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Full text of "Higher Mathematics For Engineers And Physicists"

DAMAGE BOOK 



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00 



166046 



Higher Mathematics 

for 

Engineers and Physicists 



BY 
IVAN S. SOKOLNIKOFF, Ph.D. 

Professor of Mathematics, University of Wisconsin 

AND 
ELIZABETH 8. SOKOLNIKOFF, Ph.D. 

Formerly Instructor in Mathematics, University of Wisconsin 



SECOND EDITION 
FOURTH IMPRESSION 



McGRAW-HILL BOOK COMPANY, INC. 

NEW YORK AND LONDON 
1941 



HIGHER MATHEMATICS FOB ENGINEERS AND PHYSICISTS 

COPYRIGHT, 1934, 1941, BY TUB 
McGRAw-HiLL BOOK COMPANY, INC. 

PRINTED IN THE UNITED STATES OF AMERICA 

All rights reserved. This book, or 

parts thereof, may not be reproduced 

in any form without permission of 

the publishers. 



THE MAPLE PRESS COMPANY, YORK, PA 



PREFACE 

The favorable reception of the First Edition of this volume 
appears to have sustained the authors' belief in the need of a 
book on mathematics beyond the calculus, written from the 
point of view of the student of applied science. The chief 
purpose of the book is to help to bridge the gap which separates 
many engineers from mathematics by giving them a bird's-eye 
view of those mathematical topics which are indispensable in 
the study of the physical sciences. 

It has been a common complaint of engineers and physicists 
that Ae usual courses in advanced calculus and differential 
equations place insufficient emphasis on the art of formulating 
physical problems in mathematical terms. There may also be a 
measure of truth in the criticism that many students with pro- 
nounced utilitarian leanings are obliged to depend on books 
that are more distinguished for rigor than for robust uses of 
mathematics. 

This book is an outgrowth of a course of lectures offered by 
one of the authors to students having a working knowledge of the 
elementary calculus. The keynote of the course is the practical 
utility of mathematics, and considerable effort has been made to 
select those topics which arc of most frequent and immediate use 
in applied sciences and which can be given in a course of one 
hundred lectures. The illustrative material has been chosen for 
its value in emphasizing the underlying principles rather than 
for its direct application to specific problems that may confront 
a practicing engineer. 

In preparing the revision the authors have been greatly aided 
by the reactions and suggestions of the users of this book in both 
academic and engineering circles. A considerable portion 
of the material contained in the First Edition has been rear- 
ranged and supplemented by further illustrative examples, proofs, 
and problems. The number of problems has been more than 
doubled. It was decided to omit the discussion of improper 
integrals and to absorb the chapter on Elliptic Integrals into 



vi PREFACE 

much enlarged chapters on Infinite Series and Differential 
Equations. A new chapter on Complex Variable incorporates 
some of the material that was formerly contained in the chapter 
on Conformal Representation. The original plan of making 
each chapter as nearly as possible an independent unit, in order 
to provide some flexibility and to enhance the availability of the 
book for reference purposes, has been retained. 

I. S. S. 
E. S. S. 

MADISON, WISCONSIN, 
September, 1941. 



CONTENTS 

PAGE 

PREFACE v 

CHAPTER I 
SECTION INFINITE SERIES 

1. Fundamental Concepts 1 

2. Series of Constants . ..... 6 

3. Series of Positive Terms . 9 

4. Alternating Series ' 15 

5. Series of Positive and Negative Terms . . . 16 

6. Algebra of Series . ... 21 

7. Continuity of Functions Uniform Convergence . 23 

8. Properties of Uniformly Convergent Series . . .28 

9. Power Series . 30 

10. Properties of Power Series 33 

11. Expansion of Functions in Power Series ... 35 

12. Application of Taylor's Formula . ... 41 
J3. Evaluation of Definite Integrals by Means of Power Series ... 43 

14. Rectification of Ellipse. Elliptic Integrals .47 

15. Discussion of Elliptic Integrals . . . 48 

16. Approximate Formulas in Applied Mathematics . . 55 

CHAPTER II 
FOURIER SERIES 

17. Preliminary Remarks . . . . 63 

18. Dinchlet Conditions. Derivation of Fourier Coefficients .... 65 

19 Expansion of Functions in Fourier Series . 67 

20 Sine and Cosine Series . .... 73 

21. Extension of Interval of Expansion 76 

22. Complex Form of Fourier Series . 78 

23. Differentiation and Integration of Fourier Series 80 

24. Orthogonal Functions 81 

CHAPTER III 
SOLUTION OF EQUATIONS 

25. Graphical Solutions 83 

26. Algebraic Solution of Cubic 86 

27. Some Algebraic Theorems 92 

28. Homer's Method 95 



riii CONTENTS 

JBCTION PAGE 

29. Newton's Method . . . 97 

30. Determinants of the Second and Third Order 102 

31. Determinants of the nth Order. . ... . 106 

32. Properties of Determinants . . . 107 

33. Minors . . . . . 110 

34. Matrices and Linear Dependence 114 

35. Consistent and Inconsistent Systems of Equations 117 

CHAPTER IV 
PARTIAL DIFFERENTIATION 

36. Functions of Several Variables . 123 

37. Partial Derivatives . 125 

38. Total Differential i27 

39. Total Derivatives . 130 

40. Euler's Formula 136 

41. Differentiation of Implicit Functions . 137 

42. Directional Derivatives 143 

43. Tangent Plane and Normal Line to a Surface 146 

44. Space Curves 149 

45. Directional Derivatives in Space 151 

46. Higher Partial Derivatives 153 

47. Taylor's Series for Functions of Two Variables 155 

48. Maxima and Minima of Functions of One Variable 158 

49. Maxima and Minima of Functions of Several Variables . 160 

50. Constrained Maxima and Minima . .163 

51. Differentiation under the Integral Sign 167 

CHAPTER V 

MULTIPLE INTEGRALS 

'52. Definition and Evaluation of the Double Integral 173 

53. Geometric Interpretation of the Double Integral 177 

54. Triple Integrals ... 179 

55. Jacobians. Change of Variable 183 

56. Spherical and Cylindrical Coordinates 185 

57. Surface Integrals . . 188 

58. Green's Theorem in Space . . . 191 

59. Symmetrical Form of Green's Theorem . . . 194 

CHAPTER VI 
LINE INTEGRAL 

60. Definition of Line Integral . . 197 

61. Area of a Closed Curve 199 

62. Green's Theorem for the Plane . . . 202 

63. Properties of Line Integrals 206 

64. Multiply Connected Regions . . . . . 212 

65. Line Integrals in Space . 215 

66. Illustrations of the Application of the Line Integrals . .217 



CONTENTS ix 

SECTION PAGE 

CHAPTER VII 

ORDINARY DIFFERENTIAL EQUATIONS 

67. Preliminary Remarks . ... 225 

68. Remarks on Solutions . ... 227 

69. Newtonian Laws . . 231 

70. Simple Harmonic Motion . 233 

71. Simple Pendulum . . 234 

72. Further Examples of Derivation of Differential Equations 239 

73. Hyperbolic Functions 247 
^4. First-order Differential Equations 256 
75. Equations with Separable Variables . 257 
*f6. Homogeneous Differential Equations . . 259 

77. Exact Differential Equations 262 

78. Integrating Factors . 265 

79. Equations of the First Order in Which One of the Variables Does 
Not Occur Explicitly . 267 

80. Differential Equations of the Second Order 269 

81. Gamma Functions . 272 

82. Orthogonal Trajectories 277 
'83. Singular Solutions . ... 279 

84. Linear Differential Equations . 283 

85. Linear Equations of the First Order . . 284 

86. A Non-linear Equation Reducible to Linear Form (Bernoulli's 
Equation) . . 286 

87 Linear Differential Equations of the nth Order . 287 

88 Some General Theorems . . ... 291 

89. The Meaning of the Operator 

Z> + oiJ)-i + : + a n -iD + a n f(x) ' . . 295 

90. Oscillation of a Spring and Discharge of a Condenser 299 

91. Viscous Damping 302 

92. Forced Vibrations . . ... 308 

93. Resonance . 310 

94. Simultaneous Differential Equations . . 312 

95. Linear Equations with Variable Coefficients . .... 315 

96. Variation of Parameters . . 318 

97. The Euler Equation . . 322 

98. Solution in Series . ... 325 

99. Existence of Power Series Solutions . 329 

100. BesseTs Equation 332 

101. Expansion in Series of Bessel Functions 339 

102. Legendre's Equation . . . . 342 

103. Numerical Solution of Differential Equations 346 

CHAPTER VIII 
PARTIAL DIFFERENTIAL EQUATIONS 

104. Preliminary Remarks . . .... ... 350 

105. Elimination of Arbitrary Functions . . . . . . 351 



x CONTENTS 

SECTION PAGE 

106. Integration of Partial Differential Equations. . . 353 

107. Linear Partial Differential Equations with Constant Coefficients . 357 

108. Transverse Vibration of Elastic String . . . 361 

109. Fourier Series Solution ... . 364 

110. Heat Conduction . 367 

111. Steady Heat Flow ..... 369 

112. Variable Heat Flow ... . . . 373 

113. Vibration of a Membrane . 377 

114. Laplace's Equation . . ... .... 382 

115. Flow of Electricity in a Cable . . 386 

CHAPTER IX 
VECTOR ANALYSIS 

116. Scalars and Vectors . ... 392 

117. Addition and Subtraction of Vectors 393 

118. Decomposition of Vectors. Base Vectors . 396 

119. Multiplication of Vectors . . 399 

120. Relations between Scalar and Vector Products 402 

121. Applications of Scalar and Vector Products . . . 404 

122. Differential Operators . . . 406 

123. Vector Fields . . 409 

124. Divergence of a Vector . 411 

125. Divergence Theorem . . . . . 415 

126. Curl of a Vector . 418 

127. Stokcs's Theorem . 421 

128. Two Important Theorems 422 
129 Physical Interpretation of Divergence and Curl 423 

130. Equation of Heat Flow 425 

131. Equations of Hydrodynamics 428 

132. Curvilinear Coordinates . . 433 

CHAPTER X 
COMPLEX VARIABLE 

133. Complex Numbers . ... . . 440 

134. Elementary Functions of a Complex Variable . . 444 

135. Properties of Functions of a Complex Variable 448 

136. Integration of Complex Functions .... ... 453 

137. Cauchy's Integral Theorem . ... 455 

138. Extension of Cauchy's Theorem . .... 455 

139. The Fundamental Theorem of Integral Calculus . 457 

140. Cauchy's Integral Formula . . . . . 461 

141. Taylor's Expansion. . . . ... . ... 464 

142. Conformal Mapping .... . . . . 465 

143. Method of Conjugate Functions . . 467 

144. Problems Solvable by Conjugate Functions 470 

145. Examples of Conformal Maps . ... ... 471 

146 Applications of Conformal Representation . . . 479 



CONTENTS xi 

SECTION PAGE 

CHAPTER XI 

PROBABILITY 

147. Fundamental Notions . .... 492 

148. Independent Events . . 495 

149. Mutually Exclusive Events . 497 

150. Expectation. . . . 500 

151. Repeated and Independent Trials . . . 501 

152. Distribution Curve 504 

153. Stirling's Formula . 508 

154. Probability of the Most Probable Number . . .511 

155. Approximations to Binomial Law . 512 

156. The Error Function ... 516 

157.* Precision Constant. Probable Error . .521 

CHAPTER XII 

EMPIRICAL FORMULAS AND CURVE FITTING 

158. Graphical Method . . 525 

159. Differences 527 

160. Equations That Represent Special Types of Data 528 

161. Constants Determined by Method of Averages 534 

162. Method of Least Squares . 536 

163. Method of Moments 544 

164. Harmonic Analysis . . ... 545 

165. Interpolation Formulas . . . 550 

166. Lagrange's Interpolation Formula . 552 

167. Numerical Integration 554 

168. A More General Formula 558 

ANSWERS . . 561 

INDEX . . 575 



HIGHER MATHEMATICS 

FOR ENGINEERS AND 

PHYSICISTS 



CHAPTER I 
INFINITE SERIES 

It is difficult to conceive of a single mathematical topic that 
occupies a more prominent place in applied mathematics than 
the subject of infinite series. Students of applied sciences meet 
infinite series in most of the formulas they use, and it is quite 
essential' that they acquire an intelligent understanding of the 
concepts underlying the subject. 

The first section of this chapter is intended to bring into 
sharper focus some of the basic (and hence more difficult) notions 
with which the reader became acquainted in the first course in 
calculus. It is followed by ten sections that are devoted to a 
treatment of the algebra and calculus of series and that represent 
the minimum theoretical background necessary for an intelligent 
use of series. Some of the practical uses of infinite series are 
indicated briefly in the remainder of the chapter and more fully 
in Chaps. II, VII, and VIII. 

1. Fundamental Concepts. Familiarity with the concepts 
discussed in thig section is essential to an understanding of the 
contents of this chapter. 

FUNCTION. The variable y is said to be a function of the variable 
x if to every value of x under consideration there corresponds at least 
one value of y. 

If x is the variable to which values are assigned at will, then 
it is called the independent variable. If the values of the variable 
y are determined by the assignment of values to the independent 

1 



2 MATHEM&TICS FOR ENGINEERS AND PHYSICISTS 1 

variable x, then y is called the dependent variable. The functional 
dependence of y upon x is usually denoted by the equation* 

V = /(*) 

Unless a statement to the contrary is made, it will be supposed 
in this book that the variable x is permitted to assume real 
values only and that the corresponding values of y are also real. 
In this event the function f(x) is called a real function of the real 
variable x. It will be observed that 



(1-1) y = 

does not represent a real function of x for all real values of x, for 
the values of y become imaginary if x is negative. In order that 
the symbol f(x) define a real function of x, it may be necessary to 
restrict the range of values that x may assume. Thus, (1-1) 
defines a real function of x only if x ^ 0. On the other hand, 
y \/x 2 1 defines a real function of x only if \x > I. 

SEQUENCES AND LIMITS. Let some process of construction 
yield a succession of values 

Xij 2*2, #3, , X H) , 

where it is assumed that every x t is followed by other terms. 
Such a succession of terms is called an infinite sequence. Exam- 
ples of sequences are 

(a) 1, 2, 3, , n, , 
,,,1 11 1 / iw i ! 

(6) 2' - 4 1 8' ~ 16' ' ' ' '<-*> V 
(c) 0,2,0, 2, -, ! + (-!)", . 

Sequences will be considered here only in connection with the 
theorems on infinite series, t and for this purpose it is necessary 
to have a definition of the limit of a sequence. 

DEFINITION. The sequence x\ y x%, , x n , is said to 
converge to the constant L as a limit if for any preassigned positive 
number , however small, one can find a positive integer p such that 

\x n I/I < e for all n > p. 

* Other letters are often used. In particular, if more than one function 
enters into the discussion, the functions may be denoted by /i(x), ft(x), etc.; 
by/(aO, g(x), etc.; by F(x) t G(x), etc. 

t For a somewhat more extensive treatment, see I. S, okolnikoff ? 
Advanced Calculus, pp. 3-21 f 



1 INFINITE SERIES 5 

which is convergent to the value 2. In order to establish this 
fact, note that 



is a geometric progression of ratio J^, so that 

J_ 

on 7~ ^ TvHTZTi"* 



Heftce, the absolute value of the difference between 2 and s n 
is l/2 n ~ 1 , which can bo made arbitrarily small by choosing n 
sufficiently large. 

On the other hand, if x 1, the series (1-4) becomes 



which does not converge; for s 2 n = and S2n~i = 1 for any choice 
of n and, therefore, lim s n does not exist. Moreover, if x = 2, 

n oo 

the series (1-4) becomes 

1 + 2 + 4 + + 2- 1 + , 
so that s n increases indefinitely with n and lim s n does not exist. 

n oo 

If an infinite, series does not converge for a certain value of x, 
it is said to diverge or be divergent for that value of x. It will 
be shown later that the series (1-4) is convergent for I < x < 1 
and divergent for all other values of x. 

The definition of the limit, as given above, assumes that the 
value of the limit $ is known. Frequently it is possible to infer 
the existence of S without actually knowing its value. The 
following example will serve to illustrate this point. 

Example. Consider the series 



and compare the sum of its first n terms 

_!,!,!, , JL 

Sn - l + 2! + 3! + ' ' ' + n! 

with the sum of the geometrical progression 



6 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 

S. = 1 + \ + + + ~ 



The corresponding terms of S n are never less than those of $; but, no 
matter how large n be taken, S n is less than 2. Consequently, s < 2; 
and since the successive values of s n form an increasing sequence of 
numbers, the sum of the first series must be greater than 1 and less than 
or equal to 2. A geometrical interpretation of this statement may help 
to fix the idea. If the successive values of s, 

Si = 1, 

2 = 1 + 21 = 1.5, 

ss - 1 + 5j 4- ^ = 1.667, 

*4=*l + || + ^ + jj = 1.708, 

s* = 1 + I + |j + jj + ^ = 1.717, 

are plotted as points on a straight line (Fig. 1), the points representing 
the sequence Si, $2, * , s n , - always move to the right but never 



,, 1 15 1667 2 

* - FIG. 1. 

progress as far as the point 2. It is intuitively clear that there must be 
some point s, either lying to the left of 2 or else coinciding with it, which 
the numbers s n approach as a limit. In this case the numerical value 
of the limit has not been ascertained, but its existence was established 
with the aid of what is known as the fundamental principle. 

Stated in precise form the principle reads as follows: // an infinite 
set of numbers si, 2, * * , s n , forms an increasing sequence (that is, 
SN > Sn, when N > n) and is such that every s n is less than some fixed 
number M (that is, s n < M for all values of n\ then s n approaches a limit 
s that is not greater than M (that is, lim s n = s < M). The formulation 

n oo 

of the principle for a decreasing sequence of numbers i, s 2 , * , 
s n , , which are always greater than a certain fixed number w, will 
be left to the reader. 

2. Series of Constants. The definition of the convergence of 
a series of functions evidently depends on a study of the behavior 



2 INFINITE SERIES 7 

of series of constants. The reader has had some acquaintance 
with such series in his earlier study of mathematics, but it seems 
desirable to provide a summary of some essential theorems that 
will be needed later in this chapter. The following important 
theorem gives the necessary and sufficient condition for the 
convergence of an infinite series of constants: 

00 

THEOREM. The infinite series of constants u n converges if 

n = l 

and only if there exists a positive integer n such that for all positive 
integral values of p 

\S n+p S n \ SS \U n +l + U n +2 + ' ' ' + U n+f >\ < , 

where e is any preassigned positive constant. 

The necessity of the condition can be proved immediately by recalling 
the definition of convergence. Thus, assume that the series converges, 
and let its sum be S y so that 

lim s n = S 

n > oo 

and also, for any fixed value of p, 

lim s n i P = S. 

n > oo 

Hence, 

lim (s n+p s n ) = lim (u n+} + u n +z + + u n+p ) = 0, 

n * n * * 

which is another way of saying that 



for a sufficiently large value of n. 

The proof of the sufficiency of the condition requires a fair degree 
of mathematical maturity and will not be given here.* 

This theorem is of great theoretical importance in a variety of 
investigations, but it is seldom used in any practical problem 
requiring the testing of a given series. A number of tests for 
convergence, applicable to special types of series, will be given in 
the following sections. 

It may be remarked that a sufficient condition that a series 
diverge is that the terms u n do not approach zero as a limit when 
n increases indefinitely. Thus the necessary condition for con- 
vergence of a series is that lim u n = 0, but this condition is not 

n * o 

* See SOKOLNIKOFF, I. S., Advanced Calculus, pp. 11-13. 



8 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 2 
sufficient; that is, there are series for which lim u n = but which 

n> 

are not convergent. A classical example illustrating this case 
is the harmonic series 



in which S n increases without limit as n increases. 

Despite the fact that a proof of the divergence of the harmonic 
series is given in every good course in elementary calculus, it will 
be recalled here because of its importance in subsequent con- 
siderations. Since 

_1_ + _J_ + . . ! ^ ! ! 

i "i I l r> l 



1 n + n " " 2n 2' 

it is possible, beginning with any term of the series, to add a 
definite number of terms and obtain a sum greater than V. 
If n = 2, 

3 + 4 > 2 ; 

n =' 4, 



n = 8, 

1, , J_ , , J_ 1. 

9 "*" 10 + ' ' ' + 16 > 2 ; 
n = 16, 

_1_J_ .^1 

17 + 18 + ' ' ' + 32 > 2 

Thus it is possible to group the terms of the harmonic series 



in such a way that the sum of the terms in each parenthesis 
exceeds ,; and, since the series 

1 +2 : + ! + 2 :+ ' 
is obviously divergent, the harmonic series is divergent also. 



8 INFINITE SERIES 9 

3. Series of Positive Terms. This section is concerned with 
series of the type 



a n = a! + a z + ' ' ' a n * , 

1 

where the a n are positive constants. It is evident from the 
definition of convergence and from the fundamental principle 
(see Sec. 1) that the convergence of a series of positive constants 
will be established if it is possible to demonstrate that the partial 
sums s n remain bounded. This means that there exists some 
positive number M such that s n < M for all values of n. 
The proof of the following important test is based on such a 
demonstration. 

oo 

COMPARISON TEST. Let 2 a n be a series of positive terms, 

n = l 

00 

and let 2 b n be a series of positive terms that is known to converge. 

n = l 

00 

Then the series 2 a n is convergent if there exists an integer p such 

n = l 

00 

that, for n > p, a n ^ b n . On the other hand, if 2 c n is a series of 

n-l 

positive terms that is known to be divergent and if a n ^ c n for 

oo 

n > p, then 2 #n is divergent also. 

n = l 

Since the convergence or , divergence of a series evidently is not 
affected by the addition or subtraction of a finite number of terms, the 
proof will be given on the assumption that p = 1. Let s n = i + a 2 

00 

+ - + a n , and let B denote the sum of the series 2 b n and B n its 

n = l 

nth partial sum. Then, since a ^ b n for all values of n, it follows that 
s n ^ B n for all values of n. Hence, the s n remain bounded, and the 

oo 

series 2 d n is convergent. On the other hand, if a n ^ c n for all values of 

n=l 

00 00 

n and if the series 2 c n diverges, then the series 2 a n will diverge also. 

n=l n=l 

There are two series that are frequently used as series for 
comparison. 

a. The geometric series 

(3-1) a + ar + ar 2 + + ar + , 



10 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 3 

which the reader will recall* is convergent to . _ if \r\ < 1 

and is divergent if \r\ > 1. 
6. The p series 

(3-2) l+g++ ++, 

which converges if p > 1 and diverges if p < 1. 

Consider first the case when p > 1, and write (3-2) in the form 



(3-3) 



4- 

T 



(2 n _ 



where the nth term of (3-3) contains 2 n ~ 1 terms of the series 
(3-2). Each term, after the first, of (3-3) is less than the corre- 
sponding term of the series 

1 J_ O . Jl __ L A . ___ __ L . . . 4_ On 1 

1 ^ ^ ^ ^ ^ 



2? ^ 4 p (2 n ~ 1 )p ' 

or 

(3-4) 1 + Tj^Zi + /2P-1)2 + ' ' ' + /2p-l)n-l + ' ' ' * 

Since the geometric series (3-4) has a ratio l/2 p ~ l (which is less 
than unity for p > 1), it is convergent and, by the comparison 
test, (3-2) will converge also. 

If p = 1, (3-2) becomes the harmonic series which has been 
shown to be divergent. 

If p < 1, l/n p > l/n for n > 1, so that each term of (3-2), 
after the first, is greater than the corresponding term of the 
harmonic series; hence, the series (3-2) is divergent also. 

Example 1. Test the series 

1 + I + I+ ... +!+-.. 

A T 22 T 33 T -r nn -r 

The geometric series 

1,1,1, _ 1 . 

1 + 2 2 + 2 3 + * ' ' + 2* + ' " ' 

* Since the sum of the geometric progression of n terms a -f or -f or 2 

i i -i , . a ar n a X1 N 

4- H- ar n r is equal to ^_ = __ (1 r n ). 



3 INFINITE SERIES 11 

is known to be convergent, and the terms of the geometric series are 
never less than the corresponding terms of the given series. Hence, the 
given series is convergent. 
Example 2. Test the series 



iog~2 loi~3 log! ' k^ 

Compare the terms of this series with the terms of the p series for 



given series is divergent, for its terms (after the first) are greater 
than the corresponding terms of the p series, which diverges when 
p = 1. 

00 

RATIO TEST. The series 2 n of positive terms is convergent if 



,. U'n-f-l ^ + 

lim = r < 1 

n * a n 

and divergent if 

lim ^^ > 1. 

n QO d n 

// lim -^ = 1, the series may converge or diverge. 

Consider first the case when r < 1, and let q denote some constant 
between r and 1. Then there will be some positive integer N such that 



< q for all n N. 



Hence. 



a N q, 
and 



Since q < 1, the series in the right-hand member is convergent'; there- 
fore, the series in the left-hand member converges, also. It follows that 

00 

the series S ct n is convergent*. 



n-l 



- 

If the limit of the ratio is greater than 1, then a n +i > a n for every 

00 

n ^ N so that lim a n j& 0, and hence the series S a n is divergent. 

n o n = 1 



12 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 3 

It is important to observe that this theorem makes no reference 
to the magnitude of the ratio of a n+ \/a n but deals solely with tLr 
limit of the ratio. Thus, in the case of the harmonic series the 
ratio is a n +\/a n = n/(n + 1), which remains less than 1 for all 
finite values of n, but the limit of the ratio is precisely equal 
to 1. Hence the test gives no information in this case. 

Example 1. For the series 



and, therefore, the series converges. 
Example 2. The series 

1+1L + 1L+ . . . +"j*L+ . 

10 10 2 10 3 10" 

is divergent, for t. 

r an+i ,. (n + l)!10 rt , n 
lim -^i- = lim v , ' ' -- r- = lim 



-- , -- r- ~r 

a n n-^oo 10 n+1 n! n-K 10 

Example 3. Test the series 



Here 

r fln+1 
" 



5-6 


T^ T 7 
1 


2n - l)2n ' 
. (2n - l)2n 




4n 2 


- l)(2n + 2) 
- 2n 


lim ^ X "" ^ 


4n' + 


6n + 2 , 


^"l +1-4 
2n 


1 

"2n 2 



Hence, the test fails; but if the given series be .compared with the 
p series for p = 2, it is seen to be convergent* J*- 

oo * 

CAUCHY'S INTEGRAL TEST. Let 2 a n be a series of positive 

n = l 

terms such that a n +i < a n . If there exists a positive decreasing 
function f(x), for x > 1, such that f(n) = a n , then the given series 
converges if the integral 



exists; the series diverges if the integral does not exist. 



INFINITE SERIES 



13 



The proof of this test is deduced easily from the following 
graphical considerations. Each term a n of the series may be 
thought of as representing the area of a rectangle of base unity 
and height /(n) (see Fig. 2). The sum of the areas of the first 
n inscribed rectangles is less than f" +1 f(x) dx, so that 






f(x) dx. 



But f(x) is positive, and hence 
l f(x) dx 



f(x) dx. 



If the integral on the right exists, it follows that the partial sums 
are bounded and, therefore, the series converges (see Sec. 1). 




TvC 



The sum of the areas of the circumscribed rectangles, a\ + a 2 
+ + a n , is greater than f " +1 f(x) dx] hence, the series will 
diverge if the integral does not exist. 

Example 1. Test the harmonic series 



In this case, f(x) - and 
x 



f 1 C n dx 

I - dx = lim I = lim log n 

Jl X n - oo Jl X n -> oo 

and the series is divergent. 



14 MATHEMATICS FOR EN&NEERS AND PHYSICISTS 3 

00 i 

Example 2. Apply Cauchy's test to the p series ]5 - where p > 0. 

^^ p 



Taking f(x) = i observe that 



-p 



l xf l_p- ,. "p* 1 - 

= log a;|?, if p = 1. 



/ ^XJ. 
exists if p > 1 and does not exist if p < 1. 
or* 



PROBLEMS 

1. Test for convergence 

l 



__ 

2 2 2 2 3 2 3 4 2 4 

- 



/ ^ i i 2! , 3! , 
(c) 1 + 2i + 35 + 



1 ,2,3, 

^ + v + 2"3 + ; 



21^2 3l^g~3 4Togl " 

2. Use Cauchy's integral test to investigate the convergence of 

' -- 



v 1 '/' A ~ i _j_ 2 2 ~ 2 + 3 2 

00 

3. Show that the series 2 a n of positive terms is divergent if na n 
has a limit L which is different from 0. Hint: Let lim na n = L so that 



na n > L c for n large enough. Hence, a n > 



n oo 

- 



4 INFINITE SERIES 15 

4. Test for convergence 




1 



(2n + I) 2 ' 
n 



4. Alternating Series. A series whose terms are alternately 
positive and negative is called an alternating series. There is a 
simple test, due to Leibnitz, that establishes the convergence of 
many of these series. 

TEST FOR AN ALTERNATING SERIES. // the alternating series 
a\ c&2 + a a* + * * , where the a t are positive, is such 
that a n +i < a n and lim a n = 0, then the series is convergent. 

n > oo 

Moreover, if S is the sum of the series, the numerical value of the 
difference between S and the nth partial sum is less than a n +i. 
Since 

S2n = (ai - a 2 ) + (as flu) + + (2n-i - a 2n ) 
= ai (a 2 a 3 ) (a2 n -2 a2n-i) 2n, 

it is evident that $2n is positive and also that $2n < #1 for all 
values of n. Also, $2 < $4 < *e < * * , so that these partial 



16 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 6 

sums tend to a limit S (by the fundamental principle). Since 
2n+i = 2n + 2n+i and lim a2n+i = 0, it follows that the partial 

n oo 

sums of odd order tend to this same limit. Therefore, the series 
converges. The proof of the second statement of the test will 
be left as an exercise for the reader. 

Example 1. The series 



is convergent since lim - = and - < - Moreover. 54 

n -> * n n + 1 n 

% + H y differs from the sum S by less than y*>. 
Example 2. The series 



--..-.-... 

2 "*" 3 1 4 "*" 3 2 6 "*" 3 3 
is divergent. Why? 

6. Series of Positive and Negative Terms. The alternating 
series and the series of positive constants are special types of the 
general series of constants in which the terms can be either posi- 
tive or negative. 

DEFINITION. // u\ + u<t + + u n + is an infinite 
series of terms such that the series of the absolute values of its terms, 
\Ui\ + \u%\ + ' + \u n \ + , is convergent, then the series 
Ui -f- u<i + + u n + is said to be absolutely convergent. 
If the series of absolute values is not convergent, but the given series 
is convergent, then the given series is said to be conditionally 
convergent. 

Thus, 

l-i + i_! + i ____ 
2+3 4+5 

is convergent, but the series of absolute values, 



is not, so that the original series is conditionally convergent. 

If a series is absolutely convergent, it can be shown that the 
series formed by changing the signs of any of the terms is also a 
convergent series. This is an immediate result of the following 
theorem: 



6 INFINITE SERIES 17 

THEOREM. // the series of absolute values 2 \u n \ is convergent, 

00 

then the series 2 u n is necessarily convergent. 
Let 

and 

If p n denotes the sum of the positive terms occurring in s n and q n 
denotes the sum of the negative terms, then 

(5-1) S n = Pn - qn 

and 

tn = Pn + q n > 

00 

The series 2 \u n \ is assumed to be convergent, so that 

n = l 

(5-2) lim t n = lim (p n -f q n ) ss L. 

n oo n QO 

But p n and g n are positive and increasing with n and, since (5-2) shows 
that both remain less than L, it follows from the fundamental principle 
that both the p n and q n sequences converge. If 

lim p n P and lim q n = Q, 

n > * H > oo 

then (5-1) gives 

lim s rl = lim (p n q n ) = P Q, 

00 

which establishes the convergence of 2 u n . 

Moreover, it can be shown that changing the order of the 
terms in an absolutely convergent series gives a series which is 
convergent to the same value as the original series. * However, 
conditionally convergent series do not possess this property. In 
fact, by suitably rearranging the order of the terms of a condi- 
tionally convergent series, the resulting series can be made to 
converge to any desired value. For example, it is knownf that 
the sum of the series 

111 f_n-i 

ii-i-l j_ 

1 0<0 A 1 



-.-- 

23 4 n 

* See SOKOLNIKOFF, I. S., Advanced Calculus, pp. 240-241. 
t See Example 1, Sec. 13. 



18 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 5 

p 

is log 2. The fact that the sum of this series is less than 1 and 
greater than % can be made evident by writing the series as 



which shows that the value of s n > % for n > 2; whereas, by 
writing it as 



- * O Of \ A f 

it is clear that s n < I for n > 2. Some questions might < be 
raised concerning the legitimacy of introducing parentheses in a 
convergent infinite series. The fact that the associative law 
holds unrestrictedly for convergent infinite series can be estab- 
lished easily directly from the definition of the sum of the infinite 
series. It will be shown* next that it is possible to rearrange 
the series 



so as to obtain a new series whose sum is equal to 1. The positive 
terms of this series in their original order are 



3' 5' 7' 9' 



The negative terms are 

11 



' 



g 



In order to form a series that converges to 1, first pick out, in 
order, as many positive terms as are needed to make their sum 
equal to or just greater than 1, then pick out just enough negative 
terms so that the sum of all terms so far chosen will be just less 
than 1, then more positive terms until the sum is just greater 
than 1, etc. Thus, the partial sums will be 



* 2 - - 2 = 2' 

- 1,1,1 31 
S4==1 -2 + 3 + 5 = 30' 
* General proof can be constructed along the lines of this example. 



8 INFINITE SERIES 19 

l * 1 1 47 



. . 
2 3 5-3 7 9- 1260 

1,1,11, 1,1_1_ 1093 
1 2 + 3 + 5 4 + 7 "*" 9 6 ~ 1260' 



It is clear that the series formed by this method will have a sum 
equal to 1. 

As another example, consider the conditionally convergent 
series 

(5-3) l--4= + -^--4-+---. 

' V2 -s/3 VI 

Let the order of the terms in (5-3) be rearranged to give the 
series 



LL 4. -1 L\ 



The nth term of (5-4) is 
1 



which is greater than 

j == * _|_ *__ [i . \ * , 



oo 

But the series S &n is divergent, and it follows that the series 

n = l 

(5-4) must diverge. 

00 

Inasmuch as the series S |wn| is a series of positive terms, the 

nl 

tests that were developed in Sec. 3 can be applied in establishing 

00 

the absolute convergence of the series 2 u n . In particular, the 

n = l 

ratio test can be restated in the following form: 



20 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 5 

00 

RATIO TEST. The series S u n is absolutely convergent if 



and is divergent if 



lim 



lim 



U n +l 



U n 



< 1 



> 1. 



// the limit is unity, the test gives no information. 
Example 1. In the case of the series 

. z 2 . x s . x* 



lim 



x n (n 1)! 



ft! x n ~ l 







for all values of x. Hence, the series is convergent for all values of 
x and, in particular, the series 

2 2 2 3 

1 - 2 + 21 - 3J + ' ' ' 
is absolutely convergent. 
Example 2. Consider the series 



1 



Here 



lim 

n oo 



1 - x ^ 2(1 - xY ^ 3(1 - 



. n(l - ap* 



(n + 1)(1 - x) n+l 1 



(n + 1)(1 - x) 
I 



Therefore, the series will converge if 

n-i^r < 1 or 1< |1 1 x\, 

\i x\ 

which is true for x < and for x > 2. 

For x = and for x 2 the limit is unity, but if x = the series 
becomes the divergent harmonic series 



and if x = 2 there results the convergent alternating series 



6 INFINITE SERIES . 21 

i4.i^i4... -!-( nn-i-u... 
1 ^ 2 3 ^ ^ ' n^ 

It follows that the original series converges for x < and for x ^ 2 
and diverges for < x < 2. 

6. Algebra of Series. The following important theorems are 
stated without proof:* 

THEOREM 1. Any two convergent series 

U = Ui + U 2 + ' ' + U n + ' ' ' 
V = Vi + V% + ' ' + V n + ' * ' 

can be added or subtracted term by term to give 

U + V = (Ui + Vi) + (Ma + 2 ) + ' + (tin + )+' 

or 

U - V = (m - v,) + (M* - f> 2 ) + + (u - t>0 + ' ' ' 
// /Ae original series are both absolutely convergent, then the resulting 
series will be absolutely convergent also. 

THEOREM 2. // 

U = ui + uz + + u n + 



F = fli + tf2 + ' ' + Vn + ' 

are two absolutely convergent series, then they can be multiplied like 
finite sums and the product series will converge to UV. Moreover, 
the product series will be absolutely convergent. Thus, 

UV = U\V\ + UiV% + U%Vi + U\V$ + U^Vz + UzV\ + ' ' ' . 

THEOREM 3. In an absolutely convergent series the positive 
terms by themselves form a convergent series and also the negative 
terms by themselves form a convergent series. If in a convergent 
series the positive terms form a divergent series, then the series of 
negative terms is also divergent and the original series is conditionally 
convergent. 

THEOREM 4. // u\ + u% + + u n + is an abso- 
lutely convergent series and if Mi, M^ 9 , M n , is any 
sequence of quantities whose numerical values are all less than some 
positive number N, then the series 

UiMl + UiMz + ' ' ' + U n M n + ' * 

is absolutely convergent. 

* See SOKOLNIKOFF, I. S., Advanced Calculus, pp. 212-213, 241-245. 



22 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 6 

Exarppk. Consider the series 

sin x __ sin 2x sin 3x __ 
1 2 3 + 3 3 " ' ' ' ' 

This series is absolutely convergent for all values of x> for the series 

p - 2~ 3 + 3~ 3 "" 

is absolutely convergent and (sin nx\ ^ 1. 

PROBLEMS 

1. Show that the following series are divergent: 

, . 5 7,9 11 , 2n + 3 , 

()2 - 4 + 6""8 + "" + ( ~ 1} ~2^~ + ' ' ' J 

(6) 2-22 + 3- 3*2 + 4-4$+ ; 
(c) 1 - 1 + 1 " i + ' ' ' 

2. Test for convergence, and if the series is convergent determine 
whether it is absolutely convergent. 

- 1.1 1 ..... 

1-3-5-7 . 



_ - . _ 

3 3-6 3-6-9 3-6-9-12''' 

,,2 3,4 5 

(c) i - ^ + 3 - 4 + ' ' ' 

3. For what values of x are the following series convergent? 

/v2 /y.3 /yn 

W*-l+l ---- +<-D-'^+---; 

/^2 /p4 /p6 

(6) 1 - 2! + j| - g-, + ; 
(c) 1 - + * 2 - + ; 

w; + i + 3i+---+i + ---- 

4. Determine the intervals of convergence of the following series: 



iT4 2 V^Tl 3 
% (6) * + 2b 2 + 3lx 3 + 4b 4 + ; 

, m(m - 1) m(m - 3)(m 2) 

(c) 1 + ws H -- 21 - * "! -- 31 - x + 

where w is not a positive integer. 



INFINITE SERIES 



23 



7. Continuity of Functions. Uniform Convergence. Before 
proceeding with a discussion of infinite series of functions, it is 
necessary to have a clear understanding of the concept of con- 
tinuity of functions. The reader will recall that a function 
f(x) is said to be continuous at a point x = X Q if lim f(x) = f(x Q ) 

y xo 

regardless of how x approaches XQ. From the discussion of 
the limit in Sec. 1, it appears that this concept can be defined in 
the following way : 

DEFINITION. The function f(x) is continuous at the point 
x = XQ if, corresponding to any ^reassigned positive number c, 
it is possible to find a positive number 5 such that 

(7-1) I/O) /Cr )| < whenever \x - rr | < 5. 

The foregoing analytical definition of continuity is merely a 
formulation in exact mathematical language of the intuitive 



y=f(x )-e 



-y-f(x) 



FIG. 3 

concept of continuity. If the function f(x) is represented by a 
graph and if it is continuous at the point x = #o, then it is 
possible to find a strip bounded by the two parallel lines x = x 
+ 5 and x = x 5, such that the graph of the function will lie 
entirely within the strip bounded by the parallel lines y = /(a?o) 
+ e and y = f(x Q ) e (Fig. 3). But if the function is discon- 
tinuous at some point (such as x = x\), then no interval about 
such a point can be found such that the graph of the function will 
lie entirely within the strip of width 2e, where e is arbitrarily 
small. 

DEFINITION. A function is said to be continuous in an interval 
(a, b) if it is continuous at each point of the interval 



24 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 7 

If a finite number of functions that are all continuous in an 
interval (a, 6) are added together, the sum also will be a continuous 
function in (a, fe). The question arises as to whether this 
property will be retained in the case of an infinite series of con- 
tinuous functions. Moreover, it is frequently desirable to 
obtain the derivative (or integral) of a function f(x) by means of 
term-by-term differentiation (or integration) of an infinite series 
that defines /(x). Unfortunately, such operations are not always 
valid, and many important investigations have led to erroneous 
results solely because of the improper handling of infinite series. 
A discussion of such questions requires an introduction of the 
property of uniform convergence of a series. 

It was stated in Sec. 1 that the series 



(7-2) ui(x) + u*(x) + + u n (x) + 

is convergent to the value $, when x = X Q> provided that 

(7-3) lim s n (xo) = S. 

I n> oo 

The statement embodied in (7-3) means that for any preassigned 
positive number , however small, one can find a positive number 
N such that 

|n(&o) - S\ < e for all n ^ N. 

If the series (7-2) is convergent for every value of x in the interval 
(a, 6), then the series (7-2) defines a function S(x). Let x be 
some value of x in (a, b), so that 

\s n (xo) S(xo)\ < e whenever n ;> N. 

It is important to note that, in general, the magnitude of N 

depends not only on the choice of e, but also on the value of XQ. 

This last remark may be clarified by considering the series 

(7-4) x + (x - l)x + (x - l)z 2 + 

+ (x - I)*"- 1 + . 
Since 

s(x) = x + (x - l)x + (x - l)x 2 + + (x - l)^- 1 
= x n , 

it is evident that 

lim s n (x) = lim x n = 0, if ^ x < 1. 



7 INFINITE $ERIES 25 

Thus, S(x) = for all values of x in the interval <> x < 1, and 
therefore 

MX) - S(x)\ = |* - 0| = \x\. 

Hence, the requirement that \s n (x) S(x)\ < c, for an arbitrary 
e, will be satisfied only if x n < e. This inequality leads to the 
condition 

n log x < log e. 

Since log x is negative for x between and 1, it follows that it is 
necessary to have 



logo: 

which clearly shows the dependence of N on both e and x. In fact, 
if e = 0.01 and x = 0.1, n must be greater than log 0.01/log 0.1 
= 2/ 1 = 2, so that N can be chosen as any number greater 
than 2. If e = 0.01 and x = 0.5, N must be chosen larger than 
log 0.01/log 0.5, which is greater than 6. Since the values of 
log x approach zero as x approaches unity, it appears that the 
ratio log e/log x will increase indefinitely and that it will be 
impossible to find a single value of 'TWwhich will serve for e = 0.01 
and for all values of x in ^ x < 1. 

It should be noted that the discussion applies to the interval 
(0, 1) and that it might be possible to find an N, depending on e 
only, if some other interval were chosen. If the series and the 
interval are such that it is possible to find an N, for any pre- 
assigned e, which will serve for all values of x in the interval, then 
the series is said to converge uniformly in the interval. 

00 

DEFINITION OF UNIFORM CONVERGENCE. The series S u n (x) 

nl 

is uniformly convergent in the interval (a, b) if, for any > 0, there 
exists a positive number N, independent of the value of x in (a, b), 
such that 

\S(x) - s n (x)\ < e for all n> N. 

The distinction between uniform convergence and the type of 
convergence exemplified by the discussion of the series (7-4) 
will become apparent in the discussion of the series 

(7-5) 1 + x + x* + - + x* + , 

where J^ < x ^ J^. 



26 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 7 

J %n 

Since s n (x) = -z - , it follows that 
i x 



8(x) = lira (*) = lim T-- 

n * oo n oo \1 k 

Then, 

\8(X) - (*)! = 



1 - x 
which will be less than an arbitrary e > if 

\x n \ < e(l - x). 
Hence, 

n log |x| < log e(l a;), 
or 

(7-6) n > 



. 
log |z| 

Again, it appears that the choice of AT will depend on both 
x and , but in this case it is possible to choose an N that will 
serve for all values of x in ( J^, H) Observing that the ratio 
log e(l a;) /log |#| assumes its maximum value, for a fixed e, 
when x = J^, it is evident that if N is chosen so that 



Iog2 ? 

then the inequality (7-6) will be satisfied for all n *t N. 

Upon recalling the conditions for uniform convergence, it is 
seen that the series (7-5) converges uniformly for % < x < %. 
However, it should be noted that (7-5) does not converge uni- 
formly in the interval ( 1, 1). For, in this interval, the ratio 
appearing in (7-6) will increase indefinitely as x approaches the 
values 1. The discussion given above shows that the series 
(7-5) is uniformly convergent in any interval ( a, a), where 
a < 1. 

It may be remarked that the series (7-5) does not even con- 
verge for x = 1. For x = 1, it is obviously divergent, and 
when x = 1 the series becomes 



1 - 1 + 1 - 1 + 
T-5) defin 
the value ^ when x = 1. 



If 1 < x < 1, (7-5) defines the function ^ , which takes 

1 "*-* X 



7 INFINITE SERIES 27 

As is often the case with definitions, the definition of uniform 
convergence is usually difficult to apply when the behavior of a 
particular series is to be investigated. There are available 
several tests for the uniform convergence of series, the simplest of 
which is associated with the name of the German mathematician 
Weierstrass. 

THEOREM. (WEIERSTRASS M TEST). Let 

(7-7) ui(x) + u 2 (x) + + u n (x) + 

be a series of functions of x defined in the interval (a, 6). // there 
exists a convergent series of positive constants, 

Mi + M* + + M n + , 

such that |^ t (#)| ^ M l for all values of x in (a, b), then the series 
(7-7) is uniformly and absolutely convergent in (a, 6). 

Since, by hypothesis, the series of M 's is convergent, it follows 
that for any prescribed c > there exists an N such that 

M n+ i + M n+ i +< for all n ^ N. 
But \u,(x)\ < M % for all values of x in (a, 6), so that 



for all n ^ N and for all values of x in (a, b). Therefore, the 
series (7-7) is uniformly and absolutely convergent in (a, b). 

The fact that the Weierstrass test establishes the absolute 
convergence, as well as the uniform convergence, of a series means 
that it is applicable only to series which converge absolutely. 
There are other tests that are not so restricted, but these tests are 
more complex. It should be emphasized that a series may con- 
verge uniformly but not absolutely, and vice versa. 

Example 1. Consider the series 

sin x sin 2x sin nx 

-p- + -22- + ' ' ' + -5- + ' ' ' 

Since |sin nx\ ^ 1 for all values of x, the convergent series 

15 + Ji ++ + 

will serve as an M series. It follows that the given series is uniformly 
and absolutely convergent in any interval, no matter how large. 
Example 2. As noted earlier in this section, the series 

1 + x + x 2 + - + z + 



28 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 8 

converges uniformly in any interval (a, a), where a < 1. The 
series of positive constants 

1 + a + a 2 + - + a* + 

could be used as an M series in this case, since this series converges for 
a < 1 and |a; l | ^ a* for x in ( a, a). 

PROBLEMS 

1. Show that the series (7-4) is uniformly convergent in the interval 
(0, H). 

2. By using the definition of uniform convergence, show that 

1 1 1 



(x + l)(x 4-2) (x + n- l)(x + n) 



is uniformly convergent in the interval < z < 1. 
ies to sh 

\S(X) ~ 



Hint: Rewrite the series to show that s n (x) = r- and therefore 

X ~\~ 71 



" x + n 

3. Test the following series for uniform convergence: 
, x cos x , cos 3x , cos 5x , 

(0 -p- + -3^" + 52- + ; 

.^^ sin ^cc sin QX sin u^c 

(m : ~ H 5 Tf + "= ^~ ~T ' ' ' ', 

I ' O O O * / 

(c) 1 + ^ cos ^ + x 1 cos 219 + x s cos 30 + , \x\ ^ x l < 1 ; 
. ,. cos 2x cos 3x , cos 4x 



"" 



2" "3" 4" ' 

(e) lOz + 10 2 x 2 + 10 8 OJ 3 + . 

8. Properties of Uniformly Convergent Series. As remarked 
in the preceding section, the concept of uniform convergence was 
introduced in order to allow the discussion of certain properties 
of infinite series. This section contains the statements* of three 
important theorems concerning uniformly convergent series. 

THEOREM 1. Let 



be a series such that each u l (x) is a continuous function of x in 
the interval (a, 6). // the series is uniformly convergent in (a, 6), 
then the sum of the series is also a continuous function of x in (a, 6). 

* For proofs, see I. S. Sokolnikoff, Advanced Calculus, pp. 256-262. 



8 INFINITE SERIES 29 

COROLLARY. A discontinuous function cannot be represented 
by a uniformly convergent series of continuous functions in the 
neighborhood of the point of discontinuity. 

THEOREM 2. // a series of continuous functions, 



converges uniformly to S(x) in (a, 6), then * 

S(x) dx = f ft ui(x) dx + f*u t (x) dx + + f ft u n (x) dx + 

ot Jet Jot Jot 



where a < a < b and a < < b. 
THEOREM 3. Let 



- + u n (x) + 

be a series of differentiate functions that converges to S(x) in 
(a, 6) . // the series 

u((x) + u' 2 (x) + + <(s) + 

converges uniformly in (a, 6), then it converges to S'(x). 

These theorems provide sufficient conditions only. It may be 
that the sum of the series is a continuous function when the series 
is not uniformly convergent. It is impossible to discuss neces- 
sary conditions in this brief introduction to uniform convergence. 
It may happen also that the series is differentiate or integrable 
term by term when it does not converge uniformly. In the 
chapter on Fourier scries it will be shown that a discontinuous 
function can be represented by an infinite series of continuous 
functions. In that chapter, it is established that the series 

sin 2x . sin 3x 



_- _g 

represents the function x for w < x < TT. But, if this series 
be differentiated term by term, the resulting series is 

2(cos x cos 2x + cos 3x - ), 

which does not converge in ( TT, ?r); for the necessary condition 
for convergence, namely, that lim \u n \ = 0, does not hold for any 

n * * 

value of x. 

The series used in the first example of Sec. 7, 

, . sin x , sin 2x sin 3x sin nx , 

(8-1) -jg I ^2 I 32 r ' ' H ~z r ' , 



30 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 



is uniformly convergent in any interval (a, 6) and as such defines 
a continuous function S(x). Moreover, the series can be inte- 
grated term by term to produce the integral of S(x). The 
term-by-term derivative of (8-1) is 



(8-2) 



cos x + Y 2 cos 2x + % cos 3x + 



which is convergent in (0, TT), but the M series for (8-2) cannot 
be found since I +% + }+ is divergent. This merely 
suggests that (8-2) may not converge to the derivative of S(x), 
but it does not say that it will not. 

PROBLEMS 

1. Test for uniform convergence the series obtained by term-by-term 
differentiation of the five series given in Prob. 3 of Sec. 7. 

2. Test for uniform convergence the series obtained by term-by- 
term integration of the five series given in Prob. 3, Sec. 7. 

9. Power Series. One of the most important types of infinite 
series of functions is the power series 



00 

(9-1) V a n x n ss 



a n x n 



in which the a t are independent of x. Some of the reasons for 
the usefulness of power series will become apparent in the dis- 
cussion that follows. 

Whenever a series of functions is used, the first question which 
arises is that of determining the values of the variable for which 
the series is convergent. The ratio test was applied for this 
purpose in the examples discussed in Sec. 5. In general, for a 
power series, 



lim 



u n +i 



u n 



= lim 



so that the series converges if 



and diverges if 



lim 

n * oo 

lim 

n ao 



a n -i 



\0>n-l 



< 1 



> 1. 



Therefore, the series will converge for those values of x for which 

\x < lim 



INFINITE SERIES 



31 



If lim 



On-l 



a n 



r, it follows that the series will converge when x 



lies inside the interval ( r, r), which is called the interval of con- 
vergence, the number r being called the radius of convergence. 
This discussion establishes the following theorem : 

00 

THEOREM. // the series S a n x n is such that 

n = 



lim 



then the series converges in the interval r<x<r and diverges 
outside this interval. The series may or may not converge at the 
end points of the interval. 



Example 1. Consider the series 

/v.2 /*.3 



/>n 
- 

n 



Since lim 



= lim 



n - 1 



= 1, the series converges for 



1 < x < 1 and diverges for \x\ > 1. At the end point x = 1 the 
series becomes 

.-.+S-H---- 

which is convergent. At the end point x = 1 the divergent series 



is obtained. Hence, this power series is convergent for 1 ^ x < 1. 
Example 2. The series 

1 + x + 2lx 2 + + n \x n + - - - 

will serve to demonstrate the fact that there are power series which 
converge only for the value x = 0. For 



lim 



lim 



(n - 1)! 



n! 



lim - 



0. 



Obviously, the series converges for x 0, as does every power series, 
but it diverges for every other value of x. 

* Power series in x h are frequently more useful than the 
special case in which the value of h is zero. A series of this type 
has the form 



ai(x - h) + a 2 (x - 



a n (x - 



32 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 
In this case the test ratio yields 



lim 



= lim 



\x - h\. 



If this limit is less than 1, the series is convergent; if greater than 
1, the series is divergent; and if the limit is equal to 1, the test 
fails and the values of x, which make the limit equal to 1, must be 
investigated. Thus, if the series is 



then 
lim 

n oo 


U n +l 


= lim 

n 


(x- 


D" 


(n- 


I) 2 


u n 


n< 




(*- 


I)"- 1 



= lim l - - - - |s - 1| = |x - 11- 
n _ w \ n n 2 / ' ' ' 

Therefore the series converges if x 1| < 1, orO<x<2, and 
diverges for \x 1| > 1, or x < 0, x > 2. For a; 1 = 1, or 
x = 2, the series becomes 



which is the p series for p = 2 and is therefore convergent. For 
== ) the series becomes 



which is an alternating series of decreasing terms with lim u n = 

n * oo 

and is therefore convergent. Thus the series is convergent for 
^ x <. 2. 

PROBLEM 

Find the interval of convergence for each of the following series, and 
determine its behavior at the end points of the interval: 

(a) 1 + x + x 2 + x 3 + - ; 



10 INFINITE SERIES 33 

(d) I - 2x + 3x* - 4z 3 + ' ; 



1 



_ 

2 2* 2 
(* - 2) - 1 (* - 



10. Properties of Power Series. The importance of power 
series in applied mathematics is due to the properties given in 
the theorems of this section, as will be evident from the applica- 
tions discussed in succeeding sections. 

THEOREM 1 . // r > is the radius of convergence of a power 

00 

series 2 a n x n , then the series converges absolutely and uniformly for 

n = 

every value of x in any interval a ^ x < b that is interior to 



Since the interval (a, 6) lies entirely within the interval ( r, r), it 
is possible to choose a positive number c that is less than r but greater 
than a and b. The interval (a, 6) will then lie entirely within the 
interval ( c, c); and it follows that, for a ^ x ^ 6, 

\a n x n \ < \a n c n \. 

00 

The series of positive constants S \a n c n \ is convergent, for c < r, and, 

n = 

accordingly, can be used as a Weierstrass M series establishing the 

00 

absolute and uniform convergence of S a n x n in (a, b). 

n=0 

oo 

THEOREM 2. A power series 2 a n x n defines a continuous func- 



/ton /or aK values of x in any closed interval (a, 6) 2Aa2 is interior to 
the interval of convergence of the series. 

This statement is a direct consequence of the preceding theorem 
and of Theorem 1, Sec. 8. 

THEOREM 3. // the radius of convergence of the power series 

00 _ 00 

2 a n x n is r, then the radii of convergence of the series 2 na n x n ~ 1 

n=0 n-O 

00 

and ^^ ~-^ x n+1 , obtained by term-by-term differentiation and 

n = 

integration of the given series, are also r. 



34 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 10 

If the radius of convergence can be determined from the ratio 
test, then the proof follows immediately from the fact that if 



lim 



= r. then lim 



/ * \ 




and 




na n 
lim 

n * * 


j 


(n- 


na n 






Since the series obtained by term-by-term differentiation and 
integration are also power series, these processes can be repeated 
as many times as desired and the resulting series will be power 
series that converge in the interval ( r, r). It follows from 
Theorem 1 that all these series are uniformly and absolutely 
convergent in any interval which is interior to ( r, r). How- 
ever, the behavior of these series at the end points x = r and 
x = r must be investigated in each case. 

For example, the series 

/2 <i3 /l>i 

l+x +!+!+ +*-+ 

has unity for its radius of convergence. The series converges 
for x = 1 but is divergent for a: = 1. The series obtained by 
term-by-term differentiation is 

1 + x + x* + + x* + - - , 

which has the same radius of convergence but diverges at both 
x = 1 and x = 1. On the other hand, the series obtained by 
term-by-term integration is 

/>2 /y.3 /y.4 /yrt-fl 

+ Ji> , */ . U/ - . *l/ . 

_ 1 _ L _ _1_ . . . I _ I ... 

-^-^-^ + 



which converges for both x = 1 and x = 1. 

This discussion leads to the conclusion stated in the following 
theorem : 

oo 

THEOREM 4. A power series 2 o> n x n may be differentiated and 

n0 

integrated term by term as many times as desired in any closed 
interval (a, b) that lies entirely within the interval of convergence of 
the given series. 

00 

THEOREM 5. // a power series S Q> n x n vanishes for all values 

n-p 

of x lying in a certain interval about the point x = 0, then the 



11 INFINITE SERIES 35 

coefficient of each power of x vanishes, that is, 

do = 0, ai = 0, a 2 = 0, , a* = 0, . 

The reader may attempt to construct the proof of this theorem 
with the aid of Theorem 2 of this section. 

11. Expansion of Functions in Power Series. It was stated in 
Theorem 2, Sec. 10, that a power series defines a continuous 
function of x in any interval which lies within the interval of 
convergence. This theorem suggests at once the possibility of 
using such a power series for the purpose of computation. For 
example, the values of sin x might be obtained by means of a 
power series. Accordingly, it becomes necessary to develop 
some method of obtaining such a power series, and this section 
is devoted to a derivation of Taylor's formula and a discussion of 
Taylor's series. 

One of the simplest proofs of Taylor's formula will be given 
here. * It assumes that the given function f(x) has a continuous 
nih derivative throughout the interval (a, 6). Taylor's formula 
is obtained by integrating this nth derivative n times in suc- 
cession between the limits a andjfe where x is any point in (a, 6). 
Thus, 



I f M (x)dx =/<- (a- 



) 

I I / <n) (z) (dx) 2 = I /<"-(*) dx - f /< (o) dx 

Ja Ja Ja Ja 

(x) (dx) 3 =/(- (*) _/( ( ) - (Z - a)/<"- 2) (a) 



= /(*) - /(a) -(x- a)/' (a) 



2! (n - 1)! ' 

* For other proofs, see I. 8. Sokolnikoff, Advanced Calculus, pp. 291-295. 



36 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 11 
Solving for f(x) gives 

(11-1) /(*) - /(a) + (x - a)/' (a) + (x ~ a) >(a) 



where 

(11-2) R n = f f/ (n) (*) (dx)* 

i/a Ja 

The formula given by (11-1) is known as Taylor's formula and 
the particular form of R n given in (11-2) is called the integral 
form of the remainder after n terms. The foregoing can be 
stated in the form of a theorem. 

TAYLOR'S THEOREM. Any function f(x) that possesses a con- 
tinuous derivative / (n) (x) in the interval (a, b) can be expanded in 
the form (11-1) for all values of x in (a, b). 

The term R n , which represents the difference between f(x) 
and the polynomial of degree n 1 in x a, is frequently more 
useful when expressed in a different form. Since* 

P/ (n) (z) dx - (x* '- a)/ (n) (), where a < f < x, 

Ja 

repeated integration gives 

(11-3) R n = - - - J[* /<>(*) (dxY = (X ^ q)n / (n) (e. 

The right-hand member of (11-3) is the Lagrangian form of the 
remainder after n terms. 

The special form of Taylor's formula that is obtained by 
setting a = is known as the Maclaurin formula. In this case 

(11-4) f(x) =/(0) 



where 



* The student will recall from elementary calculus that 

I <f>(x) dx == (6 a)<f>(), where a < < b. 



11 INFINITE SERIES 37 

Taylor's formula with the Lagrangian form of the remainder 
is often encountered in a somewhat different form, which results 
from setting x a = A. Since a < < x, can be written 
in the form a + Oh, where < 6 < 1. Hence, (11-1) becomes 



(11-5) /(a + h) = /(a) ~ 



' where 



In this derivation of Taylor's formula, it was assumed that 
f(x) possesses a continuous nth derivative, and as a result it 
appeared that then f(x) could be expressed as a polynomial of 
degree n in x a. It should be noted, however, that only the 
first n coefficients of this polynomial are constants, for the 
coefficient of (x a) n is a function of and the value of is 
dependent upon the choice of x. It may happen that f(x) 
possesses derivatives of all orders and that the remainder R n 
approaches zero as a limit when n - <*> regardless of the choice 
of x in (a, b). If such is the case, the infinite series 

(11-6) /(a) +f'(a)(x - a) +/"(a) (x "^ + 

--- 



is convergent and, in general,* it converges to/(x). 

The series given in (11-6) is called the Taylor's series expansion, 
or representation, of the function f(x) about the point x = a. 
The special form of (11-6) that is obtained when a = 0, namely, 



(11-7) 

is called Maclaurin's series. 
Example. Find the Taylor's series expansion of cos x in powers of 

7T 

*- 

Since 



/'(*)= -sins, /'(!)= - 1 '* 



* For a further discussion of this point, see I. S. Sokolnikoff, Advanced 
Calculus, pp. 296-298. 



38 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 11 

/"(*) = -cos*, 
/'"(*) = sin *, / 

f"(x) = cos x, f iv = 0; 



it follows that the result is 



Since it is often possible to obtain a power series expansion 
of a function f(x) by some other method, the question arises as 
to the relation of such an expansion to the Taylor's series expan- 
sion for f(x). For example, a power series expansion for the 

function ^ - is obtained easily by division, giving 

J. """* X 



The reader can check the fact that the Maclaurin expansion for 
this function is identical with the power series obtained by 
division. That this is not an exceptional case is established in 
the following theorem: 

UNIQUENESS THEOREM. There is only one possible expansion 
of a function in a power series in x a; and, therefore, if such an 
expansion be found in any manner whatsoever, it must coincide with 
Taylor's expansion about the point a. 

Suppose that f(x) could be represented by two power series in 
x a, so that 

f(x) = a + ai(x - a) + a 2 (x - a) 2 + 

+ a n (x - a) n + 
and 

f(x) = 6 + bi(x - a) + 6 2 (z - a) 2 + 

+ b n (x a) n + . 

Since both these expansions represent f(x) in the vicinity of a, 
there must be some interval about the point x = a in which 
both the expansions are valid. Then, in this interval, 



11 INFINITE SERIES 39 

00 00 

V a n (x - a) n = ] b n (x a) n , 

n-O n=0 

or 

jj (an - &)(* - a)- = 0. 

n=0 

It follows from Theorem 5, Sec. 10, that 

an ~ b n - 0, (n = 0, 1, 2, ), 
or 

a n = b n> (n = 0, 1, 2, ). 

Hence, the two power series expansions are identical. 

Taylor's formula is frequently more useful in a slightly modified 
form. Let 

x a s= h, 
so that 

x = a + h. 
Then 

/(*) = /(a) +/'(a)(z - a) +^~ ) (x - a) 2 + - 

/<->(a) - 

+ (^nyi^ a; + 

becomes 

(11-8) /(a + A) = /(a) + /'(a) A + K + ' 



(n-l)! n! ' 

in which < 8 < 1, so that a<a + 9h<a + h. 

PROBLEMS 

1. Find the expansion of each of the following functions in power 
series in x: 

(a) e x , (b) sin x, (c) cos x, (d) tan" 1 x, 

(e) sin" 1 x, (/) sec #, (g) tan x, (h) e inx . 

2. Expand 

(a) log a; in powers of x 1 ; 
(6) tan x in powers of a; T; 



(c) e* in powers of x 2; 

/j\ e v 

(a) sm a; m powers of x g; 

(e) 2 + x 2 3x* + 7x* in powers of x 1. 



40 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 11 

3. Show that sin x can be developed about any point a in a series 
(11-8) which converges for all values of h. 

4. Differentiate term by term the power series in x for sin x and thus 
obtain the power series in x for cos x. What is the interval of con- 
vergence of the resulting series? 

x 3 x 5 
6. Divide the series sin x = x ^y + ^ by the series 

x* x 4 
cos ==1 21 "t" ZT "~ "**> anc ^ ^us obtain the series for tan x. 

6. Differentiate the series for sin" 1 x to obtain the expansion in powers 
of # for (1 x 2 )~M. Find the interval of convergence. Is convergence 
absolute? Investigate the behavior of the series at the end points of the 
interval of convergence. 

7. Establish with the aid of Maclaurin's series that 

(a + &)* SE k(l + *) = k [l + mx + m(m 2 ~ 1} x* + - ], 

where m is not a positive integer. 

This series is convergent for \x\ < 1 and divergent when |a;| > 1 A 
complete discussion of this series will be found in Sokolnikoff's Advanced 
Calculus. Some facts are. 

If x I, convergence is absolute if m > 0; 

If = 1, convergence is conditional if > m > 1; 

If x = 1, convergence is absolute if m > 0; 

If x == 1, series diverges when m < 0; 

If # = 1, series diverges when m ^ 1. 

oo oo 

8. Let /(?/) = S b n y n smdy~ S a n x n be convergent power series. If 

n=0 n=0 

f(y) is a polynomial, then the powers of y in terms of x can be determined 
by repeated multiplications and thus the expansion for/(?/) in powers of x 
can be obtained. But if f(y) is an infinite series, this procedure may not 
be valid. Inasmuch as the power series in x is always convergent for x = 
and since the value of y for x is a , it is clear that the interval of con- 

oo 

vergence of S b n y n must include a if the series for f(y\ in powers of x, 

n = 

is to converge. But if a = 0, then f(y) surely can be expanded in 
power series in x by this method, for the point is contained in the 

00 

interval of convergence of S b n y n . 

n=*0 

Apply this method to deriving the series in powers of x for e*^ x by 
setting 

x s , x 6 

y = sm x = x - + - 



12 INFINITE SERIES 41 

and 



Explain why this method fails to produce the series in powers of x for 
log (1 + e*)i where e* = y. 

12. Application of Taylor's Formula. In this section two 
illustrations of the application of Taylor's formula will be given, 
and in each case the remainder will be investigated to determine 
the error made in using the sum of the first n terms of the expan- 
sion instead of the function itself. 

1. Calculate the Value of sin 10. Since 10 is closer to 
than to any other value of x for which the values of sin x and its 
derivatives are known, the Maclaurin expansion for sin x will 
be determined and evaluated for x = 10 = IT /IS radian. Then 

f(x) = /(O) + /'(O)* 



n\ 



_ 

' 



where < < 75- 

io 

Since 

f(x) = sin x, /(O) = 0; 

f'(x) = oosx, /'(0) = 1; 

/"(*) = - sin *, /"(O) = 0; 

f"(x) = - cos x, /'"(0) = -1; 

................................... y 

/<>(*) = sin (x + ~^, /<>(0) = sin ^; 
therefore, 



Here, 

R n (x) s ~ /<>($) == ~ f (n} (0x), < 6 

= . sin 
n! 

If only the terms through x 7 (or x 8 ) are used in computing 
sin 7T/18, the error will be 



42 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 18 



" 1 {** ^9 \ /T\ 9 1 fcr 

W 91 8m V Is + 2 V = Iw 9i cos ii 



9!' 
so that 



sm 18 ~ 18 U87 3! + V18/ 5! Vl8/ 7f 



with an error less than ( j^ j ^- 

2. Compute the Value of e 1 - 1 . It can be established readily, by 
expanding e x in Maclaurin's series and evaluating for = 1, that 
e = 2.71828 . In order to compute the value of e 1 - 1 , the 
expansion of e* about x = 1 will be used. The expansion is 



Since 

/(*) = e x , /(I) = ^ 

?(*) = -, /'(I) = e; 



/<>(*) = e-, /<>(!) = e; 
and 

/f'(0 = e, 1 < { < x; 
therefore 



(x - 1)* + + n J 1; (x - I) 



Here 

so that the error made in using only four terms is 

4 = || (X - 1). 
If X - 1.1, 

e 1 - 1 = 

= 1.105166e 



13 INFINITE SERIES 43 

Thus, e 1 - 1 = 1.105166e with an error of (0.0001/24)e*. Since 
lies between 1 and 1.1 and since e* is an increasing function, e* is 
certainly less than e 2 . An approximate value of e 2 is 7, and the 
error is certainly less than 0.0007/24 = 0.00003. Therefore, 

e 1 - 1 = 1.1052e, 

correct to four decimal places. 

13. Evaluation of Definite Integrals by Means of Power 
Series. One of the most important applications of infinite series 
is their use in computing numerical values of definite integrals, 
such as f J e~ x * dx, in which the indefinite integral cannot be 
found in closed form. Moreover, the values of many tran- 
scendental functions are computed most easily by this method. 
Several examples of this use of infinite series are given in this 
section. 

Example 1. Consider 

^o+^ 

Since 

(1 + 2 )-l = 1 - Z -f 2 2 _ 

for 1 3 1 < 1, it follows that 



Example 2. Since 

dz 




if | z | < 1, therefore 



It is evident that this method of obtaining the expansion of sin" 1 x is 
much less complicated than the direct application of Taylor's formula. 
Example 3. In order to evaluate the integral 



f 

Jo 



2adx 



- x)(2ax - 



44 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 13 
express it as 



dx 



A _ jL^" 5 *, 
\ 2j 



\ 
and then replace ( 1 ^- j by its expansion in powers of ~ giving 



, 1-3-5/s' 

^2-4-6 \2a 

If this integral is expressed as 

CM / x \ dx 



== + 



fh I / x 

Jo 2\2a 



- ; 2 



2-4 
and each integral evaluated, there results 



+ 



] 



This expression gives the period of the simple pendulum. By making 
the change of variable x = h sin 2 <p, the integral reduces to 



/- 2^ J.U -*'**'*>-** 

where & 2 = V2a. 

This is the form used in the discussion of the simple pendulum given 
in Sec. 71. In this illustration, h denotes the height of the pendulum 
bob and a the length of the pendulum. 

/*! e x _ e -x 

Example 4. The integral J dx cannot be evaluated by the 

usual method for evaluating a definite integral, for the indefinite integral 

cannot be obtained. Moreover, the expansion for > if obtained 

x 

directly with the aid of Maclaurin's formula, would lead to an extremely 
complicated expression for each derivative. The expansion is most 
easily obtained by using the separate expansions for e* and e~*. Thus, 

e..i+s + 5! + |? + ... 

L\ o! 



13 INFINITE SERIES 45 

and 

/ z 3 x b 
e* - e- = 2 ^ + ^ j + ^ + 

Hence, 

fl P x e-x / 1 1 

Jo i * - 2 + F3l + 5-1! +- 2 ' 1145 ' 

Example 5. In order to evaluate the integral J Q e* * dx, recall that 



so that 



. . sin 2 a; , sin 3 x 
sm JP H -- 21 --- ' -- 3"! 



Then 

C w ' f" ( ' . , sin 2 a; , sin 3 a; , 
J e***dx = Jo (,1 + sm + ~2p + -3^ + 

TT 

, f 5 / . sin 2 x sin 3 a; , . \ , 

= 2 Jo I 1 + Sln * + ~2T + ~3T + ) **, 

which can be evaluated with the aid of the Wallis formula 



. J f^ j (^ - 1)(" - 3) 2 or 1 

sin n x dx L cos n x dx -- / - ^ - ^ - : - , 
Jo n(n 2) 2 or 1 ' 

where a = 1, if n is odd, and a = ^> if n is even. 

In order to justify the term-by-term integration, it is sufficient to 
show that the series in the integrand is uniformly convergent. That 
such is the case is obvious if one considers 

l + l + i\ + wi + 

as the Weierstrass M series. 

PROBLEMS 

1. Calculate cos 10, and estimate the maximum error committed 
by neglecting terms after x 6 . 

2. Find the interval of convergence of the expansion of e* in power 
series in x. Determine the number of terms necessary to compute e 1 - 1 
accurate to four decimal places from this expansion, and compare the 
result with illustration 2, Sec. 12. 

3. Compute sin 33, correct to four decimal places. 



46 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 13 

C x dx 

4. Expand the integrand of J 1 . g g in power series in x, and 

integrate term by term. Compare the result with that of Prob. l(d), 
Sec. 11. __ 

5. Compute v/35 = 2(1 + ?^2)^ correct to five decimal places. 

6. Develop the power series in x for sin" 1 x and hence establish that 



_ , . - - 

6 ~ 2 ~*~2 3 \2/ "" 2-4 5 \2 

7. Show, by squaring and adding the power series for sin # and 
cos x, that. 

sin 2 x + cos 2 #=1. 

8. Evaluate by using series expansions of the integrands 

/ v f 1 / ON , ,,x T H sin a; efo 

(a) J Q gin (x') dx; (6) J ==; 



(e) * cos (*) dx; (/) ^ (2 - cos x)~ dx 



T 1 
= Jo 

/*l log x f - 1 log(l - z) /** 

(?) Jo 9 r-r^ <* = Jo - z *; Jo 

/*x 

(i) I &***dx.* 

9. Show, by multiplication of series, that 

(1 + x + x 2 + ) 2 = 1 + 2x + 3z 2 + 

= (1 - x)- 2 . 
10. Expand to terms in x 6 

(a) \/cos a;; 
sin g 



(c) 



cos 



11. Determine the magnitude of a, if the error in the approximation 
sin a === a is not to exceed 1 per cent. 

a ~ sin a , . a 8 a 5 

Hint: = 0.01 and sma = a -^ + -=7 . 

a oi ol 

* See form 787, B. O, Peirce, A Short Table of Integrals. 



14 INFINITE SERIES 47 

14. Rectification of Ellipse. Elliptic Integrals. In spite of its 
importance and apparent simplicity, the problem of finding the 
length of an elliptical arc is not usually considered in elementary 
calculus. This is because the integral that arises is incapable 
of evaluation in terms of elementary functions. However, the 
evaluation can be effected by means of series expansion of the 
integrand function, as will be shown in this section. 

Let the equation of the ellipse be 

r 2 7v 2 

J> + f, = l, a>b. 
The length of arc from (0, 6) to (x\, y\) is given by the integral 



Computing dy/dx and substituting its value in (14-1) gives 



r /, _!_ b * ** A - r 

~ Jo \ + rf tf^T s<to ~ Jo 






Recalling the fact that the numerical eccentricity of the ellipse 
is k = -S/& 2 W/a, the integral given above can be written as 



where k 2 < 1. 
Let x = a sin 0; then dx = a cos 6 dd, and (14-2) becomes 



(14-3) 8 = a Vl - k 2 sin 2 6 dd. 

JQ 

The series expansion of the integrand function is most easily 
obtained by writing it as (1 fc 2 sin 2 0p and expanding by 
use of the binomial theorem. Then (14-3) is replaced by 

s = a J (l - ^k* sin 2 - g fc 4 sin 4 6 - ) d8, 

and term-by-term integration* gives 

* Term-by-term integration is valid here, for the series 



serves as a Weierstrass M series. 



48 MATHEMATICS FOR ENGINEERS AND PHYSICISTS &15 

(14-4) s = a L> - ^ k* I sin 2 dO - ^ ft 4 I %in 4 0d0 - 
L ^ J o Jo 

_ l-3.5.-.(2n-3) fc2M P sin2n 9dg _ . . . 1 
2 4 6 2w Jo J 

If (14-4) is used, it is possible to evaluate s for particular 
values of k and <p. However, the integral in (14-3) is so impor- 
tant that there are extensive tables* giving its value for* many 
choices of k and <p. This integral for the value of a 1 is 
called the elliptic integral of the second kind and is denoted by 
the symbol E(k, <p). If <p = 7r/2, the integral is called the com- 
plete elliptic integral of the second kind, which is denoted by the 
symbol E. 

The elliptic integral of the second kind having been defined, it 
seems desirable to mention the elliptic integral of the first kind, 
although the latter arises in considering the motion of a simple 
pendulum and will be discussed in more detail in Sec. 71. The 
elliptic integral of the first kind, F(k, ^>), has the form 

(14-5) F(k, v) = F , d 

Jo Vl - k 2 sin 2 S 

The complete elliptic integral of the first kind, which arises 
when <p = 7T/2, is denoted by the symbol K. Values of F(k y <p) 
and of K are also tabulated, but the evaluation can be obtained 
from (14-5) by means of series expansion of the integrand. 
Thus, one has the expansion 

(14-6) F(fc, ?) = <p + i fc 2 I sin 2 6 d6 + | fc 4 | sin 4 6 d6 

"JO o JO 



16. Discussion of Elliptic Integrals. The elliptic integral of 
the first kind is a function defined by the integral 

(15-1) F(k, *) m F d . =, fc 2 < L 

Jo v 1 - k 2 sin 2 S 

* See the brief table in B. O. Peirce, A Short Table of Integrals, pp. 
121-123. 



16 INFINITE SERIES 49 

If sin 6 is replaced by z, (15-1) becomes 

(15-2) f(k, x) = (* r dz k* < 1. 

' Jo V(l - * 2 )(1 - *V) 



This is an alternative form of the elliptic integral of the first kind. 
Similarly, the same change of variable transforms the integral 
of the second kind 

(15-3) E(k, *>) = I* ^/^^~k^n ^ e dO, fc 2 < 1, 

Jo 
into 

(15-4) i(k, x) = ^^f dz, k* < 1. 

It will be recalled that any integral of the type 

\ R(x, -\/ax 2 + bx + c) dXj 
where R is a rational function of the variables x and 



\/ax' 2 + bx + c, 

is integrablb in terms of the elementary functions, i.e., power, 
trigonometric, and logarithmic functions. It can be shown that 
the integration of integrals of the type 

(15-5) J R(x, V< 3 + bx'^~cx~+~d) dx 

and 

(15-6) J R(x, \/ax* + bx 3 + ex 2 +~dx + e) dx 

requires, in general, the introduction of new functions obtained 
from the elliptic integrals. 

The evaluation of (15-5) and (15-6) can be reduced to the 
evaluation of integrals of the elementary types and the following 
new types: 
a. Elliptic integral of the first kind : 

fvi N C x dz 

F(k ' x) m va - >(! - *y>- r 

de 



/ 

VI - fc 2 sin 2 6 
b. Elliptic integral of the second kind: 



r 

Jo 



2 

E(k, x) = _ dz, or 



= I * 
Jo 



- & sin 2 6 dO. 



MATHEMATICS FOR ENGINEERS AND PHYSICISTS 15 
c. Elliptic integral of the third kind: 

dz 



or 



fi(n,*,*) = r 

Jo 

H(n, t, *) s p 
Jo 



de 



/ 

(1 + n sin 2 (?) \/l ~ & 2 sin 2 



The problem of reducing the integrals of expressions involving 
square roots of cubics and quartics to normal forms is not difficult, 
but it is tedious and will be omitted here.* Integrals involving 

y 

2.0 



1.0 




FIG. 4. 

square roots of polynomials of degree higher than the fourth lead, 
in general, to more complicated functions, the so-called hyper- 
elliptic functions. 

The graphs of the integrands of the integrals of the first and second 
kinds are of some interest (see Fig. 4). For k = 0, 



A0 == 



- k 2 sin 2 6 and -: ss - 



- A; 2 sin 2 6 



both become equal to 1, and the corresponding integrals are both equal 

to <p. For < k < 1, the curve y = 1/A0 lies entirely above the line 

t/ = 1 and the curve y = A0 lies entirely below it. As ^> increases, 

* For a detailed account see Goursat-Hedrick, Mathematical Analysis, 
vol. 1, p. 226. A monograph, Elliptic Functions by H. Hancock, may also 
be consulted. 



10 INFINITE SERIES 51 

F(kj <p) and E(k, <p) increase continuously, F being always the larger. 
As k increases, <p being fixed, the value of F(k, <p)* increases and that of 
E(k, <p) decreases. Also F(k, TT) = 2K and E(k, TT) = 2E, for the curves 
are symmetrical about = Tr/2. If ir/2 < <p < IT, it is obvious from 
the figure that 

(15-7) 

Moreover, 

(15-8) F(k, mir + <p) - 2mK + F(k, <p), 

E(k, mir + <p>) = 2mE + E(k 9 <p), 

where in is an integer. 

Since the values of K and E, and of F(k, <p) and E(k, <p) for (p <> ir/2, 
are tabulated, the relations (15-7) and (15-8) permit the evaluation of 
F(k, <p) and E(k, <p) for all values of <p. 

The discussion* above was restricted to values of k 2 < 1. If & 2 = 1, 
y = A0 becomes y = |cos 0| and ?/ = 1/A0 becomes y = |sec 0|. 

Consider 



u = r _==*===. = 

Jo ^/(i - z 2 )(\. - k'W) Jo 



- z 2 )(\. - k'W) Jo -y/i - p sin 2 

where x = sin ^>. 

For a fixed value of A, (15-9) defines u = F(z) or u = Ffo>). The 
function resulting from the solution of (15-9) for <p in terms of u is 
called the amplitude of u and is denoted by am (u, mod &), or more 
simply by <p = am u. It will be assumed that the equation u = F(<p) 
can be solved for <p. Since <p am w, 

x = sin ^> = sin am u = sn u. 
Moreover, 

cos <p = \/l ~ ^ 2 s \/l sn 2 1* = en u. 
Finally, 



The functions sn u, en u, and dn w are called the elliptic functions. 
From the definitions, it is obvious that 

am (0) = 0, sn (0) = 0, en (0) = 1, dn (0) = 1; 
am ( u) = am u, sn (u) sn u, en ( u) = en u, dn ( u) 

= dn u. 

The elliptic functions are periodic functions and in some respects 
resemble the trigonometric functions. There^ exists a complete set of 

* See Prob. 1, at the end of this section. 




52 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 15 

formulas connecting the elliptic functions analogous to the set for the 
trigonometric functions.* 

An interesting application of elliptic integrals to electrical 
problems is found in the calculation of the magnetic flux density 
in the plane of a circular loop of radius a carrying a steady 
current /. 

Upon applying the law of Biot and Savartf to a circular loop of 
radius a, the flux density B at any point P in the plane of the 
wire is given by 

(15-10) B = L f "Mr,*)* 

v ' 4?r Jc r* 

where C is the circumference of the loop, r is the radius vector 

from P to an element of arc 
ds, and (r, s) is the angle 
between r and this element 
(Fig. 5). 

If the point P is at the 
center of the loop, then (r, s) 
= 90, r = a, and the integral 
is easily evaluated to give 




= _ 

4* a 2 ~ 2a 

a familiar result. 

If, however, the point P is 
not at the center, the evalua- 
FlG * 5 * tion of the integral is not so 

easy. Consider the triangle RQS, where the side RQ r dd 
makes an angle a with ds. It is clear that ds cos a = r dQ] and, 
since a = 90 (r, s), it follows that 

cos a = sin (r, s). 
Hence, 

, rdO 

ds = ~. - f ^ 
sin (r, s) 

* See APPBL, P., and E. LACOUB, Fonctions elliptiques; PEIRCE, B. O., A 
Short Table of Integrals; GBEENHILL, A. G., The Application of Elliptic 
Functions. 

t This formula is known to engineers as 'Ampere's formula. See, for 
example, E. Bennett, Introductory Electrodynamics for Engineers. The 
system of units used here is the "rational" system of units used in M. Mason 
and W. Weaver, Electromagnetic Field. 



15 INFINITE SERIES 53 

The substitution of this value in (15-10) yields 



for the magnetic flux density at P. 

Now, from triangle OQA, it is evident that 



\/a 2 - (r sin 0) 2 = r cos + A, 
which, after squaring both sides and simplifying, becomes 

r 2 + 2rh cos 6 + (/i 2 - a 2 ) = 0. 
Solving for r gives 



r = -/i cos S \A 2 cos 2 + a 2 - /i 2 ; 

and, since r is always positive, the radical must be taken with the 
positive sign. Substituting this value of r in (15-11) gives 



B = L r 2 * 

4*- Jo -h 



cos e + \/h 2 cos 2 B + a 2 - h 2 
or, upon rationalization of the denominator, 
D I C 2ir -h cos 0'- Vh 2 cos 2 + a 2 - h 2 ,. 

B = 5 Jo P"^ ^ 

i / r 2 - r 2 * , \ 

= A / o r^ ( I ^ cos d6 + I A/a 2 h 2 sin 2 c?0 )- 

4w(a 2 - /i 2 ) \ Jo Jo / 

The first of these integrals is zero, and the second is an elliptic 
integral of the second kind, so that 



In 

B = 



A ( 
47r(a 2 



n r 2v I /?2 

MN \ I 1 - ~2 sin 2 e de 
ft 2 ) Jo \ a 2 

I 
T /2 

= f 2 , 2 , 
7r(a 2 - h 2 ) Jo 



- fc 2 sin 2 dff, 



where k = h/a. This integral can be evaluated for any k with 
the aid of the tables of elliptic integrals. 

PROBLEMS 
1. Prove that 

da , ^ . 



/V 
Jo 



- I 2 sin 2 <p ^ o A/I - ^-2 S i n 2 a 
Hint: Change the variable by setting I 2 sin 2 ^ = sin 2 a. 



54 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 15 

2. Plot, with the aid of Peirce's tables, F(k, ^>), where k = sin a, 
using a as abscissa and F(k, <p) as ordinate. Draw 10 curves on the 
same sheet of rectangular coordinate paper for <p 0, <p = 10, <p == 20, 
> = 30, <f> = 40, y> = 50, , <p = 90. 

3. Plot four curves representing F(k, p) on the same sheet of rectan- 
gular coordinate paper. Use <p as abscissa and the values of k as 0, 

K, \/3/2, and 1. 

/tp dp 

4. Plot the integrand of I .- - for the values of k = 0, 

Jo -v/1 - k 2 sin 2 <p 

%, and 1. ' Use (p as abscissa. The areas under the curves give the 
values of the elliptic integrals. 

6. Compute the value of F(0, ir/2). 

6. The major and minor axes of an elliptical arch are 200 ft. and 
50 ft., respectively. Find the length of the arch. Compute the length 
of the arch between the points where x and x = 25. Use Peirce's 
tables. 

7. Plot with the aid of Peirce's tables E(k, <p), where k = sin a. 
Use a's as abscissas and E(k, <p) as ordinates. Draw 10 curves on the 
same sheet of rectangular coordinate paper for (p = 0, 10, 20, , 
90. 

8. Plot on a sheet of rectangular coordinate paper the four curves 
representing E(k, <p). Use <p as abscissa. The four curves are for k = 0, 
H, \/3/2, and 1. 

9. Plot the integrand of I \/l~-" k 2 sin 2 p dp for the values of 

/o 

k 0, H> and 1. Use v? as abscissa. Compare the result with that of 
Prob. 4. What can be said about the relative magnitudes of F(k, (p) 
and E(k,<p} ( ? 

I*<P d(D 

10. Show that I . is an elliptic integral of the first 

Jo -y/i 4- k 2 sin 2 <p 

kind. 

Hint: Change the variable by setting sin <p = T tan x. 

11. Show that 



dx 




o \/l 



cos x o \l H sm 

Hint: Set -\/cos x = cos ^>. 

Note that the integral is improper but that it is easy to show its 
convergence. 
12. Show that 

ain*gd9 



__ 

" 2 (K - 



16 INFINITE SERIES 55 

Hint: sin 2 = j 2 - ~ 2 (1 - k* sin 2 6). 
13. Show that 

# ^ fl ^ 
Jo 



- A; 2 sin 2 



14. Find the length of one arch of the sine curve. 
16. Find the length of the portion of y = sin x lying between # 
and x = 2. 
16. Given: 



- K sin 2 
Find K and sn %K. 

17. Show that I . where a > 1, is an elliptic integral. 

J va cos 

18. Show that the length of arc of an ellipse of semiaxes a and 6 is 
given by 



s = 4a f 2 Vl ~ e 2 sin 2 d 
Jo 

= 2ira (l Z "~ 54 g4 ""'*')> where e is the eccentricity. 

16. Approximate Formulas in Applied Mathematics. It is 

frequently necessary to introduce approximations in order to 
make readily usable the results of mathematical investigations. 
For example, an engineer seldom finds it necessary to use the 
exact formula for the curvature of a curve whose equation is 
y = f(x), namely, 

dfy 

fir 2 

(16-1) K = ax 



since in most applications the slope dy/dx is small enough to 
permit the use of the approximate formula 



2 



(16-2) K - |f 2 

Many such approximations are obtained by using the first few 
terms of the Taylor's series expansion in place of the function 



56 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 16 

itself. Thus, the formula (16-2) is obtained from (16-1) by 
neglecting all except the first term in the expansion of [1 + 
(dy/dx)*\~~^ in powers of dy/dx. 

I. Small Errors. The values of physical quantities determined 
by experiment are subject to errors due to inaccuracies arising in 
the measurements of the quantities involved. It is often neces- 
sary to know the size of such errors. 

Let a capillary tube contain a column of mercury. The radius 
R of the tube can be determined by measuring the length L and 
the weight W of the column of mercury. Let L be measured in 
centimeters and W in grams. Since the density of mercury is 
P = 13.6, ^ 

fl = J-^y = 0.153 J~ 
\TTpL \ L 

The principal error arises in the measurement of L. Let L be the 
true value, and let L' = L + be the observed value. Then, 
if R is the true value of the radius, let R' = R + 77 be the com- 
puted value. The error in measuring W is negligible because of 
the high accuracy of the balance. It follows that 

fw IW 

R = 0.153 J j- and R' = 0.153 Jj- f 

or 

R + TJ = 0.15 
Therefore, 








Since is small compared with L, it follows that 77 is approxi- 

mately given by K R T* Clearly, c can be either positive or 
Z Li 

negative. 

2. Crank and Connecting Rod. If one end of a straight line 
PQ (see Fig. 6) is required to move on a circle, while the other 



10 , - INFINITE SERIES 57 

end moves on a straight line which passes through the center of 
the circle, the resulting motion is called connecting-rod motion. 
This kind of motion arises in a steam engine in which one end of 
the connecting rod is attached to the crank PB and therefore 
moves in a circle whose radius is the length of the crank, while the 
other end is attached to the crosshead and moves along a straight 
line. 

Let r be the length of the crank, I the length of the connecting 
rod, and s the displacement of the crosshead from the position A, 

D ~ 




FIG. 6 



in which the connecting rod and crank lie in a straight line. 
Then, 

AB = I + r, 
and 

AB = AQ + QD + DB = s + I cos <p + r cos 0. 

Moreover, 

PI) = Z sin <p = r sin 0, 
so that 

sin ^> = j sin 
and 

r* r * * 

cos a? = ^ / 1 77 sm z 0. 
\ I 2 

Therefore, 



Y 1 - ^ sin 2 6>J 
: 1 - ( 1 - ^ sin 2 0J + r(l - cos 6). 

--..)" 



s + Z ( 1 - - 7? sin 2 ) + r cos = I + r 
or 

s 
If 



58 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 16 
be-replaced by its expansion, it follows that 

fir 2 1 AA 2 1 

s^lj^sin^ + i^j Sin 40 + . . . J +r(1 _ cos 0) 

= fc sin 2 6 + ^ sin 4 +) + r(l - cos 0). 

\&l Ol / 

If r is small compared with Z, the displacement of the crosshead 
is given approximately by r(l cos 0) . 

3. Surveying. In railroad surveying, it is fre- 
quently useful to know the amount of difference 
between the length of a circular arc and the 
length of its corresponding chord. 

Let r be the radius of curvature of the arc 
AB (Fig. 7), and let a be the angle intercepted 
by the arc. Then, if s is the length of the arc 

AB and c is the length of chord AB, s = ra and c = 2r sin ~- 
Since 

sin x = x ^- ? + ^-j cos , 

where < < x, the error in using only the first two terms of the 




FIG. 7. 



expansion is certainly less than 
with an error less than 



5! 



Then, 



Therefore, 



1920 



o-o- ,-,-,-_ _ 

with an error that is less than ra 6 /1920. 

4. Vertical Motion under Earth's Attraction. Let it be required 
to determine the velocity of a body of mass m that is falling from 
a height s above the center of the earth and is subject to the 
earth's attraction alone. 

Let F be the attraction on the earth's surface and F f be the 
attraction at a distance h from the surface (Fig. 8). Then 



16 



F = 



INFINITE SERIES 
kmm' , , kmm' 



59 



(f 



where m' is the mass of the earth, k is the gravitational constant, 
and r is the radius of the earth. Hence, 

F (r + h) 2 
F' r 2 

Also, let g be the acceleration at the surface of the 
earth and g' be the acceleration at a distance h above 
the surface, so that F = mg and F' = mg'. It 
follows that 



F' ~" "' 



g 



and, therefore, 



But 



a> = . 
9 s 2 




Fia. 8. 



so that 



dt 2 



This equation can be solved for v = ds/dt by the following 
device: Multiplying both members by 2 ds/dt and integrating 
give 



where C is the constant of integration. If the initial velocity 
(ds/dt) 8 ^ 8Q is zero, then C = 2gr 2 /s and hence 



But s = r + A and ds/eS = v, so that the equation becomes 



( J-r 

\r + 



This formula can be used to calculate the terminal velocity 
(i.e., the velocity at the earth's surface) when the body is released 



60 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 16 
from any height. Thus, setting h = gives 

(16-3) ^ = 2^(1-1). 

Upon denoting by ho the initial height above the earth's surface, 
so that S Q = r + ho, (16-3) can be written as 



or 

(16-4) v 2 = 2gr 




Now , , = ( 1 H - J ; and if < 1, then series expansion 
is permissible, so that 



I .f If/Q 

^ r + ho = 7 " 
Hence, if ho/r < 1, (16-4) can be replaced by 



Moreover, if & is very small compared with r, then the powers 
of ho/r higher than the first Can be neglected* and 

v 2 = 2gh Q , 

which is the familiar formula for the terminal velocity of a body 
falling freely from a height ho that is not too great. 

It follows from (16-3) that the square of the terminal velocity 
will be less than 20r 2 (l/r) = 2gr. Moreover, for large values of 
s<j the terminal velocity will be very close to \/2gr. Accordingly, 
if a body falls from a very great distance it would attain a ter- 
minal velocity (air resistance being neglected) of approximately 

\/2gr = 6.95 miles per second. 

The results stated in the last paragraph may receive a different 
interpretation. Suppose a body were projected outward from 
the earth's surface with a velocity of more than \/2gr = 6.95 
miles per second. The previous discussion shows that, if air 

* Since the series is alternating, the error will be less than 2gr(h Q /r)*. 



16 INFINITE SERIES 61 

resistance is neglected, the body would travel an infinite dis- 
tance. This velocity is called the critical velocity or the velocity 
of escape. 

It may be recalled that the earth's rotation exerts a centrifugal 
force on a particle which is falling toward the earth and that this 
force diminishes the effect of the force due to the earth's attrac- 
tion. For a particle of mass m on the surface of the earth at the 
equator, this centrifugal force is 

mv 2 mw 2 r 2 mq , 

_ = ___ = ww - r = ___ dynes, 

where o> = 0.00007292 radian per second is the angular velocity 
of the earth, r = 6,370,284 m., and g = 980 cm. per second per 
second. At a distance s from the center of the earth, this force is 

o mas 

m s = m-' 

But the earth's attraction at this distance is F = mg f . Since 
9 f = 0r 2 M 

_ mgr* 
* ~ s 2 ' 

If the particle is to be in equilibrium, 

mgs _ mgr 2 
289r ~ ~s2~' 
so that 

s 3 = 289r 3 or s = 6.6r = 26,000 miles approx. 

Thus, if all other forces are neglected, a particle would be in 
equilibrium at approximately 22,000 miles above the earth's 
surface. This gives a very rough approximation to the extent 
of the earth's atmosphere. The actual thickness of the atmos- 
pheric layer is supposed to be considerably smaller. 

PROBLEMS 

1. The mass of the moon is nearly one-eighty-first that of the earth, 
and its radius is approximately three-elevenths that of the earth. 
Determine the velocity of escape for a body projected from the moon. 
Acceleration of gravity on the surface of the moon is one-sixth that on 
the surface of the earth. 



62 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 16 

2. Show that the time required for a body to reach the surface of the 
earth in Illustration 4, Sec. 16, is 



Hint: 

ds 



3. If the earth is considered as a homogeneous sphere at rest, then 
the force of attraction on a particle within the sphere can be shown to 
be proportional to the distance of the particle from the center. Let a 
hole be bored through the center of the earth, the air exhausted, and 
a stone released from rest at the surface of the earth. Show that the 
velocity of the stone at the center of the earth is about 5 miles per 
second. 

Hint: 

d*s mg 

"&---> 

where 8 is the distance of the stone from the center of the earth and r 
is the radius of the earth. 



CHAPTER III 
SOLUTION OF EQUATIONS 

Students of engineering, physics, chemistry, and other sciences 
meet the problem of the solution of equations at every stage of 
their work. This chapter gives a brief outline of some of the 
algebraic, graphical, and numerical methods of obtaining the 
real roots of equations with real coefficients, of types that occur 
frequently in the applied sciences. It also contains a short 
summary of those parts of the theory of determinants and the 
theory of matrices that are immediately applicable to the solution 
of systems of linear equations. 

26. Graphical Solutions. The subject of the solution of equa- 
tions will be introduced by considering a simple problem that 
any engineer may be called upon to solve. 

It is required to design a hollow cast-iron sphere, 1 in. in 
thickness, that will just float in water. It is assumed that the 
air in the cavity is completely exhausted. The specific gravity 
of cast iron will be denoted by p, for convenience. 

By the law of Archimedes, the weight of the sphere must 
equal the weight of the displaced water. This gives the con- 
dition on the radius of the sphere, namely, 



- (x - I) 3 ]. 
Simplifying gives 



(25-1) x 3 - Spx* + 3pz - p = 0. 

It will be convenient to remove the second-degree term in (25-1). 
To accomplish this, let x = y + k, giving 

Zyk* + & - 3 P (i/ 2 + 2yk + fc) + 3 P (y + k) - /> - 0, 



or 

y 8 -f (3fc - 3p)i/ 2 + (3fc 2 - Qpk + 3p)y + W - 3pfc 2 4- 3pfc - p - 0. 
Choosing k = p makes the equation reduce to 

(25-2) 2/ 8 + (3p ~ 3 P 2 )2/ - 2 P 3 + 3p 2 - p = 0. 

83 



84 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 26 
For cast iron, p = 7.5, and (25-2) becomes 
(25-3) y* - 146.257/ - 682.5 = 0. 

If (25-3) is solved, the solution of (25-1) is also determined, since 
x = y + 7.5. 

A graphical method of solution will be used. The solution 
of (25-3) is equivalent to the simultaneous solution of the system 

(25-4) (* = 2/ 3 > 

v ' \z = 146.25*/ + 682.5. 

The accompanying figure (Fig. 16) represents the graphs of the 
two functions of (25-4); since they inter- 
sect at y = 14.0, this value gives an 
approximate solution of (25-3). The cor- 
responding solution of (25-1) is x = 21.5. 
From the graph, it is clear that there is 
only one real solution of (25-4) and hence 



6825, 




of (25-3). 



M0 This graphical method can be applied to 

Fia - 16> any cubic equation. The general fourth- 

degree equation (quartic) can also be reduced to a form that is 
convenient for graphical methods of solution. 
Consider the quartic 

x 4 + ax 3 + bx* + ex + d = 0. 

Let x = y + k, as in the cubic equation. This substitution 
gives 

+ ?/(4fc 3 + 3ak 2 + 2bk + c) + fc 4 + ak* + bk 2 + ck + d = 0. 

In order to remove the term in y 3 , choose k = -T- This reduces 
the equation to the form 

2/ 4 + Ay* + By + C = 0. 

If A > 0, the further transformation y \/A z is made, and the 
equation is reduced to 



or 



u 4 + AW + B VA z + C = 0, 
s 4 + ** + P* + q = 0. 



26 SOLUTION OF EQUATIONS 85 

The solutions of this equation are the same as the solutions of 
the simultaneous system 

u = z 4 + z 2 , 
u = pz q. 

The graphs of these two functions are easily plotted, and the 
solutions can be read from the graph. In case A < 0, the 
transformation would be y = \/ A z, which leads to the equation 

z 4 - z 2 + pz + q = 
and the graphical solution of the system 



u = 

w = pz q. 

This method of solution for the real roots of an equation is 
also applicable to many transcendental equations. In order to 
solve 

Ax B sin x = 0, 
write it as 

ax sin x 0, 

and plot the curves of the simultaneous system 

i y = sin x, 

y = ox. 
Similarly, the equation 

a* ~ X 2 = o 

can be solved graphically by plotting the curves of the equivalent 
simultaneous system 

y = o*, 

y = x 2 . 

PROBLEMS 

1. Solve graphically 

(a) 2* - x* = 0, 

(5) a? 4 - x - 1 = 0, 

(c) x 6 - Z - 0.5 = 0, 

(d) e* + x = 0. 

2. Find, graphically, the root of 

tan x x = 
nearest %TT. 



86 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 26 

26. Algebraic Solution of Cubic. The graphical method of solution 
is perfectly general, but its accuracy depends upon the accurate con- 
struction of the graphs of the equations in the simultaneous' systems. 
This is often extremely laborious and, at most, yields only an approxi- 
mate value of the roots. 

In the case of the linear equation ax + b 0, where a 5^- 0, the solu- 
tion is x = b/a. For the quadratic equation ax 2 + bx + c = 0, 

b \/6 2 4ac 



where a 7* 0, there are two solutions given byx 

The question naturally arises as to the possibility of obtaining expres- 
sions for the roots of ajgebraic equations of degree higher than 2. 
This section will be devoted to a derivation of the -solutions of the 
general cubic equation . 

dox* + aix 2 + aw + a 3 = 0, * a 5^ 0. 
Dividing through by ao gives 
(26-1) z 3 + bx 2 +^cx + d = 0, 

and the x 2 term can be removed by making the change of variable 

6 

f tii .. . 

* ~ J 3 
The resulting equation is 

(26-2) y + py + q = 0, 

where 

b 2 
p = c - T 

and 

. be , 26 3 
- d --3 + 27- 

In order to solve (26-2), assume that 
(26-3) y = A + B, 

so that 

?/ = A 3 + B* + 3AB(A + B). 

Substitute in this last equation for A + B, from (26-3), and there is 
obtained the equation 

(26-4) ?/ - 3ABy - (A 3 + 3 ) = 0. 

A comparison of (26-4) with (26-2) shows that 

SAB = -p and A 3 + B 3 = -3, 



SOLUTION OF EQUATIONS 



87 



or 
(26-5) 



A 8 3 = - %= and A 3 + B 9 = -$. 



If B 3 is eliminated by substituting from the second of Eqs. (26-5) into 
the first, there appears the quadratic equation in A 8 , 



whose roots are 



-q 



The solution for B z yields precisely the same values. However, in 
order to satisfy Eq. (26-5), choose* 



(26-6) 



B* = 



If the values of y are to be determined from (26-3), it is necessary to 
find the cube roots of A 3 and j5 3 . Recall that if x z = a 3 , then the solu- 

tions for x are given by a, coa, and o> 2 a, where co = ^ H o~ i 



o 2 = o -- o~ ^ are 



complex roots of unity. Hence, if one cube 

root of A 3 be denoted by a and one cube root of B 3 by , the cube roots 

of A z are 

a, coa, and w 2 a, 

whereas those of J5 3 are 

|9, cop, and co 2 0. 

It would appear that there are nine choices for /, but it should be 
remembered that the values must be paired so that SAB = p. The 
only pairs that satisfy this condition are a and /3, coa and co' 2 j0, and o> 2 a 
and wjS, Hence, the values of y are 

(26-7) yi = a + ft 2/2 = coa + a> 2 ft 2/3 = co 2 a + coft 
where 



and 



* The opposite choice for the values of A 8 and J5 3 simply interchanges 
their role in what follows. 





88 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 26 



The solutions of (26-1) can be obtained from the values given in 
(26-7) by recalling that x = y - 6/3. 

The expressions for a and are quite complicated, and when the 
quantity under the square-root sign has a negative value the values of 
a and ft cannot, in general, be determined. This is the so-called 
irreducible case of the cubic, which can, however, be solved by using a 
trigonometric method. This method will be 
described later in the section, but first it is im- 
portant to find a criterion that will determine 
in advance which method should be used. 

In order to determine the character of the 
.y roots of (26-2), whose coefficients are assumed 
to be real, consider the function 



O 



f(y) == y3 + py + q 

FIG. 17. and its maximum and minimum values. 

f(y) = 3i/ 2 + p, 



Since 



it appears that, if p > 0, then f'(y) is always positive and f(y) is an 
increasing function. In this case the graph of f(y) has the form shown 
in Fig. 17, and there is evidently only one real value for which f(y) = 0. 
If p < 0, however, f(y) is zero when y = \/-p/3. Since 
f"(y) = 6i/, it follows that y = + V- P/3 gives a minimum value to 
f(y), whereas y = V' p/S furnishes a maximum value. The cor- 
responding values of f(y) are 



ffyr 




and 



The graph of f(y) will have the 
appearance of one of the curves in 
Fig. 18. 

It is evident that f(y) = will 
have only one real root if the graph 
of f(y) has the appearance shown by (1) or (5), that is, if the maximum 
and minimum values of f(y) are of the same sign. Hence, 



or 



FIG. 18. 



0, 



26 SOLUTION OF EQUATIONS .89 

is the condition that (26-2) have only one real root. It may be observed 
that this condition is automatically satisfied if p ^ 0. It should be 
noted that, if p = 0, Eq. (26-2) reduces to y* + q = 0, which obviously 
has only one real root. 

If (26-2) has three real and distinct roots, then the graph of f(y) must 
have the appearance shown in (3), and it follows that the maximum 
and minimum values must be of opposite sign. Hence, 

q 2 + 7 P 3 < 

is the condition for three real and unequal roots. 

If <f + $iiP* = 0, either the maximum or the minimum value of 
f(y) must be zero [see (2) and (4)], and (26-2) will have three real roots, 
of which two will be equal (a so-called double root). 

The expression 

(26-8) A s -27g 2 - 4p 3 

is called the discriminant of the cubic equation (26-2), for its value 
determines the character of the roots of the equation. The discriminant 
for (26-1), obtained by replacing p and q in (26-8) by their values in 
terms of 6, c, and d, is 

(26-9) A s ISbcd - 46 3 d + & 2 c 2 '- 4c 3 - 27d\ 

It may be worth noting that the discriminant of any algebraic equa- 
tion, with leading coefficient unity, is the product of the squares of 
the differences of the roots taken two at a time. Inasmuch as 

(xi - X 2 )*(x 2 - X 3 y(x 3 - XiY = (yi - 2/ 2 ) 2 (?/ 2 - 2/ 3 ) 2 (2/ 3 - 2/i) 2 , 

the discriminant has the same value for (26-1) and (26-2). 
In view of the definition of A, it follows that 

if A < 0, one root is real and two are complex; 
if A = 0, all the roots are real and two are equal; 
if A > 0, the three roots are real and unequal. 

Example. Consider the cubic equation 

x* + 3x* + 9z - 1 = 0. 

From (26-9), it follows that A = 2592, and hence there will be one real 
root and two complex roots. Setting x = y 1 yields the reduced cubic 

2/* + 6y - 8 = 0, 

and substituting p = 6 and q = 8 in (26-6) gives A 3 =4 +2 ^/Q and 
B 3 = 4 2 V6. Therefore, the solutions for y are 

^4 + 2 \/6 4- ^4 - 2 \/6, ^4 + 2 \/6 + w 2 ^4 - 2 v/6, 

and w 2 ^4 +2\/6 + w ^4-2 \/6. 



90. MATHEMATICS FOR ENGINEERS AND PHYSICISTS 26 
The solutions of the original equation can now be obtained by recalling that 

gW y - 1. 

The discussion of the solution of the cubic* equation "will be con- 
cluded by giving the derivation of the expressions for the roots in the 
case where the roots are real and unequal (that is, when A = 27 q 2 
- 4p 3 > 0). 

Let 

~ = r cos 
Jj 

and 



- + -"* 

Then* 

(A Q\ 

cos - + i sin - ) 
o o/ 

and 

(n nv 

cos - i sin - V 

If it is noted that 



and 



o 

2 



2?r . . . 2ir 

= cos ~ + i sm ~ 

o o 



27T . . 27T 



o 

co 2 = cos -r- i sm 



3 3 

it is easily checked that the expressions for 

2/i == a + 0, 7/2 = wo: + O) 2 j8, 2/3 
become 

(26-10) ?/i - 2rH cos 1 y a = 2rH cos ^ 

o 



COS 

Since 



'-v-fr 

and 

cos 8 = 

the values of 2/1, 2/2, and 2/3 can be obtained directly from the coeffi- 
cients of (26-2) or (26-1). 

* By De Moivre's theorem (cos -f- i sin 0)" = cos nO + i sin n0. 



26 SOLUTION OF EQUATIONS 91 

Example. Determine the real roots of 

x* - 3ic 2 + 3 0. 
Here 

A fc= -4(-27)(3) -27(9) > 0, 

and the roots are all real and unequal. Since p 3 and q *= 1, it follows 
that r - 1 and cos = >. Hence, 



and 



*27r Sir n STT 

i/i=2 cos > 1/2=2 cos > 2/3 = 2 cos -r- 
y y 7 



The solutions of the general quartic equation 

z 4 + kr 3 + ex* + cfc + e = 

can be found, but the methods of obtaining the expressions for the 
roots depend upon the solution of an auxiliary cubic equation. More- 
over, these expressions are, in general, so involved that they are prac- 
tically useless for computation.* It has been shown that the ordinary 
operations of algebra are, in general, insufficient for the purpose of 
obtaining ex'act solutions of algebraic equations of degree higher than 4. 
However, it is possible to obtain the expressions for the solutions of the 
general equation of the fifth degree with the aid of elliptic integrals. 

The reader should not confuse the problem of obtaining expressions 
for the exact solutions of the general algebraic equation with that of 
calculating numerical approximations to the roots of specific equations 
which have numerical coefficients. The latter problem will be dis- 
cussed in Sees. 28 and 29, and it will be shown that the real roots of 
such equations can be computed to any desired degree of accuracy. 
Moreover, if the roots are rational they can always be determined 
exactly. 

PROBLEMS 

Determine the roots of the following equations: 

(a) ?/ 3 - 2y - 1 = 0; 

(6) 7/ - 146.25y - 682.5 = 0; 

(c) x s - x 2 - 5x - 3 = 0; 

(d) x* - 2x* - x + 2 = 0; 

(e) x* - 6z 2 + 6z - 2 = 0; 
(/) x 3 + 6z 2 + 3x + 18 = 0; 
(0) 2x* + 3z 2 + 3s + 1 = 0. 

* See DICKSON, L. E., First Course in Theory of Equations, pp. 50-54; 
BURNSIDE, W. S., and A. W. PANTON, Theory of Equations, vol. 1, pp. 121-142. 



92 MATHEMATICS FOR ENGINEERS AND PHYSICISTS &27 

27. Some Algebraic Theorems. The student of any applied 
science is usually interested in obtaining numerical values, 
correct to a certain number of decimal places, for the roots of 
equations. Unless the roots are rational, the expressions for the 
exact roots, provided that they can be found at all, are usually 
complicated and the process of determining numerical values 
from them is tedious. Accordingly, it is distinctly useful to 
consider other methods of finding these numerical values. 
Plorner's method, Newton's method, and the method of inter- 
polation are the ones most frequently used; they will be dis- 
cussed in Sees. 28 and 29. However, all thc k sc methods arc 
based on the assumption that a root has first been isolated, that 
is, that there have been determined two values of the variable 
such that between them lies one and only one root. In many 
practical problems the physical setup is a guide in this isolation 
process. This section contains a review of some theorems* from 
the theory of equations that provide preliminary information as 
to the character and location of the roots. 

THEOREM 1 (Fundamental Theorem of Algebra.) Every 
algebraic equation 

f(x) == ax n + air"- 1 + + a n -iz + a n = 

has a root. 

It should be noted that this theorem does not hold for non- 
algebraic equations. For example, the equation e x has 
no root. 

THEOREM 2. (Remainder Theorem.) // the polynomial 

f(x) E= a Q x n + aix n ~ l ++ a n -ix + a n 

is divided by x I) until the remainder is independent of x, then 
this remainder has the value f(b). 

THEOREM 3. (Factor Theorem.) // f(b) = 0, then x b is 
a factor of the polynomial /(x) and b is a root off(x) = 0. 

This theorem follows directly from Theorem 2. In many 
cases the easiest way to compute the value of f(b) is to perform 
the division of f(x) by x b. This is a particularly useful 

* Those students who are not already familiar with these theorems and 
their proofs will benefit by referring to H. B. Fine, College Algebra, pp. 
425-453, and L. E. Dickson, First Course in the Theory of Equations, Chap. 
II. 



27 SOLUTION OF EQUATIONS 93 

method when the factor theorem is being used for the purpose of 
determining the roots of f(x) = 0. For if x b is a factor of 
/Or), it follows that f(x) = (# &) g(x), where g(x) is a poly- 
nomial of degree one less than that of /(#). Obviously the roots 
of g(x) = will be the remaining roots of f(jr) 0, so that only 
g(x) = need be considered in attempting to find these roots. 
Moreover, when /(>) is divided by x b the quotient is g(x). 
If synthetic division is used, the computation is usually quite 
simple. 

Example. If }(x) = z 3 + 2x 2 + 2x + 1 is divided by x + 1, the 
quotient is x 2 + x + 1 and the remainder is zero. Hence, x 1 is 
a root of f(x) and the remaining roots are determined by solving 
X 2 + x + i = o. 

THEOREM 4. Every algebraic equation of degree n has exactly 
n roots if a root of multiplicity m is counted as m roots. 

A root b of f(x) = is said to be a root of multiplicity m if 
(x b) m is a factor of /(x) but (x b) m+1 is not a factor of f(x). 

It follows from Theorems 2 and 4 that the polynomial of 
degree n can be factored into n linear factors, so that 

f(x) = a<>x n + aix n ~ l + ' + dn-ix + a n 
= a (x xi)(x x 2 ) (x x n ). 

THEOREM 5. '// 

f(x) = a x n + aix n ~ l + + a n ^x + a n 

has integral coefficients and if f(x) = has the rational root b/c, 
where b and c are integers without a common divisor, then b is an 
exact divisor of a n and c is an exact divisor of a . 

Example. Consider the equation 

f(x) = 2x* + x 2 + x - 1 = 0. 

The only possible rational roots are 1 and J. Since /(I) = 3, 
/(-I) = -3, f(- l A] = -%, and/(K) = 0, it follows that K is the 
only rational root. As a matter of fact, if f(x) is divided by x % the 
quotient is 2x 2 + 2x + 2 whose factors are 2, x co, and x o> 2 , 
where w and co 2 are the complex roots of unity.* 

THEOREM 6. Given f(x) = a; n + ^i^"" 1 + + a n ~ix + a n 
= 0. Iff(a) andf(b) are of opposite sign, then there exists at least 
* See Sec. 26 and the example following Theorem 9 of this section. 



94 MATHEMATICS FOR ENGINEERS AND PHYSICISTS ^27 

one root off(x) = between a and b. Moreover, the number of such 
roots is odd. 

Graphically this means that y = f(x) must cross the :c-axis 
an odd number of times between a and b. 

Example. If f(x) s 8x* - I2x 2 - 2x + 3 = 0, 

/(-I) = -15, /(O) = 3; /(I) - -3, /(2) = 15. 

Since /( I) is negative and /(O) is positive, there is at least one root 
between 1 and 0. Similarly, there is a root between and 1, and 
another between 1 and 2. 

THEOREM 7. (Descartes' Rule of Signs.) The number of 
positive real roots of an algebraic equation f(x) with real coeffi- 
cients is either equal to the number of variations in sign of f(x) or less 
than that number by a positive even integer. The number of negative 
real roots of f(x) = is either equal to the number of variations in 
sign of f( x) or less than that number by a positive even integer. 

Example. f(x) = 8# 3 I2x 2 2x + 3 has two changes in sign, and 
therefore there are either two or no positive roots of f(x) = 0. Also, 
f( x) s &c 3 I2x 2 + 2x + 3 has only one change in sign, and/(#) 
must have one negative root. 

THEOREM 8. Every algebraic equation of odd degree, with real 
coefficients, and leading coefficient positive has at least one real root 
whose sign is opposite to that of the constant term. 

Example. Since f(x) s= 8x* 12z 2 2x + 3 = is of odd degree 
and the constant term is positive, it follows that there must be at least 
one negative root. 

THEOREM 9. // an algebraic equation f(x) = with real coeffi- 
cients has a root a + bi, where b 9^ 0, and a and b are real, it also has 
the root a bi. 

Example. Thus, x* 1 = has the root M + K \/3 1, and there- 
fore it has the root % J \/3 i. This theorem states that imaginary 
roots always occur in pairs. 

PROBLEMS 

1. Find all the roots of the following equations: 

(a) x s + 2x* - 4z - 8 = 0; 
(6) 2x* -x 2 - 5x - 2 = 0; 

(c) 4z 4 + 4z 3 + 3z 2 - x - 1 = 0; 

(d) 2z 4 - 3z 3 - 3x - 2 0. 



28 SOLUTION OF EQUATIONS 95 

2. Isolate the roots of the following equations between consecutive 
integers : 

(a) x 3 - 2x 2 - x + 1 = 0; 
(6) 2s 3 + 4z 2 - 2x - 3 = 0; 

(c) x 3 -f 5z 2 + to + 1 = 0; 

(d) x 4 - 5z 2 + 3 = 0. 

28. Hornet's Method. Many readers are already familiar with 
Horncr's method of determining the value, to any desired number of 
decimal places, of the real roots of algebraic equations. However, the 
development given here is somewhat different from that used in the 
texts on algebra, in that it depends on Taylor's scries expansion. 

Suppose that the equation is 

(28-1) f(x) =s a x n + ttix"- 1 + + (in^x + a n = 

and that it is known that the equation has a root between c and c + 1 , 
where c is an integer. If f(x) is expanded in Taylor's series in powers of 
x c, there will result* a polynomial in x r, namely, 

i"(c\ 
/ +f'(e)(x - c) + J - (x - c) + 



Now, let x c = Xi and p~ == A ft _ r . Then (28-1) is replaced by 

(28-2) / t (xi) - A n + An-iX! + - + AiX!- 1 + Aox, - 0. 

Since (28-1) had a root between c and c + 1 and since Xi = x c, 
it is evident that (28-2) has a root between and 1. By the use of 
Theorem 6, Sec 27, this root can be isolated between d and d + 0.1, 
where d has the form a/ 10 and < a < 9. Moreover, /i(xj = f(x\ -f c) ; 
and it follows, that, if f\ has a root between d and d -f- 0.1, then / has a 
root between c + d and c + d + 0.1. It should be noted that c may 
be negative but that d will always be positive or zero. 

The function /i(^i) can be expanded in Taylor's series in powers 'of 
x\ d] and, if # 2 = #1 d, there will be obtained an equation 



/ 2 (a? 2 ) = B n + B n ^xt + ' + JW- 1 + Boxf = 0. 

But /i (0*1) = had a loot between d and d + 0.1 ; and since xz = x\ d, 
U follows that/2(x 2 ) = will have a root between and 0.1. 

This process can be continued as long as desired, each step deter- 
mining another decimal place of the root of the original equation (28-1). 

* Since /(#) is a polynomial of the nth degree, the derivatives of ordor 
higher than n are all zero. 



96 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 28 

The solution of a specific equation may help to clarify the procedure. 
Let it be required to find the values of the real roots of the equation 

F(x) ss x 4 + x* - 3z 2 - 6x - 3 = 0. 

Since there is only one variation in sign, F(x) has at most one positive 
root. F(x) has three variations, and so there will be at most three 
negative roots. The only possibilities for rational roots are 1 and 
3. Since F( 1) = 0, it follows that x = 1 is a root. Moreover, 
if F(x) is divided by x + 1, the quotient is f(x) = x 3 3x 3. Hence, 
the remaining roots of F(x) = are the three roots of 

/(x) = x s - 3x - 3 = 0. 

It is easily checked that /(x) = has no rational roots. Also, 
A = 108 243, so that'there is only one real root which, since jf(2) = 1 
and /(3) = 15, must lie between 2 and 3. Therefore, f(x) will be 
expanded in powers of x 2. Since 

/(x) = x*- 3* -3, /(2) = -1, 

f(x) = 3x 2 - 3, /' (2) - 9, 

/"(x) = Gx, /"(2) = 12, 

/'"(x) = 6, /'"(2) = 6, 

the expansion becomes 

/(x) = -1 + 9(x - 2) + ly (a; - 2) 2 + |j (x - 

Replacing a? 2 by x\ gives 

/ 1 (a? l ) s -1 + 9si + 6xi 2 + xi 3 = 0. 

Since the real root of this equation lies between and 1, the #i 2 and Xi 3 
terms do not contribute very much to the value of f\(xi). Hence, a 
first approximation to the root can be obtained by setting 9#i 1 = 0. 
This gives x\ % =0.111 , and suggests that the root probably 
lies between 0.1 and 0.2. It is easy to show that /i (0.1) = 0.039 and 
/i (0.2) = 1.048; there is thus a root between 0.1 and 0.2, and it is 
evidently closer to 0.1. Therefore, f(x) = has a root between 2.1 
and 2.2. 

Expanding /i Co; i) in powers of Xi 0.1 gives 

jfi(si) = -0.039 + 10.23(*i - 0.1) + ! |y (xi - O.I) 2 + (x, - 0.1)', 

and replacing x\ 0.1 by x 2 yields 

/ 2 (*2) = -0.039 + 10.23^2 + 6.3X2 2 + $2* - 0. 

Now 10.23x2 0.039 = gives the approximation x = 0.0038, and 
testing 0.003 and 0.004 reveals that / 2 (0.003) = -0.008253273 and 



29 SOLUTION OF EQUATIONS 97 

/ 2 (0.004) = + 0.002020864. Thus, the root lies between 0.003 and 0.004 
and is closer to 0.004. If it is desired to determine the root of f(x) = 
to three decimal places only, this value will be 2.104. If more decimal 
places are desired, the process can be continued. It should be noted 
that in each succeeding step the terms of the second and third degree 
contribute less, so that the linear approximation becomes better. 

PROBLEMS 

1. Apply Horner's method to find the cube root of 25, correct to three 
decimal places 

2. Determine the real roots of 3 2x 1 = by Horner's method. 

3. Determine the root of x 4 + x 3 7x 2 3 + 5 = 0, which lies 
bet ween 2 and 3. 

4. Determine the real root of 2x 3 '3x 2 + # 1 0. 

5. Determine the roots of x 3 3# 2 + 3 = 0. 

6. Find, correct to three decimal places, the value of the root oi 
x 5 + 3# 3 2x 2 + x + 1 =0, which lies between 1 and 0. 

7. A sphere 2 ft. in diameter is formed of wood whose specific gravity 
is M. Find to three significant figures the depth h to which the sphere 

will sink in water. The volume of a spherical segment isTT/i 2 ( r ~ Y 

The volume of the submerged segment is equal to the volume of the 
displaced water, which must weigh as much as the sphere. Since water 
weighs 62 5 Ib. per cubic foot, 



and, since r 1, 

/i 3 - 3A 2 + % = 0. 

29. Newton's Method. Horncr's method of obtaining a 
numerical solution of an equation is probably the most useful 
scheme for solving algebraic equations, but 
it is not applicable to trigonometric, ex- 
ponential, or logarithmic equations. A 
method applicable to these types as well as 
to algebraic equations was developed by Sir 
Isaac Newton sometime before 1676. 

Newton applied his method to an alge- 
braic equation, but it will be introduced 
here in the solution of a problem involving FlG 19 - 

a trigonometric function., 

Let it be required to find the angle subtended at the center of 
a circle by an arc whose length is double the length of its chord 




98 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 29 



(Fig. 19). Let the arc BCA be an arc of length 2BA. Let 2x 
be the angle (measured in radians) subtended at the center of the 
circle. Then, arc BCA = 2xr and BA = 2 DA = 2r sin x. If 
arc BCA = 2BA, then 2xr = 4r sin x, or rr 2 sin # = 0. 

The graphical solution of equations 
of this type was discussed in Sec. 25. 
A first approximation can be obtained 
by graphical means. If y = x and 
?/ = 2 sin x are plotted, it appears 
from the graph (Fig. 20) that they 
intersect for x lying between 108 and 
109, or, expressing this in radians, 




x -?$//? x 



FIG. 20. 



1.8850 < x < 1.9024. 



If xi = 1.8850 be chosen as the first approximation, the question 
of improving this value will be discussed first from the following 
graphical considerations. 

If the part of the curve y = x 2 sin x in the vicinity of the 
root be drawn on a large scale, it will have the appearance shown 
in Fig. 21. It is clear from the graph that adding to x\ the 





FIG. 21. 



distance AE, cut off by the tangent line to the curve at x\ = 
'1.8850, will give a value x% which is a better approximation to the 
actual root XQ. But AE is the subtangent at Xi and is equal to 

where /(x) = x 2 sin x. Thus,* 

Xz Xl _ /fr).. 

* See, in this connection, Prob. 8, at the end of this section. 



29 SOLUTION OF EQUATIONS 99 

Similarly, upon using 2 as the second approximation and 

observing that ff \ . is the subtangent EF, the third approxi- 



/(**) 



mation is found to bo 



and in general the nth approximation x n is given by 
(29-1) x n = Xn-.! - (r^r~y (n = 2, 3, ) 

Since x\ = 1.885, the formula gives 

_ _ f( x *) . _ - ri ~~ ^ s * n Xl 
f'(%i) l 12 cos 0:1 

= 1.8850 - ^-^ 2 = 1.8956. 
In a similar way, 



L895(i " 2 sin h895G - 1 8955 

1.8955. 



f(x) 



It follows that the angle subtended by the arc is 3.7910 radians. 
The use of Newton's method requires some preliminary 
examination of the equation. It may happen that the equation 
is of such a character that the second approximation to XQ will be 
worse than the first. A careful examination of the following 
sketches of four types of functions, 
sketched in the vicinity of their 
roots, reveals the fact that some 
care must be exercised in applying 
Newton's method. For all four 
figures, it is assumed that x has 
been isolated between Xi and x[. 
The graphical interpretation of the 
f(xi) 




correction 



as the subtan- 




Fio. 22. 



gent must be kept in mind throughout this discussion. If x\ is 
used as the first approximation, then x 2 will be obtained as the 
second approximation by using Newton's method; if x( is used, 
then 2 will be obtained. 

In Fig. 22, both x% and x'% are closer to x than x\ or x{. In this 
case the method would work regardless of which value is chosen 



100 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 2$ 



as the first approximation. In Fig. 23, x 2 is better than the 
first approximation xi, but x' 2 is worse than x{. It appears from 





FIG. 23. 

the figure that this occurs because the curve is concave down 
between xi and x[, and hence f"(x) < 0, whereas f(x i) < and 

f(x{) > 0. A similar situa- 
tion would obtain if the curve 
is concave up, so that/"(#) 
> (Fig. 24). The reader 
will readily convince himself 
from an inspection of Fig 23 
>x that caution must be ex- 
ercised in the choice of the 
first approximation if the 
curve has a maximum (or a 
minimum) in the vicinity of XQ. 

If the curve has the appearance indicated in Fig. 25, then it is 
evident that the choice of either x\ or x\ as the first approximation 
will yield a second approxima- 
tion which is worse than the 
firt one. This is due to the 
fact that the curve has a point 




FIG. 24. 




of inflection between x\ and 

r' 
Xi. 

From the foregoing discus- 
sion, it is apparent that New- 
ton's method should not be 
applied before making an investigation of the behavior of the first 
and second derivatives of f(x) in the vicinity of the root. The 



FIG. 25. 





I 

'^ 

FIG. 26. 



29 SOLUTION OF EQUATIONS 101 

conclusions drawn from this discussion can be summarized in the 
following practical rule for determining the choice of the first 
approximation: Iff'(x) andf"(x) do not vanish in the given interval 
(xi, x{) and if the signs off(xi) and f(x{} are opposite, then the first 
approximation should be chosen as that one of the two end points for 
which f(x) andf"(x) have the same sign. 

It can be proved* that if the single-valued continuous function 
f(x) is of such a nature that /( 
= has only one real root in foo 
(xi, x{) and both/'(s) and /"(a;) 
are continuous and do not 
vanish in (xi, x'^) y then repeated 
applications of Newton's 
method will determine the value 
of the root of f(x) = to any 
desired number of decimal 
places. 

The cases to which Newton's 
method does not apply can be 
treated by a method of interpolation (regula falsi) that is appli- 
cable to any equation. 

Let x be the value of x for which the chord AB intersects the 
x-axis. From similar triangles (Fig. 26), 

x Xi x( x 

Solving for x gives 

. _ 

The value x is clearly a better approximation than either xi 
or x(. 

PROBLEMS 

1. Solve Prob. 7, Sec. 28, by Newton's method. Also, apply the 
method of interpolation. 

2. Determine the angle subtended at the center of a circle by a chord 
which cuts off a segment whose area is one-quarter of that of the circle. 

3. Find the roots of e* 4x 0, correct to four decimal places. 

4. Solve x cos x 0. 

* See WEBER, H., Algebra, 2d ed. vol. 1, pp. 380-382; COATE, G. T., On the 
Convergence of Newton's Method of Approximation, Amer. Math. Monthly, 
vol. 44, pp. 464-466, 1937. 



102 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 30 

6. Solve x = tan x in the vicinity of x = %TT. 

6. Solve x + e* = 0. 

7. Solve x* - x - 1 = 0. 

8. Show with the aid of Taylor's series that, if x = x t is an approxi- 
mate root of f(x) 0, then the nth approximation is, in general, 
determined from the formula (29-1). 

f(n)( r ,\ 

Hint:f(x) = /(*0 + /'(*i)(* -*!)++ J ~r L (x - *i) + ; 
and if f(xz) == 0, then 



30. Determinants of the Second and Third Order. The solu- 
tion of systems of linear equations involves the determination of 
the particular values of two or more variables that will satisfy 
simultaneously a set of equations in those variables. Since the 
discussion is simplified by using certain properties of deter-' 
minants and matrices, the remainder of this chapter is devoted 
to some elementary theory of determinants and matrices and its 
application to the solution of systems of linear equations. 

Consider first a system composed of two linear equations in 
two unknowns, namely, 

(30-1) ( aiX + ? I2/ = J" 

\a 2 x + b 2 y = k> 2 . 

If y is eliminated between these two equations, there is obtained 
the equation 

(30-2) (aibz a 2 bi)x = kib 2 fobij 

and if x is eliminated, there results 

(30-3) (aj6 2 a 2 bi)y = ai& 2 Q^i. 

If the expression aj) 2 a 2 bi is not zero, the two equations 
(30-2) and (30-3) can be solved to give values for x and y. That 
the values so obtained are actually the solutions of the system 
(30-1) can be verified by substitution in Kqs. (30-1). 

The expression ai6 2 a z bi occurs as the coefficient for both 
x and y. Denote it by the symbol 



(30-4) 2 

#2 02 

This symbol is called a determinant of the second order. It is 
also called the determinant of the coefficients of the system 



30 



SOLUTION OF EQUATIONS 



103 



(30-1), for the elements of its first column are the coefficients of 
x and the elements of its second column are the coefficients of y. 
Using this notation, (30-2) and (30-3) become 



(30-5) 



a 2 6; 






bi 
& 2 o 2 



V = 



The definition (30-4) provides the method of evaluating the 
symbol. If 

D - 



the unique solution of (30-1) can be written as 

:fci 



x 



D 



y = 



If Z) = 0, ai6 2 = fl2&i or ai/a 2 = 61/62. But if the correspond- 
ing coefficients of the two equations are proportional, the two lines, 
whose equations are givon by (30-1), arc parallel vif a\/ai ^ 
fci//c 2 ) or coincident (if 0,1/0,% = 61/62 = fci//c 2 ). In the first 
case, the determinants appearing as the right-hand members of 
the equations in (30-5) will be different from zero and there will 
be no solution for x and y. In the second case, these deter- 
minants, as well as D, are zero and any pair of values , y that 
satisfies one equation of the system will satisfy the other equa- 
tion, also. 



Example 1. For the system 

2x - By = -4 
3x - y = 1, 
-4 -3 
1 -1 



D = 



2 -3 



= -2 

2 -4 
3 1 

7 



=* 2. 



Example 2. For the system 



but 



2 _ 

6 " 



2 -3 
6 -9 



= 0, 



The two lines whose equations are given are parallel. 



104 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 30 
Example 3. For the system 

,, 2-3 



2x - 3y = 4 

Q X - 9t, = 12, j 

2 -3 4 
6 ~ -9 ~~ 12 



-9 



= 0, 



The two lines are coincident. 

Consider next the system of three linear equations in three 
unknowns, 

+ b\y + CiZ = 
(30-6) <J d 2 x + b 2 y + c 2 z = k 2 , 



If these equations are multiplied, respectively, by 



and the resulting equations are added, the sum is 
(30-7) 



= kib-2.Cz ^i6 
The coefficient of x in (30-7) can be denoted by the symbol 



(30-8) D ^ 



0,2 62 



+ 



This symbol is called a determinant of the third order. It is 
also the determinant of the coefficients of the system (30-6). 
Using the notation of (30-8), Eq. (30-7) can be written as 



Dx s 



61 
b 2 
63 



x = 



k 2 b 2 c 2 



Similarly it can be shown that 



bi GI 
! b 2 c 2 



y = 



and 



di b\ ( 
d 2 62 < 

^3 &3 < 



^2 ^2 ^2 , 



!>!*! 

<)2k 2 . 
hk t 



30 



SOLUTION OF EQUATIONS 



105 



If D 5^ 0, the unique solutions for x f y, and z can be obtained as 

bi ki 



(30-9) x = 



0,2 ^2 C2 

fc 3 c 3 



D 



z = 



a* 62 
,b, 



In order to show that the values of x, y, and z, given in (30-9), 
actually satisfy Eqs. (30-6), these values can be substituted in 
the given equations. 

If D = 0, the three equations (30-6) are either inconsistent or 
dependent. A detailed analytic discussion of these cases will be 
given in Sec. 35. Since the three equations of (30-6) are the 
equations of three planes, a geometrical interpretation will now 
be given. 

If the three equations are inconsistent, the three planes are 
all parallel, or two are parallel and are cut by the third plane in 
two parallel lines. In either case, there is obviously no solution 
for x, yj and z. If the equations are dependent, all three planes 
intersect in the same line or all three planes coincide. In either 
case there will be an infinite number of solutions for x, y, and z. 



Example. For the system 



Therefore, 



3x- y - z = 2, 

x - 2y - 3z = 0, 

4x + y + 2z = 4. 



D = 



3 -1 -1 

100 
JL ~~~ i """" O 



2 -1 


-1 








-2 


-3 








4 1 


2 




2 




2 




2 * 








3 


-1 


2 






1 


-2 







ft 


4 


1 


4 



y = 



2 -i 

-3 
4 2 



= 2. 



2, 



PROBLEMS 



1. Evaluate 



1 2 

2 -1 



3 -1 -2 



2 -3j 

1 4 2| 

-1 1 -2! 



, and 



4 -2 1 

5 0-1 
2 3-3 



2. Find the solutions of the following systems of equations by using 
determinants: 



106 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 31 



(a) 5s - 4y = 3, 

2x + 3y = 7; 
(&) 2z + 3y - 2* = 4, 
x + y - z = 2, 
3z - 57/ + 3s = 0; 

(c) 3x - 2y = 7, 
3?/ + 2z = 6, 
2z + 3z = 1; 

(d) 3x + 2y + 2z = 3, 
z - 4y + 2z = 4, 
2x + y + z = 2. 

31. Determinants of the nth Order. Determinants of the 
second and third orders were defined in the preceding section. 
These are merely special cases of the definition of the determinants 
of any order n. Instead of a symbol with 2 2 or 3 2 elements, the 
determinant of the nth order is defined as the symbol, with n 
rows and n columns, 



#21 



D = 



#2n 



#n,l #n2 * ' * #nn 



which stands for the sum* of then! terms ( l^a^ia^ * * ' #/c n n, 
where &i, & 2 , , k n are the numbers 1, 2, , n in some 
order. The integer k is defined as the number of inversions of 
order of the subscripts fci, & 2 , * , k n from the normal order 
1, 2, , n, where a particular arrangement is said to have k 
inversions of order if it is necessary to make k successive inter- 
changes of adjacent elements f in order to make the arrangement 
assume the normal order. There are nl terms since there are n! 
permutations of the n first subscripts. Moreover, it is evident 
that each term contains as a factor one and only one clement from 
each row and one and only one element from each column. 

* This sum is sometimes called the expansion of the determinant. 

t It should be noted that it is not necessary to specify that the inter- 
changes should be of adjacent elements, for it can be proved that, if any 
particular arrangement can be obtained by k interchanges of adjacent 
elements and also by'fc' interchanges of some other type, then k and k f are 
always either both even or both odd. Hence, the sign of the term is inde- 
pendent of the particular succession of interchanges. 



32 SOLUTION OF EQUATIONS 

Example. Consider the third-order determinant 



107 



D = 



l #12 #13 
#21 #22 #23 
#31 #32 #33 



The six terms of the expansion are, apart from sign, 



#11#22#33> 
#21#32#13, 



#11#32#23, 
#31#12#23, 



#2i#12#33, 
#3l#22#13- 



The first term, in which the first subscripts have the normal order, is 
called the diagonal term, and its sign is positive. In the second term 
the arrangement 132 requires the interchange of 2 and 3 to make it 
assume the normal order; therefore, k 1, and the term has a negative 
sign. Similarly, the third term has a negative sign. The fourth term 
will have a positive sign, for the arrangement 231 requires the inter- 
change of 3 and 1 followed by the interchange of 2 and 1 in order to 
assume the normal order. Similarly, it appears that the fifth term will 
have a positive sign. In the sixth term, it is necessary to make three 
interchanges (3 and 2, 3 and 1, and 2 and 1) in order to arrive at the 
normal order; hence, this term will have a negative sign. As a result of 
this investigation, it follows that 



D = ttutt22#33 



23 #21#12#33 + #2i#,*2#13 + #31#12#23 ~ #31#22#13 



It is evident that if k is equal to zero or an even number the 
term will have a positive sign, whereas if k is odd the term will 
be negative. 

PROBLEM 

Find the signs of the six terms involving # u in the expansion of the 
determinant 

#11 #12 #13 #14 

#21 #22 #23 #24 

#31 #32 #33 #34 

#41 #12 #43 #44 

32. Properties of Determinants. 1. The value of a determi- 
nant is not changed if in the symbol the elements of corresponding 
rows and columns are interchanged. 



If 



D s 



21 



a 2n 



a n i < 



108 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 32 



then the determinant formed by interchanging the corresponding 
rows and columns is 

an 

T\l ... 



Any term ( l)*afc la* 2 a k n of D, where &i, fc 2 , , k n are 
the numbers 1, 2, , n in some order, will correspond to a 
term ( 1)^1^2 * * ' #/c n n of D', for each determinant must 
contain every possible term that is a product of one and only 
one element from each row and each column. But the number 
of inversions is the same for the term of D as it is for the term of 
ZX, owing to the fact that the corresponding first subscripts are 
the same. It follows that each term of D occurs also in )', and 
conversely each term of D f occurs also in D. 

Example. If 



then 



D' 



2 5 

1 -1 

-3 -2 

2 1 -3 
5 -1 -2 

3 4 1 



-66, 



= -66. 



2. An interchange of any two rows or of any two columns of a 
determinant will merely change the sign of the determinant. 

If D is the original determinant and D" is the determinant 
having the ith and jth rows of D interchanged, then the expansion 
of D" will have the first subscripts of each term the same as those 
of the corresponding term of D, except that i and j will be inter- 
changed. Since it requires one interchange to restore i and j to 
their original order in each term, the sign of every term will be 
changed. Thus, D" = -D. 

Example. If 



then 



2 5 
1 -1 



-3 -2 1 

2 5 
-3 -2 
i 1 



= 66. 



SOLUTION OF EQUATIONS 



109 



3. // any two rows or any two columns of a determinant are 
identical, the value of the determinant is zero. 

For, by property 2, if these two rows (or columns) were inter- 
changed, the sign of D should be changed. But since these two 
rows (or columns) are identical, D remains unchanged. There- 
fore, D = D, and hence D = 0. 

Example. If 



then 



2 -1 

3 4 

-2 5 -2 

D = 0. 



4. // each element of any row or any column be multiplied by 
m, the value of the determinant is multiplied by m. 

This follows from the definition of the determinant. Since one 
and only one element of any row or column occurs in each term, 
each term will be multiplied by m and therefore the value of the 
determinant is multiplied by m. 

Example 1. If 

5 



and 



2 

1 

-3 



D = 



-2 



-66 



2 5 

i i 

j. j. 

-6 -4 



which has each element of the last row twice the corresponding element 
of the last row of D, then 



D = -132 and D = 2D. 



Example 2. If 



6 

9 

-6 



4 
2 
3 



then 



= 2 



3 

9 

-6 



4 

1 
i 



= 2-3 



1 

3 

-2 



3 -1 



5. From properties 3 and 4, it follows that the value of a deter- 
minant is zero if any two rows or any two columns have corre- 
sponding^ elements proportional. 



110 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 33 

6. The product of two determinants D and D', both of order n, 
is the nth-order determinant D" which has as the element in its 
ith row and jth column the sum 



a in b n 



which is formed by multiplying each element a lk of the ith row of D 
by the corresponding element bk } of the jth column of D' . 
Thus, if 



D = 



an 



and 



>' = 



611 6 
621 622 



then 



D D' s D" = 



a\\b\i + 012621 #11612 + #12622 

+ 022621 #21612 H~ ^22622 



Example. The product of the following determinants is easily found 
by expanding the product determinant: 



sin 


X 


COS X 


1 




sir 


sec 


X 


tan x 


1 




CO 


CSC 


X 


cot x 


1 
















tan 


X 


+ sin 


x - 1 


cos 


X 


cot x 





sec x esc x 

tan x cot x 
- 1 - 1 - 1 

tan x sin x 1 


sec x esc x 2 



cos x cot # 

sec x esc 2 





- 2 cos 2 x(2 sec a? esc x). 



33. Minors. The method of evaluating a determinant by the 
use of the definition of Sec. 31 is exceedingly tedious, especially 
if n ^ 4. There are other schemes for this evaluation, and these 
require the definition of the minors of a determinant. The 
simplest of these schemes will be described and used here. 

'If, in the determinant Z>, the ith row and the jth column be 
suppressed, the resulting determinant A lJ (which is of order one 
less than the order of D) is called the minor of the element a l3 , 
which is in the ith row and jth column. 



Example. If 



ai2 ais 

#21 #22 a 23 024 

i 032 flss 

1 tt42 tt43 #44 



33 SOLUTION OF EQUATIONS 111 

then 

flll #12 

A23 ~ &31 &32 
fl41 42 

From the definition of a determinant, it is evident that o,,A tJ is 
composed of all the terms of D which contain the element a tj as a 
factor, except for the possibility that all the signs may be reversed. 
Then the expression ( l) &1 anAn is composed of all the terms of 
D containing an as a factor; ( l) t2 a 2 iA 2 i is composed of all the 
terms containing a 2 i as a factor; ( l)* 3 o. 3 iA 3 i is composed of all 
the terms containing a 3 i as a factor; etc. But D is composed of all 
the terms containing an, a 2i , a 3 i, , a n i as a factor, and so, 

D = ( l)*iaiiAn + ( l)* 2 a 2 iA 2 i + + ( l)* n aiAi. 

It can be proved* that fci = 1 4- 1. fc 2 = 2 + 1, & 3 = 3 + 1, 
, k n = n + 1, so that 

D = aiiAn - a 2 iA 2 i + + ( l) n+1 aniA n i. 

In the above development for D the elements an, a 2 i, , a n i 
are the elements of the first column of D. Similarly, the value 
of D can be formed by taking the elements of any other column 
or of any row. 

Using the ith column gives 

D = ( IJ^ai.Au + (-l) fc2 a 2l A 2t + + (-l) fc *a m A n ;, 

where ki = i + 1, fc 2 = i + 2, , k n = i + n. Similarly, 
using the ith row gives 

D = (-!' 



where ki = i + 1, fc 2 = i + 2, , k n = i + n. It may be 
observed that each k r is equal to the sum of the subscripts of its 
a t? and is thus equal to the sum of the number of the row and the 
number of the column in which this element occurs. This 
development is known as the expansion by minors, or the simple 
Laplace development. 

Since the term cofactor is frequently used in applications of 
this type of development, it will be defined here. The cofactor 
Ca of an element a l3 is defined as the signed minor, that is, 



* DICKSON, L. E., First Course in Theory of Equations, pp. 101-127; 
FINE, H. B., College Algebra, pp. 492-519. 



112 MATHEMATICS FOR ENGINEERS AND PHYSICISTS {33 
Thus, the expression for D can be written as 

n n 

0=(-l)'-H.^ = 



or as 



n n 

D = 2 ^A/ = 2 aijCij - 



3406 




0521 
0340 


= 3 


521 
340 


1271 




271 


= 3[- 3 



-0 



406 
340 

2 7 1 



+ 



406 
5 2 1 

2 7 1 



5 1 
2 1 



-9(2 - 7) + 12(5 - 2) - 4(0 
13. 



- 1 



2 1 



406 
5 2 1 
340 



-0 



5 1 
3 



5 2 
3 4 



4) - 6(20 - 6) 



Here, the first expansion is made by using the elements of the first 
column, for it contains two zeros (the third row is an equally good 
choice). The expansion of the first third-order determinant is made by 
using the elements of the second row, but the third column could be 
used to equal advantage. In the expansion of the last third-order 
determinant the first row was chosen, but the third row and the second 
and third columns provide equally good choices. 

The following theorem is given here because of its frequent use 
in many fields of pure and applied mathematics : 

n 

THEOREM. The sum 2 a>% 3 Ck 3 is zero, if k ^ i. 
j-i 

Each term of this sum is formed by taking the product of the 
cofactor of an element of the fcth row by the corresponding 
element of the ith row. This is the expansion of a determinant 
whose t'th and fcth rows are identical and whose value is accord- 



ingly zero. Similarly, it follows that S 



= 0, if k ^ j. 



Exampk. Let 



Then, 



D 



3-1 2 
1 2 -1 
4 -3 -2 



7, Ciz - -2, 



C 



13 



-11 



33 SOLUTION OF EQUATIONS 113 

and the sum 

3 

2) a Sj Ci, = -28 + 6 + 22 = 0. 
Similarly, 

3 

2) 02,Ci, = -7 - 4 + 11 = 0. 

By using the theory of determinants, the solution of a system 
of n non-homogeneous linear equations in n unknowns can be 
obtained. The rule for effecting the solution will be stated but 
not proved.* The proof for the cases when n = 2 and n = 3 has 
already been given in Sec. 30. 

Cramer's Rule. Let 



(33-1) 



== fo2 



1 t^nn^n 



be a system of n equations in the unknowns x\ 9 
such that the determinant 



D = 



/) 

y An 



of the coefficients is not zero. The system (33-1) has a unique 
solution given by 



D, 



x - 

Xn "" 



where Z3 t is the determinant formed by replacing the elements 
ii, a 2l , a st> , a* of the tth column of D by fci, fc 2 , fc 3 , * , 
fc n , respectively. 

Example. Solve, by Cramer's rule, the system 

3z + y + 2z = 3, 
2z - 3y - z= -3, 

a? + 2t/ + 2 = 4. 

* DICKSON, L. E., First Course in Theory of Equations, pp. 114-115. 



114 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 34 



Here 



and 





3 1 


2 




D = 2 -3 - 


1 = 8 




1 2 


1 


3 1 2 




332 


-3 -3 -1 




2 -3 -1 


4 2 1 


8 , 


1 4 1 


8 


~~ o ~ 1> 2/ 


8 




3 1 3 






2 -3 -3 






1 2 4 


-8 


z 


8 


8 ~" ' 



PROBLEMS 



1. Evaluate 






1 


-2 





3 


2 


-1 




3 










1 


4 


1 


2 


1 


4 







2 






2 


- 


-1 -3' 


1 


1 




3 


5 






3 


1 





1 


-1 







2 





2. Solve, by Cramer's rule, 


the 


systems 


(a) x -{- 2y + 3z = 


= 3, 


(6) 2x + 








2x - 


y + 


2 = 6, 3# 








3o; + 


y 


z - 


= 4. 








# - 




<<0 






i 


2y 


= 1, 


(d) 1 


2x + 








2x - 


y - 


22 


= 3, 









)X + 








/> i 


2/ + 


32 


= 2. 






a; 


2 



, and 



16 



132-1 

0432 

-3 1 I 1 

120-4 



- 3z = 2, 

- 2z = 1, 

x - y + z = 1. 
+ y + 3z + w = -2, 

2 W 1, 



34. Matrices and Linear Dependence. In order to discuss 
the systems arising in the succeeding sections, it is convenient to 
give a short introduction to the theory of matrices.* 

An m X n matrix is defined as a system of mn quantities a l} 
arranged in a rectangular array of m rows and n columns. If 
m = n, the array is called a square matrix of order n. The 
quantities a lj are called the elements of the matrix. Thus, 



(34-1) A 



an 



a 2n 



or 



an 



a 2n 



#wl Cl m 2 * * ' a 

* For detailed treatment see M. Bocher, Introduction to Higher Algebra, 
pp. 20-53; L. E, Dickson, Modern Algebraic Theories, pp. 3&~63. 



34 SOLUTION OF EQUATIONS . . 115 

where double bars or parentheses are used to enclose the array of 
elements. If the order of the elements in (34-1) is changed or if 
any element is changed, a different matrix results, Any two 
matrices A and B are said to be equal if and only if every element,. 
of A is equal to the corresponding element of J5, that is, if 
a l3 = b tj for every i and j. 

If the matrix is square, it is possible to form from the elements 
of the matrix a determinant whose elements have the same 
arrangement as those of the matrix. The determinant is called 
the determinant of the matrix. From any matrix, other matrices 
can be obtained by striking out any number of rows and columns. 
Certain of these matrices will be square matrices, and the 
determinants of these matrices arc called the determinants of the 
matrix. For an m X n matrix, there are square matrices of 
orders 1, 2, , p, where p is equal to the smaller of the 
numbers m and n. 

Example. The 2X3 matrix 



^_ / a H 
\a-2l 



contains the first-order square matrices (an), (a t2 ), (a 2 s), etc., obtained 
by striking out any two columns and any one row. It also contains 
the second-order square matrices 



obtained by striking out any column of A. 

In many applications, it is useful to employ the notion of the 
rank of a matrix A. This is defined in terms of the determinants 
of A. A matrix A is said to be of rank r if there exists at least 
one r-rowed determinant of A that is not zero, whereas all deter- 
minants of A of order higher than r are zero.* 



Example. If 



/ 1 1 3\ 
EE( 2 1 0-21 
\-l -1 1 5/ 



* In case an m X n matrix contains no determinants of order higher than r, 
obviously r is the smaller of the numbers m and n, and the matrix is said 
to be of rank r. 



M6 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 
the third-order determinants are 



34 



I 1 




1 3 




1 1 3 




2 1 


= 0, 


2 1 T-2 


= 0, 


20-2 


= 0, 


-1 -1 1 




-1 -1 5 




-1 1 5 






1 3 




1 -2 




1 1 K 
J. A c) 



0. 



Since 



1 



5*0, 



there is at least one second-order determinant different from zero, 
whereas all third-order determinants of A are zero. Therefore, the 
rank of A is 2. 

It should be observed that a matrix is said to have rank zero 
if all of its elements are zero. 

The notion of linear dependence is of importance in connection 
with the study of systems of linear equations, and it will be con- 
sidered next. 

A set of m, m > 2, quantities /i, / 2 , /a, * , f m (which may be 
constants or functions of any number of variables) is said to be 
linearly dependent if there exist m constants ci, c 2 , , c m , which 
are not all zero } such that 

(34-2) ci/i + c 2 / 2 + + c m f m s 0. 

If no such constants exist, the quantities / t are said to be linearly 
independent. 

Example. If the /, are the polynomials 

fi(x, y, z) = 2z 2 - 3xy + 4z, 
f 2 (x, y, z) 55 a?* + 2xy - 3z, 
MX, y, z) ss 4z 2 + xy- 2z, 

and if the constants are chosen as c t = 1, 02 = 2, c 3 = 1, then 

Cifi + c 2 /a + c 3 / 3 ss 0. 
Therefore, these three polynomials are linearly dependent. 

It is evident that, whenever the set of quantities is linearly 
dependent, at least one of the / t can be expressed as a linear 
combination of the others. Thus, from (34-2), if ci ^ 0, then 



35 SOLUTION OF EQUATIONS 117 

where 

a 2 = -, a 3 = -> etc. 

The definition of linear dependence requires the existence of at 
least one constant c l ^ 0, and therefore the solution for / is 
assured. 

Obviously, in most cases it would be extremely difficult to 
apply the definition in order to establish the linear dependence 
(or independence) of a given set of quantities. In case the 
quantities f t are linear functions of n variables, there is a simple 
test which will be stated without proof.* 

THEOREM. The m linear functions 

f^ z= a l \x\ + a l zXz + ' ' ' + QinXn, (i = 1, 2, , m), 

are linearly dependent if and only if the matrix of the coefficients 
is of rank r < m. Moreover, there are exactly r of the f l that form 
a linearly independent set. 

If m > n, obviously r < m, and it follows that any set of m 
linear functions in less than m unknowns must be linearly 
dependent. 

The fact that the polynomials 

/i = 2x 3y + 42, 
/ 2 = x + 2y - 82, 
/ 3 = 4s + y - 22, 

are linearly dependent can be determined by observing that the matrix 
of the coefficients, 

/2 -3 



-3 4\ 
2 -3I 
1 -2/ 



is of rank 2. 

35. Consistent and Inconsistent Systems of Equations. A set 

of equations that have at least one common solution is said to 
be a consistent set of equations. A set for which there exists no 
common solution is called an inconsistent set. 

The question of consistency is frequently of practical impor- 
tance. For example, in setting up problems in electrical net- 
works, there are often more conditions than there are variables. 

* DICKSON, L. E., Modern Algebraic Theories, pp. 55-60; BOCHER, M , 
Introduction to Higher Algebra, pp. 34-38. 



118 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 35 

This leads to a system in which there are more equations than 
there are unknowns. It is important to have a method for 
testing whether all the conditions can be satisfied simultaneously. 
THEOREM 1. Consider a system of m linear equations in n 
unknowns, 

/0-. 1 \ a%lXl ~T~ # 2 2#2 ~T~ -{- a% n Xn == ^2 

(35-1) 

; 

_ _j_ (J, mn Xn == K"mj 



where at least one k t j^ 0. If the matrix of the coefficients is of 
rank r, Eqs. (35-1) are consistent provided that the rank of 




is also r. 

The matrix K is called the augmented matrix. The proof of 
this theorem will be found in any standard work on higher 
algebra. 

Example 1. Consider the system 

2x + 3y = 1, 
x - 2y = 4, 
4x y = 9. 
Since 



/2 3\ 
(l -2) 

\4 -I/ 



is of rank 2, the equations are consistent if 
K 



/2 3 1\ 
= (1 -2 4) 

\4 -1 9/ 



is also of rank 2. This condition is satisfied; for the determinant of K 
is zero, and there exists a second-order determinant of K that is dif- 
ferent from zero. 
Example 2. The system 

2x + 3y = 1, 

x - 2y = 4, 

4s - y = 6 



35 SOLUTION OF EQUATIONS 119 

is inconsistent, because 



/2 3 1\ 

K = ( 1 -2 4 I 

\4 -1 67 



is of rank 3, whereas the matrix A is of rank 2. 

In the case in which there are n equations in n unknowns, the 
theorem on consistent equations shows that if the determinant of 
A is zero, so thjt the rank of A is r < n, then the rank of K must 
be r also, if the set of equations is to be consistent. If the rank of 
K is greater than r, the set of equations is inconsistent. This 
provides the analytic discussion that should accompany the 
geometric discussion given forn = 3 in Sec. 30. 

If the set of equations is consistent and the rank of A is r, 
then it can be shown that n r of the unknowns can be given 
arbitrary values, and the values of the remaining r unknowns are 
determined uniquely in terms of those n r arbitrary values. 
These n r unknowns cannot be chosen at random, for the 
m X T matrix of the coefficients of the remaining r unknowns must 
have rank r if these unknowns are to be uniquely determined. 

Example 3. Solve the system 

x - y + 2z = 3, 
x + y - 2z = 1, 
x + 3y - 6z = -1. 

Since A and K are botli of rank 2, the equations are consistent. If 
either y or z is chosen arbitrarily the matrix of the coefficients of the 
remaining variables will have rank 2. If z = k, the equations to 
be solved are 

x - y = 3 - 2k, 

x + y = 1 + 2/b, 

x + 3y 1 + 6k. 

Solving the first two for x and y gives x = 2 and y = 2k 1. These 
values are seen to satisfy the third equation. Therefore, the solutions 
x = 2, y 2k 1, z k satisfy the original system for all values of k. 

The preceding discussion has dealt with non-homogeneous 
linear equations. In case the fc t are all zero, the system becomes 
the set of homogeneous equations 

^11X1 + ai2#2 + * * ' + &lnn ^ 0, 

(35-2) a * lXl a X * . a * nXn ""' ' 

. . . . , 

+ _!_ 1 n ~ C\ 

(Zt2^*2 "i * * * "T~ ^*in*'f> """" U 



120 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 35 

Obviously, Zi = 2 = = x n = is a solution of (35-2). 
It may happen that there are other solutions. If a\, a%, - - - , a n 
is a solution of (35-2), it is evident that fcai, fca 2 , , ka n , where 
k is an arbitrary constant, will be a solution, also. The condition 
for solutions different from the x\ = x% = - = x n = solu- 
tion will be stated without proof. 

THEOREM 2. The system (35-2) will have a solution different 
from the solution x\ = x% = = x n = 0, if the rank of the 
matrix of the coefficients is less than n. 

It follows that if the number of equations is less than the 
number of unknowns, that is, if m < n, there are always solutions 
other than the obvious zero solution. If m = n, there exist 
other solutions if the determinant of the square matrix of the 
coefficients is zero. As in the case of the non-homogeneous 
system, if the m X n matrix of the coefficients is of rank r, then 
n r of the unknowns can be specified arbitrarily and the 
remaining r unknowns will be uniquely determined, provided that 
the rank of the matrix of the remaining unknowns is r. 

Example 4. Consider the system 

2x - y + 3z = 0, 
x + 2y z = 0, 
3x + 4y + z 0. 
Here 

2-1 3 

\A\ = 1 2 -1 = 10. 
|3 4 1 

Therefore, x - 0, y = 0, z = is the only solution. 
Example 5. Consider 

3x - 2y = 0, 

x + 4y = 0, 

2x - y = 0, 

for which the matrix of the coefficients is of rank 2. Since the number 
of unknowns is 2, x = 0, y = is the only solution. 
Example 6. Consider 

2x - y + 3z = 0, 

x + 3y - 2z 0, 
5x 4 1 V + 4* - 0. 



36 SOLUTION OF EQUATIONS 121 

Here, 

A s 




which is of rank 2. Since the number of unknowns is 3, the system 
has solutions other than x = 0, y = 0, z = 0. Let z = k, and solve 
any two of the equations for x and y. If the first two are chosen, 
x = k and y = k. By substitution, it is easily verified that x = k, 
y = k, z = k satisfies all four equations for any choice of k. 
Example 7. Consider 

2x - 4y + 2 = 0, 

3x + y - 2z = 0. 
For this system, 

A / 2 ~ 4 ^ 
A = ( 3 1 -2J' 

which is of rank 2 Since the number of unknowns is greater than the 
number of equations, there exist other solutions. Let z = k, and solve 
the two equations for x and y. There results x %k and y = %k. 
Thus, x = %k, y = J^&, 2 = k is a solution for any choice of k. 
Example 8. Consider 

x - y + 2 = 0, 
2x + 3y + 2 = 0, 
3s + 2?/ + 2z = 0. 
Here, 



\A\ = 



2 3 1 
322 



Since the determinant of A is zero, there are solutions different from 
x = 0, y = 0, z = 0. Let z = fc, and solve any two of the equations. 
If the first two are chosen, x = &, ?/ == K&> z k. It is verifiable 
by substitution that these values satisfy all three equations, whatever 
be the choice of k. 

PROBLEMS 

1. Investigate the following systems and find solutions whenever the 
systems are consistent: 

(a) x - 2y = 3, (6) 2x + y - z = 1, 
2x + y = 1, x - 2y + z = 3, 

3x y = 4. 4x 3y + z = 5. 

(c) 3x + 2y = 4, (d) 2x - y + 3z = 4, 
a? 3y = 1, s + If 3 = -1, 

2z + 6y = -1. 5s - y + 3z = 7. 



122 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 35 

2. Investigate for consistency, and obtain non-zero solutions when 
they exist. 

(a) x + 3y - 2z = 0, (b) x - 2y = 0, 

2x - y + z = 0. 3x + y = 0, 

2z - y = 0. 

(c) 3x -2y + 2 = 0, (d) 2o; - 4y + 82; = 0, 

a; + 2y - 2,3 = 0, x + 2?y - 2z = 0, 

2x y + 2z = Q. 3x - 2?/ + 2 = 0. 

(e) 4a; - 2y + 2 = 0, (/) x + 2?y + 2z = 0, 

2x - y + 3z = 0, 3x - y + z = 0, 

2x - y - 2z = Q, 2x + 3y + 2z = 0, 

Qx 3y + 42 = 0. x + 4i/ - 2z = 0. 



CHAPTER IV 
PARTIAL DIFFERENTIATION 

36. Functions of Several Variables. Most of the functions 
considered in the preceding chapters depended on a single 
independent variable. This chapter is devoted to a study of 
functions depending on more than one independent variable. 

A simple example of a function of two independent variables 
x and y is 

z = xy, 

which can be thought to represent the area of a rectangle whose 
sides are x and y. Again, the volume v of a rectangular parallel- 
epiped whose edges are x, y, and z, namely, 

v = xyz, 

is an example of a function of three independent variables x, 
y, and z. A function u of n independent variables x\, x^ , 
x n can be denoted by 

U = f(Xi, X 2 , ' ' , X n ). 

A real function of a single 
independent variable x, say 
V /(#)> can be represented 
graphically by a curve in the 
xy-pl&ne. Analogously, a real 
function z = f(x, y), of two 
independent variables x and 




Fio. 27. 



y, can be thought to represent a surface in the three-dimen- 
sional space referred to a set of coordinats axes x, y, z (Fig. 
27). However, one must not become too much dependent 
on geometric interpretations, for such interpretations may 
prove to be of more hindrance than help. For instance, 
the function v = xyz, representing the volume of a rectangular 
parallelepiped, depends on three independent variables x, y, 

123 



124 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 36 

and z and hence cannot be conveniently represented geometrically 
in a space of three dimensions. 

Corresponding to the definition of continuity of a function 
of a single independent variable x (see Sec. 7), it will be said that 
a function z = f(x, y) is continuous at the point (XQ, yo) provided 
that a small change in the values of x and y produces a small 
change in the value of z. More precisely, if the value of the 
function z = /(#, y) at the point (XQ, yd) is z , then the continuity 
of the function at the point (x , yo) means that* 

(36-1) lim f(x, y) = f(x Q) yo) = z . 



In writing the left-hand member of (36-1), it is assumed that the 
limit is independent of the mode of approach of (x, y) to (XQ, y ). 
The statement embodied in (36-1) is another way of saying 
that 

f( x > y} ~ f( x v> yo) + , 

where lim e = 0; that is, if the function /(x, y) is continuous at 

x>xo 

(XQ, yo), then its value in the neighborhood of the point (XQ, yo) can 
be made to differ from the value at the point (XQ, yo) by as little as 
desired. 

If a function is continuous at all points of some region R 
in the :n/-plane, then it is said to be continuous in the region R. 

The definition of continuity of a function of more than two 
independent variables is similar. Thus, the continuity of the 
function u = f(x, y, z) at the point (x , yo, z ) means that 



lim f(x, y, z) = /(zo, y , z ), 

z >ro 

y *i/o 

Z+ZO 



independently of the way in which (x, y, z) approaches (x , y , z ). 

PROBLEM 

Describe the surfaces represented by the following equations: 

(a) x + 2y = 3, (6) x - y + z = 1, (c) x - 2, (d) z = y, 

(e) 2x-3y + 7z = 1, (/) x 2 - ?/ = 0, (g) y* + z* = 25, 

(A) i/ 2 = 2x, (i) z 2 + ?/ - 10s = 0, (/) z 2 + 2/ 2 + z 2 = 1, 

(AO 3* + z 2 = y, (0 z 2 + 27/ 2 + z = 0, (m) z 2 + 7/ 2 = z 2 , 

* For details, see I. S. Sokolnikoff, Advanced Calculus, Chap. III. 



37 



PARTIAL DIFFERENTIATION 



125 



y 2 
T 



(r) 

37. Partial Derivatives. The analytical definition of the 
derivative of a fuhctiori y = f(x), of a single variable x, is 



. Ihn = Km 



This derivative can be interpreted geometrically as the slope of 
the curve represented by the equation y = f(x) (Fig. 28). 



z^ 




It is natural to extend the definition of the derivative to 
functions of several variables in the following way: Consider 
the function z = f(x, y) of two independent variables x and y. 
If y is held fast, z becomes a function of the single variable x 
and its derivative with respect to x can be computed in the 
usual way. Let AZ* denote the increment in the function 
z = /(x, y) when y is kept fixed and x is changed by an amount 
Ax; that is, 

Az* = f(x + Ax, y) - /(x, y). 
Then, 



lim * _ lim - 

Ax-0 AX Az-0 AX 

is called the partial derivative of z with respect to x and is denoted 
by the symbol dz/dx, or z x , or f x . 



126 MAfKEMATICS FOR ENGINEERS AND PHYSICISTS 37 
Similarly, the partial derivative of z with respect to y is defined 

by 



In general, if w = /(xi, 0:2, , x n ) is a function of n inde- 
pendent variables x\ y x 2 , * , x n , then du/dx t denotes the 
derivative of u with respect to x t when the remaining variables 
are treated as constants. Thus, if 

z = x 3 + x 2 y + 2/ 3 , 
then 

g = 3x 2 + 2xy and g = x 2 + 3</ 2 . 
Also, if u = sin (ax + by + cz), then 

= a cos (ax + by + cz) (both t/ and 2 held constant) ; 

CfX 

= 6 cos (ax + by + cz) (both x and 2 held constant) ; 

= c cos (ax + by + cz) (both x and ?/ held constant). 
dz 

In the case of z = /(x, ?/), it is easy to provide a simple geo- 
metric interpretation of partial derivatives (Fig. 27). The equa- 
tion z = /(x, T/) is the equation of a surface; and if x is given the 
fixed value x , 2 = /(x , y) is the equation of the curve AB on 
the surface, formed by the intersection of the surface and the 
plane x = XQ. Then dz/dy gives the value of the slope at any 
point of AB. Similarly, if y is given the constant value y , 
then z /(x, T/ O ) is the equation of the curve CD on the surface, 
and dz/dx gives the slope at any point of CD. 

PROBLEMS 

1. Find dz/dx and dz/dy for each of the following functions: 

(a) z = y/x\ (b) z = x 3 ?/ + tan- 1 (y/x)\ (c) z = sin xy + x; 

(d) z = e* log y; (e) z == x 2 y + sin" 1 x. 

2. Find du/dx, du/dy, and du/dz for each of the following functions: 

(a) u = x 2 y + yz - zz 2 ; (6) u = xyz + log xy; 
(c) it = z sin- 1 (x/y); (d) u = (x 2 + y 2 + z 2 )^; 

(e) it = (x* + 2/ 2 + s 2 )-^. 



38 PARTIAL DIFFERENTIATION 127 

38. Total Differential. In the case of a function of one 
variable, y = /(a:), the derivative of y with respect to x is defined 
as 



so that Ay/ Ax = /'(a;) + , where lim e = 0. Therefore, 

Ax 

/(x + Ax) - /(x) ss Ay = /'(x) Ax + e Ax, 
where e is an infinitesimal which vanishes with Ax. Then, 



is defined as the differential dy. 

For the independent variable x, the terms " increment" and 
"differential" are synonymous (that is, Ax = dx). However, 
it should be noted that the differential dy (of the dependent 
variable y) and the increment Ay differ by an amount c Ax (see 
Fig. 28). 

The differential of a function of several independent variables 
is defined similarly. Let z = /(x, y), and let x and y acquire the 
respective increments Ax and Ay. Then, 

Az = /(x + Ax, y + Ay) - /(x, y). 

If 2 = /(x, y) is a continuous function, then, as Ax and Ay 
approach zero in any manner, Az also approaches zero as a limit. 
It will be assumed here that /(x, y) is continuous and that 
df/dx and df/dy are also continuous. 

The expression for Az can be put in a more useful form by 
adding and subtracting the term/(x, y + Ay). Then, 

Az = /(x + Ax, y + Ay) - /(x, y + Ay) + /(x, y + Ay) - /(x, y). 

But 

li m /<>* + Ax, y + Ay) - /(x, y + Ay) = df(x, y + Ay) 
AZ-+O Ax dx 

so that 

/(x + Ax, v + Ay) - f(x, y + 
where lim ei = 0. Moreover, 



Ay-,0 



128 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 38 
since the derivative is continuous. Therefore, 

df(x, y + Ay) = a/Qc, y} ^ 
dx dx 2 ' 

where lim 2 = 0. 

Af/~ -0 

In like manner, 



f(x, y + Ay) - /(re, y) = 
where lim e' = 0. It follows that 



e A* 



in which = ei + c 2 . 
The expression 



is defined as the total differential of z and denoted by cfc. In 
general, if w = /Or i, # 2 , , # n ), the total differential is given by 

(38-1) du = -^~dx l + -j-dx2 + ' ' ' + ^ dx n . 

oXi 0X2 oX n 

The expression for the total differential is called the principal part 
of the increment Aw, and is a close approximation to Au for 
sufficiently small values of dx\, dx%, , and dx n . As in the 
case of a function of a single independent variable, the differential 
of each independent variable is identical with the increment of 
that variable, but the differential of the dependent variable 
differs from the increment. 

If all of the variables except one, say x t , arc considered as 
constants, the resulting differential is called the partial differ- 
ential and is denoted by 

, df . 

- 



The partial differential expresses, approximately, the change 
in u due to a change A# t = dx l in the independent variable x l . 
On the other hand, the total differential du expresses, approxi- 
mately, the change in u due to changes dx\, dx^ , dx n in all 



38 PARTIAL DIFFERENTIATION 129 

the independent variables Xi, x%, , x n . It may be noted 
that the total differential is equal to the sum of the partial 
differentials. Physically, this corresponds to the principle of 
superposition of effects. When a number of changes are taking 
place simultaneously in any system, each one proceeds as if it 
were independent of the others and the total change is the sum 
of the effects due to the independent changes. 

Example 1. A metal box without a top has inside dimensions 
6 X 4 X 2 ft. If the metal is 0.1 ft. thick, find the actual volume of 
the metal used and compare it with the approximate volume found 
by using the differential. 

The actual volume is AF, where 

AF = 6.2 X 4.2 X 2.1 - 6 X 4 X 2 = 54.684 - 48 - 6.684 cu. ft. 
Since F = xyz, where x = 6, y = 4, z = 2, 

dV = yz dx + xz dy + xy dz 

= 8(0.2) + 12(0.2) + 24(0.1) - 6.4 cu. ft. 

Example 2. Two sides of a triangular piece of 
land (Fig. 29) are measured as 100 ft. and 125 ft., FIG. 29. 

and the included angle is measured as 60. If the 
possible errors are 0.2 ft. in measuring the sides and 1 in measuring 
the angle, what is the approximate error in the area? 

Since A = %xy sin a, 




dA = %(y sin a dx + x sin a dy + xy cos a da), 
and the approximate error is therefore 



dA ~ \ t 



125 r (0 - 2) + 10 r 

+ 100(125) ~ Q = 74.0 sq. ft. 

PROBLEMS 

1. A closed cylindrical tank is 4 ft. high and 2 ft. in diameter (inside 
dimensions). What is the approximate amount of metal in the wall 
and the ends of the tank if they are 0.2 in. thick? 

2. The angle of elevation of the top of a tower is found to be 30, with 
a possible error of 0.5. The distance to the base of the tower is found 
to be 1000 ft., with a possible error of 0.1 ft. What is the possible error 
in the height of the tower as computed from these measurements? 



130 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 39 

3. What is the possible error in the length of the hypotenuse of a 
right triangle if the legs are found to be 11.5 ft. and 7.8 ft., with a 
possible error of 0.1 ft. in each measurement? 

4. The constant C in Boyle 's law pv C is calculated from the 
measurements of p and v. If p is found to be 5000 Ib. per square foot 
with a possible error of 1 per cent and v is found to be 15 cu. ft. with a 
possible error of 2 per cent, find the approximate possible error in C 
computed from these measurements. 

5. The volume v, pressure p, and absolute temperature T of a perfect 
gas are connected by the formula pv = RT, where R is a constant. If 
T = 500, p = 4000 Ib. per square foot, and v = 15.2 cu. ft., find the 
approximate change in p when T changes to 503 and v to 15.25 cu. ft. 

6. In estimating the cost of a pile of bricks measured as 6 X 50 X 4 
ft., the tape is stretched 1 per cent beyond the estimated length. If 
the count is 12 bricks to 1 cu. ft. and bricks cost $8 per thousand, find 

the error in cost. 

A 

7. In determining specific gravity by the formula s = , _ ^ 

where A is the weight in air and W is the weight in water, A can be read 
within 0.01 Ib. and W within 0.02 Ib. Find approximately the maxi- 
mum error in s if the readings are A = 1.1 Ib. and W 0.6 Ib. Find 
the maximum relative error As/s. 

8. The equation of a perfect gas is pv RT. At a certain instant 
d given amount of gas has a volume of 16 cu. ft. and is under a pressure 
of 36 Ib. per square inch. Assuming R = 10.71, find the temperature 
T. If the volume is increasing at the rate of H cu. ft. per second and 
the pressure is decreasing at the rate % Ib. per square inch per second, 
find the rate at which the temperature is changing. 

9. The period of a simple pendulum with small oscillations is 



r = 2T V^ 

If T is computed using I = 8 ft. and g 32 ft. per second per second, 
find the approximate error in T if the true values are I 8.05 ft. and 
g = 32.01 ft. per second per second. Find also the percentage error. 

10. The diameter and altitude of a can in the shape of a right circular 
cylinder are measured as 4 in. and 6 in., respectively. The possible 
error in each measurement is 0.1 in. Find approximately the maxi- 
mum possible error in the values computed for the volume and the 
lateral surface. 

39. Total Derivatives. Thus far, it has been assumed that x 
and y were independent variables. It may be that x and y 



39 PARTIAL DIFFERENTIATION 131 

are both functions of one independent variable t, so that z 
becomes a function of this single independent variable. In 
such a case, z may have a derivative with respect to t. 

Let z = f(x, y), where x = <p(t) and y = $(()] these functions 
are assumed to be differentiable. If t is given an increment A, 
then x, y, and z will have corresponding increments Az, Ay, 
and Az, which approach zero with At. As in the case when x and 
y were independent variables, 

Az = | Ax + j Ay + 61 Ax + c 2 Ay. 
Then, 



= _ Arc Ay 

A dx ~A* "*" dy A* Cl AJ "*" 2 A* 



and 

cfe 



Moreover, from (39-1) it appears that 

dz = fdx+fdy 
dx dy 

gives the expression for the differential in this case as well as 
when x and y are independent variables. 
The general case, in which 

Z = f(Xi, X 2 , * , X n ) 

with 



can be treated similarly to show that 



and 



In case ^ = a: (39-1) becomes 



cfe _ ^f < = t 

dx ~ fa Itydx ~ Ite Itydx 

This formula can be used to calculate the derivative of a func- 



132 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 39 

tion of x defined implicitly by the equation f(x, y) = 0. Let 
z /(x, y), so that 

* = # + # & 
dx dx dy dx 

Since z = f(x, y) = 0, it follows that dz/dx = and 

dy _ 



provided that d/,% ^ 0. 
As an example, let 

x*y + y*x - 1 = 

define y as an implicit function of x. Then, 



& a: 2 + 2xy 

for all values of x and y for which the denominator does not 
vanish. 

It was noted that the total differential of a function 

z = f(xi, x 2 , , Xn), where x t = <p t (), 
is given by 



It will be proved next that the same formula can be applied to 
calculate the differential even when the variables x % are functions 
of several independent variables ti, t^ * , tm. Thus, consider 

z = f(xi, X*, - , x n ), where x % = <p t (ti, t 2 , , t m ). 

In order to find the partial derivative of f(xi, x^ , x n ) with 
respect to one of the variables, say fc, the remaining variables are 
held fixed so that f(xi, x%, , x n ) becomes a function of the 
single variable t k . Then, 



(39-2) > 



oj oXi . oj dx% , , oj uX n 

df dxi , df dx% . . df dx n 



^L ^1 _i_ ^L dx * 4. . . . J_ ^/ ^n 
dxi dt m dx% dt m dx n dt m 



39 PARTIAL DIFFERENTIATION 133 

If the first equation of (39-2) is multiplied by dti, the second 
by dt%, etc., and the resulting equations are added, there results 



or 

(39-3) rf/ = ^^ + ^^+---+ 

This establishes the validity of the formula (38-1) in all cases 
where the first partial derivatives, are continuous functions, 
irrespective of whether the independent variables are Xi t x%, 
- , x n or ti, h, , t m . 

An important special case of the formula (39-8) arises in 
aerodynamics and other branches of applied^ mathematics. 
Consider a function u = f(x, y, z, t) of four variables x, y, z, and t. 
The total differential of u is 

(39-4) ^-J^ + I^ + IA+I* 

du , , du , . du , . du j. 
^*c dx + d-y dy+ te dz+ -dl dt - 

Let it be supposed that x, y, and z are not independent variables, 
but functions of the variable t. In such a case, u will depend on 
t explicitly, and also implicitly through x, y, and z. Dividing 
both members of (39-4) by dt gives 

fQo K\ du _ ^d^ .dudy .dudz du 

( ' Tt "" toT* + ^"5? + a^di "*" df 

On the other hand, if the variables x, y, and z are functions of t 
and of some other set of independent variables r, s, , one 
must replace dx/dt, dy/dt, and dz/d in the right-hand member 
of (39-5) by dx/dt, dy/dt, and dz/dt, respectively, and du/dt in 



134 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 39 

the left-hand member by du/dt. The partial derivative with 
respect to t which appears in the left-hand member differs from 
that appearing in the right-hand member, since the latter is com- 
puted from u= f(x, y, z, t) by fixing the variables x, y, and z 
and differentiating the resulting function with respect to t. In 
order to indicate the distinction between the meanings of the 
two partial derivatives with respect to t y one can write 

Du _ du dx du dy dudz du 
~~dt~~'dxdt dy'dt 'dzdt ~dt' 

The fact that the total differential of a composite function 
has the same form irrespective of whether the variables involved 
are independent or not permits one to use the same formulas for 
calculating differentials as those established for the functions 
of a single variable. Thus, 

d(u + v) = du + dv, 



= v du + u dv, 

to 

etc. 



Example 1. If u = xy + yz + zx, and x = t, y = e~', and z cos t, 

du , , . dx , , , . dy , t , N dz 

Tt = ( y + ^Tt + (x + z ^dt + (x + y) ~dt 

= (e-< + cos 0(1) + (t + cos )(-s~0 + (< + "0(- sin 
= e~* + cos t te*' e~ ( cos t t sin t e~* sin t. 

This example illustrates the fact that this method of computing du/dt 
is often shorter than the old method in which the values of x, y, and z 
in terms of t are substituted in the expression for u before the derivative 
is computed. 

Example 2. If f(x y y) = x 2 + i/ 2 , where x r cos <p and y r sin <p, 
then 

df df dx df dv 

dr = dxfr + dyfr = 2x cos ^ + 2y sin ^ = 2r a* 1 * + 2r sinV = 2r, 

^ = i^ + %l^ = 2a;( ^ r sin ^ + 22/(r cos ^ 

= 2r 2 cos v? sin <p + 2r 2 cos ^? sin <p = 0. 
Also, 

df = 2r dr or df =* 2xdx + 2y dy. 



39 PARTIAL DIFFERENTIATION 135 

Example 3. Let z = e*v, where x = log (u + v) and y = tan" 1 (u/v). 
Then, 

az az dx 1 , ay t; 

-jp-r, = *e- - and - 



Hence, 

du ~ dx du dy du ~~ u + v v 2 + u 2 
Similarly, 

az _ ye xv xe xv u 

~fo = u + V ~~ V 2 + U 2 ' 

The same results can be obtained by noting that 

dz ye xv dx + xe xv dy. 
But 

, dx , dx , 1 . , 1 , 

dx 3- du + 3- dv = : du -\ : dv 

du dv u + v u + v 



and 



dy -5- du + ^ dv = -^-n " 9 ^ r r dv. 

y du dv v 2 + u 2 v 2 + u 2 



Hence, 

. du + dv , v du u dv 

dz = ye xv : h xe xv ; 

y u + v v 

= ( i h -r~l 2 J ^ " 

\u -}- v v 2 + 'W 2 / 

But 

,1 dz j i dz j 
dz = -5- du + -~- dv: 
du dv ' 

and since dz* and dv are independent differentials, equating the coeffi- 
cients of du and dv in the two expressions for dz gives 

a^ __ 7/e*" ze^z; 

a^ u + t; * f 2 + u 2 
and 



__ 
ay ~ tT+li "" v 2 + u" 2 ' 

PROBLEMS 

1. If u = z?/z and a; = a cos 0, |/ = a sin ^, 2 == kO, find du/dQ. 

2. If ^ = a; 2 7/ 2 and ?/ = r sin 6 and # = r cos 0, find aV^ r and 



3. If u = xy ?/2J and a; = r + s, y = r s, 2; = ^, find du/dr, 
du/ds, and du/dt. 



136 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 40 

*?/ 
4. If z = e**, z = log v-w 2 + y 2 , and z/ = tan" 1 i find dz/du and 



6. If z = /(a; + M, y + *0 show that d2/dz = dz/du and 
dz/dz>. 
6. If u = x 2 y + y*z + z 2 x, verify that 

du t du t du , . , xo 



7. (a) Find du/dt, if u = e* sin yz and 2 = Z 2 , y = i 1, z = I//. 
(6) Find dw/dr and dw/d0, if ^ = x 2 - 4?/ and x = r sec 0, y = r tan 0. 

8. (a) Find dt//dz and du/dx, if ^ = x 2 + y 2 and t/ = tan x. 
(b) Given V = /(a; ; T/, 2;), where x = r cos 0, t/ = r sin 0, 2 = i. Com- 
pute dF/dr, aF/a0 ; 57/5^ in terms of dV/dx, dV/dy, and dV/dz. 

- ?y 

9. If /is a function of u and v, where u = v# 2 ~h 2/ 2 ^ n d v = tan -1 -> 



find olf/ax, a//^2/, and VW/dx) 2 + 



40. Euler's Formula. A function /(#i, a: 2 , , x n ) of n 
variables xi, x^, - , x n is said to be homogeneous of degree m 
if the function is multiplied by X m when the arguments xi, 
x z , - - , x n are replaced by \Xi, Xx 2 , , Xx n , respectively. 
For example, f(x, y) = x z /-\/x 2 + y 2 is homogeneous of degree 
1, because the substitution of \x for x and \y for y yields 

Xz 2 / V^M 7 ?- Again, /(, y) = i + log * """ log y is homo- 

y % 

geneous of degree 1, whereas /(#, 2/, z) = z*/\/x* + y 2 is 
homogeneous of degree %. 

There is an important theorem, due to Euler, concerning 
homogeneous functions. 

EULER'S THEOREM. If u = /(xi, x 2 , * , x n ) is homogeneous 
of degree m and has continuous 9 first partial derivatives, then 



The proof of the theorem follows at once upon substituting 

x'i = Xxi, #2 = ^2, ' * * , x f n ==: Xx w . 
Then, since /(rci, o: 2 , * * , x n ) is homogeneous of degree m, 
f(x(, x' 2 , , <) = 



41 PARTIAL DIFFERENTIATION 137 

Differentiating with respect to X gives 

Ij* 1 + &,** + ' ' ' + &. Xn = m * n ~ lf(Xi > ** ' ' ' ' Xn) - 

If X is set equal to 1, then x\ = x(, x* = x' 2 , - * , x n = x' n and 
the theorem follows. 

PROBLEM 

Verify Euler's theorem for each of the following functions: 
(a) f(x, y, z) = x 2 y + xy 2 + 2xyz\ 
(6) f(x, y) = vV - x* sin- 1 



N 1 , log a; - log y. 
, y) = t H -- - ; 




() 



41. Differentiation of Implicit Functions. It was noted in 
Sec. 39 that the derivative of a function of x which is defined 
implicitly by the equation f(x, y) = could be calculated by 
applying the expression for the total derivative. This section 
contains a more detailed discussion of this method. 

The equation f(x, y) = may define either x or y as an implicit 
function of the other. If the equation can be solved for y to 
give y = <p(x), then the substitution of y = <p(x) in f(x, y} = 
gives an identity. Hence, f(x, y) = may be regarded as 
a composite function of x, where x enters implicitly in y. If 



so that 



It will be observed that this discussion tacitly assumes that 
f(Xj y) = has a real solution for y for every value of x. If 
(41-1) is applied formally to z 2 + y* = 0, it is readily checked 



138 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 41 

that - = --- This result is absurd for real values of x and y, 
dx y *" 

inasmuch as the only real values of x and y that satisfy x 1 + y 2 = 
are x = and y = 0. 

Example 1. Find dy/dx, if 3x s y* + x cos / = 0. Here, 



cos y, -g~ = 6z 3 ?/ 2 sn 

so that 

rf^ __ 9o; 2 ?/ 2 + cos y 



__ 
dx ~~ ~~ tix*y x sin y 

The relation /(x, ?/, z) = may define any one of the variables as 
an implicit function of the other two. Let x and y be independent 
variables. Then /(x, y, z) = defines 3 as an implicit function 
of x and i/, and 

, dz , , dz , 
dz = ~ dx + dy. 
dx dy * 

But 



Therefore, by substitution, 



This can be written as 



Since dx and dt/ are independent differentials and the above rela- 
tion holds for all values of dx and dy, it follows that 



dx dz dx 
and 



dy dz dy 
If df/dz 9* 0, these equations can be solved to give 



Example 2. If a 2 + 2y 2 - Szz = 0, then, by (41-2), 

SL m _ 2a; - 3g - 4 ^ 

da? "" 3a; J dy ~~ -3x 



41 



PARTIAL DIFFERENTIATION 



139 



(41-3) 



Frequently, it is necessary to calculate the derivatives of a 
function that is defined implicitly by a pair of simultaneous 
equations 

>, y, z) = o, 
(*, y, = o. 

If each of these equations is solved for one of the variables, say 
Zj to yield 

z = F(x, y) and z = <(#, y), 

then one is led to consider the equation resulting from the 
elimination of 2, namely, 

F(x, y} - $(z, y} = 0. 

This equation may be thought to define y as an implicit function 
of Xj and one can apply the method discussed earlier in this 
section to calculate dy/dx. 

However, the elimination of one of the variables from the 
simultaneous equations (41-3) may prove to be difficult, and it 
is simpler to use the following procedure: The differentiation 
of (41-3) gives 



and 



These equations can be solved for the ratios to give 



dx \dy\dz 



dy dz 
dy dz 



dz dx 
d<f> d<p 
dz dx 



dx dy 
d<p d<p 
dx 'dy 



from which the derivatives can be written down at once. 
Example 3. Let 

f(x, y, z) s x 2 + 7/ 2 + z* - a 2 = 
and 

Then, 



dx:dy: dz 



2y 2z 
-2y 4z 



2z 2x 

-4z 2x 



2x 2y\ 
2x -2y\ 



4yz:12xz: 



140 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 41 
Hence, 



dy _ 



dz -* 



etc. 



dx tyz dx 

Another important case arises from a consideration of a pair 
of simultaneous equations 



(41-4) 



lf(x, y, u, v) = 0, 
\ <p(x, y, u, v) = 0, 



which may be thought to define u and v as implicit functions 
of the variables x and y. 
Differentiating (41-4) gives 

' ~ dx dy ~ " - 



(41-5) 



. 

du 



dv 



But, since u and v are regarded as functions of x and y, 

, du , . du , 
du = dx + dy 
ox oy 

and 

, dv 7 . dv 7 
ay = aa; + ai/. 
da: dy y 

Substituting for du and dv in (41-5) gives 

fdu , d/d^ , /3/ , a/aw 



du dx 



XT, ^ ) d* + 



^ v w . d/ di^ 

+ a^a^ av^ 



/^ ^ 

\dx + Tu 



dx 



du 



dv 



Since the variables a; and y are independent, the coefficients 
of dx and dt/ must vanish, and this leads to a set of four equations 
for the determination of du/dXj du/dy, dv/dx, and dv/dy. Thus, 
one obtains 

d JL i 

dx dv 
d<p d<p 
dx dv 



du 
~dx 



df_df_ 
du dv 
dtp d<p 
du dv 



41 



PARTIAL DIFFERENTIATION 



141 



and similar expressions for du/dy, dv/dx, and dv/dy. It is 
assumed in the foregoing discussion that all the derivatives 
involved are continuous and that 

df_ tf 

du dv 



Example 4. If 
then 



du dv 

+ y z + u* + v 3 = 0, 
1 + y - u* + v* = 0, 



du 
~dx 



PROBLEMS 

1. Obtain dy/dz, dw/dy, and dv/dy in Example 4, Sec. 41. 

2. Compute dy/efo, if x z + ?/ 3 3xy = 1. 

3. Find dy/dx if 



a; 2 ?/ y 2 z + 
4. Obtain du/dx and dv/dy, if 



a 3 = 0. 



tte v zt/ + t; = 0, 
ve y xv + u 0. 



5. If x = 
to x gives 



and ?/ = 



then differentiation with respect 



dx du dx dv 

du dx dv dx 

dy du dy dv 

~~ dudx dv dx 

from which du/dx and dv/dx can be computed. Consider the pair of 
equations 

x = u 2 - v 2 , 

y = uv, 

and obtain du/dx, du/dy, dv/dx, and dv/dy. 

6. Apply the method outlined in Prob. 5 to find du/dx, dv/dx, 
du/dy, and dv/dy, if 

j X = U + V, 

\y = 3u + 2v; 



(a} \, 



( 2x = v 2 w 2 , 
/ y uv. 



142 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 41 

7. If x = r cos 6 and y = r sin 0, find dr/dz and d0/d#. 

8. If w - uv and 

f N i w 2 4- r + x = 0, 
(a) ) * - u - y = 0, 

one can obtain dw/dx as follows: Differentiation of w with respect to 
x gives -3- = u T- + "r The values of du/dx and dt>/dz can be 

calculated from (a) by the method of Prob. 5. Find the expressions 
for dw/dx and dw/dy. 

9. If z uv and 

u* -f- v 2 - x - y = 0, 

W * - 02 + 3S + # = 0, 

find dz/dx. 

10. If 2 = w 2 + w 2 and 

2 = ^2 _ V 2 9 

y = w, 
find dz/dx. 

11. If 2 = n 2 + f 2 and 

u = r cos 0, 
y = r sin 0, 
find dz/dr and dz/dO. 

12. If r = (x 2 + /y 2 )' 2 ' and = tan" 1 -, find dr/dx and dO/'dx. 

x 

13. (a) Find du/dx, if x sec y + z 3 ?y 2 - 0. 

(b) Find cte/da; and dz/dy, if z 3 ?/ sin 2 + 2 3 = 0. 

14. Let u = x + y + z = and ; == s 2 + 7/ 2 + 2 2 - a 2 = 0. lir <l 
dx \dy\dz. 

15. Find du/dx, dv/dx, du/dy, and dv/dy, if 



16. Find dw/dx and dw/dy, ii w = u/v and 



17. Show that ||f = 1 and ||f | = -1, if f(x,y,z) = 0. 

Note that, in general, dz/dx and dz/dz are not reciprocals. 

18. Find du/dx, if 

u 2 - v 2 - x 3 + 3y = 0, 
-w + t; - y 2 - 2x = 0. 

19. Prove that 



42 



PARTIAL DIFFERENTIATION 



143 



if F(x, y, u, v) = and G(x, y, u, v) = 0. 

20. Show that (g) 1 + 1/r* (|)' - (|)' + ()', if . 

and y = r sin 0. 



r cos 



42. Directional Derivatives. The relation expressed in (39-1) 
has an important special case when x and y are functions of the 
distance s along some curve C, which goes through the point 
(x, y). The curve C may be thought to be represented by a pair 
of parametric equations 

x = x'(s), 

2/ = 2/0), 

where x and y are assumed to possess continuous derivatives with 
respect to the arc parameter s. 

Let P (Fig. 30) be any point of 
the curve C at which /(x, y) is 
defined and has partial deriva- 
tives df/dx and df/dy. Let 

Q(x + Ax, y + Ay) 

be ^a point close to P on this 

curve. If As is the length of the 

arc PQ and A/ is the change in / due to the increments Ax and 

Ay, then 

df r Af 

~- = hm -~ 
as A S -O As 

gives the rate of change of/ along C at the point (x, y). 




FIG 



But 



^ = y ^ a/ ^ 

ds ^x rfs ^y ^ 5 ' 



and 



dx 
-r- 
as 



,. Ax 

hm ~~ = cos a. 
A 5->o As 



dy ,. Ay 

-r = hm - = sin a. 

ds A*-O As 



Therefore, 
(42-1) 



df df . df . 

-r- = - COS a + T~ Sin a, 
as dx dy 



and it is evident that df/ds depends on the direction of the curve. 
For this reason, df/ds is called the directional derivative. It 
represents the rate of change of / in the direction of the tangent to 




144 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 42 
the particular curve chosen for the point (x, y). If a = 0, 

#= #, 

ds dx 

which is the rate of change of / in the direction of the x-axis. If 
a = r/2, 

df^V, 

ds dy 

which is the rate of change of / in the direction of the y-axis. 

Let z = f(Xj y), which can be inter- 
preted as the equation of a surface, be 
represented by drawing the contour 
lines on the zy-plane for various values 
of z. Let C (Fig. 31) be the curve in 
th e #?/-plane corresponding to the value 
2 = 7, and let C + AC be the neighbor- 
ing contour line for z = 7 + &y. 

+ x Then, A// As s= AY /As is the average 

rate of change of / with respect to the 
distance As between C and C + AC. 

Apart from infinitesimals of higher order, 

An 

= COS \f/. 

As Y ' 

where An denotes the distance from C to C + AC along the nor- 
mal to C at (x, y), and \l/ is the angle between An and As; hence, 
dn/ds = cos ^. Therefore, 

(42-2) ' 4f.*!^co8* 

' as an ds an 

This relation shows that the derivative of / in any direction may 
be found by multiplying the derivative along the normal by the 
cosine of the angle ^ between the particular direction and the 
normal. This derivative in the direction of the normal is called 
the normal derivative of /. Its numerical value obviously is the 
maximum value that df/ds can take for any direction. In applied 
mathematics the vector in the direction of the normal, of magni- 
tude df/dn, is called the gradient. 

Example. Using (42-1), find the value of a that makes df/ds a 
maximum, considering x and y to be fixed. Find the expression for 
this maximum value of df/ds. 



42 PARTIAL DIFFERENTIATION 

Since df/ds = /* cos a + / sin a, 
d /df\ 



145 



The condition for a maximum requires that 

tanai =^, or a, = tan' 1 ^- 



Using this value of i, 



j: r 



f, 



The relation (42-2) can be derived 
directly by use of this expression for 
df/dn. If a (Fig. 32) gives any di- 
rection different from the direction 
given by QJI, then 



~ = /, cos a + f y sin a. 
But a = ai \l/ 1 so that 




C+JC 



FIG. 32. 



.2. = /j.(cos i cos ^ -f sin ai sin 



cos \/> cos i sn 



Since 



cos ai = 



^ COS ^ + /a 




^fr! cos * / v/5+ ^ 

i9 I J2 

= cos ^ = Vfi +fi cos \ 



= -T 21 COS 

dn 



PROBLEMS 



1. Find the directional derivative of f(x, y) = x*y + sin xy at 
(l,7r/2), in the direction of the line making an angle of 45 with the 



146 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 43 
2. Find 



dn 



dx 



if x r cos 8, y = r sin 0, and / is a function of the variables r and 0. 

3. Find the directional derivative of f(x, y) = a; 3 ?/ + e v * in the 

direction of the curve which, at the point (1, 1), makes an angle of 

30 with the #-axis. 

4. Find the normal derivative of 
f(x, y) = x* + y*. 

43. Tangent Plane and Normal 
Line to a Surface. It will be re- 
called that 

Ax + By + Cz = D 

. 33. . . 

is the equation of a plane, where 

the coefficients A, B, and C are called the direction components 
of the normal to the plane. If a, (3, and y (Fig. 33) are the direc- 
tion angles made by the normal to the plane from the origin, then 



cos a = 



Therefore, 




A 


** cos 8 = 


B 


VA 2 + B 2 + C 2 

cos y = 7= 


c 


\/A* + B 2 + C 2 



+ B 2 + C 2 



cos a : cos ft : cos 7 = A:B:C. 



If the plane passes through the point (XQ, yo, ZQ), its equation can 
be written as 

A(x - x ) + B(y - T/O) + C(z - Z Q ) = 0. 

There is also a normal form for the equation of a plane, entirely 
analogous to the normal form for the equation of the straight 
line in the plane. This form is 



or 



a; cos a + 2/ cos /3 + 2 cos 7 = p, 
B 



+ B* + 



+ # 2 + c 2 



+ B* + C 2 

D 

+ B 2 + C 2 ' 



43 



PARTIAL DIFFERENTIATION 



147 



in which p = D/^/A 2 + J5 2 + C 2 is the distance from the origin 
to the plane. 

Consider a surface denned by z = f(x, y), in which x and y 
are considered as the independent variables. Then, 



(43-1) 



If #o and t/o arc chosen, z is determined by z = /(x, y 
Ax = x x and Ay = y y , and denote cfe by z z . 
(43-1) becomes 



. Let 
Then 



(43-2) z - z = 



(.ro, 



(X - X ) + 






Jo, ?/o) 



- 2/0), 



which is the equation of a plane. If this plane is cut by the 
plane x = x , the equation of the line of intersection is 



o, yo) 

and this is the tangent line to the curve z = /(xo, y) at the point 
(x , 2/0, z ). Similarly, the line of intersection of the plane defined 
by (43-2) and the plane y = 2/0 is the tangent line to the curve 
z = /(x, T/O) at (x , 2/0, z ). The plane defined by (43-2) is called 
the tangent plane^to the surface z = /(x, y) at (x , 2/0, z ). 

The direction cosines of the normal to this plane are propor- 
tional to 



dx 



dy 



-1. 



The equation of the normal line to the plane (43-2) at (x , 2/0, z ) 
is therefore 



(43-3) 



x x 



y ~ 



Z Zp 



d JL 

dx 



df_ 



, 2/0) 



(so, 2/0) 



This line is defined as the norrrial to the surface at (#o, 2/o, z ). 
Figure 34 shows the difference between dz = RP' and Az = RQ. 
P(%o, 2/o, z ) is the point of tangency and R(XO + Ax, t/ + Ay, z ) 
is in the plane z = z . PP 7 is the tangent plane. 



148 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 43 

In case the equation of the surface is given in the form 
F(x, y, z) = 0, 

the tangent plane and the normal line at (x 0) y , z ) have the 
respective equations 

t \ -L. dF ( \ 

(X - Xo; -f- ^-7 (y yo) 



(xo, in, zo) 



(xo, yo, z ) 



dz 



(n, 



(z - 



and 

(43-5) 



x XQ y y$ 

dF\ = dF\ 

^^|(xo, i/o, zo) y\ (%o, yo, 20) 



z 



(xo,yo,o) 



These equations follow directly from (41-2). 




FIG. 34 



Example 1. At (6, 2, 3) on the surface z 2 + ?/ 2 + ^ = 49, the 
tangent plane has the equation 



(x - 6) + 



(6, 2, 3) 



The normal line is 



(6, 2, 3) 



(y - 



(6, 2, 3) 



(z - 3) = 



6x + 2y + 3z = 49. 

s - 6 __ y - 2 __ z - 3 
~~ ~~ ~~ 



Example 2. For (2, 1, 4) on the surface z x 2 + y 2 1, the tangent 
plane is 



z - 4 = 2x 



(2, 1) 



(x - 2) + 



or 



2y - z = 6. 



44 PARTIAL DIFFERENTIATION 149 

The normal line is 

x - 2 _ y - 1 z - 4 

~T~ - 2 - -r 

PROBLEMS 

1. Find the distance from the origin to the plane x + y + z = 1. 

2. Find the equations of the tangent plane and the normal line to 

(a) 2x* + 3?/ + 4z 2 = 6 at (1, 1, K); 

(*) T + ?-S ~ lat ( 4 > 3 > 8 )' 

(<0 ^2 + |i + ^ = 1 at (*o, 2/o, *o); 
(d) z 2 + 2*/ 2 -- z 2 = at (1, 2, 3). 

3. Referring to (43-4), show that 

dF OF dF 
cosa:cos0-cos7 '^Wl? 

where cos a, cos j8, cos 7 are direction cosines of the normal line. 

4. Show that the sum of the intercepts on the coordinate axes of any 
tangent plane to x^ + yW + z^ = a 1 / 2 is constant. 

44. Space Curves. It will be recalled that a plane curve C 
whose equation is 

(44-1) y = f(x) 

can be represented in infinitely many ways by a pair of parametric 
equations 

(44-2) x = x(t), 

y = y(t) 

so chosen that when the independent variable t runs continuously 
through some set of values ti < t < t% the corresponding values 
of x and y, determined by (44-2), satisfy (44-1). 

For example, the equation of the upper half of a unit circle 
with the center at the origin of the cartesian system, 

y = 
can be represented parametrically as 

x = cos tj 

y sint, (0 ^ t ^ TT), 



150 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 44 
or 



or 

x = 2t, 

y = A/i .- 4* 2 , (0 <; i : M). 

Similarly, a space curve C can be represented by means of a 
set of equations 

x = x(t\ 
(44-3) < T/ - 



so selected that when t runs through some set of values the 
coordinates of the point P(x, y, z), defined by (44-3), trace out 
the desired curve C. 




FIG. 35. 
i 

It will be assumed that the functions in (44-2) and (44-3) 
possess continuous derivatives with respect to t, which implies 
that the curve C has a continuously turning tangent as the point 
P moves along the curve. 

Let P(x Q , yo, ZQ) (Fig. 35) be a point of the curve C defined by 
(44-3) that corresponds to some value U of the parameter t y 
and let Q be the point (XQ +' A#, yo + Ay, z + Az) that cor- 
responds to t = to + A. The direction ratios of the line PQ 



45 PARTIAL DIFFERENTIATION 151 

joining P and Q are 

Ax A?/ Az __ Ax^A^Az 
Ac Ac* Ac A* A* A2 

If A is allowed to approach zero, Ax, At/, and Az all tend to zero, 
so that the direction ratio h s of the tangent line at P(XQ, y Q , z ) 
are proportional to (dx/df)^t '(dy/dt)t^to-(dz/d()t^to- Hence, the 
equation of the tangent line to C at P is 

x XQ _ y -- j/o _ z zp 

where primes denote derivatives with respect to . 

Example. The equation of the tangent line to the circular helix 

x a cos t, 

y = a sin , 

z = a, 
at t - 7T/6, is 



a 

~ 2 ~2 a 

The element of arc c?s is given by 

(da) 2 = (dz) 2 + W + (^) 2 , 
so that the length of a space curve C can be calculated from 

rv 2 



. , 



The length of the part of the helix between the points (a, 0, 0) and 

(0, a, 7ra/2) is 



45. Directional Derivatives in Space. There is no essential 
difficulty in extending th>e results of Sec. 42 to any number of 
variables. Thus*, if u f(x, y, z) is a suitably restricted func- 
tion of the independent variables x, y, and z, then the directional 
derivative along a space curve whose tangent line at some point 
P(x, y, z) (Fig. 35) has the direction cosines cos (x, s), cos (y, s), 



152 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 45 

and cos (z, s) is 

du du f N . du f N , du f , 
^ = - cos (x, s)+- cos (i,, .) + - cos (z, ,) 



The magnitude of the normal derivative to the surface u = const. 
is given by 

' 



The vector that is normal to the surface u const, and whose 
magnitude is du/dn is called the gradient of u. 

PROBLEM 

1. Find the equation of the tangent line to the helix 

x a cos t, y a sin t, z = a, 

at the point where t = Tr/4. Find the length of the helix between the 
points t = and t 7T/4. 

2. Find the directional derivative of / = xyz at (1, 2, 3) in the 
direction of the line that makes equal angles with the coordinate axes. 

3. Find the normal derivative of / = x 2 + y 2 + z 2 at (1, 2, 3). 

4. Show that the square root of the sum of the squares of the 
directional derivatives in three perpendicular directions is equal to 
the normal derivative. 

6. Express the normal derivative (45-1) in spherical and cylindrical 
coordinates, for which the equations of transformation are 

(a) x = r sin cos <p, y r sin 6 sin <p, z = r cos 6; 

(b) x = r sin 0, y = r cos 0, z = z. 

6. What is the direction of the curve x = t, y t 2 , z 3 at the 
point (1, 1, 1)? 

7. Show that the condition that the surfaces f(x, y, z) =0 and 
g(x, y,z) =0 intersect orthogonally is that 



, . , = 

dx dx dy dy dz dz 



8. Show that the surfaces 

xyz = 1 
intersect at right angles. 



x 2 y 2 z 2 
xyz = 1 and -^ + -% - f = 



46 PARTIAL DIFFERENTIATION 153 

9. Find the angle between the normals to the tangent planes to 
the surfaces x* + y 2 + z z = 6 and 2z 2 + 3?/ 2 + z* = 9 at the point 
(1, 1, 2). 

10. Show that the direction of the tangent line to the curve of inter- 
section of the surfaces f(x, y, z) = and g(x t y, z) = is given by 



a 
cos a:cos j3:cos 7 = 



/ 



A /> 



Hint: Let (XQ, T/O, ^o) be a point on the curve of intersection, and find 
the line of intersection of the tangent planes to the surfaces at the 
point (XQ, 7/0, Zo). 

46. Higher Partial Derivatives. The partial derivatives f* i9 

/v * * * y f* ^ f( Xl > x z, ' ' ' j #n) are functions of x\, x%, - , 
x n and may have partial derivatives with respect to some or all 
of these variables. These derivatives are called second partial 
derivatives of f(x iy x z , ' ' ' , x n ). If there are only two inde- 
pendent variables x and t/, then f(x, y) may have the second 
partial derivatives 



df\ _ d 2 / _ 
Jx) ^ dx* ^ ***' 

^ JXV) 



dy \dx/ dy dx 

dx \dy/ dx dy 

d_(V\ s ^ sf 

dy \dy/ dy 2 v ' 

It should be noticed that f xy means that df/dx is jBrst found and 

then ( ) is determined, so that the subscripts indicate the 
dy \dx/ 

order in which the derivatives are taken. In 

a 2 / _ a (ef\ 

dy dx dy \dx/ 

the order is in keeping with the meaning of the symbol, so that 
the order appears as the reverse of the order in which the deriva- 
tives are taken. 

It can be proved* that, if f^ and/^ are continuous functions 
of x and y f then /^ = f yx , so that the order of differentiation is 

* See SOKOLNIKOFF, I. S., Advanced Calculus, Sec. 31. 



154 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 46 

immaterial. Similarly, when third partial derivatives are 
found, f xyx = f xxy = f yxx and f^ y = f vxy = f vyx , if these deriva- 
tives are continuous. 

Example. If f(x, y) = e**, then 

/* = ye*", f v = 6**, /, yV*, 

/* = /** = ^ xv (^ + 1), /yy = tfe*"- 

PROBLEMS 

1. Verify that ^- = ^- for 

J 60; d?/ cty d# 

(a) / = cos xy 2 , (6) / = sin 2 x cos y, (c) f = e y / x . 

2. Prove that if 

v d*f d 2 f 

(a) /(*, y) = log ( + t/ 2 ) + tan- 1 J then -^ + ^ = 0; 

(W /<*, y, ) - W + 2/ 2 + ^)~^, then g + g + g = 0. 

3. If u = x* + y* and \ X = * + 3 j' find ~ and ~- 

J (y = 2s t, ds* dt 2 

(x = r cos 0, d 2 u 3 2 u 

4. If u = /(x, y) arid j gin ^ find ^- 2 and ^- 

5. Use the results obtained in Prob. 6(6), Sec. 41, in order to show 
that ~ - u(W - u 2 )/(u* + v 2 )*. Find ^- and |~- 

2 ^ // V / l 



6. If -a? = ^(x, ?/), where x x(u, v), y = t/(w, ?;), dx/du dy/dv, 
and dx/dv = dy/dUj show that 



_ 

dw 2 " ^ 2 ~ \dx 2 

7. Show that the expressions 

dz* . /dz\* , a 2 ^ d 2 z 



upon change of variable by means of x = r cos 6 and i/ = r sin ^ become 

T/ /^Y , 1 /^A 2 A T/ d20 , 1 ^ 2 2 1 ^ 

Fl = \dr) + ^ (do) and F2 - 5? + F 2 ^ 2 



8. If F = /(a; + ct) + g(x ct} } where / and ^ are any functions 
possessing continuous second derivatives, show that 



_ , 

" c 



47 PARTIAL DIFFERENTIATION 155 

9. Show that if x e r cos and y = e r sin 8, then 
?!Z . ^!Z _ -2 /d 2]/ , d!T\ 

dx* + a?/ 2 ~' e A^ 2 a* V 

10. If 7i(x, 2/, 2) and y 2 (z, T/, 2) satisfy the equation 



show that 

satisfies the equation 

d 2 d 2 d 2 

47. Taylor's Series for Functions of Two Variables. This sec- 
tion contains a formal development of a function of two variables, 
/(x, t/), in a series analogous to the Taylor's series development 
of a function of a single variable. It is assumed that the series 
obtained here converges to the value of the function f(x t ?/), 
but the analysis of the conditions under which this convergence 
will occur is too involved to be discussed in this book. 

Consider /(x, y), which is a function of the two variables x and 
T/, and let it have continuous partial derivatives of all orders. 
Let 

(47-1) x = a + at and y = b + pt, 

where a, 6, a, and p are constants and t is a variable. Then 
(47-2) /(x, y) = /(a + at,'b + pt) = F(t). 

If F(t) is expanded in Maclaurin's series, there results 

F"(Q) F"'(Q) 

. L W *2 _L L W *3 _1_ . . . 



(47-3) 

From (47-2) and (47-1), it follows that 



= /.(,) +fy(x,y)P- 

Then, 



156 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 47 

and 

dx 
+ f yyx (x, y}p] - 



3f xyy (x, y)a 



Higher order derivatives of F(t) can be obtained by continuing 
this process, but the form is evident from those already obtained. 
Symbolically expressed, 



toa? + ^d' 

Then, 



n _ A d n f _ n o^/ 

where 



n 



-r ~ r!(n _ r)! 

Since = gives x = a and ?/ = 6, it follows that 

F(0) = /(a, 6), F'(0) = af x (a, 6) + )8/ y (a, 6), - . 
Substituting these expressions in (47-3) gives 
*XO = /(, y) = /(, fc) + W(a, &) + fify(a, &)] 

Since at = a; a and # = y 6, the expansion becomes 
(47-4) /(x, 2/) = /(a, 6) +/(a, 6)(* - a) +/,(a, 6)(y - 6) 

+ A [/*(<*, b)( - ) 2 + 2/^(a, 6)(x - a)(y - 6) 



47 PARTIAL DIFFERENTIATION 157 

This is Taylor's expansion for a function f(x, y) about the point 
(a, b). 

Another form that is frequently used is obtained by replacing 
(x a) by h and (y b) by k, so that x = a + h and y = b + fc. 
Then, 

(47-5) f(a + h,b + k) = /(a, 6) + /,(a, V)h + / v (a, 6)fc 

+ j r/~(a, &)^ 2 + 2/*,(a, b)hk + / w (a, 6)fc 2 ] + . 

This formula is frequently written symbolically as 
/(a + h, b + fc) = /(a, 6) + (h ^ + * 



Example. Obtain the expansion of tan" 1 - about (1,1) up to the 

y 

third-degree terms. Here, f(x, y) = tan" 1 -> so that 

x 

f(x, y) = tan- 1 -> /(I, 1) = tan" 1 1=7; 



/(i y) = 

xxfe y) = 



/vv(> y) = (^2 _|_ yaj 
Then, 



PROBLEMS 

1. Obtain the expansion for x^/ 2 + cos xy about (1, Tr/2) up to the 
third-degree terms. 



158 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 48 



2. Expand f(x, y) = e xv at (1, 1), obtaining three terms. 

3. Expand e x cos y at (0, 0) up to the fourth-degree terms. 

4. Show that, for small values of x and y, 



and 



e* sin y = y + xy (approx.), 

y2 

' log (1 + y) = y + xy - ^ (approx.). 



5. Expand f(x, y) = x*y + x 2 y + 1 about (0, 1). 

6. Expand (1 x 2 y 2 )^ about (0, 0) up to the third-degree 
terms. 

7. Show that the development obtained in Prob. 6 is identical with 
the binomial expansion of [1 (x 2 -f y 2 )] }/ ^. 

48. Maxima and Minima of Functions of One Variable. A 

function f(x) is said to have a maximum at x = a, if 



and 



A+ s /(a + h) - /(a) < 0, 
A- = /(a - A) - /(a) < 0, 



for all sufficiently small positive values of h. If A + and A"~ are 

both positive for all small positive values of h, then f(x) is said 

to have a minimum at # = a. 

It is shown in the elementary calculus that, if the function 

f(x) has a derivative at x = a, then the necessary condition for a 

maximum or a minimum is the 
vanishing of f'(x) at the point 
x = a. Of course, the function 
f(x) may attain a maximum 
or a minimum at x = a with- 
out having /'(a) = 0, but this 
^ x can occur only if /'(#) ceases to 
exist at the critical point (see 
Fig. 36). 
Let it be supposed that f(x) has a continuous derivative of 

order n in some interval about the point x = a. Then it follows 

from Taylor's formula that 




a 



FIG. 36. 



48 PARTIAL DIFFERENTIATION , 159 

where < 61 < 1, and 

A- B /(a - h) - /(a) 



where < 02 < 1- I^et it be assumed further that the first 
n 1 derivatives of f(x) vanish at x = a but that / (n) (a) is not 
zero. Then 



n\ 
and 



Since f (n) (x) is assumed to be continuous in some interval 
about the point x = a, f (n) (a + Bik) and f (n} (a - eji) will 
have the same sign for sufficiently small values of h. Conse- 
quently, the signs of A + and A~ will be opposite unless n is an 
even number. But if f(x) is to have a maximum or a minimum 
at x a, then A+ and A~ must be of the same sign. Accordingly, 
the necessary condition for a maximum or a minimum of f(x) 
at x = a is that the first non-vanishing derivative of f(x), at 
x a, be of even order. Moreover, since both A 4 " and A~ 
are negative if f(x) is a maximum, it follows that/ (n) (a) must be 
negative. A similar argument shows that, if /(#) has a minimum 
at x = a, then the first non-vanishing derivative of f(x) at x = a 
must be of even order and positive. 

If the first non-vanishing derivative of f(x) at x = a is of odd 
order and/" (a) = 0, then the point x = a is called a point of inflection. 

Example. Investigate f(x) x 5 5# 4 for maxima and minima. 
Now, 

f(x) - 5z 4 - 20z 3 , 

which is zero when x and x = 4. Then, 

f"(x) = 20x 3 - 60s 2 , /"(O) = 0, /"(4) * 320; 

/'"(a?) = 60s 2 - 120s, /'"(O) = 0; 
f(x) = 120s - 120, / IV (0) = -120. 

Since /"(4) > 0, /(4) = -256 is a minimum; and since / IV (0) < 0, 
/(O) = is a maximum. 



160 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 49 

PROBLEMS 

1. Examine the following for maxima and minima: 

(a) y = x* - 4x* + 1 ; 
(6) y = x*(x - 5) 2 ; 
(c) y x + cos x. 

2. Find the minimum of the function y x x , where x > 0. 
Hint: Consider the minimum of log y. 

3. Show that x = gives the minimum value of the function 

y e x + 6~ z + 2 COS . 

4. Find maxima, minima, and points of inflection, and sketch the 
curves, for the following: 

(a) y 3x -f 4 sin x + sin 2x; 

(b) y 3x 4 sin x + sin 2z; 

(c) i/ = 6z + 8 sin x + sin 2x. 

6. Find the maximum and minimum values of the function 
y x sin x + 2 cos x. 

6. Find maxima, minima, and points of inflection, and sketch the 
curves, for the following: 

(a) y = zlogz; 

(b) x 5 - (y - x 2 )* = 0. 

49. Maxima and Minima of Functions of Several Variables. 

A function of two variables f(x, y) is said to have a maximum at 
(a, >), if /(a + h, b + k) - /(a, 6) < for sufficiently small 
positive and negative values of h and ft, and a minimum, if 
/(a + h, b + k) - /(a, 6) > 0. 

Geometrically, this means that when the point (a, 6, c) on the 
surface z f(x, y) is higher than all neighboring points, then 
(a, 6, c) is a maximum; and when (a, b, c) is lower than all 
neighboring points, it is a minimum point. At a maximum or a 
minimum point (a, 6, c) the curves in which the planes x = a and 
y = b cut the surface have maxima or minima. Therefore, 
f*(a, 6) = and / y (a, 6) = 0. The conditions f x = and / y = 
can be solved simultaneously to give the critical values. 

The testing of the critical values for maxima and minima is 
more difficult than in the case of functions of one variable. 
However in many applied problems the physical interpretation 



49 PARTIAL DIFFERENTIATION 161 

will determine whether or not the critical values yield maxima 
or minima or neither. An analytical criterion can be established 
for the case of two variables in a manner analogous to the method 
used for one variable. By the use of Taylor's expansion, it can 
be shown that if / x (a, b) = and f v (a, 6) = 0, then /(a, 6) is a 
maximum if 

D ^ &(a, b) - /,(a, b}f yy (a y b) < 
with 

/**(, 6) < and f yy (a, 6) < 0, 

and a minimum if 

D**fi,,(a,b) -/,(a,6)/ w (a,6) <0 
with 

fxx(a, b) > and f yy (a, b) > 0. 
In case 

fly(a, b) - f xx (a, 6)/ w (a, 6) > 0, 

/(a, fc) is neither a maximum nor a minimum. If 

&(M) ~ /~(, &)/**(, &) =0, 

the test gives no information, just as/"(z) = gives no criterion 
in the case of one variable. 

These considerations can be extended to functions of more than 
two variables. Thus, in the case of a function f(x, y, z) of three 
variables, 

f , o, = 0, % = 

dx ' dy ' dz 

is the necessary condition for a maximum or a minimum. 



Example 1. A long piece of tin 12 in. r 
wide is made into a trough by bending up \ 
the sides to form equal angles with the 
base (Fig. 37). Find the amount to be ^t-\. 
bent up and the angle of inclination of the /2-2x 



sides that will make the carrying capacity ,_ 

Jb IG. o7. 
a maximum. 

The volume will be a maximum if the area of the trapezoidal cross 
section is a maximum. The area is 

A = I2x sin 6 - 2x 2 sin + x 2 sin 9 cos 0; 
for 12 2x is the lower base, 12 2x + 2x cos 6 is the upper base, 



162 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 49 
and x sin 6 is the altitude. Then, 

33- = I2x cos 6 - 2z 2 cos 6 + x 2 cos 2 6 - x 2 sin 2 

O0 

= z(12 cos 6 - 2z cos + z cos 2 - z sin 2 0) 
and 

|~ = 2 sin 0(6 - 2z -f x cos 0). 

dA/dx = and 3A/dB 0, if sin = and z = 0, which, from physical 
considerations, cannot give a maximum. 
There remain to be satisfied 

6 - 2x + x cos = 
and 

12 cos - 2x cos 6 -f x cos 2 6 - x sin 2 = 0. 

Solving the first equation for x and substituting in the second yield, 
upon simplification, 

cos = J-6 or = 60, and x = 4. 

Since physical considerations show that a maximum exists, x = 4 and 
= 60 must give the maximum. 
Example 2. Find the maxima and minima of the surface 



Now, 

d% x dz 



which vanish when x = y = 0. But 



dx 2 a 2 c dy 2 b 2 c dx dy 

Hence, D = l/a 2 6 2 c 2 and, consequently, there is no maximum or minimum 
at x = y = 0. The surface under consideration is a saddle-shaped 
surface called a hyperbolic paraboloid. The points for which the first 
partial derivatives vanish and D > are called minimax. The reason 
for this odd name appears from a consideration of the shape of the 
hyperbolic paraboloid near the origin of the coordinate system. The 
reader will benefit from sketching it in the vicinity of (0, 0, 0). 

PROBLEMS 

1. Divide a into three parts such that their product is a maximum. 
Test by using the second derivative criterion. 



60 PARTIAL DIFFERENTIATION 163 

2. Find the volume of the largest rectangular parallelepiped that can 
be inscribed in the ellipsoid 






4. -4- - - i 
a 2 " 1 " 6 2 "^c 2 

3. Find the dimensions of the largest rectangular parallelepiped that 
has three faces in the coordinate planes and one vertex in the plane 



4. A pentagonal frame is composed of a rectangle surmounted by an 
isosceles triangle. What are the dimensions for maximum area of the 
pentagon if the perimeter is given as P? 

5. A floating anchorage is designed with a body in the form of a right- 
circular cylinder with equal ends that are right-circular cones. If the 
volume is given, find the dimensions giving the minimum surface area. 

6. Given n points P % whose coordinates are (#, 7/ t , z t ), (i 1,2, 
, n). Show that the coordinates of the point P(x, y, z), such that 
the sum of the squares of the distances from P to the P % is a minimum, 
are given by 



(1 n 1 n ^ i n \ 
nS^S^S*)' 



50. Constrained Maxima and Minima. In a large number of 
practical and theoretical investigations, it is required that a 
maximum or minimum value of a function be found when the 
variables are connected by some relation. Thus, it may be 
required to find a maximum of u = f(x, y t z), where x, y, and z 
are connected by the relation <p(x, y, z) = 0. The resulting 
maximum is called a constrained maximum. 

The method of obtaining maxima and minima described in 
the preceding section can be used to solve a problem of con- 
strained maxima and minima, as follows: If the constraining 
relation <p(x, y, z} = can be solved for one of the variables, 
say z, in terms of the remaining two variables, and if the resulting 
expression is substituted for z in u = f(x, y, 2), there will be 
obtained a function u = F(x t y). The values of x and y that 
yield maxima and minima of u can be found by the methods of 
Sec. 49. However, the solution of <p(x, y, z) = for any one 
of the variables may be extremely difficult, and it is desirable to 
consider an ingenious device used by Lagrange. 



164 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 50 

To avoid circumlocution the maximum and minimum values 
of a function of any number of variables will be called its extremal 
values. It follows from Sec. 49 that the necessary condition 
for the existence of an extremum of a differentiable function 
f(xi, #2, , n) is the vanishing of the first partial derivatives 
of the function with respect to the independent variables Xi, 
Xz, ' ' * , %n- Inasmuch as the differential of a function is 
defined as 



it is clear that df vanishes for those values of x\, Xz, ' ' , x n 
for which the function has extremal values. Conversely, since 
the variables x t are assumed to be independent, the vanishing 
of the differential is the necessary condition for an extremum. 

It is not difficult to see that, even when some of the variables 
are not independent, the vanishing of the total differential is 
the necessary condition for an extremum. Thus, consider a 
function 

(50-1) u = f(x, y, z), 

where one of the variables, say z, is connected with x and y by 
some constraining relation 

(50-2) <p(x, y, z) = 0. 

Regarding x and y as the independent variables, the necessary 
conditions for an extremum give du/dx = and du/dy = 0, or 



du ^ df = 

dx dx "*" dz dx ' 



dy dy ~~ dz dy 
Then the total differential 

du , , du j df , . df . . dfdz , , dz 

- ~ - 



and since the expression in the parenthesis is precisely dz, it 
follows that 

(5 - 3) ^ + ^ + ^ - 



60 - PARTIAL DIFFERENTIATION 165 

The total differential of the constraining relation (50-2) is 






5* 



*- 



Let this equation be multiplied by some undetermined mul- 
tiplier X and then added to (50-3) . The result is 



Now, if X is so chosen that 



(50-5) 



f\ i^ ** \ 
dx dx 



, y, 



= o, 



then the necessary condition for an extremum of (50-1) will 
surely be satisfied. 

Thus, in order to determine the extremal values of (50-1), 
all that is necessary is to obtain the solution of the system of 
Eqs. (50-5) for the four unknowns x, y, z, and X. The multiplier 
X is called a Laqrangian multiplier. 

Example 1. Find the maximum and the minimum distances from 
the origin to the curve 

5z 2 + fay + 5?/ -8 = 0. 
The problem here is to determine the extremal values of 

f(x, y) = x z + y 2 
subject to the condition 

<p(x, y) SE 5z 2 + fay + 5i/ 2 - 8 = 0. 
Equations (50-5) in this case become 

2x + \(Wx + 6y) = 0, 
2y + X(6z + 102/) = 0, 
5z 2 + fay + 5y 2 - 8 = 0. 



Multiplying the first of these equations by y and the second by x and 



166 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 50 

then subtracting give 

6X(?/ - a: 2 ) = 0, 

so that y = x. Substituting these values of y in the third equation 
gives two equations for the determination of x, namely, 

2z 2 = 1 and x 2 = 2. 

The first of these gives / = x 1 + y 2 = 1, and the second gives/ = x 2 
+ 2/ 2 = 4. Obviously, the first value is a minimum, whereas the 
second is a maximum. The curve is an ellipse of semiaxes 2 and 1 
whose major axis makes an angle of 45 with the z-axis. 

Example 2. Find the dimensions of the rectangular box, without a 
top, of maximum capacity whose surface is 108 sq. in. 

The function to be maximized is 

f(x t y, z) 35 xyz, 
subject to the condition 

(50-6) xy + 2xz + 2yz = 108. 

The first three of Eqs. (50-5) become 

( yz + \(y + 2z) = 0, 

(50-7) 4 xz + \(x + 2z) = 0, 

(xy + \(2x + 2y) =0. 

In order to solve these equations, multiply the first by x, the second 
by ?/, and the last by z, and add. There results 

\(2xy + 4xz + 4?yz) + 3xyz = 0, 
or 

2xz + 2yz) + %xyz = 0. 



Substituting from (50-6) gives 

108X + %xy* = 0, 
or 

\ _ _ *y* 

A ~ 72' 

Substituting this value of X in (50-7) and dividing out common factors 
give 

1 - ^ (y + 2z) = 0, 

i - ^ (* + ao - o, 

1 - (2* + 2y) - 0. 



61 PARTIAL DIFFERENTIATION 167 

From the first two of these equations, it is evident that x = y. The 
substitution of a: = y in the third equation gives z = 18/y. Substitut- 
ing for y and z in the first equation yields x = 6. Thus, x = 6, y = 6, 
and 3 = 3 give the desired dimensions. 

PROBLEMS 

1. Work Probs. 1, 2, and 3, Sec 49, by using Lagrangian multipliers. 

2. Prove that the point of intersection of the medians of a triangle 
possesses the property that the sum of the squares of its distances from 
the vertices is a minimum. 

3. Find the maximum and the minimum of the sum of the angles 
made by a line from the origin with (a) the coordinate axes of a cartesian 
system; (b) the coordinate planes. 

4. Find the maximum distance from the origin to the folium of 
Descartes x 3 + y 3 3<m/ = 0. 

5. Find the shortest distance from the origin to the plane 

ax + by + cz = d. 

51. Differentiation under the Integral Sign. Integrals whose 
integrands contain a parameter have already occurred in the 
first chapter. Thus, the length of arc of an ellipse is expressi- 
ble as a definite integral containing the eccentricity of the ellipse 
as a parameter. * 

Consider a definite integral 

(51-1) *() = '/(*, ) **, 

in which the integrand contains a parameter a and where UQ and 
ui are constants. As a specific illustration, let 

7T 

/*2 

<p(a) = I sin ax dx. 
Jo 

In this case the indefinite integral 

f . cos ax ~ 

F(x, ot) I sin ax dx = h C 

is a function of both x and a ; but, upon substitution of the limits, 
there appears a function of a. alone, namely 

IT T 

, . P 2 . , cos ax\ 2 I/- TrcA 

via) = I sm ax ax = = - I 1 cos -^ I- 

Jo . a |o a \ 2 / 

* See Sec. 14. 



168 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 61 

Frequently, it becomes necessary to calculate the derivative 
of the function <p(a) when the indefinite integral is complicated 
or even cannot be written down explicitly. Inasmuch as the 
parameter a is independent of x, it appears plausible that in 
some cases it may be permissible to perform the differentiation 
under the integral sign, so that one can use the formula 

dtp _ J Ul df(Xj a) , 
da Juo da 

This formula turns out to be correct if /(x, a) and d/(z, a) /da 
are continuous functions in both x and a. Thus, forming the 
difference quotient with the aid of (51-1), 



Juo 



- f(x, ) 



v ' Aa J wo Aa 

Now the limit, as Aa > 0, of the left-hand member of (51-2) 
is precisely dy/da, whereas the limit of the expression under the 
integral sign is df/da. Hence, if it is permissible to interchange 
the order of integration and calculation of the limit, one has 



- 

da 

The restrictions imposed on the function f(x, a) can be shown 
to be sufficient to justify the inversion of the order of these 
operations. 

Suppose next that the limits of integration u\ and UQ are func- 
tions of the parameter a, so that 



In this case, one can proceed as follows : Let 

ff(x,a)dx = F(x,a) 
so that 

(51-4) ~ = f(x, a). 

Then, 

(51-5) <f>(a) = f" 1 f(x, a) dx = F(x, a) 

Juo(a) 

i, a) - F(tt , a). 



61 PARTIAL DIFFERENTIATION 169 

Assuming the continuity of all the derivatives involved, one 
can write* 

, g) 



_ _ __ __ 

da du\ da da du$ da da 

which, upon making use of (51-4) and (51-5), becomes 



~ , a)] 



- A"., ) p + f U>M f(x, a) dx. 
a da da J wo () 



The partial derivative appearing in this expression means that 
the differentiation is to be performed with respect to a, treating 
UQ and u\ as constants. Hence, making use of (51-3), 

/r , ^ d<p ., . du\ , . duo . C ui(a} df(x y a) , 
(51-6) = f( Ul , a)-- f(u , a) + J--l dx. 



This formula is known as the formula of Leibnitz, and it 
specializes to (51-3) when HI and U Q arc independent of a. The 
validity of this formula can be established under somewhat less 
restrictive hypotheses,! but the limitations imposed on the func- 
tion f(x, a) in the foregoing discussion are usually met in prob- 
lems arising in applied mathematics. 

d- r 2a - 

Example 1. Find -T-, if #(a) = J_ a2 e " 2 dx. 

Then 



/2a 9 r l> _ r ! 
- ^3* e ""' (JX 



Example 2. Formula (51-3) is frequently used for evaluating definite 
integrals. Thus, if 



(a) = J log (1 + a cos x) dx, 



* See Sec. 39. 

t See SOKOLNIKOFF, I. S., Advanced Calculus, Sec. 39, p. 121. 



170 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 51 
then 



=lxfo( l - 1+acosJ 



cos 



1 + a cos # 



1 r , 1 /.,-! . ,<* + 1\1 

= - TT H -- 7 : -- ( sin" 1 -- sm" 1 - ) 
al Vl - 2 \ 1 -a 1 + a/J 



= 

" a 



Therefore, 

(D\CX.) == 7T I I 7 :==: - 

J \a a\/l - a z 

i + vr^s 



g a + log 
\ 

or 



<p(a) = TT log (I + Vl ~ 2 ) + c. 
But, when a = 0, 



Hence, 

= TT log 2 + c and c = TT log 2, 

and ___ _ 

, . , /i + Vi - 2 \ 

^(a) - TT log ^T-^ - J. 
PROBLEMS 

J/2 
sin ao; d^ by using the Leibnitz formula, 

and check your result by direct calculation. 

2. Find dp/da, if <p(a) = J Q (1 a cos a:) 2 dx. 

r a a; 

3. Find d<p/da, if ^(a) = J tan- 1 ^ ete. 

4. Find dp/dot, if <p( = JQ tan ( x "" a ) ^ a; - 

5. Find d(p/dx y if ^(o?) = J Q V^ ^- 

/*7T fa 

6. Differentiate under the sign and thus evaluate J Q -7 - ^ 

. C* dx 7T .. . . 

by using L - = 5 - r> if a 2 > 1. 
J b Jo a cos # a 2 1 

7. Show that 

f T log (1 - 2a cos 3 + a 2 ) dx = 0, if a 2 ^ 1 

= ir log a 2 , if a a ^ 1, 



61 PARTIAL DIFFERENTIATION 171 

8. Verify that 



is a solution of the differential equation 

jg3 + k*y = /(&)- 
where A; is a constant. 



CHAPTER V 
MULTIPLE INTEGRALS 

It is assumed that the reader is somewhat familiar with the 
problem of calculating the volumes of solids with the aid of double 
and triple integrals and has some facility in setting up such 
integrals. The first three sections of this chapter contain a 
brief summary of some basic facts concerning double and triple 
integrals, preparatory to the development of the expressions for 
the volume elements in spherical and cylindrical coordinates. 
These expressions are used frequently in applied mathematics 
and are seldom included in the first course in calculus. A brief 
discussion of surface integrals is also given here. 

First, it may be well to recall the definition of the simple 
integral J* f(x) dx. Let the function f(x) be continuous and 
single valued for a < x < b. The interval (a, b) of the #-axis 
is divided into n parts by the points a z= X Q , xi, x 2 , , 
x n = b. Let A# = Xi #t_i, and let t be a value of x such that 

n 

rct-i < t ^ x^ Form the sum 23 /() A# t , and take the limit 

t==i 

of this sum as n <*> and all the Ax l > 0. Under the given 
assumptions on f(x), this limit will exist, and it is defined as the 
definite integral of f(x) over the interval (a, b) of the #-axis. 
Thus, 

lim X/({.) A*. ^ P /(*)**- 



Geometrically, this integral can be interpreted as the area between 
the curve y = f(x) and the x-axis included between the lines 
x = a and x = b. The evaluation of the integral can often be 
accomplished by the use of the following theorem. 

FUNDAMENTAL THEOREM OF INTEGRAL CALCULUS. // f(x) 
is continuous in the interval a < x ^ b and G(x) is a function such 
that dG/dx = f(x) for all values of x in this interval, then 



f b f(x) dx = O(b) - G(a). 

Ja 



172 



52 



MULTIPLE INTEGRALS 



173 



52. Definition and Evaluation of the Double Integral. The 

double integral is defined and geometrically interpreted in a 
manner entirely analogous to that sketched above for the simple 
integral. Let f(x, y) be a continuous and single-valued function 
within a region R (Fig. 38), bounded by a closed curve C, and 
upon the boundary C. Let 
the region R be subdivided 
in any manner into n sub- 
regions AjRi, A# 2 , ' ' ' , A.R n 

of areas A A i, A.A 2 , * * ,AA n . 
Let ( t , 7? t ) be any point in 
the subregion AjR t , and form 
the sum 




FIG. 38. 



The limit of this sum, as n > o and all AA t > 0, is defined as 
the double integral of f(x, y) over the region R. Thus, 



(52-1) 



Km 



f(x, y) dA. 



The region R is called the region of integration, corresponding 
to the interval of integration (a, b) in the case of the simple 
integral. The integral (52-1) is sometimes written as 



In order to evaluate the double integral, it will be simpler to 
consider first the case in which the region R (Fig. 39) is a rec- 
tangle bounded by the lines x a, x = 6, y = c, y = d. The 
extension to other types of regions will be indicated later. 
Subdivide R into mn rectangles by drawing the lines x = Xi, 
x = z 2 , , x = Zn-i, y = yi, y = 2/2, , # = y-i. 
Define Ax t = z t x t __i, where X Q = a and x n = 6, and define 
Ay, = t/ ? 2/j-i, where 2/0 = c and 7/ w = d. Let A.R t / be the rec- 
tangle bounded by the lines x = z t _i, # = x t , ?/ = y/-i, 2/ 2/?- 
Then, if the area of ARij is denoted by AAy, 

= Ax t Ay/. 



174 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 52 

t = n, j * m 

Let ( t; , 7j t ,) be any point of AB#. The sum S /(*,-, ??,) AA t ,- 
can be written as 

t = n, /"Wt 

(52-2) j) /(**" ^^ AXt A2/J '' 

This summation sign signifies that the terms can be summed for 
i and j in any manner whatsoever. Suppose that the terms of 
(52-2) are arranged so that all the rectangles A.R t i are used first, 

y- 
y=d 



Jm-l 



Vi 



->JC 



FIG. 39. 



then all the rectangles A/^ t2 , then all the rectangles A72 t3 , etc. 
This is equivalent to taking the sum of the terms for each row of 
rectangles and then adding these sums. Then (52-2) can be 
written 



(52-3) 
But 






lim 



(*.' th,) Ax.]. 

>5 

*1) Ax = J a f( x > i 



dx > 



so that 






where lim , = 0. Moreover, | /(x, tij) dx is a function of 77,, 

n * /<* 



62 MULTIPLE INTEGRALS 175 

say ^>0?,). Thus (52-3) becomes 



;1 

= f c d v(y} dy + e(d - c) + e' 

fd fb 

~ I I f($> y} dx dy + c(d c) + e , 

Jc Ja 

in which lim e = and lim e f 0. Taking the limit as n > oo 

n * m > oo 

and m > oo gives 

(52-4) J fl f(x, y) dA = J" JT* /(x, y) do; dp. 

The double integral is, therefore, evaluated by considering 
f(Xj y) as a function of x alone, but containing y as a parameter, 
and integrating it between x a and x = 6 and then integrating 
the resulting function of y between y c and y = d. The right 
member of (52-4) is known as an iterated integral, and (52-4) 
establishes the relation between the double integral over the 
rectangle R and an iterated integral over the same rectangle. 

Similarly, by taking the sum of the terms in each column and 
then adding these sums, 

(52-5) f f(x, y) dA = f f* f(x, y) dy dx. 

J K 4/O t/ C 

In case (52-5) is used, f(x, y) is first considered as a function of y 

alone and integrated between y = c 

and y = d, and then the resulting 

function of x is integrated between 

x = a and x = b. Either (52-4) or 

(52-5) can be used, but one of them 

is frequently simpler in the case of a 

particular f unction /(x, y). 

Suppose R is, not a rectangle, 
but a region bounded by a closed 
curve C (Fig. 40) that is cut by any ' FIQ 4Q 

line parallel to one of the axes in, at 
most, two points. Let Bi and JS 2 be the points of C having 
the minimum and maximum ordinates, and let AI and A 2 be 
the points of C having the minimum and maximum abscissas. 
Let x = <?i(y) be the equation of B\A\B^ and x = <p*(y) be the 




176 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 62 



equation of BiA 2 B 2 . Then, in taking the sum of the terms by 
rows and adding these sums, the limits for the first integration 
will be <pi(y) and <p2(y), instead of the constants a and b. The 
limits for the second integration will be 0i and /3 2 , in which /3i is 
the ^/-coordinate of BI and /3 2 is the y-coordinatc of B 2 . Then 
(52-4) is replaced by 



(52-6) 



dA = 



** 



Similarly, if y = fi(x) is the equation of A\BiA^ y = fz(x) is 
the equation of A\B^A^ a\ is the abscissa of Ai, and a 2 is the 
abscissa of A 2 , (52-5) is replaced by 



(52-7) 



/(*, y) dA = 



/(,, y) dy dx. 



y*b 



Tn case R is a region bounded by a closed curve C that is cut 

in more than two points by some 
parallel to one of the axes, the 
previous results can be applied to 
subregions of R whose boundaries 
satisfy the previous conditions. 
By adding algebraically the inte- 
grals over these subregions, the 
double integral over R is obtained. 

Example 1. Compute the value of 






Fio 41 



/i = f R y dA where R is the region in the first quadrant bounded by 
the ellipse 

?! 4_ yl - i 

a 2 "^ 6 2 
Upon using (52-6) and summing first by rows, 

n. . _ _ 



(Fig. 41). 



dy 



3 
Using (52-7), one has 

6 , 

/*ii /*~" v a j - 

i, = r f aV 

Jo Jo 
6 2 



* 3 / /V 2 " v a 2 -x 2 \ 

yd2/da; = Jo(2o ) da: 



53 



MULTIPLE INTEGRALS 



111 



It may be remarked that the value of /i is equal to yA, in which y is 
the ^-coordinate of the center of gravity of this quadrant of the ellipse 
and A is its area. Since A = irab/4, 

/i ab^/3 46 
$ = 1 = mb/l ** &T 

Similarly, by evaluating 7 2 = f R x dA = a 2 6/3, 



3-7T 



x = = 

A 7ra&/4 



which is the ^-coordinate of the center of gravity. 

Example 2. Moment of Inertia. It will be recalled that the moment 
of inertia of a particle about an axis is the product of its mass by the 
square of its distance from the axis. If it is desired to find the moment 
of inertia of a plane region about an axis perpendicular to the plane 
of the region, the method of Sec. 52 can 
be applied, where f(x, y) is the square of 
the distance from the point (x y y) of the ^ 
region to the axis. Then 



M = 




dA\ 



^^ 
(l >> 



For example, let it be required to find 
the moment of inertia of the area in the _ 
first quadrant (Fig. 42), bounded by the 
parabola y 2 1 x and the coordinate 
axes, about an axis perpendicular to the 

xy-p\&ne at (1,0). The distance from any point P(x t y) to (1,0) 
is r = \/(x - I) 2 + y 2 - Therefore, 



FIG. 42. 



Evaluating this integral by means of (52-6) gives 
M -- 



dy 



53. Geometric Interpretation of the Double Integral. If 

f(Xj y) is a continuous and single-valued function defined over the 
region R (Fig. 43) of the xy-pl&ne, then z = f(x, y) is the equation 
of a surface. Let C be the closed curve that is the boundary of R. 
Using R as a base, construct a cylinder having its elements parallel 



178 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 63 




to the z-axis. This cylinder intersects z = f(x, y) in a curve T, 
whose projection on the ;n/-plane is C. Denote by S the portion 
of z = f(x, y) that is enclosed by T. Let R be subdivided as in 
Sec. 52 by the lines x = x l , (i = 1, 2, , n 1), and y = y ]y 

(j = 1, 2, - - - , m - 1). 
Through each line x = x % 
pass a plane parallel to the 
2/z-plane; and through each 
line y = y/ pass a plane 
parallel to the zz-plane. 
The rectangle A72 l? -, whose 
area is AA tJ = Ax t Ay J} will 
be the base of a rectangular 
^y prism of height /(t?, r; t3 ), 
whose volume is approxi- 
mately equal to the volume 
enclosed between the surface 
and the #?/-plane by the 
planes x = x t ~i, x = ic t , 
2/ = 2/j-i, and i/ = j//. Then 
the sum 




FIG. 43. 



(53-1) 



"T" 

<-T7-i 



c* At/,- 



gives an approximate value for the volume V of the portion of the 
cylinder enclosed between z = f(x f y) and the xy-pl&ne. As 
oo and m > o ; the sum (53-1) approaches F, so that 



n 



(53-2) 



y 



The integral in (53-2) can be evaluated by (52-6) in which the 
prisms are added first in the z-direction or by (52-7) in which 
the prisms are added first in the ^-direction. 

It should be noted that formulas (52-6) and (52-7) give the 
value of the area of the region R if the function f(x, y) = 1 ; 
for the left member becomes 



which is A. A can be evaluated by 
dx dy or 



j 

dy 



54 



MULTIPLE INTEGRALS 



179 



Example. Find the volume of the tetrahedron bounded by the 

jC II 2 

plane h T + ~ = 1 and the coordinate planes (Fig. 44). Here, 
a o c 

('--: -9- 

If the prisms are summed first in the ^-direction, they will be summed 
from x = to the line a&, whose equation is 



Therefore, 



fb / X" 32/\|(l 
c I ( x H~ -- r ) V 

Jo V 2a b /|o 



, 

dy 
* 



This result was obtained by using (52-6) for the evaluation of V, but 
(52-7) could be used equally well. 

64. Triple Integrals. The triple 
integral is defined in a manner 
entirely analogous to the definition 
of the double integral. The 
function 

}(> y> z) 

is to be continuous and single 

valued over the region of space R x ' 

enclosed by the surface S. Let R 

be subdivided into subregions AR lJ k. If &V l]k is the volume of 

AR l]k , the triple integral of f(x, y, z) over R is defined by 




FIG. 44. 



(54-1) f R f(x, V ,z)dVm lim 

'** n,m,p i 



i = nj = m,k = p 



by exactly the same argument as that used in Sec. 52. 

In order to evaluate the triple integral, R is considered to be 
subdivided by planes parallel to the three coordinate planes, the 



180 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 5* 



case of the rectangular parallelepiped being treated first. In this 
case, 

A IV = &x % by, A^. 

By suitably arranging the terms of the sum 

n,j 

&;*, I**, ft,*.) 



dy dz. 



it can be shown, as in Sec. 52, that 

(54-2) f f(x, y,z)dV = F f" F f(x, y, a 

a/ ft / *0 %/ tf %/ XO 

By other arrangements of the terms of the sum, the triple integral 
can be expressed by means of iterated integrals in which the order 
of integration is any permutation of that given in (54-2). 

If R is not a rectangular parallelepiped, the triple integral 
over R will be evaluated by iterated integrals in which the limits 
for the first two integrations will be functions instead of con- 
stants. By extending the method of Sec. 53, it can be shown that 

(54-3) f R f(x, y,z)dV = ' f fj^ f(x, y, z) dx dy dz. 

Similarly, the triple integral can be evaluated by interchanging 
the order of integration in the iterated integral and suitably 

choosing the limits. 

The expression (54-3), or the similar 
expressions obtained by a different choice 
of the order of integration, gives the 
formula for the volume of R in case 
vc f( x > y> *0 = 1- Therefore, 




V = 



dx dy dz. 



Fm. 45. AlsQ? the f ormu l a (54.3) may b 

sidered as giving the total mass of the volume V that has vari- 
able density /(x, y, z). 

Example. Let it be required to find the moment of inertia I x of the 
solid bounded by the cylinder # 2 + t/ 2 = a 2 and the planes 2 = and 
z = 6 about the s-axis (Fig. 45). Assume uniform density 0. The 
function f(x, y, z) is the square of the distance of any point P(x, y, z) 
from the z-axis. Therefore, 



54 

Hence, 



MULTIPLE INTEGRALS 



181 



p (a 

Jo 



2 + 6 2 - a 2 sin 2 0) cos 2 



_ 
4 i 6 12 

PROBLEMS 



1. Evaluate 



/ >, fir Ta(l cos 0) , ,/> 

W Jo Jo pdp^, 

and describe the regions of integration in (a) and (6). 

2. Verify that f R (x 2 + y*) dy dx = f R (x* + y' 2 ) dx dy, where the 
region R is a triangle formed by the lines y = 0, y == #, and # = 1. 

3. Evaluate and describe the regions of integration for 



/ \ C a fVa 2 

() JoJ- 



4. Find the areas enclosed by the following pairs of curves: 

(a) y = x, y = z 2 ; 

(6) y = 2~x, 2/ 2 = 2(2-*); 

(c) y = 4 - x*, y = 4 - 2x- f 

(d) y 2 = 5 - x, y = x + 1; 



(e) i/ = 



a 



182 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 64 

6. Find by double integration the volume of one of the wedges cut 
off from the cylinder x 2 + y 2 = a 2 by the planes z = and z = x. 

6. Find the volume of the solid bounded by the paraboloid 
y't -f- %* = 4$ and the plane x = 5. 

7. Find the volume of the solid bounded by the plane = 0, the 
surface z = x 2 + y 2 + 2, and the cylinder x 1 + y 2 = 4. 

8. Find the smaller of the areas bounded by y = 2 x and 
x 2 + 2/ 2 = 4. 

9. Find the volume bounded by the cylinders y = # 2 , ?/ 2 = z and 
the planes 2 = and 2 = 1. 

10. Find the volume of the solid bounded by the cylinders 
x 2 + y 2 = a 2 ahd y 2 + z 2 = a 2 . 

11. Find the coordinates of the center of gravity of the area enclosed 
by y = 4 z 2 and y 4 2x. 

12. Find the moments of inertia about the x- and y- axes of the 
smaller of the areas enclosed by y a x and x 2 + y 2 a 2 . 

13. Evaluate the following: 



f< f W-2/ 2 f \/a 2 -* 2 j 7 7 

Jojo Jo ctectedy; 



14. Find by triple integration 

(a) The volume in the first octant bounded by the coordinate 
planes and the plane x + 2y + 82 4. 

(6) The volume of one of the wedges cut off from the cylinder 
x 2 + y 2 a 2 by the planes 2 = and z = x. 

(c) The volume enclosed by the cylinder x 2 + y 2 1 and the 
planes 2 = and 2 = 2 #. 

(d) The volume enclosed by the cylinders y 2 = z and x 2 + y 2 = a 2 
and by the plane 2 = 0. 

(e) The volume enclosed by the cylinders y 2 + z 2 = a 2 and 
r 2 4- ^2 ^ ^ 

(/) The volume enclosed by y 2 + 2z 2 = 4z - 8, y 2 + z 2 = 4, and 
x = 0. 

(0) The volume in the first octant bounded by the coordinate 
planes and x + 3y + 2z = 6. 

(h) The volume enclosed by the cylinder x 2 + y 2 = 9 and the 
planes 2 = 5 x and 2 = 0. 

(1) The volume of the cap cut off from y 2 + z 2 = 4# by the plane 
2 = x. 

15. Find the moments of inertia about the coordinate axes of the 
solids in Prob. 14. 



$55 MULTIPLE INTEGRALS 183 

16. Find the coordinates of the center of gravity of each of the 
volumes in Prob. 14. 

17. Find by triple integration the moment of inertia of the volume 
of a hemisphere about a diameter. 

18. Find the coordinates of the center of gravity of the volume of the 
solid in Prob. 17. 

19. Find by triple integration the moment of inertia of the volume of 
the c*. ne y 1 + z 2 = aV about its axis. 

20. imd the moment of inertia of the cone in Prob. 19 about a 
diameter of its base. 

21. Find the volume in the first octant bounded by = # + !, z = 0, 
y = 0, x = 2z, and z 2 + i/ 2 = 4. 

22. Find the coordinates of the center of gravity of the volume 
bounded by z = 2(2 x y\ z = 0, and z = 4 z 2 y 2 . 

55. Jacobians. Change of Variable. If it is desired to make 
a change of variable in a double or triple integral, the method is 
not so simple as in the case of the simple integral. It is probably 
already familiar to the reader that the element of area dA, which 
is equal to dx dyin>' otangular coordinates, is not equal to dp dd in 
polar coordinates. In order to obtain a general method for trans- 
forming the element of area or the element of volume from one 
set of coordinates to another, it is necessary to introduce the 
definition of the Jacobian, or functional determinant. 

Let u = u(xj y) and v = v(x, y) be two continuous functions 
of the independent variables x and y, such that du/dx, du/dy, 
j and dv/dy are also continuous in x and y. Then 



(55-1) 



du dv __ du dv 
'dxdy ~~ ~dy^x 



du dv 

Jx dx 



'dy dy 



is called the Jacobian, or functional determinant, of u } v with 
respect to x, y. It is usually denoted by 

J I I Or -r~, r 



In the case of three variables, let u = u(x, y, z), v = v(x, y, z), 
and w = w(x, y, z) be continuous together with their first partial 
derivatives. The Jacobian, or functional determinant, of w, v, w 



184 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 55 
with respect to x, y, z is defined by 



(55-2) 



du dv dw 
H)X 'dx ~dx 
du to dw 

dy lfy~dy 

du dv dw 
dz "dz ~dzi 



The usual symbols for it are 



or 



d(u, v f w) 
d(x, y, z) ' 



The Jacobian of any number of functions Ui, u^, , u n , 
with respect to the variables xi, x%, - - - , x n , is defined by an 
obvious extension of (55-1) and (55-2). It is denoted by 



(Ui, U 2 , ' ' ' , U n \ 
\Xi, X Z , ' ' ' , X n / 



or 



,U n ) 



The Jacobian is of great importance in mathematics.* It is 
used here in connection with the change of variable in multiple 
integrals. If it is desired to change the variable in f R f(x, y) dA 
by making x = x(u, v) and y = y(u,v), the expressionf for dA in 
terms of u and v is given by 



(55-3) 



dA = 



(*LM\ 
\U,V/ 



du dv. 



Thus, in transforming to polar coordinates by means of x = 
p cos 0, y = p sin 8, 



cos 6 



sin 6 



p sin 6 p cos 



= p cos 2 + p sin 2 6 = p. 



Therefore, 

dA = p dp d6, 

a result that is already familiar from elementary calculus. 

* Note that the Jacobian appeared in Sec. 41m connection with the differen- 
tiation of implicit functions. 

t See SOKOLNIKOFF, I. S., Advanced Calculus, Sec. 46. 



58 MULTIPLE INTEGRALS 

It follows from (55-3) that 

(55-4) f /(x, y) dA = f f f[x(u, v), y(u, v)] 
JR J JR 

The right-hand member of (55-4) can be written as 
J f R F(u,v)dudv, 



185 



du dv. 



where 



F(u, v) = f[x(u, v), y(u y v}] 



If it is desired to evaluate this double integral by means of an 
iterated integral, the limits for u and v must be determined from 
a consideration of the region R. 

Similarly, if x = x(u, v, w), y = y(u, v, w) y and z = z(u, v, w), 
then 



u v w 



du dv dw. 



(55-5) dV = J ( -^-'- ) du dv dw 

and 

(55-6) f f(x, y, z) dV 
JR 

s J J L f[x(u ' Vj w} ' 

56. Spherical and Cylindrical Coordinates. Corresponding to 
the system of polar coordinates in the plane, there are two 
systems of space coordinates that are frequently used in prac- 
tical problems. The first of these is the system of spherical, or 
polar, coordinates. Let P(x, y, z) (Fig. 46) be any point whose 
projection on the o^-plane is Q(x, y). Then the spherical 
coordinates of P are p, <p, 0, in which p is the distance OP <p, is the 
angle between OQ and the positive x-axis, and 6 is the angle 
between OP and the positive 2-axis. Then, from Fig. 46, it is 
seen that 

x = OQ cos tf> = OP cos (90 0) cos <p = p sin cos <?, 
y = OQ sin <p = p sin 6 sin <p, 
z = p cos 0. 

The element of volume in spherical coordinates can be obtained, 
by means of (55-5). Since 



186 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 



(x, V, g\ = 
\P, V* OJ 



sin cos <p sin sin <p cos 
p sin sin ^> p sin cos ^ 

p cos cos <f> p cos sin ^ p sin 
s= p 2 sin 0, 
it follows that 

(56-1) dV = p 2 sin 

This element of volume is the volume of the solid bounded by the 




Q(*,y) 



FIG. 46. 



two concentric spheres of radii p and p + dp, the two planes 

through the z-axis that make 
angles of <p and <p + d<p with 
the 2-plane, and the two cones 
of revolution whose common 
axis is the z-axis and whose 
vertical angles are 28 and 
2(8 + dB). 

The second space system cor- 
responding to polar coordinates 
in the plane is the system of 
cylindrical coordinates. Any 
point P(x t if, z), whose projection 
on the 2/-plane is Q (Fig. 47), has 
the cylindrical coordinates p, 8, z, 
where is the angle between OQ and the positive z-axis, p is the 
distance OQ, and z is the distance QP. From Fig. 47, it is evident 




FIG. 47. 



56 MULTIPLE INTEGRALS 

that x = p cos 6, y = p sin 0, and 2 = 2. Since 



187 



(x y z\ 

[ x > y> z } 

\P> 0, zj 



cos sin 

p sin p cos 6 

1 



it follows that 
(56-2) 



dV = pdp dO dz. 



This element of volume is the volume of the solid bounded by the 
two cylinders whose radii are p and p + dp y the two planes 
through the z-axis that make angles 6 and 6 + d8 with the 
zz-plane, and the two planes parallel to the :n/-plane at distances 
z and z + dz. 

Example 1. Find the re-coordinate of the center of gravity of the 
solid of uniform density <r lying in the first octant and bounded by the 
three coordinate planes and the sphere x 1 + y 2 + z* = a 2 . 

Since 



it is necessary to compute J R x dV. This integral can be calculated by 
evaluating the iterated integral 



f a /*Va 2 -z 2 f Vo 2 -?/ 2 

Jo Jo Jo 

but it is easier to transform to spherical coordinates. Then, 

7T 7T 

C xdV = f 2 f 2 / 

IT IT 

f~2 /*2 O 4 

= Jo Jo 



sn cos 



sn 



Therefore, 



oV /*| 

= TeJ 



oV 



_ _ 3a 

7ra 3 /6 " 8' 



Example 2. In the example of Sec. 54, find 7, by transforming the 
integral into cylindrical coordinates. Then, 



188 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 57 



2 V dv 



*)p dz M dp 



/V0 

T 



p 3 sin 20 , 6 2 p0\ 

+ -sv 



(8.. + 46-). 



57, Surface Integrals. Another important application of 
multiple integrals occurs in the problem of defining the area of a 
surface. Let z = f(x, y) be the equation of a surface 8 (Fig. 48). 

Let 8' be a portion of this 
surface bounded by a closed 
curve F, and such that any 
line parallel to the 2-axis cuts 
S r in only one point. If C is 
the projection of T on the xy- 
plane, let the region R, of 
which C is the boundary, be 
subdivided by lines parallel to 
the axes into subregions AB t . 
Through these subdividing 
lines pass planes parallel to 
the z-axis. These planes cut 
from S' small regions Afi of area Aov Let AA t be the area of 
AJ? t . Then, except for infinitesimals of higher order, 

AA = cos 7 t Acr t , 

where cos a, cos /3 l; and cos 7 represent the direction cosines of 
the normal to 8 at any point (x ly y^ t ) of AS(. Since (see 
Sec. 43) 

dz 

cos <: cos &: cos 7* = -r- 
ox 




FIG. 48. 



it follows that 

cos 7 = 



-1 



67 MULTIPLE INTEGRALS 

Upon using the positive value for cos 7*, 



189 



Then, 



is defined as the area of the surface S'. Since this limit is 







dz 



the value of <7 is given by 



(.7.1) 



-i 



7 



Similarly, by projecting S 
on the other coordinate 
planes, it can be shown that 



= I sec a dA 
jRi 



= I sec /3 dA. 

JR*. 



The integral of a function 
v(x, y, z) over the surface 
z = f(x, y) can now be de- 
fined by the equation 




FIG. 49. 



(57-2) 



Js 



, y, z) dcr 



-Si 



It is assumed that <p(x, y, z) is continuous and single valued for 
all points of some region T that contains S. 

Example. Find the area of that portion of the surface of the cylinder 
x* + y 2 = # 2 which lies in the first octant between the planes z = and 
z = mx (Fig. 49). 

' This surface can be projected on the zz-plane or on the t/z-plane but 
not on the xy-pl&ne (since any perpendicular to the xy-pl&ne that meets 
the surface at all will lie on the surface). The projection on the #z-plane 
is the triangle GAB. Hence, 



190 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 67 
*~foAB "****. 



But 



rr = a(a 2 - 

V a % 

Therefore, 

* "" ^"^ dz dx 



C 
I 



amx(a 2 # 2 )-^ dx = a 2 w. 

PROBLEMS 

1. Find the coordinates of the center of gravity of the area bounded 
by x^ + yM = a^, z = 0, and ?/ = 0. 

2. Find the moment of inertia of the area of one loop of p 2 = a 2 sin 26 
about an axis perpendicular to its plane at the pole. 

3. (a) Find the expression for dA in terms of u and v, if x u(l v) 
and y = uv. 

(6) Find the expression for dV in terms of u, v, and w, if x = u(l v), 
y uv(\ w), and z = mw>. 

4. Find the center of gravity of one of the wedges of uniform density 
cut from the cylinder x 2 + y 2 = a 2 by the planes 2 = mx and 2 = mx. 

6. Find the volume enclosed by the circular cylinder p ~ 2a cos 0, 
the cone z = p, and the plane 2 = (use cylindrical coordinates). 

6. Find the center of gravity of the solid of uniform density bounded 

by the four planes - 4- T + - = 1, # = 0, 2/ = 0, and 2 = 0. 

01 C 

7. Find the moment of inertia of the solid of uniform density 
bounded by the cylinder x 2 + y 2 = a 2 and the planes 2 = and 2 = 6 
about the 2-axis. 

8. Find, by the method of Sec. 57, the area of the surface of the 
sphere x 2 + y 2 + 2 2 = a 2 that lies in the first octant. 

9. Prove that 



Hint: Write out the Jacobians, and multiply. 
10. Prove that 



w y/ 

where u = u(x, y), v = v(x, y), x = z(, 17), and i/ = y(?, 17). 

11. Find the surface of the sphere x 2 + y 2 + z* = a 2 cut off by the 
cylinder a; 2 ax + y* = 0. 



58 



MULTIPLE INTEGRALS 



191 



12. Find the volume bounded by the cylinder and the sphere of 
Prob. 11. 

13. Find the surface of the cylinder x 2 + y 2 = a 2 cut off by the 
cylinder y* + z 2 = a 2 . 

14. Find the coordinates of the center of gravity of the portion of 
the surface of the sphere cut off by the right-circular cone whose vertex 
is at the center of the sphere. 

15. Use cylindrical coordinates to find the moment of inertia of the 
volume of a right-circular cylinder about its axis. 

16. Find the moments of inertia of the volume of the ellipsoid 



about its axes. 

17. Kinetic energy T is defined as T %Mv 2 , where M is the mass 
and v is the velocity of a particle. If the body is rotating with a 
constant angular velocity o>, show that 

T = 

where p is the density and / is the moment of inertia of the body about 
the axis of rotation. 

58. Green's Theorem in 
Space. An important the- 
orem that establishes the 
connection between the in- 
tegral over the volume and 
the integral over the surface 
enclosing the volume is given 
next. This theorem has 
wide applicability in numer- 
ous physical problems* and 
is frequently termed the di- 
vergence theorem. 

THEOREM. // P(x,y,z), FIG. 50. 

Q(x, y, z), R(x, y, z) and 

dP/dx, dQ/dy, dR/dz are continuous and single-valued functions in 
a region T bounded by a closed surface S, then 




It will be assumed that S (Fig. 50) is cut by any line parallel to 
* See, in this connection, Sees. 125, 130, 131. 



192 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 68 

one of the coordinate axes in at most two points. If S is not 
such a surface, then T is subdivided into regions each of which 
satisfies this condition, and the extension to more general types 
of regions is immediate. 

A parallel to the z-axis may cut S in two points (x t , y r , z t ) and 
(z t , ?/t, z t ), in which z l < z t . Let z = fi(x, y) be the equation 
satisfied by (z t , t/ t , z t ) and z = / 2 (x, y) be the equation satisfied 
by (z t , 2/ t , Zt). Thus, S is divided into two parts, Si, whose equa- 
tion is z = fi(x, y), and S 2 , whose equation is z = / 2 (#, y). Then, 



R(x, y, z) cos 7 d<r, 
taken over the exterior of S, is equal to 

XR(XJ y, z) cos 7 do- + I /?(#, ?/, z) cos 7 rfo-, 
'^ /<Si 

taken over the exteriors of the surfaces Si and S 2 . But, from 
(57-2), these surface integrals are equal to double integrals taken 
over the projection T' of T on the i/-plane. Therefore,* 

I R(x, y, z) cos 7 da = {#[>, 2/,/ 2 (z, y)] - R[x, y,fi(x, y)]} dA 

JS JT' 

R(XJ y, z) dy dx 



- dz 
or 



I R(x, y, z) cos 7 da = | -^ dV. 

Similarly, it can be shown that 

f f dP 

I P(#, t/, z) cos a dcr = I rfF 
J-s Jr ^ 

fQCr,t/,z)cos/3d<r= ( $dV. 
Js JT oy 

* The negative sign appears in the right-hand member of the equation 
because 

COS 72 dffz ** ~" COS 

where the subscripts refer to S* and Si. 



and 



68 MULTIPLE INTEGRALS 193 

Therefore, 



(58-1) (P cos a + Q cos + R cos 7) dr = + + d7. 

Since cos ad<r = dy dz, cos /3 dv = dz dx y and cos y da = dx dy, 
(58-1) can be written in the form 



dR 



(58-2) I I (P dy dz + Q dz dx + R dx dy) 

m(dP ^dQ 
\3x + ^ 

The formula (58-2) bears the name of Green.* 

Example. By transforming to a triple integral, evaluate 
/ = J J (x s dy dz + x 2 y dz dx + # 2 z da; dy\ 

where >S is the surface bounded by z = 0, z = 6, and a; 2 -f 2/ 2 = a 2 . 

Calculating the right-hand member with the aid of (58-2) and mak- 
ing use of the symmetry, one finds 

/*a / v/aT^lr* /*& 

7 = 4 Jo Jo Jo (*** + ** + **) *>dv** 

= 4-56 jT a a; 2 V~a 2 - x> dx 
= %ira*b. 

A direct calculation of the integral 7 may prove to be instructive. The 
evaluation of the integral can be carried out by calculating the sum of 
the integrals evaluated over the projections of the surface S on the 
coordinate planes. Thus, 







a ' /*\/a* y 2 

.J_ V ^M- 



which upon evaluation is seen to check with the result obtained above. 
It should be noted that the angles a, j8, 7 are made by the exterior 
normal with the positive direction of the coordinate axes. 

* The names of Gauss and Ostrogradsky are also associated with this 
theorem. 



194 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 69 

59. Symmetrical Form of Green's Theorem. One of the most 
widely used formulas in the applications of analysis to a great 
variety of problems is a form of Green's theorem obtained by 
setting 

dv ~ dv D dv 

P = u> Q = u> R = u~ 

dx dy dz 

in (58-1). The result of the substitution is 

f fdv , &> a , dv \ j 

I u I -r- cos a + cos j8 + cos 7 I ao* 
js \dx dy dz / 



C 

== 

J r 



T /^u 
Jr \dx 



5v 6u dv du , 

'dx ^ ltydy + ~dz 



But the direction cosines of the exterior normal n to the surface 
are 

dx n dy dz 

cos a = -j-i cos jS = -r-) cos 7 = 3-7 
dn an an 

so that the foregoing integral reads 

(59-1) f u ^ dff = I 11 V 2 i; dF 
' Js dn JT 



where 



f (dudv_ , du dv_ du --i.y 
Jr V^a; aa; ^ dy dy ^ dz dzj ' 



= < 

dx 2 + dy* + dz 2 ' 



Interchanging the roles of u and v in (59-1) and subtracting the 
result from (59-1) give the desired formula 



A reference to the conditions imposed upon P, Q, and R in 
the theorem of Sec. 58 shows that, in order to ensure the validity 
of this formula, it is sufficient to require the continuity of the 
functions u and v and their first and second space derivatives 
throughout a closed region T. 



59 MULTIPLE INTEGRALS 195 

PROBLEMS 

1. Evaluate, by using Green's theorem, 



J Js 



where S is the surface z 2 + y 2 + z 2 = a 2 . 

2. Show from geometrical considerations that the angle dd subtended 
at the origin by an element ds of a plane curve C is 

ds 

dd = cos (ft, r) > 

where r is the radius vector of the curve, and (n, r) is the angle between 
the radius vector and the normal to the curve. Hence, show that 



r cos (n, r) ds r I dr 

e = Jc r 



f 1 dr 

= )c~rd^ ds > 



where the integral is a line integral along the curve C. 

3. A solid angle is defined as the angle subtended at the vertex of a 
cone. The area cut out from a unit sphere by the cone, with its vertex 
at the center, is called the measure of the solid angle. The measure of 
the solid angle is clearly equal to the area cut out by the cone from any 
sphere concentric with the unit sphere divided by the square of the 
radius of this sphere. In a manner analogous to that employed in 
Prob. 2, show that the element of solid angle is 

cos(n, r) da 
&* = ~ 2 1 

where the angle between the radius vector and the exterior normal to 
the surface S is (n, r). Also, show that 

cos (n, r) do" f 1 dr 

x . _ \ > / 

CO = 



where the integral is extended over the surface S. 
4. By transforming to a triple integral, evaluate 



//, 



dy dz + y 3 dz dx + 2 3 dx dy\ 



where S is the spherical surface x 2 + y z + z* = a 2 . Also, attempt to 
calculate this integral directly. 

5. Set v 1 in Green's symmetrical formula, and assume that u 
satisfies the equation^fc Laplace, V 2 u = 0. What is the value of 

f -JT do- if S is an arDiwary closed surface? 



196 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 59 

6. The density of a square plate varies directly as the square of the 
distance from one vertex. Find the center of gravity and the moment 
of inertia of the plate about an axis perpendicular to the plate and 
passing through the center of gravity. 

7. Find the volume of a rectangular hole cut through a sphere if a 
diameter of the sphere coincides with the axis of the hole. 

8. Show that the attraction of a homogeneous sphere at a point 
exterior to the sphere is the same as though all the mass of the sphere 
were concentrated at the center of the sphere. Assume the inverse 
square law of force. 

9. The Newtonian potential V due to a body T at a point P is defined 

by the equation V(P) = j T dm/r, where dm is the element of mass of 
the body and r is the distance from the point P to the element of mass 
dm. Show that the potential of a homogeneous spherical shell of 
inner radius b and outer radius a is 

V = 27r<7(a 2 - 6 2 ), if r < 6, 

and 

4 a 3 - & 3 
= ^TTO- -- - > if r > a, 

where cr is the density. 

10. Find the Newtonian potential on the axis of a homogeneous 
circular cylinder of radius a. 

11. Show that the force of attraction of a right-circular cone upon a 
point at its vertex is 2ir<rh(l cos a), where h is the altitude of the 
cone and 2a is the angle at the vertex. 

12. Show that the force of attraction of a homogeneous right-circular 
cylinder upon a point on its axis is 



here h is altitude, a is radius, and R is the distance from the point to 
one base of the cylinder. 

13. Set up the integral representing the part of the surface of the 
sphere # 2 + y z + & 100 intercepted by the planes x = 1 and x = 4. 

14. Find the mass of a sphere whose density varies as the square of 
the distance from the center. 

15. Find the moment of inertia of the sphere in Prob. 14 about a 
diameter. 



CHAPTER VI 
LINE INTEGRAL 

The line integral, to be considered in this chapter, is as useful 
in many theoretical and practical problems as the ordinary defi- 
nite integral defined in Chap. V. The discussion of the line 
integral will be followed by several 
illustrations of its use in applied 
mathematics. 

60. Definition of Line Integral. 
Let C be any continuous curve (Fig. 
51), joining A(a,b) and B(c,d). 
Let M(x, y) and N(x, y) be two 
functions that are single- valued and 
continuous functions of x and y for 
all points of C. Choose n 1 points 
PI(XI, y^) on the curve C, which is 
thus divided into n parts. Let 

where X Q = a, y Q = 6, x n = c, y n = d. 




Jx n 



Fio. 51. 



Let { and rj t be defined by 
< 7/ t and form 



t , 170 Ax, 



The limit of this sum as n > QQ and all Ax % and 
simultaneously is defined as a line integral along C. 
Thus, 



(60-1) lim V [M(^ i/O Ax, + tf (&, nO AyJ 

n-^oo 

[M( X , y) dx + N(x, y) dy]. 



Obviously, the value of this integral depends, in general, on the 
particular choice of the curve C. If the equation of C is known 
in one of the forms y = /(x), x = <p(y) or x = /i(0, t/ = / 2 (0, the 

197 



198 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 60 

line integral may be reduced to a definite integral in one variable 
by substitution, as is indicated in the following examples. How- 
ever, it is frequently inconvenient to make this reduction, and 
thus it is desirable to consider the properties and uses of (60-1). 

Example 1. Let the points (0, 0) and (1, 1) be connected by the line 
y = x. Let M(x, y) = x y 2 and N(x t y) = 2xy. Then the line 
integral along y = x, 



becomes, on substitution of y = #, 

[(x - x 2 ) dx + 2x 2 >dx] = (x + x 2 ) dx 



If (0, 0) and (1, 1) are connected by the parabola y = x 2 , I along 
y = x 2 is 



[(x - x 4 ) dx + 2x*(2x dx)] = o (x + 3z 4 ) dx = % 

Example 2. Consider 

M(x, y) = 2x 2 + 4xy 
and 

N(x t y) = 2x 2 - ?/, 

with the curve y = x 2 connecting the points (1,1) and (2, 4). Then 

/*(2,4) /2 /*! 

J (u) (Mdx + Ndy)=J i (2x 2 + 4x-x 2 )dx+J } (2y - ?/ 2 ) dy = 13%. 
Inasmuch as dy = 2o; do:, this integral can be written as 

f 2 (2x 2 + 4x 3 ) da: + f 2 (2a; 2 - o; 4 )2x dx = 13?^. 

If the equation of the parabola in this example is written in a para- 
metric form as 

x = t, 

y = t 2 , (l<t< 2), 

then the integrand of the line integral can be expressed in terms of the 
parameter t. Substituting for x, y, dx, and dy in terms of t gives 

/*(2,4) /*2 

J (u) (Mdx + N dy)} = J t [(' + 4") + (24" - <)2] <tt 

< 2 + 8* 3 - 2 6 ) dt = 



61 LINE INTEGRAL 199 

The reader will readily verify that the value of this integral over a 
rectilinear path C joining the points (1,1) and (2, 4) is also 13%J. In 
fact, the value of this integral depends only on the end points and not 
upon the curve joining them. The reason for this remarkable behavior 
will appear in Sec. 63. 

PROBLEMS 

1. Find the value of I ' [\/y dx + (x y) dy] along the follow- 

/(0,0) 

ing curves: 

(a) Straight line x t, y = t. 

(b) Parabola x = t 2 , y = t. 

(c) Parabola x = t, y t 2 . 

(d) Cubical parabola x = t, y = t*. 

2. Find the value of J^ [x 2 y dx + (x 2 - y 2 ) dy] along (a) y = 3s 2 , 
(b) y = 3x. 

3. Find the value of J ( [o) ( x * dx + y 2 dy) along the curves of 
Prob. 1 above. 

4. Find the value of J (0 ' 0) [(x 2 + y' 2 ) dx 2xy dy] along (a) y = x\ 
(V) x = y 2 ; (c) y = x\ 

6. Find the value of J (Jo) (y s ' n x ^ x ~~ x cos y ^2/) a ^ on g y ~ x. 

6. Find the value of J(- o> (x dy + y dx) along the upper half of 
the circle x 2 + y 2 = a 2 . 

7. Evaluate the integral of Prob. 6 over the path formed by the 
lines x = a, ty a, x = a. What is the value of this integral if the 
path is a straight line joining the points (a, 0) and (a, 0)? 

8. Find the value of /|J'J) (# 2 dx + y 2 dy) along the path given by 
x = sin t, y = cos t. 

9. Evaluate the integral of Prob. 8 if the path is a straight line join- 
ing (0, 1) and (1, 0). 

10. What is the value of the integral of Prob. 8 if the path is the 
curve y = 1 x 2 ? 

61. Area of a Closed Curve. Let C be a continuous closed 
curve which nowhere crosses itself. The equation of such a 
curve, in parametric form, can be given as 



where the parameter t varies continuously from some value 



200 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 61 

t = to to t = t\ and the functions f\(t) and f 2 (t) are continuous 
and single-valued in the interval to < t <> t\. Inasmuch as the 
curve is assumed to be closed, the initial and the final points of 
the curve coincide, so that 



and 



The statement that the curve C does not cut itself implies 
that there is no other pair of values of the parameter t for which 



and 



A closed curve satisfying the condition stated above will be 
called simple. 

As t varies continuously from to to ti, the points (x, y) deter- 
mined by (61-1) will trace out the curve C in a certain sense. If 
C is described so that a man walking along the curve in the direc- 
tion of the description has the 
enclosed area always to his left, 
the curve C is said to be described 
in the positive direction^ and the 
enclosed area will be considered 
positive; but if C is described so 
that the enclosed area is to the 
right, then C is described in the 
negative direction, and the area is 
regarded as negative. 

Consider at first a simple closed 
curve C such that no line parallel 
to one of the coordinate axes, say 
the y-axis, intersects C in more 
than two points. Let C be 
, x = a 2 , y = 61, y = b 2 , which are 
and J3 2 , respectively. Clearly, C 




FIG. 52. 



bounded by the lines x 

tangent to C at Ai, A*, 

cannot be the graph of a single-valued function. Therefore, let 



61 



LINE INTEGRAL 



201 



the equation of A\B\A^ be given by y\ = fi(x), and the equation 
of AiBzAz by y* = /2(&), where /i (x) and/ 2 (z) are single-valued 
functions. Then the area enclosed by C (Fig. 52) is given by 



(61-2) 



or 



(61-3) 



= I 2/2 dz - f a2 2/1 

Jai Jai 

= - ] 2/2 da: - \ 

Jai Jai 

A = - f c ydx, 



dx, 



in which the last integral is to be taken around C in a counter- 
clockwise direction. 

Similarly, if x\ = <p\(y) is the equation of BiAiB^ and o: 2 = 
is the equation of 



x-tdy I * 

Jb\ 

= I 2 x 2 dy + C l 

Jb\ Jbi 



or 



(61-4) 



Again, the last integral is to be taken around C in a counter- 
clockwise direction. It may be 
noted that (61-3) and (61-4) both 
require that the area be to the left 
as C is described if the value of 
A is to be positive. 

By adding (61-3) and (61-4), a 
new formula for A is obtained, 
namely, 



(61-5) 



A ^ 




FIG. 53. 



This formula gives a line-integral expression for A. 

To illustrate the application of (61-5), the area between 
(1) x 2 = ty and (2) y 2 = 4z (Fig. 53) will be determined. 



202 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 62 
Then, 

A - o (~ydx + xdy} = H (-ydx + xdy) 

* JC * J(l) 

+ o (- 

* J(2) 



+ zc 



24| 24 



16 
3* 



For convenience the first integral was expressed in terms of x, 
whereas the second integral is simpler in terms of y. 

The restriction that the curve C be such that no line parallel 
to one of the coordinate axes cuts it in more than two points 
can be removed if it is possible to draw a finite number of lines 
connecting pairs of points on C, so that the area enclosed by 
the curve is subdivided into regions each of which is of the type 
considered in the foregoing. This extension is indicated in detail 
in the following sect on. 

PROBLEMS 

1. Find, by using (61-5), the area of the ellipse x = a cos <p t 
y = 6 sin <p. 

2. Find, by using (61-5), the area between y 2 = 9# and y = 3x. 

3. Find, by using (61-5), the area of the hypocycloid of four cusps 
x = a cos 3 6, y = a sin 3 6. 

4. Find, by using (61-5), the area of the triangle formed by the line 
x + y = a and the coordinate axes. 

5. Find, by using (61-5), the area enclosed by the loop of the strophoid 



62. Green's Theorem for the Plane. This remarkable theorem 
establishes the connection between a line integral and a double 
integral. 

THEOREM. // M(x, y) and N(x, y), dM/dy and dN/dx are 
continuous single-valued functions over a closed region R, bounded 
by the curve C, then 



62 LINE INTEGRAL 203 

The double integral is taken over the given region, and the curve C is 
described in the positive direction. 

The theorem will be proved first for a simple closed curve 
of the typo considered in Sec. 61 (see Fig. 52). 

Again, let y\ = fi(x) be the equation of AiBiA% and y% = / 2 (x) 
be the equation of A\E^A^ Then, 



. , , , 

-5 dx ay = I ax \ ~^-~ dy 
; dy 



[M(x, Vt ) - M(x, y,}} dx 

/*ai S*az 

= - M(XJ 2/ 2 ) da? ~ I M(x, 

Jai Jai 



or 
(62-1) 

Similarly, if xi = <pi(y) is the equation of BiAiBi and 
is the equation of B\A^.B^ 



na? 
t 



f ' tfdi, 

J^z 



or 



(62-2) J J^ g dx dy = J^ JV^, y) <fo. 

Therefore, if (62-2) is subtracted from (62-1), 

(Sf ^ S) dx dy ~ ~ fc [M(x > & dx 



It will be observed that setting M = y and N ~ x gives the 
formula (61-5). 

Now, let the region have any continuous boundary curve C, so 
long as it is possible to draw a finite number of lines that divide 
the region into subregions each of the type considered in the 
first part of this section; that is, the subregions must have 
boundary curves that are cut by any parallel to one of the 
coordinate axes in at most two points. Such a region R is shown 
in Fig. 54. 



204 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 62 

By drawing the lines AiA 2 and A 3 A 4 , the region R is divided 
into three subregions R\, Rz, and R s . The boundary curve of 
each region is of the simple type. The positive direction of 
each boundary curve is indicated by the arrows. The theorem 
can be applied to each subregion separately. When the three 
equations are added, the left-hand members add to give the 
double integral over the entire region R. The right-hand mem- 
bers give 

- f (Mdx+Ndy)- f (Mdx + Ndy)- f (Mdx + Ndy), 

JCi JCi JCs 

where 



Since each of the lines AiA 2 and A 3 A 4 is traversed once in each 
direction, the line integrals that arise from them will cancel. 

The remain ng line integrals, 
taken over the arcs of C, add 
to give the line integral over 
C. Therefore, 




nfdM _ 
\dy 

--X 



dx, 

(Mdx + N dy) 



holds for regions of the type 

R. 

Another type of region in 

which an auxiliary line is in- 

troduced is the region whose 
boundary is formed by two or more distinct curves. Thus, if R 
(Fig. 55) is the region between C\ and C 2 , the line A\A% is drawn 
in order to make the total boundary 



FIQ. 54. 



a single curve. 
integrals over A 
integrals over C 



The theorem can be applied, and the line 
^A\ and A\A% will cancel, leaving only the line 
and C 2 . 



62 



LINE INTEGRAL 



205 



If the region R is such that any closed curve drawn in it can, 
by a continuous deformation, be shrunk to a point without 
crossing the boundary of the region, then the latter is called 
simply connected. Thus, regions bounded by a circle, a rectangle, 
or an ellipse are simply connected. The region R exterior to 
C 2 and interior to Ci (Fig. 55) is not simply connected because a 
circle drawn within R and enclosing 2 cannot be shrunk to a 
point without crossing C 2 . In ordinary parlance, regions that 
have holes are not simply connected regions; they are called 





FIG. 55. 



FIG. 56. 



multiply connected regions. The importance of this classification 
will appear in the next two sections. 

Example. Evaluate by using Green's theorem 



where C is the closed path formed by y x and i/ 3 = x* from (0, 0) 
to (1, 1) (Fig. 56). Since M = x*y and N = y\ 



Then, 



dN 



.**+.*> --//,(-)** 



206 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 63 

PROBLEMS 

1. Find, by Green's theorem, the value of 



j c (x 2 y dx + y dy) 



along the closed curve C formed by y* x and y x between (0, 0) 
and (1, 1). 

2. Find, by Green's theorem, the value of 

along the closed curve C formed by y* = # 2 and y = x between (0, 0) 
and (1, 1). 

3. Use Green's theorem to find the value of 



J c [(*y - 2 ) dx + x*y dy] 



along the closed curve C formed by y = 0, x = 1, and y = x. 
4. Use Green's theorem to evaluate 



along the closed path formed by y = 1, x = 4, and y + -\/x. 

5. Check the answers of the four preceding problems by evaluating 
the line integrals directly. 

63. Properties of Line Integrals. THEOREM 1. Let M and N 

be two functions of x and y, such that M, N, dM/dy, and dN /dx are 
continuous and single-valued at every point of a simply connected 
region R. The necessary and sufficient condition that f c (M dx 
+ N dy) = around every closed curve C drawn in R is that 

dM = dN 
dy dx' 
for every point of R. 
Since 

-]>. + **>-//(-)** 

where A is the region enclosed by C, it follows that 

dM = dN 
dy dx 



LINE INTEGRAL 



207 



makes the double integral, and consequently the line integral, 
have the value zero. Conversely, let J* c (Af dx + N dy) = 
around every closed curve C drawn in R. Suppose that 

dM _ # , 
dy dx * 

at some point P of R. Since dM/dy and dN/dx are continuous 
functions of x and y, 

dM d# 



is also a continuous function of x and y. Therefore, there must 



exist some region S, about P, in which has the same 

sign as at P. Then, 

ff (-)<*'* 

J Js \ dy ox/ 

and hence f (M dx + N dy) T 
around the boundary of this re- 
gion. This contradicts the hy- 
pothesis that 

f c (M dx + N dy) = 
around every closed curve C drawn in R. 





+x 



FIG. 57. 
It follows that 



at all points of R. 
Example 1. Let 

M = 
Then, 



dM 

dy 



dN 
dx 



dM 

dy 



and 



N = 



dN 
dx 



M, N, dM/dy, and dN/dx are continuous and single-valued for all 
points of the xy-plane except (0, 0). Hence, f c (M dx + N dy) = 
around any closed curve C (Fig. 57) that does not enclose (0, 0). In 
polar coordinates, obtained by the change of variables 



208 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 63 
x = p cos B y y = p sin 6, 

f ( r? g <fo + -TT *A = f de - 
Jc \x 2 -ft/ 2 a: 2 -f ?/ V Jc 

If C does not enclose the origin, varies along C from its original value 
back to 0o. Therefore, j c dO = 0. If Ci encloses the origin, varies 

along Ci from to + 2?r, so that J Cl dO = 2?r. 
Example 2. Find, by Green's theorem, the value of 

I = f c [(x* + xy) dx + (y* + x*) dy], 

where C is the square formed by the lines y = 1 and x = 1. Since 
dM dN 



Note that the line integral has the value zero, but dM /dy ^ dN /dx. 

This does not contradict Theorem 1. 




THEOREM 2. Le^ M and ]V satisfy 
the conditions of Theorem 1. 7"/ie 
necessary and sufficient condition that 
f(a'&) (^ ^ + ^ rfy) 6e independent 
o/ i/ie citr^e connecting (a, b) and 



A(a ' b) (x, y) is that dM/dy = dN/dx at all 

~x points of the region R. In this case 



FIG. 58. the line integral is a function of the 

end points only. 

Suppose dM/dy = dN/dx. Let Ci and C 2 (Fig. 58) be any two 
curves from A to P, and let 



7i = I (M dx + N dy) | 1 

and 

J 2 = C(Mdx + N dy) 

/C2 

be the values of the line integral from A to P along Ci and C 2? 
respectively. Then /i 7 2 is the value of the integral around 
the closed path formed by Ci and C 2 . By Theorem 1, 

Jl - /2 = 0. 



63 



LINE INTEGRAL 



209 



Therefore, Ii = /2, so that the line integral taken over any two 
paths from A to P has the same value. 

Conversely, suppose that J (M dx + N dy) is independent of 
the path from A to P. Then, for any two curves Ci and C 2 , 
/! = 7 2 . It follows that / (M dx + N dy) = for the closed 
path formed by C\ and C 2 . Hence, by Theorem 1, dM/dy 
= dN/dx. 



Example. Consider 



(22) /I + 7/ 2 1+z* \ 

m \x^ dx nr~v dy r 

1,1) \ " " / 



Since dM/dy 2y/x 3 and dN/dx 2y/x s and both functions are 
continuous except at (0, 0), the line integral is independent of the 
path so long as it does not enclose the origin. Choose y 1 from 
(1, 1) to (2, 1) and x = 2 from (2, 1) to (2, 2) as the path of integration. 
Then, 

25 ,__JL 2 __5 2 2 __9 
[ x 2 \i 81 8 

THEOREM 3. Let M ami N satisfy the conditions of Theorem 1. 
The necessary and sufficient con- 



dition that there exist a function 
F(x, y} such that dF/dx = M 
and dF/dy = N is that dM/dy 
dN/dx at all points of the re- 
gion R. 

If dM/dy = dN/dx, Theorem 
2 proves that 



P(oc t y) 




j (Mdx + N dy) FlG> 59< 

is independent of the path. Therefore, 
(63*1) f ( "f (Mdx + N dy) = F(x, y}, 

a/ \u,o) 

and this function F(x, y) depends only on the coordinates of the 
end points of the path. Hence, 

F(x + Ax, y) = f * 6 +A *' y) (M dx + N dy). 

Let the path of integration be chosen as a curve C (Fig. 59) 
from A to P and the straight line PP' from P(x, y) to P'(x + Ax, 



210 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 63 
y). Then, 

F(x + Ax, y) =f^ (Mdx + N dy) + f^**'* (M dx + N dy) 

or 

(63-2) F(x + Ax, y) = F(x, y) + **** M(x, y} dx. 



The second integral reduces to the simpler form given in (63-2) 
since y is constant along PP', and therefore dy = 0. From (63-2), 



,y)1 

J 



r j p+A* -i 

= lim -r- I Jlf (a, y) rfx . 

Ax _o LAx Jx v * ^ J 

Application of the first mean-value theorem* gives 

f xH ~ Ax M(x, y) dx = Ax M(f, ), (x ^ < x + Ax). 

/# 

Therefore, 



= lim - Ax M(, 2/) = lim Jlf ({, y). 

OX Az-*0 LAX J Ax->0 

Hence, 



It can be proved similarly that 

* It may be recalled that 

> 
f(x) dx = (6 )/(), (a 



is the first mean-value theorem for definite integrals. If ff(x)dx = F(x), 
than /(a;) = /^'(x). From these relations, the mean-value theorem for 
definite integrals can be transformed into 

F(b) - F(a) = (6 - a)*"(0, 
where a ^ ^ ^i 6, or 



which is the mean-Value theorem of the differential calculus. 



63 LINE INTEGRAL 211 

The function F is really a function of both end points. Multi- 
plying dF/dx = M(x, y) by dx and dF/dy = N(x, y) by dy gives 

dF dF 

dF = ^dx + ^dy = M(x, y) dx + N(x, y) dy. 

Thus, if 

dM = dN 
dy dx 9 

the integrand in J c (M dx + N dy) is the exact differential 
of the function F(x, y), which is determined by the formula 
(63-1). 

The most general expression for a function $(x, #), whose 
total differential is d$ = M dx + N dy, is 

*(*, y) = F(x, y) + C, 

where C is an arbitrary constant. Indeed, since dF and d& 
are equal, 

d(F - $) = 0, 
so that 

F 3> = const. 

To prove the necessity of the condition of the theorem, note 
that if there exists a function F(x, y) such that 



then 



OF dF 

= M(x, y) and = N(x, y), 



= a M , d*F = dN 

dy dx ~ dy dx dy ~~ dx' 



Since dM/dy and dN '/dx are both continuous, - ^- and 
are also continuous; hence,* 



dx dy dy dx 
Therefore, 

dM = aAT 

dy d" 
* See Sec. 46. 



212 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 64 

As a corollary to Theorem 3, one can state the following: 
The necessary and sufficient condition that M(x, y) dx + N(x, y) dy 
be an exact differential is that dM/dy = dN /dx. 

PROBLEMS 

1. Show that 

.u [(x * + y ^ dx + 2xy dy] 

is independent of the path, and determine its value. 

2. Test the following for independence of path: 



(a) J (y cos x dx + sin x dy) ; 

(b) f [(x* - y*) dx + 2xy dy]; 

(c) / [(x - ?/) dx + 2xy dy]; 

(d) f[(x?-y*)dx-2(x-l)ydy]. 



3. Show that 



I ' j\ L ' \s dx + n . r % is independent of 
J(0,o) L \i ~t~ >> l-L T" 3J</ J 



the path, and find its value. 
4. Show that the line integral 



-ydx xdy 



evaluated along a square 2 units on the side and with center at the origin 
has the value 2ir. Give the reason for failure of this integral to vanish 
along this closed path. 
5. Find the values of the following line integrals: 



(y cos x dx + sin x dy) ; 




, 

(c) [(x + l)dx + (y + 1) dy]. 



64. Multiply Connected Regions. It was shown that the 
necessary and sufficient condition for the vanishing of the line 
integral J c [M(x, y) dx + N(x, y) dy] around the closed path C 
is the equality of dM/dy and dN '/dx at every point of the region 
enclosed by C. It was assumed that C was drawn in a simply 
connected region R and that the functions M(x, y) and N(x r y\ 



64 LINE INTEGRAL 213 

together with their first partial derivatives, were continuous on 
and in the interior of C. The latter condition was imposed in 
order to ensure the integrability of the functions involved. The 
reason for imposing the restriction on the connectivity of the 
region essentially lies in the type of regions permitted by Green's 
theorem. 

Thus, consider a region R containing one hole (Fig. 60). The 
region R will be assumed to consist of the exterior of C 2 and the 
interior of C\. Let a closed contour C be drawn, which lies 
entirely in R and encloses C%. 
Now, even though the functions 
M(x, y) and N(x, y) together with 
their derivatives may be continu- 
ous in R, the integral 

J c [M(x, y) dx + N(x, y) dy] 

may not vanish. For let K be 

, , . *i. 60. 

any other closed curve lying in R 

and enclosing C^ and suppose that the points A and B of K 
and C are joined by a straight line AB. Consider the integrals 




JAPA JAB JBQB JBA' 



where the subscripts on the integrals indicate the direction of 
integration along the curves K, C, and along the straight line AB, 
as is indicated in Fig. 60. Since the path AB is traversed twice, 
in opposite directions, the second and the last of the integrals 
above will annul each other, so that there will remain only the 
integral along K, traversed in the counterclockwise direction, 
and the integral along C, in the clockwise direction. Now, if M 
and N satisfy the conditions of Theorem 1, Sec. 63, then 

f (Mdx+Ndy) + f (M dx + N dy) = 0, 
JGK JQC 

where the arrows on the circles indicate the direction of integra- 
tion. Thus, 

(64-1) f (M dx + N dy) = f (M dx + N dy), 

J&K JQC 

both integrals being taken in the counterclockwise direction. 



214 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 64 



The important statement embodied in (64-1) is that the 
magnitude of the line integral evaluated over a closed path in R, 
surrounding the hole, has the same constant value whatever be 
the path enclosing C%. This value need not be zero, as is seen 

from a simple example already 
mentioned in Sec. 63. Thus, 
let the region R consist of the 
exterior of the circle of radius 
unity and with center at the ori- 
gin and of the interior of a con- 
centric circle of radius 3 (Fig. 61). 

-y 




The functions M = 



and 



N = 



x 2 + y 2 
and their deriva- 



FIG. 



x* + y* 

tivcs, obviously satisfy the con- 
ditions of continuity in R and on Ci and C 2 . Also, dM/dy 
= dN/dx. But 



X 



y dx 



\dyj, 



where C is the circle 



gives 



f 

Jo 



x = a cos 0, 
y = a sin 6, 

a 2 sin 2 + a 2 cos 2 



(1 < a < 3), 



dO = 2ir. 



The function F(x, y), of which M(x, y) dx + N(x, y) dy is 
an exact differential, is F(x, y) = tan" 1 -> which is a multiple- 



valued function. 
The function 



F(x, y) = 



[M(x, y) dx + N(x, y) dy], 



where M and N satisfy the conditions of Theorem 1, Sec. 63, will 
be single-valued if the region R is simply connected (as is required 
in Theorem 1) but not necessarily so if the region is multiply 
connected. 



65 LINE INTEGRAL 215 

65. Line Integrals in Space. 1 The line integral over a space 
curve C is defined in a way entirely analogous to that described in 
Sec. 63. 

Let C be a continuous space curve joining the points A and J5, 
and let P(x, y, z),,Q(x, y, z), and R(x, y, z) be three continuous, 
single-valued functions of the variables x, y, z. Divide the 
curve C into n arcs As t , (i = 1, 2, * , n), whose projections 
on the coordinate axes are &X{, Ay t , Az r , and form the sum 

n 

PCfc, i?., fO Az, + Q(&, *, fO At/, + B(f t , 77 t , fO AsJ, 

where (&, r; t , f t ) is a point chosen at random on the arc As t . 
The limit of this sum as n increases indefinitely in such a way 
that all As* > is called the line integral of P dx + Q dy + R dz, 
taken along C between the points A and B., It is denoted by 
the symbol 

(65-1) f c [P(x, y, z) dx + Qfo y, z) dy + R(x, y, z) dz]. 

The conditions imposed upon the functions P, Q, and R are 
sufficient to ensure the existence of the limit, provided that the 
curve C is suitably restricted. 

If the equation of the space curve C is given in parametric 
form as 

(65-2) ^ y 



where /i (0, / 2 (0, and / 8 (0 possess continuous derivatives in the 
interval to < t < fa, the line integral (65-1) can be expressed 
as a definite integral 



where P, Q, and R are expressed in terms of t with the aid of (65-2). 

It is possible to derive three theorems analogous to those given 
in Sec. 63 for line integrals in the plane. They are as follows: 

THEOREM 1. Let the region of space considered be one in which 
P(x, y, z), Q(x, y, z\ and fl(x, ?/, z) and their partial derivatives are 
continuous and single-valued functions of x, y, and z. Then the 
necessary and sufficient condition that 



216 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 66 

/ (P dx + Q dy + R dz) = 
around every closed curve in the region is that 

dP = dQ dQ^dR dR = dP 

dy dx J dz dy' dx dz' 

for every point of the region. 

THEOREM 2. Let the functions considered satisfy the conditions 
of Theorem 1. Then the necessary and sufficient condition that 

f ( 7f (Pdx + Qdy + R dz) 

J (a,b,c) 

be independent of the path from (a, fr, c) to (x, y, z) is that 
dP dQ dQ dR dR dP 



dy dx dz dy dx dz 

for every point of the region. 

THEOREM 3. Let the functions P ', Q> and R satisfy the conditions 
of Theorem 1. Then, the necessary and sufficient condition that 
there exist a function F(x, y, z) such that 



is that 



\T/1 f\pj f\Tj\ 

dx ~~ ' dy ~" ^ ^ "" 

aP^dQ dQ = dR dR = dP 

di/ " 5o: ; dz " dy' dx ~~ ^2;' 



for every point of the region. The function F(x, y, z) is given by the 
formula 

F(x, y, z) = (Pdx + Qdy + R dz). 



COROLLARY. The necessary and sufficient condition that 

Pdx + Qdy + Rdz 
be an exact differential of some function $(x, y, z) is that 

dP^dQ dQ^dR dR = <>P_ 
dy dx' dz dy' dx dz' 

for every point of the region. The function 3>(x, y, z) is determined 
from the formula 

*(x, y, z) = f ( 7f (Pdx + Qdy + R dz) + const. 

J(a,o,c) 



LINE INTEGRAL 



217 




t+1 



FIG. 62. 



These results are of particular importance in hydrodynamics and 
the theory of electromagnetism. The vector derivation and 
interpretation of these results are given in Chap. IX on Vector 
Analysis. 

66. Illustrations of the Application of the Line Integrals. 

1. Work. It will be assumed that a force F(x, y) acts at every 
point of the xi/-plane (Fig. 62). This force varies from point to 
point in magnitude and direction. 
An example of such conditions is the 
case of an electric field of force. The 
problem is to determine the work 
done on a particle moving from the 
point A (a, b) to the point B(c, d) 
along some curve C. Divide the arc 
AB of C into n segments by the -^ 
points Pi, P^ - , Pn-i, and let 
As t = PtP t+ i. Then the force acting 
at P t is F(x % , 7/r). Let it be directed along the line P t S, and let 
P t jf be the tangent to C at P t , making an angle t with P t >S. 

The component of F(x^ y l ) along PiT is F cos t and the ele- 
ment of work done on the particle in moving through the distance 
As t is approximately F(x ly y l ) cos t As t . The smaller As t , the 
better this approximation will be. Therefore, the work done in 
moving the particle from A to B along C is 

W = lim J? F(x lf 7/ t ) cos r As t = f F(x, y} cos ds. 

If a is the inclination of P t S and ft is the inclination of P t T, 
then = a /3 and cos = cos a cos + sin a sin 0, so that 

(66-1) W = f F(x, T/) (cos a cos ft + sin a sin ft} ds. 

From the definition of a, it is evident that 

F cos a = ^-component of F = X, 
F sin a = ^-component of F = F. 

Moreover, since cte/ds = cos /3 and dy/ds = sin , 
cos ft ds ^ dx and sin ft ds ~ dy. 



218 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 66 
Therefore, (66-1) becomes 



which is a line integral of the form (60-1). 

If C is a space curve, then an argument in every respect similar 
to the foregoing shows that the work done in producing a dis- 
placement along a curve C in a field of force where the components 
along the coordinate axes are Z, 7, and Z is 

W = f c (X dx + Y dy + Z dz). 

To illustrate the use of this formula, the work done in displac- 
ing a particle of mass m along some curve C, joining the points 




FIG. 63. 



A and B, will be calculated. It will be assumed that the particle 
is moving under the Newtonian law of attraction 



where k is the gravitational constant and r is the distance from 
the center of attraction (containing a unit mass) to a position 
of the particle (Fig. 63). 

The component of force in the direction of the positive #-axis 
is 



r-. / \ l\i!H> JL> 

F cos (x, r) = - -^ - 



Similarly, 



and 



Z = - 



km z 



66 LINE INTEGRAL 219 

The work done in displacing the particle from 'A to B is 



C 
= - 

i) A 



B 
W 

A ' 

But 

, x dx + y dy + zdz 

r = vz + y 2 + z 2 and rfr = - y * ' -- 

Therefore, 



= fcm I -5- = few - > 
J^ r 2 L r _M 



which depends only on the coordinates of the points A and B 
and not on the path C. Denoting the distances from to A 
and B by r\ and r2, respectively, gives 



l l \ 
) 

r 2 ri/ 



W = km 



The quantity & = Arn/r is known as the gravitational potential 
of the mass m. It is easily checked that 



-. -r ; : ; / ~ ; 

dx dy . dz 

so that the partial derivatives of the potential function $ give 
the components of force along the* coordinate axes. Moreover, 
the directional derivative of 4> in any direction s is 

cM> __ d$ dx d$ dy , ^^> dz 
ds dx ds dy ds dz ds 

= X cos (x, s) + Y cos (y, s) + Z cos (z, s) 



where F 8 is the component of force in the direction s. 

A conservative field of force is defined as a field of force in 
which the work done in producing a displacement between two 
fixed points is independent of the path. It is clear that in a 
conservative field the integral 



fa 



'dz) 



along every closed path is zero, so that the integrand is an exact 
differential. 



220 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 66 



2. Flow of a Liquid. Let C be a curve on a plane surface across 
which a liquid is flowing. The xi/-plane will be chosen to coincide 
with the surface. The lines of flow are indicated in Fig. 64 by the 
curved arrows. It will be assumed that the flow of the liquid 
takes place in planes parallel to the xy-plane and that the depth 
of the liquid is unity. The problem is to determine the amount 
of liquid that flows across C in a unit of time. 

If v t is the velocity of the liquid and OL % is the inclination of the 
tangent to the line of flow at P z , then v x \ l = v l cos <x t is the ^-com- 

ponent of v t and v v \ t = v t sin a t is 
the ?y-componcnt of v % . Let As t 
denote the segment P t P t+ i of C. 
A particle at P l will move in time 
At to P', while a particle at P l+ i will 
move to P;+i. Therefore, the 
amount of liquid crossing PJP t+ i in 
time At is equal to the volume of 
the cylinder whose altitude is unity 
and whose base is P f l+ \P( + iP(. 
Aside from infinitesimals of higher 
order, this volume is 

AF t - PJP( P*P t+1 sin t , 




FIG. 64. 

in which t denotes the angle between PJ*( and PJP t +i. But 
i = As t and, except for infinitesimals of higher order, 
^ = v l At. Therefore, AV l = v l At - As r sin 0>. The volume 
of liquid crossing C in a unit of time is 

n 

V = lim V v t sin t As t . 

n-* ^ 

If T< denotes the inclination of the tangent to C at P t , then 
r = t + a. Therefore, 

Vt sin 0i As % = y t (sin r t cos t cos r t sin a t ) As t 

== y t cos a l sin r t As t v r sin a t cos r t As t 



Hence, 
(66-2) 



V = f c ( -v y dx + v x dy) 



is the line integral which gives the amount of liquid that crosses 
C in a unit of time. 



66 LINE INTEGRAL 221 

If the contour C is a closed one and the liquid is incompressible, 
then the net amount of liquid crossing C is zero, since as much 
liquid enters the region as leaves it. This is on the assumption, 
of course, that the interior of C contains no sources or sinks. 
Thus, a steady flow of incompressible liquid is characterized by 

the equation 

/* 

v y dx + v x dy) = 0, 



over any closed contour not containing sources or sinks. This 
implies that (see Sec. 63) 

(66-3) - %* = J-", 

dy dx 

which is an important equation of hydrodynamics known as the 
equation of continuity. Moreover, from Theorem 3, Sec. 63, it is 
known that there exists a function ^ such that 



This function ^ is called the stream function, and it has a simple 
physical meaning, for 

(-v y dx +v x dy) 



represents the amount of liquid crossing, per unit time, any curve 
joining (a, 6) with (x, y). 

The function defined by the integral 



(66-5) *(x, y) = (v x dx + v y dy) 

is called the velocity potential. It is readily shown that 

(66-6) -. and f y = V 

Upon comparing (66-4) with (66-6), it is seen that 

d$ _ <W , ^5__^ 

dx " dy dy ~~ dx' 

These are the celebrated Cauchy-Riemann differential equations. 

If the integral (66-2) around a closed curve C does not vanish, 

then the region bounded by C may contain sources (if V is 



222 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 66 

positive) or sinks (if V is negative). The presence of sources 
or sinks is characterized by the singularities of the function ^, 
that is, those points for which ^ is not continuous or where its 
derivatives may cease to be continuous.* 

The foregoing discussion is readily generalized to a steady 
flow of liquids in space. Instead of the integral (66-5), one will 
have 

$(s, y, z) = f ( *'"'* (vx dx + v y dy + v z dz), 

and if the integral is independent of the path C, the equations 
corresponding to (66-3) are 

c^x__c^, = n ^_^V = A dVx dVz = 
dy dx ' dy dz ' dz dx 

In such a case the integrand is an exact differential, and the 
velocity potential $(#, y y z) gives 



3. Thermodynamics. A thermodynamical state of any sub- 
stance is found to be characterized by the following physical 
quantities: (1) pressure p, (2) volume v, and (3) absolute tempera- 
ture T. The pressure, volume, and temperature are connected 
by the equation 

(66-7) F(p, v, T) = 0, 

so that any two of the three quantities p, v, and T will suffice to 
determine completely the state of the substance. 

In the case of an ideal gas enclosed in a receptacle, Eq. (66-7) 
has the form 

pv - RT = 0, 

where R is a constant. Let p and v be chosen to determine the 
state of the gas, and consider p and v as the coordinates of a point 
P in the py-plane. As the state of the gas changes, the point P, 
which characterizes the state, will describe some curve C in the 
py-plane. If the process is cyclic, so that the substance returns 
to its original state, then the curve C will be a closed one. 

It is important to know the amount Q of heat lost or absorbed 
by the gas while the gas in the receptacle (for example, steam in 

* See in this connection Sec. 64. 



66 LINE INTEGRAL 223 

an engine cylinder) changes its state. Let Ap, Ay, and A!F be 
small changes in the pressure, volume, and temperature, respec- 
tively. Now if any two of these quantities, say p and v y do not 
change, then the amount of heat supplied is nearly proportional 
to the change in the remaining quantity. If all three quantities 
change, then the total change AQ in the amount of heat supplied 
is approximately equal to the sum of the quantities AQi, A$2, 
and AQ 3 , due to changes Ap, Ay, and AT, respectively.* 
Thus, 

AQ = AQ 1 + AQ 2 + AQ 3 
= d Ap + c 2 Ay + c 3 AT 7 , 

where ci, c%, and c 3 are constants of proportionality. Then, the 
total amount of heat supplied in the process is given by the 
equation 

(66-8) Q = (ci dp + c 2 dv + c 3 dT). 



Solving (66-7) for T in terms of p and v gives T = /(p, y), so that 

dT = ^-dp + ^dv. 
dp ^ dv 

If this expression is substituted in (66-8), one obtains 

(66-9) Q = j' c [( Cl + c |) dp + (c, + ca g) dv], 

where the integration is performed over the curve C in the 
py-plane, which is called the pv diagram. 

Consider the state of the gas in the cylinder of a steam engine, 
and let the piston be displaced through a distance As. Then, 
if the area of the piston is A, the work AW performed by the 
piston is given by 

ATF = pA As = p Aw, 

and the total work W performed during one cycle is 

W = f c pdv. 

It follows from (61-4) that this is precisely equal to the area of 
the pv diagram. 

18 This principle is called the principle of superposition of effects. 



224 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 66 

In deriving (66-9), it was assumed that p and v were the 
independent variables, and it followed, upon making use of (66-7), 
that the increment of heat is given by* 

dQ = (ci + C3~-} dp + f 2 + cz~\dv 

ss P(p, v) dp + V(p, v) dv, 

where P and V are known functions of p and v. The expression 
for dQ is not, in general, an exact differential (that is, dP/dv 
? dV/dp), for the line integral (66-9) need not vanish for a 
cyclic process. However, it is possible to show that the difference 
between dQ and the work p dv is an-oxact differential, namely, 

(66-10) dU ss dQ - pdv, 

where the function U is called the internal energy of the gas. 

It is also possible to showf that the ratio of dQ to the absolute 
temperature, namely, 

(66-11) dS^^Q 

is likewise an exact differential. The function S is called the 
entropy, and it plays a fundamental role in all investigations in 
thermodynamics. 

The formulas (66-10) and (66-11) can be used to show that 
for an isothermal process (that is, when dT = 0) 

dQ = p dv, 

so that all the heat absorbed by the gas goes into the performance 
of the work p dv. If the process is adiabatic (that is, such that 
there is no gain or loss of heat), then dQ = and, therefore, 
dS = 0. It follows that the entropy S is constant during such a 
process. 

* By making use of T and v, or T and p, as the independent variables, it is 
possible to write down two other important expressions for dQ. 

f These assertions follow from the first and second laws of thermo- 
dynamics. 



CHAPTER VII 
ORDINARY DIFFERENTIAL EQUATIONS 

67. Preliminary Remarks. The great usefulness of mathe- 
matics in the natural sciences derives from the fact that it is 
possible to formulate many laws governing natural phenomena 
with the aid of the unambiguous language of mathematics. 
Some of the natural laws, for example those dealing with the 
rates of change, are best expressed by means of equations involv- 
ing derivatives or differentials. 

Any function containing variables and their derivatives (or 
differentials) is called a differential expression, and every equation 
involving differential expressions is called a differential equation. 
Differential equations are divided into two classes, ordinary and 
partial. The former contain only one independent variable and 
derivatives with respect to it. The latter contain more than one 
independent variable. 

The order of the highest derivative contained in a differential 
equation is called the order of the differential equation. Thus, 



dx* 

is an ordinary differential equation of order 2, and 



is a partial differential equation of order 3. 

When a differential equation can be expressed as a polynomial 
in all the derivatives involved, the exponent of the highest 
derivative is called the degree of the equation. In the foregoing 
examples the degree of the ordinary equation is and that of the 
partial differential equation is 2. It should be observed that the 
degree of 



dx* + \G 



is 2, when this equation is rationalized. 

225 



226 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 67 

If an ordinary differential equation is of the first degree in 
the dependent variable and all its derivatives, it is called a 
linear differential equation. The general form for a linear 
differential equation of the nth order is 



where the PI(X) and f(x) are functions of x only. 

An explicit function y = f(x), or an equation <p(x, y) = 
which defines y as an implicit function of .r, is said to be a solution 
of the differential equation 

(67-1) F[x, y, y f , y", - - , y^] = 0, 

provided that, whenever the values of y, y', y", - - , y (n) 
are substituted in the left-hand member of (67-1), the latter 
vanishes identically. 
For example, 

(67-2) ^ + y cos x = 

has a solution 

y = e~* mx , or log y + sin x = 0, 

because the substitution of y and y' calculated from either one 
of these expressions reduces (67-2) to an identity = 0. Thus, 

differentiation of the second equation gives - -~ + cos x = 0, 

y ax 

so that y f = y cos #, and substitution in (67-2) gives = 0. 

The graph of a solution of an ordinary differential equation is 
called an integral curve of the equation. 

PROBLEM 

Classify the following differential equations, and determine their 
orders and degrees: 



ox* dx oy oy 
(c) -^ + sin y + x = 0; 




ORDINARY DIFFERENTIAL EQUATIONS 



227 



(e) y" - vT=T y' + by = 0; 

*' dt 2 ~~ dx 2 ' 
(g) y" + x*y' + xy = sin x; 



68. Remarks on Solutions. 

the first order, 

(68-1) ^ - 



Consider a differential equation of 



where f(x, y} is a single-valued and continuous function of the 
variables x and y. If a point (J Q , ?/ Q ) is chosen in the xy- 
and its coordinates are substituted in (68-1), then 

= /u 2/0) 



determines a direction associated with the point (# , 2/o), since 

dy/dx can be interpreted geometrically as the slope of an integral 

curve. If a second point (?i, yi) is chosen and its coordinates 

are substituted in (68-1), a direction is 

associated with (xi, yi). Continuing in 

this way, it is possible to find a direction 

associated with every point of the plane 

for which f(x, y) is defined. Now, sup- 

pose that a point (# , 2/o) is chosen in the 

plane (Fig. 65) and the direction associated 

with this point is determined. Let (#1, 7/1) be a point very near 

to (XQ, yo) and in the direction specified by 




FIG. G5. 



Then, 



dy 
dx 

dy 
dx 



determines a new direction. Upon proceeding a short distance 
in this new direction, a third direction given by 



228 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 68 

is determined by the selection of a point (x 2 , 1/2) which is close to 
(xi, yi). If this process is continued, there will be built up a 
curve made up of short straight-line segments. If the points 
(x , 2/o), (xi, t/i), (z 2 , 2/2), ' ' ' , (x n , 2/n) are chosen very close 
together, it becomes intuitively clear that this series of straight- 
line segments approximates a smooth curve associated with the 
initial point (X Q) t/ ). Evidently, the equation of this curve will 
be a solution of the differential equation (68-1), for the slope of 
the curve is 



In general, a different choice of (X Q , 2/o) will lead to a different 
integral curve and thus to a different solution of (68-1). 

The foregoing discussion forms the basis of one method of 
graphical solution of differential equations of the first order. 
Another important method of approximate solution of differential 
equations is the method of infinite series, which is outlined next. 

Let it be supposed that the function /(#, y) in (68-1) can be 
expanded in Taylor's series about the point (x , 2/0) ; then the 
solution of (68-1) can be obtained in the form of a power scries in 
x XQ. Indeed, denote the solution of (68-1) by 

(68-2) y = F(x). 

Then, if the integral curve defined by (68-2) is to pass through 
(XQ, 2/0), it is necessary that 

y = F(x Q ) = 2/0. 
Substituting the coordinates of (XQ, 2/0) in (68-1) gives 

J~ = /O&O, 2/o) = F'(XQ). 

Differentiating (68-1) yields 

d*y = df(x, y) , df(x, y) dy^ 
dx 2 dx dy dx 

so that the value of the second derivative of (68-2) at XQ is 



dy 



68 ORDINARY DIFFERENTIAL EQUATIONS 229 

The formula (68-3) can be used to calculate d 3 y/dx 3 , and its 
value at the point (# , 2/0) can be obtained, for the values of the 
first and second derivatives of F(x) at x = XQ are already known. 
In this manner, one can attempt to find the solution of (68-1) 
in the form of the series 



y = F(* ) + F'(x )(x - x a ) + - (x - z ) 2 + . 

In essence, this method of solution is the same as the method of 
undetermined coefficients that is discussed in Sec. 98. Another 
important method, due to the French mathematician E. Picard, 
is discussed in Sec. 103. 

Next consider a family of curves 

(68-4) y = x 2 + c, 

where c is an arbitrary constant. Differentiation of (68-4) gives 



- 

-7 AX. 

dx ' 

which is the differential equation of the family of curves (68-4), 
and which is free from arbitrary constants. If the given func- 
tional relation contains two arbitrary constants, as, for example, 

y = ci sin" 1 x + c 2 , 

then it is possible to eliminate these constants c\ and Cz by 
two differentiations. The first differentiation gives 



Solving for c\ yields 



and differentiation of this equation gives 

dy = Q 



dx 2 1 x 2 dx 

This is a differential equation of the second order, and clearly it 
has y = ci sin" 1 x + 03 as a solution. It should be observed that 
two differentiations were necessary in order to eliminate two 
arbitrary constants. 



230 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 68 

In general, if f(x 9 y, ci, c 2 , , c n ) = is a functional 
relation involving n arbitrary constants and defining y as a 
function of x, then n successive differentiations will produce n 
equations involving derivatives up to and including those of the 
nth order. These n equations together with the given equation 
/Or, t/, Ci, c 2 , , c n ) = can be used to eliminate the n con- 
stants ci, C2, , c n , and the result will be a differential equation 
of the nth order whose solution is f(x, y, Ci, c 2 , , c n ) = 0. 
It can be shown that, in general, a differential equation of the 
nth order has a solution which contains n arbitrary constants. 
Moreover, no solution of a differential equation of the nth order 
can contain more than n arbitrary constants. A solution that 
contains n arbitrary constants is called the general solution of the 
differential equation. 

The foregoing discussion does not prove these facts. It 
merely suggests that a functional relation containing n arbitrary 
constants leads to a differential equation of order n. For the 
proof of this theorem and its converse, any advanced treatise on 
differential equations* can be consulted. 

Any solution that is obtained from the general solution by 
specifying the values of the arbitrary constants is called a par- 
ticular solution. Particular solutions arc usually the ones that 
are of interest in applications of differential equations. It 
should be remarked, however, that some differential equations 
possess solutions which cannot be obtained from the general 
solution by specifying the values of the arbitrary constants. 
Some examples illustrating the existence of such solutions are 
given in Sec. 83. 

PROBLEMS 

Find the differential equations of the following families of curves : 

1. x* + cx + y = c 2 . 

2. ci sin x + c 2 cos x = y. 

3. Cix + c& x + C&-* = y. 

4. ce* - xy + e~ x = 0. 

6. (x - Cl ) 2 + (y - c 2 ) 2 = 1. 

6. y = c\e x sin x + c 2 e* cos x. 

7. c 2 x + cy + 1 = 0. 

8. cfy + ciy + c 2 = 0. 

9. y = Cix* + c& 2 + c&. 

* See INCE, E. L. Ordinary Differential Equations. 



69 ORDINARY DIFFERENTIAL EQUATIONS 231 

10. y z - 4cx = 0. 

11. y = Ci6 2aj + c 2 e 3 * + #. 

69. Newtonian Laws. In order to illustrate the prominence 
of the subject of differential equations in a study of various 
phenomena, the next four sections are primarily concerned with 
the task of setting up the differential equations from the basic 
physical principles. A systematic treatment of the problem of 
solving various typos of differential equations frequently occur- 
ring in practice will be given in the subsequent sections. 

The formulation of the basic principles from which many 
differential equations arise rests on the following fundamental 
laws of dynamics, which were enunciated by Sir Isaac Newton. 

1. Every particle persists in its state of rest or moves in a straight 
line with constant speed unless it is compelled by some force to 
change that state. 

2. The rate of change of momentum of a particle is proportional 
to the force acting on it and is in the direction of the force. 

3. Action and reaction are equal and opposite. 

The first law merely states that any change of velocity of a 
particle (that is, acceleration) is the result of some external force. 
The second law postulates that the resultant force / acting on a 
particle is proportional to the product of the mass m by its accel- 
eration a; for momentum is defined as the product of mass m and 
velocity v, and the rate of change of momentum is 

d , ^ dv 

_ (mv) = m -rr = ma. 

at at 

Thus, 

ma = kf, 

where k is the proportionality constant, which can be made equal 
to unity by a proper choice of units. 

Obviously, the second law includes the first; for if the force 
acting on a particle is zero, then its acceleration is zero and the 
particle must either remain at rest or move with constant 
velocity. 

The third law asserts that, if two particles exert forces on each 
other, then the force exerted by the first on the second is equal to 
the force exerted by the second on the first. This law can be 
used to define the mass of a body. 



232 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 69 

Frequent use of these laws will be made in the following pages. 
There is one more law, formulated by Newton, that will be found 
of cardinal importance in this study. It is the law of gravita- 
tion. Newton was led to it by his attempts to explain the 
motions of the planets. This law states that two bodies attract 
each other with a force proportional to the product of their 
masses and inversely proportional to the square of the distance 
between them, the distance being large compared with the 
dimensions of the bodies. If the force of attraction is denoted 
by F, the masses of the two bodies by mi and m%, and the distance 
between them by r, then 

(69-1) F = ^p, 

where K is the proportionality constant, called the gravitational 
constant. In the c.g.s. system the value of K is 6.664 X 10~ 8 . 

The three fundamental principles formulated by Newton in 
reality form the postulates of dynamics and furnish a definition 
of force, and the law of gravitation permits one to compare 
masses with the aid of the beam balance. 

The law of attraction (69-1) assumes a simpler form in the case 
of a small body of mass m falling to the earth from heights that 
are not too great. It can be established that a sphere attracts a 
particle at an external point as if the whole mass of the sphere 
were collected at its center. * If the height of the particle above 
the earth's surface is small compared with the radius of the earth, 
the law of attraction becomes, since r in (69-1) is sensibly con- 
stant and equal to the radius of the earth, t 

(69-2) F = mg, 

where g is a new constant called the acceleration due to gravity. 
Its value in the c.g.s. system is approximately 980 cm. per second 
per second and in the f .p.s. system 32.2 ft. per second per second. 
Thus, the differential equation of the falling body can be 
written as 

(69-3) g = g, 

where s is the distance traveled by the body and t is the time in 
seconds. Integration of (69-3) gives 
* In this connection, see Sees. 16 and 66. 



70 ORDINARY DIFFERENTIAL EQUATIONS 233 

(69-4) ~ = gt + t>o; 

and, since the velocity v is equal to ds/dt, (69-4) may be written 

v = gt + v , 

where VQ is the constant of integration so chosen as to equal the 
initial velocity, that is, the value of v at the time t = 0. 
Integrating (69-4) gives 

(69-5) s = y 2 gP + v Q t + s , 

where SQ is the distance of the body from the point of reference 
at the time t = 0. Equation (69-5) furnishes all the desired 
information about the freely falling body. 

70. Simple Harmonic Motion. Simple harmonic motion is the 
most important form of periodic motion. It represents a 
linear vibration of such a sort that the vibrating particle is 
accelerated toward the center of its path in such a way 
that the acceleration is proportional to the displacement of the 
particle from the center. If the displacement of the particle 
from its central position is denoted by #, the definition of simple 
harmonic motion demands that 

r/ 2 r 

(70-1) *Z = -,*, 

where a? 2 is a constant of proportionality and the negative sign 
signifies that the acceleration is directed oppositely to the dis- 
placement x. 

In order to find the equation of motion, that is, the displace- 
ment of the particle in terms of the time t, multiply both sides 

fir 

of (70-1) by 2 ~ and obtain 



The left-hand side of (70-2) is the derivative of (dx/dt) 2 , and 
integration yields 







()'--. 

where the constant of integration is written for convenience in 



234 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 71 

the form c 2 , for it must be positive; otherwise, the velocity 
dx/dt will be imaginary. 

Extracting the square root and solving for dt give 



dt = 



dx 



which upon integration becomes 



1 * i 

- sin" 1 = t + 



or 



(70-3) 



x = A sin 



+ 



where A = c/u and B = Cio>. The period of the motion, 

T 7 = 27T/CO, is independent of tho amplitude A. 
It will be seen in the next section that (70-3) approximately 

represents the behavior of a simple pendulum. 

71. Simple Pendulum. Let P be a position of the bob of a 

simple pendulum of mass m and of length I (Fig. 66), and let 
be the angle, measured in radians, made by OP 
with the position of equilibrium OQ. Denote the 
tangential acceleration by d*s/dt 2 , where s repre- 
sents the displacement, considered positive to the 
right of OQ. 

The acceleration d 2 s/dt* along the path of the 
bob is produced by the tangential component of 
the force of gravity mg, so that its magnitude is 
mg sin 0. Since the velocity of the bob is decreas- 
ing when the bob is moving to the right of its 
position of equilibrium OQ, the acceleration will 

be negative. Hence, since force is equal to the product of mass 

and acceleration, one can write 




FIG. 66. 



(71-1) 



dt* 



The normal component of the force of gravity acts along OP 
and is balanced by the reaction of the string (Newton's third 
law of motion). 

Remembering that s = 10, (71-1) can be written as 



(71-2) 



71 ORDINARY DIFFERENTIAL EQUATIONS 235 

and if the angle is so small that* sin can be replaced by 6, 



This equation is precisely of the form (70-1), and its general 
solution is 

(71-4) = Ci sin (ut + c 2 ), 

where Ci and c 2 are arbitrary constants and co 2 = g/l. 

However, from physical considerations it is clear that there 
is nothing arbitrary in the behavior of the pendulum. Moreover, 
it is known that, if the pendulum bob is held initially at an angle 
a and then released without receiving any impulse, the pendulum 
will vibrate in a perfectly definite manner, so that it must be 
possible to calculate the position of the bob at any later time t. 

These remarks concerning the initial position of the pendulum 
bob and the fact that the bob was released with zero velocity 
can be stated mathematically as follows: If the time at which 
the pendulum was released is denoted by t = 0, then 

SO = a when t = 0, 

^ = when t = 0. 

at 

Therefore, the general solution (71-4) of (71-3) must satisfy the 
initial conditions (71-5). Substituting the first of these initial 
conditions in (71-4) gives 

(71-6) a = Ci sin c 2 . 

Differentiation of (71-4) with respect to t shows that 

dO . . . N 

^ = Cio> cos (co + c 2 ), 

and therefore the second initial condition yields 

= Cico cos c 2 , 

which is satisfied if c 2 = ir/2. Substituting this value of c 2 in 
(71-6) gives Ci = a. Thus, the particular solution of (71-3) that 
satisfies the initial conditions is 

6 = a sin ( co + ^ J = a cos ut. 
* See Prob. 11, Sec. 13. 



236 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 71 

Naturally, a different choice of the initial conditions would lead 
to different values for c\ and 0%. 

The solution of the problem of a simple pendulum that was 
just obtained was based on the assumption that 6 was sufficiently 
small to permit the replacement of sin 6 by 6. If this is not the 
case, the problem is somewhat more difficult. In order to 

solve (71-2), multiply both sides by 2 -^> obtaining 
ddd*6 __ __?0 ^. 

*Ttw ~~ i * m dt 

Integration gives 



(dO\ __ 2g 
\dtj ~ I 



cos e + c. 
i 

Since dB/dt = when = a, 

= -~ cos a + C 



dt) I 



and 



The angular velocity is given by dQ/dt] and since the linear 

velocity is / -T:I the velocity in the path at the lowest point is 
u/t 



- (cos e cos a) I 



0=0 = \/2gl(l cos a). 

It may be observed that this is the same velocity that would 
have been acquired if the bob had fallen freely under the force 
of gravity through the same difference in level, for v = 
and h = 1(1 cos a). 

Integrating (71-7) yields 

(71-8) ' /T r de 



\/cos cos a 

which gives the formula for determining the time required for the 
bob to move from the initial position to any other. 



71 ORDINARY DIFFERENTIAL EQUATIONS 237 

If the lowest position of the bob is chosen as the initial position, 
then 6 = when t = 0, and (71-8) becomes 

(Tl-9) < - Jl "" M 



o ^/cos cos a. 
where ^ 0i ^ a. 

In order to evaluate (71-9), first reduce the integral to a more 

A 

convenient form by means of the relation cos 0=12 sin 2 = 



Then, 



Let 



then 



. . a . 
sin p: = sin -jr sin 



jn ' 

cos 2 Q ^ = sin s cos 



and 

2 sin ^ cos (p d<p 2 sin ~ cos <p d<p 



cos H x /l sin 2 ^ sin 2 ^ 

^s \ A 

Substitution of these expressions in (71-10) gives 

rr (*<p, 2 sin jj cos 

t - \ * 

\2fli Jo 



{ sin 2 ^ sin 2 ^ sin 2 ^ j ^/l sin 2 ^ sin 2 <p 



or 



I-J^ 



V 1 - sin2 1 sir 



If the time involved is the time required for the completion of 
one-quarter of the vibration, then 0i = a and hence y\ = ir/2. 



238 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 71 
The entire period is then 

" *2 

T = 



-%/!"" sin 2 ~ sin 2 <p 



9 Jo VI - W sin 2 <? 



where fc 2 = sin 2 -^ 

If (1 k 2 sin 2 <p)~W is expanded by the binomial theorem, so 
that 

T = J- Pdv(l+h 

\g Jo \ * 

term-by-term integration* gives 



It may be noted that the period is a function of the amplitude, 
which was not true in the case of simple harmonic motion. 

N 




FIG 67. 



A reference to Sec. 14 shows that the period of a simple 
pendulum is expressible as an elliptic integral of the first kind. 



e Note Wallis's formula 



(n - l)(n - 3) 



2 or 1 



P sin n e dO = f 2 cos n e dO = ^ . ^ rtx ^ :: a, 

Jo Jo n(n 2) 2 or 1 ' 

where a. 1 when n is odd, a = Tr/2 when n is even. 



72 ORDINARY DIFFERENTIAL EQUATIONS 239 

72. Further Examples of Derivation of Differential Equations. 

1. The Slipping of a Belt on a Pulley. Let T and TI be the 
tensions of the belt (Fig. 67) at the points A and B. Consider 
an element of the belt of length As, which has end points P and Q 
and subtends an angle A0 at 0. Let the tension at P be T and at 
Q be T + AT, and let the normal pressure per unit of length of the 
arc be p, so that the total normal force on the element of arc 
As is p As. If the angle A0 is assumed to be small, the normal 
pressure may be thought of as acting in the direction of the 
line ON, which bisects the angle A0. From the definition of the 
coefficient of friction /z, it follows that the frictional force is equal 
to the product of M by the normal pressure, so that the frictional 
force on PQ is up As, and, since A0 is small, this frictional force 
may be assumed to act at right angles to ON. If it is assumed 
that the belt is at the point of slipping, the components of force 
along ON must balance. Hence, 

T sin ^ + (T + AT) sin ^ = p As 
or 
(72-1) (2T + AT) sin ~? - p-As. 

Similarly, by equating the forces acting at right angles to ON, 

(T + AT) cos ^ - T cos y = M p As 
or 

(72-2) AT cos = MP As. 

Z 

Eliminating p As between (72-1) and (72-2) leads to 

, 70 o\ 2T + AT . A0 1 

(72-3) ~= tan = 

Solving (72-3) for AT gives 

*m ,. A0 2T/i 

AT = tan -^r > 

Z 1 A0 

1 M tan -^ 

and dividing both members of this equation by A0 leads to 



240 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 72 

A0 
AT 



A0 A0 A A0 

"2 1 ~ M tan "2 

The limit of this expression as A0 > is 

(72-4) f = ,r, 



snce 



,. tan a . 
lim - = 1. 



Separating the variables in (72-4) yields 

?-,*. 

which, upon integration, becomes 

log T = fjid + c 
or 

(72-5) T = c^ B . 

The arbitrary constant Ci that enters into the solution of the 
differential equation can be determined from the initial condition 
T = jT when 6 = 0. Substituting these values in (72-5) gives 



T = 
so that the tension TI corresponding to the angle of lap a is 



PROBLEM 

Find the tensions T\ in the foregoing illustration when To = 100, 
M = K> and the angles of lap are Tr/2, %TT, and TT radians. 

2. Elastic Curve. Consider a horizontal beam under the 
action of vertical loads. It is assumed that all the forces acting 
on the beam lie in the plane containing the central axis of the 
beam. Choose the #-axis along the central axis of the beam in 
undeformed state and the positive y-axis down (Fig. 68). Under 
the action of external forces F t the beam will be bent and its 
central axis deformed. The deformed central axis, shown in 



ORDINARY DIFFERENTIAL EQUATIONS 



241 



the figure by the dotted line, is known as the elastic curve, and it 
is an important problem in the theory of elasticity to determine 
its shape. 

It can be shown* that a beam made of elastic material that 
obeys Hooke's law is deformed in such a 
way that the curvature K of the elastic 
curve is proportional to the bending 
moment M. In fact, 



(72-6) K = 



M 

w 




FIG. 68. 



where E is Young's modulus, / is the moment of inertia of the 
cross section of the beam about a horizontal line passing through 
the centroid of the section and lying in the plane of the cross 
section, and y is the ordinate of the elastic curve. The important 
relation (72-6) bears the name of the Bernoulh-Euler law. 

The bending moment M in any cross section of the beam is 
equal to the algebraic sum of the moments of all the forces F, 
acting on one side of the section. The moments of the forces F, 
are taken about a horizontal line lying in the cross section in 
question. 

If the deflection of the beam is small, the slope of the elastic 
curve is also small, so that one may neglect the square of dy/dx 

in the formula for curvature. Thus, for 
small deflections the formula (72-6) can 
be written as 




(72 " 7) ~dtf = 



M_ 
El 



FIG. 69. 



As an illustration of the application of 
this formula, consider a cantilever beam 
of length I, which is built in at the left end and which carries a 
load W on its free end (Fig. 69). The weight of the beam is 
assumed negligible in comparison with the magnitude of the load 
TF, so that the moment M in any cross section at a distance x 
from the built-in end is 

M = W(l - x). 

* See TIMOSHENKO, S., Theory of Elasticity, p. 41; LOVE, A. E. H., A 
Treatise on the Mathematical Theory of Elasticity, 4th ed., pp. 129-130. 



242 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 72 
When this expression is substituted in (72-7), there results 

dtf = m (l " **' 
and integrating gives 

W fix* x* 



The constants of integration, ci and c 2 , are easily evaluated 
from the boundary conditions 

y = 0, when x = 0, 

~~ = 0, when re = 0, 

the first of which expresses the fact that the displacement at 
the built-in end is zero and the second that the slope of the 
elastic curve is zero when x = 0. It is easily checked that these 
boundary conditions require that 

W 

y " 2EI 

so that the displacement d at the free end is 

Wl 3 
3EI 

PROBLEM 

A beam of length / is freely supported at its ends and is loaded in the 
middle by a concentrated vertical load W, 
which is large in comparison with the weight 

jv^ | ^ of the beam. Show that the maximum 

deflection is one-sixteenth of that of the 
cantilever beam discussed above. Hint: 
From symmetry, it is clear that the behavior 
70 t "~ of this beam is the same as that of the 

cantilever beam of length 1/2 which is 

loaded by a concentrated load of magnitude W/2 at its free end (Fig. 

70). 

3. Cable Supporting a Horizontal Roadway. Let a cable that 
supports a horizontal roadway be suspended from two points 
A and B (Fig. 71). It will be assumed that the load on the 
roadway is so large compared with the weight of the cable that 




72 



ORDINARY DIFFERENTIAL EQUATIONS 



243 



T+AT 



the weight of the cable can be neglected. The problem is to 
determine the shape assumed by the cable. 

Denote the tension at the point P of the cable by T and that 
at the point Q by T + AT, and let w be the load per foot run. 
Since the cable is in equilibrium, the horizontal and vertical 
components of the forces acting on any portion As of the cable 
must balance. Thus, equat- 
ing the horizontal and ver- y 
tical components gives a 
system of two equations 

(72-8) T cos 

= (T + AT) cos (0 + A0) 

and 

(72-9) T sin0 = wAx 
+ (T + AT) sin (0 + A0). 

Dividing (72-9) by (72-8) ~~ 
gives 

(72-10) tan = tan (0 + A0) 




FIG. 71. 
W Ax 



But (72-8) does not depend on the magnitude of As and, since 
As is arbitrary in size, it appears that the horizontal component 
of the tension at any point of the cable is a constant, say TO. 
Substituting this value in the right-hand member of (72-10) 
and rearranging give 

1J) AT* 
tan (6 + A0) - tan = =^= f 



cr 



(72-11) 



tan (0 + A0) tan = w_ Az 
A0 ~~ To A0' 



Tho left-hand member of (72-11) is the difference quotient, and 
its limit as A0 is made to approach zero is the derivative of 

tan 0. Hence, passing to the limit gives 

,* 

w_ dx 

''Y Q Te 



(72-12) 



2 
SGC 



Recalling that tan = > so that = tan" 1 ; it follows that 
ax ax 



dx 1 + (dy/dxY 



244 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 72 
Moreover, 



sec 2 0=1+1^ 
\dx 

Substituting from these two expressions in (72-12) leads to the 
differential equation of the curve assumed by the cable, namely, 



V - w . 
d& ~ ri 

If (72-13) is integrated twice with respect to #, one obtains the 
desired equation of the curve, 

ID 
(72-14) y = _ 32 + ClX + C2; 

which is the equation of a parabola. The arbitrary constants Ci 
and C2 can be determined by substituting in (72-14) the coordi- 
nates of the points A and B. 

If the lowest point of the cable is chosen as the origin of the 
coordinate system, the equation of the parabola becomes 

<ii) 
(72-15) y = Wo X *- 

The length of any portion of the cable can easily be calculated 
with the aid of (72-15). 

PROBLEMS 

1. Find the length of the parabolic cable when the latter supports a 
roadway which is I ft. long. Express the length of the cable in an 
infinite series in powers of L Hint: Expand the integrand in the 
expression for the length of the cable. 

2. Find an approximate expression for the sag d in terms of the 
length I by using the first two terms of the infinite series expansion that 
was obtained in Prob. 1. 

4. Uniform Flexible Cable Hanging under Its Own Weight. Let 
a flexible cable (Fig. 72) be suspended from two points A and B. 
Denote the weight per \Hiit length of the cable by w, and con- 
sider the forces acting on the element of cable As. As in the 
preceding example, the horizontal and vertical components of 
force must balance, for the cable is in equilibrium. If the tension 
at P is denoted by T and that at Q by T + AT 7 , it follows that 

T cos 6 = (T + AT) cos (0 + A0) 



72 
and 



ORDINARY DIFFERENTIAL EQUATIONS 



245 



T sin B = (T + AT) sin (6 + A0) - w As. 
Dividing the second of these equations by the first gives 

w As 



tan 6 = tan (0 + A0) - 



(T + AT) cos (0 + A0) 



This equation has the structure of Eq. (72-10), and an analysis 
in every respect similar to that outlined in the preceding illus- 



T+AT 




FIG. 72. 



tration leads to the equation 



(72-16) 



w ds 



sec B = f 35' 



where To is the tension at the lowest point of the cable. Since 

ds ds/dx 



dO dd/dx 



where 



dx 



^ and 



1 + 



and since sec 2 0=1 + (dy/dx) 2 , it follows upon substitution 
in (72-16) that the differential equation of the curve assumed by 
the cable is 



(72-17) 



dx* 



246 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 72 
If dy/dx is replaced by w, (72-17) becomes 
du __ w /T~^T 2 

Tx ~ n V1 + u > 

or 

rfu ^ 7 



Integrating this equation gives 

log (u H 



This differential equation can be solved by the following 
device: Taking the reciprocal of (72-18), one obtains 



and rationalizing the denominator gives 



When (72-19) is subtracted from (72-18), there results 

rfy if +, -(f *+ 
^ . -^r. _ e Vr. 

and integration gives 



The constants c\ and eg can be determined from the condition 
that the curve passes through the points A and B, whose coordi- 
nates are assumed to be known. 

If the constants c\ and cz are chosen to be equal to zero, then 
the lowest point of the curve is at (0, TQ/W), and the equation 



73 ORDINARY DIFFERENTIAL EQUATIONS 247 

of the curve assumed by the cable has the form 

(72-20) y = 

A curve whose equation has the form (72-20) is called a catenary. 

PROBLEM 

Find the length of the catenary between the limits and v. 

73. Hyperbolic Functions. Combinations of exponential func- 
tions analogous to the one that appears in (72-20) are of such 
frequent occurrence in applied mathematics that it has been 
found convenient to give them a special 
name. The function %(e x + e~ x ) is 
called the hyperbolic cosine of x and is 
denoted by 

(73-1) cosh x - y^e* + <r*). 

The derivative of cosh x is equal to 
%(e x e~ x ) and is called the hyper- 
bolic sine of x. Thus, 

(73-2) sinh x - %(c x - e~ x ) 

These functions are called hyperbolic because they boar rela- 
tions to the rectangular hyperbola x 2 y 2 = a 2 that are very 
similar to those borne by the circular functions to the circle 

x 2 -f i/ 2 = a 2 . Thus, consider 
the equation of a circle (Fig. 73) 



whose parametric equations are 
x = a cos t 





and 



y = a sin t. 



The equation of a rectangular 
FIG. 74. hyperbola (Fig. 74) is 

(73-3) x 2 - t/ 2 = a 2 , 

and the reader can readily show with the aid of the definitions 



248 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 73 



(73-1) and (73-2) that (73-3) can be written in a parametric form 
as 

x = a cosh tj 
= a sinh t. 



(73-4) 



It will be shown next that the parameter t can be interpreted 
for the circle and the hyperbola in a similar way. 
The area u of the circular sector OAP (Fig. 73) is 



u 



2u 



so that 



On the other hand, the area of the hyperbolic sector OAP (Fig. 
74) is given by 

(73-5) u = 



~~ / *N/# 2 ~ # 2 dxj 



where the first term in (73-5) represents the area of the triangle 
OBP. 

Integrating (73-5) gives 



x + -\/x 2 a 2 Q? 1 x + y 



= 9" lo g 



2 * ~^T ) 



. 
log 



x + y 2u 
- = 



so that 

and 
(73-6) 



Also, since a; 2 y 2 = a 2 , it follows that 

r 11 - 
(73-7) --? = e a '. 



Adding and subtracting (73-6) and (73-7) lead to 
(73-8) 



x = a 



.2u 

= a cosh -r; 
a 2 



2?* 



2u 



= a 



= a sinh 



2u 



which are precisely Eqs. (73-4) with t = 2u/o 2 . 



73 ORDINARY DIFFERENTIAL EQUATIONS 249 

From (73-8), it is clear that ' 

x . 2u . y . . 2u 

- = cosh -r- and - = sinh r 
a a 2 a a 2 

and a reference to Fig. 73 shows that 

x 2u , y . 2u 

- = cos r and - = sin -=- 
a a 2 a a 2 

Therefore, the circular functions may be defined by means of 
certain ratios involving the coordinates of the point P(x, y) on 
the circle x 2 + y 2 = a 2 , whereas the hyperbolic functions are 
expressed as ratios involving the coordinates of the point P(x, y) 
on the hyperbola x 2 y 2 a 2 . 

The definitions of the hyperbolic tangent, hyperbolic cotangent, 
hyperbolic secant, and hyperbolic cosecant are as follows : 

, sinh x 

tanh x = ; > 
cosh x 

COth X ; ; 

tanh x 

sech x = ; ) 
cosh x 

csch x = 



sinh x 

The inverse hyperbolic functions are defined in a way similar 
to that used in defining the inverse circular functions. Thus, if 

y = tanh x, 
then 

x = tanh" 1 y, 

which is read " the inverse hyperbolic tangent of y." The definition 
of the remaining inverse hyperbolic functions is similar. There 
are some interesting relations that connect these inverse hyper- 
bolic functions with the logarithmic functions.* 

It will be recalled that the expansion in Maclaurin's series for e u 
is 

(73-9) ..! + + ; + ;+..., 

* See Probs. 5 and 7 at the end of this section. 



250 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 73 

so that 

(73-10) * = 1 + x + ~ + ~ + 

and 

(73-11) e- = 1 - x + - + . 

Subtracting (73-11) from (73-10) gives 



so that 

pX p X /y3 /y5 

(73-12) sinh x m $ = * + |-, + |-, + . 

On the other hand, addition of (73-10) and (73-11) shows that 

/?- I p X /v2 /y4 

(73-13) cosh x = e = 1 + |l + |l + ' ' ' 

Moreover, if it is assumed that (73-9) holds for complex num- 
bers as well as for real numbers, then 

(Yr") 2 
(73-14) e = 1 + ix + ^jf- 

and 

(73-15) ^ = l- ix + 



where i s= \/ 1. Adding (73-14) and (73-15) and simplifying 
show that 

T 2 T 4 T 6 



\ 

), 



which is recognized to be the series for cos x multiplied by 2. 
Thus, 

pix JL. p ix 

(73-16) cos x = 



It is readily verified that subtraction of (73-15) from (73-14) 
leads to the formula 

fix _. p-ix 

(73-17) sin x = - -- 



73 ORDINARY DIFFERENTIAL EQUATIONS 251 

By combining (73-16) with (73-17) there result two interesting 
relations, 

cos x + i sin x = e lx and cos x i sin x = e 



= ~ lx 



which are frequently used in various investigations in applied 
mathematics. These relations are known as the Euler formulas. 
The following table exhibits the formal analogy hat exists 
between the circular and hyperbolic functions. The relations 
that are given for hyperbolic functions can be established readily 
from the definitions for the hyperbolic sine and the hyperbolic 
cosine. 

Circular Functions Hyperbolic Functions 

sin x = 2~ (e tx e~ tar ) sinh x = ^ (e x e~ x ) 

cos z = 2 ( e ' x + "**) cosh x = ^ (e x + e~ x ) 

e lx e~* x , , e x ~* 

tan x = t - : - -v tanh a; = - - 

^(e lx + e~ %x ) e x + e~* 

cot # = r -- coth x = 



-- V/vsuii A* 7 i 

tan x tanh a; 



/v.2 /v4 /r2 /v4 

cos x = 1 - 2j + j| - cosh x = 1 + 2j + jj + 

sin 2 x + cos 2 x = 1 cosh 2 x smh 2 x = 1 

1 4- tan 2 x sec 2 x I tanh 2 x = sech 2 x 

sin 2z = 2 sin z cos x smh 2x = 2 smh # cosh x 

cos 2x = cos 2 x sin 2 # cosh 2x = cosh 2 x -f sinh 2 a; 

sin (x y) sin x cos i/ sinh (x ?/) = sinh x cosh t/ cosh x smh ?/ 
cos x sin i/ 

</ sin x d sinh x , 

cos x T - = cosh x 



, \J\JO As i 

dx dx 

d cos x _ _ . d cosh x 

dx ~ dx 



= smh a; 



d tan x d tanh x , , 

3 - = sec 2 x - -5 - = sech 2 x 
dx dx 

Example 1. A telephone wire (Fig. 75) weighing 8 Ib. per 100 ft. is 
stretched between two poles 200 ft. apart. If the sag is 1 ft., find tlu 
tension in the wire. 

Note that 



)- 



where a = T /w. 



252 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 73 

The vertical component of the tension is clearly equal to ws, where 
is the length of the wire. But the length of the catenary between the 

points whose abscissas are and x is 



s 



where y = a cosh Substituting for 



Y" 01 * 4 * dy/dx gives 







-f: 



FIG. 75. 
so that the vertical component of tension is 



1 + sinh 2 - dx 



u j - i. 

co ~ ~ a 81 "' 



T y ws = wa sinh -> 



and the total tension at any point is 



T = V2V + 



= wa J 



l + sinh 2 



wa cosh - 
a 

wy. 



At the point of support, y = a + d, so that T = w(a + d). Since d is 
usually small, the tension in the wire is nearly constant and approxi- 
mately equal to TV 
If the wire is very taut and the distance between the poles is not large, 



r-2* 



- a - 25' 



When x = Z/2, where Z is the distance between the poles, and d is the 
sa g> y a = d and 



so that 



* The symbol a = 6 is used to signify that a is approximately equal to 6. 



73 ORDINARY DIFFERENTIAL EQUATIONS 253 

or 

T = 8d' 

Substituting the numerical values for w, I, and d gives for the value 
of the tension at the lowest point 

- (O.Q8)(200) 2 _ 

To== (8)'(1) - 4001b ' 

Example 2. A parachute, supporting a mass m, is falling from a 
distance h above the ground. Determine the velocity with which it 
strikes the ground if the air resistance is proportional to the square of 
the velocity. 

If the air resistance be denoted by R, then 

R = kv 2 , 

where fc is a proportionality constant depending upon the design of 
the parachute. The force acting downward is 

d 2 s dv 

which is equal to mg kv 2 Hence, 

dv 
m -r. = mg kv* 

or 

g-r = g(\ - oV), 
where a 2 = k/gm. Integrating 

f dv r 

J i -. aV = 9J * 
gives 

1 i 1 + w 



If v = when t = 0, it follows that c\ = 0. The integrated expres- 
sion then simplifies to 

l + av 



av 



254 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 73 
or 



It is easily shown that 

lim tanh t 1, 

and it follows that the terminal velocity is v t = \/gm/k. 
But ds/dt = v, so that 

1 f 

s = - I tanh agt dt 
a J 

= - log cosh agl + c 2 ; 
and since s = when t = 0, c 2 = 0. Hence, 

m i i W * 

s = 7T log cosh A/ . 

PROBLEMS 

1. A wire is stretched between two supports 100 ft. apart. If 
the weight of the wire is 10 Ib. per 100 ft. and the tension in the wire is 
300 Ib., find the amount of sag at the middle. 

2. Newton's law of cooling states that the rate of decrease of the 
difference in temperature of a body surrounded by a medium of con- 
stant temperature is proportional to the difference between the tempera- 
ture of the body and that of the medium, that is, 

dO _ 
dt 

Find the temperature of the body at any time t, if the initial tempera- 
ture is 0i. 

3. If a wire weighing w Ib. per unit of length is stretched between two 
supports I units apart, show that the length of the wire is approximately 



where T is the tension. 

4. Show that any complex number a + U can be put in the form 

a + bi = re 6 *, where r = \/a 2 + & 2 and = tan" 1 - 



73 



ORDINARY DIFFERENTIAL EQUATIONS 



255 



5. If y = sinh x, then x is called the inverse hyperbolic sine and is de- 
noted by x = sinh- 1 y. Prove that x = sinh- 1 y = log (y + vV + 1). 

6. Establish the formulas for hyperbolic functions given in the table 
of Sec. 73. 

7. Establish the following formulas: 

(a) d sinh u = cosh u du; 

(b) d cosh u = sinh u du] 

(c) d tanh u = sech 2 u du; 

(d) d coth u csch 2 u du; 

(e) d sech u = sech u tanh u du\ 
(/) d csch i = csch u coth 



(<;) d sinh" 1 u = 

(fi) d cosh" 1 u 

(i) d tanh- 1 u = 

(j) d coth- 1 u = 



du 



du 



du 
1 - u 2 ' 



du 



u \/\ - u 2 ' 
du 



(k) d sech- 1 u = 

(I) d csch" 1 u = 

u VI + 

(w) cosh" 1 y log (y + V?/ 2 1) = sinh" 1 

(n) sinh- 1 y = log (y + Vl/ 2 H~ 1) = cosh" 1 

(o) tanh- 1 y = g log 1 _ if 2/ 2 < 1; 

(p) coth- 1 y = 5 log 

(#) seen" 1 y 

(r) csch- 1 y 



log 



+ 1; 




< 1; 



8. Plot the graphs of the hyperbolic functions. 

9. A man and a parachute, weighing w lb., fall from rest under the 
force of gravity. If the resistance of the air is assumed to be propor- 
tional to the speed v and if the limiting speed is v , find the expression for 
the speed as a function of the time t. 

Hint: 

w dv 



256 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 74 

10. A man and a parachute are falling with the speed of 100 ft. per 
second at the instant the parachute is opened. What is the speed of 
the man at the end of t sec. if the air resistance is proportional to the 
square of the speed? 

11. It can be established that the steady flow of heat across a 
large wall is proportional to the space rate of decrease of temperature 
across the wall and to the area A of the wall, that is, 



where x is the distance from one of the faces of the wall and Q is the 
constant quantity of heat passing through the wall. The constant k 
(thermal conductivity) depends on the properties of the material. 
Integrate this equation and calculate the amount of heat per square 
centimeter passing through a refrigerator wall, if the thickness of the 
wall is 6 cm. and the temperature inside the refrigerator is 0C., while 
outside it is 2(JC. Assume k = 0.0002. 

12. A tank contains initially v gaL of brine holding XQ lb. of salt in 
solution. A salt solution containing w lb. of salt per gallon enters the 
tank at the rate of r gal. per minute; and the mixture, which is kept 
uniform by stirring, leaves the tank at the same rate. What is the 
concentration of the brine at the end of t min. ? 

Hint: Let x denote the amount of salt present at the end of t min.; 
then, at a later instant t + A, the change in the quantity of salt is 
A# = wr At (x/v)r At. Hence, dx/dt wr xr/v = (r/v)(wv x). 

74. First-order Differential Equations. Generally speaking, 
the problem of solving differential equations is a very difficult 
one. There are very few types of equations whose solutions can 
be written down at once; in practice, special methods of solu- 
tion, suitable to the particular problem under consideration, 
have to be depended upon. Seeking special methods of solution 
is a difficult task, and the mathematician, at present at least, is 
almost entirely restricted to a consideration of linear differential 
equations. Very little is known concerning the solution of non- 
linear differential equations. Even such a simple-appearing 
first-order equation as 



cannot be solved in general; that is, there are no formulas avail- 
able for solving a non-linear differential equation of the first order. 
However," it is possible to classify some of the first-order non- 



76 ORDINARY DIFFERENTIAL EQUATIONS 257 

linear differential equations according to several types and to 
indicate the special methods of solution suitable for each of these 
types. The next ten sections will be concerned with the solutions 
of the special types of non-linear differential equations that are of 
common occurrence in practice. The remainder of the chapter 
will be devoted to the general methods of solution of the impor- 
tant types of linear differential equations. 

75. Equations with Separable Variables. If the given differ- 
ential equation 



can be put in the form 

/i Or) dx + / 2 (y) dy = 0, 



where /i(o;) is a function of x only and fa(y) is a function of y 
only, the equation is said to be an equation with separable 
variables. Such an equation is easily integrable, and its general 
solution is 

ffi(x) dx + f My) dy = c, 

where c is an arbitrary constant. In order to obtain an explicit 
solution, all that is necessary is to perform the indicated 
integrations. 

Example. Solve 

dy , . 

~T~ ~r & x y = & x y 

This can be written as 

~dx "*" e *^ y "" y ' ^ 
or 



Integration gives 

log . __ ^ + e* = c, 
which is the general solution required. 



258 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 75 

PROBLEMS 

Solve the following differential equations: 



1. VI - & dy + V 1 - y z dx = 0. 




dy _ sin 2 x 

dx ~~ sin y 

4. sin x cos 2 y dx + cos 2 x dy = 0. 
6. Vl + x dy - (1 + y 2 ) dx = 

0. 



6. e* VI ~ y 2 dx + - dy 



7. 



dy I + y 



dx 1 + x 

8. e ^ + y - y 2 = 0. 

9. sinh x dy + cosh y dx 0. 



dx 



dx 



12. 
13. 



og 2/ :/ + tan z sec 2 x. 
4?/ 2 ) dx + 3yx* dy = 0. 



sin 



+ (1 - e*) dy = 0. 



xy 



- 

* dx ~ x(y - 1)' 
16. (1 + x 2 ) dy - (1 + y 2 ) dx = 0. 

17 ^ - y 8 + 2y + i 

"' dx "" x 2 - 2x + T 

18. x 2 (l + y)dy + y*(x - 1) dx = 0. 

19. y(l -y)dx- (x+ 1) rfy = 0. 
dy _ x(l + y 2 ) 

^- dx " y(l + x 2 )' 

21. (y 2 - xy) dx + x 2 dy = 0. 

22. Let A be the amount of a substance at the beginning of a chemical 
reaction, and let x be the amount of the substance entered in the reaction 
after t sec. Then, the simple law of chemical reaction states that the 
rate of change of the substance is proportional to the amount of the sub- 
stance remaining; that is, dx/dt = c(A x), where c is a constant 
depending on the reaction. Show that x = A(l e~ c O. 



76 ORDINARY DIFFERENTIAL EQUATIONS 259 

23. Let a solution contain two substances whose amounts expressed 
in gram molecules, at the beginning of a reaction, are A and B. If an 
equal amount x of both substances has changed at the time t, then the 
amounts of the substances remaining are A x and B x. The basic 
law of chemical reactions states that the rate of change is proportional 
to the amounts of the substances remaining; that is, 

ft = k(A - x)(B - x). 

Solve this equation under the hypothesis that x = when t = 0. 
Discuss the case when A B. 

76. Homogeneous Differential Equations. It will be recalled* 
that a function f(x, y), of the two variables x and y, is said to be 
homogeneous of degree n provided that 

f(\x, \y) ^ \f(x, y). 
Thus, 

f(x, y) = z* + x*y + y 3 

is a homogeneous function of degree 3, and 

f(x, y) = x 2 sin ^ + xy 
y 

is a homogeneous function of degree 2. 
If the differential equation is of the form 

(76-1) /!(, y) dx + fr(x, y) dy = 0, 

where /i (x, y) and/ 2 (x, y) are homogeneous functions of the same 
degree, then (76-1) can be written in the form 



where <p(x, y) is a homogeneous function of degree zero, that is, 

<p(\x, \y) 35 \ <p(x y y) = <p(x, y). 
If X is set equal to l/x, then 

<p(x, y) = 

which shows that a homogeneous function of degree zero can 
always be expressed as a function of y/x. This suggests making- 
* See Sec. 40. 



260 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 76 
the substitution y/x = v. Then, since y = vx, 

dy dv , 
-/- = 3- a; + v. 
dx ax 

Substituting this value of dy/dx in (76-2) gives 



x 

This equation is of the type considered in Sec. 75. Separating 
the variables leads to 

dv dx 






<p(l,v) - v x 
which can be integrated at once to give 

F(v, x, c) = 0. 
Since v = y/x, the general solution of (76-1) is 



F (x' X 
An equation of the form 

dy _ a\x + a z y + 



dx bix + b^y + 63 

can be reduced to the solution of a homogeneous equation by a 
change of variable. This is indicated in detail in Prob. 11 at 
the end of this section. 

Example. Solve 

o , 9 dy dy 

v . + a .,_ = xys . . 

This equation can be put in the form 

y*dx + (x* - xy) dy = 0, 

which is of the type (76-1). By setting y = vx and dy = v dx + x dv, 
the equation becomes 

(vxY dx + (x* - vx*)(v dx + x dv) = 0. 

This reduces to 

v dx + x(l v) dv = 



76 ORDINARY DIFFERENTIAL EQUATIONS 261 

and, upon separation of the variables, to 

dx . 1 v . A 

d v = o. 

x v 

Integration yields 

log x + log v v = c 
or 



which simplifies to 



log y - jj- = c. 



PROBLEMS 

Solve the following differential equations: 
1. (x 2 + y 2 ) dy + 2xy dx = 0. 
2 ' x dx~ y " V^ 17 ?. 



4. (x + T/) = x - y. 

5. x 2 y dx - (x 3 - ?/) dy = 0. 



x 
8. x(\fxy + y) dx - x 2 dy = 0. 



dy __ y* - x Vx* - y*_ 

j 

11. Discuss the problem of transforming the differential equation 

dy _ a\x + a 2 y + a? 

into a homogeneous equation by the change of variable x = x' + h and 
y = ?/ + fc. Determine the values of h and fc for which the original 



262 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 77 
equation is transformed into 

<ty _ a\x f + a 2 y r 
dx' b&' + b*y f 

and solve this equation. If aj)z ajb\ = 0, set a\x + a^y = z. 

12. (x* - xy) dy + y* dx = 0. 

13. (y 2 - z 2 ) ch/ + 2zi/ dx = 0. 
dy 1 + 2s + y 

140 dx 1 - 2x - y 

16. (a? - y + 1) efo + (x + y - 1) dy = 0. 

16. i/ 2 dx + (xy + x*) dy = 0. 

17. (2x 3 y - ?/) dx + (Zxif - z 4 ) dy = 0. 

18. (z 2 + 2/ 2 ) rfx + Bxy dy = 0. 

19. (x 2 + y 2 ) dx - xy dy = 0. 

20. (x + y) dy - (x - y) dx = 0. 

77. Exact Differential Equations. It was shown in Sec. 63 
that the necessary and sufficient condition that the expression 

P(x, y} dx '+ Q(x, y) dy 
be an exact differential of some function F(x, y) is that 

m \\ dP dQ 

(77 - 1} ^ = ^' 

where these partial derivatives are continuous functions. 
Consider now the differential equation 

(77-2) P(x, y) dx + Q(x, y) dy = 0, 

and suppose that the functions P(x, y) and Q(x, y) satisfy the 
condition (77-1), so that there exists a function F(x, y) such that 

, v dF , , dF , 
dF = te dx + dj dy 

= P(x, y} dx + Q(x, y) dy. 

Such a differential equation is called an exact differential equation. 
It is clear that the function 



where c is an arbitrary constant, will be a solution of (77-2). 
An explicit form of the function F(x, y) will be obtained next. 



77 ORDINARY DIFFERENTIAL EQUATIONS 263 

By hypothesis the condition (77-1) is satisfied so that one can 
write 

(77-3) = P(x, y) and = Q(x, y). 

ox oy 

Now, the first of these equations will surely be satisfied by the 
expression 

(77-4) F(x, y) = / P(x, y) dx + /(</), 

where the y appearing under the integral sign is treated as a 
parameter and f(y) is an arbitrary function of y alone. The 
function f(y) will be determined next, in such a way that (77-4) 
satisfies the second of Eqs. (77-3). 

Differentiating (77-4) with respect to y and equating -the 
result to Q(x, y) give 

<W 

dy 
so that 

(77-5) j| = Q(x, y) - J P(x, y) dx. 

Hence, 

(77-6) f(y) = J [<2(x, tf) - ^ J P(, J/) dx ] dy. 
Substitution of (77-6) in (77-4) gives the explicit formula 
(77-7) F(x, y) = J P(x, y) dx + J [<3(z, y) 

y) dx \ dy. 



- J P(x, 



To illustrate the use of this formula, consider 

(2xy + 1) dx + (x* + 4y) dy = 0. 
Here, 

- ^ - 2x 
dy ~ dx ~ ZX > 

so that the formula (77-7) is applicable. The reader will verify 
that the substitution of the expressions for P and Q in (77-7) 
gives 

F(x, y) = x 2 y + x + 2y* + c. 



264 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 77 
Hence, the solution is 

x z y + x + 2y 2 = c. 

Instead of using the formula (77-7), one frequently proceeds 
as follows: Since dP/dy = dQ/dx, the existence of a function 
F(x, y) such that 

= 2xy + 1 and = x 2 + ty 
is assured. Now, if 



is integrated with respect to x } y being treated as a constant, there 
results 

F(x, y) = x*y + x + ci(y), 

where c\(y) is not a function of x but may be a function of y, 
since y was treated as a constant. Similarly, the second 
condition 

dF 

f -*' + *' 

upon integration with respect to y, gives 

F(x, y) = z 2 s/ + 2y 2 + ca(x). 

Comparison of the two expressions for F(#, #) shows that if 
F(x, y) = x 2 2/ + re + 2y*, 



then 



= 2xy + 1 and = x 2 + 



Thus, the general solution of the given equation is 
x*y + x + 2y* = c. 

PROBLEMS 

Integrate the following equations if they are exact: 

1. (y cos xy + 2x) dx + x cos xy dy = 0. 

2. (y 2 + 2z?/ + 1) dx + (2xy + x 2 ) dy = 0. 

3. (e* + 1) dx + dy = 0. 

4. (3x*y - y*) dx + (x 9 - 3y*x) dy = 0. 



78 ORDINARY DIFFERENTIAL EQUATIONS 265 

6. (3x*y - y 3 ) dx - (z 3 + 3y 2 z) dy * 0. 

& 9 cos - dx cos - dy = 0. 

a; 2 a: # x y 

x 1 

7. 2z log y dx + dy = 0. 

1 <jy2 "V/l # 2 

, do; + y . dy = 0. 

- x 2 VI- 2/ 2 

9. (2z + e* log y) da; H dy = 0. 

10. 2x sin y dx x' 2 cos y dy = Q. 

+ ( 1 , \ , 1 

12. ( 2a; + - <? x/y ) ^j ; < 

V y / y 2 

13. sin 2y dx + 2x cos 2?y dy = 0. 

14. x 2 (y + 1) dx - y*(x - 1) dy = 0. 

15. y(l + a: 2 )" 1 do; tan" 1 x dy === 0. 

78. Integrating Factors. It is not difficult to see that every 
differential equation of the type 

(78-1) M(x, y) dx + N(x, y) dy = 0, 

which has a solution F(x, y) = c, can be made exact by multi- 
plying both members by a suitable function of x and y. For 
since F(x, y) = c is a solution of (78-1), 



and it follows from a comparison of (78-1) and (78-2) that 
Therefore, 



-^T = M(S, y)M and = 



x + Ndy) =0 

is an exact equation. The function M(X, y) is termed an 
grating factor. Moreover, it is clear that there is an unlimited 
number of such functions for each equation. Despite this fact, 
it must not be concluded that an integrating factor can always 
be found easily. In simpler cases, however, the integrating 
factor can be found by inspection. 
Thus, in order to solve 

x dy y dx = 0, 



266 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 78 

which is not exact as it stands, multiply both sides by l/xy. 
Then the equation becomes 

d JL _ *5 = o, 

y x 

which is exact. Another integrating factor for this same equation 
is I/a; 2 . Similarly, multiplication by l/y 2 makes the equation 
exact. 

In Prob. 1 at the end of this section will be found a few of 
the integrable combinations that frequently occur in practice. 

Example. The differential equation 

(y 2 - x 2 ) dy + 2xy dx = 

is not an exact equation, but on rearrangement it becomes 
y 2 dy + 2xy dx x 2 dy = 0, 

which can be made exact with the aid of the integrating factor l/y 2 . 
The resulting equation is 

2xy dx x 2 dy _ 

V y2 ~ 

which integrates to 

+ - - 

y+ y~ c ' 

, PROBLEMS 

1. Verify the following: 

x dy - y dx 



(6) <*(log|) = 



(e) Hd(x* + y 2 ) = x dx + y dy; 
(/) d(xy) = x dy + y dx. 

2. Solve the following equations by finding a suitable integrating 
factor: 

(a) x dy y dx + x 2 dx = 0; 

(6) On/ 2 + y) dx + (x - x 2 y) dy = 0; 



79 ORDINARY DIFFERENTIAL EQUATIONS 267 



(c) xdy + 3ydx = xy dy, 

(d) (s 2 + 2/ 2 + 2s) <ty -2ydx = 0; 

(e) xdy y dx = xy dy, 

(/) (* 2 - 2/ 2 ) <fy - 2xy dx = 0; 

(0) x dy (y + log x) dx = 0. 

79. Equations of the First Order in Which One of the Vari- 
ables Does Not Occur Explicitly. Suppose that the dependent 
variable y does not occur explicitly in the equation. The form 
of the equation is then 

'(*)-* 

If this equation is solved for dy/dx to obtain 

^ = f(x) 
dx J(x) > 

then y is obtained by a simple quadrature as 

y = ff(*)dx + c. 
Example. Consider 



Solving for dy/dx gives 
and 



Hence, the solutions are 

2 + A/3 
y -- g - * "" c = 

and 

2 - \/3 o 
2/ --- 2^ x 2 - c = 0. 

These solutions can be combined into one equation by multiplying one 
by the other to give 

(y - c) 2 - 2x*(y - c) + Y^ = 0. 

If the independent variable is missing, the equation is of the 
form 



F(=*> 



268 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 79 
Solving for dy/dx gives 



or 

dx 1 
dy /(</)' 

Integration of this equation yields 

,= CtLji 



Occasionally, the differential equation can be solved easily by 
factoring. For example, consider 



"" ' "' dx ' 
This equation can be written in the form 



so that one is led to the solution of the differential equations 

^-2/ 2 = and 2^-* = 0. 
dx y dx 

It follows that the general solution of the given equation can be 
written as 



PROBLEMS 

Solve the following differential equations: 



*(!)'+*-' 



' dx 1 + y* dx 



80 ORDINARY DIFFERENTIAL EQUATIONS 269 

' -+- 



80. Differential Equations of the Second Order. Occasionally, 
it is possible to solve a differential equation of the second order by 
reducing the problem to that of solving first-order equations. 
Thus, if the given equation is of the form 



dy\ 
' Tx) = ' 



the substitution of p = dy/dx reduces it to 



which is an equation of the first order of the type treated in 
Sec. 79. If this equation is solved for p to give 

P = M c), 

the solution for y can be obtained at once, since p = dy/dx. 

No general rules can be given for solving non-linear differential 
equations, and the task must be left to the skill and ingenuity of 
the student. An example of the solution of a non-linear differen- 
tial equation by means of an artifice was given in Sec. 72 in 
dealing with the equation of a flexible cable. Another example 
may prove interesting and useful. 

Example. Consider the equation 



If dy/dx is replaced by p, the resulting equation is 



270 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 80 
Since 



dp dp 

*j~ P jf~> 
ax r ay 



the equation can be written as 

^2-2 2 + 2 = 

or 

dp 2p 2 y 1 
dy ~~ py 

which is a homogeneous equation. Setting p = vy gives 
dv 2yW - y 2 2v 2 - 1 



which reduces to 

Therefore, 

and 

But v = p/y, so that 

and 



dy _ v dv 

y 



* - 1 
log y = log (v 2 - 1)H + log 



Since p = dy/dx, the last equation becomes 



or 



Therefore, 




Combining these two solutions by multiplication gives the solution 

(* + C2 )'-(log C - + ^+^V = 0, 
^ y / 



80 ORDINARY DIFFERENTIAL EQUATIONS 271 

which can be written, also, as 

/ y \ 2 

(x + c 2 ) 2 - ( csch- 1 - ) = 0. 



It is seen from this example that if the given differential equar 
tion is of the form 

(80-1) F(y, </', - , y<>) = 0, 

then one can introduce the new variable 

P = V' 
and calculate the successive derivatives as follows: 



dp dp 

= L_ _ *_ 

d 



= L_ _ *_ /r) 

dx dy p> 



The substitution of these derivatives in (80-1) leads to a 
differential equation of order n 1. It may be possible to solve 
this differential equation and obtain the general solution in 
the form 

p = F(y, ci, , c n _i), 
so that 

(80-2) g = F(y, c lf , c,.!). 

Equation (80-2) is one with separable variables. 

PROBLEMS 

Solve the following differential equations: 

d*y 
1. -7-^ + y = 0. Solve by substituting dy/dx p, and also by 

using the integrating factor 2 dy/dx. 



272 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 81 
5. x'g +(*-!) =0. 

-(!)'-'- 



81. Gamma Functions. Consider a particle of mass m that 
is moving in a straight line under the influence of an attractive 
force whose intensity varies inversely as the distance of the 
particle from the center of attraction. The equation of such a 
motion is obtainable immediately from the definition of force 
(Newton's second law). Denoting the distance from the center 
of attraction by y, it follows that 

d*y k 

m 777? -- 

dt 2 y 

or 



where a = k/m. 
This is a non-linear equation of the type 

g - >. 

which can always be solved by multiplying both sides of the 
equation by 2 dy/dt and integrating. Thus, 



_ 



dt dt* dt y 

and integrating with respect to t gives 



ft) = -2alog</ + c. 

If the velocity of the particle is zero when y = y^ then c 
2a log 2/0 and 



= ^ I2(i loff 
w ^/ y 

The negative sign was chosen for the square root because y is a 



81 ORDINARY DIFFERENTIAL EQUATIONS 273 

decreasing function of t. Solving for dt and integrating yield 

*" dy 




The integral can be put in a simpler form by making the 
obvious transformation log (y$/y) = x, or y yoe~ x . If T is the 
time required to reach the center of attraction, y = 0, the integral 
becomes 



This integral cannot be evaluated in terms of a finite number of 
the elementary functions. In fact, an integral of this type led 
Euler to the discovery of the so-called Gamma functions. 

The remainder of this section will be concerned with the 
study of the improper integral 

(81-2) r(a) = f x?- l er*dx 9 where a > 0, 

which is the generalization of (81-1). It will 'be shown that 
(81-2) defines an interesting function, called the Gamma function, 
which provides a generalization of the factorial and which will 
prove useful in the study of Bessel functions. 

It is not difficult to prove* that (81-2) converges for all positive 
values of a and diverges whenever a < 0. However, it is 
possible to define the function T(a) for negative values of a with 
the aid of the recursion formula which will be developed next. 

If a > 0, then it follows from (81-2) that 

(81-3) r( + 1) = f " xe~*dx. 

/o 

Integrating the right-hand member of (81-3) by parts gives 

I xe- x dx = x a e~ x + a f x a ~ l e~- x dx 
Jo o Jo 

/ 00 

= a I x a ~ l e~ x dx 
Jo 

= r(a). 

Thus, 

(81-4) r(a + 1) = or(a). 

*See SOKOLNIKOFF, I. S., Advanced Calculus, p. 373. 



274 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 81 
But 



so that when a = 1 the formula (81-4) becomes 

T(2) = 1 T(l) = 1. 
Setting a = 2, 3, , n gives 

T(3) = 2F(2) = 1-2, 
T(4) = 3F(3) = 1-2-3, 



T(n) = (n- l)r(n- 1) = (n - 1)!, 
T(n + 1) = nT(n) = n\. 

Hence, the formula (81-4) enables one to compute the values of 
F(of) for all positive integral values of the argument a. 

If by some means (for example, by using infinite series) the 
values of F(a) are obtained for all values of a between 1 and 2, 
then, with the aid of the recursion formula (81-4), the values of 
T(a) are readily obtained when a lies between 2 and 3. These 
values being known, it is easy to obtain T(a) where 3 < a < 4, 
etc. The values of T(a) for a lying between 1 and 2 have been 
computed* to a high degree of accuracy, so that it is possible to 
find the value of F(a) for all a > 0. 

It remains to define T(a) for negative values of a. The 
recursion formula (81-4) can be written as 

(61-5) r() = <SL1>. 



The formula (81-5) becomes meaningless when a is set equal to 
zero, for 

lim F(a) = +00 and Km T(a) <*>. 

a-0-f a-0- 

It follows from (81-5) that the function F( a) is discontinuous 
when a is a positive integer. 

If any number 1 < a < is substituted in the left-hand 
side of (81-5), the right-hand side gives the value of r( a); 
for the values of a + 1 lie between and 1, and T(a) is known 

* A small table is found in B. O. Peirce, A Short Table of Integrals, p. 140. 



81 ORDINARY DIFFERENTIAL EQUATIONS 

for these values of a. Thus, 

rf-i'U 



275 



r( -o.o) = !%>, etc. 



In this manner the values of T (a) for 1 < a < can be com- 
puted. If these values are known and the recursion formula 
(81-5) is used, the values of r() for 2 < a < 1 can be 
obtained, etc. The adjoining figure represents the graph of 
r() (Fig. 76). 

It was observed that 

r(a + !) = ! 

when a is a positive integer. This 
formula may serve as the definition 
of factorials of fractional numbers. 
Thus, 



r(i) = 0! = i. 

This section will be concluded 
with an ingenious method of evaluating 




c~ x x 1 ^ dx. 

If the variable in this integral be changed by the transformation 
x = ?/ 2 , the integral becomes 

,(81-6) y 2 \ = 

Since the definite integral is independent of the variable of 
integration and is a function of the limits, 



(81-7) y 2 \ = 2 

Multiplying (81-6) by (81-7) gives 
(MO 2 = 4 f o " e-*z* 
which can be written as a double integral 
(81-8) O^!) 2 = 4 f " f e-tx'+Vy 

/0 J 



- 



Ydy, 



dy dz. 



276 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 81 

In order to evaluate (81-8), transform it into polar coordinates 
by setting z = r cos and y = r sin 6. The element of area 
dy dz becomes r dr dQ, and (81-8) becomes 



= 4 I dr I r 5 e^ 2 sin 2 cos 2 dO. 
But 

ir 

f* sin 2 cos 2 e dB = ~ 
Jo lo 

and 

/* 00 

Jo e TT r = 

The latter integral is evaluated by integration by parts. There- 
fore 

)_ . _ /~\T* f . . 
- 4 r 2" ~ 2 

It can be shown with the aid of the recursion formula that 



It follows that (81-1) has the value t/o vW(2a) sec. 

PROBLEMS 

1. Compute the values of r(a) for every integer and half integer from 
to 5 by using the relations T(l) = 1 and r(H) == V*. Plot the 
curve y = T(a) with the aid of these values. 

2. The Beta function B(m, n) is defined by the integral 



B(m, n) = a^^-Hl - x)"~ l dx. 
*/o 

If x is replaced by y 2 in T(n) I x* 1 " 1 ^"* dx, there results 

*/o 

F(n) = 2 f* e-iy^dy. 
/o 

Using this integral, form 

T(m)r(n) = 4 f x*- l e~**dx f " y*"- l e-*> dy. 
jo /o 



Express this product as a double integral, transform to polar coordi- 
nates, and show that 



<,> ,<,, 



82 



ORDINARY DIFFERENTIAL EQUATIONS 



277 



3. Show, by a suitable change of variable, that (81-2) reduces to 



4. Show that 



.cKr* 

'(a) / 
* n ~~ Jo 



82. Orthogonal Trajectories. In a variety of practical investi- 
gations, it is desirable to determine the equation of a family of 
curves that intersect the curves 
of a given family at right 
angles. For example, it is 
known that the lines of equal 
potential, due to a distribution 
of steady current flowing in a 
homogeneous conducting me- 
dium, intersect the lines of 
current flow at right angles. 
Again, the stream lines of a 
steady flow of liquid intersect 
the lines of equal velocity 
potential (see Sec. 66) at right 
angles. 

Let the equation of the given family of curves be 




FIG 77. 



(82-1) 



/O, y, c) = o, 



where c is an arbitrary parameter. By specifying the values of 
the parameter c, one obtains a family of curves (see solid curves 
in Fig. 77). Let it be required to determine the equation of a 
family of curves orthogonal to the family defined by (82-1). 

The differential equation of the family of curves (82-1) can 
be obtained by eliminating the parameter c from (82-1) and its 
derivative, 

(82 . 2 ) f + M = . 

dx dy ax 

Let the resulting differential equation be 



Now, by definition, the orthogonal family of curves cuts the 
curves of the given family (82-1) at right angles. Hence, the 



278 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 82 

slope at any point of a curve of the orthogonal family is the nega- 
tive reciprocal of the slope of the curves of the given family. 
Thus, the differential equation of the desired family of curves is 



This is a differential equation of the first order, and its general 

solution has the form 




(82-3) *(x, y, c) = 0. 

The family of curves defined 
by (82-3) is the desired family of 
curves orthogonal to the curves of 
the given family (82-1). It is 
called the family of orthogonal 
trajectories. 

Example. Let it be required to find 
the family of curves orthogonal to 
the family of circles (Fig. 78) 

(82-4) x> + 7/ 2 - ex = 0. 

The differential equation of the family (82-4) can be obtained by dif- 
ferentiating (82-4) with respect to x and eliminating the parameter c 
between (82-4) and the equation that results from the differentiation. 
The reader will check that the differential equation of the family 
(82-4) is 



FIG. 78. 



Hence, the differential equation of the family of curves orthogonal to 
(82-4) is 

This is a homogeneous differential equation whose solution is easily 
found to be 

X 2 + y2 _ cy _ 0. 

Thus, the desired family of curves is the family of circles with centers 
on the y-axis (see Fig. 78). 

PROBLEMS 

1. Find the orthogonal trajectories of the family of concentric circles 
& + y z = a 2 . 

2. Find the orthogonal trajectories of the family of hyperbolas xy = c. 



ORDINARY DIFFERENTIAL EQUATIONS 



279 



3. Find the orthogonal trajectories of the family of curves y = cx n . 
Sketch the curves of the given and the desired families for n = 1, 1,2. 

4. If the equation of a family of curves is given in polar coordinates 
as /(r, 6, c) 0, show that the tangent of the angle made by the radius 
vector and the tangent line at any point (r, 6} of a curve of the family is 

equal to r -r- Hence, show that the differential equation of the 
orthogonal trajectories of the given family of curves is obtained by 
replacing r -y- by ~ -^ in the differential equation of the given family 

of curves. 

5. Using the results of Prob. 4, show that the orthogonal trajectories 
of the family of cardioids r c(\ cos 6) is another family of cardioids. 

6. Find the orthogonal trajectories of the family of spirals r e cB . 

7. Find the orthogonal trajectories of the family of similar ellipses 

8. Find the orthogonal trajectories of the family of parabolas 
y 2 = 4px. 

9. Find the equation of the curve such that the area bounded by the 
curve, the z-axis, and an ordinate is proportional to the ordinate. 

83. Singular Solutions, It was remarked in Sec. 68 that a 
differential equation may possess 
solutions which cannot be obtained 
from the general solution by specify- 
ing the values of the arbitrary con- 
stants. Such solutions are called 
singular solutions. 

Consider a family of integral curves 
defined by 



(x>y). 




(83-1) <p(x, y, c) = 0, 

FIG. 79. 

where (83-1) is the general solution of the differential equation 



(83-2) 



Assume that the family of curves denned by (83-1) is such that it 
has an envelope* (Fig. 79). Since the slope of the envelope at 
any point (x, y) is the same as that of the integral curve which is 

* It will be recalled that an envelope of a family of curves is a fixed curve C 
such that every curve of the family is tangent to C. 



280 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 83 



tangent to the envelope at (x, y), it follows that the equation of 
the envelope must satisfy (83-2). In general, the envelope is 
not a curve belonging to the family of curves defined by (83-1), 
and hence its equation cannot be obtained from (83-1) by 
specifying the value of the arbitrary constant c. It will be 
recalled that the equation of the envelope is obtained by elimi- 
nating the parameter c between the equations 



<t>(x, y, c) = 



and = 0. 

dc 



Example. It is readily verified that the family of integral curves 

associated with the equation 



(83-3) i 

is the family of circles 
(83-4) (x - c) 2 + y- 




y--o 



FIG. 80 



The equation of the envelope of the 
family (83-4) is obtained by eliminating c between (83-4) and 



There results 

(83-5) 



-2(x - c) = 0. 



y = 



which represents the equation of a pair of lines tangent to the family of 
circles (83-4) (Fig. 80). Obviously, (83-5) is a singular solution of 
(83-3), for it cannot be obtained from (83-4) by any choice of the 
constant c. 

Inasmuch as the problem of determining the singular solutions 
of a given differential equation is relatively rare in applied work, 
the subject will not be pursued here any further. 

REVIEW PROBLEMS 

1. A particle slides down an inclined plane making an angle with 
the horizontal. If the initial velocity is zero and gravity is the only 
force acting, what are the velocity of the particle and the distance 
traveled during the time ? Compare the time of descent and the termi- 
nal velocity with those of a particle falling freely from the same height 
as that of the inclined plane. 



83 ORDINARY DIFFERENTIAL EQUATIONS 

2. A particle falls in a liquid under the action of the force of gravity. 
If the resistance to the motion is proportional to the velocity of the par- 
ticle, what is the distance traveled in t seconds when the particle starts 
from rest? 

3. A bullet is projected upward with an initial velocity of VQ ft. per 
second. If the force of gravity and a resistance that is proportional to 
the velocity are the only forces acting, find the velocity at the end of 
t sec. and the distance traveled by the bullet in t sec. 

4. The rate of decomposition of a certain chemical substance is 
proportional to the amount of the substance still unchanged. If the 
amount of the substance at the end of t hr. is x and XQ is the initial 
amount, show that x = Zoe-*', where k is the constant of propor- 
tionality. What is the constant of proportionality if x changes from 
1000 g. to 500 g. in 2 hr.? 

6. A torpedo moving in still water is retarded with a force propor- 
tional to the velocity. Find the speed at the end of t sec. and the dis- 
tance traveled in t sec., if the initial speed is 30 miles per hour. 

6. A disk is rotating about a vertical axis in an oil bath. If the 
retardation due to friction of the oil is proportional to the angular 
velocity o>, find w after t sec. The initial velocity is w . 

7. Water is flowing out through a circular hole in the side of a 
cylindrical tank 2 ft. in diameter. The velocity of the water in the 
jet is \/*2gh, where h is the height in feet of the surface of the water 
above the center of the orifice. How long will it take the water to fall 
from a height of 25 ft. to a height of 9 ft. above the orifice, if the orifice 
is 1 in. in diameter? 

8. Water is flowing out from a 2-in. horizontal pipe running full. 
Find the discharge in cubic feet pejr second if the jet of water strikes the 
ground 4 ft. beyond the end of the pipe when the pipe is 2 ft. above 
the ground. 

9. A projectile is fired, with an initial velocity ?>o, at an angle a with 
the horizontal. Find the equation of the path under the assumption 
that the force of gravity is the only force acting on the projectile. 

10. A cylindrical tumbler containing liquid is rotated with a con- 
stant angular velocity about the axis of the tumbler. Show that the 
surface of the liquid assumes the shape of a paraboloid of revolution. 

Hint: The resultant force acting on a particle of the liquid is directed 
normally to the surface. This resultant is compounded of the force of 
gravity and the centrifugal force. 

11. Two chemical substances combine in such a way as to produce 
a compound. If the rate of combination is proportional to the product 
of the unconverted amounts of the parent substances, find the amount 
of the compound produced at the end of time L The initial amounts of 
the parent substances are a and 6. 



282 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 83 

Hint: dx/dt = k(a - x)(b - x). 

12. Assume that the pressure p of the air at any height h is equal to 
the weight of the vertical column of air above it. If the density of the 
air is proportional to the pressure, what is the law connecting the 
pressure p with the height hi 

13. A particle of mass m is sliding down a rough inclined plane (the 
coefficient of friction /x = 0.2), whose height is 300 ft. and whose angle 
of inclination is 30. If the particle starts from rest, how long will it 
take to reach the foot of the plane? With what velocity will it be 

traveling then? 

Hint: The differential equation of 
motion is 

d * s , ^ 

jp = 0(sm a - AI cos a), 

where a is the angle of inclination of 
e plane. 
14. A runaway carrier in an aerial 




1000 the plane. 



_, 

r 1G. ol . 

tramway is moving along the arc of a 

second-degree parabola joining the points whose coordinates are (0, 
0) and (1000, 300) (Fig. 81). How long will it take the carrier to 
reach the lowest point if the factional resistance is neglected and if the 
carrier starts from the top with initial velocity zero? See in this con- 
nection the Engineers 1 Bulletin of the Colorado Society of Engineers, 
June, 1935. 

16. A brick is set moving in a straight line over the ice with an 
initial velocity of 20 ft. per second. If the coefficient of friction between 
the brick and the ice is 0.2, how long will it be before the brick stops? 

16. A certain radioactive salt decomposes at a rate proportional to 
the amount present at any instant t. How much of the salt will be 
left 300 years hence, if 100 mg. that was set aside 50 years ago has been 
reduced to 90 mg.? 

17. A skier weighing 150 Ib. is coasting down a 10 incline. If the 
force of friction opposing the motion is 5 Ib. and the air resistance is 
two times the speed in feet per second, what is the skier's speed after 
t sec.? 

18. A tank contains 1000 gal. of brine holding 1 Ib. of salt per gallon. 
If salt water containing 2 Ib. of salt per gallon is allowed to enter the 
tank at the rate of 1 gal. per minute and the mixture, which is kept 
uniform by stirring, is permitted to flow out at the same rate, what is 
the amount of salt in the tank at any time ? 

Hint: Let the amount of salt present at any time t be x\ then, the rate 
at which x changes is equal to the rate of gain, in pounds per minute, 



84 ORDINARY DIFFERENTIAL EQUATIONS 283 

diminished by the rate of loss. Thus, 

dx x 



dt 1000 

19. A 100-gal. tank contains pure water. If 50 per cent alcohol is 
allowed to enter the tank at the rate of 2 gal. per minute and the 
mixture of alcohol and water, which is kept uniform by stirring, leaves 
the tank at the same rate, what is the amount of alcohol in the tank 
at the end of 10 min.? 

20. The rate at which two chemical substances are combining is 
proportional to the amount of the first substance remaining unchanged. 
If initially there are 20 Ib. of this substance and 2 hr. later there are only 
10 Ib., how much of the substance will be left at the end of 4 hr.? 

21. A series circuit consists of a condenser whose capacity is c farads 
and the resistance is 72 ohms. Before the circuit was closed the con- 
denser contained a charge of # coulombs. What is the charge on the 

condenser t sec. later? (The differential equation is R -,7 + - = 0.) 

dt c 

22. The rate at which a body is cooling is proportional to the differ- 
ence in the temperatures of the body and the surrounding medium. 
It is known that the temperature of a body fell from 120 to 70C. in 
1 hr., when it was placed in air at 20C. How long will it take the 
body to cool to 40C.? 30C.? 20C.? 

23. A bullet is fired vertically down from a balloon that is 2 miles 
above the surface of the earth. On the assumption that the resistance 
is proportional to the square of the velocity, find the velocity with 
which the bullet strikes the earth if the initial velocity is 1800 ft. per 
second. 

84. Linear Differential Equations. The remainder of this 
chapter will be restricted to the treatment of linear differential 
equations, that is, equations of the type 

(84-1) Po(*)^ + ?!(*) |3+ +P-i(*)fc + P*Wv=fM, 

where the Pi(x) and/(z) are either functions of x or constants. It 
is extremely fortunate that a large number of physical phe- 
nomena are successfully described with the aid of linear differ- 
ential equations. It will be shown in the succeeding sections 
that it is possible to give a more detailed account of the treat- 
ment and solution of linear differential equations than has been 
furnished for non-linear equations. 



284 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 86 

85. Linear Equations of the First Order. A linear differential 
equation of the first order has the form 



In order to solve this equation, set y = uv, where u and v are 
functions of x that are to be determined later. With this 
substitution, (85-1) becomes 

dv . du . * , N - , N 

U dx + V ~dx +fl(x ^ UV ~ f *^ 
or 

(85-2) 

If u is suitably chosen, the bracket in (85-2) can be made equal 
to zero, thus reducing (85-2) to a simple form. In order to 
choose u so that the expression in the bracket is equal to zero, set 

du 



or 

~ 

It 

Integrating gives 

log u + J/i(x) dx = c, 

and choosing the simplest expression for u, by setting c = 0, 
produces 



With this choice of u, (85-2) becomes 

-r/jC*) dx dv _ 
dx ~~ 
or 

_ f/i() dx f 

dx "~ e J 

which integrates into 



86 ORDINARY DIFFERENTIAL EQUATIONS 285 

By hypothesis, y = uv, so that 

(85-3) y = e-W> dx / e/'' (a d %(x) dx + ce~I'> (x) dx . 
This is the general solution of (85-1). 
Example 1. Solve 

-7- + y cos x = sin 2x. 

Upon using formula (85-3), 

y = e -I COB x r** f e /cos * dx s i n 2# eta + ce~f a * dx 
= e~n * J e sia x sin 2x dx + ce~ x , 

which is easily evaluated by replacing sin 2x by 2 sin x cos x. 
Example 2. Solve 



Here, 

r2<JE 

^ = 6 ~ J x+i / 



^ J ^ (x + D 2 
wliich is easily evaluated. 

PROBLEMS 
Solve the following equations* 

1. (1 + x') dy + (xy - ) <fa - 0. 

2. (a; 2 + 1) f x + 2xy = x*. 

3. = ^' - 2xy. 



6. ^ h y cos a: = cos 3 x. 

6. x ^ + y - x 2 sin x = 0. 

dy _ y - 1 
'' dc x* + 1 

8. L -r + El = j&, given that 7 = when J = 0; L, ^, and E are 
constants. 



286 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 86 

dy 
9. T- = y + cos x sin x. 

10. -r y sec x esc x = e*(l sec x esc a?). 

11. -i \- yx = y. 

12. dx + 2xdy y dy = 0. 

13. ^ + ?/ sec 2 # = tan z sec 2 x. 

14. (x + 1) ~/ x - y = e* (x + I) 2 . 

15. ~T~ 2?/ 6 3 * = 0. 

86. A Non-linear Equation Reducible to Linear Form (Ber- 
noulli's Equation). An equation of the type 

(86-1) -~ + fi(x)y = fz(x)y n , 

in which n may be regarded different from zero and unity, can 
be reduced to linear form by the substitution z = y l ~ n . Then, 



dz ,1 . dy 
-r- = (1 - ri)y~ n 
dx '* dx 



and (86-1) becomes 



g- (n - !)/!(*)* = -(n- 

which is a linear equation in z. 
Example. Solve 

Setting z = I/y z , the equation becomes 



whose general solution is 

z 
so that 



87 ORDINARY DIFFERENTIAL EQUATIONS 287 

PROBLEMS 

Solve the following equations: 



2. - 



3 + _ =a;2 

7/ 6 dx xy 5 

~ ~ l ' 



5. x -^ + y = 2/ 2 log 

6. + xy = xV. 



_ - 

cte 1 - a; 2 ~ 1 - a; 2 ' 

87. Linear Dififerential Equations of the nth Order. No 

formulas are available for the solution of the linear differential 
equation, with variable coefficients, of order greater than 1. 
This section contains some interpretations of the symbolic 
notation that will be found useful in the solution of the linear 
differential equation 



in which the a l are constants. 

It will be convenient to introduce the new notation 



%-Dv and m Dy. 

dx y dx n y 

In this notation, (87-1) becomes 

D n y + ai D n ~ l y + a* D n ~*y + - + a n -i Dy + a n y = f(x) 
or 

(87-2) (D + oiZ)- 1 + a 2 D- 2 + + a n -iD + a)y = /(x). 

The expression in the parentheses in (87-2) is known as a linear 
differential operator of order n. Obviously, it is not an algebraic 



288 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 87 

expression multiplying y but is a symbol signifying that certain 
operations of differentiation are to be performed on the function y. 
Thus, D 2 2Z) + 5 operating on log x gives 

(D 2 - 2D + 5) log x s D 2 log x - 2D log x + 5 log x 
_ d 2 log x ~ d log # ,r i 

1 2 , , 



The gain in simplicity in using the operational notation results 
from the fact that linear differential operators with constant 
coefficients formally obey the laws which are valid for poly- 
nomials. Thus 



so that the operator D is distributive. If the symbol 
(D + ai)(D + at), 

where ai and o 2 are constants, is interpreted to mean that the 
operator D + ai is applied to (D + a^)y, then 

' (D + oi)(D + a,)y = (D 



d 



(ai + a 2 ) Dy 
= [D 2 + (ai + a 2 )Z) + 

It is readily established that operating on t/ with 
(D + a 2 )(D + ai) 

produces precisely the same result. Hence, the commutative 
law holds, or 



(D + oi)(D + a 2 )2/ s (D + a ,)(D 

s [D 2 + (ax + a 2 )D + 

It is readily established that the law of exponents also holds, 
namely, 



87 ORDINARY DIFFERENTIAL EQUATIONS 289 

so that linear operators can be multiplied like ordinary algebraic 
quantities, where the powers of D in the result are to be inter- 
preted as successive differentiations. 

The solution of (87-2) can be written in the symbolic form 

1 ,, , 

y D + a,D^ + + an-iD + a n J(X) ' 

The meaning of this symbol will be investigated next. 
Consider a simple differential equation 

(87-3) ;! = /(*) or A/ =/(*) 

The solution of (87-3), in symbolic form, is 



so that the symbol l/D must be interpreted as integration* with 
respect to x. Thus, 

1 ,, x 



The meaning of a more complicated symbol can be obtained 
from a consideration of the first-order equation 

(87-4) g + ay = f(x), 

where a is a constant. Writing this equation in the operational 
notation, it becomes 

(D + a)y = /(*). 

The symbolic solution in this case is 
(87-5) y - -f(x). 



It was established in Sec. 85 that the general solution of 

(87-4) is 

(87-6) y = ce~ ax + e~ ax J e ax f(x) dx, 

* In order to make the definition of the operator l/D unambiguous, one 
could agree that the constant of integration should be selected so that 
y = when x assumes some specific value. However, in order to avoid 
complication, the constant that arises from the integration of f(x) will be 
suppressed. 



290 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 87 

and it is desirable to give the symbolic solution (87-5) an inter- 
pretation that is consistent with the actual solution (87-6). 
Now, the solution (87-6) consists of two parts, the first of which, 
ce~ ax , if taken alone, obviously does not satisfy (87-4). The 
second part 

e -ax J c ax f(x) dx 

is a solution of (87-4), for (87-6) represents the general solution 
which reduces to 

e -ax J e a*f( X } fa 

when the arbitrary constant is taken as zero. The part of the 
solution (87-6) containing f(x) is called a particular integral of 
(87-4), and the part containing the arbitrary constant is called 
the complementary function. It may be observed that the 
complementary function cc~ ax satisfies the homogeneous linear 
differential equation* 



It is convenient to associate with the symbol (87-5) the particular 
integral of (87-4), namely, 



f(x) ES e J 



(87-7) ^~ f(x) ES e e?*f(x) dx. 

The arbitrary constant arising from the integration in (87-7) may 
be taken as zero, for the addition of this constant of integration- 
will give rise to a term that can be merged with the complemen- 
tary function. The integral operator 



as defined by (87-7), is of fundamental importance in the follow- 
ing sections. The meaning of a more complicated symbolic solu- 
tion will be given later. 

Example 1. To interpret the symbol 

1 
D + a x "> 

* The term homogeneous linear differential equation should not be confused 
with the homogeneous equation discussed in Sec. 76. The homogeneous 
linear differential equation is one of the type (84-1), where f(x) B 0. 



88 ORDINARY DIFFERENTIAL EQUATIONS 291 

write out its meaning with the aid of (87-7). Then, 

i*x m dx 

m ~ l m(m l)x m ~ 2 



. f . _ 
, if m 0, 



except when a = 0. If a = and m ^ 1, then 
Example 2. 



/vm 

D x 



c . a sin mx m cos mx 

sm mx = e~ a:c I e aa: sin mo: do? = 
J 



^r. = = - ^. - 

D + a J a 2 + m 

PROBLEMS 
1. Show that 



2. What is the meaning of jr. e mx< ! 

U ~f~ ft 

3. What is the meaning of yr r cos mx? 

D ~\~ a 

88. Some General Theorems. In Sec. 87, it was found that 
the general solution of the non-homogeneous linear differential 
equation of the first order contained as part of itself the solution 
of the homogeneous equation 



It will be shown next that a similar statement can be made 
concerning the general solution of the nth-order linear equation. 
Consider first a homogeneous linear differential equation of the 
nth order with constant coefficients, 



If y = e mx is substituted in this equation, the result is 

(m n + aim n ~ l + + a n _ira + a n )e mx = 0. 
If m is chosen so that it satisfies the equation 
(88-2) m n + cum*- 1 + + a n -im + a n = 0, 



292 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 88 

which is called the auxiliary, or characteristic, equation, then 
y = e mx will be a solution of (88-1). But (88-2) has, in general, 
n distinct roots, Wi, w 2 , , m n , so that there will be n distinct 
solutions 



Because of the linear character of (88-1), it is clear that, if y = e m f 
is a solution, then 

y = c t e w /, 

where c t is an arbitrary constant, is also a solution. Moreover, 
it is readily verified that the sum of the solutions of a homogeneous 
linear differential equation is also a solution of the equation. 
Thus, 

(88-3) y = cie m s + c 2 e m s + - - + c n e m n x 

will be a solution; and since it contains n arbitrary constants (all 
roots ra t are assumed to be distinct), it is the general solution of 
(88-1). 
Let 

(88 - 4 > +*+ +"-+*-/<'>. 



where f(x) ^ ; and assume that, by inspection or otherwise, a 
solution y = u(x) of (88-4) has been found. Then, if (88-3) 
is the general solution of the homogeneous equation (88-1), 

(88-5) y = Cie m i x + c^e m ^ x + + c n e m * x + u(x) 

will be the general solution of (88-4). This fact can be verified by 
direct substitution of (88-5). That (88-5) is the general solution 
follows from the fact that it contains n arbitrary constants 
The part of (88-5) that is denoted by u(x) is called a particular 
integral of (88-4), and the part containing the arbitrary constants 
is called the complementary function. 

Example 1. Solve 

dx* dx* dx ~~ 

The auxiliary equation is 

m 8 rn 2 2m = 0, 

and its roots are mi = 0, ra 2 = 1, m 3 = 2. Then the complementary 



88 ORDINARY DIFFERENTIAL EQUATIONS 293 

function is 

Y = d + C&-* + c&**. 

A particular integral u(x) is 

u(x) = Hxe~ x . 

Therefore, the general solution is given by 

y=Y + u(x). 

If (88-1) is written in symbolic form as 
(88-6) (D- + aJJ*- 1 + + a n ^D + a n )y = 

and the differential operator (which is of precisely the same form 
as the auxiliary equation defined above) is treated as an algebraic 
expression, then (88-6) can be written as 

(88-7) (D - mi)(D - m 2 ) (D - m n )y = 0. 
Consider the n first-order linear homogeneous equations 

(D - mi)y = 0, 
(D - m,)y = 0, 



(D - mjy = 0, 

whose solutions can be obtained at once by recalling that the 
meaning of the symbol is given by 

(D - m)y ss -| - my. 

These solutions are e m i x , e m z x , , e m x , which are precisely 
the same as the solutions obtained for (88-1) by a different 
method. 

The general solution of (88-7) was found to be (88-3) under 
the assumption that all the roots ra t were distinct. If some of the 
roots are equal, the number of arbitrary constants c t in (88-3) 
is less than n and the solution given there is not the general 
solution. Suppose that the equation 



is such that its auxiliary equation has a double root, that is, 
mi = W2 = m. Then this equation can be written as 

(D - m)(D - m)y = 0. , 



294 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 88 

If (D m)y is set equal to v, the equation becomes 

(D - m)v = 
and v = Cie mx is its solution. Since (D w)t/ = v, it follows that 

(D - m)y = cie mx , 

which is a linear equation whose solution can be found, with the 
aid of (87-6), to be 

y = e mx (c 2 



Thus, if the auxiliary equation has a double root, the solution 
corresponding to that root is 

y = e mx (Ci + CiZ). 

By an entirely similar argument, it can be established that, if the 
auxiliary equation possesses a root m of multiplicity r, then the 
solution corresponding to that root is 

y = e mx (Ci + C 2 Z + ' ' ' + CrX 1 ^ 1 ). 

Example 2. Find the solution of 

(D 3 - 3D 2 + 4)y = 0. 
The auxiliary equation is 

m 3 - 3m 2 + 4 = or (m + l)(m - 2) 2 = 0. 
Therefore the general solution is 



Example 3. Find the solution of 

(D 2 + l)y - 0. 
The auxiliary equation is 

m 2 + 1 = or (m - f)(m + i) = 0. 
Therefore, the general solution is 

y = de* + c 2 e-* = A cos 3 + B sin a?. 

PROBLEMS 

1. Find the general solutions of 



89 ORDINARY DIFFERENTIAL EQUATIONS 295 

>-!+.-. 



(d) (D 3 2Z) 2 + D)y = 0. 

(e) (D 4 + 3D 3 + 3D 2 + D)y = 0. 
(/) (Z) 4 Jb 4 )y = 0. 

to) (D 3 3D 2 + 4)y = 0. 
(h) (D 3 - 13D + 12)y = 0. 

(f) (D 3 + D 2 - D + l)y = 0. 
(j) (D* + 2D 3 + D*)y = 0. 

}. The Meaning of the Operator 

1 



D* + aiD^- 1 + + a n ~iD + a n 



f(x). 



In Sec. 87 the meaning of the operator n /(x) was given. 

AX | G/ 

Now, consider a second-order linear differential equation with 
constant coefficients, 

3 +<!+ *-"*> 

or 

(89-1) (Z> 2 + oiD + a,)y = /(x). 

It was remarked in Sec. 87 that linear operators with constant 
coefficients obey the ordinary laws of algebra and can be treated 
as polynomials. Therefore, (89-1) can be factored to read 

(D -m 1 )(D - ro,) = f(x) 
or 

(D - mOy = j^M 

= e m i x f e~ m i x f(x') dx, 
in accordance with (87-7). Hence, 

(89-2) y = j^ e m s \ <r"i*J(x) dx 

JJ niz J 

B e m ** \ e^i-*** I e~ m i*f(x) dx dx. 



296 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 89 

For mi = w 2 , (89-2) reduces to 

(89-3) y = e m i x J J e~ m i x f(x) dx dx. 

By direct substitution in (89-1), it is easy to establish the fact 
that (89-2) is a particular solution of (89-1). The general 
solution, according to Sec. 88, is made up of the sum of (89-2) and 
the general solution of the homogeneous equation 

3 +*+* 

which is known to be 

y = cie m i* + C 2 e m 2*, mi ^ m 2 , 

or 

y = (ci + Czx)e m i*, mi = m 2 . 

The interpretation of the symbol 



which represents the symbolic solution of the differential equa- 
tion 

d n y . d n ~ l y . dy . ,, . 

d + <d^+ +a n ^/ x + a n y=:f(x), 

or 

(89-4) (D + axD"- 1 + + a n _,Z) + o)y = /(*), 

can now be made easily. Write the operator in (89-4) in factored 
form, 



+ a n _x + a n 

= (D - mi)(D - m,) (D - m n ), 
so that (89-4) becomes 

V = (D - m n }(D - m n _0 (D - 

1 1 



_ _ 

D - TO n D - m n _i D - m/ > 

Successive operations on /(z) with -~ - give 

jLx ~~ tn/i 

(89-5) y = e m i x J e (w 2 ~ w i )x f e (m r m J x J e~ m n x f(x) (dx) n > 
and the result is a particular integral of (89-4). 



89 ORDINARY DIFFERENTIAL EQUATIONS 297 

It can be shown that if the operator 

_ 1 _ 
D" + axD*- 1 + - + a n _iZ) + a n 

is decomposed into partial fractions (the denominator being 
treated as a polynomial in D), then 

_ _ 1 _ ff v 

y ^ 1 + - + a n ^D + a n J(X) 



Al 4- Az An 

+ 



vD - mi D - m 2 ^ ^ D - m n 

which gives, by (87-7), 
(89-6) y = Aie m i x J e~ m i x f(x) dx + A 2 e m 2* J e~ m ff(x) dx 

which is also a particular integral of (89-4). 

Thus, there are available two methods for the determination 
of the particular integral. The first method* of finding the 
particular integral, (89-5), is known as the method of iteration, 
and the second, (89-6), as the method of partial fractions. 
Generally speaking, formula (89-6) is easier to apply. However, 
if the roots of the auxiliary equation are not all distinct, the 
decomposition of the operator into partial fractions, of the type 
considered, cannot be effected and formula (89-5) must be used. 

Example 1. Solve 

d*y dy 

-'-'- 5 i h G?/ 6 

dx 2 dx J 
or 

(Z) 2 - 5D + 6)7; = e** 
or 

The particular integral, as obtained by the method of iteration, is 



11 1 



/>x 

e ~ 



r 

I / 

J 



- - , 

D -3 D -2 ~ D - 

= e Zx C (e~* C e** dx\ dx = y- 



298 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 89 
If the method of partial fractions is used, then 

1 1_ ,4, _ /_J L_\ ,4, 

y ~D-3D-2 e \D-3 D-2) e 

f f e * x 

= e** j e **e**dx - e** J e **e** x = -y 

The complementary function is 
therefore, the general solution is 
Example 2. Solve 



or 



The particular integral is 



= x - 2, 



and the general solution is given by 

y = (ci + c 2 x)e-* + x 2. 

PROBLEMS 
1. Solve jjjj + 3j/ = a;'. 

2. 



4. The flexure y for end thrust P is given by 



where E is Young's modulus, w is the load, and I is the moment of 
inertia. Solve this equation. 

5. Solve (D 3 - 2D 2 - D + 2)y = 1 - 2*. 

6. Solve (Z) 2 + >> - H)0 = cos x - 3 sin x. 

7. Solve (D 3 - 3D + 2)y = 2 sin * - 4 cos x. 

8. Solve (D 2 - l)y = 5x - 2. 

9. Solve (D 3 - D 2 - 8D + 12)y - 1. 

10. Solve (D 4 l)y == e* cos x. 

11. Solve (D* - 2D + l)y - e. 



90 ORDINARY DIFFERENTIAL EQUATIONS 299 

12. Solve (Z) 2 + D - 2)y = sin 2z. 

13. The differential equation of the deflection y of the truss of a 
suspension bridge has the form 



where H is the horizontal tension in the cable under dead load q, h is 
the tension due to the live load p, E is Young's modulus, and I is the 
moment of inertia of the cross section of the truss about Xhe horizontal 
axis of the truss through the center of gravity of the section and per- 
pendicular to the direction of the length of the tru&=. Solve this 
equation under the assumption that p qh/H is a constant. 

14. The differential equation of the deflection y of a rotating shaft 
has the form 



where El is the flexural rigidity of the shaft, p is the mass per unit 
length of the shaft, and o> is the angular velocity of rotation. Solve 
this equation. 

15. The differential equation of the buckling of an elastically sup- 
ported beam under an axial load P has the form 

!V_,jP^_L*-n 

dx* + El dx* + El y ~ U > 

where El is the flexural rigidity and k is the modulus of the foundation. 
Solve this equation. 

90. Oscillation of a Spring and Discharge of a Condenser. 

The foregoing discussion gives all the essential facts for solving 
an nth-order linear differential equation with constant coeffi- 
cients. At this point, it is desirable to apply the methods of 
solution, outlined above, to a group of important practical 
problems. 

Suppose that it is required to determine the position of the end 
of a helical spring at any time t. It is assumed that the spring is 
set vibrating in a vacuum so that considerations of damping do 
not enter here. If a mass M (Fig. 82) is applied to the end of 
the spring, it produces an elongation s which, according to 
Hooke's law, is proportional to the applied force. Thus, 

F - ks, 
where F = Mg from the second law of motion and k represents 



300 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 90 



the stiffness of the spring. Then, 

Mg = ks. 

If at any later time t an additional force is applied to produce an 
extension y, after which this additional force is removed, the 
spring will start oscillating. The problem is to determine the 
position of the end point of the spring at any subsequent time. 

The forces acting on the mass M are the force of gravity Mg 
downward, which will be taken as the positive direction for 
the displacement y, and the tension T in the 
spring, which acts in the direction opposite 
to that of the force of gravity. Hence, from 
Newton's second law of motion, 




k(s+y) Since T is the tension in the spring when its 
elongation is s + y, Hooke's law states that 
T = k(s + y), so that 



FIG. 82. 



But Mg ks, and therefore the foregoing 
equation becomes 



Setting k/M a 1 reduces this to 



(90-1) 



dt* 



+ a*y = or (D 2 + a 2 )*/ = 0. 



Factoring gives (D ai)(D + ai)y = 0, from which it is clear 
that the general solution is 



Recalling that e tx = cos x + i sin x (Sec. 73), the solution can 
be written as 

y = Ci(cos at i sin at) + C2(cos at + i sin a) 
= A cos at + B sin at, 

where A = ci + c 2 and B = (c 2 Ci)i. The arbitrary con- 



90 ORDINARY DIFFERENTIAL EQUATIONS 301 

stants A and B can be determined from the initial conditions. 
The solution reveals the fact that the spring vibrates with a 
simple harmonic motion whose period is 



The period depends on the stiffness of the spring as would be 
expected the stiffer the spring, the greater the frequency of 
vibration. 

It is instructive to compare the solution just obtained with 
that of the corresponding electrical problem. It will be seen 
that a striking analogy exists between the mechanical and 
electrical systems. This analogy is responsible for many recent 
improvements in the design of telephone equipment. 

Let a condenser (Fig. 83) be discharged through an inductive 
coil of negligible resistance. It is known that c 

the charge Q on a condenser plate is proportional _. 
to the potential difference of the plates, that is, 

Q = CV, L 



where C is the capacity of the condenser. FIG. 83. 
Moreover, the current 7 flowing through the coil is 

_ _dQ 
~ ~dt' 

and, if the inductance be denoted by L, the e.m.f. opposing V is 
L dl/dt, since the IR drop is assumed to be negligible. Thus, 



or 

C~ J 

Simplifying gives 

d z Q I n __ n 
-575 + 77f v = 0, 

ill ' \jJU 

which is of precisely the same form as (90-1), where a 2 = 1/CL, 
and the general solution is then 



302 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 91 
The period of oscillation is 

T = 27r VOL. 

Note that the inductance L corresponds to the mass M of the 
mechanical example and that 1/C corresponds to the stiffness k 
of the spring. 

91. Viscous Damping. Let the spring of the mechanical 
example of Sec. 90 be placed in a resisting medium in which the 
damping force is proportional to the velocity. This kind of 
damping is termed viscous damping. 

Since the resisting medium opposes the displacement, the 

damping force r -~ acts in the direction opposite to that of the 
d/t 

displacement of the mass M . The force equation, in this case, 
becomes 



or, since Mg = fcs, 

^j-.L^ + Atf-o 
dt* ^ Mdt ^ M y ~ u * 

In order to solve this equation, write it in the more convenient 
form 

(91-D -g +2(> !+.,. 

In this case the auxiliary equation is 

m 2 + 26m + a 2 = 
and its roots are 

m = -6 \A 2 - a 2 , 
so that the general solution is 
(91-2) y = cie^-^^^ 1 ^^ + c 2 e ( - fe - V**^)'. 

It will be instructive to interpret the physical significance of 
the solution (91-2) corresponding to the three distinct cases that 
arise when fc 2 - a 2 > 0, fc 2 - a 2 = 0, and b* - a 2 < 0. If 
6 2 a 2 is positive, the roots of the auxiliary equation are real 
and distinct. Denote them by mi and m^ so that (91-2) is 

(91-3) y = cie m i' + c 2 e"V. 



91 



ORDINARY DIFFERENTIAL EQUATIONS 



303 



The arbitrary constants Ci and c 2 are determined from the initial 
conditions. Thus, let the spring be stretched so that y = d and 
then released without giving the mass M an initial velocity. 
The conditions are then 

y = d 
when t = and 



when t = 0. 

Substituting these values into (91-3) and the derivative of 
(91-3) gives the two equations 



d = Ci + c 2 and = 

These determine 

m^d , 

Ci = and 



+ m 2 c 2 . 
mid 



mi m 2 
Hence, the solution of (91-3) is 



mi m 2 

The graph of the displacement represented as a function of t is of 
the type shown in Fig. 84. 
Theoretically, y never becomes 
zero, although it comes arbi- 
trarily close to it. This is the 
so-called overdamped case. 
The retarding force is so great 
in this case that no vibration 
can occur. 

If b 2 a 2 = 0, the two roots 
of the auxiliary equation are equal and the general solution of 
(91-1) becomes 

y = 




FIG. 84. 



If the initial conditions are 
when t = and 



y = d 




dy 
dt 



304 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 91 

When t = 0, the solution is 

y = de~ bt (l + bf). 

This type of motion of the spring is called dead-beat. If the 
retarding force is decreased by an arbitrarily small amount, the 
motion will become oscillatory. 

The most interesting case occurs when b 2 < a 2 , so that the 
roots of the auxiliary equation are imaginary. Denote 6 2 a 2 
by a 2 , so that 

m = 6 ia 
and 

y = Cie (-b+i<x)t _|_ C2e (-6-u)l 

= e~ bt (A cos at + B sin at). 
If the initial conditions are chosen as before, 

y = d 

when t = and 



when = 0, the arbitrary constants A and B can be evaluated. 
The result is 



y = de~ bt f cos a H sin a J, 



which can be put in a more convenient form by the use of the 
identity 

A cos 6 + B sin 6 = V? 2 + B* cos (0 - tan- 1 ~Y 
The solution then appears as 






(91-4) y = - V^+T 2 <r- 6 < cos 



a 



( orf - tan- 1 -\ 
\ a/ 



The nature of the motion as described by (91-4) is seen from 
Fig. 85. It is an oscillatory motion with the amplitude decreasing 
exponentially. The period of the motion is T = 2ir/a. In the 
undamped case the period is T = 27r/a; and since 



91 ORDINARY DIFFERENTIAL EQUATIONS 305 

it follows that 

27T 2* 

a. d 

Thus the period of oscillation is increased by the damping. 

An electrical problem corresponding to the example of the 
viscous damping of a spring is the following: A condenser (Fig. 
86) of capacity C is discharged through an inductive coil whose 
resistance is not negligible. Referring to Sec. 90 and remember- 



\ ^-^ 


C 




^ 


_ 


VWWVWW- 
R 


FIG. 85. 


L 
FIG. 86. 



ing that the IR drop is not negligible, the voltage equation is 
found to be 

V -L%- 7/2 = 

at 



or 



Simplifying gives 

W r L ~dt """ CL ~~ v ' 

and this equation is of the same form as that in the mechanical 
example. The mass corresponds to the inductance L, r corre- 
sponds to the electrical resistance R y and the stiffness k corresponds 
to 1/C. Its solution is the same as that of the corresponding 
mechanical example and is obtained by setting 26 = R/L and 
a 2 = l/CL. 

PROBLEMS 

1. The force of 1000 dynes will stretch a spring 1 cm. A mass of 
100 g, is suspended at the end of the spring and set vibrating. Find the 



306 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 91 

equation of motion and the frequency of vibration if the mass is pulled 
down 2 cm. and then released. What will be the solution if the mass is 
projected down from rest with a velocity of 10 cm. per second? 

2. Two equal masses are suspended at the end of an elastic spring 
of stiffness k. One mass falls off. Describe the motion of the remain- 
ing mass. 

3. The force of 98,000 dynes extends a spring 2 cm. A mass of 
200 g. is suspended at the end, and the spring is pulled do\vn 10 cm. and 
released. Find the position of the mass at any instant t, if the resistance 
of the medium is neglected. 

4. Solve Prob. 3 under the assumption that the spring is viscously 
damped. It is given that the resistance is 2000 dynes for a velocity 
of 1 cm. per second. What must the resistance be in order that the 
motion be a dead beat? 

6. A condenser of capacity 4 microfarads is charged so that the 
potential difference of the plates is 100 volts. The condenser is then 
discharged through a coil of resistance 500 ohms and inductance 0.5 
henry. Find the potential difference at any later time t. How large 
must the resistance be in order that the discharge just fails to be 
oscillatory? Determine the potential difference for this case. Note 
that the equation in this case is 

d*V dV V 



6. Solve Prob. 5 if R = 100 ohms, C = 0.5 microfarad, and L = 
0.001 henry. 

7. A simple pendulum of length / is oscillating through a small 
angle in a medium in which the resistance is proportional to the 
velocity. Show that the differential equation of the motion is 



Discuss the motion, and show that the period is 2w -y/co 2 k* where 
co 2 = g/l 

8. An iceboat weighing 500 Ib. is driven by a wind that exerts a 
force of 25 Ib. Five pounds of this force are expended in overcoming 
frictional resistance. What speed will this boat acquire at the end 
of 30 sec. if it starts from rest? 

Hint: The force producing the motion is F = (25 5)0 = 20gr. 
Hence, 500 dv/dt = 200. 

9. A body is set sliding down an inclined plane with an initial 
velocity of v ft. per second. If the angle made by the plane with the 
horizontal is 6 and the coefficient of friction is ju, show that the distance 



91 



ORDINARY DIFFERENTIAL EQUATIONS 



307 



traveled in t sec. is 



6 



, cos 



Hint: m d*s/dt* = mg sin 6 nmg cos 6. 

10. One end of an elastic rubber band is fastened at a point P, and 
the other end supports a mass of 10 Ib. When the mass is suspended 
freely, its weight doubles the length of the band. If the original length 
of the band is 1 ft. and the weight is dropped from the point P, how far 
will the band extend? What is the equation of motion? w 

11. It is shown in books on strength of materials and 
elasticity that a long beam lying on an elastic base, the 
reaction of which is proportional to the deflection y, 
satisfies the differential equation 



Set a 4 = k/(EI), and show that the characteristic 
equation corresponding to the resulting differential equa- 
tion is m 4 + 4a 4 = 0, whose roots are m = a + ai. 
Thus show that the general solution is 



y = Cie ax cos ax + 



sin ax + c$e~ ax cos ax 



sn ax. 



FIG. 87. 



12. If a long column is subjected to an axial load P and the assump- 
tion that the curvature is small is not made, then the Bernoulli-Euler 
law gives (see Sec. 72) 



dx* 



M_ 
El' 



Since the moment M is equal to Py (Fig. 87), it follows upon setting 
dy/dx p that the differential equation of the deformed central axis is 

D^ 

dy Py 

(1 + p i)fc == ~ Ji' 

Solve this differential equation for p, and show that the length of the 
central line is given by the formula 



where k 2 = d*P/4EI, d is the maximum deflection, and F(k, tr/2) is 
the elliptic integral of the first kind. The equation of the elastic 
curve, in this case, cannot be expressed in terms of the elementary func- 
tions, for the formula for y leads to an elliptic integral. 



308 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 92 

92. Forced Vibrations. In the discussion of Sec. 91, it was 
supposed that the vibrations were free. Thus, in the case of the 
mechanical example, it was assumed that the point of support 
of the spring was stationary and, in the electrical example, that 
there was no source of e.m.f. placed in series with the coil. 

Now, suppose that the point of support of the spring is vibrating 
in accordance with some law which gives the displacement of the 
top of the spring as a function of the time t, say x /(), where x 
is measured positively downward. Just as before, the spring 
is supposed to be supporting a mass M, which produces an 

elongation s of the spring. If the 
displacement of the mass M from its 
position of rest is y, it is clear that, 
when the top of the spring is dis- 
placed through a distance x, the 

vwwwwwwmooooomooo J a tual extension of the spring is 
R 

tion is 



R L y x. If the resistance of the 

medium is neglected, the force equa- 



Jtf f = Mg - k(s + y - x) = -k(y - x), 
whereas, if the spring is viscously damped, it is 



Upon simplifying this last equation, it becomes 
(92-1) M^ + r^ + ky^kx, 

where x is supposed to be a known function of t. 

The corresponding electrical example is that of a condenser 
(Fig. 88) placed in series with the source of e.m.f. and that dis- 
charges through a coil containing inductance and resistance. 
The voltage equation is 



where f(f) is the impressed e.m.f. given as a function of t. Since 

_r-dQ _ dV 

1 ~ dt ~ C -&' 



92 ORDINARY DIFFERENTIAL EQUATIONS 309 

the equation becomes 

d~V dV 

(92-2) CL+CR + V 



An interesting case arises when the impressed e.m.f. is sinusoidal, 
for example, 

f(f) = E Q sin cot. 

Then the equation takes the form 

dW RdV 1 I,- 



Both (92-1) and (92-2) are non-homogeneous linear equations 
with constant coefficients of the type 



(92-3) ^ + 26^ + ay = o/(0. 

The solution of this equation is the sum of the complementary 
function and the particular integral (see Sec. 88). The com- 
plementary function has the form shown by (91-2), namely 

c\e m J + c 2 e m 2*, 
where 



mi = b + \/fr 2 a 2 and ra 2 = b \/b 2 a 2 . 
The particular integral, by (89-5), is 
(92-4) Y = aW J e^~ m ^ [ J er<*Sf(t) dt] dt. 

From the discussion of Sec. 91, it is clear that the part of the 
solution which is due to free vibrations is a decreasing function 
of t and will become negligibly small after sufficient time has 
elapsed. Thus the " steady-state solution" is given by the 
particular integral (92-4). 

Let it be assumed that the impressed force, x in (92-1) and 
f(t) in (92-2), is simply harmonic of period 27r/co and of amplitude 
a . Then, 

f(f) = a sin ut, 
and (92-4) becomes 

Y = aW J e (m r m ^ (J e-"Va sin wt dt) dt. 



310 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 93 
The result of integration* is 



a 2 a 



Y = = sin (wt c), 

- co 2 ) 2 + 46V 



where 

= tan" 1 



This is the steady-state solution. 

The remainder of this section will be devoted to the physical 
interpretation of the solution (92-5). It is observed that if the 
impressed frequency is very high (large w), then the amplitude 
of the sinusoid (92-5) is small, so that the effect of the impressed 
force is small. When co = a, the amplitude is a a/26, which may 
be dangerously large if b (and hence the resistance of the medium) 
is small. For a fixed 6 (resistance of the medium) and a (natural 
frequency of the system), the maximum amplitude occurs when 
(a 2 w 2 ) 2 + 4b 2 co 2 is a minimum, that is, when 

-^ [(a 2 - co 2 ) 2 + 4b 2 co 2 ] = 0. 
do) 

This is readily found to be when 

co 2 = a 2 - 26 2 . 

Upon recalling the physical significance of a and 6, these results 
can be interpreted immediately in terms of the physical quantities. 

93. Resonance. It was remarked in Sec. 92 that if the 
impressed frequency is equal to the natural frequency of vibra- 
tion, then the amplitude of (92-5) may be abnormally large. 
Stated in terms of the physical quantities of the electrical and 
mechanical examples, this means that the maximum voltage 
of the electrical system may be dangerously large or that the 
maximum displacement of the spring may be so great as to pro- 
duce rupture. 

The phenomenon of forced vibration is of profound importance 
in many engineering problems. Not so many years ago the 
collapse of a building in one of the larger American cities was 

* Integration in this case is a little tedious. For actual integration, it is 
convenient to replace sin ut by the equivalent exponential expression 



93 ORDINARY DIFFERENTIAL EQUATIONS 311 

attributed to the rhythmic swaying of the dancing couples, who 
happened to strike the natural frequency of the beam supporting 
the structure. Again, the failure of the Tacoma bridge was 
explained by some on the basis of forced vibration. It is also 
well known that soldiers are commanded to break step in crossing 
a bridge, for fear that they may strike the note of the cables. 
The walls of Jericho are reported to have fallen after seven priests 
with seven trumpets blew a long blast. 

The phenomenon of resonance occurs when the impressed 
frequency is equal to the natural frequency. Consider Eq. 
(92-3) in which 6 (resistance) is zero and f(t) = a sin at } so that 

(93-1) ~ + a*y = a 2 a sin at. 

The particular integral in this case is 

(93-2) F = a a 2 e- a " J (e 2a < J e~^ sin at dt) dt, 

since mi = ai and ra 2 = ai. If sin at is replaced by ^ > 

Zi 

(93-2) integrates into 

V 2 (t COS a ^ I 1 I * 4\ 

JL ~~"~ a$a \ ^ ~i~ o Q sin ai ~\ ^ cos ai j. 

If F is added to the complementary function c\ cos at + c 2 sin at, 
the general solution is given by 

(93-3) y = A cos at + B sin at ~- t cos at, 

where the last two terms of F have been combined with the 
complementary function. Let the initial conditions be y = 
when t = 0, and dy/dt = when t = 0. Then A = and 
B = a /2, and (93-3) will be 

(93-4) y = -~ (sin at at cos a). 

This equation represents a vibration whose amplitude increases 
with time; for the amplitude of the first term is the constant 
a /2, and the amplitude of the second term is proportional to the 
time t. In fact, if sufficient time is allowed, the amplitude may 



312 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 94 

become greater than any preassigned number. This remark 
ought not to stimulate the student to design an apparatus to 
produce an infinite amplification or an infinite force. In any 
physical case, there is some resistance b present, and a brief 
reference to (92-5) will show that b prevents the oscillations from 
becoming arbitrarily large. 

PROBLEMS 

Show that a particular integral of 

1 d *y . 2 , 1 4 

! 37s + a y sin a t 1S y = ~~ o~ t cos at* 
at 4a 

2. -jif + a z y cos at is y = -~-t sin at. 
at *j(i 

94. Simultaneous Differential Equations. In many investi- 
gations, it is necessary to consider systems of differential equa- 
tions involving several dependent variables and one independent 
variable. For example, the motion of a particle in the plane can 
be described with the aid of the variables x and y, representing 
the coordinates of the particle, each of which may depend on 
time. It will be indicated in this section how a system of n 
ordinary differential equations involving n dependent variables 
may be reduced to a study of a single differential equation of 
higher order. 

Let two dependent variables x and y be functions of an inde- 
pendent variable t, and let it be required to determine x and y 
from the simultaneous equations 

dx 
(94-1) dt 

= / 2 (0, 

where a, b, c, and d are constants. If these equations are written 
in operational form, they are 

(D + a)x + by = /i(Z), 
ex + (D + d}y = / 2 (0- 

Operating on the second of these equations with - (D + a) gives 

c 

(D + a)x + - (D + a)(D + d)y = - (D 

C C 



94 ORDINARY DIFFERENTIAL EQUATIONS 313 

and, if the first equation is subtracted from this result, 

i (D + a) CD + d)y - by = \ (D + a)/ 8 (0 - fi(t). 
c c 

This is a second-order linear differential equation which can be 
solved for y. In order to determine x, solve the second equation 
of (94-1) for x, 



-t-*} 



and substitute the value of y in terms of t. 

The reader may show in the same way that the solution of a 
system of two second-order linear differential equations can be 
reduced to the solution of a linear differential equation of the 
fourth order (see Example 2 below). ///////// 

Example 1. Consider ^ 

J 4- 2x - 2y = t, 

dy _ 

dt ~" X y ~~ e 
or 

(D + 2)x - 2y = t, 
-3* + (D + l)y = 6'. 

Operate on the second of these equations with 
+ 2) to obtain 

-(Z) + 2)x + H(D + 2)(D + l)y = y 3 (D + 2)e', 
and add this result to the first equation. The result is 

y 3 (D + 2)(D + l)y - 2y = H(D + 2)e< + t, 
which simplifies to 




This equation can be solved for y as a function of t, and the result can 
be substituted in the second of the given equations to obtain x. 

Example 2. Let the two masses MI and M 2 be suspended from two 
springs, as indicated in Fig. 89, and assume that the coefficients of 
stiffness of the springs are k\ and k% respectively. Denote the dis- 
placements of the masses from their positions of equilibrium by x 
and y. Then it can be established that the following equations must 



314 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 94 
hold: 



These equations can be simplified to read 

d*y , kt k% ^ 

w+w^-ws- ' 

d z x kz . ki + kz 

W -Ml y + MT x = "- 

By setting 

ki kz Mz 

M\ = a ' Wz = b > Mi = m ' 

the equations reduce to 

(D 2 + 6 2 )?/ - b*x = 0, 
-6 2 /ni/ + (D 2 + a 2 + 6 2 m)x = 0. 

Operating on the second of these reduced equations with 7-7- (D 2 + 6 2 ) 
and adding the result to the first of the equations give 

(D 2 + 6 2 )(Z) 2 + a 2 + b*m)x - b*mx = 
or 

[D 4 + (a 2 + & 2 + 6 2 m)Z) 2 + aV]x = 0. 

This is a fourth-order differential equation which can be solved for 
# as a function of t. It is readily checked that 

x = A sin (<ut e) 

is a solution, provided that w is suitably chosen. There will be two 
positive values of co which will satisfy the conditions. The motion of the 
spring is a combination of two simple harmonic motions of different 
frequencies. 

PROBLEMS 

1. Solve Examples 1 and 2, Sec. 94. 

2. The equations of motion of a particle of mass m are 



where a?, y, z are the coordinates of the particle and X, Y, Z are the 
components of force in the directions of the #-, T/-, and 2-axes, respec- 
tively. If the particle moves in the zy-plane under a central attractive 




95 ORDINARY DIFFERENTIAL EQUATIONS 315 

force, proportional to the distance of the particle from the origin, find 
the differential equations of motion of the particle. 

3. Find the equation of the path of a particle whose coordinates x 
and y satisfy the differential equations 

d^x dy 

(Py __ dx 

where H, E, e, and m are constants. Assume that x = y = dx/dt 
= dy/dt when t = 0. This system of differential equations occurs 
in the determination of the ratio of the charge to the mass of an electron. 

4. The currents /i and 7 2 in the two 

coupled circuits shown in Fig. 90 satisfy the - - - - ->-* 
following differential equations: 

d 2 /! d*I 2 d/ 2 , /2 

M ~ 2 + L 4-r + - 

Reduce the solution of this system to that of a single fourth-order dif- 
ferential equation. Solve the resulting equation under the assumption 
that the resistances Ri and R% are negligible. 

95. Linear Equations with Variable Coefficients. With the 
exception of linear equations with constant coefficients and such 
equations with variable coefficients as are reducible to those 
with constant coefficients by a change of variable, there are no 
general methods for solving linear differential equations of order 
higher than the first. In general, solutions of differential 
equations with variable coefficients cannot be expressed in terms 
of a finite number of elementary functions, and it was seen in a 
number of specific examples that the solutions of such equations 
lead to new functions which are defined either by definite integrals 
or by infinite series. Some of these functions are of such frequent 
occurrence in applied mathematics that it has been expedient to 
calculate their values and tabulate them, precisely as the values 
of logarithms and trigonometric functions are tabulated. It 
must be borne in mind that the term elementary function as 
applied to logarithmic and circular functions is, in a sense, a 
misnomer and that such functions as Gamma functions, Bessel 
functions, and Legendre polynomials become just as "elemen- 
tary" after their values have been tabulated. The elementari- 



316 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 96 

ness of any given function is measured by the ease with which its 
value can be ascertained. 

The remainder of this chapter contains a brief treatment of 
those linear differential equations which are of common occur- 
rence in practice. An attempt will be made to express the 
solutions in convergent power series in x. This involves the 
tacit assumption that the solutions are capable of being expanded 
in Maclaurin's series, which, of course, is not true in general, and 
it is therefore not surprising that occasionally this method fails 
to give a solution. The method consists in assuming that a 
solution of the differential equation 



is expressible in a convergent infinite series in powers of #, of the 
type 

(95-2) y = a + a& + a 2 x 2 + + a n x n + - , 

where the coefficients a t are to be determined so that the series 
will satisfy the differential equation. If the coefficients of the 
derivatives in (95-1) arc polynomials in x, then the obvious mode 
of procedure is to substitute the infinite scries (95-2) into the 
equation (95-1), expand f(x) in Maclaurin's series, combine the 
like powers of a*, and equate to zero the coefficient of each power 
of x. This leads to an infinite set of algebraic equations in the 
a t , which can sometimes be determined by algebraic means. 

It is stated without proof that a homogeneous linear differ- 
ential equation of order n, 

(95-3) g + Pl (*) g^ + + ?_!<*) | + P n(x)y = 0, 

where the p t are continuous one-valued functions of x, possesses 
n linearly independent solutions, and only n. If these solutions 
are yi(x), yi(x), , y n (x), then the general solution of the 
equation is given by 

(95-4) y = ciyi + c 2 2/ 2 + + c n y n . 



This fact can be immediately verified by substituting (95-4) in 
(95-3). It is also clear that, if u(x) is any particular solution of 



96 



ORDINARY DIFFERENTIAL EQUATIONS 



317 



(95-1), then its general solution is y c\y\ + c 2 ?_/2 + * * * 
+ c n y n + u(x), where Cit/i + c 2 t/2 + + c n y n is the solution 
of the related homogeneous equation (95-3). 

Frequently, it is of practical importance to knqw whether a 
given set of functions is linearly independent. Inasmuch as 
the definition for linear independence that is given in Sec. 34 is 
difficult to apply, a test for the linear independence of the solu- 
tions will be stated.* 

THEOREM. The necessary and sufficient condition that a given 
set of solutions y\, y%, , y n of the nth order differential equation 
(95-3) be linearly independent is that the determinant 



2/1 



W 3 



2/2 
2/2' 



- - - y n r 



This determinant is called the Wronskian. 

Example. By substitution, it can be verified that y\ = sin x, 
7/2 = cos x, and 7/3 = e lx are solutions of the differential equation 



But the Wronskian is 



c j dx* ^ dx u 

sin x cos e ix 

cos x sin ie 13 

sin x cos e tx 



= o, 



and therefore this set of solutions is not linearly independent. In other 
words, at least one of them can be expressed as a linear combination of 
the other two. It is known that 

e l * cos x + i sin x. 

It is readily verified that a linearly independent set of solutions is 
T/i = sin x, 7/2 = cos x, and 7/3 = e x , so that the general solution is 

y = ci sin x + c 2 cos x + c 3 e x . 



PROBLEMS 

Determine whether or not the following sets of functions are linearly 
independent: 

* See INCB, E. L., Ordinary Differential Equations. 



318 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 96 

1. 2/1 = sin x + x, 2/2 == e x t 2/3 ~ 3e* 2x 2 sin x. 

2. 2/1 = x 2 - 2x + 5, 2/2 = 3z - 7, 2/3 = sin x. 

3. 2/1 = 6** + x, 2/2 = cos a: + x, y z = sin x. 

4. 2/1 = (z + I) 2 , 2/2 = (x - I) 2 , 1/3 = 3*. 

5. 2/1 = log x, 2/2 = sinh x, 2/3 = e*, 2/4 = e~ 



~* 



96. Variation of Parameters. Two methods of determining a 
particular integral of a linear differential equation with constant 
coefficients were discussed and illustrated in Sec. 89. Another 
important method that is applicable to linear equations with 
either constant or variable coefficients will be described here. 
This method, due to the great French mathematician Lagrange 
(1736-1813), permits one to determine a particular integral of 

(96-1) g + Pl (x) g3 + + Pn-^x) I + p.(x)y = /(*), 
when the general solution of the related homogeneous equation 

' ' ' 



(96 " 2) n - Tx 

is known. 

Let the general solution of (96-2) be 

(96-3) y = Ciyi + c 2 yt + + c n y n , 



in which the c l are arbitrary constants, and assume that a set of 
n functions #1(0;), vi(x), , v n (x) can be so chosen that 

(96-4) y = viyi + v^y^ + + v n y n 



will satisfy (96-1). Since y\(x), y<t(x), , y n (x) are known 
functions of x, (96-1) imposes only one condition upon the v l 
in (96-4). Inasmuch as there are n functions v t , it is clear that 
n 1 further independent conditions can be imposed upon the 
v t , provided that these conditions are consistent. 
Differentiating (96-4) gives 



' ' + v n y' n ) + (v{yi + v f M* + + v' n y n ). 
As one condition to be imposed on the v^ } let 

v(y\ + v' 2 y2 + + v' n y n = 0, 
so that 



96 ORDINARY DIFFERENTIAL EQUATIONS 

Then, 



319 



and if the second condition to be satisfied by the v % is 

+ <y' = o, 



it follows that 



V" = viy[' + v&y 



+ v n y" n . 



By continuing this process a set of n 1 conditions is imposed 
on the Vt, namely, 

- + v' n y n = 0, 

h + <//: = o, . 



(96-5) 

as a consequence of which 



y f = 



+ v n y n , 



i(n-l) 



Calculating y (n} yields 



Substituting t/, y f , , y (n) in (96-1) and remembering that, 
by hypothesis, y\, t/ 2 , , y n satisfy (96-2) give the nth con- 
dition to be imposed upon the v t , namely, 

The n 1 relations (96-5) together with (96-6) give n linear 
algebraic equations which can be solved for vj, v 2 > ' * ' , v' n , pro- 
vided that the determinant of the coefficients of the v(, namely, 



(i 2/2 ' ' ' y n 

9l 2/2 ' ' ' 2/n 



is not identically zero. But this determinant is the Wronskian 
and, since yi, 2/ 2 , , y n were assumed to be linearly inde- 



320 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 96 

pendent, it is different from zero. Hence, the system of equations 
can always be solved for the v' t , and the expressions for the u t are 
obtained by integration. 

Example. As an illustration* of the application of this method of 
determining a particular integral, consider the equation 

2 = 2(sinx-2coB*). 
The general solution of the homogeneous equation is found to be 



Assume that a particular integral of the non-homogeneous equation is 
of the form 

y v\e x 



where Vi, v^, and v z are functions of x to be determined presently. 
Computing y r gives 

y 1 = vtf x + v 2 (x + \}e* 2v 3 e~' 2x + Vi'e* + v 2 xe x + v 3 f e~ 2x . 
The first condition to be imposed upon the v, is 

Vi'e' + v 2 'xe* + v 3 'e~ 2 * = 0, (1) 

so that 

y " * Vl * + V2 (x + 2)e* + 4v s e~ Zx + vi'e* + v 2 '(x + \)e* - 2v^e~ 2x 
Imposing the second condition produces 

vi'e 9 + v 2 '(x + l)e x - 2v 3 'e~** = 0, (2) 

and computing y 1 " yields 

y '" = vie x + vt(x + 3)6* - Svse-** -f vi'e* + v> 2 '(x -f 2)e x -f 4v z 'e~* x . 
Hence, the third condition to be satisfied by the v t is 

vi'e* + vj(x + 2)e* + 4z> 3 V 2 * = 2(sin x - 2 cos a). (3) 

Solving (1), (2), and (3) for Vi, v 2 f , and v s ' gives 

Vi' = ?^6"*(sin a: 2 cos 3) %e~ x (sm ^ 2 cos x), 
iV - ^^"'(sin x 2 cos x), 
Vs = %e 2a! (sin a; 2 cos x). 

The integration! of these expressions yields 

* For another illustration, see Example, Sec. 97. 

t The integration in this case is quite tedious, and generally speaking it 
is easier to solve linear equations with constant coefficients by the methods 



96 ORDINARY DIFFERENTIAL EQUATIONS 321 

Vi = Hxe~ x (3 sin x cos x) + e~* sin x + %e~* cos x, 
t>2 = }ie~ x ( 3 sin a; + cos a;), 
t>3 = %e 2x cos x, 

in which the constants of integration are omitted because a particular 
integral is desired. 

By hypothesis, a particular integral is given by 

y = Vie* + v 2 xe x + v^e~ Zx sin x, 

so that the general solution of the non-homogeneous equation is 
y = (c\ + c>2x)e* + c z e~ 2x + sin x. 

PROBLEMS 

1. Solve Probs. 1, 2, and 3, Sec. 89, by the method of variation of 
parameters. 

2. Find the solution of 



by the method of variation of parameters, and compare your result with 
that of Sec. 85. The solution of the related homogeneous equation is 
obtained easily by separation of the variables. 

3. By the method of variation of parameters, find a particular integral 
of 

d*y 3dy 5 



where the general solution of the related homogeneous equation is 



i , K 
y = ~ + czx 6 . 

4. Find the general solution of 

&y , x dy __ \_ . 

dx* "*" 1 - x dx 1 - x y ~ - 1 *' 

where the general solution of the related homogeneous equation is 
c\e x + cix. 

5. Find the general solution of 

_ x*y" - 2xy' + 2y = x log x, _ 

discussed in Sec. 89. However, the method of the present section is of 
great value when the given equation has variable coefficients, 



322 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 97 

if the general solution of the related homogeneous equation is 
y = cix 2 + C&. 

97. The Euler Equation.* Before proceeding to illustrate the 
method of solution in terms of infinite series, it will be well to 
discuss one type of differential equation with variable coefficients 
that can be reduced by a change of variable to a differential 
equation with constant coefficients. 

Consider the linear equation 

(97-1) * g + a^-> g3 + + _,* | + a n y = /(*), 

where the a t are constants. This equation can be transformed 
into one with constant coefficients by setting x = e z . For if 
x = e z , then 

dx , dz 

j- e z and -j- = e~*. 

dz dx 

Moreover, if D s= --, then 

9 dz 



dx dz dx 
and 



Similarly, 

g = e -3* (>3 _ 3 

Then, since x = e z , it follows that 

x ^ - TDjy 
X dx~ Dy ' 

x* = (D* - D)y = D(D 



x" n = D(D - 1)(D - 2) (D - n 
* Also called Cauchy's equation, 



97 ORDINARY DIFFERENTIAL EQUATIONS 323 

so that (97-1) is replaced by an equation with constant coefficients, 

[D(D - 1) (D - n + 1) + (nD(D - 1) (D - n + 2) 

+ - + a n _!Z) + a n ]y = /(*). 
Example. Consider 

d*y dy 
x W + x dx- y = xlo * x ' 

Upon making the substitution x = e e , this equation becomes 



or 

(D 3 - 3D 2 + 3D - l)y = ze*. 

The roots of the auxiliary equation are MI = w 2 = m^ = 1, so that the 
complementary function is 



The particular integral is 



so that, in terms of z, the general solution is 

and, in terms of x, 

y [ci + c 2 log x + c 3 (log x) 2 ]x - 



A particular integral for this example will be obtained by the method 
of variation of parameters in order to demonstrate the applicability 
of this method to equations with variable coefficients. Care must be 
taken first to transform the equation so that it has unity for its leading 
coefficient, for the discussion of Sec. 96 was carried through for this 
type of equation. 

Expressing the given equation in the form (96-1) gives 

d*y , 1 dy 1 1 . 



Since the general solution of the homogeneous equation was found 
to be 

CiX + C& log X + C&(\0g X)*, 



324 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 97 
the equations of condition (96-5) and (96-6) are 

Vi'x + v 2 'x log x + Va'x(log x) 2 = 0, 
vi' + fi'U + log x) + v z '[(log xY + 2 log x] = 0, 



v *'\ + v * (x log x + x) = 



Solving for vi, t> 2 ', and i; 3 ' yields 

Vl ' ** 2x (log *)*' v * = ~~x (log ^ 2> v * = 2x log ^ J 
which integrate into 

v l = H(log xY, v 2 = -K(log x) 3 , 
Hence, a particular integral is 

1 /I NO 

y = v& + v& log a; + y 3 x(log a;) 2 = 



PROBLEMS 

1. Find the general solution of Prob. 3, Sec. 96, by the method of 
Sec. 97. 

2. Find the general solution of 



Compute the particular integral by the method of variation of 
parameters. 
3. Solve 



by assuming a solution of the form y x r and determining appropriate 
values of r. 
4. Solve 



by assuming a solution to be of the form y = x r . 

5. Find the general solution of 



x*y'" - 4zV + 5xy' - 2y - 1. 

6. Find the general solution of 

y oc X *. 



98 ORDINARY DIFFERENTIAL EQUATIONS 325 

7. Find the general solution of 

x*y" - 2xy' + 2y = x log x. 

98. Solution in Series. Many differential equations occurring 
in applied mathematics cannot be solved with the aid of the 
methods described in the preceding sections, and it is natural to 
attempt to seek a solution in the form of an infinite power series. 
The method of solution of differential equations with the aid of 
infinite series is of great importance in both pure and applied 
mathematics, and there is a vast literature on the subject. This 
section and the four following sections contain only a brief 
introduction to the formal procedure used in obtaining such 
solutions. 

As an illustration of the method, consider the differential 
equation 

(98-1) y' - xy - x = 1, 

and assume that it is possible to obtain the solution of (98-1) in 
the form of a convergent power series 

(98-2) y = a + a+x + a 2 z 2 + - - - + a n x n + . 

Inasmuch as the series of derivatives of a convergent power series 
is convergent, one can write 

(98-3) y' = ai + 2a 2 z + - - + na n x n ~ l + . 

Substituting (98-2) and (98-3) in (98-1) and collecting the coeffi- 
cients of like powers of x give 

(98-4) a! + (2a 2 - a - l)x + (3o 8 - ajx* + 

+ (na n - a n -z)x n - 1 +=!. 

By hypothesis, (98-2) is a solution of (98-1), and therefore* 
equating the coefficients of like powers of x in (98-4) leads to the 
following system of equations : 

a i I (coefficient of x), 
2a 2 a 1 = (coefficient of x), 
3a 3 a\ = (coefficient of a: 2 ), 



(98-5) 



na n a n - 2 = (coefficient of a;"* 1 ), 
* See Theorem 5, Sec. 10. 



326 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 98 

The system of equations (98-5) is a system of linear equations 
in infinitely many unknowns a , ai, , a n , . Solving 
the second equation of (98-5) for a 2 in terms of a gives 



The third equation taken with the first demands that 

aa = = . 

Setting n = 4 in the coefficient of x n ~ l gives 

i -| [ -| 

&2 do ~t~ J- &0 i J- 

4 2*4 2^ 2^' 
whereas n = 5 gives 

_3_ 1 



^ 5 3-5 
In general, 



(98-6) 



2n 2*-n! 

1 



(2n + 1) 



The substitution in (98-2) of the values of a k given by (98-6) 
leads to a solution in the form 

. , OQ + 1 o , 1 n i tt O + 1 A , 1 c , 

= -- 1 4 t ' ' ' 



When the terms containing a are collected, there results 

+ 1 o , tto + 1 A , 



(98-7) y = ao + 
+ 2 n -nT X n + 



I* 



TU+-]- 



1-3-5 
If ao + 1 is set equal to c, one can write (98-7) in the more 



98 ORDINARY DIFFERENTIAL EQUATIONS 327 

compact form 



(98-8) j,- e [i+*! + 5 n + ... +-=_, .+ 



g2n 

2^2! ^ ^ 2^T! 



f x 3 x 2n + l ' 1 

+ [~~ l+X + T^ + ' ' ' + 1 - 3 (2n + 1) + ' ' '} 

The two series appearing in (98-8) are easily shown to be 
convergent for all values of x, and hence they define functions 
of x. In fact, the first of the series is recognized as the Maclaurin 



expansion of e 2 , so that (98-8) can be written as 

^ T rr3 

(98-9) y = ce 2 + 1-1 + x + ~-^ + 

T 2n-fl 

+ 1-3 . . . (2n +l) + 



} 



This is the general solution of (98-1), for it contains one arbitrary 
constant. 

Since (98-1) is a linear differential equation of the first order, 
its solution could have been obtained by using the formula 
(85-3), and it is readily verified that (85-3) gives 



X 2 



(98-10) y = ce 2 - 1 + e 2 J e 2 dx. 

The integral in (98-10) cannot be evaluated in closed form; but 
if the integrand is expanded in a power series in x, it is easy to 
show that (98-10) leads to (98-9). 

Consider next the homogeneous linear differential equation of 
the second order 

(98-11) y" - xy' + y = 0, 

and assume that the solution of (98-11) has the form 



(98-12) y = a n x n = a + a& + - + a n x n + - 

n=X) 

Then the series for y' and y" are 

00 00 

i'' = V nanX 71 " 1 and y" = V n(n l)a n z n ~ 2 . 



n>-2 



328 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 98 
If these expressions are substituted in (98-11), the result is 



n(n - l)a n a: n *" 2 - na n x n + a n x n = 0. 

n2 n-\ n = 

Combining the terms in like powers of x and setting the coeffi- 
cient of each power of x equal to zero give the system of equations 

2 12 + flo = (coefficient of x), 

3 2a 3 ai + ai = (coefficient of x), 

4 3a 4 2a 2 + a 2 = (coefficient of z 2 ), 

............................................... > 

(n + 2)(n + l)a n +2 na n + a n = (coefficient of x n ), 



Hence, 

(98-13) a n+2 = 



This recursion formula can be used to determine the coefficients 
in (98-12) in terms of a and a\. Thus, substituting n = 0, 
1, 2, in (98-13) gives 



1 
=- - o , 



1 1 

T^[ a2 = ~ 4] 

2 



4-75 

3 3 

6! 



a 2n +i = 0, 

1 3 5 (2n - 3) 

a2n = --- __ - 

Therefore, 
(98-14) y = 

where a and a\ are arbitrary. It is readily checked that the 



99 ORDINARY DIFFERENTIAL EQUATIONS 329 

series is convergent, and, since it is a power series, it defines a 
continuous function of x. 

The two linearly independent solutions of (98-11) are then 



y = 
and 



1 2 1 4 3 , 15 8 
- 



> 



These solutions are obviously linearly independent, for one of 
them defines an odd function of x and the other defines an even 
function of x. 

PROBLEMS 
Integrate in series: 



3. (x* -3 + 2) + (x-2i-l)+(*- 3)y = 0. 

4- ft - y - 1. 
cfc 2 ^ 

99. Existence of Power Series Solutions. It must be kept 
clearly in mind that the calculations performed in Sec. 98 are 
formal and depend on the assumption that the differential equa- 
tions discussed there possess power series solutions. If, for 
example, an attempt had been made to apply the method of 
solution outlined in Sec. 98 to the equation 

xy' - 1 = 0, 

it would have been futile. The general solution of this equation 
is 

y = log x + c, 

which cannot be expanded in a power series in x. 

The task of determining beforehand whether a given differen- 
tial equation possesses solutions in the form of power series 
represents one of the major problems of analysis. It will suffice 
in this introductory treatment to state, without proof, the 
conditions under which a homogeneous linear differential equa- 
tion of the second order has a power series solution. 



330 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 99 

THEOREM. Let 
(99-1) V" + fi(x)y' + f*(x)y = 

have coefficients f\(x) and f%(x) that can be expanded in power series 
in x which converge for all values of x in the interval R < x < R ; 
then there will exist two linearly independent solutions of the form 

00 

y = 2} anXn > 

n0 

which will converge for all values of x in the interval R < x < R. 

It is clear from the statement of the theorem that the differen- 
tial equation will possess power series solutions, which converge 
for all values of x, whenever f\(x) and/2(x) are polynomials in x. 

It should be noted that the coefficient of the second derivative 
term in (99-1) is unity. Frequently, the differential equation 
has the form 



and if this differential equation is put in the form (99-1), then 
/. / x Pi(x) j 

fi(x) - and 



If po(x) should vanish for some value of x in the interval within 
which the solution is desired, one must expect trouble. If 
po(x), pi(x), and pz(x) are polynomials in x and if 7>o(0) ^ 0, then 
one can surely expand fi(x) and fa(x) in power series in some 
interval, and the theorem enunciated above is applicable. 
As an illustration, consider the differential equation 

(99-2) (2 - x)y" + (x - l)y' - y = 0. " 

Inasmuch as 



and 

/ 2 (x) = (x - 2)- 1 

obviously possess power series expansions that are valid in the 
interval 2 < x < 2, it is reasonable to proceed to obtain the 
power series solution. 



99 ORDINARY DIFFERENTIAL EQUATIONS 331 

Substituting 

y = 

in (99-2) gives 



(2 - x) n(n - l)a n x-* + (x - 1) wa^"- 1 - J a n z" = 0. 

n=0 n=0 n-0 



Rearranging and combining terms give 



n-O 



(n - 



" = 0. 



Equating the coefficients of the powers of x to zero gives 



(99-3) 



2 2 Ia 2 oi a = 0, 

2 3 2a 3 - 2 2 a 2 = 0, 

2 4 3a 4 - 3 2 a 3 + a 2 = 0, 

2(w + 2)(n + l)a n+2 - (w + l) 2 a n+1 + (n - l)a n = 0, 



The coefficient of x n provides the recursion formula 



2(n + l)(n + 2) 
and setting n = 0, 1, 2, 3, gives 

#o ~f" di c 



2 2 ~ 2 - 2! 
c 



, where c = a + 



a 4 = 



3 2-3! 
3 2 a 3 a 2 c 



4! 2-4! 

4 2 a 4 - 2a 3 c 



" 2-4-5 2-5! 
It is easily shown that in general 


a " "" 2^T! ? 



332 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 100 
so that 

y = a + a\x + a 2 x 2 + + a n x w + 
= ao + ^i^ + 



/a: 2 x* 

\2i + n 



2 

+ *!/, , , * 2 



^0 + &\X + 

^& \ ^1 

flo H~ ^i 



(1 



2 2 

= Cie* + c 2 (l - x), 

where Ci = (a + ai)/2 and c 2 = (a ai)/2. 

It happens in this illustration that the series appearing in the 
solution of the differential equation represent elementary func- 
tions, so that the general solution can be written in* closed form 
Ordinarily, the infinite series arising from the solution of linear 
differential equations with variable coefficients represent func- 
tions that carniQt be expressed in terms of a finite number of 
elementary functions. This is the case discussed in the next 
section, where the series that provides the solution of the differen- 
tial equation represents a class of functions of primary importance 
in a great many problems in applied mathematics. 

It is quite obvious that the theorem of this section can be 
rephrased to include the case where the f unctions fi(x) and/ 2 '(^) 
possess series expansions in powers of x a. In this case, 
there will exist two linearly independent solutions of the form 



y = 

PROBLEM 

Solve 

y" - (x - 2)7/ = 

00 

by assuming the solution in the form y S a n (x 2) n . Also, 

n = 

00 

obtain the solution in the form y = 2 a n x n . 

n=0 

^ 100. Bessel's Equation. An important differential equation 
was encountered by a distinguished German astronomer and 



100 ORDINARY DIFFERENTIAL EQUATIONS 333 

mathematician, F. W. Bessel (1784-1846), in a study of planetary 
motion. The range of applications of the functions arising from 
the solution of this equation is partly indicated by the fact that 
these functions are indispensable in the study of vibration of 
chains, propagation of electric currents in cylindrical conductors, 
problems dealing with the flow of heat in cylinders, vibration of 
circular membranes, and in many other problems arising in every 
branch of applied mathematics. Some of the uses of this equation 
are indicated in the chapter on Partial Differential Equations. 
BesseFs equation has the form 

(100-1) x*y" + xy f + (x 2 - n 2 )y = 0, 

where n is a constant. It will be observed that this equation 
does not satisfy the conditions of the theorem of Sec. 99 because 
of the appearance of x 2 in the coefficient of y". In the notation 
of Sec. 99, 

1 n 2 

fi(x) = - and / 2 (x) = 1 - , 

JU JU 

and it is clear that /i(x) and f*(x) cannot be expanded in power 
series in x. 

In order to solve (100-1), assume that the solution can be 
obtained in the form of a generalized power series, namely, 



(100-2) y = x m 2 a rX r , 

where m is a constant to be determined later and where ao can be 
assumed to be distinct from zero because of the indeterminate 
nature of m. 

Calculating the first and second derivatives with the aid 
of (100-2) and forming the terms entering into (100-1) give 

d 2 y 

x 2 ~ = m(m I)a m + m(m + l)dix m+1 + 

CLX 

+ a k (m + k)(m + k l)x m+k + , 
dt/ 



~- Q + (m + l)ai$ m + l + ' + a k (m 



x 2 y = a Q x 
n' 2 y = ~- 



334 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 100 

Adding the left-hand members of these expressions and equating 
the result to zero give (100-1). By hypothesis, (100-2) is a 
solution, and therefore the coefficient of each power of x in the 
sum must vanish. Hence, 

(w 2 n 2 )a = (coefficient of x m ), 

[(m + I) 2 - n 2 ]ai = (coefficient of x m+l ), 

, 

[(m + fc) 2 - n 2 ]a, c + a, t _ 2 = (coefficient of x m + k ). 

The coefficient of the general term gives the recursion formula 

(100-3) a, = - r-^l-l 5- 

^ / (m + k) 2 n 2 

The equation resulting from equating to zero the coefficient of the 
term of lowest degree in x (here, x m ) is known as the indicial 
equation. In order to satisfy the indicial equation, choose 
m = n. It m is chosen as -\-n or n, a Q is arbitrary and the 
second condition requires that i = 0. 

For m = n the recursion formula becomes 



* (n + fc) 2 - n 2 fc(2n + fc) 

Setting fc = 2, 3, 4, * in turn gives 



a 2 = - 

^^rt -j- ^; 

Ch 

since ai = 



2(2n + 2)' 

PI 

3 3(2n + 3) = 

a> 
a 4 = - 



4(2n + 4) 2 4(2n + 2)(2n + 4) 

^3 = 

6 5(2n + 5) ' 

In this manner as many coefficients as desired can be computed; 
and if their values in terms of a are substituted in (100-2), there 
is obtained the following series, which converges for all values of x } 

(100-4) y = atf 



2) ' 2-4(2n + 2)(2n + 4) 
x 6 





2 4 6(2n + 2)(2n + 4)(2n + 6) 



100 ORDINARY DIFFERENTIAL EQUATIONS 335 

If m = n is chosen, ao is again arbitrary and a\ = 0, and 
the resulting series is 



*- [ 



(100-5) = *- l + + 



274(2X2^4) 

X 6 



2 4 6(2n - 2)(2n - 4)(2n - 6) 

The series (100-4) and (100-5) become identical for n = 0. For* 
positive integral values of n, (100-5) is meaningless, since some of 
the denominators of the coefficients become zero, and (100-4) is 
the only solution obtainable by this method. If n is a negative 
integer, (100-4) is meaningless and (100-5) is the only solution in 
power series in x. For n 7^ or an integer, both (100-4) and 
(100-5) are convergent and represent two distinct solutions. 
Then (100-4) multiplied by an arbitrary constant and added to 
(100-5) multiplied by an arbitrary constant gives the general solu- 
tion of the Bessel equation. 

The reason for the failure of this method to produce two dis- 
tinct solutions when n is zero or an integer is not hard to find., 
The success of this method depends upon the assumption that the 
solutions are representable in power series. The analysis leading 
to the determination of the second particular solution of (100-1) 
when n is an integer is not given here. * It is sufficient to mention 
the fact that the second solution can be obtained by assuming 
that it has the form, when n is a positive integer, 

oo 

(100-6) i/2 = Cyi(x) log x + V a k x- n + k , 

*=o 

where y\(x) is the solution (100-4) and C is a constant. Obvi- 
ously, this solution becomes infinite when x = 0. 

It will be of interest to consider the particular solution obtained 
from (100-4) by setting 

a = 

*See WATSON, G. N., Theory of Bessel Functions; GRAY, A., G. B. 
MATHEWS, and J. M. MACROBEBT, A Treatise on Bessel Functions and 
Their Applications to Physics; WHITTAKER, E. J., and G. N. WATSON, 
Modern Analysis; MCL.ACHLAN, N. W., Bessel Functions for Engineers. 



336 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 100 
The series (100-4) then becomes 

/yn /fW+2 /yn-f-4 

(100-7) J n (x) m ^ 2n+2( * n + 1}! + ^ (n + 2)! 

_ . . . + (_l)fc ^ n+2fc -4- -.. 

~ v A / 2 n + 2 *A!l^ti. 4- fc")! 

;(-!)- 



-2 



The function defined by this series is called the Bessel function 
of the first kind of order n. This series holds for non-negative 
values of n. For n = 0,* (100-7) gives 

x 2 * zk 

J Q (x) = 1 - ^2 

and for n = 1, 



2 3 2! 2 5 2!3! 



i)! 



which are called Bessel functions of the zero-th and first orders, 
respectively. 

The formula (100-7) can be generalized for non-integral 
positive values of n with the aid of the Gamma function by 
writing (n + k)\ = T(n + k + 1), so that 



(100-8) ^(') 

For n = Y<i, (100-8) becomes 



/2fc + 8\' 



For n 0, n! is defined to be unity. This is consistent with the formula 



when n = 1, as well as with the general definition of n! in Sec. 81. 



100 ORDINARY DIFFERENTIAL EQUATIONS 

It is not difficult to show that this reduces to* 

sin x. 



337 



TTX 

This formula suggests that the behavior of Bessel functions may 
be somewhat similar to that of the trigonometric functions. This 
proves to be the case, and it will be shown in the next section that 
Bessel functions can be used to represent suitably restricted 
arbitrary functions in a way similar to that in which trigonometric 
functions are used in Fourier series. 

* It is clear that JQ(X) is an even function and that Ji(x) is an 
odd function of x. Their graphs are shown in Fig. 91. For large 




FIG. 91. 



values of x the roots of JQ(X) = and J\(x) = are spaced 
approximately TT units apart. It can be shown that for large 
values of x the Values of J n (x) are given approximately by the 
formula 



It is not difficult to show with the aid of (100-6), by setting 
yi(x) = Jo(x), that the second solution of BessePs equation of 
order zero has the form 



= J,(x) log x + - 



l + + 



22.42 



(2fc) 2 



This function is called the Bessel function of the zero-th order 
* See Prob. 2 at the end of this section. 



338 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 100 
and the second kind. Thus, the general solution of the equation 



is 

y = CiJoGO + C z K Q (x), 
where C\ and Cz are arbitrary constants. 

Bessel functions of the first kind and of negative integral order 
are defined* by the relation 

/_(*) = (-1)V(*). 

The values of Bessel functions are tabulated in many books, f 
Electrical engineers frequently use the real and imaginary parts 
of J n (\/ix) J which are denoted by the symbols ber n x and 
bei n x, respectively, that is, 

~~i x) = ber n x + i bei n x. 

There are also modified Bessel functions of the first kind, which 
are denoted in the literature by I n (x). The commonly used 
notations for modified Bessel functions of the second kind are 

7 n (x), Nn(x), H n (x). 

PROBLEMS 

1. Show that 

^;/o(z) = -Ji(x). 

2. Show that 



Note that T(n + 1) = nT(n) and 

3. Show that y = J Q (kx) is a solution of the differential equation 

xy" + y' + k*xy = 0. 

4. Show that 

x*y" + xy' + (kW - n*)y = 0, k ^ 0, 

can be reduced to the form 



* See BYERLY, W. E., Fourier Series and Spherical Harmonics, Sec. 120, 
p. 219. 

t JAHNKE-EMDE, Funktionentaf cln ; BYERLY, W. E., Fourier Series and 
Spherical Harmonics; WATSON, G. N., Theory of Bessol Functions. 



101 ORDINARY DIFFERENTIAL EQUATIONS 339 

Hint: Set z = kx, and hence y' = -r- -r- = A; -7- 



^- 

/2~ 
6. Show that /_!,$ (2) = ^J cos x, so that the general solution of 

\/ 7T3J 

Bessel's equation with n % is 



6. Show that y = \/a; J n (\x) is a solution of the equation 

4&V + (4XV - 4n 2 + I)?/ = 0. 

7. Show that T/ = x n J n (x) is a solution of the equation 

xy" + (I - 2n)y' + xy = 0. 

8. Show that y = x~ n J n (x) is a solution of the equation 

xy" + (1 + 2n)i/ 7 + xy = 0. 

101. Expansion in Series of Bessel Functions. It was pointed out 
in connection with the development of arbitrary functions in trigo- 
nometric series (Sec. 24) that the Fourier series development is only a 
special case of the expansion of a suitably restricted class of functions 
in series of orthogonal functions. It will be shown in this section that 
it is possible to build up sets of orthogonal functions with the aid of 
Bessel functions, so that one can represent an arbitrary function in a 
series of Bessel functions. 

It is shown in the treatises* on Bessel functions that the equation 
Jn(x) = has infinitely many positive roots Xi, \2, , X, , 
whose values can be calculated to any desired degree of accuracy. It 
will be established next that the functions 



\/xJ n (\\x), -v/J/nCXaz), , \/xJ n (\ k x) t - 
are orthogonal in the interval from x to x = 1, so that 

(101-1) j^ 1 Vx J(X t z) V* J*(\,x) dx = 0, if i 7* j, 

= M<A/(X t )] 2 , if i = j. 

The proof of this fact depends on the following identity, the validity 
of which, for the moment, will be taken for granted: 



(101-2) (X 2 - M 2 ) xJ n (\x)J n (x) dx 

= x(iJ. 
* See the references given in the footnote, p. 335. 



340 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 101 
Let X = X, and p, = X t , where X 7^ X,; then 

J \/x J n (\x) \/z J n (\jx) dx 



Setting # = 1 and remembering that ./ n (X t ) = J n (X ; ) = give the 
first part of the formula (101-1). 

In order to establish the second part, differentiate (101-2) partially 
with respect to X, and thus obtain 



(101-3) 2X xJ n (\x)J n (x) dx + (X 2 - M 2 ) x\J n (nx)J n '(\x] dx 



Set x = 1, X = /x, and recall that if X is a root of J n (x) = then 
/ n (X) = 0. Upon simplification of (101-3), one obtains the second pait 
of the formula (101-1). 

In order to establish the identity (101-2), note that y = \/xJ n (\x) 
is a solution of the equation* 

4x*y" + (4\V - 4n 2 + l)y = 0. 

Setting u \/x / n (Xx) and v = -\/x J n (^x) gives the identities 
+ (4X 2 ^ 2 - 4n 2 + l)u = 



and 

4x*v" + (4/z 2 ^ 2 - 4n 2 + 1> = 0. 

Multiplying the first of these by v and the second by u and subtracting 
furnish the identity 

-(X 2 - fj. 2 )uv = u"v - v"u. 

The integration of both sides of this identity, between the limits and 
x, yields 



- (X 2 - /x 2 ) f* uv dx = * (u"v - v"u) dx 

- [ u \ - fo u ' vt dx \ - [Ho - K uV 

= [u'v - v'u]* Q . 

Recalling the definitions of u and v gives the desired identity (101-2). 
Since the formula (101-1) is established, it is easy to see that if f(x) 
* See Prob. 6, Sec. 100. 



101 ORDINARY DIFFERENTIAL EQUATIONS 341 

has an expansion of the form 



/(*) = 

which can be integrated term by term, then* 

2 f 1 xf(x)J n (\tX) dx 



The most common use of this formula is in connection with the expan- 
sion of functions in a series of Bessel functions of order zero. 

PROBLEMS 

1. Show that -j- [xJ\(x)\ xJ Q (x). 

2. Show that f(x) = 1, < x < 1, when expanded in a series of 
Bessel functions of order zero, gives 



Hint: Make use of the results of Prob. 1 above and Prob. 1, Sec. 100. 

3. Show that 



dx [X n < n(X)\ ~ X n , n-lW 

and 

d 

dx ' n * 

4. Show that 



[2 /sin x \ 

= V^ v"^ ~ cos v 



and 



6. Show, with the aid of the formulas of Prob. 3, that 



and 



* See Sec. 24. 



342 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 102 

*(h--} 

6. Expand e 2 v */ in a power series in h to obtain 

x (h l \ n ^? 
e^ h I) = 2) A>s 

n so 00 

and show that A n = /(#), so that 

/ A _l\ 

4. Wl (j.) + to/ l(a .) + . . . + frs/^) + . . . ] 



+ [h-V-i(x) + /r 2 J_ 2 (z) + + h-J- n (x) +]. 
102. Legendre's Equation.* The equation 

(102-1) (1 - * 2 ) g - 2* g + n(n + l)y = 0, 

where n is a constant, occurs frequently in practical investigations 
when spherical coordinates are used. One of the many uses of 
this equation in practical problems is indicated in the next 
chapter in connection with a study of the distribution of temper- 
ature in a conducting sphere. 
Assume, as in Sec. 100, that 

(102-2) y = a x m + aix m+l + + a k x m + l + 
is a solution of (102-1). Then, 

d 2 v 

-~5 = m(m I)a x m ~ 2 + (m 



+ (m + 2)(m 

+ (m + k)(m + k - 

d 2 y 

-r~ = m(m I)a x m 

> - (m + k - 2)(m + k - 3)a A; _ 2 x w+fc ~ 2 - 



~- - Q - 2(m + A? - 
ax 



n(n + l)i/ = n(n + l)a Q x m + + n(n 



Adding these expressions and equating to zero the coefficients 
of a;" 2 , x m ~ l , - - , x m+A; - 2 give the system of equations 

* A. M. Legendre (1752-1833), an outstanding French analyst who made 
many brilliant contributions to the theory of elliptic functions. , / 



102 ORDINARY DIFFERENTIAL EQUATIONS 343 

m(m I)a = 0, 
m(m + l)ai = 0, 



(m + fc)(m + fc 

+ (n m k + 2)(n + m + k l)ajb_ 2 = 0. 

In order to satisfy the first of these equations, m can be chosen as 
either or 1. If m = 1, the second equation requires that 
ai 55 0. For m = the coefficient of z m +*~ 2 gives the recursion 
formula 

(n - k + 2)(n + fc - 1) 
<** = -- k(k=-l) - a *- 2 > 

from which, in a manner analogous to that employed in Sec. 100, 
the coefficients a 2 , a 3 , a 4 , can be determined. If w = 0, 
the second of the equations of the system allows ai to be arbitrary. 
If the values of the coefficients in terms of a and ai are sub- 
stituted in (102-2), the following solution is obtained: 



(102-3) - a. l - * + ~ 1)(n + 3) 



3! 

J- < n " l ^ U - 3 ^ U + 2 )^ + 4 ) .5 _ 

~ l " 5! 

It is readily shown by means of the ratio test that for non- 
integral values of n the interval of convergence of the series in 
(102-3) is ( 1, 1). Moreover, since the first series in (102-3) 
represents an even function and the second series represents an 
odd function, the two solutions are linearly independent. The 
sum of the two series, each multiplied by an arbitrary constant, 
gives the general solution of (102-1), which is certainly valid if 
|x( < 1. It can be verified directly that the choice of m = 1 does 
not lead to a new solution but merely reproduces the solution 
(102-3) with a = 0. 

An important and interesting case arises when n is a positive 
integer. It is clear that, when n is an even integer, the first series 
in (102-3) terminates and reduces to a polynomial, whereas, when 
n is an odd integer, the second series becomes a polynomial. If 
the arbitrary constants a and a\ are so adjusted as to give these 
polynomials the value unity when x = 1, then the following set 



344 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 102 

of polynomials is obtained: 

Po(x) = 1, 
Pi(x) = x, 

/>,(*)- f *-, 

P 8 (s) s - x s - ^ x, 

r> / \ 7 5 A 5 3 9I 3*1 
2 * +' 



4 

9'7 7-5 5-3 



\ 

where the subscripts on P indicate the value of n. Clearly, each 
of these polynomials is a particular solution of the Legendre 
equation in which n has the value of the subscript on P. These 
polynomials are known as Legendre polynomials. They are 
frequently used in applied mathematics. Very often, they are 
denoted by P n (cos 0), where cos 6 = x, so that, for example. 

P 3 (cos 0) s jj cos 3 B - ^ cos 0. 

Zi Zi 

The values of the Legendre polynomials (sometimes called surface 
zonal harmonics) are tabulated* for various values of x. 

A solution that is valid for all values of x outside the interval ( ,1, 1) 
can be obtained by assuming it to have the form of a series of descending 
powers of x. A procedure analogous to that outlined above leads to 
the general solution of the form 

n(n - 1) , n(n - l)(n - 2)(n - 3) 

r __ - - rn-2 4- - - - - - r~ 4 

x 2(2n-l) x + 2 4(2n - l)(2n - 3) x 
n(n - l)(n - 2)(n - 3)(n - 4)(n - 5) n _ n 

2 4 6(2n - l)(2n - 3)(2n - 5) x ^ ' ' ' J 
(n + l)(n + 2) 

- ! - ! ' -- 



2-4(2n-h3)(2n + 5) 
(n + l)(n + 2)(n + 3)(n + 4)(n + 5)(n + 6) 



2 4 6(2n + 3)(2n + 5)(2n + 7) - 

which is valid for |x| > 1, and where n is a positive integer. 

* See JAHNKE, E., und F. EMDE, Funktionentaf eln ; BYERLY, W. E., 
Fourier Series and Spherical Harmonics. 



102 ORDINARY DIFFERENTIAL EQUATIONS 345 

It will be shown next that the Legendre polynomials are orthogonal 
in the interval from 1 to 1, so that they can be used to represent a 
suitably restricted arbitrary function defined in the interval ( 1, 1). 

Note that (102-1) can be written in an equivalent form as 

3j[(k-*Vl + ( + i)tf-o, 

and let P m (x) and P n (x) be two Legendre polynomials. Then, 

[(1 - x*}P m '(x)} + m(m + l)P m (x) s 



and 



[(1 - x*)Pn'(x)\ + n(n + l)P n (x) EE 0. 



Multiplying the first of these identities by P n (x) and the second by 
P m (x) and subtracting give 

P m (x) ^ [(1 - x*)P'(x)] - P n (x) ~ [(1 - x*)P m '(x)\ 

+ (n-m)(n + m + l)P m (x)P n (x) = 0. 

Integrating both members of this expression with respect to x between 
the limits 1 and +1 gives the formula 

Pm(x} ix [(1 ~ * 2 ) p '<*>] dx ~ S-i Pn(x} Tx l(l ~ X ^ P "'W ** 

+ (n - m)(n + m + 1) P P m (x)P n (x) dx = 0. 
J ~ i 

The application of the formula for integration by parts to the first two 
integrals reduces this formula to 



P m (x)P n (x) dx = 0. 
Therefore, 

J^ P m (x)P n (x) dx = 0, if m ^ n, 

so that the Legendre polynomials are orthogonal. 
It can be shown that 



2m + 1 

The derivation of this formula is somewhat tedious and will not be 
given here.* 

*See WHITTAKER, E. J., and G. N. WATSON, Modern Analysis, p. 305; 
MAcRoBERT, J. M., Spherical Harmonics; Byerly, W. E., Fourier Series 
and Spherical Harmonics, p. 170. 



346 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 103 

Consider next a function f(x) that is defined in the interval ( 1,1), 
and assume that it can be represented by a series of Legendre 
polynomials 

(102-4) /(*) = J arf n (x) 

n = 

that can be integrated term by term. It follows immediately from 
Sec. 24 that the coefficients in the series (102-4) are given by the 
formula 



a n = 



J 1 f( x )p n ( x ) dx, (n = 0, 1, 2, - - ). 



PROBLEMS 

1. Show that the coefficients of h n in the binomial expansion of 
(1 - 2xh + A 2 )- 1 '* are the P n (x). 

2. Verify that 

P M - (* - 1)- 



by computing P n (x) for n 0, 1, 2, 3. 

3. Expand f(x) 1 + x x 1 in a series of Legendre polynomials. 

4. Show that 

P 2 Gr) =P 2 (-aO 
and 

Pjn + lCx) = -P 2 n+l(-aO. 

6. Show that 

p rm-r iv x ' 3 ' 5 ' ' ' (2n " 1} 
P 2 n(0) - (-1)- 2 4 6 2n 

and 

P 2 n + l(0) = 0. 

6. Show, with the aid of the formula 



(1 - 2xh + h*)- = ^Pn(x) (see Prob. 1), 

n = 

that 

Pn(l) = 1 

and 

P.(-l) = (-1)-. 

103. Numerical Solution of DilGferential Equations. The 

method of infinite series solution of ordinary differential equations 
affords a powerful means of obtaining numerical approxima- 
tions to the solutions of differential equations, but its useful- 
ness is limited by the rapidity of convergence of the series. 



103 ORDINARY DIFFERENTIAL EQUATIONS 347 

Many differential equations occurring in physical problems 
cannot be solved with the aid of the methods discussed in this 
chapter, and one is obliged to resort to numerical methods. 
Only one of these methods, which was developed by the French 
mathematician E. Picard, is outlined in this section.* 

Consider the problem of finding that particular solution of the 
equation of first order, 

(103-1) f x = /(*, ), 

which assumes the value y Q when x = XQ. If both members of 
(103-1) are multiplied by dx and the result is integrated between 
the limits XQ and x, one obtains 



or 

(103-2) y = 2/0 + f x f(x, y) dx. 

JXo 

This is an integral equation, for it contains the unknown function 
y under the integral sign. 

Since the desired integral curve passes through (#o,,2/o)> assume 
as a first approximation to the solution of (103-2) that y, appear- 
ing in the right-hand member of (103-2), has the value 2/0- Then, 
the first approximation to the solution of (103-2) is 



yi(x) = 2/0 + \f(x, 2/0) dx. 

Jxo 



Performing the indicated integration gives y\ as an explicit 
function of x, and substituting y\(x) in the right-hand member 
of (103-2) gives the second approximation, namely 



dx. 
The process can be repeated to obtain 

I/ate) = 2/0 + Fflx, y z (x)] dx, 

Jxo 

and so on. It is clear that the nth approximation has the form 
Vn(x) = 2/o + C*f[x, y n -\(x)} dx. 

Jx^ 

* For other methods see Bennett, Milne, and Bateman, Numerical Inte- 
gration of Differential Equations, Bulletin of National Research Council, 
1933. 



348 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 103 

The functions 2/1(2), yi(x), , y n (x) all take on the value 
?/o when x is set equal to XQ, and it may happen that the successive 
approximations y\(x)j 2/2(2), , y n (x) improve as n increases 
indefinitely; that is, 

lim y n (x) = y(x) y 

n QO 

where y(x) is the solution of Eq. (103-1). It may be remarked 
that, in order to establish the convergence of the sequence of 
approximating functions, it is sufficient to assume that f(x, y) 
and df/dy are continuous in the neighborhood of the point (XQ, 2/0). 
Despite the fact that these conditions are usually fulfilled in 
physical problems, the convergence may be so slow as to make 
the application of the method impracticable. The usefulness of 
the method is likewise limited by the complexity of the approxi- 
mating functions. In many instances, it may be necessary to 
make use of numerical integration in order to evaluate the 
resulting integrals. * 

The method just outlined can be extended to equations of 
higher order. 

As an illustration of the application of the method to a specific prob- 
lem, let it be 'required to find the integral curve of the equation 

y' = 2x + y\ 
passing through (0, 1). 
Then (103-2) becomes 

(103-3) y = i+f* (2x + I/*) dx, 

and substituting y = 1 in the integrand of (103-3) gives 

y\ = 1 + f* (2x + 1) dx = I + x + x\ 
Then, 

2/2 = 1 + f* [2x + (1 + x + z 2 ) 2 ] dx 

= 1 + x + 2x* + x* + y^ + 
and 



f 



[2x + (l+x + 2x* + x s + %x* + Hx*Y] dx 
+ 2x* + %x* + %x* + %x* + y lQ x* + i% 5 V 
+ %o* 8 + %o* 9 + Ko* 10 + HTS*"- % 

Even though the integrations in this case are elementary, the process 
of computing the next approximation is quite tedious. As a matter of 



* See Sec. 167. 



J103 ORDINARY DIFFERENTIAL EQUATIONS 349 

fact, in this case one can obtain the desired solution more easily by the 
method of infinite series. 
Thus, assuming 

00 

<r\ 

a n x* 
and applying the method of Sec. 98 lead to the solution in the form 



I 9 \ I 1 I 1 \ 9 I *- /t *u I ****U _ 

y = O,Q -f" #o x *T \^o i L)X* i ^ x -7- 

o 



5a 3 

-h 



Since the integral curve must pass through (0, 1), it follows that a = 1, 
and the desired solution is 

y = 1 + x + 2z 2 + y^ + iy*x* + ijhsz 6 + ' . 

This agrees with the solution obtained by Picard's method up to the 
terms in x 4 . 

PROBLEM 

Find, by Picard's method, solutions of the following equations: 

(a) y' = zi/ through (1,1); 

(b) y' - x~y* through (0, K); 

(c) y' = l+2/ 2 through g, l); 

(d) y' = x + y through (1, 1). 



CHAPTER VIII 
PARTIAL DIFFERENTIAL EQUATIONS 

104. Preliminary Remarks. An equation containing partial 
derivatives has been defined in Sec. 67 as a partial differential 
equation. This chapter contains a brief introduction to the 
solution of some of the simpler types of linear partial differential 
equations which occur frequently in practice. It will be seen that 
the problem of solving partial differential equations is inherently 
more difficult than that of solving the ordinary equations and 
that Fourier series, Bessel functions, and Legendre polynomials 
play an important part in the solution of some of the practical 
problems involving partial differential equations. 

It was stated in Sec. 68 that the elimination of n arbitrary 
constants from a primitive /(x, y, Ci, c 2 , , c n ) leads, in 
general, to an ordinary differential equation of order n. Con- 
versely, the general solution of an ordinary differential equation 
has been defined to be that solution which contains n arbitrary 
constants. In the next section, it is indicated, by some examples, 
that one is led to partial differential equations by differentiating 
primitives involving arbitrary functions, and it follows that 
partial differential equations may have solutions which contain 
arbitrary functions. However, it is not always possible to 
eliminate n arbitrary functions from a given primitive by n 
successive differentiations, and the temptation to define the 
general solution of a partial differential equation as the one con- 
taining n arbitrary functions may lead to serious difficulties. 

In some important cases of linear partial differential equations 
with constant coefficients, treated in Sec. 107, it is possible to 
obtain solutions that contain the number of arbitrary functions 
equal to the order of the differential equation, and the term 
general solution is used in this chapter only in connection witfy 
such equations. With the exception of the linear partial differen- 
tial equations of the first order and of certain important types of 
linear equations of the second order, no extensive theory of the 
nature of solutions has been developed so far. 

350 



105 PARTIAL DIFFERENTIAL EQUATIONS 351 

Just as in the case of the ordinary differential equations, the 
solution of a practical problem can be obtained by eliminating 
the element of arbitrariness with the aid of the initial or boundary 
conditions. In practical problems the boundary conditions 
frequently serve as a guide in choosing a particular solution, 
which satisfies the differential equation and the boundary con- 
ditions as well. 

105. Elimination of Arbitrary Functions. Consider a family of 
surfaces defined by 

*=/(* + y), 

where / is an arbitrary function. 

If the argument of / is denoted by s, then 

*=f(x + y) -/() 

and 

dz _ df ds 

dx ~~ ds dx 

Since s = x + y, it follows that 
(105-1) g = f(x + y}, 

where f'(x + y) denotes the derivative of f(x + y) with respect to 
its argument x + y. Similarly, 

(105-2) g ==/ > (x + 2/) . 

Subtracting (105-2) from (105-1) leads to the partial differential 
equation of the first order 

^_* = o 

dx dy ' 

whose solution, clearly, is z = f(x + y). 
If 



-'GO 



then 



dx ds dx dy ds dy 

where s = y/x. 



352 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 106 

Denoting df/ds by f(y/x) and substituting the values of 
ds/dx and ds/dy give 



- f> - and - 

- ' *2/ ~ 

from which /'(y/x) can be eliminated to give 

dz . dz A 
z + 2/ = 0. 
do; d?/ 

Again the result is a partial differential equation of the first order. 
On the other hand, if 

*=/i(*)+/2(y), 

where f\(x) and fa(y) are arbitrary functions, differentiations 
with respect to x and y give 

!=/((*) and !=/;(,). 

If the first of these relations is differentiated with respect to y, 
a partial differential equation of the second order results, namely, 



dx dy 



=0. 



Differentiation of the second relation with respect to x will lead 
to the same equation, for the derivatives involved are assumed 
to be continuous. 

Another example, which is of considerable importance in the 
theory of vibrations, will be given. Let 

z =fi(x + at) +f 2 (x - at). 
If x + at ss r and x at = s, then 



and 



, 



OX uT uX uS uX 

" == J 1\*^ "i ^'/ "T~ J 2\*^ "~~ Qtt) 

Similarly, 

(105-3) d ~ = f!(z + at) + ft(x - at). 



106 PARTIAL DIFFERENTIAL EQUATIONS 353 

Also, 

dz a(/i +/ 2 ) dr a( 



dZ 3r dt ds dt 

= fi(x + a()a +ft(x - at)(-a) 
and 

(105-4) g = fi(x + a)a + K(x - a*)a*. 

Eliminating /'/(z + at) and/^'(a: - aO from (105-3) and (105-4) 
gives 

&z _ 2 M 

dt* " a dx*' 

* 

regardless of the character of /i and / 2 . This partial differential 
equation is of primary importance in the study of vibration and 
will be considered in more detail in Sees. 106, 108, and 109. 

106. Integration of Partial Differential Equations. This sec- 
tion contains two examples illustrating integration of partial 
differential equations. 

Let the differential equation be 



Integration with respect to y gives 
(106-2) ^ = /(*), 

where f(x) is arbitrary. If (106-2) is integrated with respect to 
x, then 



where ^ and <p are arbitrary functions. 
Consider next 

nnr *\ & z *&* 

(106 - 3) a?~ a 5? 

Change the variables in this equation by setting r = x + at 
and s = x at so that z becomes a function of r and s. Then 

dz = dzdr , ^^. 
dx "" dr dx ds dx' 



354 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 106 

and since dr/dx 1 and ds/dx = 1, it follows that 

d _ dz dz 
dx ~~ dr "*" ds 
and 

d 2 z /d z zdr , d 2 z ds , d 2 z ds 



Similarly, 

and since dr/dt = a and ds/d = a, it follows that 
dz _ dz __ dz 

Differentiating this with respect to t gives 

/- -\ d 2 z d 2 z dr , d 2 z ds d 2 z ds d 2 z dr 
(106-5) 2 == a 2 1" a a 2 a 



Substituting (106-4) and (106-5) in (106-3) gives the equation 

dTds = ' 

which is of the type (106-1), whose solution was found to be 
z = \[/(r) + <p(s), where ^ and <p are arbitrary. Recalling that 
r = x + at and s = x at, it is seen that 

(106-6) z = $(x + at) + <p(x - at), 

which is the solution in terms of the original variables. If in 
this solution \l/ and <p are so chosen that 

\l/(x + at) s A sin k(x + at), 
<p(x at) s A sin k(x at), 

where the variable t is thought to represent the time and x is the 
distance along the #-axis, then the first of these equations repre- 
sents a sinusoidal wave of amplitude A and wave length X = 2ir/k 
which is moving to the left with velocity a, whereas the second 



106 PARTIAL DIFFERENTIAL EQUATIONS . 355 

expression represents a similar wave moving with velocity a to 
the right (see Fig. 92). This can best be seen by recalling that the 
replacement of x by x at in z = f(x) shifts the curve at units 
in the positive direction of x and that the substitution of x + at 
for x translates the curve z = f(x) at units in the opposite direc- 




z*A$inkx 

FIG. 92. 

tion. Since t is a continuous variable representing the time, 
it is clear that the expression 

A sin k(x at) 

states that the sinusoid 

z = A sin kx 

is advancing in the positive direction of the x-axis with the 
velocity a. The period of the wave 

z = A sin k(x at) 

is defined as the time required for the wave to progress a distance 
equal to one wave length, so that 

X = aT 
or 

T = - = - 
a ka 

Consider next the wave resulting from the superposition of the 
two moving sinusoids A sin k(x at) and A sin k(x + of). 
Then, 

z = A sin k(x at) + A sin k(x + at) 
= A (sin kx cos kat cos kx sin 
+ 4 (sin kx cos fca + cos kx sin 
or 

(106-7) z = (2 A cos kat) sin to. 

The expression (106-7) is frequently referred to as a standing 
wave, because it may be thought of as representing a sinusoid 



356 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 106 

sin kx whose amplitude 2A cos kat varies with the time t in a 
simply harmonic manner. Several curves 

z = 2 A cos kat sin kx 
are drawn in Fig. 93 for various values of t. The points 



x = ' 



= o, i, 2, ), 



are stationary points of the curve and are called nodes. 

Inasmuch as (106-7) is obtained from (106-6) by making a 
particular choice of \f/ and <?, it satisfies the differential equation 



xZ2Acoskoit 




Fia. 93. 

(106-3), whatever be the values of A and k. This fact is of great 
importance in the discussion of Sec. 108. 

PROBLEMS 

1. Form partial differential equations by eliminating arbitrary 
functions. 



(a) z=f(x- 2y) 
(6) *=/(* + y 



3x + 4y. 



Note that ~ = f'(x 2 + ?/ + z 2 ) (2x + 2z 
ox \ ox 



(c) z 

(d) z = / 1 (x)/ 2 ( ?/ ). 

2. Prove that z - fi(x + iy) + f%(x iy) is a solution of 

d^" 2 + dy* = ' 

3. Form partial differential equations by eliminating the arbitrary 
functions, in which x and t are the independent variables. 

(a) z =/!(* -20 +/>(* + 20; 

(6) /(- + ); 

(c) *-/i(* + 20 +/,(* + W); 



107 PARTIAL DIFFERENTIAL EQUATIONS 357 

(d) z # 1(3 + 



(/) 2 = /i(s - + */2(* - 0- 

4. Show that z = f(a\y ax) is a solution of the equation 

dz dz 

ai dx + a *frj = "> 

where a\ and a 2 are constants. 

5. Verify that z = j\(y + 2x) and z = f(y 3ar) satisfy the equation 

**+_*'*__ 6 *!i -o 

ax 2 + dxdy a?/ 2 ~ u ' 

and hence deduce that z = fi(y + 2) + /2(?/ 3x) is also a solution of 
the equation. 

6. Show that 



is a solution of the equation 

2, 2 _*?!.. -o 

ax 4 "*" ax 2 a^/ 2 "^ a?j 4 u> 

provided that i 2 = 1. 

107. Linear Partial Differential Equations with Constant 
Coefficients. A linear partial differential equation with constant 
coefficients that often occurs in applications has the form 

(107-1) ^o + a l + a 22 



Frequently, such equations are called " homogeneous " because 
they involve only derivatives of the nth order. 

This equation can be solved by a method similar to that 
employed in solving an ordinary linear equation with constant 
coefficients. Introduce the operators 

DJ - and D, - , 

with the aid of which (107-1) can be written as 
(107-2) 



358 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 107 

It is readily established* that the operators D\ and D 2 formally 
obey the ordinary laws of algebra, so that one can deal with 
differential operators of the form 

D 2 ) s 



just as one would with polynomials in the two variables DI and 
D 2 . Accordingly, the left-hand member of (107-2) can be 
decomposed into a product of n linear factors, so that (107-2) 
reads 



(107-3) (aiDi + 0iD*)(aJ)i + jW>,) - - (aDi + /U> 2 )z = 0, 

where the quantities a, and ft, in general, are complex numbers. 
Now the system of equations 

(a l D l + ftZ> 2 )z = 0, (z = 1, 2, , n), 
or 

(107-4) ,|1 + A || = o, (i = 1, 2, - , n), 

can be readily solved. It is easy to verify that 

z = F t (a t y - ftz), 

where /*\ is an arbitrary function, is a solution of (107-4). Con- 
sequently,! the solution of (107-3) can be written in the form 

(107-5) z = 



If the linear factors appearing in (107-3) are all distinct, the 
solution (107-5) contains n arbitrary functions and will be called 
the general solution of (107-1). 

If the a t in (107-3) are all different from zero, one can write 
(107-3) as 

(107-6) (Di - miJD,)CDi - m 2 D 2 ) (Di - mj) 2 )z = 0, 

where ra = ft/at, (i = 1, 2, , n). In this case, (107-5) 
assumes the form 

(107-7) z = Ft(y + m l x) + F 2 (y + m z x) + 

+ F n (y + m n x). 

If some of the factors in (107-6) are alike, then the number of 
arbitrary functions in (107-7) will be less than n, but it is easy 

* See the corresponding discussion in Sec. 87. 
t See the corresponding case in Sec. 88. 



J107 PARTIAL DIFFERENTIAL EQUATIONS 359 

to see that the equation 

(Di - wZ> 2 ) r z = 
has the solution 

z = Fi(y + mx) + zF 2 (i/ + mx) +-,-+ x r ~ l F r (y + mx). 

Consequently, one can obtain a solution of (107-6) that contains 
the number of arbitrary functions equal to the order of the 
differential equation even in the case when some of the factors 
in the left-hand member of (107-6) are not distinct. 

As an illustration, consider the equation, which frequently 
occurs in the study of elastic plates, 

* * d*z 

* ' 



dx* dx* dy 2 dy* 
or 

(Z>i + 2D\D\ + |)z = 0. 

The decomposition into linear factors gives 

(Di + iD 2 )(D 1 - iD*)(Di + iD*)(Di - */>) = 0, 

where z 2 = 1. It follows that the general solution of this 
equation has the form 

z = Ft(y - ix) + xF*(y- ix) + F 3 (y + ix) + xF*(y + ix). 

If the right-hand member of (107-1) is a function /(x, t/), then 
the general solution of the equation is 

z = $(z, y) + u(x, y), 

where u(x, y) is a particular integral and $(#, y) is the general 
solution of the related homogeneous equation. The determina- 
tion of particular integrals of the equation 

(107-8) L(D l9 D 2 )z = f(x, y) 

can be made to depend on the calculus of operators* as was done 
in Sec. 89. In many cases the particular integral can be obtained 
by inspection. If f(x, y) is a homogeneous polynomial of degree 
k, then the particular integral has the form 

(107-9) z = c x k+n + dx k+n - l y + - + c k + n y k + n , 

in which the coefficients c l can be determined by substituting 
(107-9) into (107-8) and comparing the coefficients of the corre- 

* See, for example, M. Morris and O. Brown, Differential Equations, p. 
294; A. Cohen, Differential Equations, p. 275. 



360 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 107 

spending terms of the resulting equation. 
As an example of this, consider 



which can be written as 



or 

(/>! - 2Z) 2 )(Z>i + 3D 2 )z 

Assume the particular integral in the form 

(107-11) z = coz 5 + cix 4 y + w*y* + c*x' 2 y 3 + cay 4 + 

Substituting (107-11) in (107-10) gives 



(20c + 4ci - 12c 2 )x 8 + (12ci + 6c 2 - 3(ic 

_i_ ffj^o [ f\(* , 72c4^x?y' 2 ( (2ci I 4c^i 120f > R > )?/^ ~ fJx'^/y 

Hence, equating the coefficients of like terms on both sides of this 
equation gives 

5c -f- c t 3c 2 = 0, 

12ci + 6c 2 - 36c s = 6, 

c 2 + c 3 12c 4 = 0, 

c 3 + 2c 4 - 60c 5 = 0. 

This system of four equations in six unknowns can always be solved. 
Writing it as 

Ci 3c 2 -f- Oc 3 + Oc 4 = 5c , 
2ci + c 2 6c 3 + Oc 4 = 1, 
Oci + c 2 + c 3 12c 4 = 0, 
Oci + Oc 2 + c 3 + 2c 4 = 60c 5 , 

and solving for ci, ca, c 3 , and c 4 in terms of Co and c 5 give a two-parameter 
family of solutions, 

-65c + 6480c 6 + 21 



C 2 
C 3 = 
C4 = 



55 

70c + 2160c 5 + 7 

55 ' 

-10co + 2520c 6 - 

55 

78Qc 5 + IQco + 1 
110 



Setting c = Cs = and substituting the values of the coefficients in 
(107-11) give a particular integral of (107-10) in the form 



108 PARTIAL DIFFERENTIAL EQUATIONS 361 

(*, y) = *K*x*y + %5*V - x**V 

Therefore the general solution of (107-10) is 

* = F l (y + 2x) + F 2 (y - 3s) + u(x, y), 
where FI and F 2 are arbitrary functions. 

PROBLEMS 

1. Find the general solutions of 



2. Find the particular integrals for the following equations: 

r \ o <* . ^ ^ _ i 
W ^ Sx 2 + dx dy ~ dy* ~ 1; 

(b )**z_Vz 

^ ' dx 2 dxdy dy 2 " ' 

//zn^: Obtain the particular integral for /(a;, y) = y 2 and for/Or, ?/) = a: 
and add the solutions. 

to ^ + 3 A 22 . , 2 -^4-77- 
(c) dx 2 ^*dxdy^ Z dy 2 X ^ y ' 

(d] ^_ a ^- x * 
W dx 2 a dy 2 ~ x ' 

108. Transverse Vibration of Elastic String. Consider an 
elastic string or wire of length / stretched between two points on 
the x-axis that are I units apart, and let it be distorted into some 
curve whose equation is y f(x) (Fig. 94). At a certain instant, 
say t = 0, the string is released from rest and allowed to vibrate. 
The problem is to determine the position of any point P of the 
string at any later time t. It is assumed that the string is per- 
fectly elastic and that it does not offer any resistance to bending. 

The resulting vibration may be thought of as being composed of 
the two vibrations: 



362 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 108 



a. Transverse vibration, in which every particle of the string 
moves in the direction of the t/-axis; 

6. Longitudinal vibration, in which every particle oscillates 
in the direction of the rr-axis. 

It is tolerably clear that, if the stretching force T is large com- 
pared with the force of gravity, then the horizontal component 
of tension in the string will be sensibly constant. Therefore the 
displacement of the point P in the direction of the tf-axis can be 
neglected compared with the displacement of P in the y-direction. 
In other words, the longitudinal oscillation of the string con- 
tributes so little to the resultant vibration that the entire vibra- 
tion may be thought to be given by considering the transverse 
component-vibration. 



T+AT 




FIG. 94. 

The relation connecting the coordinates of the point P with 
the time t can best be stated in the form of a differential equation. 
Thus, denote the length of the segment of the string between the 
points P(XJ y) and Q(x + A, y + AT/) by As, and let the tension 
at P be T and at Q be T + A7 T . In view of the assumption 
stated above, the horizontal components of tension at P and Q 
are nearly equal so that the difference AT of the tensions at the 
ends of the segment As is taken as equal to the difference between 
the vertical components of tension at Q and P. The vertical 
component of tension at P is 

(A \ / ^ \ 

T lim =) = (T^J , 
As _oAs/p \ ds/ x 

and the vertical component of tension at Q is 

(V 
TIT) 
OS/x+bx 

If it is assumed that the transverse displacement of the string 
is so small that one can neglect the square of the slope of the 
string in comparison with the slope dy/dx, then the sine of the 



108 PARTIAL DIFFERENTIAL EQUATIONS 363 

angle can be replaced by the tangent,* and the resultant of 
the forces at P and Q is 



By Newton's second law, this resultant must equal the mass of 
the element of the string of length Ax multiplied by the accelera- 
tion in the direction of the ?/-axis. Hence, 



<"> "*(&),- [( 



where p is the mass per unit length of the string and 
denotes the acceleration of the element PQ of the string. 
Dividing both sides of (108-1) by p &x reduces it to 



" 



and passing to the limit as Ax gives 



where a 2 = T/p. 

The solution of (108-2) was found in Sec. 106 to be 

y = $(x + at) + v(x - at), 

where ^ and <p are arbitrary functions. These functions must 
be so chosen that, when t = 0, 



Represents the equation of the curve into which the string was 
initially distorted. Furthermore, the string was supposed to 
have been released from rest, so that dy/dt = when t = 0. It is 
beyond the scope of this book to prove that these boundary 
conditions suffice for the unique determination of the functions 

* Note that 

dy 

dy . tan 6 dx ^ dy 
~ sin "~ ' 



VI + tan 2 



364 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 109 

\// and <p. It will be shown in the next section how the solution 
of this problem is obtained with the aid of Fourier series. 

109. Fourier Series Solution. In the preceding section, it was 
established that the transverse vibrations of an elastic string are 
defined by the equation 



and in Sec. 106 it was shown that a particular solution of this 
equation is given by 

(109-2) y = 2A cos kat sin kx 

for arbitrary values of A and k. Moreover, it is clear that the 
sum of any number of solutions of the type (109-2) will satisfy 
(109-1). 

Now, suppose that the string of length I is distorted into some 
curve y = /Or), and then released without receiving any initial 
velocity. The subsequent behavior of the string is given by 
Eq. (109-1), the solution of which must be chosen so that it 
reduces to y = f(x) when t = 0. In addition to this condition, 
dy/dt when t = 0, for, by hypothesis, the string is released 
without having any initial velocity imparted to it. Furthermore, 
since the string is fixed at the ends, y = when x = and when 
x = l. 

Consider the infinite series 

/lrkrk ON vat . irx . 2irat . 2irx 

(109-3) y = di cos -y- sm -j- + #2 cos j sm =- 
II 11 

3wat . 3wx , 
+ a 3 cos -y sm -j -- 1- , 

each term of which is of the type (109-2), where k has been chosen 
so that each term reduces to zero when x = and when x = I. 
When t = 0, the series becomes 



/^ A\ . wx . . 2irx . . . 

(109-4) ai sm y + a 2 sin ~~T I" a s sin j f- 

If the coefficients a n are chosen properly, (109-4) can be made to 
represent the equation y = f(x) of the curve into which the 
string was initially distorted; for a function /(#), subject to 
certain restrictions,* can be expanded in a series of sines (109-4) 
* See Sec. 20. 



109 



PARTIAL DIFFERENTIAL EQUATIONS 



365 



and the coefficients are given by 

(109-5) n = r f /(*) sin ^r dx - 

v ' I Jo J I 

It is readily verified that the derivative of (109-3) with respect 
to t satisfies the remaining boundary condition, dy/dt = 
when t = 0. Hence the infi- 
nite series (109-3), where the 
values of a n are given by 
(109-5), gives the formal solu- 
tion of the problem. 

Illustration. If the initial dis- 
tortion of the string (Fig. 95) is 
given by 

26 




k 



FIG. 95. 



then the solution of the problem is readily found to be 



86/1 . TTX 



irat 
~ ~~2 V ?2 sm "T cos ~T~ 

7T \ 1 I I 



1 . 
7 Sin 



j cos j + 

i I 



PROBLEMS 

1. Carry out solution of the problem given in the illustration, Sec. 109. 

2. A taut string of length /, fastened at both ends, is disturbed from 
its position of equilibrium by imparting to each of its points an initial 
velocity of magnitude f(x) . Show that the solution of the problem is 



y = sm 



sm 






sm 



Hint: The schedule of conditions here is: 
(a) y = 0, when t = 0; 

(6) -| = /(*), when = 0; 

(c) T/ = 0, when x = 0; 
(rf) ?/ = 0, when a; = /. 



Observe that 



, . . 

A sm T- sm 



nirat 



366 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 109 
satisfies conditions (a), (c), and (d), and build up a solution by forming 

mrx . mrat 



S A . nirx . 
A n sm T- sir 



and utilizing condition (6). 

3. Show that the solution of the equation of a vibrating string of 
length /, satisfying the initial conditions 

y f( x )> when t = 0, and r = g(x), when t = 0, 



niral , ^\ . . mrx . nwal 
cos 7 h X o n sm j- sin T > 

, v li ^^i ' ' 

n-1 n=l 



is 

y = 2) sin -y- 
where 

= - r i - 

and 

2 

I /t^'y^ oif .^ 

/ 



4. The differential equation of a vibrating string that is viscously 
damped is 



Show that the solution of this equation, when the initial velocity is 
zero, has the form 



y = 



where 



a n e~ bt sin -7- ( cos a n l + sin a n l) , 
L * \ n /J 

70 . 2 r l ^ - mrx . 

~" ^ anc ^ a ~ 7 I f( x ) sm ~7~ ^ a '- 



6. Show that the differential equation of the transverse vibrations of 
an elastic rod carrying a load of p(x) Ib. per unit length is 

dy __ p(x) <Py 
dx* " El m dl* ' 

where E is the modulus of elasticity, / is the moment of inertia of the 
cross-sectional area of the rod about a horizontal transverse axis through 
the center of gravity, and m is the mass per unit length. 

Hint: For small deflections the bending moment M about a horizontal 
transverse axis at a distance x from the end of the rod is given by the 
Euler formula M == El d 2 y/dx*, and the shearing load p(x) is given 
by d*M/dx* = p(x). 



110 PARTIAL DIFFERENTIAL EQUATIONS 367 

6. Show that the small longitudinal vibrations of a long rod satisfy 
, the differential equation 



_ 
dt* ~ p dx 2 ' 

where u is the displacement of a point originally at a distance x from 
the end of the rod, E is the modulus of elasticity, and p is the density. 

Hint: From the definition of Young's modulus E, the force on a cross- 
sectional area q at a distance x units from the end of the rod is 
Eq(du/dx) x , for du/dx is the extension per unit length. On the other 
hand, the force on an element of the rod of length Ax is pq Ax d 2 u/dt 2 . 

1. If the rod of Prob. 6 is made of steel for which E = 22 10 8 g. 
per square centimeter and whose specific gravity is 7.8, show that the 
velocity of propagation of sound in steel is nearly 5.3 10 5 cm. per 
second, which is about 16 times as great as the velocity of sound in air. 
Note that the c.g.s. system E must be expressed in dynes per square 
centimeter. 

110. Heat Conduction. Consider the slab cut from a body 
r by two parallel planes As units apart, and suppose that the 
temperature of one of the planes is u and that of the second plane 
is u + At. It is known from the results of experiments that heat 
will flow from the plane at the higher temperature to that at the 
lower and that the amount of heat flowing across the slab, per 
unit area of the plane per second, is approximately given by 

(110-1) kg, 

where k is a constant called the thermal conductivity* of the 
substance. If the distance As between the planes is decreased, 
then the limit of (110-1), 

,. 7 Aw 7 du 
hm k = A; > 
A S ->O As as 

gives the quantity of heat flowing per second per unit area of the 
surface whose normal is directed along s, and the quantity 
du/ds gives the rate of change of temperature in the direction of 
increasing s. 

Now suppose that the initial temperature of such a body is 
given by 



and that it is required to find the temperature of the body at 
* The dimensions of k in the c.g.s. system are cal./(cm.-sec. C.). 



368 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 110 

some later instant i. It is known* that the function u, which 
gives the temperature at any later time t, must satisfy the partial 
differential equation 

(110 -2) _* 

v ' dt Cp 

where c is the specific heat of the substance, p is the density 
of the body, and k is the conductivity, f Equation (110-2) is 
derived on the assumption that k, c, and p remain independent 
of the temperature u, whereas in reality they are not constant 
but vary slowly with the temperature. Moreover, this equation 
is not true if there is any heat generated within the body.J 
This equation must be solved subject to certain boundary 
conditions. 

Thus, if the body is coated with some vsubstance which makes 
it impervious to heat so that there is no heat flow across the sur- 
face of the body, then, if the direction of the exterior normal to the 
body is denoted by n, this boundary condition can be expressed 
mathematically as 

1 du A du n 

k = or = 0. 
dn dn 

On the other hand, if the surface of the body radiates heat 
according to Newton's law of cooling, then 

i du ( \ 

*- = *(-,), 

where u is the temperature of the surrounding medium and e is 
a constant called the emissivity of the surface. It can be shown || 
that, if the initial and surface conditions are specified, then the 
problem of determining the temperature at any later time t has 
only one solution. 

It should be observed that if the flow of heat is steady, so that 
the temperature u is independent of the time t, then du/dt - 

* See derivation of this equation in Sec. 130. 

f The dimensions of c and p are, respectively, in calories per gram per 
degree centigrade and grams per cubic centimeter. The constant k/cp a 2 
sq. cm. per second is frequently called the diffusivity. 

t See Sec. 130. 

See Problem 2, Sec. 70. 

|| For detailed treatment see Carslaw, "Introduction to the Mathematical 
Theory of the Conduction of Heat in Solids." 



111 



PARTIAL DIFFERENTIAL EQUATIONS 



369 



and (110-2) reduces to 



(110-3) 



dx 2 



^ 

dy* ^ dz* 



This is known as Laplace's equation, and it occurs frequently 
in a large variety of physical problems. 

It may be remarked that the problems of diffusion and the 
drying of porous solids are governed by an equation similar to 
(110-2), so that many problems on diffusion and heat conduction 
are mathematically indistinguishable. 

111. Steady Heat Flow. Consider a large rectangular plate 
of width d, one face of which is kept at tem- 
perature u = Ui, whereas the other face is kept 
at temperature u = u^. If one face of the 
plate is placed so as to coincide with the 
2/2-plane (Fig. 96), the surface conditions can 
be expressed mathematically as 



uu, 



(111-1) 



n HI when x = 0, 
u = u<t when x rf, 



and the temperature u must satisfy Eq. - 
(110-3). In this formulation of the problem, 
it is assumed that the plate extends indefi- 
nitely in the y- and ^-directions, a condition 
that is approximated by the large rectangular 
plate if the attention is restricted to the 
middle of the plate. With these assumptions, 
it is clear that the temperature u is inde- 
pendent of the y- and ^-coordinates and that (110-3) reduces to 



U-Ug 



FIG. 96. 



(111-2) 



dx* 



= 0, 



which is to be solved subject to the conditions (111-1). 
The solution of (111-2) is easily found to be 

(111-3) u = cix + c 2 , 

where Ci and c 2 arc arbitrary constants which must be determined 
so that (111-3) satisfies (111-1). Substituting x = and x = d 
in (111-3) gives u\ = c 2 and it 2 = c\d + c 2 , so that 



u = 



-x + 



370 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 111 



gives the solution of the problem. Recalling that the amount 
of heat flowing per second per unit area of the plate is 

7 du i Uz Ui 

k = k 1 i 

dx d 



it is seen that the amount of heat flowing in t sec. over the area 

y A is given by 

_ i. 

- u^tA. 



-i 



These results can be anticipated from physical 
considerations. 

A more difficult problem will be solved next. 
Suppose that a " semi-infinite " rectangular plate 
(that is, the plate extends indefinitely in the 
* positive direction o the ?/-axis), of thickness d, 
has its faces kept at the constant temperature 
FIG. 97. u = 0, whereas its base y Q is kept at tem- 

peratujre u = f(x) (Fig. 97). It is clear physically that the tem- 
perature u at any point of the plate will be independent of 2, so 
that in this case (110-3) becomes 



uf(x) 



(111-4) 



dx* + dy* 



(111-5) 



The solution of (111-4) must be so chosen that it satisfies the 

boundary conditions: 

u = when x = 0, 

u when x = d, 

u = f(x) when y = 0, 

u = when y = oo. 

The last condition results from the observation that the tem- 
perature decreases as the point is chosen farther and farther from 
the a>axis. 

In order to solve (111-4), recourse is had to a scheme that 
often succeeds in physical problems. Assume that it is possible 
to express the solution of (111-4) as the product of two functions, 
one of which is a function of x alone and the other a function of y 
alone. Then, 

(111-6) u = X(x)Y(y). 

Substitution of (111-6) in (111-4) and simplification give 



111 PARTIAL DIFFERENTIAL EQUATIONS 371 



= _. 

j X dx 2 Y dy 2 

It will be observed that the left member of (111-7) is a function 
of x alone, whereas the right member is a function of y alone. 
Since x and y are independent variables, Eq. (111-7) can be true 
only if each member is equal to some constant, say a 2 . Hence, 
(111-7) can be -written as 

1 d 2 X 2 ,1 d 2 Y 2 

v~rr = a and 17 -FT = a 
X dx 2 Y dy 2 

or 



~ + a 2 X = and a 2 Y = 0. 

dx 2 dy 2 

The linearly independent solutions* of these equations are 

X sin ax, 

X cos ax, 

Y = e a v, 

Y = e- a ", 

and, since u = XY, the possible choices for u are 



u = 



e oj/ cos ax, 
e ay sin ao:, 



The first two of these particular solutions for u obviously do not 
satisfy the last one of the boundary conditions (111-5). The 
third particular solution e~ ay cos ax does not satisfy the first 
of the conditions (111-5). But if u is chosen as e~ ay sin ax, then 
u = when x = and u = when y = oo ; and if a is chosen 
as nir/d, where n is an integer, then 

/i 1 1 o\ ~^r y 

(111-8) u = e d sin 

/ 

satisfies all the conditions (111-5), except u = /(#) when y = 0. 
It will satisfy this condition also if f(x) = sin ~ 

Inasmuch as Eq. (111-4) is linear, any constant times, ^ solu- 
tion (111-8) will be a solution, and the sum of any number ,pf such 
* See Sec. 95. 



372 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 111 
solutions will be a solution. Hence, 

00 niry 

(111-9) = ^ e ~^" sin ^ 

n=>l 

is a formal solution. When y = 0, (111-9) becomes 

00 

2. nwx 
a n sin T- > 

n = l 

which must reduce to/(x). But, in Sec. 20, it was shown that 
the constants a n can be chosen so that the function is represented 
by a series of sines. Therefore, if 



then (111-9) will satisfy all the boundary conditions of the prob- 
lem and hence it is the required solution. 

Illustration. In the preceding problem suppose that f(x) 1 and 
d = TT. Then 

2 r* - 

a = - I sin nx ax, 

TT JO ' 

and the solution (111-9) is easily found to be 

4 / 1 1 \ * 

u = - ( e~ tf sin x + ^ e~ ?J/ sin 3z + ^ e~ 5tf sin 5x + J. 

PROBLEMS 

1. Using the result of the illustration just above, compute the tem- 
peratures at the following points: (ir/2, 1), (?r/3, 2), (?r/4, 10). 

2. Obtain the solution of the problem treated in Sec. 109 by assuming 
that y can be expressed as the product of a function of x alone by a 
function of t alone and following the arguments of Sec. 111. 

3. Compute the loss of heat per day per square meter of a large con- 
crete wall whose thickness is 25 cm., if one face is kept at 0C. and the 
other at 30C. Use k = 0.002. 

4. A refrigerator door is 10 cm. thick and has the outside dimensions 
60 cm. X 100 cm. If the temperature inside the refrigerator is 10C. 
and outside is 20C. and if k = 0.0002, find the gain of heat per day 
across the door by assuming the flow of heat to be of the same nature 
as that across an infinite plate. 

5. A semi-infinite plate 10 cm. in thickness has its faces kept at 0C. 
and its base kept at 100C. What is the steady-state temperature at 
any point of the plate? 



112 PARTIAL DIFFERENTIAL EQUATIONS 373 

112. Variable Heat Flow. Consider a rod of small uniform 
cross section and of length /. It will be assumed that the surface 
of the rod is impervious to heat and that the ends of the rod are 
kept at the constant temperature u = 0C. At a certain time 
t = 0, the distribution of tempera- 
ture along the rod is given by 
y = /(#) The problem is to find the 
temperature at any point x of the 
rod at any later time t. 

In this case the temperature u is 
a function of the distance along the FlG> 98> 

rod and the time , so that, if the rod (Fig. 98) is placed so as to 
coincide with the z-axis, (110-2) becomes 

N du n d 2 u 



where a 2 = k/cp is the diffusivity. In addition to satisfying 
(112-1) the solution u must satisfy the boundary conditions 

!' u = when x = Q) f , r . 
A , i \ f or all values of t, 

u = when x = I j 

u = f( x ) when t = 0. 

As in Sec. Ill, assume that a solution of (112-1) is given by the 
product of two functions, one a function of x alone and the other 
a function of t alone. Then, 

u = X(x)T(f), . 

and the substitution of this expression in (112-1) gives, after 
simplification, 

1 d_ T = 1 d*X 
a*T dt ~ X ~dx* ' 

This equation can hold only if each member of it is equal to some 
constant, say /3 2 . There result 



+ a^T = and + pX = 0. 

at ax* 

The linearly independent solutions of these ordinary differential 
equations are readily found to be 



X = cos fix, 
X = sin ftx. 



374 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 112 

Then, since by hypothesis u = TX, the possible choices for u are 

u = e ~ a ^ 2t cos ftx, 
u = e -*w s j n p Xm 

The first particular solution does not satisfy the first one of the 
conditions (112-2). If ft is chosen as mr/l, where n is an integer, 
then 

(112-3) u-e-'W'xin^x 

satisfies the first two conditions of (112-2) but not the last one. 

The sum of solutions of the type (112-3), each multiplied by a 
constant, will be a solution of (112-1), since the equation is linear, 
so that 



(112-4) M = 2a n e V ' / sin ^-x 

is a solution. For t = 0, (112-4) reduces to 

oo 

2. nit 
a n sin -y x, 

which can be made equal to/(x), provided that 

= ? f l /(j.) 8in !^5 da, 

/ Jo 

Then, 



(H2-5) u = 2 [r jo' /(x) sin ^ r/x j 6 ^ ' sin T 

satisfies all the conditions of the problem and is therefore the 
required formal solution. 

Next, consider an infinite slab of thickness I, whose faces are 
kept at temperature zero and whose temperature in the interior 
at the time t = is given by u = f(x). It is clear that the 
solution of this problem is independent of y and z, so that u satisfies 
the differential equation 

2 ^ u 
~di ~~ "" daT 2 " 

The boundary and initial conditions are 



u = when x = I ' ' 
u = f(x) when / = 0. 



112 



PARTIAL DIFFERENTIAL EQUATIONS 



375 



The mathematical formulation of this problem is identical with 
that of the preceding one, and therefore the solution of the 
problem is given by (112-5). 

The solutions of other important problems on heat flow are 
outlined in detail in Probs. 5 and 6 at the end of this section. 

PROBLEMS 

1. Suppose that in the first problem of Sec. 112 the ends of the rod 
are iniDervious to heat, instead of being kept at zero temperature. 
The formulation of the problem in such a case is 

du _ 2 3% 
dt ~ a dx 2 ' 



du - n 

~r~ U 

dx 



when x 



- -0 

dx ~ U 



for all values of t, 



when x I 
u = f(x) when t = 0. 
Show that the solution in this case is ' 



00 

u -- 



where 



2 f l 
= j- J 



cos 



cos 



, 
dx. 



o 



100 



01 







Fia 99. 



2. A large rectangular iron plate (Fig. 99) is 
heated throughout to 100C. and is placed in con- 
tact with and between two like plates each at 0C. The outer faces of 
these outside plates are maintained at 0C. Find the temperature of 
the inner faces of the two plates and the temperature at the midpoint 
of the inner slab 10 sec. after the plates have been put together. Given : 
a 0.2 c.g s. unit. 

Hint: The boundary conditions are 



u = 
u = 
u = f(x) 

where /(x) is when < x < 1, 
f(x) is 100 when 1 < x < 2, 
f(x) is when 2 < x < 3. 

Hence, 



when x 



when t = 0, 



C 3 j>/\ 

L f(x) s 



j f 2 ^ n<jrx j 

sm ~r- dx = Ij 1 00 sin -rr- dx. 



376 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 112 

3. An insulated metal rod 1 m. long has its ends kept at 0C., and its 
initial temperature is 50C. What is the temperature in the middle of 
the rod at any subsequent time? Use k = 1.02, c = 0.06, and p = 9.6. 

4. The faces of an infinite slab 10 cm. thick are kept at temp. 0C. 
If the initial temperature of the slab is 100C., what is the state of 
temperature at any subsequent time? 

6. Let the rod of Prob. 3 have one of its ends kept at 0C. and the 
other at 10C. If the initial temperature of the rod is 50C., find the 
temperature of the rod at any later time. 

Hint: Let the ends of the rod be at x and x = 100; then the 
conditions to be satisfied by the temperature function u(x, t) are as 
follows: w(0, t) = 0, w(100, t) = 10, u(x, 0) = 50. Denote the solu- 
tion of Prob. 3 by v(x, t) ; then if the function w(x, t) satisfies the 
conditions * 

Tt = a * d5' w(0 ' = ' (100 ' = 10 ' w(x > 0) = ' 

u(x, t) v(x, t) + w(x, t) will be the solution of the problem. Assume 
the solution w(x, t) in the form w(x, t) = x/10 + <p(x, t), and deter- 
mine the function <p(x, t). 

6. Let a rod of length / have one of its ends x maintained at a 
temperature u = 0, while the heat is dissipated from the other end 
x I according to the law 



Let the initial temperature be u(x, 0) = f(x), where f(x) is a prescribed 
function, Choose a particular solution of (112-1) in the form 

e'--'0'< sin fix, 
and show that the boundary conditions demand that 

cos 01 = -hsm/31. 
Write this transcendental equation in the form 

tan 01= -Q, 

and show that it has infinitely many positive real roots ]8i, 02, ' ' ' , 0n* 
Hence, if 

u(x,Q) = f(x) 
then the solution has the form 



n = l 



113 



PARTIAL DIFFERENTIAL EQUATIONS 



377 



The functions sin/3 n a;, (n = 1, 2, ), are easily shown to be orthog- 
onal in the interval (0, I), so that the coefficients A n in the solution are 
given by the formula 

i f(x) sin fi n x dx 
A _ J 

/In pi 

I sin 2 p n x dx 

113. Vibration of a Membrane. Consider an elastic mem- 
brane, of surface density p, which is under uniform tension T. By 
definition the tension is said to be uniform if the force exerted 
across a line of unit length in the plane of the membrane is 
independent of the orientation of the line. It will be assumed 

y 




FIG. 100 

that the plane of the membrane coincides with the 2/-plane of the 
rectangular coordinate system and that the displacement of any 
point of the membrane in the direction normal to the xy-plane is 
denoted by z. Then a consideration of the forces acting upon 
the element dA of the membrane (Fig. 100) leads to the equation 



m*n 
(113-1) 



__ 



where c 2 = T/p. The analysis leading to (113-1) is quite similar 
to that used in deriving Eq. (108-2) for the vibrating string; 
and, just as in Sec. 108, the underlying assumption here is that 
the displacement z is not too great. 

The solution of the problem of a vibrating membrane consists 
of determining the function z = f(x, y, t), which satisfies the 
differential equation (113-1) as well as the boundary and initial 
conditions characteristic of the particular physical problem under 



378 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 113 

consideration. These remarks will be illustrated by a brief 
treatment of the case in which the membrane is circular. In this 
case the shape of the membrane suggests the use of the cylindrical 
coordinate system in preference to the rectangular system. As 
will be seen presently, the choice of cylindrical coordinates is 
made because the boundary conditions assume particularly simple 
forms in these coordinates. 

The transformation of (113-1) can be accomplished readily 
with the aid of the relations* connecting cylindrical coordinates 
with rectangular, namely, 

x = r cos B y y = r sin 0, z = z 
or 

r = \fifi~+~y*, = tan" 1 -> z = z. 

X 

It will be necessary to express d 2 z/dx 2 and d 2 z/dy 2 in terms of the 
derivatives of z with respect to r and 6. Nowf 

dz = dz dr dz <90 
dx ~ dr dx 68 dx 



and 

/ \ 2 


d 2 r dz d 2 z i dO\ d 2 6 dz d 2 z dr 

. I VV I . V Vt* . ^ V ^ V ' 


a0 


dx 2 dr 2 \dx/ 
But 
dr x 


ao: 2 dr T a0 2 \dxj ] dx 2 dd ] " dr dS dx 


aa; 


dx T/X* q: y 

d 2 r y 2 


sin 2 6 d 2 6 2xy 2 sin 6 cos 6 


dr 2 (r 2 4- ?/ 2 V^ 
t/x {J, \ y ) 

Substituting these 

d * z d * z cos* e + 


r ' dx 2 (x 2 + y 2 ) 2 r 2 
values in the expression for d 2 z/dx 2 gives 
dz sin 2 6 d 2 z sin 2 dz 2 sin cos 





Similarly, 

01 - 2 fl 1. 


ar r ' a0 2 r 2 ' a0 r 2 
a 2 z ( cos sin 


1 "ara0 r 

dz cos 2 a 2 z cos 2 az ( 2 sin cos 


') 


~ o 10 olll V p 

at/ 2 ar 2 

* See Sec. 56. 
t See Sec. 39. 


ar r a0 2 r 2 . a0 r 2 
a 2 z /sin cos 


'Y 


ar a0 \ r 


/ 



113 PARTIAL DIFFERENTIAL EQUATIONS 379 

so that 

z + ^ z = ~ 4- i 4- 1 *?5 
to 2 + <ty 2 ~ dr 2 + r dr + r 2 d0 2 

and (113-1) can be written as 

(H3-2) S = 

dr 

It was remarked in Sec. 104 that the solution of such an equa- 
tion contains two arbitrary functions, and in order to make 
the problem definite it is necessary to know the initial and 
boundary conditions. Thus, suppose that the membrane is of 
radius a and is fastened at the edges. Then it is evident that the 
solution of (113-2), 

z = F(r, 6, 0, 

must satisfy the condition z = when r = a, for all values of t. 
If, moreover, the membrane is distorted initially into some 
surface whose equation is a function of the radius only (that is, the 
initial distortion is independent of 0), say z = /(r) when t = 0, 
then it is clear that the subsequent motion will preserve the 
circular symmetry and that the solution will be a function 6f 
r and t only. These conditions alone are not sufficient for the 
unique determination of the function 

z = F(r, 0, 

and it is necessary to specify the initial velocity of the membrane 
in order to make the problem perfectly definite. If the mem- 
brane is distorted and thereafter released from rest, then 

~ = when t = 0. 

ot 

Since the solution is assumed to be independent of 0, (113-2) 
becomes 



(113-3) ^ = < 

and its solution satisfying the boundary and initial conditions 

z when r = a, 

(113-4) / =/(r) when* = 0, 

-r- = when t = 
at 

will be obtained by a method similar to that used in Sec. 111. 



380 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 113 

Assume that it is possible to express the solution of (113-3) as 
the product of two functions, one of which is a function of r alone 
whereas the other is a function of t alone. Then, 



Substitution of this relation in (113-3) leads to 

R ldR\ 

+ r dr) 



dt* 
or 



(1135) 
(il6 ~ d) 

Since the left-hand member of (113-5) is, by hypothesis, a func- 
tion of t alone, whereas the right-hand member is a function of 
r alone, each member must be equal to some constant, say w 2 . 
Hence, (113-5) can be written as 



(113-6) + *r = 

and 



, 

dr 2 r dr 

where k = u/c. 

Equation (113-6) is the familiar equation of simple harmonic 
motion, and Eq (113-7) is easily reducible to the Bessel equation 
by the substitution x = kr. Thus, if x kr, 

dR = dR dr = IdR 
dx dr dx k dr' 

d*R d (ldR\ I d*R dr 1 d*R 
dx 2 dx \kdrj k dr* dx k* dr*' 

so that (113-7) assumes the form 

d*R IdR 

dx^ + xd^ +R== > 

which possesses the solution (see Sec. 100) 

R = J ( x ) = Jo(fcr). 
Therefore, 

z = RT = Jo(fer) sin co 
or 

z = Jo(kr) cos ait. 



113 PARTIAL DIFFERENTIAL EQUATIONS 381 

X 

Since the last of the boundary conditions (113-4) requires 

== when t = 0, 

it is necessary to reject the solution involving sin co. Further- 
more, the first of these conditions demands that 

2 = when r = a, 

so that 

z = Jo(ka) cos co? = 

for all values of t. This condition will be satisfied if the arbitrary 
constant k is so chosen that Jo(fca) = 0. In other words, ka 



UoOO 




FIG. 101 



must be a root of the Bessel function of order zero (Fig. 101); and 
if the r*th root of Jo(kr) is denoted by 



then 



Since k = w/c, it follows that 

co = k n c. 

Hence, a solution of (113-3) that satisfies two of the boundary 
conditions (113-4) is given by 

Jo(k n r) cos k n ct. 

The sum of any number of such solutions, each multiplied by an 
arbitrary constant, will be a solution, so that 



(113-8) z = 2^ A n Jo(k n r) cos k n ct 

will be a formal solution of (113-3). 



382 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 114 

But when t = 0, the second of the boundary conditions 
demands that z f(r). Since (113-8) becomes, for t = 0, 

00 

(113-9) z = J AJ(k n r), 

n l 

it follows that, if it is possible to choose the coefficients in the 
series (113-9) so as to make 

00 

(113-10) A n J<,(k n r) = f(r), 



then (113-7) will be the required formal solution of (113-3) which 
satisfies all the conditions (113-4). 

The problem of development of an arbitrary function in a 
series of Bessel functions has been discussed in Sec. 101, where it 
was indicated that a suitably restricted function /(r) can be 
expanded in a series (113-10), where 



114. Laplace's Equation. Let it be required to determine 
the permanent temperatures within a solid sphere of radius unity 
when one half of the surface of the sphere is kept at the constant 
temperature 0C. and the other half is kept at the constant 
temperature 1C. 

From the discussion of Sec. 110, it is evident that the tempera- 
ture u within the sphere must satisfy Laplace's equation 

d 2 u d*u d*u __ 
dx* + dy* + fc* 

The symmetry of the region within which the temperature is 
sought suggests the use of spherical coordinates. If Laplace's 
equation is transformed with the aid of the relations* (Fig. 102) 

x = r sin 6 cos <p, 
y = r sin 6 sin <p, 
z = r cos } 

in a manner similar to that employed in Sec. 113, the equation 
becomes 



* See Sec. 56. 



114 



PARTIAL DIFFERENTIAL EQUATIONS 



383 



It is necessary to seek a solution of this equation that will satisfy 
the initial conditions. 

If the plane separating the unequally heated hemispheres is 
chosen so that it coincides with 
the xy-pl&ne of the coordinate 
system and if the center of the 
sphere is taken as the origin, 
then it appears from symmetry 
that it is necessary to find the 
temperatures only for that por- 
tion of the sphere which lies to 
the right of the #2-plane (see 
Fig. 102). Moreover, it is clear 
that the temperatures will be 
independent of <p, so that (114-1) FIG 102 

becomes 




The solution of (114-2) must be chosen so as to satisfy the bound- 
ary conditions 



(114-3) 



u = 1 for < < ~ when r 1, 



u = for - 



< TT when r = 1. 



In order to solve (114-2), assume that the solution 

u = F(r, 6) 

is expressible as the product of two functions, one of which is 
independent of 6 and the other independent of r. Thus, let 

u = /2(r)6(0). 

The substitution of this expression in (114-2) leads to the two 
ordinary differential equations 

d\rR) 



dr 2 



- a*R = 



and 



1 d 



sin0^) + a*e = 0, 



where a 2 is an arbitrary constant. 



384 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 114 

The first of these equations can be expanded to read 

2 d 2 R , dR 2P A 
r 2 -T-Y + 2r -^ -- a 2 .R = 0, 
ar 2 ar 

which is an equation of the type treated in Sec. 97 and the linearly 
independent solutions of which are 

R = r m and R = l/r m +\ 
where 



so that 

a 2 = m(m + 1). 

If this value of a is substituted in (114-4), this equation becomes 



The change of the independent variable 6 to x by means of 
x = cos 6 leads to Legendre's equation 



If m is an integer, particular solutions of this equation are the 
Legendre polynomials 

P( T \ _ p ( P0 u fi\ 
m\>") -* mv^-^-'O l/y, 

and hence the particular solutions of (114-2) are 

u = r m P m (cos 6), 
_ P m (cos 6) 

lAi ~r~: 

The second of these solutions evidently cannot be used, for it 
becomes infinite when r > 0. Therefore, it will be necessary to 
build up the expression for the temperature u within the sphere 
from terms of the type r m P m (cos 0), where m is a positive integer. 

Consider the infinite series 

00 

(114-5) u = 2 A m rP m (cos 0), 

each term of which satisfies (114-2). When r = 1, (114-5) 
becomes 

00 

ra-0 



114 PARTIAL DIFFERENTIAL EQUATIONS 385 

and, if it is possible to choose the undetermined constants A m in 
such a way that (114-5) satisfies the boundary conditions (114-3), 
then (114-5) will be the desired solution of the problem. ' Now, 
it was indicated in Sec. 102 that a suitably restricted function 

y = F(x) 

can be expanded in the interval ( 1, 1) in a series of Legendre 
polynomials in the form 



F(x) = 

where the coefficients a m are given by 
(114-6) a m = ^^ J^F(aOP m (z) dx. 

In the problem under consideration, 

u = f(o) = 1 for < 6 < > 

2i 

u = f(o) = for I < 6 < TT, 

so that the problem is equivalent to expanding F(x) as 
F(x) = i A m P m (x), 



where F(x) = for -1 < x < 0, and F(x) = 1 for < x < 1. 

If formula (114-6) is used, it is readily found that the solution of 
(114-2), which satisfies the initial conditions (114-3), is 



+ 5-ri3^(coB) ---- - 

PROBLEMS 

1. Find the steady-state temperature in a circular plate of radius a 
which has one half of its circumference at 0C. and the other half at 

nrc. 

Hint: Use Laplace's equation for the plane in polar coordinates, 

<Pu ,ldu l^d^u 

6r* + r dr + r 2 d6* " U> 

and assume that u = R(r)B(6) as in Sec. 114. Hence, show that the 



386 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 115 

physically possible solution is of the form 

u = a + air cos 6 + a 2 r 2 cos 26 + a 3 r 3 cos 30 + 

+ bir sin 6 + 6 2 r 2 sin 26 + b^ sin 30 + . 

Determine the coefficients a, and 6 t so as to satisfy the boundary 
conditions. 

2. Show that Laplace's equation in cylindrical coordinates is 

d*u I du 1 d z u d*u _ 
dr* + r dr + r 2 d<p* + dz 2 ~ 

and in spherical coordinates is 

d ( z du\ . 1 d f . -du\ . 1 d*u 

~*~ \ r i~ } + ~ Q ~ \ SHI -^ J 4- . o /) ^ 5 = 0. 
dr\ dr / sin 6 00 \ SO / sin 2 6 o<p* 



3. Find the steady-state temperature at any point of a semicircular 
plate of radius a, if the bounding diameter of the plate is kept at the 
temperature 0C. and the circumference is kept at the temperature 
100C. 

Hint: Use Laplace's equation for the plane in polar coordinates, namely, 

d*u 1 du 1 d z u _ 

dr 2 + r dr + r 2 d0 2 ~~ 

4. Outline the solution of the problem of the distribution of tem- 
perature in a long cylinder whose surface is kept at the constant tem- 
perature zero and whose initial temperature in the interior is unity. 

115. Flow of Electricity in a Cable. A simple problem of 
determining the distribution of current and voltage in an elec- 

trical circuit, whose linear dimen- 
----- oc+Ax---->\ . n xi_ x 

. _____ x _.__ > i sions are so small that one can 

- 1 JL - T^ disregard the variation of the e m.f 
I > B '<dwig the circuit, has been discussed 

JE ^ in Sees. 90 and 91. This section 

^ is concerned with the more compli- 

cated problem of the flow of elec- 

tricity in linear conductors (such as telephone wires or submarine 
cables) in which the current may leak to earth. 

Let a long imperfectly insulated cable (Fig. 103) carry an 
electric current whose source is at A. The current is assumed 
to flow to the receiving end at R through the load B and to 
return through the ground. It is assumed that the leaks occur 
along the entire length of the cable because of imperfections in 
the insulating sheath. Let the distance, measured along the 



115 PARTIAL DIFFERENTIAL EQUATIONS 387 

length of the cable, be denoted by x; then both voltage and 
current will depend not only on the time t, but also on the dis- 
tance x. Accordingly, the e.m.f. F (volts) and the current 7 
(amperes) are functions of x and t. The resistance of the cable 
will be denoted by R (ohms per mile) and the conductance from 
sheath to ground by G (mhos per mile). It is known that the 
cable acts as an electrostatic condenser, and the capacitance of the 
cable to ground per unit length is assumed to be C (farads per 
mile); the inductance per mile will be denoted by L (henrys 
per mile). 

Consider an element CD of the cable of length Ax. If the 
e.m f. is F at C and F + AF at D, then the change in voltage 
across the element Ax is produced by the resistance and the 
inductance drops, so that one can write 

AF = - 

The negative sign signifies that the voltage is a decreasing func- 
tion of x. Dividing through by A:r and passing to the limit as 
Ax > gives the equation for the voltage, 

(115-1) y- = -IR - L ! 

The decrease in current, on the other hand, is due to the 
leakage and the action of the cable as a condenser. Hence, the 
drop in current, A7, across the element Ax of the cable is 

A7 = - VG Ax - ~ C Ax. 
ot 

so that 

(115-2) = VG C 

Equations (115-1) and (115-2) are simultaneous partial 
differential equations for the voltage and current. The voltage 
F can be eliminated from these equations by differentiating 
(115-2) with respect to x to obtain 

d 2 7 dV d 2 V 

'dx 2 ~~ "~ ~dx "" dx dt 

Substituting for dV /dx from (115-1) gives 



388 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 116 

527 



from which d 2 V/dx dt can be eliminated by using the expression 
for d*V/dtdx obtained from the differentiation of (115-1). 
Thus, one is led to 



(115-3) -Lc 

A similar calculation yields the equation for F, namely, 

r) 2 F <9 2 F r)F 

(115-4) - LC - - (LG + RC) d -- KGV = 0, 



which is identical in structure with (115-3). 

In general, it is impossible to neglect the capacitance C of the 
cable in practical applications of these equations to problems in 
telephony and telegraphy, but the leakage G and the inductance 
L, normally, are quite small. Neglecting the leakage arid induc- 
tance effects yields the following equations: 

(115-5) | = -IR, 

dl dV 



It is clear from (115-7) and (115-8) that the propagation of 
voltage and current, in this case, is identical with the flow of 
heat in rods. 

In order to give an indication of the use of these equations, 
consider a line / miles in length, and let the voltage at the source 
A, under steady-state conditions, be 12 volt's and at the receiving 
end R be 6 volts. At a certain instant t 0, the receiving end is 
grounded, so that its potential is reduced to zero, but the poten- 
tial at the source is maintained at its constant value of 12 volts. 
The problem is to determine the current and voltage in the line 
subsequent to the grounding of the receiving end. 

It follows that one must solve Eq. (115-8) subject to the follow- 
ing boundary conditions : 



116 PARTIAL DIFFERENTIAL EQUATIONS 389 

V = 12, at x = for all t > 0, 
7 = 0, at x = I for all t > 0. 

In addition, it is necessary to specify the initial condition that 
describes the distribution of voltage in the line at the time t = 0. 
Now, prior to the grounding of the line, the voltage V is a func- 
tion of x alone, so that (115-8) gives 



the solution of which is 

V = c\x + c%. 

Since, prior to grounding, V = 12 at x = and V = 6 at x = Z, 
it follows that c\ = 6/7 and c 2 = 12, so that 

fir 

V = - y + 12 at t = 0. 

Accordingly, it is necessary to find the solution of Eq. (115-8) 
subject to the following initial and boundary conditions: 

F(0, = 12, V(l, t) = 0, 
V(x, 0) = - * + 12. 

^ 

A reference to Sec. 112 shows that the mathematical formula- 
tion of this problem is similar to that of the problem of heat flow 
in a rod, except for the difference in the formulation of the end 
conditions. * Now, the voltage V(x, t) in the line, subsequent to 
the grounding, can be thought of as being made up of a steady- 
state distribution Vs(x) and the transient voltage VT(X, t), which 
decreases rapidly with the time. Thus, 

V(X, t) EEE V S (X) + V T (X, t). 

After the line has been grounded, the voltage at the ends of 
the line must satisfy the following conditions: 

F(0, = 12 and V(l, t} = 0. 

It was noted above that the steady-state distribution of voltage 
is a linear function of x\ and since after the lapse of some time t 
the transient effects will not be felt, it follows that 

V a (x) = - y x + 12. 
* See, however, Prob. 5, Sec. 112. 



390 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 115 

Thus, 

(115-10) V(x, t) = - y x + 12 + V T (x, t). 

The boundary conditions to be satisfied by the transient 
voltage VT(X, t) can now be determined from (115-9). Thus, 

7(0, t) = 12 = 12 + F r (0, 0, 
V(l, 0=0= Fr(Z, 0, 

fir 197* 

F(, 0) = - y + 12 = - i~ + 12 + Fr(*, 0). 

Hence, the function V T (x, t) satisfies the following initial and 
boundary conditions: 

Fr(0, = V T (l, t) = 0, 



Since it is obvious from (115-10) that VT(X, t) satisfies (115-8), 
it becomes clear that the determination of the transient voltage 
VT(X, t) is identical with the problem of determining the dis- 
tribution of the temperature in a rod when the initial distribution 
is the linear function fix /I. Referring to the solution (112-5) 
and setting a 2 = \/(RC) give 

(,\ ^^_ i / & I \j flTTX 7 \ T77H ~~i~ I ' TlTTX 

^ = >, I 7 I 7 x sm ~~T dx ) e ^ J sm -T 



A x?/ 2 r n * x j\ 

X, t) = ^ ( j I j x sin -y- dx 1 



Therefore, the problem of determining the distribution of voltage 
is solved. 

The magnitude of the current in the line is obtained from 
(115-5). It is left as an exercise for the reader to calculate 
the expression for the current /. It is easy to see that the term- 
by-term differentiation of the series for VT(X, t) is valid for all 
values of t > 0. 

From the discussion of this problem, it is clear that the deter- 
mination of the temperature of a rod whose ends are kept at 
different fixed temperatures and whose initial temperature is a 
function of the distance along the rod can be effected in a similar 
way. 

PROBLEMS 

1. On the assumption that the length I of the line in Sec. 115 is 120 
miles, R 0.1 ohm per mile, and (7 = 2- 10~ 8 farad per mile, find the 



COMPLEX VARIABLE 441 

called the plane of a complex variable, the x-axis is called the 
real axis, and the 2/-axis is called the imaginary axis. 
If v vanishes, then 

z u + - i = u 

is a number corresponding to some point on the real axis. Accord- 
ingly, this mode of representation of complex numbers (due to 
Gauss and Argand) includes as a special case the usual way of 
representing real numbers on the number axis. 
The equality of two complex numbers, 

a + ib = c + id, 

is interpreted to be equivalent to the two equations 
a c and b = d. 

In particular, a + ib = is true if, and only if , a = and 6 = 0. 
If the polar coordinates of the point (u, v) (Fig. 123) are 
(r, 6), then 

u = r cos and v = r sin 
so that 






r = \/u 2 + *> 2 and = tan" 1 - 

11 

The number r is called the modulus, or absolute value, and 6 
is called the argument, or amplitude, of the complex number 
z = u + iv. It is clear that the argument of a complex number 
is not unique ; and if one writes it as 6 + 2&?r, where TT < 6 ^ TT 
and fc = 0, 1, 2, , then 6 is called the principal argu- 
ment of z. The modulus of the complex number z is frequently 
denoted by using absolute value signs, so that 



r = \z\ = \u + iv\ = v^ 2 + ^ 2 > 
and the argument 6 is denoted by the symbol 

6 = arg z. 

The student is assumed to be familiar with the fundamental 
algebraic operations on complex numbers, and these will not be, 
entered upon in detail here. It should be recalled that 

21 + 22 = (xi + iyi) + (x 2 + iy*) = (xi + z 2 ) + i(y\ + 2/2), 
21 2 2 = G&i + iy\) - (x 2 




21 i i _ i 2 i2 , z - 



22 z 2 + **2/2 ^1 +2/1 ^i + 2/1 

provided that j^l = 



442 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 133 

It follows from the polar mode of representation that 

z\ * 82 = ri(cos 0i + i sin 0i) r 2 (cos 2 + i sin 2 ) 
= rir 2 [cos (0i + 2 ) + i sin (0i + 2 )]; 

that is, the modulus of the product is equal to the product of the 
moduli and the argument of the product is equal to the sum of the 
arguments. Moreover, 

zi ri (cos 0i + i sin 0i) 

- ~ ~ 



i r , 
= ~~ L cos (0i "" 



^ sm 



/> 
2 



7 - 2 i ~ n~\ ~~ 
z 2 r 2 (cos 2 + 1 sin 2 ) r 2 

so that ^e modulus of the quotient is the quotient of the moduli and 
the argument of the quotient is obtained by subtracting the argument 




FIG. 124. 



of the denominator from that of the numerator. If n is a positive 
integer, one obtains the formulas of De Moivre, namely, 



\/z = {r[cos (0 + 2kw) + i sin (0 + 



so that 



+ 2/C7T 

>s 

R(cos <p + i sin < 

- 



+ 2kir\ 
n ) 



(* = 0, 1, 2, 



n-1). 



This last formula can be illustrated by finding the expressions 
for the cube roots of z = 1 i. Since u = 1 and 2; = 1, it 

follows that r = -\/2 and = tan- 1 ( -y- ) Hence 



133 



COMPLEX VARIABLE 



443 



-f + 2*r 



sin 



Assigning k the values 0, 1, and 2 gives the values of the three 
roots as (Fig. 124) 



- - 



and 



* 1 = ^[ 

6/0 I 7?T 

22 = V2^C08 ^ 

23 = A/2 ( COS 



. . . 
+ i sin 



in (" s)) 

in T I 
i^/ 

inj) 




The following important inequalities will be recalled, for they 
are used frequently in this 
chapter. 

(133-1) \zi + z*\ 

N + N, 

that is, the modulus of the sum 

is less than or equal to the sum 

of the moduli. This follows 

at once upon observing (Fig. 

125) that the sum of two FlG - 125 ' 

sides of the triangle is not less than the third side. 

(133-2) \zi + z 2 | ^ M - M, 

that is, the modulus of the sum is greater than or equal to the differ- 

ence of the moduli. This follows 
from the fact that the length of 
one side of a triangle is not less 
than the difference of the other two 
sides. 

(133-3) |2i| \z%\ 




FIG. 126. 



This follows from Fig. 126. 



PROBLEMS 

1. Find the modulus and argument of 

(a) 1 + i V3, (6) 2 + 2i, 



(c) (1 + 



- i). 



444 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 134 

7T IT. 

2. If z\ = 3e* and z z e , find 3i z 2 and 21/22. Illustrate the 
results graphically. 

3. Under what conditions does one have the following relations? 

(a) \z\ -j- 22! = \z\\ -f- \Zz\t 

(&) |#1 ~f" #2! == |^l| l^l- 

4. Setting 2 = r(cos 6 -\- i sin ^), show, with the aid of the formula 
of De Moivre z n = r n (cos nd + i sin n#), that 

cos 2^ = cos 2 6 sin 2 and sin 20 = 2 sin cos 0. 

6. Find all the fifth roots of unity, and represent them graphically. 

6. Find all the values of \Xl + i and \/i, and represent them 
graphically. 

7. Find all the roots of the equation z n 1 = 0. 

8. Write the following complex numbers in the form a + bi: 

(a) 
(b) 
(c) (1 - V30 1 ; 



9. Express the following functions in the form u + 

1 (c) 2; 2 - z + 1 ; 



(a) 



(6) 



10. The conjugate of a complex number a + ib is defined as a t6. 
Prove that 

(a) The conjugate of the product of two complex numbers is equal 
to the product of the conjugates of the complex numbers. 

(6) The conjugate of the quotient of two complex numbers is equal 
to the quotient of the conjugates of the complex numbers. 

134. Elementary Functions of a Complex Variable. A com- 
plex quantity z = x + iy, where x and y are real variables, is 
called a complex variable. If the assignment of values to z 
determines corresponding values of some expression /(z), then 
f(z) is said to be & function of the complex variable z. For example, 
if 



134 COMPLEX VARIABLE 445 

the values of f(z) can be determined by recalling that, if z = x 
+ iy y then 



So long as the functions under consideration involve only the 
operations of addition, subtraction, multiplication, division, and 
root extraction, the discussion of Sec. 133 provides methods of 
determining the values of these functions when arbitrary values 
are assigned to z Thus, if /(z) is any rational function of z, 
that is, the quotient of two polynomials so that 

ft \ - apz" + a\z n ~ l + * + an 
J(Z) ~ ~"'- 1 7 



there is no difficulty in ascertaining its values. The discussion 
permits one to ascribe a meaning even to such an expression as 

1 * 



i V^-zr 

For, if z x + iy, then 



2ixy 

= [r(cos 9 + i sin 
where 



r = V(x* - y 2 - I) 2 + 4zV and = tan 2 - 

x y 1 

Applying De Moivre's formula gives 

V?'"! = r^[cos H( + 2/C7r) + i sin %(0 + 2fcx)], (fc = 0, 1), 
and therefore, 

19 + 2/C7T 






. . 

* sin -_ , (t 0| 1). 



Matters become somewhat more involved when it is necessary 
to define transcendental functions of z such as 

e z , sin z, log 2, etc. 

It is evident that it is desirable to define these functions so that 
they will include as special cases the corresponding functions 
of the real variable x. It was indicated in Sec. 73 that the series 



446 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 134 

which converges for all real values of x, can be used to define 
the function e z y where z = x + iy, so that 

Kv 2 v* \ 

i ^ I- ... 1 
2! ^ 4! / 

/ y 3 V 5 M 

+ H # ~~ 3J + 51 "" ' ' ' ) = e z (cos y + i sin i/). 

Also, from Sec. 73, 

e v _L 
COS I/ = 



sin y 



2 

e vl 



These formulas lead one to define the trigonometric functions 
for a complex variable z as 

e zl + e~ zl . e zl e~ zl 

cos z ~ -J sm z = 



2 ' ^ A ~ 2i 

sin z , cos z 

tan z = > cot z = > 

cos z sin 2 

1 1 

sec z = > esc z = 



cos z sin z 

It can be easily verified by the reader that these definitions 
permit one to use the usual relations between these functions, 
so that, for example, 

g! Q Z I = e z i + *2, 

sin 2 z + cos 2 2=1, 

sin (zi + #2) = sin z\ cos #2 + cos z\ sin 2 2 . 

The logarithm of a complex number z is defined in the same 
way as in the real variable analysis. Thus, if 

w = log 2, 
then ~~~~ "* 

z = e". 
Setting w = u + iv gives 

2 6 u+it> = e u (cos v + i sin v). 
On the other hand, z can be written as 

z = x + iy = r(cos 6 + i sin 0). 



134 COMPLEX VARIABLE 447 

Therefore, 

r(cos 6 + i sin 0) == e w (cos v + i sin v), 
which gives 

e = r, t; = + 2&7r, (fc = 0, 1, 2, ). 
Then, since u and r are real, u = log r, so that 
(134-1) w = u + iv = log z = log r + (0 + 2kir)i. 

Hence, the logarithm of a complex number has infinitely many 
values, corresponding to the different choices of the argument of 
the complex number. Setting k = 0, one obtains the principal 
argument of log z, if it is assumed that TT < 6 < TT. 

It is obvious that (134-1) provides a suitable definition of 
log z for all values of z with the exception of z = 0, for which 
log z is undefined. 

The definition (134-1) permits one to interpret the complex 
power w of a complex variable z by means of the formula 

gW __ gW log Z 

and since log z is an infinitely many-valued function, it follows 
that, in general, z w likewise has infinitely many values. 

PROBLEMS 

1. Verify the formulas 

(a) e*i e z * = e z i + **; 

(b) sin 2 z + cos 2 z = 1 ; 

-. (c) cos (zi + #2) = cos zi cos Zz sin z\ sin 3 2 ; 
/0(d) cos iz cosh z\ 

A (e) sin iz = i sinh z: 

"j } '* 

2. Represent graphically the complex numbers defined by the 
following: 

(a) logi; (d) i 1 ', 

(b) log(-l); _ (e) e*<. 

(c) log (I - V3i); 

3. Show that 

, . t I e*< - I 

(a) tan-T ?ir?1 ; 

* -^1 

(6) cot 2 = t 



448 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 135 

4. Express tan z in the form u + w. 
6. Express sin z in the form u + iv. 

6. If a and 6 are real integers, show that 

( re 0t)a+&t - r a e- be [cos (ad + b log r) + i sin (ad + b log r)]. 

7. Write in the form r(cos 6 + i sin 0) 

(a) (1 + 0*; (d) Is 

(b) (1 - O 1 -; W 2 1+ . 

(c) t' 1 -'); 

136. Properties of Functions of a Complex Variable. Let 

w = /(z) denote some functional relationship connecting w with 
z. If z is replaced by x + iy, w can be written as 

w = f(x + iy) = w(x, y) + iv(x, y), 

where u(x, y) and v(x, y) are real functions of the variables 
x and y. As an example, one may consider the simple function 

w = z 2 = (x + iy) 2 = x 2 y z + 2ixy. 

If x and ?/ are allowed to approach the values X Q and y , respec- 
tively, then it is said that the complex variable z = x + iy 
approaches z = XQ + iyo. Thus the statement 

z z , or a: + iy > o: + iyo, 
is equivalent to the two statements 

x x and y y - 

Since /(^ ) is, in general, a complex number, one extends the 
definition of continuity in the following way: The function f(z) 
is said to be continuous at the point z = z provided that 

(135-1) /(z) /(ZQ) when z > z . 

Since /(z) = u(x, y) + iv(x, y) and /(z ) = U(X Q , y Q ) + iv(x , 2/0), 
the statement (135-1) implies the continuity of the functions 
u(x, y) and v(x, y). If the function /(z) is continuous at every 
point of some region R in the z-plane, then /(z) is said to be 
continuous in the region R. 

The complex quantities z and w can be represented on separate 
complex planes, which will be called the z-plane and the w-plane, 
respectively. Thus the functional relationship w = /(z) sets 
up a correspondence between the points (x, y) of the z-plane and 
the points (u, v) of the w-plane (Figs. 127 and 128). 



135 



COMPLEX VARIABLE 



449 



If the variable z = x + iy acquires an increment Az, then 

(Fig. 127) 

z + Az = (x + Aor) + i(y + Ay) 
and 

Az Ax + iky. 

The change in w = /(z), which corresponds to the change Az in 
z, can be denoted by Aw (Fig. 128), and one defines the derivative 
of w with respect to z to be the f unction /'(z) such that 

/(z + Az) - /(z) 



(135-2) 



/'(z) = lim 

Az->0 



Az 



where the limit must exist and be independent of the mode of 
approach of Az to zero. 

v i 
z- plane w- plane 



w+Aw 





FIG. 127. 



FIG. 128. 



It should be noted that this requirement, that the limit of 
the difference quotient have the same value no matter how Az 
is allowed to approach zero, narrows down greatly the class of 
functions of a complex variable that possess derivatives. Thus, 
consider the point P in the z-plane that corresponds to z = x + iy, 
and let Q be determined by z + Az = (x + Az) + i(y + A/). 
In allowing the point Q to approach P, one can choose any one 
of infinitely many paths joining Q with P, and the definition 
(135-2) demands that the limit /'(z) be the same regardless of 
which one of the paths is chosen. 

, -Let it be assumed for the moment that w = /(z) has a unique* 
derivative at the point P; then 

* It is assumed throughout that we are concerned with single-valued 
functions; hence, the discussion of the derivatives of such functions as 
Vl 2, for example, is restricted to a study of one of the branches of the 
function. 



450 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 135 



(135-3) /'(z) = lim ^ " 2z ' 

where 

f(z) = u(x, y) + iv(x, y). 

If Q is to approach P along a straight line parallel to the z-axis, 
then Ay = 0, kz = Az, and 

dz ~dx dx dx 

On the other hand, if Q approaches P along a line parallel to the 
y-axis, then Ax = 0, Az = i Ay, and 



dz 



Since the derivative is assumed to exist, (135-4) and (135-5) 
require that the functions u(x, y) and v(x, y) satisfy the conditions 

n . ^ du dv dv du 

(IdO-O) = ) 

7 dx ^y 



These are known as the Cauchy-Riemann differential equations, 
and the foregoing discussion proves the necessity ol the condi- 
tion (135-6) if f(z) = u(x, y) + iv(x, y) is to possess a unique 
derivative. 

In order to show that the conditions (135-6) are sufficient for 
the existence of the unique derivative/' (2), one must suppose that 
the functions u(x, y) and v(x, y) possess continuous partial 
derivatives. * 

It is not difficult to show that the usual formulas for the differ- 
entiation of the elementary functions of a real variable remain 
valid, so that, for example, 

dz n n . de* d sin z 

= nz n ~~ l , -j- = c*, -5 = cos z, etc. 

dz ' dz ' dz ' 

As an illustration of the application of the formulas (135-4) anvi 
(135-5), consider the calculation of the derivative of 

y) = 0* as e x+tv 

or 

w = w + iv = 6 x (cos y + f sin y). 

* Only the existence of these derivatives was required in the proof of the 
necessity. See references at the end of Sec. 141. 



COMPLEX VARIABLE 451 

Here, u e* cos t/, v = e* sin y, and it follows that 

du du 

fa = e* cos i/, g = -e* sin y, 

av . aw 

gj - e- sin 2/, g = cos y. 

Since Eqs. (135-6) are satisfied and the partial derivatives are con- 
tinuous, dw/dz can be calculated with the aid of either (135-4) or (135-5). 
Then, 

dw ... 

-T- = e* cos y + ie* sin y = e*. 

The functions of a complex variable z that possess derivatives 
are called analytic or holomorphic. * A point at which an analytic 
function ceases to have a derivative is called a singular point. 
It is possible to provef that, if f(z) is analytic in some region R 
of the z-plane, then not only the first partial derivatives of u and 
v exist throughout the region R, but also those of all higher orders. 

This last statement leads to an important consequence of Eqs. 
(135-6). Differentiating (135-6) gives 



i 

<W ^ 
and adding gives 

Similarly, one obtains 



Hence, the real and imaginary parts of an analytic function satisfy 
Laplace's equation. 

On the other hand, if a function u(x, y) satisfying Laplace's 
equation is given, one can construct an analytic function f(z) 
whose real part is u. Multiplying the first of Eqs. (135-6) by 
dy and the second by dx and adding give 

, dv , . dv , du , . du j 

dv = dx + T- dy = dx + dy. 
dx dy dy dx 

Then, since du/dy and du/dx are known, 
(135-7) v(x, y) = f ^ ( - ^ dx + g 

J(xo,yo) \ ^2/ OX 

* The term regular is also used. 
f See Sec. 140. 



452 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 135 

where the line integral (135-7) will not be ambiguous if it is 
independent of the path joining some fixed point (X Q , t/ ) to 
the point (x, y). Applying the conditions for the independence 
of the path, * namely, 



A ( - **\ = A (<*y 

fy \ ty) " fa W 



gives 

to" 2 "*" dy* ~ U ' 

which is precisely the condition assumed to be satisfied by u(x, y}. 
Since the line integral (135-7) depends on the choice of the point 
(x , 2/0), it is clear that the function v(x, y) is determined to 
within an arbitrary real C9nstant, and hence the function 
f(z) = u + iv is determined save for a pure imaginary additive 
constant. 

It may be further remarked that the function v(x, y) may 
turn out to be multiple-valued (if the region of integration is not 
simply connected) even though u(x, y) is single-valued. The 
connection of analytic functions with Laplace's equation is one 
of the principal reasons for the great importance of the theory of 
functions of a complex variable in applied mathematics, f 

PROBLEMS 

1. Determine which of the following functions are analytic functions 
of the variable z = x + iy: 

(a) x iy; 

(b) x* ~y* + 2ixy, 

(c) Y 2 log (z 2 + y 2 ) + i tan- 1 (y/x); 



2. Verify the following formulas: 

, . d(cos z) 

(a) - = - sin 2; 



, x z) _ 

(c) dz = sec 2 z: 

K See Sec. 63. 

t See, in this connection, Sees. 66, 111, and 130. 



136 




COMPLEX VARIABLE 
1 



453 



3. Find a function w such that w = u + iv is analytic if 

(a) u = x* - y*] 



(c) t* = x\ 

(d) u = log 

(e) -u = cosh 

4. Prove that 



cos x. 



); 



(a) sinh z = K(e* ~ e ~*) is analytic; 

(6) cos (z + 2/C7T) = cos z, (/c = 0, 1, 2, 

(c) sinh (2 + 2ikir) = sinh 2, (A; = 0, 1, 2, ); 

(d) log 21 z 2 = log 0i + log z 2 ; 

(e) log z a = a log 2, where a is a complex number. 

5. Show how to construct an analytic function /(z) = u(x, y) 
+ iv (x, y) if v(a:, 2/) is given, and construct /(z) if v = 3a? 2 2/ ?/ 3 . 

6. An incompressible fluid flowing over the x7/-plane has the velocity 
potential < = x 1 y-. Find a stream function ^. 

7. Referring to Prob. 6, what is the velocity potential if the stream 
function is 



? 3 ? 



3x' 2 y y 

136. Integration of Complex Functions. 

defined by the parametric equations y 



Let C 



x 



x = 



where <p and ^ are real differentiate 

functions of the real variable t. Con- 

sider a continuous (but not necessarily _ 

analytic) f unction /(), of the complex 

variable z = x + iy, defined at all points FIG. 129. 

of C. Divide the curve C into n parts by inserting the points 

Po, PI, ' ' , Pn-i, Pn, where P coincides with the initial 

point 2o of the curve and P n with the end point z n (Fig. 129). 

Let f t be any point on the arc of the curve joining JK*-i with P t , 

and form the sum 




454 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 136 

The limit* of this sum as n > > in such a way that each element 
of arc P t _iPt approaches zero is called the line integral of f(z) 
along the contour C, that 'is, 

(136-1) (j(z) dz = lim 

*' c n 

The fact that this integral exists follows at once from the 
existence of the real line integrals into which (136-1) can be 
transformed. Indeed, separating f(z) into real and imaginary 
parts as 

/() = u(x, y) + iv(x, y) 

and noting that dz = dx + i dy give 

(136-2) f c f(z) dz = f c (udx-v dy) + i f c (v dx + u dy). 

Thus, the evaluation of the line integral of a complex function 
can be reduced to the evaluation of two line integrals of real 
functions. It follows directly from the properties of real line 
integrals that the integral of the sum of two continuous complex 
functions is equal to the sum of the integrals, that a constant 
can be taken outside the integral sign, and that the reversal of 
the 'direction of integration merely changes the sign of the 
integral. 

It follows from (136-1), upon noting that the modulus of the 
sum is not greater than the sum of the moduli, that 



\f c f(z)dz < f c |/(s) | \dz\. 



If, along C, the modulus of f(z) does not exceed in value some 
positive number M, then 



(136-3) c /(*) dz ^ M c \dz\ = M c \dx + i dy\ = M c ds = ML, 
where L is the length of C. 

* The precise meaning of the symbol lim in (136-1) is the following: Con- 
sider any particular mode of subdivision of the arc into n\ parts and denote 
the maximum value of |z, 2,_i( in this subdivision by Si, and let S Hl stand 

ni 

for S /(f) fe ~ 2t-i). A new sum, corresponding to the subdivision of the 
-l 

arc into n 2 parts, is denoted by *S>n 2 ; and the maximum value of z z,_i| 
in this new subdivision is 5 2 , etc. In this way, one forms a sequence of 
numbers Ul , n2 , , Sn m , in which the numbers n m are assumed to 
increase indefinitely in such a way that the 5 * 0. 



138 COMPLEX VARIABLE 455 

137. Cauchy's Integral Theorem. The discussion of the 
preceding section involved no assumption of the analyticity of 
the function /(z) and is applicable to any continuous complex 
function, such as for example f(z) = z = x iy, in which event 

dz 

If the integral (136-1) is to be independent of the path, then 
it immediately follows from (136-2) that 

du _____ dv_ eto _ du 

~dy ~~ "" Hix fy ~ dx 

Thus, the conditions that the integral of a complex function f(z) 
be independent of the path are precisely the Cauchy-Riemann con- 
ditions; in other words, the function f(z) must be analytic. 

Now, let R be any region of the 2-plane in which f(z) is analytic, 
and let C be a simple closed curve lying entirely within B; then 
it follows from the properties of line integrals that* the following 
important theorem holds: 

CAUCHY'S INTEGRAL THEOREM. // f(z) is analytic within 
and on a simple closed contour C, then f c f(z) dz = 0. 

It should be noted carefully that the theorem has been estab- 
lished essentially with the aid of Green's theorem, which requires 
not only the continuity of the functions u and v but also the 
continuity of the derivatives. Thus, the proof given above 
implies not merely the existence of f'(z) but its continuity as 
well.f It is possible to establish the validity of Cauchy's 
theorem under the sole hypothesis that/' (z) exists and then prove 
that the existence of the first derivative implies the existence 
of derivatives of all orders. Accordingly, the proof given above 
imposes no practical limitation on the applicability of the 
theorem. 

138. Extension of Cauchy's Theorem. In establishing 
Cauchy's theorem in Sec. 137, it was assumed that the curve C 
is a simple closed curve, so that the region bounded by C is 
simply connected. It is easy to extend the theorem of Cauchy 
to multiply connected regions in a manner indicated in Sec. 64. 
Thus, consider a doubly connected region (Fig. 130) bounded 

* See Sec. 63. 

t See (135-4) and (135-5). 



456 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 138 

by the closed contours C\ and C 2 , where C 2 lies entirely within Ci. 
It will be assumed that the function f(z) is analytic in the region 
exterior to C 2 and interior to C\ and analytic on C 2 and Ci. The 
requirement of analyticity on Ci and C 2 implies that the function 
f(z) is analytic in an extended region (indicated by the dotted 
curves KI and Jf 2 ) that contains the curves Ci and C 2 . 

If some point A of the curve Ci is joined with a point B of C 2 
by a crosscut A B, then the region becomes simply connected and 
the theorem of Cauchy is applicable. Integrating in the positive 
direction gives 

(138-1) C f(z)dz+ f f(z)dz + f f(z)dz+ f f(z) dz = 0, 
JAPAO JAB JBQB& J JBA 



where the subscripts on the integrals indicate the directions of 
integration along Ci, the crosscut AB, and C 2 . Since the second 

x . .^ and the fourth integrals in (138-1) 

are calculated over the same path 
in opposite directions, their sum is 
zero and one has 




where the integral along C\ is trav- 
ersed in the counterclockwise direc- 
FlQ ' 13 tion and that along C 2 in the 

clockwise direction. Changing the order of integration in the 
second integral in (138-2) gives* 



This important result can be extended in an obvious way to 
multiply connected regions bounded by several contours, to yield 
the following valuable theorem. 

THEOREM. // the function f(z) is analytic in a multiply con- 
nected region bounded by the exterior contour C and the interior 
contours Ci, C 2 , , C n , then the integral over the exterior contour 
C is equal to the sum of the integrals over the interior contours 
Ci, Ctj , C n - It is assumed, of course, that the integration over 

* See Sec. 64. 



139 



COMPLEX VARIABLE 



457 



all the contours is performed in the same direction and that f(z) is 
analytic on all the contours. 

139. The Fundamental Theorem of Integral Calculus. Let /() 
be analytic in some simply connected region R, and let the curve 
C join two points PO and P of R 
(Fig. 131). The coordinates of P 
and P will be determined by the 
complex numbers 2 and z. Now 
consider the function F(z) defined 
by the formula F(z) = /*,/() dz. 
The function F(z) will not depend 
upon the path j oining ZQ with z so long 
as these points lie entirely within R. 

Forming the difference quotient 
gives 




FIG. 131. 



F(z 



-F(z) _ 



/CO <fo - 



r ^ 

f(z)dz\ 

Jzo J 



In order to avoid the confusion that may occur if the variable z 
appears in the limits and also as the variable of integration, 
denote the latter by f , so that 

F(z + &z) - F(z) = J 
Az ~ A 



(139-1) 



/) rfr 



snce 
Now if 

(139-2) 



'' df = Az. 



1 /z 

lim T- I 

Az-0 A2 J z 



458 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 139 
then it follows from (139-1) that 

dF(z) 



dz 



'/GO- 



In order to prove that (139-2) holds, one merely has to make use 
of (136-3) and note that max |/(f) - f(z)\ -> as Az - 0. 
Any function F\(z) such that 



dz J ^ J 

is called a primitive or an indefinite integral of /(), and it is easy 
to show that if F\(z) and ^2(2) are any two indefinite integrals of 
f(z) then they can differ only by a constant.* Hence, if F\(z) 
is an indefinite integral of f(z), it follows that 



= f*f(z) dz = F l (z) + C. 

*fZO 



In order to evaluate the constant C, set z = 2 ; then, since 
(z) dz = Q,C = Ffa). Thus 



(139-3) F(z) = f*f(z) dz = 

JZQ 



The statement embodied in (139-3) establishes the connection 
between line and indefinite integrals and is called the fundamental 
theorem of integral calculus because of its importance in the 
evaluation of line integrals. It states that the value of the line 
integral of an analytic function is equal to the difference in the 
values of the primitive at the end points of the path of integration. 

As an example consider 

\ iri 

6z fJy pt oie\ __ 1 9 
(*<& t/ o JL ~ i 

This integral can also be evaluated by recalling that 

* Proof: Since Fi'(z) = Fj(z) = /(), it is evident that 

Fi'(z) - F 2 '(z) = d(Fi - F 2 )/dz 9 dG/dz = 0. 

But if dG/dz = 0, this means that G'(z) = ^ + i ~ = ^ - t - 0, so 

dx dx dy dy 

,, . du dv dU dv rt j jj . j j j 

that r = -r- =s: "^~ =: T~ :=I 0> an( l u an( l v do not depend on x and y. 
dx dx dy dy 



139 

Then, 



COMPLEX VARIABLE 



459 



r/*(0,ir) 
e* dz = I (e* cos y + ie* sin y)(dx + i dy) 

/*(0,7r) 

= I (e* cos ydx - e* sin y dy) 

/(o,) 
+ t J (0 (e x sin ydx + e* cos y dy). 



As a more interesting example, consider 



where n is an integer and the integral is evaluated over some curve 
joining z and z. If n 9^ 1, an indefinite integral is 



+ 1 



a) n is analytic throughout the 



For n > 0, the integrand /(z) = 
finite z-plane and hence 

(139-4) F (z - a)" dz = r-r [(z - a)" +1 - (z - a)"+ 1 ]. 
Jzo n -f- i 

If the variable point z is allowed to start from z and move along some 
closed contour C back to z , then 



(*- 



= 0. 



Of course, the latter result could have been obtained directly from 
Cauchy's integral theorem. 

Suppose next that n < 1 and that the path of integration does not 
pass through the point a. If the point a is outside the closed contour 
(7, then the integrand is analytic and it 
follows at once from Cauchy's integral 
theorem that 



(z - a)" dz = 0. 



Suppose now that the point a is within the 

contour C. Delete the point a by enclos- 

ing it in a small circle of radius p, and con- 

sider the simply connected region R shown 

in Fig. 132. Then, so long as n ^ 1 ? 

the single-valued function /(z) = (z a) n is analytic in R and (139-4) 

is applicable to any curve C joining z and z in R. Now if z is allowed 

to approach ZQ, then it follows from the right-hand member of (139-4) 

that 

(z - a)* dz = for n ^ - 1. 




FIG. 132. 



\ c 



460 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 139 



There remains to be investigated the case when n 1. For any 
path C not containing z = a, one obtains 



(139-5) 



dz . 


(z-a) 

z a 


Z Irvrr 


a 


n \ 


_ log 


log _ 


a 
a 


log 


ZQ d 

z a 


+ i arg _ 


a 


log 


ZQ a 


+ i arg (z 


a) i arg (#o ~ 


a). 



Now if the point z starts from z and describes a closed path C in such 
a way that a is within the contour, then the argument of z a changes 
by 2?r, and therefore 

./ /It 

= 2iri. 



a 



If a is outside the contour, then (z a)~ l is analytic within and on C 

and hence the line integral is zero by 
Cauchy's theorem. 

A different mode of evaluating the 
integral 




where n is an integer greater than unity 

>A: and C is a closed contour, will be given 

next. If the point a is outside CV then 
IG * 133 ' the value of the integral is zero by 

Cauchy's theorem. Accordingly, consider the case when a is inside C. 
Draw a circle 7 of radius p about the point a (Fig. 133) and, since the 
integrand is analytic in the region exterior to 7 and interior to C, it 
follows from the theorem of Sec. 138 that 

J c (z - a)~ n dz = J (z - a)~ n dz. 

But z a pe l and dz = ipe di dd on 7, so that 
dz /*27r ipe t& dO 



/* dz fir ipe l dO i r2jr 

I i -r = I ~ 5- = r I 

Jc (z a) n Jo p n e* n9 p n ~ l Jo 



i 

^~i 



= o, 



dd 



\lU7*\. 



This is the same result as that obtained above by a different method. 
The reader should apply the latter method to show that, if a is inside C, 

then J (z a)- 1 dz 



2iri. 



140 COMPLEX VARIABLE 461 

PROBLEMS 

1. Show that J* o zdz = %(z 2 2 2 ) for all paths joining 2 with z. 

2. Evaluate the integral J c (z a)" 1 dz, where C is a simple closed 
curve and a is interior to C, by expressing it as a sum of two real line 
integrals over C. 

Hint: Set z a pe 9 *; then dz = e* l (dp + i pdd). 

3. Evaluate J c z~ 2 dz where the path C is the upper half of the unit 
circle whose center is at the origin. What is the value of this integral 
if the path is the lower half of the circle? 

4. Evaluate J c z- 1 dz, where C is the path of Prob. 3. 

6. Evaluate f c (z 2 - 2z + 1) dz, where C is the circle x 1 + y* = 2. 

6. Discuss the integral J G (z + l)/z 2 dz, where C is a path enclosing 
the origin. 

7. What is the value of the integral J c (1 + 2 2 )" 1 dz, where C is the 
circle x 2 + y 2 = 9? 

8. Discuss Prob. 7 by noting that . , ^ = 2"' ( " -. '' 4^i anc ^ 

evaluating the integrals over the unit circles whose centers are at 
z i and z i. Note the theorem of Sec. 138. 

140. Cauchy's Integral Formula. The remarkable formula 
that is derived in this section permits one to calculate the value 
of an analytic function f(z), at any interior point of the region 
bounded by a simple closed curve C, from the prescribed bound- 
ary values of f(z) on C. 

Let f(z) be analytic throughout the region R enclosed by a 
simple closed curve C and also on the curve C. If a is some 
point interior to the region R (see Fig. 133), then the function 

(140-1) 



z a 

is analytic throughout the region R, with the possible exception 
of the point z = a, where the denominator of (140-1) vanishes. 
If the point a is excluded from the region R by a circle 7 of radius 
p and with center at a, then (140-1) is analytic throughout the 
region exterior to 7 and interior to C, and it follows from Cauchy's 
integral theorem that 



f M. 

Jco * - <*> 



f -M-&= f -1^-dz. 
Jco * - a JyO z - a 



462 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 140 

or 

(l4 0-2) 

The integral in the right-hand member of (140-2) can be 
written as 

(140-3) f I- d* - f M^ dz + /(a) f J*L. 
JT* - a J Y 2 - a J 7 * - a 

It was demonstrated in Sec. 139 that 



f dz 
I -- = 
J 7 z - a 



. and it will be shown next that the first integral in the right-hand 
member of (140-3) has the value zero. Set z a = pe Bl ; 
then, so long as z is on 7, dz = ipe ld dO, 
and hence 



f f(z) ~ /( ^ <fe = i f 

Jy Z a Jy 



(140-4) ~ <fe = i [/(*) - /(a)] cW. 

Jy Z a Jy 

If the maximum of \f(z) /(a) | is denoted by M, then it follows 
from Sec. 136 that 



(140-5) 



z a 



2* 



o 



Now if the circle 7 is made sufficiently small, it follows from 
the continuity of f(z) that \f(z) f(a) \ can be made as small as 
desired. On the other hand, it follows from (138-3) that the 
value of the integral (140-4) is independent of the radius p of 
the circle 7, so long as 7 is interior to R. Thus the left-hand 
member of (140-5) is independent of p; and since M *0 when 
p > 0, it follows that the value of the integral is zero. 

Accordingly, (140-2) becomes 



(140-6) -p^ = 2nf(a), 

c/C ^ a 

where a, which plays the role of a parameter, is any point interior 
to C. Denote the variable of integration in (140-6) by f, and 
let z be any point interior to C; then (140-6) can be written as 

* - 



140 COMPLEX VARIABLE 463 

The relationship stated by (140-7) is known as Cauchy's integral 
formula. 

It is not difficult to show that an integral of the form (140-7) 
can be differentiated with respect to the parameter z as many 
times as desired, * so that 

jw-. - /(r) 
(140-8) """' ~ 



In fact, if /() is any continuous (not necessarily analytic) function 
of the complex variable z, then the integral 



if? 



defines an analytic function F(2). To show this, all that is 
necessary is to form the difference quotient [F(z + Az) 
F(z)]/&z and to evaluate its limit as Az > 0. It follows from 
such a calculation that 



The assertion made in Sec. 137, concerning the fact that the 
continuity of the derivative of an analytic function follows from 
the assumption that the derivative exists, is now made clear. 

PROBLEMS 
1. If 



** 



where C is the circle of radius 2 about the origin, find the values of 



2. Apply Cauchy's integral formula to Prob. 7, Sec. 139. Use the 
integrand in the form given in Prob. 8. 

* Form the difference quotient [f(z + &z) /(2)]/Az, and investigate the 
behavior of the quotient as Az 0. 



464 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 141 
3. Evaluate with the aid of Cauchy's integral formula 



where C is the circle |f | = 2. 

4. What is the value of the integral of Prob. 3 when evaluated over 

the circle |f 1| = 1? 
5. Evaluate 

f- 2z - 1 J 
*, 




Sc 



where C is the circle \z\ = 1. 

141. Taylor's Expansion. Let/(z) 
be analytic in some region R, and 
let C be a circle lying wholly in R 
and having its center at a. If z is 

any point interior to C (Fig. 134), then it follows from Cauchy's 

integral formula that 



FIQ. 134. 



(141-1) 



/(*) 



= _! f 

2*i JC f - 



. 



But 




and substi- 



1 - "'"'*' ' " ' 1 - ' 

tuting this expression, with i = (z a)/(f a), in (141-1) 
leads to 



where 



tori 



Making use of (140-8) gives 
(141-2) /( Z )=/(a)+/'(a)(2-a) 



(f-a)-(r -)"> 

f "(a) , ., 

-77T ( 2 - ) 



142 COMPLEX VARIABLE 465 

By taking n sufficiently large, the modulus of R n may be made 
as small as desired. In order to show this fact, let the maximum 
value attained by the modulus of /() on C be M, the radius of 
the circle C be r, and the modulus of z a be p. Then |f z\ 
> r p, and 



r 

Je 



e r ~ 



p^ M27rr = Mr /pV 

27r r(r - p) " r - p \r/ 



Since p/r < 1, it follows that lim \R n \ = for every z interior 

n > oo 

to C. 

Thus, one can write the infinite series 

/(z) = /(a) +/'(a)(* - a) + ( 2 - o) + 



which converges to f(z) at every point z interior to any circle 
C that lies entirely within the region R in which f(z) is analytic. 
This series is known as the Taylor's series.* 

PROBLEMS 

1. Obtain the Taylor's series expansions, about 2 = 0, for the follow- 
ing functions : 

(a) e s , (b) sin z, (c) cos z, (d) log (1 + i). 

2. Verify the following expansions: 

(a) tan z = z + - + -r=+ ; 

(b) sinh 3 = * + fj + f]+'''; 

(c) cosh z = l+oi + T-f+*-'; 



142. Conformal Mapping. It was mentioned in Sec. 135 that 
the functional relationship w = f(z) sets up a correspondence 

* For a more extensive treatment, see D. R. Curtiss, Analytic Functions 
of a Complex Variable; E. J. Townsend, Functions of a Complex Variable, 
H. Burkhardt and S. E. Rasor, Theory of Functions of a Complex Variable; 



466 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 142 



between the points z = x + iy, of the complex z-plane, and 
w = u + iv, of the complex w-plane. If w = /(z) is analytic in 
some region R of the 2-plane, then the totality of values w belongs 
to some region R f of the w-plane, and it is said that the region R 
maps into the region R 1 . If C is some curve drawn in the region 
R and the point z is allowed to move along C, then the corre- 
sponding point w will trace a curve C' in the w-plane (Fig. 135), 
and C" is called the map of the curve C. 

The relationship of the curves C and C' is interesting. Con- 
sider a pair of points z and z + &z on C, and let the arc length 
between them be As. = PQ. The corresponding points in the 



w+Aw 




w 




FIG. 135. 

region R' are denoted by w and w + Aw, and the arc length 
between them by As 7 = P'Q' . Since the ratio of the arc lengths 
has the same limit as the ratio of the lengths of the corresponding 
chords, 

Aw 



= lim 



= lim 



lim A _ A1AXA | A | AXA-l*. . 

The function w = f(z) is assumed to be analytic, so that dw/dz 
has a unique value regardless of the manner in which Az > 0. 
Hence, the transformation causes elements of arc, passing through 
P in any direction, to experience a change in length whose 
magnitude is given by the value of the modulus of dw/dz at P. 
For example, if w = z 3 , then the linear dimensions at the point 
3=1 are stretched threefold, but at the point z = 1 + i they 
are multiplied by 6. 

It will be shown next that the argument of dw/dz determines 
the orientation of the element of arc As' relative to As. The 
argument of the complex number Az is measured by the angle 9 
made by the chord PQ with the #-axis, while arg Aw measures 



143 



COMPLEX VARIABLE 



467 



the corresponding angle 0' between the w-axis and the chord 
P'Q'. -Hence, the difference between the angles 0' and is 
equal to 

A * Aw 

arg Aw arg Az = arg > 

for the difference of the arguments of two complex numbers is 
equal to the argument of their quotient. As A 0, the vectors 
Az and Aw tend to coincidence with the tangents to C at P and 
C" at P', respectively, and hence* arg dw/dz is the angle of 
rotation of the element of arc As' relative to As. It follows 
immediately from this statement that if Ci and C^ are two curves 
which intersect at P at an angle r (Fig. 136), then the corre- 





Fio. 136. 

spending curves C[ and C in the w-planc also intersect at an 
angle r, for the tangents to these curves are rotated through the 
same angle. 

A transformation that preserves angles is called con/ormaZ, and 
thus one can state the following theorem: 

THEOREM. The mapping performed by an analytic function 
f(z) is conformal at all points of the z-plane where f'(z) 7* 0. 

143. Method of Conjugate Functions. The angle-preserving 
property of the transformations by analytic functions has many 
immediate and important physical applications. 

For example, if an incompressible fluid flows over a plane 
with a velocity potential 3>(x, y) (so that v x = d$/dx,v y d$/dy}, 
then it is known that the stream lines will be directed at right 
angles to the equipotential curves <(x, y} = const. Moreover, 
it was shown f that the functions $ and ^f satisfy the Cauchy- 

* Note that this statement assumes that dw/dz 9* at the point P. 

t See Sec. 66. 



468 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 143 



Riemann equations, and hence one can assert that the functions 
< and ^ are the real and imaginary parts, respectively, of some 
analytic f unction /(z), that is, 



/(*) - <*>(*, y) + i*(x, y). 

Now, let w = f(z) = $ + i^, and consider the two families 
of curves in the w-plane defined by 



(143-1) $(#, y} const. and 



(x, y) = const. 



The orthogonality of the curves <i> = const, and ^ = const, in 
the z-plane follows at once from the conf ormal properties of the 
transformation by the analytic function /(z). For <f> = const. 
and ^ = const, represent a net of orthogonal lines (Fig. 137) 



$= const 



- const 




-X 



FIG. 137. 



parallel to the coordinate axes in the w-plane, and they are 
transformed by the analytic function w = $ + i^ into a net of 
orthogonal curves in the z-plane. 

It is obvious then that every analytic f unction /(z) = u(x, y) + 
iv(x, y) furnishes a pair of real functions of the variables 
x and y, namely, u(x, y) and v(x, y), each of which is a solution 
of Laplace's equation. The functions u(x, y) and v(x, y) are 
called conjugate functions, and the method of obtaining solutions 
of Laplace's equation with the aid of analytic functions of a 
complex variable is called the method of conjugate functions. 

Example. The process of obtaining pairs of conjugate functions from 
analytic functions is indicated in the following example. Let 



then, 



w - u + iv = sin z = sin (x + iy) ; 

u + iv = sin x cos iy + cos x sin iy, 

= sin x cosh y + i cos x sinh ?/, 



143 COMPLEX VARIABLE 469 

so that 

u(x, y) = sin x cosh y, 
v( x > y) = cos x sinh y. 

It is not difficult to show that the inverse of an analytic func- 
tion is, in general, analytic. Thus, the solution of the equations 

u = &(x, y) and v = W(x, y) 

for x and y in terms of u and v furnishes one with a pair of 
functions 

x = <p(u, v) and y = \l/(u, v) 

that satisfy Laplace's equation in which u and v are the inde- 
pendent variables. 

The following three sections are devoted to an exposition of 
the method of conjugate functions as it is employed in solving 
important engineering problems * 

PROBLEMS 

1. Discuss the mapping properties of the transformations defined 
by the following functions. Draw the families of curves u = const, 
and v const. 

(a) w = u-\-iv = z-\-a, where a is a constant; 

(b) w = bz, where b is a constant; 

(c) w = bz + a, where a and b are constants; 

(d) w = z 2 ; 

(e) w = l/z. 

2. Obtain pairs of conjugate functions from 

(a) w = cos z-, 

(b) w = e*; 

(c) w = z 3 ; 

(d) w = log z; 

(e) w = l/z. 

* The material contained in Sees. 144 to 146 is extracted from a lecture on 
conformal representation, which was delivered by invitation at the S. P. E. E. 
Summer Session for Teachers of Mathematics to Engineering Students at 
Minneapolis, in September, 1931, by Dr. Warren Weaver, director of the 
Division of Natural Scien es of the Rockefeller Foundation, and formerly 
professor of mathematics at the University of Wisconsin. 

The authors did not feel that they could improve upon the lucidity and 
clarity of Dr. Weaver's exposition of the subject and are grateful for his kind 
permission to make use of the lecture, which was printed in the October, 
1932, issue of the American Mathematical Monthly. 



470 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 144 

144. Problems Solvable by Conjugate Functions. Specific 
examples of the method of conjugate functions will be given later, 
but it may be well to indicate here two general sorts of problems. 
Suppose that an analytic function w = u + iv = f(z) = f(x + ly) 
maps a curve C of the 2-plane (see Fig. 138), whose equation 



w-plom* 



z-plome 




FIG. 138. 



is y = <p(x), onto the entire real axis v = of the w-plane. 
will obviously occur if, and only if, 



This 



v[x, 



s 0. 



Then the function 



(z, y) s= v (x, y) 



clearly is a solution of Laplace's equation that reduces to zero 
on the curve C. In an important class of problems of applied 
mathematics, one requires a solution of Laplace's equation that 
reduces to zero, or some other constant, on some given curve. 
Thus, one may, so to speak, go at such problems backward; 
and, by plotting in the z-plane the curves u(x, y) = const, and 
v(x, y) = const., he finds for what curves C a given analytic 
function solves the above problem. Similarly, one may inter- 
change the roles of u, v and x, y and may plot in the w-plane the 
curves x(u, v) = const, and y(u, v) = const. Thus a properly 
drawn picture of the plane transformation indicates to the 
eye what problems, of this sort, are solved by a given analytic 
function. It must be emphasized that the picture must be 
" properly drawn"; that is, one requires, in one plane, the 



146 COMPLEX VARIABLE 471 

two families of curves obtained by setting equal to various con- 
stants the coordinate variables of the other plane. 

In a second and more general sort of problem, it is necessary 
to obtain a solution <(#, y) of Laplace's equation which, on a 
given curve C whose equation is y <p(x), reduces to some given 
function <>*(#, y}. The previous problem is clearly a very special 
case of this second problem. Suppose, now, that an analytic 
function w = f(z) map the curve C of the 2-plane onto the axis 
of reals v = 0, of the w-plane. Since the curve C maps onto 
v = in the w-plane, v[x, <p(x)] = 0, and the values of <* at 
points on C are equal to the values of 

**l*(u, 0), y(u, 0)] sfc.dO 

at the corresponding points on the transformed curve v 0. 
Suppose now that the function ^(w, v) be a solution of Laplace's 
equation (u and v being viewed as independent variables), such 
that 

y(u, 0) ss $>*(u). 
It is easily checked that 

&(x, y} = V[u(x, y), v(x, y}} 

is a solution of Laplace's equation, x and y being viewed as 
independent variables. Moreover,' on the curve C one has 

*[*, v(x)] = *(w, 0) = **(u) = **(*, y), 

so that $ is the solution sought. 

The chief service, in this case, of the method of conjugate 
functions, is that the form of the boundary condition is much 
simplified. Rather than seeking a function that takes on 
prescribed values on some curve C, one has rather to find a func- 
tion that takes on prescribed values on a straight line, namely, 
the axis of abscissas. This latter problem is so much simpler 
than the former that it can, indeed, be solved in general form for 
a very general function <!>,. This solution will be referred to 
later, in Sec. 146c. 

145. Examples of Conformal Maps. As a preparation for the 
consideration of applications, this section will present six specific 
instances of the conformal mapping of one plane on another. 
The examples chosen are not precisely those which one would 
select if, building up from the simplest cases, one were to study 



472 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 145 

the mathematical theory in detail. The examples are chosen 
for their characteristic features and because of their important 
and direct applications. The first case is: 

a. The Transformation w = z m j m a Positive Integer. If one 
write both z and w in polar form, so that 



then 



and 



z = re 1 *, 
w = Re 1 *, 

w == Re 1 * = z m = r m e im<p , 
R = r, 



<$> = m<p. 

Thus the curve r = const, in the 2-plane (that is, a circle about 
the origin) transforms into a curve R = const, in the w-plane 




FIG. 139. 

(also a circle about the origin), the radius of the circle in the 
w-plane being equal to the rath power of the radius of the circle 
in the 2-plane. Also, a radial line <p = const, in the z-plane 
transforms into a new radial line 3> = const., the amplitude angle 
for the transformed radial line being ra times the amplitude 
angle of the original radial line. Thus, a sector of the 2-plane 
of central angle 27r/ra is "fanned out" to cover the entire w-plane, 
this sector also being stretched radially (see Fig. 139, drawn for 
ra = 3). One notes the characteristic feature that a set of 
orthogonal curves in one plane transform into a set of orthogonal 
curves in the other plane. 



146 COMPLEX VARIABLE 473 

This example suggests several interesting questions which 
cannot be discussed here. The " angle-true " property clearly 
does not hold at the origin, which indicates that this point 
deserves special study. Further, it is clear that only a portion 
of the 2-plane maps onto the entire w-plane. In the case for 
which the figure is drawn, it would require three w-planes, so to 
speak, if the entire 2-plane were to be unambiguously mapped. 
This consideration leads to the use of many-sheeted surfaces, 
called Riemann surfaces. Such questions and apparent diffi- 
culties correctly indicate that a thorough knowledge of the mathe- 
matical theory of analytical functions is essential to a proper 
and complete understanding of even simple instances of conformal 
representation. 1 

To get a clear idea of the way in which the 2-plane maps onto 
the w-plane, one may choose various convenient families of curves 
in one plane and determine the corresponding curves in the other 
plane. The resulting picture, as was mentioned earlier, does not 
give any indication of the immediate physical applications of the 
transformation in question unless one of the sets of curves, in 
one plane or the other, consists of the straight lines parallel to the 
coordinate axes. It should thus be clear that Fig. 139 does not 
give a direct indication of the type of problem immediately 
solvable by the transformation w = z 3 . The curves in the 
w-plane obtained by setting x const, and y = const, are, in 
fact, cubic curves; and no simple physical problem is directly 
solved by this transformation. This transformation may, how- 
ever, be used to solve various physical problems for a wedge- 
shaped region, since the bounding curve C of such a wedge 
(say the line <p = and the line <p = 7r/3) is transformed into 
a curve C' of the w-plane that consists of the entire real axis. 
Thus the transformation can be used, in the way indicated in 
Sec. 144, to solve problems in which one desires a solution of 
Laplace's equation that reduces to a given function (or a con- 
stant) on the boundary of a wedge. 

H) 

6. The Transformation w = - ~ ~* This again is a trans- 

1 BIEBERBACH, L., Einfuhrung in die konforme Abbildung, Berlin, 1927; 
LEWENT, L., Conformal Representation, London, 1925; OSGOOD, W. F., 
Lehrbuch der Funktionentheorie, vol. 1, Chap. XIV. 



474 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 146 

formation that does not have immediate applicability. It has, 
however, interesting features, and subsequent discussion will 
indicate how it may be made to serve a practical purpose. 





FIG. 140. 

If, as before, one write z in the polar form re 1 *, then 



w = u + iv = 



z) 



"27 



so that 



1 ( JL l \ _L ' ! ( l \ ' 

= 2 V V C S * + Z 2 V ~ V Sm ^ 

I/ i A 

W = 2 V r/ C S 9| 



Thus, ^ and r being eliminated in turn, 



cos 2 <p sin 



2 



2 ~ 4' 



= 1. 



From these equations, it follows by inspection that the circles 
r = const, of the 2-plane transform into a family of ellipses of 
the w-plane (see Fig. 140), the ellipses being confocal, since 



146 COMPLEX VARIABLE 475 

V + r) ~ V ~~ r) " 4 = 
It is also clear that two circles of reciprocal radii transform into 
the same ellipse. Similarly, the radial lines <p = const, of the 
z-plane transform into a family of hyperbolas which, again, arc 
confocal, since 

cos 2 <f> + sin 2 <? = 1 = const. 

Thus the exterior of the unit circle of the z-plane transforms 
into the entire w-plane. The unit circle itself "flattens out" 
to form the segment from 1 to + 1 of the real axis of the w-plane. 
All larger circles are less strenuously "flattened out" and form 
ellipses, while the radial lines of the z-plane form the associated 
confocal hyperbolas of the w-plane. A similar statement can be 
made for the inside of the unit circle. 

c. The Transformation w = e z . If one set w ss Re 1 * and 
z = x + iy, then 

Re 1 * = e* + * v = e x - e lv , 
so that 

7? = e x , 
$ = y. 

It is thus clear that vertical lines of the z-plane map into circles 
of the w-plane, the radius being greater or less than 1, accord- 
ing as # is positive or negative. Horizontal lines of the z-plane, 
on the other hand, map into the radial lines of the w-plane, and 
it is clear that any horizontal strip of the z-plane of height 2ir 
will cover the entire w-plane once (see Fig. 141). 

The curves in the w-plane of Fig. 141 are drawn by setting 
equal to a constant one or the other of the coordinates of the 
z-plane. Thus these curves give direct indication of physical 
problems to which this analytic function may be applied. For 
example, one could obtain the electrostatic field due to a charged 
right circular cylinder, the lines of flow from a single line source 
of current or liquid, the circulation of a liquid around a cylindrical 
obstacle, etc. 

By considering this example in conjunction with the preceding 
example, one gives new significance to Fig. 140. In fact, if one 
starts with the z-plane of Fig. 141 and then uses the w-plane of 
Fig. 141 as the z-plane of Fig. 140, it is clear that the curves 
drawn in the w-plane of Fig. 140 then are obtainable by setting 



476 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 145 

equal to various constants the coordinates of the 2-plane of 
Fig. 141. That is to say, the w-plane curves of Fig. 140 give 
direct evidence of physical problems that can be solved by the 
pair of transformations 



d. The Transformation w = cosh z. If, in the two preceding 
equations, one eliminates the intermediate variable z\ (so he may 




' _J_L_ I L- 



w-plane 



Z-plane 



Fia. 141 

pass directly from the 2-plane of Fig. 141 to the w-plane of Fig. 
140), the result is 

J_ e ~ z 



w = 



= cosh z. 



Thus 



u -J- iv = cosh (x + iy) = cosh x cosh iy + sinh x sinh iy f 

= cosh x cos y + i sinh x sin y, 
so that 

u = cosh x cos y 
v = sinh x sin y, 
or 

^.2 ..2 

= 1, 



cosh 2 # sinh 2 x 
u 2 v* 



cos 2 



sin 2 y 



= 1. 



145 



COMPLEX VARIABLE 



477 



This transformation is shown in Fig. 142, and it may be used to 
obtain the electrostatic field due to an elliptic cylinder, the 
electrostatic field due to a charged plane from which a strip has 
been removed, the circulation of liquid around an elliptical 
cylinder, the flow of liquid through a slit in a plane, etc. 

The transformation from the z-plane to the w-plane may be 
described geometrically as follows: Consider the horizontal strip 
of the 2-plane between the lines y = and y = TT; and think 
of these lines as being broken and pivoted at the points where 
x = 0. Rotate the strip 90 counterclockwise, and at the same 



2 -plane 




FIG. 142. 



time fold each of the broken lines y = and y = ir back on itself, 
the strip thus being doubly "fanned out" so as to cover the 
entire w-plane. 

e. The Transformation w = z + e z . One has 



so that 



u + iv = x + iy + e x+lv , 

= x + iy + e x (cos y + i sin y), 

u = x + e x cos y, 
v = y + e x sin y. 



This transformation is shown in Fig. 143. If one considers the 
portion of the z-plane between the lines y = TT, then the portion 
of the strip to the right of x = 1 is to be "fanned out" by 
rotating the portion of y = +1 (to the right of x = 1) counter- 
clockwise and the portion of y = 1 (to the right of x = 1) 
clockwise until each line is folded back on itself. This trans- 



478 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 145 



formation gives the electrostatic field at the edge of a parallel 
plate condenser, the flow of liquid out of a channel into an open 
sea, etc. 

/. The Schwartz Transformation. The transformations just 
considered are simple examples and are necessarily very special 
in character. . This list of illustrations will be concluded by a 




FIG. 143. 





*Z *3 



x-p1one 



*4 5 



w-plome 



Fio. 144 

more general transformation. Suppose one has (see Fig. 144) a 
rectilinear polygon, in the w-plane, whose sides change direction 
by an angle aw when one passes the ith vertex, going around the 
boundary of the polygon so that the interior lies to the left. 
The interior of this polygon can be mapped onto the upper 
half 3-plane by the transformation 



w 



dz 



<- 



146 COMPLEX VARIABLE 479 

where i, s*, ' , Sn are the (real) points, on the #-axis of the 
2-plane, onto which map the first, second, , nth vertex 
of the polygon, and where A and B are constants which are to be 
determined to fit the scale and location of the polygon. Three 
of the points z l may be chosen at will, and the values of the 
remaining ones may be calculated. 

This theorem may be used to find, for example, the analytic 
transformation that solves the problem of determining the 
electrostatic field around a charged cylindrical conductor of any 
polygonal cross section. It should be noted, however, that one 
requires for this purpose the function y(u, v), whereas the 
theorem gives one w as a function of z. It is often exceedingly 
difficult and laborious to solve this relation for z as a function of 
w, so that one may obtain the function y. It should further be 
remarked that this theorem may be applied to polygons some 
of whose vertices are not located in the finite plane and that 
the theorem is of wide applicability and importance in connec- 
tions less direct and simple than the one just mentioned. 

146. Applications of Conformal Representation, a. Applica- 
tions to Cartography. It is natural that a mathematical theory 
which discusses the "mapping'' of one plane on another should 
have application to the problems connected with the drawing of 
geographic maps. Since the surface of a sphere cannot be made 
plane without distortion of some sort, one has to decide, when 
mapping a portion of the sphere on a plane, what type of distor- 
tion to choose and what to avoid. For some purposes, it is 
essential that areas be represented properly; for other purposes, 
it is most important that the angles on the map faithfully repre- 
sent the actual angles on the sphere. 

The first problem, in conveniently mapping a sphere on a plane, 
is to map the sphere on the plane in some fashion or other and 
then, if this fashion be unsatisfactory, to remap this plane onto a 
second plane. The first problem can be done in a wide variety 
of ways 1 which include, as important examples, stereographic 
projection and Mercator's projection. Both these examples 
are conformal projections, in that they preserve the true values 
of all angles. Having once mapped the sphere on the plane 
(or on a portion of the plane), one may now remap onto a second 

1 The Encyclopaedia Britannica article on maps lists and discusses nearly 
thirty such projections actually used in map making. 



480 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 146 

plane, and it is here that the theory of conformal representation 
finds its application ; for one can determine the analytic function 
that will conformally remap the original map onto a new region 
of any desired shape and size. Not only are all angles preserved 
in this process of conformal remapping, but the distortion in the 
neighborhood of a point is always a pure magnification. Thus 
the shapes of all small objects or regions are preserved. Such 
maps do not give a true representation of areas, and for this rea- 
son many maps are based on compromises between conformal 
transformations and area-preserving transformations. 

b. Applications to Hydrodynamics. When the velocities of all 
particles of a moving liquid lie in planes parallel to one plane 
that we may conveniently choose as the xy-p\&ue and when 
all particles having the same x and y have equal velocities, then 
the motion is said to be two-dimensional. Such cases clearly 
arise if a very thin sheet of liquid is flowing in some manner over 
a plane or if a thick layer of liquid circulates over a plane, there 
being no motion and no variation of motion normal to the plane. 
Let the x- and ^-components of velocity at any point (x, y) be u 
and y, respectively. The motion is said to be irrotational if the 
curl of the velocity vector vanishes. Analytically, this demands 
that 


dy ~~ dx 

whereas physically it states that the angular velocity of an 
infinitesimal portion of the liquid is zero. The equation just 
written assures that 

(u dx + v dy) 

is the perfect differential of some function, say 4>. This function 
is known as the velocity potential, since by a comparison of the 
two equations 

d$ = u dx v dy, 



dx + dy, 
dx dy yy 



it follows that 

/m*i\ 

(146-1) 



Now, if the liquid be incompressible, the amount of it that 
flows into any volume in a , given time must equal the amount 



146 COMPLEX VARIABLE 481 

that flows out. This demand imposes on the components of 
velocity the restriction that 



dx dy ' 

this being known as the equation of continuity. From the last 
two equations, it follows that 

d 2 <l> d 2 <l> 

f. 

d?/ 



_ 4. = v 2 $ = 

r\ 9 I f. n - v ^* ^^ V/. 

2 2 



That is, the velocity potential satisfies Laplace's equation. 

Just as the vanishing of the curl of the velocity demands that 
u dx + v dy be an exact differential, so the equation of continuity 
demands that v dx u dy be an exact differential of some func- 
tion, say ty. That is, 

d^f = v dx u dy, 

d* , , 3V j 
d* = -dx + -dy, 

so that 

(146-2) = * u = - . 

6x dy 

From (146-1) and (146-2), it follows at once that 

d$d* ,d$<M 

dx dx + dy dy 7 

which expresses the geometric fact that the curves $ = const. 
and ^ = const, intersect everywhere orthogonally. It is clear 
from (146-1) that there is no component of velocity in the 
direction of the curves on which <l> is a constant, so that the veloc- 
ity of the liquid is everywhere orthogonal to the equipotential 
curves $ = const. That is, the curves ^ = const, depict 
everywhere the direction of flow. For this reason, ^ is called 
the stream function and the curves ^ = const, are called the 
stream lines. From (146-2) and the vanishing of the curl of 
the velocity, it follows that the stream function ^ is also a solu- 
tion of Laplace's equation. 

Thus, the velocity potential $ and the stream function ^ 
in the case of the irrotational flow of a perfect incompressible 
liquid both satisfy Laplace's equation, and the curves & = const. 



482 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 146 

and ^ = const, form two orthogonal families. Every analytic 
function therefore furnishes the solution to four such problems, 
the four solutions resulting from the fact that one may choose 
the pair x, y or the pair u, v as independent variables, and that 
one may interchange the roles of the potential function and the 
stream function. Figure 142, for example, indicates two of 
the four problems solved by the analytic transformation w = 
cosh z. If one treats u and v as the independent variables and 
identifies the (solid) curves y(u, v) = const, in the w-plane 
with the curves <f> = const., then the dotted curves x(u, v) = 
ty = const, give the stream lines, and one has solved the prob- 
lem of the circulation of liquid around an elliptic cylinder. If, 
however, one sets y(u, v) = & and x(u, v) = <, then the solid 
curves of the w-plane are the stream lines, and one has solved 
the problem of the flow of liquid through a slit. The other two 
problems solved by this same function are to be obtained by 
drawing, in the 2-plane, the curves u(x, y) = const, and v(x, y) = 
const, and identifying ^ and $ with u and t;, and vice versa. 
The 2-plane curves u = const, and v = const, are very com- 
plicated and do not correspond to any simple or important 
physical problem, and hence they are not drawn on the figure. 
In fact, it is usually the case that only two of the possible four 
problems are sufficiently simple to be of any practical use. 

It should be emphasized that it is never sufficient, in obtaining 
the analytical solution of a definite physical problem, merely 
to know that certain functions satisfy Laplace's equation. 
One must also have certain boundary conditions. The graphs 
shown above disclose to the eye what physical problem has been 
solved precisely because they show what sort of boundary condi- 
tions are satisfied. For example, if the dotted curves of Fig. 142 
are stream lines, then the problem solved is the circulation around 
an elliptical obstacle just because these dotted stream lines 
satisfy the boundary condition for such a problem; namely, 
because the flow at any point on the boundary of the obstacle is 
parallel to the boundary of the obstacle. 

It is interesting to note that this same transformation w = 
cosh z (or, slightly more generally, w = a cosh z) can be used to 
solve a hydrodynamic problem of a different sort. When liquid 
seeps through a porous soil, it is found that the component in 
any direction of the velocity of the liquid is proportional to the 



146 



COMPLEX VARIABLE 



483 



negative pressure gradient in that same direction. Thus, in a 
problem of two-dimensional flow, 



If these values be inserted in the equation of continuity, namely, 
in the equation 



the result is 



0. 



Suppose, then, one considers the problem of the seepage flow 
under a gravity dam which rests on material that permits such 
seepage. One seeks (see Fig. 145) a function p that satisfies 



624/p -=-_= 



P'Po 





FIG. 145. 

Laplace's equation and that satisfies certain boundary conditions 
on the surface of the ground. That is, the pressure must be 
uniform on the surface of the ground upstream from the heel of 
the dam and zero on the surface of the ground downstream 
from the toe of the dam. If we choose a system of cartesian 
coordinates u, v with origin at the midpoint of the base of the 
dam (Fig. 145) and w-axis on the surface of the ground, then 
it is easily checked that p(u, v) = pvy(u, v)/*, where 

w = u + iv = a cosh (x + iy), 



484 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 146 

satisfies the demands of the problem. In fact, it was seen in 
Sec. 145d, where the transformation w = cosh z was studied, 
that the line y = TT of the 2-plane folds up to produce the portion 
to the left of u = 1 of the ^-axis in the w-plane, and the line 
y = of the z-plane folds up to produce the portion to the right 
of u = +1 of the u-axis. The introduction of the factor a 
in the transformation merely makes the width of the base of the 
dam 2a rather than 2. These remarks show that p(u, v) reduces 
to the constant TT on the surface of the ground upstream from the 
heel of the dam. If the head above the dam is such as to produce 
a hydrostatic pressure po, one merely has to set 

P(U, v) = 

71 

One may now easily find the distribution of uplift pressure across 
the base of the dam. In fact, the base of the dam is the repre- 
sentation, in the i^-plane, of the line x = 0, ^ y ^ TT, of the 
:n/-plane. Hence, on the base of the dam the equations 

u = a cosh x cos y, 
v = a sinh x sin y 
reduce to 

u = a cos y, 
v = 0, 
so that 

p(u, 0) = cos" 1 
TT a 

This curve is drawn in the figure. The total uplift pressure 
(per foot of dam) 



/+a 

PO I 

= I 

TT J-a 



17 

cos" 1 - du = 



which is what the uplift pressure would be if the entire base of 
the dam were subjected to a head just one-half the head above the 
dam or if the pressure decreased uniformly (linearly) from the 
static head p Q at the heel to the value zero at the toe. The point 
of application of the resultant uplift is easily calculated to be 
at a' distance b = 3a/4 from the heel of the dam. 

c. Applications to Elasticity. If opposing couples be applied 
to the ends of a right cylinder or prism of homogeneous material, 



146 COMPLEX VARIABLE 485 

the cylinder twists and shearing stresses are developed. Choose 
the axis of the prism for the z-axis of a rectangular system of 
coordinates. The angle of twist per unit length, say r, and the 
shearing stresses, due to an applied couple T, can both be cal- 
culated if one can determine a function <(#, y) satisfying Laplace's 
equation and reducing, on the boundary of a section of the prism, 
to the function 3>* = (x* + y*)/2. In fact, 1 



where 

C = 2(7 JJ ($ - $*) do: dy, 

in which (? is the modulus of rigidity of the material, whereas the 
shearing stresses are given by 

X, = G 



1 



/ \ 

\ / 

Exact analytical solutions of 

this important technical problem 

i. u i_x j r i FIG. 146. 

nave been obtained ior several 

simple sections, notably circular, elliptical, rectangular, and tri- 
angular. 2 Only recently 3 the problem was solved for an infinite 
T section (see Fig. 146). From the general discussion given in 
Sec. 144, it is clear that, to solve this latter problem, one requires 
first an analytic function that will map the boundary of this T 
section onto the entire real axis of the new w-plane. This sec- 
tion, moreover, is a rectilinear polygon, so that one can use the 
Schwartz transformation theory to produce the desired analytic 
relation. One finds that the desired mapping is carried out by 
the function 

1 LOVE, A. E. H., Theory of Elasticity, 3d ed., pp. 315-333, 1920. 

2 TRAYER, W., and H. W. MARCH, The Torsion of Members Having Sec- 
tions Common in Aircraft Construction, Bur. Aeronautics Navy Dept., 
Separate Rept. 334; also contained in Nat. Adv. Comm. Aeronautics, loth 
Ann. Rept., 1929, pp. 675-719. 

3 SOKOLNIKOFF, I. S., On a Solution of Laplace's Equation with an Appli- 
cation to the Torsion Problem for a Polygon with Reentrant Angles, Trans. 
Amer. Math. Soc., vol. 33, pp. 719-732. 



486 MATHEMATICS FOR ENGINEERS AND PHYSICISTS ?U6 

A C (W 2 - 1)*S i_ D 
Z = A I 7 9 v- O^ -H #, 

J (^ 2 - a 2 ) 

d 



. x . 
= - log (u> + 

where the first line is furnished directly by the Schwartz theorem 
and where, in the second line, the constants a, A, B have been 
evaluated so as to fit the dimensions and location of the T 
section. 

It is next necessary to break z up into its real and imaginary parts 
so as to obtain x and y as functions of u and v. These values, 
when substituted into 





give, because v = on the boundary of the section, the function 
**[*(u, 0), y(u, 0)] s *,(*). 

The remaining essential step is to obtain a function ^(u, v) 
satisfying Laplace's equation and reducing, on the axis of 
reals v = 0, to the function $#(u). Such a function is 1 



TT -oo P 2 - 2pcos0+ {*' 
where 

w = u + z'y = pe* e . 

The solution to the original problem is then given, as was earlier 
indicated in S"ec. 144, by < = ^. It is obviously a difficult and 
laborious job to carry out these calculations, but formulas have 
been obtained, in the paper referred to, from which practical 
calculations can be made. 

d. Applications to Electrostatics. The methods of complex 
variable theory are peculiarly applicable to two-dimensional 
electrical problems. In order that the problems be two-dimen- 
sional, we shall understand that the conductors under considera- 
tion are exceedingly long cylinders whose axes are normal to the 
z = x + iy plane. Under these circumstances the various 

1 SOKOLNIKOFF, I. S., On a Solution of Laplace's Equation with an Appli- 
cation to the Torsion Problem for a Polygon with Reentrant Angles, Trans. 
Amer. Math. Soc., vol. 33, pp. 71&-732. This formula is the general solution, 
spoken of in Sec. 144, of Laplace's equation subject to specified boundary 
values on the entire axis of abscissas. 



146 



COMPLEX VARIABLE 



487 



electrical quantities do not change appreciably in the direction 
normal to the 2-plane, and one has to determine these quantities 
as functions of x and y only. In certain problems, one or more 
of the conductors present will have very small cross sections and 
will be given a charge of, say e f per unit length. Such a con- 
ductor will be called a line charge of strength e f . 

The electrostatic problem for such conductors is solved when 
one has obtained a function $(x, ?/), known as the electrostatic 
potential, satisfying the following conditions: 1 



(a) 



V 2 * = ^-^ 4- -- = 
d* 2 + dw 2 



(146-3) 



at all points in free space. 

(6) <1> reduces, on the surface of the kth conductor, 
to a constant 3>fc. 

(c) In the neighborhood of a line charge of strength 
e f , $ becomes infinite as 

e' log r 

2^ ' 

where r measures distance to the line. 

(d) $ behaves at infinity as 

log R2e' 

27~~' 

where Se' is the total charge per unit length of all 
conductors present and where R measures distance 
from some reference point in the finite plane. In 
case Se' = 0, $ approaches zero as 1/R. 

It is readily shown, by standard methods, that the solution of 
such a problem is unique. This remark is of great practical 
importance, since it assures one that a function < satisfying these 
conditions is, however it may have been obtained, the correct 
solution of the physical problem. 

Physically one wishes to know the distribution of charge and 
the electrostatic force at any point. These data may be obtained 

1 See MASON, M., and W. WEAVER, The Electromagnetic Field, pp. 134, 146, 
1929; and REIMANN- WEBER, Die Differentialgleichungen der Physik, vol. 2, p. 
290, 1927. The units used in the above discussion are the rational units 
used in Mason and Weaver, loc. cit. 



488 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 146 

from the function $ in the following manner: The component E a 
in any direction s of the electrostatic force per unit charge is 
given in terms of < by the relation 



and the surface density of charge 77 on any conductor is given by 

77 = ) 

dn 

where n measures distance along the external normal to the 
conductor in question. 
Now if 

w = u + iv = f(z) = f(x + iy) 

and if the function 

$(.?, y) = u(x, y) 

satisfies condition (146-3), then 

jP _ d< du _ dv 

the last step following from the Cauchy-Riemann equations 
(135-6). Similarly 

v dy dy dx 

Thus, 



the last step resulting from the fundamental fact that the value 
of the derivative of an analytic function w is independent of the 
mode in which z approaches zero. Now the complex number 
a ib is called the " conjugate" of the complex number a + ib 
and one often denotes a conjugate by a bar, thus: 



a ib = a + bi. 

With this standard notation, the "complex electric force" 

E SE E x + iE y is given by 

(146-4) E m E, + iE y = - ?, 



146 COMPLEX VARIABLE 489 



and the magnitude \fEl + El of the electrostatic force at any 
point is given by 

*=~ 

If one chooses $ s v(x, y), then (146-4) and (146-5) are 
replaced by 

(146-6) E = E x + iE y = -i ^> 



(146-7) VW+ El = 



dw 
Tz 



Three types of electrostatic problems will now be briefly 
considered. The first and simplest two-dimensional electrostatic 
problem is that of a single long cylindrical conductor with a given 
charge. One then seeks a function that satisfies Laplace's 
equation and, in accordance with (146-3)6, reduces to a constant 
on the curve that bounds a section of the conductor. This is 
the analytical problem whose solution was indicated in Sec. 144. 
One requires a function w = u + iv = f(z) = f(x + iy) such 
that either a vertical straight line u = const, or a horizontal 
straight line v = const, of the w-plane maps into the bounding 
curve C of the conductor's section in the 2-plane. Then <(z, y) 
= u(x, y) or $(x, y) = v(x, y) solves the problem, and the 
physically important quantities are given by (146-4), (146-5) 
or by (146-6), (146-7), respectively. 

Secondly, suppose that a single long cylindrical conductor is 
in the presence of a parallel line charge of strength e f '. We 
suppose the line charge to be outside the conductor. Let C 
be the bounding curve in the 2-plane of a section of the conduc- 
tor, and let the line charge be located at z = ZQ. We may 
conveniently suppose the cylindrical conductor to be grounded, 
so that we seek a solution of Laplace's equation that reduces 
to zero on C and becomes infinite as indicated in (J46-3)c at 
z = z . Let f = f(z) transform C onto the entire axis of reals 
and the exterior of C conformally upon the upper half f -plane. 
Then if 



the function 3> ss u(x, y) is the solution sought. In fact, for 
values of z sufficiently close to 20, f(z) f(zo) behaves, except for 



490 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 146 
a constant factor, as (z ZQ). Thus, if one writes 

then, for values of z very near to 2 , 

e' 1 e' 1 e'0 

10 = 14 -|- iv = log -r -f- A = pr log i p: ~\~ A, 

2?r re ZTT r &TT 

where A remains finite as z = ZQ. Therefore, 



where B remains finite as z = Z Q . Thus u(x y y) has the proper 
type of infinity at z = 2 . Furthermore, for points z on C, 
/() is on the axis of reals in the f -plane, so that the modulus of 
/(z) fM equals the modulus of f(z) /(ZQ). Hence the 
modulus of 



/(>-/(*>) 

is unity. However, since 

log pe'*> = log p + zV, 

it is clear that the real part of the logarithm of a complex quantity 
is the logarithm of the modulus of the complex quantity. Since 
the logarithm of unity is zero, it is clear that u vanishes on C. 
As regards the behavior of u(x, y) at infinity, one notes that u 
is the logarithm of the ratio of the (real) distances of f = f(z) to 
f o = /(ZQ) and to f = /(ZQ). As z becomes infinite, this ratio 
differs from unity by an amount whose leading term is equal to 
or less than a constant times_the reciprocal distance from f(z) 
to one of the points /(z ) or/(2 ). Thus the leading term in the 
logarithm of this ratio is a constant times this reciprocal distance; 
and $ = u behaves at o in the required manner. 

'In the third type of problem there are two conductors present, 
one raised to the potential <f>o while the other is at a potential 
zero. Thus, suppose that the cross section of two long cylindrical 
conductors consists of two curves Co and Ci, such as those shown 
in Fig. 147, which do not intersect at a finite point but which, 
if one takes account of the intersection of BI and B and of AQ 
and A i at z = <, divide the extended plane into two simply 
connected regions, one of which may be called the "interior" 
and the other the "exterior" of the closed curve C + Ci. Now 
suppose that f = f(z} maps Co onto the entire negative axis of 



146 



COMPLEX VARIABLE 



491 



reals in the f -plane, with the infinitely distant point along B Q 
mapped onto f = 0, that f = f(z) also maps C\ onto the entire 
positive axis of reals with the infinitely distant point along 
BI mapped onto f = 0, and that f = /(z) maps the interior of 
Co + Ci conformally on the upper half f -plane. Then, if 

w = u + w = - log/(z), 

7T 

the function $ = v(x, y) satisfies V 2 <l> = at every point in the 
interior of C + Ci, reduces to zero on Ci, and reduces to <f> on C . 

A O 





4" plan 



FIG. 147. 

In fact, the imaginary part of the logarithm of a complex number 
is merely the amplitude of the complex number; and for points 
on Co, f(z) has an amplitude of TT, while for points on Ci, f(z) has 
an amplitude of zero. Then, 

ds 

dw ___ $o dz 

Hz ~~ 7/60' 
and the electrostatic force is given by (146-5) and (146-7). 

This third type of problem is of frequent and important 
practical occurrence. Many electrical engineering problems 
that have been solved by this method of conformal representa- 
tion are referred to in an expository article, devoted largely to 
the Schwarz transformation, by E. Weber. 1 In an earlier article 
in the same journal, 2 for instance, the theory of conformal 
representation is applied to the problem of studying the leakage 
voltage and the breakdown potential between the high- and low- 
potential portions of oil-immersed transformers. The cases 
studied come under the third type of problem discussed above. 

1 WEBER, E., Archiv fur Elektrotechnik, vol. 18, p. 174, 1926. 

2 DREYFUS, L., Archiv fUr Elektrotechnik, vol. 13, p. 123, 1924. 



CHAPTER XI 
PROBABILITY 

There is no branch of mathematics that is more intimately 
connected with everyday experiences than the theory of proba- 
bility. Recent developments in mathematical physics have 
emphasized anew the great importance of this theory in every 
branch of the physical sciences. This chapter sets forth the bare 
outline of those fundamental facts of the theory of probability 
which should form a part of the minimum equipment of every 
student of science. 

147. Fundamental Notions. Asking for the probability or 
for a measure of the happening of any event implies the possi- 
bility of the non-occurrence of this event. Unless there exists 
some ignorance concerning the happening of an event, the 
problem does not belong to the theory of probability. Thus, the 
question "What is the probability that New Year's day in 1984 
will fall on Monday?" is trivial, inasmuch as this question can 
be settled by referring to a calendar. On the other hand, the 
query "What is the probability of drawing the ace of hearts 
from a deck of 52 cards?" constitutes a problem to which the 
theory of probability gives a definite answer. In fact, one can 
reason as follows: Granting that the deck is perfect, one card is 
just as likely to appear as any other and, since there are 52 cards, 
the chance that the ace of hearts will be drawn is 1 out of 52. 
The words "just as likely," used in the preceding sentence, imply 
the existence of the ignorance that is essential to remove any 
problem 'of probability from triviality. The term "equally 
likely," or "just as likely," applied to a future event that can 
happen or fail to happen in a certain number of ways, indicates 
that the possible ways are so related that there is no reason for 
expecting the occurrence of any one of them rather than that of 
any other. 

If an event can happen in N ways, which are equally likely, 
and if, among these N ways, m are favorable, then the probability 

492 



147 PROBABILITY 493 

of the occurrence of the event in a single trial is 

m 

P-y 

Thus, the probability that the six will appear when a die is 
thrown is ^, since the total number of ways in which a die can 
fall is 6, and of these six ways only one is favorable. The proba- 
bility of drawing a heart from a deck of 52 cards is J^, since there 
are 13 hearts and the total number of equally likely ways in 
which a card can be drawn is 52. 

It is clear that, if an event is certain to happen, then the 
probability of its occurrence is 1, for all the possible ways are 
favorable. On the other hand, if an event is certain not to occur, 
the probability of its occurrence is zero. It is clear also that, if 
the probability of the happening of an event is p, then the proba- 
bility of its failure to happen is 

q = 1 - p. 

The concept of " equally likely " plays a fundamental role in the 
theory of probability. The need for caution and a careful 
analysis of the problem will be illustrated by several examples. 

Let two coins be tossed simultaneously. What will be the 
probability that they both show heads? The following reasoning 
is at fault. The total number of ways in which the coins can fall 
is three, since the possible combinations are two heads, two tails, 
and a head and a tail. Of these three ways, only one is favorable, 
and therefore the probability is J^. The fault in this reasoning 
lies in the failure to account for all the equally likely cases. 
The number of ways in which one head and one tail can fall is 
two, since the head can appear on the first coin and the tail on 
the second, or the head can appear on the second coin and the 
tail on the first. Thus, the total number of equally likely ways is 
4, and the probability of both coins showing heads is J^. The 
probability of one head and one tail showing is %, so that a head 
and a tail are twice as likely to appear as either two heads or 'two 
tails. 

Another example may prove useful. Suppose that a pair of 
dice is thrown. What is the probability that a total of eight 
shows? The total number of ways in which two dice can fall 
is 36. "(This follows from the fundamental principle of com- 



494 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 147 

binatory analysis: if one thing can be done in m different ways and 
another thing can be done in n different ways, then both things can 
be done together, or in succession, in mn different ways.) The sum 
of 8 can be obtained as follows: 2 and 6, 3 and 5, 4 and 4. 
Now, there are two ways in which 2 and 6 can fall: 2 on the 
first die and 6 on the second, and vice versa. Similarly, there are 
two ways in which 3 and 5 can fall, but there is only one way in 
which 4 and 4 can fall. Hence, the total number of equally likely 
and favorable cases is 5, so that the desired probability is %e- 

The two foregoing examples were solved simply by enumer- 
ating all the possible and all the favorable cases. Frequently, 
such enumeration is laborious and it is convenient to resprt to 
formulas. Thus, let it be required to find the probability of 
drawing 4 white balls from an urn containing 10 white, 4 black, 
and 3 red balls. The number of ways in which 4 white balls 
can be c'hosen from 10 white balls is equal to the number of com- 
binations of 10 things taken 4 at a time, namely, 

10! 

lot-4 = 



4!6! 

The total number of ways in which 4 balls can be chosen from the 
17 available is 



_ 
17/4 ~ 4T13!' 

Hence, the probability of success is 

_ 10^4 _ 10! 4! 13! _ _8_ 
P ~~ 17 C 4 ~ 4! 6! 17! ~ 34' 

Another example will illustrate further the use of formulas. 
Suppose that it is desired to find the probability of drawing 4 
white, 3 black, and 2 red balls from the urn in the preceding 
illustration. In this case the number of ways in which 4 white 
balls can be drawn is ioC4; the 3 black balls can be chosen in 
4Ca ways; and the 2 red ones in 3^2 ways. An application of the 
fundamental principle of combinatory analysis gives the required 
probability as 

ioC 4 4C 3 3 C 2 _ 252 
P 17 C 9 ~ 2431' 



148 PROBABILITY 495 

PROBLEMS 

1. What is the probability that the sum of 7 appears in a single 
throw with two dice? What is the probability of the sum of 1 1 ? Show 
that 7 is the more probable throw. 

2. An urn contains 20 balls: 10 white, 7 black, and 3 red. What is 
the probability that a ball drawn at random is red? White? Black? 
If 2 balls are drawn, what is the probability that both are white? If 
10 balls are drawn, what is the probability that 5 are white, 2 black, and 
3 red? 

148. Independent Events. A set of events is said to be 
independent if the occurrence of any one of them is not influenced 
by the occurrence of the others. On the other hand, if the 
occurrence of any one of the events affects the occurrence of the 
others, the events are said to be dependent. 

THEOREM 1. // the probabilities of occurrence of a set of n 
independent events are pi, p 2 , , p n , then the probability that 
all of the set of events will occur is p = pipz * p n . 

The proof of this theorem follows directly f eom the fundamental 
principle. Thus, let there be only two events, whose probabilities 
of success are 

mi , w 2 

PI = TT and pz = -rp 

The total number of ways in which both the events may succeed 
is mim 2 , and the total number of ways in which these events can 
succeed and fail to succeed is NiN*. Hence, the probability of 
the occurrence of both of the events is 



mi m 2 

p = NWl = Fi ' F 2 

Obviously, this proof can be extended to any number of events. 
Illustration. A coin and a die are tossed. What is the 
probability that the ace and the head will appear? The proba- 
bility that the ace will appear is J^, and the probability that the 
head will appear is J^. Therefore, the probability that both 
head and ace will appear is 



THEOREM 2. // the probability of occurrence of an event is p\ 
and if, after that event has occurred, the probability of occurrence 



496 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 148 

of a second event is pz, then the probability of occurrence of both 
events in succession is p\p%. 

The proof of this theorem is similar to that of Theorem 1 and 
will be left to the student. Obviously, the theorem can be 
extended to more than two events. 

Illustration 1. What is the probability that 2 aces be drawn 
in succession from a deck of 52 cards? The probability that an 
ace will be drawn on the first trial is ^2- After the first ace 
has been drawn, the probability of drawing another ace from the 
remaining 51 cards is %i, so that the probability of drawing 2 
aces is 

Hi = 



Illustration 2. What is the probability that the ace appears 
at least once in n throws of a die? The probability of the ace 
appearing in a single throw of the die is Y, and the probability 
that it will not appear is %. The probability that the ace will 
not appear in n successive throws is 



Hence, the probability that the ace will appear at least once is 



Illustration 3. An urn contains 30 black balls and 20 white 
balls. What is the probability that (a) A white ball and a 
black ball are drawn in succession? (6) A black ball and a white 
ball are drawn in succession? (c) Three black balls are drawn 
in succession? 

a. The probability of drawing a white ball is 2 %$. After a 
white ball is drawn, the probability of drawing a black ball is 
3 %g. Hence, the probability of drawing a white ball and a black 
ball in the order stated is 



P = 

b. The probability of drawing a black ball is 3 %o> and the 
probability that the second drawing yields a white ball is 

so that 

P = 3 %0 ' 2 %9 = % 

c. The probability of drawing 3 black balls in succession is 

P = 



149 PROBABILITY 497 

Illustration 4. The probability that Paul will solve a problem 
is Y, and the probability that John will solve it is %. What is 
the probability that the problem will be solved if Paul and John 
work independently? 

The problem will be solved unless both Paul and John fail. 
The probability of John's failure to solve it is ^ and of PauPs 
failure to solve it is %. Therefore, the probability that Paul 
and John both fail is 

M-H = H, 

and the probability that the problem will be solved is 

i - H = M- 

PROBLEMS 

1. What is the probability that 5 cards dealt from a pack of 52 cards 
are all of the same suit? 

2. Five coins are tossed simultaneously. What is the probability 
that at least one of them shows a head? All show heads? 

3. What is the probability that a monkey seated before a typewriter 
having 42 keys with 26 letters will type the word sir? 

4. If Paul hits a target 80 times out of 100 on the average and John 
hits it 90 times out of 100, what is the probability that at least one of 
them hits the target if they shoot simultaneously? 

6. The probability that Paul will be alive 10 years hence is %, and 
that John will be alive is %. What is the probability that both Paul 
and John will be dead 10 years hence? Paul alive and John dead? 
John alive and Paul dead? 

149. Mutually Exclusive Events. Events are said to be 
mutually exclusive if the occurrence of one of them prevents the 
occurrence of the others. An important theorem concerning such 
events is' the following: 

THEOREM. The probability of the occurrence of either one or 
the other of two mutually exclusive events is equal to the sum of the 
probabilities of the single events. 

The proof of this theorem follows from the definition of proba- 
bility. Consider two mutually exclusive events A and B. 
Inasmuch as the events are mutually exclusive, A and B cannot 
occur simultaneously and the possible cases are the following: (a) 
A occurs and B fails to occur, (6) B occurs and A fails to occur, 
(c) both A and B fail. Let the number of equally likely cases in 
which (a) A can occur and B fail be a, (6) B can occur and A fail 
be 0, (c) both A and B fail be y. 



498 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 

Then the total number of equally likely cases is a + ft + y. 
The probability that either A or B occurs is 

a + /3 
a + ft + y 

the probability of occurrence of A is 

a 

a + ft + y 
and the probability of occurrence of B is 



Therefore, the probability of occurrence of either A or B is equal 
to the sum of the probabilities of occurrence of A alone and of 
B alone. Obviously, this theorem can be extended to any 
number of mutually exclusive events. 

The task of determining when a given set of events is mutually 
exclusive is frequently difficult, and care must be exercised that 
this theorem is applied only to mutually exclusive events. Thus, 
in Illustration 4, Sec. 148, the probability that either Paul or 
John will solve the problem cannot be obtained by adding 
J^ and %, for solution of the problem by Paul does not elimi- 
nate the possibility of its solution by John. The events in this 
case are not mutually exclusive and the theorem of this section 
does not apply. 

Illustration 1. A bag contains 10 white balls and 15 black 
balls. Two balls are drawn in succession. What is the proba- 
bility that one of them is black and the other is white? 

The mutually exclusive events in this problem are : (a) drawing 
a white ball on the first trial and a black on the second; (6) 
drawing a black ball on the first trial and a white on the second. 
The probability of (a) is l % 5 ' ^4 and that of (b) is ^5 i% 4 , 
so that the probability of either (a) or (6) is 



Illustration 2. Paul and John are to ertgage in a game in 
which each is to draw in turn one coin at a time from a purse 
containing 3 silver and 2 gold coins. The coins are not replaced 
after being drawn. If Paul is to draw first, find the probability 
for each player that he is the first to draw a gold coin. 



149 PROBABILITY 499 

The probability that Paul succeeds in drawing a gold coin 
on the first trial is %. The probability that Paul fails and John 
succeeds in his first trial is 



The probability that Paul fails, John fails, and then Paul suc- 
ceeds is 



The probability that Paul fails, John fails, Paul fails again, and 
John succeeds is 

% M H % = Ho, 

for after three successive failures to draw a gold coin there 
remain only the two gold coins in the purse and John is certain 
to draw one of them. Therefore, Paul's total probability is 

% + H = H 

and John's probability is 

Ho + Ho = %. 

PROBLEMS 

1. One purse contains 3 silver and 7 gold coins; another purse con- 
tains 4 silver and 8 gold coins. A purse is chosen at random, and a 
coin is drawn from it. What is the probability that it is a gold coin? 

2. Paul and John are throwing alternately a pair of dice. The first 
man to throw a doublet is to win. If Paul throws first, what is his 
chance of winning on his first throw? What is the probability that 
Paul fails and John wins on his first throw? 

SfOn the average a certain student is able to solve 60 per cent of 
the problems assigned to him, If an examination contains 8 problems 
and a minimum of 5 problems is required for passing, what is the 
student's chance of passing? 

4. Two dice are thrown; what is the probability that the sum is either 
Tor 11? 

5. How many times must a die be thrown in order that the prob- 
ability that the ace appear at least once shall be greater than % ? 

6. Twenty tickets are numbered from 1 to 20, and one of them is 
drawn at random. What is the probability that the number is a 
multiple of 5 or 7? A multiple of 3 or 5? 

Note that in solving the second part of this problem, it is incorrect to 
reason as follows: The number of tickets bearing numerals that are 
multiples of 3 is 6, and the number of multiples of 5 is 4. Hence, the 



500 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 160 

probability that the number drawn is either a multiple of 3 or of 5 is 
+ ^o = . Why? 

publishing concern submits a copy of a proposed book to three 
independent critics. The odds that a book will be reviewed favorably 
by these critics are 5 to 4, 4 to 3, and 2 to 3. What is the probability 
that a majority of the three reviews will be favorable? 

8. If on the average in a shipment of 10 cases of certain goods 1 case 
is damaged, what is the probability that out of 5 cases expected at least 
4 will not be damaged? 

150. Expectation. The expectation of winning any prize is 
'defined as the value of the prize multiplied by the probability 
of winning it. Let it be required to determine the price one 
should pay for the privilege of participating in the following 
game. A purse contains 5 silver dollars and 7 fifty-cent pieces, 
and a player is to retain the two coins that he draws from the 
purse. It can be argued fallaciously as follows: The mutually 
exclusive cases are (a) 2 dollar coins, (6) 2 half-dollar coins, (c) 
1 dollar coin and 1 half-dollar coin. Therefore, the values of 
the prizes are $2, $1, and $1.50, and the fair price to pay for the 
privilege of participating is $1.50. But, the probability of 
obtaining (a) is 

^ _ 5 ^ 2 _ 5 

Pa ~ ^2 ~ 33' 



that of obtaining (b) is 
and that of obtaining (c) is 



f^ 7 

Pb = ^ = 22' 



= 5.7 = 35 
12(^2 66 
Hence, the expectation of obtaining (a) is 

6a = 2 % 3 = 0.30, 

that of obtaining (b) is 

e & = 1 % 2 = 0.32, 
and that of obtaining (c) is 

c c = 1.50 3% 6 = 0.80. 
It follows that the total expectation is 

$0.30 + $0.32 + $0.80 = $1.42, 
instead of $1.50. 



151 PROBABILITY 501 

PROBLEMS 

1. A batch of 1000 electric lamps is 5 per cent bad. If 5 lamps are 
tested, what is the probability that no defective lamps appear? What 
is the chance that a test batch of 5 lamps is 80 per cent defective? What 
is a fair price to pay for a batch of 500 such lamps if the perfect ones can 
be bought for 10 cts. each? 

2. What is a fair price to pay for a lottery ticket if there are 100 
tickets and 5 prizes of $100 each, 10 prizes of $50 each, and 20 prizes of 
$5 each? 

3. What is the expected number of throws of a coin necessary to 
produce 3 heads? 

151. Repeated and Independent Trials. Frequently it is 
required to compute the probability of the occurrence of an 
event in n trials when the probability of the occurrence of that 
event in a single trial is known. For example, it may be required 
to find the probability of throwing exactly one ace in 6 throws 
of a single die. The possible mutually exclusive cases are as 
follows: 

(1) An ace on the first throw, and none on the remaining 5 
throws. 

(2) No ace on the first throw, an ace on the second, and no 
aces on the remaining 4 throws. 

(3) No ace on the first 2 throws, an ace on the third, and no 
aces on the last 3 throws. 

(4) No aces on the first 3 throws, an ace on the fourth, and no 
aces on the last 2 throws. 

(5) No aces on the first 4 throws, an ace on the fifth, and no 
ace on the last throw. 

(6) No aces on the first 5 throws, and an ace on the last throw. 
The probability of the occurrence of (1) is 



since the probability of throwing an ace on the first trial is % 
and the probability that the ace will not appear on the succeeding 
5 throws is (%) 5 . The probability of (2) is 



the probability of (3) is 



502 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 151 

and it is clear that the probability of any one of the 6 specified 
combinations is 



Since the cases are mutually exclusive, the probability that some 
one of the 6 combinations will occur is 

P = 6 - H - W = (%) 5 

It should be observed that the probability of obtaining any 
combination of 1 ace and 5 not-aces is always the same, so that in 
order to obtain the probability of occurrence of one of the set of 
mutually exclusive cases all that is necessary is to multiply the 
probability of the occurrence of any specified combination by the 
number of distinct ways in which the events may occur. This 
leads to the formulation of an important theorem which is 
frequently termed the binomial law. 

THEOREM 1 (Binomial Law). // the probability of occurrence 
of an event in a single trial is p, then the probability that it will occur 
exactly r times in n independent trials is 

Pr = nC r p r (l ~ pY" ' 

where 

c - -iL 

nCr ~ r\(n - r)!' 

The method of proof of this theorem is obvious from the 
discussion of the specific case that precedes the theorem. The 
probability that an event will occur in a particular set of r 
trials and fail in the remaining n r trials is p r (l p) n ~ r . 
But since the number of trials is n, the number of ways in which 
the event can succeed r times and fail n r times is equal to the 
number of the combinations of n things taken r at a time, or 

n\ 

nv-T 



r\(n r)! 
Hence the probability of exactly r successes and n r failures is 

(151 - 1} P ' - r\(n-r)\ ^ ~ ">~ 

Illustration 1. What is the chance that the ace will appear 
exactly 4 times in the course of 10 throws of a die? 



161 PROBABILITY 503 

Formula (151-1) gives 

_ 10! /1V M" _ 656,250 



_ 
4 ~ 4l6! \fi ~ 60,466,170 ~ 

Illustration 2. Ten coins are tossed simultaneously. What is 
the chance that exactly 2 of them show heads? 
Here 



If in the example at the beginning of this section it had been 
required to determine the probability that the ace would appear 
at least once in the course of the 6 trials, the problem would be 
solved with the aid of the following argument: The ace will 
appear at least once if it appears exactly once, or exactly twice, 
or exactly three times, and so on. But the probability that it 
appears exactly once is 

PI = .C 

exactly twice, 

p 2 = 6 

exactly three times, 



etc. These compound events are mutually exclusive, so that the 
probability of the ace appearing at least once is the sum of the 
probabilities 

Pi, P2, Pa, ' ' ' , Pe. 

The general theorem, which includes this problem as a special 
case, is the following. 

THEOREM 2. // the probability of the occurrence of an event 
on a single trial is p, then the probability that the event will occur 
at least r times in the course of n independent trials is 

P>r = P n + nClp n ~ l q + *C 2 p- 2 g 2 + + nC n -rp r q n - r , 

where q = 1 p. 

It should be noted that n C r = n C n - r is the coefficient of p r in the 
binomial expansion for (p + g) n and that p^ r is equal to the sum 
of the first n r + 1 terms in the expansion for (p + q) n . 

Illustration 3. The probability that at least 2 of the coins 
show heads when 5 coins are tossed simultaneously is 

= PAY + 6 



504 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 162 

The first term of this sum represents the probability of exactly 5 
heads, the second represents that of exactly 4 heads, the third 
that of exactly 3 heads, and the last represents that of exactly 
2 heads. 

PROBLEMS 

1. If 5 dice are tossed simultaneously, what is the probability that 
(a) exactly 3 of them turn the ace up? (6) At least 3 turn the ace up? 

2. If the probability that a man aged 60 will live to be 70 is 0.65, what 
is the probability that out of 10 men now 60, at least 7 will live to be 70? 

3. A man is promised $1 for each ace in excess of 1 that appears in 6 
consecutive throws of a die. What is the value of his expectation? 

4. A bag contains 20 black balls and 15 white balls. What is the 
chance that at least 4 in a sample of 5 balls are black? 

5. Solve Prob. 3, Sec. 149. 

162. Distribution Curve. Some interesting and useful con- 
clusions can be deduced regarding the formula for repeated 
independent events from the consideration of an example that 
presents some features of the general case. Consider a purse 
in which are placed 2 silver and 3 gold coins, and let it be required 
to determine the probability of drawing exactly r silver coins 
in n repeated trials, the coin being replaced after each drawing. 
The probability of exactly r successes in n trials is given by the 
binomial law [see (151-1)] 

Pr = nC r p r (l ~ p)***, 

where p, the probability of drawing a silver coin on a single trial , 
is %. 

If the number of drawings is taken as n = 5, the probability 
that none of the drawings yields a silver coin is 

Po = 6Co(%) (9s) 6 = 0.07776, 
the probability that 5 trials yield exactly 1 silver coin is 

Pi = 6<7i(%)(^) 4 = 0.2592, 

and the probability that exactly 2 silver coins will appear is 
P2 = 6 C 2 (%) 2 (%) 3 = 0.3456. 

In this manner, it is possible to construct a table of the values 
that represent the probabilities of drawing exactly 0, 1, 2, 3, 4, 5 
silver coins in 5 trials. Such a table, where the values of p r are 
computed to four decimal places, is given next. 



PROBABILITY 505 

PROBABILITY OF EXACTLY r SUCCESSES IN 5 TRIALS 



r 


Pr 


r 


Pr 





0778 


3 


2304 


1 


2592 


4 


0768 


2 


3456 


5 


0102 



It will be observed that r = 2 gives the greatest, or "most 
probable," value for p r , which seems reasonable in view of the 
fact that the probability of drawing a silver coin on a single trial 
is % and one would " expect" that 2 silver coins should result 
from 5 repeated drawings. 

If the number of trials is n = 10, the formula 



Pr = lo 

gives the following set of probabilities for 0, 1, 2, 3, , 10 
successes. 

PROBABILITY OF EXACTLY r SUCCESSES IN 10 TRIALS 



r 


Pr 


r 


Pr 


r 


Pr 





0060 


4 


2508 


8 


0106 


1 


0403 


5 


2007 


9 


0016 


2 


1209 


6 


1115 


10 


0001 


3 


2150 


7 


0425 







Again it appears that the most probable number of successes in n 
trials is equal to the probability of success in a single trial 
multiplied by the number of trials. 

If a similar table is constructed for n = 30, the resulting proba- 
bilities are as shown below. 

PROBABILITY OF EXACTLY r SUCCESSES IN 30 TRIALS 



r 


Pr 


r 


Pr 


r 


Pr 


<2 


0000 


9 


0823 


16 


0489 


3 


0003 


10 


1152 


17 


0269 


4 


0012 


11 


1396 


18 


0129 


5 


0041 


12 


0.1474 


19 


0054 


6 


0115 


13 


1360 


20 


0020 


7 


0263 


14 


1100 


21 


0006 


8 


0.0505 


15 


0783 


22 


0002 










^23 


0000 



506 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 152 

In this table the entry 0.0000 is made for < r < 2 and for 
r > 23 because the values of p r were computed to four decimal 
places, and in these cases p r was found to be less than 0.00005. 
For example, the probability of drawing exactly 1 silver coin in 
30 trials is 

Pi = 30(%)(%) 29 = 0.00000442149, 

and the probability of drawing exactly 23 silver coins in 30 trials 
is 

p 23 = 3oC 23 (%) 23 (%) 7 = 0.000040128. 

Therefore, for all values 23 < r < 30, the values of p r are less 
than 0.00005 and must be recorded as 0.0000. 

Just as in the foregoing tables, the most probable number is 
equal to % n > although the probability of drawing exactly 12 

Pr 



04 
03 


~-_ 


n-5 0.4 

03 


n-10 


02 






02 




1 
( 
02 






i. 


-.1 i. 


) 


2 3 4 5 r 0123456789 10 r 


01 


: ..rill 1 



2 4 6 8 10 12 14 16 18 20 22 24 
FIG. 148. 

silver coins, 0.1474, is less than the probability of drawing the 
most probable number of silver coins in the case of 10 trials. 

These tabulated results can be more conveniently represented 
in a graphical form, where the values of r are plotted as abscissas 
and the values of p r as ordinates. Such graphs are known as 
distribution charts (Fig. 148). 

An alternative method of graphical representation is obtained 
by erecting rectangles of unit width on the ordinates which 
represent the probabilities p r of occurrence of r successes. Since 
the width of each rectangle is unity, its area is equal, numerically, 
to the probability of the value of r over which it is erected. In 
such graphs the vertical lines are not essential to the interpreta- 
tion of the graph and hence are omitted. The resulting broken 
curve constitutes what is known as a distribution curve. The 



152 



PROBABILITY 



507 



area under each step of the curve represents the value of p r ; 
and the entire area under the distribution curve is unity, for it 
represents the sum of the probabilities of 0, 1, 2, , n suc- 
cesses. Such curves, corresponding to the distribution charts 
(Fig. 148), are drawn in Fig. 149. 

It appears that, as the number of drawings, n, is increased, the 
probability of obtaining the most probable number of silver coins 
decreases. Moreover, there is a greater spread of the chart as 
the number of trials is increased, so that the probability of missing 
the most probable number by more than a specified amount 
increases with the increase in the number of drawings. 



0.4 


n5 


0.3 


- 






02 
0.1 


f 






I. 



03 
02 
1 



n-10 




012345 



01 23456789 10 



n-30 




10 12 14 16 
FIG. 149. 



18 20 2Z 24 26 r 



The following observations will serve to clarify the last state- 
ment. In the case of 5 trials the probability of missing the 
most probable number of successes by 5 is zero, for the deviation 
from the most probable number 2 cannot be greater than 3. 
But in the case of 10 trials the probability of missing the most 
probable number by this same amount has a definite non-zero 
value. Thus, in order to miss the most probable number 4 
by 5, r must be either 9 or 10. Hence, the probability of missing 
4 is 0.0016 + 0.0001 = 0.0017. In the case of 30 trials the 
most probable number of successes is 12; and, in order to miss 
12 by 5, r must be less than 8 or greater than 16. Therefore the 
probability of missing 12 by 5 is the sum 

30 



X Pr + X Pr = 0.0914. 



If the number of drawings n is made 1,000,000, the most probable 



508 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 163 

number of successes is 400,000, and the probability of missing it 
by 5 is very nearly unity. On the other hand, the probability 
of obtaining the most probable number 400,000 is a very small 
quantity. 

The important facts obtained from the foregoing considerations 
are the following: 

1. The most probable number of successes appears to be equal 
to pn. 

2. The probability of obtaining the most probable number of 
successes decreases with the increase in the number of trials n. 

3. The probability of missing the most probable number by a 
specified amount increases with the increase in the number of 
trials. 

It can be established that the last two facts, inferred from the 
special example, are true in general. The first fact, concerning 
the size of the most probable number, clearly is meaningless if 
pn is not an integer. Thus, if the number of drawings is n 24 
and p = %, then pn 4 %. It can be shown in general that the 
most probable number is pn, provided pn is an integer; otherwise, 
the most probable number is one of the two integers between 
which pn lies. In fact, the following is the complete statement 
of the theorem:* The most probable number of successes is the 
greatest integer less than np -{-p. If np + p is an integer, there 
are two most probable numbers, namely, np + P and np + p 1. 
Since p < I, it is clear that the most probable number of successes 
is approximately equal to np. This number np is called the 
"expected" number of successes. 

PROBLEM 

A penny is tossed 100 times. What is the most probable number of 
heads? What is the probability of this most probable number of 
heads? If the penny is tossed 1000 times, what is the probability of the 
most probable number of heads? 

153. Stirling's Formula. The binomial law (151-1), on which 
the major portion of the theory of probability is based, is exact, 
but it possesses the distinct disadvantage of being too com- 
plicated for purposes of computation. The labor of computing 
the values of the factorials that enter in the term n C r becomes 

* For proof and further discussion see T. C. Fry, Probability and Its Engi- 
neering Uses, Chap. IV. 



153 



PROBABILITY 



509 



prohibitive when n is a large number. Accordingly, it is desirable 
to develop an approximation formula for n!, when n is large. 

An asymptotic formula, which furnishes a good approximation 
to n!, was developed by J. Stirl- y t 
ing. By an asymptotic formula 
is meant an expression such 
that the percentage of error 
made by using the formula as 
an approximation to n \ is small 

when n is sufficiently large, j k-1 k 

whereas the error itself increases 
with the increase in n. It will be Fl0 ' 15 * 

indicated that for values of n greater than 10 the error made in 
using Stirling's formula* 

(153-1) n\ ~ n n e~ n 

is less than 1 per cent. 




Consider the function y = log x, and observe that, for k 
log x dx > HHog (k - 1) + log fc], 



2, 



since the right-hand member represents the trapezoidal area 
formed by the chord (Fig. 150) joining the points P and Q on the 
curve y log x. Denote the area between the chord and the 
curve by a*, so that 



+ a n ). 



(153-2) J ;fc _ i log x dx = ^[log (fc - 1) + log k] + a k . 
Setting k = 2, 3, , n in (153-2) and adding give 

| log x dx = K(log 1 + log 2) + H(lg 2 + log 3) + 
+ /^[log (n 1) + log n] + (a 2 + a 3 + 

Integrating the left-hand member and combining the terms of the 
right-hand member give 

n 

n log n n + 1 = log n\ Yi log n + ^ o. 
Hence, 
(153-3) log n! = (n + ^) log n - n + 1 - ^ a t . 



* The symbol ~, which is read " asymptotically equal to," is used instead 
of = to call attention to the fact that the formula is asymptotic. 



510 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 153 

Since each a t is positive, it follows that 

log nl < (n + Y^) log n - n + 1, 
and hence 
(153-4) nl < e \/n n n e~ n . 

The expression on the right of the inequality (153-4) is, therefore, 
an upper bound for nl 

To get a lower bound, solve (153-2) for &, perform the integra- 
tion, and obtain 

(153-5) a * 



Now, since the integrand is non-negative, 

f* A iY 

(153-6) ( - f ) dx > 0, 

Jk-i \x k) 

and the evaluation of (153-6) leads to the formula 
1 _*_ < 2k ~ l 

Ki J. juK/\fc L) 

By the use of this inequality, (153-5) gives 

1 1 /_! 

ak 4k(k ~- 1) 4 \A; - 1 
Hence, 



By means of this result and (153-3), one obtains 

log nl > (n + Yd log n - n + 1 - y, 
whence 
(153-7) nl > e^ \/n n n e~ n . 

Combining (153-4) and (153-7) furnishes the inequality* 

e^ \/n n n e~ n < nl < e -\/n n n e~ n , 

for all values of n > 1. Since e = 2.718, e* = 2.117, and 
\/25r = 2.507, it follows that 

n I c^ n n e~ n \/27rn. 

* The derivation of this result is given by P. M. Hummel, in Amer. Math. 
Monthly, vol. 47, p. 97, 1940. 



154 PROBABILITY 511 

It is possible to obtain a sharper lower bound for n\ by. using 
the integral* 



(153-8) 

' 



f* [-~ 

j^-i L 



instead of (153-6). 

To gain some insight into the accuracy of this formula, note 
that (153-1) gives for n = 10 the value 3,598,696, whereas the 
true value of 10! is 3,628,800. The percentage of error in this 
case is 0.8 per cent. Forn = 100, (153-1) gives 9.524847 X 10 157 , 
whereas the true value of 100! is 9.3326215 X 10 157 , so that the 
percentage of error is 0.08. It is worth noting that, even for 
n = 1, the error is under 10 per cent and that for n = 5 it is in 
the neighborhood of 2 per cent. 

PROBLEM 

Make use of (153-8) in order to show that 

e'Ma -y/^ n n e~ n < n!, if n > 1, 

and compare the value of e 1 ^ 2 with that of \/2ir. 

154. Probability of the Most Probable Number. It was 

mentioned in Sec. 152 that the most probable number of successes 
is either equal, or very nearly equal, to the expected number 
e = np. Very often it is desirable to compute the probability 
of the expected or the most probable number of successes. Of 
course, p can be computed from the exact law by substituting 
in it r = np, but formula (151-1) is cumbersome to use when 
factorials of large numbers appear in n C r . An approximate 
formula can be obtained by replacing n\ and (np)! by their 
approximate values with the aid of Stirling's formula. It is 
readily verified that, when these replacements are made, the prob- 
ability of the most probable number of successes is approximately 



where q 1 p. It must be kept in mind that (154-1) is 
subject to the same restrictions as (153-1) and gives good results 
for np ^ 10. 

* See problem at the end of this section. 



512 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 155 

Thus, if a die is tossed 100 times, the most probable number of 
aces is 16. The exact formula gives 



whereas the approximation (154-1) gives 

p " = = - 309 - 

The percentage of error is quite small. 

PROBLEMS 

1. Two hundred and fifty votes were cast for two equally likely can- 
didates for an office. What is the probability of a tie? 

2. What is the most probable number of aces in 1200 throws of a die? 
What is the probability of the most probable number? 

3. Solve, with the aid of the approximate formula, the problem at the 
end of Sec. 152. 

155. Approximations to Binomial Law. With the aid of 
formula (153-1), it is possible to devise various formulas approxi- 
mating the binomial law (151-1). One of these approximations 
is known as the Poisson formula or the law of small numbers. 
The wide range of applicability of this law can be inferred from 
the fact that it has been used successfully in dealing with such 
problems as those of beta-ray emission, telephone traffic, trans- 
mission-line surges, and the expected sales of commodities. The 
law of small numbers gives a good approximation to (151-1) in 
those problems in which r is small compared with the large number 
n, and p represents the probability of occurrence of a rare event in a 
single trial. 

Replacing n\ and (n r)! in (151-1) with the aid of Stirling's 
formula (153-1) leads to 



uss-n ~ o^i - > 

(1551) p,____p(l p) 



= , V-r+H/' ^ 

4 ~ ;) 

By hypothesis, r is small compared with n, so that 



155 PROBABILITY 513 

is very nearly equal to 



which, * for large values of n, differs little from e~ r . Similarly, 

(1 _ p )-r - (1 _ p )^ 

which in turn is nearly equal to e~ Mp , since 

(1 - p) = 1 - np + ^~^- } p - 
and 

o 2/rj2 

e - P = 1 - np + -~ 2 y - . 
The substitution of e~ r for 



and e~ np for 

in (155-1) leads to the desired law of small numbers, 

(\ tt 9"\ T> ^ n V) p np 

(LOO-*) Pr r j e ". 

Formula (155-2) is frequently written in a slightly different 
form. It will be recalled that the expected number of successes 
is e = np, so that (155-2) can be written 

= 6^ _ e 

r\ 

An application of this law to some specific cases may prove 
interesting. Suppose that it is known that, on the average, in a 
large city two persons die daily of tuberculosis. What is the 
probability that r persons will die on any day? In this case the 
expected number of deaths is c = 2, so that 

r = - e~ 2 

r\ 
* Note that lim (1 + l/n) rt = e. For a rigorous discussion, see I. S. 

n * oo 

Sokolnikoff, Advanced Calculus, pp. 28-31. 



514 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 165 
Therefore, 



r 


Pr 


r 


Pr 


r 


Pr 





136 


2 


272 


4 


091 


1 


272 


3 


181 


5 


036 



A glimpse into the accuracy of this law can be gained by con- 
sidering the following example. 

Example. What is the probability that the ace of spades will be 
drawn from a deck of cards at least once in 104 consecutive trials? This 
problem can be solved with the aid of the exact law (151-1) as follows: 
The probability that the ace will not be drawn in the 104 trials is 



Po = 



= 0.133, 



and the probability that the ace will be drawn at least once is 1 0.133 
= 0.867. On the other hand, Poisson's law (155-2) gives for the prob- 
ability of failure to draw the ace 



e ' f/ . 



/'O 

Hence, the probability of drawing at least one ace of spades is 1 e~* 
= 0.865. 

Another important approximation to the binomial law (151-1), 
namely, 

n\ 



(155-3) 



Pr = 



r\(n - r)! 



p r q n 



where q = 1 p, is obtained by assuming that r, n, and n r 
are all large enough to permit the use of the Stirling formula. 
Replacing n!, r!, and (n r)! by Stirling's approximations 
gives, upon simplification, 



r(n - r) 

Let 5 denote the deviation of r from the expected value np; 
that is, 

8 = r up. 
Then, 

n r = nq 6, 



155 PROBABILITY 515 

and (155-4) becomes 



\ f 

or 

v (np+8) 



(, \ 
i + JL) 
np/ 



where 

A = 
Then, 
log p r A = -(np + 6) log ( 1 + J - (nq - 5) log ( 1 - -V 

Assuming that \d\ < npq, so that 
< 1 and 



np 



nq 



permits one to write the two convergent series 

log (i + A) = A - * + *' 

\ n P/ n P 2n 2 p 2 3r 



and 

Hence, 
logy 



2npq 



Now, if \d\ is so small in comparison with npq that one can 
neglect all terms in this expansion beyond the first and can 
replace A by \/2irnpq, then there results the approximate 
formula 

(155-5) Pr = * e~^*, 

\f2irnpq 

which bears the name of Laplace's, or the normal, approximation. 
Since the maximum value of the exponential e~ x , for x ^ 0, is 
unity, it follows that the normal approximation gives for the 
probability that r will assume its most probable value the same 



516 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 156 



value as was obtained in Sec. 154. It is obvious that the normal 
approximation gives best results when p and q are nearly equal. 

If the mean error a is defined 
by the formula 

0- = \/npq, 
then (155-5) assumes the form 




151. 



p r = 



I 



- e 



and the graph of p r as a function of d is a bell-shaped curve 
(Fig. 151), known as the normal distribution curve.* 

PROBLEMS 

1. What is the probability of throwing an ace with a die exactly 10 
times in 1200 trials? 

2. A wholesale electrical dealer noticed that a shipment of 10,000 
electric lamps contained, on the average, 20 defective lamps. What is 
the probability that a shipment of 10,000 lamps is 1 per cent defective? 

3. In a certain large city, on the average, two persons die daily of 
cancer. What is the probability of no persons dying on any day? One 
person dying? Two? Three? Four? Five? 

4. Two dice are tossed 1000 times. What is, approximately, the 
probability of getting a sum of 4 the most probable number of times? 

5. What is the approximate probability that a sum of 4 will appear 
500 times in a set of 1000 tosses? 

156. The Error Function. Let wi, ra 2 , , m n be a set of n 
measurements, of some physical quantity, that are made inde- 
pendently and that are equally trustworthy. If the best estimate 
of the value of the measurements is m, then the " errors" in 
individual measurements are 



m, 



x n = m n m, 



Xi = mi m, #2 = 
and the sum of the errors is 
(156-1) xi + Xz + + x n = (mi + m 2 + + m n ) - mn. 

If it is assumed that on the average the positive and negative 
errors are equally balanced, then their sum is zero, and (156-1) 
becomes 

mn = mi + w 2 + + m n 

* For a detailed discussion see T. C. Fry, Probability and Its Engineering 
Uses. 



t= 1 



158 PROBABILITY 517 

or 

(156-2) m = 

IV 

It is important to note that the best value m is the arithmetic 
average of the individual measurements when it is assumed that 
the positive and negative errors are equally likely. In per- 
forming a set of measurements, not all the errors x l are equally 
likely to occur. In general, large errors are less likely to occur 
than small ones. For instance, the probability of making an 
error of 1 ft. in measuring the length of a table is less than that 
of making an error of 1 in. 

Let the probability of making an error x l be denoted by 
<p(xi). The assumption that positive and negative errors are 
equally likely to occur de- 
mands that * x 



which states that <p(x) is an 
even function. Further- 
more, the hypothesis that 
small errors are more likely 
to occur than large ones re- 
quires <p(x) to be a decreasing function for x > 0; and since 
infinitely large errors cannot occur, 

*(>) = 0. 

These observations lead to the conclusion that the function 
<p(x), which gives the probability of occurrence of the error #, 
must have the appearance shown in Fig. 152, where the errors x l 
are arranged in order of increasing magnitude. Upon recalling 
the fact that the ordinates represent the probability of occurrence 
of an error of any size x, it is clear that the area under this curve 
from oo to + oo must be unity, for all the errors are certain 
to lie in the interval ( <, oo). Hence, 




/as 
<p(x) dx = 1. 
- 



Moreover, the probability that the error lies between the limits 
Xt and t + A# is equal to the area bounded by curve y = <p(x), 



518 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 166 

the ordinates y = x l and y = x l + A#, and the x-axis, which is 
equal to the value of the integral 



r 



dx. 



By hypothesis the measurements ra t were made independently, 
so that the probability of simultaneous occurrence of the errors 
Xi, xt, , x n is equal to the product of the probabilities of 
occurrence of the individual errors, or 



(156-3) P = <P(XI}<P(X*} <p(xn) 

= <p(m\ m)(p(mz m) <p(m n m). 



The expression (156-3) is a function of the best value m, in which 
the functional form of <p is not known. Now, if it be assumed 
that the best value m is also the most probable value, that is, the 
value which makes P a maximum, then it is possible to determine 
the functional form of <p by a method due to Gauss. In other 
words, it is taken as a fundamental axiom that the probability 
(156-3) is a maximum when m is the arithmetic average of the 
measurements mi, m^ , m n > But if (156-3) is a maximum, 
its logarithm is also a maximum. Differentiating the logarithm 
of (156-3) with respect to m and setting the derivative equal to 
zero give 



M r,A A\ <i - m) <p'(m z - m) <p'(m n ~ m) 

( 1 00-4 ) , --- r H -- - f -- r- + ' ' ' H -- y -- r = 0. 

<f>(m>i ~ in) <p(mz m) <p(m n m) 

n 

This equation is subject to the condition 2 a: t = 0. 

t = i 

If 

<p'(m t m) 

<p(m t m) 

is set equal to F(x t ), (i = 1, 2, , n), Eq. (156-4) can be 
written as 



(156-5) F(XI) + F(x*) + - - + F(x n ) = 

n 

with S #t = 0. If there are only two measurements, (156-5) 
reduces to 

= 0, 



166 PROBABILITY 519 

with Xi + xz = 0, or #2 = #i. Therefore, 

F(xi)+F(-xi) = 0, 
or 

(156-6) F(x) = -F(-x). 

Similarly, if there are only three measurements, then 

Ffa) + F(x t ) + F(x 9 ) = 0, 
with xi + 2 + #3 = 0. Therefore, 



But, from (156-6), 

and since x$ x\ + #2, 

F(*0 + f (s,) = F( Xl + x,). 

Differentiating this expression partially with respect to x\ and 
leads to 

F'(xJ = F^rci + x>) and 
or 



(156-7) F'(xi) = F'(x*). 

Since x\ and x% are independent, (156-7) can be true only if 

*"(a*) = c, 



so that 

F(XI) = cxi and F(x*) = co; 2 . 

Recall that, by definition, 



so that the differential equation for <p is 



which, upon integration, gives 

(156-8) q>(x) 

where K and c are arbitrary constants. 



520 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 156 

One of these constants can be determined at once, for it is 
known that 



*(x) dx = 1. 
The substitution of <p(x) in this integral gives 
K f e~ h ^dx = 1, 

J oo 

where h 2 ^ c/2. 

This integral can be evaluated by means of a procedure similar 
to that used in Sec. 81. Set 



and then 

f >j r *s 

I* = e~* dx e-y dy 
Jo Jo 

= f f e-^^dxdy 
Jo Jo 



rl r 

I I 

Jo Jo 



w 

e~ r ~r dr d<p = 7; 
4 



where the last step results from the transformation of the double 
integral into polar coordinates and has been described in Sec. 
81. Hence, 

f* / 

I = e~* z dx = ! 

Jo * 

But 

K f* e-^dx = 2K f e~ h ^ dx = 1, 

J- oo JO 

so that 

1 J, Z, 

X =,- 



e~ h2 ** dx 2 e~ h2 * 2 d(hx) 
Thus, (156-8) can be written as 
(156-9) v(x] -** 



which is called the Gaussian law of error. The undetermined 
constant A, as will be seen in the next section, measures the 
accuracy of the observer and is known as the precision constant. 



157 



PROBABILITY 



521 



It is easy to verify the fact that the choice of <p, specified by 
(156-9), gives a maximum for the product (156-3) when the sum 
of the squares of the errors is a minimum. In fact, since # = 
m l m y (156-3) becomes 



._ f * Y 



-h* 2- 



and the maximum value of P is clearly that which makes the 
sum of the squares of the errors a minimum. 

In order to verify the assumption that the choice of the 
arithmetic average for the best value leads to the least value for 
the sum of the squares of the 
errors, all that is necessary is 
to minimize 



x, 2 - 2) (ro. 




The theory of errors based 
upon the Gaussian law (156-9) 
is often called the theory of 
least squares, 

157. Precision Constant. 
Probable Error. In the pre- 
ceding section, it was established that the probability of commit- 
ting an error of magnitude x is given by the ordinate of the curve 



FIG. 153. 



- e~ 



y = 



This curve is called the probability curve. Clearly, the proba- 
bility of an error lying in the interval between x and 
x +e is equal, numerically, to the area bounded by the proba- 
bility curve (Fig. 153), the ordinates x = and x = +, and 
the x-axis. If only the absolute value of the error is of interest, 
then the probability that the absolute value of the error does not 
exceed is 



2h / 

V^ Jo 



e~ h * x * dx. 



If hx is set equal to t, this integral assumes the form 



522 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 157 



2 /* 

= ^Jo ' 



which shows that P is a function of Ae, and, for a fixed value of 
, P increases with A. For large values of A, the probability 
curve decreases very rapidly from its maximum value, h/\/w 
at x = 0, to very small values, so that the probability of making 
large errors is very small. On the other hand, if h is small, the 
probability curve falls off very slowly so that the observer is 
almost as likely to make fairly large errors as he is to make 
small ones. For this reason the constant A is known as the pre- 
cision constant. 

That particular error which is just as likely to be exceeded as 
riot is called the probable error. More precisely, the probable 
error is that error e which makes P = }^, or 

e~ tz dt. 

An approximate solution of this equation can be obtained by 
expanding e~ tz in Maclaurin's series, integrating the result term 
by term, and retaining only the first few terms of the resulting 
series.* The solution, correct to four decimal places, found by 
this method is 

he = 0.4769, 

so that the probable error is 0.4769/A. It is commonly denoted 
by the letter r. 

In addition to the probable error, the mean absolute error and 
the mean square error are of importance in statistics. The mean 
absolute error is defined as 



XOO 
xe -hW fa _ 



0.5643 

I <vr> /*-*- fiw 

v, 



and the mean square error is defined as 

_. __ 2h 
x & -- 

V* 



C 2 fc , x , 7 _ 1 
x e ax ^Tg- 

Jo 2h 



It will be observed that the mean absolute error is the z-coordi- 
nate of the center of gravity of the area bounded by the proba- 
*See Prob. 3, at the end of this section. 



167 PROBABILITY 523 

bility curve and the positive coordinate axes, and that the 
square root of the mean square error is the radius of gyration of 
that area about the y-axis. 

The values of these mean errors can actually be computed for 
any set of observations. Thus, 



w t m 



11 n n 

so that 

(157-1) h = - 

P 

Also, 

n 

V (m, - m 



n "n 2h 2 ' 

so that h computed from this equation is 

(157-2) h = J r 

Vz 2 \/2 

These two expressions for the precision constant give a means of 
computing h for any set of observation data. The two values of 
h cannot be expected to be identical; but unless there is a fair 
agreement between them, experience indicates that the data are 
not reliable. The value of -y/x 2 = " * s commonly called the 
standard deviation, and it follows from the foregoing that the 
probable error is equal to 0.6745<r. 

PROBLEMS 

1. Evaluate the integral J Q e~^ dt by expanding the integrand in 
series, and show that 

f*X r 3 r 5 r 7 T 9 

jC^-'-^n + s^-T^ + Ai-* 

where R < a? u /1320. 

2. The expression for the probability integral given in the preceding 
problem is not suitable for computation purposes when x is large. But 



524 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 157 
C x e -t* dt = f * er< 2 dt - f er< 2 <& 

JO JO Jx 



Show, by integrating by parts, that 
f* 2J fl SJ e-*Yi ] 

Jx '-"*-J, F'-'^-W^-S-' 

and thus obtain an asymptotic expansion for the probability integral 
that can be used to compute its value when x is large. Also, show that 
the asymptotic series 

r* , i V* e-* 2 r 1,1-3 1-3-5, n 

Jo ^^^-^L 1 "^ w~ wr + ' * J 

gives a value for the integral which coffers from its true value by less 
than the last term which is used in the series. 

3. Show, with the aid of Homer's method, that the value of the 
probable error is 

0.4769 



4. Compute the probable errors for the following set of observation 
data: 

mi = 1.305, ra 2 = 1.301, w 3 = 1.295, m* = 1.286, 

m 5 = 1.318, w 6 = 1.321, ra 7 = 1.283, w 8 = 1.289, 

m 9 = 1.300, WIQ = 1.286, 

by using (157-1) and (157-2). 

5. With reference to Prob. 4, what is the probability of committing an 
error whose absolute value is less than 0.03? 

6. Two observers bring the following two sets of data, which repre- 
sent measurements of the same quantity: 



(a) mi = 105.1, 
m B = 104.8, 
(6) mj = 105.3, 
m 5 = 106.7, 


m 2 = 103.4, 
m 6 = 105.0, 
m 2 = 105.1, 
m 6 = 102.9, 


m 3 = 104.2, 
m 7 = 104.9. 
m 3 = 104.8, 
m 7 = 103.1. 


m 4 - 104.7, 
m 4 = 105.2, 



Which set of data is the more reliable? 

7. Discuss the problem of a rational way of proportioning the salaries 
of two observers whose precision constants are hi and hz. 



CHAPTER XII 
EMPIRICAL FORMULAS AND CURVE FITTING 

An empirical formula is a formula that is inferred by some 
scheme in an attempt to express the relation existing between 
quantities whose corresponding values are obtained by experi- 
ment. For example, it may be desired to obtain the relation 
connecting the load applied to a bar and the resulting elongation 
of Hie bar. Various loads are applied, and the consequent 
elongations are measured. Then, by one of the methods to be 
given in this chapter, a formula is obtained that represents the 
relationship existing between these two quantities for the 
observed values. With certain restrictions, this formula can 
then be used to predict the elongation that will result when an 
arbitrary load is applied. 

It is possible to obtain several equations of different types 
that will express the given data approximately or exactly. 
The question arises as to which of these equations will give the 
best "fit" and be most successful for use in predicting the results 
of the experiment for additional values of the quantities involved. 
If there are n sets of observed values then, theoretically at least, 
it is possible to fit the given data with an equation that involves 
n arbitrary constants. What would be the procedure if it were 
desired to obtain an equation representing these data but 
involving less than n arbitrary constants? Questions of this 
type will be considered in the succeeding sections. 

158. Graphical Method. The graphical method of obtaining 
an empirical formula and curve to represent given data is prob- 
ably already somewhat familiar to the student from elementary 
courses. It is particularly applicable when the given data can 
be represented by equations of the three types 

(1) y = mx + 6 ; (2) y = a + bx n ; (3) y = ka mx . 

If the corresponding values (x*, y v ) of the given data are 
plotted on rectangular coordinate paper and the points thus 

525 



526 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 168 

plotted lie approximately on 1 a straight line, it is assumed that 
the equation 

y = mx + b 

will represent the relationship. In order to determine the values 
of the constants m and 6, the slope and ^-intercept may be read 
from the curve, or they can be determined by solving the two 
simultaneous equations 

y\ = mxi + b, 2/2 = mx* + b 



obtained by assuming that any two suitably chosen points 
(xi, ?/i) and (#2, 2/2) lie on the line. Obviously the values of m 
and b will depend upon the judgment of the investigator regard- 
less of which method is used for their determination. 
Consider the equation 

y = a + bx n . 

If the substitution x n = t is made, then the graph of y = a + bt 
is a straight line and the determination of a and b is precisely 
the same as in the preceding case. In the special case y = bx n , 
taking logarithms on both sides gives 

log y = log b + n log x, 

which is linear in log y and log x and gives a straight line on 
logarithmic paper. The slope of this line and the intercept on the 
log i/-axis can be read from the graph. Hence, if the correspond- 
ing values (# t , ?/,), when plotted on logarithmic paper, give points 
that lie approximately on a straight line, the data can be repre- 
sented by the equation 

y = bx n , 

whose constants can be read from the graph. 

Similarly, if the data can be represented by a relation 

y = ka mx j 

the corresponding values, when plotted on semilogarithmic* 
paper, will give points that lie approximately on a straight line. 
For taking logarithms on both sides of this equation gives 

log y = log k + (m log a)x, 

* For a discussion of logarithmic and semilogarithmic paper, see C. S. 
Slichter, Elementary Mathematical Analysis. 



159 



EMPIRICAL FORMULAS AND CURVE FITTING 



527 



which is linear in log y and x and therefore plots as a straight line 
on semilogarithmic paper. 

The three types of equations cited here are, of course, not the 
only ones to which the graphic method is applicable. However, 
they are the simplest because of the fact that their graphs, on 
appropriate paper, give straight lines. When the points repre- 
senting the observed values do not approximate a straight line, 
some other method is usually preferable. 

PROBLEMS 

1. Find the equation that represents the relation connecting x and 
y if the given data are 



X 




3 


4 


5 


6 


7 


8 


9 


10 


11 


12 


y 


5 


5 6 


6 


6 4 


7 


7 5 


8 2 


8 6 


9 


9 5 



2. Find the equation of the type y bx n that represents the relation 
between x and y. 



X 


i 


2 


3 


A 


5 


6 


7 


8 


9 


y 


2 5 


3 5 


4 3 


5 


5 6 


6 2 


6 6 


7 1 


7 5 



3. From the following data, find the relation of the type y = klQ mx 
between x and y: 



X 


i 


2 


3 


4 


5 


6 


7 


8 


y 


5 


8 


1 2 


1 9 


3 


4 8 


7 5 


11 9 



169. Differences. Before proceeding to investigate rules 
for the choice of the particular type of equation that will repre- 
sent the observed values, it is advisable to define and discuss 
differences. 

Let the observed values be (# y), (i = 0, 1, 2, , n). 
The first differences are defined by 

(159-1) At/ t ss y i+1 2/ t . 

The second differences are given by 

A 2 2/ t ss A2/ l+ i Ay,. 
In general, for k > 1, the differences of order k, or the kth 



528 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 160 

differences, are defined as 

(159-2) A*2A s A^-^+i - A k ~ l y l . 

It should be noted that, if the jth differences are constant, then 
all of the differences of order higher than j will be zero. 
From (159-1) and (159-2) it follows that 

2/i = 2/o + AT/O, 

2/2 = 2/i + AS/I = (2/0 + Ai/o) + (A 2 2/ + AT/ O ) 

= 2/o + 2A2/o + A 2 2/ , 

2/3 = 2/2 + ^2/2 = (2/0 + 2A?/o + A 2 i/o) + (A 2 ?/i + A/I) 
= (2/0 + 2A2/o + A 2 2/ ) 

+ (A 3 2/ + A 2 ?/o + A 2 2/ + A2/ ) 
= 2/o + 3A2/o + 3A 2 ?/o + A 3 2/o. 

These results can be written symbolically as 

y l = (1 + A)y , 2/2 = (1 + A) 2 2/ , 2/3 = (1 + A) 3 i/o, 

in which (1 + A)' acts as an operator on 2/0, with the exponent on 
the A indicating the order of the difference. This operator is 
analogous to the differential operators discussed in Chap. VII. 
By mathematical induction, it is established easily that 

(159-3) y k = (1 + A)*y . 

160. Equations That Represent Special Types of Data. There 
are certain types of data which suggest the equation that will 
represent the relation connecting the observed values of x and y. 
Some of the more common types will be discussed in this section. 

a. Suppose that a number of pairs of observed values (x t} yl) 
have been obtained by experiment. If the x l form an arith- 
metical progression and the rth differences of the y l are constant, 
then the relation connecting the variables is 

y = a + dix + a 2 2 + + a r x r . 

For if the rth differences are constant, all differences of order 
higher than r are zero, and hence, from (159-3), 

(160-1) y* = 2/0 
where 

(160-2) 



160 EMPIRICAL FORMULAS AND CURVE FITTING 529 

is simply the coefficient of a r in the binomial expansion for 
(1 + a) k . Moreover, it was assumed that the x t are in arith- 
metical progression so that, if x\ XQ = A#, then XH x = k Ax 
and 

, _ X k Xp 

- A - * 

Ax 

Now, the expression (160-2) is a polynomial of degree r in k y and 
therefore of degree r in x fc . It follows that, upon substitution of 



7 __ 
K 



Ax 



and the collection of like powers of x k in (160-1), this equation 
assumes the form 



The relation is true for all integral values of &, and therefore 
(160-3) y = a + aix + a 2 x 2 + + a r x r 

gives the relation existing between the variables for the given 
set of observed values. 

In general, a given set of observed values will not possess con- 
stant differences of any order, but it may be that the rth differ- 
ences are sensibly constant. Then an equation of the type 
(160-3) will be a good approximation for the relation between 
the variables. 

Various modifications of (160-3) can be made. If the values of 
x? form an arithmetical progression, whereas the values of the 
rth differences of the y? are constant, then the relation connect- 
ing the variables is 

(160-4) y m = a + aix n + a 2 (z n ) 2 + + a r (x n ) r . 

Here m and n can take either positive or negative values. The 
derivation of the formula is exactly like that given above if x t n is 
replaced by X l and y? by F t . 

If, in (160-4), m = n = 1 and r = 1, the equation assumes 
the form 

IT'-o. + a^ or i = o + or y - - 



530 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 160 



Curves having this equation are frequently of use in fitting data 
to observations measuring flux density against field intensity. 
The curve is the hyperbola having the lines 

ai , 1 

x anc l y = 

for asymptotes. If the values of l/y are plotted against those 
for 1/x, the result is a straight line. A few of these curves are 
plotted in Fig. 154. 




012 4 6 8 

Fio. 154. 
Example 1. Consider the following set of observed data: 



X 


y 


&y 


A 2 ?/ 


A 3 y 


A 4 ?/ 


1 


2 105 














703 








2 


2 808 




103 










0.806 




081 




3 


3 614 




184 




-0 002 






990 




0.079 




4 


4 004 




263 




-0.001 






1 253 




078 




5 


5.857 




341 




+0 003 






1 594 




081 




6 


7 451 




422 




-0.001 






2 016 




080 




7 


9.467 




0.502 










2.518 








8 


11.985 











160 



EMPIRICAL FORMULAS AND CURVE FITTING 



531 



In this case the third differences are sensibly constant and the rela- 
tion between x and y is approximately of the form 



The question of determining the values of the constants will be con- 
sidered in a later section. 

b. If the set of pairs of observed values (# t , y l ) is such that 
the values of # t form an arithme- 
tical progression and the corre- 
sponding values of ?/ t form a 
geometrical progression, then the 
equation representing the relation 
between the variables is 

(160-5) y = ka*. 

For, taking logarithms of both 
sides, the equation becomes 

log y = log k + x log a, 

which is linear in x and log y. 
Hence, if the values x z form an 
arithmetical progression, the val- 
ues log T/t will do likewise. But 
then log y ^ log y^\ = c (for each value of z), so that 

- '- = e c = C and 




Therefore, the numbers y, form a geometrical progression. 

It can be proved that, if the values of the rth differences of 
the y % form a geometrical progression when the values of the x r 
form an arithmetical progression, then the relation between x 
and y is 

(160-6) y = a + aix + - + a r -ix r ~ l + ka x . 
If r = 1 in (160-6), the equation becomes 
y = + ka x . 

If a > 1, the values of y increase indefinitely as x is increased. 
If a < 1, the curve falls off from its value at x = and approaches 
the line y = a as an asymptote. Three of these curves are 
plotted in Fig. 155. 



532 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 160 
Example 2. Consider the data given in the following table. 



X 


1 




2 


3 






4 


5 


6 




7 


y 


2 157 


3 


519 


4 198 


4 


539 


4 708 


4 792 


4 


835 


Ay 


1 


362 





679 





341 





169 


084 


043 





Since the first differences have values very nearly equal to the num- 
bers that form the geometrical progression whose first term is 1.362 




1 



2345 
Fio 156. 



X 



and whose ratio is K, the relation between x and y is very neaily of the 
form 

y = a 4- ka x . 
c. The equation 

(160-7) y = ax n 

represents the relation existing between the variables if, when 
the x % form a geometrical progression, the y l also form a geo- 
metrical progression. For if 



x % = 



then (160-7) states that 



y l = ax? 



Hence, the g/ form a geometrical progression whose ratio is 

R SB r n . 



160 EMPIRICAL FORMULAS AND CURVE FITTING 533 



If the first differences of the 2/< form a geometrical progression 
when the Xi form a geometrical progression, then the equation 
giving the relation between x and y is 

(160-8) * y = k + ax n . 

For if x t nr t _i, then (160-8) requires that 

A^i = 2A - y^-l = k + ax? (k + asIU) 



and, similarly, 



Then 



= a#Jli(r n 1) = ar n xJL 2 (r n 1) 



r n - 1). 



and therefore the Ai/ t form a geometrical progression whose ratio 
is R = *"*. 

The curves (160-8) are parabolic if n > and hyperbolic if 
n < 0. Three of each type are plotted in Fig. 156. 

Example 3. Let the pairs of observed values be 



X 


16 


4 


1 


2 5 


6 25 


15 625 


y 


2 


2 210 


2 421 


2 661 


2 929 


3 222 



The values of x l form a geometrical progression with ratio r = 2.5, 
and the values of y % are approximately equal to the terms of the geo- 
metrical progression whose first term is 2 and whose ratio is R = 1.1. 
Hence, the relation between x and y is very nearly of the type y = ax n \ 
and since R = r n , it follows that 1.1 = (2.5) n or log 1.1 = n log 2.5 
and 

log 1.1 



n = 



log 2.5 



PROBLEMS 

1. A flat surface is exposed to a wind velocity of v miles per hour, and 
it is desired to find the relation between v and p, which is the pressure 
per square foot on the surface. By experiment the following set of 
observed values is obtained. Find the type of formula to fit them. 



V 


10 


15 


22 5 


33 75 


50 625 


75 937 


p 


3 


0.675 


1.519 


3.417 


7 689 


17 300 



534 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 161 



2. The temperature 6 of a heated body, surrounded by a medium kept 
at the constant temperature 0C., decreases with the time. Find the 
kind of formula which expresses the relation between S and t that is 
indicated by the following pairs of observed values: 



t 





1 


2 


3 


4 


5 


6 


7 


8 





60 00 


51.66 


44 46 


38 28 


32 94 


28 32 


24 42 


21 06 


18 06 



3. If C represents the number of pounds of coal burned per hour per 
square foot of grate and H represents the height of the chimney in 
feet, find the type of formula connecting H and (7, using the following 
data: 



C 


19 


20 


21 


22 


23 


24 


25 


H 


81 


90 25 


100.00 


110 25 


121 00 


132 25 


144 00 



161. Constants Determined by Method of Averages. Several 
different methods are employed in determining the constants 
which appear in the equation that expresses the relation existing 
between the variables whose observed values are given. The 
method to be described in this section is known as the method 
of averages. It is based on certain assumptions concerning the 
so-called "residuals" of the observations. Let the pairs of 
observed values be (x % , y t ), and let y = f(x) be the equation that 
represents the relation between x and y for these values. Then, 
the expressions 

> = /() - y^ 

are defined as the residuals of the observations. The method of 
averages is based on the assumption that the gum 2^ is zero. 

This assumption gives only one condition on the constants 
that appear in y = f(x). If there are r of these constants and 
if f(x) is linear in them, the further^assumption is made that, if the 
residuals are divided into r groups, then 2X = for each group. 
This second assumption leads to r equations in the r unknown 
constants. It is obvious that different methods of choosing the 
groups will lead to different values for the constants. Ordinarily, 
the groups are chosen so as to contain approximately the same 
number of residuals; and if there are to be k residuals in each 
group, the first group contains the first k residuals, the second 
group contains the succeeding k residuals, and so on. 



161 



EMPIRICAL FORMULAS AND CURVE FITTING 



535 



A modification of this method is spmetimes used when f(x) 
is not linear in its constants, but it will not be discussed here. * 

Example. Determine the constants iii the equation that represents 
the data given in Example 1, Sec. 160. 

It was shown in this example that the equation is of the type 

+ OL& 4- &2# 2 4- 3# 3 . 



Therefore. 



f(x) = 
= a + 



Vl 



a 
a 
ao 



6ai 



#2 + a s 2.105, 
4a 2 + 8a 3 - 2.808, 
9a 2 + 27a 3 - 3.614, 
16a 2 + 64a 3 - 4.604, 
25a 2 + 125a 3 - 5.857, 
36a 2 + 216a 3 - 7.451, 
49a 2 + 343a 3 - 9.467, 
64a 2 + 512a 3 - 11.985. 



Let the assumptions be that 

vi + v 2 = 0, v z + v 4 =* 0, t; 5 

Then the conditions on the constants are 



0. 



2a + 3ai + 5a 2 + 9a 3 = 4.913, 
2a + 7ai + 25a 2 + 91a 3 = 8.218, 
2a + Hoi + 61a + 341a 3 == 13.308, 
2a + 15ai + 113a 2 + 855a 3 = 21.452. 

The solution of these equations is 

a = 1.433, ai = 0.685, a 2 = -0.025, a 3 = 0.013. 

Hence, the equation, as determined by the method of averages, is 
y = 1.433 + 0.685z - 0.025z 2 + 0.013z 8 . 

PROBLEMS 
1. Use the method of averages to find the constants in the equation 

y = a + aix + a 2 x 2 , 
which is to represent the given data 



X 


i 


2 


3 


4 


5 


6 


y 


3.13 


3 76 


6 94 


12 62 


20 86 


31.53 



2. Find, by the method of averages, an equation to fit the data 
given in Prob. 3 at the end of Sec. 160. 
* See SCARBOROUGH, Numerical Analysis. 



536 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 162 

3. Find the constants in the equation of the type 

y = a + dix + a& 2 + a&*, 
which fits the data given in the table 



X 


i 


2 


3 


4 


5 


6 


7 


8 


y 


3 161 


3 073 


3 104 


3 692 


5 513 


9 089 


15 123 


24 091 



4. Find, by the method of averages, the constants in the equation 
y a + be* if it is to fit the following data: 



X 


5 


1 


1 5 


2 


2 5 


3 


y 


1 630 


1 844 


2 196 


2 778 


3 736 


5 318 



6, Use the method of averages to determine the constants in 
y = ae x +.b sin x + ex 2 so that the equation will represent the data 
in the table. 



X 


4 


6 


8 


1 


1 2 


1 4 


1 6 


1 8 


2 


y 


258 


470 


837 


1 392 


2 133 


3 069 


4 225 


5 608 


7 216 



162. Method of Least Squares. This section introduces 
another method of determining the constants that appear in the 
equation chosen to represent the given data. It is probably 
the most useful method and the one most frequently applied. 
The two methods already described give different values of the 
constants depending upon the judgment of 1^he investigator, 
either in reading from a graph or in combining the residuals into 
groups. This method has the advantage of giving a unique set 
of values to these constants. Moreover, the constants deter- 
mined by this method give the "most probable " equation in the 
sense that the values of y computed from it are the most probable 
values of the observations, it being assumed that the residuals 
follow the Gaussian law of error. In short, the principle of least 
squares asserts that the best representative curve is that for which 
the sum of the squares- of the residuals is a minimum. 

Suppose that the given set of observed values (# y), (i = 
1, 2, , n), can be represented by the equation 

V = /(*) 



162 EMPIRICAL FORMULAS AND CURVE FITTING 537 

containing the r undetermined constants ai, a^ , a r . Then, 
the n observation equations 

y^ = /(**) 

are to be solved for these r unknowns. If n = r, there are just 
enough conditions to determine the constants; if n < r, there 
are not enough conditions and the problem is indeterminate; but, 
in general, n > r, and there are more conditions than there are 
unknowns. In the general case, the values of the ak which 
satisfy any r of these equations will not satisfy the remaining 
n r equations, and the problem is to determine a set of values 
of the ak that will give the most probable values of y. Let 

v* = & - y* 

be th residuals, or deviations of the computed values from the 
observed values, where y % is the value of y obtained by sub- 
stituting x = x l in y = /(#). On the basis of the Gaussian law of 
error, the probability of obtaining the observed values y l is 



Obviously, P is a maximum when 2 ^? is a minimum. 

1 = 1 

n 

Since S = v\ is a function of the r unknowns ai, a 2 , , 

1 = 1 

a r , it follows that necessary conditions for a minimum are 



Moreover, each t; is a function of a^; therefore, 



- 2* ! + a* 1 



Equations (162-1) are called the normal equations. 

If it happens that the r equations (162-1) are linear in the r 
unknowns a 1; a 2; * , a r , then these equations can be solved 
immediately. This will certainly be the case if f(x) is a poly- 
nomial. For let 



538 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 162 

r r 

f(x) = 2 a i x '~ l so that v l = 2 ^X" 1 ~ !/ 

Then, dv x /da k = zj'- 1 , and the normal equations assume the 
form, with the aid of (162-2), 

(162-3) 2(2 aX- 1 - 2/0 tf- 1 = 0, (& = 1, 2, , r). 
It should be noted that the equation which is obtained by setting 

n 

k = 1 is S t\ = 0. Reordering the terms in (162-3), so as to 
collect the coefficients of the a,, gives 

r / n n 

Hfi2-4^ Vl V^ r '-4-fc-2\ _ ^ r /c_i ? . //, _ i o r\ 

^lU^-'-ty ^V 1 ^\ X t I Uj / X t t/t, \l\j 1, ^>, j '/ 

J=ll=l t=l c 

The r linear equations (162-4) can then be solved for the values 
of the r unknowns a\, 0,%, - - - , a r . 

This procedure may be clarified somewhat by writing out some 
of these expressions for a simple specific case. Consider the 
data given in the following table. Since the second differences 



X 


i 


2 


3 


4 


y 


1 7 


1 8 


2 3 


3 2 



of the y t are constant, the equation will have the form 

// \ i i n 

Then, v l = ai + ^x l + s^ t 2 2A, and 

dv t _ dv t _ dVi __ 2 

The normal equations 



are 




162 EMPIRICAL FORMULAS AND CURVE FITTING 539 

If the coefficients of the a, are collected and the normal equa- 
tions put in the form (162-4), one obtains the three equations 



4 4 






4 



a 2 + x a 3 = 

\-i 7 \ = i ' 

Now, 

4 

V x t = 1 + 2 + 3 + 4 = 10, ^ x? - 1 + 4 + 9 + 16 = 30, 

*=i 

t = 1.7 + 3.6 + 6.9 + 12.8 = 25, etc. 

The equations become 

4at + 10a 2 + 30a 3 = 9, 
lOai + 30a 2 + lOOa* = 25, 
30ai + 100a 2 + 354a 3 = 80.8; 
and the solutions are a x = 2, a 2 = 0.5, a 3 = 0.2. 

Even when Eqs. (162-1) are not linear in the unknowns, it may 
be possible to solve them easily. However, in most cases it is 
convenient to replace the exact residuals by approximate residuals 
which are linear in the unknowns. This is accomplished by 
expanding y = /(#), treated as a function of ai, a 2 , , a r , in 
Taylor's series in terms of a r a t = Aa t , where the a t are 
approximate values of the a t . The values of d t may be obtained 
by graphical means or by solving any r of the equations ?/, = 
f(x l ). The expansion gives 

(162-5) y = J(x, ai, , a r ) s /(^, di + Aai, , a r + Aa r ) 



/c =* 1 



where 

Af /)f /)2f ^2f 

, etc. 



fc =dfc da j 



jfc d* 



540 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 162 

Assuming that the d t are chosen so that the Aa t are small, the 
terms of degree higher than the first can be neglected and (162-5) 
becomes 

y = f(x, oi, , fir) 



The n observation equations are then replaced by the n 
approximate equations 



(162-6) 



If (162-6) is used, the residuals v ^ will be linear in the Aa/ f , and 
hence the resulting conditions, which become c 

(162-7) 

also will be linear in the Aa*,. Equations (162-7) are called the 
normal equations in this case. 

In order to illustrate the application of the method of least 
squares, two examples will be given. In the first the polynomial 
form of f(x) permits the use of (162-4), whereas the second 
requires the expansion in Taylor's series. 

Example 1. Compute the values of the constants appearing in the 
equation of Example 1, Sec. 160. 

The equation isy = a + a\x + a& 2 + a z x 3 , and from the given data 
it appears that the normal equations are 



8a + x>) 01 

V 



/ 

( 2} *') o + ( 2} ^ 

\1 7 x tl 

8 / 8 \ / 8 \ / 8 \ 8 

(2) *') o + (2) ***) i + (2) ^ 5 ) a2 + (X * 6 ) 3 = 2) *^.. 



From the form of the coefficients of the a*, it is seen that it is con- 
venient to make a table of the powers of the x t and to form the sums 



EMPIRICAL FORMULAS AND CURVE FITTING 



541 



Sav and 2z t *2/t before attempting to write down the equations in 
explicit form. 



X* 


xS 


* t 3 


z 4 


* 5 


* 


1 


1 


1 


1 


1 


1 


2 


4 


8 


16 


32 


64 


3 


9 


27 


81 


243 


729 


4 


16 


64 


256 


1,024 


4,096 


5 


25 


125 


625 


3,125 


15,625 


6 


36 


216 


1,296 


7,776 


46,656 


7 


49 


343 


2,401 


16,807 


117,649 


8 


64 


512 


4,096 


32,768 


262,144 


2*,' 36 


204 


1,296 


8,772 


61,776 


446,964 



x t 


y^ 


#t2A . 


xSy, 


x\ y\ 


1 


2 105 


2.105 


2 105 


2 105 


2 


2 808 


5 616 


11.232 


22 464 


3 


3 614 


10 842 


32.526 


97 578 


4 


4 604 


18 416 


73 664 


294 656 


5 


5 857 


29 285 


146 425 


732 125 


6 


7 451 


44 706 


268 236 


1,609 416 


7 


9 467 


66 269 


463 883 


3,247 181 


8 


11 985 


95 880 


767 040 


6,136 320 


2o^2/ t 


47 891 


273 119 


1,765 111 


12,141 845 



When the values given in the tables are inserted, the normal equa- 
tions become 



36a 

204a 

l,296ao + 

The solutions are 
a = 1.426, 



36ai 
204a! 
l,296ai 



204a 2 + I,296a 3 = 47.891, 

I,296a 2 + 8,772a 3 = 273.119, 

8,772a 2 + 61,776a 3 = 1,765.111, 

61,776a 2 + 446,964a 3 = 12,141.845. 



= 0.693, a 2 = -0.028, a 3 = 0.013. 



Therefore, the equation, as determined by the method of least squares, 

is 

y = 1.426 + 0.693z - 0.028z 2 + 0.013z 3 . 

It will be observed that these values of the constants are very nearly 
the same as those obtained by the method of averages. 



542 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 162 



Example 2. Compute the constants that appear in the equation 
that represents the following data: 



t 


i 


2 


3 


4 


e 


51 66 


44 46 


38 28 


32 94 



Since the observed values are such that the I* form an arithmetic 
progression and the t approximately form a geometric progression, the 
equation expressing the relation is of the form 

6 = ka*. 

If the points are plotted on semilogarithmic paper, it is found that 
k = 60 and a = 10- 65 = 0.86, approx. This suggests using k = 60 
and o = 0.9 as the first approximations. The first two terms of the 
expansion in Taylor's series in terms of Afc = k 60 and Aa = a 0.9 
are 



e = 60(0.9)' + ^Jjb-w AA; + l^ j*-w Aa 

= 60(0.9)' - 

If the values (, t ) are substituted in this equation, four equations 
result, namely, 

0, 60(0.9)' + (0.0)'* AA; + GO^O.g)''- 1 Aa, (i = 1, 2, 3, 4). 

The problem of obtaining from these four equations the values of Afc 
and Aa, which furnish the most probable values of t , is precisely the 
same as in the case in which the original equation is linear in its con- 
stants. The residual equations are 

v. = (0.9)' AA; + 60J,(0.9)'*- 1 Aa + 60(0.9)'* - t , (i = 1, 2, 3, 4). 
Therefore, 

4 4 

S = tv> = V [(0.9)* Afc + 60M0.9)'*- 1 Aa + 60(0.9)'* - 0J 2 , 



and the normal equations 
dS 



= and 



dS 



become 



=-- 



60(0.9)< - (9,]0. 



1-1 



162 EMPIRICAL FORMULAS AND CURVE FITTING 543 
and 

4. 

[0.9AA? + 60e,(0.9)'- l Aa + 60(0.9)'* - 0J60M0.9)''- 1 = 0. 

When these equations are written in the form 

p Afc + q Aa = r, 
with all common factors divided out, they are 



44 

(0.9) 2 <>A/c + 60 2 



4 4 

'-'Aa = T t (0.9)<> - 60 V (0.9) 



and 



- 60 2^ ,(0.9)- 1 . 

As in Example 1, the coefficients are computed most conveniently by 
the use of a table. 



* 


1 


2 


3 


4 


Totals 


(0 9)'. 


9 


81 


729 


6561 




(09) 2 <* 


81 


6561 


531441 


43046721 


2 42800821 


*,(0.9)".-* 


9 


1.458 


1 77147 


1 9131876 


6 0426576 


t? (09) 2< ." a 


1 


3 24 


5 9049 


8 503056 


18 647956 


(0t)(09)'. 


4C 494 


36 0126 


27 90612 


21.611934 


132 024654 


(^,)(09)'.- 1 


51 66 


80 028 


93.0204 


96 05304 


320 76144 



Substituting the values of the sums from the table gives 

2.42800821 AA? + 362.559456 Aa = 132.024654 - 145.6804926 
and 

6.0426576 A/c + 11 18.87736 Aa = 320.76144 - 362.559456. 

Reducing all the numbers to four decimal places gives the following 
equations to solve for Afc and Aa: 



544 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 163 

2.4280 A/c+ 362.5595 Aa = -13.6558, 
6.0427 M + 1118.8774 Aa = -41.7980. 

The solutions are 

A/c = -0.238 and Aa = -0.036. 
Hence, the required equation is 

= 59.762(0.864)'. 

PROBLEMS 

1. Find the constants in the equation for the data given in Prob. 1, 
Sec. 161. Use the method of least squares. 

2. Use the method of least squares to determine the values of the 
constants in the equation that represents the following data. 



X 


2 


4 


6 


8 


1 


y 


1 25 


1 60 


2 00 


2 50 


3 20 



3. Apply the method of least squares to the data given in Prob. 1, 
Sec. 158. 

4. Apply the method of least squares to determine the values of 
a and b in Prob. 4, Sec. 161. 

163. Method of Moments. Since the method of moments is 
one of the most popular methods in use by the statisticians and 
economists, a brief discussion of it will be presented. For certain 
types of equations, especially those which are linear in their 
constants, it provides a simple method of determining the 
constants. If the equation has the form 



r-l 



a k x K 



this method gives results identical with those obtained by the 
method of least squares. In this case the method has a theo- 
retical background that justifies its use. When the method is 
applied to other types of equations, there is, in general, no such 
justification. However, in modified forms it is convenient for 
computation and often gives very good results. 

Let the set of observed values be (z, t/ t ), (i = 1, 2, , ri), 
and the equation that represents these data be y = /(#). When 
the values x = x t are substituted in f(x), there result the corre- 
sponding computed values of y, which will be designated by 



164 EMPIRICAL FORMULAS AND CURVE FITTING 545 

?/ t . The moments of the observed values y % and of the computed 
values y t are defined, respectively, by 



n n 

= 2 x\y^ and M* = 2 
1=1 1=1 



If f(x) contains r undetermined constants, the method of moments 
is based on the assumption that 

(163-1) TA = MA, (A = 0, 1, 2, , r - 1). 

Since y, is a function of the r undetermined constants, Eqs. 
(163-1) give r simultaneous equations in these constants. 

The method of moments in this form is most useful when 
f(x) is linear in its r constants, so that the r equations (163-1) 
can be solved immediately. Various modifications and devices 
are used to simplify the computation in case f(x) is not linear in 
its constants. These will not be discussed here.* 

In the special case in which f(x) is a polynomial, that is, 

r-l 

f(x) = 2 a ^> 

j=0 

the values of y % are given by 

r-l 

& = 2 a,x't 



and therefore 

n r-l 



\ = 2 2 x ?" a (* = 0, 1, 2, , r - 1). 

1 = 1 j = j=0 i = l 

Then Eqs. (163-1) assume the form 

2 2 x * +ka < = 2 ^' (fc = 0, 1, 2, , r - 1), 

;=0 t = l i = l 

which are identical with the normal equations (162-4) obtained 
by the method of least squares. Hence, the two methods lead 
to identical results for this form of /(#). 

164. Harmonic Analysis. The problem of obtaining the 
expansion of a periodic function in an infinite trigonometric 

* For a discussion, see Frechet and Romann, Representation des lois 
empiriques; Rietz, Handbook of Mathematical Statistics. 



546 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 164 

series was considered in Chap. II. In this section will be given 
a short discussion of the problem of fitting a finite trigonometric 
sum to a set of observed values (x t , y l ) in which the values of y 
are periodic. 

Let the set of observed values 

(X , ?/o), (Zi, 2/l), ' ' ' , (.T2-l, 2/2n-l), (Z 2 n, 2/2), ' ' ' 

be such that the values of y start repeating with y^ n (that is, 
2/2n = ?/o, 2/2+i = 2/i, etc.). It will be assumed that the x l are 
equally spaced, that x = 0, and that x* 2 = 2ir. [If ^ and 
the period is c, instead of 2?r, the variable can be changed by 
setting 

t = (x, x- ). 
c 

The discussion would then be carried through for t and y % in 
place of the x v and ?/, used below.] Under these assumptions 



. 2ir ITT 

Xi = 

The equation 



i 1 ^T 

2n n 



n 

(164-1) y AO + V A k cos kx + V B k sin kx 



n l 

V 
i ffi 

contains the 2n unknown constants 



which can be determined so that (164-1) will pass through the 
2n given points (x l9 y t ) by solving the 2n simultaneous equations 

n n 1 

y, = AO + " A fc cos fcx t + V ^A, sin fcx,, 
ft-i /fe = i 

(i = 0, 1, 2, , 2n - 1). 

Since x l = zV/n, these equations become 

n n-1 

/1/>yl rtN . . ^1 . zA:7T . ^%T1 n . zfcir 

(164-2) i/ 1 = AO + >. A fc cos -- h >, #* sin , 



(t = 0, 1, 2, , 2n - 1). 

The solution of Eqs. (164-2) is much simplified by means of a 
scheme somewhat similar to that used in determining the Fourier 



164 EMPIRICAL FORMULAS AND CURVE FITTING 547 

coefficients. Multiplying both sides of each equation by its 
coefficient of A Q (that is, by unity) and adding the results give 

2n-l n 277-1 x n-1 ,271-1 



t = ( 

It can be established that 

2n-l 

^ cos ^ = 0, (k = 1, 2, , n), 
and 

277-1 

^j sin = 0, (k = 1, 2, , n 1). 
Therefore, 

2n-l 

(164-3) 2nA = ] y t . 

Multiplying both sides of each equation by its coefficient of 
A ]y (j = 1, 2, , n 1), and adding the results give 

^ y l cos - = ^, ( >, cos cos -' 1 Ak 
^J y n ^J \^J n n / 

, ^7* /%^ ikir tjir\ 

+ >, >, sm cos ) 

^-J \^J n n / 

for j = 1, 2, , n - 1. But 

cos - cos = 0, if k 5* j, 

n n ' J) 

= n, if k = j; * 

and 

2n - 1 

. ikir ijw A 
sm cos = 



>k 



n n 

for all values of k. Therefore, 



(164-4) nA, = "S- y. cos ? A (j - 1, 2 ; - , n - 1). 



/& 



548 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 164 

In order to determine the coefficient of A n the procedure is 
precisely the same, but 

2n-l ^ 

IKTT ~ . f 7 . 

cos cos ITT 0, if k 5^ n, 



= 2n, if A; = n. 

Hence, 

2n-J 

(164-5) 2nA n = y, cos zV. 

i = 

Similarly, by multiplying both sides of each equation of 
(164-2) by its coefficient of R k and adding, it can be established 
that 

2n-l 

(164-6) nB, = ^ sin V 0' = 1, 2, - , n - 1). 

t = 

Equations (164-3), (164-4), (164-5), and (164-6) give the 
solutions for the constants in (164-1). A compact schematic 
arrangement is often used to simplify the labor of evaluating 
these constants. It will be illustrated in the so-called "6-ordi- 
nate" case, that is, when 2n = 6. The method is based on the 
equations that determine the constants, together with relations 
such as 

. TT . (n - l)ir . (n + })TT . (2n - I)TT 

sin - = sin - = sin ------ - ~ = sin ----- j 

n n n n 

TT (n - !)TT (n + l)ir (2/1 - I)TT , 

cos - = ~ cos - - = cos -- = cos - - ~; etc. 
n n n n 

Six-ordinate Scheme. Here, 2n = 6, the given points are 
fo, y t ), where x l = nr/3, (i = O t 1, 2, 3, 4, 5), and Eq. (164-1) 
becomes 

y = AQ + A\ cos x + A 2 cos 2x + A 3 cos 3# + ^i sin x + B z sin 2x. 
Make the following table of definitions: 

2/o y\ 2/2 t>o #1 WQ W\ 

y\ 



_ _ _ 

Sum \VQ v\ Vz Po pi r r\ 

Difference! w Wi w 2 q\ Si 



164 



EMPIRICAL FORMULAS AND CURVE FITTING 



549 



It can be checked easily that Eqs. (164-3), (164-4), (164-5), 
and (164-6), with n = 3, become 



6A 3 = r si, 



3*! = 



In particular, suppose that the given points are 



X 





7T 

3 


27T 

3 


7T 


4r 

3 


STT 

'3 


27T 


y 


1.0 


1 4 


1 9 


1 7 


1 5 


1 2 


1 



Upon using these values of y in the table of definitions above, 





1 

1 7 


1 4 
1 5 


1 9 
1 2 




VQ = 


2 7 


vi = 29 


v 2 = 31 




WQ = 


-0 7 


MI = -0 1 


w; 2 = 07 






2 7 


2 9 


-0 7 


-0 1 






3 1 




7 


P G 


,=27 


pi = 6 


r = -0 7 


ri == 6 






qi = -0 2 




*i == -0.8 



Therefore, the equations determining the values of the con- 
stants are 



3A 2 = 



2.7 + 6.0 = 

0.7 - 0.4 = 

2.7 - 3.0 = 

0.7 + 0.8 = 



8.7 
-1.1 
-0.3 

0.1 



(0.6) = 0.3 



(-0.2) = -0.1 V3 



and 
and 
and 
and 

and 
and 



lo = 1.45, 

Li = -0.37, 

1 2 = -0.10, 

L 3 = 0.02, 

?i - 0.17, 
h = -0.06. 



Hence, the curve of type (164-1) that fits the given data is 

y = 1.45 - 37 cos x - 10 cos 2x -f 02 cos 3x + 0.17 sin x - 0.06 sin 2x. 

A convenient check upon the computations is furnished by the 
relations 



and 



+ B 2 = 



(y, - 2/5). 



550 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 166 

Substituting the values found above in the left-hand members 
gives 

1.45 - 0.37 - 0.10 + 0.02 = 1.0 and 0.17 - 0.06 -= 0.11, 

which check with the values of the right-hand members. 

Similar tables can be constructed for 8-ordinates, 12-ordinates, 
etc.* 

PROBLEMS 

1. Use the 6-ordinate scheme to fit a curve of the type (164-1) to the 
data in the following table : 



X 





7T 

3 


27T 

3 


7T 


47T 

3 


5T 

3 


2ir 


y 


8 


6 


4 


7 


9 


1 1 


8 



2. Make a suitable change of variable, and apply the 6-ordinate 
scheme to the data given in the table 



X 





7T 

6 


* 
3 


7T 

2 


27T 

~3 


5;r 
6 


7T 


y 


6 


9 


1 3 


1 


0.8 


5 


6 



165. Interpolation Formulas. When an equation has been 
obtained to represent the relation existing between x and y, as 
indicated by a given set of observed values (x ly y l ), this equation 
can be used to determine approximately the value of y corre- 
sponding to an arbitrary value of x. It would be expected that 
the equation would furnish a good approximation to the value 
of y corresponding to an x which lies within the range of the 
observed values x t . The equation may provide a good approxi- 
mation for y even if x is chosen outside this range, but this must 
not be assumed. 

Frequently, it is desired to obtain an approximation to the y 
corresponding to a certain value of x without determining the 
relation that connects the variables. Interpolation formulas 
have been developed for this purpose and for use in numerical 
integration (mechanical quadrature).! The formulas to be 

* See CAUSE and SHEARER, A Course in Fourier Analysis and Periodogram 
Analysis. 

t See Sees. 167 and 168. 



165 EMPIRICAL FORMULAS AND CURVE FITTING 551 

discussed here all assume that the desired value for y can be 
obtained from the equation 

y == a + dix + a 2 x 2 + + a m x m , 

in which the a t have been determined so that this equation is 
satisfied by m + 1 pairs of the observed values (#,, z/,). These 
m + 1 pairs may include the entire set of observed values, or 
they may be a subset chosen so that \x x l is as small as possible. 
The first interpolation formula of this discussion assumes that 
the set of m + 1 observed values #0, #1, #2, * * , x m is an 
arithmetic progression, that is, that 

(165-1) x k = xi^i + d = xo + kd, (k = 1, 2, , m). 

Since there are m + 1 pairs of observed values, there is only one 
mth difference A w t/, and all differences of order higher than m 
are zero. Hence, by (159-3), 



k A?/o H --- 2! 

*(* - 1) (fc - 



But, from (165-1), it follows that 

x k - x 
k ~ ~~d~~' 

so that the expression for yj c becomes 

/Tr.c 0\ . %k XO * , (%k 

(165-2) y k == 7/0 H -- ^ AT/O H 

So - d) ' ' ' 



..._ o . 

Relation (165-2) is satisfied by every one of the m + 1 pairs 
of observed values. Now, assume that the value of the y which 
corresponds to an arbitrary x also can be obtained from (165-2). 
Then, 



,+ n - \ i o A , Q XQ 

(165-3) 2/ = yo + J - j ^ A2/o + ^ - '-^jj 

, (x x<>)(x x d) (x XD md + d) 

' ' - - 



552 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 166 

Equation (165-3) represents the mth-degree parabola which 
passes through the m + 1 points whose coordinates are (x t , y l ). 
It assumes a more compact form and is more convenient for 

computation purposes when -. - is replaced by X. Then, 



(165-4) y = i/o + X Ay + " A 2 7/ + 



Example. Using the data given in Example 1, Sec. 160, determine 
an approximate value for the y corresponding to x = 2.2. 

First, let y be determined by using only the two neighboring observed 
values (hence, m = 1). Then, x = 2, T/ O = 2.808, AT/ O = 0.806, and 

x = M_H? = 0/2 . Hence, 

y - 2.808 + 0.2(0.806) = 2.969, 

which has been reduced to three decimal places because the observed 
data are not given more accurately. Obviously, this is simply a 
straight-line interpolation by proportional parts. 

If the three nearest values are chosen, m = 2, x = 1, y = 2.105, 
A#o = 0.703, A 2 ?/o = 0.103, and X = 2.2 - 1 = 1.2. Then, 



y = 2.105 + 1.2(0.703) + ' (0.103) = 2.961, 

correct to three decimal places. 

If the four nearest values are chosen, m = 3, X Q = 1, y 2.105, 
Ay = 0.703, A 2 ?/o = 0.103, A 3 7/ == 0.081, and X = 1.2. Therefore, 

y = 2.105 + 1.2(0.703) + (1>2) 2 ( ' 2) (0.103) 

+ (L2)( - 2 6 )( -- 8) (0.081) = 2.958, 

correct to three decimal places. 

The value obtained by substituting x 2.2 in the equation 

y = 1.426 + 0.693z - 0.028z 2 + 0.013z 3 , 

obtained by the method of least squares (see Example 1, Sec. 162) 
is 2.954. It might be expected that a better approximation to this 
value could be obtained by choosing m 4, but investigation shows 
that the additional term is too small to affect the third decimal place. 

166. Lagrange's Interpolation Formula. The interpolation 
formula developed in Sec. 165 applies only when the chosen set 



166 EMPIRICAL FORMULAS AND CURVE FITTING 553 

of Xi is an arithmetic progression. If this is not the case, some 
other type of formula must be applied. 

As in Sec. 165, select the m + 1 pairs of observed values for 
which \x x % \ is as small as possible, and denote them by (x ly 7/ t ), 
(i = 0, 1, 2, , m). Let the mth-degree polynomials Pk(x), 
(k = 0, 1, 2, , m), be defined by 

(166-1) P k (x) = (*- *o) (*-**)' ' ' (*-*) s JJ (a . _ ^ 

C .XL . -* 

Then, the coefficients ^4& of the equation 

m 

y=5^(*> 

can bedetermined so that this equation is satisfied by each of the 
m + 1 pairs of observed values (x lf y l ). For if x = XL, then 

A = ~^i- 
* P*(x*)' 

since Pk(xi) = 0, if i 5* k. Therefore, 
(166-2) y = 



is the equation of the mth-degree parabola which passes through 
the m + 1 points whose coordinates are (x ly y t ) . If x is chosen as 
any value in the range of the x,, (166-2) determines an approxi- 
mate value for the corresponding y. 

Equation (166-2) is known as Lagrange's interpolation formula. 
Obviously, it can be applied when the x, are in arithmetic progres- 
sion, but (165-4) is preferable in that it requires less tedious cal- 
culation. Since only one mth-degree parabola can be passed 
through m + 1 distinct points, it follows that (165-3), or its 
equivalent (165-4), and (166-2) are merely different forms of 
the same equation and will furnish the same value for y. 

Example. Using the data given in Prob. 1, Sec. 160, apply Lagrange's 
formula to find the value of p corresponding to v 21. 

If the two neighboring pairs of observed values are chosen, so that 
m 1, 

01 22 ^ 21 1 ^ 

P = ' 675 15^2275 + L519 22-5-=T5 = L350 ' 
correct to three decimal places. 



554 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 167 

If the three nearest values are chosen, so that m 2, 

- n .(21 - lfi)(21 -22.5) (21 - 10) (21 - 22.5) 

P - "<* (io _ I5)(io - 22.5) "*" u>0 ' (15 - 10)(15 - 22.5) 

, , , 1Q (21 - 10)(21 - 15) 
+ L519 (22.5 - 10)(22.5 - 15) = L323 ' 
correct to three decimal places. 

The value of p obtained from the equation p = 0.003z> 2 , which repre- 
sents the given data, is also 1.323. 

PROBLEMS 

1. Using the data given in Prob. 2, Sec. 160, find an approximate 
value for 6 when t = 2.3. Use m = 1, 2, and 3. 

2. Find an approximate value for the y corresponding to x 2, 
using the data given in Example 3, Sec. 160. Use m = I and m = 2. 

3. If the observed values are given by the data of Prob. 3, Se r c. 160, 
find an approximate value of H when C = 21.6. Use m 1, 2, and 3. 

4. Using the data of Prob. 1, Sec. 160, find an approximate value for 
p when v = 30. Use m 1 and m = 2. 

167. Numerical Integration.* The definite integral P f(x) dx 
is interpreted geometrically as the area under the curve y = f(x) 
between the ordinates x = a and x = b. If the function f(x) is 
such that its indefinite integral F(x) can be obtained, then from 
the fundamental theorem of the integral calculus it follows that 



However, if the function f(x) does not possess an indefinite inte- 
gral expressible in terms of known functions or if the value of 
f(x) is known only for certain isolated values of x, some kind of 
approximation formula must be used in order to secure a value 

(OT f b j(x) dx. 

A formula of numerical integration, or mechanical quadrature, 
is one that gives an approximate expression for the value of 
fa/(#) dx. The discussion given here is restricted to the case 
in which m + 1 pairs of values (x t , y l ), or [x l9 f(x % )], are given 
[either by observation or by computation from y f(x) if the 
form of f(x) is known] and where this set of given values is 
represented by (165-3) or (166-2). 

The formulas of numerical integration that are most frequently 
used are based on the assumption that the x l form an arith- 

* For discussion of the accuracy of the formulas given here, see Steffensen, 
Interpolation; and Kowalewski, Interpolation und genaherte Quadratur. 



167 EMPIRICAL FORMULAS AND CURVE FITTING 555 

metic progression, that is, that x k = XQ + kd. In that Case, all 
the m + 1 points (#, y t ), (i = 0, 1, 2, , w), lie on the 
parabola whose equation is given by (165-3). The area bounded 
by the x-axis and this parabola between x = XQ and x = x m 
is an approximation to the value of f x x "f(x) dx. 
Upon using (165-4) and recalling that 



v x 

A = ; 



it follows that 
(167 



-1) ^ y dX = JJ" [2/0 + X Ay, + 




If m = 1, (167-1) becomes 



But 

, , , v X XQ 

rf /> I 'yw/Y Q TJ /i X . 

tH/m ~~~ "H) \^ ifiAJU atH.\ji ^\. , 7 

a 
so that 

and the formula becomes 
(167-2) 

If n + 1 pairs of values are given, (167-2) can be applied 
successively to the first two pairs, the second and third pairs, 
the third and fourth pairs, etc. There results 

(167-3) t Xn ydx= I * y dx + I ** y dx + + I y dx 

Jxo Jx* Jxi Jx n -i 

_ d d 

2 2 

d 

d , 

^ -j- 2yi -f* 2t/2 "f~ * * * H" 2?/ n _i -)- ?/n)' 
2 



556 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 167 



/ 


^ 


^^ 


Y4 


^ 

Ys 


"^ 






Ye 


* 










l 












X 



X X, X 2 X 3 X4 X 5 X 6 

FIG. 157. 

If m = 2, (167-1) becomes 



Formula (167-3) is known as 
the trapezoidal rule, for it gives 
the value of the sum of the areas 
of the n trapezoids whose bases 
are the ordinates T/ O , yi, y%, ' ' ' , 
*, y n . Figure 157 shows the six 
trapezoids in the case of n = 6. 



Jo Jo L 



X 



- J/o) + 3 (2/2 - 
1 



or 

(167-4) 



r 

ty -^o 






Suppose that there are n + 1 pairs of given values, where n is 
even. If these n + 1 pairs are divided into the groups of three 

pairs with abscissas x^ x 2 i+i, 2^+2, ( i = 0, 1, , = \ 
then (167-4) can be applied to each group. Hence, 
(167-5) f xn ydx= f"ydx+ f*'ydx+ - - - 

JXO JXQ JXl 

= (t/o + 47/i + 2/2) + (2/2 + 



y dx 
+ 2/4) 



d 
3 



3 [2/0 + y n + 



+ 2/3 + ' + 2/n-l) 
+ 2(7/2 + 7/4 + + 7/ n _ 2 )]. 

Formula (167-5) is known as Simpson's rule with m = 2. 
Interpreted geometrically, it gives the value of the sum of the 
areas under the second-degree parabolas that have been passed 
through the points (x* t 7/ 2t ), (# 2l +i, 2/ 2 t+i), and (z 2l + 2 , 7/ 2l + 2 ), [i = 0, 

1, 2, - - - , (n - l)/2]. 



167 EMPIRICAL FORMULAS AND CURVE FITTINO 557 
If m = 3, (167-1) states that 



y dX = y + X Aj/o 



+ g ^~ A'yo 

27 9 . 



) 



9 9 

(2/i ~ 2/0) + (2/2 - 



o 

+ g (2/3 - 82/2 + 82/1 - 2/0) 



3 

(2/0 + 32/i + 82/2 + 2/3), 



or 



(167-6) I 3 y dx = ^ (2/0 + 3y, + 82/2 + y,). 

J^o o 

If n + 1 pairs of values are given and if n is a multiple of 3, 
then (167-6) can be applied successively to groups of four pairs 
of values to give 

C Xn ^d 

(167-7) ydx = ~[y Q + y n + 3(yi + 2/2 + 2/4 + 2/5 + 

J^o O 

+ l/n-2 + 2/n-l) + 2(2/3 + 2/6 + ' ' ' + J/n-s)]- 

Formula (167-7) is called Simpson's rule with m = 3. It is 
not encountered so frequently as (167-3) or (167-5). Other 
formulas for numerical integration can be derived by setting 
m = 4, 5, in (167-1), but the three given here are sufficient 
for ordinary purposes. In most cases, better results are obtained 
by securing a large number of observed or computed values, so 
that d will be small, and using (167-3) or (167-5). 

Example. Using the data given in Example 1, Sec. 160, find an 
approximate value for J t y dx. 

Using the trapezoidal rule (167-3) gives 



y dx = H (2.105 + 5.616 + 7.228 + 9.208 + 11.714 

+ 14.902 + 9.467) = 30.120. 



558 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 168 
Using (167-5) gives 

C 7 y dx = ^[2.105 + 9.467 + 4(2.808 + 4.604 -f 7.451) 

+ 2(3.614 + 5.857)] = 29.989. 
Using (167-7) gives 



dx = %[2.105 + 9.467 + 3(2.808 + 3.614 

+ 5.857 + 7.451) + 2(4.604)] = 29.989. 

168. A More General Formula. If numerical integration is 
to be used in a problem in which the form of f(x) is known, the 
set of values (x^ y) can usually be chosen so that the x l form an 
arithmetic progression and one of the formulas of Sec. 167 can 
be applied. Even if it is expedient to choose values closer 
together for some parts of the range than for other parts, the 
formulas of Sec. 167 can be applied successively, with appro- 
priate values of d, to those sets of values for which the x^ form an 
arithmetic progression. However, if the set of given values was 
obtained by observation, it is frequently convenient to use a 
formula that does not require that the o? t form an arithmetic 
progression. 

Suppose that a set of pairs of observed values (x % , y l ), (i = 0, 
1, 2, , m), is given. The points (x r , 7/ t ) all lie on the 
parabola whose equation is given by (166-2). The area under 
this parabola between x = XQ and x = x m is an approximation to 

XXm 
y dx. The area under the parabola (166-2) is 


C Xa ^ in C Xm 

(168-1) ydx^y^ =** P k (x) dx, 



in which the expressions for the Pk(x) are given by (166-1). 
If m = 1, (168-1) and (166-1) give 



(168-2) ydx= - (x - xO dx 



C XI 

(x - x,) dx 



XQ 



Formula (168-2) is identical with (167-2), as would be expected, 
but the formula corresponding to (167-3) is 



168 EMPIRICAL FORMULAS AND CURVE FITTING 559 
(168-3) f*" y dx = %[(xi - *o)(tfo + J/i) + (x 



If m = 2, (168-1) becomes 

(168-4) f" y dx = p^ P (* ~ *iX* - **) dx 

Jxt Jro(XQ) Jxo 

+ ETT-N G - so)(z - 

"l(^0 J^o 



J/L_H! 
i(zi) L 



3 2 

+ 

, (_ _,_ ^(,,.2 _ 3.8) 



(x 



-*)] 



Formula (168-4) reduces to (167-4) when Xi XQ = x 2 Xi = 
d. The formula that corresponds to (167-5) is too long and 
complicated to be of practical importance, and hence it is omitted 
here. It is simpler to apply (168-4) successively to groups of 
three values and then add the results. 

Example. Using the data given in Example 3, Sec. 160, find an 
approximate value of | ' y dx. 
Using (168-3) determines 

y dx = K[0.24(4.21Q) + 0.6(4.631) + 1.5(5.082) 

0.16 

+ 3.75(5.590)] = 16.187. 



560 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 168 

Applying (168-4) successively to the first three values and to the 
last three values gives 



p- 2 
JoiG 



, ^ (0.84) 2 [2(1.2 - 0.32 - 1) , 2.210(-0.84) 
yax 6 L (-0.24X-0.84) (0.24)(-0.6) 



2.421(2 + 0.16 - 1.2)1 
J 



(0.84) (0.6) 

(5.25) 2 r 2.421(7.5 - 2 - 6.25) , 2.66(-5.25) 
"*" 6 L (-1.5X-5.25) "*" (1.5)(-3.75) 



(-1.5X-5.25) " (1.5)(-3.75) 

2.929(12.5 + 1 -7.5)1 _ ? Q4 
+ - (5.25)(3.75) - J " 17 ' 194 ' 

PROBLEMS 

/7 
y dx, using the data given 

in Example 2, Sec. 160, and applying (167-3). Find the approximate 
value if (167-5) is used. 

/*50 625 

2. Apply (168-3) to determine an approximate value for I p dv, 

/io 

using the data given in Prob. 1, Sec. 160. 

3. Work the preceding problem by applying (168-4). 

4. Apply (167-3) and (167-5) to the data given in Prob. 3, Sec. 160, 

in order to determine / H dC. 
Jio 

5. Find the approximate values of f \/4 + x 3 dx obtained by using 
x = 0, 1, 2, 3, 4, 5, 6 and applying (167-3) and (167-5). 



ANSWERS 

CHAPTER I 
Pages 14-15 

1. (a) convergent; (6) divergent; (c) divergent; (d) convergent; 
(e) convergent; (/) divergent; (g) convergent; (h) divergent. 

2. (a) convergent; (6) divergent. 

4. (a) divergent; (5) convergent; (c) divergent; (d) divergent; 
(e) convergent; (/) convergent; (g) divergent; (h) convergent; 
(s) convergent; 0) divergent. 

Page 22 

3. (a) 1 < x < 1; (6) all finite values; (c) 1 < x < 1; 
(d) x > 1 and x < -1. 

4. (a) -^ < x < 4; (6) 0; (c) -I < x < I. 

Pages 39-40 

1. (a) 1 + x + ~ + ~ + ; 

,, N X 3 . X 6 X 7 , 

(&) * - 51 + 5! " 71 + ' ' ' J 
- , N , x 2 , a; 4 x 

W X - 2 + 41 " 6 + " ' " '" 
# 3 x^ x"* 

w x ~ 3" + T ~ T + ' ; 



(fc)l+x+2 i - 1 j- lir -- gr +---. 
2. (a) (x - 1) - l A(x - I) 2 + K(z ~ I) 3 - 

(re - 2)2 . (x - 2) 3 




(e) 7 -f- 29(a; - 1) + 76(a; - I) 2 + H0(* - I) 3 -h 90(z - I) 4 

+ 39(o? - I) 5 -f 7(x 

4. All finite values of x. 6. x 2 < 1. 

561 



562 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 

Pages 45-46 

1. 0.984808; 2 - 10-". 3. 0.5446. 

2. 9. 6. 2.03617. 
8. (a) 0.3103; (6) 0.0201; (c) 0.9461; 

w ._|V^ ! -^ n + ^ n -... ; 



X " + T " ~ + >(/> - 937 ; W -0-1026; 



j. E! _j_ ! 4. :L 4 4. 2_ 5 4_ 37x6 I 
v*/ * ~*~ 2! 3! "*" 4! 5! 6! ' ' * ' 
11. a <> 0.24 radian or 14. 

Pages 53-56 

6. 7T/2. 15. 1.05. 

6. 214.5 ft.; 25.1 ft. 16. 1 69; 0.881. 

14. 2 \/2 E(V2/2, 7T/2) = 3.825. 

CHAPTER II 
Pages 75-76 

16 



lirins-gg * 



1 s cos x 2 ^V 2 _ i cos nx - 

n = 2 n 



Pages 77-78 

. (2n l)wx - 



2 

2 A (-l) nH 1 4 <A 1 /0 

T i' n sm n;rx ' 2 "" 5? ^ (2ra - I) 2 C S ^ ~" ^* x ' 

n=l n 1 

18f/7T 2 4\ . 7T^ 7T 2 . 27TO; /7T 2 _ ^\ ^ SlTX 

S I I ~1~ 111 ^^ ~O~ "o" Sin o I I O O T I ^^^ O~ 

7r tt L\l IV d J d \d 6 A J 6 

in ~ ] 

,^36 v (-1)' 



cos -3- 



CHAPTER III 
Page 85 

1. (a) 2, -0.75; (6) 1.22,- -0.73; (c) 1.08, -0.55, -0.77; (d) -0.57. 

2. 4.49. 



ANSWERS 563 



Page 91 

(a) 1.618, -1, -0.618. (6) 13.968, -6.984 0.29U*. 

(c) 3, -1, -1. (d) -1,1,2. 

(e) 2 4 v/3 4 V5, 2 + V^a) 4 V^w 2 , 2 4 AX4a> 2 4- 



(/) -6, i V5, -f' VS. 



Pages 94-96 

1. (a) 2, -2, -2; (6) 2, -1, -; W M , '; W) 2, -J*, t. 

2. (a) (-1, 0), (0, 1), (2, 3); (6) (-3, -2), (-1, 0), (0, 1); 

Tc) (-4, -3), (-2, -1), (-1, 0); (d) (-3, -2), (-1, 0), (0, 1), 

(2,3) 

Page 97 

1. 2 924. 6. -0.879, 1 347, 2 532. 

2. 1618, -1, -0.618. 6. -0.418. 

3. 2061. 7. 1.226. 

4. 1.398. 

Pages 101-102 

1. 1 226; %. 6. 4.494. 

2. 2 310 radians. 6. -0.567. 

3. 0.3574, 2.1533. 7. -0.725, 1.221. 

4. 0.739. 

Pages 106-106 

1. 41; -35; 1. 

2. (a) ( 3 Ka, 2 % 3 ); (&) (1, 0, -1); (c) (5, 4, -3); (d) (1, -^, J$). 

Page 114 

1. 20; -126; -212. 

2. (a) (2, -1, 1); (6) (1, %, -K); (c) (3, -1,2); (d) (1, -1, -2,3). 

Pages 121-122 

1. (a) (1, 1); (6) inconsistent; (c) inconsistent; (d) (1, 3k 2, &). 

2. (a) (-*/7 f 5Aj/7,*);(6) (0,0) ; (c) (0,0,0); 
(d) (fc/4, 7*/8, *); (e) (fc, 2fc, 0); (/) (0, 0, 0). 

CHAPTER IV 
Page 126 

- * 2 - * 



(c) y cos xy 4 1, a cos zy; (d) e* log t/, e x /y; 

(e) 2xy 4 A * a; 2 . 
V 1 x* 

2. (a) 2xy - z\ x* + z,y - 2xz; (b) yz + 4 xz + i 



564 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 

, N z zx , x 

(c) , t sm~ x -; 

(d) , . x y * 



V* 2 



(x 2 + y 2 + 2 2 )^' (x 2 + y 2 + 2 2 )*> (x 2 + 2/ 2 -f z 2 )^ 

Pages 129-130 

1. 7T/6 cu. ft. 6. $3.46. 

2. 11.7ft. 7. 0.112; 0.054. 

3. 0.139ft. 8. 53.78; 093. 

4. 2250. 9. 0.0037T; 0.3 per cent. 
6. 10.85. 10. I.GTTJTT. 

Pages 136-136 

1. ka*(8 cos 26 + K sin 20). 2. 2r cos 20; -2r 2 sin 20. 

3. 2r - t;t 2s; s - r. 



7. (a) e^ 2 (2< sin ^ + I cos ^) ; 

(b) 2r(l - 3 tan 2 0), -6r 2 tan sec 2 0. 

8. (a) 2z, 2(x + tan x sec 2 x); 



,. N ^67 . A dV / dV dV\ dV 

(b) cos -r -- h sin - > r I cos - sin - ] 
6x 5?/ \ dy Ox ) dz 



Pages 141-143 

+ 9x 2 4w 1 4t>!/ 2 e 4uy* 1 y 
12v*(u + v)' 4w 2 (w + t;) J 4v*(u + w) ' ^ ^TZ 
4. 2/g y 

' 



xy uve v v t ve v+v x 



6. (a) -2, 3, i, -i; (6) ^rqr^ ^r^i' srjr^ snp 



-f 4u' 1 + 

13 C^ - sec y , . 

v ; 3 sec ?/ tan y + 2zV v y cos - 3-s 2 cos - 32 
14. 2(* -y):2(* -*):2(y - *). 

Pages 146-146 

3. J^[3 \/3 + 1 -f (\/3 + 1)1 or 6.811. 4. 2 \/s 2 + 2/ 2 . 



ANSWERS 565 

Page 149 



1. V3/3. 

2. (a) 2x + 3y + 2z = 6, ^LzJ = ?^J = * ~ ** ; 

(6) 6z -f 2?/ - 32; = 6, ^-TT- = ^ ^ = g ^" ; 

c ^ ^Q^ i yy. \ z j& i 2! / _ ^_^!/ _ \, 

a 2 6 2 c 2 ' Zo 2/o 

Pages 152-163 

6. dx/ds = l/\/14, rfi//c?s == 2/V14, d2/d = 3/VTi. 9. 27. 



Pages 154-155 

3. 10^20. 

4. /,* cos 2 +/* sm 20 -\-fyy sin 2 0; 

f xx r* sin 2 ^ /xi/r 2 sm 20 + /^j/r 2 cos 2 / x r cos f y r sin 0. 



Pages 157-158 



2. e 1 + (x - 1) + (y - 1) + [(x - I) 2 + 4 (3 - l)(y - 1) 



3. 1 + x + ^ (a: 2 - 7/ 2 ) + ~ (* - S^ 2 ) + ~ 



Page 160 

1. (a) (3, 26) minimum; 

(6) (3, 108) maximum, "(5, 0) minimum; 
(c) No maxima 91 minima. 

2. x = l/e. 

4. (a) cos x H and sin re = 0, inflection; 

(6) cos x = -^ and sin x = 0, inflection; 

(c) sin x = 0, inflection. 
6. (a) # = l/e; (6) x 1 %s, maximum, x = 6 ^25) inflection, 



566 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 

Pages 162-163 

1. a/3, a/3, a/3 2. 8a6c/3 \/3. 3. a/3, 6/3, r/3 

4. A/3P/(2 -v/3 + 3), (-\/3 + l)P/2(2 A/3 + 3), P/(2 V$ + 3). 

^ 

5. = /i = = v / u07r 2 V', a* = \/5l. 

OTT 

Pages 170-171 

7TC* TTOi ., 

TT sin -- cos -~- 1 

1. 5 1 4. tan a. 

.<& a: 

2. 7r. 6. 2# 2 . 

3 ' a (l ~~ 10g 2 )' 6 * 2a7r ( a2 ~ 1 )~ 2 ' 

CHAPTER V 
Pages 190-191 

1. (a/5, a/5). 8. 7ra 2 /2. 

2. 7raV16. 11. 4a 2 ^ - lY 

\2 / 

4 /7T 2\ 

3. (a) ?/ du dv; (b) u 2 v du dv dw. 12. Q a* 1 ( - - ) 

O \4 O / 

4. (37ra/16, 0, 0). 13. 8a 2 

6. 32a 3 /9 14. x = a cos 2 - 

6. (a/4, 6/4, c/4). 15. 7ra 4 fc/2 

7. <T7ra 4 6/2. 16. / a = ^{^abc(a 2 -f 6 2 ) 

Page 195 

1. 0. 4. 127ra 5 /5 

CHAPTER VI 
Page 199 

2. (a) -'%,* (b) - 2 ?/' 1 ' 

3. (a) ?^; (6) %; (c) ?; (rf) ^ 

4. (a) 0; (6) M; (c) -^5- 

Page 202 
1. 7ra6. 2. J^. 3. 37ra 2 /8. 

Page 206 

I- -Ms- 2. 0. 3. -H2 4. %. 

Page 212 

1. !%. 3. M. 

5. (a) 7T/2; (6) - x/8/4; (c) 1%. 



ANSWERS 



567 



CHAPTER VH 
Pages 230-231 

1. (y') 2 4- 5xy' - y + 5x 2 = 6. y" - 2y' + 2y = 

2. y" 4- y = o. 7. #(y') 2 yy' 4- 1 

3. zy'" y" xy' 4- y = 9. 3 y'" 3z 2 y" -f 

4. zy' 4- (1 x)y 4- 20"* = 0. - 10. 2xy' - y = 0. 
6. (y") 2 = [14- (2/') 2 l 3 



0. 



0. 



- 6y - 0. 



Pages 264-266 



1. 0.417 ft. 

2. 0o + 



9. w = vo (1 e~ 
11. 000667 cal. 



Page 268 



1. sin" 1 y 4- sin" 1 x c. 

2. (v - !)/( + 1) = ce 

3. 2 cos y sin x cos # 4- x 

4. sec # 4- tan y = c. 

5. tan" 1 y 2 \/l + x c. 

6. ^6* e x \/l y 2 = c 

13. rr sin" 1 x -f- 
14. 



16. y # log XT/ 
16. tan" 1 y tan" 1 
1 . 1 



17. - 



2/ 



a; - 1 



c. 

= c. 



7. 1 + y = c(l + x). 

8. log [(?/ - l)/y] -f -* = c. 

9. 2 tan" 1 e^ + log tanh x/2 * c. 

10 - - - - - 1( >g 2/ = c. 

11. y(2 - log y) = M tan 2 x + c. 

12. s(l + 4?/ 2 )% = c. 



8- - ~ + I + 



19. 
20. 



c(l +x)(l - ty). 
7/ 2 - c(l + ^ 2 ). 

23. (5 - X )/(A - x) = 



2. sin ] (y/x) log x c 

3. sin (y/x) + log Z = c 

4. a; 2 2zi/ y 2 = c. 

5. log y + z 3 /(3?y 3 ) = c. 



11. 



Pages 261-262 

ft __ 

7. y = 

8. a; = 

9. log a 
10. x - 



13. a: 2 + y 2 -f 



0. 



12. y 4- ce v / x = 0. 

14. + y + 2 log (2x 4- y - 3) - c. 

16. rcy 2 = c(x 4- 2y). 19. x + ce* 2 /<2*<o . o. 

17. x 3 4- y 3 = co;?/. 20. y 2 4- 2a;y x 2 = c. 
18. 



568 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 

Pages 264-265 

1. sin xy + x 2 c. 7. x 2 log t/ =* c. 

2. aty + xt/ 2 4- * - c. 8. (1 - x 2 )(l - y 2 ) - c. 

3. e* -}- x + y c. 9. e x log y + a; 2 = c. 

4. x 3 y 2/ 3 x = c. 10. Not exact. 

5. Not exact. 11. Not exact. 

6. sin (y/x) c. 13. x sin 2?/ = c. 

Pages 268-269 

1. sin" 1 y x = c. 



2. y = ^A(x VI - x 2 4- sin" 1 x) 4- c. 

3. x 4- 2/ tan" 1 y = c. 4. y ee 3 * = 0, y ce~ x 0. 

Pages 278-279 

1. y ex. 3. x 2 4- wi/ 2 = c. 

2. x 2 - y z = c. 9. y = ce*/*. 

Pages 280-283 

12. p = p e- kh . 

Pages 285-286 




cos 2 x -F 2(sin x - 1) 



6. y 2 sin x x cos x H cos x -\ --- 

XX 

7. y = 1 + ce 1 *"- 1 *. 12. x = ce- 2 " + | - |. 

8. 7 = (E/R)(\ - e-^ f /i-). 13. y = tan x - 1 + ce~** n *. 

9. y = sin x 4- ce x 14. y = (x 4- l)(e* + c). 

10. y = c tan x -f e*. 16. y = e 3 * + ce 2 *. 

11. x = 1 + c<r" 2 /2. 

Page 287 

1. ?/ = (48x~ 2 - 96x~ 4 - 4) cos x + (IGx" 1 - 96x~ 3 ) sin x + cx~ 4 . 

2. 2/~ 2 = x + ^ + ce 2x . 6. ?T 2 = 1 + x 2 4- ce x *. 

3. 2/~ 6 = %x 3 -f ex 5 . 7. x~ 2 y + 

4. x = y log ex. 8. t/~ J = 1 4- c 

5. y 1 = 1 4- log x 4- ex. 



Page 291 
2. e~ aaf f eCa+m) ^^ 3. e - ax f 



ANSWERS 569 

Pages 294-295 

1. (a) y ae-'* 4- c 2 e 8 *; (6) y - cie 3 * + c 2 e 2 *; 



(c) y (ci + C2x)e~ x ; (d) y (r i + c 2 z)e a! + c 8 ; 

(e) y = (ci 4- 02 4- c 3 z 2 )e~* 4- c 4 ; (/) y ci cos fcx + c 2 sin fcz 

+ c 3 cosh kx 4- C4 sinh 

Pages 298-299 

1 v - c e -ax , (^ 3 - * 2 ) , 2x ^ 2 

1. y - cic 4- g + y 27- 

2. y - cie- 3 * + c 2 e~ 2j; + ~- 3. y = (c t + c&)e* + x + 2. 



4. y = cie-* 1 + c 2 e** + (n)/2P)(x 2 - to + 2fc~ 2 ), where fc - 

5. y = c\e x + c 2 e~ x + Cse 2x x. 6, ?/ = cie"* + c 2 e*/ 2 4- 2 sin x. 
1. y (ci + c 2 x)e :c 4- c$e~ 2x 4- sin x. 

Pages 305-307 

1. y = 2 cos Vl%_ViO/(27r); y 2 cos \Ao* -f VlO sin VIO*. 

3. t/ = 10 cos V245. 

4. y = 10e- (cos A/220^ 4- ^r sin \722oA; /2 = 400 \/245 dynes. 

V \/220 / 

5. F - 100 \/2e- 600< cos ^500* - |); F = 100e"" 500 ^(l -f 500 \/20. 

6. 7 = 20 \/5e- 50000f (5 sinh 10000 VH>t + \/5 cosh 10000 \/50. 
0. 10 -jrj- 4- lOgrz/ = 0; maximum y = -\/3, total drop 2 4- \/3. 



Pages 314-315 

1. x - %cie' - c 2 e~ -f %te - % 5 e' - ^ - Ke, 

y = cie' -f c 2 6~ 4< 4- K& ~ %t ~ Vis- 
3. Cy cloid of radius mE/(eH 2 ). 

Pages 317-318 

1. No. 2. Yes. 3. No. 

Pages 321-322 

3. y = -(x 2 log a:)/9. 4. y c^ 4- care 4- z 2 4- 1. 

Pages 324-325 

2. y CiX- 2 4- Ca^- 1 4- M log x %. 

3. ?/ = Cix 2 4- C2X 3 . 4. ?/ = CIX OT 4- cjaj""*" 1 . 
6. y - cix 2 4- c 2 x(5+V2i)/2 4- 



6. y - dsU + VSo/a 4- c 2 x(i-\/30/2 4- *. 

3 

7. y do? 2 4- c 2 x x[(log ) f /2 4- log x]. 



570 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 

Page 329 



3. y-cil-a;+-~-h + r,(l -f x + z 2 -f 



CHAPTER VIII 
Pages 356-367 

^2! d2 _ 

^- a; ^ = 0; 

2 3 _ 1 dz dz 



Page 361 

1. (a) ^ = Fi(y -f ox) -h ^a(y - ax); 
(b) z ~Fi(y -2x) +F*(y + x); 
(e) z = Fi(y - x) -f xF*(y -* x). 

2. (a) xy; (c) x*y/2 - a;V3. 

Page 372 
1. 0.44883; 0.14922; 000004 





Pages 375-377 


2. 35.5; 41.9. 






Pages 385-386 


_ 200 -A 




* n 4i & 


n - l)a*- r 


, _ 400 ^ 1 


j.2n I <JMJ^ (^fL 


** '* i f^f) 1 




n= 1 


)(l 


4. U = 7, ^.n6~ a2 * 2 '*/o(i 


k n r), where 1=7-^. 



(2n - 1).. 



Pages 390-391 

V n=l 

00 

7. / = 0.6 -f- 1.1 2) (~ 1 ) ncos 
n-l 

CHAPTER IX 
Pages 398-399 
1. 0.5640. 2. (10 V3/3)(i -h j + k). 



ANSWERS 571 

Pages 403-404 

6. 3i + 12j + 4k. 6. 19 V3/3; (\/3/3)(8i + j - 9k). 

Page 409 

j du _ dudx dudy dudz 
' ds ~ ~dxds ~dyds dz ds' 
2. (a) jyz + jxz + kxy; (b) i2x -f j2y + k2z; 

(c) (x 2 + 2/ 2 + z 2 )-H(iz + jy + kz); 

(d) 2(x 2 + 2/ 2 -f z 2 )- ] (iz -f jy + kz). 
5. 26 A/2/5 6. 9. 

Pages 414-416 

1. (a) 3; (b) 2/r; (c) 

3 i 1 /Mf a. ^Aji . M.A j. i J^ /'Mf 4_ M _i_ M_A 
da; V dx dy ~*" dz ) ^ J dy \ dx ^~ dy "^ dz ) 



Pages 420-421 

1. (a) 0; (b) 0; (c) 

Pages 432-433 

4. ^ = cij 5. * = x 2 

Page 439 

^$ 4. 5. 1 ^? _4_ ^?. ?? i ^i^? i J^!_ 

91 a p + P ~dd + Zl a7 ; 9l a^ + 7 56 > 



fcUp ^i i r 
L~^ ""ap J "*" 'pL 

(sin gA y ) _ dA\ T 1 

~d<t> \ l \_ps 



_ _ 

psmei 08 ~d<t> l _psmO~d<f> p dp 

j_ M 1 

+ * 1 p 

10. 3p cos 0/r 4 ; 

CHAPTER X 
Pages 443-444 
1. (a) 2, 60; (6) 2 >/2, 45; (c) 1, 90. 

a (n\ x 4- * -y + ! .. . 

y ' W x 2 -f (y - I) 2 + % 2 -f(2/ -I) 2 ' 



* 2 + (2/4- I) 2 ^ & -f (y -f I) 2 



572 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 

Pages 447-448 

4 sin 2x . sinh 2y 

cos 2x -h cosh 2y cos 2x -f cosh 2y 

6. sin x cosh y -\- ^ cos x sinh y. 

7. (a) e-/[cos (log \/2) + t sin (log \/2)]j 

(b) \/2e~ r /* I cos f j ~ lg \/2 ) -h * sin ( ~" T ~ lg "N/2 ) I 

Pages 462-463 

3. (a) z*; (b) 1/z; (c) z; (d) log z; (e) cos 2. 7. x 3 - 3^ 2 . 

Page 461 
3. 2 4. Trt. 6. 0. 6. 2wi. 7. 0. 

Page 469 

2. (a) u =* cos # cosh y, v sin # sinh t/; 

(b) u = e* cos ?/, v e x sin y; 

(c) u x &xy , f : 

(d) w = log (x 2 4- i/ 2 )J 

(e) u = a:/ (a; 2 + y 2 ), i 



CHAPTER XI 
Page 496 

l. K; Ks- 2. % ; H; Ko; % 8 ; 1323/46189. 

Page 497 

1. 33/16660. 4. *%Q. 

2. 8 H 2 ;H 2 . 5. % 2 ;2;% a . 

3. M4088- 

Pages 499-600 

1. 4 Mo. 6. n > log 2/(log 6 - log 5). 

2. H; 6 . 6. ^ ; %o. 

3. 46413/78125. 7. 16 ^i 5 . 

4. %. 8. 91854/100000 

Page 601 

1. 0.775; 0.0000265; $47.50. 2. $11 3. 6. 

Page 604 

I- () 12 888; (6) 2 %48- 

2. (0.65) 10 + 10(0.65) 9 (0.35) + 45(0 65) 8 (0 35) 2 + 120(0 65) 7 (0.35) 3 . 
3. 

4- 



Page 608 
50; iooC*o(H) 100 ; 



ANSWERS 573 

Page 512 



1. l/\/125T. 2. 200; \/3/(10007r). 3. 

Page 616 

1. (200) 10 e- 200 /10! 2. (20) 100 e- 20 /100l 

3. 0.136; 0.272; 0.272; 0.181; 0.091; 0.036. 

Pages 523-524 

4. 0.00896; 0.00850. 5. 0.976; 0.983. 6. First set. 

CHAPTER XII 
Page 527 

1. y =* x/2 + %. 2. y = 2.5s 05 8. y - 0.3(10- 2a! ). 

Pages 633-634 

1. p av n . 2. = fca e . 

3. H *= a 2 C 2 -f- aiC -f a . 

Pages 536 6^6 

1. y = 4.99 - 3.13z + 1 26z 2 . 2. H - KC' 2 ~ HC' 4- Ji- 



3. y = 1.557 + 1 992o: - 751a; 2 -f 0.100s 3 . 

4. y = 1.3 + 0.2e*. 6. y = 0.3<? -1.1 sin z + 1.5s 2 . 

Page 544 

1. y = 4.99 - 3.13s + 1 26s 2 . 2. y - 10 5 *. 

Page 560 

1. ?/ = 0.75 -f- 0.10 cos s 05 cos 3s 29 sin x. 

2. ?/ = 0.85 0.25 cos 2s 0.05 cos 4s -f 0.05 cos 6s + 0.26 sin 2s 

- 0.03 sin 4s. 

Page 654 

1. 42 61; 42 50; 42.51. 3. 106.15; 106.09; 106.09. 

2. 2.581; 2.627. 4. 2.784; 2.700. 

Page 560 

1. 25.252; 25.068. 4. 666.25; 666.00. 

2. 132.137. 6. 39.30; 38.98. 

3. 128.6. 



INDEX 



Absolute convergence of series, 16, 

17, 20, 21 
Absolute value of complex number, 

441 
Addition, of series, 21 

of vectors, 393 

parallelogram law of, 394 
Adiabatic process, 224 
Aerodynamics, 133, 431 
Algebra, fundamental theorem of, 92 
Algebraic theorems, 92-94 
Alternating series, 15 
am u, 51 

Ampere's formula, 52n. 
Amplitude of complex number, 441 
Amplitude function, 51 
Analysis, harmonic, 545 
Analytic functions, 451-491 
Angle, as a line integral, 195 

direction, 146, 398 

of lap, 240 

of twist, 485 

solid, 195 
Angular velocity, 61, 191, 236, 404, 

424 

Applications, of conformal repre- 
sentation, 479-491 

of line integrals, 217-224 

of scalar and vector products, 

404-406 
Approximate formula, for n!, 509 

for probability of most probable 
number, 511 

in applied mathematics, 55 
Approximation, Laplace's or normal, 

515 

Approximations to binomial law, 512 
Arc length, 143 

of ellipse, 47 



Arc length, of sinusoid, 55 
Area, 172 

as a double integral, 178 

as a line integral, 199-202 

element of, 183, 184, 190, 437 

positive and negative, 200 

surface, 188-196 

Argument of complex number, 441 
Associative law, for series, 18 

for vectors, 394 

Asymptotic formula for n\ 509 
Asymptotic series, 524 
Atmosphere, thickness of, 61 
Attraction, law of, 218, 232 

motion under, 58, 218 

of cone, 196 

of cylinder, 196 

of sphere, 196, 232 
Augmented matrix, 118 
Auxiliary equation, 292 
Averages, method of, 534 
Axes, right- or left-handed, 397 

B 

Base vectors, 396 

Beam, 240-242, 307 

Belt on pulley, slipping of, 239 

Bending moment, 241 

Bernoulh-Euler law, 241, 307 

Bernoulli's equation, 286 

Bessel functions, 273, 336, 381 

expansion in, 339 
Bessel's equation, 332, 380 
Beta function, 276 
Binomial law, 502 

approximations to, 512 
Binomial series, 40 
Biot and Savart, law of, 52 
Boundary conditions, 242, 351, 363, 

370 
Buckling, 299 



575 



576 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 



Cable, flexible, 244 

flow of electricity in, 386 

supporting horizontal roadway, 

242 

Cartography, 479 
Catenary, 247, 252. 
Cauchy-Riemann equations, 221, 

450, 455 

Cauchy's equation, 322n. 
Cauchy's integral formula, 461 
Cauchy's integral test, 12 
Cauchy's integral theorem, 455 
Center of gravity, 177, 182, 183, 187, 

190, 191, 196, 522 

Change of variables, in derivatives, 
154 

in integrals, 183-188 
Characteristic equation, 292 
Charge, distribution of, 487 
Charts, distribution, 506 
Chemical reaction, 258 
Circular functions, 247 
Circulation, of a liquid, 475, 477 

of a vector, 418, 419 
Closed curve, area of, 199-201 

direction around, 200 

integral around, 201, 203, 206, 
216, 421 

simple, 200 
en u, 51 
Coefficients, Fourier, 65 

metric, 437 
Cofactor, 111, 112 
Combinatory analysis, fundamental 

principle of, 493 

Commutative law, 394, 399, 400 
Comparison test for series, 9 
Complementary function, 290, 292 
Complete elliptic integrals, 48 
Complex number, 440 

absolute value of, 441 

argument of, 441 

conjugate of, 444, 488 

vector representation of, 440 
Complex roots of unity, 87 



Complex variable, 440-491 
functions of, 444-491 
analytic, 451-491 
derivative of, 449 
integration of, 453 
line integral of, 454 
Taylor's expansion for, 464 
Components of force, 217 
Composite function, 134, 137 
Condenser, 283, 299, 305, 308, 387 
Conditionally convergent series, 16, 

17,21 
Conditions, Cauchy-Riemann, 221, 

450, 455 
Dinchlet, 65 

for exact differential, 212, 216 
Conductivity, 367, 426 
Conductor, 486, 489 
Conformal mapping, 465, 471 
Conformal representation, applica- 
tions of, 479-491 
Conformal transformation, 467 
Conjugate of a complex number, 

444, 488 

Conjugate functions, 468, 470 
Conservation of matter, law of, 429 
Conservative field of force, 219, 411 
Consistent systems of equations, 

117-122 
Continuity, equations of, 221, 429, 

481 

of functions, 23, 28, 124, 448 
Contour line, 144 
Convergence, absolute, 16, 17, 20, 

21, 33 

conditional, 16, 17, 21 
interval of, 31, 33 
of series, 4, 7 

tests for, 9, 11, 12, 15, 20, 27, 31, 

33 

radius of, 31, 33 
uniform, 23-30, 33 
Cooling, law of, 254 
Coordinate lines, 434 
Coordinate surfaces, 434 
Coordinates, curvilinear, 433-439 
cylindrical, 152, 185, 190, 191, 378, 
386, 434, 438 



INDEX 



577 



Coordinates, ellipsoidal, 433 

parabolic, 439 

polar, 183, 184, 276, 279, 386, 438 

spherical, 152, 185, 382, 386, 434, 

439 

cos x, 46, 250 
cosh, 247 
Cosine, hyperbolic, 247 

power series for, 38, 40 
Cosine series, 73 
Cosines, direction, 146, 147, 151, 

188, 194, 398 
coth, 249 

Cramer's rule, 113 
Cross product, 400 
Cubic equation, algebraic solution 
of,*86 

graphical solution of, 83 
Curl, 418, 422, 423, 438 
Current, 386, 427 
Curve, distribution, 504, 516 

elastic, 240, 307 

map of, 466 
Curve fitting, 525-560 
Curves, integral, 226, 228, 279 

orthogonal, 277, 468 
Curvilinear coordinates, 433-439 
Cylinder functions (see Bessel func- 
tions) 

Cylindrical coordinates, 152, 185, 
190, 191, 378, 386, 434, 438 

D 

Dam, gravity, 483 
Damping, viscous, 302 
Dead-beat motion, 304 
Decomposition of vectors, 396 
Definite integrals, 172 

change of variable in, 183-188 

evaluation of, 172 

mean-value theorem for, 210n. 
Deflection, 299 

Degree of differential equation, 225 
Del, V (see Nabla) 
Delta amplitude, dn, 51 
De Moivre's theorem, 90, 442 



Dependence, functional, 2 

linear, 116 

Dependent events, 495 
Derivation of differential equations, 

231-247 
Derivative, 125 

directional, 143, 151, 219 

normal, 144, 146, 152 

of functions of a complex variable, 
449, 452, 463 

of hyperbolic functions, 255 

of series, 29, 33 

partial, 125-143, 153 

total, 130-143 
Descartes's rule of signs, 94 
Determinants, 102-114 

cof actors of, 111 

expansion of, 106n., Ill 

functional or Jacobian, 183 

Laplace development of, 111 

minors of, 110 

of matrix, 115 

product of, 110 

properties of, 107-112 

solution of equations by, 102-114 

Wronskian, 317 
Deviation, standard, 523 
Diagonal term of determinant, 107 
Diagram, pv, 223 
Differences, 527 

Differential, exact, 211, 212, 216, 
222, 224, 262, 411, 418, 420 

of area, 184, 190 

of volume, 185, 187, 190 

partial, 128-143 

total, 127-143 
Differential equations, 225-391 

Bernoulli's, 286 

BessePs, 332, 380 

Cauchy-Riemann, 221, 450, 455 

Cauchy's, 322n. 

definition of, 225 

degree of, 225 

derivation of, 231-247 

Euler's, 322, 430 

exact, 262 

first order, 256, 267 

Fourier, 425 



578 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 



Differential equations, general solu- 
tion of, 230, 290, 292, 350, 358 
homogeneous, 259, 261 
homogeneous linear, 290 
integral curve of, 226, 228 
integrating factor of, 265 
Laplace's, 369, 382, 385, 386, 439, 

451, 470, 481 
Legendre's, 342, 384 
linear, 226, 283-349, 357 
numerical solution of, 346 
of electric circuits, 301, 305, 386 
of heat conduction, 367 
of membrane, 377 
of vibrating spring, 308 
of vibrating string, 361 
order of, 225 
ordinary, 225-349 
partial, 225, 350-391 
particular integral of, 290, 292, 

297, 318, 359 
particular solution of, 230 
second order, 269, 295 
separation of variables in, 257 
simultaneous, 312-315 
singular solution of, 279 
solution in series, 228, 325, 349, 

364 

solution of, 226 
with constant coefficients, 287- 

315, 357 
with variable coefficients, 284, 

315-349 

Differential expression, 225 
Differential operators, 287-299, 357, 

406 
Differentiation, of implicit functions, 

132-142 

of series, 29, 33, 34, 80 
partial, 123-171 
term by term, 33, 34, 80 
under integral sign, 167 
Diffusion, 369, 427 
DifTusivity, 368w. 
Direction angles, 146, 398 
Direction components, 146 
Direction cosines, 146, 147, 151, 188, 
194, 398 



Direction ratios, 150, 151 
Directional derivative, 143, 151, 219 

(See also Gradient) 
Dirichlet conditions, 65 
Discharge of condenser, 299 
Discontinuity, finite, 64 
Discriminant of cubic, 89 
Distance, element of, 435 
Distribution of charge, 487 
Distribution charts, 506 
Distribution curve, 504, 516 
Distributive law, 399, 400 
Divergence, of series, 5, 8, 20 
/ of a vector, 411, 423, 438 
Divergence theorem, 191, 415, 425, 

428 

dn w, 51 

Dot product, 399 

Double integrals, 173, 192, 202, 275 
Drying of porous solids, 369 
Dynamics, laws of, 231 



E 



e, 42 

e, 250 

Effects, superposition of, 129, 223 

E(k, <?), 48-51, 54 

Elastic curve, 240, 307 

Elasticity, 241, 422, 484-486 

Electrodynamics, 422, 423n. 

Electron, 315 

Electrostatic field, 475, 477, 479 

Electrostatic force, 487 

Electrostatic potential, 487 

Electrostatics, 486-491 

Element, of arc, 467 

of area, 184, 190, 437 

of distance, 435 

of volume, 185, 187, 190, 437 
Elementary functions, 315 

expansion of, 35-46, 65-82, 465 
Ellipse, area of, 177, 202 

center of gravity of, 177 

length of arc of, 47 
Ellipsoidal coordinates, 433 
Elliptic functions, 51 



INDEX 



579 



Elliptic integrals, 47-55 

complete, 48 

first kind, F(k, v?), 48-55, 238 

second kind, E(k, ^), 48-54 

third kind, II (n, k, <?), 50 
Empirical formulas, 525-560 
Entropy, 224 
Envelope, 279 
Equation, auxiliary, 292 

Bernoulli's, 286 

Bessel's, 332, 380 

characteristic, 292 

cubic, 86 

Euler, 322 

Fourier, 425 

mdicial, 334 

integral, 347 

Laplace's, 195, 369, 382, 385, 386, 
439, 451, 470, 481 

Legendre's, 342, 384 

of continuity, 221, 429, 481 

of plane, 147 

wave, 432 

Equations, Cauchy-Uiemann, 221, 
450, 455 

consistent, 117-122 

dependent, 105 

differential, 225-391 

Euler's, 430 

inconsistent, 105, 117-122 

normal, 537, 540 

parametric, 143, 149, 150, 199, 215 

representing special types of data, 
528 

simultaneous, 102-122, 139-141 

solution of, 83-122 

systems of, 102-122 

homogeneous linear, 119-122 
non-homogeneous linear, 113- 

119 
Error, Gaussian law of, 520, 536 

mean, 516 

mean absolute, 522 

mean square, 522 

of observation, 516 

probable, 521 

small, 56 
Error function, 516 



Euler equation, 322 
Euler formulas, 78, 251 
Euler's equations, 430 
Euler's theorem, 136 
Evaluation of integrals, by differ- 
entiation, 169 

in series, 43-46 
Even function, 68 
Events, dependent, 495 

independent, 495 

mutually exclusive, 497 
Exact differential, 211, 212, 216, 222, 

224, 262, 411, 418, 420 
Exact differential equation, 262 
Expansion, in Bessel functions, 339 

in Fourier series, 65-82 

in Legcndre polynomials, 346 

in Maclaunn's series, 37 

in power series, 37-46 

in Taylor's series, 37 

in trigonometric series, 65 

uniqueness of, 38 
Expectation, 500 

Expected number of successes, 508 
Exponential form for trigonometric 

functions, 78, 251, 446, 447 
Exponential function, expansion for, 

42, 446 

Extremal values, 164 
Extremum, 164 



F(k, *>), 48-55 

Factor, integrating, 265 

Factor theorem, 92 

Factorial, n !, approximation for, 509 

(See also Gamma functions) 
Falling body, 58, 232 
Field, 406 

conservative, 411 

electrostatic, 475, 477, 479 

irrotational, 418 
Finite discontinuity, 64 
Fitting, curve, 525-560 
Flexure, 298 
Flow, of a liquid, 220, 424, 428, 477, 

, 478, 480-484 



580 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 



Flow, of electricity in a cable, 386 
of heat, 256, 368-377, 423, 425 
seepage, 483 
Fluid motion, 220, 424, 428, 475- 

478, 480-484 
Flux, 416 

Flux density, magnetic, 52 
Force, 217, 239, 392, 406, 409 
components of, 217 
conservative field of, 219 

electrostatic, 487 
Force function, 411 
Forced vibrations, 308, 310 
Formula, asymptotic, 509 

Cauchy's integral, 461 

empirical, 525-560 

interpolation, 550 
Lagrange's, 552 

Poisson, 512 

Stirling's, 508 

Wallis's, 45 
Fourier coefficients, 65 
Fourier equation, 425 
Fourier series, 63-82 

complex form of, 78 

differentiation of, 80 

expansion in, 65-82 

integration of, 80 

solution of equations with, 364 
Functional dependence, 2 
Functional determinant, 183 
Functions, 1 

analytic, 451 

Bessel, 336 

Beta, 276 

complementary, 290 

conjugate, 468, 470 

continuous, 23, 448 

elementary, 315 

expansion of, 35, 65, 155 

Gamma, 272-277 

holomorphic, 451 

homogeneous, 136, 259 

hyperbolic, 247-256 

of a complex variable, 444-491 

of several variables, 123, 160 

orthogonal, 81, 339, 345 

periodic, 64 



Functions, potential, 219, 411 

power, 30 

real, 2, 123 

regular, 451 

scalar point, 406 

singularities of, 222 

stream, 221, 432, 453, *oi 

vector point, 406 

Fundamental principle, of combina- 
tory analysis, 493 

of sequences, 6 
Fundamental theorem, of algebra, 92 

of integral calculus, 172, 457 



G 



Gamma functions, 272-277 
Gauss- Argand diagram, 440 
Gaussian law of error, 520, 536 
Gauss's theorem, 193 
General solution of differential equa- 
tion, 230, 290, 292, 350, 358 
Geometric series, 9 
Gradient, V, 144, 152, 407, 410, 438 
Graphical method, of curve fitting, 

525 

of solution of equations, 83 
Gravitational constant, 232 
Gravitational law (see Attraction) 
Gravitational potential, 219, 408 
Gravity, center of, 177, 182, 183, 

187, 190, 191, 196, 522 
Gravity dam, 483 

Green's theorem, for the plane, 202 
in space, 191, 418 
symmetric form of, 194, 418 

H 

Harmonic analysis, 545 
Harmonic series, 8 
Heat conduction, 367 
Heat flow, 368-377 

equation of, 368, 425 

steady, 256, 368, 427 

variable, 368, 373, 425 
Helix, 151, 152 
Holomorphic functions, 451 



INDEX 



581 



Homogeneous equations, differential, 
259, 261 

linear algebraic, 119-122 

linear differential, 290 
Homogeneous function, 136, 259 

definition of, 136 

Euler's theorem on, 136 
Hooke's law, 241, 299 
Horner's method, 95 
Hydrodynamics, 221, 422, 428-433, 

480-484 
Hyperbola, 247 
Hyperbolic functions, 247-256 
Hyperbolic paraboloid, 162 



I 



Imaginary roots, 94 
Implicit functions, 132, 137-142 
Inclined plane, 280, 282, 306 
Incompressible fluid, 424, 430 
Inconsistent equations, 105, 117-122 
Independence, linear, 116, 317 

of path, 208, 216, 452, 455 
Independent events, 495 
Independent trials, 501 
Indicial equation, 334 
Infinite series, 1-62 

absolute convergence of, 16, 17, 20 

conditional convergence of, 16, 17 

definition of, 4 

expansion in, 35-46, 155-158 

of constants, 6-22 

of functions, 23-62 

of power functions, 30 

of trigonometric functions, 63-82 

operations on, 21, 29, 33-35 

sum of, 4n. 

tests for convergence of, 9, 11, 12, 
15, 20, 27, 31, 33 

theorems on, 17, 21, 27, 28, 29, 31, 
33, 34, 36, 38 

uniform convergence of, 23-30, 33 
Inflection, point of, 159 
Initial conditions, 235, 351 
Integral calculus, fundamental theo- 
rem of, 172, 457 
Integral curve, 226, 228, 279 



Integral equation, 347 
Integral formula, Cauchy's, 461 
Integral test for series, 12 
Integral theorem, Cauchy's, 455 
Integrals, around closed curve, 201, 
203, 206, 216, 421 

change of variable in, 183-188 

definite, 172 

double, 173, 192, 202, 275 

elliptic, 47-55, 238 

evaluation of, by means of series, 
43-46 

iterated, 175, 180 

line, 197-224, 410, 421, 454, 458 

mean-value theorem for, 210n. 

multiple, 172-196 

particular, 290, 292, 297, 318 

surface, 188-196, 415, 421 

transformation of (see Green's 
theorem; Stokes's theorem) 

triple, 177, 193 

volume, 180, 415 

with a parameter, 47, 167 
Integrating factor, 265 
Integration, by parts, 276 

numerical, 554-560 

of complex functions, 453 

of series, 29, 33, 34, 80 

term by term, 33, 34, 80 
Interpolation, method of, 101 
Interpolation formulas, 550-554 
Interval, 4 

of convergence, 31, 33 

of expansion, 38, 76 
Inverse hyperbolic functions, 249, 

255 

Inversions, 106 

Irrotational field, 418, 423, 430 
Isolation of roots, 92 
Isothermal process, 224 
Iterated integrals, 175, 180 
Iteration, method of, 297 



J(x\ 336 

Jacobian, 141, 183, 190 



582 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 



K 



K 9 (x), 337 



Lagrange's interpolation formula, 

552 
Lagrange's method of multipliers, 

163-167 

Lamellar field, 423 
Laplace's approximation, 515 
Laplace's equation, 195, 369, 382, 

385, 386, 439, 451, 470, 481 
Law, Bernoulli- Euler, 241 

binomial, 502, 512 

of attraction, 218 

of conservation of matter, 429 

of cooling, 254 

of dynamics, 231 

of error, 520, 536 

of gravitation, 232 

of small numbers, 512 
Least squares, method of, 536 

theory of, 521 
Legendre polynomials, 344, 384 

expansion in, 346 
^egendre's equation, 342, 384 
Leibnitz's rule (see Differentiation, 

under integral sign) 
Leibnitz's test (see Test, for alternat- 
ing series) 
Length, of arc, 143 

of ellipse, 47 

of sine curve, 55 
Level surface, 406 
Limit, 2, 124, 454 
Line, contour, 144 

coordinate, 434 

direction cosines of, 146, 147, 151 

normal, 144, 146-149 

of equal potential, 277 

of flow, 475 

stream, 277, 432, 467 

tangent, 143, 147, 151 

vector equation of, 395 
Line integrals, 197-224, 410, 421, 
454 



Line integrals, applications of, 217- 

224 
around a closed curve, 202, 206, 

216, 421 

definition of, 197, 454 
evaluation of, 202-206, 458 
for angle, 195 
for area, 201 
for work, 217 
in space, 215, 410, 421 
properties of, 206-217 
transformation of, 202, 421 
Linear dependence or independence, 

116, 317 
Linear differential equations, 288- 

349, 357 

with constant coefficients, 287 357 
with variable coefficients, 284, 

315-349 

Linear differential operator, 287-299 
Log z, 446 
Logarithmic paper, 526 

M 

M test, 27 

Maclaurin formula, 36 
Maclaurin's scries, 37, 249 
Magnitude of a vector, 393 
Map, geographic, 479 

of a curve, 466 
Mapping functions, 467 
Matrix, 114-122 

augmented, 118 

determinants of, 115 

rank of, 115 

Maxima and minima, constrained, 
163 

for functions of one variable, 158 

for functions of several variables, 

160 

Mean error, 516, 522 
Mean- value theorems, 210n. 
Measure numbers, 397 
Mechanical quadrature, 554 
Membrane, vibration of, 377 
Mercator's projection, 479 
Metric coefficients, 437 



INDEX 



583 



Minima (see Maxima and minima) 
Minimax, 162 

Modulus, of complex number, 441, 
442 

of elliptic function, k, 51 
Moment, bending, 241 
Moment of inertia, 177, 180, 182, 

183, 187, 190, 191, 196, 241 
Moments, method of, 544 
Most probable value, 505 

approximation for probability of, 

511 
Motion, dead-beat, 304 

fluid, 220 

JAWS of, 231, 234 

of a membrane, 377 

oscijktory, 304 

pendulum, 48, 234 

simple harmonic, 233, 301, 314, 
380 

under gravity, 232 
Multiple integrals, 172-196 

definition and evaluation of, 173, 
179 

geometric interpretation of, 177 
Multiplication, of complex numbers, 
442 

of determinants, 110 

of series, 21 

of vectors, 399 
Multiplicity of root, 93, 294 
Multiplier, Lagrangian, 165 
Multiply connected region, 205, 212, 

455 
Mutually exclusive events, 497 



N 



N*bla, or del, V, 194, 195, 407, 414, 

422 

Newtonian potential, 196 
Newton's law, of attraction, 218 
of cooling, 254 
of dynamics, first law, 231 
second law, 231, 272, 363 
third law, 231, 234 
of gravitation, 232 
Newton's method of solution, 97 



Normal, to a curve, 144 

to a plane, 146, 147 

to a surface, 147, 188, 407 
Normal approximation, 515 
Normal derivative, 144, 146, 152 

(See also Gradient) 
Normal distribution curve, 516 
Normal equations, 537, 540 
Normal form, 146 
Normal law (see Gaussian law of 

error) 

Normal line, 144, 146-149 
Normal orthogonal functions, 81 
Numbers, complex, 440 

measure, 397 

Numerical integration, 554-560 
Numerical solution of differential 

equations, 346 



O 



Odd function, 68 

Operator, 528 

differential, 287-299, 357, 406 
vector (see Curl; Divergence; 
Gradient; Nabla) 

Order of differential equation, 225 

Ordinary differential equations, 225- 
' 349 

(See also Differential equations) 

Ordinary discontinuity, 64 

Origin of a vector, 393 

Orthogonal curves, 277, 468 

Orthogonal functions, 81, 339, 345 

Orthogonal systems, 434 
Orthogonal trajectories, 277-279 

Orthogonal vectors, 398 

Oscillation of a spring, 299 

Oscillatory motion, 304 

Overdamped, 303 



p series, 10 

Parabola, 244 

Parabolic coordinates, 439 

Paraboloid, hyperbolic, 162 

Parachute, 253, 255 



584 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 



Parallelogram law of addition, 394 
Parameters, 277, 280 
integrals containing, 167 
variation of, 318 
Parametric equations, 143, 149, 150, 

199, 215, 247 

Partial derivatives, 125-143, 153 
Partial differential equation, 350- 

391 

derivation of, 351 
Fourier, 425 
integration of, 353 
Laplace's, 369, 382, 385, 386, 439 
linear, 357 

of elastic membrane, 377 
of electric circuits, 386 
of heat conduction, 367, 425 
of vibrating string, 361 
Partial differentials, 128-143 
Partial differentiation, 123-171 
Partial fractions, method of, 297 
Partial sum, 4 
Particular integral, 290, 292, 297, 

318, 359 

Particular solution, 230 
Path, integrals independent of, 208, 

216, 452, 455 

Pendulum, simple, 44, 234-238, 306 
Periodic function, 64 
Picard's method, 347 
Plane, equation of, 147 
inclined, 280, 282, 306 
normal form for, 146 
tangent, 146-149 
Point, of inflection, 159 

singular, 451 
Poisson formula, 512 
Polar coordinates, 183, 184, 276, 279, 

386, 438 

Polygon, rectilinear, 478, 485 
Polynomials, Legendre, 344, 384 
Porous solids, drying of, 369 
Potential, electrostatic, 487 
gravitational, 219, 408 
lines of equal, 277 
Newtonian, 196 

velocity, 221, 222, 277, 430, 432, 
453, 467, 480 



Potential function, 219, 411 
Power series, 30-62 

differentiation of, 33, 34 

evaluation of integrals by, 43-46 

expansion in, 35-46 

functions defined by, 33 

integration of, 33, 34 

interval of convergence of, 31, 33 

operations on, 33-35 

theorems on, 31-35 

uniform convergence of, 33 

uniqueness of expansion in, 38 

whose terms are infinite series, 40 
Power series solutions of differential 

equations, 325-346 
Precision constant, 520, 521 
Pressure on dam, 484 
Primitive, 458 

Principal part of increment, 128 
Probability, 492-524 
Probability curve, 521 
Probable error, 521 
Probable value, most, 505 

probability of, 511 
Product, of determinants, 110 

scalar, 399 

vector, 400 
Projection, Mercator's, 479 

stereographies, 479 
Pulloy, slipping of belt on, 239 
pv diagram, 223 



Q 



Quadrature, mechanical, 554 
Quotient, of complex numbers, 444 
of power series, 40 

R 

Radius of convergence, 31, 33 
Radius vector, 195 
Rank of matrix, 115 
Ratio test, 11, 20, 31 
Reaction, chemical, 258 
Rearrangement of series, 17 
Rectilinear polygon, 478, 485 
Recursion formula, 273, 328, 331, 
334 



INDEX 



585 



Region, multiply connected, 205, 
212, 455 

of integration, 173 

simply connected, 205 
Regulafalsi, 101 
Regular functions, 451 
Remainder in Taylor's series, 36-37 
Remainder theorem, 92 
Repeated trials, 501 
Representation, applications of con- 
formal, 479-491 
Residuals, 534, 537 
Resonance, 310 
Riemann surface, 473 
Right-handed system of axes, 397 
Rod, flow of heat in, 373 

vibrations of, 366, 367 
Roots, of equations, 83-102 
isolation of, 92 
theorems on, 92-94 

of unity, co, co 2 , 87 
Rot (see Curl) 
Rotational field, 418 
Rule, Cramer's, 113 

Simpson's, 556 

trapezoidal, W 556 



S 



Scalar field, 406, 408 

Scalar point function, 406, 418 

Scalar product, 399 

application of, 404 
Scalars, 392 
Schwartz transformation, 478, 485, 

491 

Seepage flow, 483 
Separation of variables, 257 
Sequences, 2 

fundamental principle of, 6 

limit of, 3 
Series, asymptotic, 524 

binomial, 40 

evaluation of integrals by, 43-46 

Fourier, 63-82 

infinite, 1-62 

integration and differentiation of, 
29, 33, 34 



Series, of constants, 6-22 
of functions, 23-62 
power, 30-62 
solution of differential equations 

by, 228, 325-346 
Taylor's and Maclaurin's, 37, 155, 

228, 249, 464, 539 
tests for convergence of, 9, 11, 12, 

15, 20, 27, 31, 33 
theorems on, 17, 21, 27, 28, 29, 31, 

33, 34, 36, 38 

uniform convergence of, 23-30 
Shearing stresses, 485 
Simple closed curve, 200 
Simple harmonic motion, 233, 301, 

314, 380 
equation of, 234 
period of, 234 

Simple pendulum, 44, 234-238, 306 
Simply connected region, 205 
Simpson's rule, 556 
Simultaneous differential equations, 

312-315 
Simultaneous equations, 102-122, 

139-141 
sin x, 41, 250 
sin- 1 x, 46 

Sine, hyperbolic, 247 
length of curve, 55 
power series for, 40, 41 
Sine series, 73 
Singular point, 451 
Singular solution, 279 
Singularities of function, 222 
smh x } 247 

Sink (see Source and sink) 
Six-ordinate scheme, 548 
Slipping of belt on pulley, 239 
Small numbers, law of, 512 
sn u, 51 

Solenoidal field, 423 
Solid angle, 195 
Solids, drying of porous, 369 
Solution, of cubic, 86-91 
of differential equations, 226, 228, 

325 

general, 230, 290, 292, 350, 358 
particular, 230 



586 MATHEMATICS FOR ENGINEERS AND PHYSICISTS 



Solution, of differential equations, 

singular, 279 
of equations, 8&-122 
algebraic, 86, 95 
graphical method of, 83 
transcendental, 85, 97 
of systems of linear algebraic equa- 
tions, 102-122 
steady-state, 309, 310 
Source and sink, 221, 411, 416, 424, 

427, 429 

Space curves, 149-152 
Spherical coordinates, 152, 185, 382, 

386, 434, 439 
Spring, 299, 308, 313 

oscillation of, 299 
Standard deviation, 523 
Steady heat flow, 256, 368, 369, 427 
Steady-state solution, 309, 310 
Stereographic projection, 479 
Stirling's formula, 508 
Stokes's theorem, 421 
Stream function, 221, 432, 453, 481 
Stream lines, 277, 432, 467 
Stresses, shearing, 485 
String, vibration of, 361 
Sum, of a series, 4n. 

of vectors, 393 

Superposition of effects, 129, 223 
Surface, equation of, 144 
level, 406 

normal to, 147, 407 
Surface integral, 188-196, 415, 421 
Surfaces, coordinate, 434 

Riemann, 473 
Systems of equations, consistent or 

inconsistent, 117-122 
linear algebraic, 107-122 



Taylor's theorem, 36 

Tension, 239, 243, 244, 251, 362, 377 

Test, Cauchy's integral, 12 

comparison, 9 

for alternating series, 15 

for series, 9, 11, 12, 15, 20, 27, 31, 
33 

ratio, 11, 20, 31 

Weierstrass M, 27 
Theory of least squares, 521 
Thermodynamics, 222 
Torque, 405 

Total derivatives, 130-143 
Total differential, 127-143 
Trajectories, orthogonal, 277-279 
Transformation, by analytic func- 
tions, 467 

conformal, 467 

Green's, 191, 202, 418 

of element of arc, 467 

of integrals, 202 

Schwartz, 478, 485, 491 

Stokes's, 421 
Trapezoidal rule, 556 
Trials, repeated and independent, 

501 
Trigonometric functions, 78, 251, 

446, 447 

Trigonometric series, 63-82 
Triple integrals, 177, 193 

U 

Undetermined coefficients, 229 
Uniform convergence, 23-30 

test for, 27 

Unit vectors, 394, 397 
Unity, roots of, 87 



Tangent line, 143, 147, 151 
Tangent plane, 14&-149 
tanh, 249 
Taylor's formula, 35-46, 158 

applications of, 41-46 
Taylor's series, 37, 464, 539 

for functions of two variables, 
155-158, 228 



Variable, change of, 154, 183-188 

complex, 440-491 

dependent, 2 

independent, 1 

Variable heat flow, 368, 373, 425 
Variation, of constants (see Varia- 
tion, of parameters) 



INDEX 



587 



Variation, of parameters, 318, 323 
Vector analysis, 392-439 
Vector equation of line, 395 
Vector field, 406, 408, 409, 412, 418, 

423 

Vector point function, 406 
Vector product, 400 

applications of, 404 
Vector relationships, 402 
Vectors, 144, 152, 392 

addition of, 393 

base, 396 

curl of, 418 

decomposition of, 396 

divergence of, 411 

magnitude of, 393 

multiplication of, 399 

origin of, 393 

orthogonal, 398 

radius, 195 

unit, 394, 397 

zero, 393 
Velocity, 404, 424 

angular, 61, 191, 236, 404, 424 

critical, 61 

of earth's rotation, 61 

of escape, 61 

terminal, 59, 254 



Velocity potential, 221, 222, 277, 

430, 432, 453, 467, 480 
Vibration, forced, 308, 310 
of elastic rod, 366, 367 
of membrane, 377 
of spring, 308 
of string, 361 

(See also Simple harmonic mo- 
tion) 

Viscous damping, 302, 366 
Volume, as a triple integral, 180 
element of, 185, 187, 190, 437 
Volume integral, 180, 415 

W 

Walhs's formula, 45 
Wave equation, 432 
Wedge, 473 
Weierstrass test, 27 
Work, 217, 404 
Wronskian, 317 



Z 



Zero vector, 393 

Zonal harmonics (see Legendre poly- 
nomials)