DAMAGE BOOK Pages are missing 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)