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APPENDIX.
On the nearly symmetrical solution for a nearly circular area, when w satisfies (d2/dx? + d2/dy2 + k2) w = 0 and makes div/dn = 0 on the boundary.
Starting with the true circle of radius a, we have w a function of r (the radius vector) only, and the solution is w — J^ (kr) with the condition Jo' (ka) = 0, yielding ka = 3'832, which determines & if a be given, or a if k be given. In the problem proposed the boundary is only approximately circular, so that we write r = a + p, where a is the mean value and
p — tti cos O + ft-L sin 6 + . . . + an cos nd 4- @n sin nd .......... (57)
In (57) 6 is the vectorial angle and «j etc. are quantities small relatively to a. The general solution of the differential equation being
w — AQJo (kr) + J1 (kr) { AT, cos 6 + B1 sin 0}
+ ...+Jn (kr) {An cos nd 4- Bn sin n0], . . .(58)
we are to suppose now that Al} etc., are small relatively to A0.    It remains to consider the boundary condition.
If <f> denote the small angle between r and the normal dn measured outwards,
dw    dw       .     diu   .    .
(LT       do       11 and        •          tan<£= —r^ =—^ = -(— an sin nd + /3n cos nd)   ......... (60)
with sufficient approximation, only the general term being written. In formulating the boundary condition dw/dn = Q correct to the second order of small quantities, we require dw/dr to the second order, but dw/dd to the first order only. We have
= A {^ (ka} + kpJ0" (ka) + pyJV" (ka)}
+ {Jnf (ka) + kpjn" (ka)} [An cos nd + Bn sin nO},
•—ja = - J« (ka) {— An sin w^ + Bn cos n^}
QjCk\j       CL
and for the boundary condition, setting ka = z and omitting the argument in the Bessel's functions,
A o { J0' . cos 0 + kp J0" + py Jo'"}
+ {Jn' + kpJn'} [An cos nO + 5n sin nd}
—
n sin w^ + -5n cos nd} {— an sin nd + f3n cos w-^} = 0.   (61) walls all round, with but one curve of zero elevation, between.
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