624 Prof. T. H. Havelock on an Integral Equation 



example of such a. theorem; without attempting any discus- 

 sion of the general theorem, we proceed to solve (4) directly 

 on the lines indicated in equations (9)-(13). In the nota- 

 tion of these equations, we have from (4) 



p = 0, q=-7r 2 v/h 2 , r=-2Vp/h 2 , , .. ; 



P = 2/i,/(7/i, Q = R= .... =4/k/«tA, 



Equation (9) becomes a transcendental equation, namely 



-^coth^ + l = 0, .... (14) 



The roots of this equation are negative, and it is convenient 

 to write x= — vX 2 /A 2 , then the values of X are the positive 

 roots of the equation 



\tnn\=2ph/<T. . . . . . (15) 



Using A 1? A 2 , ... for the coefficients of the solving function, 

 equations (10) become 



Ai I A 2 | A 3 , -Z-0 (ifo 



X^-nW + X 2 2 -n% 2 + X^-n'ir 2 + " * ' ' h 2 ' { } 



where n=0, 1, 2, ... and X 1} X 2 ,.^. are the positive roots 

 of (15). 



Assuming that a function /(#) can be expanded, in the 

 range — 1 < x < 1, in a series 



/■'(a?)==XC cos Xx, 

 we have 



X 



0: 



A. + sin X cos X 



f +1 



1 f(x) cosXxdx, . . (17) 



J-i 



where X is a root of (15). Taking f{x)-= cos nwsc, we obtain 

 the set of expansions 



-i 9 v* X 2 sinXcosX /,nx 



Z (X + sin X cos X)(X 2 -n 2 7r 2 ) ' ' ' { } 



Hence the solution of the set of equations in (16) is 



» 2vX, 2 sinX r cosX r _ 4//, X r 2 ... q. 



''" A 2 (X r + sin A, cos X r ) ~ o7* X, 2 + A-(l H- A-) ' ^ V 



where k = 2ph/a. These results can also be derived directly 

 by extending the forms (12) and (J 3) to include infinite 

 products. 



