624 Prof. T. H. Havelock on an Integral Equation 
example of such a theorem; without attempting any discus- 
s'on of the general theoram, 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, G= Sen, PS Srl 506 8 
P=2y/ch, Q=R=.... =4y/ch. 
Equation (9) becomes a transcendental equation, namely 
2 
g coth alia OW een tern te (OLA) 
Tro 
Ovex? 
The roots of this equation are negative, and it is convenient 
to write w= —vd*/h?, then the values of X are the positive 
roots of the equation 
NUN —=)2 010 Gan tay BE (160) 
Using Aj, Ag, ... for the coefficients of the solving function, 
equations (10) become 
Ay Ay A; 
; omatata = ar = 
Ay—n2nr? — dr —n*a? 3? —n? rr? 
Vv 
+ ++. ——=0, (16) 
where n=0,1,2,... and 24,Azs,... are the positive roots 
of (15). 
Assuming that a function f(a) can be expanded, in the 
range —1<«#<l1, ina series 
f(@)=C cos Az, 
we have 
r +1 
Oe erasrnaa ent Ned, 
A+sin A Cos \. J(2) cos rx da, (17) 
where A isa root of (15). Taking /(a@)=cosn7a, we obtain 
the set of expansions 
? sin AX cos A 
>) : oh: 
(A+sin XA cos X)(A?— n777”) 
(18) 
Hence the solution of the set of equations in (16) is 
_ _2vr,~ SiN A-COSA, __ Ap ee (19) 
"~ A2(A,+ sin A, cos Ar) GharAP+kh(1+hky’ 
where k=2ph/o. These results can also be derived directly 
by extending the forms (12) and (J3) to include infinite 
products. 
180 
