160 MR JOHN DOUGALL ON AN ANALYTICAL THEORY OF 
15. Independent symbolical solution of the general problem. 
The form of the last result suggests a method of dealing with the problem from 
the beginning, which, though not easy to develop independently with thorough rigour, 
has the advantage of conciseness, and will therefore be useful in giving a rapid account 
both of the foregoing solution and of those to be obtained in the following pages. 
We begin by observing that 
Lee sy a De EP 
ee rg ay? ‘a = sinh xzf(x, y) = sinh les + ap + a \ie, Y) 
I 
sinh Ky? + x) 7 (x, y). 
Hence sinh «zf(x, y) is a potential function, provided we regard « as an operator such 
that °=-—y*. We may, if we please, take «=zy, but it will not be necessary to 
interpret odd powers of the operator v. 
On this understanding, it is obvious from (12), (13) that we obtain a solution 
giving 
n= Ly =(i) onzg=+h 
Be =) on z= —h 
2 =fl,y)onz=h, within the area A, 
by taking - ae 
cosh «/) nich 
SS ees } wee SS } Be 
4p «(sinh he eee Kin (sinh Ihe OP ae a pe | 
37) | 
res cosh kh + 2«h sinh kh , iden sinh kh + 2«h cosh Kh 4 (37) 
pO = ~ (sinh Qh — 2xh) RMS + ~.2(cinh Qeh+ 2h) WS 
Now, taking as a specimen the first term of 4u@, we observe that the function of « 
fe oe sinh xz 
vanishes at infinity round the path W A BE of § 5. 
Hence the function is represented by the sum of its polar elements. (If «, be 
a simple pole of the function, and if in the vicinity of this pole the function 
= A,/(« —«,)+ finite, then A,/(k—«,) is the polar element at this pole, and A, is the 
residue there. The point «=0 isa multiple pole, and the polar element there has the — 
form A/«‘+B/«°’, these being the terms of negative degree in the expansion of the 
function near «= 0). 
Taking the elements belonging to + «, together we obtain 
ek 4 at A(— hg 1 ) 6 Oy 
«(sinh 2xh — 2h) Ke Ke Nk Ky KEK 
AC SB 2A,K, 
= ae ab a ab a Ke Ge Series Ree 
i Bats B Fe Vioas cosh x,h sinh «,z 1 
a ae Kh (cosh 2«,h — 1) x2 -«,? 7 
the series extending over the poles with positive imaginary part. When we put — 
«= —v’, this part of 4ud becomes 
Sek (ees cosh kh sinh xz 1 
EVI, eh Rag 2 kh(cosh 2kh — 1) y? +x? Wf 
where « is no longer an operator, but simply a root of sinh 2«h —2«h =0. 
