527 Prof. T. H. Havelock. Some Cases of 
on the sphere. We have to find X, the image of X, in the sphere ; for this 
we first put X; into a different form by using 
mJ o[« {(a+cu)?+ y?}4] = {, cos {x (x+cu) cos p} cos(xy sind) dd. (36) 
From (36) and (34), after carrying out the integration with respect to w, 
we obtain 
wX, = ca? | on *(S-2) ed ic [,t4 sin («x cos @)+ B cos («x cos d) } 
0 0 
x cos (xy sind) cospdd, (37) 
where A and B are given by (14) after writing ¢ cos ¢ for c. 
For convenience in the following analysis, we transfer the origin to the 
centre of the sphere, noting that in (37) we shall have exp.(—2«f+.«z) in 
place of exp.(—«f+«z). Also we use polar co-ordinates 
v= rcosa; y = rsinacos BP; z=rsinasin £. 
The conditions for X2 are that it must be a potential function, the 
disturbance due to it must ultimately vanish as we recede from the sphere, 
and on the sphere 
0(X1+X2)/or = 0. (38) 
To avoid repetition of expressions like (37), we take out of it a typical term 
and write 
X) = e* sin («x cos ) cos (cy sin ¢). (39) 
We know that the function 
rerwel™ sin (Kau cos p/r*) cos (Kay sin p/7?) (40) 
satisfies the first two conditions for X2, but we find it does not fulfil (38). 
An additional term is required, and it can be found in the following way. 
Suppose that on the sphere we have 
e® sin (Kx cos d) cos (Ky sin 6) = TAnYn (2, B), (41) 
where the right-hand side is an expansion in surface spherical harmonics. 
Then for the term (39), all the conditions for X2 are satisfied by 
ar ex!" sin (Ka*x cos h/7?) cos (Ka?y sin b/7?)— La"** AnY m/(m+1)7r™*1. 
(42) 
Suppose, similarly, that on the sphere we have 
e* cos (xz cos h) cos (ky sind) = YBnYn (a, 8). (43) 
126 
