inside a Hollow Unlimited Boundary. 233 
Outside the sphere or for r > a we will take for velocity-potential 
b 0 a 2 
Vr cos 6 h a 
r i 
(38). 
In case the boundary is tubular at infinity, the 1st term gives us 
V the constant velocity at infinity. If we want the boundary to 
resemble a hyperboloid or paraboloid at infinity, V will be 0. The 
2nd term in (38) represents the fictitious source at 0, which will 
ultimately be seen to represent the effect of the boundary on the 
velocity at a distance. In the 3rd term in (38) n is supposed to 
have the same values as in (35) excepting zero. Then 
dcf> 
dr 
? = y cos e + ^ - 1 {&„ (» + 1) ££ P M } (39), 
1 dfa 
r d6 
V sin 6 + X 
i 
/ b n a n+1 dP n \ 
V r n+2 d6 ) 
(40). 
It will be seen either from (35) and (38) or from (37) and (40) 
that the motion is symmetrical with respect to Ox. 
On the surface of the sphere r — a we must have 
p g hia n P n \ 
a o \ a ) 
V cos 0 + — — 2 \b n (n + 1)— 
a i a 
...(41), 
/ a n dP n \ 
U lie) 
= — V sin 6 + X 
i 
b n dP n \ 
.a dO ) 
(42). 
These must be true for all values of 6 from 0 up to a certain 
definite angle, which may be 7t/2 or any less angle. 
In (42) we must have generally a n = b n except when n = 1, in 
which case remembering that — sin 6 is dPJdO, we must have 
= a x — aV. Since dP 0 /dO is zero, a 0 is arbitrary. Putting these 
values in (41) it will become 
ii a 1 P 1 1 -r-< s \ 
1 1 S (W'&w-Pft) 
a a a 2 
= VP 1 + h ^ {a, - aV) P,-~ 
a a v 7 a 
OO 
2 (n + 1) a n P n 
2 
or (fi - b 0 ) + 3 Pi (a 1 -aF) = -2 (2 n + 1) a n P n (43). 
2 
Whether V is finite or zero (43) can be satisfied in an infinite 
number of ways, so that we get an infinite number of solutions of 
the problem proposed. The method is as follows. 
9. By Todhunter’s Laplaces Functions ( T.L.F .) Arts. 28 and 
62 it can be shewn that between the limits 6 = 0 and 7t/2 any 
