232 Mr Sharpe, Liquid Motion from a Single Source 
We notice that if in this we put p = 1 and 6 = a 2 the result is 
exactly identical with (29). The rest of the solution is now plain. 
It will be found that the equation of A 2 A X is 
cp sin 6 + u6 + ^ 
tan "(h^ tan ^ 
par . ..(32). 
Here of course p < 1. Putting p = 1 and 6 = cl x we get for deter- 
mining a x in terms of pu/c and b/c, 
par = c sin a x + pua x + 
ba x 
T 
(33), 
which is the same form as (18). Finally the equation to A X D is 
cp sin 6 + pu9 + ^ 
6 4- tan' 
= par . . .(34). 
I believe it will be found that the equations (29) and (33) are 
perfectly compatible. As a particular case we may have a x = tt/ 3, 
a 2 =57r/6. We shall then get fi/c = 3*0531 and b/c = ‘3814, and 
these values satisfy all the conditions to which these constants are 
subject. 
Part II. 
8. We will next consider the problem in three dimensions. 
Suppose a single source of liquid supply at 0 (Fig. 1) inside a 
hollow boundary DABA'D' which is a surface of revolution round 
Ox. To find the possible forms of such surfaces. The liquid 
motion is supposed to be symmetrical round the axis of x and in 
planes through that axis. As before we shall use different ex- 
pressions for the velocities within and without a sphere of radius 
a, but such that when r — a they are continuous. Let cj) x be the 
velocity-potential inside the sphere, then we may take for r < a 
(35). 
The 1st term represents the source at 0. In the 2nd term a n is 
an arbitrary constant, P n is a Legendre coefficient of 6 of the 
nth order and n has a set of integral values from 0 to oo, but 
exactly what set will be shewn presently. Then 
#i mua.S 
dr r 2 o 
, nr 
n ~df r ) 
(36), 
l#i = ? / a n r n dP , A 
r dd~o\ a n dd) 
( 37 ). 
