232 Mr Sharpe, Liquid Motion from a Single Source 



We notice that if in this we put p = 1 and = 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^ is 



cp sin + fx6 + -= 



- tan" 1 



1-p 



1+p 



- tan 



= /t*7T...(32). 



Here of course p <1. Putting p = 1 and = a ± we get for deter- 

 mining C*! in terras of fi/c and b/c, 



fjLTT = c sin a-i + fia^ + 



b<x x 



•(33), 



which is the same form as (18) 

 cp sin + fid + ■= 



+ tan -3 ' ' 



Finally the equation to A X D is 

 = /«r:..(34). 



P 2 + l 



tan 



I believe it will be found that the equations (29) and (33) are 

 perfectly compatible. As a particular case we may have a x = w/3, 

 a 2 =57r/6. We shall then get pjc = 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 (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 fa be the 

 velocity-potential inside the sphere, then we may take for r < a 



& = 



/u,a 



+ 2 







(a n r n p \ 



.(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 of the 

 ?zth order and n has a set of integral values from to oo , but 

 exactly what set will be shewn presently. Then 



dfa _fia 2! / p nr n 

 dr r 2 o V a 



1 #i = | ( a n r n - x dP n \ 

 r d0 ~ V V a n d0 J 



(36), 



(37). 



