inside a Hollow Unlimited Boundary. 
235 
we get on integration 
yjr 1 = G 1 — ya cos 6 — sin 0 
aF _f (/1 _6 0 )^P 1 ' + 
+ 2 
4 
(n + 1) 
...(51), 
where P n ' stands for dP/dd. (48) and (51) belong to points 
inside the sphere r — a. For points outside this sphere (38) 
becomes 
b 0 a 2 
r 
<f> 2 =Vr cos 6 — £ — b„)~ Pi + — bn)^ + 
2 /a n a n+1 \ 
+ f {-^r Pr.) ...(02). 
Putting in (49) for <f> the value of </> 2 from (52) and integrating we 
get 
VP 
yfr 2 = C 2 — b 0 a cos 6 — sin 6 
- 
sin 6 + 
2 
+ 3 
/ h\ a2 v h\ aZ P' S/Ona^p/V 
<** - 6.) T p. - 4 0* - W 2^ Pa - ? I P » ) 
.(53). 
10. If in (51) 1 ^! = 0 is to be the equation of that particular 
stream line which cuts 00' (fig. 1) orthogonally, it must be satis- 
fied by 0 = 7r and r = OB, so that we must have 
0 1 = — fia (54). 
OB is most easily got from (36) by putting in it dfa/dr = 0 and 
0 = 7r. We thus get from (36), (45) and (46) for finding OB 
0 = f ~ l{ a F ~ I (fi ~ M + ia^~ b ° > " &C ‘ • ' ■ '< 55 )' 
Remembering ( T.L.F . Art. 14) that P w ' vanishes when n is even 
and 0 = 7t/ 2 we get from (51) for finding OE 
0 = - (Ba + jaF— |0t — 6,)j ^- a (56) 
accurately. We see from this that if V = 0 we must for OE to be 
real have b 0 > fi. We have next to get the value of C 2 in (53). 
This may be done in two ways. We can either suppose V to be 
finite and remembering that from (35) 4yaa7r is the quantity of 
liquid given out by the source in the unit of time, and making the 
