PROFESSOR W. M. HICKS ON THE THEORY OF VORTEX RINGS. 
777 
For the pressure condition we require the stream function. Now 2nxfj is the flux 
across any diaphragm with circular boundary through the point in question. Take it 
to be the annulus v—tt, from u—u x to u = u, and then the portion of the tore u—u 
from v—tt to v=v. We need only to find the part of xfj due to (j). Now </> produces 
no motion across the annulus. Hence 
or 
r bd> dv! „ dn' 7 
2 mA = —-2 t TO—civ 
J v on dn ' dv 
., rw j 
x(j = — a\ —dv 
J i)07.l 
=4a 3 & 1 2 £(7r— v) 
= ia 3 k 2 2 r](i t—v) ■ . 
(77) 
which is many-valued, as it ought to be. 
Again, we need the flow due to y. This is 
P‘ 1 duJ _ (T 1 he c iv 
™ J Ul p dv dn' v_ v =n ~) v \_p bu dn dv _ n =n 
=rR^i 
j Wl \_p bv\ v=n ‘ )v\_p bu\ u= i, 
__ pt-l hf 
J t , l_p tl/.J, 
.,' A or 
dv 
dv to lowest order. 
'2(C~c)' I! ° 4(C-c)* SP »^ rf! ’ 
= -|{-V(2)(iL- 1 )+i}(.-0 
77— V) 
= 4 Ao 3 /cf^(77— v) = AK(dk 2 rj{TT—v) . 
_ _ A 0 / 
- as/2 [ 
■ (78) 
The pressure is given by 
v . /_ dU. 7 , 1 tv du b<f> dv \ 3 
f= F -4( U +A rfrc 6v dn) 
-iG%£+££M* + & i+ * +x ) 
=F«-lU»+A#-M'-uf»-^g+A(3t w -+*'+X 
