760 
PROFESSOR W. M. HICKS OK THE THEORY OF VORTEX RTXGS. 
Since it will be necessary to obtain the value of (f>, it will be necessary to obtain 
the value of xjj at any time ; that is when the boundaries are given by equation (51). 
This will consist of two portions, one determined by the normal motion at the boundary, 
and the other (e) by the fact that the circulation remains unaltered. 
14. To find the function y it is first necessary to know the velocity normal to the 
boundary at any time. Now since the boundary is not the circle, it is in steady motion, 
the function \p 0 will itself produce a normal motion. This must first be found. Let 9 
be the angle which the boundary at an} 7 point makes with the circle u — u, then 
~ dn du 1 dk 
clT1 dn' dv k dv 
or since 
k=k 1 (l-\-ct. cos nv- J- y sin nv) 
tan 9=n(a sin nv—y cos nv) 
Now the normal velocity outwards is 
(52) 
1 b-Jr dv n 1 t-v/r du . . 
- -A — cos 9 ---— sm 9 
p ov dn p ou dn 
where xp=:\jj 0 - J-y;, \p 0 being the stream function already found. Also, remembering 
that 6 is small, the normal velocity is 
= — cos 6— U sin 6 
dt 
= 2ak l (<x cos nv-\-y sin n?;)~wU(a sin nv—y cos nv) 
Therefore, along the boundary 
1 dv 
p dv dn 
(2ak x ct.-\-n\Jy) cos nv-\-(2ak x y —nUa) sin nv 
Then the two sets of conditions give (letters with one dash referring to the inner 
surface, and with two dashes to the outer surface) 
ART +BTT = ~« 3 U lV /(2 h )*+-(2 k^y 
'TO 
ARA+BTA = -a^V^(2k 2 )/3+-(2k,yh 
n i 
CRT +DTT = -a 2 U lv /(2^)y -^(2 k^a 
CR"„ +DT''„= -a a U„ v /(-2/;,)S 
(53) 
