186 T. H. Havelock 
Using (35) in (11) we obtain 
L/L, = k+k*prysin 6 + kp?r,B, sin? 0 — }p3r,(2kB, cos 20 — B?) sind 
—}p'r,(kB, sin 30 —2B,B,sin@)siné+. .. (44) 
Substituting from (36) and (43) and collecting the terms we obtain 
L/L) = 1+6, p+, p? +b, p? +b, pt + Me 
b, = 2rysin 0, 
by = 3rgsin?@ +7, —2r} cos 20, 
3 = 4r§ sin? 6+ 27ro7r,(2 sin 6 — sin?) —7r, sin 6 cos 20 — 8ryrj sin 0 cos 20, 
by = 5r§ sin? O + 3r27,(3 sin? @ — 2 sin’ @) + 37? — 37, cos 20 
— 3% o1o(7 sin? 6 — 12 sin* @) — 18727; sin? 6 cos 20 — 3r, rj cos 20 
+ 377? cos? 26 + 375 cos 40. 
(45) 
The integrals given in (33) and (34) give for the coefficients, 
To= 5qtan a, 
a\e 
Pa =(5) sec? a, 
Tr \3 
tT. = 2s) sec?a tana, (46) 
2d 
4 
7 ‘ 
a 25) sec? a(sec? a + 2 tan? x), 
77 , a 
"1 24g? "3 ~ 940q2” 
with a = 7(b—a)/2d. 
We may derive limiting cases from (45). If we make 6 and d infinite, we 
have a semi-infinite stream bounded by an upper plane wall; the limiting 
values of the coefficients are then 
1 . 1 . 1 . 3 . 
pa Tage Ta Tage US =e 
fh Sits = Os (47) 
447 
