Darrozes 
This relation is fulfilled for any function A,(Z), and we must solve 
the Hilbert problem for a function B, (zZ) holomorphic on the positive 
part of the real axis. This has been done in section III, 2, but the cor- 
responding solution Z% , does not fulfill the Kutta-Joukowski con- 
dition (25). 
~ (1) +(1) 
Consequently : @ =co9(s)+é€éo(s) + noe an 
Ags," = omy 81s A, (Z) + Ay (Z,)+ 3 [B, (z) - B, (z,)] 
A, (Z) and B, (z) are sectionally holomorphic functions, satisfying the 
following relations 
ale - be == 4 eS Z=0 <o (29) 
Br + By Ft 5 AP) Zep Bar eh (30) 
hy - Bo - - i cotg aes) Z=o, Y> oO (31) 
(ey eet i Gell) bE ves 
With ida Gres Wns terohech bieaieanis sue oa) 
ja Z 
The equation (29) has been solved in the preceding paragraph : 
Ng laces 
A(z) © = z Log z+ ¢) (s)% (32) 
From equation (29) 
oe Pp) henge 
By (z) = “ae tie, (s) 
and using the relation (31), the solution is 
hes 
B, (Z) = le | 6 (P) i cote ex(s) (33) 
From these results, the 7 - expansion of the flow velocity potential is 
written in the following form : 
i) (e) = (e) a 
& = G(s)-e (s)-1Logn 61 Bcos O+ np. wd ( @sin O- Logp cos 8) 
ui 
etsy esemacng Bl et S) % 
1 sin9- e) ‘(s) cotga cos + o( 772) (34) 
In this expression, we haxe assumed Bar ae vortex sheet has no 
thickness, so writing 3 ® Maz | = E32 O , we obtain 
Cie = Gee = oe. 
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