353 



of F(£;) (57) this right hand side also vanishes 

 for S ->■ 1. Because the integral is positive we 

 have found that the assumption Im sj = -Im S2 7^ 

 yields a contradiction. It follows that no other 

 residues have to be added to the resulting integrals 

 after the rotations as denoted in (58) and (59) 

 besides the two we mentioned for x > 0. 



Using some formulae from Watson (1922) we find 



Then we have to solve the following integral 

 equation 







k(x-x') f(x')dx'= &r (x) , X < 0, (73) 



-> c 



—00 



which is of the Wiener-Hopf type. 



K (+ i S) = T" tl J (5) + i '^ (5)] , ? > 0, (64) 



- 2+0 o 



Ki(+ 15)=-- [Ji(£) + i Yi(S)] , 5 > 0, 



(65) 



Adding poles to the integrals in the case x > we 

 can transform the Green function (46) into 



k(x) = h(x) + 



where 



(66) 



h(x) 



and 



e ^ dg 



2 2. 



[CJ (C)+aJi(S)] +[SYi{S)+aYi(C)] (67) 



5. THE EXPLICIT SOLUTION OF a Ki (C) -£ K (£) = 







In order to find an explicit solution of Eq. (47) 

 we first have to make some preliminary considerations. 

 Assume the following loading of the otherwise undis- 

 turbed cavity boundary 



f(x) = Ei6(x) 



(74) 



where ej is a small parameter. By (66) we find for 

 the deformation of the cavity 



Ej A sin bx 



6r (x) = e, h{x) + < 

 c '■ 



Next we consider 

 f(x) = 



-1 



X < 



f(x) 



X > 0, 



, X < 0. 



0, X > 0. 



(75) 



(76) 



A = 2? /(2c»+a^-C ) 

 



c 



(68) 



The function h(x) is symmetric, h(x) = h(-x). 

 For x -*- it has a logarithmic singularity because 

 for X = the integrand as a function of £, behaves 

 as 17/25, hence 



h (x) ~ — J.n X 



TT 



(69) 



For X ^ " the behavior of h(x) depends on the behav- 

 ior of the integrand in (67). For g ^ 0, this 

 turns out to be as 



(4a2/7i2£2)+o(53£ng) . 



(70) 



The loading given in (74) is the derivative with 

 respect to x of the loading given in (76) . Hence 

 the derivative of the deformation 6*r^{x) caused 

 by (76) has to be equal to (75) , we take 



cos bx , X > 0, (77) 



6*r (x) 



-El 



h(?)dC + 



, X < 0, 



Then it follows from Doetsch (1943) p. 233 that 



h(x) = -Trr^/a2|xP |x| ^-^ . (71) 



where we have chosen the constant of integration 

 in such a way that for x ->- +°° we have a harmonic 

 wave with mean value zero. 

 Finally consider 



Now suppose that for x < the shape of the 

 cavity is prescribed. 



FIGURE 2. Flow with swirl along hub. 



r = r + 6r (x) , -<» < x < 

 c c 



(72) 



and that the unknown pressure between hub and fluid 

 is p„ + pU^f (x) . 



f(x) =0 , X < ; f (x) = E, , X > 0. 



(78) 



The loading given in (74) is also, in this case, 

 the derivative with respect to x of the loading 

 given in (78) . Hence the same argument applies as 

 before. However, now the constant of integration 

 has to be chosen so that the disturbance tends to 

 zero for x ^ -<» , we find 



(79) 



Ej — (1-cos bx) , X > 0, 



5**r (x) = El 

 c 



h{5)df + 



, X < 0. 



Subtraction of the disturbances (77) and (79) yields 



