354 



6*r (x) - 6**r^(x) = -e^ h(5)d5 -ej - , 



Equation (87) describes the dispersion of these 

 waves when surface tension is neglected. 



-00 < X < 00 



(80) 



NUMERICAL SOLUTION OF THE INTEGRAL EQUATION 



which is constant as could be expected because 

 belongs to a constant loading of magnitude -EjpU 

 of the whole cavity. 



This displacement however can be calculated in 

 another way by using (6) where we have to replace 



Pc ^^ 



p + pU^f = p - EipU^. 



c c 



(81) 



Expanding (6) with respect to ej, we find 



r + 6*r - 5**1 

 c c 



(p -p ) 



<*> c 



Combining (80) and (82) yields 



(P -p^+eiPU^)'^ 



r EipU^ 

 T i (1 - ^^ )■ (82) 



(Poo-P. 



pU-^r /2(p -p ) 



C 00 c 



h(x)dx + A/b 



(83) 



Substituting h(C), A, and b from (67) and (68) into 

 (83) and carrying out the integration with respect 

 to X we find after some reductions 

 -1 2 



2(2a+a -C^ 



-1 



(84) 



where 



L = 



dC 



2 ^ 



^ J 5{[?J (C)+aJi(S)] +[CY (5)+ctYi(C)] } 







Solution of (84) with respect to C yields 







2L 



5 (a) = a . {1 + 

 (aL-1 



(85) 



(86) 



by which we have found the unique solution of (47) 

 for 5 real and g > 0. This derivation rests on 

 some mechanical considerations such as uniqueness 

 of the solutions in relation to radiation conditions. 

 The result however, which is interesting from the 

 point of view of zeros of transcedental equations 

 connected with Bessel functions, has been verified 

 by others in a more straight forward way and found 

 to be correct. 



By (86) it follows that an axial symmetric wave 

 moving along the cavity with velocity U has a wave- 

 length X (U) given by 



X(U) = 2TT/b =[Tir/S^(a)] [2p/(p^-p^)] 



2(P -Pc)/PU'' 



(87) 



In the left hand side of (73) the function k (x) is 

 given by (66-68) and the dimensionless quantity 

 f(x') is unknown. For x < the right hand side 

 is determined by the geometry of the hub. Let this 

 geometry be described by 



r (x) = r + 6r, (x) 

 h c h 



(88) 



where r (x) is a given function. Then the right 

 hand side is known up to an unknown shift, s, of 

 the hub along the x-axis, since the position of 

 the point of separation is a priori unknown. Hence 

 for x < we can write (73) as 



/ 



k (x - x' 



f (x- 



dx' 



(x + s) 



x < 



(89) 



where the function f(x') and s are unknown. First 

 we will describe how f(x') is computed numerically 

 from (89) for arbitrary values of s. Then s will 

 be determined by a condition to be satisfied by f 

 at X = 0. 



We make some remarks concerning the behavior of 

 f(x') for x'fO and for x' ->- -<». As will be shown 

 in the Appendix, the behavior of f(x') near the 

 origin is, for arbitrary values of s. 



-1/2 

 f(x') = 2Tr ^ B 



1/2, 



x'tO 



(90) 



where B is some constant which will be discussed 

 later. 



The hub has a constant radius far upstream, 

 hence 6r (x) tends to a finite value for x ->■ -">. 

 Since the kernel k (x) vanishes for x -+ -", the 

 perturbation due to the end part of the hub van- 

 ishes far upstream. Hence, the pressure distribu- 

 tion there is the same as that of a two-sided 

 infinitely long hub with constant radius. This 

 case was also considered in the preceding section. 

 We find f(-<=) from (82), with -Ej = f(-'»), and (6) 

 and (87); 



f (- 



a 6r, (-<«>)/r . 

 h c 



(91) 



In order to transform (89) into a discrete 

 function we choose n + 1 points on the negative 



x-axis : 



-oo<x <X ,<...<Xl<X =0, 



n n-1 



(92) 



and construct n coordinate functions, f (x) , ..., 

 ffi (x) , defined on -°> < x < as follows: For 

 m = 2, ..., n-1 the function f^ia) vanishes out- 

 side the interval (Xj^+i, Xj^_i), and inside this 

 interval its value is 



f (x) = (X 



m 



>c ^,) / (X 



m+1 m 



X ,) , X , < X < X , (93a) 

 m+1 m+1 - - m 



f(x) = (x-x ,)/(x -X ).x<x<x . (93b) 

 m m-1 m m-1 m- - m-1 



The function f j (x) vanishes for x < X2 , and: 



