351 



.6r (X) +uf- 6r (x) - '- ^'^'^^ =0 - 

 c 3x c 3r 



(33) 



We introduce the Fourier transform g(ij) of a function 

 g(x) by 



+ 00 



g(y) = -^ I g(x)e^^''dx, g(x) =-4= g (.X) e^'^^'dv. 



_oo -co (34) 



-iSx 

 j_ " Ki (g)ILe ^'^ dC 



2" J (g+ ^^"c )K^(g) - (a+e52)Ki(5) , (42) 

 where a and B are dimensionless quantities given by 



(43) 



r U'' 



Transformation of (31) yields 



ii 11- 



8r r 3r 



(U,r) = . 



(35) 



Hence, for real p 

 * (yr) =Ai(y) K^(|y|r) +A2(vi)I (|y|r), (36) 



where Kg and Iq are modified Bessel functions. 

 Because (j) ->• for r -> ■» we have 



Ao (U) = 0. 



(37) 



Substitution of (36) with (37) into (32) and (33) 

 yields 



(E-ipU) K (|y|r ) Aj (v) - (^ + p'^o)6r (y) 



c r^ r^ c 



= -u-'f(y), 



(38) 



It can be easily proved that under the assiomption 

 (4), 



a/B 



1 + [2r2{p_^ - p^)/pa^ > 1. 



(44) 



In order to find the Green function for the 

 stationary case we have to take the limit e -»■ in 

 (42). 



We now make some remarks for the case a j^ and 

 hence 6 7^ 0. 



First, the integrals in (42) are absolutely con- 

 vergent for < X < (»_ This means that when sur- 

 face tension is present Green's function k (x) is 

 finite even at the point of application of the 

 singular loading (41) . This could be expected 

 because the surface tension can be represented by 

 a membrane placed at the boundary of the cavity 

 and a membrane has the possibility to locally 

 sustend such a loading by a jump in its first 

 derivative while its deformation is still a contin- 

 uous function of x. 



Second, we consider the denominators in (42) for 

 e = and look for positive real roots of 



|y| Ki(|y|r ) Aj (u) + (£-iyU) fir (y) = 0. 



(39) 



Solving (38) and (39) for &r^ (y) and applying the 

 inverse Fourier transformation we obtain 



<5r^(x) = 



f (y) |y|K, {|y|r )e ^^''du 



,r2 a 



/27J [(£:-iyU)2K^(|y|r^) + (— - — +y2o) |y|Kj(|y|r^ 



-CO ^C "^ 



(40) 



K (C) 

 o 



Ki(?) 



(45) 



The left hand side of (45) is curved upwards for 

 C > 0, while the right hand side is curved down- 

 wards. The proof of the latter statement is rather 

 complicated and will not be given here. However, 

 taking this for granted, it means that there are 

 none or two real positive roots, which is analogous 

 to the case of ordinary gravity waves with surface 

 tension. One of the roots corresponds to a wave 

 primarily due to the swirl, the other one to 

 capillarity. [Whitham (1973), p. 446] 



We now choose 



f(x) = 6(x) 



(41) 



where 6 (x) is the delta function of Dirac, hence 

 f (y) = l//2Tr. Next we split the range of integra- 

 tion into two parts namely - «> < y < and < y 

 < »> and neglect terms of 0{£^) in the denominator, 

 then we find 



k(x)— fir (x) I 



c f (x)=6 (x) 



4. THE CASE OF ZERO SURFACE TENSION 



Green's function (42) in the stationary case for 

 zero surface tension, when we take a different 

 positive value for £ which of course is irrelevant, 

 is 



k(x) 



lim 



Sx 



2tt 



Ki (C) e 



dC 



:(S-i £)K^(?)- a Ki(C)] 



CO 



-i r 



2T7 -J 



Ki(S)e 



igx 



re 



dS 



(g-^^f^)K (5)-(a+6C^)Ki(C) 

 — r, — 



1 



271 



5x 



Ki(5) 



dC 



[(5+i e)KQ(5) - " Ki(C)]_ 



def 



+ I-. 



(46) 



