32 Mr. J. Proud man on the 



Instead of trying to solve this problem directly, let ns 

 determine a function which satisfies (3) over the area of the 

 square bounded by £= +77-, tj= + 77-, and which vanishes on 

 the sides of this square. The determination of such a func- 

 tion is known to be unique, and from considerations of 

 symmetry we see that over the triangle L'SL it will be the 

 function we require. 



Now 



4(7^ + f )t7r 9 -V) - 2 A» cosh (n + i)f cos (n + J)*?, (4) 



where A w is a constant, satisfies (3) and vanishes over 

 7] = + 7T. We shall see that we can choose the constants A„ 

 so that it will vanish on f = +7r. 



From Fourier's theorem, or otherwise, we have 



4 oa / -J\rc 



for — 7r <: r] <> it, by which we see that if we take 



( — 1 ) n 

 A„cosh (w + i)7r = 327r v 



(« + *)•' 



(4) will satisfy all the conditions for %. 

 Thus, 



0.0 3 (~l) w cosh(?i + i)£ . 



The value of % on SL, which gives the velocity on the free 

 surface, is obtained by putting ?? = £ in (5). Doing this, we 

 obtain 



in which f is connected with the distance r from the focus, 

 by £ = r*/ V'2. 



For the flux of liquid through the channel we require the 

 function 



taken over the area of the triangle L'SL (fig. 2), or, again 

 from symmetry, taken over the area of the square SL. The 

 integration is straightforward, the series for ^ being uni- 

 formly convergent over the area, and we obtain 



F 7T 6 » 1 S 1 



+ 2 2 7-ttw; -2tt 2 . , M5 tanh (n + *>, 



128tt 2 15 ~Z (n -r if Zo (n + ±)« 



