﻿4:62 Mr. G. B. Jeffery on the 2 too-Dimensional 



take O = O, so that 



dyjr 



-^ = 2v(l- m 2 - m 2 P) + 6vk 2 m 2 sn 2 (m0, k). 



= 2v(l + 2m 2 - m 2 k 2 ) - 6pm 2 dn 2 (m0, k). 



The stream-lines are all straight lines passing throuoh the 

 origin. If u be the velocity of the fluid 



1 d-dr 

 U =~rd0- 



The constants m, k can be chosen so that the velocity is zero 

 when 6 = a, and hence, since u is an even function of 

 when 6 = — a, while the values of yfr appropriate to these 

 two stream-lines differ by a given amount. Thus we have 

 .a solution for the motion of a fluid between two fixed planes 

 inclined at an angle 2« due to a line source of given strength 

 along their line of intersection, or if we exclude the origin, 

 for tbe flow along a canal with converging banks. If Q is 

 the total flux of fluid outwards from the orioin 



Q = _ 4j, a ( l + 2m 2 - m 2 k 2 ) + 12vm 2 C dn%m0, k) d0. 



Now f£ 



\ dn 2 (:,k)dz = E(t;^) 



where E denotes the elliptic integral of the second kind. 

 Hence 



Q= -±vcc(l + 2m 2 -m 2 k 2 ) +12vmE(mot, k). 



This relation, together with 



M 2 m 2 sn 2 (mcL, k) + 1—m 2 - m 2 P=0, 



is sufficient to determine m and h. (N.B. — a must be ex- 

 pressed in circular measure.) 



If the angle between the planes is small, we may write 

 sn.z=z, and we have as an approximation when the planes 

 are nearly parallel 



<fy = 2v{l - m 2 - m 2 k\l - 3m 2 2 )}. 



If the angle between the planes is 2a we have 



1 _ m * _ mPkXl - 3m 2 a 2 ) = 0, 

 or , 2 _ 1 — m 2 



m 2 (l-3mV)' 



and d-dr 6vm 2 (l — m 2 ) /z)9 

 df l-3mV ( ^- g ' ) - 



