﻿Problem of the Weir. 789 



Hence, 



I. 



= 1. . . . (e) 



o+o-)HO+<^+^} 



Proceeding in the same manner with (&) we find 



d£ 1 



J. 



It now remains to be shown that (e) and (tj) are not 

 satisfied simultaneously by any value of a. 



Put (X+*)* + \* = $h 



where 3 = £(£, o) is a function of both f and <r. 



(X) 



(e) becomes f ^ +<7)==4 

 Jo 

 and (,) „ C l dm =i 



Now '\2=u/(2gh)h. 



Also, X + (7 = u 2 /2gh + 1 - ft'/A = { u 2 + 2^(7* - &') }/ (%^), 



= (^ + U' 2 -U 2 )/(2^/i), 



= tt'*/(2^),by(3). 

 ... (\ + «r)* = u7(2^)*. 



For physical reasons, the same sign must be attached to 

 both \i and (\ + <r)*. Also \-\-a is always positive : hence, 



$=\ + o- + 2X*(\ + a-)* + X; .*. S>0. 



Also, $ + cr = 2|> + cr + \*(\ + cr)*]; .*. 3-fcr>0. 



Whether a be positive or negative the equations (X) cannot 

 be satisfied by any value of cr, except by a further value 

 0- = O, and therefore the only physical solution of the problem 

 is given by «r==l or h' = h. We conclude that a permanent 

 standing elevation, where the stream depths are different on 

 either side, cannot be maintained in an open channel of the 

 form contemplated. 



Appendix B. 

 It appears that both of the equations last written will be 

 satisfied by a further value cr = 0, provided 



J'.' 



'{/*=*, to 



and it is interesting to inquire into the physical meaning of 

 this relation. 



