﻿788 Mr. J. H. C. Searle on the 



Equation (6) of § 2, which was deduced from the con- 

 dition of " continuity " of flow and the condition that the 

 pressure over the free surface is constant, takes the following 

 form in the present case, viz., 



-J, U(o+«(i— )J rt?) • 



(«) 



for the fall of the stream-bed, d, is now zero, and therefore 

 8=l+<*/A=l. 



Again, the horizontal pressure on the weir-Pace must 

 vanish when there is no weir. Hence, neglecting atmospheric 

 pressure, we have from (11), putting 8=1, 



-i)+T [ 



i«(^-i)+ L^(0{*(»+»(i— ;}-*(»]^=o. OS) 



«/o 



(a) and (/?) have to be simultaneously satisfied by some 

 value of 37. They are both satisfied by the value -bt=1, 

 i. e., h = h\ which corresponds to a uniform flow with an 

 unbroken level surface. We have to show that no other 

 solution is possible. 



Put X(f) =$(?)/* aQ d 1 — -bt=o-. In terms of physical 

 constants these are X(f) = u 2 /\J 2 . YPfigh = u */2g7i and 

 a=l — h'/7i ; where u is the velocity of the " upper" stream 

 at a '•' distance" t ) { = z\li) above the lower stream-bed. 



(a) and (/3) now become 



^-fG&y* 



(y) 



and 



lo-(o— 2)+ f l { Vx(*. + <r)-X}<*?=0. . . (8) 



(7) is now satisfied by the value <r = 0, and (o) is satisfied 

 byo- 2 = 0. Before proceeding further these factors will be 

 removed from the respective equations. From (7) 



-J>-(x-fc) , }« 



Jo (A. + <r)s 



_ in 



Jo (*. + «r)*{(X. + a)* + \*} 



