moving through a channel with diverging boundaries 309 



Hence - = — - — --^ + C, where C is a constant. Now the lateral 

 p r^ zr^ 



stress in the liquid is pee , where 



p p r 



Hence pgg is independent of d, and is the normal stress (of the 

 nature of a tension) exerted by the liquid on the boundary. If 

 a is negative the normal pressure on the boundary decreases as 

 the channel widens, and if a is positive the normal pressure 

 increases. 



Now by substitution of the solution for u given above in the 

 differential equation satisfied by ii, we find 

 a = 4i;2 [_ 1 + m^ (1 - F + J^)] 

 = iv^ [-! + (! + uJ2vf (1 + k^)/{l + k^fl 



(!) Writing a = 0, we can immediately discriminate between 

 those cases for which the pressure on the boundary decreases and 

 those for which it increases. 



If a = 0, we have 



1 + uJ2p - (1 + k^r/{i + k^)^, 



and 1 + 2i//mo = (1 + k^f/{{l + k'^f - (1 + k^)^ 



Hence s„^ \( A + ^Y., 4 = (1 + P)* - (1 +^e)t 

 iV(l + k^y/ ) sk^ (1 + F)* 



The following diagram shows how the value of uJ2v for which 

 Pffg is independent of r varies with a, for those cases in which k is 



