moving through a channel with diverging boundaries 311 



/3 a^r" 



Thus the lateral pressure apparently decreases for all values of 

 Uq. But if 



Uq= — 2v {I — m^ — m^k^), 



and = - (1 - w2 - yn^) - Skhn^a^, 



we have Uq/2u = 3khn^a^, 



and therefore ma is not necessarily small. Hence the approxima- 

 tion is only valid for values of Uq/2v below some limiting value. 

 If this condition be satisfied the expression for pgg given above is 

 an approximation to its value for small values of a. 



Three-dimensional problem. 



Let the boundary of the channel be ^ = a, where {r, 9, (f)) are 

 polar coordinates. This problem has been considered by Prof. 

 A. H. Gibson^. In his solution Cartesian and Polar Coordinates 

 are confused, and he assumes that the stream lines are straight lines 

 diverging from the origin, a state of motion which is impossible if 

 the inertia terms are retained in the equations of motion, as he 

 retains them. One result of these errors is that in his solution p is 

 a function of 6 although the preliminary assumption is virtually 

 made that p is independent of 6. His expression for the pressure 

 appears to be quite wrong. 



Assume, in the first place, that the stream lines are straight 

 lines diverging from the origin, so that u = f{d)jr^, v = 0, w = 0. 

 The polar equations of motion reduce to 



du 1 dp 



or p or 



dht 2 du cot 6 du 1 c^u 2u 

 dr^ r dr r^ dd r^ dd^ r^ 



^ _ 1 dp 2v du 

 pr^^^dd' 



= - -^ ^ 



p d(f) 



1 Phil. Mag. (6), vol. xviii, p. 36, 1909. 



