﻿542 Mr. W. Ellis Williams on the 



Eliminating p by differentiation, we get 



Denoting, for brevity, the operator on the right-hand 

 side by ■&, this becomes : — 



vD 2 ^ = $T)ylr (7) 



We have to solve this equation subject to boundary 

 conditions of type (5). 



Let yfr=^Q + '^lr l j where -*Jr is the solution for infinitely 

 small velocities, and yjr 1 is to be looked upon as a correcting 

 term which is small compared with ^r . 



Substituting we have 



^o + vD^sdDfo + dDf (8) 



Now since yjt is a solution of: (4) ^D 2 -v^ = and 5T)yfr L 

 is of the third power of the velocity and may therefore be 

 neglected. Further in the term £D-\|r () we may neglect i^ 

 (occurring in the operator $) as compared with yjr , so that 

 $DtJt contains i/r alone and is therefore a known function 

 of the coordinates. The equation thus becomes 



z,D 2 ^=$Df (9) 



and is a differential equation to determine yfr 1 . 



Now let yjr t} = (p (r) sin 2 (9, -^^^(V) sin 2 cos 0, where 

 ^o(V) is to be determined so that ^ satisfies the boundarv 

 conditions and <pi(r) is as yet undetermined. 



Now D 2 ^ 1 (r)sin 2 6>cos6> = 



and &D<£ (» sin 2 = 



~^— (^9o W ^2 + -73— ) sm_ # cos 0. 



The term sin 2 cos thus divides out of (9) which bee 



ome 



VI 



F[>'''<r) 



This is a differential equation of: the fourth order 



to 



