1501 
solution 
f(y, «) = 5 pe gd ie Sod By (29) 
Application of the boundary conditions and equation 
(40) gives: 
Fiuje Zu) = 7 
Ale) = Gr)" f deol Ha) epfoterfar 
Api alee 
\(42) 
Mt 
0d 
Af (i, u) aAflu 
fi Ae eget . ees Me a fel 42, x) lw] opie yidv 
Integrating with respect to & , equation (43) 
becomes : 
co ot 
A to ode Het fo) lel “riot 
seine 
Eauations (/4 42) and (44) serve to define the functions 
A (4 ) andB(«). Solving for f, and f. in terms of A and B, 
we have: 
VY 
ae (45) 
4s file 4 A + 
pe ie ee 
U 
(46) 
/ 
“44 fiz My 4 - 
DE) 
aS. 
iv) 
