446 



but by (7) 



f(t,x,k,l) = —kf(t,x+l,k—l,0) 

 Therefore, 



(8) xf(t,x,k,Q) = f(t+l,x,k,Q) + kf(t,x+l,k—l,0), k = 1, 2. 3 . . 



In (5) setting t = 0: 



m 



(9) 2 r-l x l f(—m,x,k,n+m—i) = S(k,n) 



k,n = 0, 1, 2, ; m = 1, 2, 3, 



Now S(k,n) vanishes if & > n; therefore /( — m,x,k,n) satisfies the linear homo 

 geneous difference equation of order m: 



m 



(10) 2 7 x l f(—m,x,k,n+m—i) = 0. 



k > n = 0, 1, 2 . . . >» = 1, 2, 3, . . . 



of which the characteristic equation is 



(r + x) m = 

 whence the complete solution is 



(11) j{—m,x,k,n) = (c + Cin + + c m - X n m ~ ) (— xf 



m = 1, 2, 3 . . . . ; n = 0, 1.2 fc— 1; not for n~>k; 



however, the equation (10) itself will give/( — m,x,k,n) for 



n = k, fc+1, &+m — 1. 



For ??i = 1, we have 



/(— l,x,k,n) = Co (— x) n n = 0, 1, 2, 3, A-. 



and setting » = 0, we determine 



Co =/(— l,.r,A-,0). 

 setting t = — 1 in '(8) 



K—l,x,k,Q) = \ [S{k,0) + kK—l,x+l,k— 1,0)] 



= — when £ = 

 :r 



= - /(— l,x+l,k— 1,0) fc = 1, 2, 3 . . . 



