BOUNDARY PROBLEMS AND DEVELOPMENTS. 61 



represents is necessarily one with elements continuous for a ^ x ^ b. 

 Now 



X 



Y m (x) = Yo+ f {A(t)Y„Ut) + B(t)} dt 



Xo 



by definition, and since the convergence of Y m (x) to Y(x) is uniform 



X 



lim Y m (x) = l'o+ f lim { .4(0 Y m ^(t) + B(t)} dt, 



m=oo *J m= oo 



xo 



X 



namely, Y(x) = Y + J { A(t) Y(t) + B(t) } dt. 



xo 



Differentiating we see that the elements of Y'(x) are continuous, 

 a ^ x ^ b, while Y'(x) = A(x) Y(x) + -B(.r). Hence the solution 

 exists as stated. 



To prove the solution unique suppose that both Y(x) and Y(x) are 

 solutions of equation (3) each reducing to the matrix F for x = Xo, i.e. 



Y\x) = A{x) Y(x) + B(x), Y(x ) = Y , 

 T(x) = A{x) Y(x) + B(x), Y(x ) = F . 



Now Y(x) — Y(x) = D(x) is a matrix whose elements are continuous, 

 a ^ x ^ b, and which satisfies the relations 



D'(x) = A(x) D(x), D(x ) = 0. 



Let any neighborhood of the point .To be chosen, say | x — Xq | ^ 5. 

 Then, if d denotes the largest numerical maximum of any element of 

 D(x) for any x of this neighborhood so that D(x) « (d), 



A(x) D(x) « (nad), 

 and D'(x) « (nad). 



Hence D(x) « (nad | x — x \ ) « (?iadS). 



But for some x of the interval at least one element of D(x) is numeri- 

 cally equal to d (by the definition of d), and for this element it follows 

 that d ^ nadd, i.e. d { 1 - nad} ^ 0. 



