NONLINEAR BOUNDAR Y VALUE PROBLEMS 1 1 Then, if M, N are linear bounded operators in C and*/€ Ll(I, R n ), c C , we can consider the linear boundary value problem (L6) *'(0=/(0 , teI (1.7) Mx 0 +Nxx =c. If we write Ir = [-r , 1 ] (so that I0 = /), X = C(/ r * n ), Z = L 1 ^ *") * C , and if R: C(Ir, Rn)— * C(J, /?*) is the restriction operator from Ir to / (i.e . Rx is the restriction of x to /), then we can define dom L = { * Jf: Rx is absolutely continuous on / }, Lx^{(Rx)\MxQ+Nxl) so that problem (1.6)—(1.7) is equivalent to the equation Lx = (f, c) with (f, c) G Z. For each rG/, let St:L1(I t Rn)-+C, T t :C-+C be defined respectively by (s,/X*) = J 0 '+ f /M*«. i f o r +1 i f = 0 i f -r t + 5 0, s e [-r , 0] . (7X«) « *(' + *) . - r * 0 , where *(") = POO , - r u 0 , = *(0) , 0M1 , so that (3X*) = ¥' + *) i f - r r + s 0 , = tf0) i f 0 f + i lf * [-r , 0] . For each t GI, S t and r, ar e continuous linear operators and the mapping from / int o Rn defined by K/.*):fM(VX0) + (r^XO) is the unique solution of the initial value problem *'(0 = /('), x 0 = p. Consequently the boundary value problem (I.6)-(I.7) will have a solution if and only if we can find a p C such that Mp + MSJ+Trf^c. i.e. such that (1.8) (M + NTx)ip = c- NS t f. On the other hand,
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