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Consider the nonlinear, singularly perturbed, vector boundary relation problem x$\sp\prime$ = f(t,x,y,$\epsilon$), $\epsilon$y$\sp\prime$ = g(t,x,y,$\epsilon$), L(x(0),y(0),$\epsilon$) = $\alpha\sb0$, R(x(1),y(1),$\epsilon$) = $\alpha\sb1$. Suppose that there exist smooth maps $\phi$,P,A$\sb1$,A$\sb2$ such that for all appropriate (t,x): (1) g(t,x,$\phi$(t,x),0) = 0, (2) P(t,x) $\cdot$ D$\sb3$g(t,x,$\phi$(t,x),0)P(t,x)$\sp{-1}$ = Diag(A$\sb1$(t,x),A$\sb2$(t,x)), and (3) the spectrum of A$\sb1$(t,x) is bounded away and to the left of the imaginary axis and the spectrum of A$\sb2$(t,x) is bounded away and to the right of the imaginary axis. Suppose also that p$\sb0$(t) is a solution of the reduced differential equation p$\sb0\sp\prime$(t) = f(t,p$\sb0$(t),$\phi$(t,p$\sb0$(t)),0) and that L(p$\sb0$(0),$\phi$(0,p$\sb0$(0),0) = 0 and R(p$\sb0$(1),$\phi$(1,p$\sb0$(1)),0) = 0. If L and R are of a special type (projections onto complementary sets of variables) Hadlock has shown (J. Diff. Eq. 14, 498-517) that the full problem has a bounded family of solutions (x(t,$\epsilon$),y(t,$\epsilon$)) defined for all $\epsilon$ sufficiently small and $\alpha$ in some neighborhood of 0. It is also clear in this special case what reduced set of boundary conditions (cancellation law) determine p$\sb\alpha$(t) = x(t,0+). A corollary of the main result of this paper extends Hadlock's result to allow for an arbitrary set of nonlinear boundary relations L and R subject to the invertibility of a certain linear operator. The proof makes use of the local invariant manifolds of the boundary layer equation along solutions of the reduced differential equation. The vectors x and y may belong to arbitrary Banach spaces.
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