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I have a nonlinear, two point boundary value problem of the form $F(x, y(x), y'(x); \Omega ) = y''$ along with some boundary conditions of the form $y_\Omega(0) = a_\Omega, y_{\Omega}(1) = b_\Omega$. The set $\Omega$ is a set of parameters which both alters the dynamics and changes the locations of the boundary points.

Now, suppose to I have a solution to this BVP (call it $y_\Omega(x))$, and I change the parameters in $\Omega$ a small bit (resulting in new boundary points) - am I guaranteed to again have a solution $Y_{\Omega'}$ that is "close" to the original function in some sense?

Really, I am asking if there is an implicit function theorem for boundary value problems.

Alternatively, since my BVP is derived via the Euler-Lagrange equations, I would be fine if there was an implicit function theorem for the minimum of an integral, as well.

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There is an implicit function theorem for smooth nonlinear maps between Banach spaces. You can apply this to your problem by considering the mapping $$(y,\Omega)\mapsto (y''-F,y(0),y(1))$$ from $C^2[0,1]\times (your\ parameter\ set)$ to $C[0,1]\times R^2$.

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  • $\begingroup$ This answer certainly makes sense, but I think I am still a little stymied about how to show the mapping you suggest is smooth. Is this just a requirement on my operator to be nonlinear BVP operator to be non-singular? $\endgroup$ – Danny W. Feb 7 '14 at 21:54
  • $\begingroup$ In particular, unsurprisingly by BVP operator is the Euler-Lagrange operator. Are there conditions under which this operator is non-singular? $\endgroup$ – Danny W. Feb 7 '14 at 22:02
  • $\begingroup$ What you need is that F is smooth and that the linearization at your known solution is uniquely solvable. $\endgroup$ – Michael Renardy Feb 8 '14 at 12:02

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