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.


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 $C[0,1]\times R^2$.

  • $\begingroup$ 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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