Let $T \in \R$ with $T > 0$ and $Q_1, Q_2 \in N$ be given. Assume that $\gamma: [0,T] \to N$ is a solution to the boundary value problem

(BVP) $\D$ and $\gamma(0) = Q_1$ and $\gamma(T) = Q_2$.

and moreover assume that $\gamma$ is an isolated point within the set of all solutions to (BVP).

Then I would like there to be an open neighborhood $U$ of $(Q_1,Q_2) \in N$ such that for each point $(q_1,q_2) \in U$, there is a solution $\gamma_{q_1,q_2}$ to the BVP

$\D$ and $\gamma_{q_1,q_2}(0) = q_1$ and $\gamma_{q_1,q_2}(T) = q_2$

such that $\gamma_{Q_1,Q_2} = \gamma$ and the function $U \times [0,T] \to N$ given by $(q_1,q_2,t) \mapsto \gamma_{q_1,q_2}(t)$ is smooth.

Is this true? Put succinctly, if I have an isolated solution to a boundary-value problem, do nearby boundary values have yield solutions that depend smoothly on the boundary values?

If it matters, as I said above, my differential equation is the Euler-Lagrange equations for some Lagrangian $L$. Recall that the

*Action*of a path $\gamma$ is defined as the integral of $L$ over the path $\gamma$. Then the Euler-Lagrange equations are precisely the statement that the first derivative of the Action vanishes along any perturbation of the path with Derichlet boundary conditions.

For my particular problem, I know a bit more about the solution $\gamma$ than just that it is an isolated point among the set of solutions to the Euler-Lagrange equations. Just as setting the first derivative of the action to zero gives the Euler-Lagrange equations, the second derivative of the action is interesting as well. Like any second derivative, this one depends on the coordinates (it does not transform as a tensor), and for given coordinates is a symmetric bilinear form on the space of perturbations of $\gamma$. What I know is that this form has no kernel among the paths with Derichlet boundary conditions, in the sense that if $\zeta$ is a perturbation of $\gamma$ that does not perturb the endpoints, then there is some $\xi$ so that the second derivative of the action in the $\zeta\otimes \xi$ direction is non-zero. Put another way, our extra bit of knowledge about $\gamma$ is that a certain linear homogeneous second-order differential equation (with coefficients that depend on $\gamma$) along with the boundary requirements that the solution vanish at $0$ and at $T$ --- that this boundary value problem has only the trivial solution.

So, anyhoo, the condition in the previous paragraph guarantees that the original path $\gamma$ is an isolated point. I think that being an isolated point alone is enough to have the "depends smoothly on the parameters" thing that I want. But maybe this extra condition does.

Sorry if the post doesn't make any sense. It's late at night. If you might know this stuff, but want me to send you TeX with all the definitions, just drop me a line.