# Continuous variation from solution of one problem to solution of another problem

This is a differential equations question. The only thing I know about differential equations is that there can be many subtleties.

Suppose that I have a function $g:\mathbb{R}^n \times \mathbb{R} \to \mathbb{R}^n$. I have a solution $x_0$ to $g(x_0, t=0)=0$. I desire to show the existence of a solution to $g(x_1, t=1)=0$. My strategy is to take the known solution $x_0$ and vary it from $t=0$ to $t=1$.

Let $J$ be the Jacobian $J_{ij} = \partial g_i / \partial x_j$. Define the function $f(x,t) = - J^{-1} \partial g/\partial t$. Then, setting $dx/dt = f(x,t)$ gives $$\frac{dg(x,t)}{dt} = J \frac{dx}{dt} + \frac{\partial g}{\partial t} = 0.$$ So it would seem that integrating this differential equation from $t=0$ to $t=1$ will lead to the desired solution $g(x_1, t=1)=0$.

What I need then is a theorem of the following form:

Suppose $g(x_0, t=0)=0$, and suppose that $\exists f$ such that $$\frac{dx}{dt}=f(x,t) \implies \frac{dg(x,t)}{dt}=0.$$ If $f$ and $g$ are well behaved enough in the vicinity $g(x,t) \approx 0$, then $\exists x$ such that $g(x,1)=0$.

When $g(x,t)=0$, I have a lower bound on the smallest singular value of $J$, and $\partial g/\partial t$ is bounded as well. So, in this case $f$ is bounded. In fact, I think that I can pretty much prove any needed type of "well-behavedness" in the region $\left| g(x,t) \right| < \epsilon$ for some $\epsilon$. So, what properties of $f$ and $g$ are needed, and what theorem will help me here? It looks like Picard-Lindelof may help, but it seems to only give the existence of a unique solution to the differential equation, and I need to show that that solution satisfies $g(x_1, t=1)=0$. Furthermore, $f$ is not well behaved when $g$ is far from zero, and so it seems I cannot use Picard-Lindelof without prior assumption that $g(x,t)$ stays small (which is kind of assumes the fact that I am trying to prove).

NOTE: since I didn't get any answers here, I also asked on mathoverflow.

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