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2 added suggested tag; made clear that x is a function of t

I asked this question a week ago over on math.stackexchange and got no reply, so I am asking here with slightly different wording. I am trying to prove that there exists a solution to a problem. I can't solve this problem in general, but I can always continuously vary the parameters of the general problem to turn it into a specific case which is easy to solve. I believe that the solution of the easy problem can then be varied continuously to be a solution to the hard problem.

It seems to me that this would be a widely used technique, but I haven't seen it before, and I don't know how to prove that it works. To be specific, suppose that $g:\mathbb{R}^n \times \mathbb{R} \to \mathbb{R}^n$. 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 $t=1$. What I need is a theorem of the following form:

Suppose $g(x_0, t=0)=0$, and suppose that $\exists f$ such that \begin{equation} \frac{dx}{dt}=f(x,t) \implies \frac{dg(x,t)}{dt}=0. frac{dg(x,t)}{dt}=0, \end{equation} with $x$ a function of $t$. If $f$ and $g$ are "well behaved enough" in the vicinity $|g(x,t)| < \epsilon$ for some $\epsilon$, then $\exists x$ such that $g(x,1)=0$.

$f$ and $g$ in my case are "well behaved enough" that I can probably prove just about any sort of continuity condition that is needed. 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).

1

# Continuous variation from solution of easy problem to solution of hard problem

I asked this question a week ago over on math.stackexchange and got no reply, so I am asking here with slightly different wording. I am trying to prove that there exists a solution to a problem. I can't solve this problem in general, but I can always continuously vary the parameters of the general problem to turn it into a specific case which is easy to solve. I believe that the solution of the easy problem can then be varied continuously to be a solution to the hard problem.

It seems to me that this would be a widely used technique, but I haven't seen it before, and I don't know how to prove that it works. To be specific, suppose that $g:\mathbb{R}^n \times \mathbb{R} \to \mathbb{R}^n$. 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 $t=1$. What I need is a theorem of the following form:

Suppose $g(x_0, t=0)=0$, and suppose that $\exists f$ such that \begin{equation} \frac{dx}{dt}=f(x,t) \implies \frac{dg(x,t)}{dt}=0. \end{equation} If $f$ and $g$ are "well behaved enough" in the vicinity $|g(x,t)| < \epsilon$ for some $\epsilon$, then $\exists x$ such that $g(x,1)=0$.

$f$ and $g$ in my case are "well behaved enough" that I can probably prove just about any sort of continuity condition that is needed. 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).
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