0
$\begingroup$

I am trying to figure out, whether the IFT can be generalized to curves. Let's say I have a function $G(x,u)$ mapping $\mathbb{R}^{n+m}\rightarrow \mathbb{R}^n$ with invertible jacobian $\frac{\partial G}{\partial x}$. However, not only do we assume $G(x_0,u_0) = 0$, but $G(x(t),u(t)) = 0$ for a compact curve $t\mapsto (x(t),u(t))$ in $\mathbb{R}^{n+m}$.

Needless to say, for each point on the curve there is an associated neighborhood $V(t)$ where the IFT applies. I guess the main question is: Can we make it so that the neighborhoods with corresponding graphs glue? Or does this need additional assumptions?

Any help is appreciated!

Improve this question
$\endgroup$
3
  • $\begingroup$ I don't see a question here. What does your curve have to do with the function $G$? It looks like you are assuming the curve belongs to a level-set. But what do you mean about $V$ and "the IFT"? $\endgroup$ Commented Dec 6, 2022 at 16:26
  • $\begingroup$ With "the implicit function theorem" I mean the standard theorem for $\mathcal{C}^1$ functions $\mathbb{R}^{n+m}\rightarrow \mathbb{R}^n$. The curve a priori is any smooth curve satisfying $G(x(t), u(t)) = 0$ such that the corresponding Jacobian is invertible along the curve. $V(t)$ are the open neighborhoods the classic implicit function theorem provides at every point of the curve. $\endgroup$ Commented Dec 7, 2022 at 9:18
  • 2
    $\begingroup$ What do you mean by gluing neighbourhoods? Could you turn this into a precise question, i.e. is there an actual theorem you are looking to prove? $\endgroup$ Commented Dec 7, 2022 at 16:52

0

Reset to default

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.