# Sufficient condition(s) for a non-local diffeomorphism?

1. Mar 4, 2012

### dash-dot

I hope I'm posting this question in the right forum; anyway, here goes:

What would be a sufficient condition for a diffeomorphism to be non-local--specifically, for it to be valid over a given domain?

In the particular case I'm examining, the mapping I'd like to be a non-local diffeo is given by a solution to a PDE with a perturbation term. In the absence of the perturbation, I can simply take the mapping to be the identity map (which can always be shown to be a global diffeo, correct?).

Hence, my objective is to establish that for a "small enough" perturbation term, the solution to the PDE is itself a perturbation of the identity mapping, and hence valid over the domain of interest.

PS: For those who wish to see the actual PDE and obtain additional information, here it is:

Given a compact set $D\in\textbf{R}^{m+n}$ containing the origin, I would like to find a mapping
$T:D\to \textbf{R}^m; \quad T:(x,s,\phi)\mapsto q$
that is given by the solution to the PDE

$\frac{\partial T}{\partial s}+\left( \frac{\partial T}{\partial \phi}-\frac{\partial T}{\partial s}\,\frac{\partial N}{\partial \phi}\right)M\delta_a=0,$

where $x\in\textbf{R}^{n-1},\, s\in\textbf{R},\, \phi\in\textbf{R}^m,\, M=M(x,\phi)\in\textbf{R}^m,\, N=N(x,\phi)\in\textbf{R}$ and $\delta_a=\delta_a(x,s)\in\textbf{R}$.

$\delta_a(x,s)$ is the perturbation I'm talking about, and we see that when $\delta_a=0$, we can simply take $q=T(\phi)=\phi$. Hence, for small enough $\delta_a$, we expect $T$ to just be a perturbation of the identity map, and I need for it to be valid for all $(x,s,\phi)\in D$.

If anyone could recommend some books or other references that talk about when a diffeomorphism is non-local (or global, although I don't need mine to be valid globally, but if it turns out to be the case for any $T(x,s,\phi)$ satisfying the PDE above, then great), that would be awesome.

Thanks for taking the time to read this. :)

Last edited: Mar 4, 2012