Question on diffeomorphisms

Question

Suppose that we are given an autonomous ode $\dot{x} = f(x)$ where $f : \mathbb{R}^{n} \to \mathbb{R}^{n}$. My (elementary question) is that is the time one map for the ode above a local diffeomorphism around a neighborhood of a fixed point? $f \in C^1(\mathbb{R}^n)$. Thanks

Answer 1

Yes, this is true.

Let's look at a special case first. If $f$ has linear growth at infinity, then the solutions exist globally, and so does the time-one map $\Phi_1$. Since $f$ is $C^1$ smooth, so is $\Phi_1$. And the inverse of $\Phi_1$ is $\Phi_{-1}$, which is smooth for the same reason: it's just $\Phi_1$ for the system $\dot x = -f(x)$. Therefore, $\Phi_1$ is a global diffeomorphism of $\mathbb R^n$ onto itself in this case.

In general, solutions may blow up (e.g., $f(x)=x^2$), so $\Phi_1$ may not be defined everywhere. But since $f(x_0)=0$ and $f$ is locally Lipschitz, it follows that $|f(x)|\le L|x-x_0|$ in a neighborhood $U=\{x: |x-x_0|<r\}$. A solution starting near $x_0$ will move slowly and therefore will stay around $x_0$ for a long time. Specifically, we have $|x(t)-x_0|\le e^{Lt}|x(0)-x_0|$ as long as $x(t)\in U$. It follows that solutions beginning in the smaller neighborhood $V=\{x: |x-x_0|<e^{-L}r\}$ will stay in $U$ up to time $1$. Therefore, $\Phi$ and $\Phi_{-1}$ are both defined in $V$; since $\Phi_{-1}\circ \Phi_1$ is the identity, the conclusion follows.