If $T_t$ is the flow generated by the autonomous velocity $v$ and $\left.v\right|_{\partial\Omega}=0$, then $T_t(\partial\Omega)=\partial\Omega$ Let $d\in\mathbb N$ and $v\in C_c^\infty(\mathbb R^d,\mathbb R^d)$. We know that, for any $\tau>0$, there is a unique solution $X^x\in C^0([0,\tau],\mathbb R^d)$ of \begin{align}X'(t)&=v(X(t))\tag1&\text{for all }t\in[0,\tau],\\X(0)&=x\end{align} for all $x\in\mathbb R^d$. It is easy to show that $$T_t(x):=X^x(t)\;\;\;\text{for }t\in[0,\tau]$$ is a $C^1$-diffeomorphism from $\mathbb R^d$ onto $\mathbb R^d$.

Now let $\Omega\subseteq\mathbb R^d$. How can we show that,

*

*if $\left.v\right|_{\partial\Omega}=0$, then $$T_t(\partial\Omega)=\partial\Omega\tag2$$ for all $t\in[0,\tau]$?

*if $\Omega$ is closed or open, then $$T_t(\Omega)=\Omega\tag3$$ for all $t\in[0,\tau]$?


It's clear to me that any homeomorphism maps boundary (interior) points to boundary (interior) points. I guess we need to use this somehow.
EDIT: From the comments it is clear that $(2)$ holds, since it should generally hold that if $B$ is any subset of $\mathbb R^d$ with $\left.v\right|_B=0$, then $T_t(x)=x$ for all $x\in B$. But how can we prove $(3)$?
EDIT 2: If $f$ is any homeomorphism between topological spaces $E_1$ and $E_2$ and $B_1\subseteq E_1$, then we know that $f(B_1^\circ)=f(B_1)^\circ$, $f(\partial B_1)=\partial f(B_1)$ and $f(\overline{B_1})=\overline{f(B_1)}$. If $B_1$ os open, then $B_1=B_1^\circ$ and if $B_1$ is closed, then $B_1=\overline{B_1}$. I think we need to use this for $(3)$.
EDIT 3: Let $x\in\Omega^\circ$. Then there is a $\varepsilon>0$ with $B_\varepsilon(x)\subseteq\Omega$. Maybe we can at least show that there is a $t\in[0,\tau]$ (sufficiently small) such that $\left\|X^x(s)-x\right\|<\varepsilon$ for all $s\in[0,t]$. Then it would follow that $$T_s(\Omega^\circ)\subseteq\Omega^\circ\;\;\;\text{for all }s\in[0,t].\tag4$$ From pure intuition, for a sufficiently small $t$, the velocity should not be able to move the point $x$ outside of the ball $B_\varepsilon(x)$. So, $(4)$ should hold. (How would we need to argue that it must even be an equality? This seems trivial, by bijectivity though.)
 A: About the second question, you can go like this.
This is a formal argument for the more intuitive "you can't cross the boundary if the boundary is fixed, so you gotta stay inside".
Firstly, suppose that $\Omega$ is open. Take $x \in \Omega$. The map
$$ T_{()}(x): [0, \tau] \to \mathbb{R}^d $$
that sends $t$ to $T_t(x) $  is continous, thus the preimage of $\Omega$ is open. We then get that
$$A(x) =  \{t \in [0, \tau] : T_t(x) \in \Omega\} $$
is open. Suppose by contradiction that there exist $x$ such that $A(x)$ is not $[0,\tau]$. Take $t(x) = \sup \{ t: \forall s \in [0, t] , T_s(x) \in \Omega\}$.Set $y=T_{t_*(x) }(x) $.
Notice that:

*

*$y \not \in \Omega$. Indeed, $t_*(x) < \tau$ because otherwise we would have $A(x) = [0,\tau]$. If $T_{t_*(x) }(x) $ was in $\Omega$, then by openness of $A(x) $ we'd have that $T_{t_*(x) +\epsilon}(x) $ would be in $\Omega$ for all sufficiently small $\epsilon$, contradicting the sup hypothesis.


*$y \in \partial \Omega$. Indeed, we have that
$$ T_{t_*(x) }(x) = \lim_{t\to t_*(x)-} T_t(x) $$
And all the points in the limit belong to $\Omega$. Using also point 1 we get that $T_{t_*(x) }(x) \in \bar{\Omega} \setminus \Omega = \partial \Omega$.
This concludes, because $T_{t_*(x) }$ would not be injective: both $x, y$ are mapped to $y$.
The same argument applies also to negative times, yielding the equality $T_t(\Omega) = \Omega$. Indeed, take $z \in \Omega$: then $T_t (T_{-t}(z)) = z$, and $T_{-t}(z) \in \Omega$.
Finally, if we take $\Omega$ to be closed, by the previous points we get $T_t(\Omega^c) = \Omega^c$; being bijective, this yield $T_t(\Omega) = \Omega$.
A: Andrea Marino's answer is perfectly fine, I'm mainly writing down a similar attempt for my own reference.
First of all, we can show the following result:

Let $\tau>0$, $s\in[0,\tau]$, $E$ be a $\mathbb R$-Banach space and $f\in C^0([s,\tau],E)$.
Proposition 1: Let $B\subseteq E$ be closed and $$I:=f^{-1}(B)=\{t\in[s,\tau]:f(t)\in B\}.$$ If $I\ne\emptyset$, then

*

*$\sigma:=\inf I\in I$ and hence $f(\sigma)\in B$;

*if $f(0)\not\in B$, then $\sigma>s$ and $f(sigma)\in\partial B$.

Corollary 2: Let $\Omega\subseteq E$ be open and $$I:=\{t\in[s,\tau]:f(t)\not\in\Omega\}.$$ If $I\ne\emptyset$, then

*

*$\sigma:=\in I\in I$ and hence $f(\sigma)\not\in\Omega$;

*if $f(0)\in\Omega$, then $\sigma>s$ and $f(\sigma)\in\partial\Omega$.


Now, turning to the question, assume $v:[0,\tau]\times E\to E$ is uniformly Lipschitz continuous in the second argument uniformly with respect to the second and $v(\;\cdot\;,x)\in C^0([0,\tau],E)$ for all $x\in E$. Then there is a unique $X^{s,\:x}\in C^0([s,\tau],E)$ with $$X^{s,\:x}(t)=x+\int_s^tv(r,X^{s,\:x}(r))\:{\rm d}r\;\;\;\text{for all }t\in[s,\tau]\tag1$$ for all $(s,x)\in[0,\tau]\times E$. We can show that $$T_{s,\:t}(x)=X^{s,\:x}(t)\;\;\;\text{for }x\in E$$ is bijective for all $0\le s\le t\le\tau$.

Proposition 3: Let $(s,x)\in[0,\tau]\times E$. If $$v(t,x)=0\;\;\;\text{for all }t\in[s,\tau],\tag2$$ then $$X^{s,\:x}=x\tag3.$$

(This can be proven using the Lipschitz assumption and Gronwall's inequality.)

Corollary 4: Let $(s,x)\in[0,\tau]\times E$ and $\Omega\subseteq E$ be open or closed. If $$v(t,x)=0\;\;\;\text{for all }(t,x)\in[s,\tau]\times\partial\Omega\tag4,$$ then $$T_{s,\:t}(\Omega)=\Omega\;\;\;\text{for all }t\in[s,\tau].\tag5$$

Proof: We first show the following: Let $x\in\Omega$. If $\Omega$ is open, then $$T_{s,\:t}(x)\in\Omega\;\;\;\text{for all }t\in[s,\tau]\tag6.$$ In order to prove that, let $$I:=\{t\in[s,\tau]:X^{s,\:x}(t)\not\in\Omega\}.$$ Assume the claim is not true, i.e. $I\ne\emptyset$. Then, by Corollary 2, $$\sigma:=\inf I\in I\tag7$$ and $$y:=X^{s,\:x}(\sigma)\in\partial\Omega.$$ Thus, by $(2)$, $$v(t,y)=0\;\;\;\text{for all }t\in[s,\tau]\tag8$$ and hence $$T_{s,\:t}(y)=y\;\;\;\text{for all }t\in[s,\tau]\tag9$$ by Proposition 3. On the other hand, by definition, $$T_{s,\:\sigma}(x)=y\tag{10}.$$ Since $\Omega$ is open, $\Omega\cap\partial\Omega=\emptyset$ and hence $x\ne y$. But by $(9)$ and $(10)$ this implies that $T_{s,\:\sigma}$ is not injective; which is not true. So, $I=\emptyset$.

However, what we can infer from this claim is $$T_{s,\:t}(\Omega)\subseteq\Omega\;\;\;\text{for all }t\in[s,\tau],\tag{11}$$ but why is $(11)$ actually an equality?

EDIT: Couldn't we simply apply the same prove to $[s,\tau]\ni t\mapsto T_{s,\:t}^{-1}(x)$, where $x\in\Omega$ is fixed as in the first claim of my proof above? If I'm not missing something, the only relevant part was the continuity of $[s,\tau]\ni t\mapsto T_{s,\:t}(x)$ and we can show that $[s,\tau]\ni t\mapsto T_{s,\:t}^{-1}(x)$ is continuous as well. So, the proof of that claim should follow line-by-line yielding $T_{s,\:t}^{-1}(x)\in\Omega$ for all $t\in[s,\tau]$. What do you think?
