Separation property for non-injective flows

Question

Let us consider a non-injective flow $X$ on $\mathbb{R}^d$, i.e. a continuous map $X:\mathbb{R}_+\times \mathbb{R}^d \to \mathbb{R}^d$ with $X(0,\cdot)=\mathrm{id}$ and satisfying the semigroup property $X(t,X(s,\cdot))=X(t+s,\cdot)$ for all $t,s\geq 0$.

For $A\subset \mathbb{R}^d$ open, do we have $\partial X_t(A) \subset X_t(\partial A)$?

Here $\partial A$ denotes the boundary of a set $A$ and $X_t(A)$ is the set of images of points in $A$ by $X(t,\cdot)$. The property is true if $X(t,\cdot)$ is injective$^*$. To give an intuition, in fluid mechanics, this property corresponds to the idea that streamlines do not leave streamline tubes. Also, this corresponds to the fact that the property: "the boundary $\partial A$ separates $A$ from its complement" is preserved by the flow.

I am not a specialist of this type of questions, so it might be a well known property; otherwise I'd already be happy with pointers to possibly useful theories.

$^*$in case $X_t$ is injective, by the domain invariance theorem $X_t$ is actually a homeomorphism and it can be shown $\partial X_t (A)= X_t(\partial A)$ without need of the flow property (see here).

Answer 1

I came up with a proof of my claim when $A$ is bounded, I'll reproduce it here although it is likely that I am the only one interested (since I asked the question). The idea is to use facts from mapping degree theory (this might be elementary for people familiar with this theory, but I discovered it just a few days ago while thinking about this problem).

Let $T>0$ and $A\subset \mathbb{R}^d$ open and bounded. Consider the function $\tilde X: (t,x) \mapsto (t,X_t(x))$, the set $$ S = \tilde X([0,T]\times \partial A) $$ which is compact, and its (relative) complement $S^c = ([0,T]\times \mathbb{R}^d)\setminus S$ which is open in the relative topology. Since connected components of $S^c$ are path-connected, it follows, using the invariance under homotopy of the degree (and the other properties of Theorem 1 here) that $$ (t,x)\mapsto \deg(X_t,A,x) $$ is constant in each connected component of $S^c$, and more precisely, equals $1$ if the connected component intersects $\{0\}\times A$ and $0$ if it intersects $\{0\} \times (\mathbb{R}^d\setminus A)$. In particular, no connected component intersects both. Here, $\deg(f,A,z)$ is the degree of a map $f$ at a point $z$ with respect to a set $A$ (in a nutshell, it gives the "algebraic" number of solutions to $f(u)=z$ for $u\in A$ and in particular, one such solution is guaranteed to exist if $\deg(f,A,z)\neq 0$). Apparently, the use of degree theory is only permitted in bounded sets, hence our assumption on $A$ (a more detailed proof would replace $\mathbb{R}^d$ by a bounded set as an intermediate step).

So far, we have not used the semi-group property: it will be used to show that if $B$ is a connected component of $S^c$ where the above degree vanishes, then $B\cap \tilde X([0,T]\times A)=\emptyset$. To see this, let $x\in A$: if $(X_t(x))_{t\in [0,T]}$ leaves at some point the connected component of $S^c$ containing $x$, then there exists ${t_0}\in [0,T]$ such that $\tilde X_{t_0}(x)\in S$. Then the semigroup property implies $\tilde X_t(x)\in S$ for all $t\geq t_0$ and thus, for all $t\in [0,T]$, $\tilde X_t(x)\notin B$.

The final result follows by taking a section of $S^c$ at a time $t_0\in {]0,T[}$, i.e. considering $\{x \in \mathbb{R}^d \;;\; (t_0,x) \in S^c\}$: it has the property that its boundaries are contained in $X_{t_0}(\partial A)$ and that its connected component are either included in $X_{t_0}(A)$ or in its complement. Moreover, it contains $X_{t_0}(A)$. Thus $\partial X_{t_0}(A) \subset X_{t_0}(\partial A)$.

In the end, our proof tells more about the situation than just this final property. Using the vocabulary from fluid dynamics a summary of the proof would be: mapping degree theory is used to prove that stream tube walls do not have holes, and the flow property is used to assert that all stream lines are confined by these walls.