# Separation property for non-injective flows

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 $X(0,\cdot)=id$ 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).

• It is not clear to me what you understand by $\partial A$. Is this the boundary in the topology of $\mathbb{R}^d$, or in the relative topology of $\Omega$? I would rather think the latter: $\Omega$ is our "universe". Take $A = \Omega = (-1, 1) \subset \mathbb{R}$ and $X_t(x) = e^{-t}x$. $\partial(X_1(A)) = \{ -e^{-1}, e^{-1} \}$, whereas $X_1(\partial A) = \emptyset$. And if you understand by $\partial A$ its boundary in the topology of the ambient space $\mathbb{R}^d$, the semiflow should be defined on the closure of $\Omega$. – user539887 Apr 10 '18 at 20:45
• Oh I was imprecise, I indeed mean the second case, let me update my question. Thank you. – L. Chizat Apr 10 '18 at 21:55

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 $\partial X_{t_0}(A) \subset X_{t_0}(\partial A)$.
• $S$ is a subset of $\mathbb{R}^d$, while $S^c$ is a subset of $[0,T] \times \mathbb{R}^d$. Is the definition of $S$ correct? – user539887 Apr 11 '18 at 21:11
• I corrected (in three places) $X$ to $\tilde X$. It appears to me that everything is O.K. But I'm no expert on degree theory, I have to admit. – user539887 Apr 15 '18 at 20:23