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I'm looking for a proof or reference proving this theorem which was written in my ODE class:

Let $D \subseteq \mathbb{R}^d$ be open and $f:D \to \mathbb{R}^d$ continuous such that each IVP has a unique solution.

Under these conditions one can define the orbit from an initial point $x_0$ of an autonomous equation $x' = f(x)$:

The orbit passing through $x_0$ is the set $\Gamma(x_0) = \{ G(t;0,x_0). t \in ]\alpha(x_0),\omega(x_0)[\}$

One can also define the notion of invariant set for $x' = f(x)$:

$B$ is an invariant set if whenever $x_0 \in B$ then $x(t;x_0) \in O$ for all $t \ge 0$ where $x$ is the unique solution for this $x_0$.

This is the theorem I need to prove:

Let $B \subseteq D$ such that $Fr(B)$ is the union of the orbits of $x' = f(x)$.

Then $B$ is an invariant set for $x' = f(x)$

My approach

Given $x_0 \in int(B)$, if $\Gamma(x_0) \subseteq B$ we finish. Otherwise, by the continuity of $\varphi$, $\exists \tau \in ]\alpha,\omega[. \varphi(\tau) \in Fr(B)$. Then the IVP $\begin{cases}x' = f(x) \\ x(\tau) = \varphi(\tau)\end{cases}$ has two solutions: $\varphi$ and $G(\cdot;\tau,\varphi(\tau))$ which lives in the frontier. By the uniqueness of solution, $\varphi = G(\cdot;\tau,\varphi(\tau)) = \Gamma(x_1)$ for some $x_1$. Therefore, $\Gamma(x_0) = \Gamma(x_1)$ and we have that $x_0 \in Fr(B)$. This contradicts $x_0 \in int(B)$.

However if $x_0 \in Fr(B) \cap B$ is not so clear.

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  • $\begingroup$ The theorem as you stated it seems to be false. Take $D=\mathbb{R}^2$, $f(x,y)=(y,-x)$ (besides $\{(0,0)\}$, orbits are circles centered at $(0,0)$), and $B=\{\,(x,y):x^2+y^2=1\,\}\setminus\{(1,0)\}$. The boundary (not frontier) of $B$, that is, $\{\,(x,y):x^2+y^2=1\,\}$, is the union of orbits, but $B$ is not invariant. A hint: the theorem holds if one assumes $B$ to be closed. $\endgroup$ – user539887 May 15 at 7:28
  • $\begingroup$ @user539887 it appears that frontier is also accepted (en.wikipedia.org/wiki/Boundary_(topology)) Wikipedia cites Willard, which makes sense since the professor who taught me general topology learnt with that book $\endgroup$ – Javier May 26 at 19:01
  • $\begingroup$ @user539887 on the other hand is $B$ closed the most general condition? $\endgroup$ – Javier May 26 at 19:02
  • $\begingroup$ No, this is only a sufficient condition. $\endgroup$ – user539887 May 27 at 7:24

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