# If the frontier is a union of the orbits of a autonomous ODE then the set is invariant

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.

• 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. – user539887 May 15 at 7:28
• @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 – Javier May 26 at 19:01
• @user539887 on the other hand is $B$ closed the most general condition? – Javier May 26 at 19:02
• No, this is only a sufficient condition. – user539887 May 27 at 7:24