# Trajectory leaving a set

Consider the differential equation $$\dot{x}=f(x)$$, where $$f: \mathbb{R}^2 \to \mathbb{R}^2$$ is smooth. Given a set $$A \subset \mathbb{R}^2$$, are there some results saying that whenever $$x(0) \in A$$, there exists $$T>0$$, such that $$x(T) \notin A$$ (so every trajectory starting from $$A$$ will always leave $$A$$ at some time instant)?

Note: I do not specify what the set $$A$$ looks like (e.g., compact, convex, etc) because I want to know all the possibilities. This means, your answer can, of course, require that $$A$$ is convex or whatever you like, as long as the answer is correct. And I also do not specify the behavior of the trajectory after $$T$$ because it is not important in this question. The question itself is clear enough.

By the Poincaré–Bendixson theorem, if $$\overline{A}$$ is (EDIT: compact and) contained in an open set with no fixed points or periodic orbits, then every trajectory starting in $$A$$ must eventually exit $$A$$.
The utility of this is somewhat limited: it's generally easy to check for fixed points, but periodic orbits are harder to rule out, unless you have a Liapounov function or $$\text{div}(f) > 0$$.
• You need to assume that $\overline{A}$ is compact (the OP did not assume that). Otherwise, $f\equiv\mathrm{col}(1,0)$ and $A$ the open (or, for that matter, closed) upper half-plane serve as a counterexample. – user539887 Mar 12 '19 at 9:44