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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.

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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 $\text{div}(f) > 0$.

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    $\begingroup$ 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. $\endgroup$ – user539887 Mar 12 at 9:44

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