Consider a non-vanishing $C^1$ vector field $f$ on a neighbourhood of the annulus with radii $1$ and $2$ in $\mathbb{R}^2$. The vector field is transversal to the boundary of the annulus and points inward. Prove that there is a closed orbit solving $\dot x = f(x)$. Prove also that if exactly $7$ closed orbits exist, then there exist associated orbits that spiral towards one of the closed orbits from both sides.

Here is my work so far:

Because $f$ is transversal and points inwards, the orbits inside the annulus do not leave the annulus (since if they did, the vector field would have to point outward from the annulus). The Poincaré-Bendixson theorem states that any compact $\omega$-limit set of an orbit must be a periodic orbit (since no fixed points exist), so for the orbits inside the annulus there should be two possibilities: either they are periodic orbits, or their limit sets are periodic orbits. So clearly at least one periodic orbit has to exist, and if any non-periodic orbits exist, they must spiral towards a periodic orbit.

To prove that non-periodic orbits exist, use the fact that exactly $7$ periodic orbits exist. Each of these periodic orbits is a loop inside the annulus, but finitely many loops cannot cover the entire annulus. So there exist points that do not belong to any periodic orbit, and therefore have a non-periodic orbit.

These arguments feel a bit shaky. Can anyone tell me what errors I've made?


1 Answer 1


Seems fine to me. The number $7$ is a red herring. We know there must be at least $1$ (because of the transversality condition on the boundary).


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