# Periodic orbit attracts an open set of initial conditions

We have and ode $x'=f(x)$, where $x \in \mathbb{R}^2$ and $f: \mathbb{R}^2\to \mathbb{R}^2$ a $\mathcal{C}^1$ function.

Suppose that $\gamma$ is a periodic orbit for this ode.

Suppose also that there exists $y \in \mathbb{R}^2$, $y \notin \gamma$, such that $\omega(y) = \gamma$ (it means that $\gamma$ is a limit cycle).

Prove that there exist an open set U in $\mathbb{R}^2$ such that $\forall z \in U$, $\omega(z) = \gamma$.

I tried to use Poincaré Bendixson theorem and continuity of solutions with respect to initial conditions in order to prove that such $U$ exists. I couldn't see a clear way to do this.

Any ideas?

• Yes, it is a very good approach. The only problem is that the fact that for all the points on (one side of) a transversal their $\omega$-limit set is just the limit cycle is (usually) hidden in a proof. – user539887 Jun 22 '18 at 7:30