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Can someone sketch some ideas of how to use the Poincare-Bendixson Theorem to prove that there must be a fixed point contained inside a periodic orbit?

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I have a feeling that that statement is false. I do not have a counterexample though. –  picakhu Apr 6 '11 at 2:43
I cant see the question, the image link isn't showing either. –  Kate Apr 6 '11 at 3:22
This certainly doesn't hold for discrete or higher dimensional dynamical systems (ex: billiards in polygons), so I take it we can assume the system is continuous and 2 dimensional? –  Alex Becker Apr 6 '11 at 5:13
@Alex: is there even a Poincare-Bendixson theorem in more than 2 dimensions? In anycase, certain topological assumptions must be made about the domain, else just consider the rotations on $\mathbb{S}^1\times \mathbb{R}$. –  Willie Wong Apr 6 '11 at 13:21
@Brad: please flesh out your question. Your statement is presumably true under certain assumptions, none of which you stated in the question. Please put more effort into describing what exactly it is that you want to know. –  Willie Wong Apr 6 '11 at 13:22

2 Answers 2

I interpreted the question as follow:

Let $f : \mathbb{R}^2 \to \mathbb{R}^2$ be a $\mathcal{C}^1$ vector field. Suppose there exists a periodic orbit $\gamma$. According to Jordan-curve-theorem, $\mathbb{R}^2 \backslash \gamma$ admits only one bounded connected component $D_{\gamma}$. Show that $D_{\gamma}$ contains a fixed point.

In fact, you only need a weak version of Poincaré-Bendixson theorem:

Theorem: Let $f : \mathbb{R}^2 \to \mathbb{R}^2$ be a $\mathcal{C}^1$ vector field and $x \in \mathbb{R}^2$. If the $\omega$-limit $\omega(x)$ is nonempty, compact and does not contain any fixed point, then $\omega(x)$ is a periodic orbit.

Notice that it is also true for $\alpha$-limits, since by reversing time an $\alpha$-limit becomes an $\omega$-limit while the phase portrait is unchanged.

By contradiction, suppose $D_{\gamma}$ does not contain any fixed point. Then, by Poincaré-Bendixson theorem, for all $x \in D_{\gamma}$, $\omega(x)$ and $\alpha(x)$ are periodic orbits. In particular, there are infinitely many periodic orbits in $D_{\gamma}$.

For any periodic orbit $\tau$ in $D_{\gamma}$, let $K_{\tau}=\tau \cup D_{\tau}$ (where $D_{\tau}$ is defined like $D_{\gamma}$ but for $\tau$). We get a family of compacts $\{K_i : i \in I \}$ linearly ordered by inclusion, indexed by some unbounded set $I \subset \mathbb{R}_+$. Without loss of generality, we can suppose the family $\{K_i : i \in I\}$ nonincreasing (otherwise, take $\tilde{K}_i= \bigcap\limits_{j \leq i} K_j$).

Because any $x \in \mathbb{R}^2$ is between its $\omega$-limit and its $\alpha$-limit, $\bigcap\limits_{i \in I} K_i=\emptyset$.

Now, take $i_n \in I$ such that $i_n \underset{n\to + \infty}{\longrightarrow} + \infty$; then $\bigcap\limits_{n \geq 0} K_{i_n}=\emptyset$. So we find a nonincreasing sequence of compacts converging to the emptyset, a contradiction (since $\mathbb{R}^2$ is complete).

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What is meant here with inside is inside as in the Jordan-curve-theorem. The proof can be done using it as well as the Schauder-fixed-point theorem - see for example the proof in the book by G. Teschl (to be found on his homepage).

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