# AODE: Compact orbit iff periodic

I've been trying to do the following question for a while and have tried following the hint. I have done everything apart from showing that the forward orbit of $$z$$ is equal to the orbit of $$z$$. Would it be possible to get another hint for this last step?

In the question $$O$$ stands for the orbit of a element and $$Ω$$ stands for the $$\omega$$-limit set of a point.

Let $$f : E → R^N$$ be a locally Lipschitz function on an open set $$E ⊂ R^N$$ . We consider the autonomous ODE $$x' = f(x)$$. Show that if $$a ∈ E$$ has a compact orbit, then it is periodic.

Hint: The standard properties of limit sets implies $$O(a) = Ω(a) = \overline{O^+(a)} = O(z)$$ for any $$z ∈ Ω(a)$$, which yields $$O^+(z) = O(z)$$. Conclude.

• Take an element $x(-t)$ ($t>0$) in the backward orbit. As the orbit is equal to the forward orbit, this implies that there exists $T>0$ such that $x(T)=x(-t)$. Therefore, the orbit is periodic. – Severin Schraven Mar 31 at 16:44
• Thank you so much. But I think my question was a bit misleading I was able to conclude that but it was the step before i was unable to conclude. – Milos Tasic Mar 31 at 17:13

From your last comment I guess that what you need is to show that $$\begin{equation*} O(a) = \Omega(a) = \overline{O^{+}(a)} = O(z) \quad \text{for any } z \in \Omega(a). \end{equation*}$$ Denote by $$a \cdot t$$ the action of the flow of the ODE.

Since $$\Omega(a) = \bigcap_{t \ge 0} \overline{O^{+}(a \cdot t)}$$ and $$O(a)$$ is compact, hence a closed set, we have $$\Omega(a) \subset O(a).$$ But $$O(a)$$ is a minimal set (that is, a compact invariant set having no compact invariant proper subsets), so, by the invariance of $$\Omega(a)$$, $$\Omega(a) = O(a).$$ Further, $$\overline{O^{+}(a)} = O^{+}(a) \cup \Omega(a) = O(a).$$ Finally, take a $$z \in \Omega(a)$$. As $$\Omega(a) = O(a)$$, there is $$\tau \in \mathbb{R}$$ such that $$z = a \cdot \tau$$. Therefore $$O(z) = O(a)$$.

Notice that the above is indeed a proof in the theory of dynamical systems. You should add the tag ds.dynamical-systems.

EDIT: I gave a proof of the auxiliary result, but now I am not sure how it contributes toward solving the original problem: notice that we have only that $$O(a)$$ is equal to the closure of $$O^+(a)$$, not to $$O^+(a)$$ itself. I googled M. C. Irwin's Smooth Dynamical Systems: the proof on p. 45 seems to be much more complicated, using facts from general topology that are quite simple, but rather not generally known.

• Is it easy to see that the orbit is a minimal set? Anyway +1 for such a nice answer. – Severin Schraven Mar 31 at 23:04
• @SeverinSchraven Thank you. The orbit is a compact set, by assumption. For any proper subset $A$ of $O(a)$ there are $t_1,t_2\in\mathbb{R}$ such that $a \cdot t_1\in A$ and $a \cdot t_2\in O(a)\setminus A$. But $a\cdot t_2=(a \cdot t_1) \cdot (t_2-t_1)$, so $A$ cannot be invariant. – user539887 Apr 1 at 6:53
• So compactness is only needed to show $\Omega(a)\subseteq O(a)$. In the minimality we only need invariance of $\Omega(a)$, right? – Severin Schraven Apr 1 at 8:27
• A minimal set is, by definition, a compact invariant set such that there does not exist any proper compact invariant subset of it. – user539887 Apr 1 at 11:05
• @SeverinSchraven See my edit. – user539887 Apr 2 at 7:11