# Differential equations question: Follow-up on Dynamical systems?

Yesterday I asked a question on here. Unfortunately I closed off the page without fully signing up for my account so I could not comment on the answer I received, whilst the answer was very good there were a few things I did not quite get and wanted some extra clarification. I could not get access to my old account so I will have to post here sorry for any inconvenience this causes.

Here is the question for reference: How to determine which initial conditions will make the solution of a Hamiltonian system periodic?

And the answer provided by @Jonas.

Hint: Notice that the level set $H=0$ is the union of the line $y=0$ and a parabola. Simply drawing these two lines, you will find that they determine a compact invariant set containing your fixed point $(1/2,-1/4)$. So the initial conditions giving a periodic orbit are precisely those that are inside this compact invariant set, other than the fixed point.

Indeed, periodic orbits must have fixed points inside and the other other two fixed points can be ruled out because they are on the line $y=0$, which then would have to be crossed by the periodic orbit.

A few comments I want clearing up in regards to this.

Why do we look straight away for $H=0$ why is this necessary to determine the periodic orbits?

Why do periodic orbits have to contain fixed points inside?

Finally why does the fact that that the two other fixed points are on $y=0$ mean that they would be crossed by the periodic orbit? And how can we then rule them out?

In summary I don't understand the last paragraph.

I should note that my answer to the other question was only a hint, although quite complete. It is a coincidence that looking at the level set $H=0$ is all that is need for this particular system. There is no rule that can tell you how to do it general, and in particular the level set $H=0$ may have no relevance whatsoever in other problems.
A simple reasoning is the following. First observe that periodic orbits have fixed points inside (this is not easy to show, but one way that avoids the axiom of choice is to show that the index of a periodic orbit is $\pm1$, which implies that a fixed point must exist inside, in the sense of Jordan's curve theorem; see for example the good old Coddington and Levinson's book).
You have three fixed points, but two of them are on the line $y=0$, on which $H$ is constant. Since a periodic orbit having one of these two fixed points inside would go "around" at least one of them, unavoidably it would cross the line $y=0$ at least at two points. But orbits are contained in level sets of $H$ and so no orbit can cross $y=0$ like that because $H$ changes when you do it (and $H$ is a conserved quantity).
Only one fixed point remains, the one outside the line $y=0$. As I explained this line and a certain parabola determine a compact invariant set. In fact, the complement of the union of the two curves has $5$ connected components (just draw the curves) and only one of them has fixed points (actually, one fixed point, the one remaining).