How to prove that the level sets of this function are closed curves in a specific region.

Question

I need to study the Hamiltonian differential system $$ \begin{align} \dot{x} &= -2ye^{-x^2}\\ \dot{y} &= 2xe^{-x^2}(1-y^2) \end{align}$$ with Hamiltonian function $$ \begin{align} H &\colon \mathbb{R}^2 \to \mathbb{R} \\ (x,y) &\mapsto e^{-x^2}(1-y^2) \end{align}$$

Thanks to the explicit independence of $H$ from the time, $H$ is conserved along the solutions. So I need to study the level sets of $H$. In $(0,0)$ there is a local maxima, so there is a neighborhood of the constant solution $0,0$ in which the other solutions are cycles. I wanted to prove that this neighborhood is in fact the region (strictly) inside the the two lines $(1-y^2)=0$. But I don't know how to prove that in such region, the hamiltonian $H$ has "cyclic" (closed curves) level sets.

Here is the Plot and the contour lines computed by Wolfram Alpha.

Answer 1

When $H\gt0$, $1-y^2\ge H$, or $$|y|\le\sqrt{1-H}.$$

The curve has the explicit equation $$x=\pm\sqrt{\ln{H(1-y^2)}},$$ formed of two symmetric arcs (it's an oval), and $$|x|\le\sqrt{\ln H}.$$

Answer 2

Since description of cyclic orbits between lines $1-y^2 = 0 $ is given by Yves Daoust, I will give proof that you cannot get cyclic orbit when $|y| > 1$.

Lets deal with case $y>1$, case when $y<-1$ is similar.

If $y>1$ than $H<0$, so you cannot escape $\{y>1\}$, because you would have to cross line $\{y=1\}$ for which $H=0$. Also for $y>1$ we have $\dot x = -2 y e^{-x^2} < 0$, therefore if you start at any point with $y>1$ than you always drift to the left (toward $x=-\infty$).

We can show even more, you have to escape any region $\{ y > 1, |x| < L \}$ in finite time. This is because in this region $\dot x < -2 e^{-L^2} < 0 $. Upper bound for time to escape this region is $ \frac{2L}{2 e^{-L^2}}$.