Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 1 down vote accepted

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 $$|x|\le\sqrt{\ln H}.$$

share|cite|improve this answer
mmh would you mind elaborate it a little more? Can't follow your reasoning. Ok, if $H>0$, then the region is the one inside the two lines, but why level sets inside the two lines are cycles? – Riccardo Jun 1 '14 at 9:23

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}}$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.