Periodic orbit of Hamiltonian dynamical system in R^2 is stable

[MatthijsV]

Homework Statement

We are given a Hamiltonian dynamical system with a smooth Hamiltonian $$H:\mathbb{R}^2\to\mathbb{R}$$ on $$\mathbb{R}^2$$, with canonical symplectic structure. Suppose this Hamiltonian has a periodic orbit $$H^{-1}(h_0)$$. Prove that there exists an $$\epsilon>0$$ such that for all $$h\in ]h_0-\epsilon,h_0+\epsilon[$$: $$H^{-1}(h)$$ is also a periodic orbit.

Homework Equations

Liouville's theorem tells us that the flow of a Hamiltonian vector field preserves the Liouville volume form in $$\mathbb{R}^2$$.

The Attempt at a Solution

It is enough to prove that $$H^{-1}(h)$$ is bounded. It then follows from Liouville's theorem that it is also a periodic orbit. However, I have no idea how to do this.

Thanks in advance!

 

[mark726]

Thank you for your question. The proof for this statement is actually quite simple. First, note that since H is a smooth function, we can use the Implicit Function Theorem to show that for any h\in ]h_0-\epsilon,h_0+\epsilon[, the level set H^{-1}(h) is a smooth curve.

Next, since H^{-1}(h_0) is a periodic orbit, it is closed and therefore bounded. This means that there exists a constant M>0 such that for any point (x,y)\in H^{-1}(h_0), we have \| (x,y) \| \leq M. Now, consider any point (x,y)\in H^{-1}(h). Using the Implicit Function Theorem, we can write y as a smooth function of x, say y=g(x). Then, we have (x,g(x))\in H^{-1}(h) and \| (x,g(x)) \| \leq M for all x\in\mathbb{R}. Therefore, the set H^{-1}(h) is also bounded, and hence it is a periodic orbit.

I hope this helps. Please let me know if you have any further questions.