# An analytic method to prove a specific curve is closed

In my study of Hamiltonian dynamics I have come across a Hamiltonian dynamic system with a solution curve I know to be closed via computer and via intuition but I require a rigorous way to prove this, the problem simply is:

We have the curve in $\mathbb{R}^2$ described via the equation $\sin(x)\sin(\pi y)=1/2$ (of course for single period of the sines as they are periodic so this equation can describe infinitely many curves). We are to show this curve is closed. The Hamiltonian is $H(x,y)=\sin(x)\sin(\pi y)$

My intuition was regarding periodicity of the sines. My purpose is basically to show it is not a limit cycle but a closed orbit in phase space, so can someone please demonstrate rigorously why this curve is closed? I have the intuition but not the right rigorous ideas. I thank all helpers.

• Do you know about morse theory? – Tim kinsella May 26 '16 at 4:41
• No worries. I see you're an undergrad studying geometry and topology. You should definitely check out the great book "differential topology" by Guillemin and Pollock. – Tim kinsella May 26 '16 at 5:16
• A little bit of that book gives a nice solution to your problem: $f(x,y)= \sin(x)\sin(\pi y)$ is a smooth function and $1/2$ is one of its regular values (this just means the derivative of $f$ is surjective at every point $f$ maps to $1/2$). This implies that $f^{-1}(1/2)$ is an imbedded sub manifold of $\mathbb{R}^2$. If you can show that there exist no solutions on the boundary of a box around one of the solutions, then inside that box the solution set must consist of a finite union of imbedded 1-submanifolds. Now 1-submanifolds are either homeomorphic to $\mathbb{R}$ or $S^1$. – Tim kinsella May 26 '16 at 5:17
• (the classification of 1-manifolds is also done in Guillemin and Pollock). Since the solution set is closed and we are confined to a bounded subset of the plane, its not hard to see we cannot have homeomorphs of $\mathbb{R}$. This means the solution set in this region must be a finite union of homeomorphs of circles. – Tim kinsella May 26 '16 at 5:17
• Can't you simply investigate what the graph $y=\frac{1}{\pi}\arcsin\frac{1}{2\sin x}$ looks like? (It is one solution of your equation, and all the other solutions can be expressed in terms of this one too.) – Hans Lundmark May 26 '16 at 6:30