How do I calculate for what values of **H** are the solutions of the differential equation periodic?

Question

Given the differential system $$\dot{x}=p$$

$$\dot{p}=-x^3+x$$

and

$$H(x,p)=\frac{x^4}{4}-\frac{x^2}{2}+\frac{p^2}{2}+\frac{1}{2}$$

How do I calculate for what values of H are the solutions of the differential equation periodic?

Answer 1

The trajectories of the system correspond to level curves of the Hamiltonian. Closed curves that do not contain the stationary points ($(\pm1,0)$ and $(0,0)$) are periodic solutions. Rewrite the Hamiltonian as $$ H(x,p)=\frac14(x^2-1)^2+\frac12\,p^2+\frac14, $$ making it clear that level curves are bounded. When $H=1/4$ we get the stationary points $(\pm1,0)$; when $H=1/2$ the point $(0,0)$ and two homoclinic orbits joining $(0,0)$ to itself, one going around $(1,0)$ and the other around $(-1,0)$. Can you do the rest now?

The level curves $H=E$ can be drawn from the equation $$p=\pm\sqrt{2\,E-\frac12-\frac12(x^2-1)^2}$$.