I have a Hamiltonian system given by:
$$\dot{x}=x+y-x^2\\\dot{y}=2xy-y$$
I have found that the Hamiltonian function for the system is given by $$H(x,y)=xy+\frac{1}{2}y^2-x^2y$$
and I have managed to classify the equilibrium points of the system by considering the eigenvalues of the Jacobian matrix to get:
$(0,0),(1,0)$ are a saddle points of the system and $(1/2,-1/4)$ is a centre.
Now is where the problem has defeated me it asks:
Determine the set of initial conditions $(x_0,y_0) \in \Bbb{R^2}$ for which the solution to this system is periodic.
I have no idea of how to proceed from here.
I would greatly appreciate if someone could show me how to go about solving this part and explain how to do it.
Thank you very much!
Hint: Notice that the level set $H=0$ is the union of the line $y=0$ and a parabola. Simply drawing these two lines, you will find that they determine a compact invariant set containing your fixed point $(1/2,-1/4)$. So the initial conditions giving a periodic orbit are precisely those that are inside this compact invariant set, other than the fixed point.
Indeed, periodic orbits must have fixed points inside and the other other two fixed points can be ruled out because they are on the line $y=0$, which then would have to be crossed by the periodic orbit.