# Phase portrait Hamiltonian system

I have the Hamiltonian:

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

When I find the critical points associatted with the system $\dot{x}=\frac{p}{m}$ and $\dot{p}=-kx^3$, I find only one critical point in (x=0,p=0). Then I get the Jacobian matrix, evaluate it at this point and I get its eigenvalues, $\lambda_1=\lambda_2=0$.

I know that when both eigenvalues are 0, the system is unstable, but after integrating it with matlab I find ellipses around (0,0) nonetheless. What is happening?

I'm just learning dynamical systems, so a bit of help might come in handy.

Obviously there is no harmonic approximation to your system near $(0,0)$ since your potential does not have a harmonic approximation near that point. Nevertheless, the system is stable in the sense that, if you start near $(0,0)$, you will remain near $(0,0)$. Moreover, the solution is still a bounded oscillation near the fixed point, but its not a harmonic oscillation.
The potential is illustrated in the figure. The "issue" is that it is too flat to be approximated by something like $V\approx V_0+\epsilon\frac{\beta}{2}x^2$, i.e. $\beta=0$ for this potential: the usual stability theory, which is based on expansion of $H$ to $\epsilon^2$ near the fixed point, will just not work.
• But why does starting near $(0,0)$ means remaining close to it? Isn't the system unstable? Apr 21 '17 at 11:34
• The system is stable as your phase curves are closed. The potential has a minimum at $x=0$ but no osculating parabola: the motion is still bounded between two turning points; it's just not harmonic motion to any approximation.. Apr 21 '17 at 11:38
• The eigenvalues tell you nothing because your system does not have a quadratic approximation. Your Jacobian matrix contains 2nd derivatives, but all of those are $0$ at the fixed point so you get nothing out of that. Apr 21 '17 at 12:41