I have the Hamiltonian:


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.


1 Answer 1


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.

Your system is called "intrinsically non-linear" and is discussed in P.Mohazzabi, "Theory and examples of intrinsically nonlinear oscillators", Am.J.Phys. vol. 72 (2004) 492-498.

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.

enter image description here

  • $\begingroup$ But why does starting near $(0,0)$ means remaining close to it? Isn't the system unstable? $\endgroup$
    – mobzopi
    Commented Apr 21, 2017 at 11:34
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    $\begingroup$ 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.. $\endgroup$ Commented Apr 21, 2017 at 11:38
  • $\begingroup$ But what do the eigenvalues tell me about the stability then? $\endgroup$
    – mobzopi
    Commented Apr 21, 2017 at 12:37
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    $\begingroup$ 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. $\endgroup$ Commented Apr 21, 2017 at 12:41
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    $\begingroup$ @mobzopi what your instructor or textbook has done is given you an example where there the standard expansion does not work. The problem here is conceptual: simply because a system does not have a harmonic approximation does not mean the motion is unstable. $\endgroup$ Commented Apr 21, 2017 at 13:01

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