0
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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

$\endgroup$
  • $\begingroup$ But why does starting near $(0,0)$ means remaining close to it? Isn't the system unstable? $\endgroup$ – mobzopi Apr 21 '17 at 11:34
  • 1
    $\begingroup$ 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.. $\endgroup$ – ZeroTheHero Apr 21 '17 at 11:38
  • $\begingroup$ But what do the eigenvalues tell me about the stability then? $\endgroup$ – mobzopi Apr 21 '17 at 12:37
  • 1
    $\begingroup$ 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. $\endgroup$ – ZeroTheHero Apr 21 '17 at 12:41
  • 1
    $\begingroup$ @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. $\endgroup$ – ZeroTheHero Apr 21 '17 at 13:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.