Consider the second-order ODE: $$\ddot{x} + x+x^2=0,$$ here $\ddot{x}$ is the second derivative w.r.t. $t$. Take initial values $x(0)=0.5$ and $\dot{x}(0)=0.$

Question: is the solution periodic or not?

Comment: Numerical experiments seems to show that the solution is periodic when $x(0)<0.5$ and if $x(0)>0.5$ then the solution fails to be periodic, in fact, $x(t)\rightarrow -\infty.$

(Asked by Prof. J.E. Björk, Stockholm Univ.)

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    $\begingroup$ Since solution trajectories are level sets of the Hamiltonian $H(x,\dot{x}) = \frac{1}{2}\dot{x}^2 + \frac{1}{2}x^2 + \frac{1}{3}x^3$, are you asking about the level set of $H$ running through (.5,0)? $\endgroup$ – Aaron Hoffman Feb 8 '13 at 16:04
  • $\begingroup$ Aaron Hoffman: Yes, that is correct. $\endgroup$ – Per Alexandersson Feb 9 '13 at 9:37

As Aaron Hoffman pointed out, all trajectories lie on the level lines of $p^2+x^2+(2/3)x^3=c$. The LHS has two critical points: the local minimum $(0,0)$ with critical value $c=0$, and the saddle $(-1,0)$ with critical value $c=1/3$. The behavior for large $(x,p)$ is also clear. So it is easy to sketch all these level lines, and the conclusion is that the trajectory starting from $(x_0,0)$ is bounded if and only if $x_0^2+(2/3)x_0^3\leq 1/3$, and closed when this inequality is strict. For $x_0=0.5$ we obtain that the trajectory is not really closed, but tends to $(-1,0)$ as time goes to infinity.

You cannot detect this on computer because the point $(-1,0)$ is unstable. It takes infinite time to approach it on the trajectory, but once you miss, no matter how little, you will be either on a closed trajectory or escape to infty. And it will take you long time to find out, if you miss very little.

EDIT. By the way, this system is called the classical anharmonic oscillator. An explicit solution in elliptic functions exists, but physicists prefer to consider perturbative expansions. See, for example, L. Landau and E. Lifshitz, Mechanics (Course of Theoretical Physics, vol. I).

  • $\begingroup$ Thank you! J.E Bjork tells me he has some ideas as well on how to find the length of the period for initial values <0.5... $\endgroup$ – Per Alexandersson Feb 11 '13 at 9:24
  • $\begingroup$ Finding the length of the period is simple. You write your equation as $(dx/dt)^2=P(x)$. This is separable, and the period $T=\int dx/\sqrt{P},$ where the integration is over the closed trajectory. This is an elliptic integral; it can be brought to a standard form, and this gives an explicit answer. $\endgroup$ – Alexandre Eremenko Feb 11 '13 at 14:30

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