I'm generating high-resolution Poincaré sections of a double pendulum, and I'm running into some issues in creating the initial conditions for a given section.

In general, I describe my pendulum with a state vector $(\theta_{1}, P_{\theta 1}, \theta_2, P_{\theta 2})$ for angles $\theta_1$, $\theta_2$ and momenta $P_{\theta 1}$, $P_{\theta 2}$. My prescription for generating initial conditions to get my Poincaré sections is:

  1. Set the total energy $E$

  2. Set $\theta_1 = P_{\theta 2} = 0$

  3. Grid $\theta_2$ in a range from $-\pi$ to $\pi$

  4. Solve the Hamiltonian as a quadratic to determine $P_{\theta 1}$

This Hamiltonian is from this Wolfram source, and was inverted in Mathematica to give

$$P_1 = \frac{1}{2} \left(\frac{P_2 \cos (\theta_1-\theta_2)}{\sin ^2(\theta_1-\theta_2)+1}-\sqrt{4 E+\frac{P_2^2 \cos ^2(\theta_1-\theta_2)}{\left(\sin ^2(\theta_1-\theta_2)+1\right)^2}-8 P_2^2+8 \cos (\theta_1)+4 \cos (\theta_2)}\right)$$

I think this is a fine procedure? There are a few arbitrary choices (like step 2; they could really be set to anything), but I'm not sure if this is doing anything necessarily bad.

However, there are a few different kinds of problems I'm encountering with this.

The first is for low-energy sections. The quadratic only has real solutions for certain ranges of $\theta_2$, and this precludes me from getting initial conditions which show the fixed points of the poincare section:

poincare section 1

No matter what, the sections won't reduce to the fixed points because the Hamiltonian doesn't solve for those conditions. This figure is an attempt to recreate the figure from this SE article. It's getting the outermost sections, but the surfaces simply don't reduce to single points.

The second problem is with high-energy sections. I get these weird plots which look nothing like any poincare section I've seen in the literature: poincare section 2

Certainly, there's some structure there (but the convergence problem from before is still present), but there is a crazy amount of apparently chaotic behavior in the background. Shouldn't the chaotic behavior only be happening between the fixed points?

So, my questions are:

  • Does my initial condition procedure seem okay?

  • How can I generate the initial conditions for the fixed points?

  • Does the second poincare section seem sane, or should I be looking elsewhere for a problem (like maybe my integrator is doing something weird or something)?


1 Answer 1


Note that I have not checked any references or own simulations at to what Poincaré sections of the double pendulum should look like, but here are a two things that strike me as odd about your description:

  • the sections won't converge to stable points

    As the double pendulum is a Hamiltonian system, it does not have attractors such as limit cycles. Hence there is no convergence in the Poincaré sections. For a given energy, there can be infinitely many periodic solutions. Therefore, to get a meaningful Poincaré section, only take into account the solution for a single initial condition. Finally note that due to the system being Hamiltonian, you may get some energy drift, which can be addressed by using a symplectic integrator (I don’t know whether this is a relevant problem here though).

    For the same reason, you will find many distinct solutions for the chaotic case. As dense periodic orbits are a defining feature of chaos, it is not surprising that you see some of them.

  • The quadratic only has real solutions for certain ranges of θ2, and this precludes me from getting initial conditions which show the stable points of the poincare section

    If you cannot solve your equation to find an initial condition, this is because your chosen $θ_2$ already implies more positional energy than your chosen energy. There simply is no physical initial condition meeting your constraints.

  • $\begingroup$ Thanks for the response! "stable" was an error – I meant "fixed", and the "converge" was sloppy– I meant more generally how the surfaces around a fixed point on a poincare section will have smaller and smaller radii. I've edited the original question to reflect this. $\endgroup$
    – David
    Jun 28, 2019 at 21:15

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