I got really curious about ergodicity in double pendulum. The diagram below represents the time it takes for a double pendulum to "flip" when started from different initial positions. The white color on the graph corresponds to the positions where the pendulum doesn't flip during the simulation time (of about 2000 swings). While the central white region of the graph is intuitive and corresponds to states in which the pendulum does not have enough energy to flip (the region corresponds to inequality $3 \cos \theta_1+\cos\theta_2>2$), there are non-flipping states outside the region. These states do not flip, even though they have enough energy, which indicates the behavior of the system starting from such states is possibly non-ergodic. The existence of unreachable regions in the phase space would seem to imply the existence of non-trivial integrals of motion (for instance, an indicator function that is equal to 1 in the unreachable region of the phase space, and 0 everywhere else). My question is what these non-trivial integrals of motion could possibly be like? Can we find them somehow? Or will the pendulum eventually flip everywhere where it can flip, but just in an extremely long time?

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    $\begingroup$ No, the double pendulum does not have integrals of motion besides the energy. If it did it would be integrable and there would be no chaotic motion, see Non-integrability of the 2D double pendulum. And yes, it is not ergodic on energy surfaces, non-ergodicity does not imply extra integrals of motion in general. Such "partially integrable" systems are described by KAM theory, see Chaos and ergodicity in hamiltonian systems. $\endgroup$
    – Conifold
    Oct 21, 2021 at 5:18
  • $\begingroup$ I think it is important to point that there are no smooth integrals of motion. In many places (like wikipedia) smoothness is not mentioned as a requirement for the integral of motion. If smoothness is not required, non-ergodicity always implies integrals of motions. As was mentioned in the OP, an indicator function, which is equal to 1 in the accessible region of the phase space, and is equal to 0 in the inaccessible region would be an integral of motion, if there was no smoothness requirement. $\endgroup$
    – Pavlo. B.
    Oct 22, 2021 at 16:45
  • $\begingroup$ @Conifold Also, double pendulum cannot be "partially integrable" (by the definition I was able to find), because adding a single integral of motion [non-constant in any open region] would make the system completely integrable. I would appreciate if you could flash out your comment into a more complete answer, because I am quite uncertain about what definitions you use and how they answer the question. $\endgroup$
    – Pavlo. B.
    Oct 22, 2021 at 17:35


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