Double Pendulum has a very beautiful stochastic trajectory. Is there any way to calculate the distribution of probability of finding the end of pendulum at each point?

Link to formulations.

enter image description here

  • 6
    $\begingroup$ the trajectory is not "stochastic". It shows chaotic behavior but is fully deterministic given input conditions and a perfect environment. $\endgroup$
    – qwr
    Nov 13, 2016 at 18:11
  • $\begingroup$ You could use the notion of path integral in physics to obtain a probability distribution. $\endgroup$ Nov 13, 2016 at 19:33
  • 2
    $\begingroup$ This system is AFAIK not ergodic on its energy level, so the answer is subtle, and depends on your initial conditions. The Liouville measure indicated by Stephen Montgomery-Smith is somewhat too large (its support is the whole energy level, while the pendulum lives on a subset), but there are too many invariant measures to give an exact answer. $\endgroup$
    – D. Thomine
    Nov 14, 2016 at 21:07
  • $\begingroup$ @D.Thomine, what if we have the initial energy. Then, does the initial point matter? $\endgroup$
    – ar2015
    Nov 23, 2016 at 4:22

1 Answer 1


What you could try is to first write the system in Hamiltonian form. Then quote https://en.wikipedia.org/wiki/Liouville%27s_theorem_(Hamiltonian) which says that the distribution on phase space is uniform. You could assume ergodicity, that is, the solution path is dense in the energy submanifold $H(q,p) = E$. Then project this onto your 2-D space.

I think it would be a lot of work, but definitely doable.

So making a start, the potential energy is $$ V = - m_1 g L_1 \cos(\theta_1) - m_2 g L_2 \cos(\theta_2) ,$$ and the Kinetic Energy is $$ T = \tfrac12 [\dot \theta_1, \dot \theta_2] A [\dot \theta_1, \dot \theta_2]^T = \tfrac12 [\eta_1, \eta_2] A^{-1} [\eta_1, \eta_2]^T $$ where $$ A = \begin{bmatrix} m_1 L_1^2 + m_2 L_1^2 & m_2 L_1 L_2 \cos(\theta_1-\theta_2) \\ m_2 L_1 L_2 \cos(\theta_1-\theta_2) & m_2 L_2^2 \end{bmatrix} $$ and $\eta_1$ and $\eta_2$ are the generalized momentums.

See https://physics.stackexchange.com/questions/142238/non-integrability-of-the-2d-double-pendulum

The Hamiltonian $H = V + T$ is conserved, so calculate its value using initial conditions. For a given $[\theta_1,\theta_2]$, the probability density will be proportional to the volume of the ellipsoid $$ \{[\eta_1,\eta_2] \in \mathbb R^2 : \tfrac12 [\eta_1, \eta_2] A^{-1} [\eta_1, \eta_2]^T \le H - V\}$$ And this volume will be proportional to $$ \begin{cases}\det(A) (H-V)^2 & \text{if $H > V$}\\ 0 &\text{if $H \le V$}\end{cases} $$ Remember both $V$ and $A$ are functions of $\theta_1$ and $\theta_2$. Note that for any place where the pendulum can go, there will generally be two values of $\theta_1$ and $\theta_2$ for that point, where for one point the value of $\theta_1 - \theta_2$ will be exactly the negative of the value for the other position.

Remember, this will give you the density in $[\theta_1,\theta_2]$ space (the green picture on the left side of the web page you provided), so you will have to divide this by the absolute value of determinant of the Jacobian of the map that takes $[\theta_1,\theta_2]$ to $[x,y]$ (that is, $L_2L_2|\sin(\theta_1-\theta_2)|$).

Now my calculations might be all wrong because maybe there is another constant of the motion (which means I lose the ergodicity property). The web page https://physics.stackexchange.com/questions/142238/non-integrability-of-the-2d-double-pendulum suggests there isn't another constant of motion. But in the green picture in http://www.myphysicslab.com/pendulum/double-pendulum/double-pendulum-en.html, I can plainly see that for any choice of $[\theta_1, \theta_2]$ that the velocity has a particular direction (or rather a choice of two directions). Conservation of the Hamiltonian (energy) would only restrict the magnitude of the velocity going through each point!

Added later: I asked someone more knowledgeable than I about ergodicity. There are even situations where the set of points reached in phase space has positive measure on a level set of the energy function, but the double pendulum is still not ergodic. Apparently many books have been written on the subject. So my final answer is "I don't know." I'll not delete this answer, because I think the discussion is worthwhile. But I don't expect to receive any upvotes.

  • 1
    $\begingroup$ A precision : ergodicity is too much to ask for (this is a textbook example of KAM theory, so the system tends to keep some integrability). $\endgroup$
    – D. Thomine
    Nov 14, 2016 at 21:10
  • $\begingroup$ @D.Thomine Yes, I can see in the green picture that it isn't ergodic! I must admit I don't know much about the KAM theorem. But I think I can see why it destroys my answer. $\endgroup$ Nov 14, 2016 at 21:17
  • $\begingroup$ @D.Thomine Now I am rerunning the web simulation myphysicslab.com/pendulum/double-pendulum/… with larger starting values, and then it really does look like it might be ergodic. Does the KAM theory only apply to small perturbations? Or does it work for any initial conditions? $\endgroup$ Nov 15, 2016 at 21:35
  • $\begingroup$ This is not bad at all. The distribution you are talking about is invariant under the hamiltonian dynamics, so we know at least one solution for the probability distribution. This is not the most general answer, but I it is valuable for me and I think for the OP too. Though still would be great to calculate various marginal probability distributions explicitly. $\endgroup$
    – Yrogirg
    Nov 17, 2016 at 4:12

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