Suppose, we have the most simple double pendulum:

  • Both masses are equal.
  • Both limbs are equal.
  • No friction.
  • No driver.
  • Arbitrary initial conditions (no restriction to low energies)

Does this pendulum have transients for any initial conditions or is it immediately on its respective invariant set?

I have seen several time series that suggest that there are no transients, however, I could not find any general statement on this.

I here say that a system has no transients, if its trajectory comes arbitrarily close in phase space to its initial state again, i.e., for a given initial state $x(0)$ and for all $ε>0$, there is a $T>0$ such that $\left | x(T) - x(0) \right | < ε$.

  • $\begingroup$ I can't imagine why there would be transients; why do you think that there might? $\endgroup$
    – garyp
    Jul 10, 2014 at 13:41
  • $\begingroup$ @garyp: Well, for chaotic or periodic dynamical systems, the existence of transients is very much the default – actually, the only system without transients that comes to my mind is the undampened, undriven harmonic oscillator. If you have some argument that explains, why you cannot imagine that there would be transients, this might very well answer my question. $\endgroup$
    – Wrzlprmft
    Jul 10, 2014 at 14:14
  • $\begingroup$ Perhaps my use of the word "transient" is too limited. To me, transients occur in forced systems with dissipative forces. Since you specified no driver and no friction, there would be no transient. But there may be a broader definition of which I am not aware. $\endgroup$
    – garyp
    Jul 10, 2014 at 14:32
  • $\begingroup$ @garyp: I added a definition of transient. Again, if you can reason, that such transients do not occur in systems without drivers and friction, this might be the answer I am looking for. $\endgroup$
    – Wrzlprmft
    Jul 10, 2014 at 15:12
  • $\begingroup$ Is the solution of the double pendulum ultimately a periodic function or not? Is it guaranteed to have the exact same state as the initial conditions at a time >0 ? Once this is answered any characterization of the solution as transient + eigenmodes can be done. $\endgroup$ Jul 10, 2014 at 15:31

2 Answers 2


I found the (shamefully simple) answer myself:

The simple double pendulum is a conservative system, hence due to Liouville’s theorem the phase-space volume given by a given ensemble of trajectories is constant over time. However, if the system exhibited transients and thus attractors, all trajectories starting within the basin of attraction of a given attractor would eventually converge towards that attractor. As the phase-space volume of the basin of attraction is larger than that of the attractor, Liouville’s theorem would be broken. Thus transients and attractors can only occur in dissipative systems.


Simple pendulums are simple, for one, because they only have a single solution. These types of pendulums are typically single pendulum systems where the small angle approximation holds. For this reason I wouldn't classify your question as a "Simple Double Pendulum" because the "simple pendulum" is something different. (This confused me on first answer).

Transients arise because of counteracting forces (driving force and friction for example) that eventually balance each other out to a simple, steady-state solution. When you remove those forces to simplify the problem, you remove the transient solution.

In addition, in the comments you wrote "for chaotic or periodic dynamical systems, the existence of transients is very much the default". I wouldn't describe chaotic motion as having transients. Transient implies that it will at some point go away and chaotic motion, by definition, never settles down to a steady-state. You have stated a definition of the transient that, at least to my knowledge, is one that is not generally shared. To most, the "transient" is the solution that disappears with time or the solution that describes behavior for times close to t = 0.

The double pendulum typically has a chaotic solution, or one that depends heavily on initial conditions. It does not have a single, steady state solution. Thus if you define the transient as the solution that describes behavior near the starting time, you could claim that the only solution to the double pendulum is a transient solution.

This, to me, seems like a matter of definition. A transient solution normally only makes sense when it dies off and a "long-time" or steady state solution is later found. In the absence of the steady state solution, I would not classify the solution as transient.

  • $\begingroup$ Your definition of transients seems to be a different one than mine according to which chatic systems certainly can have transients, when the intial condition is not on the strange attractor but in its basin of attraction. $\endgroup$
    – Wrzlprmft
    Jul 10, 2014 at 16:06
  • $\begingroup$ Okay that may just be a matter of definition then :) I'm not sure what the "official" physics definition is. Did the rest of my answer solve your question? $\endgroup$
    – jkeuhlen
    Jul 10, 2014 at 16:39
  • $\begingroup$ I am afraid, I am not convinced. Your first paragraph is based on a wrong assumption, namely that there is only one solution to the double pendulum (I should have spotted that earlier). There might be something to your second paragraph, if you could classify the counteracting forces required (and back up that they are necessary for transients). $\endgroup$
    – Wrzlprmft
    Jul 10, 2014 at 18:33
  • 1
    $\begingroup$ 1) There is no need to tell the history of your answer and keep outdated aspects – this makes your answer very unstructured and confusing. 2) The definition I gave is at least very similar to the common use of the term by people simulating chaotic dynamics. 3) I do not see any contradiction between my definition of transients and the on you gave: E.g., if you take a Lorenz system, the behaviour close to t = 0 is qualitatively identical to the behaviour after a long time, if and only if the initial conditions are on the strange attractor – otherwise you get a transient. $\endgroup$
    – Wrzlprmft
    Jul 10, 2014 at 22:27
  • 1
    $\begingroup$ 4) I do not think there is any point in discussing about the definition of transient. I gave a (hopefully) clear definition of my criterion – how it is called does not matter for answering the question. 5) How exactly do you define steady state? Do you consider a quasiperiodic system steady? $\endgroup$
    – Wrzlprmft
    Jul 10, 2014 at 22:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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