classical mechanics - Does a simple double pendulum have transients?

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.

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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 $\left | x(T) - x(0) \right | < ε$.

Answer

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.