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.

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 $\left | x(T) - x(0) \right | < ε$.
 A: 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.
A: 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. 
