Lyapunov exponents in bounded systems This is the definition used to numerically calculate the maximal Lyapunov exponent:
$$
|\delta Z(t)| \approx e^{\lambda t} |\delta Z_0|
$$
$$
\lambda \approx \frac{1}{t} \ln\left(\frac{|\delta Z(t)|}{|\delta Z_0|}\right)
$$
Consider a system to be bounded in a region of space, like two balls moving inside a room, where we want to see if the dynamics are chaotic or not. We start recording the trajectory of a ball and after a good amount of time, we go back and start from a tiny dislocated initial position. We see that the trajectories diverge and we may conclude that the system is chaotic. Like the following:

But here is what I do not understand. If the room is bounded, then these two trajectories cannot get arbitrarily far from each other and their distance would not be larger than the size of the room. So in the formula for the Lyapunov exponent, $t$ would get larger and larger over time so $\lambda$ would become smaller and finally zero, no matter if it was positive at first.
Would this mean the Lyapunov exponent decays with time in bounded systems? A double pendulum is a bounded system, two trajectories (angles) can not be more than $\pi$ apart at any moment, so does its Lyapunov exponent decay to zero over time?
 A: TLDR: if you want to look at longer and longer time scales, you need to have smaller and smaller initial displacements so that the displacement doesn't blow up to the full system size.
You are right that in a bounded system, if you look at a fixed initial displacement $|\delta Z_0|$ and then let $t \to \infty$, the quantity you care about will have to tend to zero.  For a fixed $|\delta Z_0|$, the displacement should evolve like $|\delta Z(t)| \approx e^{\lambda t} |\delta Z_0|$ for some range of time, but eventually the displacement becomes comparable with the size of the system, and then it can't keep expanding any more.  Because of that, to calculate the Lyapunov exponent it only makes sense to estimate the displacement for a time range that is short enough so that the displacement doesn't blow up to become comparable with the size of the system.
If you want to be more precise you can define the Lyapunov exponent like this, to ensure that you only look at the displacement at time scales which are still good for the initial displacement you started with.
