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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:

enter image description here

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?

Alp Uzman
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Alireza
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1 Answers1

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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.

Adam
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  • I think I do not clearly understand your answer; even in the case of not strictly bounded systems such as the Lorenz equations, look at the figure on page 6 of this document: http://epileptologie-bonn.de/cms/upload/homepage/lehnertz/Lyapunov_Ansmann.pdf the distance between the two trajectories becomes somewhat constant after sometime but we know that the system is chaotic. in this figure where would be the short enough time range you mentioned? – Alireza Oct 07 '20 at 17:18
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    It doesn't have to be the boundedness of the system that causes the exponential divergence to stop happening, it could happen for any reason (in this case it's because the Lorenz system has an attractor, so orbits end up being "bounded" even though the system is not literally bounded). But that doesn't matter for the Lyapunov exponent. It's still true that given an initial displacement, exponential divergence happens for a while and then stops, and if the initial displacement is smaller then the exponential divergence happens for a longer time scale. – Adam Oct 07 '20 at 17:24
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    The calculation is the same regardless of the mechanism that causes exponential divergence to stop happening – Adam Oct 07 '20 at 17:24
  • So the Lyapunov exponent should be calculated up until when the distance has stopped growing? – Alireza Oct 07 '20 at 18:00
  • Yes that sounds good to me – Adam Oct 07 '20 at 18:07
  • @Alireza: So the Lyapunov exponent should be calculated up until when the distance has stopped growing? – Yes that sounds good to me – Actually, you want to stop at least an order of magnitude before it stops growing. The final distance should still be small compared to the size of the attractor. — Note that in the lecture slides you link (which happen to be mine), the answer to your question is encoded in the fact that the initial separation should be (ideally) infinitesimal. Thus, no matter how long you evolve, it will never reach the size of the attractor (but stay infinitesimal). – Wrzlprmft Oct 08 '20 at 06:29