I want to calculate the Lyapunov Exponent for a double pendulum, with a small change in the initial angle. In this study, the authors used the formula $\frac{1}{t}{ln(\frac{d}{d_0})}$ as $t$ tends to infinity, where $t$ is time, $d_0$ is the initial separation, and $d$ is the separation after a given time step. Would this work? I was worried because the differences in the trajectories are only locally exponential (physical constraints of the pendulum), and the separation (between the two trajectories, of differing angles) does not necessarily need to grow exponentially. This is mentioned in this paper, pg. 13. A lot of other studies mention taking the average of the Lyapunov Exponent along many points of the trajectory (including this source). So is $\lim_{t\to \infty}\frac{1}{t}{ln(\frac{d}{d_0})}$ a viable way to calculate the Lyapunov Exponent for a double pendulum? If not, how should I approach this problem. Any guidance would be greatly appreciated, thanks.
$\begingroup$
$\endgroup$
3
-
2$\begingroup$ Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. $\endgroup$– Community BotCommented Feb 5 at 5:43
-
2$\begingroup$ Minor comment to the post (v3): Please consider to mention explicitly author, title, etc. of link, so it is possible to reconstruct link in case of link rot. $\endgroup$– Qmechanic ♦Commented Feb 5 at 8:19
-
$\begingroup$ Would this work? – What do you mean by this and work here? Obviously, you cannot work with $t→∞$ and $d_0→0$ under any circumstances, so you practically try to approximate this. How this approximation can look like is described in your second source, for example. Do you have a more specific concern regarding this? $\endgroup$– WrzlprmftCommented Feb 5 at 14:36
Add a comment
|