Ian Jauslin

Double pendulum

The Lagrangian of the double pendulum is $$L(\theta_1,\theta_2;\dot\theta_1,\dot\theta_2)=\frac{m_1+m_2}2l_1^2\dot\theta_1^2+\frac{m_2}2l_2^2\dot\theta_2^2+m_2l_1l_2\dot\theta_2\dot\theta_2\cos(\theta_2-\theta_1)+g((m_1+m_2)l_1\cos\theta_1+m_2l_2\cos\theta_2)$$ and its Hamiltonian is $$H(\theta_1,\theta_2;p_1,p_2)=\frac{\frac{p_1^2}{2l_1^2}+\frac{\mu p_2^2}{2l_2^2}-\frac{p_1p_2}{l_1l_2}\cos(\theta_2-\theta_1)}{m_1+m_2\sin^2(\theta_2-\theta_1)}-g((m_1+m_2)l_1\cos\theta_1+m_2l_2\cos\theta_2)$$ where $\mu:=1+m_1/m_2$.

A numerical simulation of this system for $$l_1=1,\ l_2=0.9,\ m_1=m_2=1,\ g=2$$ can be found below. The double pendulum is drawn as well as the Poincare section in the plane $(\theta_2,p_2)$ with $\theta_1=0$, $p_1>p_2\frac{l_1}{l_2}\cos\theta_2$. The initial conditions can be changed below.


Below is a representation of two simultaneous trajectories the double pendulum, which start out at a distance of $10^{-6}$. This is a way to acertain the positivity of Lyapunov exponents.

Initial conditions:

In order to select an initial conditions, click on the diagrams below (tested with Firefox). The red dot gives the position of the initial $(\theta_1,\dot\theta_1)$ and $(\theta_2,\dot\theta_2)$. Then click on "Reset initial conditions".

If you are not running Firefox, then this might not work (I have been told that it does not work in Safari), in which case, you can select one of the following pre-configured initial conditions:

Reset initial conditions