Concerning this page http://scienceworld.wolfram.com/physics/DoublePendulum.html
for the double pendulum the moving equations are given by
$$ \ddot{\theta}_1=\frac{-m_2\ell_2\ddot{\theta}_2\cos(\theta_1-\theta_2)-m_2\ell_2\dot{\theta}_2^2\sin(\theta_1-\theta_2)-g(m_1+m_2)\sin\theta_1}{(m_1+m_2)\ell_1}, $$ $$ \ddot{\theta}_2=\frac{-m_2\ell_1\ddot{\theta}_1\cos(\theta_1-\theta_2)+m_2\ell_1\dot{\theta}_1^2\sin(\theta_1-\theta_2)-m_2g\sin\theta_2}{m_2\ell_2}. $$
Now I can write this as a system (*):
$$ \dot{\theta}_1=w_1, $$ $$ \dot{\theta}_2=w_2, $$ $$ \dot{w}_1=\frac{-m_2\ell_2\ddot{\theta}_2\cos(\theta_1-\theta_2)-m_2\ell_2\dot{\theta}_2^2\sin(\theta_1-\theta_2)-g(m_1+m_2)\sin\theta_1}{(m_1+m_2)\ell_1}, $$ $$ \dot{w}_2=\frac{-m_2\ell_1\ddot{\theta}_1\cos(\theta_1-\theta_2)+m_2\ell_1\dot{\theta}_1^2\sin(\theta_1-\theta_2)-m_2g\sin\theta_2}{m_2\ell_2}. $$
My question is if the vector field $f$ built by the right sides of the system (*) is divergence free.
I would say: yes, because
$$ \frac{\partial}{\partial\theta_1}f_1(\theta_1,\theta_2,w_1,w_2)=\frac{\partial}{\partial \theta_1}w_1=0, $$ $$ \frac{\partial}{\partial\theta_2}f_2(\theta_1,\theta_2,w_1,w_2)=\frac{\partial}{\partial\theta_2}w_2=0, $$ $$ \frac{\partial}{\partial w_1}f_3(\theta_1,\theta_2,w_1,w_2)=\frac{\partial}{\partial w_1}(\frac{-m_2\ell_2\ddot{\theta}_2\cos(\theta_1-\theta_2)-m_2\ell_2\dot{\theta}_2^2\sin(\theta_1-\theta_2)-g(m_1+m_2)\sin\theta_1}{(m_1+m_2)\ell_1})=0, $$ $$ \frac{\partial}{\partial w_2}f_4(\theta_1,\theta_2,w_1,w_2)=\frac{\partial}{\partial w_2}(\frac{-m_2\ell_1\ddot{\theta}_1\cos(\theta_1-\theta_2)+m_2\ell_1\dot{\theta}_1^2\sin(\theta_1-\theta_2)-m_2g\sin\theta_2}{m_2\ell_2})=0, $$ so the sum is 0, too.
But I am not sure, if I think too easy...