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Why or why not? I'm pretty sure that this isn't a Hamiltonian system because it involves a dissipation term, but using the Hamiltonian flow it gives me that the system is Hamiltonian.

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up vote 0 down vote accepted

I have a curious example :

If I take the hamiltonian $H(p,q) = (p + q)^2$, and apply Hamilton equations : I get :

$$ \dot q = \frac{\partial H}{\partial p} = 2(p + q)$$

$$ \dot p = - \frac{\partial H}{\partial q} = - 2(p + q)$$

So you have :

$$ \dot p = - \dot q$$

So it looks like very much to a dissipative system.

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@ Trimok I think your example is quite interesting. It's equivalent to a free particle, which you can see by starting with the hamiltonian $H=(p+q)^2/2$ the canonical transformation $p'=p+q, q'=q$ which leads to the hamiltonian $H'=p'^2/2$. I guess the point is that in this case the apparent dependence on $\dot{q}$ is fake, because it is canceled by the $\dot{q}$ dependence in $\dot{p}$ when going back to configuration space. – Andrew Dec 25 '13 at 5:51

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