3
$\begingroup$

I'm working on an API for simulation of port-Hamiltonian systems. As far as I understand it, a Hamiltonian system is symplectic if it is power conserving, and so including resistive elements would thus destroy the symplectic nature. My question is, if I add an extra state variable, the dissipated heat, along with its adjoint, does this restore its symplecticity?

Further, if I didn't include the extra state variable, would I still benefit from using symplectic integrators? And do systems lying on a contact manifold still benefit from symplectic integrators?

$\endgroup$
  • 1
    $\begingroup$ Could you give us a more specific operator form? Whether your addition is symplectic is hard to tell (for me) from your description. $\endgroup$ – Deathbreath Mar 6 '13 at 19:37
  • $\begingroup$ @Deathbreath, I do not, other than the for every dissipative term (e.g. for a resistor in an RLC circuit), add a term to the Hamiltonian of the extra state variable, equal to the negative of the dissipative term (e.g. a virtual resistor with negative resistance) $\endgroup$ – Sophie Taylor Mar 7 '13 at 13:40
  • $\begingroup$ In general, one cannot change a dissipative system to a conservative one. $\endgroup$ – Arnold Neumaier Mar 8 '13 at 11:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.