Let $H:P \to \mathbb{R}$ be a Hamiltonian on a symplectic manifold $(\omega,P)$ and let $X_H: P \to TP$ be the Hamiltonian vector-field. Let $F:P \to T^*P$ be a dissipative force field such that for $Y = \omega^{\sharp}(F)$ we have that $Y[H] < 0$ everywhere outside a point $x_0 \in P$. This makes $x_0$ a stable point of the dissipative Hamiltonian system $(P,\omega,H,F)$. Now let $f: \mathbb{R} \times P \to T^*P$ be a time-periodic force. My question: Does there exists a periodic orbit (near $x_0$) for the periodically forced system $(P,\omega,H, F + \epsilon f)$ for sufficiently small $\epsilon$ ? I'm sure the answer is yes, but how big can $\epsilon$ be?

  • $\begingroup$ Can you provide a reference explaining what's a "dissipative Hamiltonian system"? $\endgroup$
    – Vanessa
    Oct 20, 2011 at 23:18
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    $\begingroup$ I'm not completely familiar with all your notation, but I assume $Y$ is the vector field corresponding to $F$ and $Y[H]$ the derivative of $H$ with respect to $Y$ that indicates dissipativeness. First of all, is there a reason that the fixed point $x_0$ is stable instead of unstable? Secondly, I would look at the linearized system around $x_0$. If the eigenvalues are strictly negative, then the fixed point must persist. This can be extended to time-periodic or generally time-dependent perturbations. To estimate the size of $\epsilon$, you need to look at the proof of the stable manifold thm. $\endgroup$ Oct 21, 2011 at 8:35
  • $\begingroup$ Thankyou Jaap. I will take a look. Your interpretation of the notation is correct. Secondly, I guess you are right. There is no reason to discriminate the unstable case. For dissipative systems I'd expect to find a stable periodic orbit from a stable point. Under time-reversal a stable equilibria would become an unstable one, and the corresponding stable limit cycle would become unstable as well. $\endgroup$
    – hoj201
    Dec 8, 2011 at 22:37


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