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I would like to obtain a general formula to verify if a certain time-$t$ map of a Hamiltonian flow is twist. I have a Hamiltonian $1$ degree of freedom system $H=H(q(t),p(t))$, such that all orbits are periodic with equations of motion

$$\dot{q} = \frac{\partial H}{\partial p}, \quad \dot{p}=-\frac{\partial H}{\partial q}$$

Now the time-$t$ map of the flow generated by $H$ $$q(t)=f(q(0), p(0))$$

$$p(t) = g(q(0),p(0))$$ where $f$ and $g$ are some functions, is, in terms of $H$ and its derivatives,

$$q(t) = q(0) + \int_{0}^{t}\frac{\partial H}{\partial p} d\tilde{t} $$ $$ p(t) = p(0) - \int_{0}^{t}\frac{\partial H}{\partial q} d\tilde{t}$$

The twist condition for the time-$t$ map is $\frac{\partial f}{\partial p(0)} \neq 0$. How could I obtain an expression for $\frac{\partial f}{\partial p(0)}$ in terms of $H$? Some kind of differentiation under the integral sign?

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  • $\begingroup$ Although your notation is a bit unusual (you must at least have some $t$ in $f$ and $g$), notice that the derivative of the (whole) solution with respect to the initial condition is the solution of the linear variational equation. Depending on your specific Hamiltonian this may or may not help. $\endgroup$ – John B May 13 '16 at 20:53
  • $\begingroup$ But anyways, you say that you have a $1$ degree Hamiltonian system having only periodic orbits. It shouldn't exist something like that right? $\endgroup$ – John B May 13 '16 at 20:55
  • $\begingroup$ Why do you say that $f$ and $g$ must also depend on $t$? Surely I can absorb $t$ into $q(t)$ and $p(t)$ (say by some implicit function theorem argument). Yes, it is indeed possible to have a Hamiltonian system only with periodic orbits, say if the phase space is the cyliner $\mathbb{R} \times \mathbb{S}$ @Jonas $\endgroup$ – Alex May 13 '16 at 21:22
  • $\begingroup$ You wrote $q(t)=f(q(0),p(0))$, etc. The left-hand side has $t$, the right-hand side doesn't. Sorry on the orbits, I meant "It shouldn't exist something like that right?" including the twist property. $\endgroup$ – John B May 13 '16 at 21:32
  • $\begingroup$ Crossposted to physics.stackexchange.com/q/255713/2451 $\endgroup$ – Qmechanic May 13 '16 at 21:48

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