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Let $(M,\omega)$ be a compact symplectic manifold.

Given a Hamiltonian funciton $H:M\rightarrow \mathbb{R}$ one can defined its Hamiltonian flow $\Phi^H_t$ and we will have that given $p\in M$ then $\phi_t^H(p)\subset H^{-1}(p)$ since $\frac{d}{dt}H(\phi_t^H)=dH(X_H)=0$.

Now I am wondering what happens in the case that $H$ is a time-dependent Hamiltonian. That is suppose we have $H_t:M\rightarrow \mathbb{R}$, we can still define its Hamiltonian flow $\phi_t^{H_t}$. My question is what happens now to the level sets under the Hamiltonian flow. That is suppose that we have $c$ a regular value for all $t$ would we have that the flow $\phi_t^{H}$ takes $H_t^{-1}(c)$ into $H_{t'}^{-1}(c)$ for $t'\neq t$? Or is there another phenomen happening? I tried to prove this basically doing the same thing as in the previous case but here we don't have that $\frac{d}{dt}H_t(\phi_t^{H_t}(p))=0$, at least I don't think we do, hence I am not sure what to do. Any insight is appreciated, thanks in advance.

Someone
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  • The time derivative is not zero in general: with the same computations as in the autonomous case, you get $d/dt (H_t(\phi_t))=(\partial_tH_t)(\phi_t)$, so depending on how $H_t$ grows or decreases over time, your energy will change over time. I don't know any other interesting insights though, it's a nice question. I also want to remark that if $H$ depends on $t$, the flow should have two parameters $t,t_0$, which tell the final time and the initial time. So you should write something like $\phi_{t,t_0}$. – Lorenzo Pompili Nov 30 '23 at 22:11

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