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Let $(M,\omega)$ be a symplectic manifold, and $H : M \times [0,1] \to \mathbb{R}$ be a smooth time-dependent Hamiltonian on $M$. Then non degeneracy of $\omega$ implies the existence of a time-dependent Hamiltonian vector field $X_H^t$, uniquely defined by $$ \omega(X_H^t,.) = dH(.,t). $$ Consider the differential equation $$ \dot{x}(t) = X_H^t(x(t)). $$ If I am not mistaken, for any given $x \in M$, there exists a unique solution with initial condition $x(0) = x$ defined on an interval containing $0$ (which depends on $x$), provided that $H(.,t)$ has Lipschitz derivatives. This way, we obtain a one-parameter family $t \in I \to \phi_H^t(x) \in M$, where $I$ is the maximal interval of existence, that is the intersection of all intervals containing $0$ on which the solutions are defined. The family $\phi_H^t$ is called the Hamiltonian flow generated by $H$.

My question regarding the above is the following:

  • In the literature, one never talks about the Lipschitz condition on $H(.,t)$, even when $M$ is compact, and moreover the time-$1$ maps $\phi_H^1$ (the Hamiltonian diffeomorphisms) are always considered without specifying if they even exist.

Is there a particular reason why the Lipschitz condition would not be necessary in this situation, and why the time-$1$ maps would always be defined ?

Thanks!

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  • $\begingroup$ What do they mean by smooth in the literature? And, on a compact manifold any (sufficiently regular) ODE has solutions that can be extended to the whole time component of their domain. $\endgroup$
    – user539887
    Commented Apr 29, 2019 at 11:56
  • $\begingroup$ Smooth means infinitely differentiable. In the compact case I know that there are no problems, my questions regard the non-compact case. $\endgroup$
    – BrianT
    Commented Apr 29, 2019 at 12:02
  • $\begingroup$ So, everything is OK.: $\mathrm{d}H$ is $C^{\infty}$, too, hence locally Lipschitz, so the Picard–Lindelöf theorem applies. Incidentally, $H$ being continuous and having the space derivatives up to second order continuous does suffice. $\endgroup$
    – user539887
    Commented Apr 29, 2019 at 12:06
  • $\begingroup$ Don’t we need $dH$ to be uniformly Lipschitz ? Why does locally suffices here ? Moreover, what can we say about the interval of existence of the flow ? $\endgroup$
    – BrianT
    Commented Apr 29, 2019 at 12:40
  • $\begingroup$ Because the existence and uniqueness of a solution to an IVP is a local property. A general feature is that the right-endpoint of the interval of existence is lower semicontinuous. To get the existence for all time values the global Lipschitz property suffices. $\endgroup$
    – user539887
    Commented Apr 29, 2019 at 18:02

1 Answer 1

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As you noted, the compact case, which is the typical case, has no problems. See https://unapologetic.wordpress.com/2011/06/01/vector-fields-on-compact-manifolds-are-complete/.

For the non-compact case, there clearly needs to be conditions on $H$. If $$H(x,p) = x^\alpha - p^\alpha,$$ then on the $x = p$ line we have $$ \dot{x} \propto x^{\alpha-1}$$ so for $\alpha > 2$, we have finite time blowup.

Also, see 10.4. https://drive.google.com/file/d/1USe9SS33q1L9k3S3xsgR-oy_9VWhEz5r/view; for Floer theory on $T^*M$, you consider a "Lagrangian" $L$ defined on the tangent bundle, and you restrict to Hamiltonians that are dual to $L$ that are "Tonelli Lagrangians", a condition that explicitly stipulates that the flow exists for all $t$. You also further stipulate that $H$ grows quadratically in the $q$ variables in some sense.

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  • $\begingroup$ Notice that the OP asked about a time-dependent Hamiltonian. The linked proof is only for the autonomous case (although it can be easily modified to the time-dependent case). $\endgroup$
    – user539887
    Commented Apr 29, 2019 at 19:06
  • $\begingroup$ Yes consider the flow on $M \times [0,1]$ for $(X_H, 1)$. $\endgroup$ Commented Apr 29, 2019 at 19:15

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