= Solution
Write $z=(x,v)$ and let $\Phi(t,s,z)$ denote the <Hamiltonian flow> from time $s$ to time $t$. The <characteristic curves> satisfy <Hamilton's equations>:
$$
\boxed{\dot X_i(t)=\partial_{v_i}H(t,X(t),V(t)),\qquad
\dot V_i(t)=-\partial_{x_i}H(t,X(t),V(t)),\qquad
(X(s),V(s))=(x,v).}
$$
The <Hamiltonian Liouville equation> then reduces along each curve to
$$
\frac d{dt}f(t,X(t),V(t))=h(t,X(t),V(t)).
$$
The signs and derivative variables here are those in the PDF.
The <global characteristic flow for a Hamiltonian with bounded Hessian> follows, for example, from $H\in C^2(\mathbb R\times\mathbb R^{2d})$ and, for every finite $T$,
$$
\sup_{|t|\leq T,\ z\in\mathbb R^{2d}}\|D_z^2H(t,z)\|\leq L_T<\infty,\qquad
\sup_{|t|\leq T}|\nabla_zH(t,0)|\leq A_T<\infty.
$$
Thus the <Hamiltonian vector field> $b=(\nabla_vH,-\nabla_xH)$ is globally <Lipschitz continuous> in $z$ on each finite time interval and satisfies $|b(t,z)|\leq A_T+L_T|z|$. The <Picard-Lindelof theorem> gives local existence and uniqueness, while the <Gronwall inequality> gives, for example,
$$
|\Phi(t,s,z)|\leq (|z|+A_T|t-s|)e^{L_T|t-s|}
\qquad(|s|,|t|\leq T).
$$
This excludes finite-time escape. \b[There is a unique <Hamiltonian flow> for all finite forward and backward times], and $\Phi(s,t)$ is the inverse of $\Phi(t,s)$. These sufficient conditions are deliberately stronger than necessary.
Back to article page
