A complete Hamiltonian flow for a C2 potential preserves energy and phase volume

By [F7], $M=L{L}^T$ where $L$ is lower triangular with strictly positive diagonal. Thus $\det L$ is the positive product of its diagonal entries, and $\det M=\det L\det (L^T)=(\det L{)}^2>0$; [F7] implies that $M$ is invertible, so $M^{-1}$ is well-defined and symmetric. Since $U\in C^2$, [F9] makes $F$ a $C^1$ vector field. By [F1] it has a unique maximal $C^1$ flow, and completeness as defined in the Statement makes every time slice ${\mathrm{\Phi }}_t$ global with inverse ${\mathrm{\Phi }}_{-t}$. Differentiating ${\mathrm{\Phi }}_{-t}\circ {\mathrm{\Phi }}_t=I$ by [F2] shows $D{\mathrm{\Phi }}_t(z)$ is invertible. The derivative $A(t):=D{\mathrm{\Phi }}_t(z)$ satisfies $A^{\mathrm{\prime }}(t)=DF({\mathrm{\Phi }}_t(z))A(t)$ by [F1].

Along a solution $z(t)=(q(t),p(t))$, [F2] and the displayed Hamilton equations give $\frac{d}{dt}H(q(t),p(t))=\mathrm{\nabla }U(q(t){)}^T{M}^{-1}p(t)+(M^{-1}p(t){)}^T(-\mathrm{\nabla }U(q(t)))=0,$ where symmetry of $M^{-1}$ identifies the derivative of the kinetic energy. Hence $H({\mathrm{\Phi }}_t(z))=H(z)$.

Suppose $\mathrm{\nabla }U$ is globally $L$-Lipschitz. For $z=(q,p)$ and $w=(q^{\mathrm{\prime }},p^{\mathrm{\prime }})$, [F6] gives $\mathrm{\parallel }F(z)-F(w){\mathrm{\parallel }}_2^2\le \mathrm{\parallel }{M}^{-1}{\mathrm{\parallel }}_F^2\mathrm{\parallel }p-p^{\mathrm{\prime }}{\mathrm{\parallel }}_2^2+L^2\mathrm{\parallel }q-q^{\mathrm{\prime }}{\mathrm{\parallel }}_2^2\le C^2\mathrm{\parallel }z-w{\mathrm{\parallel }}_2^2\mathrm{.}$ Thus the time-independent field on $\mathbb{R}\times {\mathbb{R}}^{2d}$ has global state-Lipschitz constant $C$.

By [F9], the variational matrix is $DF(q,p)=\left(\begin{array}{cc}0& M^{-1}\\ -{\mathrm{\nabla }}^{2}U(q)& 0\end{array}\right),$ whose trace is zero. Since $A(t)$ is invertible by step 1.1, [F3] and the finite-sum trace identity $\mathrm{tr}(XY)={\sum }_{i,j}{X}_{ij}{Y}_{ji}=\mathrm{tr}(YX)$ give $\frac{d}{dt}\det A(t)=\det A(t)\mathrm{tr}(A(t{)}^{-1}DF({\mathrm{\Phi }}_t(z))A(t))=\det A(t)\mathrm{tr}(DF({\mathrm{\Phi }}_t(z)))=0.$ Since $A(0)=I_{2d}$, $\det A(0)=1$, so $\det D{\mathrm{\Phi }}_t(z)=1$ for all $t$.

Apply [F4] to obtain completeness and uniqueness in this globally Lipschitz case, and apply [F5] with the same $C$ to obtain the stated Gronwall stability estimate. This supplies a sufficient condition; no completeness is inferred from merely local Lipschitz continuity of $\mathrm{\nabla }U$.

For each $t\in \mathbb{R}$, step 1.1 makes ${\mathrm{\Phi }}_t:{\mathbb{R}}^{2d}\to {\mathbb{R}}^{2d}$ a global $C^1$ diffeomorphism. For any Lebesgue-measurable $E\subseteq {\mathbb{R}}^{2d}$, apply [F8] to ${\mathbf{1}}_E$ and use step 2.1 to get ${\lambda }_{2d}({\mathrm{\Phi }}_t^{-1}(E))={\int }_{{\mathbb{R}}^{2d}}{\mathbf{1}}_E({\mathrm{\Phi }}_t(z))\mathrm{\mid }\det D{\mathrm{\Phi }}_t(z)\mathrm{\mid }dz={\lambda }_{2d}(E)$. Hence the exact flow preserves phase-space Lebesgue volume.