Smooth state flow for continuous-time smooth-state force fields

If $a=b$, the only time is the prescribed initial time and the solution map is the identity, so the claims hold. Otherwise lift $G$ periodically to $\tilde{G}:J\times {\mathbb{R}}^d\to {\mathbb{R}}^d$ and write $z=(x,v)$ and $B(t,z)=(v,\tilde{G}(t,x))$. On every compact time interval and bounded state ball, $B$ is continuous, bounded and Lipschitz in state because $D_{x}G$ is bounded. Thus [F1] gives unique local solutions. Along any such solution, if $M:={\sup }_{I\times {\mathbb{T}}^d}\mathrm{\mid }G\mathrm{\mid }$, then $\mathrm{\mid }V(t)-v_0\mathrm{\mid }\le M\mathrm{\mid }t-a\mathrm{\mid }$ and $\mathrm{\mid }X(t)-x_0-(t-a)v_0\mathrm{\mid }\le M\mathrm{\mid }t-a{\mathrm{\mid }}^2\mathrm{/}2$ for lifted positions. Hence a solution on a finite subinterval of $I$ remains in one bounded state cylinder. On a slightly larger cylinder the field has one finite bound and one finite state-Lipschitz constant, so the quantitative [F1] time estimate gives a common positive continuation length for every point of the trajectory. Finitely many such equal-length continuations cover $I$. Uniqueness on overlaps from [F1] gives one solution on all of $I$.

Fix a compact subinterval and a base state $z_0$. The bound on $D_{z}B$ and [F2] give $\mathrm{\mid }z(t,z_0+h)-z(t,z_0)\mathrm{\mid }\le C\mathrm{\mid }h\mathrm{\mid }$ uniformly in $t$ for small $h$. In the integral equations, set ${\mathrm{\Delta }}_h(t):=z(t,z_0+h)-z(t,z_0)$ and $A_h(t):={\int }_0^1{D}_{z}B(t,z(t,z_0)+\theta {\mathrm{\Delta }}_h(t))d\theta \mathrm{.}$ Then ${\mathrm{\Delta }}_h(t)=h+{\int }_a^t{A}_h(s){\mathrm{\Delta }}_h(s)ds$. Uniform continuity of $D_{z}B$ on the compact time interval and torus gives $A_h\to A:=D_{z}B(t,z(t,z_0))$ uniformly in $t$. The linear integral equation $J(t)=I+{\int }_a^{t}A(s)J(s)ds$ has a unique solution by the same Picard iteration as [F1]. Subtracting it from the equation for ${\mathrm{\Delta }}_h$, applying [F2], and using ${\sup }_t\mathrm{\parallel }{A}_h-A\mathrm{\parallel }\to 0$ proves ${\sup }_t\mathrm{\mid }{\mathrm{\Delta }}_h(t)-J(t)h\mathrm{\mid }\mathrm{/}\mathrm{\mid }h\mathrm{\mid }\to 0$. Thus $D_{z_0}z(t,z_0)=J(t)$ and it satisfies $\dot{J}=D_{z}B(t,z(t,z_0))J$, $J(a)=I$. Since $D_{z}B$ is bounded uniformly in state, [F2] also gives ${\sup }_{t\in I,z_0}\mathrm{\parallel }J(t,z_0)\mathrm{\parallel }<\mathrm{\infty }$.

For each $r\ge 2$, the candidate $r$-th state derivative $J_r$ satisfies the linear variational equation ${\dot{J}}_r=D_{z}B(t,z)J_r+{\sum }_{\begin{array}{c}\pi \text{ a partition of }\{1,\dots ,r\}\\ \mathrm{\mid }\pi \mathrm{\mid }\ge 2\end{array}}{D}_z^{\mathrm{\mid }\pi \mathrm{\mid }}B(t,z)[J_{\mathrm{\mid }{C}_1\mathrm{\mid }},\dots ,J_{\mathrm{\mid }{C}_{\mathrm{\mid }\pi \mathrm{\mid }}\mathrm{\mid }}],J_r(a)=0,$ where $C_1,\dots ,C_{\mathrm{\mid }\pi \mathrm{\mid }}$ are the blocks of $\pi$ and $z=z(t,z_0)$. The coefficient and all higher derivatives of $B$ are bounded uniformly in state. By induction, the forcing term is therefore bounded uniformly over initial states; [F2] gives a unique $J_r$ with a uniform finite-time bound. Applying the difference-quotient argument of step 1.2 to the equation for $J_{r-1}$, using uniform continuity of each $D_z^{j}B$ on the compact time interval and periodic state variable for $j\le r$, shows that this $J_r$ is its derivative. Induction proves smooth dependence and the asserted uniform bounds. Solving the same system backward in time gives the inverse map, and uniqueness makes the two maps inverse. If $B$ is also smooth in time, ${\mathrm{\partial }}_{t}z=B(t,z)$ and the state-jet equations give every mixed time-state derivative by repeated chain rule.