Complete global bounded-Lipschitz mean-field characteristic evolution

Choose $h>0$ with $A{h}^2<1$; when $A=0$ any finite $h$ is allowed. For a bounded continuous displacement $u(s,x,v)$ in ${\mathbb{R}}^d$ on $[0,h]\times Z$, periodic in $x$ and with $u(0,\cdot )=0$, set $(Tu)(s,x,v):={\int }_0^s(s-r){\int }_{Z}k(x+rv+u(r,x,v)-y-rw-u(r,y,w))d{\mu }_0(y,w)dr\mathrm{.}$ with torus arguments understood modulo ${\mathbb{Z}}^d$. On the closed set $\sup \mathrm{\mid }u\mathrm{\mid }\le B{h}^2\mathrm{/}2$, the map has the same bound because $\mathrm{\mid }k\mathrm{\mid }\le B$. For every convergent sequence of $(s,x,v)$, the integrand is pointwise continuous in those variables and bounded by $B$ on $(Z,\mathcal{B}(Z),{\mu }_0)$; [F3] therefore gives joint continuity of $Tu$. Uniform limits preserve continuity, periodicity and the initial condition, so this closed function space is complete in the supremum norm.

For any two displacements $u,\tilde{u}$, the force arguments change by at most $2\sup \mathrm{\mid }u-\tilde{u}\mathrm{\mid }$. The Lipschitz bound $A$ and ${\int }_0^s(s-r)dr=s^2\mathrm{/}2$ give $\sup \mathrm{\mid }Tu-T\tilde{u}\mathrm{\mid }\le A{h}^2\sup \mathrm{\mid }u-\tilde{u}\mathrm{\mid }$. Starting with $u_0=0$ and iterating $u_{n+1}=T{u}_n$, successive differences are bounded by a geometric sequence with ratio $A{h}^2<1$; hence $(u_n)$ is uniformly Cauchy. Its limit $u$ belongs to the closed complete set, and the Lipschitz estimate gives $Tu=u$. The same estimate makes this fixed point unique.

Put $X_s(x,v)=x+sv+u(s,x,v)$ modulo the torus and $V_s(x,v)=v+{\mathrm{\partial }}_{s}u(s,x,v)$. The Volterra formula and continuity of the integrand imply ${\mathrm{\partial }}_s^2{X}_s={\int }_{Z}k(X_s(x,v)-X_s(y,w))d{\mu }_0(y,w)$, with $X_0=x$ and ${\dot{X}}_0=v$; thus $(X_s,V_s)$ solves the displayed characteristic system. The bound $\mathrm{\mid }{V}_s-v\mathrm{\mid }\le B\mathrm{\mid }s\mathrm{\mid }$ gives a finite first moment for every finite $s$. For bounded continuous $\psi$, the integrand $\psi (X_s,V_s)$ is pointwise continuous in $s$ and bounded by $\mathrm{\parallel }\psi {\mathrm{\parallel }}_{\mathrm{\infty }}$, so [F3] gives narrow continuity of ${\mu }_s$.

The force $F_{{\mu }_s}(x)$ is continuous in $(s,x)$: for $(s_n,x_n)\to (s,x)$, its integrand converges pointwise and is bounded by $B$, so [F3] applies on $(Z,{\mu }_0)$. Along each characteristic, the chain rule gives $\frac{d}{ds}\phi (s,X_s,V_s)=({\mathrm{\partial }}_s\phi +V_s\cdot {\mathrm{\nabla }}_x\phi +F_{{\mu }_s}(X_s)\cdot {\mathrm{\nabla }}_v\phi )(s,X_s,V_s)\mathrm{.}$ On any bounded time interval the right side is bounded by $C(1+\mathrm{\mid }v\mathrm{\mid })$, which is integrable by the assumed first moment. Difference quotients have the same common bound; [F3] therefore permits differentiation under the ${\mu }_0$ integral and gives exactly the weak identity in the Statement.

The fixed-point estimate and force bounds do not depend on the starting time or probability. Repeating the construction on successive intervals of length $h$ therefore gives a unique characteristic solution on every bounded positive interval. For negative times let $R(x,v)=(x,-v)$ and solve the same positive-time problem from $R_{\mathrm{\#}}{\mu }_0$. If its characteristics are $({\tilde{X}}_r,{\tilde{V}}_r)$, define, for $s\le 0$, $X_s(x,v)={\tilde{X}}_{-s}(x,-v)$ and $V_s(x,v)=-{\tilde{V}}_{-s}(x,-v)$. Then $X_0=x$, $V_0=v$, ${\dot{X}}_s=V_s$, and ${\dot{V}}_s=F_{{\mu }_s}(X_s)$: the transformed law has the same position marginal as ${\mu }_s$, and the force depends only on position. Thus these are characteristics from ${\mu }_0$ for negative times. Uniqueness on overlapping intervals follows from step 2.1 and continuation, so the positive and negative solutions form one global family.

Suppose ${\mu }_0$ has the smooth density and all moments in the Statement. Its characteristic field is $b(s,x,v)=(v,F_{{\mu }_s}(x))$. Every state derivative of $F_{{\mu }_s}$ is bounded on a finite time slab because it is an integral of a bounded derivative of $k$. Step 4.1 makes $b$ continuous in time; [F2] gives a smooth state flow $S_{s,0}$ and its smooth backward inverse. Its state derivative $J_s=D{S}_{s,0}$ solves ${\dot{J}}_s=Db(s,S_{s,0})J_s$, $J_0=I$, by [F2]. The divergence of $b$ is zero: ${\mathrm{\nabla }}_x\cdot v=0$ and $F_{{\mu }_s}$ is independent of $v$. Since the inverse flow makes $J_s$ invertible, multilinear differentiation of its determinant gives $\frac{d}{ds}\det J_s=\det J_s\mathrm{tr}(J_s^{-1}Db{J}_s)=\det J_s\mathrm{tr}(Db)=0$; with $\det J_0=1$ this yields $\det J_s=1$. Define ${\rho }_s={\rho }_0\circ S_{0,s}$ relative to $dxdv$. It is smooth in state, and the chain rule gives the classical transport equation.

Orient $Z$ by its standard translation-chart volume form $\mathrm{v}\mathrm{o}\mathrm{l}=d{x}^1\wedge \cdots \wedge d{x}^d\wedge d{v}^1\wedge \cdots \wedge d{v}^d$. By [F9] and [F10], its associated volume measure is the product $dxdv$ specified in the statement. Since $\det J_s=1$, $S_{s,0}$ preserves this orientation. For $g\in C_c^{\mathrm{\infty }}(Z)$ the top form $g{\rho }_s\mathrm{v}\mathrm{o}\mathrm{l}$ has compact support, so [F4] and step 5.1 give ${\int }_{Z}g(z){\rho }_s(z)dxdv={\int }_{Z}g(S_{s,0}z){\rho }_s(S_{s,0}z)dxdv={\int }_{Z}g(S_{s,0}z){\rho }_0(z)dxdv\mathrm{.}$ Take the cutoffs ${\chi }_R(v)$ of [F5] for positive integers $R$. The identity with $g={\chi }_R$ gives ${\int }_{\mathrm{\mid }v\mathrm{\mid }\le R}{\rho }_{s}dxdv\le {\int }_Z{\chi }_R{\rho }_{s}dxdv={\int }_Z{\chi }_R(V_s(z)){\rho }_0(z)dxdv\le 1$, since ${\rho }_s\ge 0$ and $0\le {\chi }_R\le 1$. The indicators of the balls $\{\mathrm{\mid }v\mathrm{\mid }\le R\}$ increase to $1$, so [F6] implies ${\int }_Z{\rho }_{s}dxdv\le 1$. Now ${\rho }_s$ is integrable and ${\chi }_R(v)\to 1$ pointwise. Apply [F3] to both sides of the cutoff identity, with majorants ${\rho }_s$ and ${\rho }_0$, to obtain ${\int }_Z{\rho }_{s}dxdv={\int }_Z{\rho }_{0}dxdv=1$. Thus ${\rho }_{s}dxdv$ is a probability. The identity also says ${\rho }_{s}dxdv$ and the characteristic law ${\mu }_s$ agree on all smooth compactly supported tests. Apply [F7] to conclude ${\rho }_{s}dxdv={\mu }_s$, so ${\rho }_s$ is the density of ${\mu }_s$.

For every time-derivative order $r$, differentiating $F_{{\mu }_s}(x)=\int k(x-X_s(z)){\rho }_0(z)dz$ gives finite sums of bounded derivatives of $k$ times products of characteristic time derivatives. The base bounds $\mathrm{\mid }{V}_s-v\mathrm{\mid }\le B\mathrm{\mid }s\mathrm{\mid }$ and $\mathrm{\mid }{\dot{V}}_s\mathrm{\mid }\le B$, followed by induction in the differentiated characteristic equation, bound each order-$r$ integrand by $C(r,T)(1+\mathrm{\mid }v{\mathrm{\mid }}^r)$ on $\mathrm{\mid }s\mathrm{\mid }\le T$. Every such envelope is integrable by the all-moment assumption. Dominated convergence [F3] therefore justifies each time differentiation under the integral, so $F_{{\mu }_s}$ is jointly smooth in $(s,x)$. The joint-smooth-time clause of [F2] then makes $S_{s,0}$ and ${\rho }_s$ jointly smooth, proving the classical branch.