Smooth periodic finite-N and scaled mean-field phase data

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.

Lift the torus positions periodically to ${\mathbb{R}}^{dN}$. The force bound $K_{\ast }:=\mathrm{\parallel }K{\mathrm{\parallel }}_{\mathrm{\infty }}$ gives $\mathrm{\mid }{p}_i(t)-p_i(0)\mathrm{\mid }\le K_{\ast }\mathrm{\mid }t\mathrm{\mid }$ and $\mathrm{\mid }{q}_i(t)-q_i(0)-(t\mathrm{/}m)p_i(0)\mathrm{\mid }\le K_{\ast }{t}^2\mathrm{/}(2m)$ for each particle. Thus on every finite time interval the lifted trajectory stays in a bounded state cylinder. The lifted vector field from [F1] is smooth with bounded first derivatives there, so [F3] gives a common local continuation interval; repeating this fixed extension across the finite interval gives a global lifted solution, and periodicity makes it a well-defined torus solution. Reversing time gives existence on negative intervals. Local smooth dependence from [F3], composed over finitely many local intervals, makes every global finite-time map smooth; uniqueness makes ${\mathrm{\Phi }}_{-t}^N$ its inverse.

With the symplectic form in [F1] and Hamiltonian sign convention in [F2], the Hamiltonian in [F1] defines the canonical Hamiltonian vector field on this cotangent phase space. Apply [F4] on each local time interval of step 1.1 and compose the volume-preserving local maps; this gives $({\mathrm{\Phi }}_t^N{)}^{\ast }{\mathrm{\Omega }}_N={\mathrm{\Omega }}_N$ for every real $t$. The Hamiltonian vector field is equivariant under every permutation $\sigma$ of particle labels because the kinetic sum and even pair-potential sum are invariant. Hence $\sigma {\mathrm{\Phi }}_t^N(z)$ and ${\mathrm{\Phi }}_t^N(\sigma z)$ solve the same initial-value problem; uniqueness from [F3] gives equality.

Along characteristics $z(t)={\mathrm{\Phi }}_t^N(z_0)$, the derivative of $F_N(t,z(t))=F_N^0(z_0)$ is zero. The chain rule therefore gives ${\mathrm{\partial }}_t{F}_N+X_{H_N}\cdot \mathrm{\nabla }{F}_N=0$. Substituting the Hamilton equations from [F1] and grouping the two terms for each pair $i<j$, using $K(q_j-q_i)=-K(q_i-q_j)$, yields ${\mathrm{\partial }}_t{F}_N+{\sum }_{i=1}^N\frac{p_i}{m}\cdot {\mathrm{\nabla }}_{q_i}{F}_N+\frac{1}{N}{\sum }_{1\le i<j\le N}K(q_i-q_j)\cdot ({\mathrm{\nabla }}_{p_i}-{\mathrm{\nabla }}_{p_j})F_N=0.$ The rapid tails justify the displayed derivatives as classical ones on every finite time slab.

On each finite time slab the rapid bounds on all phase derivatives make ${\mathrm{\partial }}_t{F}_N$ integrable and supply one integrable majorant for its time difference quotients; [F6] therefore permits differentiation under the $d{\mathrm{\Omega }}_N$ integral. Integrate the equation in step 3.1 over all phase variables. The $q_i$-derivative terms have zero integral by periodicity. For each force term, $K(q_i-q_j)$ is independent of momenta, so integration by parts in $p_i$ and $p_j$ has zero boundary terms by the rapid tails. Fubini [F5] justifies these coordinate integrations and their order. Hence $\frac{d}{dt}\int F_N(t)d{\mathrm{\Omega }}_N=0$; its value is $1$ at $t=0$ by [A2], so $F_N(t)$ remains normalized. The same permutation identity from step 2.1 and [A2] give $F_N(t,\sigma z)=F_N(t,z)$. Finally, the vector field derivatives of every order are bounded globally: its first derivative has entries $1\mathrm{/}m$ and $DK$, and higher derivatives are bounded derivatives of the periodic force. The variational equations and [F3] bound every finite state derivative of ${\mathrm{\Phi }}_{-t}^N$ on the time slab. Since $\mathrm{\mid }{p}_i(t)-p_i(0)\mathrm{\mid }\le K_{\ast }\mathrm{\mid }t\mathrm{\mid }$, the momentum norms differ by at most $\sqrt{N}{K}_{\ast }\mathrm{\mid }t\mathrm{\mid }$; the chain rule transfers every rapid-decay bound of $F_N^0$ to $F_N(t)$ uniformly on the slab.

Use the Hamilton equations and odd force $K$ from [F3], with the classical Liouville solution from [F1] under the countable-choice density measure assumption [A1]. Substituting these equations into ${\mathrm{\partial }}_t{F}_N+{\sum }_i({\dot{q}}_i\cdot {\mathrm{\nabla }}_{q_i}{F}_N+{\dot{p}}_i\cdot {\mathrm{\nabla }}_{p_i}{F}_N)=0$ and grouping the two terms for each unordered pair $i<j$ yields ${\mathrm{\partial }}_t{F}_N+{\sum }_{i=1}^N\frac{p_i}{m}\cdot {\mathrm{\nabla }}_{q_i}{F}_N+\frac{1}{N}{\sum }_{1\le i<j\le N}K(q_i-q_j)\cdot ({\mathrm{\nabla }}_{p_i}-{\mathrm{\nabla }}_{p_j})F_N=0.$ The sign in the second paired derivative uses $K(q_j-q_i)=-K(q_i-q_j)$.

Fix $k<N$ and integrate step 1.1 over $z_{k+1},\dots ,z_N$. The integrations and time differentiation are legitimate by the all-phase-derivative rapid tails (the time derivative from step 1.1 adds at most one factor of momentum), compact periodic position variables, and [A1, F2]. A position derivative in an integrated variable has zero integral by periodicity; a momentum derivative in an integrated variable has zero boundary term by the stated decay. Thus one-body terms for $i\le k$ and interaction pairs entirely inside $\{1,\dots ,k\}$ give the first three terms of the target equation. Pairs with both labels greater than $k$ integrate to zero by the same momentum integration-by-parts argument.

For a cross pair $i\le k<j$, the term containing ${\mathrm{\nabla }}_{p_j}{F}_N$ integrates to zero because $K(q_i-q_j)$ is independent of $p_j$ and the $p_j$-tail vanishes at infinity. The term containing ${\mathrm{\nabla }}_{p_i}{F}_N$ remains. By permutation symmetry of $F_N$, for fixed $i\le k$ every label $j>k$ has the same integrated contribution, equal to ${\mathrm{\nabla }}_{p_i}\cdot \int K(q_i-q_{k+1})f_{k+1}(t,z_1,\dots ,z_k,z_{k+1})d{z}_{k+1}\mathrm{.}$ There are exactly $N-k$ such labels, each carrying coefficient $1\mathrm{/}N$. This gives the factor $(N-k)\mathrm{/}N$ and all terms of the displayed hierarchy.

If $k=N$, there are no cross pairs, so the same integration argument leaves only the one-body transport and internal-pair terms; the formula's cross coefficient is $(N-N)\mathrm{/}N=0$. Thus the statement holds for every $1\le k\le N$, with no factorization assumption at any step.

Fix a bounded Lipschitz function $f:Z\to \mathbb{R}$. Use ${\lambda }_d$ and $m_{{\mathbb{T}}^d}$ from [F5]. Choose the smooth cutoff $\chi$ of [F2] and set $\eta (u)=c\chi (2u)$ on ${\mathbb{R}}^d$, where $c=({\int }_{{\mathbb{R}}^d}\chi (2u)d{\lambda }_d(u){)}^{-1}$; its integral is positive and finite because $\chi =1$ on the unit ball and is supported in the radius-two ball. Thus $\eta$ is a nonnegative smooth probability density supported in the unit ball. For $0<ϵ<1\mathrm{/}4$, define ${\eta }_{ϵ}^{\mathbb{R}}(v)={ϵ}^{-d}\eta (v\mathrm{/}ϵ)$ and periodize it to ${\mathbb{T}}^d$ by ${\eta }_{ϵ}^{\mathbb{T}}(x)={\sum }_{n\in {\mathbb{Z}}^d}{ϵ}^{-d}\eta ((\tilde{x}+n)\mathrm{/}ϵ)$, where $\tilde{x}$ is any representative. The sum is locally finite and independent of the representative; the fundamental-domain formula for $m_{{\mathbb{T}}^d}$ in [F5], followed by the change of variables $u=(\tilde{x}+n)\mathrm{/}ϵ$, gives integral one, and its support satisfies $d_{{\mathbb{T}}^d}(x,0)\le ϵ$. Set $f_{ϵ}(x,v)={\int }_{{\mathbb{T}}^d}{\int }_{{\mathbb{R}}^d}{\eta }_{ϵ}^{\mathbb{T}}(x-y){\eta }_{ϵ}^{\mathbb{R}}(v-w)f(y,w)d{\lambda }_d(w)d{m}_{{\mathbb{T}}^d}(y)$. Differentiating the compactly supported smooth kernels under the integral shows that $f_{ϵ}$ is smooth, and their nonnegativity and unit integrals give $\mathrm{\parallel }{f}_{ϵ}{\mathrm{\parallel }}_{\mathrm{\infty }}\le \mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$. On their support, $d_{{\mathbb{T}}^d}(x,y)\le ϵ$ and $\mathrm{\mid }v-w\mathrm{\mid }\le ϵ$, so $\mathrm{\parallel }{f}_{ϵ}-f{\mathrm{\parallel }}_{\mathrm{\infty }}\le 2ϵ\mathrm{Lip}(f)$. For each $R>0$, ${\chi }_R(v)f_{ϵ}(x,v)$ is in $C_c^{\mathrm{\infty }}(Z)$, so its integrals agree by the hypothesis. Let $R\to \mathrm{\infty }$ and apply [F3] to each probability measure; then let $ϵ\downarrow 0$ and use uniform convergence. Thus $\mu$ and $\nu$ have equal integrals for every bounded Lipschitz $f$.

If $C\subseteq Z$ is a nonempty closed set, distance to $C$ is 1-Lipschitz by the triangle inequality. Hence $h_n(z)=\max (0,1-n{d}_Z(z,C))$ is bounded Lipschitz and decreases pointwise to ${\mathbf{1}}_C$. Step 1.1 gives equal integrals of $h_n$, and [F3] gives $\mu (C)=\nu (C)$. The empty closed set has mass zero for both. Let $\mathcal{D}:=\{E\in \mathcal{B}(Z):\mu (E)=\nu (E)\}$. Since $\mu$ and $\nu$ are probabilities, $Z\in \mathcal{D}$; equality of total mass makes $\mathcal{D}$ closed under complements in $Z$, and countable additivity makes it closed under countable disjoint unions. Thus $\mathcal{D}$ is a lambda-system. The $d_Z$ sum metric induces the product topology on $Z$, so its closed sets form a pi-system containing $Z$ and generating $\mathcal{B}(Z)$. Apply [F4] to this pi-system and $\mathcal{D}$ to get $\mu =\nu$.

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.