= Solution
\b[A finite first moment suffices for the velocity, but may not suffice for the printed potential.] For example, $W(z)=|z|^2/2$ has all the stated properties. If $\mu$ is the law of $(S,0)$ with density $2s^{-3}$ on $s\geq1$, then $\mathbb ES=2$ but $\mathbb ES^2=\infty$, and $(W*\mu)(x)=\infty$ for every $x$. Thus the literal definition of $V$ needs a finite second <absolute moment>, or another hypothesis ensuring integrability of $W$.
For the specified class $\mathcal P_1$, define the <velocity fields> directly by $b_\mu=K*\mu$ and $b_{\widehat\mu}=K*\widehat\mu$. Equivalently, use the <renormalized interaction potential>
$$
V_{\mathrm{ren}}(t,x)=\int\left[W(x-z)-W(-z)\right]d\mu_t(z).
$$
The bounded <Hessian matrix> and $\nabla W(0)=0$ imply
$$
|W(x-z)-W(-z)|\leq L|x||z|+\frac L2|x|^2.
$$
Thus $V_{\mathrm{ren}}$ is finite, is twice continuously differentiable in space, and has $\nabla V_{\mathrm{ren}}=\nabla W*\mu$, by dominated <differentiation under the integral sign>. Its rotated negative <gradient> is exactly $b_\mu$. Whenever $W*\mu$ is finite, the two potentials differ only by an $x$-independent constant and give the same <transport equation>.
We check the hypotheses for <characteristic representation of a measure-valued continuity equation>. For every $x,y$,
$$
|b_\mu(t,x)-b_\mu(t,y)|\leq L|x-y|,\qquad |b_\mu(t,x)|\leq L\left(|x|+m_1(\mu_t)\right),\qquad m_1(\mu_t)=\int|z|\,d\mu_t(z).
$$
For any <coupling> $\gamma\in\Gamma(\mu_t,\widehat\mu_t)$, the <Lipschitz condition> on $K$ also gives
$$
\begin{aligned}
|b_\mu(t,x)-b_{\widehat\mu}(t,x)|&=\left|\int[K(x-z)-K(x-\widehat z)]\,d\gamma(z,\widehat z)\right|\\
&\leq L\int|z-\widehat z|\,d\gamma(z,\widehat z).
\end{aligned}
$$
Taking the infimum over <couplings> proves the uniform bound
$$
\boxed{\sup_x|b_\mu(t,x)-b_{\widehat\mu}(t,x)|\leq Ld_1(\mu_t,\widehat\mu_t).}
$$
Applying the same estimate at two times of one curve proves time continuity of $b_\mu$, uniformly in $x$. Also $|m_1(\mu_t)-m_1(\mu_s)|\leq d_1(\mu_t,\mu_s)$, so $m_1(\mu_t)$ is bounded on compact time intervals. Spatial <partial derivatives> of $b_\mu$ are continuous and bounded by $L$, using dominated convergence in space and <weak convergence of probability measures> in time for the bounded continuous kernel $DK$. All corresponding facts hold for $\widehat\mu$.
The <characteristic representation of a measure-valued continuity equation> says that a continuous <velocity field>, continuously differentiable and globally <Lipschitz> in space and with locally uniform linear growth, has a global <characteristic flow map> $X_\mu(t;s,a)$ solving
$$
\frac d{dt}X_\mu(t;s,a)=b_\mu(t,X_\mu(t;s,a)),\qquad X_\mu(s;s,a)=a.
$$
For each $f\in\mathcal P_1$, its <pushforward measure> is the unique <weak solution> in $C([0,\infty),\mathcal P_1)$ of the associated <continuity equation>. The hypotheses have all been verified above, and the fields have zero <divergence>, so these are also the two required <transport equations>. Explicitly,
$$
\boxed{\rho_t=X_\mu(t;0,\cdot)_\#f,\qquad\widehat\rho_t=X_{\widehat\mu}(t;0,\cdot)_\#f.}
$$
For completeness, the <Gronwall inequality> gives
$$
|X_\mu(t;0,a)|\leq e^{Lt}|a|+L\int_0^te^{L(t-s)}m_1(\mu_s)\,ds,
$$
so these <pushforward measures> have finite first <absolute moments>. Integrating $|X_\mu(t;0,a)-X_\mu(s;0,a)|$ against $f$ proves continuity in the <Wasserstein distance>. The <chain rule> proves the weak identity. For uniqueness at a fixed terminal time $t$, use the backward <test function> $\varphi(s,x)=\psi(X_\mu(t;s,x))$, where $\psi$ is continuously differentiable with compact <support>. It satisfies $\partial_s\varphi+b_\mu\cdot\nabla\varphi=0$ and has compact spatial <support> uniformly for $0\leq s\leq t$, by the backward linear-growth bound. The weak identity with temporal cutoffs therefore gives $\int\psi\,d\rho_t=\int\psi(X_\mu(t;0,a))\,df(a)$, identifying every <weak solution> with the displayed <pushforward measure>.
Finally compare the two <characteristic curves> with the same initial point $a$. Put $\Delta_t=X_\mu(t;0,a)-X_{\widehat\mu}(t;0,a)$. The spatial <Lipschitz condition> and the boxed velocity estimate give
$$
\frac d{dt}|\Delta_t|\leq L|\Delta_t|+Ld_1(\mu_t,\widehat\mu_t),\qquad\Delta_0=0,
$$
for almost every time; the integral form gives the same conclusion at zeros of $\Delta$. The <Gronwall inequality> implies
$$
|\Delta_t|\leq L\int_0^te^{L(t-s)}d_1(\mu_s,\widehat\mu_s)\,ds.
$$
The <pushforward measure> $(X_\mu(t;0,\cdot),X_{\widehat\mu}(t;0,\cdot))_\#f$ is a <coupling> of $\rho_t,\widehat\rho_t$. Integrating the last bound against the <probability measure> $f$ proves
$$
\boxed{d_1(\rho_t,\widehat\rho_t)\leq L\int_0^te^{L(t-s)}d_1(\mu_s,\widehat\mu_s)\,ds.}
$$
The same constant $L$ depends only on the uniform bound for the <Hessian matrix> of $W$, and is independent of the prescribed curves and their common initial data.
Back to article page
