Probability-flow ODE has the same one-time marginals

For symmetric $a_t$, the product rule gives ${\sum }_{i,j}{\mathrm{\partial }}_i{\mathrm{\partial }}_j(a_{t,ij}{p}_t)=\mathrm{\nabla }\text{\hspace{0.17em}⁣}\cdot ((\mathrm{\nabla }\text{\hspace{0.17em}⁣}\cdot a_t)p_t+a_t\mathrm{\nabla }{p}_t)$, since ${\sum }_j{\mathrm{\partial }}_j(a_{t,ij}{p}_t)=(\mathrm{\nabla }\text{\hspace{0.17em}⁣}\cdot a_t{)}_i{p}_t+{\sum }_j{a}_{t,ij}{\mathrm{\partial }}_j{p}_t$. By [F1], $\mathrm{\nabla }{p}_t=p_t{s}_t$. Substitution into the Fokker–Planck equation therefore yields ${\mathrm{\partial }}_t{p}_t+\mathrm{\nabla }\text{\hspace{0.17em}⁣}\cdot (v_t{p}_t)=0$. For a time-dependent test $\phi$ with common compact spatial support, differentiate $\int \phi (t,x)p_t(x)dx$, substitute this equation, and integrate by parts; this gives exactly the weak continuity identity stated above.

For $Z_t={\mathrm{\Phi }}_{t,0}(Z_0)$ and any continuously differentiable time-dependent test $\phi$ with common compact spatial support, the pathwise chain rule gives $\frac{d}{dt}\phi (t,Z_t)={\mathrm{\partial }}_t\phi (t,Z_t)+{\mathrm{\nabla }}_x\phi (t,Z_t)\cdot v_t(Z_t)$. The derivative is bounded on the compact support by continuity of $v$ and $\phi$, so it is integrable; taking expectations and integrating in time proves that $Q_t:=\mathrm{Law}(Z_t)$ satisfies the same weak continuity equation and has $Q_0=P_0$.

Fix $t\in [0,T]$ and $\psi \in C_c^1({\mathbb{R}}^d)$, and set $\phi (r,x)=\psi ({\mathrm{\Phi }}_{t,r}(x))$ for $0\le r\le t$. The flow identities in [A1] imply ${\mathrm{\partial }}_r\phi +v_r\cdot {\mathrm{\nabla }}_x\phi =0$; the support stays compact because it is the image of $\mathrm{supp}\psi$ under the continuous maps ${\mathrm{\Phi }}_{r,t}$ on the compact interval $[0,t]$. Applying the weak equation from steps 1.1 and 1.2 gives $\int \psi d{P}_t=\int \psi ({\mathrm{\Phi }}_{t,0}(x))P_0(dx)=\int \psi d{Q}_t$. Thus $P_t$ and $Q_t$ agree on every $C_c^1$ test.

For a bounded rational open box $R={\prod }_{i=1}^d(a_i,b_i)$, let $h(z)$ be $0$ for $z\le 0$, $3z^2-2z^3$ for $0<z<1$, and $1$ for $z\ge 1$. Its derivative is $6z(1-z)\ge 0$ on $(0,1)$ and is $0$ at the joins, so $h$ is nondecreasing and $C^1$. Set ${\theta }_n(u)=h(nu)$ and $g_n(x)={\prod }_i{\theta }_n(x_i-a_i){\theta }_n(b_i-x_i)$. Then $g_n\in C_c^1$, $0\le g_n\le 1$, and $g_n\uparrow {\mathbf{1}}_R$ pointwise. Step 2.1 and [F2] give $P_t(R)=Q_t(R)$. Bounded rational open boxes, together with $\mathrm{\varnothing }$ and ${\mathbb{R}}^d$, form a $\pi$-system. They generate the Borel sets because every point of an open set lies in a bounded rational open box contained in that set. The class $\mathcal{D}=\{A\text{ Borel}:P_t(A)=Q_t(A)\}$ contains ${\mathbb{R}}^d$, is closed under complements by equality of the total masses, and is closed under disjoint countable unions by countable additivity; hence it is a $\lambda$-system. By [F3], [F4], and the definition of law, $P_t=Q_t$ on all Borel sets, proving equality of the one-time marginals.