I'm reading section Continuity equation and Benamou–Brenier in these notes, i.e.,
Let $\Omega \subset \mathbb{R}^d$ be a convex set $\left(\Omega=\mathbb{R}^d\right.$ is admissible), let $\bar{\rho}_0 \in \mathcal{P}_2(\Omega)$ be a probability density with finite second moment, and let $v:[0, T] \times \Omega \rightarrow \mathbb{R}^d$ be a smooth bounded vector field tangent to the boundary of $\Omega$. Let $X(t, x)$ denote the flow of $v$, namely $$ (5.1) \quad \left\{\begin{array}{l} \dot{X}(t, x)=v(t, X(t, x)), \\ X(0, x)=x \end{array}\right. $$ and set $\rho_t=(X(t))_{\#} \bar{\rho}_0$. Note that, since $v$ is tangent to the boundary, the flow remains inside $\Omega$, hence $\rho_t \in \mathcal{P}(\Omega)$.
Lemma 5.1. Let $v_t(\cdot):=v(t, \cdot)$. Then $\left(\rho_t, v_t\right)$ solves continuity equation $$ (5.2) \quad \partial_t \rho_t+\operatorname{div}\left(v_t \rho_t\right)=0 $$ in the distributional sense.
Proof. Let $\psi \in C_c^{\infty}(\Omega)$, and consider the function $t \mapsto \int_{\Omega} \rho_t(x) \psi(x) d x$. Then, using the definitions of $X$ and $\rho_t$, we get $$ \begin{aligned} \int_{\Omega} \partial_t \rho_t(x) \psi(x) d x & =\frac{d}{d t} \int_{\Omega} \rho_t(x) \psi(x) d x \stackrel{(i)}{=} \frac{d}{d t} \int_{\Omega} \psi(X(t, x)) \bar{\rho}_0(x) d x \\ & =\int_{\Omega} \nabla \psi(X(t, x)) \cdot \dot{X}(t, x) \bar{\rho}_0(x) d x \\ & \stackrel{(i i)}{=} \int_{\Omega} \nabla \psi(X(t, x)) \cdot v_t(X(t, x)) \bar{\rho}_0(x) d x \\ & \stackrel{(i i i)}{=} \int_{\Omega} \nabla \psi(x) \cdot v_t(x) \rho_t(x) d x=-\int_{\Omega} \psi(x) \operatorname{div}\left(v_t \rho_t\right) d x \end{aligned} $$ where (i) and (iii) follow from $\rho_t=(X(t))_{\#} \bar{\rho}_0$, while (ii) follows from (5.1).
I'm not familiar with distribution theory. Could you confirm if I understand the statement of Lemma 5.1. correctly? Thank you so much for your help!
My understanding By assumption, $v$ is a bounded smooth vector field. On the other hand, no smoothness condition is imposed on $\bar{\rho}_0$. However, $\rho_t$ is differentiable. Then $\partial_t \rho_t$ and $\operatorname{div}\left(v_t \rho_t\right)$ makes sense as measurable functions. So $\partial_t \rho_t+\operatorname{div}\left(v_t \rho_t\right)=0$ in the distributional sense if and only if $\partial_t \rho_t (x)+\operatorname{div}\left(v_t \rho_t\right) (x)=0$ for a.e. $(t, x) \in [0, T] \times \Omega$.