From the paper of [Ambrosio-Crippa](https://www.cambridge.org/core/journals/proceedings-of-the-royal-society-of-edinburgh-section-a-mathematics/article/abs/continuity-equations-and-ode-flows-with-nonsmooth-velocity/DD2A02C300757F63F7F0A96FAD8A70C5), it is known that if $\beta:\mathbb R^d\times[0, T[\longrightarrow\mathbb R^d$ is suitably regular, then the system
$$
\begin{cases}
\dfrac{\partial\mu}{\partial t}(x, t)+\operatorname{div}(\beta(x, t)\mu(x, t))=0,&(x, t)\in\mathbb R^d\times]0, T]\\
\mu(\cdot, 0)=\mu\_0
\end{cases}
$$
is well-posed. In the case when $\mu\_0$ is absolutely continuous, that is $\mu\_0=m\_0\mathcal L^d$, where $m\_0:\mathbb R^d\longrightarrow\mathbb R$ is the density, all the measures $\mu(\cdot, t)$ are absolutely continuous too, and their density $m(\cdot, t)$ can be explicitly computed as
$$
m(\cdot, t)=\frac{m\_0(\cdot)}{\operatorname{det}J\Phi\_t(\cdot)}\ \circ\ \Phi\_t^{-1}(\cdot),\label{1}\tag{$\triangle$}
$$
where $\Phi\_t$ is the flow associated to
$$
\begin{cases}
y'(s)=\beta(y(s), s),&s\in]0, T[\\
y(0)=x
\end{cases}.
$$


**Question.** Is it true that $m$ satisfies (in the sense of distributions) the continuity equation
$$
\begin{cases}
\dfrac{\partial m}{\partial t}(x, t)+\operatorname{div}(\beta(x, t)m(x, t))=0,&(x, t)\in\mathbb R^d\times]0, T]\\
m(\cdot, 0)=m\_0
\end{cases}\quad?\label{2}\tag{$\star$}
$$
I proved that, with suitable regularity hypotheses on the field $\beta$, if $m\_0\in H^1(\mathbb R^d)\cap W^{1, \infty}(\mathbb R^d)$ then $m(\cdot, t)$ belongs to the same space for every $t$. I think that \eqref{2} is true but, to be honest, i cannot prove that using directly \eqref{1}. Can you help me or give me some suggestion?



Continuity equation for a density of a measure
I am not so sure to understand the problem, maybe I am missing something. You should not use $(\triangle)$ but instead go back to the equation satisfied by the measure. Indeed, since $\mu$ is solution of your conservative transport equation, you have for any test function $\varphi\in\mathscr{D}(\mathbf{R}\_+\times\mathbf{R}^d)$, noting $\mu\_t:=\mu(t,\cdot)$
$$
\int\_0^{+\infty} \int\_{\mathbf{R}^d} (\partial\_t \varphi+\beta\cdot\nabla\_x \varphi) \mathrm{d}\mu\_t \,\mathrm{d}t = -\int\_{\mathbf{R}^d} \varphi(0)\mathrm{d}\mu\_0. 
$$
If you did prove what you claim then you get directly your formulation.


