From the paper of Ambrosio-Crippa, it is known that if $\beta:\mathbb R^d\times[0, T[\longrightarrow\mathbb R^d$ is suitably regular, then the system $$ \begin{cases} \frac{\partial\mu}{\partial t}(x, t)+\operatorname{div}(\beta(x, t)\mu(x, t))=0,&(x, t)\in\mathbb R^d\times]0, T]\\ \mu(\cdot, 0)=\mu_0 \end{cases} $$ 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 $$ m(\cdot, t)=\frac{m_0(\cdot)}{\operatorname{det}J\Phi_t(\cdot)}\ \circ\ \Phi_t^{-1}(\cdot),\qquad(\triangle) $$ 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 $$ (\star)\ \ \begin{cases} \frac{\partial m}{\partial t}(x, t)+\operatorname{div}(\beta(x, t)m(x, t))=0,&(x, t)\in\mathbb R^d\times]0, T]\\ m(\cdot, 0)=m_0 \end{cases}? $$ 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 $(\star)$ is true but, to be honest, i cannot prove that using directly $(\triangle)$. Can you help me or give me some suggestion?