# Continuity equation for a density of a measure

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 $$\begin{cases} \dfrac{\partial\mu}{\partial t}(x, t)+\operatorname{div}(\beta(x, t)\mu(x, t))=0,&(x, t)\in\mathbb R^d\times]0, T]\\ \mu(\cdot, 0)=\mu_0 \end{cases}$$ 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?

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.
• Thank you for replying. Yes, indeed i proved the claim $(\star)$ just using the weak formulation. But i was asking myself if we can obtain $(\star)$ via $(\triangle)$, but maybe i'm wrong.. Oct 31, 2022 at 14:51
• Yes, i mean that. Of course it should be true, but I cannot prove them. In particular, i can compute all the derivatives but I cannot turn out to an equality between $\partial_tm$ and $-\operatorname{div}(\beta m)$. Btw, thank you anyway. Oct 31, 2022 at 15:32