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?


1 Answer 1


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.

  • $\begingroup$ 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.. $\endgroup$
    – Redeldio
    Oct 31, 2022 at 14:51
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    $\begingroup$ Oh you mean by computing the derivatives via chain rule or so ? It is obviously true but like that, without more context, I find it a bit cumbersome to try this way ! $\endgroup$ Oct 31, 2022 at 15:11
  • $\begingroup$ 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. $\endgroup$
    – Redeldio
    Oct 31, 2022 at 15:32

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