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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.