6
$\begingroup$

On page 3 of this preprint, after recalling the definition of flow generated by a vector field, the authors remark that "a necessary condition for a flow $\varphi_t(\cdot)$ generated by $a(t, \cdot)$ to be measure preserving is: $\mathrm{div}\, a = 0$ in a suitable sense".

Assuming $a(t,\cdot) \in BV(\mathbb{R}^N; \mathbb{R}^N)$, is the "suitable sense" the sense of distributions? If yes, how can one prove that $\mathrm{div}\, a = 0$ in the sense of distributions implies $\phi_t$ measure preserving?

Improve this question
$\endgroup$
1
  • 1
    $\begingroup$ It seems to me that the other are talking about the other implication? $\endgroup$
    – user2520938
    Commented Apr 12, 2019 at 7:51

1 Answer 1

Reset to default
6
$\begingroup$

Let $\mu_t = (\varphi_t)_{\sharp} \mu$ denote the image of the measure $\mu$ by the flow of $a$ (where $\mu$ can be Lebesgue measure). It is well-known that the family of measures $\{\mu_t\}_{t\in \mathbb R}$ satisfies Liouville equation (aka continuity equation) $$ \partial_t \mu_t + \operatorname{div\,} (a \mu_t) = 0 \tag{1} $$ in the sense of distributions. Indeed, for any smooth compactly supported function $f=f(t,x)$ $$ \iint (\partial_t f(t,y) + a(t,y) \nabla f(t,y)) \,d\mu_t(y) \,dt = \iint \bigl((\partial_t f)(t,\varphi_t(x)) + a(t, \varphi_t(x)) (\nabla f)(t, \varphi_t(x))\bigr) \, d\mu(x) \,dt = \iint \partial_t \bigl( f(t, \varphi_t(x)) \bigr) \,dt \,d\mu(x) = \int 0 \,d\mu = 0, $$ where we have used the chain rule and the fact that $\partial_t\varphi_t(x) = a(t, \varphi_t(x))$ for a.e. $t$.

Claim 1. If $\varphi_t$ preserves the measure $\mu$ then $\operatorname{div} (a\mu) = 0$.

Indeed, if the flow of $\varphi_t$ preserves the measure $\mu$, i.e. $\mu_t = \mu$ for all $t$, then by (1) $$ \operatorname{div} (a \mu) = 0. \tag{2} $$ In particular, if $\mu$ is the Lebesgue measure then $\operatorname{div} a = 0$.

Claim 2. Suppose that $\mu_t$ is the unique solution of (2) with the initial condition $\mu_t|_{t=0} = \mu$ (e.g. in the class of non-negative measure-valued solutions, absolutely continuous with respect to Lebesgue measure). Then $\operatorname{div} (a\mu) = 0$ implies that the flow $\varphi_t$ preserves the measure $\mu$.

Indeed, uniqueness implies $\mu_t = \mu$ for a.e. $t$, that is $\mu$ is preserved by the flow of $a$.

(I learned this trick in some paper by E.O. Stepanov, but don't remember precisely which one.)

Improve this answer
$\endgroup$
3
  • 1
    $\begingroup$ Thank you. 1) How do you prove that the family of measures $\{\mu_t\}_{t\in \mathbb{R}}$ satisfies the continuity equation? 2) Why if $\mu$ is the Lebesgue measure then $\mathrm{div}\,a=0$? 3) Are you claiming that $\mathrm{div} \, a = 0$ is sufficient and necessary if (1) has a unique solution? So, for example, if $a \in L^1_tBV_x$? $\endgroup$
    – Riku
    Commented Apr 11, 2019 at 19:41
  • $\begingroup$ I've also put up a question on a related topic here: mathoverflow.net/questions/327789/… $\endgroup$
    – Riku
    Commented Apr 12, 2019 at 9:39
  • $\begingroup$ @Riku I've added the proof of (1). (2) is by definition ($\operatorname{div} (a \mu) = 0$ in the sense of distributions means $\int a \nabla f d \mu = 0$ for any appropriate test function $f$). Concerning (3) I don't claim uniqueness. But note that in the paper you cited only Claim 1 is stated. $\endgroup$
    – Skeeve
    Commented Apr 14, 2019 at 10:29

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .