# Prove that the flow of a divergence-free vector field is measure preserving

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?

• It seems to me that the other are talking about the other implication? – user2520938 Apr 12 '19 at 7:51

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.)

• 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 $a \in L^1_tBV_x$? – Riku Apr 11 '19 at 19:41
• I've also put up a question on a related topic here: mathoverflow.net/questions/327789/… – Riku Apr 12 '19 at 9:39
• @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. – Skeeve Apr 14 '19 at 10:29