Consider the continuity equation

$$\partial_t u(t,x) + \mathrm{div}(a(t,x)u(t,x)) = 0,$$ where $u: [0,T]\times \mathbb{R}^N \to \mathbb{R}$ is the solution and $a:[0,T]\times \mathbb{R}^N \to \mathbb{R}^N$ is a vector field.

I've head many times that assumptions on $\mathrm{div}\,a$, for example that it is bounded, amount to asking some requirements on the "concentration of mass" transported by the equation.

I'm not sure what that means (neither heuristically, nor rigorously) and I would appreciate some insight (or detailed references) on this point.

Previous posts with closely related question and framework are Role of absolute continuity of divergence of BV function in proof of renormalization property and Prove that the flow of a divergence-free vector field is measure preserving.

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    $\begingroup$ are you sure it's "concentration of mass" and not "conservation of mass" that you are referring to? (the former makes no sense to me, the latter does, that's why I ask) $\endgroup$ – Carlo Beenakker Apr 12 '19 at 11:17
  • $\begingroup$ @CarloBeenakker I've heard it both ways actually, but I'm completely unclear what it means in this context. Could you write down what sense you can make of the phrase? One thing I can think about is that $\mathrm{div}\, a = 0$ is related to measure preserving flows for the associated ODE (as in the post mathoverflow.net/questions/327781/…). $\endgroup$ – Riku Apr 12 '19 at 11:19
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    $\begingroup$ you could start from continuummechanics.org/continuityequation.html $\endgroup$ – Carlo Beenakker Apr 12 '19 at 11:24
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    $\begingroup$ In the smooth setting the jacobian $J(t,x) = \det(\nabla_x \Phi(t,x))$ of the flow $\Phi$ of $a$ satisfies $\partial_t J(t,x) = (\operatorname{div} a)(t,\Phi(t,x)) J(t,x)$. From here one can deduce that $J(t,x)$ is bounded from below and above once $\operatorname{div} a$ is bounded $\endgroup$ – Skeeve Apr 12 '19 at 12:14
  • $\begingroup$ @Skeeve And what does that imply? $\endgroup$ – Riku Apr 12 '19 at 12:25

It is instructive to think about 1 dimensional case. Take $a(x)=b-\alpha x$, ($\alpha\geq 0$) then the divergence is simply $-\alpha$.

Case 1: $b=0,\alpha\geq 0$: A trajectory starting at $x(0)=x_0$ on the real line will follow the dynamics $x(t)=x_0e^{-\alpha t}$. Hence, each point exponentially attracted to the origin $x=0$ as $t\rightarrow\infty$ (hence the origin acts like a sink). This means that this flow will concentrate all the "mass" to the neighborhood of origin at an exponential rate given by $\alpha$. So if alpha is large, faster is this concentration.

Case 2: $b\neq 0,\alpha = 0$: Divergence is 0. Hence there is no mass concentration, the initial mass/measure just translates with velocity $b$.

Multi-dimensional case: Let $\phi(x,t)$ denote the flow map of $a(x)$ (i.e., the solution of ODE $\dot{x}=a(x)$), and $D(x,t)=\det(\nabla_x \phi(x,t))$, where $x\in\mathbb{R}^n$. Then the following one dimensional ODE holds (as already mentioned by Skeeve):


The jacobian of the flow map $J(x,t)=\nabla_x \phi(x,t)$ is intuitively the linear approximation of the flow map between neighborhood of starting position $x$, and the endpoint $\phi(x,t)$. The determinant of this linear map describes the expansion or concentration of this neighborhood as it is mapped by the flow. (See here for geometric interpretation of determinant: https://math.stackexchange.com/questions/250534/geometric-meaning-of-the-determinant-of-a-matrix). By the above equation, the rate of change of this this determinant is determined by $\nabla.a$

  • $\begingroup$ Thank you. This is very instructive indeed. Why "each point exponentially attracted to the origin"? I see that as $t \to \infty \ x(t) \to 0$, but why is the origin ($t=0$, $x=0$) attractive? $\endgroup$ – Riku Apr 12 '19 at 16:36
  • $\begingroup$ Also, what happens if $\alpha <0$? More in general, what does this insight give? In general, do we have to consider separately $\mathrm{div} a$ greater smaller or equal to $0$? $\endgroup$ – Riku Apr 12 '19 at 16:38
  • $\begingroup$ In general (including multi-dimensional case), $\nabla. a$ gives the local "expansion" or "concentration" depending upon whether it is positive or negative. $\endgroup$ – Piyush Grover Apr 12 '19 at 16:51
  • $\begingroup$ I see. How can you make the argument heuristically and rigorously in general? $\endgroup$ – Riku Apr 12 '19 at 17:12
  • $\begingroup$ See my edit that explains this point. $\endgroup$ – Piyush Grover Apr 12 '19 at 17:25

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