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. 




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Previous posts with closely related question and framework are [Role of absolute continuity of divergence of BV function in proof of renormalization property](https://mathoverflow.net/questions/327789/role-of-absolute-continuity-of-divergence-of-bv-function-in-proof-of-renormaliza) and [Prove that the flow of a divergence-free vector field is measure preserving](https://mathoverflow.net/questions/327781/prove-that-the-flow-of-a-divergence-free-vector-field-is-measure-preserving).



Role of the divergence of the vector field in transport equations: mass concentration?
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):


$\frac{dD(x,t)}{dt}=(\nabla.a(\phi(x,t)))D(x,t)$.


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 $\nabla.a$


