Continuity equation for velocity Jacobian I heard in class that for a given velocity field $u$, $$f := \det J_u$$ (the determinant of the Jacobian of $u$) satisfies the following:
$$f_t + div(f u) = 0$$
with $f(0)=1$.
I cannot convince myself that this is true, or find any resource that proves it. Can someone provide a derivation of this?
 A: I do not know what you mean by "the determinant of the Jacobian of $u$". If $$J_{u \:ij} := \frac{\partial u_i}{\partial x_j}$$ then the statement is trivially false. Take in $\mathbb R^n$ the field of velocities $u_i =x_i$, for $i=1,\ldots, n$. In this case $J_{u \:ij} = \delta_{ij}$,  $\det J =1$, $div u =1$ and $\partial_t \det J_u=0$ and so that
$$f_t +div(f u) = 0 + div u =1$$
However there is a geometric quantity  related with yours satisfying a conservation equation.
For every time $t$, the particle of fluid with position $x_0$ at $t=0$ has the position $x = x(t,x_0)$ and the map $\mathbb R^n \ni x_0 \mapsto x \in \mathbb R^n$ is a diffeomorphism on its image. The field of velocities is obtained by the derivative 
$$u(t,x) = (\partial_t x(t,x_0))|_{x_0(t,x)}$$
Next take a volume of fluid  $V_0$ at time $t=0$, consider its time evolution $V_t = \{x=x(t,x_0)\:|\: x_0 \in V_0\} $. 
Let us finally denote by  $J$ the Jacobian matrix $\partial x_0/ \partial x$ and $f := \det J$.
It holds
$$\int_{V_t} f  dx  =\int_{V_0} \det J  (\det(J))^{-1} dx_0 = \int_{V_0} 1 dx_0  = const \tag{1}$$
We have established that, for every choice of the initial volume $V_0$, the right hand side of (1) is  constant in time. Passing to the corresponding local conservation rule with the usual procedure, you have the identity you pointed out:
$$f_t +div(f u) = 0 \tag{2}$$
where 
$f = \det J$, because (2) is the local form of the global identity
$$\frac{d}{dt}\int_{V_t} f  dx=0$$
when it is valid for every choice of initial volumes $V_0$ (it is sufficient that it is valid for every ball with finite radius centered at every point).
