I'm studying the transport equation $$ \partial_tu+b \cdot \nabla u = 0. $$
In some notes I found the following: If $u(t,\cdot)$ is compactly supported for every $t$ then $\int u \, dx$ is conserved if the divergence of $b$ is zero. I have trouble to understand what the word "conserved" means in this context.
Any idea?
As pointed out by Jan Bohr in a comment to the question itself, the quantity
$E(t) = \displaystyle \int_{\Bbb R^n} u(x, t) \; d^nx \tag 1$
is said to be conserved if and only if it is constant in the time variable $t$; the following calculation shows that this is so:
$\dot E(t) = \displaystyle \int_{\Bbb R^n} \partial_t u(x, t) \; d^nx; \tag 2$
$\partial_t u + b \cdot \nabla u = 0; \tag 3$
$\dot E(t) = \displaystyle -\int_{\Bbb R^n} b \cdot \nabla u \; d^nx; \tag 4$
$\nabla \cdot (ub) = b \cdot \nabla u + u \nabla \cdot b = b \cdot \nabla u, \tag 5$
since
$\nabla \cdot b = 0; \tag 6$
thus (4) becomes
$\dot E(t) = \displaystyle -\int_{\Bbb R^n} \nabla \cdot (ub) \; d^nx; \tag 7$
let $S(R) \subset \Bbb R^n$ be the $n - 1$ sphere of radius $R$ centered at $0$; then since $u(x, t)$ is compactly supported in $\Bbb R^n$ for every $t$, for any $t$ we may choose $R$ sufficiently large that $u(x, t)$ vanishes on and outside of $S(R)$; for such $R$, we apply the divergence theorem to (7) and find
$\dot E(t) = \displaystyle -\int_{\Bbb R^n} \nabla \cdot (ub) \; d^nx = -\int_{S(R)} (ub) \cdot n \; dS, \tag 8$
where $n$ is the outward-pointing unit normal vector field, and $dS$ is the standard measure, on $S(R)$; since $u$ vanishes on $S(R)$, (8) yields
$\dot E(t) = \displaystyle -\int_{S(R)} (ub) \cdot n \; dS = 0 \tag 9$
and thus $E(t)$ is "conserved".