How to prove a PDE preserves mass? My general question is: suppose you are given a PDE (possibly with boundary conditions). What does it mean to say the PDE "conserves mass"?
Specifically, if you are given the PDE
$$- \nabla \cdot (a(x) \nabla u(x)) = f(x) \quad \text{in a domain $\Omega$,}$$
$$u = 0 \quad \text{ on } \partial \Omega,$$
how would you determine whether "mass is conserved"?
My guess would be that the mass is the integral $\int_\Omega u(x) \, dx$. But if this is the case, I don't know how to prove whether it is constant or not.
Edit: In this specific problem, there is no time involved, so you can't actually use a time derivative. However, it is the starting problem of a homogenization procedure, so actually it reads
$$- \nabla \cdot (a_\epsilon(x) \nabla u_\epsilon(x)) = f(x) \quad \text{in a domain $\Omega$,}$$
$$u_\epsilon = 0 \quad \text{ on } \partial \Omega.$$
So the mass conservation might be related to the $\epsilon$, but I don't think so (because from the context I gather that determining whether the problem conserves mass is supposed to be done for a fixed $\epsilon$).
Any help is appreciated!
 A: The concept of conservation of mass (or any other meaningful quantity) is usually applied to time dependent equations, like the heat equation
$$
u_t-\Delta u=0,
$$
or the wave equation
$$
u_{tt}-\Delta u=0.
$$
For the heat equation conservation of mass (or heat, to be more precise) means that $\int_\Omega u(x,t)\,dx$ is constant (i.e.  it does not depend on $t$.)
A: The time independent PDE you gave:
\begin{align}
- \nabla \cdot (a(x) \nabla u(x)) &= f(x) \quad \text{in  }\Omega,
\\
u &= 0 \quad \text{ on } \partial \Omega,
\end{align}
is normally the limit equilibrium of the time dependent problem:
$$
\frac{\partial u}{\partial t}- \nabla \cdot (a(x) \nabla u(x,t)) = f(x),
$$
where we let $t\to \infty$, and 
$$
\lim_{t\to \infty}\frac{\partial u}{\partial t} = 0,
$$
this roughly means $u$ won't be changed along with time anymore in the equilibrium state. If $u$ is the mass density in $\Omega$, then it won't change anymore with respect to time, so I am guessing the "conservation of mass" in OP means this. 
We can let 
$$
u(x)= \lim_{t\to \infty}u(x,t)
$$
which solves the time independent problem. We assume everything is nice and smooth we can interchange the limit and the integral, then
$$
\lim_{t\to \infty}\frac{d}{dt}\int_{\Omega} u = \lim_{t\to \infty}\int_{\Omega}\frac{\partial u}{\partial t} = \int_{\Omega} \nabla \cdot (a \nabla u) + \int_{\Omega} f = 0.
$$

In the equation,
\begin{align}
- \nabla \cdot (a_{\epsilon}(x) \nabla u_{\epsilon}(x)) &= f(x) \quad \text{in  }\Omega,
\\
u_{\epsilon} &= 0 \quad \text{ on } \partial \Omega,
\end{align}
the mean of "conservation" is rather straightforward: For any open subset $M\subset \Omega$
$$
\text{Constant} = \int_{M}f = -\int_{M}\nabla \cdot (a_{\epsilon} \nabla u_{\epsilon}) = \int_{\partial M} a_{\epsilon} \nabla u_{\epsilon}\cdot \nu \,dS,
$$
where $-a_{\epsilon} \nabla u_{\epsilon}$ is the flux of this mass density $u_{\epsilon}$. In an arbitrary region $M$ within this domain, The amount of total mass flux on the boundary of $M$ is constant independent of $\epsilon$ for any $\epsilon$. 
A: Conservation laws imply some notion of 'time' in which a quantity $M(t)$ like the mass $$\int_\Omega u(x,t) \,\mathrm d x$$ is conserved.
This is equivalent to proving that
$$\frac {\mathrm d}{\mathrm d t}\int_\Omega u(x,t) \,\mathrm d x = \int_\Omega \frac {\partial}{\partial t} u(x,t) \,\mathrm d x = 0$$
For example, an equation of the form
$$\frac {\partial u}{\partial t} = \nabla \cdot (F(u,\nabla u,x,t)) \quad \text{and}\quad G(u,\nabla u,x,t)|_{\partial \Omega} = 0$$
has the property that
$$\int_\Omega \frac {\partial}{\partial t} u(x,t) \,\mathrm d x = \int_\Omega \nabla \cdot F \,\mathrm d x = \int_{\partial\Omega} F \cdot \mathrm d S $$
so that mass is conserved if e.g. $G=0 \implies F=0$.
So if $F=u^2 \nabla u$ and $G=u$ then this last integral is zero as the boundary conditions imply $F$ vanishes on the boundary.
