How to prove a PDE preserves mass?

Question

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!

Answer 1

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 $t$.)

Answer 2

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 $$ \lim_{t\to \infty}\frac{d}{dt}\int_{\Omega} u = \lim_{t\to \infty}\int_{\Omega}\frac{\partial u}{\partial t} = \int_{\Omega} \nabla \cdot (a \nabla u) + \int_{\Omega} f = 0. $$

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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 $\epsilon$.

Answer 3

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.