Why is the equation $u_t + (a u)_x = 0$ conservative but $u_t + a u_x = 0$ isn't?

Question

Consider the following advection equations: $$u_t + a u_x = 0$$

$$u_t + (au)_x = 0$$

I have heard the second one described as being conservative, yet the first one isn't. Why is this?

Answer 1

Physically speaking, ''conservative'' means that the total mass of $u$ is conserved in time. To see this, suppose $u : [0,T] \times [0,1] \to \mathbb R$ with $u(t,0) = u(t,1)=0$ for all $t \in [0,T]$ and define the mass of $u$ at time $t$ by

$$M(t)= \int_0^1 u(t,x)dx.$$

If $u$ satisfies the second (conservative) equation, you get

$$ M'(t) = \int_0^1 \partial_t u(t,x)dx = -\int_0^1 \partial_x(a(t,x)u(t,x))dx = u(t,0)a(t,0)-u(t,1)a(t,1) = 0$$

Therefore the mass of $u$ is constant in time, so $u$ is conserved. In the first equation things go wrong because you get, after an integration by parts

$$M'(t) = -\int_0^1 a(t,x) \partial_x u(t,x)dx = \int_0^1 u(t,x) \partial_x a(t,x)dx + u(t,0)a(t,0)-u(t,1)a(t,1)$$

$$= \int_0^1 u(t,x) \partial_x a(t,x)dx.$$

and so $M$ is in general not constant in time: equation non conservative.

Answer 2

Essentially, as @Stefano has shown, writing a PDE in conservative form leads to something being constant in time, specifically an integral. It can be shown (and I am happy to do this if you wish), that

$$\frac{d}{dt}\int_{a(t)}^{b(t)} F(t,x)\,dx \;=\; \int_{a(t)}^{b(t)}\left[ \frac{\partial F}{\partial t} + \frac{\partial}{\partial x}\left(\phantom{\frac{}{}} F(t,x)v(t,x)\,\right)\right]\,dx$$

where $v(t,x)=\frac{\partial x}{\partial t}$ is the instantaneous velocity. Clearly if the term under the integral on the right-hand side vanishes, then the integral on the left-hand side is constant in time. This is just a one-dimensional version of the more general Reynold's transport theorem. The joy of putting PDEs in conservative form (if it can be done) is that this integral form allows for discontinuous solutions (like shocks), whereas non-conservative PDEs only admit smooth solutions.