Explaining the one-dimensional continuity equation with respect to density evolution I've got a rather abstract question
So the continuity equation for a one-dimensional continuum is:
$$
\frac{\partial \rho}{\partial t} + \frac{\partial}{\partial x}(\rho v)=0
$$
and we can expand and rearrange that to give:
$$
\frac{\partial \rho}{\partial t} + v\frac{\partial \rho}{\partial x} = -\rho \frac{\partial v}{\partial x}
$$
which gives the material derivative on the left.   I'm cool with all this.
But when it comes to explaining it, why does the density $\rho$ evolve according to the above equation?
I can see that we take the derivative of density with respect to time, and the velocity of the density derivative with respect to position for the material derivative side... and that equals the inverse of density multiplied by acceleration... but I'm lacking the intuition to take it the step further and also put it into plain english :(
Can anyone give me a bit of a hint? Textbooks have failed me.
Cheers
 A: The modelling assumption is that `mass' of $\rho$ is conserved in each interval which is transported by the velocity $v(x,t)$. You can think that all the mass is transported (or advected) by the velocity $v$.
Consider an interval moving with the velocity $v(x,t)$. That is $(a(t),b(t))$ with $a'(t) = v(a(t),t)$ and $b'(t) = v(b(t),t)$. Then
$$
\frac{\mathrm{d}}{\mathrm{d}t} \int_{a(t)}^{b(t)} \rho(x,t) \, \mathrm{d}x = 0.
$$
The Reynolds transport theorem tells you how to swap the time derivative with the integral sign:
$$
\frac{\mathrm{d}}{\mathrm{d}t} \int_{a(t)}^{b(t)} \rho(x,t) \, \mathrm{d}x 
= \int_{a(t)}^{b(t)} \frac{\partial}{\partial t} \rho(x,t) \, \mathrm{d}x + b'(t) \rho(b(t),t) - a'(t) \rho( a(t),t )
= 0.
$$
Finally, use the Fundamental Theorem of Calculus on the boundary terms to get
$$
  \int_{a(t)}^{b(t)} \frac{\partial}{\partial t} \rho(x,t) + \frac{\partial}{\partial x} ( v(x,t) \rho(x,t) ) \, \mathrm{d} x = 0
$$
Since the choice of interval was arbitrary, this shows the pointwise relation. 
