I am trying to understand the advection equation with variable velocity coefficient a bit better. In particular I don't understand how the equation can be conservative.

The advection equation,

$$ \frac{\partial u}{\partial t} + \frac{\partial}{\partial x}(\boldsymbol{v}u) = 0 $$

Let's interpret $u(x,t)$ as being the concentration of some physical species ($cm^{-3}$) or some other physical quantity which cannot be created or destroyed. If we integrate $u(x,t)$ over our domain then we should get constant,

$$ \int_{x_{\text{min}}}^{x_{\text{max}}} u(x,t) dx = \text{constant} $$

(This is what I mean by being conservative.)

If we now let the velocity be a function of space (and time), $\boldsymbol{v}(x,t)$, then the chain rule must be applied to give,

$$ \frac{\partial u}{\partial t} + \boldsymbol{v}\frac{\partial u}{\partial x} + \underbrace{u\frac{\partial \boldsymbol{v}}{\partial x}}_{\text{?}} = 0 $$

The final term "looks" like a source term and this is what I find confusing. It will increase or decrease the quantity $u$ depending on the divergence of the velocity field.

Following this question, I understand how to impose conservation boundary conditions. However, for the variable velocity advection equation I don't understand how conservation boundary conditions can be derived because of the additional "source term" that is introduced by applying the chain rule. Can this equation be conservative? If so, how can correct boundary conditions be applied?


1 Answer 1


The fundamental quantity in transport is the flux, $\mathbf v u$ for advection. The divergence theorem states that

$$\int_\Omega \nabla\cdot (\mathbf v u) = \int_{\partial \Omega} (\mathbf v u) \cdot \mathbf n . $$

An equation is conservative when it is preserves this equality. Dropping to 1D with $\Omega = (a,b)$ and using the equation $u_t + (\mathbf v u)_x = 0$, we have

$$ \Big(\int_a^b u\Big)_t = \int_a^b u_t = -\int_a^b (\mathbf v u)_x = -\mathbf v u |_a^b $$

where the term on the right is just the difference in flux between the left and right boundaries.

Regarding your second observation, the non-conservative (non-divergence) form is misleading (and only justified for smooth solutions). The product $\mathbf v \cdot \nabla u$ is not conservative transport if $\mathbf v$ is not divergence-free (i.e., constant in 1D). You should stick with the conservative form and resist the urge to apply the chain rule when evaluating conservation properties.

  • $\begingroup$ Thank you for a really clear answer, yet again, Jed! I think I will ask a follow-up question to this, but first need to try to implement your suggestion. $\endgroup$
    – boyfarrell
    Commented May 20, 2013 at 0:39

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