2
$\begingroup$

I have always come across the terms "conservative" as opposed to "non-conservative" forms of equations in fluid mechanics. Is there a good reference that someone can share to clearly distinguish the two or shed some light on what they mean and what are their advantages/disadvantages?

$\endgroup$
6
$\begingroup$

In short, you can recognize a conservative formulation if a divergence operator is involved in the equation. For instance, the mass conservation equation is naturally written in conservative form :

$$\frac{\partial \rho}{\partial t} + div (\rho \mathbf{u})=0$$

By integrating over $\Omega$ (a reference volume) and with the flux-divergence theorem :

$$\int_{\Omega} \frac{\partial \rho}{\partial t} \mathrm{d}\Omega + \int_{\partial \Omega} \rho \mathbf{u} \cdot \mathbf{n} \mathrm{d}S = 0$$

you can understand that the quantity of mass in the reference volume is equal to the variation of $\rho$ in time (first term) + the sum of the fluxes ($\rho \mathbf{u}$) across the surface boundary S with respect to the normal $\mathbf{n}$ pointing outward, that is the difference between what is going in and what is leaving $\Omega$. It is conservative in the sense that if we sum all the contributions acting on the volume $\Omega$, the result will be constant at each time, it means that mass is not lost or created.

For a pure advection equation (non-conservative) : $$\frac{\partial \phi}{\partial t} + \mathbf{u} \cdot \nabla\phi=0$$ the theorem below does not apply (no divergence operator). It means that the quantity $\phi$ is advected through the computational domain so if we place ourselves in an eulerian formalism looking at the quantity $\phi$ in a reference volume $\Omega$, the result can be 1 at t=dt and 0.5 at t=t+dt => non-conservative.

It is recommended to solve equations in conservative form as they respect the conservation principle but sometimes an additional advection equation is required (multifluid simulations), that one can be very tedious to discretize without tremendous numerical problems.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.