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I am trying to solve an equation for chemical advection of a fluid in a porous media with this equation:

$\begin{align} \frac{\partial (\phi(x,t) C(x,t))}{\partial t} = - \nabla . (\vec{q(\phi, t, x)} C(x,t)) \end{align} $

with $\phi$ beeing the porosity, C the concentration of an element, and $\vec{q}$ the fluid flux.

I have another system of equations to compute $\phi$ and $\vec{q}$ at the next time step, so I only have 1 unknown for this equation: C.

What I first did was to simplify the system that way:

$\begin{align} \frac{\partial C}{\partial t} = - \frac{\vec{q}}{\phi} \nabla ( C) \end{align} $

Which is a quasi-linear advection equation and easy to solve. The thing is that I would like to solve this using the conservative form without simplifying my system because mass conservation is important.

I would like to use a WENO scheme for the spatial discretization and a SSP Runge-kutta method for the time because I have sharp gradients in composition. I've already did it with the classical Burgers' equation, but I don't know what to do with $\phi$ in the time derivative.

Here are my 2 ideas:

either developing the time derivative that way:

$\begin{align} \frac{\partial (\phi(x,t) C(x,t))}{\partial t} = \phi(x,t) \frac{\partial C(x,t)}{\partial t} + C(x,t) \frac{\partial \phi(x,t)}{\partial t} \end{align} $

where I can approximate the second term because I know the value of $\phi$ at the next timestep. My gut feeling tells me this is a bad idea.

My second idea is to make a change of variable:

$\begin{align} \phi C = \tilde{C} \end{align} $

knowing that the flux is equal to this, with $\vec{v}$ the velocity of the fluid: $ \vec{q} = \vec{v} \times \phi $

My equation then would becomes:

$\begin{align} \frac{\partial (\tilde{C}(x,t))}{\partial t} = - \nabla . (\vec{v} \tilde{C}) \end{align} $

That I should know how to solve. Then I could retrieve $C$ by dividing $\tilde{C}$ by $\phi$ at the next timestep.

Would that works? Or do you have other suggestions?

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    $\begingroup$ The second approach is the right way to go. $\endgroup$ Jan 28, 2022 at 16:18
  • $\begingroup$ Thanks @DavidKetcheson that's the confirmation I was looking for :) $\endgroup$
    – Iddingsite
    Jan 28, 2022 at 16:35

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