Hello Jonathan,
Once again, thank you for your answer.
> How does $v^*$ appear in your other equations? Is it possible that one
> of them is more naturally solved for velocity?
I definitely don't think so. The other equations, in cartesian coordinates
and without coordinate change have the following form
\[
\frac{\partial}{\partial t} (cx)
+ \nabla \cdot (cv^*x)
= \nabla \cdot (c D \nabla x)
\]
and they are necessary to calculate $x$, the mole fraction.
The legend has it that the velocity must be calculated from continuity
\[
\frac{\partial}{\partial t} (cx)
+ \nabla \cdot (cv^*x)
= \nabla \cdot (c D \nabla x)
\]
But last night, I reread a thesis I have and I realized that the
convection can be neglected. Which of course solves the problem with
velocity. I will go with this approach now because I need results for a
conference in early spring and I'm confident it will give satisfactory
results.
I thought perhaps there is an obvious solution that I wasn't seeing, but
now you confirmed that accounting for convection is no trivial task.
Many thanks!
Etienne
