2
$\begingroup$

Does the CFL condition play any role in a pure Stokes flow, i.e. convective term is neglibile, or vanishing? If not, what is the "equivalent" condition for stability? I have read something about the diffusional time scale but it's quite vague. Anyone with in-depth insight?

And a quotation from a paper [1]:
"The time step is $3.10^-3\dot\gamma^-1$, corresponding to a CFL number based on the diffusional time scale $CFL_d=v\Delta t/\Delta^2$ of 50."

[1] Gallier, S., Lemaire, E., Lobry, L., & Peters, F. (2014). A fictitious domain approach for the simulation of dense suspensions. Journal of Computational Physics, 256, 367-387.

Improve this question
$\endgroup$
13
  • $\begingroup$ Do you mean the time-dependent Stokes equation? It would be useful to state the exact problem you are considering. $\endgroup$
    – Wolfgang Bangerth
    Commented Oct 3, 2017 at 0:06
  • $\begingroup$ @WolfgangBangerth I will clarify! I mean in the flow regime that the Navier stokes equations reduce to the Stokes equation: $\mu\nabla^2\mathbf{u}-\nabla p=0$ $\endgroup$
    – nabber
    Commented Oct 3, 2017 at 4:44
  • $\begingroup$ But there is no time dependence here. The CFL condition relates the time step size to the mesh size -- but you don't have a time step here. $\endgroup$
    – Wolfgang Bangerth
    Commented Oct 3, 2017 at 17:21
  • $\begingroup$ @WolfgangBangerth Although it is evident that this not apply in the steady case, does it apply in the unsteady case? $$\mu \nabla^2 \mathbf{u}- \nabla p = \frac{\partial \rho \mathbf{u}} {\partial{t}}$$ Generally we would solve the diffusion operator with an implicit scheme, so I don't think you would get a CFL in this condition no? $\endgroup$
    – BlaB
    Commented Oct 3, 2017 at 20:11
  • $\begingroup$ In the time dependent case, if you use an implicit time stepping scheme, then the CFL stability condition does not apply. If you use an explicit time stepping scheme, then a variation of the condition applies indeed. $\endgroup$
    – Wolfgang Bangerth
    Commented Oct 3, 2017 at 21:25

1 Answer 1

Reset to default
3
$\begingroup$

Essentially, the time dependent Stokes equation looks like the heat equation: $$ \frac{\partial u}{\partial t} - \nu\Delta u = f-\nabla p, $$ plus the incompressibility condition $\nabla \cdot u=0$ that for the current discussion is immaterial. Thus, the same considerations for time step choice apply as for the heat equation.

Consequently, using an implicit method such as the backward Euler method generally yields an unconditionally stable method for which you can choose the time step as large as you want. (Although you may want to choose it not too large for accuracy reasons -- stability does not imply accuracy.)

On the other hand, for an explicit method, you need to choose the time step $\Delta t$ subject to some CFL-like condition that says that $$ \Delta t \le C \frac{1}{\nu} \Delta x^2 $$ where $\Delta x$ is the mesh size. This is not practical: it requires you to choose the time step four times smaller for each mesh refinement. These time steps would be so small that the time discretization error is vastly smaller than the spatial discretization error, and that's not useful. As a consequence, practical implementations do not choose explicit methods for the Stokes equation.

[For an explanation of the why this condition arises, see lecture 27 at http://www.math.colostate.edu/~bangerth/videos.html .]

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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