To compute the evolution of a free surface between two incompressible, immiscible liquids, two tightly coupled equations have to be solved, the volume fraction advection and the Navier-Stokes equations:

$$\frac{\partial\alpha}{\partial t}=\nabla \cdot (U\alpha)$$ $$\frac{\partial \rho U}{\partial t} + \nabla \cdot (\rho U U)-\nabla\cdot (\mu \nabla U) = -\nabla p + \sigma\kappa\nabla\alpha$$ with $\kappa = -\nabla \cdot \frac{\nabla \alpha}{|\nabla \alpha|}$.

These equations are tightly coupled, and by far the most common approach is to couple them explicitly using a very small time step, i.e. solve for $\alpha$ once, then once for $U$ and then proceed to the next time step. Regardless of the time marching scheme used to solve $\alpha$ and $U$ very small time steps are required to maintain the coupling of the two equations.

Now this is probably a naive question, but why is it not helpful to wrap the solution in an outer loop, relax U and $\alpha$ and solve with larger time steps?

  • $\begingroup$ I see no reason not to do so. It's more difficult to compute, though. It may also be the case that you are bound to small time steps by accuracy considerations (just because you can take a large time step with implicit methods without losing stability doesn't mean that you actually want to do it since you may still lose accuracy). $\endgroup$ – Wolfgang Bangerth Apr 2 '14 at 11:39
  • $\begingroup$ @WolfgangBangerth Would convergence be guaranteed though? Perhaps the system would only converge with strong underrelaxation beating the purpose? $\endgroup$ – akid Apr 2 '14 at 12:18
  • $\begingroup$ By the way, my question might be a bit ambiguous. When saying explicit solution, I'm not talking about solving $\frac{\partial \alpha}{\partial t}$ or $\frac{\partial \rho U}{\partial t}$, but the coupling between the two equations. $\endgroup$ – akid Apr 2 '14 at 12:21
  • $\begingroup$ @akid: You should probably make it clear in the question that you don't mean explicit time-stepping since you use the word "explicit" in the same sentence with the phrase "time step". $\endgroup$ – Bill Barth Apr 2 '14 at 12:49
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    $\begingroup$ Ah, what you're talking about then is "operator splitting". Is the time step people take dominated by ensuring that the coupling works, or by what the Navier-Stokes equations dictate? $\endgroup$ – Wolfgang Bangerth Apr 2 '14 at 14:35

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