I am working on a project involving steady solutions for the Navier-Stokes Equations. In the past I've only worked with the unsteady Navier-Stokes, so some of this is new to me.

In particular, at each iteration for some problems, one may need to resort to a relaxation parameter at each iteration, so letting $\tilde{\boldsymbol{u}}^n$ be the velocity field at an iteration $n$, we have:

$$ \boldsymbol{u}^n = \gamma \tilde{\boldsymbol{u}}^n + (1 - \gamma) \boldsymbol{u}^{n-1} $$ For some value of $\gamma$ between zero and one. This seems to work alright for my purposes, but I have a couple of questions:

  1. Is it necessary to relax both pressure and velocity? It seems for my purposes relaxing only velocity seems to be fine, and the pressure still converges to the desired physical solution. I do worry, however, that there are problems with consistency if the pressure is not also relaxed.
  2. If the pressure must also be relaxed, how does the pressure relxation parameter relate to the velocity relaxation parameter? Would they simply be the same value of $\gamma$?

Potentially relevant details:

  • I am using a finite element code with P2 for velocity and P1 for pressure.
  • I am prescribing a Dirichlet inflow with a parabolic profile and a Neumann outflow.
  • I am solving the coupled monolithic system (and not using the pressure Schur Complement approach).

Thanks in advance to anyone who has any answers for me!

  • 1
    $\begingroup$ With stable FEM don't you have a zero block for the pressure in the continuity equation? If so how do you solve this system without Schur complement approach? $\endgroup$
    – chris
    Commented Sep 20, 2016 at 14:53
  • $\begingroup$ It is actually possible to solve the saddle-point system monolithically (though this can become unfeasible when the number of unknowns is very high). There is a zero block for pressure in the continuity equation, but the pressure can still be uniquely determined as the Lagrange multplier enforcing the incompressibility constraint in the momentum equation (with the proper boundary conditions). When solved monolithically the splitting error incurred near the boundary is eliminated. $\endgroup$
    – A.Vigs
    Commented Sep 22, 2016 at 0:30

1 Answer 1

  1. No, it's not necessary to apply relaxation as long as the process converges to steady-state. Think of it as a correction: \begin{equation*} u^n = u^{n-1} + \gamma (\bar{u}^n-u^{n-1}) \end{equation*} As you converge to steady-state where $u^n=u^{n-1}$, the correction $u^\prime=\bar{u}^n-u^{n-1}$ becomes zero and so the dependence on $\gamma$ drops out. Same holds for the pressure, no need to relax for consistency of the converged solution. Of course, with $\gamma=1$ you might not get convergence.
  2. There is no obvious relation. You may use a lower value for the relaxation parameter $\gamma_p$ for the pressure to compensate for a higher value of $\gamma_u$ or vice-versa.

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