[CIG-SHORT] Convergence problem with Neumann BCs

Brad Aagaard baagaard at usgs.gov
Thu Oct 31 08:36:16 PDT 2013


On 10/30/2013 03:43 PM, Eric Lindsey wrote:
> Hi,
>
> I'd like to apply background stresses to a simple elastic 2D plane strain
> model in pylith, then add fault slip or other quasistatic deformation;
> later I'll modify the material properties etc. Of course the stresses also
> cause some deformation in the volume that I'd like to ignore. So I'm
> imposing the slip at the second time step, with the hopes of subtracting
> out the deformation from the first time step. (Is this the right way to do
> this?)

If you want to apply a background stress field over the domain without 
causing deformation, use initial stresses in the materials. This 
approach requires that you can analytically calculate the stress field 
over the domain. See examples/3d/hex8/step16.cfg for an illustration of 
how to do this.

If due to spatial variations in physical properties it is not possible 
to calculate the initial (background) stress field over the domain, then 
you can use your approach. As noted below, you will need some Dirichlet 
BC to remove rigid body motions.

> However, I noticed that the effects from the boundary conditions continue
> to increase, and then oscillate, and don't stabilize until I let it run to
> the 8th step or so (ideally I'd only use 2 time steps, I think). I think
> the problem is with the solver's convergence; I see the message
>
> ...
> 499 KSP Residual norm 9.952574507238e-04
> 500 KSP Residual norm 9.952555275971e-04
> Linear solve did not converge due to DIVERGED_ITS iterations 500
>
> I've taken most of the configuration straight from
> examples/2d/greensfns/strikeslip, and just modified the boundary conditions
> from Dirichlet to Neumann. Config file is attached; any advice would be
> great.

You do not have any Dirichlet BC, so the problem is ill-posed. You need 
sufficient Dirichlet BC to remove rigid body motion (translations and 
rotations). Usually this is done by constraining the DOF perpendicular 
to the boundary.

Regards,
Brad



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