[CIG-SHORT] difficulty with implicit formulation and rate-and-state fault slip

[Matthew Knepley]

On Thu, Aug 23, 2012 at 2:27 PM, Lucas Abraham Willemsen
<lawillem at mit.edu>wrote:

> Thanks Brad,
>
> It seems like I (falsely) assumed that the nonlinear solver was enabled
> automatically when an implicit solution strategy was used. Thanks for your
> help.
>

But we should never have to remember. Always run with --petsc.snes_view and
then if something is wrong, you
can always check the solver setup. This is the used in all our examples.

  Thanks,

     Matt


> best,
> Lucas
> ________________________________________
> From: cig-short-bounces at geodynamics.org [cig-short-bounces at geodynamics.org]
> on behalf of Brad Aagaard [baagaard at usgs.gov]
> Sent: Thursday, August 23, 2012 15:20
> To: cig-short at geodynamics.org
> Subject: Re: [CIG-SHORT] difficulty with implicit formulation and
> rate-and-state fault slip
>
> Lucas,
>
> You are going to feel a bit embarrassed about this one. You forgot to
> use a nonlinear solver. Using just the linear solver is almost
> equivalent to the first iteration of the nonlinear solver in which
> PyLith assumes the fault is locked. This is why you don't get any slip.
>
> Brad
>
>
> On 08/23/2012 09:15 AM, Lucas Abraham Willemsen wrote:
> > Hello,
> >
> > I'm currently investigating the behaviour of a 2D-block with an internal
> fault with rate-and-state friction. The idea is to compare the final
> displacement calculated with a dynamic and a quasi-static simulation.
> > I first do a dynamic simulation using a quasi-static
> nondimensionalization object (equivalent to using a dynamic object). The
> min_slip_rate parameter in rate-and-state friction is nondimensional, so
> the length and time scales I choose will influence the frictional
> coefficient and therefore the strength at the start of the simulation when
> the slip_rate is still 0. With all the values I use the coefficient of
> friction is exactly 0.4 at zero slip rate. This is easily verified in the
> dynamic project. I impose an initial normal stress on the fault of 190 MPa.
> With a frictional coefficient of 0.4 the strength of the fault should be
> 76.0 MPa. This is verified by the locked behavior for initial shear
> stresses below 76.0 MPa and fault movement for initial shear stresses above
> 76.0 MPa.
> >
> > I then copied the project and changed the formulation from explicit
> (dynamic) to implicit (quasi-static), keeping everything else unchanged. To
> my surprise the fault did not rupture anymore for shear stresses above 76.0
> MPa. I then changed the project slightly:
> >
> > -shorter total simulation time: to reduce computation time
> > -change absorbing boundary conditions (not very natural in
> quasi-static): Instead I fix the x and y displacement at the bottom
> > -reduce zero-tolerance significantly to 1.0e-15, making sure it is still
> larger than the absolute KSP tolerance.
> > -impose a very large shear stress rate on the fault (10 MPa per second
> for 10 seconds)
> >
> > With the new shear stress rate, the final shear stress on the fault is
> 180 MPa, while the normal stress is still 190 MPa. Even with such large
> shear stresses, the slip on the fault is zero according to paraview. The
> same would happen if I would increase the shear stress even 10 times more
> to 1800 MPa. Am I doing something wrong with the solver parameters? I did
> choose very conservative solver parameters resulting in many linear
> iterations, just in case. I left the nonlinear parameters untouched. I
> currently only simulate 10 seconds, since I also did this in the dynamic
> simulations. I tried smearing out the shear traction rate over a year
> instead, but the slip on the fault remained zero.
> >
> > My project can be found at the following location:
> http://web.mit.edu/lawillem/www/send.zip . It contains both the dynamic
> simulation (with the quasi nondim object) and the quasi-dynamic simulation.
> > It runs using the following command: pylith tri3_80m.cfg
> >
> > I'd be very grateful if someone could explain to my why the fault does
> not rupture in the quasi-static case, even though it does in the dynamic
> case (the strength should be the same since I left the parameters related
> to the friction unchanged).
> >
> > with kind regards,
> > Lucas
> >
> >
> >
> >
> >
> > _______________________________________________
> > CIG-SHORT mailing list
> > CIG-SHORT at geodynamics.org
> > http://geodynamics.org/cgi-bin/mailman/listinfo/cig-short
> >
>
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-- 
What most experimenters take for granted before they begin their
experiments is infinitely more interesting than any results to which their
experiments lead.
-- Norbert Wiener
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