[CIG-SHORT] Slippery Nodes in Pylith

[Nick Wogan]

Charles,

I couldn't get the SNES solver to converge with friction and cohesion both
zero. It converged after ~100 SNES iterations for a coefficient of friction
of 0.1 for the particular problem I am looking at.

The slippery fault is suppose to represent a fault that has had many years
to evolve and almost completely relax the shear stress on the fault plane.
I suppose 0.1 coefficient of friction is representative of this?

Thanks,
Nick



On Thu, Mar 16, 2017 at 6:24 PM, Charles Williams <willic3 at gmail.com> wrote:

> Hi Nick,
>
> It’s only linear if you already know the slip direction.  In the more
> general case, it’s a nonlinear problem.  Also, I suspect your solution will
> be unstable if the friction and cohesion are both zero (I might be wrong —
> you can experiment).  You might need just a bit of friction and/or
> cohesion.  What is the slippery fault meant to represent?
>
> Cheers,
> Charles
>
>
> On 17/03/2017, at 1:42 PM, Nick Wogan <nicholaswogan at gmail.com> wrote:
>
> Brad,
>
> Is this the best way to do slippery faults? I thought slippery faults
> could be solved linearly.
>
> [pylithapp.timedependent.interfaces.fault]
> friction = pylith.friction.StaticFriction
> friction.db_properties.values = [friction-coefficient,cohesion]
> friction.db_properties.data = [0.0,0.0*Pa]
>
> What am i missing?
> Nick
>
> On Thu, Mar 16, 2017 at 4:32 PM, Brad Aagaard <baagaard at usgs.gov> wrote:
>
>> On 03/16/2017 04:29 PM, Nick Wogan wrote:
>>
>>> Hi,
>>>
>>> Is there a slippery nodes option for faults available?
>>>
>>> Thanks,
>>> Nick
>>>
>>
>> See Section 6.4.5 Dynamic Earthquake Rupture in the PyLith v2.1.4 manual.
>>
>> Regards,
>> Brad
>>
>>
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