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

[Brad Aagaard]

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
>
>
>
>
>
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