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some modifications to the iterative solver to make them easier to use. 
There are also improved versions of the Darcy flux solver and the incompressible solver.


1
2 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
3 %
4 % Copyright (c) 2003-2008 by University of Queensland
5 % Earth Systems Science Computational Center (ESSCC)
6 % http://www.uq.edu.au/esscc
7 %
8 % Primary Business: Queensland, Australia
9 % Licensed under the Open Software License version 3.0
10 % http://www.opensource.org/licenses/osl-3.0.php
11 %
12 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
13
14
15 \chapter{Models}
16 \label{MODELS CHAPTER}
17
18 The following sections give a breif overview of the model classes and their corresponding methods.
19
20 \section{Stokes Problem}
21 The velocity \index{velocity} field $v$ and pressure $p$ of an incompressible fluid \index{incompressible fluid} is given as the solution of the Stokes problem\index{Stokes problem}
22 \begin{equation}\label{Stokes 1}
23 -\left(\eta(v\hackscore{i,j}+ v\hackscore{i,j})\right)\hackscore{,j}+p\hackscore{,i}=f\hackscore{i}-\sigma\hackscore{ij,j}
24 \end{equation}
25 where $\eta$ is the viscosity, $F\hackscore{i}$ defines an internal force \index{force, internal} and $\sigma\hackscore{ij}$ is an intial stress \index{stress, initial}. We assume an incompressible media:
26 \begin{equation}\label{Stokes 2}
27 -v\hackscore{i,i}=0
28 \end{equation}
29 Natural boundary conditions are taken in the form
30 \begin{equation}\label{Stokes Boundary}
31 \left(\eta(v\hackscore{i,j}+ v\hackscore{i,j})\right)n\hackscore{j}-n\hackscore{i}p=s\hackscore{i}+\sigma\hackscore{ij} n\hackscore{i}
32 \end{equation}
33 which can be overwritten by constraints of the form
34 \begin{equation}\label{Stokes Boundary0}
35 v\hackscore{i}(x)=v^D\hackscore{i}(x)
36 \end{equation}
37 at some locations $x$ at the boundary of the domain. The index $i$ may depend on the location $x$ on the boundary.
38 $v^D$ is a given function on the domain.
39
40 \subsection{Solution Method \label{STOKES SOLVE}}
41 In block form equation equations~\ref{Stokes 1} and~\ref{Stokes 2} takes the form of a saddle point problem
42 \index{saddle point problem}
43 \begin{equation}
44 \left[ \begin{array}{cc}
45 A & B^{*} \\
46 B & 0 \\
47 \end{array} \right]
48 \left[ \begin{array}{c}
49 v \\
50 p \\
51 \end{array} \right]
52 =\left[ \begin{array}{c}
53 G \\
54 0 \\
55 \end{array} \right]
56 \label{SADDLEPOINT}
57 \end{equation}
58 where $A$ is coercive, self-adjoint linear operator in a suitable Hilbert space, $B$ is the $(-1) \cdot$ divergence operator and $B^{*}$ is it adjoint operator (=gradient operator). For more details on the mathematics see references \cite{AAMIRBERKYAN2008,MBENZI2005}.
59 We use iterative techniques to solve this problem. To make sure that the incomressibilty condition holds
60 with sufficient accuracy we check for
61 \begin{equation}
62 \|v\hackscore{k,k}\| \hackscore \le \epsilon
63 \|\sqrt{v\hackscore{j,k}v\hackscore{j,k}}\|
64 \end{equation}
65 where $\epsilon$ is the desired relative accuracy and
66 \begin{equation}
67 \|p\|^2= \int\hackscore{\Omega} p^2 \; dx
68 \label{PRESSURE NORM}
69 \end{equation}
70 defines the $L^2$-norm.
71 There are two approaches to solve this problem. The first approach, called the Uzawa scheme \index{Uzawa scheme}
72 eliminates the velocity $v$ from the problem. The second approach solves the equation in coupled form after the application of a preconditioner.
73
74 \subsubsection{Uzawa scheme}
75 The first eqution in~\ref{SADDLEPOINT} gives $v=A^{-1}(G-B^{*}p)$ assuming $p$ is known. This is inserted into the
76 second eqution which leads to
77 \begin{equation}
78 S p = B A^{-1} G
79 \end{equation}
80 with the Schur complement \index{Schur complement} $S=BA^{-1}B^{*}$. This problem can be solved iteratively using the reconditioned Conjugate Gradient Method (PCG)~\index{PCG!Preconditioned Conjugate Gradient Method}
81 with the preconditioner $\hat{S}$ defined as $q=\hat{S}^{-1}p$ by solving
82 \begin{equation}
83 \frac{1}{\eta}q = p
84 \end{equation}
85 see~\cite{ELMAN} for more details. The evaluation of $w=Sp$ is done in the form
86 \begin{equation}
87 \begin{array}{rcl}
88 A v & = & B^{*}p \\
89 w & = & Bv \\
90 \end{array}
91 \label{EVAL PCG}
92 \end{equation}
93 The residual \index{residual} $r=B A^{-1} G - S p$ is given as
94 \begin{equation}
95 r=B A^{-1} (G - B^* p) = Bv \mbox{ with } v = A^{-1}(G-B^{*}p)
96 \end{equation}
97 Therefore one uses the tuple $(v,Bv)$ to represent the residual of the current pressure $p$. Notice that before the iteration is started the right hand side $B A^{-1} G$ needs to be calculated. The bilinear form $(.,.)$ used is defined as
98 \begin{equation}
99 (p,(v,Bv))=\int\hackscore{\Omega} p \cdot Bv \; dx
100 \end{equation}
101 where $p$ is the pressure increment and $(v,Bv)$ represents an increment in the residual.
102
103 \subsubsection{Coupled Solver}
104 An alternative approach to solve the saddle point problem~\ref{SADDLEPOINT} directly using an iterative such as
105 the generalized minimal residual method (GMRES) \index{generalized minimal residual method!GMRES} with a suitable
106 preconditioner. Here we use the operator
107 \begin{equation}
108 \left[ \begin{array}{cc}
109 A^{-1} & 0 \\
110 S^{-1} B A^{-1} & -S^{-1} \\
111 \end{array} \right]
112 \label{SADDLEPOINT PRECODITIONER}
113 \end{equation}
114 where again $S$ is the Schur complement~\cite{ELMAN}. In partice we will use an approximation $\hat{S}$ for $S$. The evaluation $(w,q)$ of the iteration operator for a given $(v,p)$ is done as
115 \begin{equation}
116 \begin{array}{rcl}
117 A w & = & Av+B^{*}p \\
118 \hat{S} q & = & B(w-v) \\
119 \end{array}
120 \label{COUPLES SADDLEPOINT iteration}
121 \end{equation}
122 We use the inner product induced by the norm
123 \begin{equation}
124 \|(v,p)\|^2= \int\hackscore{\Omega} v\hackscore{i,j} v\hackscore{i,j} + \left( \frac{p}{\eta}\right)^2\; dx
125 \label{COUPLES NORM}
126 \end{equation}
127 In PDE form~\ref{COUPLES SADDLEPOINT iteration} takes the form
128 \begin{equation}
129 \begin{array}{rcl}
130 -\left(\eta(w\hackscore{i,j}+ w\hackscore{i,j})\right)\hackscore{,j} & = & -\left(\eta(v\hackscore{i,j}+ v\hackscore{i,j})\right)\hackscore{,j}+p\hackscore{,i} \\
131 \frac{1}{\eta} q & = & - (w-v)\hackscore{i,i} \\
132 \end{array}
133 \label{SADDLEPOINT iteration 2}
134 \end{equation}
135
136
137 \subsection{Functions}
138
139 \begin{classdesc}{StokesProblemCartesian}{domain}
140 opens the Stokes problem\index{Stokes problem} on the \Domain domain. The approximation
141 order needs to be two.
142 \end{classdesc}
143
144 \begin{methoddesc}[StokesProblemCartesian]{initialize}{\optional{f=Data(), \optional{fixed_u_mask=Data(), \optional{eta=1, \optional{surface_stress=Data(), \optional{stress=Data()}}}}}}
145 assigns values to the model parameters. In any call all values must be set.
146 \var{f} defines the external force $f$, \var{eta} the viscosity $\eta$,
147 \var{surface_stress} the surface stress $s$ and \var{stress} the initial stress $\sigma$.
148 The locations and compontents where the velocity is fixed are set by
149 the values of \var{fixed_u_mask}. The method will try to cast the given values to appropriate
150 \Data class objects.
151 \end{methoddesc}
152
153 \begin{methoddesc}[StokesProblemCartesian]{solve}{v,p,
154 \optional{max_iter=20, \optional{verbose=False, \optional{useUzawa=True}}}}
155 solves the problem and return approximations for velocity and pressure.
156 The arguments \var{v} and \var{p} define initial guess. The values of \var{v} marked
157 by \var{fixed_u_mask} remain unchanged.
158 If \var{useUzawa} is set to \True
159 the Uzawa\index{Uszwa} scheme is used. Otherwise the problem is solved in coupled form. In most cases
160 the Uzawa scheme is more efficient.
161 \var{max_iter} defines the maximum number of iteration steps.
162 If \var{verbose} is set to \True informations on the progress of of the solver are printed.
163 \end{methoddesc}
164
165
166 \begin{methoddesc}[StokesProblemCartesian]{setTolerance}{\optional{tolerance=1.e-8}}
167 sets the tolerance in an appropriate norm relative to the right hand side. The tolerance must be non-negative and less than 1.
168 \end{methoddesc}
169 \begin{methoddesc}[StokesProblemCartesian]{getTolerance}{}
170 returns the current relative tolerance.
171 \end{methoddesc}
172 \begin{methoddesc}[StokesProblemCartesian]{setAbsoluteTolerance}{\optional{tolerance=0.}}
173 sets the absolute tolerance for the error in the relevant norm. The tolerance must be non-negative. Typically the
174 absolute talerance is set to 0.
175 \end{methoddesc}
176 \begin{methoddesc}[StokesProblemCartesian]{getAbsoluteTolerance}{}
177 sreturns the current absolute tolerance.
178 \end{methoddesc}
179 \begin{methoddesc}[StokesProblemCartesian]{setSubToleranceReductionFactor}{\optional{reduction=None}}
180 sets the reduction factor for the tolerance used to solve the PDEs. A reduction factor
181 in the order of one will minimize compute time per iteration step but my slow down convergence or even lead to
182 divergency. On the other hand a very small value for the PDE tolerance could result in a wast of compute time.
183 If \var{reduction} is set to \var{None} the sub-tolerance is solved adaptively but
184 in cases a very small tolerance is set ($<10^{-6}$) it is recommended to set the
185 reduction factor by hand. This may require some experiments.
186 \end{methoddesc}
187 \begin{methoddesc}[StokesProblemCartesian]{getSubToleranceReductionFactor}{}
188 return the current reduction factor for the sub-problem tolerance.
189 \end{methoddesc}
190
191 \subsection{Example: Lit Driven Cavity}
192 The following script \file{lit\hackscore driven\hackscore cavity.py}
193 \index{scripts!\file{helmholtz.py}} which is available in the \ExampleDirectory
194 illustrates the usage of the \class{StokesProblemCartesian} class to solve
195 the lit driven cavity problem~\cite{LITDRIVENCAVITY}:
196 \begin{python}
197 from esys.escript import *
198 from esys.finley import Rectangle
199 from esys.escript.models import StokesProblemCartesian
200 NE=25
201 dom = Rectangle(NE,NE,order=2)
202 x = dom.getX()
203 sc=StokesProblemCartesian(dom)
204 mask= (whereZero(x[0])*[1.,0]+whereZero(x[0]-1))*[1.,0] + \
205 (whereZero(x[1])*[0.,1.]+whereZero(x[1]-1))*[1.,1]
206 sc.initialize(eta=.1, fixed_u_mask= mask)
207 v=Vector(0.,Solution(dom))
208 v[0]+=whereZero(x[1]-1.)
209 p=Scalar(0.,ReducedSolution(dom))
210 v,p=sc.solve(v,p, verbose=True)
211 saveVTK("u.xml",velocity=v,pressure=p)
212 \end{python}
213
214 \section{Darcy Flux}
215 We want to calculate the velocity $u$ and pressure $p$ on a domain $\Omega$ solving
216 the Darcy flux problem \index{Darcy flux}\index{Darcy flow}
217 \begin{equation}\label{DARCY PROBLEM}
218 \begin{array}{rcl}
219 u\hackscore{i} + \kappa\hackscore{ij} p\hackscore{,j} & = & g\hackscore{i} \\
220 u\hackscore{k,k} & = & f
221 \end{array}
222 \end{equation}
223 with the boundary conditions
224 \begin{equation}\label{DARCY BOUNDARY}
225 \begin{array}{rcl}
226 u\hackscore{i} \; n\hackscore{i} = u^{N}\hackscore{i} \; n\hackscore{i} & \mbox{ on } & \Gamma\hackscore{N} \\
227 p = p^{D} & \mbox{ on } & \Gamma\hackscore{D} \\
228 \end{array}
229 \end{equation}
230 where $\Gamma\hackscore{N}$ and $\Gamma\hackscore{D}$ are a partition of the boundary of $\Omega$ with $\Gamma\hackscore{D}$ non empty, $n\hackscore{i}$ is the outer normal field of the boundary of $\Omega$, $u^{N}\hackscore{i}$ and $p^{D}$ are given functions on $\Omega$, $g\hackscore{i}$ and $f$ are given source terms and $\kappa\hackscore{ij}$ is the given permability. We assume that $\kappa\hackscore{ij}$ is symmetric (which is not really required) and positive definite, i.e there are positive constants $\alpha\hackscore{0}$ and $\alpha\hackscore{1}$ wich are independent from the location in $\Omega$ such that
231 \begin{equation}
232 \alpha\hackscore{0} \; x\hackscore{i} x\hackscore{i} \le \kappa\hackscore{ij} x\hackscore{i} x\hackscore{j} \le \alpha\hackscore{1} \; x\hackscore{i} x\hackscore{i}
233 \end{equation}
234 for all $x\hackscore{i}$.
235
236
237 \subsection{Solution Method \label{DARCY SOLVE}}
238 In practical applications it is an advantage to calculate the pressure $p$ as a correction of a 'static' pressure $p^{ref}$ which is the solution of
239 \begin{equation}
240 -(\kappa\hackscore{ki}\kappa\hackscore{kj} p\hackscore{,j}^{ref})\hackscore{,i} = - (\kappa\hackscore{ki} (g\hackscore{k}- u^{N}\hackscore{k}))\hackscore{,i}
241 \mbox{ with }
242 p^{ref} = p^{D} \mbox{ on } \Gamma\hackscore{D}
243 \end{equation}
244 With setting $u \leftarrow u-u^{N}$ and $p \leftarrow p-p^{ref}$ and
245 \begin{equation}
246 \begin{array}{rcl}
247 g\hackscore{i} & \leftarrow & g\hackscore{i} - u^{N}\hackscore{i} - \kappa\hackscore{ij} p^{ref}\hackscore{,j }\\
248 f & \leftarrow & f - u^{N}\hackscore{k,k}
249 \end{array}
250 \end{equation}
251 we can assume that $u^{N}\hackscore{i} \; n\hackscore{i}=0$ and
252 $p^{D}=0$. Notice that with this setting
253 \begin{equation}\label{DIV FREE DARCY FLUX}
254 (\kappa\hackscore{ki} g\hackscore{k})\hackscore{,i} = 0
255 \end{equation}
256 We set
257 \begin{equation}
258 V = \{ q \in H^{1}(\Omega) : q=0 \mbox{ on } \Gamma\hackscore{D} \}
259 \end{equation}
260 and
261 \begin{equation}
262 W = \{ v \in (L^2(\Omega))^{d} : v\hackscore{k,k} \in L^2(\Omega) \mbox{ and } u\hackscore{i} \; n\hackscore{i} =0 \mbox{ on } \Gamma\hackscore{N} \}
263 \end{equation}
264 and define the operator $Q: V \rightarrow (L^2(\Omega))^{d}$ defined by
265 \begin{equation}
266 (Qp)\hackscore{i} = \kappa\hackscore{ij} p\hackscore{,j}
267 \end{equation}
268 and the operator $D: W \rightarrow L^2(\Omega)$ defined by
269 \begin{equation}
270 Dv = v\hackscore{k,k}
271 \end{equation}
272 In operator notation the Darcy problem~\ref{DARCY PROBLEM} is written in the form
273 \begin{equation}
274 \begin{array}{rcl}
275 u + Qp & = & g \\
276 Du & = & f
277 \end{array}
278 \end{equation}
279 Notice that because of~\ref{DIV FREE DARCY FLUX} we have
280 \begin{equation} \label{ABSTRACT DIV FREE DARCY FLUX}
281 Q^*g =0
282 \end{equation}
283 where $Q^*$ denote the adjoint operators of $Q$.
284 We solve this equation by minimising the functional
285 \begin{equation}
286 J(u,p):=\|u + Qp - g\|^2\hackscore{0} + \|Du-f\|\hackscore{0}^2
287 \end{equation}
288 over $W \times V$ where $\|.\|\hackscore{0}$ denotes the norm in $L^2(\Omega)$. A simple calculation shows that
289 one has to solve
290 \begin{equation}
291 ( v + Qq , u + Qp - g) + (Dv,Du-f) =0
292 \end{equation}
293 for all $v\in W$ and $q \in V$.which translates back into operator notation
294 \begin{equation}
295 \begin{array}{rcl}
296 (I+D^*D)u + Qp & = & D^*f + g \\
297 Q^*u + Q^*Q p & = & 0 \\
298 \end{array}
299 \end{equation}
300 where $D^*$ and $Q^*$ denote the adjoint operators. We use the fact that $Q^*g=$.
301 In~\cite{XXX} it has been shown that this problem is continuous and coercive in $W \times V$ and therefore has a unique solution. Also standart FEM methods can be used for discretization. It is also possible
302 to solve the problem is coupled form, however this approach leads in some cases to a very ill-conditioned stiffness matrix in particular in the case of a very small or large permability ($\alpha\hackscore{1} \ll 1$ or $\alpha\hackscore{0} \gg 1$)
303
304 The approach we are taking is to eliminate the velocity $u$ from the problem. Assuming that $p$ is known we have
305 \begin{equation}
306 v= (I+D^*D)^{-1} (D^*f + g - Qp)
307 \end{equation}
308 (notice that $(I+D^*D)$ is coercive in $W$) which is inserted into the second equation
309 \begin{equation}
310 Q^* (I+D^*D)^{-1} (D^*f + g - Qp) + Q^* Q p = 0
311 \end{equation}
312 which is
313 \begin{equation}
314 Q^* ( I - (I+D^*D)^{-1} ) Q p = - Q^* (I+D^*D)^{-1} (D^*f + g) )
315 \end{equation}
316 We use the PCG method to solve this. The residual $r$ ($\in V^*$) is given as
317 \begin{equation}
318 \begin{array}{rcl}
319 r & = & Q^* \left( -(I+D^*D)^{-1} (D^*f + g) - Qp + (I+D^*D)^{-1}Q p \right)\\
320 & =& Q^* \left( - Qp - (I+D^*D)^{-1} (D^*f + g - Qp) \right) \\
321 & =& - Q^* \left( Qp + v \right)
322 \end{array}
323 \end{equation}
324 So in a partical implementation we use the pair $(Qp,v)$ to represent the residual. This will save the
325 reconstruction of the velocity $v$. In this notation the right hand side is given as
326 $(0,(I+D^*D)^{-1} (D^*f + g))$. The evaluation of the iteration operator for a given $p$ is then
327 returning $(Qp,w)$ where $w$ is the solution of
328 \begin{equation}\label{UPDATE W}
329 (I+D^*D)w = Qp
330 \end{equation}
331 We use $Q^*Q$ as a a preconditioner for the iteration operator $Q^* ( I - (I+D^*D)^{-1} ) Q$.
332
333 \subsection{Functions}
334 \begin{classdesc}{DarcyFlow}{domain}
335 opens the Darcy flux problem\index{Darcy flux} on the \Domain domain.
336 \end{classdesc}
337
338 \begin{methoddesc}[DarcyFlow]{initialize}{\optional{f=Data(), \optional{fixed_u_mask=Data(), \optional{eta=1, \optional{surface_stress=Data(), \optional{stress=Data()}}}}}}
339 assigns values to the model parameters. In any call all values must be set.
340 \var{f} defines the external force $f$, \var{eta} the viscosity $\eta$,
341 \var{surface_stress} the surface stress $s$ and \var{stress} the initial stress $\sigma$.
342 The locations and compontents where the velocity is fixed are set by
343 the values of \var{fixed_u_mask}. The method will try to cast the given values to appropriate
344 \Data class objects.
345 \end{methoddesc}
346
347
348 \subsection{Example: Gravity Flow}
349
350 %================================================
351 \section{Temperature Advection Diffusion\label{TEMP ADV DIFF}}
352
353 \begin{equation}
354 \rho c\hackscore{p} \left (\frac{\partial T}{\partial t} + \vec{v} \cdot \nabla T \right ) = k \nabla^{2}T
355 \label{HEAT EQUATION}
356 \end{equation}
357
358 where $\vec{v}$ is the velocity vector, $T$ is the temperature, $\rho$ is the density, $\eta$ is the viscosity, $c\hackscore{p}$ is the specific heat at constant pressure and $k$ is the thermal conductivity.
359
360 \subsection{Description}
361
362 \subsection{Method}
363
364 \begin{classdesc}{TemperatureCartesian}{dom,theta=THETA,useSUPG=SUPG}
365 \end{classdesc}
366
367 \subsection{Benchmark Problem}
368 %===============================================================================================================
369
370 %=========================================================
371 \input{levelsetmodel}
372
373 % \section{Drucker Prager Model}
374
375 \section{Isotropic Kelvin Material \label{IKM}}
376 As proposed by Kelvin~\ref{KELVN} material strain $D\hackscore{ij}=v\hackscore{i,j}+v\hackscore{j,i}$ can be decomposed into
377 an elastic part $D\hackscore{ij}^{el}$ and visco-plastic part $D\hackscore{ij}^{vp}$:
378 \begin{equation}\label{IKM-EQU-2}
379 D\hackscore{ij}=D\hackscore{ij}^{el}+D\hackscore{ij}^{vp}
380 \end{equation}
381 with the elastic strain given as
382 \begin{equation}\label{IKM-EQU-3}
383 D\hackscore{ij}'^{el}=\frac{1}{2 \mu} \dot{\sigma}\hackscore{ij}'
384 \end{equation}
385 where $\sigma'\hackscore{ij}$ is the deviatoric stress (Notice that $\sigma'\hackscore{ii}=0$).
386 If the material is composed by materials $q$ the visco-plastic strain can be decomposed as
387 \begin{equation}\label{IKM-EQU-4}
388 D\hackscore{ij}'^{vp}=\sum\hackscore{q} D\hackscore{ij}'^{q}
389 \end{equation}
390 where $D\hackscore{ij}^{q}$ is the strain in material $q$ given as
391 \begin{equation}\label{IKM-EQU-5}
392 D\hackscore{ij}'^{q}=\frac{1}{2 \eta^{q}} \sigma'\hackscore{ij}
393 \end{equation}
394 where $\eta^{q}$ is the viscosity of material $q$. We assume the following
395 betwee the the strain in material $q$
396 \begin{equation}\label{IKM-EQU-5b}
397 \eta^{q}=\eta^{q}\hackscore{N} \left(\frac{\tau}{\tau\hackscore{t}^q}\right)^{\frac{1}{n^{q}}-1}
398 \mbox{ with } \tau=\sqrt{\frac{1}{2}\sigma'\hackscore{ij} \sigma'\hackscore{ij}}
399 \end{equation}
400 for a given power law coefficients $n^{q}$ and transition stresses $\tau\hackscore{t}^q$, see~\ref{POERLAW}.
401 Notice that $n^{q}=1$ gives a constant viscosity.
402 After inserting equation~\ref{IKM-EQU-5} into equation \ref{IKM-EQU-4} one gets:
403 \begin{equation}\label{IKM-EQU-6}
404 D\hackscore{ij}'^{vp}=\frac{1}{2 \eta^{vp}} \sigma'\hackscore{ij} \mbox{ with } \frac{1}{\eta^{vp}} = \sum\hackscore{q} \frac{1}{\eta^{q}} \;.
405 \end{equation}
406 With
407 \begin{equation}\label{IKM-EQU-8}
408 \dot{\gamma}=\sqrt{2 D\hackscore{ij} D\hackscore{ij}}
409 \end{equation}
410 one gets
411 \begin{equation}\label{IKM-EQU-8b}
412 \tau = \eta^{vp} \dot{\gamma}^{vp} \;.
413 \end{equation}
414 With the Drucker-Prager cohesion factor $\tau\hackscore{Y}$, Drucker-Prager friction $\beta$ and total pressure $p$ we want to achieve
415 \begin{equation}\label{IKM-EQU-8c}
416 \tau \le \tau\hackscore{Y} + \beta \; p
417 \end{equation}
418 which leads to the condition
419 \begin{equation}\label{IKM-EQU-8d}
420 \eta^{vp} \le \frac{\tau\hackscore{Y} + \beta \; p}{ \dot{\gamma}^{vp}} \; .
421 \end{equation}
422 Therefore we modify the definition of $\eta^{vp}$ to the form
423 \begin{equation}\label{IKM-EQU-6b}
424 \frac{1}{\eta^{vp}}=\max(\sum\hackscore{q} \frac{1}{\eta^{q}}, \frac{\dot{\gamma}^{vp}} {\tau\hackscore{Y} + \beta \; p})
425 \end{equation}
426 The deviatoric stress needs to fullfill the equilibrion equation
427 \begin{equation}\label{IKM-EQU-1}
428 -\sigma'\hackscore{ij,j}+p\hackscore{,i}=F\hackscore{i}
429 \end{equation}
430 where $F\hackscore{j}$ is a given external fource. We assume an incompressible media:
431 \begin{equation}\label{IKM-EQU-2}
432 -v\hackscore{i,i}=0
433 \end{equation}
434 Natural boundary conditions are taken in the form
435 \begin{equation}\label{IKM-EQU-Boundary}
436 \sigma'\hackscore{ij}n\hackscore{j}-n\hackscore{i}p=f
437 \end{equation}
438 which can be overwritten by a constraint
439 \begin{equation}\label{IKM-EQU-Boundary0}
440 v\hackscore{i}(x)=0
441 \end{equation}
442 where the index $i$ may depend on the location $x$ on the bondary.
443
444 \subsection{Solution Method \label{IKM-SOLVE}}
445 By using a first order finite difference approximation wit step size $dt>0$~\ref{IKM-EQU-3} get the form
446 \begin{equation}\label{IKM-EQU-3b}
447 D\hackscore{ij}'^{el}=\frac{1}{2 \mu dt } \left( \sigma\hackscore{ij}' - \sigma\hackscore{ij}^{'-} \right)
448 \end{equation}
449 where $\sigma\hackscore{ij}^{'-}$ is the deviatoric stress at the precious time step.
450 Now we can combine equations~\ref{IKM-EQU-2}, \ref{IKM-EQU-3b} and~\ref{IKM-EQU-6b} to get
451 \begin{equation}\label{IKM-EQU-10}
452 \sigma\hackscore{ij}' = 2 \eta\hackscore{eff} \left( D\hackscore{ij}' +
453 \frac{1}{ 2 \mu \; dt} \sigma\hackscore{ij}^{'-}\right) \mbox{ with }
454 \frac{1}{\eta\hackscore{eff}}=\frac{1}{\mu \; dt}+\frac{1}{\eta^{vp}}
455 \end{equation}
456 Notice that $\eta\hackscore{eff}$ is a function of diatoric stress $\sigma\hackscore{ij}'$.
457 After inserting~\ref{IKM-EQU-10} into~\ref{IKM-EQU-1} we get
458 \begin{equation}\label{IKM-EQU-1ib}
459 -\left(\eta\hackscore{eff} (v\hackscore{i,j}+ v\hackscore{i,j})
460 \right)\hackscore{,j}+p\hackscore{,i}=F\hackscore{i}+
461 \frac{\eta\hackscore{eff}}{\mu dt } \sigma\hackscore{ij,j}^{'-}
462 \end{equation}
463 Together with the incomressibilty condition~\ref{IKM-EQU-2} we need to solve a problem with a form almost identical
464 to the Stokes problem discussed in section~\ref{STOKES SOLVE} but with the difference that $\eta\hackscore{eff}$ is depending on the solution. Analog to the iteration scheme~\ref{SADDLEPOINT iteration 2} we can run
465 \begin{equation}
466 \begin{array}{rcl}
467 -\left(\eta\hackscore{eff}(dv\hackscore{i,j}+ dv\hackscore{i,j}
468 )\right)\hackscore{,j} & = & F\hackscore{i}+ \sigma\hackscore{ij,j}'-p\hackscore{,i} \\
469 \frac{1}{\eta\hackscore{eff}} dp & = & - v\hackscore{i,i}^+
470 \end{array}
471 \label{IKM iteration 2}
472 \end{equation}
473 where $v^+=v+dv$. As this problem is non-linear the Jacobi-free Newton-GMRES method is used with the norm
474 \begin{equation}
475 \|(v, p)\|^2= \int\hackscore{\Omega} v\hackscore{i,j}^2 + \frac{1}{\bar{\eta}^2\hackscore{eff}} p^2 \; dx
476 \label{IKM iteration 3}
477 \end{equation}
478 where $\bar{\eta}\hackscore{eff}$ is the caracteristic viscosity, for instance:
479 \begin{equation}
480 \frac{1}{\bar{\eta}\hackscore{eff}} = \frac{1}{\tau^{-}}+\sum\hackscore{q} \frac{1}{\eta^{q}\hackscore{N}}
481 \label{IKM iteration 4}
482 \end{equation}
483 In oder to perform step~\ref{IKM iteration 2} we need to calculate the $\eta\hackscore{eff}$ as well as $\sigma\hackscore{ij}'$ while via $\tau$ the first is a function of the latter. The priority is the
484 calculation of $\eta\hackscore{eff}$ with the Newton-Raphson scheme. This value can then be used to calculate
485 $\sigma\hackscore{ij}'$ via~\ref{IKM-EQU-10}. We need to solve
486 \begin{equation}
487 \tau = \eta\hackscore{eff} \cdot \epsilon \mbox{ with }
488 \epsilon = \sqrt{ 2 \left( D\hackscore{ij}' +
489 \frac{1}{ 2 \mu \; dt} \sigma\hackscore{ij}^{'-}\right)^2}
490 \label{IKM iteration 5}
491 \end{equation}
492 The Newton scheme takes the form
493 \begin{equation}
494 \tau\hackscore{n+1} = \min(\tau\hackscore{n} - \frac{\tau\hackscore{n} - \eta\hackscore{eff} \cdot \epsilon}{1 - \eta\hackscore{eff}' \cdot \epsilon}, \tau\hackscore{Y} + \beta \; p)
495 = \min(\frac{\eta\hackscore{eff} - \tau\hackscore{n} \eta\hackscore{eff}'}
496 {1 - \eta\hackscore{eff}' \cdot \epsilon}, \frac{\tau\hackscore{Y} + \beta \; p}{\epsilon}) \epsilon
497 \label{IKM iteration 6}
498 \end{equation}
499 where $\eta\hackscore{eff}'$ denotes the derivative of $\eta\hackscore{eff}$ with respect of $\tau$. The second term in $\min$ is droped of $\tau\hackscore{Y} + \beta \; p<0$ or $\epsilon=0$. In fact we have
500 \begin{equation}
501 \eta\hackscore{eff}' = - \eta\hackscore{eff}^2 \left(\frac{1}{\eta\hackscore{eff}}\right)'
502 \mbox{ with }
503 \left(\frac{1}{\eta\hackscore{eff}}\right)' = \sum\hackscore{q} \left(\frac{1}{\eta^{q}} \right)'
504 \label{IKM iteration 7}
505 \end{equation}
506 \begin{equation}\label{IKM-EQU-5XX}
507 \left(\frac{1}{\eta^{q}} \right)'
508 = \frac{1-\frac{1}{n^{q}}}{\eta^{q}\hackscore{N}} \frac{\tau^{-\frac{1}{n^{q}}}}{(\tau\hackscore{t}^q)^{1-\frac{1}{n^{q}}}}
509 = \frac{1-\frac{1}{n^{q}}}{ \tau \eta^{q}}
510 \end{equation}
511 Notice that allways $\eta\hackscore{eff}'\le 0$ which makes the denomionator in~\ref{IKM iteration 6}
512 positive.
513
514
515

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