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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 \label{STOKES STOP}
65 \end{equation}
66 where $\epsilon$ is the desired relative accuracy and
67 \begin{equation}
68 \|p\|^2= \int\hackscore{\Omega} p^2 \; dx
69 \label{PRESSURE NORM}
70 \end{equation}
71 defines the $L^2$-norm. We use the Uzawa scheme \index{Uzawa scheme} to solve the problem.
72
73 In fact the first equation in~\ref{SADDLEPOINT} gives for a known pressure
74 \begin{equation}
75 v=A^{-1}(G-B^{*}p)
76 \label{V CALC}
77 \end{equation}
78 which is inserted into the second equation leading to
79 \begin{equation}
80 S p = B A^{-1} G
81 \end{equation}
82 with the Schur complement \index{Schur complement} $S=BA^{-1}B^{*}$. This problem can be solved iteratively
83 with the preconditioner $\hat{S}$ defined as $q=\hat{S}^{-1}p$ by solving
84 \begin{equation}
85 \frac{1}{\eta}q = p
86 \end{equation}
87 see~\cite{ELMAN} for more details. Note that the residual for the current approximation $p$ is given as
88 \begin{equation}
89 r=B A^{-1} (G - B^* p) = Bv
90 \end{equation}
91 where $v$ is given by~\ref{V CALC}.
92
93 If one uses the generalized minimal residual method (GMRES) \index{generalized minimal residual method!GMRES}
94 the method is directly applied to the preconditioned system
95 \begin{equation}
96 \hat{S}^{-1} S p = \hat{S}^{-1} B A^{-1} G
97 \end{equation}
98 We use the norm
99 \begin{equation}
100 \|p\|\hackscore{GMRES} = \|\hat{S} p \|
101 \end{equation}
102 Notice that for the residual $\hat{r}=\hat{S}^{-1} r$ one has
103 \begin{equation}
104 \
105 \end{equation}
106 If $p^{0}$ provides an initial guess for the pressure we use~\ref{V CALC} to get a first initial guess for the
107 velocity $v^{0}$ which we use to set an absolute tolerance $ATOL =\epsilon \|\sqrt{v^{0}\hackscore{j,k}v^{0}\hackscore{j,k}}\|$.
108 The GMRES is terminated when
109 \begin{equation}
110 \|\hat{r}\|\hackscore{GMRES} \le ATOL
111 \end{equation}
112 Notice that $\|\hat{r}\|\hackscore{GMRES}= \|r \| = \|Bv\| = \|v\hackscore{k,k}\|$ so we we can expect that
113 the target stopping criterion~\ref{STOKES STOP} is fullfilled. However, if $v$ is very different from the
114 initial choice of $v^{0}$ the value of $ATOL$ is corrected and GMRES is restarted with a new tolerance. For time dependend problems this apprach works well as value for $p$ form a previous time step provides a good initial guess.
115
116 Alternatively, as $S$ is symmetric and positive definite one can apply the preconditioned conjugate gradient method (PCG) \index{preconditioned conjugate gradient method!PCG}. PCG use the norm
117 \begin{equation}
118 \|r\|\hackscore{PCG}^2 = \int\hackscore{\Omega} r \hat{S}^{-1}r \; dx = \int\hackscore{\Omega} \eta r^2 \; dx
119 \end{equation}
120 To take the extra factor $\eta$ into consideration when checking the stopping criterion we use the following
121 definition for $ATOL$:
122 \begin{equation}
123 ATOL = \epsilon \frac{\|\sqrt{v^{0}\hackscore{j,k}v^{0}\hackscore{j,k}}\| }{\|v^{0}\hackscore{k,k}\|}
124 \|v^{0}\hackscore{k,k}\|\hackscore{PCG}
125 \end{equation}
126
127
128
129 \subsection{Functions}
130
131 \begin{classdesc}{StokesProblemCartesian}{domain}
132 opens the Stokes problem\index{Stokes problem} on the \Domain domain. The approximation
133 order needs to be two.
134 \end{classdesc}
135
136 \begin{methoddesc}[StokesProblemCartesian]{initialize}{\optional{f=Data(), \optional{fixed_u_mask=Data(), \optional{eta=1, \optional{surface_stress=Data(), \optional{stress=Data()}}}}}}
137 assigns values to the model parameters. In any call all values must be set.
138 \var{f} defines the external force $f$, \var{eta} the viscosity $\eta$,
139 \var{surface_stress} the surface stress $s$ and \var{stress} the initial stress $\sigma$.
140 The locations and compontents where the velocity is fixed are set by
141 the values of \var{fixed_u_mask}. The method will try to cast the given values to appropriate
142 \Data class objects.
143 \end{methoddesc}
144
145 \begin{methoddesc}[StokesProblemCartesian]{solve}{v,p,
146 \optional{max_iter=20, \optional{verbose=False, \optional{usePCG=True}}}}
147 solves the problem and return approximations for velocity and pressure.
148 The arguments \var{v} and \var{p} define initial guess. The values of \var{v} marked
149 by \var{fixed_u_mask} remain unchanged.
150 If \var{usePCG} is set to \True
151 reconditioned conjugate gradient method (PCG) \index{preconditioned conjugate gradient method!PCG} scheme is used. Otherwise the problem is solved generalized minimal residual method (GMRES) \index{generalized minimal residual method!GMRES}. In most cases
152 the PCG scheme is more efficient.
153 \var{max_iter} defines the maximum number of iteration steps.
154 If \var{verbose} is set to \True informations on the progress of of the solver are printed.
155 \end{methoddesc}
156
157
158 \begin{methoddesc}[StokesProblemCartesian]{setTolerance}{\optional{tolerance=1.e-4}}
159 sets the tolerance in an appropriate norm relative to the right hand side. The tolerance must be non-negative and less than 1.
160 \end{methoddesc}
161 \begin{methoddesc}[StokesProblemCartesian]{getTolerance}{}
162 returns the current relative tolerance.
163 \end{methoddesc}
164 \begin{methoddesc}[StokesProblemCartesian]{setAbsoluteTolerance}{\optional{tolerance=0.}}
165 sets the absolute tolerance for the error in the relevant norm. The tolerance must be non-negative. Typically the
166 absolute talerance is set to 0.
167 \end{methoddesc}
168 \begin{methoddesc}[StokesProblemCartesian]{getAbsoluteTolerance}{}
169 sreturns the current absolute tolerance.
170 \end{methoddesc}
171 \begin{methoddesc}[StokesProblemCartesian]{setSubProblemTolerance}{\optional{rtol=None}}
172 sets the tolerance to solve the involved PDEs. The subtolerance \var{rtol} should not be choosen to large
173 in order to avoid feed back of errors in the subproblem solution into the outer iteration.
174 On the otherhand is choosen to small compute time is wasted.
175 If \var{rtol} is set to \var{None} the sub-tolerance is set automatically depending on the
176 tolerance choosen for the oter iteration.
177 \end{methoddesc}
178 \begin{methoddesc}[StokesProblemCartesian]{getSubProblemTolerance}{}
179 return the tolerance for the involved PDEs.
180 \end{methoddesc}
181
182 \subsection{Example: Lit Driven Cavity}
183 The following script \file{lit\hackscore driven\hackscore cavity.py}
184 \index{scripts!\file{helmholtz.py}} which is available in the \ExampleDirectory
185 illustrates the usage of the \class{StokesProblemCartesian} class to solve
186 the lit driven cavity problem:
187 \begin{python}
188 from esys.escript import *
189 from esys.finley import Rectangle
190 from esys.escript.models import StokesProblemCartesian
191 NE=25
192 dom = Rectangle(NE,NE,order=2)
193 x = dom.getX()
194 sc=StokesProblemCartesian(dom)
195 mask= (whereZero(x[0])*[1.,0]+whereZero(x[0]-1))*[1.,0] + \
196 (whereZero(x[1])*[0.,1.]+whereZero(x[1]-1))*[1.,1]
197 sc.initialize(eta=.1, fixed_u_mask= mask)
198 v=Vector(0.,Solution(dom))
199 v[0]+=whereZero(x[1]-1.)
200 p=Scalar(0.,ReducedSolution(dom))
201 v,p=sc.solve(v,p, verbose=True)
202 saveVTK("u.xml",velocity=v,pressure=p)
203 \end{python}
204
205 \section{Darcy Flux}
206 We want to calculate the velocity $u$ and pressure $p$ on a domain $\Omega$ solving
207 the Darcy flux problem \index{Darcy flux}\index{Darcy flow}
208 \begin{equation}\label{DARCY PROBLEM}
209 \begin{array}{rcl}
210 u\hackscore{i} + \kappa\hackscore{ij} p\hackscore{,j} & = & g\hackscore{i} \\
211 u\hackscore{k,k} & = & f
212 \end{array}
213 \end{equation}
214 with the boundary conditions
215 \begin{equation}\label{DARCY BOUNDARY}
216 \begin{array}{rcl}
217 u\hackscore{i} \; n\hackscore{i} = u^{N}\hackscore{i} \; n\hackscore{i} & \mbox{ on } & \Gamma\hackscore{N} \\
218 p = p^{D} & \mbox{ on } & \Gamma\hackscore{D} \\
219 \end{array}
220 \end{equation}
221 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
222 \begin{equation}
223 \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}
224 \end{equation}
225 for all $x\hackscore{i}$.
226
227
228 \subsection{Solution Method \label{DARCY SOLVE}}
229 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
230 \begin{equation}
231 -(\kappa\hackscore{ki}\kappa\hackscore{kj} p\hackscore{,j}^{ref})\hackscore{,i} = - (\kappa\hackscore{ki} (g\hackscore{k}- u^{N}\hackscore{k}))\hackscore{,i}
232 \mbox{ with }
233 p^{ref} = p^{D} \mbox{ on } \Gamma\hackscore{D}
234 \end{equation}
235 With setting $u \leftarrow u-u^{N}$ and $p \leftarrow p-p^{ref}$ and
236 \begin{equation}
237 \begin{array}{rcl}
238 g\hackscore{i} & \leftarrow & g\hackscore{i} - u^{N}\hackscore{i} - \kappa\hackscore{ij} p^{ref}\hackscore{,j }\\
239 f & \leftarrow & f - u^{N}\hackscore{k,k}
240 \end{array}
241 \end{equation}
242 we can assume that $u^{N}\hackscore{i} \; n\hackscore{i}=0$ and
243 $p^{D}=0$.
244 We set
245 \begin{equation}
246 V = \{ q \in H^{1}(\Omega) : q=0 \mbox{ on } \Gamma\hackscore{D} \}
247 \end{equation}
248 and
249 \begin{equation}
250 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} \}
251 \end{equation}
252 and define the operator $Q: V \rightarrow (L^2(\Omega))^{d}$ defined by
253 \begin{equation}
254 (Qp)\hackscore{i} = \kappa\hackscore{ij} p\hackscore{,j}
255 \end{equation}
256 and the operator $D: W \rightarrow L^2(\Omega)$ defined by
257 \begin{equation}
258 Dv = v\hackscore{k,k}
259 \end{equation}
260 In operator notation the Darcy problem~\ref{DARCY PROBLEM} is written in the form
261 \begin{equation}
262 \begin{array}{rcl}
263 u + Qp & = & g \\
264 Du & = & f
265 \end{array}
266 \end{equation}
267 We solve this equation by minimising the functional
268 \begin{equation}
269 J(u,p):=\|u + Qp - g\|^2\hackscore{0} + \|Du-f\|\hackscore{0}^2
270 \end{equation}
271 over $W \times V$ where $\|.\|\hackscore{0}$ denotes the norm in $L^2(\Omega)$. A simple calculation shows that
272 one has to solve
273 \begin{equation}
274 ( v + Qq , u + Qp - g) + (Dv,Du-f) =0
275 \end{equation}
276 for all $v\in W$ and $q \in V$.which translates back into operator notation
277 \begin{equation}
278 \begin{array}{rcl}
279 (I+D^*D)u + Qp & = & D^*f + g \\
280 Q^*u + Q^*Q p & = & Q^*g \\
281 \end{array}
282 \end{equation}
283 where $D^*$ and $Q^*$ denote the adjoint operators.
284 In~\cite{LEASTSQUARESFEM1994} 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
285 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 $\alpha\hackscore{0} \gg 1$)
286
287 The approach we are taking is to eliminate the velocity $u$ from the problem. Assuming that $p$ is known we have
288 \begin{equation}\label{DARCY V FORM}
289 v= (I+D^*D)^{-1} (D^*f + g - Qp)
290 \end{equation}
291 (notice that $(I+D^*D)$ is coercive in $W$) which is inserted into the second equation
292 \begin{equation}
293 Q^* (I+D^*D)^{-1} (D^*f + g - Qp) + Q^* Q p = Q^*g
294 \end{equation}
295 which is
296 \begin{equation}
297 Q^* ( I - (I+D^*D)^{-1} ) Q p = Q^* (g-(I+D^*D)^{-1} (D^*f + g) )
298 \end{equation}
299 We use the PCG \index{linear solver!PCG}\index{PCG} method to solve this. The residual $r$ ($\in V^*$) is given as
300 \begin{equation}
301 \begin{array}{rcl}
302 r & = & Q^* \left( g -(I+D^*D)^{-1} (D^*f + g) - Qp + (I+D^*D)^{-1}Q p \right)\\
303 & =& Q^* \left( g- Qp - (I+D^*D)^{-1} (D^*f + g - Qp) \right) \\
304 & =& Q^* \left( g - Qp - v \right)
305 \end{array}
306 \end{equation}
307 So in a partical implementation we use $\hat{r}=g-Qp-v$ to represent the residual.
308 The evaluation of the iteration operator for a given $p$ is then
309 returning $Qp+v$ where $v$ is the solution of
310 \begin{equation}\label{UPDATE W}
311 (I+D^*D)v = Qp
312 \end{equation}
313 We use $(Q^*Q)^{-1}$ as a preconditioner for the iteration operator $Q^* ( I - (I+D^*D)^{-1} ) Q$. So the application of the preconditioner to $\hat{r}$ representing the residual is given by solving
314 implemented by solving
315 \begin{equation}\label{UPDATE P}
316 Q^*Q q = Q^*\hat{r}
317 \end{equation}
318 The residual norm used in the PCG is given as
319 \begin{equation}\label{DARCY R NORM}
320 \|r\|\hackscore{PCG}^2 = \int r \cdot (Q^*Q)^{-1} r \; dx =\int \hat{r} \cdot Q (Q^*Q)^{-1} Q^* \hat{r} \; dx \approx
321 \|\hat{r}\|\hackscore{0}^2
322 \end{equation}
323 The iteration is terminated if
324 \begin{equation}\label{DARCY STOP}
325 \|r\|\hackscore{PCG} \le \mbox{ATOL}
326 \end{equation}
327 where we set
328 \begin{equation}\label{DARCY ATOL DEF}
329 \mbox{ATOL} = \mbox{atol} + \mbox{rtol} \cdot \left(\frac{1}{\|v\|\hackscore{0}} + \frac{1}{\|Qp\|\hackscore{0}} \right)^{-1}
330 \end{equation}
331 where rtol is a given relative tolerance and $\mbox{atol}$ is a given absolute tolerance (typically $=0$).
332 Notice that if $Qp$ and $v$ both are zero, the pair $(0,p)$ is a solution.
333 The problem is that ATOL is depending on the solution $p$ (and $v$ calculated form~\ref{DARCY V FORM}). In partcice one use the initial guess for $p$
334 to get a first value for ATOL. If the stopping crierion is met in the PCG iteration, a new $v$ is calculated from the current pressure approximation and ATOL is recalculated. If \ref{DARCY STOP} is still fullfilled the calculation is terminated and $(v,p)$ is returned. Otherwise PCG is restarted with a new ATOL.
335
336 \subsection{Functions}
337 \begin{classdesc}{DarcyFlow}{domain}
338 opens the Darcy flux problem\index{Darcy flux} on the \Domain domain.
339 \end{classdesc}
340 \begin{methoddesc}[DarcyFlow]{setValue}{\optional{f=None, \optional{g=None, \optional{location_of_fixed_pressure=None, \optional{location_of_fixed_flux=None, \optional{permeability=None}}}}}}
341 assigns values to the model parameters. Values can be assigned using various calls - in particular
342 in a time dependend problem only values that change over time needs to be reset. The permability can be defined as scalar (isotropic), a vector (orthotropic) or a matrix (anisotropic).
343 \var{f} and \var{g} are the corresponding parameters in~\ref{DARCY PROBLEM}.
344 The locations and compontents where the flux is prescribed are set by positive values in
345 \var{location_of_fixed_flux}.
346 The locations where the pressure is prescribed are set by
347 by positive values of \var{location_of_fixed_pressure}.
348 The values of the pressure and flux are defined by the initial guess.
349 Notice that at any point on the boundary of the domain the pressure or the normal component of
350 the flux must be defined. There must be at least one point where the pressure is prescribed.
351 The method will try to cast the given values to appropriate
352 \Data class objects.
353 \end{methoddesc}
354
355 \begin{methoddesc}[DarcyFlow]{setTolerance}{\optional{rtol=1e-4}}
356 sets the relative tolerance \mbox{rtol} in \ref{DARCY ATOL DEF}.
357 \end{methoddesc}
358
359 \begin{methoddesc}[DarcyFlow]{setAbsoluteTolerance}{\optional{atol=0.}}
360 sets the absolute tolerance \mbox{atol} in \ref{DARCY ATOL DEF}.
361 \end{methoddesc}
362
363 \begin{methoddesc}[DarcyFlow]{setSubProblemTolerance}{\optional{rtol=None}}
364 sets the relative tolerance used to solve the involved PDEs. If no argument is given,
365 the square of the current relative tolerance is used. The sub-problem tolerance should be choosen as large as possible to minimize the compute time. However, a too large value for the sub-problem tolerance may lead to slow convergence or even dibergence in the outer iteration.
366 \end{methoddesc}
367
368 \begin{methoddesc}[DarcyFlow]{solve}{u0,p0, \optional{max_iter=100, \optional{verbose=False \optional{sub_rtol=1.e-8}}}}
369 solves the problem. and returns approximations for the flux $v$ and the pressure $p$.
370 \var{u0} and \var{p0} define initial guess for flux and pressure. Values marked
371 by positive values \var{location_of_fixed_flux} and \var{location_of_fixed_pressure}, respectively, are kept unchanged.
372 \end{methoddesc}
373
374
375 \subsection{Example: Gravity Flow}
376 later
377
378 %================================================
379 % \section{Temperature Advection Diffusion\label{TEMP ADV DIFF}}
380
381 %\begin{equation}
382 % \rho c\hackscore{p} \left (\frac{\partial T}{\partial t} + \vec{v} \cdot \nabla T \right ) = k \nabla^{2}T
383 % \label{HEAT EQUATION}
384 % \end{equation}
385
386 % 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.
387
388 % \subsection{Description}
389
390 % \subsection{Method}
391 %
392 % \begin{classdesc}{TemperatureCartesian}{dom,theta=THETA,useSUPG=SUPG}
393 % \end{classdesc}
394
395 % \subsection{Benchmark Problem}
396 %===============================================================================================================
397
398 %=========================================================
399 \input{levelsetmodel}
400
401 % \section{Drucker Prager Model}
402
403 \section{Isotropic Kelvin Material \label{IKM}}
404 As proposed by Kelvin~\cite{Muhlhaus2005} material strain $D\hackscore{ij}=\frac{1}{2}(v\hackscore{i,j}+v\hackscore{j,i})$ can be decomposed into
405 an elastic part $D\hackscore{ij}^{el}$ and visco-plastic part $D\hackscore{ij}^{vp}$:
406 \begin{equation}\label{IKM-EQU-2}
407 D\hackscore{ij}=D\hackscore{ij}^{el}+D\hackscore{ij}^{vp}
408 \end{equation}
409 with the elastic strain given as
410 \begin{equation}\label{IKM-EQU-3}
411 D\hackscore{ij}^{el'}=\frac{1}{2 \mu} \dot{\sigma}\hackscore{ij}'
412 \end{equation}
413 where $\sigma'\hackscore{ij}$ is the deviatoric stress (Notice that $\sigma'\hackscore{ii}=0$).
414 If the material is composed by materials $q$ the visco-plastic strain can be decomposed as
415 \begin{equation}\label{IKM-EQU-4}
416 D\hackscore{ij}^{vp'}=\sum\hackscore{q} D\hackscore{ij}^{q'}
417 \end{equation}
418 where $D\hackscore{ij}^{q}$ is the strain in material $q$ given as
419 \begin{equation}\label{IKM-EQU-5}
420 D\hackscore{ij}^{q'}=\frac{1}{2 \eta^{q}} \sigma'\hackscore{ij}
421 \end{equation}
422 where $\eta^{q}$ is the viscosity of material $q$. We assume the following
423 betwee the the strain in material $q$
424 \begin{equation}\label{IKM-EQU-5b}
425 \eta^{q}=\eta^{q}\hackscore{N} \left(\frac{\tau}{\tau\hackscore{t}^q}\right)^{\frac{1}{n^{q}}-1}
426 \mbox{ with } \tau=\sqrt{\frac{1}{2}\sigma'\hackscore{ij} \sigma'\hackscore{ij}}
427 \end{equation}
428 for a given power law coefficients $n^{q}\ge1$ and transition stresses $\tau\hackscore{t}^q$, see~\cite{Muhlhaus2005}.
429 Notice that $n^{q}=1$ gives a constant viscosity.
430 After inserting equation~\ref{IKM-EQU-5} into equation \ref{IKM-EQU-4} one gets:
431 \begin{equation}\label{IKM-EQU-6}
432 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}} \;.
433 \end{equation}
434 and finally with~\ref{IKM-EQU-2}
435 \begin{equation}\label{IKM-EQU-2bb}
436 D\hackscore{ij}'=\frac{1}{2 \eta^{vp}} \sigma'\hackscore{ij}+\frac{1}{2 \mu} \dot{\sigma}\hackscore{ij}'
437 \end{equation}
438 The total stress $\tau$ needs to fullfill the yield condition \index{yield condition}
439 \begin{equation}\label{IKM-EQU-8c}
440 \tau \le \tau\hackscore{Y} + \beta \; p
441 \end{equation}
442 with the Drucker-Prager \index{Druck-Prager} cohesion factor \index{cohesion factor} $\tau\hackscore{Y}$, Drucker-Prager friction $\beta$ and total pressure $p$.
443 The deviatoric stress needs to fullfill the equilibrion equation
444 \begin{equation}\label{IKM-EQU-1}
445 -\sigma'\hackscore{ij,j}+p\hackscore{,i}=F\hackscore{i}
446 \end{equation}
447 where $F\hackscore{j}$ is a given external fource. We assume an incompressible media:
448 \begin{equation}\label{IKM-EQU-2bbb}
449 -v\hackscore{i,i}=0
450 \end{equation}
451 Natural boundary conditions are taken in the form
452 \begin{equation}\label{IKM-EQU-Boundary}
453 \sigma'\hackscore{ij}n\hackscore{j}-n\hackscore{i}p=f
454 \end{equation}
455 which can be overwritten by a constraint
456 \begin{equation}\label{IKM-EQU-Boundary0}
457 v\hackscore{i}(x)=0
458 \end{equation}
459 where the index $i$ may depend on the location $x$ on the bondary.
460
461 \subsection{Solution Method \label{IKM-SOLVE}}
462 By using a first order finite difference approximation wit step size $dt>0$~\ref{IKM-EQU-3} get the form
463 \begin{equation}\label{IKM-EQU-3b}
464 \dot{\sigma}\hackscore{ij}=\frac{1}{dt } \left( \sigma\hackscore{ij} - \sigma\hackscore{ij}^{-} \right)
465 \end{equation}
466 and
467 \begin{equation}\label{IKM-EQU-2b}
468 D\hackscore{ij}'=\left(\frac{1}{2 \eta^{vp}} + \frac{1}{2 \mu dt}\right) \sigma\hackscore{ij}'-\frac{1}{2 \mu dt } \sigma\hackscore{ij}^{-'}
469 \end{equation}
470 where $\sigma\hackscore{ij}^{-}$ is the stress at the precious time step. With
471 \begin{equation}\label{IKM-EQU-2c}
472 \dot{\gamma} = \sqrt{ 2 \left( D\hackscore{ij}' +
473 \frac{1}{ 2 \mu \; dt} \sigma\hackscore{ij}^{-'}\right)^2}
474 \end{equation}
475 we have
476 \begin{equation}
477 \tau = \eta\hackscore{eff} \cdot \dot{\gamma}
478 \end{equation}
479 where
480 \begin{equation}
481 \eta\hackscore{eff}= min( \left(\frac{1}{\mu \; dt}+\frac{1}{\eta^{vp}}\right)^{-1}
482 , \eta\hackscore{max}) \mbox{ with }
483 \eta\hackscore{max} = \left\{
484 \begin{array}{rcl}
485 \frac{\tau\hackscore{Y} + \beta \; p}{\dot{\gamma}} & & \dot{\gamma}>0 \\
486 &\mbox{ if } \\
487 \infty & & \mbox{otherwise}
488 \end{array}
489 \right.
490 \end{equation}
491 The upper bound $\eta\hackscore{max}$ makes sure that yield condtion~\ref{IKM-EQU-8c} holds. With this setting the eqaution \ref{IKM-EQU-2b} takes the form
492 \begin{equation}\label{IKM-EQU-10}
493 \sigma\hackscore{ij}' = 2 \eta\hackscore{eff} \left( D\hackscore{ij}' +
494 \frac{1}{ 2 \mu \; dt} \sigma\hackscore{ij}^{'-}\right)
495 \end{equation}
496 After inserting~\ref{IKM-EQU-10} into~\ref{IKM-EQU-1} we get
497 \begin{equation}\label{IKM-EQU-1ib}
498 -\left(\eta\hackscore{eff} (v\hackscore{i,j}+ v\hackscore{i,j})
499 \right)\hackscore{,j}+p\hackscore{,i}=F\hackscore{i}+
500 \left(\frac{\eta\hackscore{eff}}{\mu dt } \sigma\hackscore{ij}^{'-} \right)\hackscore{,j}
501 \end{equation}
502 Combining this with the incomressibilty condition~\ref{IKM-EQU-2} we need to solve a
503 Stokes problem as discussed in section~\ref{STOKES SOLVE} in each time step.
504
505 If we set
506 \begin{equation}\label{IKM-EQU-44}
507 \frac{1}{\eta(\tau)}= \frac{1}{\mu \; dt}+\frac{1}{\eta^{vp}}
508 \end{equation}
509 we need to solve the nonlinear problem
510 \begin{equation}
511 \eta\hackscore{eff} - min(\eta( \dot{\gamma} \cdot \eta\hackscore{eff})
512 , \eta\hackscore{max}) =0
513 \end{equation}
514 We use the Newton-Raphson Scheme \index{Newton-Raphson scheme} to solve this problem
515 \begin{equation}\label{IKM-EQU-45}
516 \eta\hackscore{eff}^{(n+1)} = \min(\eta\hackscore{max},
517 \eta\hackscore{eff}^{(n)} -
518 \frac{\eta\hackscore{eff}^{(n)} - \eta( \tau^{(n)}) }
519 {1-\dot{\gamma} \cdot \eta'( \tau^{(n)} )} )
520 =\min(\eta\hackscore{max},
521 \frac{\eta( \tau^{(n)}) -\tau^{(n)} \cdot \eta'( \tau^{(n)} ) }
522 {1-\dot{\gamma} \cdot \eta'( \tau^{(n)} )} )
523 \end{equation}
524 where $\eta'$ denotes the derivative of $\eta$ with respect of $\tau$
525 and $\tau^{(n)} = \dot{\gamma} \cdot \eta\hackscore{eff}^{(n)}$.
526
527 Looking at the evaluation of $\eta$ in~\ref{IKM-EQU-44} it makes sense formulate
528 the iteration~\ref{IKM-EQU-45} using $\Theta=\eta^{-1}$.
529 In fact we have
530 \begin{equation}
531 \eta' = - \frac{\Theta'}{\Theta^2}
532 \mbox{ with }
533 \Theta' = \sum\hackscore{q} \left(\frac{1}{\eta^{q}} \right)'
534 \label{IKM iteration 7}
535 \end{equation}
536 As
537 \begin{equation}\label{IKM-EQU-47}
538 \left(\frac{1}{\eta^{q}} \right)'
539 = \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}}}}
540 = \frac{1-\frac{1}{n^{q}}}{\eta^{q}}\frac{1}{\tau}
541 \end{equation}
542 we have
543 \begin{equation}\label{IKM-EQU-48}
544 \Theta' = \frac{1}{\tau} \omega \mbox{ with } \omega = \sum\hackscore{q}\frac{1-\frac{1}{n^{q}}}{\eta^{q}}
545 \end{equation}
546 which leads to
547 \begin{equation}\label{IKM-EQU-49}
548 \eta\hackscore{eff}^{(n+1)} = \min(\eta\hackscore{max},
549 \eta\hackscore{eff}^{(n)}
550 \frac{\Theta^{(n)} + \omega^{(n)} }
551 {\eta\hackscore{eff}^{(n)} \Theta^{(n)^2}+\omega^{(n)} })
552 \end{equation}

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