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starting a section for the Level Set class in the Models chapter.

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 Without loss of generality we can assume that $u^{N}\hackscore{i} \; n\hackscore{i}=0$ and
239 $p^{D}$. Otherewise one solves for $u-u^{N}$ and $p-p^{D}$ and sets
240 \begin{equation}
241 \begin{array}{rcl}
242 g\hackscore{i} & \leftarrow & g\hackscore{i} - u^{N}\hackscore{i} - \kappa\hackscore{ij} p^{D}\hackscore{,j }\\
243 f & \leftarrow & f - u^{N}\hackscore{k,k}
244 \end{array}
245 \end{equation}
246 We set
247 \begin{equation}
248 V = \{ q \in H^{1}(\Omega) : q=0 \mbox{ on } \Gamma\hackscore{D} \}
249 \end{equation}
250 and
251 \begin{equation}
252 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} \}
253 \end{equation}
254 and define the operator $Q: V \rightarrow (L^2(\Omega))^{d}$ defined by
255 \begin{equation}
256 (Qp)\hackscore{i} = \kappa\hackscore{ij} p\hackscore{,j}
257 \end{equation}
258 and the operator $D: W \rightarrow L^2(\Omega)$ defined by
259 \begin{equation}
260 Dv = v\hackscore{k,k}
261 \end{equation}
262 In operator notation the Darcy problem~\ref{DARCY PROBLEM} is written in the form
263 \begin{equation}
264 \begin{array}{rcl}
265 u + Qp & = & g \\
266 Du & = & f
267 \end{array}
268 \end{equation}
269 We solve this equation by minimising the functional
270 \begin{equation}
271 J(u,p):=\|u + Qp - g\|^2\hackscore{0} + \|Du-f\|\hackscore{0}^2
272 \end{equation}
273 over $W \times V$ where $\|.\|\hackscore{0}$ denotes the norm in $L^2(\Omega)$. A simple calculation shows that
274 one has to solve
275 \begin{equation}
276 ( v + Qq , u + Qp - g) + (Dv,Du-f) =0
277 \end{equation}
278 for all $v\in W$ and $q \in V$.which translates back into operator notation
279 \begin{equation}
280 \begin{array}{rcl}
281 (I+D^*D)u + Qp & = & D^*f + g \\
282 Q^*u + Q^*Q p & = & Q^* g \\
283 \end{array}
284 \end{equation}
285 where $D^*$ and $Q^*$ denote the adjoint operators.
286 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
287 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 $\alpha\hackscore{0} \gg 1$)
288
289 The approach we are taking is to eliminate the velocity $u$ from the problem. Assuming that $p$ is known we have
290 \begin{equation}
291 v= (I+D^*D)^{-1} (D^*f + g - Qp)
292 \end{equation}
293 (notice that $(I+D^*D)$ is coercive in $W$) which is inserted into the second equation
294 \begin{equation}
295 Q^* (I+D^*D)^{-1} (D^*f + g - Qp) + Q^* Q p = Q^* g
296 \end{equation}
297 which is
298 \begin{equation}
299 Q^* ( I - (I+D^*D)^{-1} ) Q p = Q^* ( g -(I+D^*D)^{-1} (D^*f + g) )
300 \end{equation}
301 We use the PCG method to solve this. The residual $r$ ($\in V^*$) is given as
302 \begin{equation}
303 \begin{array}{rcl}
304 r & = & Q^* ( g -(I+D^*D)^{-1} (D^*f + g) - Qp + (I+D^*D)^{-1}Q p \\
305 & =& Q^* \left( (g-Qp) - (I+D^*D)^{-1} (D^*f + g - Qp) \right) \\
306 & =& Q^* \left( (g-Qp) - v \right)
307 \end{array}
308 \end{equation}
309 So in a partical implementation we use the pair $(g-Qp,v)$ to represent the residual. This will save the
310 reconstruction of the velocity $v$. In this notation the right hand side is given as
311 $(g,(I+D^*D)^{-1} (D^*f + g))$. The evaluation of the iteration operator for a given $p$ is then
312 returning $(Qp,w)$ where $w$ is the solution of
313 \begin{equation}\label{UPDATE W}
314 (I+D^*D)w = Qp
315 \end{equation}
316 We use $Q^*Q$ as a a preconditioner for the iteration operator $Q^* ( I - (I+D^*D)^{-1} ) Q$.
317
318 \subsection{Functions}
319 \begin{classdesc}{DarcyFlow}{domain}
320 opens the Darcy flux problem\index{Darcy flux} on the \Domain domain.
321 \end{classdesc}
322
323 \begin{methoddesc}[DarcyFlow]{initialize}{\optional{f=Data(), \optional{fixed_u_mask=Data(), \optional{eta=1, \optional{surface_stress=Data(), \optional{stress=Data()}}}}}}
324 assigns values to the model parameters. In any call all values must be set.
325 \var{f} defines the external force $f$, \var{eta} the viscosity $\eta$,
326 \var{surface_stress} the surface stress $s$ and \var{stress} the initial stress $\sigma$.
327 The locations and compontents where the velocity is fixed are set by
328 the values of \var{fixed_u_mask}. The method will try to cast the given values to appropriate
329 \Data class objects.
330 \end{methoddesc}
331
332
333 \subsection{Example: Gravity Flow}
334
335 %================================================
336 \section{Temperature Advection Diffusion\label{TEMP ADV DIFF}}
337
338 \begin{equation}
339 \rho c\hackscore{p} \left (\frac{\partial T}{\partial t} + \vec{v} \cdot \nabla T \right ) = k \nabla^{2}T
340 \label{HEAT EQUATION}
341 \end{equation}
342
343 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.
344
345 \subsection{Description}
346
347 \subsection{Method}
348
349 \begin{classdesc}{TemperatureCartesian}{dom,theta=THETA,useSUPG=SUPG}
350 \end{classdesc}
351
352 \subsection{Benchmark Problem}
353 %===============================================================================================================
354
355 %=========================================================
356 \input{levelsetmodel}
357 % \section{Level Set Method}
358
359 %\subsection{Description}
360
361 %\subsection{Method}
362
363 %Advection and Reinitialisation
364
365 %\begin{classdesc}{LevelSet}{mesh, func\_new, reinit\_max, reinit\_each, tolerance, smooth}
366 %\end{classdesc}
367
368 %example usage:
369
370 %levelset = LevelSet(mesh, func\_new, reinit\_max, reinit\_each, tolerance, smooth)
371
372 %\begin{methoddesc}[LevelSet]{update\_parameter}{parameter}
373 %Update the parameter.
374 %\end{methoddesc}
375
376 %\begin{methoddesc}[LevelSet]{update\_phi}{paramter}{velocity}{dt}{t\_step}
377 %Update level set function; advection and reinitialization
378 %\end{methoddesc}
379
380 %\subsection{Benchmark Problem}
381
382 %Rayleigh-Taylor instability problem
383
384
385 % \section{Drucker Prager Model}
386
387 \section{Isotropic Kelvin Material \label{IKM}}
388 As proposed by Kelvin~\ref{KELVN} material strain $D\hackscore{ij}=v\hackscore{i,j}+v\hackscore{j,i}$ can be decomposed into
389 an elastic part $D\hackscore{ij}^{el}$ and visco-plastic part $D\hackscore{ij}^{vp}$:
390 \begin{equation}\label{IKM-EQU-2}
391 D\hackscore{ij}=D\hackscore{ij}^{el}+D\hackscore{ij}^{vp}
392 \end{equation}
393 with the elastic strain given as
394 \begin{equation}\label{IKM-EQU-3}
395 D\hackscore{ij}'^{el}=\frac{1}{2 \mu} \dot{\sigma}\hackscore{ij}'
396 \end{equation}
397 where $\sigma'\hackscore{ij}$ is the deviatoric stress (Notice that $\sigma'\hackscore{ii}=0$).
398 If the material is composed by materials $q$ the visco-plastic strain can be decomposed as
399 \begin{equation}\label{IKM-EQU-4}
400 D\hackscore{ij}'^{vp}=\sum\hackscore{q} D\hackscore{ij}'^{q}
401 \end{equation}
402 where $D\hackscore{ij}^{q}$ is the strain in material $q$ given as
403 \begin{equation}\label{IKM-EQU-5}
404 D\hackscore{ij}'^{q}=\frac{1}{2 \eta^{q}} \sigma'\hackscore{ij}
405 \end{equation}
406 where $\eta^{q}$ is the viscosity of material $q$. We assume the following
407 betwee the the strain in material $q$
408 \begin{equation}\label{IKM-EQU-5b}
409 \eta^{q}=\eta^{q}\hackscore{N} \left(\frac{\tau}{\tau\hackscore{t}^q}\right)^{\frac{1}{n^{q}}-1}
410 \mbox{ with } \tau=\sqrt{\frac{1}{2}\sigma'\hackscore{ij} \sigma'\hackscore{ij}}
411 \end{equation}
412 for a given power law coefficients $n^{q}$ and transition stresses $\tau\hackscore{t}^q$, see~\ref{POERLAW}.
413 Notice that $n^{q}=1$ gives a constant viscosity.
414 After inserting equation~\ref{IKM-EQU-5} into equation \ref{IKM-EQU-4} one gets:
415 \begin{equation}\label{IKM-EQU-6}
416 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}} \;.
417 \end{equation}
418 With
419 \begin{equation}\label{IKM-EQU-8}
420 \dot{\gamma}=\sqrt{2 D\hackscore{ij} D\hackscore{ij}}
421 \end{equation}
422 one gets
423 \begin{equation}\label{IKM-EQU-8b}
424 \tau = \eta^{vp} \dot{\gamma}^{vp} \;.
425 \end{equation}
426 With the Drucker-Prager cohesion factor $\tau\hackscore{Y}$, Drucker-Prager friction $\beta$ and total pressure $p$ we want to achieve
427 \begin{equation}\label{IKM-EQU-8c}
428 \tau \le \tau\hackscore{Y} + \beta \; p
429 \end{equation}
430 which leads to the condition
431 \begin{equation}\label{IKM-EQU-8d}
432 \eta^{vp} \le \frac{\tau\hackscore{Y} + \beta \; p}{ \dot{\gamma}^{vp}} \; .
433 \end{equation}
434 Therefore we modify the definition of $\eta^{vp}$ to the form
435 \begin{equation}\label{IKM-EQU-6b}
436 \frac{1}{\eta^{vp}}=\max(\sum\hackscore{q} \frac{1}{\eta^{q}}, \frac{\dot{\gamma}^{vp}} {\tau\hackscore{Y} + \beta \; p})
437 \end{equation}
438 The deviatoric stress needs to fullfill the equilibrion equation
439 \begin{equation}\label{IKM-EQU-1}
440 -\sigma'\hackscore{ij,j}+p\hackscore{,i}=F\hackscore{i}
441 \end{equation}
442 where $F\hackscore{j}$ is a given external fource. We assume an incompressible media:
443 \begin{equation}\label{IKM-EQU-2}
444 -v\hackscore{i,i}=0
445 \end{equation}
446 Natural boundary conditions are taken in the form
447 \begin{equation}\label{IKM-EQU-Boundary}
448 \sigma'\hackscore{ij}n\hackscore{j}-n\hackscore{i}p=f
449 \end{equation}
450 which can be overwritten by a constraint
451 \begin{equation}\label{IKM-EQU-Boundary0}
452 v\hackscore{i}(x)=0
453 \end{equation}
454 where the index $i$ may depend on the location $x$ on the bondary.
455
456 \subsection{Solution Method \label{IKM-SOLVE}}
457 By using a first order finite difference approximation wit step size $dt>0$~\ref{IKM-EQU-3} get the form
458 \begin{equation}\label{IKM-EQU-3b}
459 D\hackscore{ij}'^{el}=\frac{1}{2 \mu dt } \left( \sigma\hackscore{ij}' - \sigma\hackscore{ij}^{'-} \right)
460 \end{equation}
461 where $\sigma\hackscore{ij}^{'-}$ is the deviatoric stress at the precious time step.
462 Now we can combine equations~\ref{IKM-EQU-2}, \ref{IKM-EQU-3b} and~\ref{IKM-EQU-6b} to get
463 \begin{equation}\label{IKM-EQU-10}
464 \sigma\hackscore{ij}' = 2 \eta\hackscore{eff} \left( D\hackscore{ij}' +
465 \frac{1}{ 2 \mu \; dt} \sigma\hackscore{ij}^{'-}\right) \mbox{ with }
466 \frac{1}{\eta\hackscore{eff}}=\frac{1}{\mu \; dt}+\frac{1}{\eta^{vp}}
467 \end{equation}
468 Notice that $\eta\hackscore{eff}$ is a function of diatoric stress $\sigma\hackscore{ij}'$.
469 After inserting~\ref{IKM-EQU-10} into~\ref{IKM-EQU-1} we get
470 \begin{equation}\label{IKM-EQU-1ib}
471 -\left(\eta\hackscore{eff} (v\hackscore{i,j}+ v\hackscore{i,j})
472 \right)\hackscore{,j}+p\hackscore{,i}=F\hackscore{i}+
473 \frac{\eta\hackscore{eff}}{\mu dt } \sigma\hackscore{ij,j}^{'-}
474 \end{equation}
475 Together with the incomressibilty condition~\ref{IKM-EQU-2} we need to solve a problem with a form almost identical
476 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
477 \begin{equation}
478 \begin{array}{rcl}
479 -\left(\eta\hackscore{eff}(dv\hackscore{i,j}+ dv\hackscore{i,j}
480 )\right)\hackscore{,j} & = & F\hackscore{i}+ \sigma\hackscore{ij,j}'-p\hackscore{,i} \\
481 \frac{1}{\eta\hackscore{eff}} dp & = & - v\hackscore{i,i}^+
482 \end{array}
483 \label{IKM iteration 2}
484 \end{equation}
485 where $v^+=v+dv$. As this problem is non-linear the Jacobi-free Newton-GMRES method is used with the norm
486 \begin{equation}
487 \|(v, p)\|^2= \int\hackscore{\Omega} v\hackscore{i,j}^2 + \frac{1}{\bar{\eta}^2\hackscore{eff}} p^2 \; dx
488 \label{IKM iteration 3}
489 \end{equation}
490 where $\bar{\eta}\hackscore{eff}$ is the caracteristic viscosity, for instance:
491 \begin{equation}
492 \frac{1}{\bar{\eta}\hackscore{eff}} = \frac{1}{\tau^{-}}+\sum\hackscore{q} \frac{1}{\eta^{q}\hackscore{N}}
493 \label{IKM iteration 4}
494 \end{equation}
495 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
496 calculation of $\eta\hackscore{eff}$ with the Newton-Raphson scheme. This value can then be used to calculate
497 $\sigma\hackscore{ij}'$ via~\ref{IKM-EQU-10}. We need to solve
498 \begin{equation}
499 \tau = \eta\hackscore{eff} \cdot \epsilon \mbox{ with }
500 \epsilon = \sqrt{ 2 \left( D\hackscore{ij}' +
501 \frac{1}{ 2 \mu \; dt} \sigma\hackscore{ij}^{'-}\right)^2}
502 \label{IKM iteration 5}
503 \end{equation}
504 The Newton scheme takes the form
505 \begin{equation}
506 \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)
507 = \min(\frac{\eta\hackscore{eff} - \tau\hackscore{n} \eta\hackscore{eff}'}
508 {1 - \eta\hackscore{eff}' \cdot \epsilon}, \frac{\tau\hackscore{Y} + \beta \; p}{\epsilon}) \epsilon
509 \label{IKM iteration 6}
510 \end{equation}
511 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
512 \begin{equation}
513 \eta\hackscore{eff}' = - \eta\hackscore{eff}^2 \left(\frac{1}{\eta\hackscore{eff}}\right)'
514 \mbox{ with }
515 \left(\frac{1}{\eta\hackscore{eff}}\right)' = \sum\hackscore{q} \left(\frac{1}{\eta^{q}} \right)'
516 \label{IKM iteration 7}
517 \end{equation}
518 \begin{equation}\label{IKM-EQU-5XX}
519 \left(\frac{1}{\eta^{q}} \right)'
520 = \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}}}}
521 = \frac{1-\frac{1}{n^{q}}}{ \tau \eta^{q}}
522 \end{equation}
523 Notice that allways $\eta\hackscore{eff}'\le 0$ which makes the denomionator in~\ref{IKM iteration 6}
524 positive.
525
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527

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