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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
17 The following sections give a breif overview of the model classes and their corresponding methods.
18
19 \section{Stokes Problem}
20 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}
21 \begin{equation}\label{Stokes 1}
22 -\left(\eta(v\hackscore{i,j}+ v\hackscore{i,j})\right)\hackscore{,j}+p\hackscore{,i}=f\hackscore{i}-\sigma\hackscore{ij,j}
23 \end{equation}
24 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:
25 \begin{equation}\label{Stokes 2}
26 -v\hackscore{i,i}=0
27 \end{equation}
28 Natural boundary conditions are taken in the form
29 \begin{equation}\label{Stokes Boundary}
30 \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}
31 \end{equation}
32 which can be overwritten by constraints of the form
33 \begin{equation}\label{Stokes Boundary0}
34 v\hackscore{i}(x)=v^D\hackscore{i}(x)
35 \end{equation}
36 at some locations $x$ at the boundary of the domain. The index $i$ may depend on the location $x$ on the boundary.
37 $v^D$ is a given function on the domain.
38
39 \subsection{Solution Method \label{STOKES SOLVE}}
40 In block form equation equations~\ref{Stokes 1} and~\ref{Stokes 2} takes the form of a saddle point problem
41 \index{saddle point problem}
42 \begin{equation}
43 \left[ \begin{array}{cc}
44 A & B^{*} \\
45 B & 0 \\
46 \end{array} \right]
47 \left[ \begin{array}{c}
48 v \\
49 p \\
50 \end{array} \right]
51 =\left[ \begin{array}{c}
52 G \\
53 0 \\
54 \end{array} \right]
55 \label{SADDLEPOINT}
56 \end{equation}
57 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}.
58 We use iterative techniques to solve this problem. To make sure that the incomressibilty condition holds
59 with sufficient accuracy we check for
60 \begin{equation}
61 \|v\hackscore{k,k}\| \hackscore \le \epsilon
62 \|\sqrt{v\hackscore{j,k}v\hackscore{j,k}}\|
63 \end{equation}
64 where $\epsilon$ is the desired relative accuracy and
65 \begin{equation}
66 \|p\|^2= \int\hackscore{\Omega} p^2 \; dx
67 \label{PRESSURE NORM}
68 \end{equation}
69 defines the $L^2$-norm.
70 There are two approaches to solve this problem. The first approach, called the Uzawa scheme \index{Uzawa scheme}
71 eliminates the velocity $v$ from the problem. The second approach solves the equation in coupled form after the application of a preconditioner.
72
73 \subsubsection{Uzawa scheme}
74 The first eqution in~\ref{SADDLEPOINT} gives $v=A^{-1}(G-B^{*}p)$ assuming $p$ is known. This is inserted into the
75 second eqution which leads to
76 \begin{equation}
77 S p = B A^{-1} G
78 \end{equation}
79 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}
80 with the preconditioner $\hat{S}$ defined as $q=\hat{S}^{-1}p$ by solving
81 \begin{equation}
82 \frac{1}{\eta}q = p
83 \end{equation}
84 see~\cite{ELMAN} for more details. The evaluation of $w=Sp$ is done in the form
85 \begin{equation}
86 \begin{array}{rcl}
87 A v & = & B^{*}p \\
88 w & = & Bv \\
89 \end{array}
90 \label{EVAL PCG}
91 \end{equation}
92 The residual \index{residual} $r=B A^{-1} G - S p$ is given as
93 \begin{equation}
94 r=B A^{-1} (G - B^* p) = Bv \mbox{ with } v = A^{-1}(G-B^{*}p)
95 \end{equation}
96 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
97 \begin{equation}
98 (p,(v,Bv))=\int\hackscore{\Omega} p \cdot Bv \; dx
99 \end{equation}
100 where $p$ is the pressure increment and $(v,Bv)$ represents an increment in the residual.
101
102 \subsubsection{Coupled Solver}
103 An alternative approach to solve the saddle point problem~\ref{SADDLEPOINT} directly using an iterative such as
104 the generalized minimal residual method (GMRES) \index{generalized minimal residual method!GMRES} with a suitable
105 preconditioner. Here we use the operator
106 \begin{equation}
107 \left[ \begin{array}{cc}
108 A^{-1} & 0 \\
109 S^{-1} B A^{-1} & -S^{-1} \\
110 \end{array} \right]
111 \label{SADDLEPOINT PRECODITIONER}
112 \end{equation}
113 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
114 \begin{equation}
115 \begin{array}{rcl}
116 A w & = & Av+B^{*}p \\
117 \hat{S} q & = & B(w-v) \\
118 \end{array}
119 \label{COUPLES SADDLEPOINT iteration}
120 \end{equation}
121 We use the inner product induced by the norm
122 \begin{equation}
123 \|(v,p)\|^2= \int\hackscore{\Omega} v\hackscore{i,j} v\hackscore{i,j} + \left( \frac{p}{\eta}\right)^2\; dx
124 \label{COUPLES NORM}
125 \end{equation}
126 In PDE form~\ref{COUPLES SADDLEPOINT iteration} takes the form
127 \begin{equation}
128 \begin{array}{rcl}
129 -\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} \\
130 \frac{1}{\eta} q & = & - (w-v)\hackscore{i,i} \\
131 \end{array}
132 \label{SADDLEPOINT iteration 2}
133 \end{equation}
134
135
136 \subsection{Functions}
137
138 \begin{classdesc}{StokesProblemCartesian}{domain}
139 opens the Stokes problem\index{Stokes problem} on the \Domain domain. The approximation
140 order needs to be two.
141 \end{classdesc}
142
143 \begin{methoddesc}[StokesProblemCartesian]{initialize}{\optional{f=Data(), \optional{fixed_u_mask=Data(), \optional{eta=1, \optional{surface_stress=Data(), \optional{stress=Data()}}}}}}
144 assigns values to the model parameters. In any call all values must be set.
145 \var{f} defines the external force $f$, \var{eta} the viscosity $\eta$,
146 \var{surface_stress} the surface stress $s$ and \var{stress} the initial stress $\sigma$.
147 The locations and compontents where the velocity is fixed are set by
148 the values of \var{fixed_u_mask}. The method will try to cast the given values to appropriate
149 \Data class objects.
150 \end{methoddesc}
151
152 \begin{methoddesc}[StokesProblemCartesian]{solve}{v,p,
153 \optional{max_iter=20, \optional{verbose=False, \optional{useUzawa=True}}}}
154 solves the problem and return approximations for velocity and pressure.
155 The arguments \var{v} and \var{p} define initial guess. The values of \var{v} marked
156 by \var{fixed_u_mask} remain unchanged.
157 If \var{useUzawa} is set to \True
158 the Uzawa\index{Uszwa} scheme is used. Otherwise the problem is solved in coupled form. In most cases
159 the Uzawa scheme is more efficient.
160 \var{max_iter} defines the maximum number of iteration steps.
161 If \var{verbose} is set to \True informations on the progress of of the solver are printed.
162 \end{methoddesc}
163
164
165 \begin{methoddesc}[StokesProblemCartesian]{setTolerance}{\optional{tolerance=1.e-8}}
166 sets the tolerance in an appropriate norm relative to the right hand side. The tolerance must be non-negative and less than 1.
167 \end{methoddesc}
168 \begin{methoddesc}[StokesProblemCartesian]{getTolerance}{}
169 returns the current relative tolerance.
170 \end{methoddesc}
171 \begin{methoddesc}[StokesProblemCartesian]{setAbsoluteTolerance}{\optional{tolerance=0.}}
172 sets the absolute tolerance for the error in the relevant norm. The tolerance must be non-negative. Typically the
173 absolute talerance is set to 0.
174 \end{methoddesc}
175 \begin{methoddesc}[StokesProblemCartesian]{getAbsoluteTolerance}{}
176 sreturns the current absolute tolerance.
177 \end{methoddesc}
178 \begin{methoddesc}[StokesProblemCartesian]{setSubToleranceReductionFactor}{\optional{reduction=None}}
179 sets the reduction factor for the tolerance used to solve the PDEs. A reduction factor
180 in the order of one will minimize compute time per iteration step but my slow down convergence or even lead to
181 divergency. On the other hand a very small value for the PDE tolerance could result in a wast of compute time.
182 If \var{reduction} is set to \var{None} the sub-tolerance is solved adaptively but
183 in cases a very small tolerance is set ($<10^{-6}$) it is recommended to set the
184 reduction factor by hand. This may require some experiments.
185 \end{methoddesc}
186 \begin{methoddesc}[StokesProblemCartesian]{getSubToleranceReductionFactor}{}
187 return the current reduction factor for the sub-problem tolerance.
188 \end{methoddesc}
189
190 \subsection{Example: Lit Driven Cavity}
191 The following script \file{lit\hackscore driven\hackscore cavity.py}
192 \index{scripts!\file{helmholtz.py}} which is available in the \ExampleDirectory
193 illustrates the usage of the \class{StokesProblemCartesian} class to solve
194 the lit driven cavity problem~\cite{LITDRIVENCAVITY}:
195 \begin{python}
196 from esys.escript import *
197 from esys.finley import Rectangle
198 from esys.escript.models import StokesProblemCartesian
199 NE=25
200 dom = Rectangle(NE,NE,order=2)
201 x = dom.getX()
202 sc=StokesProblemCartesian(dom)
203 mask= (whereZero(x[0])*[1.,0]+whereZero(x[0]-1))*[1.,0] + \
204 (whereZero(x[1])*[0.,1.]+whereZero(x[1]-1))*[1.,1]
205 sc.initialize(eta=.1, fixed_u_mask= mask)
206 v=Vector(0.,Solution(dom))
207 v[0]+=whereZero(x[1]-1.)
208 p=Scalar(0.,ReducedSolution(dom))
209 v,p=sc.solve(v,p, verbose=True)
210 saveVTK("u.xml",velocity=v,pressure=p)
211 \end{python}
212
213 \section{Darcy Flux}
214 We want to calculate the velocity $u$ and pressure $p$ on a domain $\Omega$ solving
215 the Darcy flux problem \index{Darcy flux}\index{Darcy flow}
216 \begin{equation}\label{DARCY PROBLEM}
217 \begin{array}{rcl}
218 u\hackscore{i} + \kappa\hackscore{ij} p\hackscore{,j} & = & g\hackscore{i} \\
219 u\hackscore{k,k} & = & f
220 \end{array}
221 \end{equation}
222 with the boundary conditions
223 \begin{equation}\label{DARCY BOUNDARY}
224 \begin{array}{rcl}
225 u\hackscore{i} \; n\hackscore{i} = u^{N}\hackscore{i} \; n\hackscore{i} & \mbox{ on } & \Gamma\hackscore{N} \\
226 p = p^{D} & \mbox{ on } & \Gamma\hackscore{D} \\
227 \end{array}
228 \end{equation}
229 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
230 \begin{equation}
231 \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}
232 \end{equation}
233 for all $x\hackscore{i}$.
234
235
236 \subsection{Solution Method \label{DARCY SOLVE}}
237 Without loss of generality we can assume that $u^{N}\hackscore{i} \; n\hackscore{i}=0$ and
238 $p^{D}$. Otherewise one solves for $u-u^{N}$ and $p-p^{D}$ and sets
239 \begin{equation}
240 \begin{array}{rcl}
241 g\hackscore{i} & \leftarrow & g\hackscore{i} - u^{N}\hackscore{i} - \kappa\hackscore{ij} p^{D}\hackscore{,j }\\
242 f & \leftarrow & f - u^{N}\hackscore{k,k}
243 \end{array}
244 \end{equation}
245 We set
246 \begin{equation}
247 V = \{ q \in H^{1}(\Omega) : q=0 \mbox{ on } \Gamma\hackscore{D} \}
248 \end{equation}
249 and
250 \begin{equation}
251 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} \}
252 \end{equation}
253 and define the operator $Q: V \rightarrow (L^2(\Omega))^{d}$ defined by
254 \begin{equation}
255 (Qp)\hackscore{i} = \kappa\hackscore{ij} p\hackscore{,j}
256 \end{equation}
257 and the operator $D: W \rightarrow L^2(\Omega)$ defined by
258 \begin{equation}
259 Dv = v\hackscore{k,k}
260 \end{equation}
261 In operator notation the Darcy problem~\ref{DARCY PROBLEM} is written in the form
262 \begin{equation}
263 \begin{array}{rcl}
264 u + Qp & = & g \\
265 Du & = & f
266 \end{array}
267 \end{equation}
268 We solve this equation by minimising the functional
269 \begin{equation}
270 J(u,p):=\|u + Qp - g\|^2\hackscore{0} + \|Du-f\|\hackscore{0}^2
271 \end{equation}
272 over $W \times V$ where $\|.\|\hackscore{0}$ denotes the norm in $L^2(\Omega)$. A simple calculation shows that
273 one has to solve
274 \begin{equation}
275 ( v + Qq , u + Qp - g) + (Dv,Du-f) =0
276 \end{equation}
277 for all $v\in W$ and $q \in V$.which translates back into operator notation
278 \begin{equation}
279 \begin{array}{rcl}
280 (I+D^*D)u + Qp & = & D^*f + g \\
281 Q^*u + Q^*Q p & = & Q^* g \\
282 \end{array}
283 \end{equation}
284 where $D^*$ and $Q^*$ denote the adjoint operators.
285 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
286 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 $\alpha\hackscore{0} \gg 1$)
287
288 The approach we are taking is to eliminate the velocity $u$ from the problem. Assuming that $p$ is known we have
289 \begin{equation}
290 v= (I+D^*D)^{-1} (D^*f + g - Qp)
291 \end{equation}
292 (notice that $(I+D^*D)$ is coercive in $W$) which is inserted into the second equation
293 \begin{equation}
294 Q^* (I+D^*D)^{-1} (D^*f + g - Qp) + Q^* Q p = Q^* g
295 \end{equation}
296 which is
297 \begin{equation}
298 Q^* ( I - (I+D^*D)^{-1} ) Q p = Q^* ( g -(I+D^*D)^{-1} (D^*f + g) )
299 \end{equation}
300 We use the PCG method to solve this. The residual $r$ ($\in V^*$) is given as
301 \begin{equation}
302 \begin{array}{rcl}
303 r & = & Q^* ( g -(I+D^*D)^{-1} (D^*f + g) - Qp + (I+D^*D)^{-1}Q p \\
304 & =& Q^* \left( (g-Qp) - (I+D^*D)^{-1} (D^*f + g - Qp) \right) \\
305 & =& Q^* \left( (g-Qp) - v \right)
306 \end{array}
307 \end{equation}
308 So in a partical implementation we use the pair $(g-Qp,v)$ to represent the residual. This will save the
309 reconstruction of the velocity $v$. In this notation the right hand side is given as
310 $(g,(I+D^*D)^{-1} (D^*f + g))$. The evaluation of the iteration operator for a given $p$ is then
311 returning $(Qp,w)$ where $w$ is the solution of
312 \begin{equation}\label{UPDATE W}
313 (I+D^*D)w = Qp
314 \end{equation}
315 We use $Q^*Q$ as a a preconditioner for the iteration operator $Q^* ( I - (I+D^*D)^{-1} ) Q$.
316
317 \subsection{Functions}
318 \begin{classdesc}{DarcyFlow}{domain}
319 opens the Darcy flux problem\index{Darcy flux} on the \Domain domain.
320 \end{classdesc}
321
322 \begin{methoddesc}[DarcyFlow]{initialize}{\optional{f=Data(), \optional{fixed_u_mask=Data(), \optional{eta=1, \optional{surface_stress=Data(), \optional{stress=Data()}}}}}}
323 assigns values to the model parameters. In any call all values must be set.
324 \var{f} defines the external force $f$, \var{eta} the viscosity $\eta$,
325 \var{surface_stress} the surface stress $s$ and \var{stress} the initial stress $\sigma$.
326 The locations and compontents where the velocity is fixed are set by
327 the values of \var{fixed_u_mask}. The method will try to cast the given values to appropriate
328 \Data class objects.
329 \end{methoddesc}
330
331
332 \subsection{Example: Gravity Flow}
333
334 %================================================
335 \section{Temperature Advection Diffusion\label{TEMP ADV DIFF}}
336
337 \begin{equation}
338 \rho c\hackscore{p} \left (\frac{\partial T}{\partial t} + \vec{v} \cdot \nabla T \right ) = k \nabla^{2}T
339 \label{HEAT EQUATION}
340 \end{equation}
341
342 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.
343
344 \subsection{Description}
345
346 \subsection{Method}
347
348 \begin{classdesc}{TemperatureCartesian}{dom,theta=THETA,useSUPG=SUPG}
349 \end{classdesc}
350
351 \subsection{Benchmark Problem}
352 %===============================================================================================================
353
354 %=========================================================
355 % \section{Level Set Method}
356
357 %\subsection{Description}
358
359 %\subsection{Method}
360
361 %Advection and Reinitialisation
362
363 %\begin{classdesc}{LevelSet}{mesh, func\_new, reinit\_max, reinit\_each, tolerance, smooth}
364 %\end{classdesc}
365
366 %example usage:
367
368 %levelset = LevelSet(mesh, func\_new, reinit\_max, reinit\_each, tolerance, smooth)
369
370 %\begin{methoddesc}[LevelSet]{update\_parameter}{parameter}
371 %Update the parameter.
372 %\end{methoddesc}
373
374 %\begin{methoddesc}[LevelSet]{update\_phi}{paramter}{velocity}{dt}{t\_step}
375 %Update level set function; advection and reinitialization
376 %\end{methoddesc}
377
378 %\subsection{Benchmark Problem}
379
380 %Rayleigh-Taylor instability problem
381
382
383 % \section{Drucker Prager Model}
384
385 \section{Isotropic Kelvin Material \label{IKM}}
386 As proposed by Kelvin~\ref{KELVN} material strain $D\hackscore{ij}=v\hackscore{i,j}+v\hackscore{j,i}$ can be decomposed into
387 an elastic part $D\hackscore{ij}^{el}$ and visco-plastic part $D\hackscore{ij}^{vp}$:
388 \begin{equation}\label{IKM-EQU-2}
389 D\hackscore{ij}=D\hackscore{ij}^{el}+D\hackscore{ij}^{vp}
390 \end{equation}
391 with the elastic strain given as
392 \begin{equation}\label{IKM-EQU-3}
393 D\hackscore{ij}'^{el}=\frac{1}{2 \mu} \dot{\sigma}\hackscore{ij}'
394 \end{equation}
395 where $\sigma'\hackscore{ij}$ is the deviatoric stress (Notice that $\sigma'\hackscore{ii}=0$).
396 If the material is composed by materials $q$ the visco-plastic strain can be decomposed as
397 \begin{equation}\label{IKM-EQU-4}
398 D\hackscore{ij}'^{vp}=\sum\hackscore{q} D\hackscore{ij}'^{q}
399 \end{equation}
400 where $D\hackscore{ij}^{q}$ is the strain in material $q$ given as
401 \begin{equation}\label{IKM-EQU-5}
402 D\hackscore{ij}'^{q}=\frac{1}{2 \eta^{q}} \sigma'\hackscore{ij}
403 \end{equation}
404 where $\eta^{q}$ is the viscosity of material $q$. We assume the following
405 betwee the the strain in material $q$
406 \begin{equation}\label{IKM-EQU-5b}
407 \eta^{q}=\eta^{q}\hackscore{N} \left(\frac{\tau}{\tau\hackscore{t}^q}\right)^{\frac{1}{n^{q}}-1}
408 \mbox{ with } \tau=\sqrt{\frac{1}{2}\sigma'\hackscore{ij} \sigma'\hackscore{ij}}
409 \end{equation}
410 for a given power law coefficients $n^{q}$ and transition stresses $\tau\hackscore{t}^q$, see~\ref{POERLAW}.
411 Notice that $n^{q}=1$ gives a constant viscosity.
412 After inserting equation~\ref{IKM-EQU-5} into equation \ref{IKM-EQU-4} one gets:
413 \begin{equation}\label{IKM-EQU-6}
414 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}} \;.
415 \end{equation}
416 With
417 \begin{equation}\label{IKM-EQU-8}
418 \dot{\gamma}=\sqrt{2 D\hackscore{ij} D\hackscore{ij}}
419 \end{equation}
420 one gets
421 \begin{equation}\label{IKM-EQU-8b}
422 \tau = \eta^{vp} \dot{\gamma}^{vp} \;.
423 \end{equation}
424 With the Drucker-Prager cohesion factor $\tau\hackscore{Y}$, Drucker-Prager friction $\beta$ and total pressure $p$ we want to achieve
425 \begin{equation}\label{IKM-EQU-8c}
426 \tau \le \tau\hackscore{Y} + \beta \; p
427 \end{equation}
428 which leads to the condition
429 \begin{equation}\label{IKM-EQU-8d}
430 \eta^{vp} \le \frac{\tau\hackscore{Y} + \beta \; p}{ \dot{\gamma}^{vp}} \; .
431 \end{equation}
432 Therefore we modify the definition of $\eta^{vp}$ to the form
433 \begin{equation}\label{IKM-EQU-6b}
434 \frac{1}{\eta^{vp}}=\max(\sum\hackscore{q} \frac{1}{\eta^{q}}, \frac{\dot{\gamma}^{vp}} {\tau\hackscore{Y} + \beta \; p})
435 \end{equation}
436 The deviatoric stress needs to fullfill the equilibrion equation
437 \begin{equation}\label{IKM-EQU-1}
438 -\sigma'\hackscore{ij,j}+p\hackscore{,i}=F\hackscore{i}
439 \end{equation}
440 where $F\hackscore{j}$ is a given external fource. We assume an incompressible media:
441 \begin{equation}\label{IKM-EQU-2}
442 -v\hackscore{i,i}=0
443 \end{equation}
444 Natural boundary conditions are taken in the form
445 \begin{equation}\label{IKM-EQU-Boundary}
446 \sigma'\hackscore{ij}n\hackscore{j}-n\hackscore{i}p=f
447 \end{equation}
448 which can be overwritten by a constraint
449 \begin{equation}\label{IKM-EQU-Boundary0}
450 v\hackscore{i}(x)=0
451 \end{equation}
452 where the index $i$ may depend on the location $x$ on the bondary.
453
454 \subsection{Solution Method \label{IKM-SOLVE}}
455 By using a first order finite difference approximation wit step size $dt>0$~\ref{IKM-EQU-3} get the form
456 \begin{equation}\label{IKM-EQU-3b}
457 D\hackscore{ij}'^{el}=\frac{1}{2 \mu dt } \left( \sigma\hackscore{ij}' - \sigma\hackscore{ij}^{'-} \right)
458 \end{equation}
459 where $\sigma\hackscore{ij}^{'-}$ is the deviatoric stress at the precious time step.
460 Now we can combine equations~\ref{IKM-EQU-2}, \ref{IKM-EQU-3b} and~\ref{IKM-EQU-6b} to get
461 \begin{equation}\label{IKM-EQU-10}
462 \sigma\hackscore{ij}' = 2 \eta\hackscore{eff} \left( D\hackscore{ij}' +
463 \frac{1}{ 2 \mu \; dt} \sigma\hackscore{ij}^{'-}\right) \mbox{ with }
464 \frac{1}{\eta\hackscore{eff}}=\frac{1}{\mu \; dt}+\frac{1}{\eta^{vp}}
465 \end{equation}
466 Notice that $\eta\hackscore{eff}$ is a function of diatoric stress $\sigma\hackscore{ij}'$.
467 After inserting~\ref{IKM-EQU-10} into~\ref{IKM-EQU-1} we get
468 \begin{equation}\label{IKM-EQU-1ib}
469 -\left(\eta\hackscore{eff} (v\hackscore{i,j}+ v\hackscore{i,j})
470 \right)\hackscore{,j}+p\hackscore{,i}=F\hackscore{i}+
471 \frac{\eta\hackscore{eff}}{\mu dt } \sigma\hackscore{ij,j}^{'-}
472 \end{equation}
473 Together with the incomressibilty condition~\ref{IKM-EQU-2} we need to solve a problem with a form almost identical
474 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
475 \begin{equation}
476 \begin{array}{rcl}
477 -\left(\eta\hackscore{eff}(dv\hackscore{i,j}+ dv\hackscore{i,j}
478 )\right)\hackscore{,j} & = & F\hackscore{i}+ \sigma\hackscore{ij,j}'-p\hackscore{,i} \\
479 \frac{1}{\eta\hackscore{eff}} dp & = & - v\hackscore{i,i}^+
480 \end{array}
481 \label{IKM iteration 2}
482 \end{equation}
483 where $v^+=v+dv$. As this problem is non-linear the Jacobi-free Newton-GMRES method is used with the norm
484 \begin{equation}
485 \|(v, p)\|^2= \int\hackscore{\Omega} v\hackscore{i,j}^2 + \frac{1}{\bar{\eta}^2\hackscore{eff}} p^2 \; dx
486 \label{IKM iteration 3}
487 \end{equation}
488 where $\bar{\eta}\hackscore{eff}$ is the caracteristic viscosity, for instance:
489 \begin{equation}
490 \frac{1}{\bar{\eta}\hackscore{eff}} = \frac{1}{\tau^{-}}+\sum\hackscore{q} \frac{1}{\eta^{q}\hackscore{N}}
491 \label{IKM iteration 4}
492 \end{equation}
493 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
494 calculation of $\eta\hackscore{eff}$ with the Newton-Raphson scheme. This value can then be used to calculate
495 $\sigma\hackscore{ij}'$ via~\ref{IKM-EQU-10}. We need to solve
496 \begin{equation}
497 \tau = \eta\hackscore{eff} \cdot \epsilon \mbox{ with }
498 \epsilon = \sqrt{ 2 \left( D\hackscore{ij}' +
499 \frac{1}{ 2 \mu \; dt} \sigma\hackscore{ij}^{'-}\right)^2}
500 \label{IKM iteration 5}
501 \end{equation}
502 The Newton scheme takes the form
503 \begin{equation}
504 \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)
505 = \min(\frac{\eta\hackscore{eff} - \tau\hackscore{n} \eta\hackscore{eff}'}
506 {1 - \eta\hackscore{eff}' \cdot \epsilon}, \frac{\tau\hackscore{Y} + \beta \; p}{\epsilon}) \epsilon
507 \label{IKM iteration 6}
508 \end{equation}
509 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
510 \begin{equation}
511 \eta\hackscore{eff}' = - \eta\hackscore{eff}^2 \left(\frac{1}{\eta\hackscore{eff}}\right)'
512 \mbox{ with }
513 \left(\frac{1}{\eta\hackscore{eff}}\right)' = \sum\hackscore{q} \left(\frac{1}{\eta^{q}} \right)'
514 \label{IKM iteration 7}
515 \end{equation}
516 \begin{equation}\label{IKM-EQU-5XX}
517 \left(\frac{1}{\eta^{q}} \right)'
518 = \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}}}}
519 = \frac{1-\frac{1}{n^{q}}}{ \tau \eta^{q}}
520 \end{equation}
521 Notice that allways $\eta\hackscore{eff}'\le 0$ which makes the denomionator in~\ref{IKM iteration 6}
522 positive.
523
524
525

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