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1 ksteube 1811
2     %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
3 lgraham 1702 %
4 ksteube 1811 % Copyright (c) 2003-2008 by University of Queensland
5     % Earth Systems Science Computational Center (ESSCC)
6     % http://www.uq.edu.au/esscc
7 lgraham 1702 %
8 ksteube 1811 % 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 lgraham 1702 %
12 ksteube 1811 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
13 lgraham 1702
14 ksteube 1811
15 lgraham 1702 \chapter{Models}
16    
17     The following sections give a breif overview of the model classes and their corresponding methods.
18    
19 gross 1878 \section{Stokes Problem}
20 gross 2100 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 gross 1878 \begin{equation}\label{Stokes 1}
22 gross 2100 -\left(\eta(v\hackscore{i,j}+ v\hackscore{i,j})\right)\hackscore{,j}+p\hackscore{,i}=f\hackscore{i}-\sigma\hackscore{ij,j}
23 gross 1878 \end{equation}
24 gross 2100 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 gross 1878 \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 gross 2100 \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 gross 1878 \end{equation}
32 gross 2100 which can be overwritten by constraints of the form
33 gross 1878 \begin{equation}\label{Stokes Boundary0}
34 gross 2100 v\hackscore{i}(x)=v^D\hackscore{i}(x)
35 gross 1878 \end{equation}
36 gross 2100 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 lgraham 1702
39 gross 1878 \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 gross 2100 \index{saddle point problem}
42 lgraham 1702 \begin{equation}
43     \left[ \begin{array}{cc}
44 gross 1878 A & B^{*} \\
45     B & 0 \\
46 lgraham 1702 \end{array} \right]
47     \left[ \begin{array}{c}
48 gross 2100 v \\
49 lgraham 1702 p \\
50     \end{array} \right]
51     =\left[ \begin{array}{c}
52 gross 2100 G \\
53 gross 1878 0 \\
54 lgraham 1702 \end{array} \right]
55     \label{SADDLEPOINT}
56     \end{equation}
57 gross 2100 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 gross 1878 \begin{equation}
61 gross 2100 \|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 gross 1878 \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 gross 2100 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 gross 1878 \begin{equation}
115     \begin{array}{rcl}
116 gross 2100 A w & = & Av+B^{*}p \\
117     \hat{S} q & = & B(w-v) \\
118 gross 1878 \end{array}
119 gross 2100 \label{COUPLES SADDLEPOINT iteration}
120 gross 1878 \end{equation}
121 gross 2100 We use the inner product induced by the norm
122 gross 1878 \begin{equation}
123 gross 2100 \|(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 gross 1878 \begin{array}{rcl}
129 gross 2100 -\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 gross 1878 \end{array}
132     \label{SADDLEPOINT iteration 2}
133     \end{equation}
134 lgraham 1702
135    
136 gross 2100 \subsection{Functions}
137 gross 1878
138 gross 2100 \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 gross 1878
143 gross 2100 \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 gross 1878
152 gross 2100 \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 lgraham 1702
164    
165 gross 2100 \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 lgraham 1702
190 gross 2100 \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 jfenwick 2104 the lit driven cavity problem~\cite{LITDRIVENCAVITY}:
195 gross 2100 \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 lgraham 1702
213 gross 2100 \section{Darcy Flux}
214 gross 2108 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 gross 2100 \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 lgraham 1702
235    
236 gross 2100 \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 lgraham 1702 \begin{equation}
240 gross 2100 \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 gross 2108 \begin{classdesc}{DarcyFlow}{domain}
319     opens the Darcy flux problem\index{Darcy flux} on the \Domain domain.
320     \end{classdesc}
321 gross 2100
322 gross 2108 \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 gross 2100 \subsection{Example: Gravity Flow}
333    
334     %================================================
335     \section{Temperature Advection Diffusion\label{TEMP ADV DIFF}}
336    
337     \begin{equation}
338 lgraham 1709 \rho c\hackscore{p} \left (\frac{\partial T}{\partial t} + \vec{v} \cdot \nabla T \right ) = k \nabla^{2}T
339 lgraham 1702 \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 gross 2100 %===============================================================================================================
353 lgraham 1702
354 gross 2100 %=========================================================
355     % \section{Level Set Method}
356 lgraham 1702
357 gross 2100 %\subsection{Description}
358 lgraham 1702
359 gross 2100 %\subsection{Method}
360 lgraham 1702
361 gross 2100 %Advection and Reinitialisation
362 lgraham 1702
363 gross 2100 %\begin{classdesc}{LevelSet}{mesh, func\_new, reinit\_max, reinit\_each, tolerance, smooth}
364     %\end{classdesc}
365 lgraham 1702
366     %example usage:
367    
368     %levelset = LevelSet(mesh, func\_new, reinit\_max, reinit\_each, tolerance, smooth)
369    
370 gross 2100 %\begin{methoddesc}[LevelSet]{update\_parameter}{parameter}
371     %Update the parameter.
372     %\end{methoddesc}
373 lgraham 1702
374 gross 2100 %\begin{methoddesc}[LevelSet]{update\_phi}{paramter}{velocity}{dt}{t\_step}
375     %Update level set function; advection and reinitialization
376     %\end{methoddesc}
377 lgraham 1702
378 gross 2100 %\subsection{Benchmark Problem}
379 lgraham 1702
380 gross 2100 %Rayleigh-Taylor instability problem
381 lgraham 1702
382    
383 gross 1859 % \section{Drucker Prager Model}
384 lgraham 1702
385 gross 1841 \section{Isotropic Kelvin Material \label{IKM}}
386 gross 1859 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 gross 1841 \begin{equation}\label{IKM-EQU-2}
389 gross 1859 D\hackscore{ij}=D\hackscore{ij}^{el}+D\hackscore{ij}^{vp}
390 gross 1841 \end{equation}
391 gross 1859 with the elastic strain given as
392 gross 1841 \begin{equation}\label{IKM-EQU-3}
393 gross 1878 D\hackscore{ij}'^{el}=\frac{1}{2 \mu} \dot{\sigma}\hackscore{ij}'
394 gross 1841 \end{equation}
395 gross 1859 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 gross 1841 \begin{equation}\label{IKM-EQU-4}
398 gross 1859 D\hackscore{ij}'^{vp}=\sum\hackscore{q} D\hackscore{ij}'^{q}
399 gross 1841 \end{equation}
400 gross 1859 where $D\hackscore{ij}^{q}$ is the strain in material $q$ given as
401 gross 1841 \begin{equation}\label{IKM-EQU-5}
402 gross 1859 D\hackscore{ij}'^{q}=\frac{1}{2 \eta^{q}} \sigma'\hackscore{ij}
403 gross 1841 \end{equation}
404 gross 1859 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 gross 1841 After inserting equation~\ref{IKM-EQU-5} into equation \ref{IKM-EQU-4} one gets:
413 gross 1859 \begin{equation}\label{IKM-EQU-6}
414 gross 2100 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 gross 1841 \end{equation}
416 gross 1859 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 gross 1841 \begin{equation}\label{IKM-EQU-1}
438 gross 1859 -\sigma'\hackscore{ij,j}+p\hackscore{,i}=F\hackscore{i}
439 gross 1841 \end{equation}
440 gross 1859 where $F\hackscore{j}$ is a given external fource. We assume an incompressible media:
441 gross 1841 \begin{equation}\label{IKM-EQU-2}
442 gross 1859 -v\hackscore{i,i}=0
443 gross 1841 \end{equation}
444 gross 1878 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 gross 2100 \sigma\hackscore{ij}' = 2 \eta\hackscore{eff} \left( D\hackscore{ij}' +
463     \frac{1}{ 2 \mu \; dt} \sigma\hackscore{ij}^{'-}\right) \mbox{ with }
464 gross 1878 \frac{1}{\eta\hackscore{eff}}=\frac{1}{\mu \; dt}+\frac{1}{\eta^{vp}}
465     \end{equation}
466 gross 2100 Notice that $\eta\hackscore{eff}$ is a function of diatoric stress $\sigma\hackscore{ij}'$.
467 gross 1859 After inserting~\ref{IKM-EQU-10} into~\ref{IKM-EQU-1} we get
468     \begin{equation}\label{IKM-EQU-1ib}
469 gross 2100 -\left(\eta\hackscore{eff} (v\hackscore{i,j}+ v\hackscore{i,j})
470     \right)\hackscore{,j}+p\hackscore{,i}=F\hackscore{i}+
471 gross 1878 \frac{\eta\hackscore{eff}}{\mu dt } \sigma\hackscore{ij,j}^{'-}
472 gross 1859 \end{equation}
473 gross 1878 Together with the incomressibilty condition~\ref{IKM-EQU-2} we need to solve a problem with a form almost identical
474 jfenwick 1966 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 gross 1878 \begin{equation}
476     \begin{array}{rcl}
477 gross 2100 -\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 gross 1878 \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 gross 2100 \|(v, p)\|^2= \int\hackscore{\Omega} v\hackscore{i,j}^2 + \frac{1}{\bar{\eta}^2\hackscore{eff}} p^2 \; dx
486 gross 1878 \label{IKM iteration 3}
487     \end{equation}
488 gross 2100 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 gross 1841
524 gross 1878
525 gross 2100

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