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


1 ksteube 1811
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3 lgraham 1702 %
4 ksteube 1811 % Copyright (c) 2003-2008 by University of Queensland
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7 lgraham 1702 %
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11 lgraham 1702 %
12 ksteube 1811 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
13 lgraham 1702
14 ksteube 1811
15 lgraham 1702 \chapter{Models}
16 lgraham 2128 \label{MODELS CHAPTER}
17 lgraham 1702
18     The following sections give a breif overview of the model classes and their corresponding methods.
19    
20 gross 1878 \section{Stokes Problem}
21 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}
22 gross 1878 \begin{equation}\label{Stokes 1}
23 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}
24 gross 1878 \end{equation}
25 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:
26 gross 1878 \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 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}
32 gross 1878 \end{equation}
33 gross 2100 which can be overwritten by constraints of the form
34 gross 1878 \begin{equation}\label{Stokes Boundary0}
35 gross 2100 v\hackscore{i}(x)=v^D\hackscore{i}(x)
36 gross 1878 \end{equation}
37 gross 2100 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 lgraham 1702
40 gross 1878 \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 gross 2100 \index{saddle point problem}
43 lgraham 1702 \begin{equation}
44     \left[ \begin{array}{cc}
45 gross 1878 A & B^{*} \\
46     B & 0 \\
47 lgraham 1702 \end{array} \right]
48     \left[ \begin{array}{c}
49 gross 2100 v \\
50 lgraham 1702 p \\
51     \end{array} \right]
52     =\left[ \begin{array}{c}
53 gross 2100 G \\
54 gross 1878 0 \\
55 lgraham 1702 \end{array} \right]
56     \label{SADDLEPOINT}
57     \end{equation}
58 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}.
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 gross 1878 \begin{equation}
62 gross 2100 \|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 gross 1878 \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 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
115 gross 1878 \begin{equation}
116     \begin{array}{rcl}
117 gross 2100 A w & = & Av+B^{*}p \\
118     \hat{S} q & = & B(w-v) \\
119 gross 1878 \end{array}
120 gross 2100 \label{COUPLES SADDLEPOINT iteration}
121 gross 1878 \end{equation}
122 gross 2100 We use the inner product induced by the norm
123 gross 1878 \begin{equation}
124 gross 2100 \|(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 gross 1878 \begin{array}{rcl}
130 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} \\
131     \frac{1}{\eta} q & = & - (w-v)\hackscore{i,i} \\
132 gross 1878 \end{array}
133     \label{SADDLEPOINT iteration 2}
134     \end{equation}
135 lgraham 1702
136    
137 gross 2100 \subsection{Functions}
138 gross 1878
139 gross 2100 \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 gross 1878
144 gross 2100 \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 gross 1878
153 gross 2100 \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 lgraham 1702
165    
166 gross 2100 \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 lgraham 1702
191 gross 2100 \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 jfenwick 2104 the lit driven cavity problem~\cite{LITDRIVENCAVITY}:
196 gross 2100 \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 lgraham 1702
214 gross 2100 \section{Darcy Flux}
215 gross 2108 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 gross 2100 \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 lgraham 1702
236    
237 gross 2100 \subsection{Solution Method \label{DARCY SOLVE}}
238 gross 2156 In practical applications it is an advantage to calculate the pressure $p$ as a correction of a 'static' pressure $p^{ref}$ which is the solution of
239 lgraham 1702 \begin{equation}
240 gross 2156 -(\kappa\hackscore{ki}\kappa\hackscore{kj} p\hackscore{,j}^{ref})\hackscore{,i} = - (\kappa\hackscore{ki} (g\hackscore{k}- u^{N}\hackscore{k}))\hackscore{,i}
241     \mbox{ with }
242     p^{ref} = p^{D} \mbox{ on } \Gamma\hackscore{D}
243     \end{equation}
244     With setting $u \leftarrow u-u^{N}$ and $p \leftarrow p-p^{ref}$ and
245     \begin{equation}
246 gross 2100 \begin{array}{rcl}
247 gross 2156 g\hackscore{i} & \leftarrow & g\hackscore{i} - u^{N}\hackscore{i} - \kappa\hackscore{ij} p^{ref}\hackscore{,j }\\
248 gross 2100 f & \leftarrow & f - u^{N}\hackscore{k,k}
249     \end{array}
250     \end{equation}
251 gross 2156 we can assume that $u^{N}\hackscore{i} \; n\hackscore{i}=0$ and
252     $p^{D}=0$. Notice that with this setting
253     \begin{equation}\label{DIV FREE DARCY FLUX}
254     (\kappa\hackscore{ki} g\hackscore{k})\hackscore{,i} = 0
255     \end{equation}
256 gross 2100 We set
257     \begin{equation}
258     V = \{ q \in H^{1}(\Omega) : q=0 \mbox{ on } \Gamma\hackscore{D} \}
259     \end{equation}
260     and
261     \begin{equation}
262     W = \{ v \in (L^2(\Omega))^{d} : v\hackscore{k,k} \in L^2(\Omega) \mbox{ and } u\hackscore{i} \; n\hackscore{i} =0 \mbox{ on } \Gamma\hackscore{N} \}
263     \end{equation}
264     and define the operator $Q: V \rightarrow (L^2(\Omega))^{d}$ defined by
265     \begin{equation}
266     (Qp)\hackscore{i} = \kappa\hackscore{ij} p\hackscore{,j}
267     \end{equation}
268     and the operator $D: W \rightarrow L^2(\Omega)$ defined by
269     \begin{equation}
270     Dv = v\hackscore{k,k}
271     \end{equation}
272     In operator notation the Darcy problem~\ref{DARCY PROBLEM} is written in the form
273     \begin{equation}
274     \begin{array}{rcl}
275     u + Qp & = & g \\
276     Du & = & f
277     \end{array}
278     \end{equation}
279 gross 2156 Notice that because of~\ref{DIV FREE DARCY FLUX} we have
280     \begin{equation} \label{ABSTRACT DIV FREE DARCY FLUX}
281     Q^*g =0
282     \end{equation}
283     where $Q^*$ denote the adjoint operators of $Q$.
284 gross 2100 We solve this equation by minimising the functional
285     \begin{equation}
286     J(u,p):=\|u + Qp - g\|^2\hackscore{0} + \|Du-f\|\hackscore{0}^2
287     \end{equation}
288     over $W \times V$ where $\|.\|\hackscore{0}$ denotes the norm in $L^2(\Omega)$. A simple calculation shows that
289     one has to solve
290     \begin{equation}
291     ( v + Qq , u + Qp - g) + (Dv,Du-f) =0
292     \end{equation}
293     for all $v\in W$ and $q \in V$.which translates back into operator notation
294     \begin{equation}
295     \begin{array}{rcl}
296     (I+D^*D)u + Qp & = & D^*f + g \\
297 gross 2156 Q^*u + Q^*Q p & = & 0 \\
298 gross 2100 \end{array}
299     \end{equation}
300 gross 2156 where $D^*$ and $Q^*$ denote the adjoint operators. We use the fact that $Q^*g=$.
301 gross 2100 In~\cite{XXX} it has been shown that this problem is continuous and coercive in $W \times V$ and therefore has a unique solution. Also standart FEM methods can be used for discretization. It is also possible
302     to solve the problem is coupled form, however this approach leads in some cases to a very ill-conditioned stiffness matrix in particular in the case of a very small or large permability ($\alpha\hackscore{1} \ll 1$ or $\alpha\hackscore{0} \gg 1$)
303    
304     The approach we are taking is to eliminate the velocity $u$ from the problem. Assuming that $p$ is known we have
305     \begin{equation}
306     v= (I+D^*D)^{-1} (D^*f + g - Qp)
307     \end{equation}
308     (notice that $(I+D^*D)$ is coercive in $W$) which is inserted into the second equation
309     \begin{equation}
310 gross 2156 Q^* (I+D^*D)^{-1} (D^*f + g - Qp) + Q^* Q p = 0
311 gross 2100 \end{equation}
312     which is
313     \begin{equation}
314 gross 2156 Q^* ( I - (I+D^*D)^{-1} ) Q p = - Q^* (I+D^*D)^{-1} (D^*f + g) )
315 gross 2100 \end{equation}
316     We use the PCG method to solve this. The residual $r$ ($\in V^*$) is given as
317     \begin{equation}
318     \begin{array}{rcl}
319 gross 2156 r & = & Q^* \left( -(I+D^*D)^{-1} (D^*f + g) - Qp + (I+D^*D)^{-1}Q p \right)\\
320     & =& Q^* \left( - Qp - (I+D^*D)^{-1} (D^*f + g - Qp) \right) \\
321     & =& - Q^* \left( Qp + v \right)
322 gross 2100 \end{array}
323     \end{equation}
324 gross 2156 So in a partical implementation we use the pair $(Qp,v)$ to represent the residual. This will save the
325 gross 2100 reconstruction of the velocity $v$. In this notation the right hand side is given as
326 gross 2156 $(0,(I+D^*D)^{-1} (D^*f + g))$. The evaluation of the iteration operator for a given $p$ is then
327 gross 2100 returning $(Qp,w)$ where $w$ is the solution of
328     \begin{equation}\label{UPDATE W}
329     (I+D^*D)w = Qp
330     \end{equation}
331     We use $Q^*Q$ as a a preconditioner for the iteration operator $Q^* ( I - (I+D^*D)^{-1} ) Q$.
332    
333     \subsection{Functions}
334 gross 2108 \begin{classdesc}{DarcyFlow}{domain}
335     opens the Darcy flux problem\index{Darcy flux} on the \Domain domain.
336     \end{classdesc}
337 gross 2100
338 gross 2108 \begin{methoddesc}[DarcyFlow]{initialize}{\optional{f=Data(), \optional{fixed_u_mask=Data(), \optional{eta=1, \optional{surface_stress=Data(), \optional{stress=Data()}}}}}}
339     assigns values to the model parameters. In any call all values must be set.
340     \var{f} defines the external force $f$, \var{eta} the viscosity $\eta$,
341     \var{surface_stress} the surface stress $s$ and \var{stress} the initial stress $\sigma$.
342     The locations and compontents where the velocity is fixed are set by
343     the values of \var{fixed_u_mask}. The method will try to cast the given values to appropriate
344     \Data class objects.
345     \end{methoddesc}
346    
347    
348 gross 2100 \subsection{Example: Gravity Flow}
349    
350     %================================================
351     \section{Temperature Advection Diffusion\label{TEMP ADV DIFF}}
352    
353     \begin{equation}
354 lgraham 1709 \rho c\hackscore{p} \left (\frac{\partial T}{\partial t} + \vec{v} \cdot \nabla T \right ) = k \nabla^{2}T
355 lgraham 1702 \label{HEAT EQUATION}
356     \end{equation}
357    
358     where $\vec{v}$ is the velocity vector, $T$ is the temperature, $\rho$ is the density, $\eta$ is the viscosity, $c\hackscore{p}$ is the specific heat at constant pressure and $k$ is the thermal conductivity.
359    
360     \subsection{Description}
361    
362     \subsection{Method}
363    
364     \begin{classdesc}{TemperatureCartesian}{dom,theta=THETA,useSUPG=SUPG}
365     \end{classdesc}
366    
367     \subsection{Benchmark Problem}
368 gross 2100 %===============================================================================================================
369 lgraham 1702
370 gross 2100 %=========================================================
371 lgraham 2138 \input{levelsetmodel}
372 lgraham 1702
373 gross 1859 % \section{Drucker Prager Model}
374 lgraham 1702
375 gross 1841 \section{Isotropic Kelvin Material \label{IKM}}
376 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
377     an elastic part $D\hackscore{ij}^{el}$ and visco-plastic part $D\hackscore{ij}^{vp}$:
378 gross 1841 \begin{equation}\label{IKM-EQU-2}
379 gross 1859 D\hackscore{ij}=D\hackscore{ij}^{el}+D\hackscore{ij}^{vp}
380 gross 1841 \end{equation}
381 gross 1859 with the elastic strain given as
382 gross 1841 \begin{equation}\label{IKM-EQU-3}
383 gross 1878 D\hackscore{ij}'^{el}=\frac{1}{2 \mu} \dot{\sigma}\hackscore{ij}'
384 gross 1841 \end{equation}
385 gross 1859 where $\sigma'\hackscore{ij}$ is the deviatoric stress (Notice that $\sigma'\hackscore{ii}=0$).
386     If the material is composed by materials $q$ the visco-plastic strain can be decomposed as
387 gross 1841 \begin{equation}\label{IKM-EQU-4}
388 gross 1859 D\hackscore{ij}'^{vp}=\sum\hackscore{q} D\hackscore{ij}'^{q}
389 gross 1841 \end{equation}
390 gross 1859 where $D\hackscore{ij}^{q}$ is the strain in material $q$ given as
391 gross 1841 \begin{equation}\label{IKM-EQU-5}
392 gross 1859 D\hackscore{ij}'^{q}=\frac{1}{2 \eta^{q}} \sigma'\hackscore{ij}
393 gross 1841 \end{equation}
394 gross 1859 where $\eta^{q}$ is the viscosity of material $q$. We assume the following
395     betwee the the strain in material $q$
396     \begin{equation}\label{IKM-EQU-5b}
397     \eta^{q}=\eta^{q}\hackscore{N} \left(\frac{\tau}{\tau\hackscore{t}^q}\right)^{\frac{1}{n^{q}}-1}
398     \mbox{ with } \tau=\sqrt{\frac{1}{2}\sigma'\hackscore{ij} \sigma'\hackscore{ij}}
399     \end{equation}
400     for a given power law coefficients $n^{q}$ and transition stresses $\tau\hackscore{t}^q$, see~\ref{POERLAW}.
401     Notice that $n^{q}=1$ gives a constant viscosity.
402 gross 1841 After inserting equation~\ref{IKM-EQU-5} into equation \ref{IKM-EQU-4} one gets:
403 gross 1859 \begin{equation}\label{IKM-EQU-6}
404 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}} \;.
405 gross 1841 \end{equation}
406 gross 1859 With
407     \begin{equation}\label{IKM-EQU-8}
408     \dot{\gamma}=\sqrt{2 D\hackscore{ij} D\hackscore{ij}}
409     \end{equation}
410     one gets
411     \begin{equation}\label{IKM-EQU-8b}
412     \tau = \eta^{vp} \dot{\gamma}^{vp} \;.
413     \end{equation}
414     With the Drucker-Prager cohesion factor $\tau\hackscore{Y}$, Drucker-Prager friction $\beta$ and total pressure $p$ we want to achieve
415     \begin{equation}\label{IKM-EQU-8c}
416     \tau \le \tau\hackscore{Y} + \beta \; p
417     \end{equation}
418     which leads to the condition
419     \begin{equation}\label{IKM-EQU-8d}
420     \eta^{vp} \le \frac{\tau\hackscore{Y} + \beta \; p}{ \dot{\gamma}^{vp}} \; .
421     \end{equation}
422     Therefore we modify the definition of $\eta^{vp}$ to the form
423     \begin{equation}\label{IKM-EQU-6b}
424     \frac{1}{\eta^{vp}}=\max(\sum\hackscore{q} \frac{1}{\eta^{q}}, \frac{\dot{\gamma}^{vp}} {\tau\hackscore{Y} + \beta \; p})
425     \end{equation}
426     The deviatoric stress needs to fullfill the equilibrion equation
427 gross 1841 \begin{equation}\label{IKM-EQU-1}
428 gross 1859 -\sigma'\hackscore{ij,j}+p\hackscore{,i}=F\hackscore{i}
429 gross 1841 \end{equation}
430 gross 1859 where $F\hackscore{j}$ is a given external fource. We assume an incompressible media:
431 gross 1841 \begin{equation}\label{IKM-EQU-2}
432 gross 1859 -v\hackscore{i,i}=0
433 gross 1841 \end{equation}
434 gross 1878 Natural boundary conditions are taken in the form
435     \begin{equation}\label{IKM-EQU-Boundary}
436     \sigma'\hackscore{ij}n\hackscore{j}-n\hackscore{i}p=f
437     \end{equation}
438     which can be overwritten by a constraint
439     \begin{equation}\label{IKM-EQU-Boundary0}
440     v\hackscore{i}(x)=0
441     \end{equation}
442     where the index $i$ may depend on the location $x$ on the bondary.
443    
444     \subsection{Solution Method \label{IKM-SOLVE}}
445     By using a first order finite difference approximation wit step size $dt>0$~\ref{IKM-EQU-3} get the form
446     \begin{equation}\label{IKM-EQU-3b}
447     D\hackscore{ij}'^{el}=\frac{1}{2 \mu dt } \left( \sigma\hackscore{ij}' - \sigma\hackscore{ij}^{'-} \right)
448     \end{equation}
449     where $\sigma\hackscore{ij}^{'-}$ is the deviatoric stress at the precious time step.
450     Now we can combine equations~\ref{IKM-EQU-2}, \ref{IKM-EQU-3b} and~\ref{IKM-EQU-6b} to get
451     \begin{equation}\label{IKM-EQU-10}
452 gross 2100 \sigma\hackscore{ij}' = 2 \eta\hackscore{eff} \left( D\hackscore{ij}' +
453     \frac{1}{ 2 \mu \; dt} \sigma\hackscore{ij}^{'-}\right) \mbox{ with }
454 gross 1878 \frac{1}{\eta\hackscore{eff}}=\frac{1}{\mu \; dt}+\frac{1}{\eta^{vp}}
455     \end{equation}
456 gross 2100 Notice that $\eta\hackscore{eff}$ is a function of diatoric stress $\sigma\hackscore{ij}'$.
457 gross 1859 After inserting~\ref{IKM-EQU-10} into~\ref{IKM-EQU-1} we get
458     \begin{equation}\label{IKM-EQU-1ib}
459 gross 2100 -\left(\eta\hackscore{eff} (v\hackscore{i,j}+ v\hackscore{i,j})
460     \right)\hackscore{,j}+p\hackscore{,i}=F\hackscore{i}+
461 gross 1878 \frac{\eta\hackscore{eff}}{\mu dt } \sigma\hackscore{ij,j}^{'-}
462 gross 1859 \end{equation}
463 gross 1878 Together with the incomressibilty condition~\ref{IKM-EQU-2} we need to solve a problem with a form almost identical
464 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
465 gross 1878 \begin{equation}
466     \begin{array}{rcl}
467 gross 2100 -\left(\eta\hackscore{eff}(dv\hackscore{i,j}+ dv\hackscore{i,j}
468     )\right)\hackscore{,j} & = & F\hackscore{i}+ \sigma\hackscore{ij,j}'-p\hackscore{,i} \\
469     \frac{1}{\eta\hackscore{eff}} dp & = & - v\hackscore{i,i}^+
470 gross 1878 \end{array}
471     \label{IKM iteration 2}
472     \end{equation}
473     where $v^+=v+dv$. As this problem is non-linear the Jacobi-free Newton-GMRES method is used with the norm
474     \begin{equation}
475 gross 2100 \|(v, p)\|^2= \int\hackscore{\Omega} v\hackscore{i,j}^2 + \frac{1}{\bar{\eta}^2\hackscore{eff}} p^2 \; dx
476 gross 1878 \label{IKM iteration 3}
477     \end{equation}
478 gross 2100 where $\bar{\eta}\hackscore{eff}$ is the caracteristic viscosity, for instance:
479     \begin{equation}
480     \frac{1}{\bar{\eta}\hackscore{eff}} = \frac{1}{\tau^{-}}+\sum\hackscore{q} \frac{1}{\eta^{q}\hackscore{N}}
481     \label{IKM iteration 4}
482     \end{equation}
483     In oder to perform step~\ref{IKM iteration 2} we need to calculate the $\eta\hackscore{eff}$ as well as $\sigma\hackscore{ij}'$ while via $\tau$ the first is a function of the latter. The priority is the
484     calculation of $\eta\hackscore{eff}$ with the Newton-Raphson scheme. This value can then be used to calculate
485     $\sigma\hackscore{ij}'$ via~\ref{IKM-EQU-10}. We need to solve
486     \begin{equation}
487     \tau = \eta\hackscore{eff} \cdot \epsilon \mbox{ with }
488     \epsilon = \sqrt{ 2 \left( D\hackscore{ij}' +
489     \frac{1}{ 2 \mu \; dt} \sigma\hackscore{ij}^{'-}\right)^2}
490     \label{IKM iteration 5}
491     \end{equation}
492     The Newton scheme takes the form
493     \begin{equation}
494     \tau\hackscore{n+1} = \min(\tau\hackscore{n} - \frac{\tau\hackscore{n} - \eta\hackscore{eff} \cdot \epsilon}{1 - \eta\hackscore{eff}' \cdot \epsilon}, \tau\hackscore{Y} + \beta \; p)
495     = \min(\frac{\eta\hackscore{eff} - \tau\hackscore{n} \eta\hackscore{eff}'}
496     {1 - \eta\hackscore{eff}' \cdot \epsilon}, \frac{\tau\hackscore{Y} + \beta \; p}{\epsilon}) \epsilon
497     \label{IKM iteration 6}
498     \end{equation}
499     where $\eta\hackscore{eff}'$ denotes the derivative of $\eta\hackscore{eff}$ with respect of $\tau$. The second term in $\min$ is droped of $\tau\hackscore{Y} + \beta \; p<0$ or $\epsilon=0$. In fact we have
500     \begin{equation}
501     \eta\hackscore{eff}' = - \eta\hackscore{eff}^2 \left(\frac{1}{\eta\hackscore{eff}}\right)'
502     \mbox{ with }
503     \left(\frac{1}{\eta\hackscore{eff}}\right)' = \sum\hackscore{q} \left(\frac{1}{\eta^{q}} \right)'
504     \label{IKM iteration 7}
505     \end{equation}
506     \begin{equation}\label{IKM-EQU-5XX}
507     \left(\frac{1}{\eta^{q}} \right)'
508     = \frac{1-\frac{1}{n^{q}}}{\eta^{q}\hackscore{N}} \frac{\tau^{-\frac{1}{n^{q}}}}{(\tau\hackscore{t}^q)^{1-\frac{1}{n^{q}}}}
509     = \frac{1-\frac{1}{n^{q}}}{ \tau \eta^{q}}
510     \end{equation}
511     Notice that allways $\eta\hackscore{eff}'\le 0$ which makes the denomionator in~\ref{IKM iteration 6}
512     positive.
513 gross 1841
514 gross 1878
515 gross 2100

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