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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
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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     We want to calculate the velocity $u$ and pressure $p$ on a domain $\Omega$ such that
215     \begin{equation}\label{DARCY PROBLEM}
216     \begin{array}{rcl}
217     u\hackscore{i} + \kappa\hackscore{ij} p\hackscore{,j} & = & g\hackscore{i} \\
218     u\hackscore{k,k} & = & f
219     \end{array}
220     \end{equation}
221     with the boundary conditions
222     \begin{equation}\label{DARCY BOUNDARY}
223     \begin{array}{rcl}
224     u\hackscore{i} \; n\hackscore{i} = u^{N}\hackscore{i} \; n\hackscore{i} & \mbox{ on } & \Gamma\hackscore{N} \\
225     p = p^{D} & \mbox{ on } & \Gamma\hackscore{D} \\
226     \end{array}
227     \end{equation}
228     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
229     \begin{equation}
230     \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}
231     \end{equation}
232     for all $x\hackscore{i}$.
233 lgraham 1702
234    
235 gross 2100 \subsection{Solution Method \label{DARCY SOLVE}}
236     Without loss of generality we can assume that $u^{N}\hackscore{i} \; n\hackscore{i}=0$ and
237     $p^{D}$. Otherewise one solves for $u-u^{N}$ and $p-p^{D}$ and sets
238 lgraham 1702 \begin{equation}
239 gross 2100 \begin{array}{rcl}
240     g\hackscore{i} & \leftarrow & g\hackscore{i} - u^{N}\hackscore{i} - \kappa\hackscore{ij} p^{D}\hackscore{,j }\\
241     f & \leftarrow & f - u^{N}\hackscore{k,k}
242     \end{array}
243     \end{equation}
244     We set
245     \begin{equation}
246     V = \{ q \in H^{1}(\Omega) : q=0 \mbox{ on } \Gamma\hackscore{D} \}
247     \end{equation}
248     and
249     \begin{equation}
250     W = \{ v \in (L^2(\Omega))^{d} : v\hackscore{k,k} \in L^2(\Omega) \mbox{ and } u\hackscore{i} \; n\hackscore{i} =0 \mbox{ on } \Gamma\hackscore{N} \}
251     \end{equation}
252     and define the operator $Q: V \rightarrow (L^2(\Omega))^{d}$ defined by
253     \begin{equation}
254     (Qp)\hackscore{i} = \kappa\hackscore{ij} p\hackscore{,j}
255     \end{equation}
256     and the operator $D: W \rightarrow L^2(\Omega)$ defined by
257     \begin{equation}
258     Dv = v\hackscore{k,k}
259     \end{equation}
260     In operator notation the Darcy problem~\ref{DARCY PROBLEM} is written in the form
261     \begin{equation}
262     \begin{array}{rcl}
263     u + Qp & = & g \\
264     Du & = & f
265     \end{array}
266     \end{equation}
267     We solve this equation by minimising the functional
268     \begin{equation}
269     J(u,p):=\|u + Qp - g\|^2\hackscore{0} + \|Du-f\|\hackscore{0}^2
270     \end{equation}
271     over $W \times V$ where $\|.\|\hackscore{0}$ denotes the norm in $L^2(\Omega)$. A simple calculation shows that
272     one has to solve
273     \begin{equation}
274     ( v + Qq , u + Qp - g) + (Dv,Du-f) =0
275     \end{equation}
276     for all $v\in W$ and $q \in V$.which translates back into operator notation
277     \begin{equation}
278     \begin{array}{rcl}
279     (I+D^*D)u + Qp & = & D^*f + g \\
280     Q^*u + Q^*Q p & = & Q^* g \\
281     \end{array}
282     \end{equation}
283     where $D^*$ and $Q^*$ denote the adjoint operators.
284     In~\cite{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
285     to solve the problem is coupled form, however this approach leads in some cases to a very ill-conditioned stiffness matrix in particular in the case of a very small or large permability ($\alpha\hackscore{1} \ll 1$ or $\alpha\hackscore{0} \gg 1$)
286    
287     The approach we are taking is to eliminate the velocity $u$ from the problem. Assuming that $p$ is known we have
288     \begin{equation}
289     v= (I+D^*D)^{-1} (D^*f + g - Qp)
290     \end{equation}
291     (notice that $(I+D^*D)$ is coercive in $W$) which is inserted into the second equation
292     \begin{equation}
293     Q^* (I+D^*D)^{-1} (D^*f + g - Qp) + Q^* Q p = Q^* g
294     \end{equation}
295     which is
296     \begin{equation}
297     Q^* ( I - (I+D^*D)^{-1} ) Q p = Q^* ( g -(I+D^*D)^{-1} (D^*f + g) )
298     \end{equation}
299     We use the PCG method to solve this. The residual $r$ ($\in V^*$) is given as
300     \begin{equation}
301     \begin{array}{rcl}
302     r & = & Q^* ( g -(I+D^*D)^{-1} (D^*f + g) - Qp + (I+D^*D)^{-1}Q p \\
303     & =& Q^* \left( (g-Qp) - (I+D^*D)^{-1} (D^*f + g - Qp) \right) \\
304     & =& Q^* \left( (g-Qp) - v \right)
305     \end{array}
306     \end{equation}
307     So in a partical implementation we use the pair $(g-Qp,v)$ to represent the residual. This will save the
308     reconstruction of the velocity $v$. In this notation the right hand side is given as
309     $(g,(I+D^*D)^{-1} (D^*f + g))$. The evaluation of the iteration operator for a given $p$ is then
310     returning $(Qp,w)$ where $w$ is the solution of
311     \begin{equation}\label{UPDATE W}
312     (I+D^*D)w = Qp
313     \end{equation}
314     We use $Q^*Q$ as a a preconditioner for the iteration operator $Q^* ( I - (I+D^*D)^{-1} ) Q$.
315    
316     \subsection{Functions}
317    
318     \subsection{Example: Gravity Flow}
319    
320     %================================================
321     \section{Temperature Advection Diffusion\label{TEMP ADV DIFF}}
322    
323     \begin{equation}
324 lgraham 1709 \rho c\hackscore{p} \left (\frac{\partial T}{\partial t} + \vec{v} \cdot \nabla T \right ) = k \nabla^{2}T
325 lgraham 1702 \label{HEAT EQUATION}
326     \end{equation}
327    
328     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.
329    
330     \subsection{Description}
331    
332     \subsection{Method}
333    
334     \begin{classdesc}{TemperatureCartesian}{dom,theta=THETA,useSUPG=SUPG}
335     \end{classdesc}
336    
337     \subsection{Benchmark Problem}
338 gross 2100 %===============================================================================================================
339 lgraham 1702
340 gross 2100 %=========================================================
341     % \section{Level Set Method}
342 lgraham 1702
343 gross 2100 %\subsection{Description}
344 lgraham 1702
345 gross 2100 %\subsection{Method}
346 lgraham 1702
347 gross 2100 %Advection and Reinitialisation
348 lgraham 1702
349 gross 2100 %\begin{classdesc}{LevelSet}{mesh, func\_new, reinit\_max, reinit\_each, tolerance, smooth}
350     %\end{classdesc}
351 lgraham 1702
352     %example usage:
353    
354     %levelset = LevelSet(mesh, func\_new, reinit\_max, reinit\_each, tolerance, smooth)
355    
356 gross 2100 %\begin{methoddesc}[LevelSet]{update\_parameter}{parameter}
357     %Update the parameter.
358     %\end{methoddesc}
359 lgraham 1702
360 gross 2100 %\begin{methoddesc}[LevelSet]{update\_phi}{paramter}{velocity}{dt}{t\_step}
361     %Update level set function; advection and reinitialization
362     %\end{methoddesc}
363 lgraham 1702
364 gross 2100 %\subsection{Benchmark Problem}
365 lgraham 1702
366 gross 2100 %Rayleigh-Taylor instability problem
367 lgraham 1702
368    
369 gross 1859 % \section{Drucker Prager Model}
370 lgraham 1702
371 gross 1841 \section{Isotropic Kelvin Material \label{IKM}}
372 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
373     an elastic part $D\hackscore{ij}^{el}$ and visco-plastic part $D\hackscore{ij}^{vp}$:
374 gross 1841 \begin{equation}\label{IKM-EQU-2}
375 gross 1859 D\hackscore{ij}=D\hackscore{ij}^{el}+D\hackscore{ij}^{vp}
376 gross 1841 \end{equation}
377 gross 1859 with the elastic strain given as
378 gross 1841 \begin{equation}\label{IKM-EQU-3}
379 gross 1878 D\hackscore{ij}'^{el}=\frac{1}{2 \mu} \dot{\sigma}\hackscore{ij}'
380 gross 1841 \end{equation}
381 gross 1859 where $\sigma'\hackscore{ij}$ is the deviatoric stress (Notice that $\sigma'\hackscore{ii}=0$).
382     If the material is composed by materials $q$ the visco-plastic strain can be decomposed as
383 gross 1841 \begin{equation}\label{IKM-EQU-4}
384 gross 1859 D\hackscore{ij}'^{vp}=\sum\hackscore{q} D\hackscore{ij}'^{q}
385 gross 1841 \end{equation}
386 gross 1859 where $D\hackscore{ij}^{q}$ is the strain in material $q$ given as
387 gross 1841 \begin{equation}\label{IKM-EQU-5}
388 gross 1859 D\hackscore{ij}'^{q}=\frac{1}{2 \eta^{q}} \sigma'\hackscore{ij}
389 gross 1841 \end{equation}
390 gross 1859 where $\eta^{q}$ is the viscosity of material $q$. We assume the following
391     betwee the the strain in material $q$
392     \begin{equation}\label{IKM-EQU-5b}
393     \eta^{q}=\eta^{q}\hackscore{N} \left(\frac{\tau}{\tau\hackscore{t}^q}\right)^{\frac{1}{n^{q}}-1}
394     \mbox{ with } \tau=\sqrt{\frac{1}{2}\sigma'\hackscore{ij} \sigma'\hackscore{ij}}
395     \end{equation}
396     for a given power law coefficients $n^{q}$ and transition stresses $\tau\hackscore{t}^q$, see~\ref{POERLAW}.
397     Notice that $n^{q}=1$ gives a constant viscosity.
398 gross 1841 After inserting equation~\ref{IKM-EQU-5} into equation \ref{IKM-EQU-4} one gets:
399 gross 1859 \begin{equation}\label{IKM-EQU-6}
400 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}} \;.
401 gross 1841 \end{equation}
402 gross 1859 With
403     \begin{equation}\label{IKM-EQU-8}
404     \dot{\gamma}=\sqrt{2 D\hackscore{ij} D\hackscore{ij}}
405     \end{equation}
406     one gets
407     \begin{equation}\label{IKM-EQU-8b}
408     \tau = \eta^{vp} \dot{\gamma}^{vp} \;.
409     \end{equation}
410     With the Drucker-Prager cohesion factor $\tau\hackscore{Y}$, Drucker-Prager friction $\beta$ and total pressure $p$ we want to achieve
411     \begin{equation}\label{IKM-EQU-8c}
412     \tau \le \tau\hackscore{Y} + \beta \; p
413     \end{equation}
414     which leads to the condition
415     \begin{equation}\label{IKM-EQU-8d}
416     \eta^{vp} \le \frac{\tau\hackscore{Y} + \beta \; p}{ \dot{\gamma}^{vp}} \; .
417     \end{equation}
418     Therefore we modify the definition of $\eta^{vp}$ to the form
419     \begin{equation}\label{IKM-EQU-6b}
420     \frac{1}{\eta^{vp}}=\max(\sum\hackscore{q} \frac{1}{\eta^{q}}, \frac{\dot{\gamma}^{vp}} {\tau\hackscore{Y} + \beta \; p})
421     \end{equation}
422     The deviatoric stress needs to fullfill the equilibrion equation
423 gross 1841 \begin{equation}\label{IKM-EQU-1}
424 gross 1859 -\sigma'\hackscore{ij,j}+p\hackscore{,i}=F\hackscore{i}
425 gross 1841 \end{equation}
426 gross 1859 where $F\hackscore{j}$ is a given external fource. We assume an incompressible media:
427 gross 1841 \begin{equation}\label{IKM-EQU-2}
428 gross 1859 -v\hackscore{i,i}=0
429 gross 1841 \end{equation}
430 gross 1878 Natural boundary conditions are taken in the form
431     \begin{equation}\label{IKM-EQU-Boundary}
432     \sigma'\hackscore{ij}n\hackscore{j}-n\hackscore{i}p=f
433     \end{equation}
434     which can be overwritten by a constraint
435     \begin{equation}\label{IKM-EQU-Boundary0}
436     v\hackscore{i}(x)=0
437     \end{equation}
438     where the index $i$ may depend on the location $x$ on the bondary.
439    
440     \subsection{Solution Method \label{IKM-SOLVE}}
441     By using a first order finite difference approximation wit step size $dt>0$~\ref{IKM-EQU-3} get the form
442     \begin{equation}\label{IKM-EQU-3b}
443     D\hackscore{ij}'^{el}=\frac{1}{2 \mu dt } \left( \sigma\hackscore{ij}' - \sigma\hackscore{ij}^{'-} \right)
444     \end{equation}
445     where $\sigma\hackscore{ij}^{'-}$ is the deviatoric stress at the precious time step.
446     Now we can combine equations~\ref{IKM-EQU-2}, \ref{IKM-EQU-3b} and~\ref{IKM-EQU-6b} to get
447     \begin{equation}\label{IKM-EQU-10}
448 gross 2100 \sigma\hackscore{ij}' = 2 \eta\hackscore{eff} \left( D\hackscore{ij}' +
449     \frac{1}{ 2 \mu \; dt} \sigma\hackscore{ij}^{'-}\right) \mbox{ with }
450 gross 1878 \frac{1}{\eta\hackscore{eff}}=\frac{1}{\mu \; dt}+\frac{1}{\eta^{vp}}
451     \end{equation}
452 gross 2100 Notice that $\eta\hackscore{eff}$ is a function of diatoric stress $\sigma\hackscore{ij}'$.
453 gross 1859 After inserting~\ref{IKM-EQU-10} into~\ref{IKM-EQU-1} we get
454     \begin{equation}\label{IKM-EQU-1ib}
455 gross 2100 -\left(\eta\hackscore{eff} (v\hackscore{i,j}+ v\hackscore{i,j})
456     \right)\hackscore{,j}+p\hackscore{,i}=F\hackscore{i}+
457 gross 1878 \frac{\eta\hackscore{eff}}{\mu dt } \sigma\hackscore{ij,j}^{'-}
458 gross 1859 \end{equation}
459 gross 1878 Together with the incomressibilty condition~\ref{IKM-EQU-2} we need to solve a problem with a form almost identical
460 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
461 gross 1878 \begin{equation}
462     \begin{array}{rcl}
463 gross 2100 -\left(\eta\hackscore{eff}(dv\hackscore{i,j}+ dv\hackscore{i,j}
464     )\right)\hackscore{,j} & = & F\hackscore{i}+ \sigma\hackscore{ij,j}'-p\hackscore{,i} \\
465     \frac{1}{\eta\hackscore{eff}} dp & = & - v\hackscore{i,i}^+
466 gross 1878 \end{array}
467     \label{IKM iteration 2}
468     \end{equation}
469     where $v^+=v+dv$. As this problem is non-linear the Jacobi-free Newton-GMRES method is used with the norm
470     \begin{equation}
471 gross 2100 \|(v, p)\|^2= \int\hackscore{\Omega} v\hackscore{i,j}^2 + \frac{1}{\bar{\eta}^2\hackscore{eff}} p^2 \; dx
472 gross 1878 \label{IKM iteration 3}
473     \end{equation}
474 gross 2100 where $\bar{\eta}\hackscore{eff}$ is the caracteristic viscosity, for instance:
475     \begin{equation}
476     \frac{1}{\bar{\eta}\hackscore{eff}} = \frac{1}{\tau^{-}}+\sum\hackscore{q} \frac{1}{\eta^{q}\hackscore{N}}
477     \label{IKM iteration 4}
478     \end{equation}
479     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
480     calculation of $\eta\hackscore{eff}$ with the Newton-Raphson scheme. This value can then be used to calculate
481     $\sigma\hackscore{ij}'$ via~\ref{IKM-EQU-10}. We need to solve
482     \begin{equation}
483     \tau = \eta\hackscore{eff} \cdot \epsilon \mbox{ with }
484     \epsilon = \sqrt{ 2 \left( D\hackscore{ij}' +
485     \frac{1}{ 2 \mu \; dt} \sigma\hackscore{ij}^{'-}\right)^2}
486     \label{IKM iteration 5}
487     \end{equation}
488     The Newton scheme takes the form
489     \begin{equation}
490     \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)
491     = \min(\frac{\eta\hackscore{eff} - \tau\hackscore{n} \eta\hackscore{eff}'}
492     {1 - \eta\hackscore{eff}' \cdot \epsilon}, \frac{\tau\hackscore{Y} + \beta \; p}{\epsilon}) \epsilon
493     \label{IKM iteration 6}
494     \end{equation}
495     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
496     \begin{equation}
497     \eta\hackscore{eff}' = - \eta\hackscore{eff}^2 \left(\frac{1}{\eta\hackscore{eff}}\right)'
498     \mbox{ with }
499     \left(\frac{1}{\eta\hackscore{eff}}\right)' = \sum\hackscore{q} \left(\frac{1}{\eta^{q}} \right)'
500     \label{IKM iteration 7}
501     \end{equation}
502     \begin{equation}\label{IKM-EQU-5XX}
503     \left(\frac{1}{\eta^{q}} \right)'
504     = \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}}}}
505     = \frac{1-\frac{1}{n^{q}}}{ \tau \eta^{q}}
506     \end{equation}
507     Notice that allways $\eta\hackscore{eff}'\le 0$ which makes the denomionator in~\ref{IKM iteration 6}
508     positive.
509 gross 1841
510 gross 1878
511 gross 2100

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