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4  % Copyright (c) 2003-2008 by University of Queensland  % Copyright (c) 2003-2009 by University of Queensland
5  % Earth Systems Science Computational Center (ESSCC)  % Earth Systems Science Computational Center (ESSCC)
6  % http://www.uq.edu.au/esscc  % http://www.uq.edu.au/esscc
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15  \chapter{Models}  \chapter{Models}
16    \label{MODELS CHAPTER}
17
18  The following sections give a breif overview of the model classes and their corresponding methods.  The following sections give a breif overview of the model classes and their corresponding methods.
19
20  \section{Stokes Problem}  \section{Stokes Problem}
21  The velocity field $v$ and pressure $p$ of an incompressible fluid is given as the solution of their  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}
Stokes problem
22  \begin{equation}\label{Stokes 1}  \begin{equation}\label{Stokes 1}
23  -\left(\eta(v\hackscore{i,j}+ v\hackscore{i,j})\right)\hackscore{,j}+p\hackscore{,i}=F\hackscore{i}  -\left(\eta(v\hackscore{i,j}+ v\hackscore{i,j})\right)\hackscore{,j}+p\hackscore{,i}=f\hackscore{i}-\sigma\hackscore{ij,j}
24  \end{equation}  \end{equation}
25  where $\eta$ is the viscosity and $F_i$ defines an internal force. We assume an incompressible media:  where $\eta$ is the viscosity, $F\hackscore{i}$ defines an internal force \index{force, internal} and $\sigma\hackscore{ij}$ is an intial stress \index{stress, initial}. We assume an incompressible media:
26  \begin{equation}\label{Stokes 2}  \begin{equation}\label{Stokes 2}
27  -v\hackscore{i,i}=0  -v\hackscore{i,i}=0
28  \end{equation}  \end{equation}
29  Natural boundary conditions are taken in the form  Natural boundary conditions are taken in the form
30  \begin{equation}\label{Stokes Boundary}  \begin{equation}\label{Stokes Boundary}
31  \left(\eta(v\hackscore{i,j}+ v\hackscore{i,j})\right)n\hackscore{j}-n\hackscore{i}p=f  \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{j}
32  \end{equation}  \end{equation}
33  which can be overwritten by a constraint  which can be overwritten by constraints of the form
34  \begin{equation}\label{Stokes Boundary0}  \begin{equation}\label{Stokes Boundary0}
35  v\hackscore{i}(x)=0  v\hackscore{i}(x)=v^D\hackscore{i}(x)
36  \end{equation}  \end{equation}
37  where the index $i$ may depend on the location $x$ on the bondary.  at some locations $x$ at the boundary of the domain. The index $i$ may depend on the location $x$ on the boundary.
38    $v^D$ is a given function on the domain.
39
40  \subsection{Solution Method \label{STOKES SOLVE}}  \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  In block form equation equations~\ref{Stokes 1} and~\ref{Stokes 2} takes the form of a saddle point problem
43  \begin{equation}  \begin{equation}
44  \left[ \begin{array}{cc}  \left[ \begin{array}{cc}
45  A     & B^{*} \\  A     & B^{*} \\
46  B & 0 \\  B & 0 \\
47  \end{array} \right]  \end{array} \right]
48  \left[ \begin{array}{c}  \left[ \begin{array}{c}
49  u \\  v \\
50  p \\  p \\
51  \end{array} \right]  \end{array} \right]
52  =\left[ \begin{array}{c}  =\left[ \begin{array}{c}
53  F \\  G \\
54  0 \\  0 \\
55  \end{array} \right]  \end{array} \right]
57  \end{equation}  \end{equation}
58  where $A$ is coercive, self-adjoint linear operator in $V$, $B$ is a divergence operator and $B^{*}$ is it adjoint operator (=gradient operator)). For more details on the mathematics see references \cite{AAMIRBERKYAN2008,MBENZI2005}.  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 apply the preconditioner matrix  We use iterative techniques to solve this problem. To make sure that the incomressibilty condition holds
60    with sufficient accuracy we check for
61  \begin{equation}  \begin{equation}
62  \left[ \begin{array}{cc}  \|v\hackscore{k,k}\| \hackscore \le  \epsilon
63  A^{-1}     & 0 \\  \|\sqrt{v\hackscore{j,k}v\hackscore{j,k}}\|
64  S^{-1} B A^{-1}  & -S^{-1} \\  \label{STOKES STOP}
\end{array} \right]
65  \end{equation}  \end{equation}
66  with the Schur complement $S=BA^{-1}B^{*}$ to solve the problem iteratively. The updates $dv$ and $dp$ for  where $\epsilon$ is the desired relative accuracy and
given velocity $v$ and pressure $p$ is given as
67  \begin{equation}  \begin{equation}
68  \begin{array}{rcl}  \|p\|^2= \int\hackscore{\Omega} p^2 \; dx
69  A dv & = & F-Av-B^{*}p \\  \label{PRESSURE NORM}
S dp & = & B(v+dv) \\
\end{array}
70  \end{equation}  \end{equation}
71  This scheme is called the Uzawa scheme.  defines the $L^2$-norm. We use the Uzawa scheme \index{Uzawa scheme} to solve the problem.
72
73  In the case of the Stokes problem it can be shown that $S$ can be approximated by $\frac{1}{\eta}$. With this the iteration scheme can be implemented as  In fact the first equation in~\ref{SADDLEPOINT} gives for a known pressure
74  \begin{equation}  \begin{equation}
75  \begin{array}{rcl}  v=A^{-1}(G-B^{*}p)
76  -\left(\eta(dv\hackscore{i,j}+ dv\hackscore{i,j})\right)\hackscore{,j} & = & F\hackscore{i}+\left(\eta(v\hackscore{i,j}+ v\hackscore{i,j})\right)\hackscore{,j}-p\hackscore{,i} \\  \label{V CALC}
77  \frac{1}{\eta} dp & = & - \left(v\hackscore{,i}+dv\hackscore{,i} \right)\hackscore{,i} \\  \end{equation}
78  \end{array}  which is inserted into the second equation leading to
80    S p =  B A^{-1} G
81  \end{equation}  \end{equation}
82  To accelerate the convergence we are using the restarted $GMRES$ method using the norm  with the Schur complement \index{Schur complement} $S=BA^{-1}B^{*}$. This problem can be solved iteratively
83    with the preconditioner $\hat{S}$ defined as $q=\hat{S}^{-1}p$ by solving
84    \begin{equation} \label{P PREC}
85    \frac{1}{\eta}q = p
86    \end{equation}
87    see~\cite{ELMAN} for more details. Note that the residual for the current approximation $p$ is given as
88  \begin{equation}  \begin{equation}
89  \|(v,p)\|^2= \int\hackscore{\Omega} \eta^2 v\hackscore{i,j}^2 + p^2 \; dx  r=B A^{-1} (G - B^* p) = Bv
90  \end{equation}  \end{equation}
91  or alternatively the $PCG$ method on the pressure only using the norm  where $v$ is given by~\ref{V CALC}.
92
93    If one uses the generalized minimal residual method (GMRES) \index{generalized minimal residual method!GMRES}
94    the method is directly applied to the preconditioned system
95  \begin{equation}  \begin{equation}
96  \|p\|^2= \int\hackscore{\Omega} \frac{1}{\eta^2} p^2 \; dx  \hat{S}^{-1} S p =  \hat{S}^{-1} B A^{-1} G
97  \end{equation}  \end{equation}
98    We use the norm
99    \begin{equation}
100    \|p\|\hackscore{GMRES} = \|\hat{S} p \|
101    \end{equation}
102    Notice that for the residual $\hat{r}=\hat{S}^{-1} r$ one has
103    \begin{equation}
104    \
105    \end{equation}
106    If $p^{0}$ provides an initial guess for the pressure we use~\ref{V CALC} to get a first initial guess for the
107    velocity $v^{0}$ which we use to set an absolute tolerance $ATOL =\epsilon \|\sqrt{v^{0}\hackscore{j,k}v^{0}\hackscore{j,k}}\|$.
108    The GMRES is terminated when
109    \begin{equation}
110    \|\hat{r}\|\hackscore{GMRES} \le ATOL
111    \end{equation}
112    Notice that $\|\hat{r}\|\hackscore{GMRES}= \|r \| = \|Bv\| = \|v\hackscore{k,k}\|$ so we we can expect that
113    the target stopping criterion~\ref{STOKES STOP} is fullfilled. However, if $v$ is very different from the
114    initial choice of $v^{0}$ the value of $ATOL$ is corrected and GMRES is restarted with a new tolerance. For time dependend problems this apprach works well as value for $p$ form a previous time step provides a good initial guess.
115
116    Alternatively, as $S$ is symmetric and positive definite one can apply the preconditioned conjugate gradient method (PCG) \index{preconditioned conjugate gradient method!PCG}. PCG use the norm
117    \begin{equation}
118    \|r\|\hackscore{PCG}^2 = \int\hackscore{\Omega} r \hat{S}^{-1}r \; dx = \int\hackscore{\Omega}  \eta r^2 \; dx
119    \end{equation}
120    To take the extra factor $\eta$ into consideration when checking the stopping criterion we use the following
121    definition for $ATOL$:
122    \begin{equation}
123    ATOL = \epsilon \frac{\|\sqrt{v^{0}\hackscore{j,k}v^{0}\hackscore{j,k}}\|  }{\|v^{0}\hackscore{k,k}\|}
124    \|v^{0}\hackscore{k,k}\|\hackscore{PCG}
125    \end{equation}
126
127
128
129    \subsection{Functions}
130
132  opens the stokes equations on the \Domain domain. Setting debug=True switches the debug mode to on.  opens the Stokes problem\index{Stokes problem} on the \Domain domain. The approximation
133    order needs to be two.
135    the tolerances for all subproblems are set automatically.
136  \end{classdesc}  \end{classdesc}
137
138  example usage:  \begin{methoddesc}[StokesProblemCartesian]{initialize}{\optional{f=Data(), \optional{fixed_u_mask=Data(), \optional{eta=1, \optional{surface_stress=Data(), \optional{stress=Data()}}}}}}
139    assigns values to the model parameters. In any call all values must be set.
140  solution=StokesProblemCartesian(mesh) \\  \var{f} defines the external force $f$, \var{eta} the viscosity $\eta$,
141  solution.setTolerance(TOL) \\  \var{surface_stress} the surface stress $s$ and \var{stress} the initial stress $\sigma$.
142  solution.initialize(fixed\_u\_mask=b\_c,eta=eta,f=Y) \\  The locations and compontents where the velocity is fixed are set by
143  velocity,pressure=solution.solve(velocity,pressure,max\_iter=max\_iter,solver=solver) \\  the values of \var{fixed_u_mask}. The method will try to cast the given values to appropriate
144    \Data class objects.
145    \end{methoddesc}
146
147  % \subsection{Benchmark Problem}  \begin{methoddesc}[StokesProblemCartesian]{solve}{v,p
148  %  \optional{, max_iter=100 \optional{, verbose=False \optional{, usePCG=True }}}}
149  % Convection problem  solves the problem and return approximations for velocity and pressure.
150    The arguments \var{v} and \var{p} define initial guess. The values of \var{v} marked
152    If \var{usePCG} is set to \True
153    reconditioned conjugate gradient method (PCG) \index{preconditioned conjugate gradient method!PCG}  scheme is used. Otherwise the problem is solved generalized minimal residual method (GMRES) \index{generalized minimal residual method!GMRES}. In most cases
154    the PCG scheme is more efficient.
155    \var{max_iter} defines the maximum number of iteration steps.
156
157    If \var{verbose} is set to \True informations on the progress of of the solver are printed.
158    \end{methoddesc}
159
\section{Temperature Cartesian}
160
161  \begin{equation}  \begin{methoddesc}[StokesProblemCartesian]{setTolerance}{\optional{tolerance=1.e-4}}
162  \rho c\hackscore{p} \left (\frac{\partial T}{\partial t} + \vec{v} \cdot \nabla T \right ) = k \nabla^{2}T  sets the tolerance in an appropriate norm relative to the right hand side. The tolerance must be non-negative and less than 1.
163  \label{HEAT EQUATION}  \end{methoddesc}
164  \end{equation}  \begin{methoddesc}[StokesProblemCartesian]{getTolerance}{}
165    returns the current relative tolerance.
166  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.  \end{methoddesc}
167    \begin{methoddesc}[StokesProblemCartesian]{setAbsoluteTolerance}{\optional{tolerance=0.}}
168    sets the absolute tolerance for the error in the relevant norm. The tolerance must be non-negative. Typically the
169    absolute talerance is set to 0.
170    \end{methoddesc}
171
172  \subsection{Description}  \begin{methoddesc}[StokesProblemCartesian]{getAbsoluteTolerance}{}
173    sreturns the current absolute tolerance.
174    \end{methoddesc}
175
176  \subsection{Method}  \begin{methoddesc}[StokesProblemCartesian]{getSolverOptionsVelocity}{}
177    returns the solver options used  solve the equations~(\ref{V CALC}) for velocity.
178    \end{methoddesc}
179
180  \begin{classdesc}{TemperatureCartesian}{dom,theta=THETA,useSUPG=SUPG}  \begin{methoddesc}[StokesProblemCartesian]{getSolverOptionsPressure}{}
181  \end{classdesc}  returns the solver options used  solve the equation~(\ref{P PREC}) for pressure.
182    \end{methoddesc}
183
184  \subsection{Benchmark Problem}  \begin{methoddesc}[StokesProblemCartesian]{getSolverOptionsDiv}{}
185    set the solver options for solving the equation to project the divergence of the velocity onto the function space of pressure.
186    \end{methoddesc}
187
188
189  \section{Level Set Method}  \subsection{Example: Lit Driven Cavity}
190     The following script \file{lit\hackscore driven\hackscore cavity.py}
191    \index{scripts!\file{helmholtz.py}} which is available in the \ExampleDirectory
192    illustrates the usage of the \class{StokesProblemCartesian} class to solve
193    the lit driven cavity problem:
194    \begin{python}
195    from esys.escript import *
196    from esys.finley import Rectangle
197    from esys.escript.models import StokesProblemCartesian
198    NE=25
199    dom = Rectangle(NE,NE,order=2)
200    x = dom.getX()
201    sc=StokesProblemCartesian(dom)
203          (whereZero(x)*[0.,1.]+whereZero(x-1))*[1.,1]
205    v=Vector(0.,Solution(dom))
206    v+=whereZero(x-1.)
207    p=Scalar(0.,ReducedSolution(dom))
208    v,p=sc.solve(v,p, verbose=True)
209    saveVTK("u.xml",velocity=v,pressure=p)
210    \end{python}
211
212    \section{Darcy Flux}
213    We want to calculate the velocity $u$ and pressure $p$ on a domain $\Omega$ solving
214    the Darcy flux problem \index{Darcy flux}\index{Darcy flow}
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
\subsection{Description}
234
235  \subsection{Method}  \subsection{Solution Method \label{DARCY SOLVE}}
236    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
237    \begin{equation}
238    -(\kappa\hackscore{ki}\kappa\hackscore{kj} p\hackscore{,j}^{ref})\hackscore{,i} =  - (\kappa\hackscore{ki} (g\hackscore{k}- u^{N}\hackscore{k}))\hackscore{,i}
239    \mbox{ with }
240    p^{ref} = p^{D} \mbox{ on } \Gamma\hackscore{D}
241    \end{equation}
242    With setting $u \leftarrow u-u^{N}$ and $p \leftarrow p-p^{ref}$ and
243    \begin{equation}
244    \begin{array}{rcl}
245    g\hackscore{i} & \leftarrow & g\hackscore{i} - u^{N}\hackscore{i} -  \kappa\hackscore{ij} p^{ref}\hackscore{,j }\\
246    f & \leftarrow & f - u^{N}\hackscore{k,k}
247    \end{array}
248    \end{equation}
249    we can assume that $u^{N}\hackscore{i} \; n\hackscore{i}=0$ and
250    $p^{D}=0$.
251    We set
252    \begin{equation}
253    V = \{ q \in H^{1}(\Omega) : q=0 \mbox{ on } \Gamma\hackscore{D} \}
254    \end{equation}
255    and
256    \begin{equation}
257    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} \}
258    \end{equation}
259    and define the operator $Q: V \rightarrow (L^2(\Omega))^{d}$ defined by
260    \begin{equation}
261    (Qp)\hackscore{i} = \kappa\hackscore{ij} p\hackscore{,j}
262    \end{equation}
263    and the operator $D: W \rightarrow L^2(\Omega)$ defined by
264    \begin{equation}
265    Dv = v\hackscore{k,k}
266    \end{equation}
267    In operator notation the Darcy problem~\ref{DARCY PROBLEM} is written in the form
268    \begin{equation}
269    \begin{array}{rcl}
270    u + Qp & = & g \\
271    Du & = & f
272    \end{array}
273    \end{equation}
274    We solve this equation by minimising the functional
275    \begin{equation}
276    J(u,p):=\|u + Qp - g\|^2\hackscore{0} + \|Du-f\|\hackscore{0}^2
277    \end{equation}
278    over $W \times V$ where $\|.\|\hackscore{0}$ denotes the norm in $L^2(\Omega)$. A simple calculation shows that
279    one has to solve
280    \begin{equation}
281    ( v + Qq , u + Qp - g) + (Dv,Du-f) =0
282    \end{equation}
283    for all $v\in W$ and $q \in V$.which translates back into operator notation
284    \begin{equation}
285    \begin{array}{rcl}
286    (I+D^*D)u + Qp & = & D^*f + g \\
287    Q^*u  + Q^*Q p & = & Q^*g \\
288    \end{array}
289    \end{equation}
290    where $D^*$ and $Q^*$ denote the adjoint operators.
291    In~\cite{LEASTSQUARESFEM1994} 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
292    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 $\alpha\hackscore{0} \gg 1$)
293
294    The approach we are taking is to eliminate the velocity $u$ from the problem. Assuming that $p$ is known we have
295    \begin{equation}\label{DARCY V FORM}
296    v= (I+D^*D)^{-1} (D^*f + g - Qp)
297    \end{equation}
298    (notice that $(I+D^*D)$ is coercive in $W$) which is inserted into the second equation
299    \begin{equation}
300    Q^* (I+D^*D)^{-1} (D^*f + g - Qp) + Q^* Q p = Q^*g
301    \end{equation}
302    which is
303    \begin{equation}
304    Q^* ( I - (I+D^*D)^{-1} ) Q p = Q^* (g-(I+D^*D)^{-1} (D^*f + g) )
305    \end{equation}
306    We use the PCG \index{linear solver!PCG}\index{PCG} method to solve this. The residual $r$ ($\in V^*$) is given as
307    \begin{equation}
308    \begin{array}{rcl}
309    r & = & Q^*  \left( g -(I+D^*D)^{-1} (D^*f + g) - Qp + (I+D^*D)^{-1}Q p \right)\\
310    & =&  Q^* \left( g- Qp - (I+D^*D)^{-1} (D^*f + g - Qp) \right) \\
311    & =&  Q^* \left( g - Qp - v \right)
312    \end{array}
313    \end{equation}
314    So in a partical implementation we use $\hat{r}=g-Qp-v$ to represent the residual.
315    The evaluation of the iteration operator for a given $p$ is then
316    returning $Qp+v$ where $v$ is the solution of
317    \begin{equation}\label{UPDATE W}
318    (I+D^*D)v = Qp
319    \end{equation}
320    We use $(Q^*Q)^{-1}$ as a preconditioner for the iteration operator $Q^* ( I - (I+D^*D)^{-1} ) Q$. So the application of the preconditioner to $\hat{r}$ representing the residual is given by solving
321    implemented by solving
322    \begin{equation}\label{UPDATE P}
323    Q^*Q q  = Q^*\hat{r}
324    \end{equation}
325    The residual norm used in the PCG is given as
326    \begin{equation}\label{DARCY R NORM}
327    \|r\|\hackscore{PCG}^2 = \int r \cdot (Q^*Q)^{-1} r \; dx =\int \hat{r} \cdot Q (Q^*Q)^{-1} Q^* \hat{r} \; dx \approx
328    \|\hat{r}\|\hackscore{0}^2
329    \end{equation}
330    The iteration is terminated if
331    \begin{equation}\label{DARCY STOP}
332    \|r\|\hackscore{PCG} \le \mbox{ATOL}
333    \end{equation}
334    where we set
335    \begin{equation}\label{DARCY ATOL DEF}
336    \mbox{ATOL} = \mbox{atol} + \mbox{rtol} \cdot \left(\frac{1}{\|v\|\hackscore{0}} + \frac{1}{\|Qp\|\hackscore{0}} \right)^{-1}
337    \end{equation}
338    where rtol is a given relative tolerance and $\mbox{atol}$ is a given absolute tolerance (typically $=0$).
339    Notice that if $Qp$ and $v$ both are zero, the pair $(0,p)$ is a solution.
340    The problem is that ATOL is depending on the solution $p$ (and $v$ calculated form~\ref{DARCY V FORM}). In partcice one use the initial guess for $p$
341    to get a first value for ATOL. If the stopping crierion is met in the PCG iteration, a new $v$ is calculated from the current pressure approximation and ATOL is recalculated. If \ref{DARCY STOP} is still fullfilled the calculation is terminated and $(v,p)$ is returned. Otherwise PCG is restarted with a new ATOL.
342
343    \subsection{Functions}
345    opens the Darcy flux problem\index{Darcy flux} on the \Domain domain.
346    If \var{adaptSubTolerance} is set to \True,
347    the relative tolerances for solving~(\ref{DARCY V FORM}),~(\ref{UPDATE W})
348    and~(\ref{UPDATE P}) are set automatically.
349    \end{classdesc}
350
351  Advection and Reinitialisation  \begin{methoddesc}[DarcyFlow]{setValue}{\optional{f=None, \optional{g=None, \optional{location_of_fixed_pressure=None, \optional{location_of_fixed_flux=None, \optional{permeability=None}}}}}}
352    assigns values to the model parameters. Values can be assigned using various calls - in particular
353    in a time dependend problem only values that change over time needs to be reset. The permability can be defined as scalar (isotropic), a vector (orthotropic) or a matrix (anisotropic).
354    \var{f} and \var{g} are the corresponding parameters in~\ref{DARCY PROBLEM}.
355    The locations and compontents where the flux is prescribed are set by positive values in
356    \var{location_of_fixed_flux}.
357    The locations where the pressure is prescribed are set by
358    by positive values of \var{location_of_fixed_pressure}.
359    The values of the pressure and flux are defined by the initial guess.
360    Notice that at any point on the boundary of the domain the pressure or the normal component of
361    the flux must be defined. There must be at least one point where the pressure is prescribed.
362    The method will try to cast the given values to appropriate
363    \Data class objects.
364    \end{methoddesc}
365
366  \begin{classdesc}{LevelSet}{mesh, func\_new, reinit\_max, reinit\_each, tolerance, smooth}  \begin{methoddesc}[DarcyFlow]{setTolerance}{\optional{rtol=1e-4}}
367  \end{classdesc}  sets the relative tolerance \mbox{rtol} in \ref{DARCY ATOL DEF}.
368    \end{methoddesc}
369
370  %example usage:  \begin{methoddesc}[DarcyFlow]{setAbsoluteTolerance}{\optional{atol=0.}}
371    sets the absolute tolerance \mbox{atol} in \ref{DARCY ATOL DEF}.
372    \end{methoddesc}
373
374  %levelset = LevelSet(mesh, func\_new, reinit\_max, reinit\_each, tolerance, smooth)  \begin{methoddesc}[DarcyFlow]{getSolverOptionsFlux}{}
375    Returns the solver options used to solve the flux problems~(\ref{DARCY V FORM}) and~(\ref{UPDATE W}). Use the returned \SolverOptions object to control the solution algorithms. If the adaption of subtolerance is choosen, the tolerance will
376    be overwritten before the solver is called.
377    \end{methoddesc}
378
379  \begin{methoddesc}[LevelSet]{update\_parameter}{parameter}  \begin{methoddesc}[DarcyFlow]{getSolverOptionsPressure}{}
380  Update the parameter.  Returns the solver options used to solve the pressure problems~(\ref{UPDATE P}).
381    Use the returned \SolverOptions object to control the solution algorithms. If the adaption of subtolerance is choosen, the tolerance will
382    be overwritten before the solver is called.
383  \end{methoddesc}  \end{methoddesc}
384
385  \begin{methoddesc}[LevelSet]{update\_phi}{paramter}{velocity}{dt}{t\_step}  \begin{methoddesc}[DarcyFlow]{solve}{u0,p0, \optional{max_iter=100, \optional{verbose=False}}}
386  Update level set function; advection and reinitialization  solves the problem. and returns approximations for the flux $v$ and the pressure $p$.
387    \var{u0} and \var{p0} define initial guess for flux and pressure. Values marked
388    by positive values \var{location_of_fixed_flux} and \var{location_of_fixed_pressure}, respectively, are kept unchanged. \var{max_iter} sets the maximum number of iterations steps allowed for solving the coupled problem.
389  \end{methoddesc}  \end{methoddesc}
390
\subsection{Benchmark Problem}
391
392  Rayleigh-Taylor instability problem  \subsection{Example: Gravity Flow}
393    later
394
395    %\input{levelsetmodel}
396
397
% \section{Drucker Prager Model}
398
399  \section{Isotropic Kelvin Material \label{IKM}}  \section{Isotropic Kelvin Material \label{IKM}}
400  As proposed by Kelvin~\ref{KELVN} material strain $D\hackscore{ij}=v\hackscore{i,j}+v\hackscore{j,i}$ can be decomposed into  As proposed by Kelvin~\cite{Muhlhaus2005} material strain $D\hackscore{ij}=\frac{1}{2}(v\hackscore{i,j}+v\hackscore{j,i})$ can be decomposed into
401  an elastic part $D\hackscore{ij}^{el}$ and visco-plastic part $D\hackscore{ij}^{vp}$:  an elastic part $D\hackscore{ij}^{el}$ and visco-plastic part $D\hackscore{ij}^{vp}$:
402  \begin{equation}\label{IKM-EQU-2}  \begin{equation}\label{IKM-EQU-2}
403  D\hackscore{ij}=D\hackscore{ij}^{el}+D\hackscore{ij}^{vp}  D\hackscore{ij}=D\hackscore{ij}^{el}+D\hackscore{ij}^{vp}
404  \end{equation}  \end{equation}
405  with the elastic strain given as  with the elastic strain given as
406  \begin{equation}\label{IKM-EQU-3}  \begin{equation}\label{IKM-EQU-3}
407  D\hackscore{ij}'^{el}=\frac{1}{2 \mu} \dot{\sigma}\hackscore{ij}'  D\hackscore{ij}^{el'}=\frac{1}{2 \mu} \dot{\sigma}\hackscore{ij}'
408  \end{equation}  \end{equation}
409  where $\sigma'\hackscore{ij}$ is the deviatoric stress (Notice that $\sigma'\hackscore{ii}=0$).  where $\sigma'\hackscore{ij}$ is the deviatoric stress (Notice that $\sigma'\hackscore{ii}=0$).
410  If the material is composed by materials $q$ the visco-plastic strain can be decomposed as  If the material is composed by materials $q$ the visco-plastic strain can be decomposed as
411  \begin{equation}\label{IKM-EQU-4}  \begin{equation}\label{IKM-EQU-4}
412  D\hackscore{ij}'^{vp}=\sum\hackscore{q} D\hackscore{ij}'^{q}  D\hackscore{ij}^{vp'}=\sum\hackscore{q} D\hackscore{ij}^{q'}
413  \end{equation}  \end{equation}
414  where $D\hackscore{ij}^{q}$ is the strain in material $q$ given as  where $D\hackscore{ij}^{q}$ is the strain in material $q$ given as
415  \begin{equation}\label{IKM-EQU-5}  \begin{equation}\label{IKM-EQU-5}
416  D\hackscore{ij}'^{q}=\frac{1}{2 \eta^{q}} \sigma'\hackscore{ij}  D\hackscore{ij}^{q'}=\frac{1}{2 \eta^{q}} \sigma'\hackscore{ij}
417  \end{equation}  \end{equation}
418  where $\eta^{q}$ is the viscosity of material $q$. We assume the following  where $\eta^{q}$ is the viscosity of material $q$. We assume the following
419  betwee the the strain in material $q$  betwee the the strain in material $q$
420  \begin{equation}\label{IKM-EQU-5b}  \begin{equation}\label{IKM-EQU-5b}
421  \eta^{q}=\eta^{q}\hackscore{N} \left(\frac{\tau}{\tau\hackscore{t}^q}\right)^{\frac{1}{n^{q}}-1}  \eta^{q}=\eta^{q}\hackscore{N} \left(\frac{\tau}{\tau\hackscore{t}^q}\right)^{1-n^{q}}
422  \mbox{ with } \tau=\sqrt{\frac{1}{2}\sigma'\hackscore{ij} \sigma'\hackscore{ij}}  \mbox{ with } \tau=\sqrt{\frac{1}{2}\sigma'\hackscore{ij} \sigma'\hackscore{ij}}
423  \end{equation}  \end{equation}
424  for a given power law coefficients $n^{q}$ and transition stresses $\tau\hackscore{t}^q$, see~\ref{POERLAW}.  for a given power law coefficients $n^{q}\ge1$ and transition stresses $\tau\hackscore{t}^q$, see~\cite{Muhlhaus2005}.
425  Notice that $n^{q}=1$ gives a constant viscosity.  Notice that $n^{q}=1$ gives a constant viscosity.
426  After inserting equation~\ref{IKM-EQU-5} into equation \ref{IKM-EQU-4} one gets:  After inserting equation~\ref{IKM-EQU-5} into equation \ref{IKM-EQU-4} one gets:
427  \begin{equation}\label{IKM-EQU-6}  \begin{equation}\label{IKM-EQU-6}
428  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}} \;.  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}} \;.
\end{equation}
With
\begin{equation}\label{IKM-EQU-8}
\dot{\gamma}=\sqrt{2 D\hackscore{ij} D\hackscore{ij}}
429  \end{equation}  \end{equation}
430  one gets  and finally with~\ref{IKM-EQU-2}
431  \begin{equation}\label{IKM-EQU-8b}  \begin{equation}\label{IKM-EQU-2bb}
432  \tau = \eta^{vp} \dot{\gamma}^{vp} \;.  D\hackscore{ij}'=\frac{1}{2 \eta^{vp}} \sigma'\hackscore{ij}+\frac{1}{2 \mu} \dot{\sigma}\hackscore{ij}'
433  \end{equation}  \end{equation}
434  With the Drucker-Prager cohesion factor $\tau\hackscore{Y}$, Drucker-Prager friction $\beta$ and total pressure $p$ we want to achieve  The total stress $\tau$ needs to fullfill the yield condition \index{yield condition}
435  \begin{equation}\label{IKM-EQU-8c}  \begin{equation}\label{IKM-EQU-8c}
436  \tau \le \tau\hackscore{Y} + \beta \; p  \tau \le \tau\hackscore{Y} + \beta \; p
437  \end{equation}  \end{equation}
438  which leads to the condition  with the Drucker-Prager \index{Druck-Prager} cohesion factor \index{cohesion factor} $\tau\hackscore{Y}$, Drucker-Prager friction $\beta$ and total pressure $p$.
\begin{equation}\label{IKM-EQU-8d}
\eta^{vp} \le \frac{\tau\hackscore{Y} + \beta \; p}{ \dot{\gamma}^{vp}} \; .
\end{equation}
Therefore we modify the definition of $\eta^{vp}$ to the form
\begin{equation}\label{IKM-EQU-6b}
\frac{1}{\eta^{vp}}=\max(\sum\hackscore{q} \frac{1}{\eta^{q}}, \frac{\dot{\gamma}^{vp}} {\tau\hackscore{Y} + \beta \; p})
\end{equation}
439  The deviatoric stress needs to fullfill the equilibrion equation  The deviatoric stress needs to fullfill the equilibrion equation
440  \begin{equation}\label{IKM-EQU-1}  \begin{equation}\label{IKM-EQU-1}
441  -\sigma'\hackscore{ij,j}+p\hackscore{,i}=F\hackscore{i}  -\sigma'\hackscore{ij,j}+p\hackscore{,i}=F\hackscore{i}
442  \end{equation}  \end{equation}
443  where $F\hackscore{j}$ is a given external fource. We assume an incompressible media:  where $F\hackscore{j}$ is a given external fource. We assume an incompressible media:
444  \begin{equation}\label{IKM-EQU-2}  \begin{equation}\label{IKM-EQU-2bbb}
445  -v\hackscore{i,i}=0  -v\hackscore{i,i}=0
446  \end{equation}  \end{equation}
447  Natural boundary conditions are taken in the form  Natural boundary conditions are taken in the form
# Line 234  where the index $i$ may depend on the lo Line 457  where the index $i$ may depend on the lo
457  \subsection{Solution Method \label{IKM-SOLVE}}  \subsection{Solution Method \label{IKM-SOLVE}}
458  By using a first order finite difference approximation wit step size $dt>0$~\ref{IKM-EQU-3} get the form  By using a first order finite difference approximation wit step size $dt>0$~\ref{IKM-EQU-3} get the form
459  \begin{equation}\label{IKM-EQU-3b}  \begin{equation}\label{IKM-EQU-3b}
460  D\hackscore{ij}'^{el}=\frac{1}{2 \mu dt } \left( \sigma\hackscore{ij}' - \sigma\hackscore{ij}^{'-} \right)  \dot{\sigma}\hackscore{ij}=\frac{1}{dt } \left( \sigma\hackscore{ij} - \sigma\hackscore{ij}^{-} \right)
461    \end{equation}
462    and
463    \begin{equation}\label{IKM-EQU-2b}
464    D\hackscore{ij}'=\left(\frac{1}{2 \eta^{vp}} + \frac{1}{2 \mu dt}\right) \sigma\hackscore{ij}'-\frac{1}{2 \mu dt } \sigma\hackscore{ij}^{-'}
465    \end{equation}
466    where $\sigma\hackscore{ij}^{-}$ is the stress at the precious time step. With
467    \begin{equation}\label{IKM-EQU-2c}
468    \dot{\gamma} = \sqrt{ 2 \left( D\hackscore{ij}' +
469    \frac{1}{  2 \mu \; dt} \sigma\hackscore{ij}^{-'}\right)^2}
470    \end{equation}
471    we have
472    \begin{equation}
473    \tau = \eta\hackscore{eff} \cdot \dot{\gamma}
474    \end{equation}
475    where
476    \begin{equation}
477    \eta\hackscore{eff}= min( \left(\frac{1}{\mu \; dt}+\frac{1}{\eta^{vp}}\right)^{-1}
478    , \eta\hackscore{max}) \mbox{ with }
479    \eta\hackscore{max} = \left\{
480    \begin{array}{rcl}
481    \frac{\tau\hackscore{Y} + \beta \; p}{\dot{\gamma}} & & \dot{\gamma}>0 \\
482    &\mbox{ if } \\
483    \infty & & \mbox{otherwise}
484    \end{array}
485    \right.
486  \end{equation}  \end{equation}
487  where $\sigma\hackscore{ij}^{'-}$ is the deviatoric stress at the precious time step.  The upper bound $\eta\hackscore{max}$ makes sure that yield condtion~\ref{IKM-EQU-8c} holds. With this setting the eqaution \ref{IKM-EQU-2b} takes the form
Now we can combine equations~\ref{IKM-EQU-2}, \ref{IKM-EQU-3b} and~\ref{IKM-EQU-6b} to get
488  \begin{equation}\label{IKM-EQU-10}  \begin{equation}\label{IKM-EQU-10}
489  \sigma\hackscore{ij}' =  2 \eta\hackscore{eff} D\hackscore{ij}' +  \sigma\hackscore{ij}' =  2 \eta\hackscore{eff}  \left( D\hackscore{ij}' +
490  \frac{\eta\hackscore{eff}}{\mu \; dt}  \frac{1}{  2 \mu \; dt} \sigma\hackscore{ij}^{'-}\right)
\sigma\hackscore{ij}^{'-} \mbox{ with }
\frac{1}{\eta\hackscore{eff}}=\frac{1}{\mu \; dt}+\frac{1}{\eta^{vp}}
491  \end{equation}  \end{equation}
Notice that $\eta\hackscore{eff}$ is a function of stress.
492  After inserting~\ref{IKM-EQU-10} into~\ref{IKM-EQU-1} we get  After inserting~\ref{IKM-EQU-10} into~\ref{IKM-EQU-1} we get
493  \begin{equation}\label{IKM-EQU-1ib}  \begin{equation}\label{IKM-EQU-1ib}
494  -\left(\eta\hackscore{eff} (v\hackscore{i,j}+ v\hackscore{i,j})\right)\hackscore{,j}+p\hackscore{,i}=F\hackscore{i}+  -\left(\eta\hackscore{eff} (v\hackscore{i,j}+ v\hackscore{i,j})
495  \frac{\eta\hackscore{eff}}{\mu dt } \sigma\hackscore{ij,j}^{'-}  \right)\hackscore{,j}+p\hackscore{,i}=F\hackscore{i}+
496     \left(\frac{\eta\hackscore{eff}}{\mu dt } \sigma\hackscore{ij}^{'-} \right)\hackscore{,j}
497    \end{equation}
498    Combining this with the incomressibilty condition~\ref{IKM-EQU-2} we need to solve a
499    Stokes problem as discussed in section~\ref{STOKES SOLVE} in each time step.
500
501    If we set
502    \begin{equation}\label{IKM-EQU-44}
503    \frac{1}{\eta(\tau)}= \frac{1}{\mu \; dt}+\frac{1}{\eta^{vp}}
504  \end{equation}  \end{equation}
505  Together with the incomressibilty condition~\ref{IKM-EQU-2} we need to solve a problem with a form almost identical  we need to solve the nonlinear problem
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
506  \begin{equation}  \begin{equation}
507  \begin{array}{rcl}  \eta\hackscore{eff} -  min(\eta( \dot{\gamma} \cdot \eta\hackscore{eff})
508  -\left(\eta\hackscore{eff}(dv\hackscore{i,j}+ dv\hackscore{i,j})\right)\hackscore{,j} & = & F\hackscore{i}+ \sigma\hackscore{ij,j}' \\  , \eta\hackscore{max}) =0
\frac{1}{\eta\hackscore{eff}} dp & = & - v\hackscore{i,i}^+ \\
d\sigma\hackscore{ij}' & = & 2 \eta\hackscore{eff} D\hackscore{ij}^{+'} + \frac{\eta\hackscore{eff}}{\mu \; dt} \sigma\hackscore{ij}' -  \sigma\hackscore{ij}\\
\end{array}
\label{IKM iteration 2}
509  \end{equation}  \end{equation}
510  where $v^+=v+dv$. As this problem is non-linear the Jacobi-free Newton-GMRES method is used with the norm  We use the Newton-Raphson Scheme \index{Newton-Raphson scheme} to solve this problem
511    \begin{equation}\label{IKM-EQU-45}
512    \eta\hackscore{eff}^{(n+1)} = \min(\eta\hackscore{max},
513    \eta\hackscore{eff}^{(n)} -
514    \frac{\eta\hackscore{eff}^{(n)} - \eta( \tau^{(n)}) }
515    {1-\dot{\gamma} \cdot \eta'( \tau^{(n)} )} )
516    =\min(\eta\hackscore{max},
517    \frac{\eta( \tau^{(n)}) -\tau^{(n)} \cdot  \eta'( \tau^{(n)} )  }
518    {1-\dot{\gamma} \cdot \eta'( \tau^{(n)} )} )
519    \end{equation}
520    where $\eta'$ denotes the derivative of $\eta$ with respect of $\tau$
521    and $\tau^{(n)} = \dot{\gamma} \cdot \eta\hackscore{eff}^{(n)}$.
522
523    Looking at the evaluation of $\eta$ in~\ref{IKM-EQU-44} it makes sense formulate
524    the iteration~\ref{IKM-EQU-45} using $\Theta=\eta^{-1}$.
525    In fact we have
526  \begin{equation}  \begin{equation}
527  \|(\sigma, p)\|^2= \int\hackscore{\Omega} \sigma\hackscore{i,j}^2 + p^2 \; dx  \eta' = - \frac{\Theta'}{\Theta^2}
528  \label{IKM iteration 3}  \mbox{ with }
529  \end{equation}  \Theta' = \sum\hackscore{q} \left(\frac{1}{\eta^{q}} \right)'
530    \label{IKM iteration 7}
531    \end{equation}
532    As
533    \begin{equation}\label{IKM-EQU-47}
534    \left(\frac{1}{\eta^{q}} \right)'
535    = \frac{n^{q}-1}{\eta^{q}\hackscore{N}} \cdot \frac{\tau^{n^{q}-2}}{(\tau\hackscore{t}^q)^{n^{q}-1}}
536    = \frac{n^{q}-1}{\eta^{q}}\cdot\frac{1}{\tau}
537    \end{equation}
538    we have
539    \begin{equation}\label{IKM-EQU-48}
540    \Theta' = \frac{1}{\tau} \omega \mbox{ with } \omega = \sum\hackscore{q}\frac{n^{q}-1}{\eta^{q}}
541    \end{equation}
543    \begin{equation}\label{IKM-EQU-49}
544    \eta\hackscore{eff}^{(n+1)} = \min(\eta\hackscore{max},
545    \eta\hackscore{eff}^{(n)}
546    \frac{\Theta^{(n)}  + \omega^{(n)}  }
547    {\eta\hackscore{eff}^{(n)} \Theta^{(n)^2}+\omega^{(n)} })
548    \end{equation}
549
550
551    \subsection{Functions}
552
553    \begin{classdesc}{IncompressibleIsotropicFlowCartesian}{
554    domain
555    \optional{, stress=0
556    \optional{, v=0
557    \optional{, p=0
558    \optional{, t=0
559    \optional{, numMaterials=1
560    \optional{, verbose=True
562    }}}}}}}}
563    opens an incompressible, isotropic flow problem in Cartesian cooridninates
564    on the domain \var{domain}.
565    \var{stress},
566    \var{v},
567    \var{p}, and
568    \var{t} set the initial deviatoric stress, velocity, pressure and time.
569    \var{numMaterials} specifies the number of materials used in the power law
570    model. Some progress information are printed if \var{verbose} is set to
571    \True. If \var{adaptSubTolerance} is equal to True the tolerances for subproblems are set automatically.
572    \end{classdesc}
573
574    \begin{methoddesc}[IncompressibleIsotropicFlowCartesian]{getDomain}{}
575    returns the domain.
576    \end{methoddesc}
577
578    \begin{methoddesc}[IncompressibleIsotropicFlowCartesian]{getTime}{}
579    Returns current time.
580    \end{methoddesc}
581
582    \begin{methoddesc}[IncompressibleIsotropicFlowCartesian]{getStress}{}
583    Returns current stress.
584    \end{methoddesc}
585
586    \begin{methoddesc}[IncompressibleIsotropicFlowCartesian]{getDeviatoricStress}{}
587    Returns current deviatoric stress.
588    \end{methoddesc}
589
590    \begin{methoddesc}[IncompressibleIsotropicFlowCartesian]{getPressure}{}
591    Returns current pressure.
592    \end{methoddesc}
593
594    \begin{methoddesc}[IncompressibleIsotropicFlowCartesian]{getVelocity}{}
595    Returns current velocity.
596    \end{methoddesc}
597
598    \begin{methoddesc}[IncompressibleIsotropicFlowCartesian]{getDeviatoricStrain}{}
599    Returns deviatoric strain of current velocity
600    \end{methoddesc}
601
602    \begin{methoddesc}[IncompressibleIsotropicFlowCartesian]{getTau}{}
603    Returns current second invariant of deviatoric stress
604    \end{methoddesc}
605
607    Returns current second invariant of deviatoric strain
608    \end{methoddesc}
609
610
611    \begin{methoddesc}[IncompressibleIsotropicFlowCartesian]{setTolerance}{tol=1.e-4}
612    Sets the tolerance used to terminate the iteration on a time step.
613    \end{methoddesc}
614
615    \begin{methoddesc}[IncompressibleIsotropicFlowCartesian]{setFlowTolerance}{tol=1.e-4}
616    Sets the relative tolerance for the incompressible solver, see \class{StokesProblemCartesian} for details.
617    \end{methoddesc}
618
619    \begin{methoddesc}[IncompressibleIsotropicFlowCartesian]{setElasticShearModulus}{mu=None}
620    Sets the elastic shear modulus $\mu$. If \var{mu} is set to None (default) elasticity is not applied.
621    \end{methoddesc}
622
623    \begin{methoddesc}[IncompressibleIsotropicFlowCartesian]{setEtaTolerance=}{rtol=1.e-8}
624    sets the relative tolerance for the effectice viscosity. Iteration on a time step is completed if the realtive of the effective viscosity is less than \var{rtol}.
625    \end{methoddesc}
626
627    \begin{methoddesc}[IncompressibleIsotropicFlowCartesian]{setDruckerPragerLaw}
628    {\optional{tau_Y=None, \optional{friction=None}}}
629    Sets the parameters $\tau\hackscore{Y}$ and $\beta$ for the Drucker-Prager model in condition~\ref{IKM-EQU-8c}. If \var{tau_Y} is set to None (default) Drucker-Prager
630    condition is not applied.
631    \end{methoddesc}
632
633    \begin{methoddesc}[IncompressibleIsotropicFlowCartesian]{setElasticShearModulus}{mu=None}
634    Sets the elastic shear modulus $\mu$. If \var{mu} is set to None (default) elasticity is not applied.
635    \end{methoddesc}
636
637
638    \begin{methoddesc}[IncompressibleIsotropicFlowCartesian]{setPowerLaws}{eta_N, tau_t, power}
639    Sets the parameters of the power-law for all materials as defined in
640    equation~\ref{IKM-EQU-5b}.
641    \var{eta_N} is the list of viscosities $\eta^{q}\hackscore{N}$,
642    \var{tau_t} is the list of reference stresses  $\tau\hackscore{t}^q$,
643    and \var{power} is the list of  power law coefficients $n^{q}$.
644    \end{methoddesc}
645
646
647    \begin{methoddesc}[IncompressibleIsotropicFlowCartesian]{update}{dt
648    \optional{, iter_max=100
649    \optional{, inner_iter_max=20
650    }}}
651    Updates stress, velocity and pressure for time increment \var{dt}.
652    where \var{iter_max} is the maximum number of iteration steps on a time step to
653    update the effective viscosity and \var{inner_iter_max} is the maximum
654    number of itertion steps in the incompressible solver.
655    \end{methoddesc}
656
657    \subsection{Example}
658    later
659
660    % \section{Drucker Prager Model}

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