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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  \label{Stokes 1}  \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
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  \label{Stokes 2}  \label{Stokes 2}
27  -v\hackscore{i,i}=0  -v\hackscore{i,i}=0
28
29  Natural boundary conditions are taken in the form  Natural boundary conditions are taken in the form
30  \label{Stokes Boundary}  \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{i}
32
33  which can be overwritten by a constraint  which can be overwritten by constraints of the form
34  \label{Stokes Boundary0}  \label{Stokes Boundary0}
35  v\hackscore{i}(x)=0  v\hackscore{i}(x)=v^D\hackscore{i}(x)
36
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
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
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
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
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
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
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
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} \\
78  \end{array}  which is inserted into the second equation leading to
80    S p =  B A^{-1} G
81
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
85  \|(v,p)\|^2= \int\hackscore{\Omega} \eta^2 v\hackscore{i,j}^2 + p^2 \; dx  \frac{1}{\eta}q = p
86
87  or alternatively the $PCG$ method on the pressure only using the norm  see~\cite{ELMAN} for more details. Note that the residual for the current approximation $p$ is given as
88
89  \|p\|^2= \int\hackscore{\Omega} \frac{1}{\eta^2} p^2 \; dx  r=B A^{-1} (G - B^* p) = Bv
90
91    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
96    \hat{S}^{-1} S p =  \hat{S}^{-1} B A^{-1} G
97
98    We use the norm
99
100    \|p\|\hackscore{GMRES} = \|\hat{S} p \|
101
102    Notice that for the residual $\hat{r}=\hat{S}^{-1} r$ one has
103
104    \
105
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
110    \|\hat{r}\|\hackscore{GMRES} \le ATOL
111
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
118    \|r\|\hackscore{PCG}^2 = \int\hackscore{\Omega} r \hat{S}^{-1}r \; dx = \int\hackscore{\Omega}  \eta r^2 \; dx
119
120    To take the extra factor $\eta$ into consideration when checking the stopping criterion we use the following
121    definition for $ATOL$:
122
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
126
127
128
129    \subsection{Functions}
130
131  \begin{classdesc}{StokesProblemCartesian}{domain,debug}  \begin{classdesc}{StokesProblemCartesian}{domain}
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.
134  \end{classdesc}  \end{classdesc}
135
136  example usage:  \begin{methoddesc}[StokesProblemCartesian]{initialize}{\optional{f=Data(), \optional{fixed_u_mask=Data(), \optional{eta=1, \optional{surface_stress=Data(), \optional{stress=Data()}}}}}}
137    assigns values to the model parameters. In any call all values must be set.
138  solution=StokesProblemCartesian(mesh) \\  \var{f} defines the external force $f$, \var{eta} the viscosity $\eta$,
139  solution.setTolerance(TOL) \\  \var{surface_stress} the surface stress $s$ and \var{stress} the initial stress $\sigma$.
140  solution.initialize(fixed\_u\_mask=b\_c,eta=eta,f=Y) \\  The locations and compontents where the velocity is fixed are set by
141  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
142    \Data class objects.
143    \end{methoddesc}
144
145  % \subsection{Benchmark Problem}  \begin{methoddesc}[StokesProblemCartesian]{solve}{v,p,
146  %  \optional{max_iter=20, \optional{verbose=False, \optional{usePCG=True}}}}
147  % Convection problem  solves the problem and return approximations for velocity and pressure.
148    The arguments \var{v} and \var{p} define initial guess. The values of \var{v} marked
150    If \var{usePCG} is set to \True
151    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
152    the PCG scheme is more efficient.
153    \var{max_iter} defines the maximum number of iteration steps.
154    If \var{verbose} is set to \True informations on the progress of of the solver are printed.
155    \end{methoddesc}
156
157
158  \section{Temperature Cartesian}  \begin{methoddesc}[StokesProblemCartesian]{setTolerance}{\optional{tolerance=1.e-4}}
159    sets the tolerance in an appropriate norm relative to the right hand side. The tolerance must be non-negative and less than 1.
160    \end{methoddesc}
161    \begin{methoddesc}[StokesProblemCartesian]{getTolerance}{}
162    returns the current relative tolerance.
163    \end{methoddesc}
164    \begin{methoddesc}[StokesProblemCartesian]{setAbsoluteTolerance}{\optional{tolerance=0.}}
165    sets the absolute tolerance for the error in the relevant norm. The tolerance must be non-negative. Typically the
166    absolute talerance is set to 0.
167    \end{methoddesc}
168    \begin{methoddesc}[StokesProblemCartesian]{getAbsoluteTolerance}{}
169    sreturns the current absolute tolerance.
170    \end{methoddesc}
171    \begin{methoddesc}[StokesProblemCartesian]{setSubProblemTolerance}{\optional{rtol=None}}
172    sets the tolerance to solve the involved PDEs. The subtolerance \var{rtol} should not be choosen to large
173    in order to avoid feed back of errors in the subproblem solution into the outer iteration.
174    On the otherhand is choosen to small compute time is wasted.
175    If \var{rtol} is set to \var{None} the sub-tolerance is set automatically depending on the
176    tolerance choosen for the oter iteration.
177    \end{methoddesc}
178    \begin{methoddesc}[StokesProblemCartesian]{getSubProblemTolerance}{}
179    return the tolerance for the involved PDEs.
180    \end{methoddesc}
181
182    \subsection{Example: Lit Driven Cavity}
183     The following script \file{lit\hackscore driven\hackscore cavity.py}
184    \index{scripts!\file{helmholtz.py}} which is available in the \ExampleDirectory
185    illustrates the usage of the \class{StokesProblemCartesian} class to solve
186    the lit driven cavity problem~\cite{LITDRIVENCAVITY}:
187    \begin{python}
188    from esys.escript import *
189    from esys.finley import Rectangle
190    from esys.escript.models import StokesProblemCartesian
191    NE=25
192    dom = Rectangle(NE,NE,order=2)
193    x = dom.getX()
194    sc=StokesProblemCartesian(dom)
196          (whereZero(x[1])*[0.,1.]+whereZero(x[1]-1))*[1.,1]
198    v=Vector(0.,Solution(dom))
199    v[0]+=whereZero(x[1]-1.)
200    p=Scalar(0.,ReducedSolution(dom))
201    v,p=sc.solve(v,p, verbose=True)
202    saveVTK("u.xml",velocity=v,pressure=p)
203    \end{python}
204
205    \section{Darcy Flux}
206    We want to calculate the velocity $u$ and pressure $p$ on a domain $\Omega$ solving
207    the Darcy flux problem \index{Darcy flux}\index{Darcy flow}
208    \label{DARCY PROBLEM}
209    \begin{array}{rcl}
210    u\hackscore{i} + \kappa\hackscore{ij} p\hackscore{,j} & = & g\hackscore{i} \\
211    u\hackscore{k,k} & = & f
212    \end{array}
213
214    with the boundary conditions
215    \label{DARCY BOUNDARY}
216    \begin{array}{rcl}
217    u\hackscore{i} \; n\hackscore{i}  = u^{N}\hackscore{i}  \; n\hackscore{i} & \mbox{ on } & \Gamma\hackscore{N} \\
218    p = p^{D} &  \mbox{ on } & \Gamma\hackscore{D} \\
219    \end{array}
220
221    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
222
223  \rho c\hackscore{p} \left (\frac{\partial T}{\partial t} + \vec{v} \cdot \nabla T \right ) = k \nabla^{2}T  \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}
\label{HEAT EQUATION}
224
225    for all $x\hackscore{i}$.
226
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.
227
228  \subsection{Description}  \subsection{Solution Method \label{DARCY SOLVE}}
229    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
230  \subsection{Method}
231    -(\kappa\hackscore{ki}\kappa\hackscore{kj} p\hackscore{,j}^{ref})\hackscore{,i} =  - (\kappa\hackscore{ki} (g\hackscore{k}- u^{N}\hackscore{k}))\hackscore{,i}
232  \begin{classdesc}{TemperatureCartesian}{dom,theta=THETA,useSUPG=SUPG}  \mbox{ with }
233    p^{ref} = p^{D} \mbox{ on } \Gamma\hackscore{D}
234
235    With setting $u \leftarrow u-u^{N}$ and $p \leftarrow p-p^{ref}$ and
236
237    \begin{array}{rcl}
238    g\hackscore{i} & \leftarrow & g\hackscore{i} - u^{N}\hackscore{i} -  \kappa\hackscore{ij} p^{ref}\hackscore{,j }\\
239    f & \leftarrow & f - u^{N}\hackscore{k,k}
240    \end{array}
241
242    we can assume that $u^{N}\hackscore{i} \; n\hackscore{i}=0$ and
243    $p^{D}=0$.
244    We set
245
246    V = \{ q \in H^{1}(\Omega) : q=0 \mbox{ on } \Gamma\hackscore{D} \}
247
248    and
249
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
252    and define the operator $Q: V \rightarrow (L^2(\Omega))^{d}$ defined by
253
254    (Qp)\hackscore{i} = \kappa\hackscore{ij} p\hackscore{,j}
255
256    and the operator $D: W \rightarrow L^2(\Omega)$ defined by
257
258    Dv = v\hackscore{k,k}
259
260    In operator notation the Darcy problem~\ref{DARCY PROBLEM} is written in the form
261
262    \begin{array}{rcl}
263    u + Qp & = & g \\
264    Du & = & f
265    \end{array}
266
267    We solve this equation by minimising the functional
268
269    J(u,p):=\|u + Qp - g\|^2\hackscore{0} + \|Du-f\|\hackscore{0}^2
270
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
274    ( v + Qq , u + Qp - g) + (Dv,Du-f) =0
275
276    for all $v\in W$ and $q \in V$.which translates back into operator notation
277
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
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    \label{DARCY V FORM}
289    v= (I+D^*D)^{-1} (D^*f + g - Qp)
290
291    (notice that $(I+D^*D)$ is coercive in $W$) which is inserted into the second equation
292
293    Q^* (I+D^*D)^{-1} (D^*f + g - Qp) + Q^* Q p = Q^*g
294
295    which is
296
297    Q^* ( I - (I+D^*D)^{-1} ) Q p = Q^* (g-(I+D^*D)^{-1} (D^*f + g) )
298
299    We use the PCG \index{linear solver!PCG}\index{PCG} method to solve this. The residual $r$ ($\in V^*$) is given as
300
301    \begin{array}{rcl}
302    r & = & Q^*  \left( g -(I+D^*D)^{-1} (D^*f + g) - Qp + (I+D^*D)^{-1}Q p \right)\\
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
307    So in a partical implementation we use $\hat{r}=g-Qp-v$ to represent the residual.
308    The evaluation of the iteration operator for a given $p$ is then
309    returning $Qp+v$ where $v$ is the solution of
310    \label{UPDATE W}
311    (I+D^*D)v = Qp
312
313    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
314    implemented by solving
315    \label{UPDATE P}
316    Q^*Q q  = Q^*\hat{r}
317
318    The residual norm used in the PCG is given as
319    \label{DARCY R NORM}
320    \|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
321    \|\hat{r}\|\hackscore{0}^2
322
323    The iteration is terminated if
324    \label{DARCY STOP}
325    \|r\|\hackscore{PCG} \le \mbox{ATOL}
326
327    where we set
328    \label{DARCY ATOL DEF}
329    \mbox{ATOL} = \mbox{atol} + \mbox{rtol} \cdot \left(\frac{1}{\|v\|\hackscore{0}} + \frac{1}{\|Qp\|\hackscore{0}} \right)^{-1}
330
331    where rtol is a given relative tolerance and $\mbox{atol}$ is a given absolute tolerance (typically $=0$).
332    Notice that if $Qp$ and $v$ both are zero, the pair $(0,p)$ is a solution.
333    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 $p$
334    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.
335
336    \subsection{Functions}
337    \begin{classdesc}{DarcyFlow}{domain}
338    opens the Darcy flux problem\index{Darcy flux} on the \Domain domain.
339  \end{classdesc}  \end{classdesc}
340    \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}}}}}}
341    assigns values to the model parameters. Values can be assigned using various calls - in particular
342    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).
343    \var{f} and \var{g} are the corresponding parameters in~\ref{DARCY PROBLEM}.
344    The locations and compontents where the flux is prescribed are set by positive values in
345    \var{location_of_fixed_flux}.
346    The locations where the pressure is prescribed are set by
347    by positive values of \var{location_of_fixed_pressure}.
348    The values of the pressure and flux are defined by the initial guess.
349    Notice that at any point on the boundary of the domain the pressure or the normal component of
350    the flux must be defined. There must be at least one point where the pressure is prescribed.
351    The method will try to cast the given values to appropriate
352    \Data class objects.
353    \end{methoddesc}
354
355  \subsection{Benchmark Problem}  \begin{methoddesc}[DarcyFlow]{setTolerance}{\optional{rtol=1e-4}}
356    sets the relative tolerance \mbox{rtol} in \ref{DARCY ATOL DEF}.
357    \end{methoddesc}
358
359    \begin{methoddesc}[DarcyFlow]{setAbsoluteTolerance}{\optional{atol=0.}}
360    sets the absolute tolerance \mbox{atol} in \ref{DARCY ATOL DEF}.
361    \end{methoddesc}
362
363  \section{Level Set Method}  \begin{methoddesc}[DarcyFlow]{setSubProblemTolerance}{\optional{rtol=None}}
364    sets the relative tolerance used to solve the involved PDEs. If no argument is given,
365    the square of the current relative tolerance is used. The sub-problem tolerance should be choosen as large as possible to minimize the compute time. However, a too large value for the sub-problem tolerance may lead to slow convergence or even dibergence in the outer iteration.
366    \end{methoddesc}
367
368  \subsection{Description}  \begin{methoddesc}[DarcyFlow]{solve}{u0,p0, \optional{max_iter=100, \optional{verbose=False \optional{sub_rtol=1.e-8}}}}
369    solves the problem. and returns approximations for the flux $v$ and the pressure $p$.
370    \var{u0} and \var{p0} define initial guess for flux and pressure. Values marked
371    by positive values \var{location_of_fixed_flux} and \var{location_of_fixed_pressure}, respectively, are kept unchanged.
372    \end{methoddesc}
373
\subsection{Method}
374
375  Advection and Reinitialisation  \subsection{Example: Gravity Flow}
376    later
377
378  \begin{classdesc}{LevelSet}{mesh, func\_new, reinit\_max, reinit\_each, tolerance, smooth}  %================================================
380
381  %example usage:  %
382    % \rho c\hackscore{p} \left (\frac{\partial T}{\partial t} + \vec{v} \cdot \nabla T \right ) = k \nabla^{2}T
383    % \label{HEAT EQUATION}
384    %
385
386  %levelset = LevelSet(mesh, func\_new, reinit\_max, reinit\_each, tolerance, smooth)  % 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.
387
388  \begin{methoddesc}[LevelSet]{update\_parameter}{parameter}  % \subsection{Description}
Update the parameter.
\end{methoddesc}

\begin{methoddesc}[LevelSet]{update\_phi}{paramter}{velocity}{dt}{t\_step}
Update level set function; advection and reinitialization
\end{methoddesc}
389
390  \subsection{Benchmark Problem}  % \subsection{Method}
391    %
392    % \begin{classdesc}{TemperatureCartesian}{dom,theta=THETA,useSUPG=SUPG}
393    % \end{classdesc}
394
395  Rayleigh-Taylor instability problem  % \subsection{Benchmark Problem}
396    %===============================================================================================================
397
398    %=========================================================
399    \input{levelsetmodel}
400
401  % \section{Drucker Prager Model}  % \section{Drucker Prager Model}
402
# Line 170  D\hackscore{ij}=D\hackscore{ij}^{el}+D\h Line 408  D\hackscore{ij}=D\hackscore{ij}^{el}+D\h
408
409  with the elastic strain given as  with the elastic strain given as
410  \label{IKM-EQU-3}  \label{IKM-EQU-3}
411  D\hackscore{ij}'^{el}=\frac{1}{2 \mu} \dot{\sigma}\hackscore{ij}'  D\hackscore{ij}^{el'}=\frac{1}{2 \mu} \dot{\sigma}\hackscore{ij}'
412
413  where $\sigma'\hackscore{ij}$ is the deviatoric stress (Notice that $\sigma'\hackscore{ii}=0$).  where $\sigma'\hackscore{ij}$ is the deviatoric stress (Notice that $\sigma'\hackscore{ii}=0$).
414  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
415  \label{IKM-EQU-4}  \label{IKM-EQU-4}
416  D\hackscore{ij}'^{vp}=\sum\hackscore{q} D\hackscore{ij}'^{q}  D\hackscore{ij}^{vp'}=\sum\hackscore{q} D\hackscore{ij}^{q'}
417
418  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
419  \label{IKM-EQU-5}  \label{IKM-EQU-5}
420  D\hackscore{ij}'^{q}=\frac{1}{2 \eta^{q}} \sigma'\hackscore{ij}  D\hackscore{ij}^{q'}=\frac{1}{2 \eta^{q}} \sigma'\hackscore{ij}
421
422  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
423  betwee the the strain in material $q$  betwee the the strain in material $q$
# Line 187  betwee the the strain in material $q$ Line 425  betwee the the strain in material $q$
425  \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)^{\frac{1}{n^{q}}-1}
426  \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}}
427
428  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~\ref{POERLAW}.
429  Notice that $n^{q}=1$ gives a constant viscosity.  Notice that $n^{q}=1$ gives a constant viscosity.
430  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:
431  \label{IKM-EQU-6}  \label{IKM-EQU-6}
432  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}} \;.
433
434  With  With
435  \label{IKM-EQU-8}  \label{IKM-EQU-8}
# Line 239  D\hackscore{ij}'^{el}=\frac{1}{2 \mu dt Line 477  D\hackscore{ij}'^{el}=\frac{1}{2 \mu dt
477  where $\sigma\hackscore{ij}^{'-}$ is the deviatoric stress at the precious time step.  where $\sigma\hackscore{ij}^{'-}$ is the deviatoric stress at the precious time step.
478  Now we can combine equations~\ref{IKM-EQU-2}, \ref{IKM-EQU-3b} and~\ref{IKM-EQU-6b} to get  Now we can combine equations~\ref{IKM-EQU-2}, \ref{IKM-EQU-3b} and~\ref{IKM-EQU-6b} to get
479  \label{IKM-EQU-10}  \label{IKM-EQU-10}
480  \sigma\hackscore{ij}' =  2 \eta\hackscore{eff} D\hackscore{ij}' +  \sigma\hackscore{ij}' =  2 \eta\hackscore{eff}  \left( D\hackscore{ij}' +
481  \frac{\eta\hackscore{eff}}{\mu \; dt}  \frac{1}{  2 \mu \; dt} \sigma\hackscore{ij}^{'-}\right)  \mbox{ with }
\sigma\hackscore{ij}^{'-} \mbox{ with }
482  \frac{1}{\eta\hackscore{eff}}=\frac{1}{\mu \; dt}+\frac{1}{\eta^{vp}}  \frac{1}{\eta\hackscore{eff}}=\frac{1}{\mu \; dt}+\frac{1}{\eta^{vp}}
483
484  Notice that $\eta\hackscore{eff}$ is a function of stress.  Notice that $\eta\hackscore{eff}$ is a function of diatoric stress $\sigma\hackscore{ij}'$.
485  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
486  \label{IKM-EQU-1ib}  \label{IKM-EQU-1ib}
487  -\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})
488    \right)\hackscore{,j}+p\hackscore{,i}=F\hackscore{i}+
489  \frac{\eta\hackscore{eff}}{\mu dt } \sigma\hackscore{ij,j}^{'-}  \frac{\eta\hackscore{eff}}{\mu dt } \sigma\hackscore{ij,j}^{'-}
490
491  Together with the incomressibilty condition~\ref{IKM-EQU-2} we need to solve a problem with a form almost identical  Combining this with the incomressibilty condition~\ref{IKM-EQU-2} we need to solve a
492  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  Stokes problem as discussed in section~\ref{STOKES SOLVE} in each time step.
493    In oder to perform step~\ref{IKM iteration 2} we need to calculate the $\eta\hackscore{eff}$ which
494    is a function of $\sigma\hackscore{ij}$ via $\tau$.  To get $\tau$ and $\eta\hackscore{eff}$ we need to solve the
495    non-linear equation
496
497  \begin{array}{rcl}  \tau = \eta\hackscore{eff} \cdot \dot{\gamma}\hackscore{total} \mbox{ with }
498  -\left(\eta\hackscore{eff}(dv\hackscore{i,j}+ dv\hackscore{i,j})\right)\hackscore{,j} & = & F\hackscore{i}+ \sigma\hackscore{ij,j}' \\  \dot{\gamma}\hackscore{total} = \sqrt{ 2 \left( D\hackscore{ij}' +
499  \frac{1}{\eta\hackscore{eff}} dp & = & - v\hackscore{i,i}^+ \\  \frac{1}{  2 \mu \; dt} \sigma\hackscore{ij}^{'-}\right)^2}
500  d\sigma\hackscore{ij}' & = & 2 \eta\hackscore{eff} D\hackscore{ij}^{+'} + \frac{\eta\hackscore{eff}}{\mu \; dt} \sigma\hackscore{ij}' -  \sigma\hackscore{ij}\\  \label{IKM iteration 5}
501  \end{array}
502  \label{IKM iteration 2}  The Newton scheme takes the form

where $v^+=v+dv$. As this problem is non-linear the Jacobi-free Newton-GMRES method is used with the norm
503
504  \|(\sigma, p)\|^2= \int\hackscore{\Omega} \sigma\hackscore{i,j}^2 + p^2 \; dx  \tau\hackscore{n+1} = \min(\tau\hackscore{n} - \frac{\tau\hackscore{n} - \eta\hackscore{eff}  \cdot \dot{\gamma}\hackscore{total}}{1 - \eta\hackscore{eff}'  \cdot  \dot{\gamma}\hackscore{total}}, \tau\hackscore{Y} + \beta \; p)
505  \label{IKM iteration 3}  \label{IKM iteration 6}
506
507    where $\eta\hackscore{eff}'$ denotes the derivative of $\eta\hackscore{eff}$ with respect of $\tau$. The second term in $\min$ is droped of $\tau\hackscore{Y} + \beta \; p<0$ (?)). We have
508
509    \eta\hackscore{eff}' = - \eta\hackscore{eff}^2 \left(\frac{1}{\eta\hackscore{eff}}\right)'
510    \mbox{ with }
511    \left(\frac{1}{\eta\hackscore{eff}}\right)' = \sum\hackscore{q} \left(\frac{1}{\eta^{q}} \right)'
512    \label{IKM iteration 7}
513
514    \label{IKM-EQU-5XX}
515    \left(\frac{1}{\eta^{q}} \right)'
516    = \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}}}}
517    = \frac{1-\frac{1}{n^{q}}}{\tau}\frac{1}{\eta^{q}}
518
519    Notice that allways $\eta\hackscore{eff}'\le 0$ which makes the denomionator in~\ref{IKM iteration 6}
520    positive.
521
522
523

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