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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
42    \index{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]
56  \label{SADDLEPOINT}  \label{SADDLEPOINT}
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]
\label{SADDLEPOINT PRECODITIONER}
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}
\label{SADDLEPOINT iteration}
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
79  \label{SADDLEPOINT iteration 2}
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
\label{SADDLEPOINT iteration 3}
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
\label{SADDLEPOINT iteration 4}
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  \begin{classdesc}{StokesProblemCartesian}{domain,debug}  \subsection{Functions}
opens the stokes equations on the \Domain domain. Setting debug=True switches the debug mode to on.
\end{classdesc}
130
131  example usage:  \begin{classdesc}{StokesProblemCartesian}{domain}
132    opens the Stokes problem\index{Stokes problem} on the \Domain domain. The approximation
133    order needs to be two.
134    \end{classdesc}
135
136  solution=StokesProblemCartesian(mesh) \\  \begin{methoddesc}[StokesProblemCartesian]{initialize}{\optional{f=Data(), \optional{fixed_u_mask=Data(), \optional{eta=1, \optional{surface_stress=Data(), \optional{stress=Data()}}}}}}
137  solution.setTolerance(TOL) \\  assigns values to the model parameters. In any call all values must be set.
138  solution.initialize(fixed\_u\_mask=b\_c,eta=eta,f=Y) \\  \var{f} defines the external force $f$, \var{eta} the viscosity $\eta$,
139  velocity,pressure=solution.solve(velocity,pressure,max\_iter=max\_iter,solver=solver) \\  \var{surface_stress} the surface stress $s$ and \var{stress} the initial stress $\sigma$.
140    The locations and compontents where the velocity is fixed are set by
141    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
149    by \var{fixed_u_mask} remain unchanged.
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)
195    mask= (whereZero(x[0])*[1.,0]+whereZero(x[0]-1))*[1.,0] + \
196          (whereZero(x[1])*[0.,1.]+whereZero(x[1]-1))*[1.,1]
197    sc.initialize(eta=.1, fixed_u_mask= mask)
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.

\subsection{Description}
227
228  \subsection{Method}  \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
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    \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  \begin{classdesc}{TemperatureCartesian}{dom,theta=THETA,useSUPG=SUPG}  The approach we are taking is to eliminate the velocity $u$ from the problem. Assuming that $p$ is known we have
288
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( - 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 the pair $(Qp,v)$ 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    $(0,(I+D^*D)^{-1} (D^*f + g))$. The evaluation of the iteration operator for a given $p$ is then
310    returning $(Qp,w)$ where $w$ is the solution of
311    \label{UPDATE W}
312    (I+D^*D)w = Qp
313
314    We use $Q^*Q$ as a a preconditioner for the iteration operator $Q^* ( I - (I+D^*D)^{-1} ) Q$.
315
316    The iteration PCG \index{linear solver!PCG}\index{PCG} is terminated if
317    \label{DARCY STOP}
318    \int r \cdot (Q^*Q)^{-1} r \; dx \le \mbox{ATOL}^2
319
320    where ATOL is a given absolute tolerance.
321    The initial residual $r\hackscore{0}$ is
322    \label{DARCY STOP 2}
323    r\hackscore{0}=Q^* \left( g - v\hackscore{ref} \right) \mbox{ with } v\hackscore{ref} = (I+D^*D)^{-1} (D^*f + g)
324
325    so the
326    \label{DARCY NORM 0}
327    \int r\hackscore0 \cdot (Q^*Q)^{-1} r\hackscore0 \; dx = \int \left( g - v\hackscore{ref} \right)  \cdot  Q  p\hackscore{ref} \; dx \mbox{ with }p\hackscore{ref} = (Q^*Q)^{-1} Q^* \left( g - v\hackscore{ref} \right)
328
329    So we set
330    \label{DARCY NORM 1}
331    ATOL = atol + rtol \cdot \max(|g - v\hackscore{ref}|\hackscore{0}, |Q p\hackscore{ref} |\hackscore{0} )
332
333    where atol and rtol a given absolute and relative tolerances, respectively. The reference flux $v\hackscore{ref}$
334    and reference pressure $p\hackscore{ref}$ may be calcualated from their definition which would require to solve to
335    PDEs but in a practical application estimates can be used for instance solutions from previous time steps or for simplified scenarious (e.g. constant permability).
336
337    \subsection{Functions}
338    \begin{classdesc}{DarcyFlow}{domain}
339    opens the Darcy flux problem\index{Darcy flux} on the \Domain domain.
340  \end{classdesc}  \end{classdesc}
341    \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}}}}}}
342    assigns values to the model parameters. Values can be assigned using various calls - in particular
343    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).
344    \var{f} and \var{g} are the corresponding parameters in~\ref{DARCY PROBLEM}.
345    The locations and compontents where the flux is prescribed are set by positive values in
346    \var{location_of_fixed_flux}.
347    The locations where the pressure is prescribed are set by
348    by positive values of \var{location_of_fixed_pressure}.
349    The values of the pressure and flux are defined by the initial guess.
350    Notice that at any point on the boundary of the domain the pressure or the normal component of
351    the flux must be defined. There must be at least one point where the pressure is prescribed.
352    The method will try to cast the given values to appropriate
353    \Data class objects.
354    \end{methoddesc}
355
356  \subsection{Benchmark Problem}  \begin{methoddesc}[DarcyFlow]{setTolerance}{\optional{atol=0,\optional{rtol=1e-8,\optional{p_ref=None,\optional{v_ref=None}}}}}
357    sets the absolute tolerance ATOL according to~\ref{DARCY NORM 1}. If \var{p_ref} is not present $0$ is used.
358    If \var{v_ref} is not present $0$ is used. If the final result ATOL is not positive an exception is thrown.
359    \end{methoddesc}
360
\section{Level Set Method}
361
\subsection{Description}
362
363  \subsection{Method}  \begin{methoddesc}[DarcyFlow]{solve}{u0,p0, \optional{max_iter=100, \optional{verbose=False \optional{sub_rtol=1.e-8}}}}
364    solves the problem. and returns approximations for the flux $v$ and the pressure $p$.
365    \var{u0} and \var{p0} define initial guess for flux and pressure. Values marked
366    by positive values \var{location_of_fixed_flux} and \var{location_of_fixed_pressure}, respectively, are kept unchanged.
367    \var{sub_rtol} defines the tolerance used to solve the involved PDEs. \var{sub_rtol} needs to be choosen sufficiently small to ensure convergence but users need to keep in mind that a very small value for \var{sub_rtol} will result in a long compute time. Typically  $\var{sub_rtol}=\var{rtol}^2$ is a good choice if $\var{rtol}$ is not choosen too small.
368    \end{methoddesc}
369
Advection and Reinitialisation
370
371  \begin{classdesc}{LevelSet}{mesh, func\_new, reinit\_max, reinit\_each, tolerance, smooth}  \subsection{Example: Gravity Flow}
372  \end{classdesc}  later
373
374  %example usage:  %================================================
375    % \section{Temperature Advection Diffusion\label{TEMP ADV DIFF}}
376
377  %levelset = LevelSet(mesh, func\_new, reinit\_max, reinit\_each, tolerance, smooth)  %
378    % \rho c\hackscore{p} \left (\frac{\partial T}{\partial t} + \vec{v} \cdot \nabla T \right ) = k \nabla^{2}T
379    % \label{HEAT EQUATION}
380    %
381
382  \begin{methoddesc}[LevelSet]{update\_parameter}{parameter}  % 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.
Update the parameter.
\end{methoddesc}
383
384  \begin{methoddesc}[LevelSet]{update\_phi}{paramter}{velocity}{dt}{t\_step}  % \subsection{Description}
Update level set function; advection and reinitialization
\end{methoddesc}
385
386  \subsection{Benchmark Problem}  % \subsection{Method}
387    %
388    % \begin{classdesc}{TemperatureCartesian}{dom,theta=THETA,useSUPG=SUPG}
389    % \end{classdesc}
390
391  Rayleigh-Taylor instability problem  % \subsection{Benchmark Problem}
392    %===============================================================================================================
393
394    %=========================================================
395    \input{levelsetmodel}
396
397  % \section{Drucker Prager Model}  % \section{Drucker Prager Model}
398
# Line 170  D\hackscore{ij}=D\hackscore{ij}^{el}+D\h Line 404  D\hackscore{ij}=D\hackscore{ij}^{el}+D\h
404
405  with the elastic strain given as  with the elastic strain given as
406  \label{IKM-EQU-3}  \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
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  \label{IKM-EQU-4}  \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
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  \label{IKM-EQU-5}  \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
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 $q$
# Line 187  betwee the the strain in material $q$ Line 421  betwee the the strain in material $q$
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)^{\frac{1}{n^{q}}-1}
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
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~\ref{POERLAW}.
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  \label{IKM-EQU-6}  \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}} \;.
429
430  With  With
431  \label{IKM-EQU-8}  \label{IKM-EQU-8}
# Line 239  D\hackscore{ij}'^{el}=\frac{1}{2 \mu dt Line 473  D\hackscore{ij}'^{el}=\frac{1}{2 \mu dt
473  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.
474  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
475  \label{IKM-EQU-10}  \label{IKM-EQU-10}
476  \sigma\hackscore{ij}' =  2 \eta\hackscore{eff} D\hackscore{ij}' +  \sigma\hackscore{ij}' =  2 \eta\hackscore{eff}  \left( D\hackscore{ij}' +
477  \frac{\eta\hackscore{eff}}{\mu \; dt}  \frac{1}{  2 \mu \; dt} \sigma\hackscore{ij}^{'-}\right)  \mbox{ with }
\sigma\hackscore{ij}^{'-} \mbox{ with }
478  \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}}
479
480  Notice that $\eta\hackscore{eff}$ is a function of stress.  Notice that $\eta\hackscore{eff}$ is a function of diatoric stress $\sigma\hackscore{ij}'$.
481  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
482  \label{IKM-EQU-1ib}  \label{IKM-EQU-1ib}
483  -\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})
484    \right)\hackscore{,j}+p\hackscore{,i}=F\hackscore{i}+
485  \frac{\eta\hackscore{eff}}{\mu dt } \sigma\hackscore{ij,j}^{'-}  \frac{\eta\hackscore{eff}}{\mu dt } \sigma\hackscore{ij,j}^{'-}
486
487  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
488  to the Stokes problem discussed in section~\label{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.
489    In oder to perform step~\ref{IKM iteration 2} we need to calculate the $\eta\hackscore{eff}$ which
490    is a function of $\sigma\hackscore{ij}$ via $\tau$.  To get $\tau$ and $\eta\hackscore{eff}$ we need to solve the
491    non-linear equation
492
493  \begin{array}{rcl}  \tau = \eta\hackscore{eff} \cdot \dot{\gamma}\hackscore{total} \mbox{ with }
494  -\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}' +
495  \frac{1}{\eta\hackscore{eff}} dp & = & - v\hackscore{i,i}^+ \\  \frac{1}{  2 \mu \; dt} \sigma\hackscore{ij}^{'-}\right)^2}
496  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}
497  \end{array}
498  \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
499
500  \|(\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)
501  \label{IKM iteration 3}  \label{IKM iteration 6}
502
503    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
504
505    \eta\hackscore{eff}' = - \eta\hackscore{eff}^2 \left(\frac{1}{\eta\hackscore{eff}}\right)'
506    \mbox{ with }
507    \left(\frac{1}{\eta\hackscore{eff}}\right)' = \sum\hackscore{q} \left(\frac{1}{\eta^{q}} \right)'
508    \label{IKM iteration 7}
509
510    \label{IKM-EQU-5XX}
511    \left(\frac{1}{\eta^{q}} \right)'
512    = \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}}}}
513    = \frac{1-\frac{1}{n^{q}}}{\tau}\frac{1}{\eta^{q}}
514
515    Notice that allways $\eta\hackscore{eff}'\le 0$ which makes the denomionator in~\ref{IKM iteration 6}
516    positive.
517
518
519

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