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13    
14    
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{i}
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
42    \index{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]
56  \label{SADDLEPOINT}  \label{SADDLEPOINT}
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]  
 \label{SADDLEPOINT PRECODITIONER}  
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}  
 \label{SADDLEPOINT iteration}  
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
79  \label{SADDLEPOINT iteration 2}  \begin{equation}
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}  \begin{equation}
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  \end{equation}  \end{equation}
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  \begin{equation}  \begin{equation}
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  \end{equation}  \end{equation}
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    \begin{equation}
96    \hat{S}^{-1} S p =  \hat{S}^{-1} B A^{-1} G
97    \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  \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    \begin{equation}\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    \end{equation}
214    with the boundary conditions
215    \begin{equation}\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    \end{equation}
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  \begin{equation}  \begin{equation}
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  \end{equation}  \end{equation}
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}  
   
 \subsection{Method}  
227    
228  \begin{classdesc}{TemperatureCartesian}{dom,theta=THETA,useSUPG=SUPG}  \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    \begin{equation}
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    \end{equation}
235    With setting $u \leftarrow u-u^{N}$ and $p \leftarrow p-p^{ref}$ and
236    \begin{equation}
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    \end{equation}
242    we can assume that $u^{N}\hackscore{i}  \; n\hackscore{i}=0$ and
243    $p^{D}=0$.
244    We set
245    \begin{equation}
246    V = \{ q \in H^{1}(\Omega) : q=0 \mbox{ on } \Gamma\hackscore{D} \}
247    \end{equation}
248    and
249    \begin{equation}
250    W = \{ v \in (L^2(\Omega))^{d} : v\hackscore{k,k} \in L^2(\Omega) \mbox{ and } u\hackscore{i} \; n\hackscore{i} =0  \mbox{ on } \Gamma\hackscore{N} \}
251    \end{equation}
252    and define the operator $Q: V \rightarrow (L^2(\Omega))^{d}$ defined by
253    \begin{equation}
254    (Qp)\hackscore{i} = \kappa\hackscore{ij} p\hackscore{,j}
255    \end{equation}
256    and the operator $D: W \rightarrow L^2(\Omega)$ defined by
257    \begin{equation}
258    Dv = v\hackscore{k,k}
259    \end{equation}
260    In operator notation the Darcy problem~\ref{DARCY PROBLEM} is written in the form
261    \begin{equation}
262    \begin{array}{rcl}
263    u + Qp & = & g \\
264    Du & = & f
265    \end{array}
266    \end{equation}
267    We solve this equation by minimising the functional
268    \begin{equation}
269    J(u,p):=\|u + Qp - g\|^2\hackscore{0} + \|Du-f\|\hackscore{0}^2
270    \end{equation}
271    over $W \times V$ where $\|.\|\hackscore{0}$ denotes the norm in $L^2(\Omega)$. A simple calculation shows that
272    one has to solve
273    \begin{equation}
274    ( v + Qq , u + Qp - g) + (Dv,Du-f) =0
275    \end{equation}
276    for all $v\in W$ and $q \in V$.which translates back into operator notation
277    \begin{equation}
278    \begin{array}{rcl}
279    (I+D^*D)u + Qp & = & D^*f + g \\
280    Q^*u  + Q^*Q p & = & Q^*g \\
281    \end{array}
282    \end{equation}
283    where $D^*$ and $Q^*$ denote the adjoint operators.
284    In~\cite{XXX} it has been shown that this problem is continuous and coercive in $W \times V$ and therefore has a unique solution. Also standart FEM methods can be used for discretization. It is also possible
285    to solve the problem is coupled form, however this approach leads in some cases to a very ill-conditioned stiffness matrix in particular in the case of a very small or large permability ($\alpha\hackscore{1} \ll 1$ or $\alpha\hackscore{0} \gg 1$)  
286    
287    The approach we are taking is to eliminate the velocity $u$ from the problem. Assuming that $p$ is known we have
288    \begin{equation}\label{DARCY V FORM}
289    v= (I+D^*D)^{-1} (D^*f + g - Qp)
290    \end{equation}
291    (notice that $(I+D^*D)$ is coercive in $W$) which is inserted into the second equation
292    \begin{equation}
293    Q^* (I+D^*D)^{-1} (D^*f + g - Qp) + Q^* Q p = Q^*g
294    \end{equation}
295    which is
296    \begin{equation}
297    Q^* ( I - (I+D^*D)^{-1} ) Q p = Q^* (g-(I+D^*D)^{-1} (D^*f + g) )
298    \end{equation}
299    We use the PCG \index{linear solver!PCG}\index{PCG} method to solve this. The residual $r$ ($\in V^*$) is given as
300    \begin{equation}
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    \end{equation}
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    \begin{equation}\label{UPDATE W}
311    (I+D^*D)v = Qp
312    \end{equation}
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    \begin{equation}\label{UPDATE P}
316    Q^*Q q  = Q^*\hat{r}
317    \end{equation}
318    The residual norm used in the PCG is given as
319    \begin{equation}\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    \end{equation}
323    The iteration is terminated if
324    \begin{equation}\label{DARCY STOP}
325    \|r\|\hackscore{PCG} \le \mbox{ATOL}
326    \end{equation}
327    where we set
328    \begin{equation}\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    \end{equation}
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  \section{Level Set Method}  \begin{methoddesc}[DarcyFlow]{setAbsoluteTolerance}{\optional{atol=0.}}
360    sets the absolute tolerance \mbox{atol} in \ref{DARCY ATOL DEF}.
361    \end{methoddesc}
362    
363  \subsection{Description}  \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{Method}  \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    
 Advection and Reinitialisation  
374    
375  \begin{classdesc}{LevelSet}{mesh, func\_new, reinit\_max, reinit\_each, tolerance, smooth}  \subsection{Example: Gravity Flow}
376  \end{classdesc}  later
377    
378  %example usage:  %================================================
379    % \section{Temperature Advection Diffusion\label{TEMP ADV DIFF}}
380    
381  %levelset = LevelSet(mesh, func\_new, reinit\_max, reinit\_each, tolerance, smooth)  %\begin{equation}
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    % \end{equation}
385    
386  \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}  
387    
388  \begin{methoddesc}[LevelSet]{update\_phi}{paramter}{velocity}{dt}{t\_step}  % \subsection{Description}
 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  \end{equation}  \end{equation}
409  with the elastic strain given as  with the elastic strain given as
410  \begin{equation}\label{IKM-EQU-3}  \begin{equation}\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  \end{equation}  \end{equation}
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  \begin{equation}\label{IKM-EQU-4}  \begin{equation}\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  \end{equation}  \end{equation}
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  \begin{equation}\label{IKM-EQU-5}  \begin{equation}\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  \end{equation}  \end{equation}
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  \end{equation}  \end{equation}
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  \begin{equation}\label{IKM-EQU-6}  \begin{equation}\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}} \;.
 \end{equation}  
 With  
 \begin{equation}\label{IKM-EQU-8}  
 \dot{\gamma}=\sqrt{2 D\hackscore{ij} D\hackscore{ij}}  
433  \end{equation}  \end{equation}
434  one gets  and finally with~\ref{IKM-EQU-2}
435  \begin{equation}\label{IKM-EQU-8b}  \begin{equation}\label{IKM-EQU-2b}
436  \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}'
437  \end{equation}  \end{equation}
438  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}
439  \begin{equation}\label{IKM-EQU-8c}  \begin{equation}\label{IKM-EQU-8c}
440  \tau \le \tau\hackscore{Y} + \beta \; p  \tau \le \tau\hackscore{Y} + \beta \; p
441  \end{equation}  \end{equation}
442  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}  
443  The deviatoric stress needs to fullfill the equilibrion equation  The deviatoric stress needs to fullfill the equilibrion equation
444  \begin{equation}\label{IKM-EQU-1}  \begin{equation}\label{IKM-EQU-1}
445  -\sigma'\hackscore{ij,j}+p\hackscore{,i}=F\hackscore{i}  -\sigma'\hackscore{ij,j}+p\hackscore{,i}=F\hackscore{i}
# Line 234  where the index $i$ may depend on the lo Line 461  where the index $i$ may depend on the lo
461  \subsection{Solution Method \label{IKM-SOLVE}}  \subsection{Solution Method \label{IKM-SOLVE}}
462  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
463  \begin{equation}\label{IKM-EQU-3b}  \begin{equation}\label{IKM-EQU-3b}
464  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)
465  \end{equation}  \end{equation}
466  where $\sigma\hackscore{ij}^{'-}$ is the deviatoric stress at the precious time step.  and
467  Now we can combine equations~\ref{IKM-EQU-2}, \ref{IKM-EQU-3b} and~\ref{IKM-EQU-6b} to get  \begin{equation}\label{IKM-EQU-2b}
468    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}^{-'}
469    \end{equation}
470    where $\sigma\hackscore{ij}^{-}$ is the stress at the precious time step. With
471    \begin{equation}\label{IKM-EQU-2c}
472    \dot{\gamma} = \sqrt{ 2 \left( D\hackscore{ij}' -
473    \frac{1}{  2 \mu \; dt} \sigma\hackscore{ij}^{'-}\right)^2}
474    \end{equation}
475    we have
476    \begin{equation}
477    \tau = \eta\hackscore{eff} \cdot \dot{\gamma}
478    \end{equation}
479    where
480    \begin{equation}
481    \eta\hackscore{eff}= min( \left(\frac{1}{\mu \; dt}+\frac{1}{\eta^{vp}}\right)^{-1}
482    , \eta\hackscore{max}) \mbox{ with }
483    \eta\hackscore{max} = \left\{
484    \begin{array}{rcl}
485    \frac{\tau\hackscore{Y} + \beta \; p}{\dot{\gamma}} & & \dot{\gamma}>0 \\
486    &\mbox{ if } \\
487    \infty & & \mbox{otherwise}
488    \end{array}
489    \right.
490    \end{equation}
491    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
492  \begin{equation}\label{IKM-EQU-10}  \begin{equation}\label{IKM-EQU-10}
493  \sigma\hackscore{ij}' =  2 \eta\hackscore{eff} D\hackscore{ij}' +  \sigma\hackscore{ij}' =  2 \eta\hackscore{eff}  \left( D\hackscore{ij}' -
494  \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}}  
495  \end{equation}  \end{equation}
 Notice that $\eta\hackscore{eff}$ is a function of stress.  
496  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
497  \begin{equation}\label{IKM-EQU-1ib}  \begin{equation}\label{IKM-EQU-1ib}
498  -\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})
499    \right)\hackscore{,j}+p\hackscore{,i}=F\hackscore{i}-
500  \frac{\eta\hackscore{eff}}{\mu dt } \sigma\hackscore{ij,j}^{'-}  \frac{\eta\hackscore{eff}}{\mu dt } \sigma\hackscore{ij,j}^{'-}
501  \end{equation}  \end{equation}
502  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
503  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.
504  \begin{equation}  
505  \begin{array}{rcl}  If we set
506  -\left(\eta\hackscore{eff}(dv\hackscore{i,j}+ dv\hackscore{i,j})\right)\hackscore{,j} & = & F\hackscore{i}+ \sigma\hackscore{ij,j}' \\  \begin{equation}\label{IKM-EQU-44}
507  \frac{1}{\eta\hackscore{eff}} dp & = & - v\hackscore{i,i}^+ \\  \frac{1}{\eta(\tau)}= \frac{1}{\mu \; dt}+\frac{1}{\eta^{vp}}
 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}  
508  \end{equation}  \end{equation}
509  where $v^+=v+dv$. As this problem is non-linear the Jacobi-free Newton-GMRES method is used with the norm  we need to solve the nonlinear problem
510  \begin{equation}  \begin{equation}
511  \|(\sigma, p)\|^2= \int\hackscore{\Omega} \sigma\hackscore{i,j}^2 + p^2 \; dx  \eta\hackscore{eff} -  min(\eta( \dot{\gamma} \cdot \eta\hackscore{eff})
512  \label{IKM iteration 3}  , \eta\hackscore{max}) =0
513  \end{equation}  \end{equation}
514    We use the Newton-Raphson Scheme \index{Newton-Raphson scheme} to solve this problem
515    \begin{equation}\label{IKM-EQU-45}
516    \eta\hackscore{eff}^{(n+1)} = \min(\eta\hackscore{max},
517    \eta\hackscore{eff}^{(n)} -
518    \frac{\eta\hackscore{eff}^{(n)} - \eta( \tau^{(n)}) }
519    {1-\dot{\gamma} \cdot \eta'( \tau^{(n)} )} )
520    =\min(\eta\hackscore{max},
521    \frac{\eta( \tau^{(n)}) -\tau^{(n)} \cdot  \eta'( \tau^{(n)} )  }
522    {1-\dot{\gamma} \cdot \eta'( \tau^{(n)} )} )
523    \end{equation}
524    where $\eta'$ denotes the derivative of $\eta$ with respect of $\tau$
525    and $\tau^{(n)} =  \dot{\gamma} \cdot \eta\hackscore{eff}^{(n)}$.
526    
527    Looking at the evaluation of $\eta$ in~\ref{IKM-EQU-44} it makes sense formulate
528    the iteration~\ref{IKM-EQU-45} using $\Theta=\eta^{-1}$.
529    In fact we have
530    \begin{equation}
531    \eta' = - \frac{\Theta'}{\Theta^2}
532    \mbox{ with }
533    \Theta' = \sum\hackscore{q} \left(\frac{1}{\eta^{q}} \right)'
534    \label{IKM iteration 7}
535    \end{equation}
536    As
537    \begin{equation}\label{IKM-EQU-47}
538    \left(\frac{1}{\eta^{q}} \right)'
539    = \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}}}}
540    = \frac{1-\frac{1}{n^{q}}}{\eta^{q}}\frac{1}{\tau}
541    \end{equation}
542    we have
543    \begin{equation}\label{IKM-EQU-48}
544    \Theta' = \frac{1}{\tau} \omega \mbox{ with } \omega = \sum\hackscore{q}\frac{1-\frac{1}{n^{q}}}{\eta^{q}}
545    \end{equation}
546    which leads to
547    \begin{equation}\label{IKM-EQU-45}
548    \eta\hackscore{eff}^{(n+1)} = \min(\eta\hackscore{max},
549    \eta\hackscore{eff}^{(n)}
550    \frac{\Theta^{(n)}  + \omega^{(n)}  }
551    {\eta\hackscore{eff}^{(n)} \Theta^{(n)^2}+\omega^{(n)} })
552    \end{equation}

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