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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}
62    \|v\hackscore{k,k}\| \hackscore \le  \epsilon
63    \|\sqrt{v\hackscore{j,k}v\hackscore{j,k}}\|
64    \end{equation}
65    where $\epsilon$ is the desired relative accuracy and
66    \begin{equation}
67    \|p\|^2= \int\hackscore{\Omega} p^2 \; dx
68    \label{PRESSURE NORM}
69    \end{equation}
70    defines the $L^2$-norm.
71    There are two approaches to solve this problem. The first approach, called the Uzawa scheme \index{Uzawa scheme}
72    eliminates the velocity $v$ from the problem. The second approach solves the equation in coupled form after the application of a preconditioner.
73    
74    \subsubsection{Uzawa scheme}
75    The first eqution in~\ref{SADDLEPOINT} gives $v=A^{-1}(G-B^{*}p)$ assuming $p$ is known. This is inserted into the
76    second eqution which leads to
77    \begin{equation}
78    S p =  B A^{-1} G
79    \end{equation}
80    with the Schur complement \index{Schur complement} $S=BA^{-1}B^{*}$. This problem can be solved iteratively using the reconditioned Conjugate Gradient Method (PCG)~\index{PCG!Preconditioned Conjugate Gradient Method}
81    with the preconditioner $\hat{S}$ defined as $q=\hat{S}^{-1}p$ by solving
82    \begin{equation}
83    \frac{1}{\eta}q = p
84    \end{equation}
85    see~\cite{ELMAN} for more details. The evaluation of $w=Sp$ is done in the form
86    \begin{equation}
87    \begin{array}{rcl}
88    A v & = & B^{*}p \\
89    w & = & Bv \\
90    \end{array}
91    \label{EVAL PCG}
92    \end{equation}
93    The residual \index{residual}  $r=B A^{-1} G - S p$ is given as
94    \begin{equation}
95    r=B A^{-1} (G - B^* p) = Bv \mbox{ with } v = A^{-1}(G-B^{*}p)
96    \end{equation}
97    Therefore one uses the tuple $(v,Bv)$ to represent the residual of the current pressure $p$. Notice that before the iteration is started the right hand side $B A^{-1} G$ needs to be calculated. The bilinear form $(.,.)$ used is defined as
98    \begin{equation}
99    (p,(v,Bv))=\int\hackscore{\Omega} p \cdot Bv \; dx
100    \end{equation}
101    where $p$ is the pressure increment and $(v,Bv)$ represents an increment in the residual.
102    
103    \subsubsection{Coupled Solver}
104    An alternative approach to solve the saddle point problem~\ref{SADDLEPOINT} directly using an iterative such as
105    the generalized minimal residual method (GMRES) \index{generalized minimal residual method!GMRES} with a suitable
106    preconditioner. Here we use the operator
107  \begin{equation}  \begin{equation}
108  \left[ \begin{array}{cc}  \left[ \begin{array}{cc}
109  A^{-1}     & 0 \\  A^{-1}     & 0 \\
# Line 63  S^{-1} B A^{-1}  & -S^{-1} \\ Line 111  S^{-1} B A^{-1}  & -S^{-1} \\
111  \end{array} \right]  \end{array} \right]
112  \label{SADDLEPOINT PRECODITIONER}  \label{SADDLEPOINT PRECODITIONER}
113  \end{equation}  \end{equation}
114  with the Schur complement $S=BA^{-1}B^{*}$ to solve the problem iteratively. The updates $dv$ and $dp$ for  where again $S$ is the Schur complement~\cite{ELMAN}. In partice we will use an approximation $\hat{S}$ for $S$. The evaluation $(w,q)$ of the iteration operator for a given $(v,p)$ is done as
 given velocity $v$ and pressure $p$ is given as  
115  \begin{equation}  \begin{equation}
116  \begin{array}{rcl}  \begin{array}{rcl}
117  A dv & = & F-Av-B^{*}p \\  A w & = & Av+B^{*}p \\
118  S dp & = & B(v+dv) \\  \hat{S} q & = & B(w-v) \\
119  \end{array}  \end{array}
120  \label{SADDLEPOINT iteration}  \label{COUPLES SADDLEPOINT iteration}
121  \end{equation}  \end{equation}
122  This scheme is called the Uzawa scheme.  We use the inner product induced by the norm
123    \begin{equation}
124  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  \|(v,p)\|^2= \int\hackscore{\Omega}  v\hackscore{i,j}  v\hackscore{i,j} + \left( \frac{p}{\eta}\right)^2\; dx
125    \label{COUPLES NORM}
126    \end{equation}
127    In PDE form~\ref{COUPLES SADDLEPOINT iteration} takes the form
128  \begin{equation}  \begin{equation}
129  \begin{array}{rcl}  \begin{array}{rcl}
130  -\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} \\  -\left(\eta(w\hackscore{i,j}+ w\hackscore{i,j})\right)\hackscore{,j} & = & -\left(\eta(v\hackscore{i,j}+ v\hackscore{i,j})\right)\hackscore{,j}+p\hackscore{,i} \\
131  \frac{1}{\eta} dp & = & - \left(v\hackscore{,i}+dv\hackscore{,i} \right)\hackscore{,i} \\  \frac{1}{\eta}  q & = & - (w-v)\hackscore{i,i} \\
132  \end{array}  \end{array}
133  \label{SADDLEPOINT iteration 2}  \label{SADDLEPOINT iteration 2}
134  \end{equation}  \end{equation}
 To accelerate the convergence we are using the restarted $GMRES$ method using the norm  
 \begin{equation}  
 \|(v,p)\|^2= \int\hackscore{\Omega} \eta^2 v\hackscore{i,j}^2 + p^2 \; dx  
 \label{SADDLEPOINT iteration 3}  
 \end{equation}  
 or alternatively the $PCG$ method on the pressure only using the norm  
 \begin{equation}  
 \|p\|^2= \int\hackscore{\Omega} \frac{1}{\eta^2} p^2 \; dx  
 \label{SADDLEPOINT iteration 4}  
 \end{equation}  
135    
136    
137    \subsection{Functions}
138    
139    \begin{classdesc}{StokesProblemCartesian}{domain}
140    opens the Stokes problem\index{Stokes problem} on the \Domain domain. The approximation
141  \begin{classdesc}{StokesProblemCartesian}{domain,debug}  order needs to be two.
 opens the stokes equations on the \Domain domain. Setting debug=True switches the debug mode to on.  
142  \end{classdesc}  \end{classdesc}
143    
144  example usage:  \begin{methoddesc}[StokesProblemCartesian]{initialize}{\optional{f=Data(), \optional{fixed_u_mask=Data(), \optional{eta=1, \optional{surface_stress=Data(), \optional{stress=Data()}}}}}}
145    assigns values to the model parameters. In any call all values must be set.
146  solution=StokesProblemCartesian(mesh) \\  \var{f} defines the external force $f$, \var{eta} the viscosity $\eta$,
147  solution.setTolerance(TOL) \\  \var{surface_stress} the surface stress $s$ and \var{stress} the initial stress $\sigma$.
148  solution.initialize(fixed\_u\_mask=b\_c,eta=eta,f=Y) \\  The locations and compontents where the velocity is fixed are set by
149  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
150    \Data class objects.
151    \end{methoddesc}
152    
153  % \subsection{Benchmark Problem}  \begin{methoddesc}[StokesProblemCartesian]{solve}{v,p,
154  %  \optional{max_iter=20, \optional{verbose=False, \optional{useUzawa=True}}}}
155  % Convection problem  solves the problem and return approximations for velocity and pressure.
156    The arguments \var{v} and \var{p} define initial guess. The values of \var{v} marked
157    by \var{fixed_u_mask} remain unchanged.
158    If \var{useUzawa} is set to \True
159    the Uzawa\index{Uszwa} scheme is used. Otherwise the problem is solved in coupled form. In most cases
160    the Uzawa scheme is more efficient.
161    \var{max_iter} defines the maximum number of iteration steps.
162    If \var{verbose} is set to \True informations on the progress of of the solver are printed.
163    \end{methoddesc}
164    
165    
166  \section{Temperature Cartesian}  \begin{methoddesc}[StokesProblemCartesian]{setTolerance}{\optional{tolerance=1.e-8}}
167    sets the tolerance in an appropriate norm relative to the right hand side. The tolerance must be non-negative and less than 1.
168    \end{methoddesc}
169    \begin{methoddesc}[StokesProblemCartesian]{getTolerance}{}
170    returns the current relative tolerance.
171    \end{methoddesc}
172    \begin{methoddesc}[StokesProblemCartesian]{setAbsoluteTolerance}{\optional{tolerance=0.}}
173    sets the absolute tolerance for the error in the relevant norm. The tolerance must be non-negative. Typically the
174    absolute talerance is set to 0.
175    \end{methoddesc}
176    \begin{methoddesc}[StokesProblemCartesian]{getAbsoluteTolerance}{}
177    sreturns the current absolute tolerance.
178    \end{methoddesc}
179    \begin{methoddesc}[StokesProblemCartesian]{setSubToleranceReductionFactor}{\optional{reduction=None}}
180    sets the reduction factor for the tolerance used to solve the PDEs. A reduction factor
181    in the order of one will minimize compute time per iteration step but my slow down convergence or even lead to
182    divergency. On the other hand a very small value for the PDE tolerance could result in a wast of compute time.
183    If \var{reduction} is set to \var{None} the sub-tolerance is solved adaptively but
184    in cases a very small tolerance is set ($<10^{-6}$) it is recommended to set the
185    reduction factor by hand. This may require some experiments.
186    \end{methoddesc}
187    \begin{methoddesc}[StokesProblemCartesian]{getSubToleranceReductionFactor}{}
188    return the current reduction factor for the sub-problem tolerance.
189    \end{methoddesc}
190    
191    \subsection{Example: Lit Driven Cavity}
192     The following script \file{lit\hackscore driven\hackscore cavity.py}
193    \index{scripts!\file{helmholtz.py}} which is available in the \ExampleDirectory
194    illustrates the usage of the \class{StokesProblemCartesian} class to solve
195    the lit driven cavity problem~\cite{LITDRIVENCAVITY}:
196    \begin{python}
197    from esys.escript import *
198    from esys.finley import Rectangle
199    from esys.escript.models import StokesProblemCartesian
200    NE=25
201    dom = Rectangle(NE,NE,order=2)
202    x = dom.getX()
203    sc=StokesProblemCartesian(dom)
204    mask= (whereZero(x[0])*[1.,0]+whereZero(x[0]-1))*[1.,0] + \
205          (whereZero(x[1])*[0.,1.]+whereZero(x[1]-1))*[1.,1]
206    sc.initialize(eta=.1, fixed_u_mask= mask)
207    v=Vector(0.,Solution(dom))
208    v[0]+=whereZero(x[1]-1.)
209    p=Scalar(0.,ReducedSolution(dom))
210    v,p=sc.solve(v,p, verbose=True)
211    saveVTK("u.xml",velocity=v,pressure=p)
212    \end{python}
213    
214    \section{Darcy Flux}
215    We want to calculate the velocity $u$ and pressure $p$ on a domain $\Omega$ solving
216    the Darcy flux problem \index{Darcy flux}\index{Darcy flow}
217    \begin{equation}\label{DARCY PROBLEM}
218    \begin{array}{rcl}
219    u\hackscore{i} + \kappa\hackscore{ij} p\hackscore{,j} & = & g\hackscore{i} \\
220    u\hackscore{k,k} & = & f
221    \end{array}
222    \end{equation}
223    with the boundary conditions
224    \begin{equation}\label{DARCY BOUNDARY}
225    \begin{array}{rcl}
226    u\hackscore{i} \; n\hackscore{i}  = u^{N}\hackscore{i}  \; n\hackscore{i} & \mbox{ on } & \Gamma\hackscore{N} \\
227    p = p^{D} &  \mbox{ on } & \Gamma\hackscore{D} \\
228    \end{array}
229    \end{equation}
230    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
231  \begin{equation}  \begin{equation}
232  \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}  
233  \end{equation}  \end{equation}
234    for all $x\hackscore{i}$.
235    
 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}  
236    
237  \subsection{Method}  \subsection{Solution Method \label{DARCY SOLVE}}
238    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
239    \begin{equation}
240    -(\kappa\hackscore{ki}\kappa\hackscore{kj} p\hackscore{,j}^{ref})\hackscore{,i} =  - (\kappa\hackscore{ki} (g\hackscore{k}- u^{N}\hackscore{k}))\hackscore{,i}
241    \mbox{ with }
242    p^{ref} = p^{D} \mbox{ on } \Gamma\hackscore{D}
243    \end{equation}
244    With setting $u \leftarrow u-u^{N}$ and $p \leftarrow p-p^{ref}$ and
245    \begin{equation}
246    \begin{array}{rcl}
247    g\hackscore{i} & \leftarrow & g\hackscore{i} - u^{N}\hackscore{i} -  \kappa\hackscore{ij} p^{ref}\hackscore{,j }\\
248    f & \leftarrow & f - u^{N}\hackscore{k,k}
249    \end{array}
250    \end{equation}
251    we can assume that $u^{N}\hackscore{i}  \; n\hackscore{i}=0$ and
252    $p^{D}=0$.
253    We set
254    \begin{equation}
255    V = \{ q \in H^{1}(\Omega) : q=0 \mbox{ on } \Gamma\hackscore{D} \}
256    \end{equation}
257    and
258    \begin{equation}
259    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} \}
260    \end{equation}
261    and define the operator $Q: V \rightarrow (L^2(\Omega))^{d}$ defined by
262    \begin{equation}
263    (Qp)\hackscore{i} = \kappa\hackscore{ij} p\hackscore{,j}
264    \end{equation}
265    and the operator $D: W \rightarrow L^2(\Omega)$ defined by
266    \begin{equation}
267    Dv = v\hackscore{k,k}
268    \end{equation}
269    In operator notation the Darcy problem~\ref{DARCY PROBLEM} is written in the form
270    \begin{equation}
271    \begin{array}{rcl}
272    u + Qp & = & g \\
273    Du & = & f
274    \end{array}
275    \end{equation}
276    We solve this equation by minimising the functional
277    \begin{equation}
278    J(u,p):=\|u + Qp - g\|^2\hackscore{0} + \|Du-f\|\hackscore{0}^2
279    \end{equation}
280    over $W \times V$ where $\|.\|\hackscore{0}$ denotes the norm in $L^2(\Omega)$. A simple calculation shows that
281    one has to solve
282    \begin{equation}
283    ( v + Qq , u + Qp - g) + (Dv,Du-f) =0
284    \end{equation}
285    for all $v\in W$ and $q \in V$.which translates back into operator notation
286    \begin{equation}
287    \begin{array}{rcl}
288    (I+D^*D)u + Qp & = & D^*f + g \\
289    Q^*u  + Q^*Q p & = & Q^*g \\
290    \end{array}
291    \end{equation}
292    where $D^*$ and $Q^*$ denote the adjoint operators.
293    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
294    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 $\alpha\hackscore{0} \gg 1$)  
295    
296  \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
297    \begin{equation}
298    v= (I+D^*D)^{-1} (D^*f + g - Qp)
299    \end{equation}
300    (notice that $(I+D^*D)$ is coercive in $W$) which is inserted into the second equation
301    \begin{equation}
302    Q^* (I+D^*D)^{-1} (D^*f + g - Qp) + Q^* Q p = Q^*g
303    \end{equation}
304    which is
305    \begin{equation}
306    Q^* ( I - (I+D^*D)^{-1} ) Q p = Q^* (g-(I+D^*D)^{-1} (D^*f + g) )
307    \end{equation}
308    We use the PCG \index{linear solver!PCG}\index{PCG} method to solve this. The residual $r$ ($\in V^*$) is given as
309    \begin{equation}
310    \begin{array}{rcl}
311    r & = & Q^*  \left( g -(I+D^*D)^{-1} (D^*f + g) - Qp + (I+D^*D)^{-1}Q p \right)\\
312    & =&  Q^* \left( - Qp - (I+D^*D)^{-1} (D^*f + g - Qp) \right) \\
313    & =&  Q^* \left( g - Qp - v \right)
314    \end{array}
315    \end{equation}
316    So in a partical implementation we use the pair $(Qp,v)$ to represent the residual. This will save the
317    reconstruction of the velocity $v$. In this notation the right hand side is given as
318    $(0,(I+D^*D)^{-1} (D^*f + g))$. The evaluation of the iteration operator for a given $p$ is then
319    returning $(Qp,w)$ where $w$ is the solution of
320    \begin{equation}\label{UPDATE W}
321    (I+D^*D)w = Qp
322    \end{equation}
323    We use $Q^*Q$ as a a preconditioner for the iteration operator $Q^* ( I - (I+D^*D)^{-1} ) Q$.
324    
325    The iteration PCG \index{linear solver!PCG}\index{PCG} is terminated if
326    \begin{equation}\label{DARCY STOP}
327    \int r \cdot (Q^*Q)^{-1} r \; dx \le \mbox{ATOL}^2
328    \end{equation}
329    where ATOL is a given absolute tolerance.
330    The initial residual $r\hackscore{0}$ is
331    \begin{equation}\label{DARCY STOP 2}
332    r\hackscore{0}=Q^* \left( g - v\hackscore{ref} \right) \mbox{ with } v\hackscore{ref} = (I+D^*D)^{-1} (D^*f + g)
333    \end{equation}
334    so the
335    \begin{equation}\label{DARCY NORM 0}
336    \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)
337    \end{equation}
338    So we set
339    \begin{equation}\label{DARCY NORM 1}
340    ATOL = atol + rtol \cdot \max(|g - v\hackscore{ref}|\hackscore{0}, |Q p\hackscore{ref} |\hackscore{0} )
341    \end{equation}
342    where atol and rtol a given absolute and relative tolerances, respectively. The reference flux $v\hackscore{ref}$
343    and reference pressure $p\hackscore{ref}$ may be calcualated from their definition which would require to solve to
344    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).
345    
346    \subsection{Functions}
347    \begin{classdesc}{DarcyFlow}{domain}
348    opens the Darcy flux problem\index{Darcy flux} on the \Domain domain.
349  \end{classdesc}  \end{classdesc}
350    \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}}}}}}
351    assigns values to the model parameters. Values can be assigned using various calls - in particular
352    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).
353    \var{f} and \var{g} are the corresponding parameters in~\ref{DARCY PROBLEM}.
354    The locations and compontents where the flux is prescribed are set by positive values in
355    \var{location_of_fixed_flux}.
356    The locations where the pressure is prescribed are set by
357    by positive values of \var{location_of_fixed_pressure}.
358    The values of the pressure and flux are defined by the initial guess.
359    Notice that at any point on the boundary of the domain the pressure or the normal component of
360    the flux must be defined. There must be at least one point where the pressure is prescribed.
361    The method will try to cast the given values to appropriate
362    \Data class objects.
363    \end{methoddesc}
364    
365  \subsection{Benchmark Problem}  \begin{methoddesc}[DarcyFlow]{setTolerance}{\optional{atol=0,\optional{rtol=1e-8,\optional{p_ref=None,\optional{v_ref=None}}}}}
366    sets the absolute tolerance ATOL according to~\ref{DARCY NORM 1}. If \var{p_ref} is not present $0$ is used.
367    If \var{v_ref} is not present $0$ is used. If the final result ATOL is not positive an exception is thrown.
368    \end{methoddesc}
369    
370    
 \section{Level Set Method}  
371    
372  \subsection{Description}  \begin{methoddesc}[DarcyFlow]{solve}{u0,p0, \optional{max_iter=100, \optional{verbose=False \optional{sub_rtol=1.e-8}}}}
373    solves the problem. and returns approximations for the flux $v$ and the pressure $p$.
374    \var{u0} and \var{p0} define initial guess for flux and pressure. Values marked
375    by positive values \var{location_of_fixed_flux} and \var{location_of_fixed_pressure}, respectively, are kept unchanged.
376    \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.
377    \end{methoddesc}
378    
 \subsection{Method}  
379    
380  Advection and Reinitialisation  \subsection{Example: Gravity Flow}
381    later
382    
383  \begin{classdesc}{LevelSet}{mesh, func\_new, reinit\_max, reinit\_each, tolerance, smooth}  %================================================
384  \end{classdesc}  % \section{Temperature Advection Diffusion\label{TEMP ADV DIFF}}
385    
386  %example usage:  %\begin{equation}
387    % \rho c\hackscore{p} \left (\frac{\partial T}{\partial t} + \vec{v} \cdot \nabla T \right ) = k \nabla^{2}T
388    % \label{HEAT EQUATION}
389    % \end{equation}
390    
391  %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.
392    
393  \begin{methoddesc}[LevelSet]{update\_parameter}{parameter}  % \subsection{Description}
 Update the parameter.  
 \end{methoddesc}  
394    
395  \begin{methoddesc}[LevelSet]{update\_phi}{paramter}{velocity}{dt}{t\_step}  % \subsection{Method}
396  Update level set function; advection and reinitialization  %
397  \end{methoddesc}  % \begin{classdesc}{TemperatureCartesian}{dom,theta=THETA,useSUPG=SUPG}
398    % \end{classdesc}
 \subsection{Benchmark Problem}  
399    
400  Rayleigh-Taylor instability problem  % \subsection{Benchmark Problem}
401    %===============================================================================================================
402    
403    %=========================================================
404    \input{levelsetmodel}
405    
406  % \section{Drucker Prager Model}  % \section{Drucker Prager Model}
407    
# Line 191  for a given power law coefficients $n^{q Line 434  for a given power law coefficients $n^{q
434  Notice that $n^{q}=1$ gives a constant viscosity.  Notice that $n^{q}=1$ gives a constant viscosity.
435  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:
436  \begin{equation}\label{IKM-EQU-6}  \begin{equation}\label{IKM-EQU-6}
437  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}} \;.
438  \end{equation}  \end{equation}
439  With  With
440  \begin{equation}\label{IKM-EQU-8}  \begin{equation}\label{IKM-EQU-8}
# Line 239  D\hackscore{ij}'^{el}=\frac{1}{2 \mu dt Line 482  D\hackscore{ij}'^{el}=\frac{1}{2 \mu dt
482  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.
483  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
484  \begin{equation}\label{IKM-EQU-10}  \begin{equation}\label{IKM-EQU-10}
485  \sigma\hackscore{ij}' =  2 \eta\hackscore{eff} D\hackscore{ij}' +  \sigma\hackscore{ij}' =  2 \eta\hackscore{eff}  \left( D\hackscore{ij}' +
486  \frac{\eta\hackscore{eff}}{\mu \; dt}  \frac{1}{  2 \mu \; dt} \sigma\hackscore{ij}^{'-}\right)  \mbox{ with }
 \sigma\hackscore{ij}^{'-} \mbox{ with }  
487  \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}}
488  \end{equation}  \end{equation}
489  Notice that $\eta\hackscore{eff}$ is a function of stress.  Notice that $\eta\hackscore{eff}$ is a function of diatoric stress $\sigma\hackscore{ij}'$.
490  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
491  \begin{equation}\label{IKM-EQU-1ib}  \begin{equation}\label{IKM-EQU-1ib}
492  -\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})
493    \right)\hackscore{,j}+p\hackscore{,i}=F\hackscore{i}+
494  \frac{\eta\hackscore{eff}}{\mu dt } \sigma\hackscore{ij,j}^{'-}  \frac{\eta\hackscore{eff}}{\mu dt } \sigma\hackscore{ij,j}^{'-}
495  \end{equation}  \end{equation}
496  Together with the incomressibilty condition~\ref{IKM-EQU-2} we need to solve a problem with a form almost identical  Together with the incomressibilty condition~\ref{IKM-EQU-2} we need to solve a problem with a form almost identical
497  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  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
498  \begin{equation}  \begin{equation}
499  \begin{array}{rcl}  \begin{array}{rcl}
500  -\left(\eta\hackscore{eff}(dv\hackscore{i,j}+ dv\hackscore{i,j})\right)\hackscore{,j} & = & F\hackscore{i}+ \sigma\hackscore{ij,j}' \\  -\left(\eta\hackscore{eff}(dv\hackscore{i,j}+ dv\hackscore{i,j}
501  \frac{1}{\eta\hackscore{eff}} dp & = & - v\hackscore{i,i}^+ \\  )\right)\hackscore{,j} & = & F\hackscore{i}+ \sigma\hackscore{ij,j}'-p\hackscore{,i} \\
502  d\sigma\hackscore{ij}' & = & 2 \eta\hackscore{eff} D\hackscore{ij}^{+'} + \frac{\eta\hackscore{eff}}{\mu \; dt} \sigma\hackscore{ij}' -  \sigma\hackscore{ij}\\  \frac{1}{\eta\hackscore{eff}} dp & = & - v\hackscore{i,i}^+
503  \end{array}  \end{array}
504  \label{IKM iteration 2}  \label{IKM iteration 2}
505  \end{equation}  \end{equation}
506  where $v^+=v+dv$. As this problem is non-linear the Jacobi-free Newton-GMRES method is used with the norm  where $v^+=v+dv$. As this problem is non-linear the Jacobi-free Newton-GMRES method is used with the norm
507  \begin{equation}  \begin{equation}
508  \|(\sigma, p)\|^2= \int\hackscore{\Omega} \sigma\hackscore{i,j}^2 + p^2 \; dx  \|(v, p)\|^2= \int\hackscore{\Omega} v\hackscore{i,j}^2 + \frac{1}{\bar{\eta}^2\hackscore{eff}} p^2 \; dx
509  \label{IKM iteration 3}  \label{IKM iteration 3}
510  \end{equation}  \end{equation}
511    where  $\bar{\eta}\hackscore{eff}$ is the caracteristic viscosity, for instance:
512    \begin{equation}
513    \frac{1}{\bar{\eta}\hackscore{eff}} = \frac{1}{\tau^{-}}+\sum\hackscore{q}  \frac{1}{\eta^{q}\hackscore{N}}
514    \label{IKM iteration 4}
515    \end{equation}
516    In oder to perform step~\ref{IKM iteration 2} we need to calculate the $\eta\hackscore{eff}$ as well as $\sigma\hackscore{ij}'$ while via $\tau$ the first is a function of the latter. The priority is the
517    calculation of $\eta\hackscore{eff}$ with the Newton-Raphson scheme. This value can then be used to calculate
518    $\sigma\hackscore{ij}'$ via~\ref{IKM-EQU-10}. We need to solve
519    \begin{equation}
520    \tau = \eta\hackscore{eff} \cdot \epsilon \mbox{ with }
521    \epsilon = \sqrt{ 2 \left( D\hackscore{ij}' +
522    \frac{1}{  2 \mu \; dt} \sigma\hackscore{ij}^{'-}\right)^2}
523    \label{IKM iteration 5}
524    \end{equation}
525    The Newton scheme takes the form
526    \begin{equation}
527    \tau\hackscore{n+1} = \min(\tau\hackscore{n} - \frac{\tau\hackscore{n} - \eta\hackscore{eff}  \cdot \epsilon}{1 - \eta\hackscore{eff}'  \cdot  \epsilon}, \tau\hackscore{Y} + \beta \; p)
528    = \min(\frac{\eta\hackscore{eff} - \tau\hackscore{n}  \eta\hackscore{eff}'}
529    {1 - \eta\hackscore{eff}'  \cdot  \epsilon}, \frac{\tau\hackscore{Y} + \beta \; p}{\epsilon}) \epsilon
530    \label{IKM iteration 6}
531    \end{equation}
532    where $\eta\hackscore{eff}'$ denotes the derivative of $\eta\hackscore{eff}$ with respect of $\tau$. The second term in $\min$ is droped of $\tau\hackscore{Y} + \beta \; p<0$ or $\epsilon=0$. In fact we have
533    \begin{equation}
534    \eta\hackscore{eff}' = - \eta\hackscore{eff}^2 \left(\frac{1}{\eta\hackscore{eff}}\right)'
535    \mbox{ with }
536    \left(\frac{1}{\eta\hackscore{eff}}\right)' = \sum\hackscore{q} \left(\frac{1}{\eta^{q}} \right)'
537    \label{IKM iteration 7}
538    \end{equation}
539    \begin{equation}\label{IKM-EQU-5XX}
540    \left(\frac{1}{\eta^{q}} \right)'
541    = \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}}}}
542    = \frac{1-\frac{1}{n^{q}}}{ \tau \eta^{q}}
543    \end{equation}
544    Notice that allways $\eta\hackscore{eff}'\le 0$ which makes the denomionator in~\ref{IKM iteration 6}
545    positive.
546    
547    
548    

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