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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,debug}  \begin{classdesc}{StokesProblemCartesian}{domain}
140  opens the stokes equations on the \Domain domain. Setting debug=True switches the debug mode to on.  opens the Stokes problem\index{Stokes problem} on the \Domain domain. The approximation
141    order needs to be two.
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    \var{f} defines the external force $f$, \var{eta} the viscosity $\eta$,
147    \var{surface_stress} the surface stress $s$ and \var{stress} the initial stress $\sigma$.
148    The locations and compontents where the velocity is fixed are set by
149    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  solution=StokesProblemCartesian(mesh) \\  \begin{methoddesc}[StokesProblemCartesian]{solve}{v,p,
154  solution.setTolerance(TOL) \\  \optional{max_iter=20, \optional{verbose=False, \optional{useUzawa=True}}}}
155  solution.initialize(fixed\_u\_mask=b\_c,eta=eta,f=Y) \\  solves the problem and return approximations for velocity and pressure.
156  velocity,pressure=solution.solve(velocity,pressure,max\_iter=max\_iter,solver=solver) \\  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    
 % \subsection{Benchmark Problem}  
 %  
 % Convection problem  
165    
166    \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}
232    \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}
233    \end{equation}
234    for all $x\hackscore{i}$.
235    
 \section{Temperature Cartesian}  
236    
237    \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}  \begin{equation}
240  \rho c\hackscore{p} \left (\frac{\partial T}{\partial t} + \vec{v} \cdot \nabla T \right ) = k \nabla^{2}T  -(\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  \label{HEAT EQUATION}  \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    $p^{D}=0$. Notice that with this setting
253    \begin{equation}\label{DIV FREE DARCY FLUX}
254    (\kappa\hackscore{ki} g\hackscore{k})\hackscore{,i} = 0
255    \end{equation}
256    We set
257    \begin{equation}
258    V = \{ q \in H^{1}(\Omega) : q=0 \mbox{ on } \Gamma\hackscore{D} \}
259    \end{equation}
260    and
261    \begin{equation}
262    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} \}
263    \end{equation}
264    and define the operator $Q: V \rightarrow (L^2(\Omega))^{d}$ defined by
265    \begin{equation}
266    (Qp)\hackscore{i} = \kappa\hackscore{ij} p\hackscore{,j}
267    \end{equation}
268    and the operator $D: W \rightarrow L^2(\Omega)$ defined by
269    \begin{equation}
270    Dv = v\hackscore{k,k}
271  \end{equation}  \end{equation}
272    In operator notation the Darcy problem~\ref{DARCY PROBLEM} is written in the form
273    \begin{equation}
274    \begin{array}{rcl}
275    u + Qp & = & g \\
276    Du & = & f
277    \end{array}
278    \end{equation}
279    Notice that because of~\ref{DIV FREE DARCY FLUX} we have
280    \begin{equation} \label{ABSTRACT DIV FREE DARCY FLUX}
281    Q^*g =0
282    \end{equation}
283    where $Q^*$ denote the adjoint operators of $Q$.
284    We solve this equation by minimising the functional
285    \begin{equation}
286    J(u,p):=\|u + Qp - g\|^2\hackscore{0} + \|Du-f\|\hackscore{0}^2
287    \end{equation}
288    over $W \times V$ where $\|.\|\hackscore{0}$ denotes the norm in $L^2(\Omega)$. A simple calculation shows that
289    one has to solve
290    \begin{equation}
291    ( v + Qq , u + Qp - g) + (Dv,Du-f) =0
292    \end{equation}
293    for all $v\in W$ and $q \in V$.which translates back into operator notation
294    \begin{equation}
295    \begin{array}{rcl}
296    (I+D^*D)u + Qp & = & D^*f + g \\
297    Q^*u  + Q^*Q p & = & 0 \\
298    \end{array}
299    \end{equation}
300    where $D^*$ and $Q^*$ denote the adjoint operators. We use the fact that $Q^*g=$.
301    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
302    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 $\alpha\hackscore{0} \gg 1$)  
303    
304  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.  The approach we are taking is to eliminate the velocity $u$ from the problem. Assuming that $p$ is known we have
305    \begin{equation}
306    v= (I+D^*D)^{-1} (D^*f + g - Qp)
307    \end{equation}
308    (notice that $(I+D^*D)$ is coercive in $W$) which is inserted into the second equation
309    \begin{equation}
310    Q^* (I+D^*D)^{-1} (D^*f + g - Qp) + Q^* Q p = 0
311    \end{equation}
312    which is
313    \begin{equation}
314    Q^* ( I - (I+D^*D)^{-1} ) Q p = - Q^* (I+D^*D)^{-1} (D^*f + g) )
315    \end{equation}
316    We use the PCG method to solve this. The residual $r$ ($\in V^*$) is given as
317    \begin{equation}
318    \begin{array}{rcl}
319    r & = & Q^*  \left( -(I+D^*D)^{-1} (D^*f + g) - Qp + (I+D^*D)^{-1}Q p \right)\\
320    & =&  Q^* \left( - Qp - (I+D^*D)^{-1} (D^*f + g - Qp) \right) \\
321    & =&  - Q^* \left( Qp + v \right)
322    \end{array}
323    \end{equation}
324    So in a partical implementation we use the pair $(Qp,v)$ to represent the residual. This will save the
325    reconstruction of the velocity $v$. In this notation the right hand side is given as
326    $(0,(I+D^*D)^{-1} (D^*f + g))$. The evaluation of the iteration operator for a given $p$ is then
327    returning $(Qp,w)$ where $w$ is the solution of
328    \begin{equation}\label{UPDATE W}
329    (I+D^*D)w = Qp
330    \end{equation}
331    We use $Q^*Q$ as a a preconditioner for the iteration operator $Q^* ( I - (I+D^*D)^{-1} ) Q$.
332    
333    \subsection{Functions}
334    \begin{classdesc}{DarcyFlow}{domain}
335    opens the Darcy flux problem\index{Darcy flux} on the \Domain domain.
336    \end{classdesc}
337    
338  \subsection{Description}  \begin{methoddesc}[DarcyFlow]{initialize}{\optional{f=Data(), \optional{fixed_u_mask=Data(), \optional{eta=1, \optional{surface_stress=Data(), \optional{stress=Data()}}}}}}
339    assigns values to the model parameters. In any call all values must be set.
340    \var{f} defines the external force $f$, \var{eta} the viscosity $\eta$,
341    \var{surface_stress} the surface stress $s$ and \var{stress} the initial stress $\sigma$.
342    The locations and compontents where the velocity is fixed are set by
343    the values of \var{fixed_u_mask}. The method will try to cast the given values to appropriate
344    \Data class objects.
345    \end{methoddesc}
346    
 \subsection{Method}  
347    
348  \begin{classdesc}{TemperatureCartesian}{dom,theta=THETA,useSUPG=SUPG}  \subsection{Example: Gravity Flow}
 \end{classdesc}  
349    
350  \subsection{Benchmark Problem}  %================================================
351    \section{Temperature Advection Diffusion\label{TEMP ADV DIFF}}
352    
353    \begin{equation}
354    \rho c\hackscore{p} \left (\frac{\partial T}{\partial t} + \vec{v} \cdot \nabla T \right ) = k \nabla^{2}T
355    \label{HEAT EQUATION}
356    \end{equation}
357    
358  \section{Level Set Method}  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.
359    
360  \subsection{Description}  \subsection{Description}
361    
362  \subsection{Method}  \subsection{Method}
363    
364  Advection and Reinitialisation  \begin{classdesc}{TemperatureCartesian}{dom,theta=THETA,useSUPG=SUPG}
   
 \begin{classdesc}{LevelSet}{mesh, func\_new, reinit\_max, reinit\_each, tolerance, smooth}  
365  \end{classdesc}  \end{classdesc}
366    
 %example usage:  
   
 %levelset = LevelSet(mesh, func\_new, reinit\_max, reinit\_each, tolerance, smooth)  
   
 \begin{methoddesc}[LevelSet]{update\_parameter}{parameter}  
 Update the parameter.  
 \end{methoddesc}  
   
 \begin{methoddesc}[LevelSet]{update\_phi}{paramter}{velocity}{dt}{t\_step}  
 Update level set function; advection and reinitialization  
 \end{methoddesc}  
   
367  \subsection{Benchmark Problem}  \subsection{Benchmark Problem}
368    %===============================================================================================================
369    
370  Rayleigh-Taylor instability problem  %=========================================================
371    \input{levelsetmodel}
372    
373  % \section{Drucker Prager Model}  % \section{Drucker Prager Model}
374    
# Line 191  for a given power law coefficients $n^{q Line 401  for a given power law coefficients $n^{q
401  Notice that $n^{q}=1$ gives a constant viscosity.  Notice that $n^{q}=1$ gives a constant viscosity.
402  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:
403  \begin{equation}\label{IKM-EQU-6}  \begin{equation}\label{IKM-EQU-6}
404  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}} \;.
405  \end{equation}  \end{equation}
406  With  With
407  \begin{equation}\label{IKM-EQU-8}  \begin{equation}\label{IKM-EQU-8}
# Line 239  D\hackscore{ij}'^{el}=\frac{1}{2 \mu dt Line 449  D\hackscore{ij}'^{el}=\frac{1}{2 \mu dt
449  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.
450  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
451  \begin{equation}\label{IKM-EQU-10}  \begin{equation}\label{IKM-EQU-10}
452  \sigma\hackscore{ij}' =  2 \eta\hackscore{eff} D\hackscore{ij}' +  \sigma\hackscore{ij}' =  2 \eta\hackscore{eff}  \left( D\hackscore{ij}' +
453  \frac{\eta\hackscore{eff}}{\mu \; dt}  \frac{1}{  2 \mu \; dt} \sigma\hackscore{ij}^{'-}\right)  \mbox{ with }
 \sigma\hackscore{ij}^{'-} \mbox{ with }  
454  \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}}
455  \end{equation}  \end{equation}
456  Notice that $\eta\hackscore{eff}$ is a function of stress.  Notice that $\eta\hackscore{eff}$ is a function of diatoric stress $\sigma\hackscore{ij}'$.
457  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
458  \begin{equation}\label{IKM-EQU-1ib}  \begin{equation}\label{IKM-EQU-1ib}
459  -\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})
460    \right)\hackscore{,j}+p\hackscore{,i}=F\hackscore{i}+
461  \frac{\eta\hackscore{eff}}{\mu dt } \sigma\hackscore{ij,j}^{'-}  \frac{\eta\hackscore{eff}}{\mu dt } \sigma\hackscore{ij,j}^{'-}
462  \end{equation}  \end{equation}
463  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
464  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
465  \begin{equation}  \begin{equation}
466  \begin{array}{rcl}  \begin{array}{rcl}
467  -\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}
468  \frac{1}{\eta\hackscore{eff}} dp & = & - v\hackscore{i,i}^+ \\  )\right)\hackscore{,j} & = & F\hackscore{i}+ \sigma\hackscore{ij,j}'-p\hackscore{,i} \\
469  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}^+
470  \end{array}  \end{array}
471  \label{IKM iteration 2}  \label{IKM iteration 2}
472  \end{equation}  \end{equation}
473  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
474  \begin{equation}  \begin{equation}
475  \|(\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
476  \label{IKM iteration 3}  \label{IKM iteration 3}
477  \end{equation}  \end{equation}
478    where  $\bar{\eta}\hackscore{eff}$ is the caracteristic viscosity, for instance:
479    \begin{equation}
480    \frac{1}{\bar{\eta}\hackscore{eff}} = \frac{1}{\tau^{-}}+\sum\hackscore{q}  \frac{1}{\eta^{q}\hackscore{N}}
481    \label{IKM iteration 4}
482    \end{equation}
483    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
484    calculation of $\eta\hackscore{eff}$ with the Newton-Raphson scheme. This value can then be used to calculate
485    $\sigma\hackscore{ij}'$ via~\ref{IKM-EQU-10}. We need to solve
486    \begin{equation}
487    \tau = \eta\hackscore{eff} \cdot \epsilon \mbox{ with }
488    \epsilon = \sqrt{ 2 \left( D\hackscore{ij}' +
489    \frac{1}{  2 \mu \; dt} \sigma\hackscore{ij}^{'-}\right)^2}
490    \label{IKM iteration 5}
491    \end{equation}
492    The Newton scheme takes the form
493    \begin{equation}
494    \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)
495    = \min(\frac{\eta\hackscore{eff} - \tau\hackscore{n}  \eta\hackscore{eff}'}
496    {1 - \eta\hackscore{eff}'  \cdot  \epsilon}, \frac{\tau\hackscore{Y} + \beta \; p}{\epsilon}) \epsilon
497    \label{IKM iteration 6}
498    \end{equation}
499    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
500    \begin{equation}
501    \eta\hackscore{eff}' = - \eta\hackscore{eff}^2 \left(\frac{1}{\eta\hackscore{eff}}\right)'
502    \mbox{ with }
503    \left(\frac{1}{\eta\hackscore{eff}}\right)' = \sum\hackscore{q} \left(\frac{1}{\eta^{q}} \right)'
504    \label{IKM iteration 7}
505    \end{equation}
506    \begin{equation}\label{IKM-EQU-5XX}
507    \left(\frac{1}{\eta^{q}} \right)'
508    = \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}}}}
509    = \frac{1-\frac{1}{n^{q}}}{ \tau \eta^{q}}
510    \end{equation}
511    Notice that allways $\eta\hackscore{eff}'\le 0$ which makes the denomionator in~\ref{IKM iteration 6}
512    positive.
513    
514    
515    

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