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revision 2138 by lgraham, Mon Dec 8 02:27:07 2008 UTC revision 2252 by gross, Fri Feb 6 07:51:28 2009 UTC
# Line 61  with sufficient accuracy we check for Line 61  with sufficient accuracy we check for
61  \begin{equation}  \begin{equation}
62  \|v\hackscore{k,k}\| \hackscore \le  \epsilon  \|v\hackscore{k,k}\| \hackscore \le  \epsilon
63  \|\sqrt{v\hackscore{j,k}v\hackscore{j,k}}\|  \|\sqrt{v\hackscore{j,k}v\hackscore{j,k}}\|
64    \label{STOKES STOP}
65  \end{equation}  \end{equation}
66  where $\epsilon$ is the desired relative accuracy and  where $\epsilon$ is the desired relative accuracy and
67  \begin{equation}  \begin{equation}
68  \|p\|^2= \int\hackscore{\Omega} p^2 \; dx  \|p\|^2= \int\hackscore{\Omega} p^2 \; dx
69  \label{PRESSURE NORM}  \label{PRESSURE NORM}
70  \end{equation}  \end{equation}
71  defines the $L^2$-norm.  defines the $L^2$-norm. We use the Uzawa scheme \index{Uzawa scheme} to solve the problem.
72  There are two approaches to solve this problem. The first approach, called the Uzawa scheme \index{Uzawa scheme}  
73  eliminates the velocity $v$ from the problem. The second approach solves the equation in coupled form after the application of a preconditioner.  In fact the first equation in~\ref{SADDLEPOINT} gives for a known pressure
74    \begin{equation}
75  \subsubsection{Uzawa scheme}  v=A^{-1}(G-B^{*}p)
76  The first eqution in~\ref{SADDLEPOINT} gives $v=A^{-1}(G-B^{*}p)$ assuming $p$ is known. This is inserted into the  \label{V CALC}
77  second eqution which leads to  \end{equation}
78    which is inserted into the second equation leading to
79  \begin{equation}  \begin{equation}
80  S p =  B A^{-1} G  S p =  B A^{-1} G
81  \end{equation}  \end{equation}
82  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}  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  with the preconditioner $\hat{S}$ defined as $q=\hat{S}^{-1}p$ by solving
84  \begin{equation}  \begin{equation}
85  \frac{1}{\eta}q = p  \frac{1}{\eta}q = p
86  \end{equation}  \end{equation}
87  see~\cite{ELMAN} for more details. The evaluation of $w=Sp$ is done in the form  see~\cite{ELMAN} for more details. Note that the residual for the current approximation $p$ is given as
88  \begin{equation}  \begin{equation}
89  \begin{array}{rcl}  r=B A^{-1} (G - B^* p) = Bv
 A v & = & B^{*}p \\  
 w & = & Bv \\  
 \end{array}  
 \label{EVAL PCG}  
90  \end{equation}  \end{equation}
91  The residual \index{residual}  $r=B A^{-1} G - S p$ is given as  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}  \begin{equation}
96  r=B A^{-1} (G - B^* p) = Bv \mbox{ with } v = A^{-1}(G-B^{*}p)  \hat{S}^{-1} S p =  \hat{S}^{-1} B A^{-1} G
97  \end{equation}  \end{equation}
98  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  We use the norm
99  \begin{equation}  \begin{equation}
100  (p,(v,Bv))=\int\hackscore{\Omega} p \cdot Bv \; dx  \|p\|\hackscore{GMRES} = \|\hat{S} p \|
101  \end{equation}  \end{equation}
102  where $p$ is the pressure increment and $(v,Bv)$ represents an increment in the residual.  Notice that for the residual $\hat{r}=\hat{S}^{-1} r$ one has
   
 \subsubsection{Coupled Solver}  
 An alternative approach to solve the saddle point problem~\ref{SADDLEPOINT} directly using an iterative such as  
 the generalized minimal residual method (GMRES) \index{generalized minimal residual method!GMRES} with a suitable  
 preconditioner. Here we use the operator  
103  \begin{equation}  \begin{equation}
104  \left[ \begin{array}{cc}  \
105  A^{-1}     & 0 \\  \end{equation}
106  S^{-1} B A^{-1}  & -S^{-1} \\  If $p^{0}$ provides an initial guess for the pressure we use~\ref{V CALC} to get a first initial guess for the
107  \end{array} \right]  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  \label{SADDLEPOINT PRECODITIONER}  The GMRES is terminated when
 \end{equation}  
 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  
109  \begin{equation}  \begin{equation}
110  \begin{array}{rcl}  \|\hat{r}\|\hackscore{GMRES} \le ATOL
111  A w & = & Av+B^{*}p \\  \end{equation}
112  \hat{S} q & = & B(w-v) \\  Notice that $\|\hat{r}\|\hackscore{GMRES}= \|r \| = \|Bv\|  = \|v\hackscore{k,k}\|$ so we we can expect that
113  \end{array}  the target stopping criterion~\ref{STOKES STOP} is fullfilled. However, if $v$ is very different from the
114  \label{COUPLES SADDLEPOINT iteration}  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  \end{equation}  
116  We use the inner product induced by the norm  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}  \begin{equation}
118  \|(v,p)\|^2= \int\hackscore{\Omega}  v\hackscore{i,j}  v\hackscore{i,j} + \left( \frac{p}{\eta}\right)^2\; dx  \|r\|\hackscore{PCG}^2 = \int\hackscore{\Omega} r \hat{S}^{-1}r \; dx = \int\hackscore{\Omega}  \eta r^2 \; dx
119  \label{COUPLES NORM}  \end{equation}
120  \end{equation}  To take the extra factor $\eta$ into consideration when checking the stopping criterion we use the following
121  In PDE form~\ref{COUPLES SADDLEPOINT iteration} takes the form  definition for $ATOL$:
122  \begin{equation}  \begin{equation}
123  \begin{array}{rcl}  ATOL = \epsilon \frac{\|\sqrt{v^{0}\hackscore{j,k}v^{0}\hackscore{j,k}}\|  }{\|v^{0}\hackscore{k,k}\|}
124  -\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} \\  \|v^{0}\hackscore{k,k}\|\hackscore{PCG}
125  \frac{1}{\eta}  q & = & - (w-v)\hackscore{i,i} \\  \end{equation}
126  \end{array}  
 \label{SADDLEPOINT iteration 2}  
 \end{equation}  
127    
128    
129  \subsection{Functions}  \subsection{Functions}
# Line 151  the values of \var{fixed_u_mask}. The me Line 143  the values of \var{fixed_u_mask}. The me
143  \end{methoddesc}  \end{methoddesc}
144    
145  \begin{methoddesc}[StokesProblemCartesian]{solve}{v,p,  \begin{methoddesc}[StokesProblemCartesian]{solve}{v,p,
146  \optional{max_iter=20, \optional{verbose=False, \optional{useUzawa=True}}}}  \optional{max_iter=20, \optional{verbose=False, \optional{usePCG=True}}}}
147  solves the problem and return approximations for velocity and pressure.  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  The arguments \var{v} and \var{p} define initial guess. The values of \var{v} marked
149  by \var{fixed_u_mask} remain unchanged.  by \var{fixed_u_mask} remain unchanged.
150  If \var{useUzawa} is set to \True  If \var{usePCG} is set to \True
151  the Uzawa\index{Uszwa} scheme is used. Otherwise the problem is solved in coupled form. In most cases  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 Uzawa scheme is more efficient.  the PCG scheme is more efficient.
153  \var{max_iter} defines the maximum number of iteration steps.  \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.  If \var{verbose} is set to \True informations on the progress of of the solver are printed.
155  \end{methoddesc}  \end{methoddesc}
156    
157    
158  \begin{methoddesc}[StokesProblemCartesian]{setTolerance}{\optional{tolerance=1.e-8}}  \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.  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}  \end{methoddesc}
161  \begin{methoddesc}[StokesProblemCartesian]{getTolerance}{}  \begin{methoddesc}[StokesProblemCartesian]{getTolerance}{}
# Line 176  absolute talerance is set to 0. Line 168  absolute talerance is set to 0.
168  \begin{methoddesc}[StokesProblemCartesian]{getAbsoluteTolerance}{}  \begin{methoddesc}[StokesProblemCartesian]{getAbsoluteTolerance}{}
169  sreturns the current absolute tolerance.  sreturns the current absolute tolerance.
170  \end{methoddesc}  \end{methoddesc}
171  \begin{methoddesc}[StokesProblemCartesian]{setSubToleranceReductionFactor}{\optional{reduction=None}}  \begin{methoddesc}[StokesProblemCartesian]{setSubProblemTolerance}{\optional{rtol=None}}
172  sets the reduction factor for the tolerance used to solve the PDEs. A reduction factor  sets the tolerance to solve the involved PDEs. The subtolerance \var{rtol} should not be choosen to large
173  in the order of one will minimize compute time per iteration step but my slow down convergence or even lead to  in order to avoid feed back of errors in the subproblem solution into the outer iteration.
174  divergency. On the other hand a very small value for the PDE tolerance could result in a wast of compute time.  On the otherhand is choosen to small compute time is wasted.
175  If \var{reduction} is set to \var{None} the sub-tolerance is solved adaptively but  If \var{rtol} is set to \var{None} the sub-tolerance is set automatically depending on the
176  in cases a very small tolerance is set ($<10^{-6}$) it is recommended to set the  tolerance choosen for the oter iteration.
 reduction factor by hand. This may require some experiments.  
177  \end{methoddesc}  \end{methoddesc}
178  \begin{methoddesc}[StokesProblemCartesian]{getSubToleranceReductionFactor}{}  \begin{methoddesc}[StokesProblemCartesian]{getSubProblemTolerance}{}
179  return the current reduction factor for the sub-problem tolerance.  return the tolerance for the involved PDEs.
180  \end{methoddesc}  \end{methoddesc}
181    
182  \subsection{Example: Lit Driven Cavity}  \subsection{Example: Lit Driven Cavity}
# Line 235  for all $x\hackscore{i}$. Line 226  for all $x\hackscore{i}$.
226    
227    
228  \subsection{Solution Method \label{DARCY SOLVE}}  \subsection{Solution Method \label{DARCY SOLVE}}
229  Without loss of generality we can assume that $u^{N}\hackscore{i}  \; n\hackscore{i}=0$ and  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  $p^{D}$. Otherewise one solves for $u-u^{N}$ and $p-p^{D}$ and sets  \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}  \begin{equation}
237  \begin{array}{rcl}  \begin{array}{rcl}
238  g\hackscore{i} & \leftarrow & g\hackscore{i} - u^{N}\hackscore{i} -  \kappa\hackscore{ij} p^{D}\hackscore{,j }\\  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}  f & \leftarrow & f - u^{N}\hackscore{k,k}
240  \end{array}  \end{array}
241  \end{equation}  \end{equation}
242    we can assume that $u^{N}\hackscore{i}  \; n\hackscore{i}=0$ and
243    $p^{D}=0$.
244  We set  We set
245  \begin{equation}  \begin{equation}
246  V = \{ q \in H^{1}(\Omega) : q=0 \mbox{ on } \Gamma\hackscore{D} \}  V = \{ q \in H^{1}(\Omega) : q=0 \mbox{ on } \Gamma\hackscore{D} \}
# Line 279  for all $v\in W$ and $q \in V$.which tra Line 277  for all $v\in W$ and $q \in V$.which tra
277  \begin{equation}  \begin{equation}
278  \begin{array}{rcl}  \begin{array}{rcl}
279  (I+D^*D)u + Qp & = & D^*f + g \\  (I+D^*D)u + Qp & = & D^*f + g \\
280  Q^*u  + Q^*Q p & = & Q^* g \\  Q^*u  + Q^*Q p & = & Q^*g \\
281  \end{array}  \end{array}
282  \end{equation}  \end{equation}
283  where $D^*$ and $Q^*$ denote the adjoint operators.  where $D^*$ and $Q^*$ denote the adjoint operators.
# Line 292  v= (I+D^*D)^{-1} (D^*f + g - Qp) Line 290  v= (I+D^*D)^{-1} (D^*f + g - Qp)
290  \end{equation}  \end{equation}
291  (notice that $(I+D^*D)$ is coercive in $W$) which is inserted into the second equation  (notice that $(I+D^*D)$ is coercive in $W$) which is inserted into the second equation
292  \begin{equation}  \begin{equation}
293  Q^* (I+D^*D)^{-1} (D^*f + g - Qp) + Q^* Q p = Q^* g  Q^* (I+D^*D)^{-1} (D^*f + g - Qp) + Q^* Q p = Q^*g
294  \end{equation}  \end{equation}
295  which is  which is
296  \begin{equation}  \begin{equation}
297  Q^* ( I - (I+D^*D)^{-1} ) Q p = Q^* ( g -(I+D^*D)^{-1} (D^*f + g) )  Q^* ( I - (I+D^*D)^{-1} ) Q p = Q^* (g-(I+D^*D)^{-1} (D^*f + g) )
298  \end{equation}  \end{equation}
299  We use the PCG method to solve this. The residual $r$ ($\in V^*$) is given as  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}  \begin{equation}
301  \begin{array}{rcl}  \begin{array}{rcl}
302  r & = & Q^* ( g -(I+D^*D)^{-1} (D^*f + g) - Qp + (I+D^*D)^{-1}Q p \\  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) \\  & =&  Q^* \left( - Qp - (I+D^*D)^{-1} (D^*f + g - Qp) \right) \\
304  & =&  Q^* \left( (g-Qp) - v \right)  & =&  Q^* \left( g - Qp - v \right)
305  \end{array}  \end{array}
306  \end{equation}  \end{equation}
307  So in a partical implementation we use the pair $(g-Qp,v)$ to represent the residual. This will save the  So in a partical implementation we use the pair $(Qp,v)$ to represent the residual. This will save the
308  reconstruction of the velocity $v$. In this notation the right hand side is given as  reconstruction of the velocity $v$. In this notation the right hand side is given as
309  $(g,(I+D^*D)^{-1} (D^*f + g))$. The evaluation of the iteration operator for a given $p$ is then  $(0,(I+D^*D)^{-1} (D^*f + g))$. The evaluation of the iteration operator for a given $p$ is then
310  returning $(Qp,w)$ where $w$ is the solution of  returning $(Qp,w)$ where $w$ is the solution of
311  \begin{equation}\label{UPDATE W}  \begin{equation}\label{UPDATE W}
312  (I+D^*D)w = Qp  (I+D^*D)w = Qp
313  \end{equation}  \end{equation}
314  We use $Q^*Q$ as a a preconditioner for the iteration operator $Q^* ( I - (I+D^*D)^{-1} ) Q$.  We use $Q^*Q$ as a a preconditioner for the iteration operator $Q^* ( I - (I+D^*D)^{-1} ) Q$.
315    
316    The iteration PCG \index{linear solver!PCG}\index{PCG} is terminated if
317    \begin{equation}\label{DARCY STOP}
318    \int r \cdot (Q^*Q)^{-1} r \; dx \le \mbox{ATOL}^2
319    \end{equation}
320    where ATOL is a given absolute tolerance.
321    The initial residual $r\hackscore{0}$ is
322    \begin{equation}\label{DARCY STOP 2}
323    r\hackscore{0}=Q^* \left( g - v\hackscore{ref} \right) \mbox{ with } v\hackscore{ref} = (I+D^*D)^{-1} (D^*f + g)
324    \end{equation}
325    so the
326    \begin{equation}\label{DARCY NORM 0}
327    \int r\hackscore0 \cdot (Q^*Q)^{-1} r\hackscore0 \; dx = \int \left( g - v\hackscore{ref} \right)  \cdot  Q  p\hackscore{ref} \; dx \mbox{ with }p\hackscore{ref} = (Q^*Q)^{-1} Q^* \left( g - v\hackscore{ref} \right)
328    \end{equation}
329    So we set
330    \begin{equation}\label{DARCY NORM 1}
331    ATOL = atol + rtol \cdot \max(|g - v\hackscore{ref}|\hackscore{0}, |Q p\hackscore{ref} |\hackscore{0} )
332    \end{equation}
333    where atol and rtol a given absolute and relative tolerances, respectively. The reference flux $v\hackscore{ref}$
334    and reference pressure $p\hackscore{ref}$ may be calcualated from their definition which would require to solve to
335    PDEs but in a practical application estimates can be used for instance solutions from previous time steps or for simplified scenarious (e.g. constant permability).
336    
337  \subsection{Functions}  \subsection{Functions}
338  \begin{classdesc}{DarcyFlow}{domain}  \begin{classdesc}{DarcyFlow}{domain}
339  opens the Darcy flux problem\index{Darcy flux} on the \Domain domain.  opens the Darcy flux problem\index{Darcy flux} on the \Domain domain.
340  \end{classdesc}  \end{classdesc}
341    \begin{methoddesc}[DarcyFlow]{setValue}{\optional{f=None, \optional{g=None, \optional{location_of_fixed_pressure=None, \optional{location_of_fixed_flux=None, \optional{permeability=None}}}}}}
342  \begin{methoddesc}[DarcyFlow]{initialize}{\optional{f=Data(), \optional{fixed_u_mask=Data(), \optional{eta=1, \optional{surface_stress=Data(), \optional{stress=Data()}}}}}}  assigns values to the model parameters. Values can be assigned using various calls - in particular
343  assigns values to the model parameters. In any call all values must be set.  in a time dependend problem only values that change over time needs to be reset. The permability can be defined as scalar (isotropic), a vector (orthotropic) or a matrix (anisotropic).
344  \var{f} defines the external force $f$, \var{eta} the viscosity $\eta$,  \var{f} and \var{g} are the corresponding parameters in~\ref{DARCY PROBLEM}.
345  \var{surface_stress} the surface stress $s$ and \var{stress} the initial stress $\sigma$.  The locations and compontents where the flux is prescribed are set by positive values in
346  The locations and compontents where the velocity is fixed are set by  \var{location_of_fixed_flux}.
347  the values of \var{fixed_u_mask}. The method will try to cast the given values to appropriate  The locations where the pressure is prescribed are set by
348    by positive values of \var{location_of_fixed_pressure}.
349    The values of the pressure and flux are defined by the initial guess.
350    Notice that at any point on the boundary of the domain the pressure or the normal component of
351    the flux must be defined. There must be at least one point where the pressure is prescribed.
352    The method will try to cast the given values to appropriate
353  \Data class objects.  \Data class objects.
354  \end{methoddesc}  \end{methoddesc}
355    
356    \begin{methoddesc}[DarcyFlow]{setTolerance}{\optional{atol=0,\optional{rtol=1e-8,\optional{p_ref=None,\optional{v_ref=None}}}}}
357    sets the absolute tolerance ATOL according to~\ref{DARCY NORM 1}. If \var{p_ref} is not present $0$ is used.
358    If \var{v_ref} is not present $0$ is used. If the final result ATOL is not positive an exception is thrown.
359    \end{methoddesc}
360    
361    
362    
363    \begin{methoddesc}[DarcyFlow]{solve}{u0,p0, \optional{max_iter=100, \optional{verbose=False \optional{sub_rtol=1.e-8}}}}
364    solves the problem. and returns approximations for the flux $v$ and the pressure $p$.
365    \var{u0} and \var{p0} define initial guess for flux and pressure. Values marked
366    by positive values \var{location_of_fixed_flux} and \var{location_of_fixed_pressure}, respectively, are kept unchanged.
367    \var{sub_rtol} defines the tolerance used to solve the involved PDEs. \var{sub_rtol} needs to be choosen sufficiently small to ensure convergence but users need to keep in mind that a very small value for \var{sub_rtol} will result in a long compute time. Typically  $\var{sub_rtol}=\var{rtol}^2$ is a good choice if $\var{rtol}$ is not choosen too small.
368    \end{methoddesc}
369    
370    
371  \subsection{Example: Gravity Flow}  \subsection{Example: Gravity Flow}
372    later
373    
374  %================================================  %================================================
375  \section{Temperature Advection Diffusion\label{TEMP ADV DIFF}}  % \section{Temperature Advection Diffusion\label{TEMP ADV DIFF}}
376    
377  \begin{equation}  %\begin{equation}
378  \rho c\hackscore{p} \left (\frac{\partial T}{\partial t} + \vec{v} \cdot \nabla T \right ) = k \nabla^{2}T  % \rho c\hackscore{p} \left (\frac{\partial T}{\partial t} + \vec{v} \cdot \nabla T \right ) = k \nabla^{2}T
379  \label{HEAT EQUATION}  % \label{HEAT EQUATION}
380  \end{equation}  % \end{equation}
   
 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.  
381    
382  \subsection{Description}  % 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.
383    
384  \subsection{Method}  % \subsection{Description}
385    
386  \begin{classdesc}{TemperatureCartesian}{dom,theta=THETA,useSUPG=SUPG}  % \subsection{Method}
387  \end{classdesc}  %
388    % \begin{classdesc}{TemperatureCartesian}{dom,theta=THETA,useSUPG=SUPG}
389    % \end{classdesc}
390    
391  \subsection{Benchmark Problem}  % \subsection{Benchmark Problem}
392  %===============================================================================================================  %===============================================================================================================
393    
394  %=========================================================  %=========================================================
395  \input{levelsetmodel}  \input{levelsetmodel}
 % \section{Level Set Method}  
   
 %\subsection{Description}  
   
 %\subsection{Method}  
   
 %Advection and Reinitialisation  
   
 %\begin{classdesc}{LevelSet}{mesh, func\_new, reinit\_max, reinit\_each, tolerance, smooth}  
 %\end{classdesc}  
   
 %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}  
   
 %\subsection{Benchmark Problem}  
   
 %Rayleigh-Taylor instability problem  
   
396    
397  % \section{Drucker Prager Model}  % \section{Drucker Prager Model}
398    
# Line 392  D\hackscore{ij}=D\hackscore{ij}^{el}+D\h Line 404  D\hackscore{ij}=D\hackscore{ij}^{el}+D\h
404  \end{equation}  \end{equation}
405  with the elastic strain given as  with the elastic strain given as
406  \begin{equation}\label{IKM-EQU-3}  \begin{equation}\label{IKM-EQU-3}
407  D\hackscore{ij}'^{el}=\frac{1}{2 \mu} \dot{\sigma}\hackscore{ij}'  D\hackscore{ij}^{el'}=\frac{1}{2 \mu} \dot{\sigma}\hackscore{ij}'
408  \end{equation}  \end{equation}
409  where $\sigma'\hackscore{ij}$ is the deviatoric stress (Notice that $\sigma'\hackscore{ii}=0$).  where $\sigma'\hackscore{ij}$ is the deviatoric stress (Notice that $\sigma'\hackscore{ii}=0$).
410  If the material is composed by materials $q$ the visco-plastic strain can be decomposed as  If the material is composed by materials $q$ the visco-plastic strain can be decomposed as
411  \begin{equation}\label{IKM-EQU-4}  \begin{equation}\label{IKM-EQU-4}
412  D\hackscore{ij}'^{vp}=\sum\hackscore{q} D\hackscore{ij}'^{q}  D\hackscore{ij}^{vp'}=\sum\hackscore{q} D\hackscore{ij}^{q'}
413  \end{equation}  \end{equation}
414  where $D\hackscore{ij}^{q}$ is the strain in material $q$ given as  where $D\hackscore{ij}^{q}$ is the strain in material $q$ given as
415  \begin{equation}\label{IKM-EQU-5}  \begin{equation}\label{IKM-EQU-5}
416  D\hackscore{ij}'^{q}=\frac{1}{2 \eta^{q}} \sigma'\hackscore{ij}  D\hackscore{ij}^{q'}=\frac{1}{2 \eta^{q}} \sigma'\hackscore{ij}
417  \end{equation}  \end{equation}
418  where $\eta^{q}$ is the viscosity of material $q$. We assume the following  where $\eta^{q}$ is the viscosity of material $q$. We assume the following
419  betwee the the strain in material $q$  betwee the the strain in material $q$
# Line 409  betwee the the strain in material $q$ Line 421  betwee the the strain in material $q$
421  \eta^{q}=\eta^{q}\hackscore{N} \left(\frac{\tau}{\tau\hackscore{t}^q}\right)^{\frac{1}{n^{q}}-1}  \eta^{q}=\eta^{q}\hackscore{N} \left(\frac{\tau}{\tau\hackscore{t}^q}\right)^{\frac{1}{n^{q}}-1}
422  \mbox{ with } \tau=\sqrt{\frac{1}{2}\sigma'\hackscore{ij} \sigma'\hackscore{ij}}  \mbox{ with } \tau=\sqrt{\frac{1}{2}\sigma'\hackscore{ij} \sigma'\hackscore{ij}}
423  \end{equation}  \end{equation}
424  for a given power law coefficients $n^{q}$ and transition stresses $\tau\hackscore{t}^q$, see~\ref{POERLAW}.  for a given power law coefficients $n^{q}\ge1$ and transition stresses $\tau\hackscore{t}^q$, see~\ref{POERLAW}.
425  Notice that $n^{q}=1$ gives a constant viscosity.  Notice that $n^{q}=1$ gives a constant viscosity.
426  After inserting equation~\ref{IKM-EQU-5} into equation \ref{IKM-EQU-4} one gets:  After inserting equation~\ref{IKM-EQU-5} into equation \ref{IKM-EQU-4} one gets:
427  \begin{equation}\label{IKM-EQU-6}  \begin{equation}\label{IKM-EQU-6}
# Line 472  After inserting~\ref{IKM-EQU-10} into~\r Line 484  After inserting~\ref{IKM-EQU-10} into~\r
484  \right)\hackscore{,j}+p\hackscore{,i}=F\hackscore{i}+  \right)\hackscore{,j}+p\hackscore{,i}=F\hackscore{i}+
485  \frac{\eta\hackscore{eff}}{\mu dt } \sigma\hackscore{ij,j}^{'-}  \frac{\eta\hackscore{eff}}{\mu dt } \sigma\hackscore{ij,j}^{'-}
486  \end{equation}  \end{equation}
487  Together with the incomressibilty condition~\ref{IKM-EQU-2} we need to solve a problem with a form almost identical  Combining this with the incomressibilty condition~\ref{IKM-EQU-2} we need to solve a
488  to the Stokes problem discussed in section~\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.
489  \begin{equation}  In oder to perform step~\ref{IKM iteration 2} we need to calculate the $\eta\hackscore{eff}$ which
490  \begin{array}{rcl}  is a function of $\sigma\hackscore{ij}$ via $\tau$.  To get $\tau$ and $\eta\hackscore{eff}$ we need to solve the
491  -\left(\eta\hackscore{eff}(dv\hackscore{i,j}+ dv\hackscore{i,j}  non-linear equation
 )\right)\hackscore{,j} & = & F\hackscore{i}+ \sigma\hackscore{ij,j}'-p\hackscore{,i} \\  
 \frac{1}{\eta\hackscore{eff}} dp & = & - v\hackscore{i,i}^+  
 \end{array}  
 \label{IKM iteration 2}  
 \end{equation}  
 where $v^+=v+dv$. As this problem is non-linear the Jacobi-free Newton-GMRES method is used with the norm  
 \begin{equation}  
 \|(v, p)\|^2= \int\hackscore{\Omega} v\hackscore{i,j}^2 + \frac{1}{\bar{\eta}^2\hackscore{eff}} p^2 \; dx  
 \label{IKM iteration 3}  
 \end{equation}  
 where  $\bar{\eta}\hackscore{eff}$ is the caracteristic viscosity, for instance:  
 \begin{equation}  
 \frac{1}{\bar{\eta}\hackscore{eff}} = \frac{1}{\tau^{-}}+\sum\hackscore{q}  \frac{1}{\eta^{q}\hackscore{N}}  
 \label{IKM iteration 4}  
 \end{equation}  
 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  
 calculation of $\eta\hackscore{eff}$ with the Newton-Raphson scheme. This value can then be used to calculate  
 $\sigma\hackscore{ij}'$ via~\ref{IKM-EQU-10}. We need to solve  
492  \begin{equation}  \begin{equation}
493  \tau = \eta\hackscore{eff} \cdot \epsilon \mbox{ with }  \tau = \eta\hackscore{eff} \cdot \dot{\gamma}\hackscore{total} \mbox{ with }
494  \epsilon = \sqrt{ 2 \left( D\hackscore{ij}' +  \dot{\gamma}\hackscore{total} = \sqrt{ 2 \left( D\hackscore{ij}' +
495  \frac{1}{  2 \mu \; dt} \sigma\hackscore{ij}^{'-}\right)^2}  \frac{1}{  2 \mu \; dt} \sigma\hackscore{ij}^{'-}\right)^2}
496  \label{IKM iteration 5}  \label{IKM iteration 5}
497  \end{equation}  \end{equation}
498  The Newton scheme takes the form  The Newton scheme takes the form
499  \begin{equation}  \begin{equation}
500  \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)  \tau\hackscore{n+1} = \min(\tau\hackscore{n} - \frac{\tau\hackscore{n} - \eta\hackscore{eff}  \cdot \dot{\gamma}\hackscore{total}}{1 - \eta\hackscore{eff}'  \cdot  \dot{\gamma}\hackscore{total}}, \tau\hackscore{Y} + \beta \; p)
 = \min(\frac{\eta\hackscore{eff} - \tau\hackscore{n}  \eta\hackscore{eff}'}  
 {1 - \eta\hackscore{eff}'  \cdot  \epsilon}, \frac{\tau\hackscore{Y} + \beta \; p}{\epsilon}) \epsilon  
501  \label{IKM iteration 6}  \label{IKM iteration 6}
502  \end{equation}  \end{equation}
503  where $\eta\hackscore{eff}'$ denotes the derivative of $\eta\hackscore{eff}$ with respect of $\tau$. The second term in $\min$ is droped of $\tau\hackscore{Y} + \beta \; p<0$ or $\epsilon=0$. In fact we have  where $\eta\hackscore{eff}'$ denotes the derivative of $\eta\hackscore{eff}$ with respect of $\tau$. The second term in $\min$ is droped of $\tau\hackscore{Y} + \beta \; p<0$ (?)). We have
504  \begin{equation}  \begin{equation}
505  \eta\hackscore{eff}' = - \eta\hackscore{eff}^2 \left(\frac{1}{\eta\hackscore{eff}}\right)'  \eta\hackscore{eff}' = - \eta\hackscore{eff}^2 \left(\frac{1}{\eta\hackscore{eff}}\right)'
506  \mbox{ with }  \mbox{ with }
# Line 518  where $\eta\hackscore{eff}'$ denotes the Line 510  where $\eta\hackscore{eff}'$ denotes the
510  \begin{equation}\label{IKM-EQU-5XX}  \begin{equation}\label{IKM-EQU-5XX}
511  \left(\frac{1}{\eta^{q}} \right)'  \left(\frac{1}{\eta^{q}} \right)'
512  = \frac{1-\frac{1}{n^{q}}}{\eta^{q}\hackscore{N}} \frac{\tau^{-\frac{1}{n^{q}}}}{(\tau\hackscore{t}^q)^{1-\frac{1}{n^{q}}}}  = \frac{1-\frac{1}{n^{q}}}{\eta^{q}\hackscore{N}} \frac{\tau^{-\frac{1}{n^{q}}}}{(\tau\hackscore{t}^q)^{1-\frac{1}{n^{q}}}}
513  = \frac{1-\frac{1}{n^{q}}}{ \tau \eta^{q}}  = \frac{1-\frac{1}{n^{q}}}{\tau}\frac{1}{\eta^{q}}
514  \end{equation}  \end{equation}
515  Notice that allways $\eta\hackscore{eff}'\le 0$ which makes the denomionator in~\ref{IKM iteration 6}  Notice that allways $\eta\hackscore{eff}'\le 0$ which makes the denomionator in~\ref{IKM iteration 6}
516  positive.  positive.

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