/[escript]/trunk/doc/user/Models.tex
ViewVC logotype

Annotation of /trunk/doc/user/Models.tex

Parent Directory Parent Directory | Revision Log Revision Log


Revision 2415 - (hide annotations)
Wed May 13 02:48:39 2009 UTC (10 years, 5 months ago) by gross
File MIME type: application/x-tex
File size: 26441 byte(s)
FileWriter added: this class takes care of writing data which are global in MPI to a file. It is recommended to use this class rather then the build in open as it takes care of the case of many processors.
1 ksteube 1811
2     %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
3 lgraham 1702 %
4 ksteube 1811 % Copyright (c) 2003-2008 by University of Queensland
5     % Earth Systems Science Computational Center (ESSCC)
6     % http://www.uq.edu.au/esscc
7 lgraham 1702 %
8 ksteube 1811 % Primary Business: Queensland, Australia
9     % Licensed under the Open Software License version 3.0
10     % http://www.opensource.org/licenses/osl-3.0.php
11 lgraham 1702 %
12 ksteube 1811 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
13 lgraham 1702
14 ksteube 1811
15 lgraham 1702 \chapter{Models}
16 lgraham 2128 \label{MODELS CHAPTER}
17 lgraham 1702
18     The following sections give a breif overview of the model classes and their corresponding methods.
19    
20 gross 1878 \section{Stokes Problem}
21 gross 2100 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}
22 gross 1878 \begin{equation}\label{Stokes 1}
23 gross 2100 -\left(\eta(v\hackscore{i,j}+ v\hackscore{i,j})\right)\hackscore{,j}+p\hackscore{,i}=f\hackscore{i}-\sigma\hackscore{ij,j}
24 gross 1878 \end{equation}
25 gross 2100 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 gross 1878 \begin{equation}\label{Stokes 2}
27     -v\hackscore{i,i}=0
28     \end{equation}
29     Natural boundary conditions are taken in the form
30     \begin{equation}\label{Stokes Boundary}
31 gross 2100 \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 gross 1878 \end{equation}
33 gross 2100 which can be overwritten by constraints of the form
34 gross 1878 \begin{equation}\label{Stokes Boundary0}
35 gross 2100 v\hackscore{i}(x)=v^D\hackscore{i}(x)
36 gross 1878 \end{equation}
37 gross 2100 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 lgraham 1702
40 gross 1878 \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
42 gross 2100 \index{saddle point problem}
43 lgraham 1702 \begin{equation}
44     \left[ \begin{array}{cc}
45 gross 1878 A & B^{*} \\
46     B & 0 \\
47 lgraham 1702 \end{array} \right]
48     \left[ \begin{array}{c}
49 gross 2100 v \\
50 lgraham 1702 p \\
51     \end{array} \right]
52     =\left[ \begin{array}{c}
53 gross 2100 G \\
54 gross 1878 0 \\
55 lgraham 1702 \end{array} \right]
56     \label{SADDLEPOINT}
57     \end{equation}
58 gross 2100 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 use iterative techniques to solve this problem. To make sure that the incomressibilty condition holds
60     with sufficient accuracy we check for
61 gross 1878 \begin{equation}
62 gross 2100 \|v\hackscore{k,k}\| \hackscore \le \epsilon
63     \|\sqrt{v\hackscore{j,k}v\hackscore{j,k}}\|
64 gross 2251 \label{STOKES STOP}
65 gross 2100 \end{equation}
66     where $\epsilon$ is the desired relative accuracy and
67     \begin{equation}
68     \|p\|^2= \int\hackscore{\Omega} p^2 \; dx
69     \label{PRESSURE NORM}
70     \end{equation}
71 gross 2251 defines the $L^2$-norm. We use the Uzawa scheme \index{Uzawa scheme} to solve the problem.
72    
73     In fact the first equation in~\ref{SADDLEPOINT} gives for a known pressure
74 gross 2100 \begin{equation}
75 gross 2251 v=A^{-1}(G-B^{*}p)
76     \label{V CALC}
77     \end{equation}
78     which is inserted into the second equation leading to
79     \begin{equation}
80 gross 2100 S p = B A^{-1} G
81     \end{equation}
82 gross 2251 with the Schur complement \index{Schur complement} $S=BA^{-1}B^{*}$. This problem can be solved iteratively
83 gross 2100 with the preconditioner $\hat{S}$ defined as $q=\hat{S}^{-1}p$ by solving
84     \begin{equation}
85     \frac{1}{\eta}q = p
86     \end{equation}
87 gross 2251 see~\cite{ELMAN} for more details. Note that the residual for the current approximation $p$ is given as
88 gross 2100 \begin{equation}
89 gross 2251 r=B A^{-1} (G - B^* p) = Bv
90 gross 2100 \end{equation}
91 gross 2251 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 gross 2100 \begin{equation}
96 gross 2251 \hat{S}^{-1} S p = \hat{S}^{-1} B A^{-1} G
97 gross 2100 \end{equation}
98 gross 2251 We use the norm
99 gross 2100 \begin{equation}
100 gross 2251 \|p\|\hackscore{GMRES} = \|\hat{S} p \|
101     \end{equation}
102     Notice that for the residual $\hat{r}=\hat{S}^{-1} r$ one has
103 gross 2100 \begin{equation}
104 gross 2251 \
105     \end{equation}
106     If $p^{0}$ provides an initial guess for the pressure we use~\ref{V CALC} to get a first initial guess for the
107     velocity $v^{0}$ which we use to set an absolute tolerance $ATOL =\epsilon \|\sqrt{v^{0}\hackscore{j,k}v^{0}\hackscore{j,k}}\|$.
108     The GMRES is terminated when
109 gross 1878 \begin{equation}
110 gross 2251 \|\hat{r}\|\hackscore{GMRES} \le ATOL
111     \end{equation}
112     Notice that $\|\hat{r}\|\hackscore{GMRES}= \|r \| = \|Bv\| = \|v\hackscore{k,k}\|$ so we we can expect that
113     the target stopping criterion~\ref{STOKES STOP} is fullfilled. However, if $v$ is very different from the
114     initial choice of $v^{0}$ the value of $ATOL$ is corrected and GMRES is restarted with a new tolerance. For time dependend problems this apprach works well as value for $p$ form a previous time step provides a good initial guess.
115    
116     Alternatively, as $S$ is symmetric and positive definite one can apply the preconditioned conjugate gradient method (PCG) \index{preconditioned conjugate gradient method!PCG}. PCG use the norm
117 gross 1878 \begin{equation}
118 gross 2251 \|r\|\hackscore{PCG}^2 = \int\hackscore{\Omega} r \hat{S}^{-1}r \; dx = \int\hackscore{\Omega} \eta r^2 \; dx
119     \end{equation}
120     To take the extra factor $\eta$ into consideration when checking the stopping criterion we use the following
121     definition for $ATOL$:
122 gross 2100 \begin{equation}
123 gross 2251 ATOL = \epsilon \frac{\|\sqrt{v^{0}\hackscore{j,k}v^{0}\hackscore{j,k}}\| }{\|v^{0}\hackscore{k,k}\|}
124     \|v^{0}\hackscore{k,k}\|\hackscore{PCG}
125     \end{equation}
126 lgraham 1702
127    
128 gross 2251
129 gross 2100 \subsection{Functions}
130 gross 1878
131 gross 2100 \begin{classdesc}{StokesProblemCartesian}{domain}
132     opens the Stokes problem\index{Stokes problem} on the \Domain domain. The approximation
133     order needs to be two.
134     \end{classdesc}
135 gross 1878
136 gross 2100 \begin{methoddesc}[StokesProblemCartesian]{initialize}{\optional{f=Data(), \optional{fixed_u_mask=Data(), \optional{eta=1, \optional{surface_stress=Data(), \optional{stress=Data()}}}}}}
137     assigns values to the model parameters. In any call all values must be set.
138     \var{f} defines the external force $f$, \var{eta} the viscosity $\eta$,
139     \var{surface_stress} the surface stress $s$ and \var{stress} the initial stress $\sigma$.
140     The locations and compontents where the velocity is fixed are set by
141     the values of \var{fixed_u_mask}. The method will try to cast the given values to appropriate
142     \Data class objects.
143     \end{methoddesc}
144 gross 1878
145 gross 2100 \begin{methoddesc}[StokesProblemCartesian]{solve}{v,p,
146 gross 2251 \optional{max_iter=20, \optional{verbose=False, \optional{usePCG=True}}}}
147 gross 2100 solves the problem and return approximations for velocity and pressure.
148     The arguments \var{v} and \var{p} define initial guess. The values of \var{v} marked
149     by \var{fixed_u_mask} remain unchanged.
150 gross 2251 If \var{usePCG} is set to \True
151     reconditioned conjugate gradient method (PCG) \index{preconditioned conjugate gradient method!PCG} scheme is used. Otherwise the problem is solved generalized minimal residual method (GMRES) \index{generalized minimal residual method!GMRES}. In most cases
152     the PCG scheme is more efficient.
153 gross 2100 \var{max_iter} defines the maximum number of iteration steps.
154     If \var{verbose} is set to \True informations on the progress of of the solver are printed.
155     \end{methoddesc}
156 lgraham 1702
157    
158 gross 2251 \begin{methoddesc}[StokesProblemCartesian]{setTolerance}{\optional{tolerance=1.e-4}}
159 gross 2100 sets the tolerance in an appropriate norm relative to the right hand side. The tolerance must be non-negative and less than 1.
160     \end{methoddesc}
161     \begin{methoddesc}[StokesProblemCartesian]{getTolerance}{}
162     returns the current relative tolerance.
163     \end{methoddesc}
164     \begin{methoddesc}[StokesProblemCartesian]{setAbsoluteTolerance}{\optional{tolerance=0.}}
165     sets the absolute tolerance for the error in the relevant norm. The tolerance must be non-negative. Typically the
166     absolute talerance is set to 0.
167     \end{methoddesc}
168     \begin{methoddesc}[StokesProblemCartesian]{getAbsoluteTolerance}{}
169     sreturns the current absolute tolerance.
170     \end{methoddesc}
171 gross 2251 \begin{methoddesc}[StokesProblemCartesian]{setSubProblemTolerance}{\optional{rtol=None}}
172     sets the tolerance to solve the involved PDEs. The subtolerance \var{rtol} should not be choosen to large
173     in order to avoid feed back of errors in the subproblem solution into the outer iteration.
174     On the otherhand is choosen to small compute time is wasted.
175     If \var{rtol} is set to \var{None} the sub-tolerance is set automatically depending on the
176     tolerance choosen for the oter iteration.
177 gross 2100 \end{methoddesc}
178 gross 2251 \begin{methoddesc}[StokesProblemCartesian]{getSubProblemTolerance}{}
179     return the tolerance for the involved PDEs.
180 gross 2100 \end{methoddesc}
181 lgraham 1702
182 gross 2100 \subsection{Example: Lit Driven Cavity}
183     The following script \file{lit\hackscore driven\hackscore cavity.py}
184     \index{scripts!\file{helmholtz.py}} which is available in the \ExampleDirectory
185     illustrates the usage of the \class{StokesProblemCartesian} class to solve
186 gross 2371 the lit driven cavity problem:
187 gross 2100 \begin{python}
188     from esys.escript import *
189     from esys.finley import Rectangle
190     from esys.escript.models import StokesProblemCartesian
191     NE=25
192     dom = Rectangle(NE,NE,order=2)
193     x = dom.getX()
194     sc=StokesProblemCartesian(dom)
195     mask= (whereZero(x[0])*[1.,0]+whereZero(x[0]-1))*[1.,0] + \
196     (whereZero(x[1])*[0.,1.]+whereZero(x[1]-1))*[1.,1]
197     sc.initialize(eta=.1, fixed_u_mask= mask)
198     v=Vector(0.,Solution(dom))
199     v[0]+=whereZero(x[1]-1.)
200     p=Scalar(0.,ReducedSolution(dom))
201     v,p=sc.solve(v,p, verbose=True)
202     saveVTK("u.xml",velocity=v,pressure=p)
203     \end{python}
204 lgraham 1702
205 gross 2100 \section{Darcy Flux}
206 gross 2108 We want to calculate the velocity $u$ and pressure $p$ on a domain $\Omega$ solving
207     the Darcy flux problem \index{Darcy flux}\index{Darcy flow}
208 gross 2100 \begin{equation}\label{DARCY PROBLEM}
209     \begin{array}{rcl}
210     u\hackscore{i} + \kappa\hackscore{ij} p\hackscore{,j} & = & g\hackscore{i} \\
211     u\hackscore{k,k} & = & f
212     \end{array}
213     \end{equation}
214     with the boundary conditions
215     \begin{equation}\label{DARCY BOUNDARY}
216     \begin{array}{rcl}
217     u\hackscore{i} \; n\hackscore{i} = u^{N}\hackscore{i} \; n\hackscore{i} & \mbox{ on } & \Gamma\hackscore{N} \\
218     p = p^{D} & \mbox{ on } & \Gamma\hackscore{D} \\
219     \end{array}
220     \end{equation}
221     where $\Gamma\hackscore{N}$ and $\Gamma\hackscore{D}$ are a partition of the boundary of $\Omega$ with $\Gamma\hackscore{D}$ non empty, $n\hackscore{i}$ is the outer normal field of the boundary of $\Omega$, $u^{N}\hackscore{i}$ and $p^{D}$ are given functions on $\Omega$, $g\hackscore{i}$ and $f$ are given source terms and $\kappa\hackscore{ij}$ is the given permability. We assume that $\kappa\hackscore{ij}$ is symmetric (which is not really required) and positive definite, i.e there are positive constants $\alpha\hackscore{0}$ and $\alpha\hackscore{1}$ wich are independent from the location in $\Omega$ such that
222     \begin{equation}
223     \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}
224     \end{equation}
225     for all $x\hackscore{i}$.
226 lgraham 1702
227    
228 gross 2100 \subsection{Solution Method \label{DARCY SOLVE}}
229 gross 2156 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 lgraham 1702 \begin{equation}
231 gross 2156 -(\kappa\hackscore{ki}\kappa\hackscore{kj} p\hackscore{,j}^{ref})\hackscore{,i} = - (\kappa\hackscore{ki} (g\hackscore{k}- u^{N}\hackscore{k}))\hackscore{,i}
232     \mbox{ with }
233     p^{ref} = p^{D} \mbox{ on } \Gamma\hackscore{D}
234     \end{equation}
235     With setting $u \leftarrow u-u^{N}$ and $p \leftarrow p-p^{ref}$ and
236     \begin{equation}
237 gross 2100 \begin{array}{rcl}
238 gross 2156 g\hackscore{i} & \leftarrow & g\hackscore{i} - u^{N}\hackscore{i} - \kappa\hackscore{ij} p^{ref}\hackscore{,j }\\
239 gross 2100 f & \leftarrow & f - u^{N}\hackscore{k,k}
240     \end{array}
241     \end{equation}
242 gross 2156 we can assume that $u^{N}\hackscore{i} \; n\hackscore{i}=0$ and
243 gross 2208 $p^{D}=0$.
244 gross 2100 We set
245     \begin{equation}
246     V = \{ q \in H^{1}(\Omega) : q=0 \mbox{ on } \Gamma\hackscore{D} \}
247     \end{equation}
248     and
249     \begin{equation}
250     W = \{ v \in (L^2(\Omega))^{d} : v\hackscore{k,k} \in L^2(\Omega) \mbox{ and } u\hackscore{i} \; n\hackscore{i} =0 \mbox{ on } \Gamma\hackscore{N} \}
251     \end{equation}
252     and define the operator $Q: V \rightarrow (L^2(\Omega))^{d}$ defined by
253     \begin{equation}
254     (Qp)\hackscore{i} = \kappa\hackscore{ij} p\hackscore{,j}
255     \end{equation}
256     and the operator $D: W \rightarrow L^2(\Omega)$ defined by
257     \begin{equation}
258     Dv = v\hackscore{k,k}
259     \end{equation}
260     In operator notation the Darcy problem~\ref{DARCY PROBLEM} is written in the form
261     \begin{equation}
262     \begin{array}{rcl}
263     u + Qp & = & g \\
264     Du & = & f
265     \end{array}
266     \end{equation}
267     We solve this equation by minimising the functional
268     \begin{equation}
269     J(u,p):=\|u + Qp - g\|^2\hackscore{0} + \|Du-f\|\hackscore{0}^2
270     \end{equation}
271     over $W \times V$ where $\|.\|\hackscore{0}$ denotes the norm in $L^2(\Omega)$. A simple calculation shows that
272     one has to solve
273     \begin{equation}
274     ( v + Qq , u + Qp - g) + (Dv,Du-f) =0
275     \end{equation}
276     for all $v\in W$ and $q \in V$.which translates back into operator notation
277     \begin{equation}
278     \begin{array}{rcl}
279     (I+D^*D)u + Qp & = & D^*f + g \\
280 gross 2208 Q^*u + Q^*Q p & = & Q^*g \\
281 gross 2100 \end{array}
282     \end{equation}
283 gross 2208 where $D^*$ and $Q^*$ denote the adjoint operators.
284 gross 2371 In~\cite{LEASTSQUARESFEM1994} it has been shown that this problem is continuous and coercive in $W \times V$ and therefore has a unique solution. Also standart FEM methods can be used for discretization. It is also possible
285 gross 2100 to solve the problem is coupled form, however this approach leads in some cases to a very ill-conditioned stiffness matrix in particular in the case of a very small or large permability ($\alpha\hackscore{1} \ll 1$ or $\alpha\hackscore{0} \gg 1$)
286    
287     The approach we are taking is to eliminate the velocity $u$ from the problem. Assuming that $p$ is known we have
288 gross 2264 \begin{equation}\label{DARCY V FORM}
289 gross 2100 v= (I+D^*D)^{-1} (D^*f + g - Qp)
290     \end{equation}
291     (notice that $(I+D^*D)$ is coercive in $W$) which is inserted into the second equation
292     \begin{equation}
293 gross 2208 Q^* (I+D^*D)^{-1} (D^*f + g - Qp) + Q^* Q p = Q^*g
294 gross 2100 \end{equation}
295     which is
296     \begin{equation}
297 gross 2208 Q^* ( I - (I+D^*D)^{-1} ) Q p = Q^* (g-(I+D^*D)^{-1} (D^*f + g) )
298 gross 2100 \end{equation}
299 gross 2208 We use the PCG \index{linear solver!PCG}\index{PCG} method to solve this. The residual $r$ ($\in V^*$) is given as
300 gross 2100 \begin{equation}
301     \begin{array}{rcl}
302 gross 2208 r & = & Q^* \left( g -(I+D^*D)^{-1} (D^*f + g) - Qp + (I+D^*D)^{-1}Q p \right)\\
303 gross 2264 & =& Q^* \left( g- Qp - (I+D^*D)^{-1} (D^*f + g - Qp) \right) \\
304 gross 2208 & =& Q^* \left( g - Qp - v \right)
305 gross 2100 \end{array}
306     \end{equation}
307 gross 2264 So in a partical implementation we use $\hat{r}=g-Qp-v$ to represent the residual.
308     The evaluation of the iteration operator for a given $p$ is then
309     returning $Qp+v$ where $v$ is the solution of
310 gross 2100 \begin{equation}\label{UPDATE W}
311 gross 2264 (I+D^*D)v = Qp
312 gross 2100 \end{equation}
313 gross 2264 We use $(Q^*Q)^{-1}$ as a preconditioner for the iteration operator $Q^* ( I - (I+D^*D)^{-1} ) Q$. So the application of the preconditioner to $\hat{r}$ representing the residual is given by solving
314     implemented by solving
315     \begin{equation}\label{UPDATE P}
316     Q^*Q q = Q^*\hat{r}
317 gross 2208 \end{equation}
318 gross 2264 The residual norm used in the PCG is given as
319     \begin{equation}\label{DARCY R NORM}
320     \|r\|\hackscore{PCG}^2 = \int r \cdot (Q^*Q)^{-1} r \; dx =\int \hat{r} \cdot Q (Q^*Q)^{-1} Q^* \hat{r} \; dx \approx
321     \|\hat{r}\|\hackscore{0}^2
322 gross 2208 \end{equation}
323 gross 2264 The iteration is terminated if
324     \begin{equation}\label{DARCY STOP}
325     \|r\|\hackscore{PCG} \le \mbox{ATOL}
326 gross 2208 \end{equation}
327 gross 2264 where we set
328     \begin{equation}\label{DARCY ATOL DEF}
329     \mbox{ATOL} = \mbox{atol} + \mbox{rtol} \cdot \left(\frac{1}{\|v\|\hackscore{0}} + \frac{1}{\|Qp\|\hackscore{0}} \right)^{-1}
330 gross 2208 \end{equation}
331 gross 2264 where rtol is a given relative tolerance and $\mbox{atol}$ is a given absolute tolerance (typically $=0$).
332     Notice that if $Qp$ and $v$ both are zero, the pair $(0,p)$ is a solution.
333     The problem is that ATOL is depending on the solution $p$ (and $v$ calculated form~\ref{DARCY V FORM}). In partcice one use the initial guess for $p$
334     to get a first value for ATOL. If the stopping crierion is met in the PCG iteration, a new $v$ is calculated from the current pressure approximation and ATOL is recalculated. If \ref{DARCY STOP} is still fullfilled the calculation is terminated and $(v,p)$ is returned. Otherwise PCG is restarted with a new ATOL.
335 gross 2208
336 gross 2100 \subsection{Functions}
337 gross 2108 \begin{classdesc}{DarcyFlow}{domain}
338     opens the Darcy flux problem\index{Darcy flux} on the \Domain domain.
339     \end{classdesc}
340 gross 2208 \begin{methoddesc}[DarcyFlow]{setValue}{\optional{f=None, \optional{g=None, \optional{location_of_fixed_pressure=None, \optional{location_of_fixed_flux=None, \optional{permeability=None}}}}}}
341     assigns values to the model parameters. Values can be assigned using various calls - in particular
342     in a time dependend problem only values that change over time needs to be reset. The permability can be defined as scalar (isotropic), a vector (orthotropic) or a matrix (anisotropic).
343     \var{f} and \var{g} are the corresponding parameters in~\ref{DARCY PROBLEM}.
344     The locations and compontents where the flux is prescribed are set by positive values in
345     \var{location_of_fixed_flux}.
346     The locations where the pressure is prescribed are set by
347     by positive values of \var{location_of_fixed_pressure}.
348     The values of the pressure and flux are defined by the initial guess.
349     Notice that at any point on the boundary of the domain the pressure or the normal component of
350     the flux must be defined. There must be at least one point where the pressure is prescribed.
351     The method will try to cast the given values to appropriate
352 gross 2108 \Data class objects.
353     \end{methoddesc}
354    
355 gross 2264 \begin{methoddesc}[DarcyFlow]{setTolerance}{\optional{rtol=1e-4}}
356     sets the relative tolerance \mbox{rtol} in \ref{DARCY ATOL DEF}.
357 gross 2208 \end{methoddesc}
358 gross 2108
359 gross 2264 \begin{methoddesc}[DarcyFlow]{setAbsoluteTolerance}{\optional{atol=0.}}
360     sets the absolute tolerance \mbox{atol} in \ref{DARCY ATOL DEF}.
361     \end{methoddesc}
362 gross 2208
363 gross 2264 \begin{methoddesc}[DarcyFlow]{setSubProblemTolerance}{\optional{rtol=None}}
364     sets the relative tolerance used to solve the involved PDEs. If no argument is given,
365     the square of the current relative tolerance is used. The sub-problem tolerance should be choosen as large as possible to minimize the compute time. However, a too large value for the sub-problem tolerance may lead to slow convergence or even dibergence in the outer iteration.
366     \end{methoddesc}
367 gross 2208
368     \begin{methoddesc}[DarcyFlow]{solve}{u0,p0, \optional{max_iter=100, \optional{verbose=False \optional{sub_rtol=1.e-8}}}}
369     solves the problem. and returns approximations for the flux $v$ and the pressure $p$.
370     \var{u0} and \var{p0} define initial guess for flux and pressure. Values marked
371     by positive values \var{location_of_fixed_flux} and \var{location_of_fixed_pressure}, respectively, are kept unchanged.
372     \end{methoddesc}
373    
374    
375 gross 2100 \subsection{Example: Gravity Flow}
376 gross 2208 later
377 gross 2100
378     %================================================
379 gross 2208 % \section{Temperature Advection Diffusion\label{TEMP ADV DIFF}}
380 gross 2100
381 gross 2208 %\begin{equation}
382     % \rho c\hackscore{p} \left (\frac{\partial T}{\partial t} + \vec{v} \cdot \nabla T \right ) = k \nabla^{2}T
383     % \label{HEAT EQUATION}
384     % \end{equation}
385 lgraham 1702
386 gross 2208 % 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.
387 lgraham 1702
388 gross 2208 % \subsection{Description}
389 lgraham 1702
390 gross 2208 % \subsection{Method}
391     %
392     % \begin{classdesc}{TemperatureCartesian}{dom,theta=THETA,useSUPG=SUPG}
393     % \end{classdesc}
394 lgraham 1702
395 gross 2208 % \subsection{Benchmark Problem}
396 gross 2100 %===============================================================================================================
397 lgraham 1702
398 gross 2100 %=========================================================
399 lgraham 2138 \input{levelsetmodel}
400 lgraham 1702
401 gross 1859 % \section{Drucker Prager Model}
402 lgraham 1702
403 gross 1841 \section{Isotropic Kelvin Material \label{IKM}}
404 gross 2415 As proposed by Kelvin~\cite{Muhlhaus2005} material strain $D\hackscore{ij}=\frac{1}{2}(v\hackscore{i,j}+v\hackscore{j,i})$ can be decomposed into
405 gross 1859 an elastic part $D\hackscore{ij}^{el}$ and visco-plastic part $D\hackscore{ij}^{vp}$:
406 gross 1841 \begin{equation}\label{IKM-EQU-2}
407 gross 1859 D\hackscore{ij}=D\hackscore{ij}^{el}+D\hackscore{ij}^{vp}
408 gross 1841 \end{equation}
409 gross 1859 with the elastic strain given as
410 gross 1841 \begin{equation}\label{IKM-EQU-3}
411 gross 2252 D\hackscore{ij}^{el'}=\frac{1}{2 \mu} \dot{\sigma}\hackscore{ij}'
412 gross 1841 \end{equation}
413 gross 1859 where $\sigma'\hackscore{ij}$ is the deviatoric stress (Notice that $\sigma'\hackscore{ii}=0$).
414     If the material is composed by materials $q$ the visco-plastic strain can be decomposed as
415 gross 1841 \begin{equation}\label{IKM-EQU-4}
416 gross 2252 D\hackscore{ij}^{vp'}=\sum\hackscore{q} D\hackscore{ij}^{q'}
417 gross 1841 \end{equation}
418 gross 1859 where $D\hackscore{ij}^{q}$ is the strain in material $q$ given as
419 gross 1841 \begin{equation}\label{IKM-EQU-5}
420 gross 2252 D\hackscore{ij}^{q'}=\frac{1}{2 \eta^{q}} \sigma'\hackscore{ij}
421 gross 1841 \end{equation}
422 gross 1859 where $\eta^{q}$ is the viscosity of material $q$. We assume the following
423     betwee the the strain in material $q$
424     \begin{equation}\label{IKM-EQU-5b}
425     \eta^{q}=\eta^{q}\hackscore{N} \left(\frac{\tau}{\tau\hackscore{t}^q}\right)^{\frac{1}{n^{q}}-1}
426     \mbox{ with } \tau=\sqrt{\frac{1}{2}\sigma'\hackscore{ij} \sigma'\hackscore{ij}}
427     \end{equation}
428 gross 2371 for a given power law coefficients $n^{q}\ge1$ and transition stresses $\tau\hackscore{t}^q$, see~\cite{Muhlhaus2005}.
429 gross 1859 Notice that $n^{q}=1$ gives a constant viscosity.
430 gross 1841 After inserting equation~\ref{IKM-EQU-5} into equation \ref{IKM-EQU-4} one gets:
431 gross 1859 \begin{equation}\label{IKM-EQU-6}
432 gross 2100 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}} \;.
433 gross 1841 \end{equation}
434 gross 2300 and finally with~\ref{IKM-EQU-2}
435 gross 2371 \begin{equation}\label{IKM-EQU-2bb}
436 gross 2300 D\hackscore{ij}'=\frac{1}{2 \eta^{vp}} \sigma'\hackscore{ij}+\frac{1}{2 \mu} \dot{\sigma}\hackscore{ij}'
437 gross 1859 \end{equation}
438 gross 2300 The total stress $\tau$ needs to fullfill the yield condition \index{yield condition}
439 gross 1859 \begin{equation}\label{IKM-EQU-8c}
440     \tau \le \tau\hackscore{Y} + \beta \; p
441     \end{equation}
442 gross 2300 with the Drucker-Prager \index{Druck-Prager} cohesion factor \index{cohesion factor} $\tau\hackscore{Y}$, Drucker-Prager friction $\beta$ and total pressure $p$.
443 gross 1859 The deviatoric stress needs to fullfill the equilibrion equation
444 gross 1841 \begin{equation}\label{IKM-EQU-1}
445 gross 1859 -\sigma'\hackscore{ij,j}+p\hackscore{,i}=F\hackscore{i}
446 gross 1841 \end{equation}
447 gross 1859 where $F\hackscore{j}$ is a given external fource. We assume an incompressible media:
448 gross 2371 \begin{equation}\label{IKM-EQU-2bbb}
449 gross 1859 -v\hackscore{i,i}=0
450 gross 1841 \end{equation}
451 gross 1878 Natural boundary conditions are taken in the form
452     \begin{equation}\label{IKM-EQU-Boundary}
453     \sigma'\hackscore{ij}n\hackscore{j}-n\hackscore{i}p=f
454     \end{equation}
455     which can be overwritten by a constraint
456     \begin{equation}\label{IKM-EQU-Boundary0}
457     v\hackscore{i}(x)=0
458     \end{equation}
459     where the index $i$ may depend on the location $x$ on the bondary.
460    
461     \subsection{Solution Method \label{IKM-SOLVE}}
462     By using a first order finite difference approximation wit step size $dt>0$~\ref{IKM-EQU-3} get the form
463     \begin{equation}\label{IKM-EQU-3b}
464 gross 2300 \dot{\sigma}\hackscore{ij}=\frac{1}{dt } \left( \sigma\hackscore{ij} - \sigma\hackscore{ij}^{-} \right)
465 gross 1878 \end{equation}
466 gross 2300 and
467     \begin{equation}\label{IKM-EQU-2b}
468 gross 2415 D\hackscore{ij}'=\left(\frac{1}{2 \eta^{vp}} + \frac{1}{2 \mu dt}\right) \sigma\hackscore{ij}'-\frac{1}{2 \mu dt } \sigma\hackscore{ij}^{-'}
469 gross 2300 \end{equation}
470     where $\sigma\hackscore{ij}^{-}$ is the stress at the precious time step. With
471     \begin{equation}\label{IKM-EQU-2c}
472 gross 2415 \dot{\gamma} = \sqrt{ 2 \left( D\hackscore{ij}' +
473     \frac{1}{ 2 \mu \; dt} \sigma\hackscore{ij}^{-'}\right)^2}
474 gross 2300 \end{equation}
475     we have
476     \begin{equation}
477     \tau = \eta\hackscore{eff} \cdot \dot{\gamma}
478     \end{equation}
479     where
480     \begin{equation}
481     \eta\hackscore{eff}= min( \left(\frac{1}{\mu \; dt}+\frac{1}{\eta^{vp}}\right)^{-1}
482     , \eta\hackscore{max}) \mbox{ with }
483     \eta\hackscore{max} = \left\{
484     \begin{array}{rcl}
485     \frac{\tau\hackscore{Y} + \beta \; p}{\dot{\gamma}} & & \dot{\gamma}>0 \\
486     &\mbox{ if } \\
487     \infty & & \mbox{otherwise}
488     \end{array}
489     \right.
490     \end{equation}
491     The upper bound $\eta\hackscore{max}$ makes sure that yield condtion~\ref{IKM-EQU-8c} holds. With this setting the eqaution \ref{IKM-EQU-2b} takes the form
492 gross 1878 \begin{equation}\label{IKM-EQU-10}
493 gross 2415 \sigma\hackscore{ij}' = 2 \eta\hackscore{eff} \left( D\hackscore{ij}' +
494 gross 2300 \frac{1}{ 2 \mu \; dt} \sigma\hackscore{ij}^{'-}\right)
495 gross 1878 \end{equation}
496 gross 1859 After inserting~\ref{IKM-EQU-10} into~\ref{IKM-EQU-1} we get
497     \begin{equation}\label{IKM-EQU-1ib}
498 gross 2100 -\left(\eta\hackscore{eff} (v\hackscore{i,j}+ v\hackscore{i,j})
499 gross 2415 \right)\hackscore{,j}+p\hackscore{,i}=F\hackscore{i}+
500     \left(\frac{\eta\hackscore{eff}}{\mu dt } \sigma\hackscore{ij}^{'-} \right)\hackscore{,j}
501 gross 1859 \end{equation}
502 gross 2252 Combining this with the incomressibilty condition~\ref{IKM-EQU-2} we need to solve a
503     Stokes problem as discussed in section~\ref{STOKES SOLVE} in each time step.
504 gross 2300
505     If we set
506     \begin{equation}\label{IKM-EQU-44}
507     \frac{1}{\eta(\tau)}= \frac{1}{\mu \; dt}+\frac{1}{\eta^{vp}}
508     \end{equation}
509     we need to solve the nonlinear problem
510 gross 1878 \begin{equation}
511 gross 2300 \eta\hackscore{eff} - min(\eta( \dot{\gamma} \cdot \eta\hackscore{eff})
512     , \eta\hackscore{max}) =0
513     \end{equation}
514     We use the Newton-Raphson Scheme \index{Newton-Raphson scheme} to solve this problem
515     \begin{equation}\label{IKM-EQU-45}
516     \eta\hackscore{eff}^{(n+1)} = \min(\eta\hackscore{max},
517     \eta\hackscore{eff}^{(n)} -
518     \frac{\eta\hackscore{eff}^{(n)} - \eta( \tau^{(n)}) }
519     {1-\dot{\gamma} \cdot \eta'( \tau^{(n)} )} )
520     =\min(\eta\hackscore{max},
521     \frac{\eta( \tau^{(n)}) -\tau^{(n)} \cdot \eta'( \tau^{(n)} ) }
522     {1-\dot{\gamma} \cdot \eta'( \tau^{(n)} )} )
523 gross 2100 \end{equation}
524 gross 2300 where $\eta'$ denotes the derivative of $\eta$ with respect of $\tau$
525     and $\tau^{(n)} = \dot{\gamma} \cdot \eta\hackscore{eff}^{(n)}$.
526    
527     Looking at the evaluation of $\eta$ in~\ref{IKM-EQU-44} it makes sense formulate
528     the iteration~\ref{IKM-EQU-45} using $\Theta=\eta^{-1}$.
529     In fact we have
530 gross 2100 \begin{equation}
531 gross 2300 \eta' = - \frac{\Theta'}{\Theta^2}
532 gross 2100 \mbox{ with }
533 gross 2300 \Theta' = \sum\hackscore{q} \left(\frac{1}{\eta^{q}} \right)'
534 gross 2100 \label{IKM iteration 7}
535     \end{equation}
536 gross 2300 As
537     \begin{equation}\label{IKM-EQU-47}
538 gross 2100 \left(\frac{1}{\eta^{q}} \right)'
539     = \frac{1-\frac{1}{n^{q}}}{\eta^{q}\hackscore{N}} \frac{\tau^{-\frac{1}{n^{q}}}}{(\tau\hackscore{t}^q)^{1-\frac{1}{n^{q}}}}
540 gross 2300 = \frac{1-\frac{1}{n^{q}}}{\eta^{q}}\frac{1}{\tau}
541 gross 2100 \end{equation}
542 gross 2300 we have
543     \begin{equation}\label{IKM-EQU-48}
544     \Theta' = \frac{1}{\tau} \omega \mbox{ with } \omega = \sum\hackscore{q}\frac{1-\frac{1}{n^{q}}}{\eta^{q}}
545     \end{equation}
546     which leads to
547 gross 2371 \begin{equation}\label{IKM-EQU-49}
548 gross 2300 \eta\hackscore{eff}^{(n+1)} = \min(\eta\hackscore{max},
549     \eta\hackscore{eff}^{(n)}
550     \frac{\Theta^{(n)} + \omega^{(n)} }
551     {\eta\hackscore{eff}^{(n)} \Theta^{(n)^2}+\omega^{(n)} })
552     \end{equation}

  ViewVC Help
Powered by ViewVC 1.1.26