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

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

Parent Directory Parent Directory | Revision Log Revision Log


Revision 2548 - (hide annotations)
Mon Jul 20 06:20:06 2009 UTC (10 years, 7 months ago) by jfenwick
File MIME type: application/x-tex
File size: 30253 byte(s)
Updating copyright notices
1 ksteube 1811
2     %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
3 lgraham 1702 %
4 jfenwick 2548 % Copyright (c) 2003-2009 by University of Queensland
5 ksteube 1811 % 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 2502 \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{j}
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 gross 2474 \begin{equation} \label{P PREC}
85 gross 2100 \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 2474 \begin{classdesc}{StokesProblemCartesian}{domain\optional{, adaptSubTolerance=\True}}
132 gross 2100 opens the Stokes problem\index{Stokes problem} on the \Domain domain. The approximation
133     order needs to be two.
134 gross 2474 If \var{adaptSubTolerance} is \True
135     the tolerances for all subproblems are set automatically.
136 gross 2100 \end{classdesc}
137 gross 1878
138 gross 2100 \begin{methoddesc}[StokesProblemCartesian]{initialize}{\optional{f=Data(), \optional{fixed_u_mask=Data(), \optional{eta=1, \optional{surface_stress=Data(), \optional{stress=Data()}}}}}}
139     assigns values to the model parameters. In any call all values must be set.
140     \var{f} defines the external force $f$, \var{eta} the viscosity $\eta$,
141     \var{surface_stress} the surface stress $s$ and \var{stress} the initial stress $\sigma$.
142     The locations and compontents where the velocity is fixed are set by
143     the values of \var{fixed_u_mask}. The method will try to cast the given values to appropriate
144     \Data class objects.
145     \end{methoddesc}
146 gross 1878
147 gross 2474 \begin{methoddesc}[StokesProblemCartesian]{solve}{v,p
148     \optional{, max_iter=100 \optional{, verbose=False \optional{, usePCG=True }}}}
149 gross 2100 solves the problem and return approximations for velocity and pressure.
150     The arguments \var{v} and \var{p} define initial guess. The values of \var{v} marked
151     by \var{fixed_u_mask} remain unchanged.
152 gross 2251 If \var{usePCG} is set to \True
153     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
154     the PCG scheme is more efficient.
155 gross 2100 \var{max_iter} defines the maximum number of iteration steps.
156 gross 2474
157 gross 2100 If \var{verbose} is set to \True informations on the progress of of the solver are printed.
158     \end{methoddesc}
159 lgraham 1702
160    
161 gross 2251 \begin{methoddesc}[StokesProblemCartesian]{setTolerance}{\optional{tolerance=1.e-4}}
162 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.
163     \end{methoddesc}
164     \begin{methoddesc}[StokesProblemCartesian]{getTolerance}{}
165     returns the current relative tolerance.
166     \end{methoddesc}
167     \begin{methoddesc}[StokesProblemCartesian]{setAbsoluteTolerance}{\optional{tolerance=0.}}
168     sets the absolute tolerance for the error in the relevant norm. The tolerance must be non-negative. Typically the
169     absolute talerance is set to 0.
170     \end{methoddesc}
171 gross 2474
172 gross 2100 \begin{methoddesc}[StokesProblemCartesian]{getAbsoluteTolerance}{}
173     sreturns the current absolute tolerance.
174     \end{methoddesc}
175 gross 2474
176     \begin{methoddesc}[StokesProblemCartesian]{getSolverOptionsVelocity}{}
177     returns the solver options used solve the equations~(\ref{V CALC}) for velocity.
178 gross 2100 \end{methoddesc}
179 gross 2474
180     \begin{methoddesc}[StokesProblemCartesian]{getSolverOptionsPressure}{}
181 gross 2513 returns the solver options used solve the equation~(\ref{P PREC}) for pressure.
182 gross 2100 \end{methoddesc}
183 lgraham 1702
184 gross 2474 \begin{methoddesc}[StokesProblemCartesian]{getSolverOptionsDiv}{}
185     set the solver options for solving the equation to project the divergence of the velocity onto the function space of pressure.
186     \end{methoddesc}
187    
188    
189 gross 2100 \subsection{Example: Lit Driven Cavity}
190     The following script \file{lit\hackscore driven\hackscore cavity.py}
191     \index{scripts!\file{helmholtz.py}} which is available in the \ExampleDirectory
192     illustrates the usage of the \class{StokesProblemCartesian} class to solve
193 gross 2371 the lit driven cavity problem:
194 gross 2100 \begin{python}
195     from esys.escript import *
196     from esys.finley import Rectangle
197     from esys.escript.models import StokesProblemCartesian
198     NE=25
199     dom = Rectangle(NE,NE,order=2)
200     x = dom.getX()
201     sc=StokesProblemCartesian(dom)
202     mask= (whereZero(x[0])*[1.,0]+whereZero(x[0]-1))*[1.,0] + \
203     (whereZero(x[1])*[0.,1.]+whereZero(x[1]-1))*[1.,1]
204     sc.initialize(eta=.1, fixed_u_mask= mask)
205     v=Vector(0.,Solution(dom))
206     v[0]+=whereZero(x[1]-1.)
207     p=Scalar(0.,ReducedSolution(dom))
208     v,p=sc.solve(v,p, verbose=True)
209     saveVTK("u.xml",velocity=v,pressure=p)
210     \end{python}
211 lgraham 1702
212 gross 2100 \section{Darcy Flux}
213 gross 2108 We want to calculate the velocity $u$ and pressure $p$ on a domain $\Omega$ solving
214     the Darcy flux problem \index{Darcy flux}\index{Darcy flow}
215 gross 2100 \begin{equation}\label{DARCY PROBLEM}
216     \begin{array}{rcl}
217     u\hackscore{i} + \kappa\hackscore{ij} p\hackscore{,j} & = & g\hackscore{i} \\
218     u\hackscore{k,k} & = & f
219     \end{array}
220     \end{equation}
221     with the boundary conditions
222     \begin{equation}\label{DARCY BOUNDARY}
223     \begin{array}{rcl}
224     u\hackscore{i} \; n\hackscore{i} = u^{N}\hackscore{i} \; n\hackscore{i} & \mbox{ on } & \Gamma\hackscore{N} \\
225     p = p^{D} & \mbox{ on } & \Gamma\hackscore{D} \\
226     \end{array}
227     \end{equation}
228     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
229     \begin{equation}
230     \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}
231     \end{equation}
232     for all $x\hackscore{i}$.
233 lgraham 1702
234    
235 gross 2100 \subsection{Solution Method \label{DARCY SOLVE}}
236 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
237 lgraham 1702 \begin{equation}
238 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}
239     \mbox{ with }
240     p^{ref} = p^{D} \mbox{ on } \Gamma\hackscore{D}
241     \end{equation}
242     With setting $u \leftarrow u-u^{N}$ and $p \leftarrow p-p^{ref}$ and
243     \begin{equation}
244 gross 2100 \begin{array}{rcl}
245 gross 2156 g\hackscore{i} & \leftarrow & g\hackscore{i} - u^{N}\hackscore{i} - \kappa\hackscore{ij} p^{ref}\hackscore{,j }\\
246 gross 2100 f & \leftarrow & f - u^{N}\hackscore{k,k}
247     \end{array}
248     \end{equation}
249 gross 2156 we can assume that $u^{N}\hackscore{i} \; n\hackscore{i}=0$ and
250 gross 2208 $p^{D}=0$.
251 gross 2100 We set
252     \begin{equation}
253     V = \{ q \in H^{1}(\Omega) : q=0 \mbox{ on } \Gamma\hackscore{D} \}
254     \end{equation}
255     and
256     \begin{equation}
257     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} \}
258     \end{equation}
259     and define the operator $Q: V \rightarrow (L^2(\Omega))^{d}$ defined by
260     \begin{equation}
261     (Qp)\hackscore{i} = \kappa\hackscore{ij} p\hackscore{,j}
262     \end{equation}
263     and the operator $D: W \rightarrow L^2(\Omega)$ defined by
264     \begin{equation}
265     Dv = v\hackscore{k,k}
266     \end{equation}
267     In operator notation the Darcy problem~\ref{DARCY PROBLEM} is written in the form
268     \begin{equation}
269     \begin{array}{rcl}
270     u + Qp & = & g \\
271     Du & = & f
272     \end{array}
273     \end{equation}
274     We solve this equation by minimising the functional
275     \begin{equation}
276     J(u,p):=\|u + Qp - g\|^2\hackscore{0} + \|Du-f\|\hackscore{0}^2
277     \end{equation}
278     over $W \times V$ where $\|.\|\hackscore{0}$ denotes the norm in $L^2(\Omega)$. A simple calculation shows that
279     one has to solve
280     \begin{equation}
281     ( v + Qq , u + Qp - g) + (Dv,Du-f) =0
282     \end{equation}
283     for all $v\in W$ and $q \in V$.which translates back into operator notation
284     \begin{equation}
285     \begin{array}{rcl}
286     (I+D^*D)u + Qp & = & D^*f + g \\
287 gross 2208 Q^*u + Q^*Q p & = & Q^*g \\
288 gross 2100 \end{array}
289     \end{equation}
290 gross 2208 where $D^*$ and $Q^*$ denote the adjoint operators.
291 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
292 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 $\alpha\hackscore{0} \gg 1$)
293    
294     The approach we are taking is to eliminate the velocity $u$ from the problem. Assuming that $p$ is known we have
295 gross 2264 \begin{equation}\label{DARCY V FORM}
296 gross 2100 v= (I+D^*D)^{-1} (D^*f + g - Qp)
297     \end{equation}
298     (notice that $(I+D^*D)$ is coercive in $W$) which is inserted into the second equation
299     \begin{equation}
300 gross 2208 Q^* (I+D^*D)^{-1} (D^*f + g - Qp) + Q^* Q p = Q^*g
301 gross 2100 \end{equation}
302     which is
303     \begin{equation}
304 gross 2208 Q^* ( I - (I+D^*D)^{-1} ) Q p = Q^* (g-(I+D^*D)^{-1} (D^*f + g) )
305 gross 2100 \end{equation}
306 gross 2208 We use the PCG \index{linear solver!PCG}\index{PCG} method to solve this. The residual $r$ ($\in V^*$) is given as
307 gross 2100 \begin{equation}
308     \begin{array}{rcl}
309 gross 2208 r & = & Q^* \left( g -(I+D^*D)^{-1} (D^*f + g) - Qp + (I+D^*D)^{-1}Q p \right)\\
310 gross 2264 & =& Q^* \left( g- Qp - (I+D^*D)^{-1} (D^*f + g - Qp) \right) \\
311 gross 2208 & =& Q^* \left( g - Qp - v \right)
312 gross 2100 \end{array}
313     \end{equation}
314 gross 2264 So in a partical implementation we use $\hat{r}=g-Qp-v$ to represent the residual.
315     The evaluation of the iteration operator for a given $p$ is then
316     returning $Qp+v$ where $v$ is the solution of
317 gross 2100 \begin{equation}\label{UPDATE W}
318 gross 2264 (I+D^*D)v = Qp
319 gross 2100 \end{equation}
320 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
321     implemented by solving
322     \begin{equation}\label{UPDATE P}
323     Q^*Q q = Q^*\hat{r}
324 gross 2208 \end{equation}
325 gross 2264 The residual norm used in the PCG is given as
326     \begin{equation}\label{DARCY R NORM}
327     \|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
328     \|\hat{r}\|\hackscore{0}^2
329 gross 2208 \end{equation}
330 gross 2264 The iteration is terminated if
331     \begin{equation}\label{DARCY STOP}
332     \|r\|\hackscore{PCG} \le \mbox{ATOL}
333 gross 2208 \end{equation}
334 gross 2264 where we set
335     \begin{equation}\label{DARCY ATOL DEF}
336     \mbox{ATOL} = \mbox{atol} + \mbox{rtol} \cdot \left(\frac{1}{\|v\|\hackscore{0}} + \frac{1}{\|Qp\|\hackscore{0}} \right)^{-1}
337 gross 2208 \end{equation}
338 gross 2264 where rtol is a given relative tolerance and $\mbox{atol}$ is a given absolute tolerance (typically $=0$).
339     Notice that if $Qp$ and $v$ both are zero, the pair $(0,p)$ is a solution.
340     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 $p$
341     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.
342 gross 2208
343 gross 2100 \subsection{Functions}
344 gross 2474 \begin{classdesc}{DarcyFlow}{domain \optional{, adaptSubTolerance=\True}}
345 gross 2108 opens the Darcy flux problem\index{Darcy flux} on the \Domain domain.
346 gross 2474 If \var{adaptSubTolerance} is set to \True,
347     the relative tolerances for solving~(\ref{DARCY V FORM}),~(\ref{UPDATE W})
348     and~(\ref{UPDATE P}) are set automatically.
349 gross 2108 \end{classdesc}
350 gross 2474
351 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}}}}}}
352     assigns values to the model parameters. Values can be assigned using various calls - in particular
353     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).
354     \var{f} and \var{g} are the corresponding parameters in~\ref{DARCY PROBLEM}.
355     The locations and compontents where the flux is prescribed are set by positive values in
356     \var{location_of_fixed_flux}.
357     The locations where the pressure is prescribed are set by
358     by positive values of \var{location_of_fixed_pressure}.
359     The values of the pressure and flux are defined by the initial guess.
360     Notice that at any point on the boundary of the domain the pressure or the normal component of
361     the flux must be defined. There must be at least one point where the pressure is prescribed.
362     The method will try to cast the given values to appropriate
363 gross 2108 \Data class objects.
364     \end{methoddesc}
365    
366 gross 2264 \begin{methoddesc}[DarcyFlow]{setTolerance}{\optional{rtol=1e-4}}
367     sets the relative tolerance \mbox{rtol} in \ref{DARCY ATOL DEF}.
368 gross 2208 \end{methoddesc}
369 gross 2108
370 gross 2264 \begin{methoddesc}[DarcyFlow]{setAbsoluteTolerance}{\optional{atol=0.}}
371     sets the absolute tolerance \mbox{atol} in \ref{DARCY ATOL DEF}.
372     \end{methoddesc}
373 gross 2208
374 gross 2474 \begin{methoddesc}[DarcyFlow]{getSolverOptionsFlux}{}
375     Returns the solver options used to solve the flux problems~(\ref{DARCY V FORM}) and~(\ref{UPDATE W}). Use the returned \SolverOptions object to control the solution algorithms. If the adaption of subtolerance is choosen, the tolerance will
376     be overwritten before the solver is called.
377 gross 2264 \end{methoddesc}
378 gross 2208
379 gross 2474 \begin{methoddesc}[DarcyFlow]{getSolverOptionsPressure}{}
380     Returns the solver options used to solve the pressure problems~(\ref{UPDATE P}).
381     Use the returned \SolverOptions object to control the solution algorithms. If the adaption of subtolerance is choosen, the tolerance will
382     be overwritten before the solver is called.
383     \end{methoddesc}
384    
385     \begin{methoddesc}[DarcyFlow]{solve}{u0,p0, \optional{max_iter=100, \optional{verbose=False}}}
386 gross 2208 solves the problem. and returns approximations for the flux $v$ and the pressure $p$.
387     \var{u0} and \var{p0} define initial guess for flux and pressure. Values marked
388 gross 2474 by positive values \var{location_of_fixed_flux} and \var{location_of_fixed_pressure}, respectively, are kept unchanged. \var{max_iter} sets the maximum number of iterations steps allowed for solving the coupled problem.
389 gross 2208 \end{methoddesc}
390    
391    
392 gross 2100 \subsection{Example: Gravity Flow}
393 gross 2208 later
394 gross 2100
395 gross 2432 %\input{levelsetmodel}
396 gross 2100
397 lgraham 1702
398    
399 gross 1841 \section{Isotropic Kelvin Material \label{IKM}}
400 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
401 gross 1859 an elastic part $D\hackscore{ij}^{el}$ and visco-plastic part $D\hackscore{ij}^{vp}$:
402 gross 1841 \begin{equation}\label{IKM-EQU-2}
403 gross 1859 D\hackscore{ij}=D\hackscore{ij}^{el}+D\hackscore{ij}^{vp}
404 gross 1841 \end{equation}
405 gross 1859 with the elastic strain given as
406 gross 1841 \begin{equation}\label{IKM-EQU-3}
407 gross 2252 D\hackscore{ij}^{el'}=\frac{1}{2 \mu} \dot{\sigma}\hackscore{ij}'
408 gross 1841 \end{equation}
409 gross 1859 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
411 gross 1841 \begin{equation}\label{IKM-EQU-4}
412 gross 2252 D\hackscore{ij}^{vp'}=\sum\hackscore{q} D\hackscore{ij}^{q'}
413 gross 1841 \end{equation}
414 gross 1859 where $D\hackscore{ij}^{q}$ is the strain in material $q$ given as
415 gross 1841 \begin{equation}\label{IKM-EQU-5}
416 gross 2252 D\hackscore{ij}^{q'}=\frac{1}{2 \eta^{q}} \sigma'\hackscore{ij}
417 gross 1841 \end{equation}
418 gross 1859 where $\eta^{q}$ is the viscosity of material $q$. We assume the following
419     betwee the the strain in material $q$
420     \begin{equation}\label{IKM-EQU-5b}
421 gross 2438 \eta^{q}=\eta^{q}\hackscore{N} \left(\frac{\tau}{\tau\hackscore{t}^q}\right)^{1-n^{q}}
422 gross 1859 \mbox{ with } \tau=\sqrt{\frac{1}{2}\sigma'\hackscore{ij} \sigma'\hackscore{ij}}
423     \end{equation}
424 gross 2371 for a given power law coefficients $n^{q}\ge1$ and transition stresses $\tau\hackscore{t}^q$, see~\cite{Muhlhaus2005}.
425 gross 1859 Notice that $n^{q}=1$ gives a constant viscosity.
426 gross 1841 After inserting equation~\ref{IKM-EQU-5} into equation \ref{IKM-EQU-4} one gets:
427 gross 1859 \begin{equation}\label{IKM-EQU-6}
428 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}} \;.
429 gross 1841 \end{equation}
430 gross 2300 and finally with~\ref{IKM-EQU-2}
431 gross 2371 \begin{equation}\label{IKM-EQU-2bb}
432 gross 2300 D\hackscore{ij}'=\frac{1}{2 \eta^{vp}} \sigma'\hackscore{ij}+\frac{1}{2 \mu} \dot{\sigma}\hackscore{ij}'
433 gross 1859 \end{equation}
434 gross 2300 The total stress $\tau$ needs to fullfill the yield condition \index{yield condition}
435 gross 1859 \begin{equation}\label{IKM-EQU-8c}
436     \tau \le \tau\hackscore{Y} + \beta \; p
437     \end{equation}
438 gross 2300 with the Drucker-Prager \index{Druck-Prager} cohesion factor \index{cohesion factor} $\tau\hackscore{Y}$, Drucker-Prager friction $\beta$ and total pressure $p$.
439 gross 1859 The deviatoric stress needs to fullfill the equilibrion equation
440 gross 1841 \begin{equation}\label{IKM-EQU-1}
441 gross 1859 -\sigma'\hackscore{ij,j}+p\hackscore{,i}=F\hackscore{i}
442 gross 1841 \end{equation}
443 gross 1859 where $F\hackscore{j}$ is a given external fource. We assume an incompressible media:
444 gross 2371 \begin{equation}\label{IKM-EQU-2bbb}
445 gross 1859 -v\hackscore{i,i}=0
446 gross 1841 \end{equation}
447 gross 1878 Natural boundary conditions are taken in the form
448     \begin{equation}\label{IKM-EQU-Boundary}
449     \sigma'\hackscore{ij}n\hackscore{j}-n\hackscore{i}p=f
450     \end{equation}
451     which can be overwritten by a constraint
452     \begin{equation}\label{IKM-EQU-Boundary0}
453     v\hackscore{i}(x)=0
454     \end{equation}
455     where the index $i$ may depend on the location $x$ on the bondary.
456    
457     \subsection{Solution Method \label{IKM-SOLVE}}
458     By using a first order finite difference approximation wit step size $dt>0$~\ref{IKM-EQU-3} get the form
459     \begin{equation}\label{IKM-EQU-3b}
460 gross 2300 \dot{\sigma}\hackscore{ij}=\frac{1}{dt } \left( \sigma\hackscore{ij} - \sigma\hackscore{ij}^{-} \right)
461 gross 1878 \end{equation}
462 gross 2300 and
463     \begin{equation}\label{IKM-EQU-2b}
464 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}^{-'}
465 gross 2300 \end{equation}
466     where $\sigma\hackscore{ij}^{-}$ is the stress at the precious time step. With
467     \begin{equation}\label{IKM-EQU-2c}
468 gross 2415 \dot{\gamma} = \sqrt{ 2 \left( D\hackscore{ij}' +
469     \frac{1}{ 2 \mu \; dt} \sigma\hackscore{ij}^{-'}\right)^2}
470 gross 2300 \end{equation}
471     we have
472     \begin{equation}
473     \tau = \eta\hackscore{eff} \cdot \dot{\gamma}
474     \end{equation}
475     where
476     \begin{equation}
477     \eta\hackscore{eff}= min( \left(\frac{1}{\mu \; dt}+\frac{1}{\eta^{vp}}\right)^{-1}
478     , \eta\hackscore{max}) \mbox{ with }
479     \eta\hackscore{max} = \left\{
480     \begin{array}{rcl}
481     \frac{\tau\hackscore{Y} + \beta \; p}{\dot{\gamma}} & & \dot{\gamma}>0 \\
482     &\mbox{ if } \\
483     \infty & & \mbox{otherwise}
484     \end{array}
485     \right.
486     \end{equation}
487     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
488 gross 1878 \begin{equation}\label{IKM-EQU-10}
489 gross 2415 \sigma\hackscore{ij}' = 2 \eta\hackscore{eff} \left( D\hackscore{ij}' +
490 gross 2300 \frac{1}{ 2 \mu \; dt} \sigma\hackscore{ij}^{'-}\right)
491 gross 1878 \end{equation}
492 gross 1859 After inserting~\ref{IKM-EQU-10} into~\ref{IKM-EQU-1} we get
493     \begin{equation}\label{IKM-EQU-1ib}
494 gross 2100 -\left(\eta\hackscore{eff} (v\hackscore{i,j}+ v\hackscore{i,j})
495 gross 2415 \right)\hackscore{,j}+p\hackscore{,i}=F\hackscore{i}+
496     \left(\frac{\eta\hackscore{eff}}{\mu dt } \sigma\hackscore{ij}^{'-} \right)\hackscore{,j}
497 gross 1859 \end{equation}
498 gross 2252 Combining this with the incomressibilty condition~\ref{IKM-EQU-2} we need to solve a
499     Stokes problem as discussed in section~\ref{STOKES SOLVE} in each time step.
500 gross 2300
501     If we set
502     \begin{equation}\label{IKM-EQU-44}
503     \frac{1}{\eta(\tau)}= \frac{1}{\mu \; dt}+\frac{1}{\eta^{vp}}
504     \end{equation}
505     we need to solve the nonlinear problem
506 gross 1878 \begin{equation}
507 gross 2300 \eta\hackscore{eff} - min(\eta( \dot{\gamma} \cdot \eta\hackscore{eff})
508     , \eta\hackscore{max}) =0
509     \end{equation}
510     We use the Newton-Raphson Scheme \index{Newton-Raphson scheme} to solve this problem
511     \begin{equation}\label{IKM-EQU-45}
512     \eta\hackscore{eff}^{(n+1)} = \min(\eta\hackscore{max},
513     \eta\hackscore{eff}^{(n)} -
514     \frac{\eta\hackscore{eff}^{(n)} - \eta( \tau^{(n)}) }
515     {1-\dot{\gamma} \cdot \eta'( \tau^{(n)} )} )
516     =\min(\eta\hackscore{max},
517     \frac{\eta( \tau^{(n)}) -\tau^{(n)} \cdot \eta'( \tau^{(n)} ) }
518     {1-\dot{\gamma} \cdot \eta'( \tau^{(n)} )} )
519 gross 2100 \end{equation}
520 gross 2300 where $\eta'$ denotes the derivative of $\eta$ with respect of $\tau$
521     and $\tau^{(n)} = \dot{\gamma} \cdot \eta\hackscore{eff}^{(n)}$.
522    
523     Looking at the evaluation of $\eta$ in~\ref{IKM-EQU-44} it makes sense formulate
524     the iteration~\ref{IKM-EQU-45} using $\Theta=\eta^{-1}$.
525     In fact we have
526 gross 2100 \begin{equation}
527 gross 2300 \eta' = - \frac{\Theta'}{\Theta^2}
528 gross 2100 \mbox{ with }
529 gross 2300 \Theta' = \sum\hackscore{q} \left(\frac{1}{\eta^{q}} \right)'
530 gross 2100 \label{IKM iteration 7}
531     \end{equation}
532 gross 2300 As
533     \begin{equation}\label{IKM-EQU-47}
534 gross 2100 \left(\frac{1}{\eta^{q}} \right)'
535 gross 2438 = \frac{n^{q}-1}{\eta^{q}\hackscore{N}} \cdot \frac{\tau^{n^{q}-2}}{(\tau\hackscore{t}^q)^{n^{q}-1}}
536     = \frac{n^{q}-1}{\eta^{q}}\cdot\frac{1}{\tau}
537 gross 2100 \end{equation}
538 gross 2300 we have
539     \begin{equation}\label{IKM-EQU-48}
540 gross 2438 \Theta' = \frac{1}{\tau} \omega \mbox{ with } \omega = \sum\hackscore{q}\frac{n^{q}-1}{\eta^{q}}
541 gross 2300 \end{equation}
542     which leads to
543 gross 2371 \begin{equation}\label{IKM-EQU-49}
544 gross 2300 \eta\hackscore{eff}^{(n+1)} = \min(\eta\hackscore{max},
545     \eta\hackscore{eff}^{(n)}
546     \frac{\Theta^{(n)} + \omega^{(n)} }
547     {\eta\hackscore{eff}^{(n)} \Theta^{(n)^2}+\omega^{(n)} })
548 gross 2432 \end{equation}
549    
550    
551     \subsection{Functions}
552    
553     \begin{classdesc}{IncompressibleIsotropicFlowCartesian}{
554     domain
555     \optional{, stress=0
556     \optional{, v=0
557     \optional{, p=0
558     \optional{, t=0
559     \optional{, numMaterials=1
560 gross 2474 \optional{, verbose=True
561     \optional{, adaptSubTolerance=True
562     }}}}}}}}
563 gross 2433 opens an incompressible, isotropic flow problem in Cartesian cooridninates
564     on the domain \var{domain}.
565 gross 2432 \var{stress},
566     \var{v},
567     \var{p}, and
568     \var{t} set the initial deviatoric stress, velocity, pressure and time.
569     \var{numMaterials} specifies the number of materials used in the power law
570     model. Some progress information are printed if \var{verbose} is set to
571 gross 2474 \True. If \var{adaptSubTolerance} is equal to True the tolerances for subproblems are set automatically.
572 gross 2432 \end{classdesc}
573    
574 gross 2433 \begin{methoddesc}[IncompressibleIsotropicFlowCartesian]{getDomain}{}
575     returns the domain.
576     \end{methoddesc}
577 gross 2432
578 gross 2433 \begin{methoddesc}[IncompressibleIsotropicFlowCartesian]{getTime}{}
579     Returns current time.
580 gross 2432 \end{methoddesc}
581    
582 gross 2433 \begin{methoddesc}[IncompressibleIsotropicFlowCartesian]{getStress}{}
583     Returns current stress.
584 gross 2432 \end{methoddesc}
585    
586 gross 2433 \begin{methoddesc}[IncompressibleIsotropicFlowCartesian]{getDeviatoricStress}{}
587     Returns current deviatoric stress.
588     \end{methoddesc}
589 gross 2432
590 gross 2433 \begin{methoddesc}[IncompressibleIsotropicFlowCartesian]{getPressure}{}
591     Returns current pressure.
592 gross 2432 \end{methoddesc}
593 gross 2433
594     \begin{methoddesc}[IncompressibleIsotropicFlowCartesian]{getVelocity}{}
595     Returns current velocity.
596 gross 2432 \end{methoddesc}
597 gross 2433
598     \begin{methoddesc}[IncompressibleIsotropicFlowCartesian]{getDeviatoricStrain}{}
599     Returns deviatoric strain of current velocity
600 gross 2432 \end{methoddesc}
601 gross 2433
602     \begin{methoddesc}[IncompressibleIsotropicFlowCartesian]{getTau}{}
603     Returns current second invariant of deviatoric stress
604 gross 2432 \end{methoddesc}
605 gross 2433
606     \begin{methoddesc}[IncompressibleIsotropicFlowCartesian]{getGammaDot}{}
607     Returns current second invariant of deviatoric strain
608 gross 2432 \end{methoddesc}
609 gross 2433
610    
611     \begin{methoddesc}[IncompressibleIsotropicFlowCartesian]{setTolerance}{tol=1.e-4}
612     Sets the tolerance used to terminate the iteration on a time step.
613 gross 2432 \end{methoddesc}
614    
615 gross 2433 \begin{methoddesc}[IncompressibleIsotropicFlowCartesian]{setFlowTolerance}{tol=1.e-4}
616     Sets the relative tolerance for the incompressible solver, see \class{StokesProblemCartesian} for details.
617     \end{methoddesc}
618 gross 2432
619 gross 2474 \begin{methoddesc}[IncompressibleIsotropicFlowCartesian]{setElasticShearModulus}{mu=None}
620 gross 2433 Sets the elastic shear modulus $\mu$. If \var{mu} is set to None (default) elasticity is not applied.
621     \end{methoddesc}
622    
623     \begin{methoddesc}[IncompressibleIsotropicFlowCartesian]{setEtaTolerance=}{rtol=1.e-8}
624     sets the relative tolerance for the effectice viscosity. Iteration on a time step is completed if the realtive of the effective viscosity is less than \var{rtol}.
625     \end{methoddesc}
626    
627     \begin{methoddesc}[IncompressibleIsotropicFlowCartesian]{setDruckerPragerLaw}
628     {\optional{tau_Y=None, \optional{friction=None}}}
629     Sets the parameters $\tau\hackscore{Y}$ and $\beta$ for the Drucker-Prager model in condition~\ref{IKM-EQU-8c}. If \var{tau_Y} is set to None (default) Drucker-Prager
630     condition is not applied.
631     \end{methoddesc}
632    
633     \begin{methoddesc}[IncompressibleIsotropicFlowCartesian]{setElasticShearModulus}{mu=None}
634     Sets the elastic shear modulus $\mu$. If \var{mu} is set to None (default) elasticity is not applied.
635     \end{methoddesc}
636    
637    
638     \begin{methoddesc}[IncompressibleIsotropicFlowCartesian]{setPowerLaws}{eta_N, tau_t, power}
639     Sets the parameters of the power-law for all materials as defined in
640     equation~\ref{IKM-EQU-5b}.
641     \var{eta_N} is the list of viscosities $\eta^{q}\hackscore{N}$,
642     \var{tau_t} is the list of reference stresses $\tau\hackscore{t}^q$,
643     and \var{power} is the list of power law coefficients $n^{q}$.
644     \end{methoddesc}
645    
646    
647     \begin{methoddesc}[IncompressibleIsotropicFlowCartesian]{update}{dt
648     \optional{, iter_max=100
649     \optional{, inner_iter_max=20
650     }}}
651     Updates stress, velocity and pressure for time increment \var{dt}.
652     where \var{iter_max} is the maximum number of iteration steps on a time step to
653     update the effective viscosity and \var{inner_iter_max} is the maximum
654     number of itertion steps in the incompressible solver.
655     \end{methoddesc}
656    
657     \subsection{Example}
658     later
659    
660 gross 2432 % \section{Drucker Prager Model}

  ViewVC Help
Powered by ViewVC 1.1.26