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 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 7 lgraham 1702 % 8 ksteube 1811 % Primary Business: Queensland, Australia 9 % Licensed under the Open Software License version 3.0 10 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}