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

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starting a section for the Level Set class in the Models chapter.


 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 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 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 \end{equation} 65 where $\epsilon$ is the desired relative accuracy and 66 \begin{equation} 67 \|p\|^2= \int\hackscore{\Omega} p^2 \; dx 68 \label{PRESSURE NORM} 69 \end{equation} 70 defines the $L^2$-norm. 71 There are two approaches to solve this problem. The first approach, called the Uzawa scheme \index{Uzawa scheme} 72 eliminates the velocity $v$ from the problem. The second approach solves the equation in coupled form after the application of a preconditioner. 73 74 \subsubsection{Uzawa scheme} 75 The first eqution in~\ref{SADDLEPOINT} gives $v=A^{-1}(G-B^{*}p)$ assuming $p$ is known. This is inserted into the 76 second eqution which leads to 77 \begin{equation} 78 S p = B A^{-1} G 79 \end{equation} 80 with the Schur complement \index{Schur complement} $S=BA^{-1}B^{*}$. This problem can be solved iteratively using the reconditioned Conjugate Gradient Method (PCG)~\index{PCG!Preconditioned Conjugate Gradient Method} 81 with the preconditioner $\hat{S}$ defined as $q=\hat{S}^{-1}p$ by solving 82 \begin{equation} 83 \frac{1}{\eta}q = p 84 \end{equation} 85 see~\cite{ELMAN} for more details. The evaluation of $w=Sp$ is done in the form 86 \begin{equation} 87 \begin{array}{rcl} 88 A v & = & B^{*}p \\ 89 w & = & Bv \\ 90 \end{array} 91 \label{EVAL PCG} 92 \end{equation} 93 The residual \index{residual} $r=B A^{-1} G - S p$ is given as 94 \begin{equation} 95 r=B A^{-1} (G - B^* p) = Bv \mbox{ with } v = A^{-1}(G-B^{*}p) 96 \end{equation} 97 Therefore one uses the tuple $(v,Bv)$ to represent the residual of the current pressure $p$. Notice that before the iteration is started the right hand side $B A^{-1} G$ needs to be calculated. The bilinear form $(.,.)$ used is defined as 98 \begin{equation} 99 (p,(v,Bv))=\int\hackscore{\Omega} p \cdot Bv \; dx 100 \end{equation} 101 where $p$ is the pressure increment and $(v,Bv)$ represents an increment in the residual. 102 103 \subsubsection{Coupled Solver} 104 An alternative approach to solve the saddle point problem~\ref{SADDLEPOINT} directly using an iterative such as 105 the generalized minimal residual method (GMRES) \index{generalized minimal residual method!GMRES} with a suitable 106 preconditioner. Here we use the operator 107 \begin{equation} 108 gross 1878 \left[ \begin{array}{cc} 109 A^{-1} & 0 \\ 110 S^{-1} B A^{-1} & -S^{-1} \\ 111 \end{array} \right] 112 \label{SADDLEPOINT PRECODITIONER} 113 \end{equation} 114 gross 2100 where again $S$ is the Schur complement~\cite{ELMAN}. In partice we will use an approximation $\hat{S}$ for $S$. The evaluation $(w,q)$ of the iteration operator for a given $(v,p)$ is done as 115 gross 1878 \begin{equation} 116 \begin{array}{rcl} 117 gross 2100 A w & = & Av+B^{*}p \\ 118 \hat{S} q & = & B(w-v) \\ 119 gross 1878 \end{array} 120 gross 2100 \label{COUPLES SADDLEPOINT iteration} 121 gross 1878 \end{equation} 122 gross 2100 We use the inner product induced by the norm 123 gross 1878 \begin{equation} 124 gross 2100 \|(v,p)\|^2= \int\hackscore{\Omega} v\hackscore{i,j} v\hackscore{i,j} + \left( \frac{p}{\eta}\right)^2\; dx 125 \label{COUPLES NORM} 126 \end{equation} 127 In PDE form~\ref{COUPLES SADDLEPOINT iteration} takes the form 128 \begin{equation} 129 gross 1878 \begin{array}{rcl} 130 gross 2100 -\left(\eta(w\hackscore{i,j}+ w\hackscore{i,j})\right)\hackscore{,j} & = & -\left(\eta(v\hackscore{i,j}+ v\hackscore{i,j})\right)\hackscore{,j}+p\hackscore{,i} \\ 131 \frac{1}{\eta} q & = & - (w-v)\hackscore{i,i} \\ 132 gross 1878 \end{array} 133 \label{SADDLEPOINT iteration 2} 134 \end{equation} 135 lgraham 1702 136 137 gross 2100 \subsection{Functions} 138 gross 1878 139 gross 2100 \begin{classdesc}{StokesProblemCartesian}{domain} 140 opens the Stokes problem\index{Stokes problem} on the \Domain domain. The approximation 141 order needs to be two. 142 \end{classdesc} 143 gross 1878 144 gross 2100 \begin{methoddesc}[StokesProblemCartesian]{initialize}{\optional{f=Data(), \optional{fixed_u_mask=Data(), \optional{eta=1, \optional{surface_stress=Data(), \optional{stress=Data()}}}}}} 145 assigns values to the model parameters. In any call all values must be set. 146 \var{f} defines the external force $f$, \var{eta} the viscosity $\eta$, 147 \var{surface_stress} the surface stress $s$ and \var{stress} the initial stress $\sigma$. 148 The locations and compontents where the velocity is fixed are set by 149 the values of \var{fixed_u_mask}. The method will try to cast the given values to appropriate 150 \Data class objects. 151 \end{methoddesc} 152 gross 1878 153 gross 2100 \begin{methoddesc}[StokesProblemCartesian]{solve}{v,p, 154 \optional{max_iter=20, \optional{verbose=False, \optional{useUzawa=True}}}} 155 solves the problem and return approximations for velocity and pressure. 156 The arguments \var{v} and \var{p} define initial guess. The values of \var{v} marked 157 by \var{fixed_u_mask} remain unchanged. 158 If \var{useUzawa} is set to \True 159 the Uzawa\index{Uszwa} scheme is used. Otherwise the problem is solved in coupled form. In most cases 160 the Uzawa scheme is more efficient. 161 \var{max_iter} defines the maximum number of iteration steps. 162 If \var{verbose} is set to \True informations on the progress of of the solver are printed. 163 \end{methoddesc} 164 lgraham 1702 165 166 gross 2100 \begin{methoddesc}[StokesProblemCartesian]{setTolerance}{\optional{tolerance=1.e-8}} 167 sets the tolerance in an appropriate norm relative to the right hand side. The tolerance must be non-negative and less than 1. 168 \end{methoddesc} 169 \begin{methoddesc}[StokesProblemCartesian]{getTolerance}{} 170 returns the current relative tolerance. 171 \end{methoddesc} 172 \begin{methoddesc}[StokesProblemCartesian]{setAbsoluteTolerance}{\optional{tolerance=0.}} 173 sets the absolute tolerance for the error in the relevant norm. The tolerance must be non-negative. Typically the 174 absolute talerance is set to 0. 175 \end{methoddesc} 176 \begin{methoddesc}[StokesProblemCartesian]{getAbsoluteTolerance}{} 177 sreturns the current absolute tolerance. 178 \end{methoddesc} 179 \begin{methoddesc}[StokesProblemCartesian]{setSubToleranceReductionFactor}{\optional{reduction=None}} 180 sets the reduction factor for the tolerance used to solve the PDEs. A reduction factor 181 in the order of one will minimize compute time per iteration step but my slow down convergence or even lead to 182 divergency. On the other hand a very small value for the PDE tolerance could result in a wast of compute time. 183 If \var{reduction} is set to \var{None} the sub-tolerance is solved adaptively but 184 in cases a very small tolerance is set ($<10^{-6}$) it is recommended to set the 185 reduction factor by hand. This may require some experiments. 186 \end{methoddesc} 187 \begin{methoddesc}[StokesProblemCartesian]{getSubToleranceReductionFactor}{} 188 return the current reduction factor for the sub-problem tolerance. 189 \end{methoddesc} 190 lgraham 1702 191 gross 2100 \subsection{Example: Lit Driven Cavity} 192 The following script \file{lit\hackscore driven\hackscore cavity.py} 193 \index{scripts!\file{helmholtz.py}} which is available in the \ExampleDirectory 194 illustrates the usage of the \class{StokesProblemCartesian} class to solve 195 jfenwick 2104 the lit driven cavity problem~\cite{LITDRIVENCAVITY}: 196 gross 2100 \begin{python} 197 from esys.escript import * 198 from esys.finley import Rectangle 199 from esys.escript.models import StokesProblemCartesian 200 NE=25 201 dom = Rectangle(NE,NE,order=2) 202 x = dom.getX() 203 sc=StokesProblemCartesian(dom) 204 mask= (whereZero(x[0])*[1.,0]+whereZero(x[0]-1))*[1.,0] + \ 205 (whereZero(x[1])*[0.,1.]+whereZero(x[1]-1))*[1.,1] 206 sc.initialize(eta=.1, fixed_u_mask= mask) 207 v=Vector(0.,Solution(dom)) 208 v[0]+=whereZero(x[1]-1.) 209 p=Scalar(0.,ReducedSolution(dom)) 210 v,p=sc.solve(v,p, verbose=True) 211 saveVTK("u.xml",velocity=v,pressure=p) 212 \end{python} 213 lgraham 1702 214 gross 2100 \section{Darcy Flux} 215 gross 2108 We want to calculate the velocity $u$ and pressure $p$ on a domain $\Omega$ solving 216 the Darcy flux problem \index{Darcy flux}\index{Darcy flow} 217 gross 2100 \begin{equation}\label{DARCY PROBLEM} 218 \begin{array}{rcl} 219 u\hackscore{i} + \kappa\hackscore{ij} p\hackscore{,j} & = & g\hackscore{i} \\ 220 u\hackscore{k,k} & = & f 221 \end{array} 222 \end{equation} 223 with the boundary conditions 224 \begin{equation}\label{DARCY BOUNDARY} 225 \begin{array}{rcl} 226 u\hackscore{i} \; n\hackscore{i} = u^{N}\hackscore{i} \; n\hackscore{i} & \mbox{ on } & \Gamma\hackscore{N} \\ 227 p = p^{D} & \mbox{ on } & \Gamma\hackscore{D} \\ 228 \end{array} 229 \end{equation} 230 where $\Gamma\hackscore{N}$ and $\Gamma\hackscore{D}$ are a partition of the boundary of $\Omega$ with $\Gamma\hackscore{D}$ non empty, $n\hackscore{i}$ is the outer normal field of the boundary of $\Omega$, $u^{N}\hackscore{i}$ and $p^{D}$ are given functions on $\Omega$, $g\hackscore{i}$ and $f$ are given source terms and $\kappa\hackscore{ij}$ is the given permability. We assume that $\kappa\hackscore{ij}$ is symmetric (which is not really required) and positive definite, i.e there are positive constants $\alpha\hackscore{0}$ and $\alpha\hackscore{1}$ wich are independent from the location in $\Omega$ such that 231 \begin{equation} 232 \alpha\hackscore{0} \; x\hackscore{i} x\hackscore{i} \le \kappa\hackscore{ij} x\hackscore{i} x\hackscore{j} \le \alpha\hackscore{1} \; x\hackscore{i} x\hackscore{i} 233 \end{equation} 234 for all $x\hackscore{i}$. 235 lgraham 1702 236 237 gross 2100 \subsection{Solution Method \label{DARCY SOLVE}} 238 Without loss of generality we can assume that $u^{N}\hackscore{i} \; n\hackscore{i}=0$ and 239 $p^{D}$. Otherewise one solves for $u-u^{N}$ and $p-p^{D}$ and sets 240 lgraham 1702 \begin{equation} 241 gross 2100 \begin{array}{rcl} 242 g\hackscore{i} & \leftarrow & g\hackscore{i} - u^{N}\hackscore{i} - \kappa\hackscore{ij} p^{D}\hackscore{,j }\\ 243 f & \leftarrow & f - u^{N}\hackscore{k,k} 244 \end{array} 245 \end{equation} 246 We set 247 \begin{equation} 248 V = \{ q \in H^{1}(\Omega) : q=0 \mbox{ on } \Gamma\hackscore{D} \} 249 \end{equation} 250 and 251 \begin{equation} 252 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} \} 253 \end{equation} 254 and define the operator $Q: V \rightarrow (L^2(\Omega))^{d}$ defined by 255 \begin{equation} 256 (Qp)\hackscore{i} = \kappa\hackscore{ij} p\hackscore{,j} 257 \end{equation} 258 and the operator $D: W \rightarrow L^2(\Omega)$ defined by 259 \begin{equation} 260 Dv = v\hackscore{k,k} 261 \end{equation} 262 In operator notation the Darcy problem~\ref{DARCY PROBLEM} is written in the form 263 \begin{equation} 264 \begin{array}{rcl} 265 u + Qp & = & g \\ 266 Du & = & f 267 \end{array} 268 \end{equation} 269 We solve this equation by minimising the functional 270 \begin{equation} 271 J(u,p):=\|u + Qp - g\|^2\hackscore{0} + \|Du-f\|\hackscore{0}^2 272 \end{equation} 273 over $W \times V$ where $\|.\|\hackscore{0}$ denotes the norm in $L^2(\Omega)$. A simple calculation shows that 274 one has to solve 275 \begin{equation} 276 ( v + Qq , u + Qp - g) + (Dv,Du-f) =0 277 \end{equation} 278 for all $v\in W$ and $q \in V$.which translates back into operator notation 279 \begin{equation} 280 \begin{array}{rcl} 281 (I+D^*D)u + Qp & = & D^*f + g \\ 282 Q^*u + Q^*Q p & = & Q^* g \\ 283 \end{array} 284 \end{equation} 285 where $D^*$ and $Q^*$ denote the adjoint operators. 286 In~\cite{XXX} it has been shown that this problem is continuous and coercive in $W \times V$ and therefore has a unique solution. Also standart FEM methods can be used for discretization. It is also possible 287 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 $\alpha\hackscore{0} \gg 1$) 288 289 The approach we are taking is to eliminate the velocity $u$ from the problem. Assuming that $p$ is known we have 290 \begin{equation} 291 v= (I+D^*D)^{-1} (D^*f + g - Qp) 292 \end{equation} 293 (notice that $(I+D^*D)$ is coercive in $W$) which is inserted into the second equation 294 \begin{equation} 295 Q^* (I+D^*D)^{-1} (D^*f + g - Qp) + Q^* Q p = Q^* g 296 \end{equation} 297 which is 298 \begin{equation} 299 Q^* ( I - (I+D^*D)^{-1} ) Q p = Q^* ( g -(I+D^*D)^{-1} (D^*f + g) ) 300 \end{equation} 301 We use the PCG method to solve this. The residual $r$ ($\in V^*$) is given as 302 \begin{equation} 303 \begin{array}{rcl} 304 r & = & Q^* ( g -(I+D^*D)^{-1} (D^*f + g) - Qp + (I+D^*D)^{-1}Q p \\ 305 & =& Q^* \left( (g-Qp) - (I+D^*D)^{-1} (D^*f + g - Qp) \right) \\ 306 & =& Q^* \left( (g-Qp) - v \right) 307 \end{array} 308 \end{equation} 309 So in a partical implementation we use the pair $(g-Qp,v)$ to represent the residual. This will save the 310 reconstruction of the velocity $v$. In this notation the right hand side is given as 311 $(g,(I+D^*D)^{-1} (D^*f + g))$. The evaluation of the iteration operator for a given $p$ is then 312 returning $(Qp,w)$ where $w$ is the solution of 313 \begin{equation}\label{UPDATE W} 314 (I+D^*D)w = Qp 315 \end{equation} 316 We use $Q^*Q$ as a a preconditioner for the iteration operator $Q^* ( I - (I+D^*D)^{-1} ) Q$. 317 318 \subsection{Functions} 319 gross 2108 \begin{classdesc}{DarcyFlow}{domain} 320 opens the Darcy flux problem\index{Darcy flux} on the \Domain domain. 321 \end{classdesc} 322 gross 2100 323 gross 2108 \begin{methoddesc}[DarcyFlow]{initialize}{\optional{f=Data(), \optional{fixed_u_mask=Data(), \optional{eta=1, \optional{surface_stress=Data(), \optional{stress=Data()}}}}}} 324 assigns values to the model parameters. In any call all values must be set. 325 \var{f} defines the external force $f$, \var{eta} the viscosity $\eta$, 326 \var{surface_stress} the surface stress $s$ and \var{stress} the initial stress $\sigma$. 327 The locations and compontents where the velocity is fixed are set by 328 the values of \var{fixed_u_mask}. The method will try to cast the given values to appropriate 329 \Data class objects. 330 \end{methoddesc} 331 332 333 gross 2100 \subsection{Example: Gravity Flow} 334 335 %================================================ 336 \section{Temperature Advection Diffusion\label{TEMP ADV DIFF}} 337 338 \begin{equation} 339 lgraham 1709 \rho c\hackscore{p} \left (\frac{\partial T}{\partial t} + \vec{v} \cdot \nabla T \right ) = k \nabla^{2}T 340 lgraham 1702 \label{HEAT EQUATION} 341 \end{equation} 342 343 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. 344 345 \subsection{Description} 346 347 \subsection{Method} 348 349 \begin{classdesc}{TemperatureCartesian}{dom,theta=THETA,useSUPG=SUPG} 350 \end{classdesc} 351 352 \subsection{Benchmark Problem} 353 gross 2100 %=============================================================================================================== 354 lgraham 1702 355 gross 2100 %========================================================= 356 lgraham 2138 \input{levelsetmodel} 357 gross 2100 % \section{Level Set Method} 358 lgraham 1702 359 gross 2100 %\subsection{Description} 360 lgraham 1702 361 gross 2100 %\subsection{Method} 362 lgraham 1702 363 gross 2100 %Advection and Reinitialisation 364 lgraham 1702 365 gross 2100 %\begin{classdesc}{LevelSet}{mesh, func\_new, reinit\_max, reinit\_each, tolerance, smooth} 366 %\end{classdesc} 367 lgraham 1702 368 %example usage: 369 370 %levelset = LevelSet(mesh, func\_new, reinit\_max, reinit\_each, tolerance, smooth) 371 372 gross 2100 %\begin{methoddesc}[LevelSet]{update\_parameter}{parameter} 373 %Update the parameter. 374 %\end{methoddesc} 375 lgraham 1702 376 gross 2100 %\begin{methoddesc}[LevelSet]{update\_phi}{paramter}{velocity}{dt}{t\_step} 377 %Update level set function; advection and reinitialization 378 %\end{methoddesc} 379 lgraham 1702 380 gross 2100 %\subsection{Benchmark Problem} 381 lgraham 1702 382 gross 2100 %Rayleigh-Taylor instability problem 383 lgraham 1702 384 385 gross 1859 % \section{Drucker Prager Model} 386 lgraham 1702 387 gross 1841 \section{Isotropic Kelvin Material \label{IKM}} 388 gross 1859 As proposed by Kelvin~\ref{KELVN} material strain $D\hackscore{ij}=v\hackscore{i,j}+v\hackscore{j,i}$ can be decomposed into 389 an elastic part $D\hackscore{ij}^{el}$ and visco-plastic part $D\hackscore{ij}^{vp}$: 390 gross 1841 \begin{equation}\label{IKM-EQU-2} 391 gross 1859 D\hackscore{ij}=D\hackscore{ij}^{el}+D\hackscore{ij}^{vp} 392 gross 1841 \end{equation} 393 gross 1859 with the elastic strain given as 394 gross 1841 \begin{equation}\label{IKM-EQU-3} 395 gross 1878 D\hackscore{ij}'^{el}=\frac{1}{2 \mu} \dot{\sigma}\hackscore{ij}' 396 gross 1841 \end{equation} 397 gross 1859 where $\sigma'\hackscore{ij}$ is the deviatoric stress (Notice that $\sigma'\hackscore{ii}=0$). 398 If the material is composed by materials $q$ the visco-plastic strain can be decomposed as 399 gross 1841 \begin{equation}\label{IKM-EQU-4} 400 gross 1859 D\hackscore{ij}'^{vp}=\sum\hackscore{q} D\hackscore{ij}'^{q} 401 gross 1841 \end{equation} 402 gross 1859 where $D\hackscore{ij}^{q}$ is the strain in material $q$ given as 403 gross 1841 \begin{equation}\label{IKM-EQU-5} 404 gross 1859 D\hackscore{ij}'^{q}=\frac{1}{2 \eta^{q}} \sigma'\hackscore{ij} 405 gross 1841 \end{equation} 406 gross 1859 where $\eta^{q}$ is the viscosity of material $q$. We assume the following 407 betwee the the strain in material $q$ 408 \begin{equation}\label{IKM-EQU-5b} 409 \eta^{q}=\eta^{q}\hackscore{N} \left(\frac{\tau}{\tau\hackscore{t}^q}\right)^{\frac{1}{n^{q}}-1} 410 \mbox{ with } \tau=\sqrt{\frac{1}{2}\sigma'\hackscore{ij} \sigma'\hackscore{ij}} 411 \end{equation} 412 for a given power law coefficients $n^{q}$ and transition stresses $\tau\hackscore{t}^q$, see~\ref{POERLAW}. 413 Notice that $n^{q}=1$ gives a constant viscosity. 414 gross 1841 After inserting equation~\ref{IKM-EQU-5} into equation \ref{IKM-EQU-4} one gets: 415 gross 1859 \begin{equation}\label{IKM-EQU-6} 416 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}} \;. 417 gross 1841 \end{equation} 418 gross 1859 With 419 \begin{equation}\label{IKM-EQU-8} 420 \dot{\gamma}=\sqrt{2 D\hackscore{ij} D\hackscore{ij}} 421 \end{equation} 422 one gets 423 \begin{equation}\label{IKM-EQU-8b} 424 \tau = \eta^{vp} \dot{\gamma}^{vp} \;. 425 \end{equation} 426 With the Drucker-Prager cohesion factor $\tau\hackscore{Y}$, Drucker-Prager friction $\beta$ and total pressure $p$ we want to achieve 427 \begin{equation}\label{IKM-EQU-8c} 428 \tau \le \tau\hackscore{Y} + \beta \; p 429 \end{equation} 430 which leads to the condition 431 \begin{equation}\label{IKM-EQU-8d} 432 \eta^{vp} \le \frac{\tau\hackscore{Y} + \beta \; p}{ \dot{\gamma}^{vp}} \; . 433 \end{equation} 434 Therefore we modify the definition of $\eta^{vp}$ to the form 435 \begin{equation}\label{IKM-EQU-6b} 436 \frac{1}{\eta^{vp}}=\max(\sum\hackscore{q} \frac{1}{\eta^{q}}, \frac{\dot{\gamma}^{vp}} {\tau\hackscore{Y} + \beta \; p}) 437 \end{equation} 438 The deviatoric stress needs to fullfill the equilibrion equation 439 gross 1841 \begin{equation}\label{IKM-EQU-1} 440 gross 1859 -\sigma'\hackscore{ij,j}+p\hackscore{,i}=F\hackscore{i} 441 gross 1841 \end{equation} 442 gross 1859 where $F\hackscore{j}$ is a given external fource. We assume an incompressible media: 443 gross 1841 \begin{equation}\label{IKM-EQU-2} 444 gross 1859 -v\hackscore{i,i}=0 445 gross 1841 \end{equation} 446 gross 1878 Natural boundary conditions are taken in the form 447 \begin{equation}\label{IKM-EQU-Boundary} 448 \sigma'\hackscore{ij}n\hackscore{j}-n\hackscore{i}p=f 449 \end{equation} 450 which can be overwritten by a constraint 451 \begin{equation}\label{IKM-EQU-Boundary0} 452 v\hackscore{i}(x)=0 453 \end{equation} 454 where the index $i$ may depend on the location $x$ on the bondary. 455 456 \subsection{Solution Method \label{IKM-SOLVE}} 457 By using a first order finite difference approximation wit step size $dt>0$~\ref{IKM-EQU-3} get the form 458 \begin{equation}\label{IKM-EQU-3b} 459 D\hackscore{ij}'^{el}=\frac{1}{2 \mu dt } \left( \sigma\hackscore{ij}' - \sigma\hackscore{ij}^{'-} \right) 460 \end{equation} 461 where $\sigma\hackscore{ij}^{'-}$ is the deviatoric stress at the precious time step. 462 Now we can combine equations~\ref{IKM-EQU-2}, \ref{IKM-EQU-3b} and~\ref{IKM-EQU-6b} to get 463 \begin{equation}\label{IKM-EQU-10} 464 gross 2100 \sigma\hackscore{ij}' = 2 \eta\hackscore{eff} \left( D\hackscore{ij}' + 465 \frac{1}{ 2 \mu \; dt} \sigma\hackscore{ij}^{'-}\right) \mbox{ with } 466 gross 1878 \frac{1}{\eta\hackscore{eff}}=\frac{1}{\mu \; dt}+\frac{1}{\eta^{vp}} 467 \end{equation} 468 gross 2100 Notice that $\eta\hackscore{eff}$ is a function of diatoric stress $\sigma\hackscore{ij}'$. 469 gross 1859 After inserting~\ref{IKM-EQU-10} into~\ref{IKM-EQU-1} we get 470 \begin{equation}\label{IKM-EQU-1ib} 471 gross 2100 -\left(\eta\hackscore{eff} (v\hackscore{i,j}+ v\hackscore{i,j}) 472 \right)\hackscore{,j}+p\hackscore{,i}=F\hackscore{i}+ 473 gross 1878 \frac{\eta\hackscore{eff}}{\mu dt } \sigma\hackscore{ij,j}^{'-} 474 gross 1859 \end{equation} 475 gross 1878 Together with the incomressibilty condition~\ref{IKM-EQU-2} we need to solve a problem with a form almost identical 476 jfenwick 1966 to the Stokes problem discussed in section~\ref{STOKES SOLVE} but with the difference that $\eta\hackscore{eff}$ is depending on the solution. Analog to the iteration scheme~\ref{SADDLEPOINT iteration 2} we can run 477 gross 1878 \begin{equation} 478 \begin{array}{rcl} 479 gross 2100 -\left(\eta\hackscore{eff}(dv\hackscore{i,j}+ dv\hackscore{i,j} 480 )\right)\hackscore{,j} & = & F\hackscore{i}+ \sigma\hackscore{ij,j}'-p\hackscore{,i} \\ 481 \frac{1}{\eta\hackscore{eff}} dp & = & - v\hackscore{i,i}^+ 482 gross 1878 \end{array} 483 \label{IKM iteration 2} 484 \end{equation} 485 where $v^+=v+dv$. As this problem is non-linear the Jacobi-free Newton-GMRES method is used with the norm 486 \begin{equation} 487 gross 2100 \|(v, p)\|^2= \int\hackscore{\Omega} v\hackscore{i,j}^2 + \frac{1}{\bar{\eta}^2\hackscore{eff}} p^2 \; dx 488 gross 1878 \label{IKM iteration 3} 489 \end{equation} 490 gross 2100 where $\bar{\eta}\hackscore{eff}$ is the caracteristic viscosity, for instance: 491 \begin{equation} 492 \frac{1}{\bar{\eta}\hackscore{eff}} = \frac{1}{\tau^{-}}+\sum\hackscore{q} \frac{1}{\eta^{q}\hackscore{N}} 493 \label{IKM iteration 4} 494 \end{equation} 495 In oder to perform step~\ref{IKM iteration 2} we need to calculate the $\eta\hackscore{eff}$ as well as $\sigma\hackscore{ij}'$ while via $\tau$ the first is a function of the latter. The priority is the 496 calculation of $\eta\hackscore{eff}$ with the Newton-Raphson scheme. This value can then be used to calculate 497 $\sigma\hackscore{ij}'$ via~\ref{IKM-EQU-10}. We need to solve 498 \begin{equation} 499 \tau = \eta\hackscore{eff} \cdot \epsilon \mbox{ with } 500 \epsilon = \sqrt{ 2 \left( D\hackscore{ij}' + 501 \frac{1}{ 2 \mu \; dt} \sigma\hackscore{ij}^{'-}\right)^2} 502 \label{IKM iteration 5} 503 \end{equation} 504 The Newton scheme takes the form 505 \begin{equation} 506 \tau\hackscore{n+1} = \min(\tau\hackscore{n} - \frac{\tau\hackscore{n} - \eta\hackscore{eff} \cdot \epsilon}{1 - \eta\hackscore{eff}' \cdot \epsilon}, \tau\hackscore{Y} + \beta \; p) 507 = \min(\frac{\eta\hackscore{eff} - \tau\hackscore{n} \eta\hackscore{eff}'} 508 {1 - \eta\hackscore{eff}' \cdot \epsilon}, \frac{\tau\hackscore{Y} + \beta \; p}{\epsilon}) \epsilon 509 \label{IKM iteration 6} 510 \end{equation} 511 where $\eta\hackscore{eff}'$ denotes the derivative of $\eta\hackscore{eff}$ with respect of $\tau$. The second term in $\min$ is droped of $\tau\hackscore{Y} + \beta \; p<0$ or $\epsilon=0$. In fact we have 512 \begin{equation} 513 \eta\hackscore{eff}' = - \eta\hackscore{eff}^2 \left(\frac{1}{\eta\hackscore{eff}}\right)' 514 \mbox{ with } 515 \left(\frac{1}{\eta\hackscore{eff}}\right)' = \sum\hackscore{q} \left(\frac{1}{\eta^{q}} \right)' 516 \label{IKM iteration 7} 517 \end{equation} 518 \begin{equation}\label{IKM-EQU-5XX} 519 \left(\frac{1}{\eta^{q}} \right)' 520 = \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}}}} 521 = \frac{1-\frac{1}{n^{q}}}{ \tau \eta^{q}} 522 \end{equation} 523 Notice that allways $\eta\hackscore{eff}'\le 0$ which makes the denomionator in~\ref{IKM iteration 6} 524 positive. 525 gross 1841 526 gross 1878 527 gross 2100