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

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a power law model. tests still need to be added and needs to be linked with an incompressible solver.

 1 2 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 3 % 4 % Copyright (c) 2003-2008 by University of Queensland 5 % Earth Systems Science Computational Center (ESSCC) 6 7 % 8 % Primary Business: Queensland, Australia 9 % Licensed under the Open Software License version 3.0 10 11 % 12 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 13 14 15 \chapter{Models} 16 \label{MODELS CHAPTER} 17 18 The following sections give a breif overview of the model classes and their corresponding methods. 19 20 \section{Stokes Problem} 21 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 \begin{equation}\label{Stokes 1} 23 -\left(\eta(v\hackscore{i,j}+ v\hackscore{i,j})\right)\hackscore{,j}+p\hackscore{,i}=f\hackscore{i}-\sigma\hackscore{ij,j} 24 \end{equation} 25 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 \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 \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 \end{equation} 33 which can be overwritten by constraints of the form 34 \begin{equation}\label{Stokes Boundary0} 35 v\hackscore{i}(x)=v^D\hackscore{i}(x) 36 \end{equation} 37 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 40 \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 \index{saddle point problem} 43 \begin{equation} 44 \left[ \begin{array}{cc} 45 A & B^{*} \\ 46 B & 0 \\ 47 \end{array} \right] 48 \left[ \begin{array}{c} 49 v \\ 50 p \\ 51 \end{array} \right] 52 =\left[ \begin{array}{c} 53 G \\ 54 0 \\ 55 \end{array} \right] 56 \label{SADDLEPOINT} 57 \end{equation} 58 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 \begin{equation} 62 \|v\hackscore{k,k}\| \hackscore \le \epsilon 63 \|\sqrt{v\hackscore{j,k}v\hackscore{j,k}}\| 64 \label{STOKES STOP} 65 \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 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 \begin{equation} 75 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 S p = B A^{-1} G 81 \end{equation} 82 with the Schur complement \index{Schur complement} $S=BA^{-1}B^{*}$. This problem can be solved iteratively 83 with the preconditioner $\hat{S}$ defined as $q=\hat{S}^{-1}p$ by solving 84 \begin{equation} 85 \frac{1}{\eta}q = p 86 \end{equation} 87 see~\cite{ELMAN} for more details. Note that the residual for the current approximation $p$ is given as 88 \begin{equation} 89 r=B A^{-1} (G - B^* p) = Bv 90 \end{equation} 91 where $v$ is given by~\ref{V CALC}. 92 93 If one uses the generalized minimal residual method (GMRES) \index{generalized minimal residual method!GMRES} 94 the method is directly applied to the preconditioned system 95 \begin{equation} 96 \hat{S}^{-1} S p = \hat{S}^{-1} B A^{-1} G 97 \end{equation} 98 We use the norm 99 \begin{equation} 100 \|p\|\hackscore{GMRES} = \|\hat{S} p \| 101 \end{equation} 102 Notice that for the residual $\hat{r}=\hat{S}^{-1} r$ one has 103 \begin{equation} 104 \ 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 \begin{equation} 110 \|\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 \begin{equation} 118 \|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 \begin{equation} 123 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} 126 127 128 129 \subsection{Functions} 130 131 \begin{classdesc}{StokesProblemCartesian}{domain} 132 opens the Stokes problem\index{Stokes problem} on the \Domain domain. The approximation 133 order needs to be two. 134 \end{classdesc} 135 136 \begin{methoddesc}[StokesProblemCartesian]{initialize}{\optional{f=Data(), \optional{fixed_u_mask=Data(), \optional{eta=1, \optional{surface_stress=Data(), \optional{stress=Data()}}}}}} 137 assigns values to the model parameters. In any call all values must be set. 138 \var{f} defines the external force $f$, \var{eta} the viscosity $\eta$, 139 \var{surface_stress} the surface stress $s$ and \var{stress} the initial stress $\sigma$. 140 The locations and compontents where the velocity is fixed are set by 141 the values of \var{fixed_u_mask}. The method will try to cast the given values to appropriate 142 \Data class objects. 143 \end{methoddesc} 144 145 \begin{methoddesc}[StokesProblemCartesian]{solve}{v,p, 146 \optional{max_iter=20, \optional{verbose=False, \optional{usePCG=True}}}} 147 solves the problem and return approximations for velocity and pressure. 148 The arguments \var{v} and \var{p} define initial guess. The values of \var{v} marked 149 by \var{fixed_u_mask} remain unchanged. 150 If \var{usePCG} is set to \True 151 reconditioned conjugate gradient method (PCG) \index{preconditioned conjugate gradient method!PCG} scheme is used. Otherwise the problem is solved generalized minimal residual method (GMRES) \index{generalized minimal residual method!GMRES}. In most cases 152 the PCG scheme is more efficient. 153 \var{max_iter} defines the maximum number of iteration steps. 154 If \var{verbose} is set to \True informations on the progress of of the solver are printed. 155 \end{methoddesc} 156 157 158 \begin{methoddesc}[StokesProblemCartesian]{setTolerance}{\optional{tolerance=1.e-4}} 159 sets the tolerance in an appropriate norm relative to the right hand side. The tolerance must be non-negative and less than 1. 160 \end{methoddesc} 161 \begin{methoddesc}[StokesProblemCartesian]{getTolerance}{} 162 returns the current relative tolerance. 163 \end{methoddesc} 164 \begin{methoddesc}[StokesProblemCartesian]{setAbsoluteTolerance}{\optional{tolerance=0.}} 165 sets the absolute tolerance for the error in the relevant norm. The tolerance must be non-negative. Typically the 166 absolute talerance is set to 0. 167 \end{methoddesc} 168 \begin{methoddesc}[StokesProblemCartesian]{getAbsoluteTolerance}{} 169 sreturns the current absolute tolerance. 170 \end{methoddesc} 171 \begin{methoddesc}[StokesProblemCartesian]{setSubProblemTolerance}{\optional{rtol=None}} 172 sets the tolerance to solve the involved PDEs. The subtolerance \var{rtol} should not be choosen to large 173 in order to avoid feed back of errors in the subproblem solution into the outer iteration. 174 On the otherhand is choosen to small compute time is wasted. 175 If \var{rtol} is set to \var{None} the sub-tolerance is set automatically depending on the 176 tolerance choosen for the oter iteration. 177 \end{methoddesc} 178 \begin{methoddesc}[StokesProblemCartesian]{getSubProblemTolerance}{} 179 return the tolerance for the involved PDEs. 180 \end{methoddesc} 181 182 \subsection{Example: Lit Driven Cavity} 183 The following script \file{lit\hackscore driven\hackscore cavity.py} 184 \index{scripts!\file{helmholtz.py}} which is available in the \ExampleDirectory 185 illustrates the usage of the \class{StokesProblemCartesian} class to solve 186 the lit driven cavity problem~\cite{LITDRIVENCAVITY}: 187 \begin{python} 188 from esys.escript import * 189 from esys.finley import Rectangle 190 from esys.escript.models import StokesProblemCartesian 191 NE=25 192 dom = Rectangle(NE,NE,order=2) 193 x = dom.getX() 194 sc=StokesProblemCartesian(dom) 195 mask= (whereZero(x[0])*[1.,0]+whereZero(x[0]-1))*[1.,0] + \ 196 (whereZero(x[1])*[0.,1.]+whereZero(x[1]-1))*[1.,1] 197 sc.initialize(eta=.1, fixed_u_mask= mask) 198 v=Vector(0.,Solution(dom)) 199 v[0]+=whereZero(x[1]-1.) 200 p=Scalar(0.,ReducedSolution(dom)) 201 v,p=sc.solve(v,p, verbose=True) 202 saveVTK("u.xml",velocity=v,pressure=p) 203 \end{python} 204 205 \section{Darcy Flux} 206 We want to calculate the velocity $u$ and pressure $p$ on a domain $\Omega$ solving 207 the Darcy flux problem \index{Darcy flux}\index{Darcy flow} 208 \begin{equation}\label{DARCY PROBLEM} 209 \begin{array}{rcl} 210 u\hackscore{i} + \kappa\hackscore{ij} p\hackscore{,j} & = & g\hackscore{i} \\ 211 u\hackscore{k,k} & = & f 212 \end{array} 213 \end{equation} 214 with the boundary conditions 215 \begin{equation}\label{DARCY BOUNDARY} 216 \begin{array}{rcl} 217 u\hackscore{i} \; n\hackscore{i} = u^{N}\hackscore{i} \; n\hackscore{i} & \mbox{ on } & \Gamma\hackscore{N} \\ 218 p = p^{D} & \mbox{ on } & \Gamma\hackscore{D} \\ 219 \end{array} 220 \end{equation} 221 where $\Gamma\hackscore{N}$ and $\Gamma\hackscore{D}$ are a partition of the boundary of $\Omega$ with $\Gamma\hackscore{D}$ non empty, $n\hackscore{i}$ is the outer normal field of the boundary of $\Omega$, $u^{N}\hackscore{i}$ and $p^{D}$ are given functions on $\Omega$, $g\hackscore{i}$ and $f$ are given source terms and $\kappa\hackscore{ij}$ is the given permability. We assume that $\kappa\hackscore{ij}$ is symmetric (which is not really required) and positive definite, i.e there are positive constants $\alpha\hackscore{0}$ and $\alpha\hackscore{1}$ wich are independent from the location in $\Omega$ such that 222 \begin{equation} 223 \alpha\hackscore{0} \; x\hackscore{i} x\hackscore{i} \le \kappa\hackscore{ij} x\hackscore{i} x\hackscore{j} \le \alpha\hackscore{1} \; x\hackscore{i} x\hackscore{i} 224 \end{equation} 225 for all $x\hackscore{i}$. 226 227 228 \subsection{Solution Method \label{DARCY SOLVE}} 229 In practical applications it is an advantage to calculate the pressure $p$ as a correction of a 'static' pressure $p^{ref}$ which is the solution of 230 \begin{equation} 231 -(\kappa\hackscore{ki}\kappa\hackscore{kj} p\hackscore{,j}^{ref})\hackscore{,i} = - (\kappa\hackscore{ki} (g\hackscore{k}- u^{N}\hackscore{k}))\hackscore{,i} 232 \mbox{ with } 233 p^{ref} = p^{D} \mbox{ on } \Gamma\hackscore{D} 234 \end{equation} 235 With setting $u \leftarrow u-u^{N}$ and $p \leftarrow p-p^{ref}$ and 236 \begin{equation} 237 \begin{array}{rcl} 238 g\hackscore{i} & \leftarrow & g\hackscore{i} - u^{N}\hackscore{i} - \kappa\hackscore{ij} p^{ref}\hackscore{,j }\\ 239 f & \leftarrow & f - u^{N}\hackscore{k,k} 240 \end{array} 241 \end{equation} 242 we can assume that $u^{N}\hackscore{i} \; n\hackscore{i}=0$ and 243 $p^{D}=0$. 244 We set 245 \begin{equation} 246 V = \{ q \in H^{1}(\Omega) : q=0 \mbox{ on } \Gamma\hackscore{D} \} 247 \end{equation} 248 and 249 \begin{equation} 250 W = \{ v \in (L^2(\Omega))^{d} : v\hackscore{k,k} \in L^2(\Omega) \mbox{ and } u\hackscore{i} \; n\hackscore{i} =0 \mbox{ on } \Gamma\hackscore{N} \} 251 \end{equation} 252 and define the operator $Q: V \rightarrow (L^2(\Omega))^{d}$ defined by 253 \begin{equation} 254 (Qp)\hackscore{i} = \kappa\hackscore{ij} p\hackscore{,j} 255 \end{equation} 256 and the operator $D: W \rightarrow L^2(\Omega)$ defined by 257 \begin{equation} 258 Dv = v\hackscore{k,k} 259 \end{equation} 260 In operator notation the Darcy problem~\ref{DARCY PROBLEM} is written in the form 261 \begin{equation} 262 \begin{array}{rcl} 263 u + Qp & = & g \\ 264 Du & = & f 265 \end{array} 266 \end{equation} 267 We solve this equation by minimising the functional 268 \begin{equation} 269 J(u,p):=\|u + Qp - g\|^2\hackscore{0} + \|Du-f\|\hackscore{0}^2 270 \end{equation} 271 over $W \times V$ where $\|.\|\hackscore{0}$ denotes the norm in $L^2(\Omega)$. A simple calculation shows that 272 one has to solve 273 \begin{equation} 274 ( v + Qq , u + Qp - g) + (Dv,Du-f) =0 275 \end{equation} 276 for all $v\in W$ and $q \in V$.which translates back into operator notation 277 \begin{equation} 278 \begin{array}{rcl} 279 (I+D^*D)u + Qp & = & D^*f + g \\ 280 Q^*u + Q^*Q p & = & Q^*g \\ 281 \end{array} 282 \end{equation} 283 where $D^*$ and $Q^*$ denote the adjoint operators. 284 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 285 to solve the problem is coupled form, however this approach leads in some cases to a very ill-conditioned stiffness matrix in particular in the case of a very small or large permability ($\alpha\hackscore{1} \ll 1$ or $\alpha\hackscore{0} \gg 1$) 286 287 The approach we are taking is to eliminate the velocity $u$ from the problem. Assuming that $p$ is known we have 288 \begin{equation}\label{DARCY V FORM} 289 v= (I+D^*D)^{-1} (D^*f + g - Qp) 290 \end{equation} 291 (notice that $(I+D^*D)$ is coercive in $W$) which is inserted into the second equation 292 \begin{equation} 293 Q^* (I+D^*D)^{-1} (D^*f + g - Qp) + Q^* Q p = Q^*g 294 \end{equation} 295 which is 296 \begin{equation} 297 Q^* ( I - (I+D^*D)^{-1} ) Q p = Q^* (g-(I+D^*D)^{-1} (D^*f + g) ) 298 \end{equation} 299 We use the PCG \index{linear solver!PCG}\index{PCG} method to solve this. The residual $r$ ($\in V^*$) is given as 300 \begin{equation} 301 \begin{array}{rcl} 302 r & = & Q^* \left( g -(I+D^*D)^{-1} (D^*f + g) - Qp + (I+D^*D)^{-1}Q p \right)\\ 303 & =& Q^* \left( g- Qp - (I+D^*D)^{-1} (D^*f + g - Qp) \right) \\ 304 & =& Q^* \left( g - Qp - v \right) 305 \end{array} 306 \end{equation} 307 So in a partical implementation we use $\hat{r}=g-Qp-v$ to represent the residual. 308 The evaluation of the iteration operator for a given $p$ is then 309 returning $Qp+v$ where $v$ is the solution of 310 \begin{equation}\label{UPDATE W} 311 (I+D^*D)v = Qp 312 \end{equation} 313 We use $(Q^*Q)^{-1}$ as a preconditioner for the iteration operator $Q^* ( I - (I+D^*D)^{-1} ) Q$. So the application of the preconditioner to $\hat{r}$ representing the residual is given by solving 314 implemented by solving 315 \begin{equation}\label{UPDATE P} 316 Q^*Q q = Q^*\hat{r} 317 \end{equation} 318 The residual norm used in the PCG is given as 319 \begin{equation}\label{DARCY R NORM} 320 \|r\|\hackscore{PCG}^2 = \int r \cdot (Q^*Q)^{-1} r \; dx =\int \hat{r} \cdot Q (Q^*Q)^{-1} Q^* \hat{r} \; dx \approx 321 \|\hat{r}\|\hackscore{0}^2 322 \end{equation} 323 The iteration is terminated if 324 \begin{equation}\label{DARCY STOP} 325 \|r\|\hackscore{PCG} \le \mbox{ATOL} 326 \end{equation} 327 where we set 328 \begin{equation}\label{DARCY ATOL DEF} 329 \mbox{ATOL} = \mbox{atol} + \mbox{rtol} \cdot \left(\frac{1}{\|v\|\hackscore{0}} + \frac{1}{\|Qp\|\hackscore{0}} \right)^{-1} 330 \end{equation} 331 where rtol is a given relative tolerance and $\mbox{atol}$ is a given absolute tolerance (typically $=0$). 332 Notice that if $Qp$ and $v$ both are zero, the pair $(0,p)$ is a solution. 333 The problem is that ATOL is depending on the solution $p$ (and $v$ calculated form~\ref{DARCY V FORM}). In partcice one use the initial guess for $p$ 334 to get a first value for ATOL. If the stopping crierion is met in the PCG iteration, a new $v$ is calculated from the current pressure approximation and ATOL is recalculated. If \ref{DARCY STOP} is still fullfilled the calculation is terminated and $(v,p)$ is returned. Otherwise PCG is restarted with a new ATOL. 335 336 \subsection{Functions} 337 \begin{classdesc}{DarcyFlow}{domain} 338 opens the Darcy flux problem\index{Darcy flux} on the \Domain domain. 339 \end{classdesc} 340 \begin{methoddesc}[DarcyFlow]{setValue}{\optional{f=None, \optional{g=None, \optional{location_of_fixed_pressure=None, \optional{location_of_fixed_flux=None, \optional{permeability=None}}}}}} 341 assigns values to the model parameters. Values can be assigned using various calls - in particular 342 in a time dependend problem only values that change over time needs to be reset. The permability can be defined as scalar (isotropic), a vector (orthotropic) or a matrix (anisotropic). 343 \var{f} and \var{g} are the corresponding parameters in~\ref{DARCY PROBLEM}. 344 The locations and compontents where the flux is prescribed are set by positive values in 345 \var{location_of_fixed_flux}. 346 The locations where the pressure is prescribed are set by 347 by positive values of \var{location_of_fixed_pressure}. 348 The values of the pressure and flux are defined by the initial guess. 349 Notice that at any point on the boundary of the domain the pressure or the normal component of 350 the flux must be defined. There must be at least one point where the pressure is prescribed. 351 The method will try to cast the given values to appropriate 352 \Data class objects. 353 \end{methoddesc} 354 355 \begin{methoddesc}[DarcyFlow]{setTolerance}{\optional{rtol=1e-4}} 356 sets the relative tolerance \mbox{rtol} in \ref{DARCY ATOL DEF}. 357 \end{methoddesc} 358 359 \begin{methoddesc}[DarcyFlow]{setAbsoluteTolerance}{\optional{atol=0.}} 360 sets the absolute tolerance \mbox{atol} in \ref{DARCY ATOL DEF}. 361 \end{methoddesc} 362 363 \begin{methoddesc}[DarcyFlow]{setSubProblemTolerance}{\optional{rtol=None}} 364 sets the relative tolerance used to solve the involved PDEs. If no argument is given, 365 the square of the current relative tolerance is used. The sub-problem tolerance should be choosen as large as possible to minimize the compute time. However, a too large value for the sub-problem tolerance may lead to slow convergence or even dibergence in the outer iteration. 366 \end{methoddesc} 367 368 \begin{methoddesc}[DarcyFlow]{solve}{u0,p0, \optional{max_iter=100, \optional{verbose=False \optional{sub_rtol=1.e-8}}}} 369 solves the problem. and returns approximations for the flux $v$ and the pressure $p$. 370 \var{u0} and \var{p0} define initial guess for flux and pressure. Values marked 371 by positive values \var{location_of_fixed_flux} and \var{location_of_fixed_pressure}, respectively, are kept unchanged. 372 \end{methoddesc} 373 374 375 \subsection{Example: Gravity Flow} 376 later 377 378 %================================================ 379 % \section{Temperature Advection Diffusion\label{TEMP ADV DIFF}} 380 381 % 382 % \rho c\hackscore{p} \left (\frac{\partial T}{\partial t} + \vec{v} \cdot \nabla T \right ) = k \nabla^{2}T 383 % \label{HEAT EQUATION} 384 % 385 386 % where $\vec{v}$ is the velocity vector, $T$ is the temperature, $\rho$ is the density, $\eta$ is the viscosity, % % $c\hackscore{p}$ is the specific heat at constant pressure and $k$ is the thermal conductivity. 387 388 % \subsection{Description} 389 390 % \subsection{Method} 391 % 392 % \begin{classdesc}{TemperatureCartesian}{dom,theta=THETA,useSUPG=SUPG} 393 % \end{classdesc} 394 395 % \subsection{Benchmark Problem} 396 %=============================================================================================================== 397 398 %========================================================= 399 \input{levelsetmodel} 400 401 % \section{Drucker Prager Model} 402 403 \section{Isotropic Kelvin Material \label{IKM}} 404 As proposed by Kelvin~\ref{KELVN} material strain $D\hackscore{ij}=v\hackscore{i,j}+v\hackscore{j,i}$ can be decomposed into 405 an elastic part $D\hackscore{ij}^{el}$ and visco-plastic part $D\hackscore{ij}^{vp}$: 406 \begin{equation}\label{IKM-EQU-2} 407 D\hackscore{ij}=D\hackscore{ij}^{el}+D\hackscore{ij}^{vp} 408 \end{equation} 409 with the elastic strain given as 410 \begin{equation}\label{IKM-EQU-3} 411 D\hackscore{ij}^{el'}=\frac{1}{2 \mu} \dot{\sigma}\hackscore{ij}' 412 \end{equation} 413 where $\sigma'\hackscore{ij}$ is the deviatoric stress (Notice that $\sigma'\hackscore{ii}=0$). 414 If the material is composed by materials $q$ the visco-plastic strain can be decomposed as 415 \begin{equation}\label{IKM-EQU-4} 416 D\hackscore{ij}^{vp'}=\sum\hackscore{q} D\hackscore{ij}^{q'} 417 \end{equation} 418 where $D\hackscore{ij}^{q}$ is the strain in material $q$ given as 419 \begin{equation}\label{IKM-EQU-5} 420 D\hackscore{ij}^{q'}=\frac{1}{2 \eta^{q}} \sigma'\hackscore{ij} 421 \end{equation} 422 where $\eta^{q}$ is the viscosity of material $q$. We assume the following 423 betwee the the strain in material $q$ 424 \begin{equation}\label{IKM-EQU-5b} 425 \eta^{q}=\eta^{q}\hackscore{N} \left(\frac{\tau}{\tau\hackscore{t}^q}\right)^{\frac{1}{n^{q}}-1} 426 \mbox{ with } \tau=\sqrt{\frac{1}{2}\sigma'\hackscore{ij} \sigma'\hackscore{ij}} 427 \end{equation} 428 for a given power law coefficients $n^{q}\ge1$ and transition stresses $\tau\hackscore{t}^q$, see~\ref{POERLAW}. 429 Notice that $n^{q}=1$ gives a constant viscosity. 430 After inserting equation~\ref{IKM-EQU-5} into equation \ref{IKM-EQU-4} one gets: 431 \begin{equation}\label{IKM-EQU-6} 432 D\hackscore{ij}'^{vp}=\frac{1}{2 \eta^{vp}} \sigma'\hackscore{ij} \mbox{ with } \frac{1}{\eta^{vp}} = \sum\hackscore{q} \frac{1}{\eta^{q}} \;. 433 \end{equation} 434 and finally with~\ref{IKM-EQU-2} 435 \begin{equation}\label{IKM-EQU-2b} 436 D\hackscore{ij}'=\frac{1}{2 \eta^{vp}} \sigma'\hackscore{ij}+\frac{1}{2 \mu} \dot{\sigma}\hackscore{ij}' 437 \end{equation} 438 The total stress $\tau$ needs to fullfill the yield condition \index{yield condition} 439 \begin{equation}\label{IKM-EQU-8c} 440 \tau \le \tau\hackscore{Y} + \beta \; p 441 \end{equation} 442 with the Drucker-Prager \index{Druck-Prager} cohesion factor \index{cohesion factor} $\tau\hackscore{Y}$, Drucker-Prager friction $\beta$ and total pressure $p$. 443 The deviatoric stress needs to fullfill the equilibrion equation 444 \begin{equation}\label{IKM-EQU-1} 445 -\sigma'\hackscore{ij,j}+p\hackscore{,i}=F\hackscore{i} 446 \end{equation} 447 where $F\hackscore{j}$ is a given external fource. We assume an incompressible media: 448 \begin{equation}\label{IKM-EQU-2} 449 -v\hackscore{i,i}=0 450 \end{equation} 451 Natural boundary conditions are taken in the form 452 \begin{equation}\label{IKM-EQU-Boundary} 453 \sigma'\hackscore{ij}n\hackscore{j}-n\hackscore{i}p=f 454 \end{equation} 455 which can be overwritten by a constraint 456 \begin{equation}\label{IKM-EQU-Boundary0} 457 v\hackscore{i}(x)=0 458 \end{equation} 459 where the index $i$ may depend on the location $x$ on the bondary. 460 461 \subsection{Solution Method \label{IKM-SOLVE}} 462 By using a first order finite difference approximation wit step size $dt>0$~\ref{IKM-EQU-3} get the form 463 \begin{equation}\label{IKM-EQU-3b} 464 \dot{\sigma}\hackscore{ij}=\frac{1}{dt } \left( \sigma\hackscore{ij} - \sigma\hackscore{ij}^{-} \right) 465 \end{equation} 466 and 467 \begin{equation}\label{IKM-EQU-2b} 468 D\hackscore{ij}'=\left(\frac{1}{2 \eta^{vp}} + \frac{1}{2 \mu dt}\right) \sigma\hackscore{ij}'+\frac{1}{2 \mu dt } \sigma\hackscore{ij}^{-'} 469 \end{equation} 470 where $\sigma\hackscore{ij}^{-}$ is the stress at the precious time step. With 471 \begin{equation}\label{IKM-EQU-2c} 472 \dot{\gamma} = \sqrt{ 2 \left( D\hackscore{ij}' - 473 \frac{1}{ 2 \mu \; dt} \sigma\hackscore{ij}^{'-}\right)^2} 474 \end{equation} 475 we have 476 \begin{equation} 477 \tau = \eta\hackscore{eff} \cdot \dot{\gamma} 478 \end{equation} 479 where 480 \begin{equation} 481 \eta\hackscore{eff}= min( \left(\frac{1}{\mu \; dt}+\frac{1}{\eta^{vp}}\right)^{-1} 482 , \eta\hackscore{max}) \mbox{ with } 483 \eta\hackscore{max} = \left\{ 484 \begin{array}{rcl} 485 \frac{\tau\hackscore{Y} + \beta \; p}{\dot{\gamma}} & & \dot{\gamma}>0 \\ 486 &\mbox{ if } \\ 487 \infty & & \mbox{otherwise} 488 \end{array} 489 \right. 490 \end{equation} 491 The upper bound $\eta\hackscore{max}$ makes sure that yield condtion~\ref{IKM-EQU-8c} holds. With this setting the eqaution \ref{IKM-EQU-2b} takes the form 492 \begin{equation}\label{IKM-EQU-10} 493 \sigma\hackscore{ij}' = 2 \eta\hackscore{eff} \left( D\hackscore{ij}' - 494 \frac{1}{ 2 \mu \; dt} \sigma\hackscore{ij}^{'-}\right) 495 \end{equation} 496 After inserting~\ref{IKM-EQU-10} into~\ref{IKM-EQU-1} we get 497 \begin{equation}\label{IKM-EQU-1ib} 498 -\left(\eta\hackscore{eff} (v\hackscore{i,j}+ v\hackscore{i,j}) 499 \right)\hackscore{,j}+p\hackscore{,i}=F\hackscore{i}- 500 \frac{\eta\hackscore{eff}}{\mu dt } \sigma\hackscore{ij,j}^{'-} 501 \end{equation} 502 Combining this with the incomressibilty condition~\ref{IKM-EQU-2} we need to solve a 503 Stokes problem as discussed in section~\ref{STOKES SOLVE} in each time step. 504 505 If we set 506 \begin{equation}\label{IKM-EQU-44} 507 \frac{1}{\eta(\tau)}= \frac{1}{\mu \; dt}+\frac{1}{\eta^{vp}} 508 \end{equation} 509 we need to solve the nonlinear problem 510 \begin{equation} 511 \eta\hackscore{eff} - min(\eta( \dot{\gamma} \cdot \eta\hackscore{eff}) 512 , \eta\hackscore{max}) =0 513 \end{equation} 514 We use the Newton-Raphson Scheme \index{Newton-Raphson scheme} to solve this problem 515 \begin{equation}\label{IKM-EQU-45} 516 \eta\hackscore{eff}^{(n+1)} = \min(\eta\hackscore{max}, 517 \eta\hackscore{eff}^{(n)} - 518 \frac{\eta\hackscore{eff}^{(n)} - \eta( \tau^{(n)}) } 519 {1-\dot{\gamma} \cdot \eta'( \tau^{(n)} )} ) 520 =\min(\eta\hackscore{max}, 521 \frac{\eta( \tau^{(n)}) -\tau^{(n)} \cdot \eta'( \tau^{(n)} ) } 522 {1-\dot{\gamma} \cdot \eta'( \tau^{(n)} )} ) 523 \end{equation} 524 where $\eta'$ denotes the derivative of $\eta$ with respect of $\tau$ 525 and $\tau^{(n)} = \dot{\gamma} \cdot \eta\hackscore{eff}^{(n)}$. 526 527 Looking at the evaluation of $\eta$ in~\ref{IKM-EQU-44} it makes sense formulate 528 the iteration~\ref{IKM-EQU-45} using $\Theta=\eta^{-1}$. 529 In fact we have 530 \begin{equation} 531 \eta' = - \frac{\Theta'}{\Theta^2} 532 \mbox{ with } 533 \Theta' = \sum\hackscore{q} \left(\frac{1}{\eta^{q}} \right)' 534 \label{IKM iteration 7} 535 \end{equation} 536 As 537 \begin{equation}\label{IKM-EQU-47} 538 \left(\frac{1}{\eta^{q}} \right)' 539 = \frac{1-\frac{1}{n^{q}}}{\eta^{q}\hackscore{N}} \frac{\tau^{-\frac{1}{n^{q}}}}{(\tau\hackscore{t}^q)^{1-\frac{1}{n^{q}}}} 540 = \frac{1-\frac{1}{n^{q}}}{\eta^{q}}\frac{1}{\tau} 541 \end{equation} 542 we have 543 \begin{equation}\label{IKM-EQU-48} 544 \Theta' = \frac{1}{\tau} \omega \mbox{ with } \omega = \sum\hackscore{q}\frac{1-\frac{1}{n^{q}}}{\eta^{q}} 545 \end{equation} 546 which leads to 547 \begin{equation}\label{IKM-EQU-45} 548 \eta\hackscore{eff}^{(n+1)} = \min(\eta\hackscore{max}, 549 \eta\hackscore{eff}^{(n)} 550 \frac{\Theta^{(n)} + \omega^{(n)} } 551 {\eta\hackscore{eff}^{(n)} \Theta^{(n)^2}+\omega^{(n)} }) 552 \end{equation}

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