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

Revision 2156 - (show annotations)
Mon Dec 15 05:09:02 2008 UTC (11 years, 2 months ago) by gross
File MIME type: application/x-tex
File size: 24635 byte(s)
some modifications to the iterative solver to make them easier to use.
There are also improved versions of the Darcy flux solver and the 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 \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 \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 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 \begin{equation} 116 \begin{array}{rcl} 117 A w & = & Av+B^{*}p \\ 118 \hat{S} q & = & B(w-v) \\ 119 \end{array} 120 \label{COUPLES SADDLEPOINT iteration} 121 \end{equation} 122 We use the inner product induced by the norm 123 \begin{equation} 124 \|(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 \begin{array}{rcl} 130 -\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 \end{array} 133 \label{SADDLEPOINT iteration 2} 134 \end{equation} 135 136 137 \subsection{Functions} 138 139 \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} 143 144 \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} 152 153 \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} 164 165 166 \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} 190 191 \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 the lit driven cavity problem~\cite{LITDRIVENCAVITY}: 196 \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)*[1.,0]+whereZero(x-1))*[1.,0] + \ 205 (whereZero(x)*[0.,1.]+whereZero(x-1))*[1.,1] 206 sc.initialize(eta=.1, fixed_u_mask= mask) 207 v=Vector(0.,Solution(dom)) 208 v+=whereZero(x-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} 213 214 \section{Darcy Flux} 215 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 \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 $x\hackscore{i}$. 235 236 237 \subsection{Solution Method \label{DARCY SOLVE}} 238 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 239 \begin{equation} 240 -(\kappa\hackscore{ki}\kappa\hackscore{kj} p\hackscore{,j}^{ref})\hackscore{,i} = - (\kappa\hackscore{ki} (g\hackscore{k}- u^{N}\hackscore{k}))\hackscore{,i} 241 \mbox{ with } 242 p^{ref} = p^{D} \mbox{ on } \Gamma\hackscore{D} 243 \end{equation} 244 With setting $u \leftarrow u-u^{N}$ and $p \leftarrow p-p^{ref}$ and 245 \begin{equation} 246 \begin{array}{rcl} 247 g\hackscore{i} & \leftarrow & g\hackscore{i} - u^{N}\hackscore{i} - \kappa\hackscore{ij} p^{ref}\hackscore{,j }\\ 248 f & \leftarrow & f - u^{N}\hackscore{k,k} 249 \end{array} 250 \end{equation} 251 we can assume that $u^{N}\hackscore{i} \; n\hackscore{i}=0$ and 252 $p^{D}=0$. Notice that with this setting 253 \begin{equation}\label{DIV FREE DARCY FLUX} 254 (\kappa\hackscore{ki} g\hackscore{k})\hackscore{,i} = 0 255 \end{equation} 256 We set 257 \begin{equation} 258 V = \{ q \in H^{1}(\Omega) : q=0 \mbox{ on } \Gamma\hackscore{D} \} 259 \end{equation} 260 and 261 \begin{equation} 262 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} \} 263 \end{equation} 264 and define the operator $Q: V \rightarrow (L^2(\Omega))^{d}$ defined by 265 \begin{equation} 266 (Qp)\hackscore{i} = \kappa\hackscore{ij} p\hackscore{,j} 267 \end{equation} 268 and the operator $D: W \rightarrow L^2(\Omega)$ defined by 269 \begin{equation} 270 Dv = v\hackscore{k,k} 271 \end{equation} 272 In operator notation the Darcy problem~\ref{DARCY PROBLEM} is written in the form 273 \begin{equation} 274 \begin{array}{rcl} 275 u + Qp & = & g \\ 276 Du & = & f 277 \end{array} 278 \end{equation} 279 Notice that because of~\ref{DIV FREE DARCY FLUX} we have 280 \begin{equation} \label{ABSTRACT DIV FREE DARCY FLUX} 281 Q^*g =0 282 \end{equation} 283 where $Q^*$ denote the adjoint operators of $Q$. 284 We solve this equation by minimising the functional 285 \begin{equation} 286 J(u,p):=\|u + Qp - g\|^2\hackscore{0} + \|Du-f\|\hackscore{0}^2 287 \end{equation} 288 over $W \times V$ where $\|.\|\hackscore{0}$ denotes the norm in $L^2(\Omega)$. A simple calculation shows that 289 one has to solve 290 \begin{equation} 291 ( v + Qq , u + Qp - g) + (Dv,Du-f) =0 292 \end{equation} 293 for all $v\in W$ and $q \in V$.which translates back into operator notation 294 \begin{equation} 295 \begin{array}{rcl} 296 (I+D^*D)u + Qp & = & D^*f + g \\ 297 Q^*u + Q^*Q p & = & 0 \\ 298 \end{array} 299 \end{equation} 300 where $D^*$ and $Q^*$ denote the adjoint operators. We use the fact that $Q^*g=$. 301 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 302 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 $\alpha\hackscore{0} \gg 1$) 303 304 The approach we are taking is to eliminate the velocity $u$ from the problem. Assuming that $p$ is known we have 305 \begin{equation} 306 v= (I+D^*D)^{-1} (D^*f + g - Qp) 307 \end{equation} 308 (notice that $(I+D^*D)$ is coercive in $W$) which is inserted into the second equation 309 \begin{equation} 310 Q^* (I+D^*D)^{-1} (D^*f + g - Qp) + Q^* Q p = 0 311 \end{equation} 312 which is 313 \begin{equation} 314 Q^* ( I - (I+D^*D)^{-1} ) Q p = - Q^* (I+D^*D)^{-1} (D^*f + g) ) 315 \end{equation} 316 We use the PCG method to solve this. The residual $r$ ($\in V^*$) is given as 317 \begin{equation} 318 \begin{array}{rcl} 319 r & = & Q^* \left( -(I+D^*D)^{-1} (D^*f + g) - Qp + (I+D^*D)^{-1}Q p \right)\\ 320 & =& Q^* \left( - Qp - (I+D^*D)^{-1} (D^*f + g - Qp) \right) \\ 321 & =& - Q^* \left( Qp + v \right) 322 \end{array} 323 \end{equation} 324 So in a partical implementation we use the pair $(Qp,v)$ to represent the residual. This will save the 325 reconstruction of the velocity $v$. In this notation the right hand side is given as 326 $(0,(I+D^*D)^{-1} (D^*f + g))$. The evaluation of the iteration operator for a given $p$ is then 327 returning $(Qp,w)$ where $w$ is the solution of 328 \begin{equation}\label{UPDATE W} 329 (I+D^*D)w = Qp 330 \end{equation} 331 We use $Q^*Q$ as a a preconditioner for the iteration operator $Q^* ( I - (I+D^*D)^{-1} ) Q$. 332 333 \subsection{Functions} 334 \begin{classdesc}{DarcyFlow}{domain} 335 opens the Darcy flux problem\index{Darcy flux} on the \Domain domain. 336 \end{classdesc} 337 338 \begin{methoddesc}[DarcyFlow]{initialize}{\optional{f=Data(), \optional{fixed_u_mask=Data(), \optional{eta=1, \optional{surface_stress=Data(), \optional{stress=Data()}}}}}} 339 assigns values to the model parameters. In any call all values must be set. 340 \var{f} defines the external force $f$, \var{eta} the viscosity $\eta$, 341 \var{surface_stress} the surface stress $s$ and \var{stress} the initial stress $\sigma$. 342 The locations and compontents where the velocity is fixed are set by 343 the values of \var{fixed_u_mask}. The method will try to cast the given values to appropriate 344 \Data class objects. 345 \end{methoddesc} 346 347 348 \subsection{Example: Gravity Flow} 349 350 %================================================ 351 \section{Temperature Advection Diffusion\label{TEMP ADV DIFF}} 352 353 \begin{equation} 354 \rho c\hackscore{p} \left (\frac{\partial T}{\partial t} + \vec{v} \cdot \nabla T \right ) = k \nabla^{2}T 355 \label{HEAT EQUATION} 356 \end{equation} 357 358 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. 359 360 \subsection{Description} 361 362 \subsection{Method} 363 364 \begin{classdesc}{TemperatureCartesian}{dom,theta=THETA,useSUPG=SUPG} 365 \end{classdesc} 366 367 \subsection{Benchmark Problem} 368 %=============================================================================================================== 369 370 %========================================================= 371 \input{levelsetmodel} 372 373 % \section{Drucker Prager Model} 374 375 \section{Isotropic Kelvin Material \label{IKM}} 376 As proposed by Kelvin~\ref{KELVN} material strain $D\hackscore{ij}=v\hackscore{i,j}+v\hackscore{j,i}$ can be decomposed into 377 an elastic part $D\hackscore{ij}^{el}$ and visco-plastic part $D\hackscore{ij}^{vp}$: 378 \begin{equation}\label{IKM-EQU-2} 379 D\hackscore{ij}=D\hackscore{ij}^{el}+D\hackscore{ij}^{vp} 380 \end{equation} 381 with the elastic strain given as 382 \begin{equation}\label{IKM-EQU-3} 383 D\hackscore{ij}'^{el}=\frac{1}{2 \mu} \dot{\sigma}\hackscore{ij}' 384 \end{equation} 385 where $\sigma'\hackscore{ij}$ is the deviatoric stress (Notice that $\sigma'\hackscore{ii}=0$). 386 If the material is composed by materials $q$ the visco-plastic strain can be decomposed as 387 \begin{equation}\label{IKM-EQU-4} 388 D\hackscore{ij}'^{vp}=\sum\hackscore{q} D\hackscore{ij}'^{q} 389 \end{equation} 390 where $D\hackscore{ij}^{q}$ is the strain in material $q$ given as 391 \begin{equation}\label{IKM-EQU-5} 392 D\hackscore{ij}'^{q}=\frac{1}{2 \eta^{q}} \sigma'\hackscore{ij} 393 \end{equation} 394 where $\eta^{q}$ is the viscosity of material $q$. We assume the following 395 betwee the the strain in material $q$ 396 \begin{equation}\label{IKM-EQU-5b} 397 \eta^{q}=\eta^{q}\hackscore{N} \left(\frac{\tau}{\tau\hackscore{t}^q}\right)^{\frac{1}{n^{q}}-1} 398 \mbox{ with } \tau=\sqrt{\frac{1}{2}\sigma'\hackscore{ij} \sigma'\hackscore{ij}} 399 \end{equation} 400 for a given power law coefficients $n^{q}$ and transition stresses $\tau\hackscore{t}^q$, see~\ref{POERLAW}. 401 Notice that $n^{q}=1$ gives a constant viscosity. 402 After inserting equation~\ref{IKM-EQU-5} into equation \ref{IKM-EQU-4} one gets: 403 \begin{equation}\label{IKM-EQU-6} 404 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}} \;. 405 \end{equation} 406 With 407 \begin{equation}\label{IKM-EQU-8} 408 \dot{\gamma}=\sqrt{2 D\hackscore{ij} D\hackscore{ij}} 409 \end{equation} 410 one gets 411 \begin{equation}\label{IKM-EQU-8b} 412 \tau = \eta^{vp} \dot{\gamma}^{vp} \;. 413 \end{equation} 414 With the Drucker-Prager cohesion factor $\tau\hackscore{Y}$, Drucker-Prager friction $\beta$ and total pressure $p$ we want to achieve 415 \begin{equation}\label{IKM-EQU-8c} 416 \tau \le \tau\hackscore{Y} + \beta \; p 417 \end{equation} 418 which leads to the condition 419 \begin{equation}\label{IKM-EQU-8d} 420 \eta^{vp} \le \frac{\tau\hackscore{Y} + \beta \; p}{ \dot{\gamma}^{vp}} \; . 421 \end{equation} 422 Therefore we modify the definition of $\eta^{vp}$ to the form 423 \begin{equation}\label{IKM-EQU-6b} 424 \frac{1}{\eta^{vp}}=\max(\sum\hackscore{q} \frac{1}{\eta^{q}}, \frac{\dot{\gamma}^{vp}} {\tau\hackscore{Y} + \beta \; p}) 425 \end{equation} 426 The deviatoric stress needs to fullfill the equilibrion equation 427 \begin{equation}\label{IKM-EQU-1} 428 -\sigma'\hackscore{ij,j}+p\hackscore{,i}=F\hackscore{i} 429 \end{equation} 430 where $F\hackscore{j}$ is a given external fource. We assume an incompressible media: 431 \begin{equation}\label{IKM-EQU-2} 432 -v\hackscore{i,i}=0 433 \end{equation} 434 Natural boundary conditions are taken in the form 435 \begin{equation}\label{IKM-EQU-Boundary} 436 \sigma'\hackscore{ij}n\hackscore{j}-n\hackscore{i}p=f 437 \end{equation} 438 which can be overwritten by a constraint 439 \begin{equation}\label{IKM-EQU-Boundary0} 440 v\hackscore{i}(x)=0 441 \end{equation} 442 where the index $i$ may depend on the location $x$ on the bondary. 443 444 \subsection{Solution Method \label{IKM-SOLVE}} 445 By using a first order finite difference approximation wit step size $dt>0$~\ref{IKM-EQU-3} get the form 446 \begin{equation}\label{IKM-EQU-3b} 447 D\hackscore{ij}'^{el}=\frac{1}{2 \mu dt } \left( \sigma\hackscore{ij}' - \sigma\hackscore{ij}^{'-} \right) 448 \end{equation} 449 where $\sigma\hackscore{ij}^{'-}$ is the deviatoric stress at the precious time step. 450 Now we can combine equations~\ref{IKM-EQU-2}, \ref{IKM-EQU-3b} and~\ref{IKM-EQU-6b} to get 451 \begin{equation}\label{IKM-EQU-10} 452 \sigma\hackscore{ij}' = 2 \eta\hackscore{eff} \left( D\hackscore{ij}' + 453 \frac{1}{ 2 \mu \; dt} \sigma\hackscore{ij}^{'-}\right) \mbox{ with } 454 \frac{1}{\eta\hackscore{eff}}=\frac{1}{\mu \; dt}+\frac{1}{\eta^{vp}} 455 \end{equation} 456 Notice that $\eta\hackscore{eff}$ is a function of diatoric stress $\sigma\hackscore{ij}'$. 457 After inserting~\ref{IKM-EQU-10} into~\ref{IKM-EQU-1} we get 458 \begin{equation}\label{IKM-EQU-1ib} 459 -\left(\eta\hackscore{eff} (v\hackscore{i,j}+ v\hackscore{i,j}) 460 \right)\hackscore{,j}+p\hackscore{,i}=F\hackscore{i}+ 461 \frac{\eta\hackscore{eff}}{\mu dt } \sigma\hackscore{ij,j}^{'-} 462 \end{equation} 463 Together with the incomressibilty condition~\ref{IKM-EQU-2} we need to solve a problem with a form almost identical 464 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 465 \begin{equation} 466 \begin{array}{rcl} 467 -\left(\eta\hackscore{eff}(dv\hackscore{i,j}+ dv\hackscore{i,j} 468 )\right)\hackscore{,j} & = & F\hackscore{i}+ \sigma\hackscore{ij,j}'-p\hackscore{,i} \\ 469 \frac{1}{\eta\hackscore{eff}} dp & = & - v\hackscore{i,i}^+ 470 \end{array} 471 \label{IKM iteration 2} 472 \end{equation} 473 where $v^+=v+dv$. As this problem is non-linear the Jacobi-free Newton-GMRES method is used with the norm 474 \begin{equation} 475 \|(v, p)\|^2= \int\hackscore{\Omega} v\hackscore{i,j}^2 + \frac{1}{\bar{\eta}^2\hackscore{eff}} p^2 \; dx 476 \label{IKM iteration 3} 477 \end{equation} 478 where $\bar{\eta}\hackscore{eff}$ is the caracteristic viscosity, for instance: 479 \begin{equation} 480 \frac{1}{\bar{\eta}\hackscore{eff}} = \frac{1}{\tau^{-}}+\sum\hackscore{q} \frac{1}{\eta^{q}\hackscore{N}} 481 \label{IKM iteration 4} 482 \end{equation} 483 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 484 calculation of $\eta\hackscore{eff}$ with the Newton-Raphson scheme. This value can then be used to calculate 485 $\sigma\hackscore{ij}'$ via~\ref{IKM-EQU-10}. We need to solve 486 \begin{equation} 487 \tau = \eta\hackscore{eff} \cdot \epsilon \mbox{ with } 488 \epsilon = \sqrt{ 2 \left( D\hackscore{ij}' + 489 \frac{1}{ 2 \mu \; dt} \sigma\hackscore{ij}^{'-}\right)^2} 490 \label{IKM iteration 5} 491 \end{equation} 492 The Newton scheme takes the form 493 \begin{equation} 494 \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) 495 = \min(\frac{\eta\hackscore{eff} - \tau\hackscore{n} \eta\hackscore{eff}'} 496 {1 - \eta\hackscore{eff}' \cdot \epsilon}, \frac{\tau\hackscore{Y} + \beta \; p}{\epsilon}) \epsilon 497 \label{IKM iteration 6} 498 \end{equation} 499 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 500 \begin{equation} 501 \eta\hackscore{eff}' = - \eta\hackscore{eff}^2 \left(\frac{1}{\eta\hackscore{eff}}\right)' 502 \mbox{ with } 503 \left(\frac{1}{\eta\hackscore{eff}}\right)' = \sum\hackscore{q} \left(\frac{1}{\eta^{q}} \right)' 504 \label{IKM iteration 7} 505 \end{equation} 506 \begin{equation}\label{IKM-EQU-5XX} 507 \left(\frac{1}{\eta^{q}} \right)' 508 = \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}}}} 509 = \frac{1-\frac{1}{n^{q}}}{ \tau \eta^{q}} 510 \end{equation} 511 Notice that allways $\eta\hackscore{eff}'\le 0$ which makes the denomionator in~\ref{IKM iteration 6} 512 positive. 513 514 515