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 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 17 The following sections give a breif overview of the model classes and their corresponding methods. 18 19 \section{Stokes Problem} 20 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} 21 \begin{equation}\label{Stokes 1} 22 -\left(\eta(v\hackscore{i,j}+ v\hackscore{i,j})\right)\hackscore{,j}+p\hackscore{,i}=f\hackscore{i}-\sigma\hackscore{ij,j} 23 \end{equation} 24 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: 25 \begin{equation}\label{Stokes 2} 26 -v\hackscore{i,i}=0 27 \end{equation} 28 Natural boundary conditions are taken in the form 29 \begin{equation}\label{Stokes Boundary} 30 \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} 31 \end{equation} 32 which can be overwritten by constraints of the form 33 \begin{equation}\label{Stokes Boundary0} 34 v\hackscore{i}(x)=v^D\hackscore{i}(x) 35 \end{equation} 36 at some locations $x$ at the boundary of the domain. The index $i$ may depend on the location $x$ on the boundary. 37 $v^D$ is a given function on the domain. 38 39 \subsection{Solution Method \label{STOKES SOLVE}} 40 In block form equation equations~\ref{Stokes 1} and~\ref{Stokes 2} takes the form of a saddle point problem 41 \index{saddle point problem} 42 \begin{equation} 43 \left[ \begin{array}{cc} 44 A & B^{*} \\ 45 B & 0 \\ 46 \end{array} \right] 47 \left[ \begin{array}{c} 48 v \\ 49 p \\ 50 \end{array} \right] 51 =\left[ \begin{array}{c} 52 G \\ 53 0 \\ 54 \end{array} \right] 55 \label{SADDLEPOINT} 56 \end{equation} 57 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}. 58 We use iterative techniques to solve this problem. To make sure that the incomressibilty condition holds 59 with sufficient accuracy we check for 60 \begin{equation} 61 \|v\hackscore{k,k}\| \hackscore \le \epsilon 62 \|\sqrt{v\hackscore{j,k}v\hackscore{j,k}}\| 63 \end{equation} 64 where $\epsilon$ is the desired relative accuracy and 65 \begin{equation} 66 \|p\|^2= \int\hackscore{\Omega} p^2 \; dx 67 \label{PRESSURE NORM} 68 \end{equation} 69 defines the $L^2$-norm. 70 There are two approaches to solve this problem. The first approach, called the Uzawa scheme \index{Uzawa scheme} 71 eliminates the velocity $v$ from the problem. The second approach solves the equation in coupled form after the application of a preconditioner. 72 73 \subsubsection{Uzawa scheme} 74 The first eqution in~\ref{SADDLEPOINT} gives $v=A^{-1}(G-B^{*}p)$ assuming $p$ is known. This is inserted into the 75 second eqution which leads to 76 \begin{equation} 77 S p = B A^{-1} G 78 \end{equation} 79 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} 80 with the preconditioner $\hat{S}$ defined as $q=\hat{S}^{-1}p$ by solving 81 \begin{equation} 82 \frac{1}{\eta}q = p 83 \end{equation} 84 see~\cite{ELMAN} for more details. The evaluation of $w=Sp$ is done in the form 85 \begin{equation} 86 \begin{array}{rcl} 87 A v & = & B^{*}p \\ 88 w & = & Bv \\ 89 \end{array} 90 \label{EVAL PCG} 91 \end{equation} 92 The residual \index{residual} $r=B A^{-1} G - S p$ is given as 93 \begin{equation} 94 r=B A^{-1} (G - B^* p) = Bv \mbox{ with } v = A^{-1}(G-B^{*}p) 95 \end{equation} 96 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 97 \begin{equation} 98 (p,(v,Bv))=\int\hackscore{\Omega} p \cdot Bv \; dx 99 \end{equation} 100 where $p$ is the pressure increment and $(v,Bv)$ represents an increment in the residual. 101 102 \subsubsection{Coupled Solver} 103 An alternative approach to solve the saddle point problem~\ref{SADDLEPOINT} directly using an iterative such as 104 the generalized minimal residual method (GMRES) \index{generalized minimal residual method!GMRES} with a suitable 105 preconditioner. Here we use the operator 106 \begin{equation} 107 \left[ \begin{array}{cc} 108 A^{-1} & 0 \\ 109 S^{-1} B A^{-1} & -S^{-1} \\ 110 \end{array} \right] 111 \label{SADDLEPOINT PRECODITIONER} 112 \end{equation} 113 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 114 \begin{equation} 115 \begin{array}{rcl} 116 A w & = & Av+B^{*}p \\ 117 \hat{S} q & = & B(w-v) \\ 118 \end{array} 119 \label{COUPLES SADDLEPOINT iteration} 120 \end{equation} 121 We use the inner product induced by the norm 122 \begin{equation} 123 \|(v,p)\|^2= \int\hackscore{\Omega} v\hackscore{i,j} v\hackscore{i,j} + \left( \frac{p}{\eta}\right)^2\; dx 124 \label{COUPLES NORM} 125 \end{equation} 126 In PDE form~\ref{COUPLES SADDLEPOINT iteration} takes the form 127 \begin{equation} 128 \begin{array}{rcl} 129 -\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} \\ 130 \frac{1}{\eta} q & = & - (w-v)\hackscore{i,i} \\ 131 \end{array} 132 \label{SADDLEPOINT iteration 2} 133 \end{equation} 134 135 136 \subsection{Functions} 137 138 \begin{classdesc}{StokesProblemCartesian}{domain} 139 opens the Stokes problem\index{Stokes problem} on the \Domain domain. The approximation 140 order needs to be two. 141 \end{classdesc} 142 143 \begin{methoddesc}[StokesProblemCartesian]{initialize}{\optional{f=Data(), \optional{fixed_u_mask=Data(), \optional{eta=1, \optional{surface_stress=Data(), \optional{stress=Data()}}}}}} 144 assigns values to the model parameters. In any call all values must be set. 145 \var{f} defines the external force $f$, \var{eta} the viscosity $\eta$, 146 \var{surface_stress} the surface stress $s$ and \var{stress} the initial stress $\sigma$. 147 The locations and compontents where the velocity is fixed are set by 148 the values of \var{fixed_u_mask}. The method will try to cast the given values to appropriate 149 \Data class objects. 150 \end{methoddesc} 151 152 \begin{methoddesc}[StokesProblemCartesian]{solve}{v,p, 153 \optional{max_iter=20, \optional{verbose=False, \optional{useUzawa=True}}}} 154 solves the problem and return approximations for velocity and pressure. 155 The arguments \var{v} and \var{p} define initial guess. The values of \var{v} marked 156 by \var{fixed_u_mask} remain unchanged. 157 If \var{useUzawa} is set to \True 158 the Uzawa\index{Uszwa} scheme is used. Otherwise the problem is solved in coupled form. In most cases 159 the Uzawa scheme is more efficient. 160 \var{max_iter} defines the maximum number of iteration steps. 161 If \var{verbose} is set to \True informations on the progress of of the solver are printed. 162 \end{methoddesc} 163 164 165 \begin{methoddesc}[StokesProblemCartesian]{setTolerance}{\optional{tolerance=1.e-8}} 166 sets the tolerance in an appropriate norm relative to the right hand side. The tolerance must be non-negative and less than 1. 167 \end{methoddesc} 168 \begin{methoddesc}[StokesProblemCartesian]{getTolerance}{} 169 returns the current relative tolerance. 170 \end{methoddesc} 171 \begin{methoddesc}[StokesProblemCartesian]{setAbsoluteTolerance}{\optional{tolerance=0.}} 172 sets the absolute tolerance for the error in the relevant norm. The tolerance must be non-negative. Typically the 173 absolute talerance is set to 0. 174 \end{methoddesc} 175 \begin{methoddesc}[StokesProblemCartesian]{getAbsoluteTolerance}{} 176 sreturns the current absolute tolerance. 177 \end{methoddesc} 178 \begin{methoddesc}[StokesProblemCartesian]{setSubToleranceReductionFactor}{\optional{reduction=None}} 179 sets the reduction factor for the tolerance used to solve the PDEs. A reduction factor 180 in the order of one will minimize compute time per iteration step but my slow down convergence or even lead to 181 divergency. On the other hand a very small value for the PDE tolerance could result in a wast of compute time. 182 If \var{reduction} is set to \var{None} the sub-tolerance is solved adaptively but 183 in cases a very small tolerance is set ($<10^{-6}$) it is recommended to set the 184 reduction factor by hand. This may require some experiments. 185 \end{methoddesc} 186 \begin{methoddesc}[StokesProblemCartesian]{getSubToleranceReductionFactor}{} 187 return the current reduction factor for the sub-problem tolerance. 188 \end{methoddesc} 189 190 \subsection{Example: Lit Driven Cavity} 191 The following script \file{lit\hackscore driven\hackscore cavity.py} 192 \index{scripts!\file{helmholtz.py}} which is available in the \ExampleDirectory 193 illustrates the usage of the \class{StokesProblemCartesian} class to solve 194 the lit driven cavity problem~\cite{LITDRIVENCAVITY}: 195 \begin{python} 196 from esys.escript import * 197 from esys.finley import Rectangle 198 from esys.escript.models import StokesProblemCartesian 199 NE=25 200 dom = Rectangle(NE,NE,order=2) 201 x = dom.getX() 202 sc=StokesProblemCartesian(dom) 203 mask= (whereZero(x)*[1.,0]+whereZero(x-1))*[1.,0] + \ 204 (whereZero(x)*[0.,1.]+whereZero(x-1))*[1.,1] 205 sc.initialize(eta=.1, fixed_u_mask= mask) 206 v=Vector(0.,Solution(dom)) 207 v+=whereZero(x-1.) 208 p=Scalar(0.,ReducedSolution(dom)) 209 v,p=sc.solve(v,p, verbose=True) 210 saveVTK("u.xml",velocity=v,pressure=p) 211 \end{python} 212 213 \section{Darcy Flux} 214 We want to calculate the velocity $u$ and pressure $p$ on a domain $\Omega$ such that 215 \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 $x\hackscore{i}$. 233 234 235 \subsection{Solution Method \label{DARCY SOLVE}} 236 Without loss of generality we can assume that $u^{N}\hackscore{i} \; n\hackscore{i}=0$ and 237 $p^{D}$. Otherewise one solves for $u-u^{N}$ and $p-p^{D}$ and sets 238 \begin{equation} 239 \begin{array}{rcl} 240 g\hackscore{i} & \leftarrow & g\hackscore{i} - u^{N}\hackscore{i} - \kappa\hackscore{ij} p^{D}\hackscore{,j }\\ 241 f & \leftarrow & f - u^{N}\hackscore{k,k} 242 \end{array} 243 \end{equation} 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} 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 method to solve this. The residual $r$ ($\in V^*$) is given as 300 \begin{equation} 301 \begin{array}{rcl} 302 r & = & Q^* ( g -(I+D^*D)^{-1} (D^*f + g) - Qp + (I+D^*D)^{-1}Q p \\ 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 the pair $(g-Qp,v)$ to represent the residual. This will save the 308 reconstruction of the velocity $v$. In this notation the right hand side is given as 309 $(g,(I+D^*D)^{-1} (D^*f + g))$. The evaluation of the iteration operator for a given $p$ is then 310 returning $(Qp,w)$ where $w$ is the solution of 311 \begin{equation}\label{UPDATE W} 312 (I+D^*D)w = Qp 313 \end{equation} 314 We use $Q^*Q$ as a a preconditioner for the iteration operator $Q^* ( I - (I+D^*D)^{-1} ) Q$. 315 316 \subsection{Functions} 317 318 \subsection{Example: Gravity Flow} 319 320 %================================================ 321 \section{Temperature Advection Diffusion\label{TEMP ADV DIFF}} 322 323 \begin{equation} 324 \rho c\hackscore{p} \left (\frac{\partial T}{\partial t} + \vec{v} \cdot \nabla T \right ) = k \nabla^{2}T 325 \label{HEAT EQUATION} 326 \end{equation} 327 328 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. 329 330 \subsection{Description} 331 332 \subsection{Method} 333 334 \begin{classdesc}{TemperatureCartesian}{dom,theta=THETA,useSUPG=SUPG} 335 \end{classdesc} 336 337 \subsection{Benchmark Problem} 338 %=============================================================================================================== 339 340 %========================================================= 341 % \section{Level Set Method} 342 343 %\subsection{Description} 344 345 %\subsection{Method} 346 347 %Advection and Reinitialisation 348 349 %\begin{classdesc}{LevelSet}{mesh, func\_new, reinit\_max, reinit\_each, tolerance, smooth} 350 %\end{classdesc} 351 352 %example usage: 353 354 %levelset = LevelSet(mesh, func\_new, reinit\_max, reinit\_each, tolerance, smooth) 355 356 %\begin{methoddesc}[LevelSet]{update\_parameter}{parameter} 357 %Update the parameter. 358 %\end{methoddesc} 359 360 %\begin{methoddesc}[LevelSet]{update\_phi}{paramter}{velocity}{dt}{t\_step} 361 %Update level set function; advection and reinitialization 362 %\end{methoddesc} 363 364 %\subsection{Benchmark Problem} 365 366 %Rayleigh-Taylor instability problem 367 368 369 % \section{Drucker Prager Model} 370 371 \section{Isotropic Kelvin Material \label{IKM}} 372 As proposed by Kelvin~\ref{KELVN} material strain $D\hackscore{ij}=v\hackscore{i,j}+v\hackscore{j,i}$ can be decomposed into 373 an elastic part $D\hackscore{ij}^{el}$ and visco-plastic part $D\hackscore{ij}^{vp}$: 374 \begin{equation}\label{IKM-EQU-2} 375 D\hackscore{ij}=D\hackscore{ij}^{el}+D\hackscore{ij}^{vp} 376 \end{equation} 377 with the elastic strain given as 378 \begin{equation}\label{IKM-EQU-3} 379 D\hackscore{ij}'^{el}=\frac{1}{2 \mu} \dot{\sigma}\hackscore{ij}' 380 \end{equation} 381 where $\sigma'\hackscore{ij}$ is the deviatoric stress (Notice that $\sigma'\hackscore{ii}=0$). 382 If the material is composed by materials $q$ the visco-plastic strain can be decomposed as 383 \begin{equation}\label{IKM-EQU-4} 384 D\hackscore{ij}'^{vp}=\sum\hackscore{q} D\hackscore{ij}'^{q} 385 \end{equation} 386 where $D\hackscore{ij}^{q}$ is the strain in material $q$ given as 387 \begin{equation}\label{IKM-EQU-5} 388 D\hackscore{ij}'^{q}=\frac{1}{2 \eta^{q}} \sigma'\hackscore{ij} 389 \end{equation} 390 where $\eta^{q}$ is the viscosity of material $q$. We assume the following 391 betwee the the strain in material $q$ 392 \begin{equation}\label{IKM-EQU-5b} 393 \eta^{q}=\eta^{q}\hackscore{N} \left(\frac{\tau}{\tau\hackscore{t}^q}\right)^{\frac{1}{n^{q}}-1} 394 \mbox{ with } \tau=\sqrt{\frac{1}{2}\sigma'\hackscore{ij} \sigma'\hackscore{ij}} 395 \end{equation} 396 for a given power law coefficients $n^{q}$ and transition stresses $\tau\hackscore{t}^q$, see~\ref{POERLAW}. 397 Notice that $n^{q}=1$ gives a constant viscosity. 398 After inserting equation~\ref{IKM-EQU-5} into equation \ref{IKM-EQU-4} one gets: 399 \begin{equation}\label{IKM-EQU-6} 400 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}} \;. 401 \end{equation} 402 With 403 \begin{equation}\label{IKM-EQU-8} 404 \dot{\gamma}=\sqrt{2 D\hackscore{ij} D\hackscore{ij}} 405 \end{equation} 406 one gets 407 \begin{equation}\label{IKM-EQU-8b} 408 \tau = \eta^{vp} \dot{\gamma}^{vp} \;. 409 \end{equation} 410 With the Drucker-Prager cohesion factor $\tau\hackscore{Y}$, Drucker-Prager friction $\beta$ and total pressure $p$ we want to achieve 411 \begin{equation}\label{IKM-EQU-8c} 412 \tau \le \tau\hackscore{Y} + \beta \; p 413 \end{equation} 414 which leads to the condition 415 \begin{equation}\label{IKM-EQU-8d} 416 \eta^{vp} \le \frac{\tau\hackscore{Y} + \beta \; p}{ \dot{\gamma}^{vp}} \; . 417 \end{equation} 418 Therefore we modify the definition of $\eta^{vp}$ to the form 419 \begin{equation}\label{IKM-EQU-6b} 420 \frac{1}{\eta^{vp}}=\max(\sum\hackscore{q} \frac{1}{\eta^{q}}, \frac{\dot{\gamma}^{vp}} {\tau\hackscore{Y} + \beta \; p}) 421 \end{equation} 422 The deviatoric stress needs to fullfill the equilibrion equation 423 \begin{equation}\label{IKM-EQU-1} 424 -\sigma'\hackscore{ij,j}+p\hackscore{,i}=F\hackscore{i} 425 \end{equation} 426 where $F\hackscore{j}$ is a given external fource. We assume an incompressible media: 427 \begin{equation}\label{IKM-EQU-2} 428 -v\hackscore{i,i}=0 429 \end{equation} 430 Natural boundary conditions are taken in the form 431 \begin{equation}\label{IKM-EQU-Boundary} 432 \sigma'\hackscore{ij}n\hackscore{j}-n\hackscore{i}p=f 433 \end{equation} 434 which can be overwritten by a constraint 435 \begin{equation}\label{IKM-EQU-Boundary0} 436 v\hackscore{i}(x)=0 437 \end{equation} 438 where the index $i$ may depend on the location $x$ on the bondary. 439 440 \subsection{Solution Method \label{IKM-SOLVE}} 441 By using a first order finite difference approximation wit step size $dt>0$~\ref{IKM-EQU-3} get the form 442 \begin{equation}\label{IKM-EQU-3b} 443 D\hackscore{ij}'^{el}=\frac{1}{2 \mu dt } \left( \sigma\hackscore{ij}' - \sigma\hackscore{ij}^{'-} \right) 444 \end{equation} 445 where $\sigma\hackscore{ij}^{'-}$ is the deviatoric stress at the precious time step. 446 Now we can combine equations~\ref{IKM-EQU-2}, \ref{IKM-EQU-3b} and~\ref{IKM-EQU-6b} to get 447 \begin{equation}\label{IKM-EQU-10} 448 \sigma\hackscore{ij}' = 2 \eta\hackscore{eff} \left( D\hackscore{ij}' + 449 \frac{1}{ 2 \mu \; dt} \sigma\hackscore{ij}^{'-}\right) \mbox{ with } 450 \frac{1}{\eta\hackscore{eff}}=\frac{1}{\mu \; dt}+\frac{1}{\eta^{vp}} 451 \end{equation} 452 Notice that $\eta\hackscore{eff}$ is a function of diatoric stress $\sigma\hackscore{ij}'$. 453 After inserting~\ref{IKM-EQU-10} into~\ref{IKM-EQU-1} we get 454 \begin{equation}\label{IKM-EQU-1ib} 455 -\left(\eta\hackscore{eff} (v\hackscore{i,j}+ v\hackscore{i,j}) 456 \right)\hackscore{,j}+p\hackscore{,i}=F\hackscore{i}+ 457 \frac{\eta\hackscore{eff}}{\mu dt } \sigma\hackscore{ij,j}^{'-} 458 \end{equation} 459 Together with the incomressibilty condition~\ref{IKM-EQU-2} we need to solve a problem with a form almost identical 460 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 461 \begin{equation} 462 \begin{array}{rcl} 463 -\left(\eta\hackscore{eff}(dv\hackscore{i,j}+ dv\hackscore{i,j} 464 )\right)\hackscore{,j} & = & F\hackscore{i}+ \sigma\hackscore{ij,j}'-p\hackscore{,i} \\ 465 \frac{1}{\eta\hackscore{eff}} dp & = & - v\hackscore{i,i}^+ 466 \end{array} 467 \label{IKM iteration 2} 468 \end{equation} 469 where $v^+=v+dv$. As this problem is non-linear the Jacobi-free Newton-GMRES method is used with the norm 470 \begin{equation} 471 \|(v, p)\|^2= \int\hackscore{\Omega} v\hackscore{i,j}^2 + \frac{1}{\bar{\eta}^2\hackscore{eff}} p^2 \; dx 472 \label{IKM iteration 3} 473 \end{equation} 474 where $\bar{\eta}\hackscore{eff}$ is the caracteristic viscosity, for instance: 475 \begin{equation} 476 \frac{1}{\bar{\eta}\hackscore{eff}} = \frac{1}{\tau^{-}}+\sum\hackscore{q} \frac{1}{\eta^{q}\hackscore{N}} 477 \label{IKM iteration 4} 478 \end{equation} 479 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 480 calculation of $\eta\hackscore{eff}$ with the Newton-Raphson scheme. This value can then be used to calculate 481 $\sigma\hackscore{ij}'$ via~\ref{IKM-EQU-10}. We need to solve 482 \begin{equation} 483 \tau = \eta\hackscore{eff} \cdot \epsilon \mbox{ with } 484 \epsilon = \sqrt{ 2 \left( D\hackscore{ij}' + 485 \frac{1}{ 2 \mu \; dt} \sigma\hackscore{ij}^{'-}\right)^2} 486 \label{IKM iteration 5} 487 \end{equation} 488 The Newton scheme takes the form 489 \begin{equation} 490 \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) 491 = \min(\frac{\eta\hackscore{eff} - \tau\hackscore{n} \eta\hackscore{eff}'} 492 {1 - \eta\hackscore{eff}' \cdot \epsilon}, \frac{\tau\hackscore{Y} + \beta \; p}{\epsilon}) \epsilon 493 \label{IKM iteration 6} 494 \end{equation} 495 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 496 \begin{equation} 497 \eta\hackscore{eff}' = - \eta\hackscore{eff}^2 \left(\frac{1}{\eta\hackscore{eff}}\right)' 498 \mbox{ with } 499 \left(\frac{1}{\eta\hackscore{eff}}\right)' = \sum\hackscore{q} \left(\frac{1}{\eta^{q}} \right)' 500 \label{IKM iteration 7} 501 \end{equation} 502 \begin{equation}\label{IKM-EQU-5XX} 503 \left(\frac{1}{\eta^{q}} \right)' 504 = \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}}}} 505 = \frac{1-\frac{1}{n^{q}}}{ \tau \eta^{q}} 506 \end{equation} 507 Notice that allways $\eta\hackscore{eff}'\le 0$ which makes the denomionator in~\ref{IKM iteration 6} 508 positive. 509 510 511