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

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some notes for the kelvin model added. Implementation is still missing

 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 gross 2251 \label{STOKES STOP} 65 gross 2100 \end{equation} 66 where $\epsilon$ is the desired relative accuracy and 67 \begin{equation} 68 \|p\|^2= \int\hackscore{\Omega} p^2 \; dx 69 \label{PRESSURE NORM} 70 \end{equation} 71 gross 2251 defines the $L^2$-norm. We use the Uzawa scheme \index{Uzawa scheme} to solve the problem. 72 73 In fact the first equation in~\ref{SADDLEPOINT} gives for a known pressure 74 gross 2100 \begin{equation} 75 gross 2251 v=A^{-1}(G-B^{*}p) 76 \label{V CALC} 77 \end{equation} 78 which is inserted into the second equation leading to 79 \begin{equation} 80 gross 2100 S p = B A^{-1} G 81 \end{equation} 82 gross 2251 with the Schur complement \index{Schur complement} $S=BA^{-1}B^{*}$. This problem can be solved iteratively 83 gross 2100 with the preconditioner $\hat{S}$ defined as $q=\hat{S}^{-1}p$ by solving 84 \begin{equation} 85 \frac{1}{\eta}q = p 86 \end{equation} 87 gross 2251 see~\cite{ELMAN} for more details. Note that the residual for the current approximation $p$ is given as 88 gross 2100 \begin{equation} 89 gross 2251 r=B A^{-1} (G - B^* p) = Bv 90 gross 2100 \end{equation} 91 gross 2251 where $v$ is given by~\ref{V CALC}. 92 93 If one uses the generalized minimal residual method (GMRES) \index{generalized minimal residual method!GMRES} 94 the method is directly applied to the preconditioned system 95 gross 2100 \begin{equation} 96 gross 2251 \hat{S}^{-1} S p = \hat{S}^{-1} B A^{-1} G 97 gross 2100 \end{equation} 98 gross 2251 We use the norm 99 gross 2100 \begin{equation} 100 gross 2251 \|p\|\hackscore{GMRES} = \|\hat{S} p \| 101 \end{equation} 102 Notice that for the residual $\hat{r}=\hat{S}^{-1} r$ one has 103 gross 2100 \begin{equation} 104 gross 2251 \ 105 \end{equation} 106 If $p^{0}$ provides an initial guess for the pressure we use~\ref{V CALC} to get a first initial guess for the 107 velocity $v^{0}$ which we use to set an absolute tolerance $ATOL =\epsilon \|\sqrt{v^{0}\hackscore{j,k}v^{0}\hackscore{j,k}}\|$. 108 The GMRES is terminated when 109 gross 1878 \begin{equation} 110 gross 2251 \|\hat{r}\|\hackscore{GMRES} \le ATOL 111 \end{equation} 112 Notice that $\|\hat{r}\|\hackscore{GMRES}= \|r \| = \|Bv\| = \|v\hackscore{k,k}\|$ so we we can expect that 113 the target stopping criterion~\ref{STOKES STOP} is fullfilled. However, if $v$ is very different from the 114 initial choice of $v^{0}$ the value of $ATOL$ is corrected and GMRES is restarted with a new tolerance. For time dependend problems this apprach works well as value for $p$ form a previous time step provides a good initial guess. 115 116 Alternatively, as $S$ is symmetric and positive definite one can apply the preconditioned conjugate gradient method (PCG) \index{preconditioned conjugate gradient method!PCG}. PCG use the norm 117 gross 1878 \begin{equation} 118 gross 2251 \|r\|\hackscore{PCG}^2 = \int\hackscore{\Omega} r \hat{S}^{-1}r \; dx = \int\hackscore{\Omega} \eta r^2 \; dx 119 \end{equation} 120 To take the extra factor $\eta$ into consideration when checking the stopping criterion we use the following 121 definition for $ATOL$: 122 gross 2100 \begin{equation} 123 gross 2251 ATOL = \epsilon \frac{\|\sqrt{v^{0}\hackscore{j,k}v^{0}\hackscore{j,k}}\| }{\|v^{0}\hackscore{k,k}\|} 124 \|v^{0}\hackscore{k,k}\|\hackscore{PCG} 125 \end{equation} 126 lgraham 1702 127 128 gross 2251 129 gross 2100 \subsection{Functions} 130 gross 1878 131 gross 2100 \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 gross 1878 136 gross 2100 \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 gross 1878 145 gross 2100 \begin{methoddesc}[StokesProblemCartesian]{solve}{v,p, 146 gross 2251 \optional{max_iter=20, \optional{verbose=False, \optional{usePCG=True}}}} 147 gross 2100 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 gross 2251 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 gross 2100 \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 lgraham 1702 157 158 gross 2251 \begin{methoddesc}[StokesProblemCartesian]{setTolerance}{\optional{tolerance=1.e-4}} 159 gross 2100 sets the tolerance in an appropriate norm relative to the right hand side. The tolerance must be non-negative and less than 1. 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 gross 2251 \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 gross 2100 \end{methoddesc} 178 gross 2251 \begin{methoddesc}[StokesProblemCartesian]{getSubProblemTolerance}{} 179 return the tolerance for the involved PDEs. 180 gross 2100 \end{methoddesc} 181 lgraham 1702 182 gross 2100 \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 jfenwick 2104 the lit driven cavity problem~\cite{LITDRIVENCAVITY}: 187 gross 2100 \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 lgraham 1702 205 gross 2100 \section{Darcy Flux} 206 gross 2108 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 gross 2100 \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 lgraham 1702 227 228 gross 2100 \subsection{Solution Method \label{DARCY SOLVE}} 229 gross 2156 In practical applications it is an advantage to calculate the pressure $p$ as a correction of a 'static' pressure $p^{ref}$ which is the solution of 230 lgraham 1702 \begin{equation} 231 gross 2156 -(\kappa\hackscore{ki}\kappa\hackscore{kj} p\hackscore{,j}^{ref})\hackscore{,i} = - (\kappa\hackscore{ki} (g\hackscore{k}- u^{N}\hackscore{k}))\hackscore{,i} 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 gross 2100 \begin{array}{rcl} 238 gross 2156 g\hackscore{i} & \leftarrow & g\hackscore{i} - u^{N}\hackscore{i} - \kappa\hackscore{ij} p^{ref}\hackscore{,j }\\ 239 gross 2100 f & \leftarrow & f - u^{N}\hackscore{k,k} 240 \end{array} 241 \end{equation} 242 gross 2156 we can assume that $u^{N}\hackscore{i} \; n\hackscore{i}=0$ and 243 gross 2208 $p^{D}=0$. 244 gross 2100 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 gross 2208 Q^*u + Q^*Q p & = & Q^*g \\ 281 gross 2100 \end{array} 282 \end{equation} 283 gross 2208 where $D^*$ and $Q^*$ denote the adjoint operators. 284 gross 2100 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 gross 2208 Q^* (I+D^*D)^{-1} (D^*f + g - Qp) + Q^* Q p = Q^*g 294 gross 2100 \end{equation} 295 which is 296 \begin{equation} 297 gross 2208 Q^* ( I - (I+D^*D)^{-1} ) Q p = Q^* (g-(I+D^*D)^{-1} (D^*f + g) ) 298 gross 2100 \end{equation} 299 gross 2208 We use the PCG \index{linear solver!PCG}\index{PCG} method to solve this. The residual $r$ ($\in V^*$) is given as 300 gross 2100 \begin{equation} 301 \begin{array}{rcl} 302 gross 2208 r & = & Q^* \left( g -(I+D^*D)^{-1} (D^*f + g) - Qp + (I+D^*D)^{-1}Q p \right)\\ 303 gross 2156 & =& Q^* \left( - Qp - (I+D^*D)^{-1} (D^*f + g - Qp) \right) \\ 304 gross 2208 & =& Q^* \left( g - Qp - v \right) 305 gross 2100 \end{array} 306 \end{equation} 307 gross 2156 So in a partical implementation we use the pair $(Qp,v)$ to represent the residual. This will save the 308 gross 2100 reconstruction of the velocity $v$. In this notation the right hand side is given as 309 gross 2156 $(0,(I+D^*D)^{-1} (D^*f + g))$. The evaluation of the iteration operator for a given $p$ is then 310 gross 2100 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 gross 2208 The iteration PCG \index{linear solver!PCG}\index{PCG} is terminated if 317 \begin{equation}\label{DARCY STOP} 318 \int r \cdot (Q^*Q)^{-1} r \; dx \le \mbox{ATOL}^2 319 \end{equation} 320 where ATOL is a given absolute tolerance. 321 The initial residual $r\hackscore{0}$ is 322 \begin{equation}\label{DARCY STOP 2} 323 r\hackscore{0}=Q^* \left( g - v\hackscore{ref} \right) \mbox{ with } v\hackscore{ref} = (I+D^*D)^{-1} (D^*f + g) 324 \end{equation} 325 so the 326 \begin{equation}\label{DARCY NORM 0} 327 \int r\hackscore0 \cdot (Q^*Q)^{-1} r\hackscore0 \; dx = \int \left( g - v\hackscore{ref} \right) \cdot Q p\hackscore{ref} \; dx \mbox{ with }p\hackscore{ref} = (Q^*Q)^{-1} Q^* \left( g - v\hackscore{ref} \right) 328 \end{equation} 329 So we set 330 \begin{equation}\label{DARCY NORM 1} 331 ATOL = atol + rtol \cdot \max(|g - v\hackscore{ref}|\hackscore{0}, |Q p\hackscore{ref} |\hackscore{0} ) 332 \end{equation} 333 where atol and rtol a given absolute and relative tolerances, respectively. The reference flux $v\hackscore{ref}$ 334 and reference pressure $p\hackscore{ref}$ may be calcualated from their definition which would require to solve to 335 PDEs but in a practical application estimates can be used for instance solutions from previous time steps or for simplified scenarious (e.g. constant permability). 336 337 gross 2100 \subsection{Functions} 338 gross 2108 \begin{classdesc}{DarcyFlow}{domain} 339 opens the Darcy flux problem\index{Darcy flux} on the \Domain domain. 340 \end{classdesc} 341 gross 2208 \begin{methoddesc}[DarcyFlow]{setValue}{\optional{f=None, \optional{g=None, \optional{location_of_fixed_pressure=None, \optional{location_of_fixed_flux=None, \optional{permeability=None}}}}}} 342 assigns values to the model parameters. Values can be assigned using various calls - in particular 343 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). 344 \var{f} and \var{g} are the corresponding parameters in~\ref{DARCY PROBLEM}. 345 The locations and compontents where the flux is prescribed are set by positive values in 346 \var{location_of_fixed_flux}. 347 The locations where the pressure is prescribed are set by 348 by positive values of \var{location_of_fixed_pressure}. 349 The values of the pressure and flux are defined by the initial guess. 350 Notice that at any point on the boundary of the domain the pressure or the normal component of 351 the flux must be defined. There must be at least one point where the pressure is prescribed. 352 The method will try to cast the given values to appropriate 353 gross 2108 \Data class objects. 354 \end{methoddesc} 355 356 gross 2208 \begin{methoddesc}[DarcyFlow]{setTolerance}{\optional{atol=0,\optional{rtol=1e-8,\optional{p_ref=None,\optional{v_ref=None}}}}} 357 sets the absolute tolerance ATOL according to~\ref{DARCY NORM 1}. If \var{p_ref} is not present $0$ is used. 358 If \var{v_ref} is not present $0$ is used. If the final result ATOL is not positive an exception is thrown. 359 \end{methoddesc} 360 gross 2108 361 gross 2208 362 363 \begin{methoddesc}[DarcyFlow]{solve}{u0,p0, \optional{max_iter=100, \optional{verbose=False \optional{sub_rtol=1.e-8}}}} 364 solves the problem. and returns approximations for the flux $v$ and the pressure $p$. 365 \var{u0} and \var{p0} define initial guess for flux and pressure. Values marked 366 by positive values \var{location_of_fixed_flux} and \var{location_of_fixed_pressure}, respectively, are kept unchanged. 367 \var{sub_rtol} defines the tolerance used to solve the involved PDEs. \var{sub_rtol} needs to be choosen sufficiently small to ensure convergence but users need to keep in mind that a very small value for \var{sub_rtol} will result in a long compute time. Typically $\var{sub_rtol}=\var{rtol}^2$ is a good choice if $\var{rtol}$ is not choosen too small. 368 \end{methoddesc} 369 370 371 gross 2100 \subsection{Example: Gravity Flow} 372 gross 2208 later 373 gross 2100 374 %================================================ 375 gross 2208 % \section{Temperature Advection Diffusion\label{TEMP ADV DIFF}} 376 gross 2100 377 gross 2208 % 378 % \rho c\hackscore{p} \left (\frac{\partial T}{\partial t} + \vec{v} \cdot \nabla T \right ) = k \nabla^{2}T 379 % \label{HEAT EQUATION} 380 % 381 lgraham 1702 382 gross 2208 % 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. 383 lgraham 1702 384 gross 2208 % \subsection{Description} 385 lgraham 1702 386 gross 2208 % \subsection{Method} 387 % 388 % \begin{classdesc}{TemperatureCartesian}{dom,theta=THETA,useSUPG=SUPG} 389 % \end{classdesc} 390 lgraham 1702 391 gross 2208 % \subsection{Benchmark Problem} 392 gross 2100 %=============================================================================================================== 393 lgraham 1702 394 gross 2100 %========================================================= 395 lgraham 2138 \input{levelsetmodel} 396 lgraham 1702 397 gross 1859 % \section{Drucker Prager Model} 398 lgraham 1702 399 gross 1841 \section{Isotropic Kelvin Material \label{IKM}} 400 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 401 an elastic part $D\hackscore{ij}^{el}$ and visco-plastic part $D\hackscore{ij}^{vp}$: 402 gross 1841 \begin{equation}\label{IKM-EQU-2} 403 gross 1859 D\hackscore{ij}=D\hackscore{ij}^{el}+D\hackscore{ij}^{vp} 404 gross 1841 \end{equation} 405 gross 1859 with the elastic strain given as 406 gross 1841 \begin{equation}\label{IKM-EQU-3} 407 gross 2252 D\hackscore{ij}^{el'}=\frac{1}{2 \mu} \dot{\sigma}\hackscore{ij}' 408 gross 1841 \end{equation} 409 gross 1859 where $\sigma'\hackscore{ij}$ is the deviatoric stress (Notice that $\sigma'\hackscore{ii}=0$). 410 If the material is composed by materials $q$ the visco-plastic strain can be decomposed as 411 gross 1841 \begin{equation}\label{IKM-EQU-4} 412 gross 2252 D\hackscore{ij}^{vp'}=\sum\hackscore{q} D\hackscore{ij}^{q'} 413 gross 1841 \end{equation} 414 gross 1859 where $D\hackscore{ij}^{q}$ is the strain in material $q$ given as 415 gross 1841 \begin{equation}\label{IKM-EQU-5} 416 gross 2252 D\hackscore{ij}^{q'}=\frac{1}{2 \eta^{q}} \sigma'\hackscore{ij} 417 gross 1841 \end{equation} 418 gross 1859 where $\eta^{q}$ is the viscosity of material $q$. We assume the following 419 betwee the the strain in material $q$ 420 \begin{equation}\label{IKM-EQU-5b} 421 \eta^{q}=\eta^{q}\hackscore{N} \left(\frac{\tau}{\tau\hackscore{t}^q}\right)^{\frac{1}{n^{q}}-1} 422 \mbox{ with } \tau=\sqrt{\frac{1}{2}\sigma'\hackscore{ij} \sigma'\hackscore{ij}} 423 \end{equation} 424 gross 2252 for a given power law coefficients $n^{q}\ge1$ and transition stresses $\tau\hackscore{t}^q$, see~\ref{POERLAW}. 425 gross 1859 Notice that $n^{q}=1$ gives a constant viscosity. 426 gross 1841 After inserting equation~\ref{IKM-EQU-5} into equation \ref{IKM-EQU-4} one gets: 427 gross 1859 \begin{equation}\label{IKM-EQU-6} 428 gross 2100 D\hackscore{ij}'^{vp}=\frac{1}{2 \eta^{vp}} \sigma'\hackscore{ij} \mbox{ with } \frac{1}{\eta^{vp}} = \sum\hackscore{q} \frac{1}{\eta^{q}} \;. 429 gross 1841 \end{equation} 430 gross 1859 With 431 \begin{equation}\label{IKM-EQU-8} 432 \dot{\gamma}=\sqrt{2 D\hackscore{ij} D\hackscore{ij}} 433 \end{equation} 434 one gets 435 \begin{equation}\label{IKM-EQU-8b} 436 \tau = \eta^{vp} \dot{\gamma}^{vp} \;. 437 \end{equation} 438 With the Drucker-Prager cohesion factor $\tau\hackscore{Y}$, Drucker-Prager friction $\beta$ and total pressure $p$ we want to achieve 439 \begin{equation}\label{IKM-EQU-8c} 440 \tau \le \tau\hackscore{Y} + \beta \; p 441 \end{equation} 442 which leads to the condition 443 \begin{equation}\label{IKM-EQU-8d} 444 \eta^{vp} \le \frac{\tau\hackscore{Y} + \beta \; p}{ \dot{\gamma}^{vp}} \; . 445 \end{equation} 446 Therefore we modify the definition of $\eta^{vp}$ to the form 447 \begin{equation}\label{IKM-EQU-6b} 448 \frac{1}{\eta^{vp}}=\max(\sum\hackscore{q} \frac{1}{\eta^{q}}, \frac{\dot{\gamma}^{vp}} {\tau\hackscore{Y} + \beta \; p}) 449 \end{equation} 450 The deviatoric stress needs to fullfill the equilibrion equation 451 gross 1841 \begin{equation}\label{IKM-EQU-1} 452 gross 1859 -\sigma'\hackscore{ij,j}+p\hackscore{,i}=F\hackscore{i} 453 gross 1841 \end{equation} 454 gross 1859 where $F\hackscore{j}$ is a given external fource. We assume an incompressible media: 455 gross 1841 \begin{equation}\label{IKM-EQU-2} 456 gross 1859 -v\hackscore{i,i}=0 457 gross 1841 \end{equation} 458 gross 1878 Natural boundary conditions are taken in the form 459 \begin{equation}\label{IKM-EQU-Boundary} 460 \sigma'\hackscore{ij}n\hackscore{j}-n\hackscore{i}p=f 461 \end{equation} 462 which can be overwritten by a constraint 463 \begin{equation}\label{IKM-EQU-Boundary0} 464 v\hackscore{i}(x)=0 465 \end{equation} 466 where the index $i$ may depend on the location $x$ on the bondary. 467 468 \subsection{Solution Method \label{IKM-SOLVE}} 469 By using a first order finite difference approximation wit step size $dt>0$~\ref{IKM-EQU-3} get the form 470 \begin{equation}\label{IKM-EQU-3b} 471 D\hackscore{ij}'^{el}=\frac{1}{2 \mu dt } \left( \sigma\hackscore{ij}' - \sigma\hackscore{ij}^{'-} \right) 472 \end{equation} 473 where $\sigma\hackscore{ij}^{'-}$ is the deviatoric stress at the precious time step. 474 Now we can combine equations~\ref{IKM-EQU-2}, \ref{IKM-EQU-3b} and~\ref{IKM-EQU-6b} to get 475 \begin{equation}\label{IKM-EQU-10} 476 gross 2100 \sigma\hackscore{ij}' = 2 \eta\hackscore{eff} \left( D\hackscore{ij}' + 477 \frac{1}{ 2 \mu \; dt} \sigma\hackscore{ij}^{'-}\right) \mbox{ with } 478 gross 1878 \frac{1}{\eta\hackscore{eff}}=\frac{1}{\mu \; dt}+\frac{1}{\eta^{vp}} 479 \end{equation} 480 gross 2100 Notice that $\eta\hackscore{eff}$ is a function of diatoric stress $\sigma\hackscore{ij}'$. 481 gross 1859 After inserting~\ref{IKM-EQU-10} into~\ref{IKM-EQU-1} we get 482 \begin{equation}\label{IKM-EQU-1ib} 483 gross 2100 -\left(\eta\hackscore{eff} (v\hackscore{i,j}+ v\hackscore{i,j}) 484 \right)\hackscore{,j}+p\hackscore{,i}=F\hackscore{i}+ 485 gross 1878 \frac{\eta\hackscore{eff}}{\mu dt } \sigma\hackscore{ij,j}^{'-} 486 gross 1859 \end{equation} 487 gross 2252 Combining this with the incomressibilty condition~\ref{IKM-EQU-2} we need to solve a 488 Stokes problem as discussed in section~\ref{STOKES SOLVE} in each time step. 489 In oder to perform step~\ref{IKM iteration 2} we need to calculate the $\eta\hackscore{eff}$ which 490 is a function of $\sigma\hackscore{ij}$ via $\tau$. To get $\tau$ and $\eta\hackscore{eff}$ we need to solve the 491 non-linear equation 492 gross 1878 \begin{equation} 493 gross 2252 \tau = \eta\hackscore{eff} \cdot \dot{\gamma}\hackscore{total} \mbox{ with } 494 \dot{\gamma}\hackscore{total} = \sqrt{ 2 \left( D\hackscore{ij}' + 495 gross 2100 \frac{1}{ 2 \mu \; dt} \sigma\hackscore{ij}^{'-}\right)^2} 496 \label{IKM iteration 5} 497 \end{equation} 498 The Newton scheme takes the form 499 \begin{equation} 500 gross 2252 \tau\hackscore{n+1} = \min(\tau\hackscore{n} - \frac{\tau\hackscore{n} - \eta\hackscore{eff} \cdot \dot{\gamma}\hackscore{total}}{1 - \eta\hackscore{eff}' \cdot \dot{\gamma}\hackscore{total}}, \tau\hackscore{Y} + \beta \; p) 501 gross 2100 \label{IKM iteration 6} 502 \end{equation} 503 gross 2252 where $\eta\hackscore{eff}'$ denotes the derivative of $\eta\hackscore{eff}$ with respect of $\tau$. The second term in $\min$ is droped of $\tau\hackscore{Y} + \beta \; p<0$ (?)). We have 504 gross 2100 \begin{equation} 505 \eta\hackscore{eff}' = - \eta\hackscore{eff}^2 \left(\frac{1}{\eta\hackscore{eff}}\right)' 506 \mbox{ with } 507 \left(\frac{1}{\eta\hackscore{eff}}\right)' = \sum\hackscore{q} \left(\frac{1}{\eta^{q}} \right)' 508 \label{IKM iteration 7} 509 \end{equation} 510 \begin{equation}\label{IKM-EQU-5XX} 511 \left(\frac{1}{\eta^{q}} \right)' 512 = \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}}}} 513 gross 2252 = \frac{1-\frac{1}{n^{q}}}{\tau}\frac{1}{\eta^{q}} 514 gross 2100 \end{equation} 515 Notice that allways $\eta\hackscore{eff}'\le 0$ which makes the denomionator in~\ref{IKM iteration 6} 516 positive. 517 gross 1841 518 gross 1878 519 gross 2100