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more work on the dary solver


 1 ksteube 1811 2 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 3 lgraham 1702 % 4 ksteube 1811 % Copyright (c) 2003-2008 by University of Queensland 5 % Earth Systems Science Computational Center (ESSCC) 6 7 lgraham 1702 % 8 ksteube 1811 % Primary Business: Queensland, Australia 9 % Licensed under the Open Software License version 3.0 10 11 lgraham 1702 % 12 ksteube 1811 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 13 lgraham 1702 14 ksteube 1811 15 lgraham 1702 \chapter{Models} 16 lgraham 2128 \label{MODELS CHAPTER} 17 lgraham 1702 18 The following sections give a breif overview of the model classes and their corresponding methods. 19 20 gross 1878 \section{Stokes Problem} 21 gross 2100 The velocity \index{velocity} field $v$ and pressure $p$ of an incompressible fluid \index{incompressible fluid} is given as the solution of the Stokes problem\index{Stokes problem} 22 gross 1878 \begin{equation}\label{Stokes 1} 23 gross 2100 -\left(\eta(v\hackscore{i,j}+ v\hackscore{i,j})\right)\hackscore{,j}+p\hackscore{,i}=f\hackscore{i}-\sigma\hackscore{ij,j} 24 gross 1878 \end{equation} 25 gross 2100 where $\eta$ is the viscosity, $F\hackscore{i}$ defines an internal force \index{force, internal} and $\sigma\hackscore{ij}$ is an intial stress \index{stress, initial}. We assume an incompressible media: 26 gross 1878 \begin{equation}\label{Stokes 2} 27 -v\hackscore{i,i}=0 28 \end{equation} 29 Natural boundary conditions are taken in the form 30 \begin{equation}\label{Stokes Boundary} 31 gross 2100 \left(\eta(v\hackscore{i,j}+ v\hackscore{i,j})\right)n\hackscore{j}-n\hackscore{i}p=s\hackscore{i}+\sigma\hackscore{ij} n\hackscore{i} 32 gross 1878 \end{equation} 33 gross 2100 which can be overwritten by constraints of the form 34 gross 1878 \begin{equation}\label{Stokes Boundary0} 35 gross 2100 v\hackscore{i}(x)=v^D\hackscore{i}(x) 36 gross 1878 \end{equation} 37 gross 2100 at some locations $x$ at the boundary of the domain. The index $i$ may depend on the location $x$ on the boundary. 38 $v^D$ is a given function on the domain. 39 lgraham 1702 40 gross 1878 \subsection{Solution Method \label{STOKES SOLVE}} 41 In block form equation equations~\ref{Stokes 1} and~\ref{Stokes 2} takes the form of a saddle point problem 42 gross 2100 \index{saddle point problem} 43 lgraham 1702 \begin{equation} 44 \left[ \begin{array}{cc} 45 gross 1878 A & B^{*} \\ 46 B & 0 \\ 47 lgraham 1702 \end{array} \right] 48 \left[ \begin{array}{c} 49 gross 2100 v \\ 50 lgraham 1702 p \\ 51 \end{array} \right] 52 =\left[ \begin{array}{c} 53 gross 2100 G \\ 54 gross 1878 0 \\ 55 lgraham 1702 \end{array} \right] 56 \label{SADDLEPOINT} 57 \end{equation} 58 gross 2100 where $A$ is coercive, self-adjoint linear operator in a suitable Hilbert space, $B$ is the $(-1) \cdot$ divergence operator and $B^{*}$ is it adjoint operator (=gradient operator). For more details on the mathematics see references \cite{AAMIRBERKYAN2008,MBENZI2005}. 59 We use iterative techniques to solve this problem. To make sure that the incomressibilty condition holds 60 with sufficient accuracy we check for 61 gross 1878 \begin{equation} 62 gross 2100 \|v\hackscore{k,k}\| \hackscore \le \epsilon 63 \|\sqrt{v\hackscore{j,k}v\hackscore{j,k}}\| 64 \end{equation} 65 where $\epsilon$ is the desired relative accuracy and 66 \begin{equation} 67 \|p\|^2= \int\hackscore{\Omega} p^2 \; dx 68 \label{PRESSURE NORM} 69 \end{equation} 70 defines the $L^2$-norm. 71 There are two approaches to solve this problem. The first approach, called the Uzawa scheme \index{Uzawa scheme} 72 eliminates the velocity $v$ from the problem. The second approach solves the equation in coupled form after the application of a preconditioner. 73 74 \subsubsection{Uzawa scheme} 75 The first eqution in~\ref{SADDLEPOINT} gives $v=A^{-1}(G-B^{*}p)$ assuming $p$ is known. This is inserted into the 76 second eqution which leads to 77 \begin{equation} 78 S p = B A^{-1} G 79 \end{equation} 80 with the Schur complement \index{Schur complement} $S=BA^{-1}B^{*}$. This problem can be solved iteratively using the reconditioned Conjugate Gradient Method (PCG)~\index{PCG!Preconditioned Conjugate Gradient Method} 81 with the preconditioner $\hat{S}$ defined as $q=\hat{S}^{-1}p$ by solving 82 \begin{equation} 83 \frac{1}{\eta}q = p 84 \end{equation} 85 see~\cite{ELMAN} for more details. The evaluation of $w=Sp$ is done in the form 86 \begin{equation} 87 \begin{array}{rcl} 88 A v & = & B^{*}p \\ 89 w & = & Bv \\ 90 \end{array} 91 \label{EVAL PCG} 92 \end{equation} 93 The residual \index{residual} $r=B A^{-1} G - S p$ is given as 94 \begin{equation} 95 r=B A^{-1} (G - B^* p) = Bv \mbox{ with } v = A^{-1}(G-B^{*}p) 96 \end{equation} 97 Therefore one uses the tuple $(v,Bv)$ to represent the residual of the current pressure $p$. Notice that before the iteration is started the right hand side $B A^{-1} G$ needs to be calculated. The bilinear form $(.,.)$ used is defined as 98 \begin{equation} 99 (p,(v,Bv))=\int\hackscore{\Omega} p \cdot Bv \; dx 100 \end{equation} 101 where $p$ is the pressure increment and $(v,Bv)$ represents an increment in the residual. 102 103 \subsubsection{Coupled Solver} 104 An alternative approach to solve the saddle point problem~\ref{SADDLEPOINT} directly using an iterative such as 105 the generalized minimal residual method (GMRES) \index{generalized minimal residual method!GMRES} with a suitable 106 preconditioner. Here we use the operator 107 \begin{equation} 108 gross 1878 \left[ \begin{array}{cc} 109 A^{-1} & 0 \\ 110 S^{-1} B A^{-1} & -S^{-1} \\ 111 \end{array} \right] 112 \label{SADDLEPOINT PRECODITIONER} 113 \end{equation} 114 gross 2100 where again $S$ is the Schur complement~\cite{ELMAN}. In partice we will use an approximation $\hat{S}$ for $S$. The evaluation $(w,q)$ of the iteration operator for a given $(v,p)$ is done as 115 gross 1878 \begin{equation} 116 \begin{array}{rcl} 117 gross 2100 A w & = & Av+B^{*}p \\ 118 \hat{S} q & = & B(w-v) \\ 119 gross 1878 \end{array} 120 gross 2100 \label{COUPLES SADDLEPOINT iteration} 121 gross 1878 \end{equation} 122 gross 2100 We use the inner product induced by the norm 123 gross 1878 \begin{equation} 124 gross 2100 \|(v,p)\|^2= \int\hackscore{\Omega} v\hackscore{i,j} v\hackscore{i,j} + \left( \frac{p}{\eta}\right)^2\; dx 125 \label{COUPLES NORM} 126 \end{equation} 127 In PDE form~\ref{COUPLES SADDLEPOINT iteration} takes the form 128 \begin{equation} 129 gross 1878 \begin{array}{rcl} 130 gross 2100 -\left(\eta(w\hackscore{i,j}+ w\hackscore{i,j})\right)\hackscore{,j} & = & -\left(\eta(v\hackscore{i,j}+ v\hackscore{i,j})\right)\hackscore{,j}+p\hackscore{,i} \\ 131 \frac{1}{\eta} q & = & - (w-v)\hackscore{i,i} \\ 132 gross 1878 \end{array} 133 \label{SADDLEPOINT iteration 2} 134 \end{equation} 135 lgraham 1702 136 137 gross 2100 \subsection{Functions} 138 gross 1878 139 gross 2100 \begin{classdesc}{StokesProblemCartesian}{domain} 140 opens the Stokes problem\index{Stokes problem} on the \Domain domain. The approximation 141 order needs to be two. 142 \end{classdesc} 143 gross 1878 144 gross 2100 \begin{methoddesc}[StokesProblemCartesian]{initialize}{\optional{f=Data(), \optional{fixed_u_mask=Data(), \optional{eta=1, \optional{surface_stress=Data(), \optional{stress=Data()}}}}}} 145 assigns values to the model parameters. In any call all values must be set. 146 \var{f} defines the external force $f$, \var{eta} the viscosity $\eta$, 147 \var{surface_stress} the surface stress $s$ and \var{stress} the initial stress $\sigma$. 148 The locations and compontents where the velocity is fixed are set by 149 the values of \var{fixed_u_mask}. The method will try to cast the given values to appropriate 150 \Data class objects. 151 \end{methoddesc} 152 gross 1878 153 gross 2100 \begin{methoddesc}[StokesProblemCartesian]{solve}{v,p, 154 \optional{max_iter=20, \optional{verbose=False, \optional{useUzawa=True}}}} 155 solves the problem and return approximations for velocity and pressure. 156 The arguments \var{v} and \var{p} define initial guess. The values of \var{v} marked 157 by \var{fixed_u_mask} remain unchanged. 158 If \var{useUzawa} is set to \True 159 the Uzawa\index{Uszwa} scheme is used. Otherwise the problem is solved in coupled form. In most cases 160 the Uzawa scheme is more efficient. 161 \var{max_iter} defines the maximum number of iteration steps. 162 If \var{verbose} is set to \True informations on the progress of of the solver are printed. 163 \end{methoddesc} 164 lgraham 1702 165 166 gross 2100 \begin{methoddesc}[StokesProblemCartesian]{setTolerance}{\optional{tolerance=1.e-8}} 167 sets the tolerance in an appropriate norm relative to the right hand side. The tolerance must be non-negative and less than 1. 168 \end{methoddesc} 169 \begin{methoddesc}[StokesProblemCartesian]{getTolerance}{} 170 returns the current relative tolerance. 171 \end{methoddesc} 172 \begin{methoddesc}[StokesProblemCartesian]{setAbsoluteTolerance}{\optional{tolerance=0.}} 173 sets the absolute tolerance for the error in the relevant norm. The tolerance must be non-negative. Typically the 174 absolute talerance is set to 0. 175 \end{methoddesc} 176 \begin{methoddesc}[StokesProblemCartesian]{getAbsoluteTolerance}{} 177 sreturns the current absolute tolerance. 178 \end{methoddesc} 179 \begin{methoddesc}[StokesProblemCartesian]{setSubToleranceReductionFactor}{\optional{reduction=None}} 180 sets the reduction factor for the tolerance used to solve the PDEs. A reduction factor 181 in the order of one will minimize compute time per iteration step but my slow down convergence or even lead to 182 divergency. On the other hand a very small value for the PDE tolerance could result in a wast of compute time. 183 If \var{reduction} is set to \var{None} the sub-tolerance is solved adaptively but 184 in cases a very small tolerance is set ($<10^{-6}$) it is recommended to set the 185 reduction factor by hand. This may require some experiments. 186 \end{methoddesc} 187 \begin{methoddesc}[StokesProblemCartesian]{getSubToleranceReductionFactor}{} 188 return the current reduction factor for the sub-problem tolerance. 189 \end{methoddesc} 190 lgraham 1702 191 gross 2100 \subsection{Example: Lit Driven Cavity} 192 The following script \file{lit\hackscore driven\hackscore cavity.py} 193 \index{scripts!\file{helmholtz.py}} which is available in the \ExampleDirectory 194 illustrates the usage of the \class{StokesProblemCartesian} class to solve 195 jfenwick 2104 the lit driven cavity problem~\cite{LITDRIVENCAVITY}: 196 gross 2100 \begin{python} 197 from esys.escript import * 198 from esys.finley import Rectangle 199 from esys.escript.models import StokesProblemCartesian 200 NE=25 201 dom = Rectangle(NE,NE,order=2) 202 x = dom.getX() 203 sc=StokesProblemCartesian(dom) 204 mask= (whereZero(x[0])*[1.,0]+whereZero(x[0]-1))*[1.,0] + \ 205 (whereZero(x[1])*[0.,1.]+whereZero(x[1]-1))*[1.,1] 206 sc.initialize(eta=.1, fixed_u_mask= mask) 207 v=Vector(0.,Solution(dom)) 208 v[0]+=whereZero(x[1]-1.) 209 p=Scalar(0.,ReducedSolution(dom)) 210 v,p=sc.solve(v,p, verbose=True) 211 saveVTK("u.xml",velocity=v,pressure=p) 212 \end{python} 213 lgraham 1702 214 gross 2100 \section{Darcy Flux} 215 gross 2108 We want to calculate the velocity $u$ and pressure $p$ on a domain $\Omega$ solving 216 the Darcy flux problem \index{Darcy flux}\index{Darcy flow} 217 gross 2100 \begin{equation}\label{DARCY PROBLEM} 218 \begin{array}{rcl} 219 u\hackscore{i} + \kappa\hackscore{ij} p\hackscore{,j} & = & g\hackscore{i} \\ 220 u\hackscore{k,k} & = & f 221 \end{array} 222 \end{equation} 223 with the boundary conditions 224 \begin{equation}\label{DARCY BOUNDARY} 225 \begin{array}{rcl} 226 u\hackscore{i} \; n\hackscore{i} = u^{N}\hackscore{i} \; n\hackscore{i} & \mbox{ on } & \Gamma\hackscore{N} \\ 227 p = p^{D} & \mbox{ on } & \Gamma\hackscore{D} \\ 228 \end{array} 229 \end{equation} 230 where $\Gamma\hackscore{N}$ and $\Gamma\hackscore{D}$ are a partition of the boundary of $\Omega$ with $\Gamma\hackscore{D}$ non empty, $n\hackscore{i}$ is the outer normal field of the boundary of $\Omega$, $u^{N}\hackscore{i}$ and $p^{D}$ are given functions on $\Omega$, $g\hackscore{i}$ and $f$ are given source terms and $\kappa\hackscore{ij}$ is the given permability. We assume that $\kappa\hackscore{ij}$ is symmetric (which is not really required) and positive definite, i.e there are positive constants $\alpha\hackscore{0}$ and $\alpha\hackscore{1}$ wich are independent from the location in $\Omega$ such that 231 \begin{equation} 232 \alpha\hackscore{0} \; x\hackscore{i} x\hackscore{i} \le \kappa\hackscore{ij} x\hackscore{i} x\hackscore{j} \le \alpha\hackscore{1} \; x\hackscore{i} x\hackscore{i} 233 \end{equation} 234 for all $x\hackscore{i}$. 235 lgraham 1702 236 237 gross 2100 \subsection{Solution Method \label{DARCY SOLVE}} 238 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 239 lgraham 1702 \begin{equation} 240 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} 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 gross 2100 \begin{array}{rcl} 247 gross 2156 g\hackscore{i} & \leftarrow & g\hackscore{i} - u^{N}\hackscore{i} - \kappa\hackscore{ij} p^{ref}\hackscore{,j }\\ 248 gross 2100 f & \leftarrow & f - u^{N}\hackscore{k,k} 249 \end{array} 250 \end{equation} 251 gross 2156 we can assume that $u^{N}\hackscore{i} \; n\hackscore{i}=0$ and 252 gross 2208 $p^{D}=0$. 253 gross 2100 We set 254 \begin{equation} 255 V = \{ q \in H^{1}(\Omega) : q=0 \mbox{ on } \Gamma\hackscore{D} \} 256 \end{equation} 257 and 258 \begin{equation} 259 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} \} 260 \end{equation} 261 and define the operator $Q: V \rightarrow (L^2(\Omega))^{d}$ defined by 262 \begin{equation} 263 (Qp)\hackscore{i} = \kappa\hackscore{ij} p\hackscore{,j} 264 \end{equation} 265 and the operator $D: W \rightarrow L^2(\Omega)$ defined by 266 \begin{equation} 267 Dv = v\hackscore{k,k} 268 \end{equation} 269 In operator notation the Darcy problem~\ref{DARCY PROBLEM} is written in the form 270 \begin{equation} 271 \begin{array}{rcl} 272 u + Qp & = & g \\ 273 Du & = & f 274 \end{array} 275 \end{equation} 276 We solve this equation by minimising the functional 277 \begin{equation} 278 J(u,p):=\|u + Qp - g\|^2\hackscore{0} + \|Du-f\|\hackscore{0}^2 279 \end{equation} 280 over $W \times V$ where $\|.\|\hackscore{0}$ denotes the norm in $L^2(\Omega)$. A simple calculation shows that 281 one has to solve 282 \begin{equation} 283 ( v + Qq , u + Qp - g) + (Dv,Du-f) =0 284 \end{equation} 285 for all $v\in W$ and $q \in V$.which translates back into operator notation 286 \begin{equation} 287 \begin{array}{rcl} 288 (I+D^*D)u + Qp & = & D^*f + g \\ 289 gross 2208 Q^*u + Q^*Q p & = & Q^*g \\ 290 gross 2100 \end{array} 291 \end{equation} 292 gross 2208 where $D^*$ and $Q^*$ denote the adjoint operators. 293 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 294 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 $\alpha\hackscore{0} \gg 1$) 295 296 The approach we are taking is to eliminate the velocity $u$ from the problem. Assuming that $p$ is known we have 297 \begin{equation} 298 v= (I+D^*D)^{-1} (D^*f + g - Qp) 299 \end{equation} 300 (notice that $(I+D^*D)$ is coercive in $W$) which is inserted into the second equation 301 \begin{equation} 302 gross 2208 Q^* (I+D^*D)^{-1} (D^*f + g - Qp) + Q^* Q p = Q^*g 303 gross 2100 \end{equation} 304 which is 305 \begin{equation} 306 gross 2208 Q^* ( I - (I+D^*D)^{-1} ) Q p = Q^* (g-(I+D^*D)^{-1} (D^*f + g) ) 307 gross 2100 \end{equation} 308 gross 2208 We use the PCG \index{linear solver!PCG}\index{PCG} method to solve this. The residual $r$ ($\in V^*$) is given as 309 gross 2100 \begin{equation} 310 \begin{array}{rcl} 311 gross 2208 r & = & Q^* \left( g -(I+D^*D)^{-1} (D^*f + g) - Qp + (I+D^*D)^{-1}Q p \right)\\ 312 gross 2156 & =& Q^* \left( - Qp - (I+D^*D)^{-1} (D^*f + g - Qp) \right) \\ 313 gross 2208 & =& Q^* \left( g - Qp - v \right) 314 gross 2100 \end{array} 315 \end{equation} 316 gross 2156 So in a partical implementation we use the pair $(Qp,v)$ to represent the residual. This will save the 317 gross 2100 reconstruction of the velocity $v$. In this notation the right hand side is given as 318 gross 2156 $(0,(I+D^*D)^{-1} (D^*f + g))$. The evaluation of the iteration operator for a given $p$ is then 319 gross 2100 returning $(Qp,w)$ where $w$ is the solution of 320 \begin{equation}\label{UPDATE W} 321 (I+D^*D)w = Qp 322 \end{equation} 323 We use $Q^*Q$ as a a preconditioner for the iteration operator $Q^* ( I - (I+D^*D)^{-1} ) Q$. 324 325 gross 2208 The iteration PCG \index{linear solver!PCG}\index{PCG} is terminated if 326 \begin{equation}\label{DARCY STOP} 327 \int r \cdot (Q^*Q)^{-1} r \; dx \le \mbox{ATOL}^2 328 \end{equation} 329 where ATOL is a given absolute tolerance. 330 The initial residual $r\hackscore{0}$ is 331 \begin{equation}\label{DARCY STOP 2} 332 r\hackscore{0}=Q^* \left( g - v\hackscore{ref} \right) \mbox{ with } v\hackscore{ref} = (I+D^*D)^{-1} (D^*f + g) 333 \end{equation} 334 so the 335 \begin{equation}\label{DARCY NORM 0} 336 \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) 337 \end{equation} 338 So we set 339 \begin{equation}\label{DARCY NORM 1} 340 ATOL = atol + rtol \cdot \max(|g - v\hackscore{ref}|\hackscore{0}, |Q p\hackscore{ref} |\hackscore{0} ) 341 \end{equation} 342 where atol and rtol a given absolute and relative tolerances, respectively. The reference flux $v\hackscore{ref}$ 343 and reference pressure $p\hackscore{ref}$ may be calcualated from their definition which would require to solve to 344 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). 345 346 gross 2100 \subsection{Functions} 347 gross 2108 \begin{classdesc}{DarcyFlow}{domain} 348 opens the Darcy flux problem\index{Darcy flux} on the \Domain domain. 349 \end{classdesc} 350 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}}}}}} 351 assigns values to the model parameters. Values can be assigned using various calls - in particular 352 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). 353 \var{f} and \var{g} are the corresponding parameters in~\ref{DARCY PROBLEM}. 354 The locations and compontents where the flux is prescribed are set by positive values in 355 \var{location_of_fixed_flux}. 356 The locations where the pressure is prescribed are set by 357 by positive values of \var{location_of_fixed_pressure}. 358 The values of the pressure and flux are defined by the initial guess. 359 Notice that at any point on the boundary of the domain the pressure or the normal component of 360 the flux must be defined. There must be at least one point where the pressure is prescribed. 361 The method will try to cast the given values to appropriate 362 gross 2108 \Data class objects. 363 \end{methoddesc} 364 365 gross 2208 \begin{methoddesc}[DarcyFlow]{setTolerance}{\optional{atol=0,\optional{rtol=1e-8,\optional{p_ref=None,\optional{v_ref=None}}}}} 366 sets the absolute tolerance ATOL according to~\ref{DARCY NORM 1}. If \var{p_ref} is not present $0$ is used. 367 If \var{v_ref} is not present $0$ is used. If the final result ATOL is not positive an exception is thrown. 368 \end{methoddesc} 369 gross 2108 370 gross 2208 371 372 \begin{methoddesc}[DarcyFlow]{solve}{u0,p0, \optional{max_iter=100, \optional{verbose=False \optional{sub_rtol=1.e-8}}}} 373 solves the problem. and returns approximations for the flux $v$ and the pressure $p$. 374 \var{u0} and \var{p0} define initial guess for flux and pressure. Values marked 375 by positive values \var{location_of_fixed_flux} and \var{location_of_fixed_pressure}, respectively, are kept unchanged. 376 \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. 377 \end{methoddesc} 378 379 380 gross 2100 \subsection{Example: Gravity Flow} 381 gross 2208 later 382 gross 2100 383 %================================================ 384 gross 2208 % \section{Temperature Advection Diffusion\label{TEMP ADV DIFF}} 385 gross 2100 386 gross 2208 % 387 % \rho c\hackscore{p} \left (\frac{\partial T}{\partial t} + \vec{v} \cdot \nabla T \right ) = k \nabla^{2}T 388 % \label{HEAT EQUATION} 389 % 390 lgraham 1702 391 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. 392 lgraham 1702 393 gross 2208 % \subsection{Description} 394 lgraham 1702 395 gross 2208 % \subsection{Method} 396 % 397 % \begin{classdesc}{TemperatureCartesian}{dom,theta=THETA,useSUPG=SUPG} 398 % \end{classdesc} 399 lgraham 1702 400 gross 2208 % \subsection{Benchmark Problem} 401 gross 2100 %=============================================================================================================== 402 lgraham 1702 403 gross 2100 %========================================================= 404 lgraham 2138 \input{levelsetmodel} 405 lgraham 1702 406 gross 1859 % \section{Drucker Prager Model} 407 lgraham 1702 408 gross 1841 \section{Isotropic Kelvin Material \label{IKM}} 409 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 410 an elastic part $D\hackscore{ij}^{el}$ and visco-plastic part $D\hackscore{ij}^{vp}$: 411 gross 1841 \begin{equation}\label{IKM-EQU-2} 412 gross 1859 D\hackscore{ij}=D\hackscore{ij}^{el}+D\hackscore{ij}^{vp} 413 gross 1841 \end{equation} 414 gross 1859 with the elastic strain given as 415 gross 1841 \begin{equation}\label{IKM-EQU-3} 416 gross 1878 D\hackscore{ij}'^{el}=\frac{1}{2 \mu} \dot{\sigma}\hackscore{ij}' 417 gross 1841 \end{equation} 418 gross 1859 where $\sigma'\hackscore{ij}$ is the deviatoric stress (Notice that $\sigma'\hackscore{ii}=0$). 419 If the material is composed by materials $q$ the visco-plastic strain can be decomposed as 420 gross 1841 \begin{equation}\label{IKM-EQU-4} 421 gross 1859 D\hackscore{ij}'^{vp}=\sum\hackscore{q} D\hackscore{ij}'^{q} 422 gross 1841 \end{equation} 423 gross 1859 where $D\hackscore{ij}^{q}$ is the strain in material $q$ given as 424 gross 1841 \begin{equation}\label{IKM-EQU-5} 425 gross 1859 D\hackscore{ij}'^{q}=\frac{1}{2 \eta^{q}} \sigma'\hackscore{ij} 426 gross 1841 \end{equation} 427 gross 1859 where $\eta^{q}$ is the viscosity of material $q$. We assume the following 428 betwee the the strain in material $q$ 429 \begin{equation}\label{IKM-EQU-5b} 430 \eta^{q}=\eta^{q}\hackscore{N} \left(\frac{\tau}{\tau\hackscore{t}^q}\right)^{\frac{1}{n^{q}}-1} 431 \mbox{ with } \tau=\sqrt{\frac{1}{2}\sigma'\hackscore{ij} \sigma'\hackscore{ij}} 432 \end{equation} 433 for a given power law coefficients $n^{q}$ and transition stresses $\tau\hackscore{t}^q$, see~\ref{POERLAW}. 434 Notice that $n^{q}=1$ gives a constant viscosity. 435 gross 1841 After inserting equation~\ref{IKM-EQU-5} into equation \ref{IKM-EQU-4} one gets: 436 gross 1859 \begin{equation}\label{IKM-EQU-6} 437 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}} \;. 438 gross 1841 \end{equation} 439 gross 1859 With 440 \begin{equation}\label{IKM-EQU-8} 441 \dot{\gamma}=\sqrt{2 D\hackscore{ij} D\hackscore{ij}} 442 \end{equation} 443 one gets 444 \begin{equation}\label{IKM-EQU-8b} 445 \tau = \eta^{vp} \dot{\gamma}^{vp} \;. 446 \end{equation} 447 With the Drucker-Prager cohesion factor $\tau\hackscore{Y}$, Drucker-Prager friction $\beta$ and total pressure $p$ we want to achieve 448 \begin{equation}\label{IKM-EQU-8c} 449 \tau \le \tau\hackscore{Y} + \beta \; p 450 \end{equation} 451 which leads to the condition 452 \begin{equation}\label{IKM-EQU-8d} 453 \eta^{vp} \le \frac{\tau\hackscore{Y} + \beta \; p}{ \dot{\gamma}^{vp}} \; . 454 \end{equation} 455 Therefore we modify the definition of $\eta^{vp}$ to the form 456 \begin{equation}\label{IKM-EQU-6b} 457 \frac{1}{\eta^{vp}}=\max(\sum\hackscore{q} \frac{1}{\eta^{q}}, \frac{\dot{\gamma}^{vp}} {\tau\hackscore{Y} + \beta \; p}) 458 \end{equation} 459 The deviatoric stress needs to fullfill the equilibrion equation 460 gross 1841 \begin{equation}\label{IKM-EQU-1} 461 gross 1859 -\sigma'\hackscore{ij,j}+p\hackscore{,i}=F\hackscore{i} 462 gross 1841 \end{equation} 463 gross 1859 where $F\hackscore{j}$ is a given external fource. We assume an incompressible media: 464 gross 1841 \begin{equation}\label{IKM-EQU-2} 465 gross 1859 -v\hackscore{i,i}=0 466 gross 1841 \end{equation} 467 gross 1878 Natural boundary conditions are taken in the form 468 \begin{equation}\label{IKM-EQU-Boundary} 469 \sigma'\hackscore{ij}n\hackscore{j}-n\hackscore{i}p=f 470 \end{equation} 471 which can be overwritten by a constraint 472 \begin{equation}\label{IKM-EQU-Boundary0} 473 v\hackscore{i}(x)=0 474 \end{equation} 475 where the index $i$ may depend on the location $x$ on the bondary. 476 477 \subsection{Solution Method \label{IKM-SOLVE}} 478 By using a first order finite difference approximation wit step size $dt>0$~\ref{IKM-EQU-3} get the form 479 \begin{equation}\label{IKM-EQU-3b} 480 D\hackscore{ij}'^{el}=\frac{1}{2 \mu dt } \left( \sigma\hackscore{ij}' - \sigma\hackscore{ij}^{'-} \right) 481 \end{equation} 482 where $\sigma\hackscore{ij}^{'-}$ is the deviatoric stress at the precious time step. 483 Now we can combine equations~\ref{IKM-EQU-2}, \ref{IKM-EQU-3b} and~\ref{IKM-EQU-6b} to get 484 \begin{equation}\label{IKM-EQU-10} 485 gross 2100 \sigma\hackscore{ij}' = 2 \eta\hackscore{eff} \left( D\hackscore{ij}' + 486 \frac{1}{ 2 \mu \; dt} \sigma\hackscore{ij}^{'-}\right) \mbox{ with } 487 gross 1878 \frac{1}{\eta\hackscore{eff}}=\frac{1}{\mu \; dt}+\frac{1}{\eta^{vp}} 488 \end{equation} 489 gross 2100 Notice that $\eta\hackscore{eff}$ is a function of diatoric stress $\sigma\hackscore{ij}'$. 490 gross 1859 After inserting~\ref{IKM-EQU-10} into~\ref{IKM-EQU-1} we get 491 \begin{equation}\label{IKM-EQU-1ib} 492 gross 2100 -\left(\eta\hackscore{eff} (v\hackscore{i,j}+ v\hackscore{i,j}) 493 \right)\hackscore{,j}+p\hackscore{,i}=F\hackscore{i}+ 494 gross 1878 \frac{\eta\hackscore{eff}}{\mu dt } \sigma\hackscore{ij,j}^{'-} 495 gross 1859 \end{equation} 496 gross 1878 Together with the incomressibilty condition~\ref{IKM-EQU-2} we need to solve a problem with a form almost identical 497 jfenwick 1966 to the Stokes problem discussed in section~\ref{STOKES SOLVE} but with the difference that $\eta\hackscore{eff}$ is depending on the solution. Analog to the iteration scheme~\ref{SADDLEPOINT iteration 2} we can run 498 gross 1878 \begin{equation} 499 \begin{array}{rcl} 500 gross 2100 -\left(\eta\hackscore{eff}(dv\hackscore{i,j}+ dv\hackscore{i,j} 501 )\right)\hackscore{,j} & = & F\hackscore{i}+ \sigma\hackscore{ij,j}'-p\hackscore{,i} \\ 502 \frac{1}{\eta\hackscore{eff}} dp & = & - v\hackscore{i,i}^+ 503 gross 1878 \end{array} 504 \label{IKM iteration 2} 505 \end{equation} 506 where $v^+=v+dv$. As this problem is non-linear the Jacobi-free Newton-GMRES method is used with the norm 507 \begin{equation} 508 gross 2100 \|(v, p)\|^2= \int\hackscore{\Omega} v\hackscore{i,j}^2 + \frac{1}{\bar{\eta}^2\hackscore{eff}} p^2 \; dx 509 gross 1878 \label{IKM iteration 3} 510 \end{equation} 511 gross 2100 where $\bar{\eta}\hackscore{eff}$ is the caracteristic viscosity, for instance: 512 \begin{equation} 513 \frac{1}{\bar{\eta}\hackscore{eff}} = \frac{1}{\tau^{-}}+\sum\hackscore{q} \frac{1}{\eta^{q}\hackscore{N}} 514 \label{IKM iteration 4} 515 \end{equation} 516 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 517 calculation of $\eta\hackscore{eff}$ with the Newton-Raphson scheme. This value can then be used to calculate 518 $\sigma\hackscore{ij}'$ via~\ref{IKM-EQU-10}. We need to solve 519 \begin{equation} 520 \tau = \eta\hackscore{eff} \cdot \epsilon \mbox{ with } 521 \epsilon = \sqrt{ 2 \left( D\hackscore{ij}' + 522 \frac{1}{ 2 \mu \; dt} \sigma\hackscore{ij}^{'-}\right)^2} 523 \label{IKM iteration 5} 524 \end{equation} 525 The Newton scheme takes the form 526 \begin{equation} 527 \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) 528 = \min(\frac{\eta\hackscore{eff} - \tau\hackscore{n} \eta\hackscore{eff}'} 529 {1 - \eta\hackscore{eff}' \cdot \epsilon}, \frac{\tau\hackscore{Y} + \beta \; p}{\epsilon}) \epsilon 530 \label{IKM iteration 6} 531 \end{equation} 532 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 533 \begin{equation} 534 \eta\hackscore{eff}' = - \eta\hackscore{eff}^2 \left(\frac{1}{\eta\hackscore{eff}}\right)' 535 \mbox{ with } 536 \left(\frac{1}{\eta\hackscore{eff}}\right)' = \sum\hackscore{q} \left(\frac{1}{\eta^{q}} \right)' 537 \label{IKM iteration 7} 538 \end{equation} 539 \begin{equation}\label{IKM-EQU-5XX} 540 \left(\frac{1}{\eta^{q}} \right)' 541 = \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}}}} 542 = \frac{1-\frac{1}{n^{q}}}{ \tau \eta^{q}} 543 \end{equation} 544 Notice that allways $\eta\hackscore{eff}'\le 0$ which makes the denomionator in~\ref{IKM iteration 6} 545 positive. 546 gross 1841 547 gross 1878 548 gross 2100