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

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first version of testing for transport 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 17 The following sections give a breif overview of the model classes and their corresponding methods. 18 19 \section{Stokes Cartesian (Saddle Point Problem)} 20 21 \subsection{Description} 22 23 Saddle point type problems emerge in a number of applications throughout physics and engineering. Finite element discretisation of the Navier-Stokes (momentum) equations for incompressible flow leads to equations of a saddle point type, which can be formulated as a solution of the following operator problem for $u \in V$ and $p \in Q$ with suitable Hilbert spaces $V$ and $Q$: 24 25 \begin{equation} 26 \left[ \begin{array}{cc} 27 A & B \\ 28 b^{*} & 0 \\ 29 \end{array} \right] 30 \left[ \begin{array}{c} 31 u \\ 32 p \\ 33 \end{array} \right] 34 =\left[ \begin{array}{c} 35 f \\ 36 g \\ 37 \end{array} \right] 38 \label{SADDLEPOINT} 39 \end{equation} 40 41 where $A$ is coercive, self-adjoint linear operator in $V$, $B$ is a linear operator from $Q$ into $V$ and $B^{*}$ is the adjoint operator of $B$. $f$ and $g$ are given elements from $V$ and $Q$ respectivitly. For more details on the mathematics see references \cite{AAMIRBERKYAN2008,MBENZI2005}. 42 43 The Uzawa scheme scheme is used to solve the momentum equation with the secondary condition of incompressibility \cite{GROSS2006,AAMIRBERKYAN2008}. 44 45 \begin{classdesc}{StokesProblemCartesian}{domain,debug} 46 opens the stokes equations on the \Domain domain. Setting debug=True switches the debug mode to on. 47 \end{classdesc} 48 49 example usage: 50 51 solution=StokesProblemCartesian(mesh) \\ 52 solution.setTolerance(TOL) \\ 53 solution.initialize(fixed\_u\_mask=b\_c,eta=eta,f=Y) \\ 54 velocity,pressure=solution.solve(velocity,pressure,max\_iter=max\_iter,solver=solver) \\ 55 56 % \subsection{Benchmark Problem} 57 % 58 % Convection problem 59 60 61 \section{Temperature Cartesian} 62 63 \begin{equation} 64 \rho c\hackscore{p} \left (\frac{\partial T}{\partial t} + \vec{v} \cdot \nabla T \right ) = k \nabla^{2}T 65 \label{HEAT EQUATION} 66 \end{equation} 67 68 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. 69 70 \subsection{Description} 71 72 \subsection{Method} 73 74 \begin{classdesc}{TemperatureCartesian}{dom,theta=THETA,useSUPG=SUPG} 75 \end{classdesc} 76 77 \subsection{Benchmark Problem} 78 79 80 \section{Level Set Method} 81 82 \subsection{Description} 83 84 \subsection{Method} 85 86 Advection and Reinitialisation 87 88 \begin{classdesc}{LevelSet}{mesh, func\_new, reinit\_max, reinit\_each, tolerance, smooth} 89 \end{classdesc} 90 91 %example usage: 92 93 %levelset = LevelSet(mesh, func\_new, reinit\_max, reinit\_each, tolerance, smooth) 94 95 \begin{methoddesc}[LevelSet]{update\_parameter}{parameter} 96 Update the parameter. 97 \end{methoddesc} 98 99 \begin{methoddesc}[LevelSet]{update\_phi}{paramter}{velocity}{dt}{t\_step} 100 Update level set function; advection and reinitialization 101 \end{methoddesc} 102 103 \subsection{Benchmark Problem} 104 105 Rayleigh-Taylor instability problem 106 107 108 % \section{Drucker Prager Model} 109 110 \section{Isotropic Kelvin Material \label{IKM}} 111 As proposed by Kelvin~\ref{KELVN} material strain $D\hackscore{ij}=v\hackscore{i,j}+v\hackscore{j,i}$ can be decomposed into 112 an elastic part $D\hackscore{ij}^{el}$ and visco-plastic part $D\hackscore{ij}^{vp}$: 113 \begin{equation}\label{IKM-EQU-2} 114 D\hackscore{ij}=D\hackscore{ij}^{el}+D\hackscore{ij}^{vp} 115 \end{equation} 116 with the elastic strain given as 117 \begin{equation}\label{IKM-EQU-3} 118 D\hackscore{ij}'^{el}=\frac{1}{2 \mu} \sigma\hackscore{ij}' 119 \end{equation} 120 where $\sigma'\hackscore{ij}$ is the deviatoric stress (Notice that $\sigma'\hackscore{ii}=0$). 121 If the material is composed by materials $q$ the visco-plastic strain can be decomposed as 122 \begin{equation}\label{IKM-EQU-4} 123 D\hackscore{ij}'^{vp}=\sum\hackscore{q} D\hackscore{ij}'^{q} 124 \end{equation} 125 where $D\hackscore{ij}^{q}$ is the strain in material $q$ given as 126 \begin{equation}\label{IKM-EQU-5} 127 D\hackscore{ij}'^{q}=\frac{1}{2 \eta^{q}} \sigma'\hackscore{ij} 128 \end{equation} 129 where $\eta^{q}$ is the viscosity of material $q$. We assume the following 130 betwee the the strain in material $q$ 131 \begin{equation}\label{IKM-EQU-5b} 132 \eta^{q}=\eta^{q}\hackscore{N} \left(\frac{\tau}{\tau\hackscore{t}^q}\right)^{\frac{1}{n^{q}}-1} 133 \mbox{ with } \tau=\sqrt{\frac{1}{2}\sigma'\hackscore{ij} \sigma'\hackscore{ij}} 134 \end{equation} 135 for a given power law coefficients $n^{q}$ and transition stresses $\tau\hackscore{t}^q$, see~\ref{POERLAW}. 136 Notice that $n^{q}=1$ gives a constant viscosity. 137 After inserting equation~\ref{IKM-EQU-5} into equation \ref{IKM-EQU-4} one gets: 138 \begin{equation}\label{IKM-EQU-6} 139 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}} \;. 140 \end{equation} 141 With 142 \begin{equation}\label{IKM-EQU-8} 143 \dot{\gamma}=\sqrt{2 D\hackscore{ij} D\hackscore{ij}} 144 \end{equation} 145 one gets 146 \begin{equation}\label{IKM-EQU-8b} 147 \tau = \eta^{vp} \dot{\gamma}^{vp} \;. 148 \end{equation} 149 With the Drucker-Prager cohesion factor $\tau\hackscore{Y}$, Drucker-Prager friction $\beta$ and total pressure $p$ we want to achieve 150 \begin{equation}\label{IKM-EQU-8c} 151 \tau \le \tau\hackscore{Y} + \beta \; p 152 \end{equation} 153 which leads to the condition 154 \begin{equation}\label{IKM-EQU-8d} 155 \eta^{vp} \le \frac{\tau\hackscore{Y} + \beta \; p}{ \dot{\gamma}^{vp}} \; . 156 \end{equation} 157 Therefore we modify the definition of $\eta^{vp}$ to the form 158 \begin{equation}\label{IKM-EQU-6b} 159 \frac{1}{\eta^{vp}}=\max(\sum\hackscore{q} \frac{1}{\eta^{q}}, \frac{\dot{\gamma}^{vp}} {\tau\hackscore{Y} + \beta \; p}) 160 \end{equation} 161 Now we can combine equations~\ref{IKM-EQU-2}, \ref{IKM-EQU-3} and~\ref{IKM-EQU-6b} to get 162 \begin{equation}\label{IKM-EQU-10} 163 D\hackscore{ij}'=\frac{1}{2 \eta\hackscore{eff}} \sigma\hackscore{ij}' \mbox{ with } 164 \frac{1}{\eta\hackscore{eff}}=\frac{1}{\mu}+\frac{1}{\eta^{vp}} 165 \end{equation} 166 The deviatoric stress needs to fullfill the equilibrion equation 167 \begin{equation}\label{IKM-EQU-1} 168 -\sigma'\hackscore{ij,j}+p\hackscore{,i}=F\hackscore{i} 169 \end{equation} 170 where $F\hackscore{j}$ is a given external fource. We assume an incompressible media: 171 \begin{equation}\label{IKM-EQU-2} 172 -v\hackscore{i,i}=0 173 \end{equation} 174 After inserting~\ref{IKM-EQU-10} into~\ref{IKM-EQU-1} we get 175 \begin{equation}\label{IKM-EQU-1ib} 176 -\left(\eta\hackscore{eff} (v\hackscore{i,j}+ v\hackscore{i,j})\right)\hackscore{,j}+p\hackscore{,i}=F\hackscore{i} 177 \end{equation} 178