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some notes for the kelvin model added. Implementation is still missing
1 ksteube 1811
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3 lgraham 1702 %
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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 %\begin{equation}
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     % \end{equation}
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

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