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4 ksteube 1811 % Copyright (c) 2003-2008 by University of Queensland
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6     % http://www.uq.edu.au/esscc
7 gross 625 %
8 ksteube 1811 % Primary Business: Queensland, Australia
9     % Licensed under the Open Software License version 3.0
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13 gross 625
14 ksteube 1811
15 jgs 121 \section{The Diffusion Problem}
16 jgs 102 \label{DIFFUSION CHAP}
17    
18     \begin{figure}
19 gross 599 \centerline{\includegraphics[width=\figwidth]{figures/DiffusionDomain.eps}}
20 jgs 102 \caption{Temperature Diffusion Problem with Circular Heat Source}
21     \label{DIFFUSION FIG 1}
22     \end{figure}
23    
24 jgs 121 \subsection{\label{DIFFUSION OUT SEC}Outline}
25 ksteube 1316 In this chapter we will discuss how to solve a time-dependent temperature diffusion\index{diffusion equation} PDE for
26     a given block of material. Within the block there is a heat source which drives the temperature diffusion.
27 jgs 107 On the surface, energy can radiate into the surrounding environment.
28 jgs 102 \fig{DIFFUSION FIG 1} shows the configuration.
29    
30     In the next \Sec{DIFFUSION TEMP SEC} we will present the relevant model. A
31     time integration scheme is introduced to calculate the temperature at given time nodes $t^{(n)}$.
32 jgs 107 We will see that at each time step a Helmholtz equation \index{Helmholtz equation}
33     must be solved.
34     The implementation of a Helmholtz equation solver will be discussed in \Sec{DIFFUSION HELM SEC}.
35 jgs 102 In Section~\ref{DIFFUSION TRANS SEC} the solver of the Helmholtz equation is used to build a
36     solver for the temperature diffusion problem.
37    
38 jgs 121 \subsection{\label{DIFFUSION TEMP SEC}Temperature Diffusion}
39 jgs 107 The unknown temperature $T$ is a function of its location in the domain and time $t>0$. The governing equation
40 jgs 102 in the interior of the domain is given by
41     \begin{equation}
42 lkettle 575 \rho c\hackscore p T\hackscore{,t} - (\kappa T\hackscore{,i})\hackscore{,i} = q\hackscore H
43 jgs 102 \label{DIFFUSION TEMP EQ 1}
44     \end{equation}
45     where $\rho c\hackscore p$ and $\kappa$ are given material constants. In case of a composite
46 lkettle 575 material the parameters depend on their location in the domain. $q\hackscore H$ is
47 ksteube 1316 a heat source (or sink) within the domain. We are using the Einstein summation convention \index{summation convention}
48 lkettle 575 as introduced in \Chap{FirstSteps}. In our case we assume $q\hackscore H$ to be equal to a constant heat production rate
49 jgs 107 $q^{c}$ on a circle or sphere with center $x^c$ and radius $r$ and $0$ elsewhere:
50 jgs 102 \begin{equation}
51 lkettle 575 q\hackscore H(x,t)=
52 jgs 102 \left\{
53     \begin{array}{lcl}
54     q^c & & \|x-x^c\| \le r \\
55     & \mbox{if} \\
56     0 & & \mbox{else} \\
57     \end{array}
58     \right.
59     \label{DIFFUSION TEMP EQ 1b}
60     \end{equation}
61     for all $x$ in the domain and all time $t>0$.
62    
63     On the surface of the domain we are
64 jgs 107 specifying a radiation condition
65 ksteube 1316 which prescribes the normal component of the flux $\kappa T\hackscore{,i}$ to be proportional
66 jgs 102 to the difference of the current temperature to the surrounding temperature $T\hackscore{ref}$:
67     \begin{equation}
68     \kappa T\hackscore{,i} n\hackscore i = \eta (T\hackscore{ref}-T)
69     \label{DIFFUSION TEMP EQ 2}
70     \end{equation}
71 jgs 107 $\eta$ is a given material coefficient depending on the material of the block and the surrounding medium.
72 ksteube 1316 $n\hackscore i$ is the $i$-th component of the outer normal field \index{outer normal field}
73 jgs 102 at the surface of the domain.
74    
75 ksteube 1316 To solve the time-dependent \eqn{DIFFUSION TEMP EQ 1} the initial temperature at time
76 jgs 102 $t=0$ has to be given. Here we assume that the initial temperature is the surrounding temperature:
77     \begin{equation}
78     T(x,0)=T\hackscore{ref}
79     \label{DIFFUSION TEMP EQ 4}
80     \end{equation}
81     for all $x$ in the domain. It is pointed out that
82 jgs 107 the initial conditions satisfy the
83 jgs 102 boundary condition defined by \eqn{DIFFUSION TEMP EQ 2}.
84    
85 jgs 107 The temperature is calculated at discrete time nodes $t^{(n)}$ where
86 jgs 102 $t^{(0)}=0$ and $t^{(n)}=t^{(n-1)}+h$ where $h>0$ is the step size which is assumed to be constant.
87 jgs 107 In the following the upper index ${(n)}$ refers to a value at time $t^{(n)}$. The simplest
88     and most robust scheme to approximate the time derivative of the the temperature is
89     the backward Euler
90 gross 660 \index{backward Euler} scheme. The backward Euler
91 jgs 107 scheme is based
92     on the Taylor expansion of $T$ at time $t^{(n)}$:
93 jgs 102 \begin{equation}
94 lkettle 573 T^{(n)}\approx T^{(n-1)}+T\hackscore{,t}^{(n)}(t^{(n)}-t^{(n-1)})
95     =T^{(n-1)} + h \cdot T\hackscore{,t}^{(n)}
96 jgs 102 \label{DIFFUSION TEMP EQ 6}
97     \end{equation}
98 jgs 107 This is inserted into \eqn{DIFFUSION TEMP EQ 1}. By separating the terms at
99 jgs 102 $t^{(n)}$ and $t^{(n-1)}$ one gets for $n=1,2,3\ldots$
100     \begin{equation}
101 lkettle 575 \frac{\rho c\hackscore p}{h} T^{(n)} - (\kappa T^{(n)}\hackscore{,i})\hackscore{,i} = q\hackscore H + \frac{\rho c\hackscore p}{h} T^{(n-1)}
102 jgs 102 \label{DIFFUSION TEMP EQ 7}
103     \end{equation}
104     where $T^{(0)}=T\hackscore{ref}$ is taken form the initial condition given by \eqn{DIFFUSION TEMP EQ 4}.
105 jgs 107 Together with the natural boundary condition
106     \begin{equation}
107     \kappa T\hackscore{,i}^{(n)} n\hackscore i = \eta (T\hackscore{ref}-T^{(n)})
108     \label{DIFFUSION TEMP EQ 2222}
109     \end{equation}
110     taken from \eqn{DIFFUSION TEMP EQ 2}
111 jgs 102 this forms a boundary value problem that has to be solved for each time step.
112     As a first step to implement a solver for the temperature diffusion problem we will
113     first implement a solver for the boundary value problem that has to be solved at each time step.
114    
115 jgs 121 \subsection{\label{DIFFUSION HELM SEC}Helmholtz Problem}
116 jgs 102 The partial differential equation to be solved for $T^{(n)}$ has the form
117     \begin{equation}
118 lkettle 575 \omega T^{(n)} - (\kappa T^{(n)}\hackscore{,i})\hackscore{,i} = f
119 jgs 102 \label{DIFFUSION HELM EQ 1}
120     \end{equation}
121 lkettle 575 and we set
122 jgs 102 \begin{equation}
123 lkettle 575 \omega=\frac{\rho c\hackscore p}{h} \mbox{ and } f=q\hackscore H +\frac{\rho c\hackscore p}{h}T^{(n-1)} \;.
124 jgs 102 \label{DIFFUSION HELM EQ 1b}
125     \end{equation}
126 jgs 107 With $g=\eta T\hackscore{ref}$ the radiation condition defined by \eqn{DIFFUSION TEMP EQ 2222}
127 jgs 102 takes the form
128     \begin{equation}
129 lkettle 575 \kappa T^{(n)}\hackscore{,i} n\hackscore{i} = g - \eta T^{(n)}\mbox{ on } \Gamma
130 jgs 102 \label{DIFFUSION HELM EQ 2}
131     \end{equation}
132 gross 568 The partial differential \eqn{DIFFUSION HELM EQ 1} together with boundary conditions of \eqn{DIFFUSION HELM EQ 2}
133 jgs 107 is called the Helmholtz equation \index{Helmholtz equation}.
134 jgs 102
135 gross 568 We want to use the \LinearPDE class provided by \escript to define and solve a general linear,steady, second order PDE such as the
136     Helmholtz equation. For a single PDE the \LinearPDE class supports the following form:
137 gross 625 \begin{equation}\label{LINEARPDE.SINGLE.1 TUTORIAL}
138     -(A\hackscore{jl} u\hackscore{,l})\hackscore{,j}+D u = Y \; .
139 jgs 102 \end{equation}
140 gross 625 where we show only the coefficients relevant for the problem discussed here. For the general form of
141     single PDE see \eqn{LINEARPDE.SINGLE.1}.
142     The coefficients $A$, and $Y$ have to be specified through \Data objects in the
143 gross 568 \Function on the PDE or objects that can be converted into such \Data objects.
144 gross 625 $A$ is a \RankTwo and $D$ and $Y$ are scalar.
145 gross 568 The following natural
146     boundary conditions are considered \index{boundary condition!natural} on $\Gamma$:
147 gross 625 \begin{equation}\label{LINEARPDE.SINGLE.2 TUTORIAL}
148     n\hackscore{j}A\hackscore{jl} u\hackscore{,l}+d u= y \;.
149 jgs 102 \end{equation}
150 gross 625 Notice that the coefficient $A$ is the same like in the PDE~\eqn{LINEARPDE.SINGLE.1 TUTORIAL}.
151     The coefficients $d$ and $y$ are
152 gross 568 each a \Scalar in the \FunctionOnBoundary. Constraints \index{constraint} for the solution prescribing the value of the
153     solution at certain locations in the domain. They have the form
154 gross 625 \begin{equation}\label{LINEARPDE.SINGLE.3 TUTORIAL}
155 gross 568 u=r \mbox{ where } q>0
156     \end{equation}
157     $r$ and $q$ are each \Scalar where $q$ is the characteristic function
158     \index{characteristic function} defining where the constraint is applied.
159 gross 625 The constraints defined by \eqn{LINEARPDE.SINGLE.3 TUTORIAL} override any other condition set by
160     \eqn{LINEARPDE.SINGLE.1 TUTORIAL} or \eqn{LINEARPDE.SINGLE.2 TUTORIAL}.
161 gross 568 The \Poisson class of the \linearPDEs module,
162     which we have already used in \Chap{FirstSteps}, is in fact a subclass of the more general
163     \LinearPDE class. The \linearPDEs module provides a \Helmholtz class but
164     we will make direct use of the general \LinearPDE class.
165 jgs 102
166 jgs 107 By inspecting the Helmholtz equation \index{Helmholtz equation}
167 lkettle 575 (\ref{DIFFUSION HELM EQ 1}) and boundary condition (\ref{DIFFUSION HELM EQ 2}) and
168     substituting $u$ for $T^{(n)}$
169 jgs 102 we can easily assign values to the coefficients in the
170     general PDE of the \LinearPDE class:
171     \begin{equation}\label{DIFFUSION HELM EQ 3}
172     \begin{array}{llllll}
173     A\hackscore{ij}=\kappa \delta\hackscore{ij} & D=\omega & Y=f \\
174     d=\eta & y= g & \\
175     \end{array}
176     \end{equation}
177 jgs 107 $\delta\hackscore{ij}$ is the Kronecker symbol \index{Kronecker symbol} defined by $\delta\hackscore{ij}=1$ for
178 lkettle 575 $i=j$ and $0$ otherwise. Undefined coefficients are assumed to be not present.\footnote{There is a difference
179 gross 568 in \escript of being not present and set to zero. As not present coefficients are not processed,
180 lkettle 575 it is more efficient to leave a coefficient undefined instead of assigning zero to it.}
181     In this diffusion example we do not need to define a characteristic function $q$ because the
182     boundary conditions we consider in \eqn{DIFFUSION HELM EQ 2} are just the natural boundary
183 gross 625 conditions which are already defined in the \LinearPDE class (shown in \eqn{LINEARPDE.SINGLE.2 TUTORIAL}).
184 jgs 102
185 gross 568 Defining and solving the Helmholtz equation is very easy now:
186     \begin{python}
187     from esys.escript import *
188     from linearPDEs import LinearPDE
189     mypde=LinearPDE(mydomain)
190     mypde.setValue(A=kappa*kronecker(mydomain),D=omega,Y=f,d=eta,y=g)
191     u=mypde.getSolution()
192     \end{python}
193     where we assume that \code{mydomain} is a \Domain object and
194     \code{kappa} \code{omega} \code{eta} and \code{g} are given scalar values
195     typically \code{float} or \Data objects. The \method{setValue} method
196     assigns values to the coefficients of the general PDE. The \method{getSolution} method solves
197 gross 569 the PDE and returns the solution \code{u} of the PDE. \function{kronecker} is \escript function
198 gross 568 returning the Kronecker symbol.
199    
200     The coefficients can set by several calls of \method{setValue} where the order can be chosen arbitrarily.
201 lkettle 573 If a value is assigned to a coefficient several times, the last assigned value is used when
202 gross 568 the solution is calculated:
203     \begin{python}
204     mypde=LinearPDE(mydomain)
205     mypde.setValue(A=kappa*kronecker(mydomain),d=eta)
206     mypde.setValue(D=omega,Y=f,y=g)
207     mypde.setValue(d=2*eta) # overwrites d=eta
208     u=mypde.getSolution()
209     \end{python}
210     In some cases the solver of the PDE can make use of the fact that the PDE is symmetric\index{symmetric PDE} where the
211     PDE is called symmetric if
212 gross 625 \begin{equation}\label{LINEARPDE.SINGLE.4 TUTORIAL}
213     A\hackscore{jl}=A\hackscore{lj}\;.
214 gross 568 \end{equation}
215 gross 625 Note that $D$ and $d$ may have any value and the right hand sides $Y$, $y$ as well as the constraints
216 gross 568 are not relevant. The Helmholtz problem is symmetric.
217 gross 569 The \LinearPDE class provides the method \method{checkSymmetry} method to check if the given PDE is symmetric.
218 gross 568 \begin{python}
219     mypde=LinearPDE(mydomain)
220     mypde.setValue(A=kappa*kronecker(mydomain),d=eta)
221     print mypde.checkSymmetry() # returns True
222     mypde.setValue(B=kronecker(mydomain)[0])
223     print mypde.checkSymmetry() # returns False
224     mypde.setValue(C=kronecker(mydomain)[0])
225     print mypde.checkSymmetry() # returns True
226     \end{python}
227 gross 569 Unfortunately, a \method{checkSymmetry} is very expensive and is recommendable to use for
228     testing and debugging purposes only. The \method{setSymmetryOn} method is used to
229 gross 568 declare a PDE symmetric:
230     \begin{python}
231     mypde = LinearPDE(mydomain)
232     mypde.setValue(A=kappa*kronecker(mydomain))
233     mypde.setSymmetryOn()
234     \end{python}
235 gross 569 Now we want to see how we actually solve the Helmholtz equation.
236     on a rectangular domain
237     of length $l\hackscore{0}=5$ and height $l\hackscore{1}=1$. We choose a simple test solution such that we
238     can verify the returned solution against the exact answer. Actually, we
239 lkettle 575 take $T=x\hackscore{0}$ (here $q\hackscore H = 0$) and then calculate the right hand side terms $f$ and $g$ such that
240 jgs 107 the test solution becomes the solution of the problem. If we assume $\kappa$ as being constant,
241 jgs 102 an easy calculation shows that we have to choose $f=\omega \cdot x\hackscore{0}$. On the boundary we get
242 jgs 107 $\kappa n\hackscore{i} u\hackscore{,i}=\kappa n\hackscore{0}$.
243 gross 569 So we have to set $g=\kappa n\hackscore{0}+\eta x\hackscore{0}$. The following script \file{helmholtz.py}
244 gross 568 \index{scripts!\file{helmholtz.py}} which is available in the \ExampleDirectory
245 jgs 102 implements this test problem using the \finley PDE solver:
246     \begin{python}
247 gross 568 from esys.escript import *
248 gross 569 from esys.escript.linearPDEs import LinearPDE
249 jgs 107 from esys.finley import Rectangle
250 jgs 102 #... set some parameters ...
251 jgs 107 kappa=1.
252 jgs 102 omega=0.1
253     eta=10.
254     #... generate domain ...
255 jgs 107 mydomain = Rectangle(l0=5.,l1=1.,n0=50, n1=10)
256 jgs 102 #... open PDE and set coefficients ...
257 gross 568 mypde=LinearPDE(mydomain)
258     mypde.setSymmetryOn()
259 jgs 102 n=mydomain.getNormal()
260     x=mydomain.getX()
261 gross 569 mypde.setValue(A=kappa*kronecker(mydomain),D=omega,Y=omega*x[0], \
262     d=eta,y=kappa*n[0]+eta*x[0])
263 jgs 102 #... calculate error of the PDE solution ...
264     u=mypde.getSolution()
265     print "error is ",Lsup(u-x[0])
266 lkettle 575 saveVTK("x0.xml",sol=u)
267 jgs 102 \end{python}
268 lgraham 1700 To visualize the solution `x0.~xml' just use the command
269 lkettle 575 \begin{python}
270     mayavi -d u.xml -m SurfaceMap &
271     \end{python}
272     and it is easy to see that the solution $T=x\hackscore{0}$ is calculated.
273    
274 lgraham 1700 The script is similar to the script \file{poisson.py} discussed in \Chap{FirstSteps}.
275 jgs 102 \code{mydomain.getNormal()} returns the outer normal field on the surface of the domain. The function \function{Lsup}
276 lkettle 573 imported by the \code{from esys.escript import *} statement and returns the maximum absolute value of its argument.
277 jgs 107 The error shown by the print statement should be in the order of $10^{-7}$. As piecewise bi-linear interpolation is
278 gross 569 used by \finley approximate the solution and our solution is a linear function of the spatial coordinates one might
279 jgs 107 expect that the error would be zero or in the order of machine precision (typically $\approx 10^{-15}$).
280     However most PDE packages use an iterative solver which is terminated
281     when a given tolerance has been reached. The default tolerance is $10^{-8}$. This value can be altered by using the
282 jgs 102 \method{setTolerance} of the \LinearPDE class.
283    
284 jgs 121 \subsection{The Transition Problem}
285 jgs 102 \label{DIFFUSION TRANS SEC}
286 ksteube 1316 Now we are ready to solve the original time-dependent problem. The main
287 jgs 107 part of the script is the loop over time $t$ which takes the following form:
288 jgs 102 \begin{python}
289 jgs 107 t=0
290     T=Tref
291 gross 569 mypde=LinearPDE(mydomain)
292     mypde.setValue(A=kappa*kronecker(mydomain),D=rhocp/h,d=eta,y=eta*Tref)
293 jgs 102 while t<t_end:
294 gross 569 mypde.setValue(Y=q+rhocp/h*T)
295 jgs 102 T=mypde.getSolution()
296     t+=h
297     \end{python}
298     \var{kappa}, \var{rhocp}, \var{eta} and \var{Tref} are input parameters of the model. \var{q} is the heat source
299 gross 569 in the domain and \var{h} is the time step size.
300     The variable \var{T}
301 jgs 102 holds the current temperature. It is used to calculate the right hand side coefficient \var{f} in the
302 jgs 107 Helmholtz equation in \eqn{DIFFUSION HELM EQ 1}. The statement \code{T=mypde.getSolution()} overwrites \var{T} with the
303     temperature of the new time step $\var{t}+\var{h}$. To get this iterative process going we need to specify the
304 jgs 102 initial temperature distribution, which equal to $T\hackscore{ref}$.
305 gross 569 The \LinearPDE class object \var{mypde}
306     and coefficients not changing over time are set up before the loop over time is entered. In each time step only the coefficient
307     $Y$ is reset as it depends on the temperature of the previous time step. This allows the PDE solver to reuse information
308     from previous time steps as much as possible.
309 jgs 102
310 lkettle 575 The heat source $q\hackscore H$ which is defined in \eqn{DIFFUSION TEMP EQ 1b} is \var{qc}
311 jgs 107 in an area defined as a circle of radius \var{r} and center \var{xc} and zero outside this circle.
312 lkettle 575 \var{q0} is a fixed constant. The following script defines $q\hackscore H$ as desired:
313 jgs 102 \begin{python}
314 lkettle 573 from esys.escript import length,whereNegative
315 jgs 102 xc=[0.02,0.002]
316     r=0.001
317     x=mydomain.getX()
318 lkettle 575 qH=q0*whereNegative(length(x-xc)-r)
319 jgs 102 \end{python}
320     \var{x} is a \Data class object of
321 jgs 107 the \escript module defining locations in the \Domain \var{mydomain}.
322     The \function{length()} imported from the \escript module returns the
323     Euclidean norm:
324     \begin{equation}\label{DIFFUSION HELM EQ 3aba}
325     \|y\|=\sqrt{
326     y\hackscore{i}
327     y\hackscore{i}
328     } = \function{esys.escript.length}(y)
329     \end{equation}
330     So \code{length(x-xc)} calculates the distances
331     of the location \var{x} to the center of the circle \var{xc} where the heat source is acting.
332 ksteube 1316 Note that the coordinates of \var{xc} are defined as a list of floating point numbers. It is automatically
333 gross 569 converted into a \Data class object before being subtracted from \var{x}. The function \function{whereNegative}
334     applied to
335 ksteube 1316 \code{length(x-xc)-r}, returns a \Data object which is equal to one where the object is negative (inside the circle) and
336     zero elsewhere. After multiplication with \var{qc} we get a function with the desired property of having value \var{qc} inside
337     the circle and zero elsewhere.
338 jgs 102
339 jgs 107 Now we can put the components together to create the script \file{diffusion.py} which is available in the \ExampleDirectory:
340 jgs 102 \index{scripts!\file{diffusion.py}}:
341     \begin{python}
342 gross 569 from esys.escript import *
343     from esys.escript.linearPDEs import LinearPDE
344 jgs 107 from esys.finley import Rectangle
345 jgs 102 #... set some parameters ...
346 jgs 107 xc=[0.02,0.002]
347 jgs 102 r=0.001
348 jgs 107 qc=50.e6
349 jgs 102 Tref=0.
350     rhocp=2.6e6
351     eta=75.
352     kappa=240.
353 jgs 107 tend=5.
354     # ... time, time step size and counter ...
355     t=0
356 jgs 102 h=0.1
357     i=0
358     #... generate domain ...
359 jgs 107 mydomain = Rectangle(l0=0.05,l1=0.01,n0=250, n1=50)
360 jgs 102 #... open PDE ...
361 gross 569 mypde=LinearPDE(mydomain)
362     mypde.setSymmetryOn()
363     mypde.setValue(A=kappa*kronecker(mydomain),D=rhocp/h,d=eta,y=eta*Tref)
364 jgs 102 # ... set heat source: ....
365     x=mydomain.getX()
366 lkettle 575 qH=qc*whereNegative(length(x-xc)-r)
367 jgs 102 # ... set initial temperature ....
368     T=Tref
369     # ... start iteration:
370 jgs 107 while t<tend:
371 jgs 102 i+=1
372     t+=h
373     print "time step :",t
374 lkettle 575 mypde.setValue(Y=qH+rhocp/h*T)
375 jgs 102 T=mypde.getSolution()
376 gross 569 saveVTK("T.%d.xml"%i,temp=T)
377 jgs 102 \end{python}
378 gross 569 The script will create the files \file{T.1.xml},
379     \file{T.2.xml}, $\ldots$, \file{T.50.xml} in the directory where the script has been started. The files give the
380     temperature distributions at time steps $1$, $2$, $\ldots$, $50$ in the \VTK file format.
381 jgs 107
382     \begin{figure}
383 gross 599 \centerline{\includegraphics[width=\figwidth]{figures/DiffusionRes1.eps}}
384     \centerline{\includegraphics[width=\figwidth]{figures/DiffusionRes16.eps}}
385     \centerline{\includegraphics[width=\figwidth]{figures/DiffusionRes32.eps}}
386     \centerline{\includegraphics[width=\figwidth]{figures/DiffusionRes48.eps}}
387 lkettle 581 \caption{Results of the Temperature Diffusion Problem for Time Steps $1$ $16$, $32$ and $48$.}
388 jgs 107 \label{DIFFUSION FIG 2}
389     \end{figure}
390 ksteube 1316 \fig{DIFFUSION FIG 2} shows the result for some selected time steps.
391 jgs 102 An easy way to visualize the results is the command
392     \begin{verbatim}
393 gross 569 mayavi -d T.1.xml -m SurfaceMap &
394 jgs 102 \end{verbatim}
395 ksteube 1316 Use the \texttt{Configure Data} window in mayavi
396 gross 569 to move forward and and backwards in time.

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