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Tue Sep 25 03:18:30 2007 UTC (12 years ago) by ksteube
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Quickly edited chapters 1 and 2 of the User Guide, but it needs more work.
Ran entire document through spell checker.

1 ksteube 1316 %
2 jgs 102 % $Id$
3 gross 625 %
4 ksteube 1316 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
5 gross 625 %
6 ksteube 1316 % Copyright 2003-2007 by ACceSS MNRF
7     % Copyright 2007 by University of Queensland
8     %
9     % http://esscc.uq.edu.au
10     % Primary Business: Queensland, Australia
11     % Licensed under the Open Software License version 3.0
12     % http://www.opensource.org/licenses/osl-3.0.php
13     %
14     %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
15     %
16 gross 625
17 jgs 121 \section{The Diffusion Problem}
18 jgs 102 \label{DIFFUSION CHAP}
19    
20     \begin{figure}
21 gross 599 \centerline{\includegraphics[width=\figwidth]{figures/DiffusionDomain.eps}}
22 jgs 102 \caption{Temperature Diffusion Problem with Circular Heat Source}
23     \label{DIFFUSION FIG 1}
24     \end{figure}
25    
26 jgs 121 \subsection{\label{DIFFUSION OUT SEC}Outline}
27 ksteube 1316 In this chapter we will discuss how to solve a time-dependent temperature diffusion\index{diffusion equation} PDE for
28     a given block of material. Within the block there is a heat source which drives the temperature diffusion.
29 jgs 107 On the surface, energy can radiate into the surrounding environment.
30 jgs 102 \fig{DIFFUSION FIG 1} shows the configuration.
31    
32     In the next \Sec{DIFFUSION TEMP SEC} we will present the relevant model. A
33     time integration scheme is introduced to calculate the temperature at given time nodes $t^{(n)}$.
34 jgs 107 We will see that at each time step a Helmholtz equation \index{Helmholtz equation}
35     must be solved.
36     The implementation of a Helmholtz equation solver will be discussed in \Sec{DIFFUSION HELM SEC}.
37 jgs 102 In Section~\ref{DIFFUSION TRANS SEC} the solver of the Helmholtz equation is used to build a
38     solver for the temperature diffusion problem.
39    
40 jgs 121 \subsection{\label{DIFFUSION TEMP SEC}Temperature Diffusion}
41 jgs 107 The unknown temperature $T$ is a function of its location in the domain and time $t>0$. The governing equation
42 jgs 102 in the interior of the domain is given by
43     \begin{equation}
44 lkettle 575 \rho c\hackscore p T\hackscore{,t} - (\kappa T\hackscore{,i})\hackscore{,i} = q\hackscore H
45 jgs 102 \label{DIFFUSION TEMP EQ 1}
46     \end{equation}
47     where $\rho c\hackscore p$ and $\kappa$ are given material constants. In case of a composite
48 lkettle 575 material the parameters depend on their location in the domain. $q\hackscore H$ is
49 ksteube 1316 a heat source (or sink) within the domain. We are using the Einstein summation convention \index{summation convention}
50 lkettle 575 as introduced in \Chap{FirstSteps}. In our case we assume $q\hackscore H$ to be equal to a constant heat production rate
51 jgs 107 $q^{c}$ on a circle or sphere with center $x^c$ and radius $r$ and $0$ elsewhere:
52 jgs 102 \begin{equation}
53 lkettle 575 q\hackscore H(x,t)=
54 jgs 102 \left\{
55     \begin{array}{lcl}
56     q^c & & \|x-x^c\| \le r \\
57     & \mbox{if} \\
58     0 & & \mbox{else} \\
59     \end{array}
60     \right.
61     \label{DIFFUSION TEMP EQ 1b}
62     \end{equation}
63     for all $x$ in the domain and all time $t>0$.
64    
65     On the surface of the domain we are
66 jgs 107 specifying a radiation condition
67 ksteube 1316 which prescribes the normal component of the flux $\kappa T\hackscore{,i}$ to be proportional
68 jgs 102 to the difference of the current temperature to the surrounding temperature $T\hackscore{ref}$:
69     \begin{equation}
70     \kappa T\hackscore{,i} n\hackscore i = \eta (T\hackscore{ref}-T)
71     \label{DIFFUSION TEMP EQ 2}
72     \end{equation}
73 jgs 107 $\eta$ is a given material coefficient depending on the material of the block and the surrounding medium.
74 ksteube 1316 $n\hackscore i$ is the $i$-th component of the outer normal field \index{outer normal field}
75 jgs 102 at the surface of the domain.
76    
77 ksteube 1316 To solve the time-dependent \eqn{DIFFUSION TEMP EQ 1} the initial temperature at time
78 jgs 102 $t=0$ has to be given. Here we assume that the initial temperature is the surrounding temperature:
79     \begin{equation}
80     T(x,0)=T\hackscore{ref}
81     \label{DIFFUSION TEMP EQ 4}
82     \end{equation}
83     for all $x$ in the domain. It is pointed out that
84 jgs 107 the initial conditions satisfy the
85 jgs 102 boundary condition defined by \eqn{DIFFUSION TEMP EQ 2}.
86    
87 jgs 107 The temperature is calculated at discrete time nodes $t^{(n)}$ where
88 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.
89 jgs 107 In the following the upper index ${(n)}$ refers to a value at time $t^{(n)}$. The simplest
90     and most robust scheme to approximate the time derivative of the the temperature is
91     the backward Euler
92 gross 660 \index{backward Euler} scheme. The backward Euler
93 jgs 107 scheme is based
94     on the Taylor expansion of $T$ at time $t^{(n)}$:
95 jgs 102 \begin{equation}
96 lkettle 573 T^{(n)}\approx T^{(n-1)}+T\hackscore{,t}^{(n)}(t^{(n)}-t^{(n-1)})
97     =T^{(n-1)} + h \cdot T\hackscore{,t}^{(n)}
98 jgs 102 \label{DIFFUSION TEMP EQ 6}
99     \end{equation}
100 jgs 107 This is inserted into \eqn{DIFFUSION TEMP EQ 1}. By separating the terms at
101 jgs 102 $t^{(n)}$ and $t^{(n-1)}$ one gets for $n=1,2,3\ldots$
102     \begin{equation}
103 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)}
104 jgs 102 \label{DIFFUSION TEMP EQ 7}
105     \end{equation}
106     where $T^{(0)}=T\hackscore{ref}$ is taken form the initial condition given by \eqn{DIFFUSION TEMP EQ 4}.
107 jgs 107 Together with the natural boundary condition
108     \begin{equation}
109     \kappa T\hackscore{,i}^{(n)} n\hackscore i = \eta (T\hackscore{ref}-T^{(n)})
110     \label{DIFFUSION TEMP EQ 2222}
111     \end{equation}
112     taken from \eqn{DIFFUSION TEMP EQ 2}
113 jgs 102 this forms a boundary value problem that has to be solved for each time step.
114     As a first step to implement a solver for the temperature diffusion problem we will
115     first implement a solver for the boundary value problem that has to be solved at each time step.
116    
117 jgs 121 \subsection{\label{DIFFUSION HELM SEC}Helmholtz Problem}
118 jgs 102 The partial differential equation to be solved for $T^{(n)}$ has the form
119     \begin{equation}
120 lkettle 575 \omega T^{(n)} - (\kappa T^{(n)}\hackscore{,i})\hackscore{,i} = f
121 jgs 102 \label{DIFFUSION HELM EQ 1}
122     \end{equation}
123 lkettle 575 and we set
124 jgs 102 \begin{equation}
125 lkettle 575 \omega=\frac{\rho c\hackscore p}{h} \mbox{ and } f=q\hackscore H +\frac{\rho c\hackscore p}{h}T^{(n-1)} \;.
126 jgs 102 \label{DIFFUSION HELM EQ 1b}
127     \end{equation}
128 jgs 107 With $g=\eta T\hackscore{ref}$ the radiation condition defined by \eqn{DIFFUSION TEMP EQ 2222}
129 jgs 102 takes the form
130     \begin{equation}
131 lkettle 575 \kappa T^{(n)}\hackscore{,i} n\hackscore{i} = g - \eta T^{(n)}\mbox{ on } \Gamma
132 jgs 102 \label{DIFFUSION HELM EQ 2}
133     \end{equation}
134 gross 568 The partial differential \eqn{DIFFUSION HELM EQ 1} together with boundary conditions of \eqn{DIFFUSION HELM EQ 2}
135 jgs 107 is called the Helmholtz equation \index{Helmholtz equation}.
136 jgs 102
137 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
138     Helmholtz equation. For a single PDE the \LinearPDE class supports the following form:
139 gross 625 \begin{equation}\label{LINEARPDE.SINGLE.1 TUTORIAL}
140     -(A\hackscore{jl} u\hackscore{,l})\hackscore{,j}+D u = Y \; .
141 jgs 102 \end{equation}
142 gross 625 where we show only the coefficients relevant for the problem discussed here. For the general form of
143     single PDE see \eqn{LINEARPDE.SINGLE.1}.
144     The coefficients $A$, and $Y$ have to be specified through \Data objects in the
145 gross 568 \Function on the PDE or objects that can be converted into such \Data objects.
146 gross 625 $A$ is a \RankTwo and $D$ and $Y$ are scalar.
147 gross 568 The following natural
148     boundary conditions are considered \index{boundary condition!natural} on $\Gamma$:
149 gross 625 \begin{equation}\label{LINEARPDE.SINGLE.2 TUTORIAL}
150     n\hackscore{j}A\hackscore{jl} u\hackscore{,l}+d u= y \;.
151 jgs 102 \end{equation}
152 gross 625 Notice that the coefficient $A$ is the same like in the PDE~\eqn{LINEARPDE.SINGLE.1 TUTORIAL}.
153     The coefficients $d$ and $y$ are
154 gross 568 each a \Scalar in the \FunctionOnBoundary. Constraints \index{constraint} for the solution prescribing the value of the
155     solution at certain locations in the domain. They have the form
156 gross 625 \begin{equation}\label{LINEARPDE.SINGLE.3 TUTORIAL}
157 gross 568 u=r \mbox{ where } q>0
158     \end{equation}
159     $r$ and $q$ are each \Scalar where $q$ is the characteristic function
160     \index{characteristic function} defining where the constraint is applied.
161 gross 625 The constraints defined by \eqn{LINEARPDE.SINGLE.3 TUTORIAL} override any other condition set by
162     \eqn{LINEARPDE.SINGLE.1 TUTORIAL} or \eqn{LINEARPDE.SINGLE.2 TUTORIAL}.
163 gross 568 The \Poisson class of the \linearPDEs module,
164     which we have already used in \Chap{FirstSteps}, is in fact a subclass of the more general
165     \LinearPDE class. The \linearPDEs module provides a \Helmholtz class but
166     we will make direct use of the general \LinearPDE class.
167 jgs 102
168 jgs 107 By inspecting the Helmholtz equation \index{Helmholtz equation}
169 lkettle 575 (\ref{DIFFUSION HELM EQ 1}) and boundary condition (\ref{DIFFUSION HELM EQ 2}) and
170     substituting $u$ for $T^{(n)}$
171 jgs 102 we can easily assign values to the coefficients in the
172     general PDE of the \LinearPDE class:
173     \begin{equation}\label{DIFFUSION HELM EQ 3}
174     \begin{array}{llllll}
175     A\hackscore{ij}=\kappa \delta\hackscore{ij} & D=\omega & Y=f \\
176     d=\eta & y= g & \\
177     \end{array}
178     \end{equation}
179 jgs 107 $\delta\hackscore{ij}$ is the Kronecker symbol \index{Kronecker symbol} defined by $\delta\hackscore{ij}=1$ for
180 lkettle 575 $i=j$ and $0$ otherwise. Undefined coefficients are assumed to be not present.\footnote{There is a difference
181 gross 568 in \escript of being not present and set to zero. As not present coefficients are not processed,
182 lkettle 575 it is more efficient to leave a coefficient undefined instead of assigning zero to it.}
183     In this diffusion example we do not need to define a characteristic function $q$ because the
184     boundary conditions we consider in \eqn{DIFFUSION HELM EQ 2} are just the natural boundary
185 gross 625 conditions which are already defined in the \LinearPDE class (shown in \eqn{LINEARPDE.SINGLE.2 TUTORIAL}).
186 jgs 102
187 gross 568 Defining and solving the Helmholtz equation is very easy now:
188     \begin{python}
189     from esys.escript import *
190     from linearPDEs import LinearPDE
191     mypde=LinearPDE(mydomain)
192     mypde.setValue(A=kappa*kronecker(mydomain),D=omega,Y=f,d=eta,y=g)
193     u=mypde.getSolution()
194     \end{python}
195     where we assume that \code{mydomain} is a \Domain object and
196     \code{kappa} \code{omega} \code{eta} and \code{g} are given scalar values
197     typically \code{float} or \Data objects. The \method{setValue} method
198     assigns values to the coefficients of the general PDE. The \method{getSolution} method solves
199 gross 569 the PDE and returns the solution \code{u} of the PDE. \function{kronecker} is \escript function
200 gross 568 returning the Kronecker symbol.
201    
202     The coefficients can set by several calls of \method{setValue} where the order can be chosen arbitrarily.
203 lkettle 573 If a value is assigned to a coefficient several times, the last assigned value is used when
204 gross 568 the solution is calculated:
205     \begin{python}
206     mypde=LinearPDE(mydomain)
207     mypde.setValue(A=kappa*kronecker(mydomain),d=eta)
208     mypde.setValue(D=omega,Y=f,y=g)
209     mypde.setValue(d=2*eta) # overwrites d=eta
210     u=mypde.getSolution()
211     \end{python}
212     In some cases the solver of the PDE can make use of the fact that the PDE is symmetric\index{symmetric PDE} where the
213     PDE is called symmetric if
214 gross 625 \begin{equation}\label{LINEARPDE.SINGLE.4 TUTORIAL}
215     A\hackscore{jl}=A\hackscore{lj}\;.
216 gross 568 \end{equation}
217 gross 625 Note that $D$ and $d$ may have any value and the right hand sides $Y$, $y$ as well as the constraints
218 gross 568 are not relevant. The Helmholtz problem is symmetric.
219 gross 569 The \LinearPDE class provides the method \method{checkSymmetry} method to check if the given PDE is symmetric.
220 gross 568 \begin{python}
221     mypde=LinearPDE(mydomain)
222     mypde.setValue(A=kappa*kronecker(mydomain),d=eta)
223     print mypde.checkSymmetry() # returns True
224     mypde.setValue(B=kronecker(mydomain)[0])
225     print mypde.checkSymmetry() # returns False
226     mypde.setValue(C=kronecker(mydomain)[0])
227     print mypde.checkSymmetry() # returns True
228     \end{python}
229 gross 569 Unfortunately, a \method{checkSymmetry} is very expensive and is recommendable to use for
230     testing and debugging purposes only. The \method{setSymmetryOn} method is used to
231 gross 568 declare a PDE symmetric:
232     \begin{python}
233     mypde = LinearPDE(mydomain)
234     mypde.setValue(A=kappa*kronecker(mydomain))
235     mypde.setSymmetryOn()
236     \end{python}
237 gross 569 Now we want to see how we actually solve the Helmholtz equation.
238     on a rectangular domain
239     of length $l\hackscore{0}=5$ and height $l\hackscore{1}=1$. We choose a simple test solution such that we
240     can verify the returned solution against the exact answer. Actually, we
241 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
242 jgs 107 the test solution becomes the solution of the problem. If we assume $\kappa$ as being constant,
243 jgs 102 an easy calculation shows that we have to choose $f=\omega \cdot x\hackscore{0}$. On the boundary we get
244 jgs 107 $\kappa n\hackscore{i} u\hackscore{,i}=\kappa n\hackscore{0}$.
245 gross 569 So we have to set $g=\kappa n\hackscore{0}+\eta x\hackscore{0}$. The following script \file{helmholtz.py}
246 gross 568 \index{scripts!\file{helmholtz.py}} which is available in the \ExampleDirectory
247 jgs 102 implements this test problem using the \finley PDE solver:
248     \begin{python}
249 gross 568 from esys.escript import *
250 gross 569 from esys.escript.linearPDEs import LinearPDE
251 jgs 107 from esys.finley import Rectangle
252 jgs 102 #... set some parameters ...
253 jgs 107 kappa=1.
254 jgs 102 omega=0.1
255     eta=10.
256     #... generate domain ...
257 jgs 107 mydomain = Rectangle(l0=5.,l1=1.,n0=50, n1=10)
258 jgs 102 #... open PDE and set coefficients ...
259 gross 568 mypde=LinearPDE(mydomain)
260     mypde.setSymmetryOn()
261 jgs 102 n=mydomain.getNormal()
262     x=mydomain.getX()
263 gross 569 mypde.setValue(A=kappa*kronecker(mydomain),D=omega,Y=omega*x[0], \
264     d=eta,y=kappa*n[0]+eta*x[0])
265 jgs 102 #... calculate error of the PDE solution ...
266     u=mypde.getSolution()
267     print "error is ",Lsup(u-x[0])
268 lkettle 575 saveVTK("x0.xml",sol=u)
269 jgs 102 \end{python}
270 lkettle 575 To visualise the solution `x0.~xml' just use the command
271     \begin{python}
272     mayavi -d u.xml -m SurfaceMap &
273     \end{python}
274     and it is easy to see that the solution $T=x\hackscore{0}$ is calculated.
275    
276 gross 569 The script is similar to the script \file{poisson.py} dicussed in \Chap{FirstSteps}.
277 jgs 102 \code{mydomain.getNormal()} returns the outer normal field on the surface of the domain. The function \function{Lsup}
278 lkettle 573 imported by the \code{from esys.escript import *} statement and returns the maximum absolute value of its argument.
279 jgs 107 The error shown by the print statement should be in the order of $10^{-7}$. As piecewise bi-linear interpolation is
280 gross 569 used by \finley approximate the solution and our solution is a linear function of the spatial coordinates one might
281 jgs 107 expect that the error would be zero or in the order of machine precision (typically $\approx 10^{-15}$).
282     However most PDE packages use an iterative solver which is terminated
283     when a given tolerance has been reached. The default tolerance is $10^{-8}$. This value can be altered by using the
284 jgs 102 \method{setTolerance} of the \LinearPDE class.
285    
286 jgs 121 \subsection{The Transition Problem}
287 jgs 102 \label{DIFFUSION TRANS SEC}
288 ksteube 1316 Now we are ready to solve the original time-dependent problem. The main
289 jgs 107 part of the script is the loop over time $t$ which takes the following form:
290 jgs 102 \begin{python}
291 jgs 107 t=0
292     T=Tref
293 gross 569 mypde=LinearPDE(mydomain)
294     mypde.setValue(A=kappa*kronecker(mydomain),D=rhocp/h,d=eta,y=eta*Tref)
295 jgs 102 while t<t_end:
296 gross 569 mypde.setValue(Y=q+rhocp/h*T)
297 jgs 102 T=mypde.getSolution()
298     t+=h
299     \end{python}
300     \var{kappa}, \var{rhocp}, \var{eta} and \var{Tref} are input parameters of the model. \var{q} is the heat source
301 gross 569 in the domain and \var{h} is the time step size.
302     The variable \var{T}
303 jgs 102 holds the current temperature. It is used to calculate the right hand side coefficient \var{f} in the
304 jgs 107 Helmholtz equation in \eqn{DIFFUSION HELM EQ 1}. The statement \code{T=mypde.getSolution()} overwrites \var{T} with the
305     temperature of the new time step $\var{t}+\var{h}$. To get this iterative process going we need to specify the
306 jgs 102 initial temperature distribution, which equal to $T\hackscore{ref}$.
307 gross 569 The \LinearPDE class object \var{mypde}
308     and coefficients not changing over time are set up before the loop over time is entered. In each time step only the coefficient
309     $Y$ is reset as it depends on the temperature of the previous time step. This allows the PDE solver to reuse information
310     from previous time steps as much as possible.
311 jgs 102
312 lkettle 575 The heat source $q\hackscore H$ which is defined in \eqn{DIFFUSION TEMP EQ 1b} is \var{qc}
313 jgs 107 in an area defined as a circle of radius \var{r} and center \var{xc} and zero outside this circle.
314 lkettle 575 \var{q0} is a fixed constant. The following script defines $q\hackscore H$ as desired:
315 jgs 102 \begin{python}
316 lkettle 573 from esys.escript import length,whereNegative
317 jgs 102 xc=[0.02,0.002]
318     r=0.001
319     x=mydomain.getX()
320 lkettle 575 qH=q0*whereNegative(length(x-xc)-r)
321 jgs 102 \end{python}
322     \var{x} is a \Data class object of
323 jgs 107 the \escript module defining locations in the \Domain \var{mydomain}.
324     The \function{length()} imported from the \escript module returns the
325     Euclidean norm:
326     \begin{equation}\label{DIFFUSION HELM EQ 3aba}
327     \|y\|=\sqrt{
328     y\hackscore{i}
329     y\hackscore{i}
330     } = \function{esys.escript.length}(y)
331     \end{equation}
332     So \code{length(x-xc)} calculates the distances
333     of the location \var{x} to the center of the circle \var{xc} where the heat source is acting.
334 ksteube 1316 Note that the coordinates of \var{xc} are defined as a list of floating point numbers. It is automatically
335 gross 569 converted into a \Data class object before being subtracted from \var{x}. The function \function{whereNegative}
336     applied to
337 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
338     zero elsewhere. After multiplication with \var{qc} we get a function with the desired property of having value \var{qc} inside
339     the circle and zero elsewhere.
340 jgs 102
341 jgs 107 Now we can put the components together to create the script \file{diffusion.py} which is available in the \ExampleDirectory:
342 jgs 102 \index{scripts!\file{diffusion.py}}:
343     \begin{python}
344 gross 569 from esys.escript import *
345     from esys.escript.linearPDEs import LinearPDE
346 jgs 107 from esys.finley import Rectangle
347 jgs 102 #... set some parameters ...
348 jgs 107 xc=[0.02,0.002]
349 jgs 102 r=0.001
350 jgs 107 qc=50.e6
351 jgs 102 Tref=0.
352     rhocp=2.6e6
353     eta=75.
354     kappa=240.
355 jgs 107 tend=5.
356     # ... time, time step size and counter ...
357     t=0
358 jgs 102 h=0.1
359     i=0
360     #... generate domain ...
361 jgs 107 mydomain = Rectangle(l0=0.05,l1=0.01,n0=250, n1=50)
362 jgs 102 #... open PDE ...
363 gross 569 mypde=LinearPDE(mydomain)
364     mypde.setSymmetryOn()
365     mypde.setValue(A=kappa*kronecker(mydomain),D=rhocp/h,d=eta,y=eta*Tref)
366 jgs 102 # ... set heat source: ....
367     x=mydomain.getX()
368 lkettle 575 qH=qc*whereNegative(length(x-xc)-r)
369 jgs 102 # ... set initial temperature ....
370     T=Tref
371     # ... start iteration:
372 jgs 107 while t<tend:
373 jgs 102 i+=1
374     t+=h
375     print "time step :",t
376 lkettle 575 mypde.setValue(Y=qH+rhocp/h*T)
377 jgs 102 T=mypde.getSolution()
378 gross 569 saveVTK("T.%d.xml"%i,temp=T)
379 jgs 102 \end{python}
380 gross 569 The script will create the files \file{T.1.xml},
381     \file{T.2.xml}, $\ldots$, \file{T.50.xml} in the directory where the script has been started. The files give the
382     temperature distributions at time steps $1$, $2$, $\ldots$, $50$ in the \VTK file format.
383 jgs 107
384     \begin{figure}
385 gross 599 \centerline{\includegraphics[width=\figwidth]{figures/DiffusionRes1.eps}}
386     \centerline{\includegraphics[width=\figwidth]{figures/DiffusionRes16.eps}}
387     \centerline{\includegraphics[width=\figwidth]{figures/DiffusionRes32.eps}}
388     \centerline{\includegraphics[width=\figwidth]{figures/DiffusionRes48.eps}}
389 lkettle 581 \caption{Results of the Temperature Diffusion Problem for Time Steps $1$ $16$, $32$ and $48$.}
390 jgs 107 \label{DIFFUSION FIG 2}
391     \end{figure}
392 ksteube 1316 \fig{DIFFUSION FIG 2} shows the result for some selected time steps.
393 jgs 102 An easy way to visualize the results is the command
394     \begin{verbatim}
395 gross 569 mayavi -d T.1.xml -m SurfaceMap &
396 jgs 102 \end{verbatim}
397 ksteube 1316 Use the \texttt{Configure Data} window in mayavi
398 gross 569 to move forward and and backwards in time.

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