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Some initial corrections to the user's guide.

1 ksteube 1811
2     %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
3 ksteube 1316 %
4 jfenwick 2881 % Copyright (c) 2003-2010 by University of Queensland
5 ksteube 1811 % Earth Systems Science Computational Center (ESSCC)
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
10     % http://www.opensource.org/licenses/osl-3.0.php
11 gross 625 %
12 ksteube 1811 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
13 jgs 102
14 ksteube 1811
15 caltinay 3274 \section{The First Steps}\label{FirstSteps}
16 ksteube 1316 In this chapter we give an introduction how to use \escript to solve
17 caltinay 3274 a partial differential equation\index{partial differential equation} (PDE\index{partial differential equation!PDE}).
18     We assume you are at least a little familiar with Python.
19     The knowledge presented in the Python tutorial at \url{http://docs.python.org/tut/tut.html} is more than sufficient.
20 jgs 102
21 caltinay 3274 The PDE\index{partial differential equation} we wish to solve is the Poisson equation\index{Poisson equation}
22 jgs 102 \begin{equation}
23 caltinay 3274 -\Delta u=f
24     \label{eq:FirstSteps.1}
25 jgs 102 \end{equation}
26 ksteube 1316 for the solution $u$. The function $f$ is the given right hand side. The domain of interest, denoted by $\Omega$,
27 jgs 102 is the unit square
28     \begin{equation}
29     \Omega=[0,1]^2=\{ (x\hackscore 0;x\hackscore 1) | 0\le x\hackscore{0} \le 1 \mbox{ and } 0\le x\hackscore{1} \le 1 \}
30     \label{eq:FirstSteps.1b}
31     \end{equation}
32 jgs 107 The domain is shown in \fig{fig:FirstSteps.1}.
33 caltinay 3274 \begin{figure}[ht]
34     \centerline{\includegraphics{figures/FirstStepDomain}}
35     \caption{Domain $\Omega=[0,1]^2$ with outer normal field $n$.}
36     \label{fig:FirstSteps.1}
37 artak 1971 \end{figure}
38 jgs 102
39 ksteube 1316 $\Delta$ denotes the Laplace operator\index{Laplace operator}, which is defined by
40 jgs 102 \begin{equation}
41     \Delta u = (u\hackscore {,0})\hackscore{,0}+(u\hackscore{,1})\hackscore{,1}
42     \label{eq:FirstSteps.1.1}
43     \end{equation}
44 ksteube 1316 where, for any function $u$ and any direction $i$, $u\hackscore{,i}$
45 caltinay 3274 denotes the partial derivative \index{partial derivative} of $u$ with respect
46     to $i$.\footnote{You may be more familiar with the Laplace
47     operator\index{Laplace operator} being written as $\nabla^2$, and written in
48     the form
49 jgs 102 \begin{equation*}
50 caltinay 3274 \nabla^2 u = \nabla^t \cdot \nabla u = \frac{\partial^2 u}{\partial x\hackscore 0^2}
51     + \frac{\partial^2 u}{\partial x\hackscore 1^2}
52 jgs 102 \end{equation*}
53     and \eqn{eq:FirstSteps.1} as
54     \begin{equation*}
55 caltinay 3274 -\nabla^2 u = f
56 jgs 102 \end{equation*}
57     }
58 jgs 107 Basically, in the subindex of a function, any index to the left of the comma denotes a spatial derivative with respect
59 artak 1971 to the index. To get a more compact form we will write $u\hackscore{,ij}=(u\hackscore {,i})\hackscore{,j}$
60 jgs 102 which leads to
61     \begin{equation}
62     \Delta u = u\hackscore{,00}+u\hackscore{,11}=\sum\hackscore{i=0}^2 u\hackscore{,ii}
63     \label{eq:FirstSteps.1.1b}
64     \end{equation}
65 ksteube 1316 We often find that use
66     of nested $\sum$ symbols makes formulas cumbersome, and we use the more
67 caltinay 3274 convenient Einstein summation convention\index{summation convention}. This
68 ksteube 1316 drops the $\sum$ sign and assumes that a summation is performed over any repeated index.
69     For instance we write
70 jgs 102 \begin{eqnarray}
71     x\hackscore{i}y\hackscore{i}=\sum\hackscore{i=0}^2 x\hackscore{i}y\hackscore{i} \\
72     x\hackscore{i}u\hackscore{,i}=\sum\hackscore{i=0}^2 x\hackscore{i}u\hackscore{,i} \\
73     u\hackscore{,ii}=\sum\hackscore{i=0}^2 u\hackscore{,ii} \\
74 jgs 107 x\hackscore{ij}u\hackscore{i,j}=\sum\hackscore{j=0}^2\sum\hackscore{i=0}^2 x\hackscore{ij}u\hackscore{i,j} \\
75 jgs 102 \label{eq:FirstSteps.1.1c}
76     \end{eqnarray}
77     With the summation convention we can write the Poisson equation \index{Poisson equation} as
78     \begin{equation}
79     - u\hackscore{,ii} =1
80     \label{eq:FirstSteps.1.sum}
81     \end{equation}
82 lkettle 575 where $f=1$ in this example.
83    
84 jgs 102 On the boundary of the domain $\Omega$ the normal derivative $n\hackscore{i} u\hackscore{,i}$
85 caltinay 3274 of the solution $u$ shall be zero, i.e. $u$ shall fulfill
86 jgs 102 the homogeneous Neumann boundary condition\index{Neumann
87     boundary condition!homogeneous}
88     \begin{equation}
89     n\hackscore{i} u\hackscore{,i}= 0 \;.
90     \label{eq:FirstSteps.2}
91     \end{equation}
92     $n=(n\hackscore{i})$ denotes the outer normal field
93     of the domain, see \fig{fig:FirstSteps.1}. Remember that we
94 caltinay 3274 are applying the Einstein summation convention \index{summation convention}, i.e. $n\hackscore{i} u\hackscore{,i}= n\hackscore{0} u\hackscore{,0} +%
95     n\hackscore{1} u\hackscore{,1}$.\footnote{Some readers may familiar with the
96     notation $\frac{\partial u}{\partial n} = n\hackscore{i} u\hackscore{,i}$
97 jgs 102 for the normal derivative.}
98     The Neumann boundary condition of \eqn{eq:FirstSteps.2} should be fulfilled on the
99     set $\Gamma^N$ which is the top and right edge of the domain:
100     \begin{equation}
101 caltinay 3274 \Gamma^N=\{(x\hackscore 0;x\hackscore 1) \in \Omega | x\hackscore{0}=1 \mbox{ or } x\hackscore{1}=1 \}
102     \label{eq:FirstSteps.2b}
103 jgs 102 \end{equation}
104 jgs 107 On the bottom and the left edge of the domain which is defined
105 jgs 102 as
106     \begin{equation}
107 caltinay 3274 \Gamma^D=\{(x\hackscore 0;x\hackscore 1) \in \Omega | x\hackscore{0}=0 \mbox{ or } x\hackscore{1}=0 \}
108     \label{eq:FirstSteps.2c}
109 jgs 102 \end{equation}
110 caltinay 3274 the solution shall be identical to zero:
111 jgs 102 \begin{equation}
112 caltinay 3274 u=0 \; .
113     \label{eq:FirstSteps.2d}
114 jgs 102 \end{equation}
115 caltinay 3274 This kind of boundary condition is called a homogeneous Dirichlet boundary
116     condition\index{Dirichlet boundary condition!homogeneous}.
117     The partial differential equation in \eqn{eq:FirstSteps.1.sum} together
118 jgs 102 with the Neumann boundary condition \eqn{eq:FirstSteps.2} and
119 caltinay 3274 Dirichlet boundary condition in \eqn{eq:FirstSteps.2d} form a so-called
120     boundary value
121     problem\index{boundary value problem} (BVP\index{boundary value problem!BVP})
122     for the unknown function~$u$.
123 jgs 102
124 gross 2371 \begin{figure}[ht]
125 caltinay 3274 \centerline{\includegraphics{figures/FirstStepMesh}}
126     \caption{Mesh of $4 \times 4$ elements on a rectangular domain. Here
127     each element is a quadrilateral and described by four nodes, namely
128     the corner points. The solution is interpolated by a bi-linear
129     polynomial.}
130     \label{fig:FirstSteps.2}
131 jgs 102 \end{figure}
132    
133 caltinay 3274 In general the BVP\index{boundary value problem!BVP} cannot be solved
134     analytically and numerical methods have to be used to construct an
135     approximation of the solution $u$.
136     Here we will use the finite element method\index{finite element method}
137     (FEM\index{finite element method!FEM}).
138     The basic idea is to fill the domain with a set of points called nodes.
139     The solution is approximated by its values on the nodes\index{finite element method!nodes}.
140     Moreover, the domain is subdivided into smaller sub-domains called
141     elements\index{finite element method!element}.
142     On each element the solution is represented by a polynomial of a certain
143     degree through its values at the nodes located in the element.
144     The nodes and their connection through elements is called a
145     mesh\index{finite element method!mesh}. \fig{fig:FirstSteps.2} shows an
146 jgs 102 example of a FEM mesh with four elements in the $x_0$ and four elements
147 caltinay 3274 in the $x_1$ direction over the unit square.
148 gross 2370 For more details we refer the reader to the literature, for instance \Ref{Zienc,NumHand}.
149 jgs 102
150 caltinay 3274 The \escript solver we want to use to solve this problem is embedded into the python interpreter language.
151     So you can solve the problem interactively but you will learn quickly that it
152     is more efficient to use scripts which you can edit with your favorite editor.
153     To enter the escript environment, use the \program{run-escript}
154     command\footnote{\program{run-escript} is not available under Windows yet.
155     If you run under Windows you can just use the \program{python} command and the
156     \env{OMP_NUM_THREADS} environment variable to control the number of threads.}:
157 gross 2370 \begin{verbatim}
158 caltinay 3274 run-escript
159 gross 2370 \end{verbatim}
160     which will pass you on to the python prompt
161     \begin{verbatim}
162     Python 2.5.2 (r252:60911, Oct 5 2008, 19:29:17)
163     [GCC 4.3.2] on linux2
164     Type "help", "copyright", "credits" or "license" for more information.
165     >>>
166     \end{verbatim}
167     Here you can use all available python commands and language features, for instance
168     \begin{python}
169 caltinay 3274 >>> x=2+3
170     >>> print "2+3=",x
171     2+3= 5
172 gross 2370 \end{python}
173 caltinay 3274 We refer to the python user's guide if you not familiar with python.
174 gross 2370
175 caltinay 3274 \escript provides the class \Poisson to define a Poisson equation\index{Poisson equation}.
176     (We will discuss a more general form of a PDE\index{partial differential equation!PDE}
177     that can be defined through the \LinearPDE class later.)
178     The instantiation of a \Poisson class object requires the specification of the domain $\Omega$.
179     In \escript the \Domain class objects are used to describe the geometry of a
180     domain but it also contains information about the discretization methods and
181     the actual solver which is used to solve the PDE.
182     Here we are using the FEM library \finley\index{finite element method}.
183     The following statements create the \Domain object \var{mydomain} from the
184     \finley method \method{Rectangle}:
185 jgs 102 \begin{python}
186 ksteube 1316 from esys.finley import Rectangle
187     mydomain = Rectangle(l0=1.,l1=1.,n0=40, n1=20)
188 jgs 102 \end{python}
189     In this case the domain is a rectangle with the lower, left corner at point $(0,0)$ and
190     the right, upper corner at $(\var{l0},\var{l1})=(1,1)$.
191 jgs 107 The arguments \var{n0} and \var{n1} define the number of elements in $x\hackscore{0}$ and
192 jgs 102 $x\hackscore{1}$-direction respectively. For more details on \method{Rectangle} and
193     other \Domain generators within the \finley module,
194     see \Chap{CHAPTER ON FINLEY}.
195    
196 jgs 107 The following statements define the \Poisson class object \var{mypde} with domain \var{mydomain} and
197 jgs 102 the right hand side $f$ of the PDE to constant $1$:
198     \begin{python}
199 ksteube 1316 from esys.escript.linearPDEs import Poisson
200     mypde = Poisson(mydomain)
201     mypde.setValue(f=1)
202 jgs 102 \end{python}
203     We have not specified any boundary condition but the
204     \Poisson class implicitly assumes homogeneous Neuman boundary conditions \index{Neumann
205     boundary condition!homogeneous} defined by \eqn{eq:FirstSteps.2}. With this boundary
206     condition the BVP\index{boundary value problem!BVP} we have defined has no unique solution. In fact, with any solution $u$
207     and any constant $C$ the function $u+C$ becomes a solution as well. We have to add
208     a Dirichlet boundary condition \index{Dirichlet boundary condition}. This is done
209 jgs 107 by defining a characteristic function \index{characteristic function}
210     which has positive values at locations $x=(x\hackscore{0},x\hackscore{1})$ where Dirichlet boundary condition is set
211 jgs 102 and $0$ elsewhere. In our case of $\Gamma^D$ defined by \eqn{eq:FirstSteps.2c},
212 gross 565 we need to construct a function \var{gammaD} which is positive for the cases $x\hackscore{0}=0$ or $x\hackscore{1}=0$. To get
213 ksteube 1316 an object \var{x} which contains the coordinates of the nodes in the domain use
214 jgs 102 \begin{python}
215 ksteube 1316 x=mydomain.getX()
216 jgs 102 \end{python}
217 gross 660 The method \method{getX} of the \Domain \var{mydomain}
218 jgs 107 gives access to locations
219 ksteube 1316 in the domain defined by \var{mydomain}. The object \var{x} is actually a \Data object which will be
220     discussed in \Chap{ESCRIPT CHAP} in more detail. What we need to know here is that
221 gross 660
222 ksteube 1316 \var{x} has \Rank (number of dimensions) and a \Shape (list of dimensions) which can be viewed by
223 gross 565 calling the \method{getRank} and \method{getShape} methods:
224     \begin{python}
225 ksteube 1316 print "rank ",x.getRank(),", shape ",x.getShape()
226 gross 565 \end{python}
227 ksteube 1316 This will print something like
228 gross 565 \begin{python}
229 ksteube 1316 rank 1, shape (2,)
230 gross 565 \end{python}
231     The \Data object also maintains type information which is represented by the
232     \FunctionSpace of the object. For instance
233     \begin{python}
234 ksteube 1316 print x.getFunctionSpace()
235 gross 565 \end{python}
236     will print
237     \begin{python}
238 ksteube 1316 Function space type: Finley_Nodes on FinleyMesh
239 gross 565 \end{python}
240 ksteube 1316 which tells us that the coordinates are stored on the nodes of (rather than on points in the interior of) a \finley mesh.
241 gross 565 To get the $x\hackscore{0}$ coordinates of the locations we use the
242     statement
243     \begin{python}
244 ksteube 1316 x0=x[0]
245 gross 565 \end{python}
246     Object \var{x0}
247     is again a \Data object now with \Rank $0$ and
248     \Shape $()$. It inherits the \FunctionSpace from \var{x}:
249     \begin{python}
250 ksteube 1316 print x0.getRank(),x0.getShape(),x0.getFunctionSpace()
251 gross 565 \end{python}
252     will print
253     \begin{python}
254 ksteube 1316 0 () Function space type: Finley_Nodes on FinleyMesh
255 gross 565 \end{python}
256 ksteube 1316 We can now construct a function \var{gammaD} which is only non-zero on the bottom and left edges
257     of the domain with
258 gross 565 \begin{python}
259 ksteube 1316 from esys.escript import whereZero
260     gammaD=whereZero(x[0])+whereZero(x[1])
261 gross 565 \end{python}
262 ksteube 1316
263     \code{whereZero(x[0])} creates function which equals $1$ where \code{x[0]} is (almost) equal to zero
264 jgs 107 and $0$ elsewhere.
265 gross 565 Similarly, \code{whereZero(x[1])} creates function which equals $1$ where \code{x[1]} is
266 jgs 107 equal to zero and $0$ elsewhere.
267 gross 565 The sum of the results of \code{whereZero(x[0])} and \code{whereZero(x[1])}
268 artak 1971 gives a function on the domain \var{mydomain} which is strictly positive where $x\hackscore{0}$ or $x\hackscore{1}$ is equal to zero.
269 gross 565 Note that \var{gammaD} has the same \Rank, \Shape and \FunctionSpace like \var{x0} used to define it. So from
270     \begin{python}
271 ksteube 1316 print gammaD.getRank(),gammaD.getShape(),gammaD.getFunctionSpace()
272 gross 565 \end{python}
273     one gets
274     \begin{python}
275 ksteube 1316 0 () Function space type: Finley_Nodes on FinleyMesh
276 gross 565 \end{python}
277 ksteube 1316 An additional parameter \var{q} of the \code{setValue} method of the \Poisson class defines the
278 jgs 102 characteristic function \index{characteristic function} of the locations
279     of the domain where homogeneous Dirichlet boundary condition \index{Dirichlet boundary condition!homogeneous}
280     are set. The complete definition of our example is now:
281     \begin{python}
282 ksteube 1316 from esys.linearPDEs import Poisson
283     x = mydomain.getX()
284     gammaD = whereZero(x[0])+whereZero(x[1])
285     mypde = Poisson(domain=mydomain)
286     mypde.setValue(f=1,q=gammaD)
287 jgs 102 \end{python}
288 lkettle 573 The first statement imports the \Poisson class definition from the \linearPDEs module \escript package.
289 jgs 107 To get the solution of the Poisson equation defined by \var{mypde} we just have to call its
290 jgs 102 \method{getSolution}.
291    
292 ksteube 1316 Now we can write the script to solve our Poisson problem
293 jgs 102 \begin{python}
294 ksteube 1316 from esys.escript import *
295     from esys.escript.linearPDEs import Poisson
296     from esys.finley import Rectangle
297     # generate domain:
298     mydomain = Rectangle(l0=1.,l1=1.,n0=40, n1=20)
299     # define characteristic function of Gamma^D
300     x = mydomain.getX()
301     gammaD = whereZero(x[0])+whereZero(x[1])
302     # define PDE and get its solution u
303     mypde = Poisson(domain=mydomain)
304     mypde.setValue(f=1,q=gammaD)
305     u = mypde.getSolution()
306 jgs 102 \end{python}
307 gross 2574 The question is what we do with the calculated solution \var{u}. Besides postprocessing, eg. calculating the gradient or the average value, which will be discussed later, plotting the solution is one one things you might want to do. \escript offers two ways to do this, both base on external modules or packages and so data need to converted
308     to hand over the solution. The first option is using the \MATPLOTLIB module which allows plotting 2D results relatively quickly, see~\cite{matplotlib}. However, there are limitations when using this tool, eg. in problem size and when solving 3D problems. Therefore \escript provides a second options based on \VTK files which is especially
309     designed for large scale and 3D problem and which can be read by a variety of software packages such as \mayavi \cite{mayavi}, \VisIt~\cite{VisIt}.
310 jgs 102
311 gross 2574 \subsection{Plotting Using \MATPLOTLIB}
312 gross 2580 The \MATPLOTLIB module provides a simple and easy to use way to visualize PDE solutions (or other \Data objects).
313     To hand over data from \escript to \MATPLOTLIB the values need to mapped onto a rectangular grid
314     \footnote{Users of Debian 5(Lenny) please note: this example makes use of the \function{griddata} method in \module{matplotlib.mlab}.
315 jfenwick 2575 This method is not part of version 0.98.1 which is available with Lenny.
316 jfenwick 2583 If you wish to use contour plots, you may need to install a later version.
317     Users of Ubuntu 8.10 or later should be fine.}. We will make use
318 gross 2574 of the \numpy module.
319    
320     First we need to create a rectangular grid. We use the following statements:
321     \begin{python}
322     import numpy
323     x_grid = numpy.linspace(0.,1.,50)
324     y_grid = numpy.linspace(0.,1.,50)
325     \end{python}
326     \var{x_grid} is an array defining the x coordinates of the grids while
327     \var{y_grid} defines the y coordinates of the grid. In this case we use $50$ points over the interval $[0,1]$
328     in both directions.
329    
330 jfenwick 2575 Now the values created by \escript need to be interpolated to this grid. We will use the \MATPLOTLIB
331 gross 2574 \function{mlab.griddata} function to do this. We can easily extract spatial coordinates as a \var{list} by
332     \begin{python}
333     x=mydomain.getX()[0].toListOfTuples()
334     y=mydomain.getX()[1].toListOfTuples()
335     \end{python}
336     In principle we can apply the same \member{toListOfTuples} method to extract the values from the
337     PDE solution \var{u}. However, we have to make sure that the \Data object we extract the values from
338 gross 2580 uses the same \FunctionSpace as we have us when extracting \var{x} and \var{y}. We apply the
339     \function{interpolation} to \var{u} before extraction to achieve this:
340 gross 2574 \begin{python}
341     z=interpolate(u,mydomain.getX().getFunctionSpace())
342     \end{python}
343     The values in \var{z} are now the values at the points with the coordinates given by \var{x} and \var{y}. These
344     values are now interpolated to the grid defined by \var{x_grid} and \var{y_grid} by using
345     \begin{python}
346     import matplotlib
347     z_grid = matplotlib.mlab.griddata(x,y,z,xi=x_grid,yi=y_grid )
348     \end{python}
349     \var{z_grid} gives now the values of the PDE solution \var{u} at the grid. The values can be plotted now
350     using the \function{contourf}:
351     \begin{python}
352     matplotlib.pyplot.contourf(x_grid, y_grid, z_grid, 5)
353     matplotlib.pyplot.savefig("u.png")
354     \end{python}
355     Here we use $5$ contours. The last statement writes the plot to the file \file{u.png} in the PNG format. Alternatively, one can use
356     \begin{python}
357     matplotlib.pyplot.contourf(x_grid, y_grid, z_grid, 5)
358     matplotlib.pyplot.show()
359     \end{python}
360     which gives an interactive browser window.
361    
362     \begin{figure}
363     \centerline{\includegraphics[width=\figwidth]{figures/FirstStepResultMATPLOTLIB}}
364     \caption{Visualization of the Poisson Equation Solution for $f=1$ using \MATPLOTLIB.}
365     \label{fig:FirstSteps.3b}
366     \end{figure}
367    
368     Now we can write the script to solve our Poisson problem
369     \begin{python}
370     from esys.escript import *
371     from esys.escript.linearPDEs import Poisson
372     from esys.finley import Rectangle
373     import numpy
374     import matplotlib
375     import pylab
376     # generate domain:
377     mydomain = Rectangle(l0=1.,l1=1.,n0=40, n1=20)
378     # define characteristic function of Gamma^D
379     x = mydomain.getX()
380     gammaD = whereZero(x[0])+whereZero(x[1])
381     # define PDE and get its solution u
382     mypde = Poisson(domain=mydomain)
383     mypde.setValue(f=1,q=gammaD)
384     u = mypde.getSolution()
385     # interpolate u to a matplotlib grid:
386     x_grid = numpy.linspace(0.,1.,50)
387     y_grid = numpy.linspace(0.,1.,50)
388     x=mydomain.getX()[0].toListOfTuples()
389     y=mydomain.getX()[1].toListOfTuples()
390     z=interpolate(u,mydomain.getX().getFunctionSpace())
391     z_grid = matplotlib.mlab.griddata(x,y,z,xi=x_grid,yi=y_grid )
392     # interpolate u to a rectangular grid:
393     matplotlib.pyplot.contourf(x_grid, y_grid, z_grid, 5)
394     matplotlib.pyplot.savefig("u.png")
395     \end{python}
396     The entire code is available as \file{poisson\hackscore matplotlib.py} in the \ExampleDirectory.
397     You can run the script using the {\it escript} environment
398 ksteube 1316 \begin{verbatim}
399 jfenwick 2923 run-escript poisson_matplotlib.py
400 ksteube 1316 \end{verbatim}
401 gross 2574 This will create the \file{u.png}, see Figure~\fig{fig:FirstSteps.3b}.
402     For details on the usage of the \MATPLOTLIB module we refer to the documentation~\cite{matplotlib}.
403 ksteube 1316
404 gross 2574 As pointed out, \MATPLOTLIB is restricted to the two-dimensional case and
405     should be used for small problems only. It can not be used under \MPI as the \member{toListOfTuples} method is
406 jfenwick 2575 not safe under \MPI\footnote{The phrase 'safe under \MPI' means that a program will produce correct results when run on more than one processor under \MPI.}.
407 gross 2574
408 gross 2580 \begin{figure}
409     \centerline{\includegraphics[width=\figwidth]{figures/FirstStepResult}}
410     \caption{Visualization of the Poisson Equation Solution for $f=1$}
411     \label{fig:FirstSteps.3}
412     \end{figure}
413    
414 gross 2574 \subsection{Visualization using \VTK}
415    
416     As an alternative {\it escript} supports the usage of visualization tools which base on \VTK, eg. mayavi \cite{mayavi}, \VisIt~\cite{VisIt}. In this case the solution is written to a file in the \VTK format. This file the can read by the tool of choice. Using \VTK file is \MPI safe.
417    
418     To write the solution \var{u} in Poisson problem to the file \file{u.xml} one need to add the line
419     \begin{python}
420     saveVTK("u.xml",sol=u)
421     \end{python}
422     The solution \var{u} is now available in the \file{u.xml} tagged with the name "sol".
423    
424     The Poisson problem script is now
425     \begin{python}
426     from esys.escript import *
427     from esys.escript.linearPDEs import Poisson
428     from esys.finley import Rectangle
429     # generate domain:
430     mydomain = Rectangle(l0=1.,l1=1.,n0=40, n1=20)
431     # define characteristic function of Gamma^D
432     x = mydomain.getX()
433     gammaD = whereZero(x[0])+whereZero(x[1])
434     # define PDE and get its solution u
435     mypde = Poisson(domain=mydomain)
436     mypde.setValue(f=1,q=gammaD)
437     u = mypde.getSolution()
438     # write u to an external file
439     saveVTK("u.xml",sol=u)
440     \end{python}
441 gross 2683 The entire code is available as \file{poisson\hackscore VTK.py} in the \ExampleDirectory.
442 gross 2574
443     You can run the script using the {\it escript} environment
444     and visualize the solution using \mayavi:
445     \begin{verbatim}
446 jfenwick 2955 run-escript poisson_VTK.py
447 gross 2574 mayavi2 -d u.xml -m SurfaceMap
448     \end{verbatim}
449     The result is shown in Figure~\fig{fig:FirstSteps.3}.

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