ViewVC logotype

Contents of /branches/stage3.0/doc/user/firststep.tex

Parent Directory Parent Directory | Revision Log Revision Log

Revision 2584 - (show annotations)
Wed Aug 5 02:44:51 2009 UTC (9 years, 6 months ago) by jfenwick
File MIME type: application/x-tex
File size: 21089 byte(s)
Merging Lutz' doco changes to release
2 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
3 %
4 % Copyright (c) 2003-2009 by University of Queensland
5 % Earth Systems Science Computational Center (ESSCC)
6 % http://www.uq.edu.au/esscc
7 %
8 % 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 %
12 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
15 \section{The First Steps}
16 \label{FirstSteps}
20 In this chapter we give an introduction how to use \escript to solve
21 a partial differential equation \index{partial differential equation} (PDE \index{partial differential equation!PDE}). We assume you are at least a little familiar with Python. The knowledge presented at the Python tutorial at \url{http://docs.python.org/tut/tut.html}
22 is more than sufficient.
24 The PDE \index{partial differential equation} we wish to solve is the Poisson equation \index{Poisson equation}
25 \begin{equation}
26 -\Delta u =f
27 \label{eq:FirstSteps.1}
28 \end{equation}
29 for the solution $u$. The function $f$ is the given right hand side. The domain of interest, denoted by $\Omega$,
30 is the unit square
31 \begin{equation}
32 \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 \}
33 \label{eq:FirstSteps.1b}
34 \end{equation}
35 The domain is shown in \fig{fig:FirstSteps.1}.
36 \begin{figure} [ht]
37 \centerline{\includegraphics[width=\figwidth]{figures/FirstStepDomain}}
38 \caption{Domain $\Omega=[0,1]^2$ with outer normal field $n$.}
39 \label{fig:FirstSteps.1}
40 \end{figure}
42 $\Delta$ denotes the Laplace operator\index{Laplace operator}, which is defined by
43 \begin{equation}
44 \Delta u = (u\hackscore {,0})\hackscore{,0}+(u\hackscore{,1})\hackscore{,1}
45 \label{eq:FirstSteps.1.1}
46 \end{equation}
47 where, for any function $u$ and any direction $i$, $u\hackscore{,i}$
48 denotes the partial derivative \index{partial derivative} of $u$ with respect to $i$.
49 \footnote{You
50 may be more familiar with the Laplace operator\index{Laplace operator} being written
51 as $\nabla^2$, and written in the form
52 \begin{equation*}
53 \nabla^2 u = \nabla^t \cdot \nabla u = \frac{\partial^2 u}{\partial x\hackscore 0^2}
54 + \frac{\partial^2 u}{\partial x\hackscore 1^2}
55 \end{equation*}
56 and \eqn{eq:FirstSteps.1} as
57 \begin{equation*}
58 -\nabla^2 u = f
59 \end{equation*}
60 }
61 Basically, in the subindex of a function, any index to the left of the comma denotes a spatial derivative with respect
62 to the index. To get a more compact form we will write $u\hackscore{,ij}=(u\hackscore {,i})\hackscore{,j}$
63 which leads to
64 \begin{equation}
65 \Delta u = u\hackscore{,00}+u\hackscore{,11}=\sum\hackscore{i=0}^2 u\hackscore{,ii}
66 \label{eq:FirstSteps.1.1b}
67 \end{equation}
68 We often find that use
69 of nested $\sum$ symbols makes formulas cumbersome, and we use the more
70 convenient Einstein summation convention \index{summation convention}. This
71 drops the $\sum$ sign and assumes that a summation is performed over any repeated index.
72 For instance we write
73 \begin{eqnarray}
74 x\hackscore{i}y\hackscore{i}=\sum\hackscore{i=0}^2 x\hackscore{i}y\hackscore{i} \\
75 x\hackscore{i}u\hackscore{,i}=\sum\hackscore{i=0}^2 x\hackscore{i}u\hackscore{,i} \\
76 u\hackscore{,ii}=\sum\hackscore{i=0}^2 u\hackscore{,ii} \\
77 x\hackscore{ij}u\hackscore{i,j}=\sum\hackscore{j=0}^2\sum\hackscore{i=0}^2 x\hackscore{ij}u\hackscore{i,j} \\
78 \label{eq:FirstSteps.1.1c}
79 \end{eqnarray}
80 With the summation convention we can write the Poisson equation \index{Poisson equation} as
81 \begin{equation}
82 - u\hackscore{,ii} =1
83 \label{eq:FirstSteps.1.sum}
84 \end{equation}
85 where $f=1$ in this example.
87 On the boundary of the domain $\Omega$ the normal derivative $n\hackscore{i} u\hackscore{,i}$
88 of the solution $u$ shall be zero, ie. $u$ shall fulfill
89 the homogeneous Neumann boundary condition\index{Neumann
90 boundary condition!homogeneous}
91 \begin{equation}
92 n\hackscore{i} u\hackscore{,i}= 0 \;.
93 \label{eq:FirstSteps.2}
94 \end{equation}
95 $n=(n\hackscore{i})$ denotes the outer normal field
96 of the domain, see \fig{fig:FirstSteps.1}. Remember that we
97 are applying the Einstein summation convention \index{summation convention}, i.e
98 $n\hackscore{i} u\hackscore{,i}= n\hackscore{0} u\hackscore{,0} +
99 n\hackscore{1} u\hackscore{,1}$.
100 \footnote{Some readers may familiar with the notation
101 $
102 \frac{\partial u}{\partial n} = n\hackscore{i} u\hackscore{,i}
103 $
104 for the normal derivative.}
105 The Neumann boundary condition of \eqn{eq:FirstSteps.2} should be fulfilled on the
106 set $\Gamma^N$ which is the top and right edge of the domain:
107 \begin{equation}
108 \Gamma^N=\{(x\hackscore 0;x\hackscore 1) \in \Omega | x\hackscore{0}=1 \mbox{ or } x\hackscore{1}=1 \}
109 \label{eq:FirstSteps.2b}
110 \end{equation}
111 On the bottom and the left edge of the domain which is defined
112 as
113 \begin{equation}
114 \Gamma^D=\{(x\hackscore 0;x\hackscore 1) \in \Omega | x\hackscore{0}=0 \mbox{ or } x\hackscore{1}=0 \}
115 \label{eq:FirstSteps.2c}
116 \end{equation}
117 the solution shall be identically zero:
118 \begin{equation}
119 u=0 \; .
120 \label{eq:FirstSteps.2d}
121 \end{equation}
122 This kind of boundary condition is called a homogeneous Dirichlet boundary condition
123 \index{Dirichlet boundary condition!homogeneous}. The partial differential equation in \eqn{eq:FirstSteps.1.sum} together
124 with the Neumann boundary condition \eqn{eq:FirstSteps.2} and
125 Dirichlet boundary condition in \eqn{eq:FirstSteps.2d} form a so
126 called boundary value
127 problem\index{boundary value problem} (BVP\index{boundary value problem!BVP}) for
128 the unknown function~$u$.
131 \begin{figure}[ht]
132 \centerline{\includegraphics[width=\figwidth]{figures/FirstStepMesh}}
133 \caption{Mesh of $4 \time 4$ elements on a rectangular domain. Here
134 each element is a quadrilateral and described by four nodes, namely
135 the corner points. The solution is interpolated by a bi-linear
136 polynomial.}
137 \label{fig:FirstSteps.2}
138 \end{figure}
140 In general the BVP\index{boundary value problem!BVP} cannot be solved analytically and numerical
141 methods have to be used construct an approximation of the solution
142 $u$. Here we will use the finite element method\index{finite element
143 method} (FEM\index{finite element
144 method!FEM}). The basic idea is to fill the domain with a
145 set of points called nodes. The solution is approximated by its
146 values on the nodes\index{finite element
147 method!nodes}. Moreover, the domain is subdivided into smaller
148 sub-domains called elements \index{finite element
149 method!element}. On each element the solution is
150 represented by a polynomial of a certain degree through its values at
151 the nodes located in the element. The nodes and its connection through
152 elements is called a mesh\index{finite element
153 method!mesh}. \fig{fig:FirstSteps.2} shows an
154 example of a FEM mesh with four elements in the $x_0$ and four elements
155 in the $x_1$ direction over the unit square.
156 For more details we refer the reader to the literature, for instance \Ref{Zienc,NumHand}.
158 The \escript solver we want to use to solve this problem is embedded into the python interpreter language. So you can solve the problem interactively but you will learn quickly
159 that is more efficient to use scripts which you can edit with your favorite editor.
160 To enter the escript environment you use \program{escript} command\footnote{\program{escript} is not available under Windows yet. If you run under windows you can just use the
161 \program{python} command and the \env{OMP_NUM_THREADS} environment variable to control the number
162 of threads.}:
163 \begin{verbatim}
164 escript
165 \end{verbatim}
166 which will pass you on to the python prompt
167 \begin{verbatim}
168 Python 2.5.2 (r252:60911, Oct 5 2008, 19:29:17)
169 [GCC 4.3.2] on linux2
170 Type "help", "copyright", "credits" or "license" for more information.
171 >>>
172 \end{verbatim}
173 Here you can use all available python commands and language features, for instance
174 \begin{python}
175 >>> x=2+3
176 >>> print "2+3=",x
177 2+3= 5
178 \end{python}
179 We refer to the python users guide if you not familiar with python.
181 \escript provides the class \Poisson to define a Poisson equation \index{Poisson equation}.
182 (We will discuss a more general form of a PDE \index{partial differential equation!PDE}
183 that can be defined through the \LinearPDE class later). The instantiation of
184 a \Poisson class object requires the specification of the domain $\Omega$. In \escript
185 the \Domain class objects are used to describe the geometry of a domain but it also
186 contains information about the discretization methods and the actual solver which is used
187 to solve the PDE. Here we are using the FEM library \finley \index{finite element
188 method}. The following statements create the \Domain object \var{mydomain} from the
189 \finley method \method{Rectangle}
190 \begin{python}
191 from esys.finley import Rectangle
192 mydomain = Rectangle(l0=1.,l1=1.,n0=40, n1=20)
193 \end{python}
194 In this case the domain is a rectangle with the lower, left corner at point $(0,0)$ and
195 the right, upper corner at $(\var{l0},\var{l1})=(1,1)$.
196 The arguments \var{n0} and \var{n1} define the number of elements in $x\hackscore{0}$ and
197 $x\hackscore{1}$-direction respectively. For more details on \method{Rectangle} and
198 other \Domain generators within the \finley module,
199 see \Chap{CHAPTER ON FINLEY}.
201 The following statements define the \Poisson class object \var{mypde} with domain \var{mydomain} and
202 the right hand side $f$ of the PDE to constant $1$:
203 \begin{python}
204 from esys.escript.linearPDEs import Poisson
205 mypde = Poisson(mydomain)
206 mypde.setValue(f=1)
207 \end{python}
208 We have not specified any boundary condition but the
209 \Poisson class implicitly assumes homogeneous Neuman boundary conditions \index{Neumann
210 boundary condition!homogeneous} defined by \eqn{eq:FirstSteps.2}. With this boundary
211 condition the BVP\index{boundary value problem!BVP} we have defined has no unique solution. In fact, with any solution $u$
212 and any constant $C$ the function $u+C$ becomes a solution as well. We have to add
213 a Dirichlet boundary condition \index{Dirichlet boundary condition}. This is done
214 by defining a characteristic function \index{characteristic function}
215 which has positive values at locations $x=(x\hackscore{0},x\hackscore{1})$ where Dirichlet boundary condition is set
216 and $0$ elsewhere. In our case of $\Gamma^D$ defined by \eqn{eq:FirstSteps.2c},
217 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
218 an object \var{x} which contains the coordinates of the nodes in the domain use
219 \begin{python}
220 x=mydomain.getX()
221 \end{python}
222 The method \method{getX} of the \Domain \var{mydomain}
223 gives access to locations
224 in the domain defined by \var{mydomain}. The object \var{x} is actually a \Data object which will be
225 discussed in \Chap{ESCRIPT CHAP} in more detail. What we need to know here is that
227 \var{x} has \Rank (number of dimensions) and a \Shape (list of dimensions) which can be viewed by
228 calling the \method{getRank} and \method{getShape} methods:
229 \begin{python}
230 print "rank ",x.getRank(),", shape ",x.getShape()
231 \end{python}
232 This will print something like
233 \begin{python}
234 rank 1, shape (2,)
235 \end{python}
236 The \Data object also maintains type information which is represented by the
237 \FunctionSpace of the object. For instance
238 \begin{python}
239 print x.getFunctionSpace()
240 \end{python}
241 will print
242 \begin{python}
243 Function space type: Finley_Nodes on FinleyMesh
244 \end{python}
245 which tells us that the coordinates are stored on the nodes of (rather than on points in the interior of) a \finley mesh.
246 To get the $x\hackscore{0}$ coordinates of the locations we use the
247 statement
248 \begin{python}
249 x0=x[0]
250 \end{python}
251 Object \var{x0}
252 is again a \Data object now with \Rank $0$ and
253 \Shape $()$. It inherits the \FunctionSpace from \var{x}:
254 \begin{python}
255 print x0.getRank(),x0.getShape(),x0.getFunctionSpace()
256 \end{python}
257 will print
258 \begin{python}
259 0 () Function space type: Finley_Nodes on FinleyMesh
260 \end{python}
261 We can now construct a function \var{gammaD} which is only non-zero on the bottom and left edges
262 of the domain with
263 \begin{python}
264 from esys.escript import whereZero
265 gammaD=whereZero(x[0])+whereZero(x[1])
266 \end{python}
268 \code{whereZero(x[0])} creates function which equals $1$ where \code{x[0]} is (almost) equal to zero
269 and $0$ elsewhere.
270 Similarly, \code{whereZero(x[1])} creates function which equals $1$ where \code{x[1]} is
271 equal to zero and $0$ elsewhere.
272 The sum of the results of \code{whereZero(x[0])} and \code{whereZero(x[1])}
273 gives a function on the domain \var{mydomain} which is strictly positive where $x\hackscore{0}$ or $x\hackscore{1}$ is equal to zero.
274 Note that \var{gammaD} has the same \Rank, \Shape and \FunctionSpace like \var{x0} used to define it. So from
275 \begin{python}
276 print gammaD.getRank(),gammaD.getShape(),gammaD.getFunctionSpace()
277 \end{python}
278 one gets
279 \begin{python}
280 0 () Function space type: Finley_Nodes on FinleyMesh
281 \end{python}
282 An additional parameter \var{q} of the \code{setValue} method of the \Poisson class defines the
283 characteristic function \index{characteristic function} of the locations
284 of the domain where homogeneous Dirichlet boundary condition \index{Dirichlet boundary condition!homogeneous}
285 are set. The complete definition of our example is now:
286 \begin{python}
287 from esys.linearPDEs import Poisson
288 x = mydomain.getX()
289 gammaD = whereZero(x[0])+whereZero(x[1])
290 mypde = Poisson(domain=mydomain)
291 mypde.setValue(f=1,q=gammaD)
292 \end{python}
293 The first statement imports the \Poisson class definition from the \linearPDEs module \escript package.
294 To get the solution of the Poisson equation defined by \var{mypde} we just have to call its
295 \method{getSolution}.
297 Now we can write the script to solve our Poisson problem
298 \begin{python}
299 from esys.escript import *
300 from esys.escript.linearPDEs import Poisson
301 from esys.finley import Rectangle
302 # generate domain:
303 mydomain = Rectangle(l0=1.,l1=1.,n0=40, n1=20)
304 # define characteristic function of Gamma^D
305 x = mydomain.getX()
306 gammaD = whereZero(x[0])+whereZero(x[1])
307 # define PDE and get its solution u
308 mypde = Poisson(domain=mydomain)
309 mypde.setValue(f=1,q=gammaD)
310 u = mypde.getSolution()
311 \end{python}
312 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
313 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
314 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}.
316 \subsection{Plotting Using \MATPLOTLIB}
317 The \MATPLOTLIB module provides a simple and easy to use way to visualize PDE solutions (or other \Data objects).
318 To hand over data from \escript to \MATPLOTLIB the values need to mapped onto a rectangular grid
319 \footnote{Users of Debian 5(Lenny) please note: this example makes use of the \function{griddata} method in \module{matplotlib.mlab}.
320 This method is not part of version 0.98.1 which is available with Lenny.
321 If you wish to use contour plots, you may need to install a later version.
322 Users of Ubuntu 8.10 or later should be fine.}. We will make use
323 of the \numpy module.
325 First we need to create a rectangular grid. We use the following statements:
326 \begin{python}
327 import numpy
328 x_grid = numpy.linspace(0.,1.,50)
329 y_grid = numpy.linspace(0.,1.,50)
330 \end{python}
331 \var{x_grid} is an array defining the x coordinates of the grids while
332 \var{y_grid} defines the y coordinates of the grid. In this case we use $50$ points over the interval $[0,1]$
333 in both directions.
335 Now the values created by \escript need to be interpolated to this grid. We will use the \MATPLOTLIB
336 \function{mlab.griddata} function to do this. We can easily extract spatial coordinates as a \var{list} by
337 \begin{python}
338 x=mydomain.getX()[0].toListOfTuples()
339 y=mydomain.getX()[1].toListOfTuples()
340 \end{python}
341 In principle we can apply the same \member{toListOfTuples} method to extract the values from the
342 PDE solution \var{u}. However, we have to make sure that the \Data object we extract the values from
343 uses the same \FunctionSpace as we have us when extracting \var{x} and \var{y}. We apply the
344 \function{interpolation} to \var{u} before extraction to achieve this:
345 \begin{python}
346 z=interpolate(u,mydomain.getX().getFunctionSpace())
347 \end{python}
348 The values in \var{z} are now the values at the points with the coordinates given by \var{x} and \var{y}. These
349 values are now interpolated to the grid defined by \var{x_grid} and \var{y_grid} by using
350 \begin{python}
351 import matplotlib
352 z_grid = matplotlib.mlab.griddata(x,y,z,xi=x_grid,yi=y_grid )
353 \end{python}
354 \var{z_grid} gives now the values of the PDE solution \var{u} at the grid. The values can be plotted now
355 using the \function{contourf}:
356 \begin{python}
357 matplotlib.pyplot.contourf(x_grid, y_grid, z_grid, 5)
358 matplotlib.pyplot.savefig("u.png")
359 \end{python}
360 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
361 \begin{python}
362 matplotlib.pyplot.contourf(x_grid, y_grid, z_grid, 5)
363 matplotlib.pyplot.show()
364 \end{python}
365 which gives an interactive browser window.
367 \begin{figure}
368 \centerline{\includegraphics[width=\figwidth]{figures/FirstStepResultMATPLOTLIB}}
369 \caption{Visualization of the Poisson Equation Solution for $f=1$ using \MATPLOTLIB.}
370 \label{fig:FirstSteps.3b}
371 \end{figure}
373 Now we can write the script to solve our Poisson problem
374 \begin{python}
375 from esys.escript import *
376 from esys.escript.linearPDEs import Poisson
377 from esys.finley import Rectangle
378 import numpy
379 import matplotlib
380 import pylab
381 # generate domain:
382 mydomain = Rectangle(l0=1.,l1=1.,n0=40, n1=20)
383 # define characteristic function of Gamma^D
384 x = mydomain.getX()
385 gammaD = whereZero(x[0])+whereZero(x[1])
386 # define PDE and get its solution u
387 mypde = Poisson(domain=mydomain)
388 mypde.setValue(f=1,q=gammaD)
389 u = mypde.getSolution()
390 # interpolate u to a matplotlib grid:
391 x_grid = numpy.linspace(0.,1.,50)
392 y_grid = numpy.linspace(0.,1.,50)
393 x=mydomain.getX()[0].toListOfTuples()
394 y=mydomain.getX()[1].toListOfTuples()
395 z=interpolate(u,mydomain.getX().getFunctionSpace())
396 z_grid = matplotlib.mlab.griddata(x,y,z,xi=x_grid,yi=y_grid )
397 # interpolate u to a rectangular grid:
398 matplotlib.pyplot.contourf(x_grid, y_grid, z_grid, 5)
399 matplotlib.pyplot.savefig("u.png")
400 \end{python}
401 The entire code is available as \file{poisson\hackscore matplotlib.py} in the \ExampleDirectory.
402 You can run the script using the {\it escript} environment
403 \begin{verbatim}
404 escript poisson_matplotlib.py
405 \end{verbatim}
406 This will create the \file{u.png}, see Figure~\fig{fig:FirstSteps.3b}.
407 For details on the usage of the \MATPLOTLIB module we refer to the documentation~\cite{matplotlib}.
409 As pointed out, \MATPLOTLIB is restricted to the two-dimensional case and
410 should be used for small problems only. It can not be used under \MPI as the \member{toListOfTuples} method is
411 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.}.
413 \begin{figure}
414 \centerline{\includegraphics[width=\figwidth]{figures/FirstStepResult}}
415 \caption{Visualization of the Poisson Equation Solution for $f=1$}
416 \label{fig:FirstSteps.3}
417 \end{figure}
419 \subsection{Visualization using \VTK}
421 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.
423 To write the solution \var{u} in Poisson problem to the file \file{u.xml} one need to add the line
424 \begin{python}
425 saveVTK("u.xml",sol=u)
426 \end{python}
427 The solution \var{u} is now available in the \file{u.xml} tagged with the name "sol".
429 The Poisson problem script is now
430 \begin{python}
431 from esys.escript import *
432 from esys.escript.linearPDEs import Poisson
433 from esys.finley import Rectangle
434 # generate domain:
435 mydomain = Rectangle(l0=1.,l1=1.,n0=40, n1=20)
436 # define characteristic function of Gamma^D
437 x = mydomain.getX()
438 gammaD = whereZero(x[0])+whereZero(x[1])
439 # define PDE and get its solution u
440 mypde = Poisson(domain=mydomain)
441 mypde.setValue(f=1,q=gammaD)
442 u = mypde.getSolution()
443 # write u to an external file
444 saveVTK("u.xml",sol=u)
445 \end{python}
446 The entire code is available as \file{poisson\hackscore VTK.py} in the \ExampleDirectory
448 You can run the script using the {\it escript} environment
449 and visualize the solution using \mayavi:
450 \begin{verbatim}
451 escript poisson\hackscore VTK.py
452 mayavi2 -d u.xml -m SurfaceMap
453 \end{verbatim}
454 The result is shown in Figure~\fig{fig:FirstSteps.3}.


Name Value
svn:eol-style native
svn:keywords Author Date Id Revision

  ViewVC Help
Powered by ViewVC 1.1.26