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Some initial corrections to the user's guide.

1
2 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
3 %
4 % Copyright (c) 2003-2010 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 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
13
14
15 \section{The First Steps}\label{FirstSteps}
16 In this chapter we give an introduction how to use \escript to solve
17 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
21 The PDE\index{partial differential equation} we wish to solve is the Poisson equation\index{Poisson equation}
22 \begin{equation}
23 -\Delta u=f
24 \label{eq:FirstSteps.1}
25 \end{equation}
26 for the solution $u$. The function $f$ is the given right hand side. The domain of interest, denoted by $\Omega$,
27 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 The domain is shown in \fig{fig:FirstSteps.1}.
33 \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 \end{figure}
38
39 $\Delta$ denotes the Laplace operator\index{Laplace operator}, which is defined by
40 \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 where, for any function $u$ and any direction $i$, $u\hackscore{,i}$
45 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 \begin{equation*}
50 \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 \end{equation*}
53 and \eqn{eq:FirstSteps.1} as
54 \begin{equation*}
55 -\nabla^2 u = f
56 \end{equation*}
57 }
58 Basically, in the subindex of a function, any index to the left of the comma denotes a spatial derivative with respect
59 to the index. To get a more compact form we will write $u\hackscore{,ij}=(u\hackscore {,i})\hackscore{,j}$
60 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 We often find that use
66 of nested $\sum$ symbols makes formulas cumbersome, and we use the more
67 convenient Einstein summation convention\index{summation convention}. This
68 drops the $\sum$ sign and assumes that a summation is performed over any repeated index.
69 For instance we write
70 \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 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 \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 where $f=1$ in this example.
83
84 On the boundary of the domain $\Omega$ the normal derivative $n\hackscore{i} u\hackscore{,i}$
85 of the solution $u$ shall be zero, i.e. $u$ shall fulfill
86 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 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 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 \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 \end{equation}
104 On the bottom and the left edge of the domain which is defined
105 as
106 \begin{equation}
107 \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 \end{equation}
110 the solution shall be identical to zero:
111 \begin{equation}
112 u=0 \; .
113 \label{eq:FirstSteps.2d}
114 \end{equation}
115 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 with the Neumann boundary condition \eqn{eq:FirstSteps.2} and
119 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
124 \begin{figure}[ht]
125 \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 \end{figure}
132
133 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 example of a FEM mesh with four elements in the $x_0$ and four elements
147 in the $x_1$ direction over the unit square.
148 For more details we refer the reader to the literature, for instance \Ref{Zienc,NumHand}.
149
150 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 \begin{verbatim}
158 run-escript
159 \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 >>> x=2+3
170 >>> print "2+3=",x
171 2+3= 5
172 \end{python}
173 We refer to the python user's guide if you not familiar with python.
174
175 \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 \begin{python}
186 from esys.finley import Rectangle
187 mydomain = Rectangle(l0=1.,l1=1.,n0=40, n1=20)
188 \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 The arguments \var{n0} and \var{n1} define the number of elements in $x\hackscore{0}$ and
192 $x\hackscore{1}$-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 The following statements define the \Poisson class object \var{mypde} with domain \var{mydomain} and
197 the right hand side $f$ of the PDE to constant $1$:
198 \begin{python}
199 from esys.escript.linearPDEs import Poisson
200 mypde = Poisson(mydomain)
201 mypde.setValue(f=1)
202 \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 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 and $0$ elsewhere. In our case of $\Gamma^D$ defined by \eqn{eq:FirstSteps.2c},
212 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 an object \var{x} which contains the coordinates of the nodes in the domain use
214 \begin{python}
215 x=mydomain.getX()
216 \end{python}
217 The method \method{getX} of the \Domain \var{mydomain}
218 gives access to locations
219 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
222 \var{x} has \Rank (number of dimensions) and a \Shape (list of dimensions) which can be viewed by
223 calling the \method{getRank} and \method{getShape} methods:
224 \begin{python}
225 print "rank ",x.getRank(),", shape ",x.getShape()
226 \end{python}
227 This will print something like
228 \begin{python}
229 rank 1, shape (2,)
230 \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 print x.getFunctionSpace()
235 \end{python}
236 will print
237 \begin{python}
238 Function space type: Finley_Nodes on FinleyMesh
239 \end{python}
240 which tells us that the coordinates are stored on the nodes of (rather than on points in the interior of) a \finley mesh.
241 To get the $x\hackscore{0}$ coordinates of the locations we use the
242 statement
243 \begin{python}
244 x0=x[0]
245 \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 print x0.getRank(),x0.getShape(),x0.getFunctionSpace()
251 \end{python}
252 will print
253 \begin{python}
254 0 () Function space type: Finley_Nodes on FinleyMesh
255 \end{python}
256 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 \begin{python}
259 from esys.escript import whereZero
260 gammaD=whereZero(x[0])+whereZero(x[1])
261 \end{python}
262
263 \code{whereZero(x[0])} creates function which equals $1$ where \code{x[0]} is (almost) equal to zero
264 and $0$ elsewhere.
265 Similarly, \code{whereZero(x[1])} creates function which equals $1$ where \code{x[1]} is
266 equal to zero and $0$ elsewhere.
267 The sum of the results of \code{whereZero(x[0])} and \code{whereZero(x[1])}
268 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 Note that \var{gammaD} has the same \Rank, \Shape and \FunctionSpace like \var{x0} used to define it. So from
270 \begin{python}
271 print gammaD.getRank(),gammaD.getShape(),gammaD.getFunctionSpace()
272 \end{python}
273 one gets
274 \begin{python}
275 0 () Function space type: Finley_Nodes on FinleyMesh
276 \end{python}
277 An additional parameter \var{q} of the \code{setValue} method of the \Poisson class defines the
278 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 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 \end{python}
288 The first statement imports the \Poisson class definition from the \linearPDEs module \escript package.
289 To get the solution of the Poisson equation defined by \var{mypde} we just have to call its
290 \method{getSolution}.
291
292 Now we can write the script to solve our Poisson problem
293 \begin{python}
294 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 \end{python}
307 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
311 \subsection{Plotting Using \MATPLOTLIB}
312 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 This method is not part of version 0.98.1 which is available with Lenny.
316 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 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 Now the values created by \escript need to be interpolated to this grid. We will use the \MATPLOTLIB
331 \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 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 \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 \begin{verbatim}
399 run-escript poisson_matplotlib.py
400 \end{verbatim}
401 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
404 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 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
408 \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 \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 The entire code is available as \file{poisson\hackscore VTK.py} in the \ExampleDirectory.
442
443 You can run the script using the {\it escript} environment
444 and visualize the solution using \mayavi:
445 \begin{verbatim}
446 run-escript poisson_VTK.py
447 mayavi2 -d u.xml -m SurfaceMap
448 \end{verbatim}
449 The result is shown in Figure~\fig{fig:FirstSteps.3}.

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