ViewVC logotype

Contents of /trunk/doc/user/firststep.tex

Parent Directory Parent Directory | Revision Log Revision Log

Revision 1316 - (show annotations)
Tue Sep 25 03:18:30 2007 UTC (11 years, 6 months ago) by ksteube
File MIME type: application/x-tex
File size: 13727 byte(s)
Quickly edited chapters 1 and 2 of the User Guide, but it needs more work.
Ran entire document through spell checker.

1 %
2 % $Id$
3 %
4 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
5 %
6 % 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 %
17 \section{The First Steps}
18 \label{FirstSteps}
20 \begin{figure}
21 \centerline{\includegraphics[width=\figwidth]{figures/FirstStepDomain}}
22 \caption{Domain $\Omega=[0,1]^2$ with outer normal field $n$.}
23 \label{fig:FirstSteps.1}
24 \end{figure}
26 In this chapter we give an introduction how to use \escript to solve
27 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}
28 is more than sufficient.
30 The PDE \index{partial differential equation} we wish to solve is the Poisson equation \index{Poisson equation}
31 \begin{equation}
32 -\Delta u =f
33 \label{eq:FirstSteps.1}
34 \end{equation}
35 for the solution $u$. The function $f$ is the given right hand side. The domain of interest, denoted by $\Omega$,
36 is the unit square
37 \begin{equation}
38 \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 \}
39 \label{eq:FirstSteps.1b}
40 \end{equation}
41 The domain is shown in \fig{fig:FirstSteps.1}.
43 $\Delta$ denotes the Laplace operator\index{Laplace operator}, which is defined by
44 \begin{equation}
45 \Delta u = (u\hackscore {,0})\hackscore{,0}+(u\hackscore{,1})\hackscore{,1}
46 \label{eq:FirstSteps.1.1}
47 \end{equation}
48 where, for any function $u$ and any direction $i$, $u\hackscore{,i}$
49 denotes the partial derivative \index{partial derivative} of $u$ with respect to $i$.
50 \footnote{You
51 may be more familiar with the Laplace operator\index{Laplace operator} being written
52 as $\nabla^2$, and written in the form
53 \begin{equation*}
54 \nabla^2 u = \nabla^t \cdot \nabla u = \frac{\partial^2 u}{\partial x\hackscore 0^2}
55 + \frac{\partial^2 u}{\partial x\hackscore 1^2}
56 \end{equation*}
57 and \eqn{eq:FirstSteps.1} as
58 \begin{equation*}
59 -\nabla^2 u = f
60 \end{equation*}
61 }
62 Basically, in the subindex of a function, any index to the left of the comma denotes a spatial derivative with respect
63 to the index. To get a more compact form we will write $w\hackscore{,ij}=(w\hackscore {,i})\hackscore{,j}$
64 which leads to
65 \begin{equation}
66 \Delta u = u\hackscore{,00}+u\hackscore{,11}=\sum\hackscore{i=0}^2 u\hackscore{,ii}
67 \label{eq:FirstSteps.1.1b}
68 \end{equation}
69 We often find that use
70 of nested $\sum$ symbols makes formulas cumbersome, and we use the more
71 convenient Einstein summation convention \index{summation convention}. This
72 drops the $\sum$ sign and assumes that a summation is performed over any repeated index.
73 For instance we write
74 \begin{eqnarray}
75 x\hackscore{i}y\hackscore{i}=\sum\hackscore{i=0}^2 x\hackscore{i}y\hackscore{i} \\
76 x\hackscore{i}u\hackscore{,i}=\sum\hackscore{i=0}^2 x\hackscore{i}u\hackscore{,i} \\
77 u\hackscore{,ii}=\sum\hackscore{i=0}^2 u\hackscore{,ii} \\
78 x\hackscore{ij}u\hackscore{i,j}=\sum\hackscore{j=0}^2\sum\hackscore{i=0}^2 x\hackscore{ij}u\hackscore{i,j} \\
79 \label{eq:FirstSteps.1.1c}
80 \end{eqnarray}
81 With the summation convention we can write the Poisson equation \index{Poisson equation} as
82 \begin{equation}
83 - u\hackscore{,ii} =1
84 \label{eq:FirstSteps.1.sum}
85 \end{equation}
86 where $f=1$ in this example.
88 On the boundary of the domain $\Omega$ the normal derivative $n\hackscore{i} u\hackscore{,i}$
89 of the solution $u$ shall be zero, ie. $u$ shall fulfill
90 the homogeneous Neumann boundary condition\index{Neumann
91 boundary condition!homogeneous}
92 \begin{equation}
93 n\hackscore{i} u\hackscore{,i}= 0 \;.
94 \label{eq:FirstSteps.2}
95 \end{equation}
96 $n=(n\hackscore{i})$ denotes the outer normal field
97 of the domain, see \fig{fig:FirstSteps.1}. Remember that we
98 are applying the Einstein summation convention \index{summation convention}, i.e
99 $n\hackscore{i} u\hackscore{,i}= n\hackscore{0} u\hackscore{,0} +
100 n\hackscore{1} u\hackscore{,1}$.
101 \footnote{Some readers may familiar with the notation
102 \begin{equation*}
103 \frac{\partial u}{\partial n} = n\hackscore{i} u\hackscore{,i}
104 \end{equation*}
105 for the normal derivative.}
106 The Neumann boundary condition of \eqn{eq:FirstSteps.2} should be fulfilled on the
107 set $\Gamma^N$ which is the top and right edge of the domain:
108 \begin{equation}
109 \Gamma^N=\{(x\hackscore 0;x\hackscore 1) \in \Omega | x\hackscore{0}=1 \mbox{ or } x\hackscore{1}=1 \}
110 \label{eq:FirstSteps.2b}
111 \end{equation}
112 On the bottom and the left edge of the domain which is defined
113 as
114 \begin{equation}
115 \Gamma^D=\{(x\hackscore 0;x\hackscore 1) \in \Omega | x\hackscore{0}=0 \mbox{ or } x\hackscore{1}=0 \}
116 \label{eq:FirstSteps.2c}
117 \end{equation}
118 the solution shall be identically zero:
119 \begin{equation}
120 u=0 \; .
121 \label{eq:FirstSteps.2d}
122 \end{equation}
123 This kind of boundary condition is called a homogeneous Dirichlet boundary condition
124 \index{Dirichlet boundary condition!homogeneous}. The partial differential equation in \eqn{eq:FirstSteps.1.sum} together
125 with the Neumann boundary condition \eqn{eq:FirstSteps.2} and
126 Dirichlet boundary condition in \eqn{eq:FirstSteps.2d} form a so
127 called boundary value
128 problem\index{boundary value problem} (BVP\index{boundary value problem!BVP}) for
129 the unknown
130 function $u$.
133 \begin{figure}
134 \centerline{\includegraphics[width=\figwidth]{figures/FirstStepMesh.eps}}
135 \caption{Mesh of $4 \time 4$ elements on a rectangular domain. Here
136 each element is a quadrilateral and described by four nodes, namely
137 the corner points. The solution is interpolated by a bi-linear
138 polynomial.}
139 \label{fig:FirstSteps.2}
140 \end{figure}
142 In general the BVP\index{boundary value problem!BVP} cannot be solved analytically and numerical
143 methods have to be used construct an approximation of the solution
144 $u$. Here we will use the finite element method\index{finite element
145 method} (FEM\index{finite element
146 method!FEM}). The basic idea is to fill the domain with a
147 set of points called nodes. The solution is approximated by its
148 values on the nodes\index{finite element
149 method!nodes}. Moreover, the domain is subdivided into smaller
150 sub-domains called elements \index{finite element
151 method!element}. On each element the solution is
152 represented by a polynomial of a certain degree through its values at
153 the nodes located in the element. The nodes and its connection through
154 elements is called a mesh\index{finite element
155 method!mesh}. \fig{fig:FirstSteps.2} shows an
156 example of a FEM mesh with four elements in the $x_0$ and four elements
157 in the $x_1$ direction over the unit square.
158 For more details we refer the reader to the literature, for instance
159 \Ref{Zienc,NumHand}.
161 \escript provides the class \Poisson to define a Poisson equation \index{Poisson equation}.
162 (We will discuss a more general form of a PDE \index{partial differential equation!PDE}
163 that can be defined through the \LinearPDE class later). The instantiation of
164 a \Poisson class object requires the specification of the domain $\Omega$. In \escript
165 the \Domain class objects are used to describe the geometry of a domain but it also
166 contains information about the discretization methods and the actual solver which is used
167 to solve the PDE. Here we are using the FEM library \finley \index{finite element
168 method}. The following statements create the \Domain object \var{mydomain} from the
169 \finley method \method{Rectangle}
170 \begin{python}
171 from esys.finley import Rectangle
172 mydomain = Rectangle(l0=1.,l1=1.,n0=40, n1=20)
173 \end{python}
174 In this case the domain is a rectangle with the lower, left corner at point $(0,0)$ and
175 the right, upper corner at $(\var{l0},\var{l1})=(1,1)$.
176 The arguments \var{n0} and \var{n1} define the number of elements in $x\hackscore{0}$ and
177 $x\hackscore{1}$-direction respectively. For more details on \method{Rectangle} and
178 other \Domain generators within the \finley module,
179 see \Chap{CHAPTER ON FINLEY}.
181 The following statements define the \Poisson class object \var{mypde} with domain \var{mydomain} and
182 the right hand side $f$ of the PDE to constant $1$:
183 \begin{python}
184 from esys.escript.linearPDEs import Poisson
185 mypde = Poisson(mydomain)
186 mypde.setValue(f=1)
187 \end{python}
188 We have not specified any boundary condition but the
189 \Poisson class implicitly assumes homogeneous Neuman boundary conditions \index{Neumann
190 boundary condition!homogeneous} defined by \eqn{eq:FirstSteps.2}. With this boundary
191 condition the BVP\index{boundary value problem!BVP} we have defined has no unique solution. In fact, with any solution $u$
192 and any constant $C$ the function $u+C$ becomes a solution as well. We have to add
193 a Dirichlet boundary condition \index{Dirichlet boundary condition}. This is done
194 by defining a characteristic function \index{characteristic function}
195 which has positive values at locations $x=(x\hackscore{0},x\hackscore{1})$ where Dirichlet boundary condition is set
196 and $0$ elsewhere. In our case of $\Gamma^D$ defined by \eqn{eq:FirstSteps.2c},
197 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
198 an object \var{x} which contains the coordinates of the nodes in the domain use
199 \begin{python}
200 x=mydomain.getX()
201 \end{python}
202 The method \method{getX} of the \Domain \var{mydomain}
203 gives access to locations
204 in the domain defined by \var{mydomain}. The object \var{x} is actually a \Data object which will be
205 discussed in \Chap{ESCRIPT CHAP} in more detail. What we need to know here is that
207 \var{x} has \Rank (number of dimensions) and a \Shape (list of dimensions) which can be viewed by
208 calling the \method{getRank} and \method{getShape} methods:
209 \begin{python}
210 print "rank ",x.getRank(),", shape ",x.getShape()
211 \end{python}
212 This will print something like
213 \begin{python}
214 rank 1, shape (2,)
215 \end{python}
216 The \Data object also maintains type information which is represented by the
217 \FunctionSpace of the object. For instance
218 \begin{python}
219 print x.getFunctionSpace()
220 \end{python}
221 will print
222 \begin{python}
223 Function space type: Finley_Nodes on FinleyMesh
224 \end{python}
225 which tells us that the coordinates are stored on the nodes of (rather than on points in the interior of) a \finley mesh.
226 To get the $x\hackscore{0}$ coordinates of the locations we use the
227 statement
228 \begin{python}
229 x0=x[0]
230 \end{python}
231 Object \var{x0}
232 is again a \Data object now with \Rank $0$ and
233 \Shape $()$. It inherits the \FunctionSpace from \var{x}:
234 \begin{python}
235 print x0.getRank(),x0.getShape(),x0.getFunctionSpace()
236 \end{python}
237 will print
238 \begin{python}
239 0 () Function space type: Finley_Nodes on FinleyMesh
240 \end{python}
241 We can now construct a function \var{gammaD} which is only non-zero on the bottom and left edges
242 of the domain with
243 \begin{python}
244 from esys.escript import whereZero
245 gammaD=whereZero(x[0])+whereZero(x[1])
246 \end{python}
248 \code{whereZero(x[0])} creates function which equals $1$ where \code{x[0]} is (almost) equal to zero
249 and $0$ elsewhere.
250 Similarly, \code{whereZero(x[1])} creates function which equals $1$ where \code{x[1]} is
251 equal to zero and $0$ elsewhere.
252 The sum of the results of \code{whereZero(x[0])} and \code{whereZero(x[1])}
253 gives a function on the domain \var{mydomain} which is exactly positive where $x\hackscore{0}$ or $x\hackscore{1}$ is equal to zero.
254 Note that \var{gammaD} has the same \Rank, \Shape and \FunctionSpace like \var{x0} used to define it. So from
255 \begin{python}
256 print gammaD.getRank(),gammaD.getShape(),gammaD.getFunctionSpace()
257 \end{python}
258 one gets
259 \begin{python}
260 0 () Function space type: Finley_Nodes on FinleyMesh
261 \end{python}
262 An additional parameter \var{q} of the \code{setValue} method of the \Poisson class defines the
263 characteristic function \index{characteristic function} of the locations
264 of the domain where homogeneous Dirichlet boundary condition \index{Dirichlet boundary condition!homogeneous}
265 are set. The complete definition of our example is now:
266 \begin{python}
267 from esys.linearPDEs import Poisson
268 x = mydomain.getX()
269 gammaD = whereZero(x[0])+whereZero(x[1])
270 mypde = Poisson(domain=mydomain)
271 mypde.setValue(f=1,q=gammaD)
272 \end{python}
273 The first statement imports the \Poisson class definition from the \linearPDEs module \escript package.
274 To get the solution of the Poisson equation defined by \var{mypde} we just have to call its
275 \method{getSolution}.
277 Now we can write the script to solve our Poisson problem
278 \begin{python}
279 from esys.escript import *
280 from esys.escript.linearPDEs import Poisson
281 from esys.finley import Rectangle
282 # generate domain:
283 mydomain = Rectangle(l0=1.,l1=1.,n0=40, n1=20)
284 # define characteristic function of Gamma^D
285 x = mydomain.getX()
286 gammaD = whereZero(x[0])+whereZero(x[1])
287 # define PDE and get its solution u
288 mypde = Poisson(domain=mydomain)
289 mypde.setValue(f=1,q=gammaD)
290 u = mypde.getSolution()
291 # write u to an external file
292 saveVTK("u.xml",sol=u)
293 \end{python}
294 The entire code is available as \file{poisson.py} in the \ExampleDirectory
296 The last statement writes the solution (tagged with the name "sol") to a file named \file{u.xml} in
297 \VTK file format.
298 Now you may run the script and visualize the solution using \mayavi:
299 \begin{verbatim}
300 python poisson.py
301 mayavi -d u.xml -m SurfaceMap
302 \end{verbatim}
303 See \fig{fig:FirstSteps.3}.
305 \begin{figure}
306 \centerline{\includegraphics[width=\figwidth]{figures/FirstStepResult.eps}}
307 \caption{Visualization of the Poisson Equation Solution for $f=1$}
308 \label{fig:FirstSteps.3}
309 \end{figure}


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

  ViewVC Help
Powered by ViewVC 1.1.26