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that does not look to bad now although more stuff could be added.
1 jgs 102 % $Id$
2 gross 625 %
3     % Copyright © 2006 by ACcESS MNRF
4     % \url{http://www.access.edu.au
5     % Primary Business: Queensland, Australia.
6     % Licensed under the Open Software License version 3.0
7     % http://www.opensource.org/licenses/osl-3.0.php
8     %
9 jgs 102
10 gross 625
11 jgs 121 \section{The First Steps}
12 jgs 102 \label{FirstSteps}
13    
14     \begin{figure}
15 gross 599 \centerline{\includegraphics[width=\figwidth]{figures/FirstStepDomain}}
16 jgs 102 \caption{Domain $\Omega=[0,1]^2$ with outer normal field $n$.}
17     \label{fig:FirstSteps.1}
18     \end{figure}
19    
20 jgs 107 In this chapter we will give an introduction how to use \escript to solve
21     a partial differential equation \index{partial differential equation} (PDE \index{partial differential equation!PDE}). The reader should be familiar with Python. The knowledge presented at the Python tutorial at \url{http://docs.python.org/tut/tut.html}
22     is sufficient. It is helpful if the reader has some basic knowledge of PDEs \index{partial differential equation}.
23 jgs 102
24 jgs 107 The PDE \index{partial differential equation} we wish to solve is the Poisson equation \index{Poisson equation}
25 jgs 102 \begin{equation}
26     -\Delta u =f
27     \label{eq:FirstSteps.1}
28     \end{equation}
29 jgs 107 for the solution $u$. The function $f$ is the given right hand side. The domain of interest, denoted by $\Omega$
30 jgs 102 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 jgs 107 The domain is shown in \fig{fig:FirstSteps.1}.
36 jgs 102
37     $\Delta$ denotes the Laplace operator\index{Laplace operator} which is defined by
38     \begin{equation}
39     \Delta u = (u\hackscore {,0})\hackscore{,0}+(u\hackscore{,1})\hackscore{,1}
40     \label{eq:FirstSteps.1.1}
41     \end{equation}
42 jgs 107 where, for any function $w$ and any direction $i$, $u\hackscore{,i}$
43     denotes the partial derivative \index{partial derivative} of $u$ with respect to $i$.
44 jgs 102 \footnote{Some readers
45     may be more familiar with the Laplace operator\index{Laplace operator} being written
46     as $\nabla^2$, and written in the form
47     \begin{equation*}
48 jgs 110 \nabla^2 u = \nabla^t \cdot \nabla u = \frac{\partial^2 u}{\partial x\hackscore 0^2}
49 jgs 102 + \frac{\partial^2 u}{\partial x\hackscore 1^2}
50     \end{equation*}
51     and \eqn{eq:FirstSteps.1} as
52     \begin{equation*}
53     -\nabla^2 u = f
54     \end{equation*}
55     }
56 jgs 107 Basically, in the subindex of a function, any index to the left of the comma denotes a spatial derivative with respect
57 jgs 102 to the index. To get a more compact form we will write $w\hackscore{,ij}=(w\hackscore {,i})\hackscore{,j}$
58     which leads to
59     \begin{equation}
60     \Delta u = u\hackscore{,00}+u\hackscore{,11}=\sum\hackscore{i=0}^2 u\hackscore{,ii}
61     \label{eq:FirstSteps.1.1b}
62     \end{equation}
63     In some cases, and we will see examples for this in the next chapter,
64     the usage of the nested $\sum$ symbols blows up the formulas and therefore
65 jgs 107 it is convenient to use the Einstein summation convention \index{summation convention}. This
66     drops the $\sum$ sign and assumes that a summation over a repeated index is performed
67 jgs 102 ("repeated index means summation"). For instance we write
68     \begin{eqnarray}
69     x\hackscore{i}y\hackscore{i}=\sum\hackscore{i=0}^2 x\hackscore{i}y\hackscore{i} \\
70     x\hackscore{i}u\hackscore{,i}=\sum\hackscore{i=0}^2 x\hackscore{i}u\hackscore{,i} \\
71     u\hackscore{,ii}=\sum\hackscore{i=0}^2 u\hackscore{,ii} \\
72 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} \\
73 jgs 102 \label{eq:FirstSteps.1.1c}
74     \end{eqnarray}
75     With the summation convention we can write the Poisson equation \index{Poisson equation} as
76     \begin{equation}
77     - u\hackscore{,ii} =1
78     \label{eq:FirstSteps.1.sum}
79     \end{equation}
80 lkettle 575 where $f=1$ in this example.
81    
82 jgs 102 On the boundary of the domain $\Omega$ the normal derivative $n\hackscore{i} u\hackscore{,i}$
83     of the solution $u$ shall be zero, ie. $u$ shall fulfill
84     the homogeneous Neumann boundary condition\index{Neumann
85     boundary condition!homogeneous}
86     \begin{equation}
87     n\hackscore{i} u\hackscore{,i}= 0 \;.
88     \label{eq:FirstSteps.2}
89     \end{equation}
90     $n=(n\hackscore{i})$ denotes the outer normal field
91     of the domain, see \fig{fig:FirstSteps.1}. Remember that we
92     are applying the Einstein summation convention \index{summation convention}, i.e
93 jgs 107 $n\hackscore{i} u\hackscore{,i}= n\hackscore{0} u\hackscore{,0} +
94     n\hackscore{1} u\hackscore{,1}$.
95 jgs 102 \footnote{Some readers may familiar with the notation
96     \begin{equation*}
97     \frac{\partial u}{\partial n} = n\hackscore{i} u\hackscore{,i}
98     \end{equation*}
99     for the normal derivative.}
100     The Neumann boundary condition of \eqn{eq:FirstSteps.2} should be fulfilled on the
101     set $\Gamma^N$ which is the top and right edge of the domain:
102     \begin{equation}
103     \Gamma^N=\{(x\hackscore 0;x\hackscore 1) \in \Omega | x\hackscore{0}=1 \mbox{ or } x\hackscore{1}=1 \}
104     \label{eq:FirstSteps.2b}
105     \end{equation}
106 jgs 107 On the bottom and the left edge of the domain which is defined
107 jgs 102 as
108     \begin{equation}
109     \Gamma^D=\{(x\hackscore 0;x\hackscore 1) \in \Omega | x\hackscore{0}=0 \mbox{ or } x\hackscore{1}=0 \}
110     \label{eq:FirstSteps.2c}
111     \end{equation}
112     the solution shall be identically zero:
113     \begin{equation}
114     u=0 \; .
115     \label{eq:FirstSteps.2d}
116     \end{equation}
117 jgs 107 This kind of boundary condition is called a homogeneous Dirichlet boundary condition
118 jgs 102 \index{Dirichlet boundary condition!homogeneous}. The partial differential equation in \eqn{eq:FirstSteps.1.sum} together
119     with the Neumann boundary condition \eqn{eq:FirstSteps.2} and
120     Dirichlet boundary condition in \eqn{eq:FirstSteps.2d} form a so
121     called boundary value
122 jgs 107 problem\index{boundary value problem} (BVP\index{boundary value problem!BVP}) for
123     the unknown
124 jgs 102 function $u$.
125    
126    
127     \begin{figure}
128 gross 599 \centerline{\includegraphics[width=\figwidth]{figures/FirstStepMesh.eps}}
129 jgs 102 \caption{Mesh of $4 \time 4$ elements on a rectangular domain. Here
130     each element is a quadrilateral and described by four nodes, namely
131     the corner points. The solution is interpolated by a bi-linear
132     polynomial.}
133     \label{fig:FirstSteps.2}
134     \end{figure}
135    
136     In general the BVP\index{boundary value problem!BVP} cannot be solved analytically and numerical
137     methods have to be used construct an approximation of the solution
138     $u$. Here we will use the finite element method\index{finite element
139     method} (FEM\index{finite element
140     method!FEM}). The basic idea is to fill the domain with a
141 jgs 107 set of points called nodes. The solution is approximated by its
142 jgs 102 values on the nodes\index{finite element
143 lkettle 573 method!nodes}. Moreover, the domain is subdivided into smaller
144     sub-domains called elements \index{finite element
145 jgs 102 method!element}. On each element the solution is
146     represented by a polynomial of a certain degree through its values at
147     the nodes located in the element. The nodes and its connection through
148     elements is called a mesh\index{finite element
149 jgs 107 method!mesh}. \fig{fig:FirstSteps.2} shows an
150 jgs 102 example of a FEM mesh with four elements in the $x_0$ and four elements
151     in the $x_1$ direction over the unit square.
152     For more details we refer the reader to the literature, for instance
153     \Ref{Zienc,NumHand}.
154    
155     \escript provides the class \Poisson to define a Poisson equation \index{Poisson equation}.
156     (We will discuss a more general form of a PDE \index{partial differential equation!PDE}
157     that can be defined through the \LinearPDE class later). The instantiation of
158     a \Poisson class object requires the specification of the domain $\Omega$. In \escript
159     the \Domain class objects are used to describe the geometry of a domain but it also
160     contains information about the discretization methods and the actual solver which is used
161     to solve the PDE. Here we are using the FEM library \finley \index{finite element
162     method}. The following statements create the \Domain object \var{mydomain} from the
163     \finley method \method{Rectangle}
164     \begin{python}
165 jgs 107 from esys.finley import Rectangle
166     mydomain = Rectangle(l0=1.,l1=1.,n0=40, n1=20)
167 jgs 102 \end{python}
168     In this case the domain is a rectangle with the lower, left corner at point $(0,0)$ and
169     the right, upper corner at $(\var{l0},\var{l1})=(1,1)$.
170 jgs 107 The arguments \var{n0} and \var{n1} define the number of elements in $x\hackscore{0}$ and
171 jgs 102 $x\hackscore{1}$-direction respectively. For more details on \method{Rectangle} and
172     other \Domain generators within the \finley module,
173     see \Chap{CHAPTER ON FINLEY}.
174    
175 jgs 107 The following statements define the \Poisson class object \var{mypde} with domain \var{mydomain} and
176 jgs 102 the right hand side $f$ of the PDE to constant $1$:
177     \begin{python}
178 gross 565 from esys.escript.linearPDEs import Poisson
179 jgs 107 mypde = Poisson(mydomain)
180     mypde.setValue(f=1)
181 jgs 102 \end{python}
182     We have not specified any boundary condition but the
183     \Poisson class implicitly assumes homogeneous Neuman boundary conditions \index{Neumann
184     boundary condition!homogeneous} defined by \eqn{eq:FirstSteps.2}. With this boundary
185     condition the BVP\index{boundary value problem!BVP} we have defined has no unique solution. In fact, with any solution $u$
186     and any constant $C$ the function $u+C$ becomes a solution as well. We have to add
187     a Dirichlet boundary condition \index{Dirichlet boundary condition}. This is done
188 jgs 107 by defining a characteristic function \index{characteristic function}
189     which has positive values at locations $x=(x\hackscore{0},x\hackscore{1})$ where Dirichlet boundary condition is set
190 jgs 102 and $0$ elsewhere. In our case of $\Gamma^D$ defined by \eqn{eq:FirstSteps.2c},
191 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
192     an object \var{x} which represents locations in the domain one uses
193 jgs 102 \begin{python}
194 lkettle 573 x=mydomain.getX()
195 jgs 102 \end{python}
196 gross 660 The method \method{getX} of the \Domain \var{mydomain}
197 jgs 107 gives access to locations
198 gross 565 in the domain defined by \var{mydomain}. The object \var{x} is actually a \Data object which is
199 gross 660 discussed in \Chap{ESCRIPT CHAP} in more details. What we need to know here is that
200    
201 gross 565 \var{x} has \Rank (=number of dimensions) and a \Shape (=tuple of dimensions) which can be checked by
202     calling the \method{getRank} and \method{getShape} methods:
203     \begin{python}
204     print "rank ",x.getRank(),", shape ",x.getShape()
205     \end{python}
206     will print something like
207     \begin{python}
208     rank 1, shape (2,)
209     \end{python}
210     The \Data object also maintains type information which is represented by the
211     \FunctionSpace of the object. For instance
212     \begin{python}
213     print x.getFunctionSpace()
214     \end{python}
215     will print
216     \begin{python}
217     Function space type: Finley_Nodes on FinleyMesh
218     \end{python}
219     which tells us that the coordinates are stored on the nodes of a \finley mesh.
220     To get the $x\hackscore{0}$ coordinates of the locations we use the
221     statement
222     \begin{python}
223     x0=x[0]
224     \end{python}
225     Object \var{x0}
226     is again a \Data object now with \Rank $0$ and
227     \Shape $()$. It inherits the \FunctionSpace from \var{x}:
228     \begin{python}
229     print x0.getRank(),x0.getShape(),x0.getFunctionSpace()
230     \end{python}
231     will print
232     \begin{python}
233     0 () Function space type: Finley_Nodes on FinleyMesh
234     \end{python}
235     We can now construct the function \var{gammaD} by
236     \begin{python}
237 lkettle 573 from esys.escript import whereZero
238 gross 565 gammaD=whereZero(x[0])+whereZero(x[1])
239     \end{python}
240     where
241     \code{whereZero(x[0])} creates function which equals $1$ where \code{x[0]} is (allmost) equal to zero
242 jgs 107 and $0$ elsewhere.
243 gross 565 Similarly, \code{whereZero(x[1])} creates function which equals $1$ where \code{x[1]} is
244 jgs 107 equal to zero and $0$ elsewhere.
245 gross 565 The sum of the results of \code{whereZero(x[0])} and \code{whereZero(x[1])}
246     gives a function on the domain \var{mydomain} which is exactly positive where $x\hackscore{0}$ or $x\hackscore{1}$ is equal to zero.
247     Note that \var{gammaD} has the same \Rank, \Shape and \FunctionSpace like \var{x0} used to define it. So from
248     \begin{python}
249     print gammaD.getRank(),gammaD.getShape(),gammaD.getFunctionSpace()
250     \end{python}
251     one gets
252     \begin{python}
253     0 () Function space type: Finley_Nodes on FinleyMesh
254     \end{python}
255 jgs 107 The additional parameter \var{q} of the \code{setValue} method of the \Poisson class defines the
256 jgs 102 characteristic function \index{characteristic function} of the locations
257     of the domain where homogeneous Dirichlet boundary condition \index{Dirichlet boundary condition!homogeneous}
258     are set. The complete definition of our example is now:
259     \begin{python}
260 jgs 107 from esys.linearPDEs import Poisson
261 jgs 102 x = mydomain.getX()
262 gross 565 gammaD = whereZero(x[0])+whereZero(x[1])
263 jgs 107 mypde = Poisson(domain=mydomain)
264 lkettle 573 mypde.setValue(f=1,q=gammaD)
265 jgs 102 \end{python}
266 lkettle 573 The first statement imports the \Poisson class definition from the \linearPDEs module \escript package.
267 jgs 107 To get the solution of the Poisson equation defined by \var{mypde} we just have to call its
268 jgs 102 \method{getSolution}.
269    
270     Now we can write the script to solve our test problem (Remember that
271 gross 569 lines starting with '\#' are comment lines in Python) (available as \file{poisson.py}
272 jgs 102 in the \ExampleDirectory):
273     \begin{python}
274 gross 565 from esys.escript import *
275 gross 569 from esys.escript.linearPDEs import Poisson
276 jgs 107 from esys.finley import Rectangle
277 jgs 102 # generate domain:
278 jgs 107 mydomain = Rectangle(l0=1.,l1=1.,n0=40, n1=20)
279 jgs 102 # define characteristic function of Gamma^D
280     x = mydomain.getX()
281 gross 565 gammaD = whereZero(x[0])+whereZero(x[1])
282 jgs 102 # define PDE and get its solution u
283 gross 565 mypde = Poisson(domain=mydomain)
284     mypde.setValue(f=1,q=gammaD)
285 jgs 102 u = mypde.getSolution()
286     # write u to an external file
287 gross 565 saveVTK("u.xml",sol=u)
288 jgs 102 \end{python}
289 gross 565 The last statement writes the solution tagged with the name "sol" to the external file \file{u.xml} in
290     \VTK file format. \VTK is a software library
291 jgs 102 for the visualization of scientific, engineering and analytical data and is freely available
292 lkettle 573 from \url{http://www.vtk.org}. There are a variety of graphical user interfaces
293 gross 565 for \VTK available, for instance \mayavi which can be downloaded from \url{http://mayavi.sourceforge.net/} but is also available on most
294     \LINUX distributions.
295 jgs 102
296     \begin{figure}
297 gross 599 \centerline{\includegraphics[width=\figwidth]{figures/FirstStepResult.eps}}
298 lkettle 573 \caption{Visualization of the Poisson Equation Solution for $f=1$}
299 jgs 102 \label{fig:FirstSteps.3}
300     \end{figure}
301    
302 jgs 107 You can edit the script file using your favourite text editor (or the Integrated DeveLopment Environment IDLE
303 gross 569 for Python, see \url{http://idlefork.sourceforge.net}). If the script file has the name \file{poisson.py} \index{scripts!\file{poisson.py}} you can run the
304 jgs 102 script from any shell using the command:
305 gross 565 \begin{python}
306 gross 569 python poisson.py
307 gross 565 \end{python}
308 gross 569 After the script has (hopefully successfully) been completed you will find the file \file{u.xml} in the current
309 jgs 102 directory. An easy way to visualize the results is the command
310 gross 565 \begin{python}
311     mayavi -d u.xml -m SurfaceMap &
312     \end{python}
313     to show the results, see \fig{fig:FirstSteps.3}.

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