 # Diff of /trunk/doc/user/firststep.tex

trunk/esys2/doc/user/firststep.tex revision 102 by jgs, Wed Dec 15 07:08:39 2004 UTC trunk/doc/user/firststep.tex revision 2548 by jfenwick, Mon Jul 20 06:20:06 2009 UTC
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% $Id$
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2  \chapter{The First Steps}  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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
11    %
12    %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
13
14
15    \section{The First Steps}
16  \label{FirstSteps}  \label{FirstSteps}
17
\begin{figure}
\centerline{\includegraphics[width=\figwidth]{FirstStepDomain}}
\caption{Domain $\Omega=[0,1]^2$ with outer normal field $n$.}
\label{fig:FirstSteps.1}
\end{figure}
18
In this chapter we will show the basics of how to use \escript to solve
a partial differential equation \index{partial differential equation} (PDE \index{partial differential equation!PDE}). The reader should be familiar with
the basics of Python. The knowledge presented the Python tutorial at \url{http://docs.python.org/tut/tut.html}
is sufficient. It is helpful if the reader has some basic knowledge on PDEs \index{PDE}.
19
20  The \index{PDE} we want to solve is the Poisson equation \index{Poisson equation}  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.
23
24    The PDE \index{partial differential equation} we wish to solve is the Poisson equation \index{Poisson equation}
25  \begin{equation}  \begin{equation}
26  -\Delta u =f  -\Delta u =f
27  \label{eq:FirstSteps.1}  \label{eq:FirstSteps.1}
28  \end{equation}  \end{equation}
29  for the solution $u$. The domain of interest which we denote by $\Omega$  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  is the unit square
31  \begin{equation}  \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 \}  \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}  \label{eq:FirstSteps.1b}
34  \end{equation}  \end{equation}
35  The domain is shown in Figure~\fig{fig:FirstSteps.1}.  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}
41
42  $\Delta$ denotes the Laplace operator\index{Laplace operator} which is defined by  $\Delta$ denotes the Laplace operator\index{Laplace operator}, which is defined by
43  \begin{equation}  \begin{equation}
44  \Delta u = (u\hackscore {,0})\hackscore{,0}+(u\hackscore{,1})\hackscore{,1}  \Delta u = (u\hackscore {,0})\hackscore{,0}+(u\hackscore{,1})\hackscore{,1}
45  \label{eq:FirstSteps.1.1}  \label{eq:FirstSteps.1.1}
46  \end{equation}  \end{equation}
47  where for any function $w$ and any direction $i$ $u\hackscore{,i}$  where, for any function $u$ and any direction $i$, $u\hackscore{,i}$
48  denotes the partial derivative \index{partial derivative} of $w$ with respect to $i$.    denotes the partial derivative \index{partial derivative} of $u$ with respect to $i$.
50  may be more familiar with the Laplace operator\index{Laplace operator} being written  may be more familiar with the Laplace operator\index{Laplace operator} being written
51  as $\nabla^2$, and written in the form  as $\nabla^2$, and written in the form
52  \begin{equation*}  \begin{equation*}
53  \nabla^2 u = \frac{\partial^2 u}{\partial x\hackscore 0^2}  \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}  + \frac{\partial^2 u}{\partial  x\hackscore 1^2}
55  \end{equation*}  \end{equation*}
56  and \eqn{eq:FirstSteps.1} as  and \eqn{eq:FirstSteps.1} as
# Line 46  and \eqn{eq:FirstSteps.1} as Line 58  and \eqn{eq:FirstSteps.1} as
58  -\nabla^2 u = f  -\nabla^2 u = f
59  \end{equation*}  \end{equation*}
60  }  }
61  Basically in the subindex of a function any index left to the comma denotes a spatial derivative with respect  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 $w\hackscore{,ij}=(w\hackscore {,i})\hackscore{,j}$  to the index. To get a more compact form we will write $u\hackscore{,ij}=(u\hackscore {,i})\hackscore{,j}$
64  \begin{equation}  \begin{equation}
65  \Delta u = u\hackscore{,00}+u\hackscore{,11}=\sum\hackscore{i=0}^2 u\hackscore{,ii}  \Delta u = u\hackscore{,00}+u\hackscore{,11}=\sum\hackscore{i=0}^2 u\hackscore{,ii}
66  \label{eq:FirstSteps.1.1b}  \label{eq:FirstSteps.1.1b}
67  \end{equation}  \end{equation}
68  In some cases, and we will see examples for this in the next chapter,  We often find that use
69  the usage of the nested $\sum$ symbols blows up the formulas and therefore  of nested $\sum$ symbols makes formulas cumbersome, and we use the more
70  it is convenient to use Einstein summation convention \index{summation convention} which  convenient Einstein summation convention \index{summation convention}. This
71  says that $\sum$ sign is dropped and a summation over a repeated index is performed  drops the $\sum$ sign and assumes that a summation is performed over any repeated index.
72  ("repeated index means summation"). For instance we write  For instance we write
73  \begin{eqnarray}  \begin{eqnarray}
74  x\hackscore{i}y\hackscore{i}=\sum\hackscore{i=0}^2 x\hackscore{i}y\hackscore{i}   \\  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}   \\  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} \\  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}  \label{eq:FirstSteps.1.1c}
79  \end{eqnarray}  \end{eqnarray}
80  With the summation convention we can write the Poisson equation \index{Poisson equation} as  With the summation convention we can write the Poisson equation \index{Poisson equation} as
# Line 69  With the summation convention we can wri Line 82  With the summation convention we can wri
82  - u\hackscore{,ii} =1  - u\hackscore{,ii} =1
83  \label{eq:FirstSteps.1.sum}  \label{eq:FirstSteps.1.sum}
84  \end{equation}  \end{equation}
85    where $f=1$ in this example.
86
87  On the boundary of the domain $\Omega$ the normal derivative $n\hackscore{i} u\hackscore{,i}$  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  of the solution $u$ shall be zero, ie. $u$ shall fulfill
89  the homogeneous Neumann boundary condition\index{Neumann  the homogeneous Neumann boundary condition\index{Neumann
# Line 80  n\hackscore{i} u\hackscore{,i}= 0 \;. Line 95  n\hackscore{i} u\hackscore{,i}= 0 \;.
95  $n=(n\hackscore{i})$ denotes the outer normal field  $n=(n\hackscore{i})$ denotes the outer normal field
96  of the domain, see \fig{fig:FirstSteps.1}. Remember that we  of the domain, see \fig{fig:FirstSteps.1}. Remember that we
97  are applying the Einstein summation convention \index{summation convention}, i.e  are applying the Einstein summation convention \index{summation convention}, i.e
98  $n\hackscore{i} u\hackscore{,i}= n\hackscore1 u\hackscore{,1} +$n\hackscore{i} u\hackscore{,i}= n\hackscore{0} u\hackscore{,0} +
99  n\hackscore2 u\hackscore{,2}$. n\hackscore{1} u\hackscore{,1}$.
100  \footnote{Some readers may familiar with the notation  \footnote{Some readers may familiar with the notation
101  \begin{equation*}  $102 \frac{\partial u}{\partial n} = n\hackscore{i} u\hackscore{,i} \frac{\partial u}{\partial n} = n\hackscore{i} u\hackscore{,i} 103 \end{equation*}$
104  for the normal derivative.}  for the normal derivative.}
105  The Neumann boundary condition of \eqn{eq:FirstSteps.2} should be fulfilled on the  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:  set $\Gamma^N$ which is the top and right edge of the domain:
# Line 93  set $\Gamma^N$ which is the top and righ Line 108  set $\Gamma^N$ which is the top and righ
108  \Gamma^N=\{(x\hackscore 0;x\hackscore 1) \in \Omega | x\hackscore{0}=1 \mbox{ or } x\hackscore{1}=1  \}  \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}  \label{eq:FirstSteps.2b}
110  \end{equation}  \end{equation}
111  On the bottom and the left edge of the domain which defined  On the bottom and the left edge of the domain which is defined
112  as  as
113  \begin{equation}  \begin{equation}
114  \Gamma^D=\{(x\hackscore 0;x\hackscore 1) \in \Omega | x\hackscore{0}=0 \mbox{ or } x\hackscore{1}=0  \}  \Gamma^D=\{(x\hackscore 0;x\hackscore 1) \in \Omega | x\hackscore{0}=0 \mbox{ or } x\hackscore{1}=0  \}
# Line 104  the solution shall be identically zero: Line 119  the solution shall be identically zero:
119  u=0 \; .  u=0 \; .
120  \label{eq:FirstSteps.2d}  \label{eq:FirstSteps.2d}
121  \end{equation}  \end{equation}
122  The kind of boundary condition is called a homogeneous Dirichlet boundary condition  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  \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  with the Neumann boundary condition \eqn{eq:FirstSteps.2} and
125  Dirichlet boundary condition in \eqn{eq:FirstSteps.2d} form a so  Dirichlet boundary condition in \eqn{eq:FirstSteps.2d} form a so
126  called boundary value  called boundary value
127  problem\index{boundary value problem} (BVP\index{boundary value problem!BVP}) for unknown  problem\index{boundary value problem} (BVP\index{boundary value problem!BVP}) for
128  function $u$.  the unknown function~$u$.
129
130
131  \begin{figure}  \begin{figure}[ht]
132  \centerline{\includegraphics[width=\figwidth]{FirstStepMesh}}  \centerline{\includegraphics[width=\figwidth]{figures/FirstStepMesh}}
133  \caption{Mesh of $4 \time 4$ elements on a rectangular domain.  Here  \caption{Mesh of $4 \time 4$ elements on a rectangular domain.  Here
134  each element is a quadrilateral and described by four nodes, namely  each element is a quadrilateral and described by four nodes, namely
135  the corner points. The solution is interpolated by a bi-linear  the corner points. The solution is interpolated by a bi-linear
# Line 127  methods have to be used construct an app Line 142  methods have to be used construct an app
142  $u$. Here we will use the finite element method\index{finite element  $u$. Here we will use the finite element method\index{finite element
143  method} (FEM\index{finite element  method} (FEM\index{finite element
144  method!FEM}). The basic idea is to fill the domain with a  method!FEM}). The basic idea is to fill the domain with a
145  set of points, so called nodes. The solution is approximated by its  set of points called nodes. The solution is approximated by its
146  values on the nodes\index{finite element  values on the nodes\index{finite element
147  method!nodes}. Moreover, the domain is subdivide into small  method!nodes}. Moreover, the domain is subdivided into smaller
148  sub-domain, so-called elements \index{finite element  sub-domains called elements \index{finite element
149  method!element}. On each element the solution is  method!element}. On each element the solution is
150  represented by a polynomial of a certain degree through its values at  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  the nodes located in the element. The nodes and its connection through
152  elements is called a mesh\index{finite element  elements is called a mesh\index{finite element
153  method!mesh}. Figure~\fig{fig:FirstSteps.2} shows an  method!mesh}. \fig{fig:FirstSteps.2} shows an
154  example of a FEM mesh with four elements in the $x_0$ and four elements  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.    in the $x_1$ direction over the unit square.
156  For more details we refer the reader to the literature, for instance  For more details we refer the reader to the literature, for instance \Ref{Zienc,NumHand}.
157  \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
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
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.
180
181  \escript provides the class \Poisson to define a Poisson equation \index{Poisson equation}.  \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}  (We will discuss a more general form of a PDE \index{partial differential equation!PDE}
# Line 151  to solve the PDE. Here we are using the Line 188  to solve the PDE. Here we are using the
188  method}. The following statements create the \Domain object \var{mydomain} from the  method}. The following statements create the \Domain object \var{mydomain} from the
189  \finley method \method{Rectangle}  \finley method \method{Rectangle}
190  \begin{python}  \begin{python}
191  import finley    from esys.finley import Rectangle
192  mydomain = finley.Rectangle(l0=1.,l1=1.,n0=40, n1=20)    mydomain = Rectangle(l0=1.,l1=1.,n0=40, n1=20)
193  \end{python}  \end{python}
194  In this case the domain is a rectangle with the lower, left corner at point $(0,0)$ and  In this case the domain is a rectangle with the lower, left corner at point $(0,0)$ and
195  the right, upper corner at $(\var{l0},\var{l1})=(1,1)$.  the right, upper corner at $(\var{l0},\var{l1})=(1,1)$.
196  The arguments \var{l0} and \var{l1} define the number of elements in $x\hackscore{0}$ and  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  $x\hackscore{1}$-direction respectively. For more details on \method{Rectangle} and
198  other \Domain generators within the \finley module,  other \Domain generators within the \finley module,
199  see \Chap{CHAPTER ON FINLEY}.  see \Chap{CHAPTER ON FINLEY}.
200
201  The following statements define the \Poisson object \var{mypde} with domain var{mydomain} and  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 $1$:  the right hand side $f$ of the PDE to constant $1$:
203  \begin{python}  \begin{python}
204  import escript    from esys.escript.linearPDEs import Poisson
205  mypde = escript.Poisson(domain=mydomain,f=1)    mypde = Poisson(mydomain)
206      mypde.setValue(f=1)
207  \end{python}  \end{python}
208  We have not specified any boundary condition but the  We have not specified any boundary condition but the
209  \Poisson class implicitly assumes homogeneous Neuman boundary conditions \index{Neumann  \Poisson class implicitly assumes homogeneous Neuman boundary conditions \index{Neumann
# Line 173  boundary condition!homogeneous} defined Line 211  boundary condition!homogeneous} defined
211  condition the BVP\index{boundary value problem!BVP} we have defined has no unique solution. In fact, with any solution $u$  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  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  a Dirichlet boundary condition \index{Dirichlet boundary condition}. This is done
214  by defines a characteristic function \index{characteristic function}  by defining a characteristic function \index{characteristic function}
215  which has a positive values at locations $(x_0,x_1)$ where Dirichlet boundary condition is set  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},  and $0$ elsewhere. In our case of $\Gamma^D$ defined by \eqn{eq:FirstSteps.2c},
217  we need a function which is positive for the cases $x_0=0$ or $x_1=0$:    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}  \begin{python}
220  x=mydomain.getX()    x=mydomain.getX()
221  \end{python}  \end{python}
222  In first statement returns, the method \method{getX} of the \Domain \var{mydomain} access to the locations  The method \method{getX} of the \Domain \var{mydomain}
223  in the domain defined by \var{mydomain}. The object \var{x} is actually an \Data object  gives access to locations
224  which we will learn more about later. \code{x} returns the $x_0$ coordinates of the locations and  in the domain defined by \var{mydomain}. The object \var{x} is actually a \Data object which will be
225  \code{x.whereZero()} creates function which equals $1$ where \code{x} is (nearly) equal to zero  discussed in \Chap{ESCRIPT CHAP} in more detail. What we need to know here is that
and $0$ elsewhere. The sum of the results of \code{x.whereZero()} and \code{x.whereZero()} gives a function on the domain \var{mydomain} which is exactly positive where $x_0$ or $x_1$ is equal to zero.
226
227  The additional parameter \var{q} of the \Poisson object creater defines the  \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
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
266    \end{python}
267
268    \code{whereZero(x)} creates function which equals $1$ where \code{x} is (almost) equal to zero
269    and $0$ elsewhere.
270    Similarly, \code{whereZero(x)} creates function which equals $1$ where \code{x} is
271    equal to zero and $0$ elsewhere.
272    The sum of the results of \code{whereZero(x)} and \code{whereZero(x)}
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}
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  characteristic function \index{characteristic function} of the locations
284  of the domain where homogeneous Dirichlet boundary condition \index{Dirichlet boundary condition!homogeneous}  of the domain where homogeneous Dirichlet boundary condition \index{Dirichlet boundary condition!homogeneous}
285  are set. The complete definition of our example is now:  are set. The complete definition of our example is now:
286  \begin{python}  \begin{python}
287  from linearPDEs import Poisson    from esys.linearPDEs import Poisson
288  x = mydomain.getX()    x = mydomain.getX()
290  mypde = Poisson(domain=mydomain,f=1,q=gammaD)    mypde = Poisson(domain=mydomain)
292  \end{python}  \end{python}
293  The first statement imports the \Poisson class definition form the \linearPDEsPack module which is part of the \escript module.  The first statement imports the \Poisson class definition from the \linearPDEs module \escript package.
294  To get the solution of the Poisson equation defines by \var{mypde} we just have to call its  To get the solution of the Poisson equation defined by \var{mypde} we just have to call its
295  \method{getSolution}.  \method{getSolution}.
296
297  Now we can write the script to solve our test problem (Remember that  Now we can write the script to solve our Poisson problem
298  lines starting with '\#' are commend lines in Python) (available as \file{mypoisson.py}  \begin{python}
299  in the \ExampleDirectory):    from esys.escript import *
300  \begin{python}    from esys.escript.linearPDEs import Poisson
301  import esys.finley    from esys.finley import Rectangle
302  from esys.linearPDEs import Poisson    # generate domain:
303  # generate domain:    mydomain = Rectangle(l0=1.,l1=1.,n0=40, n1=20)
304  mydomain = esys.finley.Rectangle(l0=1.,l1=1.,n0=40, n1=20)    # define characteristic function of Gamma^D
305  # define characteristic function of Gamma^D    x = mydomain.getX()
306  x = mydomain.getX()    gammaD = whereZero(x)+whereZero(x)
307  gammaD = x.whereZero()+x.whereZero()    # define PDE and get its solution u
308  # define PDE and get its solution u    mypde = Poisson(domain=mydomain)
310  u = mypde.getSolution()    u = mypde.getSolution()
311  # write u to an external file    # write u to an external file
312  u.saveDX("u.dx")    saveVTK("u.xml",sol=u)
313  \end{python}  \end{python}
314  The last statement writes the solution to the external file \file{u.dx} in  The entire code is available as \file{poisson.py} in the \ExampleDirectory
315  \OpenDX file format. \OpenDX is a software package
316  for the visualization of scientific, engineering and analytical data and is freely available  The last statement writes the solution (tagged with the name "sol") to a file named \file{u.xml} in
317  from \url{http://www.opendx.org}.  \VTK file format.
318    Now you may run the script using the \escript environment
319    and visualize the solution using \mayavi:
320    \begin{verbatim}
321      escript poisson.py
322      mayavi -d u.xml -m SurfaceMap
323    \end{verbatim}
324    See \fig{fig:FirstSteps.3}.
325
326  \begin{figure}  \begin{figure}
327  \centerline{\includegraphics[width=\figwidth]{FirstStepResult.eps}}  \centerline{\includegraphics[width=\figwidth]{figures/FirstStepResult}}
328  \caption{\OpenDX visualization of the Possion equation soluition for $f=1$}  \caption{Visualization of the Poisson Equation Solution for $f=1$}
329  \label{fig:FirstSteps.3}  \label{fig:FirstSteps.3}
330  \end{figure}  \end{figure}
331
You can edit this script using your favourite text editor (or the Integrated DeveLopment Environment IDLE
for Python). If the script file has the name \file{mypoisson.py} \index{scripts!\file{mypoisson.py}} you can run the
script from any shell using the command:
\begin{verbatim}
python mypoisson.py
\end{verbatim}
After the script has (hopefully successfully) been completed you will find the file \file{u.dx} in the current
directory. An easy way to visualize the results is the command
\begin{verbatim}
dx -prompter
\end{verbatim}
to start the generic data visualization interface of \OpenDX. \fig{fig:FirstSteps.3} shows the result.

Legend:
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