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

revision 993 by gross, Fri Feb 23 06:39:38 2007 UTC revision 2548 by jfenwick, Mon Jul 20 06:20:06 2009 UTC
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1  % $Id$
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  %  %
9  %               \url{http://www.access.edu.au  % Licensed under the Open Software License version 3.0
11  %  %
12    %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
13
14
15  \section{The First Steps}  \section{The First Steps}
16  \label{FirstSteps}  \label{FirstSteps}
17
\begin{figure}
\centerline{\includegraphics[width=\figwidth]{figures/FirstStepDomain}}
\caption{Domain $\Omega=[0,1]^2$ with outer normal field $n$.}
\label{fig:FirstSteps.1}
\end{figure}
18
19  In this chapter we will give an introduction how to use \escript to solve
20  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}  In this chapter we give an introduction how to use \escript to solve
21  is sufficient. It is helpful if the reader has some basic knowledge of PDEs \index{partial differential equation}.  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}  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 function $f$ is the given right hand side. The domain of interest, denoted 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 \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 $u$ 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*}
# Line 54  and \eqn{eq:FirstSteps.1} as Line 59  and \eqn{eq:FirstSteps.1} as
59  \end{equation*}  \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  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 the Einstein summation convention \index{summation convention}. This  convenient Einstein summation convention \index{summation convention}. This
71  drops the $\sum$ sign and assumes that 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}   \\
# Line 93  are applying the Einstein summation conv Line 98  are applying the Einstein summation conv
98  $n\hackscore{i} u\hackscore{,i}= n\hackscore{0} u\hackscore{,0} +$n\hackscore{i} u\hackscore{,i}= n\hackscore{0} u\hackscore{,0} +
99  n\hackscore{1} u\hackscore{,1}$. n\hackscore{1} u\hackscore{,1}$.
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 120  with the Neumann boundary condition \eqn Line 125  with the Neumann boundary condition \eqn
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  problem\index{boundary value problem} (BVP\index{boundary value problem!BVP}) for
128  the unknown  the unknown function~$u$.
function $u$.
129
130
131  \begin{figure}  \begin{figure}[ht]
132  \centerline{\includegraphics[width=\figwidth]{figures/FirstStepMesh.eps}}  \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 149  elements is called a mesh\index{finite e Line 153  elements is called a mesh\index{finite e
153  method!mesh}. \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 162  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  from esys.finley import Rectangle    from esys.finley import Rectangle
192  mydomain = 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 $(\var{l0},\var{l1})=(1,1)$.
# Line 175  see \Chap{CHAPTER ON FINLEY}. Line 201  see \Chap{CHAPTER ON FINLEY}.
201  The following statements define the \Poisson class 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  from esys.escript.linearPDEs import Poisson    from esys.escript.linearPDEs import Poisson
205  mypde = Poisson(mydomain)    mypde = Poisson(mydomain)
206  mypde.setValue(f=1)    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 189  by defining a characteristic function \i Line 215  by defining a characteristic function \i
215  which has positive values at locations $x=(x\hackscore{0},x\hackscore{1})$ 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 to construct a function \var{gammaD} which is positive for the cases $x\hackscore{0}=0$ or $x\hackscore{1}=0$. To get  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 represents locations in the domain one uses  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  The method \method{getX} of the \Domain \var{mydomain}  The method \method{getX} of the \Domain \var{mydomain}
224  in the domain defined by \var{mydomain}. The object \var{x} is actually a \Data object which is  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 details. What we need to know here is that  discussed in \Chap{ESCRIPT CHAP} in more detail. What we need to know here is that
226
227  \var{x} has \Rank (=number of dimensions) and a \Shape (=tuple of dimensions) which can be checked by  \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:  calling the \method{getRank} and \method{getShape} methods:
229  \begin{python}  \begin{python}
230  print "rank ",x.getRank(),", shape ",x.getShape()    print "rank ",x.getRank(),", shape ",x.getShape()
231  \end{python}  \end{python}
232  will print something like  This will print something like
233  \begin{python}  \begin{python}
234  rank 1, shape (2,)    rank 1, shape (2,)
235  \end{python}  \end{python}
236  The \Data object also maintains type information which is represented by the  The \Data object also maintains type information which is represented by the
237  \FunctionSpace of the object. For instance  \FunctionSpace of the object. For instance
238  \begin{python}  \begin{python}
239  print x.getFunctionSpace()    print x.getFunctionSpace()
240  \end{python}  \end{python}
241  will print  will print
242  \begin{python}  \begin{python}
243  Function space type: Finley_Nodes on FinleyMesh    Function space type: Finley_Nodes on FinleyMesh
244  \end{python}  \end{python}
245  which tells us that the coordinates are stored on the nodes of a \finley mesh.  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  To get the  $x\hackscore{0}$ coordinates of the locations we use the
247  statement  statement
248  \begin{python}  \begin{python}
249  x0=x    x0=x
250  \end{python}  \end{python}
251  Object \var{x0}  Object \var{x0}
252  is again a \Data object now with \Rank $0$ and  is again a \Data object now with \Rank $0$ and
253  \Shape $()$. It inherits the \FunctionSpace from \var{x}:  \Shape $()$. It inherits the \FunctionSpace from \var{x}:
254  \begin{python}  \begin{python}
255  print x0.getRank(),x0.getShape(),x0.getFunctionSpace()    print x0.getRank(),x0.getShape(),x0.getFunctionSpace()
256  \end{python}  \end{python}
257  will print  will print
258  \begin{python}  \begin{python}
259  0 () Function space type: Finley_Nodes on FinleyMesh    0 () Function space type: Finley_Nodes on FinleyMesh
260  \end{python}  \end{python}
261  We can now construct the function \var{gammaD} by  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}  \begin{python}
264  from esys.escript import whereZero    from esys.escript import whereZero
266  \end{python}  \end{python}
267  where
268  \code{whereZero(x)} creates function which equals $1$ where \code{x} is (allmost) equal to zero  \code{whereZero(x)} creates function which equals $1$ where \code{x} is (almost) equal to zero
269  and $0$ elsewhere.  and $0$ elsewhere.
270  Similarly, \code{whereZero(x)} creates function which equals $1$ where \code{x} is  Similarly, \code{whereZero(x)} creates function which equals $1$ where \code{x} is
271  equal to zero and $0$ elsewhere.  equal to zero and $0$ elsewhere.
272  The sum of the results of \code{whereZero(x)} and \code{whereZero(x)}  The sum of the results of \code{whereZero(x)} and \code{whereZero(x)}
273  gives a function on the domain \var{mydomain} which is exactly positive where $x\hackscore{0}$ or $x\hackscore{1}$ is equal to zero.  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  Note that \var{gammaD} has the same \Rank, \Shape and \FunctionSpace like \var{x0} used to define it. So from
275  \begin{python}  \begin{python}
277  \end{python}  \end{python}
278  one gets  one gets
279  \begin{python}  \begin{python}
280  0 () Function space type: Finley_Nodes on FinleyMesh    0 () Function space type: Finley_Nodes on FinleyMesh
281  \end{python}  \end{python}
282  The additional parameter \var{q} of the \code{setValue} method of the \Poisson class defines the  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 esys.linearPDEs import Poisson    from esys.linearPDEs import Poisson
288  x = mydomain.getX()    x = mydomain.getX()
290  mypde = Poisson(domain=mydomain)    mypde = Poisson(domain=mydomain)
292  \end{python}  \end{python}
293  The first statement imports the \Poisson class definition from the \linearPDEs module \escript package.  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  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 comment lines in Python) (available as \file{poisson.py}  \begin{python}
299  in the \ExampleDirectory):    from esys.escript import *
300  \begin{python}    from esys.escript.linearPDEs import Poisson
301  from esys.escript import *    from esys.finley import Rectangle
302  from esys.escript.linearPDEs import Poisson    # generate domain:
303  from esys.finley import Rectangle    mydomain = Rectangle(l0=1.,l1=1.,n0=40, n1=20)
304  # generate domain:    # define characteristic function of Gamma^D
305  mydomain = Rectangle(l0=1.,l1=1.,n0=40, n1=20)    x = mydomain.getX()
306  # define characteristic function of Gamma^D    gammaD = whereZero(x)+whereZero(x)
307  x = mydomain.getX()    # define PDE and get its solution u
308  gammaD = whereZero(x)+whereZero(x)    mypde = Poisson(domain=mydomain)
309  # define PDE and get its solution u    mypde.setValue(f=1,q=gammaD)
310  mypde = Poisson(domain=mydomain)    u = mypde.getSolution()
311  mypde.setValue(f=1,q=gammaD)    # write u to an external file
312  u = mypde.getSolution()    saveVTK("u.xml",sol=u)
313  # write u to an external file  \end{python}
314  saveVTK("u.xml",sol=u)  The entire code is available as \file{poisson.py} in the \ExampleDirectory
315  \end{python}
316  The last statement writes the solution tagged with the name "sol" to the external file \file{u.xml} in  The last statement writes the solution (tagged with the name "sol") to a file named \file{u.xml} in
317  \VTK file format. \VTK is a software library  \VTK file format.
318  for the visualization of scientific, engineering and analytical data and is freely available  Now you may run the script using the \escript environment
319  from \url{http://www.vtk.org}. There are a variety of graphical user interfaces  and visualize the solution using \mayavi:
320  for \VTK available, for instance \mayavi which can be downloaded from \url{http://mayavi.sourceforge.net/} but is also available on most  \begin{verbatim}
321  \LINUX distributions.    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]{figures/FirstStepResult.eps}}  \centerline{\includegraphics[width=\figwidth]{figures/FirstStepResult}}
328  \caption{Visualization of the Poisson Equation Solution 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 the script file using your favourite text editor (or the Integrated DeveLopment Environment IDLE
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
script from any shell using the command:
\begin{python}
python poisson.py
\end{python}
After the script has (hopefully successfully) been completed you will find the file \file{u.xml} in the current
directory. An easy way to visualize the results is the command
\begin{python}
mayavi -d u.xml -m SurfaceMap &
\end{python}
to show the results, see \fig{fig:FirstSteps.3}.

\subsection{Visualization with \pyvisi}