1 

2 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 
3 
% 
4 
% Copyright (c) 20032008 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/osl3.0.php 
11 
% 
12 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 
13 

14 

15 
\section{The First Steps} 
16 
\label{FirstSteps} 
17 

18 

19 

20 
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} 
26 
\Delta u =f 
27 
\label{eq:FirstSteps.1} 
28 
\end{equation} 
29 
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 
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 
The domain is shown in \fig{fig:FirstSteps.1}. 
36 
\begin{figure} [h!] 
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 
43 
\begin{equation} 
44 
\Delta u = (u\hackscore {,0})\hackscore{,0}+(u\hackscore{,1})\hackscore{,1} 
45 
\label{eq:FirstSteps.1.1} 
46 
\end{equation} 
47 
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 $i$. 
49 
\footnote{You 
50 
may be more familiar with the Laplace operator\index{Laplace operator} being written 
51 
as $\nabla^2$, and written in the form 
52 
\begin{equation*} 
53 
\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} 
55 
\end{equation*} 
56 
and \eqn{eq:FirstSteps.1} as 
57 
\begin{equation*} 
58 
\nabla^2 u = f 
59 
\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 
62 
to the index. To get a more compact form we will write $u\hackscore{,ij}=(u\hackscore {,i})\hackscore{,j}$ 
63 
which leads to 
64 
\begin{equation} 
65 
\Delta u = u\hackscore{,00}+u\hackscore{,11}=\sum\hackscore{i=0}^2 u\hackscore{,ii} 
66 
\label{eq:FirstSteps.1.1b} 
67 
\end{equation} 
68 
We often find that use 
69 
of nested $\sum$ symbols makes formulas cumbersome, and we use the more 
70 
convenient Einstein summation convention \index{summation convention}. This 
71 
drops the $\sum$ sign and assumes that a summation is performed over any repeated index. 
72 
For instance we write 
73 
\begin{eqnarray} 
74 
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} \\ 
76 
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} 
79 
\end{eqnarray} 
80 
With the summation convention we can write the Poisson equation \index{Poisson equation} as 
81 
\begin{equation} 
82 
 u\hackscore{,ii} =1 
83 
\label{eq:FirstSteps.1.sum} 
84 
\end{equation} 
85 
where $f=1$ in this example. 
86 

87 
On the boundary of the domain $\Omega$ the normal derivative $n\hackscore{i} u\hackscore{,i}$ 
88 
of the solution $u$ shall be zero, ie. $u$ shall fulfill 
89 
the homogeneous Neumann boundary condition\index{Neumann 
90 
boundary condition!homogeneous} 
91 
\begin{equation} 
92 
n\hackscore{i} u\hackscore{,i}= 0 \;. 
93 
\label{eq:FirstSteps.2} 
94 
\end{equation} 
95 
$n=(n\hackscore{i})$ denotes the outer normal field 
96 
of the domain, see \fig{fig:FirstSteps.1}. Remember that we 
97 
are applying the Einstein summation convention \index{summation convention}, i.e 
98 
$n\hackscore{i} u\hackscore{,i}= n\hackscore{0} u\hackscore{,0} + 
99 
n\hackscore{1} u\hackscore{,1}$. 
100 
\footnote{Some readers may familiar with the notation 
101 
$ 
102 
\frac{\partial u}{\partial n} = n\hackscore{i} u\hackscore{,i} 
103 
$ 
104 
for the normal derivative.} 
105 
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: 
107 
\begin{equation} 
108 
\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} 
110 
\end{equation} 
111 
On the bottom and the left edge of the domain which is defined 
112 
as 
113 
\begin{equation} 
114 
\Gamma^D=\{(x\hackscore 0;x\hackscore 1) \in \Omega  x\hackscore{0}=0 \mbox{ or } x\hackscore{1}=0 \} 
115 
\label{eq:FirstSteps.2c} 
116 
\end{equation} 
117 
the solution shall be identically zero: 
118 
\begin{equation} 
119 
u=0 \; . 
120 
\label{eq:FirstSteps.2d} 
121 
\end{equation} 
122 
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 
124 
with the Neumann boundary condition \eqn{eq:FirstSteps.2} and 
125 
Dirichlet boundary condition in \eqn{eq:FirstSteps.2d} form a so 
126 
called boundary value 
127 
problem\index{boundary value problem} (BVP\index{boundary value problem!BVP}) for 
128 
the unknown function~$u$. 
129 

130 

131 
\begin{figure}[h] 
132 
\centerline{\includegraphics[width=\figwidth]{figures/FirstStepMesh.eps}} 
133 
\caption{Mesh of $4 \time 4$ elements on a rectangular domain. Here 
134 
each element is a quadrilateral and described by four nodes, namely 
135 
the corner points. The solution is interpolated by a bilinear 
136 
polynomial.} 
137 
\label{fig:FirstSteps.2} 
138 
\end{figure} 
139 

140 
In general the BVP\index{boundary value problem!BVP} cannot be solved analytically and numerical 
141 
methods have to be used construct an approximation of the solution 
142 
$u$. Here we will use the finite element method\index{finite element 
143 
method} (FEM\index{finite element 
144 
method!FEM}). The basic idea is to fill the domain with a 
145 
set of points called nodes. The solution is approximated by its 
146 
values on the nodes\index{finite element 
147 
method!nodes}. Moreover, the domain is subdivided into smaller 
148 
subdomains called elements \index{finite element 
149 
method!element}. On each element the solution is 
150 
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 
152 
elements is called a mesh\index{finite element 
153 
method!mesh}. \fig{fig:FirstSteps.2} shows an 
154 
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. 
156 
For more details we refer the reader to the literature, for instance 
157 
\Ref{Zienc,NumHand}. 
158 

159 
\escript provides the class \Poisson to define a Poisson equation \index{Poisson equation}. 
160 
(We will discuss a more general form of a PDE \index{partial differential equation!PDE} 
161 
that can be defined through the \LinearPDE class later). The instantiation of 
162 
a \Poisson class object requires the specification of the domain $\Omega$. In \escript 
163 
the \Domain class objects are used to describe the geometry of a domain but it also 
164 
contains information about the discretization methods and the actual solver which is used 
165 
to solve the PDE. Here we are using the FEM library \finley \index{finite element 
166 
method}. The following statements create the \Domain object \var{mydomain} from the 
167 
\finley method \method{Rectangle} 
168 
\begin{python} 
169 
from esys.finley import Rectangle 
170 
mydomain = Rectangle(l0=1.,l1=1.,n0=40, n1=20) 
171 
\end{python} 
172 
In this case the domain is a rectangle with the lower, left corner at point $(0,0)$ and 
173 
the right, upper corner at $(\var{l0},\var{l1})=(1,1)$. 
174 
The arguments \var{n0} and \var{n1} define the number of elements in $x\hackscore{0}$ and 
175 
$x\hackscore{1}$direction respectively. For more details on \method{Rectangle} and 
176 
other \Domain generators within the \finley module, 
177 
see \Chap{CHAPTER ON FINLEY}. 
178 

179 
The following statements define the \Poisson class object \var{mypde} with domain \var{mydomain} and 
180 
the right hand side $f$ of the PDE to constant $1$: 
181 
\begin{python} 
182 
from esys.escript.linearPDEs import Poisson 
183 
mypde = Poisson(mydomain) 
184 
mypde.setValue(f=1) 
185 
\end{python} 
186 
We have not specified any boundary condition but the 
187 
\Poisson class implicitly assumes homogeneous Neuman boundary conditions \index{Neumann 
188 
boundary condition!homogeneous} defined by \eqn{eq:FirstSteps.2}. With this boundary 
189 
condition the BVP\index{boundary value problem!BVP} we have defined has no unique solution. In fact, with any solution $u$ 
190 
and any constant $C$ the function $u+C$ becomes a solution as well. We have to add 
191 
a Dirichlet boundary condition \index{Dirichlet boundary condition}. This is done 
192 
by defining a characteristic function \index{characteristic function} 
193 
which has positive values at locations $x=(x\hackscore{0},x\hackscore{1})$ where Dirichlet boundary condition is set 
194 
and $0$ elsewhere. In our case of $\Gamma^D$ defined by \eqn{eq:FirstSteps.2c}, 
195 
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 
196 
an object \var{x} which contains the coordinates of the nodes in the domain use 
197 
\begin{python} 
198 
x=mydomain.getX() 
199 
\end{python} 
200 
The method \method{getX} of the \Domain \var{mydomain} 
201 
gives access to locations 
202 
in the domain defined by \var{mydomain}. The object \var{x} is actually a \Data object which will be 
203 
discussed in \Chap{ESCRIPT CHAP} in more detail. What we need to know here is that 
204 

205 
\var{x} has \Rank (number of dimensions) and a \Shape (list of dimensions) which can be viewed by 
206 
calling the \method{getRank} and \method{getShape} methods: 
207 
\begin{python} 
208 
print "rank ",x.getRank(),", shape ",x.getShape() 
209 
\end{python} 
210 
This will print something like 
211 
\begin{python} 
212 
rank 1, shape (2,) 
213 
\end{python} 
214 
The \Data object also maintains type information which is represented by the 
215 
\FunctionSpace of the object. For instance 
216 
\begin{python} 
217 
print x.getFunctionSpace() 
218 
\end{python} 
219 
will print 
220 
\begin{python} 
221 
Function space type: Finley_Nodes on FinleyMesh 
222 
\end{python} 
223 
which tells us that the coordinates are stored on the nodes of (rather than on points in the interior of) a \finley mesh. 
224 
To get the $x\hackscore{0}$ coordinates of the locations we use the 
225 
statement 
226 
\begin{python} 
227 
x0=x[0] 
228 
\end{python} 
229 
Object \var{x0} 
230 
is again a \Data object now with \Rank $0$ and 
231 
\Shape $()$. It inherits the \FunctionSpace from \var{x}: 
232 
\begin{python} 
233 
print x0.getRank(),x0.getShape(),x0.getFunctionSpace() 
234 
\end{python} 
235 
will print 
236 
\begin{python} 
237 
0 () Function space type: Finley_Nodes on FinleyMesh 
238 
\end{python} 
239 
We can now construct a function \var{gammaD} which is only nonzero on the bottom and left edges 
240 
of the domain with 
241 
\begin{python} 
242 
from esys.escript import whereZero 
243 
gammaD=whereZero(x[0])+whereZero(x[1]) 
244 
\end{python} 
245 

246 
\code{whereZero(x[0])} creates function which equals $1$ where \code{x[0]} is (almost) equal to zero 
247 
and $0$ elsewhere. 
248 
Similarly, \code{whereZero(x[1])} creates function which equals $1$ where \code{x[1]} is 
249 
equal to zero and $0$ elsewhere. 
250 
The sum of the results of \code{whereZero(x[0])} and \code{whereZero(x[1])} 
251 
gives a function on the domain \var{mydomain} which is strictly positive where $x\hackscore{0}$ or $x\hackscore{1}$ is equal to zero. 
252 
Note that \var{gammaD} has the same \Rank, \Shape and \FunctionSpace like \var{x0} used to define it. So from 
253 
\begin{python} 
254 
print gammaD.getRank(),gammaD.getShape(),gammaD.getFunctionSpace() 
255 
\end{python} 
256 
one gets 
257 
\begin{python} 
258 
0 () Function space type: Finley_Nodes on FinleyMesh 
259 
\end{python} 
260 
An additional parameter \var{q} of the \code{setValue} method of the \Poisson class defines the 
261 
characteristic function \index{characteristic function} of the locations 
262 
of the domain where homogeneous Dirichlet boundary condition \index{Dirichlet boundary condition!homogeneous} 
263 
are set. The complete definition of our example is now: 
264 
\begin{python} 
265 
from esys.linearPDEs import Poisson 
266 
x = mydomain.getX() 
267 
gammaD = whereZero(x[0])+whereZero(x[1]) 
268 
mypde = Poisson(domain=mydomain) 
269 
mypde.setValue(f=1,q=gammaD) 
270 
\end{python} 
271 
The first statement imports the \Poisson class definition from the \linearPDEs module \escript package. 
272 
To get the solution of the Poisson equation defined by \var{mypde} we just have to call its 
273 
\method{getSolution}. 
274 

275 
Now we can write the script to solve our Poisson problem 
276 
\begin{python} 
277 
from esys.escript import * 
278 
from esys.escript.linearPDEs import Poisson 
279 
from esys.finley import Rectangle 
280 
# generate domain: 
281 
mydomain = Rectangle(l0=1.,l1=1.,n0=40, n1=20) 
282 
# define characteristic function of Gamma^D 
283 
x = mydomain.getX() 
284 
gammaD = whereZero(x[0])+whereZero(x[1]) 
285 
# define PDE and get its solution u 
286 
mypde = Poisson(domain=mydomain) 
287 
mypde.setValue(f=1,q=gammaD) 
288 
u = mypde.getSolution() 
289 
# write u to an external file 
290 
saveVTK("u.xml",sol=u) 
291 
\end{python} 
292 
The entire code is available as \file{poisson.py} in the \ExampleDirectory 
293 

294 
The last statement writes the solution (tagged with the name "sol") to a file named \file{u.xml} in 
295 
\VTK file format. 
296 
Now you may run the script and visualize the solution using \mayavi: 
297 
\begin{verbatim} 
298 
python poisson.py 
299 
mayavi d u.xml m SurfaceMap 
300 
\end{verbatim} 
301 
See \fig{fig:FirstSteps.3}. 
302 

303 
\begin{figure} 
304 
\centerline{\includegraphics[width=\figwidth]{figures/FirstStepResult.eps}} 
305 
\caption{Visualization of the Poisson Equation Solution for $f=1$} 
306 
\label{fig:FirstSteps.3} 
307 
\end{figure} 
308 
