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1
2 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
3 %
4 % Copyright (c) 2003-2008 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/osl-3.0.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}}
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 bi-linear
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 sub-domains 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 \Ref{Zienc,NumHand}.
157
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 favourite 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
162 of threads.}:
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
170 Type "help", "copyright", "credits" or "license" for more information.
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 familar with python.
180
181 \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}
183 that can be defined through the \LinearPDE class later). The instantiation of
184 a \Poisson class object requires the specification of the domain $\Omega$. In \escript
185 the \Domain class objects are used to describe the geometry of a domain but it also
186 contains information about the discretization methods and the actual solver which is used
187 to solve the PDE. Here we are using the FEM library \finley \index{finite element
188 method}. The following statements create the \Domain object \var{mydomain} from the
189 \finley method \method{Rectangle}
190 \begin{python}
191 from esys.finley import Rectangle
192 mydomain = Rectangle(l0=1.,l1=1.,n0=40, n1=20)
193 \end{python}
194 In this case the domain is a rectangle with the lower, left corner at point $(0,0)$ and
195 the right, upper corner at $(\var{l0},\var{l1})=(1,1)$.
196 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
198 other \Domain generators within the \finley module,
199 see \Chap{CHAPTER ON FINLEY}.
200
201 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 $1$:
203 \begin{python}
204 from esys.escript.linearPDEs import Poisson
205 mypde = Poisson(mydomain)
206 mypde.setValue(f=1)
207 \end{python}
208 We have not specified any boundary condition but the
209 \Poisson class implicitly assumes homogeneous Neuman boundary conditions \index{Neumann
210 boundary condition!homogeneous} defined by \eqn{eq:FirstSteps.2}. With this boundary
211 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
213 a Dirichlet boundary condition \index{Dirichlet boundary condition}. This is done
214 by defining a characteristic function \index{characteristic function}
215 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},
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
218 an object \var{x} which contains the coordinates of the nodes in the domain use
219 \begin{python}
220 x=mydomain.getX()
221 \end{python}
222 The method \method{getX} of the \Domain \var{mydomain}
223 gives access to locations
224 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 detail. What we need to know here is that
226
227 \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[0]
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
265 gammaD=whereZero(x[0])+whereZero(x[1])
266 \end{python}
267
268 \code{whereZero(x[0])} creates function which equals $1$ where \code{x[0]} is (almost) equal to zero
269 and $0$ elsewhere.
270 Similarly, \code{whereZero(x[1])} creates function which equals $1$ where \code{x[1]} is
271 equal to zero and $0$ elsewhere.
272 The sum of the results of \code{whereZero(x[0])} and \code{whereZero(x[1])}
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}
276 print gammaD.getRank(),gammaD.getShape(),gammaD.getFunctionSpace()
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
284 of the domain where homogeneous Dirichlet boundary condition \index{Dirichlet boundary condition!homogeneous}
285 are set. The complete definition of our example is now:
286 \begin{python}
287 from esys.linearPDEs import Poisson
288 x = mydomain.getX()
289 gammaD = whereZero(x[0])+whereZero(x[1])
290 mypde = Poisson(domain=mydomain)
291 mypde.setValue(f=1,q=gammaD)
292 \end{python}
293 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
295 \method{getSolution}.
296
297 Now we can write the script to solve our Poisson problem
298 \begin{python}
299 from esys.escript import *
300 from esys.escript.linearPDEs import Poisson
301 from esys.finley import Rectangle
302 # generate domain:
303 mydomain = Rectangle(l0=1.,l1=1.,n0=40, n1=20)
304 # define characteristic function of Gamma^D
305 x = mydomain.getX()
306 gammaD = whereZero(x[0])+whereZero(x[1])
307 # define PDE and get its solution u
308 mypde = Poisson(domain=mydomain)
309 mypde.setValue(f=1,q=gammaD)
310 u = mypde.getSolution()
311 # write u to an external file
312 saveVTK("u.xml",sol=u)
313 \end{python}
314 The entire code is available as \file{poisson.py} in the \ExampleDirectory
315
316 The last statement writes the solution (tagged with the name "sol") to a file named \file{u.xml} in
317 \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}
327 \centerline{\includegraphics[width=\figwidth]{figures/FirstStepResult}}
328 \caption{Visualization of the Poisson Equation Solution for $f=1$}
329 \label{fig:FirstSteps.3}
330 \end{figure}
331

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