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Transport class uses now useBackwardEuler rather than the mysterious theta.
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 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
15 \section{The First Steps}
16 \label{FirstSteps}
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.
24 The PDE \index{partial differential equation} we wish to solve i s 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}
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.
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~$u$.
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}
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
157 \Ref{Zienc,NumHand}.
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}.
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
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 non-zero 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}
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}.
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
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}.
303 \begin{figure}
304 \centerline{\includegraphics[width=\figwidth]{figures/FirstStepResult}}
305 \caption{Visualization of the Poisson Equation Solution for $f=1$}
306 \label{fig:FirstSteps.3}
307 \end{figure}


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