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

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 1 % $Id$ 2 3 \chapter{The First Steps} 4 \label{FirstSteps} 5 6 \begin{figure} 7 \centerline{\includegraphics[width=\figwidth]{FirstStepDomain}} 8 \caption{Domain $\Omega=[0,1]^2$ with outer normal field $n$.} 9 \label{fig:FirstSteps.1} 10 \end{figure} 11 12 In this chapter we will show the basics of how to use \escript to solve 13 a partial differential equation \index{partial differential equation} (PDE \index{partial differential equation!PDE}). The reader should be familiar with 14 the basics of Python. The knowledge presented the Python tutorial at \url{http://docs.python.org/tut/tut.html} 15 is sufficient. It is helpful if the reader has some basic knowledge on PDEs \index{PDE}. 16 17 The \index{PDE} we want to solve is the Poisson equation \index{Poisson equation} 18 \begin{equation} 19 -\Delta u =f 20 \label{eq:FirstSteps.1} 21 \end{equation} 22 for the solution $u$. The domain of interest which we denote by $\Omega$ 23 is the unit square 24 \begin{equation} 25 \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 \} 26 \label{eq:FirstSteps.1b} 27 \end{equation} 28 The domain is shown in Figure~\fig{fig:FirstSteps.1}. 29 30 $\Delta$ denotes the Laplace operator\index{Laplace operator} which is defined by 31 \begin{equation} 32 \Delta u = (u\hackscore {,0})\hackscore{,0}+(u\hackscore{,1})\hackscore{,1} 33 \label{eq:FirstSteps.1.1} 34 \end{equation} 35 where for any function $w$ and any direction $i$ $u\hackscore{,i}$ 36 denotes the partial derivative \index{partial derivative} of $w$ with respect to $i$. 37 \footnote{Some readers 38 may be more familiar with the Laplace operator\index{Laplace operator} being written 39 as $\nabla^2$, and written in the form 40 \begin{equation*} 41 \nabla^2 u = \frac{\partial^2 u}{\partial x\hackscore 0^2} 42 + \frac{\partial^2 u}{\partial x\hackscore 1^2} 43 \end{equation*} 44 and \eqn{eq:FirstSteps.1} as 45 \begin{equation*} 46 -\nabla^2 u = f 47 \end{equation*} 48 } 49 Basically in the subindex of a function any index left to the comma denotes a spatial derivative with respect 50 to the index. To get a more compact form we will write $w\hackscore{,ij}=(w\hackscore {,i})\hackscore{,j}$ 51 which leads to 52 \begin{equation} 53 \Delta u = u\hackscore{,00}+u\hackscore{,11}=\sum\hackscore{i=0}^2 u\hackscore{,ii} 54 \label{eq:FirstSteps.1.1b} 55 \end{equation} 56 In some cases, and we will see examples for this in the next chapter, 57 the usage of the nested $\sum$ symbols blows up the formulas and therefore 58 it is convenient to use Einstein summation convention \index{summation convention} which 59 says that $\sum$ sign is dropped and a summation over a repeated index is performed 60 ("repeated index means summation"). For instance we write 61 \begin{eqnarray} 62 x\hackscore{i}y\hackscore{i}=\sum\hackscore{i=0}^2 x\hackscore{i}y\hackscore{i} \\ 63 x\hackscore{i}u\hackscore{,i}=\sum\hackscore{i=0}^2 x\hackscore{i}u\hackscore{,i} \\ 64 u\hackscore{,ii}=\sum\hackscore{i=0}^2 u\hackscore{,ii} \\ 65 \label{eq:FirstSteps.1.1c} 66 \end{eqnarray} 67 With the summation convention we can write the Poisson equation \index{Poisson equation} as 68 \begin{equation} 69 - u\hackscore{,ii} =1 70 \label{eq:FirstSteps.1.sum} 71 \end{equation} 72 On the boundary of the domain $\Omega$ the normal derivative $n\hackscore{i} u\hackscore{,i}$ 73 of the solution $u$ shall be zero, ie. $u$ shall fulfill 74 the homogeneous Neumann boundary condition\index{Neumann 75 boundary condition!homogeneous} 76 \begin{equation} 77 n\hackscore{i} u\hackscore{,i}= 0 \;. 78 \label{eq:FirstSteps.2} 79 \end{equation} 80 $n=(n\hackscore{i})$ denotes the outer normal field 81 of the domain, see \fig{fig:FirstSteps.1}. Remember that we 82 are applying the Einstein summation convention \index{summation convention}, i.e 83 $n\hackscore{i} u\hackscore{,i}= n\hackscore1 u\hackscore{,1} + 84 n\hackscore2 u\hackscore{,2}$. 85 \footnote{Some readers may familiar with the notation 86 \begin{equation*} 87 \frac{\partial u}{\partial n} = n\hackscore{i} u\hackscore{,i} 88 \end{equation*} 89 for the normal derivative.} 90 The Neumann boundary condition of \eqn{eq:FirstSteps.2} should be fulfilled on the 91 set $\Gamma^N$ which is the top and right edge of the domain: 92 \begin{equation} 93 \Gamma^N=\{(x\hackscore 0;x\hackscore 1) \in \Omega | x\hackscore{0}=1 \mbox{ or } x\hackscore{1}=1 \} 94 \label{eq:FirstSteps.2b} 95 \end{equation} 96 On the bottom and the left edge of the domain which defined 97 as 98 \begin{equation} 99 \Gamma^D=\{(x\hackscore 0;x\hackscore 1) \in \Omega | x\hackscore{0}=0 \mbox{ or } x\hackscore{1}=0 \} 100 \label{eq:FirstSteps.2c} 101 \end{equation} 102 the solution shall be identically zero: 103 \begin{equation} 104 u=0 \; . 105 \label{eq:FirstSteps.2d} 106 \end{equation} 107 The kind of boundary condition is called a homogeneous Dirichlet boundary condition 108 \index{Dirichlet boundary condition!homogeneous}. The partial differential equation in \eqn{eq:FirstSteps.1.sum} together 109 with the Neumann boundary condition \eqn{eq:FirstSteps.2} and 110 Dirichlet boundary condition in \eqn{eq:FirstSteps.2d} form a so 111 called boundary value 112 problem\index{boundary value problem} (BVP\index{boundary value problem!BVP}) for unknown 113 function $u$. 114 115 116 \begin{figure} 117 \centerline{\includegraphics[width=\figwidth]{FirstStepMesh}} 118 \caption{Mesh of $4 \time 4$ elements on a rectangular domain. Here 119 each element is a quadrilateral and described by four nodes, namely 120 the corner points. The solution is interpolated by a bi-linear 121 polynomial.} 122 \label{fig:FirstSteps.2} 123 \end{figure} 124 125 In general the BVP\index{boundary value problem!BVP} cannot be solved analytically and numerical 126 methods have to be used construct an approximation of the solution 127 $u$. Here we will use the finite element method\index{finite element 128 method} (FEM\index{finite element 129 method!FEM}). The basic idea is to fill the domain with a 130 set of points, so called nodes. The solution is approximated by its 131 values on the nodes\index{finite element 132 method!nodes}. Moreover, the domain is subdivide into small 133 sub-domain, so-called elements \index{finite element 134 method!element}. On each element the solution is 135 represented by a polynomial of a certain degree through its values at 136 the nodes located in the element. The nodes and its connection through 137 elements is called a mesh\index{finite element 138 method!mesh}. Figure~\fig{fig:FirstSteps.2} shows an 139 example of a FEM mesh with four elements in the $x_0$ and four elements 140 in the $x_1$ direction over the unit square. 141 For more details we refer the reader to the literature, for instance 142 \Ref{Zienc,NumHand}. 143 144 \escript provides the class \Poisson to define a Poisson equation \index{Poisson equation}. 145 (We will discuss a more general form of a PDE \index{partial differential equation!PDE} 146 that can be defined through the \LinearPDE class later). The instantiation of 147 a \Poisson class object requires the specification of the domain $\Omega$. In \escript 148 the \Domain class objects are used to describe the geometry of a domain but it also 149 contains information about the discretization methods and the actual solver which is used 150 to solve the PDE. Here we are using the FEM library \finley \index{finite element 151 method}. The following statements create the \Domain object \var{mydomain} from the 152 \finley method \method{Rectangle} 153 \begin{python} 154 import finley 155 mydomain = finley.Rectangle(l0=1.,l1=1.,n0=40, n1=20) 156 \end{python} 157 In this case the domain is a rectangle with the lower, left corner at point $(0,0)$ and 158 the right, upper corner at $(\var{l0},\var{l1})=(1,1)$. 159 The arguments \var{l0} and \var{l1} define the number of elements in $x\hackscore{0}$ and 160 $x\hackscore{1}$-direction respectively. For more details on \method{Rectangle} and 161 other \Domain generators within the \finley module, 162 see \Chap{CHAPTER ON FINLEY}. 163 164 The following statements define the \Poisson object \var{mypde} with domain var{mydomain} and 165 the right hand side $f$ of the PDE to constant $1$: 166 \begin{python} 167 import escript 168 mypde = escript.Poisson(domain=mydomain,f=1) 169 \end{python} 170 We have not specified any boundary condition but the 171 \Poisson class implicitly assumes homogeneous Neuman boundary conditions \index{Neumann 172 boundary condition!homogeneous} defined by \eqn{eq:FirstSteps.2}. With this boundary 173 condition the BVP\index{boundary value problem!BVP} we have defined has no unique solution. In fact, with any solution $u$ 174 and any constant $C$ the function $u+C$ becomes a solution as well. We have to add 175 a Dirichlet boundary condition \index{Dirichlet boundary condition}. This is done 176 by defines a characteristic function \index{characteristic function} 177 which has a positive values at locations $(x_0,x_1)$ where Dirichlet boundary condition is set 178 and $0$ elsewhere. In our case of $\Gamma^D$ defined by \eqn{eq:FirstSteps.2c}, 179 we need a function which is positive for the cases $x_0=0$ or $x_1=0$: 180 \begin{python} 181 x=mydomain.getX() 182 gammaD=x[0].whereZero()+x[1].whereZero() 183 \end{python} 184 In first statement returns, the method \method{getX} of the \Domain \var{mydomain} access to the locations 185 in the domain defined by \var{mydomain}. The object \var{x} is actually an \Data object 186 which we will learn more about later. \code{x[0]} returns the $x_0$ coordinates of the locations and 187 \code{x[0].whereZero()} creates function which equals $1$ where \code{x[0]} is (nearly) equal to zero 188 and $0$ elsewhere. The sum of the results of \code{x[0].whereZero()} and \code{x[1].whereZero()} gives a function on the domain \var{mydomain} which is exactly positive where $x_0$ or $x_1$ is equal to zero. 189 190 The additional parameter \var{q} of the \Poisson object creater defines the 191 characteristic function \index{characteristic function} of the locations 192 of the domain where homogeneous Dirichlet boundary condition \index{Dirichlet boundary condition!homogeneous} 193 are set. The complete definition of our example is now: 194 \begin{python} 195 from linearPDEs import Poisson 196 x = mydomain.getX() 197 gammaD = x[0].whereZero()+x[1].whereZero() 198 mypde = Poisson(domain=mydomain,f=1,q=gammaD) 199 \end{python} 200 The first statement imports the \Poisson class definition form the \linearPDEsPack module which is part of the \escript module. 201 To get the solution of the Poisson equation defines by \var{mypde} we just have to call its 202 \method{getSolution}. 203 204 Now we can write the script to solve our test problem (Remember that 205 lines starting with '\#' are commend lines in Python) (available as \file{mypoisson.py} 206 in the \ExampleDirectory): 207 \begin{python} 208 import esys.finley 209 from esys.linearPDEs import Poisson 210 # generate domain: 211 mydomain = esys.finley.Rectangle(l0=1.,l1=1.,n0=40, n1=20) 212 # define characteristic function of Gamma^D 213 x = mydomain.getX() 214 gammaD = x[0].whereZero()+x[1].whereZero() 215 # define PDE and get its solution u 216 mypde = Poisson(domain=mydomain,f=1,q=gammaD) 217 u = mypde.getSolution() 218 # write u to an external file 219 u.saveDX("u.dx") 220 \end{python} 221 The last statement writes the solution to the external file \file{u.dx} in 222 \OpenDX file format. \OpenDX is a software package 223 for the visualization of scientific, engineering and analytical data and is freely available 224 from \url{http://www.opendx.org}. 225 226 \begin{figure} 227 \centerline{\includegraphics[width=\figwidth]{FirstStepResult.eps}} 228 \caption{\OpenDX visualization of the Possion equation soluition for $f=1$} 229 \label{fig:FirstSteps.3} 230 \end{figure} 231 232 You can edit this script using your favourite text editor (or the Integrated DeveLopment Environment IDLE 233 for Python). If the script file has the name \file{mypoisson.py} \index{scripts!\file{mypoisson.py}} you can run the 234 script from any shell using the command: 235 \begin{verbatim} 236 python mypoisson.py 237 \end{verbatim} 238 After the script has (hopefully successfully) been completed you will find the file \file{u.dx} in the current 239 directory. An easy way to visualize the results is the command 240 \begin{verbatim} 241 dx -prompter 242 \end{verbatim} 243 to start the generic data visualization interface of \OpenDX. \fig{fig:FirstSteps.3} shows the result. 244 245 246 247

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