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% $Id$ 
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\chapter{The First Steps} 
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\label{FirstSteps} 
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\begin{figure} 
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\centerline{\includegraphics[width=\figwidth]{FirstStepDomain}} 
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\caption{Domain $\Omega=[0,1]^2$ with outer normal field $n$.} 
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\label{fig:FirstSteps.1} 
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\end{figure} 
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In this chapter we will give an introduction how to use \escript to solve 
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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} 
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is sufficient. It is helpful if the reader has some basic knowledge of PDEs \index{partial differential equation}. 
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The PDE \index{partial differential equation} we wish to solve is the Poisson equation \index{Poisson equation} 
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\begin{equation} 
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\Delta u =f 
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\label{eq:FirstSteps.1} 
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\end{equation} 
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for the solution $u$. The function $f$ is the given right hand side. The domain of interest, denoted by $\Omega$ 
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is the unit square 
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\begin{equation} 
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\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 \} 
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\label{eq:FirstSteps.1b} 
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\end{equation} 
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The domain is shown in \fig{fig:FirstSteps.1}. 
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$\Delta$ denotes the Laplace operator\index{Laplace operator} which is defined by 
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\begin{equation} 
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\Delta u = (u\hackscore {,0})\hackscore{,0}+(u\hackscore{,1})\hackscore{,1} 
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\label{eq:FirstSteps.1.1} 
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\end{equation} 
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where, for any function $w$ and any direction $i$, $u\hackscore{,i}$ 
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denotes the partial derivative \index{partial derivative} of $u$ with respect to $i$. 
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\footnote{Some readers 
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may be more familiar with the Laplace operator\index{Laplace operator} being written 
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as $\nabla^2$, and written in the form 
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\begin{equation*} 
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\nabla^2 u = \nabla^t \cdot \nabla u = \frac{\partial^2 u}{\partial x\hackscore 0^2} 
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+ \frac{\partial^2 u}{\partial x\hackscore 1^2} 
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\end{equation*} 
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and \eqn{eq:FirstSteps.1} as 
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\begin{equation*} 
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\nabla^2 u = f 
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\end{equation*} 
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} 
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Basically, in the subindex of a function, any index to the left of the comma denotes a spatial derivative with respect 
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to the index. To get a more compact form we will write $w\hackscore{,ij}=(w\hackscore {,i})\hackscore{,j}$ 
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which leads to 
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\begin{equation} 
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\Delta u = u\hackscore{,00}+u\hackscore{,11}=\sum\hackscore{i=0}^2 u\hackscore{,ii} 
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\label{eq:FirstSteps.1.1b} 
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\end{equation} 
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In some cases, and we will see examples for this in the next chapter, 
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the usage of the nested $\sum$ symbols blows up the formulas and therefore 
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it is convenient to use the Einstein summation convention \index{summation convention}. This 
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drops the $\sum$ sign and assumes that a summation over a repeated index is performed 
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("repeated index means summation"). For instance we write 
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\begin{eqnarray} 
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x\hackscore{i}y\hackscore{i}=\sum\hackscore{i=0}^2 x\hackscore{i}y\hackscore{i} \\ 
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x\hackscore{i}u\hackscore{,i}=\sum\hackscore{i=0}^2 x\hackscore{i}u\hackscore{,i} \\ 
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u\hackscore{,ii}=\sum\hackscore{i=0}^2 u\hackscore{,ii} \\ 
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x\hackscore{ij}u\hackscore{i,j}=\sum\hackscore{j=0}^2\sum\hackscore{i=0}^2 x\hackscore{ij}u\hackscore{i,j} \\ 
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\label{eq:FirstSteps.1.1c} 
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\end{eqnarray} 
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With the summation convention we can write the Poisson equation \index{Poisson equation} as 
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\begin{equation} 
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 u\hackscore{,ii} =1 
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\label{eq:FirstSteps.1.sum} 
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\end{equation} 
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On the boundary of the domain $\Omega$ the normal derivative $n\hackscore{i} u\hackscore{,i}$ 
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of the solution $u$ shall be zero, ie. $u$ shall fulfill 
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the homogeneous Neumann boundary condition\index{Neumann 
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boundary condition!homogeneous} 
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\begin{equation} 
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n\hackscore{i} u\hackscore{,i}= 0 \;. 
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\label{eq:FirstSteps.2} 
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\end{equation} 
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$n=(n\hackscore{i})$ denotes the outer normal field 
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of the domain, see \fig{fig:FirstSteps.1}. Remember that we 
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are applying the Einstein summation convention \index{summation convention}, i.e 
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$n\hackscore{i} u\hackscore{,i}= n\hackscore{0} u\hackscore{,0} + 
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n\hackscore{1} u\hackscore{,1}$. 
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\footnote{Some readers may familiar with the notation 
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\begin{equation*} 
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\frac{\partial u}{\partial n} = n\hackscore{i} u\hackscore{,i} 
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\end{equation*} 
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for the normal derivative.} 
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The Neumann boundary condition of \eqn{eq:FirstSteps.2} should be fulfilled on the 
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set $\Gamma^N$ which is the top and right edge of the domain: 
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\begin{equation} 
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\Gamma^N=\{(x\hackscore 0;x\hackscore 1) \in \Omega  x\hackscore{0}=1 \mbox{ or } x\hackscore{1}=1 \} 
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\label{eq:FirstSteps.2b} 
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\end{equation} 
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On the bottom and the left edge of the domain which is defined 
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as 
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\begin{equation} 
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\Gamma^D=\{(x\hackscore 0;x\hackscore 1) \in \Omega  x\hackscore{0}=0 \mbox{ or } x\hackscore{1}=0 \} 
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\label{eq:FirstSteps.2c} 
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\end{equation} 
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the solution shall be identically zero: 
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\begin{equation} 
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u=0 \; . 
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\label{eq:FirstSteps.2d} 
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\end{equation} 
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This kind of boundary condition is called a homogeneous Dirichlet boundary condition 
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\index{Dirichlet boundary condition!homogeneous}. The partial differential equation in \eqn{eq:FirstSteps.1.sum} together 
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with the Neumann boundary condition \eqn{eq:FirstSteps.2} and 
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Dirichlet boundary condition in \eqn{eq:FirstSteps.2d} form a so 
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called boundary value 
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problem\index{boundary value problem} (BVP\index{boundary value problem!BVP}) for 
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the unknown 
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function $u$. 
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\begin{figure} 
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\centerline{\includegraphics[width=\figwidth]{FirstStepMesh}} 
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\caption{Mesh of $4 \time 4$ elements on a rectangular domain. Here 
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each element is a quadrilateral and described by four nodes, namely 
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the corner points. The solution is interpolated by a bilinear 
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polynomial.} 
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\label{fig:FirstSteps.2} 
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\end{figure} 
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In general the BVP\index{boundary value problem!BVP} cannot be solved analytically and numerical 
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methods have to be used construct an approximation of the solution 
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$u$. Here we will use the finite element method\index{finite element 
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method} (FEM\index{finite element 
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method!FEM}). The basic idea is to fill the domain with a 
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set of points called nodes. The solution is approximated by its 
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values on the nodes\index{finite element 
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method!nodes}. Moreover, the domain is subdivided into small, 
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subdomain called elements \index{finite element 
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method!element}. On each element the solution is 
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represented by a polynomial of a certain degree through its values at 
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the nodes located in the element. The nodes and its connection through 
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elements is called a mesh\index{finite element 
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method!mesh}. \fig{fig:FirstSteps.2} shows an 
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example of a FEM mesh with four elements in the $x_0$ and four elements 
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in the $x_1$ direction over the unit square. 
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For more details we refer the reader to the literature, for instance 
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\Ref{Zienc,NumHand}. 
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\escript provides the class \Poisson to define a Poisson equation \index{Poisson equation}. 
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(We will discuss a more general form of a PDE \index{partial differential equation!PDE} 
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that can be defined through the \LinearPDE class later). The instantiation of 
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a \Poisson class object requires the specification of the domain $\Omega$. In \escript 
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the \Domain class objects are used to describe the geometry of a domain but it also 
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contains information about the discretization methods and the actual solver which is used 
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to solve the PDE. Here we are using the FEM library \finley \index{finite element 
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method}. The following statements create the \Domain object \var{mydomain} from the 
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\finley method \method{Rectangle} 
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\begin{python} 
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from esys.finley import Rectangle 
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mydomain = Rectangle(l0=1.,l1=1.,n0=40, n1=20) 
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\end{python} 
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In this case the domain is a rectangle with the lower, left corner at point $(0,0)$ and 
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the right, upper corner at $(\var{l0},\var{l1})=(1,1)$. 
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The arguments \var{n0} and \var{n1} define the number of elements in $x\hackscore{0}$ and 
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$x\hackscore{1}$direction respectively. For more details on \method{Rectangle} and 
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other \Domain generators within the \finley module, 
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see \Chap{CHAPTER ON FINLEY}. 
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The following statements define the \Poisson class object \var{mypde} with domain \var{mydomain} and 
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the right hand side $f$ of the PDE to constant $1$: 
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\begin{python} 
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from esys.escript import Poisson 
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mypde = Poisson(mydomain) 
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mypde.setValue(f=1) 
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\end{python} 
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We have not specified any boundary condition but the 
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\Poisson class implicitly assumes homogeneous Neuman boundary conditions \index{Neumann 
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boundary condition!homogeneous} defined by \eqn{eq:FirstSteps.2}. With this boundary 
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condition the BVP\index{boundary value problem!BVP} we have defined has no unique solution. In fact, with any solution $u$ 
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and any constant $C$ the function $u+C$ becomes a solution as well. We have to add 
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a Dirichlet boundary condition \index{Dirichlet boundary condition}. This is done 
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by defining a characteristic function \index{characteristic function} 
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which has positive values at locations $x=(x\hackscore{0},x\hackscore{1})$ where Dirichlet boundary condition is set 
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and $0$ elsewhere. In our case of $\Gamma^D$ defined by \eqn{eq:FirstSteps.2c}, 
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we need a function which is positive for the cases $x\hackscore{0}=0$ or $x\hackscore{1}=0$: 
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\begin{python} 
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x=mydomain.getX() 
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gammaD=x[0].whereZero()+x[1].whereZero() 
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\end{python} 
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In the first statement, the method \method{getX} of the \Domain \var{mydomain} 
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gives access to locations 
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in the domain defined by \var{mydomain}. The object \var{x} is actually a \Data object 
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which we will learn more about later. \code{x[0]} returns the $x\hackscore{0}$ coordinates of the locations and 
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\code{x[0].whereZero()} creates function which equals $1$ where \code{x[0]} is (nearly) equal to zero 
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and $0$ elsewhere. 
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Similarly, \code{x[1].whereZero()} creates function which equals $1$ where \code{x[1]} is 
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equal to zero and $0$ elsewhere. 
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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 $x\hackscore{0}$ or $x\hackscore{1}$ is equal to zero. 
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The additional parameter \var{q} of the \code{setValue} method of the \Poisson class defines the 
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characteristic function \index{characteristic function} of the locations 
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of the domain where homogeneous Dirichlet boundary condition \index{Dirichlet boundary condition!homogeneous} 
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are set. The complete definition of our example is now: 
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\begin{python} 
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from esys.linearPDEs import Poisson 
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x = mydomain.getX() 
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gammaD = x[0].whereZero()+x[1].whereZero() 
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mypde = Poisson(domain=mydomain) 
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mypde = setValue(f=1,q=gammaD) 
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\end{python} 
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The first statement imports the \Poisson class definition form the \linearPDEsPack module which is part of the \ESyS package. 
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To get the solution of the Poisson equation defined by \var{mypde} we just have to call its 
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\method{getSolution}. 
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Now we can write the script to solve our test problem (Remember that 
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lines starting with '\#' are comment lines in Python) (available as \file{mypoisson.py} 
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in the \ExampleDirectory): 
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\begin{python} 
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from esys.finley import Rectangle 
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from esys.linearPDEs import Poisson 
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# generate domain: 
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mydomain = Rectangle(l0=1.,l1=1.,n0=40, n1=20) 
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# define characteristic function of Gamma^D 
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x = mydomain.getX() 
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gammaD = x[0].whereZero()+x[1].whereZero() 
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# define PDE and get its solution u 
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mypde = Poisson(domain=mydomain,f=1,q=gammaD) 
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u = mypde.getSolution() 
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# write u to an external file 
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u.saveDX("u.dx") 
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\end{python} 
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The last statement writes the solution to the external file \file{u.dx} in 
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\OpenDX file format. \OpenDX is a software package 
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for the visualization of scientific, engineering and analytical data and is freely available 
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from \url{http://www.opendx.org}. 
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\begin{figure} 
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\centerline{\includegraphics[width=\figwidth]{FirstStepResult.eps}} 
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\caption{\OpenDX Visualization of the Possion Equation Solution for $f=1$} 
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\label{fig:FirstSteps.3} 
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\end{figure} 
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You can edit the script file using your favourite text editor (or the Integrated DeveLopment Environment IDLE 
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for Python, see \url{http://idlefork.sourceforge.net}). If the script file has the name \file{mypoisson.py} \index{scripts!\file{mypoisson.py}} you can run the 
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script from any shell using the command: 
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\begin{verbatim} 
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python mypoisson.py 
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\end{verbatim} 
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After the script has (hopefully successfully) been completed you will find the file \file{u.dx} in the current 
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directory. An easy way to visualize the results is the command 
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\begin{verbatim} 
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dx prompter & 
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\end{verbatim} 
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to start the generic data visualization interface of \OpenDX. \fig{fig:FirstSteps.3} shows the result. 