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 1 jgs 82 % $Id$ 2 3 \chapter{The First Steps} 4 5 \section{Introduction} 6 7 \subsection{Getting the software} 8 9 \escript, \ESyS, all freely available. Where do people get \finley from? 10 11 \begin{enumerate} 12 \item how to get the software 13 \item a few words about the general structure 14 \item installation 15 \end{enumerate} 16 17 \section{How to solve a linear PDE} 18 19 \begin{figure} 20 \centerline{\includegraphics[width=\figwidth]{FirstStepDomain}} 21 \caption{Domain $\Omega$ with outer normal field $n$.} 22 \label{fig:FirstSteps.1} 23 \end{figure} 24 25 \begin{figure} 26 \centerline{\includegraphics[width=\figwidth]{FirstStepMesh}} 27 \caption{Mesh of $4 \time 4$ elements on a rectangular domain. Here 28 each element is a quadrilateral and described by four nodes, namely 29 the corner points. The solution is interpolated by a bi-linear 30 polynomial.} 31 \label{fig:FirstSteps.2} 32 \end{figure} 33 34 We want to solve the \index{partial differential equation}(\index{PDE}) 35 \begin{equation} 36 -\Delta u + \alpha u =f 37 \label{eq:FirstSteps.1} 38 \end{equation} 39 for the solution $u$ on the domain $\Omega$. Here we assume that the 40 domain is the rectangle of length $1$ and height $2$, see 41 \fig{fig:FirstSteps.1}. $\Delta$ denotes the \index{Laplace 42 operator} which is defined by 43 \begin{equation} 44 \Delta u = (u\hackscore {,1})\hackscore{,1}+(u\hackscore{,2})\hackscore{,2} 45 \label{eq:FirstSteps.1.1} 46 \end{equation} 47 where for any function $w$ and any direction $i$ $u\hackscore{,i}$ 48 denotes the derivative of $w$ with respect to $i$. $\alpha$ is a 49 known constant (we will set $\alpha=10$) and $f$ is a given function 50 which may depend on the location in the domain. On the boundary of 51 the domain $\Omega$ the solution $u$ shall fulfill the so-called 52 homogeneous \index{Neumann boundary condition} 53 \begin{equation} 54 \frac{\partial u}{\partial n} = 0 55 \label{eq:FirstSteps.2} 56 \end{equation} 57 where $n=(n\hackscore1,n\hackscore2)$ denotes the outer normal field 58 of the domain, see Figure~\ref{fig:FirstSteps.1} and 59 \begin{equation} 60 \frac{\partial u}{\partial n} = n\hackscore1 u\hackscore{,1} + 61 n\hackscore2 u\hackscore{,2} 62 \label{eq:FirstSteps.2.1} 63 \end{equation} 64 denotes the normal derivative on the boundary. The partial 65 differential \eqn{eq:FirstSteps.1}) together with the 66 boundary condition~(\eqn{eq:FirstSteps.2}) forms a boundary value 67 problem (\index{BVP}) for unknown function $u$. 68 69 In most cases, the BVP cannot be solved analytically and numerical 70 methods have to be used construct an approximation of the solution 71 $u$. Here we will use the \index{finite element method} 72 (\index{FEM}). The basic idea is to fill the domain with a set of 73 points, so called nodes. The solution is approximated by its values on 74 the nodes. Moreover, the domain is subdivide into small subdomain, 75 so-called elements. On each element the solution is represented by a 76 polynomial of a certain degree through its values at the nodes located 77 in the element. The nodes and its connection through elements is 78 called a \index{mesh}. \fig{fig:FirstSteps.2} shows an example 79 of a FEM mesh with four elements in the $x_0$ and for elements in the 80 $x_1$ direction over a rectangular domain. On more details we referring 81 to the literature, for instance \cite{X1,X2,X3}. 82 83 \escript provides the class \linearPDE to define a 84 general linear, steady differential partial differential equation of 85 second order. We will discuss the most general form that can be 86 defined through \class{linearPDE} later. The components which are 87 relevant for us here is as follows: 88 \begin{equation} 89 -\sum\hackscore{i,j=0}^k (A\hackscore{ij} 90 u\hackscore{,j})\hackscore{,i} + D u = Y 91 \label{eq:FirstSteps.3} 92 \end{equation} 93 In this form $D$ and $Y$ are scalars and $A$ is a $k \times k$ matrix 94 where $k$ denotes the spatial dimension (in our example we have 95 $k=2$). By comparing the template~(\ref{eq:FirstSteps.3}) with the 96 differential equation~(\ref{eq:FirstSteps.1}) we want to solve we can 97 immediately identify the appropriate values for $A$, $D$ and $Y$: 98 \begin{equation} 99 \begin{array}{lcccc} 100 A\hackscore{ij} & = & \delta\hackscore{ij}& =& 101 \left[ 102 \begin{array}{cc} 103 1 & 0\\ 104 0 & 1 105 \end{array} 106 \right] \\ 107 D & =& \alpha \\ 108 Y & = & f \\ 109 \end{array} 110 \label{eq:FirstSteps.3.1} 111 \end{equation} 112 When the PDE is defined via~(\eqn{eq:FirstSteps.3}), \class{linearPDE} 113 makes a particular assumptions about the boundary conditions: 114 \begin{equation} 115 -\sum\hackscore{i,j=0}^k n\hackscore{i} A\hackscore{ij} 116 u\hackscore{,j} = 0 117 \label{eq:FirstSteps.4} 118 \end{equation} 119 Note that this boundary condition does not require extra information 120 as it only refers to the coefficient $A$ which already appears in the 121 PDE and so natural for it. Therefore this boundary condition is called 122 a \index{natural boundary condition}. 123 124 We will set $\alpha=10$ and $f=10$ such that $u=1$ becomes the exact 125 solution of the boundary value problem. We make this very simple 126 choice to be able to test our program as we can compare the result 127 with the exact solution. Later we will set $\alpha$ and $f$ to 128 functions of their locations in the domain in which case we will not 129 be able to give an analytic solution. However, after testing our 130 program on this very simple case, we can be confident that it working 131 correctly before we apply it is a more complicated situation. 132 133 This is the program to solve the boundary value problem: (Remember that 134 lines starting with '\#' are commend lines in Python) 135 %\verbatiminput{examples/FirstSteps1.py} 136 \begin{python} 137 # import ESyS and finley 138 from ESyS import * 139 import finley 140 # set a value for alpha: 141 alpha=10 142 # generate mesh: 143 mydomain=finley.Rectangle(n0=40,n1=20,l0=2.,l1=1.) 144 # generate a system: 145 mypde=linearPDE(A=[[1,0],[0,1]],D=alpha,Y=10,domain=mydomain) 146 # generate a test solution: 147 u=mypde.getSolution() 148 # calculate the error of the solution 149 error=u-1. 150 print "norm of the approximation error is ",Lsup(error) 151 \end{python} 152 Line 2 import \escript and a few other tools for \ESyS. In Line 3 is 153 importing \finley which is used to solve the partial differential 154 equation. In line 7, a rectangular domain of length $l\hackscore 0=2$ 155 and height $l\hackscore 1=1$ is generated and subdivided in 156 $n\hackscore 0=40$ and $n\hackscore 1=20$ elements in $x\hackscore 0$ 157 and $x\hackscore 1$ direction, respectively. 158 159 We are using a function of \finley. This determines later in the code 160 which solver for the PDE is actually being used to solve the PDE. 161 162 The solution is done in three steps: 163 \begin{enumerate} 164 \item generate a finite element mesh subdividing the domain into elements. 165 \item assemble the system of linear equations $Mu=b$ from the BVP 166 \item solve the linear system to get $u$ 167 \end{enumerate} 168 The returned $u$ is given an approximation of the solution of the BVP 169 at the nodes of the finite element mesh. The quality of the 170 approximation depends on the size of the elements of the finite 171 element mesh: As smaller the element size the better the 172 approximation. In our example we know the solution of the BVP so we 173 can compare the returned approximation with the true solution. In 174 fact, as the true solution is simple, we can expect that the 175 approximation is exact. 176 177 The first step imports the package \ESyS which includes among 178 others the module \finley. 179 180 \section{A Time Dependent Problem} 181 182 % \verbatiminput{exam/finley\hackscoretime.py} 183 184 \section{With Dirichlet Conditions} 185 186 % \verbatiminput{exam/finley\hackscoredirichlet.py} 187 188 \section{Systems of PDEs} 189 190 % \verbatiminput{exam/system\hackscoretime.py} 191 192 \section{Explicit Schemes} 193 % \verbatiminput{exam/explicit\hackscoretime.py}

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