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

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The user guide now builds under pdflatex (if you have converted the figures). Unfortunately, it has a really ugly title.

 1 2 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 3 % 4 % Copyright (c) 2003-2008 by University of Queensland 5 % Earth Systems Science Computational Center (ESSCC) 6 7 % 8 % Primary Business: Queensland, Australia 9 % Licensed under the Open Software License version 3.0 10 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 157 \Ref{Zienc,NumHand}. 158 159 \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}. 178 179 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 204 205 \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 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)+whereZero(x) 244 \end{python} 245 246 \code{whereZero(x)} creates function which equals $1$ where \code{x} is (almost) equal to zero 247 and $0$ elsewhere. 248 Similarly, \code{whereZero(x)} creates function which equals $1$ where \code{x} is 249 equal to zero and $0$ elsewhere. 250 The sum of the results of \code{whereZero(x)} and \code{whereZero(x)} 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)+whereZero(x) 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}. 274 275 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)+whereZero(x) 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 293 294 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}. 302 303 \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} 308

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