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 1 ksteube 1316 % 2 jgs 102 % $Id$ 3 gross 625 % 4 ksteube 1316 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 5 gross 625 % 6 ksteube 1316 % Copyright 2003-2007 by ACceSS MNRF 7 % Copyright 2007 by University of Queensland 8 % 9 10 % Primary Business: Queensland, Australia 11 % Licensed under the Open Software License version 3.0 12 13 % 14 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 15 % 16 jgs 102 17 jgs 121 \section{The First Steps} 18 jgs 102 \label{FirstSteps} 19 20 \begin{figure} 21 gross 599 \centerline{\includegraphics[width=\figwidth]{figures/FirstStepDomain}} 22 jgs 102 \caption{Domain $\Omega=[0,1]^2$ with outer normal field $n$.} 23 \label{fig:FirstSteps.1} 24 \end{figure} 25 26 ksteube 1316 In this chapter we give an introduction how to use \escript to solve 27 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} 28 is more than sufficient. 29 jgs 102 30 jgs 107 The PDE \index{partial differential equation} we wish to solve is the Poisson equation \index{Poisson equation} 31 jgs 102 \begin{equation} 32 -\Delta u =f 33 \label{eq:FirstSteps.1} 34 \end{equation} 35 ksteube 1316 for the solution $u$. The function $f$ is the given right hand side. The domain of interest, denoted by $\Omega$, 36 jgs 102 is the unit square 37 \begin{equation} 38 \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 \} 39 \label{eq:FirstSteps.1b} 40 \end{equation} 41 jgs 107 The domain is shown in \fig{fig:FirstSteps.1}. 42 jgs 102 43 ksteube 1316 $\Delta$ denotes the Laplace operator\index{Laplace operator}, which is defined by 44 jgs 102 \begin{equation} 45 \Delta u = (u\hackscore {,0})\hackscore{,0}+(u\hackscore{,1})\hackscore{,1} 46 \label{eq:FirstSteps.1.1} 47 \end{equation} 48 ksteube 1316 where, for any function $u$ and any direction $i$, $u\hackscore{,i}$ 49 jgs 107 denotes the partial derivative \index{partial derivative} of $u$ with respect to $i$. 50 ksteube 1316 \footnote{You 51 jgs 102 may be more familiar with the Laplace operator\index{Laplace operator} being written 52 as $\nabla^2$, and written in the form 53 \begin{equation*} 54 jgs 110 \nabla^2 u = \nabla^t \cdot \nabla u = \frac{\partial^2 u}{\partial x\hackscore 0^2} 55 jgs 102 + \frac{\partial^2 u}{\partial x\hackscore 1^2} 56 \end{equation*} 57 and \eqn{eq:FirstSteps.1} as 58 \begin{equation*} 59 -\nabla^2 u = f 60 \end{equation*} 61 } 62 jgs 107 Basically, in the subindex of a function, any index to the left of the comma denotes a spatial derivative with respect 63 jgs 102 to the index. To get a more compact form we will write $w\hackscore{,ij}=(w\hackscore {,i})\hackscore{,j}$ 64 which leads to 65 \begin{equation} 66 \Delta u = u\hackscore{,00}+u\hackscore{,11}=\sum\hackscore{i=0}^2 u\hackscore{,ii} 67 \label{eq:FirstSteps.1.1b} 68 \end{equation} 69 ksteube 1316 We often find that use 70 of nested $\sum$ symbols makes formulas cumbersome, and we use the more 71 convenient Einstein summation convention \index{summation convention}. This 72 drops the $\sum$ sign and assumes that a summation is performed over any repeated index. 73 For instance we write 74 jgs 102 \begin{eqnarray} 75 x\hackscore{i}y\hackscore{i}=\sum\hackscore{i=0}^2 x\hackscore{i}y\hackscore{i} \\ 76 x\hackscore{i}u\hackscore{,i}=\sum\hackscore{i=0}^2 x\hackscore{i}u\hackscore{,i} \\ 77 u\hackscore{,ii}=\sum\hackscore{i=0}^2 u\hackscore{,ii} \\ 78 jgs 107 x\hackscore{ij}u\hackscore{i,j}=\sum\hackscore{j=0}^2\sum\hackscore{i=0}^2 x\hackscore{ij}u\hackscore{i,j} \\ 79 jgs 102 \label{eq:FirstSteps.1.1c} 80 \end{eqnarray} 81 With the summation convention we can write the Poisson equation \index{Poisson equation} as 82 \begin{equation} 83 - u\hackscore{,ii} =1 84 \label{eq:FirstSteps.1.sum} 85 \end{equation} 86 lkettle 575 where $f=1$ in this example. 87 88 jgs 102 On the boundary of the domain $\Omega$ the normal derivative $n\hackscore{i} u\hackscore{,i}$ 89 of the solution $u$ shall be zero, ie. $u$ shall fulfill 90 the homogeneous Neumann boundary condition\index{Neumann 91 boundary condition!homogeneous} 92 \begin{equation} 93 n\hackscore{i} u\hackscore{,i}= 0 \;. 94 \label{eq:FirstSteps.2} 95 \end{equation} 96 $n=(n\hackscore{i})$ denotes the outer normal field 97 of the domain, see \fig{fig:FirstSteps.1}. Remember that we 98 are applying the Einstein summation convention \index{summation convention}, i.e 99 jgs 107 $n\hackscore{i} u\hackscore{,i}= n\hackscore{0} u\hackscore{,0} + 100 n\hackscore{1} u\hackscore{,1}$. 101 jgs 102 \footnote{Some readers may familiar with the notation 102 \begin{equation*} 103 \frac{\partial u}{\partial n} = n\hackscore{i} u\hackscore{,i} 104 \end{equation*} 105 for the normal derivative.} 106 The Neumann boundary condition of \eqn{eq:FirstSteps.2} should be fulfilled on the 107 set $\Gamma^N$ which is the top and right edge of the domain: 108 \begin{equation} 109 \Gamma^N=\{(x\hackscore 0;x\hackscore 1) \in \Omega | x\hackscore{0}=1 \mbox{ or } x\hackscore{1}=1 \} 110 \label{eq:FirstSteps.2b} 111 \end{equation} 112 jgs 107 On the bottom and the left edge of the domain which is defined 113 jgs 102 as 114 \begin{equation} 115 \Gamma^D=\{(x\hackscore 0;x\hackscore 1) \in \Omega | x\hackscore{0}=0 \mbox{ or } x\hackscore{1}=0 \} 116 \label{eq:FirstSteps.2c} 117 \end{equation} 118 the solution shall be identically zero: 119 \begin{equation} 120 u=0 \; . 121 \label{eq:FirstSteps.2d} 122 \end{equation} 123 jgs 107 This kind of boundary condition is called a homogeneous Dirichlet boundary condition 124 jgs 102 \index{Dirichlet boundary condition!homogeneous}. The partial differential equation in \eqn{eq:FirstSteps.1.sum} together 125 with the Neumann boundary condition \eqn{eq:FirstSteps.2} and 126 Dirichlet boundary condition in \eqn{eq:FirstSteps.2d} form a so 127 called boundary value 128 jgs 107 problem\index{boundary value problem} (BVP\index{boundary value problem!BVP}) for 129 the unknown 130 jgs 102 function $u$. 131 132 133 \begin{figure} 134 gross 599 \centerline{\includegraphics[width=\figwidth]{figures/FirstStepMesh.eps}} 135 jgs 102 \caption{Mesh of $4 \time 4$ elements on a rectangular domain. Here 136 each element is a quadrilateral and described by four nodes, namely 137 the corner points. The solution is interpolated by a bi-linear 138 polynomial.} 139 \label{fig:FirstSteps.2} 140 \end{figure} 141 142 In general the BVP\index{boundary value problem!BVP} cannot be solved analytically and numerical 143 methods have to be used construct an approximation of the solution 144 $u$. Here we will use the finite element method\index{finite element 145 method} (FEM\index{finite element 146 method!FEM}). The basic idea is to fill the domain with a 147 jgs 107 set of points called nodes. The solution is approximated by its 148 jgs 102 values on the nodes\index{finite element 149 lkettle 573 method!nodes}. Moreover, the domain is subdivided into smaller 150 sub-domains called elements \index{finite element 151 jgs 102 method!element}. On each element the solution is 152 represented by a polynomial of a certain degree through its values at 153 the nodes located in the element. The nodes and its connection through 154 elements is called a mesh\index{finite element 155 jgs 107 method!mesh}. \fig{fig:FirstSteps.2} shows an 156 jgs 102 example of a FEM mesh with four elements in the $x_0$ and four elements 157 in the $x_1$ direction over the unit square. 158 For more details we refer the reader to the literature, for instance 159 \Ref{Zienc,NumHand}. 160 161 \escript provides the class \Poisson to define a Poisson equation \index{Poisson equation}. 162 (We will discuss a more general form of a PDE \index{partial differential equation!PDE} 163 that can be defined through the \LinearPDE class later). The instantiation of 164 a \Poisson class object requires the specification of the domain $\Omega$. In \escript 165 the \Domain class objects are used to describe the geometry of a domain but it also 166 contains information about the discretization methods and the actual solver which is used 167 to solve the PDE. Here we are using the FEM library \finley \index{finite element 168 method}. The following statements create the \Domain object \var{mydomain} from the 169 \finley method \method{Rectangle} 170 \begin{python} 171 ksteube 1316 from esys.finley import Rectangle 172 mydomain = Rectangle(l0=1.,l1=1.,n0=40, n1=20) 173 jgs 102 \end{python} 174 In this case the domain is a rectangle with the lower, left corner at point $(0,0)$ and 175 the right, upper corner at $(\var{l0},\var{l1})=(1,1)$. 176 jgs 107 The arguments \var{n0} and \var{n1} define the number of elements in $x\hackscore{0}$ and 177 jgs 102 $x\hackscore{1}$-direction respectively. For more details on \method{Rectangle} and 178 other \Domain generators within the \finley module, 179 see \Chap{CHAPTER ON FINLEY}. 180 181 jgs 107 The following statements define the \Poisson class object \var{mypde} with domain \var{mydomain} and 182 jgs 102 the right hand side $f$ of the PDE to constant $1$: 183 \begin{python} 184 ksteube 1316 from esys.escript.linearPDEs import Poisson 185 mypde = Poisson(mydomain) 186 mypde.setValue(f=1) 187 jgs 102 \end{python} 188 We have not specified any boundary condition but the 189 \Poisson class implicitly assumes homogeneous Neuman boundary conditions \index{Neumann 190 boundary condition!homogeneous} defined by \eqn{eq:FirstSteps.2}. With this boundary 191 condition the BVP\index{boundary value problem!BVP} we have defined has no unique solution. In fact, with any solution $u$ 192 and any constant $C$ the function $u+C$ becomes a solution as well. We have to add 193 a Dirichlet boundary condition \index{Dirichlet boundary condition}. This is done 194 jgs 107 by defining a characteristic function \index{characteristic function} 195 which has positive values at locations $x=(x\hackscore{0},x\hackscore{1})$ where Dirichlet boundary condition is set 196 jgs 102 and $0$ elsewhere. In our case of $\Gamma^D$ defined by \eqn{eq:FirstSteps.2c}, 197 gross 565 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 198 ksteube 1316 an object \var{x} which contains the coordinates of the nodes in the domain use 199 jgs 102 \begin{python} 200 ksteube 1316 x=mydomain.getX() 201 jgs 102 \end{python} 202 gross 660 The method \method{getX} of the \Domain \var{mydomain} 203 jgs 107 gives access to locations 204 ksteube 1316 in the domain defined by \var{mydomain}. The object \var{x} is actually a \Data object which will be 205 discussed in \Chap{ESCRIPT CHAP} in more detail. What we need to know here is that 206 gross 660 207 ksteube 1316 \var{x} has \Rank (number of dimensions) and a \Shape (list of dimensions) which can be viewed by 208 gross 565 calling the \method{getRank} and \method{getShape} methods: 209 \begin{python} 210 ksteube 1316 print "rank ",x.getRank(),", shape ",x.getShape() 211 gross 565 \end{python} 212 ksteube 1316 This will print something like 213 gross 565 \begin{python} 214 ksteube 1316 rank 1, shape (2,) 215 gross 565 \end{python} 216 The \Data object also maintains type information which is represented by the 217 \FunctionSpace of the object. For instance 218 \begin{python} 219 ksteube 1316 print x.getFunctionSpace() 220 gross 565 \end{python} 221 will print 222 \begin{python} 223 ksteube 1316 Function space type: Finley_Nodes on FinleyMesh 224 gross 565 \end{python} 225 ksteube 1316 which tells us that the coordinates are stored on the nodes of (rather than on points in the interior of) a \finley mesh. 226 gross 565 To get the $x\hackscore{0}$ coordinates of the locations we use the 227 statement 228 \begin{python} 229 ksteube 1316 x0=x 230 gross 565 \end{python} 231 Object \var{x0} 232 is again a \Data object now with \Rank $0$ and 233 \Shape $()$. It inherits the \FunctionSpace from \var{x}: 234 \begin{python} 235 ksteube 1316 print x0.getRank(),x0.getShape(),x0.getFunctionSpace() 236 gross 565 \end{python} 237 will print 238 \begin{python} 239 ksteube 1316 0 () Function space type: Finley_Nodes on FinleyMesh 240 gross 565 \end{python} 241 ksteube 1316 We can now construct a function \var{gammaD} which is only non-zero on the bottom and left edges 242 of the domain with 243 gross 565 \begin{python} 244 ksteube 1316 from esys.escript import whereZero 245 gammaD=whereZero(x)+whereZero(x) 246 gross 565 \end{python} 247 ksteube 1316 248 \code{whereZero(x)} creates function which equals $1$ where \code{x} is (almost) equal to zero 249 jgs 107 and $0$ elsewhere. 250 gross 565 Similarly, \code{whereZero(x)} creates function which equals $1$ where \code{x} is 251 jgs 107 equal to zero and $0$ elsewhere. 252 gross 565 The sum of the results of \code{whereZero(x)} and \code{whereZero(x)} 253 gives a function on the domain \var{mydomain} which is exactly positive where $x\hackscore{0}$ or $x\hackscore{1}$ is equal to zero. 254 Note that \var{gammaD} has the same \Rank, \Shape and \FunctionSpace like \var{x0} used to define it. So from 255 \begin{python} 256 ksteube 1316 print gammaD.getRank(),gammaD.getShape(),gammaD.getFunctionSpace() 257 gross 565 \end{python} 258 one gets 259 \begin{python} 260 ksteube 1316 0 () Function space type: Finley_Nodes on FinleyMesh 261 gross 565 \end{python} 262 ksteube 1316 An additional parameter \var{q} of the \code{setValue} method of the \Poisson class defines the 263 jgs 102 characteristic function \index{characteristic function} of the locations 264 of the domain where homogeneous Dirichlet boundary condition \index{Dirichlet boundary condition!homogeneous} 265 are set. The complete definition of our example is now: 266 \begin{python} 267 ksteube 1316 from esys.linearPDEs import Poisson 268 x = mydomain.getX() 269 gammaD = whereZero(x)+whereZero(x) 270 mypde = Poisson(domain=mydomain) 271 mypde.setValue(f=1,q=gammaD) 272 jgs 102 \end{python} 273 lkettle 573 The first statement imports the \Poisson class definition from the \linearPDEs module \escript package. 274 jgs 107 To get the solution of the Poisson equation defined by \var{mypde} we just have to call its 275 jgs 102 \method{getSolution}. 276 277 ksteube 1316 Now we can write the script to solve our Poisson problem 278 jgs 102 \begin{python} 279 ksteube 1316 from esys.escript import * 280 from esys.escript.linearPDEs import Poisson 281 from esys.finley import Rectangle 282 # generate domain: 283 mydomain = Rectangle(l0=1.,l1=1.,n0=40, n1=20) 284 # define characteristic function of Gamma^D 285 x = mydomain.getX() 286 gammaD = whereZero(x)+whereZero(x) 287 # define PDE and get its solution u 288 mypde = Poisson(domain=mydomain) 289 mypde.setValue(f=1,q=gammaD) 290 u = mypde.getSolution() 291 # write u to an external file 292 saveVTK("u.xml",sol=u) 293 jgs 102 \end{python} 294 ksteube 1316 The entire code is available as \file{poisson.py} in the \ExampleDirectory 295 jgs 102 296 ksteube 1316 The last statement writes the solution (tagged with the name "sol") to a file named \file{u.xml} in 297 \VTK file format. 298 Now you may run the script and visualize the solution using \mayavi: 299 \begin{verbatim} 300 python poisson.py 301 mayavi -d u.xml -m SurfaceMap 302 \end{verbatim} 303 See \fig{fig:FirstSteps.3}. 304 305 jgs 102 \begin{figure} 306 gross 599 \centerline{\includegraphics[width=\figwidth]{figures/FirstStepResult.eps}} 307 lkettle 573 \caption{Visualization of the Poisson Equation Solution for $f=1$} 308 jgs 102 \label{fig:FirstSteps.3} 309 \end{figure} 310

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