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

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 1 2 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 3 % 4 % Copyright (c) 2003-2009 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} [ht] 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}[ht] 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 \Ref{Zienc,NumHand}. 157 158 The \escript solver we want to use to solve this problem is embedded into the python interpreter language. So you can solve the problem interactively but you will learn quickly 159 that is more efficient to use scripts which you can edit with your favorite editor. 160 To enter the escript environment you use \program{escript} command\footnote{\program{escript} is not available under Windows yet. If you run under windows you can just use the 161 \program{python} command and the \env{OMP_NUM_THREADS} environment variable to control the number 162 of threads.}: 163 \begin{verbatim} 164 escript 165 \end{verbatim} 166 which will pass you on to the python prompt 167 \begin{verbatim} 168 Python 2.5.2 (r252:60911, Oct 5 2008, 19:29:17) 169 [GCC 4.3.2] on linux2 170 Type "help", "copyright", "credits" or "license" for more information. 171 >>> 172 \end{verbatim} 173 Here you can use all available python commands and language features, for instance 174 \begin{python} 175 >>> x=2+3 176 >>> print "2+3=",x 177 2+3= 5 178 \end{python} 179 We refer to the python users guide if you not familiar with python. 180 181 \escript provides the class \Poisson to define a Poisson equation \index{Poisson equation}. 182 (We will discuss a more general form of a PDE \index{partial differential equation!PDE} 183 that can be defined through the \LinearPDE class later). The instantiation of 184 a \Poisson class object requires the specification of the domain $\Omega$. In \escript 185 the \Domain class objects are used to describe the geometry of a domain but it also 186 contains information about the discretization methods and the actual solver which is used 187 to solve the PDE. Here we are using the FEM library \finley \index{finite element 188 method}. The following statements create the \Domain object \var{mydomain} from the 189 \finley method \method{Rectangle} 190 \begin{python} 191 from esys.finley import Rectangle 192 mydomain = Rectangle(l0=1.,l1=1.,n0=40, n1=20) 193 \end{python} 194 In this case the domain is a rectangle with the lower, left corner at point $(0,0)$ and 195 the right, upper corner at $(\var{l0},\var{l1})=(1,1)$. 196 The arguments \var{n0} and \var{n1} define the number of elements in $x\hackscore{0}$ and 197 $x\hackscore{1}$-direction respectively. For more details on \method{Rectangle} and 198 other \Domain generators within the \finley module, 199 see \Chap{CHAPTER ON FINLEY}. 200 201 The following statements define the \Poisson class object \var{mypde} with domain \var{mydomain} and 202 the right hand side $f$ of the PDE to constant $1$: 203 \begin{python} 204 from esys.escript.linearPDEs import Poisson 205 mypde = Poisson(mydomain) 206 mypde.setValue(f=1) 207 \end{python} 208 We have not specified any boundary condition but the 209 \Poisson class implicitly assumes homogeneous Neuman boundary conditions \index{Neumann 210 boundary condition!homogeneous} defined by \eqn{eq:FirstSteps.2}. With this boundary 211 condition the BVP\index{boundary value problem!BVP} we have defined has no unique solution. In fact, with any solution $u$ 212 and any constant $C$ the function $u+C$ becomes a solution as well. We have to add 213 a Dirichlet boundary condition \index{Dirichlet boundary condition}. This is done 214 by defining a characteristic function \index{characteristic function} 215 which has positive values at locations $x=(x\hackscore{0},x\hackscore{1})$ where Dirichlet boundary condition is set 216 and $0$ elsewhere. In our case of $\Gamma^D$ defined by \eqn{eq:FirstSteps.2c}, 217 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 218 an object \var{x} which contains the coordinates of the nodes in the domain use 219 \begin{python} 220 x=mydomain.getX() 221 \end{python} 222 The method \method{getX} of the \Domain \var{mydomain} 223 gives access to locations 224 in the domain defined by \var{mydomain}. The object \var{x} is actually a \Data object which will be 225 discussed in \Chap{ESCRIPT CHAP} in more detail. What we need to know here is that 226 227 \var{x} has \Rank (number of dimensions) and a \Shape (list of dimensions) which can be viewed by 228 calling the \method{getRank} and \method{getShape} methods: 229 \begin{python} 230 print "rank ",x.getRank(),", shape ",x.getShape() 231 \end{python} 232 This will print something like 233 \begin{python} 234 rank 1, shape (2,) 235 \end{python} 236 The \Data object also maintains type information which is represented by the 237 \FunctionSpace of the object. For instance 238 \begin{python} 239 print x.getFunctionSpace() 240 \end{python} 241 will print 242 \begin{python} 243 Function space type: Finley_Nodes on FinleyMesh 244 \end{python} 245 which tells us that the coordinates are stored on the nodes of (rather than on points in the interior of) a \finley mesh. 246 To get the $x\hackscore{0}$ coordinates of the locations we use the 247 statement 248 \begin{python} 249 x0=x[0] 250 \end{python} 251 Object \var{x0} 252 is again a \Data object now with \Rank $0$ and 253 \Shape $()$. It inherits the \FunctionSpace from \var{x}: 254 \begin{python} 255 print x0.getRank(),x0.getShape(),x0.getFunctionSpace() 256 \end{python} 257 will print 258 \begin{python} 259 0 () Function space type: Finley_Nodes on FinleyMesh 260 \end{python} 261 We can now construct a function \var{gammaD} which is only non-zero on the bottom and left edges 262 of the domain with 263 \begin{python} 264 from esys.escript import whereZero 265 gammaD=whereZero(x[0])+whereZero(x[1]) 266 \end{python} 267 268 \code{whereZero(x[0])} creates function which equals $1$ where \code{x[0]} is (almost) equal to zero 269 and $0$ elsewhere. 270 Similarly, \code{whereZero(x[1])} creates function which equals $1$ where \code{x[1]} is 271 equal to zero and $0$ elsewhere. 272 The sum of the results of \code{whereZero(x[0])} and \code{whereZero(x[1])} 273 gives a function on the domain \var{mydomain} which is strictly positive where $x\hackscore{0}$ or $x\hackscore{1}$ is equal to zero. 274 Note that \var{gammaD} has the same \Rank, \Shape and \FunctionSpace like \var{x0} used to define it. So from 275 \begin{python} 276 print gammaD.getRank(),gammaD.getShape(),gammaD.getFunctionSpace() 277 \end{python} 278 one gets 279 \begin{python} 280 0 () Function space type: Finley_Nodes on FinleyMesh 281 \end{python} 282 An additional parameter \var{q} of the \code{setValue} method of the \Poisson class defines the 283 characteristic function \index{characteristic function} of the locations 284 of the domain where homogeneous Dirichlet boundary condition \index{Dirichlet boundary condition!homogeneous} 285 are set. The complete definition of our example is now: 286 \begin{python} 287 from esys.linearPDEs import Poisson 288 x = mydomain.getX() 289 gammaD = whereZero(x[0])+whereZero(x[1]) 290 mypde = Poisson(domain=mydomain) 291 mypde.setValue(f=1,q=gammaD) 292 \end{python} 293 The first statement imports the \Poisson class definition from the \linearPDEs module \escript package. 294 To get the solution of the Poisson equation defined by \var{mypde} we just have to call its 295 \method{getSolution}. 296 297 Now we can write the script to solve our Poisson problem 298 \begin{python} 299 from esys.escript import * 300 from esys.escript.linearPDEs import Poisson 301 from esys.finley import Rectangle 302 # generate domain: 303 mydomain = Rectangle(l0=1.,l1=1.,n0=40, n1=20) 304 # define characteristic function of Gamma^D 305 x = mydomain.getX() 306 gammaD = whereZero(x[0])+whereZero(x[1]) 307 # define PDE and get its solution u 308 mypde = Poisson(domain=mydomain) 309 mypde.setValue(f=1,q=gammaD) 310 u = mypde.getSolution() 311 \end{python} 312 The question is what we do with the calculated solution \var{u}. Besides postprocessing, eg. calculating the gradient or the average value, which will be discussed later, plotting the solution is one one things you might want to do. \escript offers two ways to do this, both base on external modules or packages and so data need to converted 313 to hand over the solution. The first option is using the \MATPLOTLIB module which allows plotting 2D results relatively quickly, see~\cite{matplotlib}. However, there are limitations when using this tool, eg. in problem size and when solving 3D problems. Therefore \escript provides a second options based on \VTK files which is especially 314 designed for large scale and 3D problem and which can be read by a variety of software packages such as \mayavi \cite{mayavi}, \VisIt~\cite{VisIt}. 315 316 \subsection{Plotting Using \MATPLOTLIB} 317 318 Users of Debian 5(Lenny) please note: this example makes use of the griddata method in \module{matplotlib.mlab}. 319 This method is not part of version 0.98.1 which is available with Lenny. 320 If you wish to use \MATPLOTLIB, you may need to install a later version. 321 Users of Ubuntu 8.10 or later and people should be fine. 322 323 The \MATPLOTLIB module provides a simple and easy to use way to visualize PDE solutions (or other \Data objects). 324 To hand over data from \escript to \MATPLOTLIB the values need to mapped onto a rectangular grid. We will make use 325 of the \numpy module. 326 327 First we need to create a rectangular grid. We use the following statements: 328 \begin{python} 329 import numpy 330 x_grid = numpy.linspace(0.,1.,50) 331 y_grid = numpy.linspace(0.,1.,50) 332 \end{python} 333 \var{x_grid} is an array defining the x coordinates of the grids while 334 \var{y_grid} defines the y coordinates of the grid. In this case we use $50$ points over the interval $[0,1]$ 335 in both directions. 336 337 Now the values created by \escript need to be interpolated to this grid. We will use the \MATPLOTLIB 338 \function{mlab.griddata} function to do this. We can easily extract spatial coordinates as a \var{list} by 339 \begin{python} 340 x=mydomain.getX()[0].toListOfTuples() 341 y=mydomain.getX()[1].toListOfTuples() 342 \end{python} 343 In principle we can apply the same \member{toListOfTuples} method to extract the values from the 344 PDE solution \var{u}. However, we have to make sure that the \Data object we extract the values from 345 uses the same \FunctionSpace as we have us when extracting \var{x} and \var{y}. By applying the 346 \function{interpolation} to \var{u} before extraction we can achieve this: 347 \begin{python} 348 z=interpolate(u,mydomain.getX().getFunctionSpace()) 349 \end{python} 350 The values in \var{z} are now the values at the points with the coordinates given by \var{x} and \var{y}. These 351 values are now interpolated to the grid defined by \var{x_grid} and \var{y_grid} by using 352 \begin{python} 353 import matplotlib 354 z_grid = matplotlib.mlab.griddata(x,y,z,xi=x_grid,yi=y_grid ) 355 \end{python} 356 \var{z_grid} gives now the values of the PDE solution \var{u} at the grid. The values can be plotted now 357 using the \function{contourf}: 358 \begin{python} 359 matplotlib.pyplot.contourf(x_grid, y_grid, z_grid, 5) 360 matplotlib.pyplot.savefig("u.png") 361 \end{python} 362 Here we use $5$ contours. The last statement writes the plot to the file \file{u.png} in the PNG format. Alternatively, one can use 363 \begin{python} 364 matplotlib.pyplot.contourf(x_grid, y_grid, z_grid, 5) 365 matplotlib.pyplot.show() 366 \end{python} 367 which gives an interactive browser window. 368 369 \begin{figure} 370 \centerline{\includegraphics[width=\figwidth]{figures/FirstStepResultMATPLOTLIB}} 371 \caption{Visualization of the Poisson Equation Solution for $f=1$ using \MATPLOTLIB.} 372 \label{fig:FirstSteps.3b} 373 \end{figure} 374 375 Now we can write the script to solve our Poisson problem 376 \begin{python} 377 from esys.escript import * 378 from esys.escript.linearPDEs import Poisson 379 from esys.finley import Rectangle 380 import numpy 381 import matplotlib 382 import pylab 383 # generate domain: 384 mydomain = Rectangle(l0=1.,l1=1.,n0=40, n1=20) 385 # define characteristic function of Gamma^D 386 x = mydomain.getX() 387 gammaD = whereZero(x[0])+whereZero(x[1]) 388 # define PDE and get its solution u 389 mypde = Poisson(domain=mydomain) 390 mypde.setValue(f=1,q=gammaD) 391 u = mypde.getSolution() 392 # interpolate u to a matplotlib grid: 393 x_grid = numpy.linspace(0.,1.,50) 394 y_grid = numpy.linspace(0.,1.,50) 395 x=mydomain.getX()[0].toListOfTuples() 396 y=mydomain.getX()[1].toListOfTuples() 397 z=interpolate(u,mydomain.getX().getFunctionSpace()) 398 z_grid = matplotlib.mlab.griddata(x,y,z,xi=x_grid,yi=y_grid ) 399 # interpolate u to a rectangular grid: 400 matplotlib.pyplot.contourf(x_grid, y_grid, z_grid, 5) 401 matplotlib.pyplot.savefig("u.png") 402 \end{python} 403 The entire code is available as \file{poisson\hackscore matplotlib.py} in the \ExampleDirectory. 404 You can run the script using the {\it escript} environment 405 \begin{verbatim} 406 escript poisson_matplotlib.py 407 \end{verbatim} 408 This will create the \file{u.png}, see Figure~\fig{fig:FirstSteps.3b}. 409 For details on the usage of the \MATPLOTLIB module we refer to the documentation~\cite{matplotlib}. 410 411 As pointed out, \MATPLOTLIB is restricted to the two-dimensional case and 412 should be used for small problems only. It can not be used under \MPI as the \member{toListOfTuples} method is 413 not safe under \MPI\footnote{The phrase 'safe under \MPI' means that a program will produce correct results when run on more than one processor under \MPI.}. 414 415 \subsection{Visualization using \VTK} 416 417 As an alternative {\it escript} supports the usage of visualization tools which base on \VTK, eg. mayavi \cite{mayavi}, \VisIt~\cite{VisIt}. In this case the solution is written to a file in the \VTK format. This file the can read by the tool of choice. Using \VTK file is \MPI safe. 418 419 To write the solution \var{u} in Poisson problem to the file \file{u.xml} one need to add the line 420 \begin{python} 421 saveVTK("u.xml",sol=u) 422 \end{python} 423 The solution \var{u} is now available in the \file{u.xml} tagged with the name "sol". 424 425 \begin{figure} 426 \centerline{\includegraphics[width=\figwidth]{figures/FirstStepResult}} 427 \caption{Visualization of the Poisson Equation Solution for $f=1$} 428 \label{fig:FirstSteps.3} 429 \end{figure} 430 431 The Poisson problem script is now 432 \begin{python} 433 from esys.escript import * 434 from esys.escript.linearPDEs import Poisson 435 from esys.finley import Rectangle 436 # generate domain: 437 mydomain = Rectangle(l0=1.,l1=1.,n0=40, n1=20) 438 # define characteristic function of Gamma^D 439 x = mydomain.getX() 440 gammaD = whereZero(x[0])+whereZero(x[1]) 441 # define PDE and get its solution u 442 mypde = Poisson(domain=mydomain) 443 mypde.setValue(f=1,q=gammaD) 444 u = mypde.getSolution() 445 # write u to an external file 446 saveVTK("u.xml",sol=u) 447 \end{python} 448 The entire code is available as \file{poisson\hackscore VTK.py} in the \ExampleDirectory 449 450 You can run the script using the {\it escript} environment 451 and visualize the solution using \mayavi: 452 \begin{verbatim} 453 escript poisson\hackscore VTK.py 454 mayavi2 -d u.xml -m SurfaceMap 455 \end{verbatim} 456 The result is shown in Figure~\fig{fig:FirstSteps.3}.

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