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

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Updated section 1.2 of user's guide.


 1 2 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 3 % 4 % Copyright (c) 2003-2010 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}\label{FirstSteps} 16 In this chapter we give an introduction how to use \escript to solve 17 a partial differential equation\index{partial differential equation} (PDE\index{partial differential equation!PDE}). 18 We assume you are at least a little familiar with Python. 19 The knowledge presented in the Python tutorial at \url{http://docs.python.org/tut/tut.html} is more than sufficient. 20 21 The PDE\index{partial differential equation} we wish to solve is the Poisson equation\index{Poisson equation} 22 \begin{equation} 23 -\Delta u=f 24 \label{eq:FirstSteps.1} 25 \end{equation} 26 for the solution $u$. The function $f$ is the given right hand side. The domain of interest, denoted by $\Omega$, 27 is the unit square 28 \begin{equation} 29 \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 \} 30 \label{eq:FirstSteps.1b} 31 \end{equation} 32 The domain is shown in \fig{fig:FirstSteps.1}. 33 \begin{figure}[ht] 34 \centerline{\includegraphics{figures/FirstStepDomain}} 35 \caption{Domain $\Omega=[0,1]^2$ with outer normal field $n$.} 36 \label{fig:FirstSteps.1} 37 \end{figure} 38 39 $\Delta$ denotes the Laplace operator\index{Laplace operator}, which is defined by 40 \begin{equation} 41 \Delta u = (u\hackscore {,0})\hackscore{,0}+(u\hackscore{,1})\hackscore{,1} 42 \label{eq:FirstSteps.1.1} 43 \end{equation} 44 where, for any function $u$ and any direction $i$, $u\hackscore{,i}$ 45 denotes the partial derivative \index{partial derivative} of $u$ with respect 46 to $i$.\footnote{You may be more familiar with the Laplace 47 operator\index{Laplace operator} being written as $\nabla^2$, and written in 48 the form 49 \begin{equation*} 50 \nabla^2 u = \nabla^t \cdot \nabla u = \frac{\partial^2 u}{\partial x\hackscore 0^2} 51 + \frac{\partial^2 u}{\partial x\hackscore 1^2} 52 \end{equation*} 53 and \eqn{eq:FirstSteps.1} as 54 \begin{equation*} 55 -\nabla^2 u = f 56 \end{equation*} 57 } 58 Basically, in the subindex of a function, any index to the left of the comma denotes a spatial derivative with respect 59 to the index. To get a more compact form we will write $u\hackscore{,ij}=(u\hackscore {,i})\hackscore{,j}$ 60 which leads to 61 \begin{equation} 62 \Delta u = u\hackscore{,00}+u\hackscore{,11}=\sum\hackscore{i=0}^2 u\hackscore{,ii} 63 \label{eq:FirstSteps.1.1b} 64 \end{equation} 65 We often find that use 66 of nested $\sum$ symbols makes formulas cumbersome, and we use the more 67 convenient Einstein summation convention\index{summation convention}. This 68 drops the $\sum$ sign and assumes that a summation is performed over any repeated index. 69 For instance we write 70 \begin{eqnarray} 71 x\hackscore{i}y\hackscore{i}=\sum\hackscore{i=0}^2 x\hackscore{i}y\hackscore{i} \\ 72 x\hackscore{i}u\hackscore{,i}=\sum\hackscore{i=0}^2 x\hackscore{i}u\hackscore{,i} \\ 73 u\hackscore{,ii}=\sum\hackscore{i=0}^2 u\hackscore{,ii} \\ 74 x\hackscore{ij}u\hackscore{i,j}=\sum\hackscore{j=0}^2\sum\hackscore{i=0}^2 x\hackscore{ij}u\hackscore{i,j} \\ 75 \label{eq:FirstSteps.1.1c} 76 \end{eqnarray} 77 With the summation convention we can write the Poisson equation \index{Poisson equation} as 78 \begin{equation} 79 - u\hackscore{,ii} =1 80 \label{eq:FirstSteps.1.sum} 81 \end{equation} 82 where $f=1$ in this example. 83 84 On the boundary of the domain $\Omega$ the normal derivative $n\hackscore{i} u\hackscore{,i}$ 85 of the solution $u$ shall be zero, i.e. $u$ shall fulfill 86 the homogeneous Neumann boundary condition\index{Neumann 87 boundary condition!homogeneous} 88 \begin{equation} 89 n\hackscore{i} u\hackscore{,i}= 0 \;. 90 \label{eq:FirstSteps.2} 91 \end{equation} 92 $n=(n\hackscore{i})$ denotes the outer normal field 93 of the domain, see \fig{fig:FirstSteps.1}. Remember that we 94 are applying the Einstein summation convention \index{summation convention}, i.e. $n\hackscore{i} u\hackscore{,i}= n\hackscore{0} u\hackscore{,0} +% 95 n\hackscore{1} u\hackscore{,1}$.\footnote{Some readers may familiar with the 96 notation $\frac{\partial u}{\partial n} = n\hackscore{i} u\hackscore{,i}$ 97 for the normal derivative.} 98 The Neumann boundary condition of \eqn{eq:FirstSteps.2} should be fulfilled on the 99 set $\Gamma^N$ which is the top and right edge of the domain: 100 \begin{equation} 101 \Gamma^N=\{(x\hackscore 0;x\hackscore 1) \in \Omega | x\hackscore{0}=1 \mbox{ or } x\hackscore{1}=1 \} 102 \label{eq:FirstSteps.2b} 103 \end{equation} 104 On the bottom and the left edge of the domain which is defined 105 as 106 \begin{equation} 107 \Gamma^D=\{(x\hackscore 0;x\hackscore 1) \in \Omega | x\hackscore{0}=0 \mbox{ or } x\hackscore{1}=0 \} 108 \label{eq:FirstSteps.2c} 109 \end{equation} 110 the solution shall be identical to zero: 111 \begin{equation} 112 u=0 \; . 113 \label{eq:FirstSteps.2d} 114 \end{equation} 115 This kind of boundary condition is called a homogeneous Dirichlet boundary 116 condition\index{Dirichlet boundary condition!homogeneous}. 117 The partial differential equation in \eqn{eq:FirstSteps.1.sum} together 118 with the Neumann boundary condition \eqn{eq:FirstSteps.2} and 119 Dirichlet boundary condition in \eqn{eq:FirstSteps.2d} form a so-called 120 boundary value 121 problem\index{boundary value problem} (BVP\index{boundary value problem!BVP}) 122 for the unknown function~$u$. 123 124 \begin{figure}[ht] 125 \centerline{\includegraphics{figures/FirstStepMesh}} 126 \caption{Mesh of $4 \times 4$ elements on a rectangular domain. Here 127 each element is a quadrilateral and described by four nodes, namely 128 the corner points. The solution is interpolated by a bi-linear 129 polynomial.} 130 \label{fig:FirstSteps.2} 131 \end{figure} 132 133 In general the BVP\index{boundary value problem!BVP} cannot be solved 134 analytically and numerical methods have to be used to construct an 135 approximation of the solution $u$. 136 Here we will use the finite element method\index{finite element method} 137 (FEM\index{finite element method!FEM}). 138 The basic idea is to fill the domain with a set of points called nodes. 139 The solution is approximated by its values on the nodes\index{finite element method!nodes}. 140 Moreover, the domain is subdivided into smaller sub-domains called 141 elements\index{finite element method!element}. 142 On each element the solution is represented by a polynomial of a certain 143 degree through its values at the nodes located in the element. 144 The nodes and their connection through elements is called a 145 mesh\index{finite element method!mesh}. \fig{fig:FirstSteps.2} shows an 146 example of a FEM mesh with four elements in the $x_0$ and four elements 147 in the $x_1$ direction over the unit square. 148 For more details we refer the reader to the literature, for instance \Ref{Zienc,NumHand}. 149 150 The \escript solver we want to use to solve this problem is embedded into the python interpreter language. 151 So you can solve the problem interactively but you will learn quickly that it 152 is more efficient to use scripts which you can edit with your favorite editor. 153 To enter the escript environment, use the \program{run-escript} 154 command\footnote{\program{run-escript} is not available under Windows yet. 155 If you run under Windows you can just use the \program{python} command and the 156 \env{OMP_NUM_THREADS} environment variable to control the number of threads.}: 157 \begin{verbatim} 158 run-escript 159 \end{verbatim} 160 which will pass you on to the python prompt 161 \begin{verbatim} 162 Python 2.5.2 (r252:60911, Oct 5 2008, 19:29:17) 163 [GCC 4.3.2] on linux2 164 Type "help", "copyright", "credits" or "license" for more information. 165 >>> 166 \end{verbatim} 167 Here you can use all available python commands and language features, for instance 168 \begin{python} 169 >>> x=2+3 170 >>> print "2+3=",x 171 2+3= 5 172 \end{python} 173 We refer to the python user's guide if you not familiar with python. 174 175 \escript provides the class \Poisson to define a Poisson equation\index{Poisson equation}. 176 (We will discuss a more general form of a PDE\index{partial differential equation!PDE} 177 that can be defined through the \LinearPDE class later.) 178 The instantiation of a \Poisson class object requires the specification of the domain $\Omega$. 179 In \escript the \Domain class objects are used to describe the geometry of a 180 domain but it also contains information about the discretization methods and 181 the actual solver which is used to solve the PDE. 182 Here we are using the FEM library \finley\index{finite element method}. 183 The following statements create the \Domain object \var{mydomain} from the 184 \finley method \method{Rectangle}: 185 \begin{python} 186 from esys.finley import Rectangle 187 mydomain = Rectangle(l0=1.,l1=1.,n0=40, n1=20) 188 \end{python} 189 In this case the domain is a rectangle with the lower, left corner at point $(0,0)$ and 190 the right, upper corner at $(\var{l0},\var{l1})=(1,1)$. 191 The arguments \var{n0} and \var{n1} define the number of elements in $x\hackscore{0}$ and 192 $x\hackscore{1}$-direction respectively. For more details on \method{Rectangle} and 193 other \Domain generators within the \finley module, 194 see \Chap{CHAPTER ON FINLEY}. 195 196 The following statements define the \Poisson class object \var{mypde} with domain \var{mydomain} and 197 the right hand side $f$ of the PDE to constant $1$: 198 \begin{python} 199 from esys.escript.linearPDEs import Poisson 200 mypde = Poisson(mydomain) 201 mypde.setValue(f=1) 202 \end{python} 203 We have not specified any boundary condition but the \Poisson class implicitly 204 assumes homogeneous Neuman boundary conditions\index{Neumann boundary condition!homogeneous} defined by \eqn{eq:FirstSteps.2}. 205 With this boundary condition the BVP\index{boundary value problem!BVP} we have 206 defined has no unique solution. 207 In fact, with any solution $u$ and any constant $C$ the function $u+C$ becomes 208 a solution as well. 209 We have to add a Dirichlet boundary condition\index{Dirichlet boundary condition}. 210 This is done by defining a characteristic function\index{characteristic function} 211 which has positive values at locations $x=(x\hackscore{0},x\hackscore{1})$ 212 where Dirichlet boundary condition is set and $0$ elsewhere. 213 In our case of $\Gamma^D$ defined by \eqn{eq:FirstSteps.2c}, we need to 214 construct a function \var{gammaD} which is positive for the cases $x\hackscore{0}=0$ or $x\hackscore{1}=0$. 215 To get an object \var{x} which contains the coordinates of the nodes in the domain use 216 \begin{python} 217 x=mydomain.getX() 218 \end{python} 219 The method \method{getX} of the \Domain \var{mydomain} gives access to locations 220 in the domain defined by \var{mydomain}. 221 The object \var{x} is actually a \Data object which will be discussed in 222 \Chap{ESCRIPT CHAP} in more detail. 223 What we need to know here is that \var{x} has \Rank (number of dimensions) and 224 a \Shape (list of dimensions) which can be viewed by calling the \method{getRank} and \method{getShape} methods: 225 \begin{python} 226 print "rank ",x.getRank(),", shape ",x.getShape() 227 \end{python} 228 This will print something like 229 \begin{python} 230 rank 1, shape (2,) 231 \end{python} 232 The \Data object also maintains type information which is represented by the 233 \FunctionSpace of the object. For instance 234 \begin{python} 235 print x.getFunctionSpace() 236 \end{python} 237 will print 238 \begin{python} 239 Function space type: Finley_Nodes on FinleyMesh 240 \end{python} 241 which tells us that the coordinates are stored on the nodes of (rather than on 242 points in the interior of) a \finley mesh. 243 To get the $x\hackscore{0}$ coordinates of the locations we use the statement 244 \begin{python} 245 x0=x 246 \end{python} 247 Object \var{x0} is again a \Data object now with \Rank $0$ and \Shape $()$. 248 It inherits the \FunctionSpace from \var{x}: 249 \begin{python} 250 print x0.getRank(), x0.getShape(), x0.getFunctionSpace() 251 \end{python} 252 will print 253 \begin{python} 254 0 () Function space type: Finley_Nodes on FinleyMesh 255 \end{python} 256 We can now construct a function \var{gammaD} which is only non-zero on the 257 bottom and left edges of the domain with 258 \begin{python} 259 from esys.escript import whereZero 260 gammaD=whereZero(x)+whereZero(x) 261 \end{python} 262 263 \code{whereZero(x)} creates a function which equals $1$ where \code{x} is (almost) equal to zero and $0$ elsewhere. 264 Similarly, \code{whereZero(x)} creates a function which equals $1$ where \code{x} is equal to zero and $0$ elsewhere. 265 The sum of the results of \code{whereZero(x)} and \code{whereZero(x)} 266 gives a function on the domain \var{mydomain} which is strictly positive where $x\hackscore{0}$ or $x\hackscore{1}$ is equal to zero. 267 Note that \var{gammaD} has the same \Rank, \Shape and \FunctionSpace like \var{x0} used to define it. 268 So from 269 \begin{python} 270 print gammaD.getRank(), gammaD.getShape(), gammaD.getFunctionSpace() 271 \end{python} 272 one gets 273 \begin{python} 274 0 () Function space type: Finley_Nodes on FinleyMesh 275 \end{python} 276 An additional parameter \var{q} of the \code{setValue} method of the \Poisson 277 class defines the characteristic function\index{characteristic function} of 278 the locations of the domain where the homogeneous Dirichlet boundary condition\index{Dirichlet boundary condition!homogeneous} is set. 279 The complete definition of our example is now: 280 \begin{python} 281 from esys.escript.linearPDEs import Poisson 282 x = mydomain.getX() 283 gammaD = whereZero(x)+whereZero(x) 284 mypde = Poisson(domain=mydomain) 285 mypde.setValue(f=1,q=gammaD) 286 \end{python} 287 The first statement imports the \Poisson class definition from the \linearPDEs module. 288 To get the solution of the Poisson equation defined by \var{mypde} we just have to call its \method{getSolution} method. 289 290 Now we can write the script to solve our Poisson problem 291 \begin{python} 292 from esys.escript import * 293 from esys.escript.linearPDEs import Poisson 294 from esys.finley import Rectangle 295 # generate domain: 296 mydomain = Rectangle(l0=1.,l1=1.,n0=40, n1=20) 297 # define characteristic function of Gamma^D 298 x = mydomain.getX() 299 gammaD = whereZero(x)+whereZero(x) 300 # define PDE and get its solution u 301 mypde = Poisson(domain=mydomain) 302 mypde.setValue(f=1,q=gammaD) 303 u = mypde.getSolution() 304 \end{python} 305 The question is what we do with the calculated solution \var{u}. 306 Besides postprocessing, e.g. calculating the gradient or the average value, which will be discussed later, plotting the solution is one of the things you might want to do. 307 \escript offers two ways to do this, both based on external modules or packages and so data need to converted to hand over the solution. 308 The first option is using the \MATPLOTLIB module which allows plotting 2D results relatively quickly from within the Python script, see~\cite{matplotlib}. 309 However, there are limitations when using this tool, especially for large problems and when solving 3-dimensional problems. 310 Therefore, \escript provides functionality to export data as files which can subsequently be read by third-party software packages such as \mayavi\cite{mayavi} or \VisIt~\cite{VisIt}. 311 312 \subsection{Plotting Using \MATPLOTLIB} 313 The \MATPLOTLIB module provides a simple and easy-to-use way to visualize PDE solutions (or other \Data objects). 314 To hand over data from \escript to \MATPLOTLIB the values need to mapped onto 315 a rectangular grid\footnote{Users of Debian 5 (Lenny) please note: this example 316 makes use of the \function{griddata} method in \module{matplotlib.mlab}. 317 This method is not part of version 0.98.1 which is available with Lenny. 318 If you wish to use contour plots, you may need to install a later version. 319 Users of Ubuntu 8.10 or later should be fine.}. We will make use of the \numpy module. 320 321 First we need to create a rectangular grid which is accomplished by the following statements: 322 \begin{python} 323 import numpy 324 x_grid = numpy.linspace(0., 1., 50) 325 y_grid = numpy.linspace(0., 1., 50) 326 \end{python} 327 \var{x_grid} is an array defining the x coordinates of the grid while 328 \var{y_grid} defines the y coordinates of the grid. 329 In this case we use $50$ points over the interval $[0,1]$ in both directions. 330 331 Now the values created by \escript need to be interpolated to this grid. 332 We will use the \MATPLOTLIB \function{mlab.griddata} function to do this. 333 Spatial coordinates are easily extracted as a \var{list} by 334 \begin{python} 335 x=mydomain.getX().toListOfTuples() 336 y=mydomain.getX().toListOfTuples() 337 \end{python} 338 In principle we can apply the same \member{toListOfTuples} method to extract the values from the PDE solution \var{u}. 339 However, we have to make sure that the \Data object we extract the values from 340 uses the same \FunctionSpace as we have used when extracting \var{x} and \var{y}. 341 We apply the \function{interpolation} to \var{u} before extraction to achieve this: 342 \begin{python} 343 z=interpolate(u, mydomain.getX().getFunctionSpace()) 344 \end{python} 345 The values in \var{z} are the values at the points with the coordinates given by \var{x} and \var{y}. 346 These values are interpolated to the grid defined by \var{x_grid} and \var{y_grid} by using 347 \begin{python} 348 import matplotlib 349 z_grid = matplotlib.mlab.griddata(x, y, z, xi=x_grid, yi=y_grid) 350 \end{python} 351 Now \var{z_grid} gives the values of the PDE solution \var{u} at the grid which can be plotted using \function{contourf}: 352 \begin{python} 353 matplotlib.pyplot.contourf(x_grid, y_grid, z_grid, 5) 354 matplotlib.pyplot.savefig("u.png") 355 \end{python} 356 Here we use 5 contours. The last statement writes the plot to the file \file{u.png} in the PNG format. 357 Alternatively, one can use 358 \begin{python} 359 matplotlib.pyplot.contourf(x_grid, y_grid, z_grid, 5) 360 matplotlib.pyplot.show() 361 \end{python} 362 which gives an interactive browser window. 363 364 \begin{figure} 365 \centerline{\includegraphics[width=\figwidth]{figures/FirstStepResultMATPLOTLIB}} 366 \caption{Visualization of the Poisson Equation Solution for $f=1$ using \MATPLOTLIB} 367 \label{fig:FirstSteps.3b} 368 \end{figure} 369 370 Now we can write the script to solve our Poisson problem 371 \begin{python} 372 from esys.escript import * 373 from esys.escript.linearPDEs import Poisson 374 from esys.finley import Rectangle 375 import numpy 376 import matplotlib 377 import pylab 378 # generate domain: 379 mydomain = Rectangle(l0=1.,l1=1.,n0=40, n1=20) 380 # define characteristic function of Gamma^D 381 x = mydomain.getX() 382 gammaD = whereZero(x)+whereZero(x) 383 # define PDE and get its solution u 384 mypde = Poisson(domain=mydomain) 385 mypde.setValue(f=1,q=gammaD) 386 u = mypde.getSolution() 387 # interpolate u to a matplotlib grid: 388 x_grid = numpy.linspace(0.,1.,50) 389 y_grid = numpy.linspace(0.,1.,50) 390 x=mydomain.getX().toListOfTuples() 391 y=mydomain.getX().toListOfTuples() 392 z=interpolate(u,mydomain.getX().getFunctionSpace()) 393 z_grid = matplotlib.mlab.griddata(x,y,z,xi=x_grid,yi=y_grid ) 394 # interpolate u to a rectangular grid: 395 matplotlib.pyplot.contourf(x_grid, y_grid, z_grid, 5) 396 matplotlib.pyplot.savefig("u.png") 397 \end{python} 398 The entire code is available as \file{poisson\hackscore matplotlib.py} in the \ExampleDirectory. 399 You can run the script using the {\it escript} environment 400 \begin{verbatim} 401 run-escript poisson_matplotlib.py 402 \end{verbatim} 403 This will create the \file{u.png}, see \fig{fig:FirstSteps.3b}. 404 For details on the usage of the \MATPLOTLIB module we refer to the documentation~\cite{matplotlib}. 405 406 As pointed out, \MATPLOTLIB is restricted to the two-dimensional case and 407 should be used for small problems only. 408 It can not be used under \MPI as the \member{toListOfTuples} method is not 409 safe under \MPI\footnote{The phrase 'safe under \MPI' means that a program 410 will produce correct results when run on more than one processor under \MPI.}. 411 412 \begin{figure} 413 \centerline{\includegraphics[width=\figwidth]{figures/FirstStepResult}} 414 \caption{Visualization of the Poisson Equation Solution for $f=1$} 415 \label{fig:FirstSteps.3} 416 \end{figure} 417 418 \subsection{Visualization using export files} 419 420 As an alternative to \MATPLOTLIB, {\it escript} supports exporting data to 421 \VTK and \SILO files which can be read by visualization tools such as 422 mayavi\cite{mayavi} and \VisIt~\cite{VisIt}. This method is \MPI safe and 423 works with large 2D and 3D problems. 424 425 To write the solution \var{u} of the Poisson problem in the \VTK file format 426 to the file \file{u.vtu} one needs to add: 427 \begin{python} 428 saveVTK("u.vtu", sol=u) 429 \end{python} 430 This file can then be opened in a \VTK compatible visualization tool where the 431 solution is accessible by the name {\it sol}. 432 433 The Poisson problem script is now 434 \begin{python} 435 from esys.escript import * 436 from esys.escript.linearPDEs import Poisson 437 from esys.finley import Rectangle 438 # generate domain: 439 mydomain = Rectangle(l0=1.,l1=1.,n0=40, n1=20) 440 # define characteristic function of Gamma^D 441 x = mydomain.getX() 442 gammaD = whereZero(x)+whereZero(x) 443 # define PDE and get its solution u 444 mypde = Poisson(domain=mydomain) 445 mypde.setValue(f=1,q=gammaD) 446 u = mypde.getSolution() 447 # write u to an external file 448 saveVTK("u.vtu",sol=u) 449 \end{python} 450 The entire code is available as \file{poisson\hackscore vtk.py} in the \ExampleDirectory. 451 452 You can run the script using the {\it escript} environment and visualize the 453 solution using \mayavi: 454 \begin{verbatim} 455 run-escript poisson_VTK.py 456 mayavi2 -d u.vtu -m SurfaceMap 457 \end{verbatim} 458 The result is shown in \fig{fig:FirstSteps.3}. 459

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