# Contents of /trunk/doc/cookbook/example01.tex

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more review in the cookbook

 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 We will start by examining a simple one dimensional heat diffusion example. This problem will provide a good launch pad to build our knowledge of \esc and demonstrate how to solve simple partial differential equations (PDEs)\footnote{Wikipedia provides an excellent and comprehensive introduction to \textit{Partial Differential Equations} \url{http://en.wikipedia.org/wiki/Partial_differential_equation}, however their relevance to \esc and implementation should become a clearer as we develop our understanding further into the cookbook.} 15 16 \section{One Dimensional Heat Diffusion in an Iron Rod} 17 \sslist{onedheatdiff001.py and cblib.py} 18 %\label{Sec:1DHDv0} 19 The first model consists of a simple cold iron bar at a constant temperature of zero \reffig{fig:onedhdmodel}. The bar is perfectly insulated on all sides with a heating element at one end. Intuition tells us that as heat is applied; energy will disperse along the bar via conduction. With time the bar will reach a constant temperature equivalent to that of the heat source. 20 \begin{figure}[h!] 21 \centerline{\includegraphics[width=4.in]{figures/onedheatdiff}} 22 \caption{One dimensional model of an Iron bar.} 23 \label{fig:onedhdmodel} 24 \end{figure} 25 \subsection{1D Heat Diffusion Equation} 26 We can model the heat distribution of this problem over time using the one dimensional heat diffusion equation\footnote{A detailed discussion on how the heat diffusion equation is derived can be found at \url{http://online.redwoods.edu/instruct/darnold/DEProj/sp02/AbeRichards/paper.pdf}}; 27 which is defined as: 28 \begin{equation} 29 \rho c\hackscore p \frac{\partial T}{\partial t} - \kappa \frac{\partial^{2} T}{\partial x^{2}} = q\hackscore H 30 \label{eqn:hd} 31 \end{equation} 32 where $\rho$ is the material density, $c\hackscore p$ is the specific heat and $\kappa$ is the thermal 33 conductivity\footnote{A list of some common thermal conductivities is available from Wikipedia \url{http://en.wikipedia.org/wiki/List_of_thermal_conductivities}}. Here we assume that these material 34 parameters are \textbf{constant}. 35 The heat source is defined by the right hand side of \refEq{eqn:hd} as $q\hackscore{H}$; this can take the form of a constant or a function of time and space. For example $q\hackscore{H} = q\hackscore{0}e^{-\gamma t}$ where we have the output of our heat source decaying with time. There are also two partial derivatives in \refEq{eqn:hd}; $\frac{\partial T}{\partial t}$ describes the change in temperature with time while $\frac{\partial ^2 T}{\partial x^2}$ is the spatial change of temperature. As there is only a single spatial dimension to our problem, our temperature solution $T$ is only dependent on the time $t$ and our position along the iron bar $x$. 36 37 \subsection{PDEs and the General Form} 38 Potentially, it is now possible to solve PDE \refEq{eqn:hd} analytically and this would produce an exact solution to our problem. However, it is not always possible or practical to solve a problem this way. Alternatively, computers can be used to solve these kinds of problems. To do this, a numerical approach is required to discretised 39 the PDE \refEq{eqn:hd} in time and space so finally we are left with a finite number of equations for a finite number of spatial and time steps in the model. While discretization introduces approximations and a degree of error, we find that a sufficiently sampled model is generally accurate enough for the requirements of the modeler. 40 41 Firstly, we will discretise the PDE \refEq{eqn:hd} in the time direction which will 42 leave as with a steady linear PDE which is involving spatial derivatives only and needs to be solved in each time 43 step to progress in time - \esc can help us here. 44 45 For the discretization in time we will use is the Backwards Euler approximation scheme\footnote{see \url{http://en.wikipedia.org/wiki/Euler_method}}. It bases on the 46 approximation 47 \begin{equation} 48 \frac{\partial T(t)}{\partial t} \approx \frac{T(t)-T(t-h)}{h} 49 \label{eqn:beuler} 50 \end{equation} 51 for $\frac{\partial T}{\partial t}$ at time $t$ 52 where $h$ is the time step size. This can also be written as; 53 \begin{equation} 54 \frac{\partial T}{\partial t}(t^{(n)}) \approx \frac{T^{(n)} - T^{(n-1)}}{h} 55 \label{eqn:Tbeuler} 56 \end{equation} 57 where the upper index $n$ denotes the n\textsuperscript{th} time step. So one has 58 \begin{equation} 59 \begin{array}{rcl} 60 t^{(n)} & = & t^{(n-1)}+h \\ 61 T^{(n)} & = & T(t^{(n-1)}) \\ 62 \end{array} 63 \label{eqn:Neuler} 64 \end{equation} 65 Substituting \refEq{eqn:Tbeuler} into \refEq{eqn:hd} we get; 66 \begin{equation} 67 \frac{\rho c\hackscore p}{h} (T^{(n)} - T^{(n-1)}) - \kappa \frac{\partial^{2} T^{(n)}}{\partial x^{2}} = q\hackscore H 68 \label{eqn:hddisc} 69 \end{equation} 70 Notice that we evaluate the spatial derivative term at current time $t^{(n)}$ - therefore the name \textbf{backward Euler} scheme. Alternatively, one can use evaluate the spatial derivative term at the previous time $t^{(n-1)}$. This 71 approach is called the \textbf{forward Euler} scheme. This scheme can provide some computational advantages which 72 we are not discussed here but has the major disadvantage that depending on the 73 material parameter as well as the discretiztion of the spatial derivative term the time step size $h$ needs to be chosen sufficiently small to achieve a stable temperature when progressing in time. The term \textit{stable} means 74 that the approximation of the temperature will not grow beyond its initial bounds and becomes unphysical. 75 The backward Euler which we use here is unconditionally stable meaning that under the assumption of 76 physically correct problem set-up the temperature approximation remains physical for all times. 77 The user needs to keep in mind that the discretization error introduced by \refEq{eqn:beuler} 78 is sufficiently small so a good approximation of the true temperature is calculated. It is 79 therefore crucial that the user remains critical about his/her results and for instance compares 80 the results for different time and spatial step sizes. 81 82 To get the temperature $T^{(n)}$ at time $t^{(n)}$ we need to solve the linear 83 differential equation \refEq{eqn:hddisc} which is only including spatial derivatives. To solve this problem 84 we want to to use \esc. 85 86 \esc interfaces with any given PDE via a general form. For the purpose of this introduction we will illustrate a simpler version of the full linear PDE general form which is available in the \esc user's guide. A simplified form that suits our heat diffusion problem\footnote{In the form of the \esc users guide which using the Einstein convention is written as 87 $-(A\hackscore{jl} u\hackscore{,l})\hackscore{,j}+D u =Y$} 88 is described by; 89 \begin{equation}\label{eqn:commonform nabla} 90 -\nabla\cdot(A\cdot\nabla u) + Du = f 91 \end{equation} 92 where $A$, $D$ and $f$ are known values and $u$ is the unknown solution. The symbol $\nabla$ which is called the \textit{Nabla operator} or \textit{del operator} represents 93 the spatial derivative of its subject - in this case $u$. Lets assume for a moment that we deal with a one-dimensional problem then ; 94 \begin{equation} 95 \nabla = \frac{\partial}{\partial x} 96 \end{equation} 97 and we can write \refEq{eqn:commonform nabla} as; 98 \begin{equation}\label{eqn:commonform} 99 -A\frac{\partial^{2}u}{\partial x^{2}} + Du = f 100 \end{equation} 101 if $A$ is constant. To match this simplified general form to our problem \refEq{eqn:hddisc} 102 we rearrange \refEq{eqn:hddisc}; 103 \begin{equation} 104 \frac{\rho c\hackscore p}{h} T^{(n)} - \kappa \frac{\partial^2 T^{(n)}}{\partial x^2} = q\hackscore H + \frac{\rho c\hackscore p}{h} T^{(n-1)} 105 \label{eqn:hdgenf} 106 \end{equation} 107 The PDE is now in a form that satisfies \refEq{eqn:commonform nabla} which is required for \esc to solve our PDE. This can be done by generating a solution for successive increments in the time nodes $t^{(n)}$ where 108 $t^{(0)}=0$ and $t^{(n)}=t^{(n-1)}+h$ where $h>0$ is the step size and assumed to be constant. 109 In the following the upper index ${(n)}$ refers to a value at time $t^{(n)}$. Finally, by comparing \refEq{eqn:hdgenf} with \refEq{eqn:commonform} it can be seen that; 110 \begin{equation}\label{ESCRIPT SET} 111 u=T^{(n)}; 112 A = \kappa; D = \frac{\rho c \hackscore{p}}{h}; f = q \hackscore{H} + \frac{\rho c\hackscore p}{h} T^{(n-1)} 113 \end{equation} 114 115 Now that the general form has been established, it can be submitted to \esc. Note that it is necessary to establish the state of our system at time zero or $T^{(n=0)}$. This is due to the time derivative approximation we have used from \refEq{eqn:Tbeuler}. Our model stipulates a starting temperature in the iron bar of 0\textcelsius. Thus the temperature distribution is simply; 116 \begin{equation} 117 T(x,0) = T\hackscore{ref} = 0 118 \end{equation} 119 for all $x$ in the domain. 120 121 \subsection{Boundary Conditions} 122 With the PDE sufficiently modified, consideration must now be given to the boundary conditions of our model. Typically there are two main types of boundary conditions known as \textbf{Neumann} and \textbf{Dirichlet} boundary conditions\footnote{More information on Boundary Conditions is available at Wikipedia \url{http://en.wikipedia.org/wiki/Boundary_conditions}}, respectively. 123 A \textbf{Dirichlet boundary condition} is conceptually simpler and is used to prescribe a known value to the unknown - in our example the temperature - on parts of or the entire boundary of the region of interest. 124 For our model problem we want to keep the initial temperature setting on the left side of the 125 iron bar over time. This defines a Dirichlet boundary condition for the PDE \refEq{eqn:hddisc} to be solved at each time step. 126 127 On the other end of the iron rod we want to add an appropriate boundary condition to define insolation to prevent 128 any loss or inflow of energy at the right end of the rod. Mathematically this is expressed by prescribing 129 the heat flux $\kappa \frac{\partial T}{\partial x}$ to zero on the right end of the rod 130 In our simplified one dimensional model this is expressed 131 in the form; 132 \begin{equation} 133 \kappa \frac{\partial T}{\partial x} = 0 134 \end{equation} 135 or in a more general case as 136 \begin{equation}\label{NEUMAN 1} 137 \kappa \nabla T \cdot n = 0 138 \end{equation} 139 where $n$ is the outer normal field \index{outer normal field} at the surface of the domain. 140 For the iron rod the outer normal field on the right hand side is the vector $(1,0)$. The $\cdot$ (dot) refers to the 141 dot product of the vectors $\nabla T$ and $n$. In fact, the term $\nabla T \cdot n$ is the normal derivative of 142 the temperature $T$. Other notations which are used are\footnote{The \esc notation for the normal 143 derivative is $T\hackscore{,i} n\hackscore i$.}; 144 \begin{equation} 145 \nabla T \cdot n = \frac{\partial T}{\partial n} \; . 146 \end{equation} 147 A condition of the type \refEq{NEUMAN 1} defines a \textbf{Neuman boundary condition} for the PDE. 148 149 The PDE \refEq{eqn:hdgenf} together with the Dirichlet boundary condition set on the left face of the rod 150 and the Neuman boundary condition~\ref{eqn:hdgenf} define a \textbf{boundary value problem}. 151 It is a nature of a boundary value problem that it allows to make statements on the solution in the 152 interior of the domain from information known on the boundary only. In most cases 153 we use the term partial differential equation but in fact mean a boundary value problem. 154 It is important to keep in mind that boundary conditions need to be complete and consistent in the sense that 155 at any point on the boundary either a Dirichlet or a Neuman boundary condition must be set. 156 157 Conviniently, \esc makes default assumption on the boundary conditions which the user may modify where appropriate. 158 For a problem of the form in~\refEq{eqn:commonform nabla} the default condition\footnote{In the form of the \esc users guide which is using the Einstein convention is written as 159 $n\hackscore{j}A\hackscore{jl} u\hackscore{,l}=0$.} is; 160 \begin{equation}\label{NEUMAN 2} 161 n\cdot A \cdot\nabla u = 0 162 \end{equation} 163 which is used everywhere on the boundary. Again $n$ denotes the outer normal field. 164 Notice that the coefficient $A$ is the same as in the \esc PDE~\ref{eqn:commonform nabla}. 165 With the settings for the coefficients we have already identified in \refEq{ESCRIPT SET} this 166 condition translates into 167 \begin{equation}\label{NEUMAN 2} 168 \kappa \frac{\partial T}{\partial x} = 0 169 \end{equation} 170 for the right hand side of the rod. This is identical to the Neuman boundary condition we want to set. \esc will take care of this condition for us. We will discuss the Dirichlet boundary condition later. 171 172 173 174 175 \subsection{A \textit{1D} Clarification} 176 It is necessary for clarification that we revisit the general PDE from \refeq{eqn:commonform nabla} under the light of a two dimensional domain. \esc is inherently designed to solve problems that are greater than one dimension and so \refEq{eqn:commonform nabla} needs to be read as a higher dimensional problem. In the case of two spatial dimensions the \textit{Nabla operator} has in fact two components $\nabla = (\frac{\partial}{\partial x}, \frac{\partial}{\partial y})$. In full, \refEq{eqn:commonform nabla} assuming a constant coefficient $A$, takes the form; 177 \begin{equation}\label{eqn:commonform2D} 178 -A\hackscore{00}\frac{\partial^{2}u}{\partial x^{2}} 179 -A\hackscore{01}\frac{\partial^{2}u}{\partial x\partial y} 180 -A\hackscore{10}\frac{\partial^{2}u}{\partial y\partial x} 181 -A\hackscore{11}\frac{\partial^{2}u}{\partial y^{2}} 182 + Du = f 183 \end{equation} 184 Notice that for the higher dimensional case $A$ becomes a matrix. It is also 185 important to notice that the usage of the Nabla operator creates 186 a compact formulation which is also independent from the spatial dimension. 187 So to make the general PDE \refEq{eqn:commonform2D} one dimensional as 188 shown in \refEq{eqn:commonform} we need to set 189 \begin{equation} 190 A\hackscore{00}=A; A\hackscore{01}=A\hackscore{10}=A\hackscore{11}=0 191 \end{equation} 192 193 \subsection{Developing a PDE Solution Script} 194 \label{sec:key} 195 To solve the heat diffusion equation (equation \refEq{eqn:hd}) we will write a simple \pyt script which uses the \modescript, \modfinley and \modmpl modules. At this point we assume that you have some basic understanding of the \pyt programming language. If not there are some pointers and links available in Section \ref{sec:escpybas} . 196 197 By developing a script for \esc, the heat diffusion equation can be solved at successive time steps for a predefined period using our general form \refEq{eqn:hdgenf}. Firstly it is necessary to import all the libraries\footnote{The libraries contain predefined scripts that are required to solve certain problems, these can be simple like $sine$ and $cosine$ functions or more complicated like those from our \esc library.} 198 that we will require. 199 \begin{python} 200 from esys.escript import * 201 # This defines the LinearPDE module as LinearPDE 202 from esys.escript.linearPDEs import LinearPDE 203 # This imports the rectangle domain function from finley. 204 from esys.finley import Rectangle 205 # A useful unit handling package which will make sure all our units 206 # match up in the equations under SI. 207 from esys.escript.unitsSI import * 208 import pylab as pl #Plotting package. 209 import numpy as np #Array package. 210 import os #This package is necessary to handle saving our data. 211 \end{python} 212 It is generally a good idea to import all of the \modescript library, although if the functions and classes required are known they can be specified individually. The function \verb|LinearPDE| has been imported explicitly for ease of use later in the script. \verb|Rectangle| is going to be our type of model. The module \verb unitsSI provides support for SI unit definitions with our variables; and the \verb|os| module is needed to handle file outputs once our PDE has been solved. \verb pylab and \verb numpy are modules developed independently of \esc. They are used because they have efficient plotting and array handling capabilities. 213 214 Once our library dependencies have been established, defining the problem specific variables is the next step. In general the number of variables needed will vary between problems. These variables belong to two categories. They are either directly related to the PDE and can be used as inputs into the \esc solver, or they are script variables used to control internal functions and iterations in our problem. For this PDE there are a number of constants which will need values. Firstly, the model upon which we wish to solve our problem needs to be defined. There are many different types of models in \modescript which we will demonstrate in later tutorials but for our iron rod, we will simply use a rectangular model. 215 216 Using a rectangular model simplifies our rod which would be a \textit{3D} object, into a single dimension. The iron rod will have a lengthways cross section that looks like a rectangle. As a result we do not need to model the volume of the rod because a cylinder is symmetrical about its centre. There are four arguments we must consider when we decide to create a rectangular model, the model \textit{length}, \textit{width} and \textit{step size} in each direction. When defining the size of our problem it will help us determine appropriate values for our model arguments. If we make our dimensions large but our step sizes very small we will to a point, increase the accuracy of our solution. Unfortunately we also increase the number of calculations that must be solved per time step. This means more computational time is required to produce a solution. In this \textit{1D} problem, the bar is defined as being 1 metre long. An appropriate step size \verb|ndx| would be 1 to 10\% of the length. Our \verb|ndy| need only be 1, this is because our problem stipulates no partial derivatives in the $y$ direction. Thus the temperature does not vary with $y$. Hence, the model parameters can be defined as follows; note we have used the \verb unitsSI convention to make sure all our input units are converted to SI. 217 \begin{python} 218 #Domain related. 219 mx = 1*m #meters - model length 220 my = .1*m #meters - model width 221 ndx = 100 # mesh steps in x direction 222 ndy = 1 # mesh steps in y direction - one dimension means one element 223 \end{python} 224 The material constants and the temperature variables must also be defined. For the iron rod in the model they are defined as: 225 \begin{python} 226 #PDE related 227 q=200. * Celsius #Kelvin - our heat source temperature 228 Tref = 0. * Celsius #Kelvin - starting temp of iron bar 229 rho = 7874. *kg/m**3 #kg/m^{3} density of iron 230 cp = 449.*J/(kg*K) #j/Kg.K thermal capacity 231 rhocp = rho*cp 232 kappa = 80.*W/m/K #watts/m.Kthermal conductivity 233 \end{python} 234 Finally, to control our script we will have to specify our timing controls and where we would like to save the output from the solver. This is simple enough: 235 \begin{python} 236 t=0 #our start time, usually zero 237 tend=5.*minute #seconds - time to end simulation 238 outputs = 200 # number of time steps required. 239 h=(tend-t)/outputs #size of time step 240 #user warning statement 241 print "Expected Number of time outputs is: ", (tend-t)/h 242 i=0 #loop counter 243 #the folder to put our outputs in, leave blank "" for script path 244 save_path="data/onedheatdiff001" 245 \end{python} 246 Now that we know our inputs we will build a domain using the \verb Rectangle() function from \verb finley . The four arguments allow us to define our domain \verb rod as: 247 \begin{python} 248 #generate domain using rectangle 249 rod = Rectangle(l0=mx,l1=my,n0=ndx, n1=ndy) 250 \end{python} 251 \verb rod now describes a domain in the manner of Section \ref{ss:domcon}. As we define our variables, various function spaces will be created to accommodate them. There is an easy way to extract finite points from the domain \verb|rod| using the domain property function \verb|getX()| . This function sets the vertices of each cell as finite points to solve in the solution. If we let \verb|x| be these finite points, then; 252 \begin{python} 253 #extract finite points - the solution points 254 x=rod.getX() 255 \end{python} 256 The data locations of specific function spaces can be returned in a similar manner by extracting the relevant function space from the domain followed by the \verb .getX() operator. 257 258 With a domain and all our required variables established, it is now possible to set up our PDE so that it can be solved by \esc. The first step is to define the type of PDE that we are trying to solve in each time step. In this example it is a single linear PDE\footnote{in comparison to a system of PDEs which will be discussed later.}. We also need to state the values of our general form variables. 259 \begin{python} 260 mypde=LinearSinglePDE(rod) 261 mypde.setValue(A=kappa*kronecker(rod),D=rhocp/h) 262 \end{python} 263 264 In a few special cases it may be possible to decrease the computational time of the solver if our PDE is symmetric. Symmetry of a PDE is defined by; 265 \begin{equation}\label{eqn:symm} 266 A\hackscore{jl}=A\hackscore{lj} 267 \end{equation} 268 Symmetry is only dependent on the $A$ coefficient in the general form and the other coefficients $D$ and $d$ as well as the RHS $Y$ and $y$ may take any value. From the above definition we can see that our PDE is symmetric. The \verb LinearPDE class provides the method \method{checkSymmetry} to check if the given PDE is symmetric. As our PDE is symmetrical we will enable symmetry via; 269 \begin{python} 270 myPDE.setSymmetryOn() 271 \end{python} 272 273 We now need to specify our boundary conditions and initial values. The initial values required to solve this PDE are temperatures for each discrete point in our model. We will set our bar to: 274 \begin{python} 275 T = Tref 276 \end{python} 277 Boundary conditions are a little more difficult. Fortunately the \esc solver will handle our insulated boundary conditions by default with a zero flux operator. However, we will need to apply our heat source $q_{H}$ to the end of the bar at $x=0$ . \esc makes this easy by letting us define areas in our model. The finite points in the model were previously defined as \verb x and it is possible to set all of points that satisfy $x=0$ to \verb q via the \verb whereZero() function. There are a few \verb where functions available in \esc. They will return a value \verb 1 where they are satisfied and \verb 0 where they are not. In this case our \verb qH is only applied to the far LHS of our model as required. 278 \begin{python} 279 # ... set heat source: .... 280 qH=q*whereZero(x[0]) 281 \end{python} 282 283 Finally we will initialise an iteration loop to solve our PDE for all the time steps we specified in the variable section. As the RHS of the general form is dependent on the previous values for temperature \verb T across the bar this must be updated in the loop. Our output at each time step is \verb T the heat distribution and \verb totT the total heat in the system. 284 \begin{python} 285 while t<=tend: 286 i+=1 #increment the counter 287 t+=h #increment the current time 288 mypde.setValue(Y=qH+rhocp/h*T) #set variable PDE coefficients 289 T=mypde.getSolution() #get the PDE solution 290 totT = rhocp*T #get the total heat solution in the system 291 \end{python} 292 293 \subsection{Plotting the heat solutions} 294 Visualisation of the solution can be achieved using \mpl a module contained within \pylab. We start by modifying our solution script from before. Prior to the \verb while loop we will need to extract our finite solution points to a data object that is compatible with \mpl. First it is necessary to convert \verb x to a list of tuples. These are then converted to a \numpy array and the $x$ locations extracted via an array slice to the variable \verb plx . 295 \begin{python} 296 #convert solution points for plotting 297 plx = x.toListOfTuples() 298 plx = np.array(plx) #convert to tuple to numpy array 299 plx = plx[:,0] #extract x locations 300 \end{python} 301 As there are two solution outputs, we will generate two plots and save each to a file for every time step in the solution. The following is appended to the end of the \verb while loop and creates two figures. The first figure is for the temperature distribution, and the second the total temperature in the bar. Both cases are similar with a few minor changes for scale and labelling. We start by converting the solution to a tuple and then plotting this against our \textit{x coordinates} \verb plx from before. The axis is then standardised and a title applied. Finally, the figure is saved to a *.png file and cleared for the following iteration. 302 \begin{python} 303 #establish figure 1 for temperature vs x plots 304 tempT = T.toListOfTuples(scalarastuple=False) 305 pl.figure(1) #current figure 306 pl.plot(plx,tempT) #plot solution 307 #define axis extents and title 308 pl.axis([0,1.0,273.14990+0.00008,0.004+273.1499]) 309 pl.title("Temperature accross Rod") 310 #save figure to file 311 pl.savefig(os.path.join(save_path+"/tempT","rodpyplot%03d.png") %i) 312 pl.clf() #clear figure 313 314 #establish figure 2 for total temperature vs x plots and repeat 315 tottempT = totT.toListOfTuples(scalarastuple=False) 316 pl.figure(2) 317 pl.plot(plx,tottempT) 318 pl.axis([0,1.0,9.657E08,12000+9.657E08]) 319 pl.title("Total temperature across Rod") 320 pl.savefig(os.path.join(save_path+"/totT","ttrodpyplot%03d.png")%i) 321 pl.clf() 322 \end{python} 323 \begin{figure} 324 \begin{center} 325 \includegraphics[width=4in]{figures/ttrodpyplot150} 326 \caption{Total temperature ($T$) distribution in rod at $t=150$} 327 \label{fig:onedheatout} 328 \end{center} 329 \end{figure} 330 331 \subsubsection{Parallel scripts (MPI)} 332 In some of the example files for this cookbook the plotting commands are a little different. 333 For example, 334 \begin{python} 335 if getMPIRankWorld() == 0: 336 pl.savefig(os.path.join(save_path+"/totT","ttrodpyplot%03d.png")%i) 337 pl.clf() 338 \end{python} 339 340 The additional \verb if statement is not necessary for normal desktop use. 341 It becomes important for scripts run on parallel computers. 342 Its purpose is to ensure that only one copy of the file is written. 343 For more details on writing scripts for parallel computing please consult the \emph{user's guide}. 344 345 \subsection{Make a video} 346 Our saved plots from the previous section can be cast into a video using the following command appended to the end of the script. \verb mencoder is linux only however, and other platform users will need to use an alternative video encoder. 347 \begin{python} 348 # compile the *.png files to create two *.avi videos that show T change 349 # with time. This operation uses linux mencoder. For other operating 350 # systems it is possible to use your favourite video compiler to 351 # convert image files to videos. 352 353 os.system("mencoder mf://"+save_path+"/tempT"+"/*.png -mf type=png:\ 354 w=800:h=600:fps=25 -ovc lavc -lavcopts vcodec=mpeg4 -oac copy -o \ 355 onedheatdiff001tempT.avi") 356 357 os.system("mencoder mf://"+save_path+"/totT"+"/*.png -mf type=png:\ 358 w=800:h=600:fps=25 -ovc lavc -lavcopts vcodec=mpeg4 -oac copy -o \ 359 onedheatdiff001totT.avi") 360 \end{python} 361