1 
ahallam 
2401 

2 


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 
3 


% 
4 


% Copyright (c) 20032009 by University of Queensland 
5 


% Earth Systems Science Computational Center (ESSCC) 
6 


% http://www.uq.edu.au/esscc 
7 


% 
8 


% Primary Business: Queensland, Australia 
9 


% Licensed under the Open Software License version 3.0 
10 


% http://www.opensource.org/licenses/osl3.0.php 
11 


% 
12 


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 
13 



14 
ahallam 
2801 
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 
ahallam 
2401 
\section{One Dimensional Heat Diffusion in an Iron Rod} 
17 
ahallam 
2658 
\sslist{onedheatdiff001.py and cblib.py} 
18 
ahallam 
2401 
%\label{Sec:1DHDv0} 
19 
ahallam 
2801 
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 
ahallam 
2494 
\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 
ahallam 
2495 
\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 
ahallam 
2494 
which is defined as: 
28 
ahallam 
2401 
\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 
gross 
2861 
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 
ahallam 
2401 

37 
gross 
2861 
\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 
gross 
2477 

41 
gross 
2861 
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(th)}{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^{(n1)}}{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^{(n1)}+h \\ 
61 


T^{(n)} & = & T(t^{(n1)}) \\ 
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^{(n1)})  \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 $t^{(n1)}$. 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 setup 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 
gross 
2477 
$(A\hackscore{jl} u\hackscore{,l})\hackscore{,j}+D u =Y$} 
88 
ahallam 
2606 
is described by; 
89 
gross 
2477 
\begin{equation}\label{eqn:commonform nabla} 
90 
jfenwick 
2657 
\nabla\cdot(A\cdot\nabla u) + Du = f 
91 
ahallam 
2411 
\end{equation} 
92 
gross 
2861 
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 
ahallam 
2495 
the spatial derivative of its subject  in this case $u$. Lets assume for a moment that we deal with a onedimensional problem then ; 
94 
gross 
2477 
\begin{equation} 
95 


\nabla = \frac{\partial}{\partial x} 
96 


\end{equation} 
97 
gross 
2861 
and we can write \refEq{eqn:commonform nabla} as; 
98 
ahallam 
2411 
\begin{equation}\label{eqn:commonform} 
99 
gross 
2477 
A\frac{\partial^{2}u}{\partial x^{2}} + Du = f 
100 
ahallam 
2411 
\end{equation} 
101 
gross 
2861 
if $A$ is constant. To match this simplified general form to our problem \refEq{eqn:hddisc} 
102 


we rearrange \refEq{eqn:hddisc}; 
103 
ahallam 
2411 
\begin{equation} 
104 
ahallam 
2645 
\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^{(n1)} 
105 
ahallam 
2494 
\label{eqn:hdgenf} 
106 


\end{equation} 
107 
ahallam 
2775 
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 
ahallam 
2495 
$t^{(0)}=0$ and $t^{(n)}=t^{(n1)}+h$ where $h>0$ is the step size and assumed to be constant. 
109 
gross 
2861 
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 
gross 
2862 
\begin{equation}\label{ESCRIPT SET} 
111 
gross 
2861 
u=T^{(n)}; 
112 
ahallam 
2494 
A = \kappa; D = \frac{\rho c \hackscore{p}}{h}; f = q \hackscore{H} + \frac{\rho c\hackscore p}{h} T^{(n1)} 
113 


\end{equation} 
114 



115 
gross 
2861 
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 
ahallam 
2495 
\begin{equation} 
117 


T(x,0) = T\hackscore{ref} = 0 
118 


\end{equation} 
119 


for all $x$ in the domain. 
120 
ahallam 
2494 

121 
ahallam 
2495 
\subsection{Boundary Conditions} 
122 
gross 
2862 
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 
ahallam 
2495 

127 
gross 
2862 
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 
ahallam 
2494 
\begin{equation} 
133 
gross 
2862 
\kappa \frac{\partial T}{\partial x} = 0 
134 
ahallam 
2494 
\end{equation} 
135 
gross 
2862 
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 
ahallam 
2645 
\begin{equation} 
145 
gross 
2862 
\nabla T \cdot n = \frac{\partial T}{\partial n} \; . 
146 
ahallam 
2645 
\end{equation} 
147 
gross 
2862 
A condition of the type \refEq{NEUMAN 1} defines a \textbf{Neuman boundary condition} for the PDE. 
148 
ahallam 
2494 

149 
gross 
2862 
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 
ahallam 
2495 
\subsection{A \textit{1D} Clarification} 
176 
gross 
2861 
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 
gross 
2477 
\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 
ahallam 
2606 
Notice that for the higher dimensional case $A$ becomes a matrix. It is also 
185 
ahallam 
2495 
important to notice that the usage of the Nabla operator creates 
186 


a compact formulation which is also independent from the spatial dimension. 
187 
gross 
2861 
So to make the general PDE \refEq{eqn:commonform2D} one dimensional as 
188 


shown in \refEq{eqn:commonform} we need to set 
189 
ahallam 
2606 
\begin{equation} 
190 
ahallam 
2494 
A\hackscore{00}=A; A\hackscore{01}=A\hackscore{10}=A\hackscore{11}=0 
191 
gross 
2477 
\end{equation} 
192 



193 
ahallam 
2495 
\subsection{Developing a PDE Solution Script} 
194 
ahallam 
2801 
\label{sec:key} 
195 
gross 
2861 
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 
gross 
2477 

197 
gross 
2861 
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 
ahallam 
2495 
that we will require. 
199 
ahallam 
2775 
\begin{python} 
200 
ahallam 
2495 
from esys.escript import * 
201 
ahallam 
2606 
# 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 
ahallam 
2775 
\end{python} 
212 
ahallam 
2801 
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 \verbLinearPDE has been imported explicitly for ease of use later in the script. \verbRectangle is going to be our type of model. The module \verb unitsSI provides support for SI unit definitions with our variables; and the \verbos 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 
gross 
2477 

214 
ahallam 
2801 
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 
ahallam 
2401 

216 
ahallam 
2801 
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 \verbndx would be 1 to 10\% of the length. Our \verbndy 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 
ahallam 
2775 
\begin{python} 
218 
ahallam 
2495 
#Domain related. 
219 
ahallam 
2606 
mx = 1*m #meters  model length 
220 
ahallam 
2495 
my = .1*m #meters  model width 
221 
ahallam 
2606 
ndx = 100 # mesh steps in x direction 
222 


ndy = 1 # mesh steps in y direction  one dimension means one element 
223 
ahallam 
2775 
\end{python} 
224 
ahallam 
2495 
The material constants and the temperature variables must also be defined. For the iron rod in the model they are defined as: 
225 
ahallam 
2775 
\begin{python} 
226 
ahallam 
2495 
#PDE related 
227 


q=200. * Celsius #Kelvin  our heat source temperature 
228 
ahallam 
2606 
Tref = 0. * Celsius #Kelvin  starting temp of iron bar 
229 
ahallam 
2495 
rho = 7874. *kg/m**3 #kg/m^{3} density of iron 
230 
ahallam 
2606 
cp = 449.*J/(kg*K) #j/Kg.K thermal capacity 
231 
ahallam 
2495 
rhocp = rho*cp 
232 
ahallam 
2606 
kappa = 80.*W/m/K #watts/m.Kthermal conductivity 
233 
ahallam 
2775 
\end{python} 
234 
ahallam 
2495 
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 
ahallam 
2775 
\begin{python} 
236 
ahallam 
2495 
t=0 #our start time, usually zero 
237 
ahallam 
2606 
tend=5.*minute #seconds  time to end simulation 
238 
ahallam 
2495 
outputs = 200 # number of time steps required. 
239 


h=(tendt)/outputs #size of time step 
240 
ahallam 
2606 
#user warning statement 
241 


print "Expected Number of time outputs is: ", (tendt)/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 
ahallam 
2775 
\end{python} 
246 
ahallam 
2606 
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 
ahallam 
2775 
\begin{python} 
248 
ahallam 
2606 
#generate domain using rectangle 
249 


rod = Rectangle(l0=mx,l1=my,n0=ndx, n1=ndy) 
250 
ahallam 
2775 
\end{python} 
251 
ahallam 
2801 
\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 \verbrod using the domain property function \verbgetX() . This function sets the vertices of each cell as finite points to solve in the solution. If we let \verbx be these finite points, then; 
252 
ahallam 
2775 
\begin{python} 
253 
ahallam 
2606 
#extract finite points  the solution points 
254 


x=rod.getX() 
255 
ahallam 
2775 
\end{python} 
256 
ahallam 
2801 
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 
ahallam 
2658 

258 
ahallam 
2775 
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 
ahallam 
2495 
mypde=LinearSinglePDE(rod) 
261 


mypde.setValue(A=kappa*kronecker(rod),D=rhocp/h) 
262 
ahallam 
2775 
\end{python} 
263 
ahallam 
2401 

264 
ahallam 
2495 
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 
ahallam 
2801 
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 
ahallam 
2775 
\begin{python} 
270 
ahallam 
2495 
myPDE.setSymmetryOn() 
271 
ahallam 
2775 
\end{python} 
272 
ahallam 
2401 

273 
ahallam 
2801 
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 
ahallam 
2775 
\begin{python} 
275 
ahallam 
2401 
T = Tref 
276 
ahallam 
2775 
\end{python} 
277 
ahallam 
2801 
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 
ahallam 
2775 
\begin{python} 
279 
ahallam 
2495 
# ... set heat source: .... 
280 


qH=q*whereZero(x[0]) 
281 
ahallam 
2775 
\end{python} 
282 
ahallam 
2401 

283 
ahallam 
2801 
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 
ahallam 
2775 
\begin{python} 
285 
ahallam 
2495 
while t<=tend: 
286 
ahallam 
2606 
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 
ahallam 
2775 
\end{python} 
292 
ahallam 
2401 

293 
ahallam 
2606 
\subsection{Plotting the heat solutions} 
294 
ahallam 
2801 
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 
ahallam 
2775 
\begin{python} 
296 
ahallam 
2606 
#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 
ahallam 
2775 
\end{python} 
301 
ahallam 
2801 
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 
ahallam 
2775 
\begin{python} 
303 
ahallam 
2606 
#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 
ahallam 
2801 
pl.title("Total temperature across Rod") 
320 
ahallam 
2606 
pl.savefig(os.path.join(save_path+"/totT","ttrodpyplot%03d.png")%i) 
321 


pl.clf() 
322 
ahallam 
2775 
\end{python} 
323 
ahallam 
2645 
\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 
jfenwick 
2657 
\subsubsection{Parallel scripts (MPI)} 
332 
ahallam 
2801 
In some of the example files for this cookbook the plotting commands are a little different. 
333 
jfenwick 
2657 
For example, 
334 
ahallam 
2775 
\begin{python} 
335 
jfenwick 
2657 
if getMPIRankWorld() == 0: 
336 


pl.savefig(os.path.join(save_path+"/totT","ttrodpyplot%03d.png")%i) 
337 


pl.clf() 
338 
ahallam 
2775 
\end{python} 
339 
jfenwick 
2657 

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 
ahallam 
2801 
For more details on writing scripts for parallel computing please consult the \emph{user's guide}. 
344 
jfenwick 
2657 

345 
ahallam 
2606 
\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 
ahallam 
2775 
\begin{python} 
348 
ahallam 
2606 
# compile the *.png files to create two *.avi videos that show T change 
349 
jfenwick 
2657 
# with time. This operation uses linux mencoder. For other operating 
350 
ahallam 
2606 
# systems it is possible to use your favourite video compiler to 
351 


# convert image files to videos. 
352 
gross 
2477 

353 
ahallam 
2606 
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 
gross 
2477 

357 
ahallam 
2606 
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 
ahallam 
2775 
\end{python} 
361 
gross 
2477 
