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 
gross 
2878 

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 
2870 
% 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) = \left 
118 


% \end{equation} 
119 


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

121 
ahallam 
2495 
\subsection{Boundary Conditions} 
122 
gross 
2878 
\label{SEC BOUNDARY COND} 
123 
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. 
124 


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. 
125 


For our model problem we want to keep the initial temperature setting on the left side of the 
126 


iron bar over time. This defines a Dirichlet boundary condition for the PDE \refEq{eqn:hddisc} to be solved at each time step. 
127 
ahallam 
2495 

128 
gross 
2878 
On the other end of the iron rod we want to add an appropriate boundary condition to define insulation to prevent 
129 
gross 
2862 
any loss or inflow of energy at the right end of the rod. Mathematically this is expressed by prescribing 
130 


the heat flux $\kappa \frac{\partial T}{\partial x}$ to zero on the right end of the rod 
131 


In our simplified one dimensional model this is expressed 
132 


in the form; 
133 
ahallam 
2494 
\begin{equation} 
134 
gross 
2862 
\kappa \frac{\partial T}{\partial x} = 0 
135 
ahallam 
2494 
\end{equation} 
136 
gross 
2862 
or in a more general case as 
137 


\begin{equation}\label{NEUMAN 1} 
138 


\kappa \nabla T \cdot n = 0 
139 


\end{equation} 
140 


where $n$ is the outer normal field \index{outer normal field} at the surface of the domain. 
141 


For the iron rod the outer normal field on the right hand side is the vector $(1,0)$. The $\cdot$ (dot) refers to the 
142 


dot product of the vectors $\nabla T$ and $n$. In fact, the term $\nabla T \cdot n$ is the normal derivative of 
143 


the temperature $T$. Other notations which are used are\footnote{The \esc notation for the normal 
144 


derivative is $T\hackscore{,i} n\hackscore i$.}; 
145 
ahallam 
2645 
\begin{equation} 
146 
gross 
2862 
\nabla T \cdot n = \frac{\partial T}{\partial n} \; . 
147 
ahallam 
2645 
\end{equation} 
148 
gross 
2862 
A condition of the type \refEq{NEUMAN 1} defines a \textbf{Neuman boundary condition} for the PDE. 
149 
ahallam 
2494 

150 
gross 
2862 
The PDE \refEq{eqn:hdgenf} together with the Dirichlet boundary condition set on the left face of the rod 
151 


and the Neuman boundary condition~\ref{eqn:hdgenf} define a \textbf{boundary value problem}. 
152 


It is a nature of a boundary value problem that it allows to make statements on the solution in the 
153 


interior of the domain from information known on the boundary only. In most cases 
154 


we use the term partial differential equation but in fact mean a boundary value problem. 
155 


It is important to keep in mind that boundary conditions need to be complete and consistent in the sense that 
156 


at any point on the boundary either a Dirichlet or a Neuman boundary condition must be set. 
157 



158 
gross 
2878 
Conveniently, \esc makes default assumption on the boundary conditions which the user may modify where appropriate. 
159 
gross 
2862 
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 
160 


$n\hackscore{j}A\hackscore{jl} u\hackscore{,l}=0$.} is; 
161 


\begin{equation}\label{NEUMAN 2} 
162 


n\cdot A \cdot\nabla u = 0 
163 


\end{equation} 
164 


which is used everywhere on the boundary. Again $n$ denotes the outer normal field. 
165 


Notice that the coefficient $A$ is the same as in the \esc PDE~\ref{eqn:commonform nabla}. 
166 


With the settings for the coefficients we have already identified in \refEq{ESCRIPT SET} this 
167 


condition translates into 
168 
gross 
2867 
\begin{equation}\label{NEUMAN 2b} 
169 
gross 
2862 
\kappa \frac{\partial T}{\partial x} = 0 
170 


\end{equation} 
171 


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. 
172 



173 
gross 
2870 
\subsection{Outline of the Implementation} 
174 


\label{sec:outline} 
175 


To solve the heat diffusion equation (equation \refEq{eqn:hd}) we will write a simple \pyt script. 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}. The script we will discuss later in details will have four major steps. Firstly we need to define the domain where we want to 
176 


calculate the temperature. For our problem this is the iron rod which has a rectangular shape. Secondly we need to define the PDE 
177 
gross 
2878 
we need to solve in each time step to get the updated temperature. Thirdly we need to define the the coefficients of the PDE and finally we need to solve the PDE. The last two steps need to be repeated until the final time marker has been reached. As a work flow this takes the form; 
178 


\begin{enumerate} 
179 


\item create domain 
180 


\item create PDE 
181 


\item while end time not reached: 
182 


\begin{enumerate} 
183 


\item set PDE coefficients 
184 


\item solve PDE 
185 


\item update time marker 
186 


\end{enumerate} 
187 


\item end of calculation 
188 


\end{enumerate} 
189 
gross 
2870 
In the terminology of \pyt the domain and PDE are represented by \textbf{objects}. The nice feature of an object is that it defined by it usage and features 
190 
gross 
2878 
rather than its actual representation. So we will create a domain object to describe the geometry of our iron rod. The main feature 
191 
gross 
2870 
of the object we will use is the fact that we can define PDEs and spatially distributed values such as the temperature 
192 


on a domain. In fact the domain object has many more features  most of them you will 
193 


never use and do not need to understand. Similar a PDE object is defined by the fact that we can define the coefficients of the PDE and solve the PDE. At a 
194 
gross 
2878 
later stage you may use more advanced features of the PDE class but you need to worry about them only at the point when you use them. 
195 
gross 
2870 

196 



197 


\begin{figure}[t] 
198 


\centering 
199 


\includegraphics[width=6in]{figures/functionspace.pdf} 
200 


\label{fig:fs} 
201 


\caption{\esc domain construction overview} 
202 


\end{figure} 
203 



204 


\subsection{The Domain Constructor in \esc} 
205 


\label{ss:domcon} 
206 


It is helpful to have a better understanding how spatially distributed value such as the temperature or PDE coefficients are interpreted in \esc. Again 
207 


from the user's point of view the representation of these spatially distributed values is not relevant. 
208 



209 


There are various ways to construct domain objects. The simplest form is as rectangular shaped region with a length and height. There is 
210 


a ready to use function call for this. Besides the spatial dimensions the function call will require you to specify the number 
211 
gross 
2878 
elements or cells to be used along the length and height, see Figure~\ref{fig:fs}. Any specially distributed value 
212 
gross 
2870 
and the PDE is represented in discrete form using this element representation\footnote{We will use the finite element method (FEM), see \url{http://en.wikipedia.org/wiki/Finite_element_method}i for details.}. Therefore we will have access to an approximation of the true PDE solution only. 
213 


The quality of the approximation depends  besides other factors mainly on the number of elements being used. In fact, the 
214 


approximation becomes better the more elements are used. However, computational costs and compute time grow with the number of 
215 
gross 
2878 
elements being used. It therefore important that you find the right balance between the demand in accuracy and acceptable resource usage. 
216 
gross 
2870 

217 


In general, one can thinks about a domain object as a composition of nodes and elements. 
218 


As shown in Figure~\ref{fig:fs}, an element is defined by the nodes used to describe its vertices. 
219 


To represent spatial distributed values the user can use 
220 


the values at the nodes, at the elements in the interior of the domain or at elements located at the surface of the domain. 
221 


The different approach used to represent values is called \textbf{function space} and is attached to all objects 
222 


in \esc representing a spatial distributed value such as the solution of a PDE. The three 
223 


function spaces we will use at the moment are; 
224 


\begin{enumerate} 
225 


\item the nodes, called by \verbContinuousFunction(domain) ; 
226 


\item the elements/cells, called by \verbFunction(domain) ; and 
227 


\item the boundary, called by \verbFunctionOnBoundary(domain) . 
228 


\end{enumerate} 
229 


A function space object such as \verbContinuousFunction(domain) has the method \verbgetX attached to it. This method returns the 
230 


location of the socalled \textbf{sample points} used to represent values with the particular function space attached to it. So the 
231 


call \verbContinuousFunction(domain).getX() will return the coordinates of the nodes used to describe the domain while 
232 


the \verbFunction(domain).getX() returns the coordinates of numerical integration points within elements, see 
233 


Figure~\ref{fig:fs}. 
234 



235 
gross 
2878 
This distinction between different representations of spatial distributed values 
236 
gross 
2870 
is important in order to be able to vary the degrees of smoothness in a PDE problem. 
237 


The coefficients of a PDE need not be continuous thus this qualifies as a \verbFunction() type. 
238 


On the other hand a temperature distribution must be continuous and needs to be represented with a \verbContinuousFunction() function space. 
239 


An influx may only be defined at the boundary and is therefore a \verb FunctionOnBoundary() object. 
240 


\esc allows certain transformations of the function spaces. A \verb ContinuousFunction() can be transformed into a \verbFunctionOnBoundary() 
241 


or \verbFunction(). On the other hand there is not enough information in a \verb FunctionOnBoundary() to transform it to a \verb ContinuousFunction() . 
242 
gross 
2878 
These transformations, which are called \textbf{interpolation} are invoked automatically by \esc if needed. 
243 
gross 
2870 

244 


Later in this introduction we will discuss how 
245 


to define specific areas of geometry with different materials which are represented by different material coefficients such the 
246 
gross 
2878 
thermal conductivities $kappa$. A very powerful technique to define these types of PDE 
247 
gross 
2870 
coefficients is tagging. Blocks of materials and boundaries can be named and values can be defined on subregions based on their names. 
248 
gross 
2878 
This is simplifying PDE coefficient and flux definitions. It makes for much easier scripting. We will discuss this technique in Section~\ref{STEADYSTATE HEAT REFRACTION}. 
249 
gross 
2870 

250 



251 


\subsection{A Clarification for the 1D Case} 
252 
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; 
253 
gross 
2477 
\begin{equation}\label{eqn:commonform2D} 
254 


A\hackscore{00}\frac{\partial^{2}u}{\partial x^{2}} 
255 


A\hackscore{01}\frac{\partial^{2}u}{\partial x\partial y} 
256 


A\hackscore{10}\frac{\partial^{2}u}{\partial y\partial x} 
257 


A\hackscore{11}\frac{\partial^{2}u}{\partial y^{2}} 
258 


+ Du = f 
259 


\end{equation} 
260 
ahallam 
2606 
Notice that for the higher dimensional case $A$ becomes a matrix. It is also 
261 
ahallam 
2495 
important to notice that the usage of the Nabla operator creates 
262 


a compact formulation which is also independent from the spatial dimension. 
263 
gross 
2861 
So to make the general PDE \refEq{eqn:commonform2D} one dimensional as 
264 


shown in \refEq{eqn:commonform} we need to set 
265 
ahallam 
2606 
\begin{equation} 
266 
ahallam 
2494 
A\hackscore{00}=A; A\hackscore{01}=A\hackscore{10}=A\hackscore{11}=0 
267 
gross 
2477 
\end{equation} 
268 



269 
gross 
2867 

270 
ahallam 
2495 
\subsection{Developing a PDE Solution Script} 
271 
ahallam 
2801 
\label{sec:key} 
272 
gross 
2878 
\sslist{onedheatdiffbase.py} 
273 
gross 
2870 
We will write a simple \pyt script which uses the \modescript, \modfinley and \modmpl modules. 
274 
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.} 
275 
ahallam 
2495 
that we will require. 
276 
ahallam 
2775 
\begin{python} 
277 
ahallam 
2495 
from esys.escript import * 
278 
ahallam 
2606 
# This defines the LinearPDE module as LinearPDE 
279 


from esys.escript.linearPDEs import LinearPDE 
280 


# This imports the rectangle domain function from finley. 
281 


from esys.finley import Rectangle 
282 


# A useful unit handling package which will make sure all our units 
283 


# match up in the equations under SI. 
284 


from esys.escript.unitsSI import * 
285 
ahallam 
2775 
\end{python} 
286 
gross 
2878 
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. 
287 
gross 
2477 

288 
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. 
289 
ahallam 
2401 

290 
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. 
291 
ahallam 
2775 
\begin{python} 
292 
ahallam 
2495 
#Domain related. 
293 
ahallam 
2606 
mx = 1*m #meters  model length 
294 
ahallam 
2495 
my = .1*m #meters  model width 
295 
ahallam 
2606 
ndx = 100 # mesh steps in x direction 
296 


ndy = 1 # mesh steps in y direction  one dimension means one element 
297 
ahallam 
2775 
\end{python} 
298 
ahallam 
2495 
The material constants and the temperature variables must also be defined. For the iron rod in the model they are defined as: 
299 
ahallam 
2775 
\begin{python} 
300 
ahallam 
2495 
#PDE related 
301 


rho = 7874. *kg/m**3 #kg/m^{3} density of iron 
302 
gross 
2878 
cp = 449.*J/(kg*K) # J/Kg.K thermal capacity 
303 


rhocp = rho*cp 
304 


kappa = 80.*W/m/K # watts/m.Kthermal conductivity 
305 


qH=0 * J/(sec*m**3) # J/(sec.m^{3}) no heat source 
306 


T0=100 * Celsius # initial temperature at left end of rod 
307 


Tref=20 * Celsius # base temperature 
308 
ahallam 
2775 
\end{python} 
309 
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: 
310 
ahallam 
2775 
\begin{python} 
311 
gross 
2878 
t=0 * day #our start time, usually zero 
312 


tend=1. * day #  time to end simulation 
313 
ahallam 
2495 
outputs = 200 # number of time steps required. 
314 


h=(tendt)/outputs #size of time step 
315 
ahallam 
2606 
#user warning statement 
316 


print "Expected Number of time outputs is: ", (tendt)/h 
317 


i=0 #loop counter 
318 
ahallam 
2775 
\end{python} 
319 
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: 
320 
ahallam 
2775 
\begin{python} 
321 
ahallam 
2606 
#generate domain using rectangle 
322 


rod = Rectangle(l0=mx,l1=my,n0=ndx, n1=ndy) 
323 
ahallam 
2775 
\end{python} 
324 
gross 
2878 
\verb rod now describes a domain in the manner of Section \ref{ss:domcon}. There is an easy way to extract 
325 


the coordinates of the nodes used to describe the domain \verbrod using the 
326 


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; 
327 
ahallam 
2775 
\begin{python} 
328 
gross 
2878 
#extract data points  the solution points 
329 
ahallam 
2606 
x=rod.getX() 
330 
ahallam 
2775 
\end{python} 
331 
gross 
2878 
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() method. 
332 
ahallam 
2658 

333 
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. 
334 


\begin{python} 
335 
ahallam 
2495 
mypde=LinearSinglePDE(rod) 
336 
gross 
2878 
A=zeros((2,2))) 
337 


A[0,0]=kappa 
338 


q=whereZero(x[0]) 
339 


mypde.setValue(A=A, D=rhocp/h, q=q, r=T0) 
340 
ahallam 
2775 
\end{python} 
341 
gross 
2878 
The argument \verbq has not been discussed yet: In fact the arguments \verbq and \verbr are used to define 
342 


Dirichlet boundary condition as discussed in Section~\ref{SEC BOUNDARY COND}. In the \esc 
343 


PDE from the argument \verbq indicates by a positive value for which nodes we want to apply a 
344 


Dirichlet boundary condition, ie. where we want to prescribe the value of the PDE solution 
345 


rather then using the PDE. The actually value for the solution to be taken is set by the argument \verbr. 
346 


In our case we want to keep the initial temperature $T0$ on the left face of the rode for all times. Notice, 
347 


that as set to a constant value \verbr is assumed to have the same value 
348 


at all nodes, however only the value at those nodes marked by a positive value by \verbq are actually used. 
349 
ahallam 
2401 

350 
gross 
2878 
In order to set \verbq we use 
351 


\verbwhereZero function. The function returns the value (positive) one for those data points (=nodes) where the argument is equal to zero and otherwise returns (nonpositive) value zero. 
352 


As \verbx[0] given the $x$coordinates of the nodes for the domain, 
353 


\verbwhereZero(x[0]) gives the value $1$ for the nodes at the left end of the rod $x=x_0=0$ and 
354 


zero elsewhere which is exactly what we need. 
355 



356 


In a many cases it may be possible to decrease the computational time of the solver if the PDE is symmetric. 
357 


Symmetry of a PDE is defined by; 
358 
ahallam 
2495 
\begin{equation}\label{eqn:symm} 
359 


A\hackscore{jl}=A\hackscore{lj} 
360 


\end{equation} 
361 
gross 
2878 
Symmetry is only dependent on the $A$ coefficient in the general form and the other coefficients $D$ as well as the right hand side $Y$ 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; 
362 
ahallam 
2775 
\begin{python} 
363 
ahallam 
2495 
myPDE.setSymmetryOn() 
364 
ahallam 
2775 
\end{python} 
365 
gross 
2878 
Next we need to establish the initial temperature distribution \verbT. We want to have this initial 
366 


value to be \verbTref except at the left end of the rod $x=0$ where we have the temperature \verbT0. We use; 
367 


\begin{python} 
368 


# ... set initial temperature .... 
369 


T = T0*whereZero(x[0])+Tref*(1whereZero(x[0])) 
370 


\end{python} 
371 


Finally we will initialize an iteration loop to solve our PDE for all the time steps we specified in the variable section. As the right hand side 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. 
372 


\begin{python} 
373 


while t < tend: 
374 


i+=1 #increment the counter 
375 


t+=h #increment the current time 
376 


mypde.setValue(Y=qH+rhocp/h*T) # set variable PDE coefficients 
377 


T=mypde.getSolution() #get the PDE solution 
378 


totE = integrate(rhocp*T) #get the total heat (energy) in the system 
379 


\end{python} 
380 


The last statement in this script calculates the total energy in the system as volume integral 
381 


of $\rho \c_p T$ over the rod. 
382 
ahallam 
2401 

383 
gross 
2878 
\subsection{Plotting the Total Energy} 
384 


\sslist{onedheatdiff001.py} 
385 



386 


\esc does not include its own plotting capabilities. However, it is possible to use a variety of free \pyt packages for visualization. 
387 


Two types will be demonstrated in this cookbook; \mpl\footnote{\url{http://matplotlib.sourceforge.net/}} and \verb VTK \footnote{\url{http://www.vtk.org/}} visualisation. 
388 


The \mpl package is a component of SciPy\footnote{\url{http://www.scipy.org}} and is good for basic graphs and plots. 
389 


For more complex visualisation tasks in particular when it comes to two and three dimensional problems it is recommended to us more advanced tools for instance \mayavi \footnote{\url{http://code.enthought.com/projects/mayavi/}} 
390 


which bases the \verbVTK toolkit. We will discuss the usage of \verbVTK based 
391 


visualization in Chapter~\ref{Sec:2DHD} where will discuss a two dimensional PDE. 
392 



393 


For our simple problem we have two plotting tasks: Firstly we are interested in showing the 
394 


behavior of the total energy over time and secondly in how the temperature distribution within the rod is 
395 


developing over time. Lets start with the first task. 
396 



397 


\begin{figure} 
398 


\begin{center} 
399 


\includegraphics[width=4in]{figures/ttrodpyplot150} 
400 


\caption{Total Energy in Rod over Time (in seconds).} 
401 


\label{fig:onedheatout1} 
402 


\end{center} 
403 


\end{figure} 
404 



405 


The trick is to create a record of the time marks and the corresponding total energies observed. 
406 


\pyt provides the concept of lists for this. Before 
407 


the time loop is opened we create empty lists for the time marks \verbt_list and the total energies \verbE_list. 
408 


After the new temperature as been calculated by solving the PDE we append the new time marker and total energy 
409 


to the corresponding list using the \verbappend method. With these modifications the script looks as follows: 
410 
ahallam 
2775 
\begin{python} 
411 
gross 
2878 
t_list=[] 
412 


E_list=[] 
413 


# ... start iteration: 
414 


while t<tend: 
415 


t+=h 
416 


mypde.setValue(Y=qH+rhocp/h*T) # set variable PDE coefficients 
417 


T=mypde.getSolution() #get the PDE solution 
418 


totE=integrate(rhocp*T) 
419 


t_list.append(t) # add current time mark to record 
420 


E_list.append(totE) # add current total energy to record 
421 
ahallam 
2775 
\end{python} 
422 
gross 
2878 
To plot $t$ over $totE$ we use the \mpl a module contained within \pylab which needs to be loaded before used; 
423 
ahallam 
2775 
\begin{python} 
424 
gross 
2878 
import pylab as pl # plotting package. 
425 
ahallam 
2775 
\end{python} 
426 
gross 
2878 
Here we are not using the \verbfrom pylab import * in order to name clashes for function names 
427 


with \esc. 
428 
ahallam 
2401 

429 
gross 
2878 
The following statements are added to the script after the time loop has been completed; 
430 
ahallam 
2775 
\begin{python} 
431 
gross 
2878 
pl.plot(t_list,E_list) 
432 


pl.title("Total Energy") 
433 


pl.savefig("totE.png") 
434 
ahallam 
2775 
\end{python} 
435 
gross 
2878 
The first statement hands over the time marks and corresponding total energies to the plotter. 
436 


The second statement is setting the title for the plot. The last statement renders the plot and writes the 
437 


result into the file \verbtotE.png which can be displayed by (almost) any image viewer. Your result should look 
438 


similar to Figure~\ref{fig:onedheatout1}. 
439 
ahallam 
2401 

440 
gross 
2878 
\subsection{Plotting the Temperature Distribution} 
441 


\sslist{onedheatdiff001b.py} 
442 


For plotting the spatial distribution of the temperature we need to modify the strategy we have used 
443 


for the total energy. Instead of producing a final plot at the end we will generate a 
444 


picture at each time step which can be browsed as slide show or composed to a movie. 
445 


The first problem we encounter is that if we produce an image in each time step we need 
446 


to make sure that the images previously generated are not overwritten. 
447 



448 


To develop an incrementing file name we can use the following convention. It is convenient to 
449 


put all image file showing the same variable  in our case the temperature distribution  
450 


into a separate directory. As part of the \verbos module\footnote{The \texttt{os} module provides 
451 


a powerful interface to interact with the operating system, see \url{http://docs.python.org/library/os.html}.} \pyt 
452 


provides the \verbos.path.join command to build file and 
453 


directory names in a platform independent way. Assuming that 
454 


\verbsave_path is name of directory we want to put the results the command is; 
455 
ahallam 
2775 
\begin{python} 
456 
gross 
2878 
import os 
457 


os.path.join(save_path, "tempT%03d.png"%i ) 
458 
ahallam 
2775 
\end{python} 
459 
gross 
2878 
where \verbi is the time step counter. 
460 


There are two arguments to the \verb join command. The \verb save_path variable is a predefined string pointing to the directory we want to save our data in, for example a single subfolder called \verb data would be defined by; 
461 


\begin{verbatim} 
462 


save_path = "data" 
463 


\end{verbatim} 
464 


while a subfolder of \verb data called \verb onedheatdiff001 would be defined by; 
465 


\begin{verbatim} 
466 


save_path = os.path.join("data","onedheatdiff001") 
467 


\end{verbatim} 
468 


The second argument of \verb join \xspace contains a string which is the filename or subdirectory name. We can use the operator \verb% to increment our file names with the value \verbi denoting a incrementing counter. The substring \verb %03d does this by defining the following parameters; 
469 


\begin{itemize} 
470 


\item \verb 0 becomes the padding number; 
471 


\item \verb 3 tells us the amount of padding numbers that are required; and 
472 


\item \verb d indicates the end of the \verb % operator. 
473 


\end{itemize} 
474 


To increment the file name a \verb %i is required directly after the operation the string is involved in. When correctly implemented the output files from this command would be place in the directory defined by \verb save_path as; 
475 


\begin{verbatim} 
476 


rodpyplot.png 
477 


rodpyplot.png 
478 


rodpyplot.png 
479 


... 
480 


\end{verbatim} 
481 


and so on. 
482 



483 


A subfolder check/constructor is available in \esc. The command; 
484 


\begin{verbatim} 
485 


mkDir(save_path) 
486 


\end{verbatim} 
487 


will check for the existence of \verb save_path and if missing, make the required directories. 
488 



489 


We start by modifying our solution script from before. 
490 


Prior to the \verbwhile loop we will need to extract our finite solution points to a data object that is compatible with \mpl. First we create the node coordinates of the data points used to represent 
491 


the temperature as a \pyt list of tuples. As a solution of a PDE 
492 


the temperature has the \verbSolution(rod) function space attribute. We use 
493 


the \verbgetX() method to get the coordinates of the data points as an \esc object 
494 


which is then converted to a \numpy array. The $x$ component is then extracted via an array slice to the variable \verbplx; 
495 
ahallam 
2775 
\begin{python} 
496 
gross 
2878 
import numpy as np # array package. 
497 


#convert solution points for plotting 
498 


plx = Solution(rod).getX().toListOfTuples() 
499 


plx = np.array(plx) # convert to tuple to numpy array 
500 


plx = plx[:,0] # extract x locations 
501 
ahallam 
2775 
\end{python} 
502 
gross 
2878 

503 
ahallam 
2645 
\begin{figure} 
504 


\begin{center} 
505 
gross 
2878 
\includegraphics[width=4in]{figures/rodpyplot001} 
506 


\includegraphics[width=4in]{figures/rodpyplot050} 
507 


\includegraphics[width=4in]{figures/rodpyplot200} 
508 


\caption{Temperature ($T$) distribution in rod at time steps $1$, $50$ and $200$.} 
509 
ahallam 
2645 
\label{fig:onedheatout} 
510 


\end{center} 
511 


\end{figure} 
512 



513 
gross 
2878 
For each time step we will generate a plot of the temperature distribution and save each to a file. We use the same 
514 


techniques provided by \mpl as we have used to plot the total energy over time. 
515 


The following is appended to the end of the \verb while loop and creates one figure of the temperature distribution. We start by converting the solution to a tuple and then plotting this against our \textit{x coordinates} \verb plx we have generated before. We add a title to the diagram before it is rendered into a file. 
516 


Finally, the figure is saved to a \verb*.png file and cleared for the following iteration. 
517 
ahallam 
2775 
\begin{python} 
518 
gross 
2878 
# ... start iteration: 
519 


while t<tend: 
520 


.... 
521 


T=mypde.getSolution() #get the PDE solution 
522 


tempT = T.toListOfTuples() # convert to a tuple 
523 


pl.plot(plx,tempT) # plot solution 
524 


# set scale (Temperature should be between Tref and T0) 
525 


pl.axis([0,mx,Tref*.9,T0*1.1]) 
526 


# add title 
527 


pl.title("Temperature across rod at time %e minutes"%(t/minutes)) 
528 


#save figure to file 
529 


pl.savefig(os.path.join(save_path,"tempT","rodpyplot%03d.png") %i) 
530 
ahallam 
2775 
\end{python} 
531 
gross 
2878 
Some results are shown in Figure~\ref{fig:onedheatout}. 
532 
jfenwick 
2657 

533 
ahallam 
2606 
\subsection{Make a video} 
534 
gross 
2878 
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. 
535 
ahallam 
2775 
\begin{python} 
536 
gross 
2878 
# compile the *.png files to create a *.avi videos that show T change 
537 


# with time. This operation uses Linux mencoder. For other operating 
538 


# systems it is possible to use your favorite video compiler to 
539 
ahallam 
2606 
# convert image files to videos. 
540 
gross 
2477 

541 
ahallam 
2606 
os.system("mencoder mf://"+save_path+"/tempT"+"/*.png mf type=png:\ 
542 
gross 
2878 
w=800:h=600:fps=25 ovc lavc lavcopts vcodec=mpeg4 oac copy o \ 
543 


onedheatdiff001tempT.avi") 
544 
ahallam 
2775 
\end{python} 
545 
gross 
2477 
