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% Copyright (c) 20032018 by The University of Queensland 
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% Licensed under the Apache License, version 2.0 
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% Development until 2012 by Earth Systems Science Computational Center (ESSCC) 
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% Development 20122013 by School of Earth Sciences 
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% Development from 2014 by Centre for Geoscience Computing (GeoComp) 
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 
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\section{Example 1: One Dimensional Heat Diffusion in Granite} 
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\label{Sec:1DHDv00} 
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The first model consists of two blocks of isotropic material, for instance 
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granite, sitting next to each other (\autoref{fig:onedgbmodel}). 
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Initial temperature in \textit{Block 1} is \verbT1 and in \textit{Block 2} is 
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\verbT2. 
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We assume that the system is insulated. 
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What would happen to the temperature distribution in each block over time? 
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Intuition tells us that heat will be transported from the hotter block to the 
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cooler one until both 
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blocks have the same temperature. 
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\begin{figure}[ht] 
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\centerline{\includegraphics[width=4.in]{figures/onedheatdiff001}} 
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\caption{Example 1: Temperature differential along a single interface between 
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two granite blocks.} 
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\label{fig:onedgbmodel} 
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\end{figure} 
35 

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\subsection{1D Heat Diffusion Equation} 
37 
We can model the heat distribution of this problem over time using the one 
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dimensional heat diffusion equation\footnote{A detailed discussion on how the 
39 
heat diffusion equation is derived can be found at 
40 
\url{ 
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http://online.redwoods.edu/instruct/darnold/DEProj/sp02/AbeRichards/paper.pdf}}; 
42 
which is defined as: 
43 
\begin{equation} 
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\rho c_p \frac{\partial T}{\partial t}  \kappa \frac{\partial^{2} 
45 
T}{\partial x^{2}} = q_H 
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\label{eqn:hd} 
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\end{equation} 
48 
where $\rho$ is the material density, $c_p$ is the specific heat and 
49 
$\kappa$ is the thermal 
50 
conductivity\footnote{A list of some common thermal conductivities is available 
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from Wikipedia 
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\url{http://en.wikipedia.org/wiki/List_of_thermal_conductivities}}. Here we 
53 
assume that these material 
54 
parameters are \textbf{constant}. 
55 
The heat source is defined by the right hand side of \refEq{eqn:hd} as 
56 
$q_{H}$; this can take the form of a constant or a function of time and 
57 
space. For example $q_{H} = q_{0}e^{\gamma t}$ where we have 
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the output of our heat source decaying with time. There are also two partial 
59 
derivatives in \refEq{eqn:hd}; $\frac{\partial T}{\partial t}$ describes the 
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change in temperature with time while $\frac{\partial ^2 T}{\partial x^2}$ is 
61 
the spatial change of temperature. As there is only a single spatial dimension 
62 
to our problem, our temperature solution $T$ is only dependent on the time $t$ 
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and our signed distance from the blockblock interface $x$. 
64 

65 
\subsection{PDEs and the General Form} 
66 
It is possible to solve PDE \refEq{eqn:hd} analytically and obtain an exact 
67 
solution to our problem. However, it is not always practical to solve the 
68 
problem this way. Alternatively, computers can be used to find the solution. To 
69 
do this, a numerical approach is required to discretise 
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the PDE \refEq{eqn:hd} across time and space, this reduces the problem to a 
71 
finite number of equations for a finite number of spatial points and time steps. 
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These parameters together define the model. While discretisation introduces 
73 
approximations and a degree of error, a sufficiently sampled model is generally 
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accurate enough to satisfy the accuracy requirements for the final solution. 
75 

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Firstly, we discretise the PDE \refEq{eqn:hd} in time. This leaves us with a 
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steady linear PDE which involves spatial derivatives only and needs to be solved 
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in each time step to progress in time. \esc can help us here. 
79 

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For time discretisation we use the Backward Euler approximation 
81 
scheme\footnote{see \url{http://en.wikipedia.org/wiki/Euler_method}}. It is 
82 
based on the approximation 
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\begin{equation} 
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\frac{\partial T(t)}{\partial t} \approx \frac{T(t)T(th)}{h} 
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\label{eqn:beuler} 
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\end{equation} 
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for $\frac{\partial T}{\partial t}$ at time $t$ 
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where $h$ is the time step size. This can also be written as; 
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\begin{equation} 
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\frac{\partial T}{\partial t}(t^{(n)}) \approx \frac{T^{(n)}  T^{(n1)}}{h} 
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\label{eqn:Tbeuler} 
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\end{equation} 
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where the upper index $n$ denotes the n\textsuperscript{th} time step. So one 
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has 
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\begin{equation} 
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\begin{array}{rcl} 
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t^{(n)} & = & t^{(n1)}+h \\ 
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T^{(n)} & = & T(t^{(n1)}) \\ 
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\end{array} 
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\label{eqn:Neuler} 
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\end{equation} 
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Substituting \refEq{eqn:Tbeuler} into \refEq{eqn:hd} we get; 
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\begin{equation} 
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\frac{\rho c_p}{h} (T^{(n)}  T^{(n1)})  \kappa \frac{\partial^{2} 
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T^{(n)}}{\partial x^{2}} = q_H 
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\label{eqn:hddisc} 
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\end{equation} 
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Notice that we evaluate the spatial derivative term at the current time 
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$t^{(n)}$  therefore the name \textbf{backward Euler} scheme. Alternatively, 
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one can evaluate the spatial derivative term at the previous time $t^{(n1)}$. 
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This approach is called the \textbf{forward Euler} scheme. This scheme can 
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provide some computational advantages, which 
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are not discussed here. However, the \textbf{forward Euler} scheme has a major 
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disadvantage. Namely, depending on the 
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material parameters as well as the domain discretization of the spatial 
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derivative term, the time step size $h$ needs to be chosen sufficiently small to 
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achieve a stable temperature when progressing in time. Stability is achieved if 
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the temperature does not grow beyond its initial bounds and becomes 
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nonphysical. 
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The backward Euler scheme, which we use here, is unconditionally stable meaning 
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that under the assumption of a 
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physically correct problem setup the temperature approximation remains physical 
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for all time steps. 
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The user needs to keep in mind that the discretisation error introduced by 
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\refEq{eqn:beuler} 
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is sufficiently small, thus a good approximation of the true temperature is 
127 
computed. It is therefore very important that any results are viewed with 
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caution. For example, one may compare the results for different time and 
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spatial step sizes. 
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131 
To get the temperature $T^{(n)}$ at time $t^{(n)}$ we need to solve the linear 
132 
differential equation \refEq{eqn:hddisc} which only includes spatial 
133 
derivatives. To solve this problem we want to use \esc. 
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In \esc any given PDE can be described by the general form. For the purpose of 
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this introduction we illustrate a simpler version of the general form for full 
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linear PDEs which is available in the \esc user's guide. A simplified form that 
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suits our heat diffusion problem\footnote{The form in the \esc users guide which 
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uses the Einstein convention is written as 
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$(A_{jl} u_{,l})_{,j}+D u =Y$} 
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is described by; 
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\begin{equation}\label{eqn:commonform nabla} 
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\nabla\cdot(A\cdot\nabla u) + Du = f 
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\end{equation} 
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where $A$, $D$ and $f$ are known values and $u$ is the unknown solution. The 
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symbol $\nabla$ which is called the \textit{Nabla operator} or \textit{del 
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operator} represents 
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the spatial derivative of its subject  in this case $u$. Lets assume for a 
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moment that we deal with a onedimensional problem then ; 
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\begin{equation} 
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\nabla = \frac{\partial}{\partial x} 
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\end{equation} 
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and we can write \refEq{eqn:commonform nabla} as; 
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\begin{equation}\label{eqn:commonform} 
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A\frac{\partial^{2}u}{\partial x^{2}} + Du = f 
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\end{equation} 
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if $A$ is constant. To match this simplified general form to our problem 
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\refEq{eqn:hddisc} 
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we rearrange \refEq{eqn:hddisc}; 
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\begin{equation} 
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\frac{\rho c_p}{h} T^{(n)}  \kappa \frac{\partial^2 T^{(n)}}{\partial 
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x^2} = q_H + \frac{\rho c_p}{h} T^{(n1)} 
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\label{eqn:hdgenf} 
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\end{equation} 
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The PDE is now in a form that satisfies \refEq{eqn:commonform nabla} which is 
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required for \esc to solve our PDE. This can be done by generating a solution 
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for successive increments in the time nodes $t^{(n)}$ where 
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$t^{(0)}=0$ and $t^{(n)}=t^{(n1)}+h$ where $h>0$ is the step size and assumed 
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to be constant. 
170 
In the following the upper index ${(n)}$ refers to a value at time $t^{(n)}$. 
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Finally, by comparing \refEq{eqn:hdgenf} with \refEq{eqn:commonform} one can see 
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that; 
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\begin{equation}\label{ESCRIPT SET} 
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u=T^{(n)}; 
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A = \kappa; D = \frac{\rho c _{p}}{h}; f = q _{H} + \frac{\rho 
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c_p}{h} T^{(n1)} 
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\end{equation} 
178 

179 
\subsection{Boundary Conditions} 
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\label{SEC BOUNDARY COND} 
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With the PDE sufficiently modified, consideration must now be given to the 
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boundary conditions of our model. Typically there are two main types of boundary 
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conditions known as \textbf{Neumann} and \textbf{Dirichlet} boundary 
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conditions\footnote{More information on Boundary Conditions is available at 
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Wikipedia \url{http://en.wikipedia.org/wiki/Boundary_conditions}}, 
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respectively. 
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A \textbf{Dirichlet boundary condition} is conceptually simpler and is used to 
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prescribe a known value to the unknown solution (in our example the temperature) 
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on parts of the boundary or on the entire boundary of the region of interest. 
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We discuss the Dirichlet boundary condition in our second example presented in 
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Section~\ref{Sec:1DHDv0}. 
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However, for this example we have made the model assumption that the system is 
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insulated, so we need to add an appropriate boundary condition to prevent 
195 
any loss or inflow of energy at the boundary of our domain. Mathematically this 
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is expressed by prescribing 
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the heat flux $\kappa \frac{\partial T}{\partial x}$ to zero. In our simplified 
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one dimensional model this is expressed 
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in the form; 
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\begin{equation} 
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\kappa \frac{\partial T}{\partial x} = 0 
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\end{equation} 
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or in a more general case as 
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\begin{equation}\label{NEUMAN 1} 
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\kappa \nabla T \cdot n = 0 
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\end{equation} 
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where $n$ is the outer normal field \index{outer normal field} at the surface 
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of the domain. 
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The $\cdot$ (dot) refers to the dot product of the vectors $\nabla T$ and $n$. 
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In fact, the term $\nabla T \cdot n$ is the normal derivative of 
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the temperature $T$. Other notations used here are\footnote{The \esc notation 
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for the normal 
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derivative is $T_{,i} n_i$.}; 
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\begin{equation} 
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\nabla T \cdot n = \frac{\partial T}{\partial n} \; . 
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\end{equation} 
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A condition of the type \refEq{NEUMAN 1} defines a \textbf{Neumann boundary 
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condition} for the PDE. 
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The PDE \refEq{eqn:hdgenf} 
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and the Neumann boundary condition~\ref{eqn:hdgenf} (potentially together with 
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the Dirichlet boundary conditions) define a \textbf{boundary value problem}. 
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It is the nature of a boundary value problem to allow making statements about 
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the solution in the 
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interior of the domain from information known on the boundary only. In most 
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cases we use the term partial differential equation but in fact it is a 
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boundary value problem. 
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It is important to keep in mind that boundary conditions need to be complete and 
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consistent in the sense that 
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at any point on the boundary either a Dirichlet or a Neumann boundary condition 
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must be set. 
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Conveniently, \esc makes a default assumption on the boundary conditions which 
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the user may modify where appropriate. 
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For a problem of the form in~\refEq{eqn:commonform nabla} the default 
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condition\footnote{In the \esc user guide which uses the Einstein convention 
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this is written as 
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$n_{j}A_{jl} u_{,l}=0$.} is; 
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\begin{equation}\label{NEUMAN 2} 
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n\cdot A \cdot\nabla u = 0 
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\end{equation} 
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which is used everywhere on the boundary. Again $n$ denotes the outer normal 
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field. 
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Notice that the coefficient $A$ is the same as in the \esc 
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PDE~\ref{eqn:commonform nabla}. 
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With the settings for the coefficients we have already identified in 
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\refEq{ESCRIPT SET} this 
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condition translates into 
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\begin{equation}\label{NEUMAN 2b} 
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\kappa \frac{\partial T}{\partial x} = 0 
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\end{equation} 
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for the boundary of the domain. This is identical to the Neumann boundary 
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condition we want to set. \esc will take care of this condition for us. We 
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discuss the Dirichlet boundary condition later. 
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\subsection{Outline of the Implementation} 
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\label{sec:outline} 
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To solve the heat diffusion equation (\refEq{eqn:hd}) we write a simple \pyt 
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script. At this point we assume that you have some basic understanding of the 
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\pyt programming language. If not, there are some pointers and links available 
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in Section \ref{sec:escpybas}. The script (discussed in \refSec{sec:key}) has 
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four major steps. Firstly, we need to define the domain where we want to 
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calculate the temperature. For our problem this is the joint blocks of granite 
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which has a rectangular shape. Secondly, we need to define the PDE to solve in 
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each time step to get the updated temperature. Thirdly, we need to define the 
266 
coefficients of the PDE and finally we need to solve the PDE. The last two steps 
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need to be repeated until the final time marker has been reached. The work flow 
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is described in \reffig{fig:wf}. 
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% \begin{enumerate} 
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% \item create domain 
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% \item create PDE 
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% \item while end time not reached: 
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% \begin{enumerate} 
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% \item set PDE coefficients 
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% \item solve PDE 
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% \item update time marker 
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% \end{enumerate} 
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% \item end of calculation 
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% \end{enumerate} 
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\begin{figure}[h!] 
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\centering 
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\includegraphics[width=1in]{figures/workflow.png} 
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\caption{Workflow for developing an \esc model and solution} 
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\label{fig:wf} 
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\end{figure} 
287 

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In the terminology of \pyt, the domain and PDE are represented by 
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\textbf{objects}. The nice feature of an object is that it is defined by its 
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usage and features 
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rather than its actual representation. So we will create a domain object to 
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describe the geometry of the two 
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granite blocks. Then we define PDEs and spatially distributed values such as the 
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temperature 
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on this domain. Similarly, to define a PDE object we use the fact that one needs 
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only to define the coefficients of the PDE and solve the PDE. The PDE object has 
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advanced features, but these are not required in simple cases. 
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299 

300 
\begin{figure}[htp] 
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\centering 
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\includegraphics[width=6in]{figures/functionspace.pdf} 
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\caption{\esc domain construction overview} 
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\label{fig:fs} 
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\end{figure} 
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\subsection{The Domain Constructor in \esc} 
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\label{ss:domcon} 
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Whilst it is not strictly relevant or necessary, a better understanding of 
310 
how values are spatially distributed (\textit{e.g.} Temperature) and how PDE 
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coefficients are interpreted in \esc can be helpful. 
312 

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There are various ways to construct domain objects. The simplest form is a 
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rectangular shaped region with a length and height. There is 
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a ready to use function for this named \verb rectangle(). Besides the spatial 
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dimensions this function requires to specify the number of 
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elements or cells to be used along the length and height, see \reffig{fig:fs}. 
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Any spatially distributed value 
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and the PDE is represented in discrete form using this element 
320 
representation\footnote{We use the finite element method (FEM), see 
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\url{http://en.wikipedia.org/wiki/Finite_element_method} for details.}. 
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Therefore we will have access to an approximation of the true PDE solution 
323 
only. 
324 
The quality of the approximation depends  besides other factors  mainly on the 
325 
number of elements being used. In fact, the 
326 
approximation becomes better when more elements are used. However, computational 
327 
cost grows with the number of 
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elements being used. It is therefore important that you find the right balance 
329 
between the demand in accuracy and acceptable resource usage. 
330 

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In general, one can think about a domain object as a composition of nodes and 
332 
elements. 
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As shown in \reffig{fig:fs}, an element is defined by the nodes that are used to 
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describe its vertices. 
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To represent spatially distributed values the user can use 
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the values at the nodes, at the elements in the interior of the domain or at the 
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elements located on the surface of the domain. 
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The different approach used to represent values is called \textbf{function 
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space} and is attached to all objects 
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in \esc representing a spatially distributed value such as the solution of 
341 
a PDE. The three function spaces we use at the moment are; 
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\begin{enumerate} 
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\item the nodes, called by \verbContinuousFunction(domain) ; 
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\item the elements/cells, called by \verbFunction(domain) ; and 
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\item the boundary, called by \verbFunctionOnBoundary(domain). 
346 
\end{enumerate} 
347 
A function space object such as \verbContinuousFunction(domain) has the method 
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\verbgetX attached to it. This method returns the 
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location of the socalled \textbf{sample points} used to represent values of the 
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particular function space. So the 
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call \verbContinuousFunction(domain).getX() will return the coordinates of the 
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nodes used to describe the domain while 
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\verbFunction(domain).getX() returns the coordinates of numerical 
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integration points within elements, see \reffig{fig:fs}. 
355 

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This distinction between different representations of spatially distributed 
357 
values 
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is important in order to be able to vary the degrees of smoothness in a PDE 
359 
problem. 
360 
The coefficients of a PDE do not need to be continuous, thus this qualifies as a 
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\verbFunction() type. 
362 
On the other hand a temperature distribution must be continuous and needs to be 
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represented with a \verbContinuousFunction() function space. 
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An influx may only be defined at the boundary and is therefore a 
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\verbFunctionOnBoundary() object. 
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\esc allows certain transformations of the function spaces. A 
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\verbContinuousFunction() can be transformed into a 
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\verbFunctionOnBoundary() or \verbFunction(). On the other hand there is 
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not enough information in a \verbFunctionOnBoundary() to transform it to a 
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\verbContinuousFunction(). 
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These transformations, which are called \textbf{interpolation} are invoked 
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automatically by \esc if needed. 
373 

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Later in this introduction we discuss how 
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to define specific areas of geometry with different materials which are 
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represented by different material coefficients such as the 
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thermal conductivities $\kappa$. A very powerful technique to define these types 
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of PDE 
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coefficients is tagging. Blocks of materials and boundaries can be named and 
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values can be defined on subregions based on their names. 
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This is a method for simplifying PDE coefficient and flux definitions. It makes 
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scripting much easier and we will discuss this technique in 
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Section~\ref{STEADYSTATE HEAT REFRACTION}. 
384 

385 

386 
\subsection{A Clarification for the 1D Case} 
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\label{SEC: 1D CLARIFICATION} 
388 
It is necessary for clarification that we revisit our general PDE from 
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\refeq{eqn:commonform nabla} for a two dimensional domain. \esc is inherently 
390 
designed to solve problems that are multidimensional and so 
391 
\refEq{eqn:commonform nabla} needs to be read as a higher dimensional problem. 
392 
In the case of two spatial dimensions the \textit{Nabla operator} has in fact 
393 
two components $\nabla = (\frac{\partial}{\partial x}, \frac{\partial}{\partial 
394 
y})$. Assuming the coefficient $A$ is constant, the \refEq{eqn:commonform nabla} 
395 
takes the following form; 
396 
\begin{equation}\label{eqn:commonform2D} 
397 
A_{00}\frac{\partial^{2}u}{\partial x^{2}} 
398 
A_{01}\frac{\partial^{2}u}{\partial x\partial y} 
399 
A_{10}\frac{\partial^{2}u}{\partial y\partial x} 
400 
A_{11}\frac{\partial^{2}u}{\partial y^{2}} 
401 
+ Du = f 
402 
\end{equation} 
403 
Notice that for the higher dimensional case $A$ becomes a matrix. It is also 
404 
important to notice that the usage of the Nabla operator creates 
405 
a compact formulation which is also independent from the spatial dimension. 
406 
To make the general PDE \refEq{eqn:commonform2D} one dimensional as 
407 
shown in \refEq{eqn:commonform} we need to set 
408 
\begin{equation} 
409 
A_{00}=A; A_{01}=A_{10}=A_{11}=0 
410 
\end{equation} 
411 

412 

413 
\subsection{Developing a PDE Solution Script} 
414 
\label{sec:key} 
415 
\sslist{example01a.py} 
416 
We write a simple \pyt script which uses the \modescript, \modfinley and \modmpl 
417 
modules. 
418 
By developing a script for \esc, the heat diffusion equation can be solved at 
419 
successive time steps for a predefined period using our general form 
420 
\refEq{eqn:hdgenf}. Firstly it is necessary to import all the 
421 
libraries\footnote{The libraries contain predefined scripts that are required to 
422 
solve certain problems, these can be simple like sine and cosine functions or 
423 
more complicated like those from our \esc library.} 
424 
that we will require. 
425 
\begin{python} 
426 
from esys.escript import * 
427 
# This defines the LinearPDE module as LinearPDE 
428 
from esys.escript.linearPDEs import LinearPDE 
429 
# This imports the rectangle domain function from finley. 
430 
from esys.finley import Rectangle 
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# A useful unit handling package which will make sure all our units 
432 
# match up in the equations under SI. 
433 
from esys.escript.unitsSI import * 
434 
\end{python} 
435 
It is generally a good idea to import all of the \modescript library, although 
436 
if the functions and classes required are known they can be specified 
437 
individually. The function \verbLinearPDE has been imported explicitly for 
438 
ease of use later in the script. \verbRectangle is going to be our type of 
439 
domain. The module \verbunitsSI provides support for SI unit definitions with 
440 
our variables. 
441 

442 
Once our library dependencies have been established, defining the problem 
443 
specific variables is the next step. In general the number of variables needed 
444 
will vary between problems. These variables belong to two categories. They are 
445 
either directly related to the PDE and can be used as inputs into the \esc 
446 
solver, or they are script variables used to control internal functions and 
447 
iterations in our problem. For this PDE there are a number of constants which 
448 
need values. Firstly, the domain upon which we wish to solve our problem needs 
449 
to be defined. There are different types of domains in \modescript which we 
450 
demonstrate in later tutorials but for our granite blocks, we simply use a 
451 
rectangular domain. 
452 

453 
Using a rectangular domain simplifies our granite blocks (which would in reality 
454 
be a \textit{3D} object) into a single dimension. The granite blocks will have a 
455 
lengthways cross section that looks like a rectangle. As a result we do not 
456 
need to model the volume of the block due to symmetry. There are four arguments 
457 
we must consider when we decide to create a rectangular domain, the domain 
458 
\textit{length}, \textit{width} and \textit{step size} in each direction. When 
459 
defining the size of our problem it will help us determine appropriate values 
460 
for our model arguments. If we make our dimensions large but our step sizes very 
461 
small we increase the accuracy of our solution. Unfortunately we also increase 
462 
the number of calculations that must be solved per time step. This means more 
463 
computational time is required to produce a solution. In this \textit{1D} 
464 
problem, the bar is defined as being 1 metre long. An appropriate step size 
465 
\verbndx would be 1 to 10\% of the length. Our \verbndy needs only be 1, 
466 
this is because our problem stipulates no partial derivatives in the $y$ 
467 
direction. 
468 
Thus the temperature does not vary with $y$. Hence, the model parameters can be 
469 
defined as follows; note we have used the \verbunitsSI convention to make sure 
470 
all our input units are converted to SI. 
471 
\begin{python} 
472 
mx = 500.*m #meters  model length 
473 
my = 100.*m #meters  model width 
474 
ndx = 50 # mesh steps in x direction 
475 
ndy = 1 # mesh steps in y direction 
476 
boundloc = mx/2 # location of boundary between the two blocks 
477 
\end{python} 
478 
The material constants and the temperature variables must also be defined. For 
479 
the granite in the model they are defined as: 
480 
\begin{python} 
481 
#PDE related 
482 
rho = 2750. *kg/m**3 #kg/m^{3} density of iron 
483 
cp = 790.*J/(kg*K) # J/Kg.K thermal capacity 
484 
rhocp = rho*cp 
485 
kappa = 2.2*W/m/K # watts/m.Kthermal conductivity 
486 
qH=0 * J/(sec*m**3) # J/(sec.m^{3}) no heat source 
487 
T1=20 * Celsius # initial temperature at Block 1 
488 
T2=2273. * Celsius # base temperature at Block 2 
489 
\end{python} 
490 
Finally, to control our script we will have to specify our timing controls and 
491 
where we would like to save the output from the solver. This is simple enough: 
492 
\begin{python} 
493 
t=0 * day # our start time, usually zero 
494 
tend=50 * yr #  time to end simulation 
495 
outputs = 200 # number of time steps required. 
496 
h=(tendt)/outputs #size of time step 
497 
#user warning statement 
498 
print("Expected Number of time outputs is: ", (tendt)/h) 
499 
i=0 #loop counter 
500 
\end{python} 
501 
Now that we know our inputs we will build a domain using the 
502 
\verbRectangle() function from \FINLEY. The four arguments allow us to 
503 
define our domain \verbmodel as: 
504 
\begin{python} 
505 
#generate domain using rectangle 
506 
blocks = Rectangle(l0=mx,l1=my,n0=ndx, n1=ndy) 
507 
\end{python} 
508 
\verbblocks now describes a domain in the manner of Section \ref{ss:domcon}. 
509 

510 
With a domain and all the required variables established, it is now possible to 
511 
set up our PDE so that it can be solved by \esc. The first step is to define the 
512 
type of PDE that we are trying to solve in each time step. In this example it is 
513 
a single linear PDE\footnote{in contrast to a system of PDEs which we discuss 
514 
later.}. We also need to state the values of our general form variables. 
515 
\begin{python} 
516 
mypde=LinearPDE(blocks) 
517 
A=zeros((2,2))) 
518 
A[0,0]=kappa 
519 
mypde.setValue(A=A, D=rhocp/h) 
520 
\end{python} 
521 
In many cases it may be possible to decrease the computational time of the 
522 
solver if the PDE is symmetric. 
523 
Symmetry of a PDE is defined by; 
524 
\begin{equation}\label{eqn:symm} 
525 
A_{jl}=A_{lj} 
526 
\end{equation} 
527 
Symmetry is only dependent on the $A$ coefficient in the general form and the 
528 
other coefficients $D$ as well as the right hand side $Y$. From the above 
529 
definition we can see that our PDE is symmetric. The \verbLinearPDE class 
530 
provides the method \method{checkSymmetry} to check if the given PDE is 
531 
symmetric. As our PDE is symmetrical we enable symmetry via; 
532 
\begin{python} 
533 
myPDE.setSymmetryOn() 
534 
\end{python} 
535 
Next we need to establish the initial temperature distribution \verbT. We need 
536 
to 
537 
assign the value \verbT1 to all sample points left to the contact interface at 
538 
$x_{0}=\frac{mx}{2}$ 
539 
and the value \verbT2 right to the contact interface. \esc 
540 
provides the \verbwhereNegative function to construct this. More 
541 
specifically, \verbwhereNegative returns the value $1$ at those sample points 
542 
where the argument has a negative value. Otherwise zero is returned. 
543 
If \verbx are the $x_{0}$ 
544 
coordinates of the sample points used to represent the temperature distribution 
545 
then \verbx[0]boundloc gives us a negative value for 
546 
all sample points left to the interface and nonnegative value to 
547 
the right of the interface. So with; 
548 
\begin{python} 
549 
# ... set initial temperature .... 
550 
T= T1*whereNegative(x[0]boundloc)+T2*(1whereNegative(x[0]boundloc)) 
551 
\end{python} 
552 
we get the desired temperature distribution. To get the actual sample points 
553 
\verbx we use the \verbgetX() method of the function space 
554 
\verbSolution(blocks) which is used to represent the solution of a PDE; 
555 
\begin{python} 
556 
x=Solution(blocks).getX() 
557 
\end{python} 
558 
As \verbx are the sample points for the function space 
559 
\verbSolution(blocks) 
560 
the initial temperature \verbT is using these sample points for 
561 
representation. 
562 
Although \esc is trying to be forgiving with the choice of sample points and to 
563 
convert 
564 
where necessary the adjustment of the function space is not always possible. So 
565 
it is advisable to make a careful choice on the function space used. 
566 

567 
Finally we initialise an iteration loop to solve our PDE for all the time steps 
568 
we specified in the variable section. As the right hand side of the general form 
569 
is dependent on the previous values for temperature \verb T across the bar this 
570 
must be updated in the loop. Our output at each time step is \verb T the heat 
571 
distribution and \verb totT the total heat in the system. 
572 
\begin{python} 
573 
while t < tend: 
574 
i+=1 #increment the counter 
575 
t+=h #increment the current time 
576 
mypde.setValue(Y=qH+rhocp/h*T) # set variable PDE coefficients 
577 
T=mypde.getSolution() #get the PDE solution 
578 
totE = integrate(rhocp*T) #get the total heat (energy) in the system 
579 
\end{python} 
580 
The last statement in this script calculates the total energy in the system as 
581 
the volume integral of $\rho c_{p} T$ over the block. 
582 
As the blocks are insulated no energy should be lost or added. 
583 
The total energy should stay constant for the example discussed here. 
584 

585 
\subsection{Running the Script} 
586 
The script presented so far is available under 
587 
\verbexample01a.py. You can edit this file with your favourite text editor. 
588 
On most operating systems\footnote{The \texttt{runescript} launcher is not 
589 
supported under {\it MS Windows}.} you can use the 
590 
\program{runescript} command 
591 
to launch {\it escript} scripts. For the example script use; 
592 
\begin{verbatim} 
593 
runescript example01a.py 
594 
\end{verbatim} 
595 
The program will print a progress report. Alternatively, you can use 
596 
the python interpreter directly; 
597 
\begin{verbatim} 
598 
python example01a.py 
599 
\end{verbatim} 
600 
if the system is configured correctly (please talk to your system 
601 
administrator). 
602 

603 
\subsection{Plotting the Total Energy} 
604 
\sslist{example01b.py} 
605 

606 
\esc does not include its own plotting capabilities. However, it is possible to 
607 
use a variety of free \pyt packages for visualisation. 
608 
Two types will be demonstrated in this cookbook; 
609 
\mpl\footnote{\url{http://matplotlib.sourceforge.net/}} and 
610 
\verbVTK\footnote{\url{http://www.vtk.org/}}. 
611 
The \mpl package is a component of SciPy\footnote{\url{http://www.scipy.org}} 
612 
and is good for basic graphs and plots. 
613 
For more complex visualisation tasks, in particular two and three dimensional 
614 
problems we recommend the use of more advanced tools. For instance, \mayavi 
615 
\footnote{\url{http://code.enthought.com/projects/mayavi/}} 
616 
which is based upon the \verbVTK toolkit. The usage of \verbVTK based 
617 
visualisation is discussed in Chapter~\ref{Sec:2DHD} which focuses on a two 
618 
dimensional PDE. 
619 

620 
For our simple granite block problem, we have two plotting tasks. Firstly, we 
621 
are interested in showing the 
622 
behaviour of the total energy over time and secondly, how the temperature 
623 
distribution within the block is developing over time. 
624 
Let us start with the first task. 
625 

626 
The idea is to create a record of the time marks and the corresponding total 
627 
energies observed. 
628 
\pyt provides the concept of lists for this. Before 
629 
the time loop is opened we create empty lists for the time marks \verbt_list 
630 
and the total energies \verbE_list. 
631 
After the new temperature has been calculated by solving the PDE we append the 
632 
new time marker and the total energy value for that time 
633 
to the corresponding list using the \verbappend method. With these 
634 
modifications our script looks as follows: 
635 
\begin{python} 
636 
t_list=[] 
637 
E_list=[] 
638 
# ... start iteration: 
639 
while t<tend: 
640 
t+=h 
641 
mypde.setValue(Y=qH+rhocp/h*T) # set variable PDE coefficients 
642 
T=mypde.getSolution() #get the PDE solution 
643 
totE=integrate(rhocp*T) 
644 
t_list.append(t) # add current time mark to record 
645 
E_list.append(totE) # add current total energy to record 
646 
\end{python} 
647 
To plot $t$ over $totE$ we use \mpl a module contained within \pylab which needs 
648 
to be loaded before use; 
649 
\begin{python} 
650 
import pylab as pl # plotting package. 
651 
\end{python} 
652 
Here we are not using \verbfrom pylab import * in order to avoid name 
653 
clashes for function names within \esc. 
654 

655 
The following statements are added to the script after the time loop has been 
656 
completed; 
657 
\begin{python} 
658 
pl.plot(t_list,E_list) 
659 
pl.title("Total Energy") 
660 
pl.axis([0,max(t_list),0,max(E_list)*1.1]) 
661 
pl.savefig("totE.png") 
662 
\end{python} 
663 
The first statement hands over the time marks and corresponding total energies 
664 
to the plotter. 
665 
The second statement sets the title for the plot. The third statement 
666 
sets the axis ranges. In most cases these are set appropriately by the plotter. 
667 

668 
The last statement generates the plot and writes the result into the file 
669 
\verbtotE.png which can be displayed by (almost) any image viewer. 
670 
As expected the total energy is constant over time, see 
671 
\reffig{fig:onedheatout1}. 
672 

673 
\begin{figure}[ht] 
674 
\begin{center} 
675 
\includegraphics[width=4in]{figures/ttblockspyplot150} 
676 
\caption{Example 1b: Total Energy in the Blocks over Time (in seconds)} 
677 
\label{fig:onedheatout1} 
678 
\end{center} 
679 
\end{figure} 
680 
\clearpage 
681 

682 
\subsection{Plotting the Temperature Distribution} 
683 
\label{sec: plot T} 
684 
\sslist{example01c.py} 
685 
For plotting the spatial distribution of the temperature we need to modify the 
686 
strategy we have used for the total energy. 
687 
Instead of producing a final plot at the end we will generate a 
688 
picture at each time step which can be browsed as a slide show or composed into 
689 
a movie. 
690 
The first problem we encounter is that if we produce an image at each time step 
691 
we need to make sure that the images previously generated are not overwritten. 
692 

693 
To develop an incrementing file name we can use the following convention. It is 
694 
convenient to put all image files showing the same variable  in our case the 
695 
temperature distribution  into a separate directory. 
696 
As part of the \verbos module\footnote{The \texttt{os} module provides 
697 
a powerful interface to interact with the operating system, see 
698 
\url{http://docs.python.org/library/os.html}.} \pyt 
699 
provides the \verbos.path.join command to build file and directory names in a 
700 
platform independent way. Assuming that 
701 
\verbsave_path is the name of the directory we want to put the results in the 
702 
command is; 
703 
\begin{python} 
704 
import os 
705 
os.path.join(save_path, "tempT%03d.png"%i ) 
706 
\end{python} 
707 
where \verbi is the time step counter. 
708 
There are two arguments to the \verbjoin command. The \verbsave_path 
709 
variable is a predefined string pointing to the directory we want to save our 
710 
data, for example a single subfolder called \verbdata would be defined by; 
711 
\begin{verbatim} 
712 
save_path = "data" 
713 
\end{verbatim} 
714 
while a subfolder of \verbdata called \verbexample01 would be defined by; 
715 
\begin{verbatim} 
716 
save_path = os.path.join("data","example01") 
717 
\end{verbatim} 
718 
The second argument of \verbjoin contains a string which is the file 
719 
name or subdirectory name. We can use the operator \verb% to use the value of 
720 
\verbi as part of our filename. The substring \verb%03d indicates that we 
721 
want to substitute a value into the name; 
722 
\begin{itemize} 
723 
\item \verb 0 means that small numbers should have leading zeroes; 
724 
\item \verb 3 means that numbers should be written using at least 3 digits; 
725 
and 
726 
\item \verb d means that the value to substitute will be a decimal integer. 
727 
\end{itemize} 
728 

729 
To actually substitute the value of \verbi into the name write \verb%i after 
730 
the string. 
731 
When done correctly, the output files from this command will be placed in the 
732 
directory defined by \verb save_path as; 
733 
\begin{verbatim} 
734 
blockspyplot001.png 
735 
blockspyplot002.png 
736 
blockspyplot003.png 
737 
... 
738 
\end{verbatim} 
739 
and so on. 
740 

741 
A subfolder check/constructor is available in \esc. The command; 
742 
\begin{verbatim} 
743 
mkDir(save_path) 
744 
\end{verbatim} 
745 
will check for the existence of \verb save_path and if missing, create the 
746 
required directories. 
747 

748 
We start by modifying our solution script. 
749 
Prior to the \verbwhile loop we need to extract our finite solution 
750 
points to a data object that is compatible with \mpl. First we create the node 
751 
coordinates of the sample points used to represent 
752 
the temperature as a \pyt list of tuples or a \numpy array as requested by the 
753 
plotting function. 
754 
We need to convert the array \verbx previously set as 
755 
\verbSolution(blocks).getX() into a \pyt list 
756 
and then to a \numpy array. The $x_{0}$ component is then extracted via 
757 
an array slice to the variable \verbplx; 
758 
\begin{python} 
759 
import numpy as np # array package. 
760 
#convert solution points for plotting 
761 
plx = x.toListOfTuples() 
762 
plx = np.array(plx) # convert to tuple to numpy array 
763 
plx = plx[:,0] # extract x locations 
764 
\end{python} 
765 

766 
\begin{figure} 
767 
\begin{center} 
768 
\includegraphics[width=4in]{figures/blockspyplot001} 
769 
\includegraphics[width=4in]{figures/blockspyplot050} 
770 
\includegraphics[width=4in]{figures/blockspyplot200} 
771 
\caption{Example 1c: Temperature ($T$) distribution in the blocks at time steps 
772 
$1$, $50$ and $200$} 
773 
\label{fig:onedheatout} 
774 
\end{center} 
775 
\end{figure} 
776 
\clearpage 
777 

778 
We use the same techniques provided by \mpl as we have used to plot the total 
779 
energy over time. 
780 
For each time step we generate a plot of the temperature distribution and save 
781 
each to a file. 
782 
The following is appended to the end of the \verbwhile loop and creates one 
783 
figure of the temperature distribution. We start by converting the solution to a 
784 
tuple and then plotting this against our \textit{x coordinates} \verbplx we 
785 
have generated before. We add a title to the diagram before it is rendered into 
786 
a file. 
787 
Finally, the figure is saved to a \verb*.png file and cleared for the 
788 
following iteration. 
789 
\begin{python} 
790 
# ... start iteration: 
791 
while t<tend: 
792 
i+=1 
793 
t+=h 
794 
mypde.setValue(Y=qH+rhocp/h*T) 
795 
T=mypde.getSolution() 
796 
totE=integrate(rhocp*T) 
797 
print("time step %s at t=%e days completed. total energy = %e."%(i,t/day,totE)) 
798 
t_list.append(t) 
799 
E_list.append(totE) 
800 

801 
#establish figure 1 for temperature vs x plots 
802 
tempT = T.toListOfTuples() 
803 
pl.figure(1) #current figure 
804 
pl.plot(plx,tempT) #plot solution 
805 
# add title 
806 
pl.axis([0,mx,T1*.9,T2*1.1]) 
807 
pl.title("Temperature across blocks at time %d days"%(t/day)) 
808 
#save figure to file 
809 
pl.savefig(os.path.join(save_path,"tempT", "blockspyplot%03d.png"%i)) 
810 
pl.clf() #clear figure 
811 
\end{python} 
812 
Some results are shown in \reffig{fig:onedheatout}. 
813 

814 
\subsection{Making a Video} 
815 
Our saved plots from the previous section can be cast into a video using the 
816 
following command appended to the end of the script. The \verb mencoder command 
817 
is not available on every platform, so some users need to use an alternative 
818 
video encoder. 
819 
\begin{python} 
820 
# compile the *.png files to create a *.avi video that shows T change 
821 
# with time. This operation uses Linux mencoder. For other operating 
822 
# systems it is possible to use your favourite video compiler to 
823 
# convert image files to videos. 
824 

825 
os.system("mencoder mf://"+save_path+"/tempT"+"/*.png mf type=png:\ 
826 
w=800:h=600:fps=25 ovc lavc lavcopts vcodec=mpeg4 oac copy o \ 
827 
example01tempT.avi") 
828 
\end{python} 
829 
