13 


14 
\section{One Dimensional Heat Diffusion in an Iron Rod} 
\section{One Dimensional Heat Diffusion in an Iron Rod} 
15 
%\label{Sec:1DHDv0} 
%\label{Sec:1DHDv0} 
16 
We will start by examining a simple one dimensional heat diffusion example. While this exact problem is not strictly relevant to earth sciences; it will provide a good launch pad to build our knowledge of \ESCRIPT and how to solve simple partial differential equations (PDEs) 
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 \ESCRIPT and 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 \ESCRIPT and implementation should become a clearer as we develop our understanding further into the cookbook.} 

\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 \ESCRIPT and implementation should become a clearer as we develop our understanding further into the cookbook.} 

17 



The first model consists of a simple cold iron bar at a constant temperature of zero Figure \ref{fig:onedhdmodel}. The bar is perfectly insulated on all sides with a heating element of some description at one end. Intuition tells us that as heat is applied; energy will disperse along the bar via convection. With time the bar will reach a constant temperature equivalent to the heat source. 

18 
\begin{figure}[h!] 
\begin{figure}[h!] 
19 
\centerline{\includegraphics[width=4.in]{figures/onedheatdiff}} 
\centerline{\includegraphics[width=4.in]{figures/onedheatdiff}} 
20 
\caption{One dimensional model of an Iron bar.} 
\caption{One dimensional model of an Iron bar.} 
21 
\label{fig:onedhdmodel} 
\label{fig:onedhdmodel} 
22 
\end{figure} 
\end{figure} 
23 

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 the heat source. 
24 


25 
We can model the heat distribution of this problem in time using the one dimensional heat diffusion equation 
\subsection{1D Heat Diffusion Equation} 
26 
\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}}; 
We can model the heat distribution of this problem over time using the one dimensional heat diffusion equation\footnote{A detailed discussion on how the heat diffusion equation is derived can be found at \url{http://online.redwoods.edu/instruct/darnold/DEProj/sp02/AbeRichards/paper.pdf}}; 
27 
which is defined as: 
which is defined as: 
28 
\begin{equation} 
\begin{equation} 
29 
\rho c\hackscore p \frac{\partial T}{\partial t}  \kappa \frac{\partial^{2} T}{\partial x^{2}} = q\hackscore H 
\rho c\hackscore p \frac{\partial T}{\partial t}  \kappa \frac{\partial^{2} T}{\partial x^{2}} = q\hackscore H 
30 
\label{eqn:hd} 
\label{eqn:hd} 
31 
\end{equation} 
\end{equation} 
32 
where $\rho$ is the material density, $c\hackscore p$ is the specific heat and $\kappa$ is the thermal conductivity constant for a given material 
where $\rho$ is the material density, $c\hackscore p$ is the specific heat and $\kappa$ is the thermal conductivity constant for a given material\footnote{A list of some common thermal conductivities is available from Wikipedia \url{http://en.wikipedia.org/wiki/List_of_thermal_conductivities}}. 
33 
\footnote{A list of some common thermal conductivities is available from Wikipedia \url{http://en.wikipedia.org/wiki/List_of_thermal_conductivities}}. 
The heat source is defined by the right hand side of \ref{eqn:hd} as $q\hackscore{H}$; this can take the form of a constant or a function of time and space. For example $q\hackscore{H} = Te^{\gamma t}$ where we have the output of our heat source decaying with time. There are also two partial derivatives in \ref{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 $x$. 

The heatsource is defined by the right hand side of \eqref{eqn:hd} as $q\hackscore{H}$, this could be a constant or a function of time. For example $q\hackscore{H} = Te^{\gamma t}$ where we have the output of our heatsource decaying with time. There are also two partial derivatives in \eqref{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 $x$. 

34 


35 
When solving a PDE analytically we are finding an exact solution to our equation, however it is not always possible to solve a problem this way. Computers can be used in these cases when a large number of sums or visualisation is required. Computers require a numerical approach to solving problems  \ESCRIPT is one example  and it becomes necessary to discretize the equation so that we are left with a finite number of equations for a finite number of spatial and time steps in the model. While this is of course an approximation and introduces a degree of error, we find that a sufficiently sampled model is generally accurate enough for the requirements of the modeller. 
\subsection{Escript, PDEs and The General Form} 
36 

Potentially, it is now possible to solve \ref{eqn:hd} analytically. 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 when a large number of sums or visualisation is required. To do this, a numerical approach is required  \ESCRIPT can help us here  and it becomes necessary to discretize the equation so that 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 modeller. 
37 


38 
To solve this equation we will write a simple python script which uses \ESCRIPT and \FINLEY of the \ESYS module. At this point we assume that you have some basic understanding of the python programming language. If not there are some pointers and links available in Section \ref{sec:escpybas} . 
\ESCRIPT interfaces with any generic PDE via a general form. In this example we will illustrate a simpler version of the full linear PDE general form which is availabe in the user's guide. The first step is to define the coefficients of the general form using our PDE \ref{eqn:hd}. This simplified form suits our heat diffusion problem\footnote{In the form of the \ESCRIPT users guide which using the Einstein convention is written as 




In the following section we will develop a script for \ESCRIPT to solve the heat equation stepbystep. 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 sin and cos functions or more complicated like those from our \ESCRIPT library.} 


that we will require. 


\begin{verbatim} 


from esys.escript import * 


from esys.escript.linearPDEs import SingleLinearPDE 


from esys.finley import Rectangle 


from esys.escript.unitsSI import * 


import os 


\end{verbatim} 


It is generally a good idea to import all of the \verb escript library, although if you know the packages you need you can specify them individually. The function \verbLinearPDE has been imported for ease of use later in the script. \verbRectangle is going to be our type of domain. The package \verb unitsSI is a module of \esc that provides support for units definitions with our variables; and the \verbos package is needed to handle file outputs once our PDE has been solved. 





Once our libraries dependancies 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 escript 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 domain upon which we wish to solve our problem needs to be defined. There are many different types of domains in escript. We will demonstrate a few in later tutorials but for our iron rod we will simply use a rectangular domain. 





Using a rectangular domain simplifies 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 center. There are four arguments we must consider when we decide to create a rectangular domain, the model length, width and step size in each direction. When defining the size of our problem it will help us determine appropriate values for our domain 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 our \textit{1D} problem we will define our bar as being 1 metre long. An appropriate \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 so the temperature does not vary with $y$. Thus the domain perameters can be defined as follows; note we have used the \verb unitsSI convention to make sure all our input units are converted to SI. 


\begin{verbatim} 


#Domain related. 


mx = 1*m #meters  model lenght 


my = .1*m #meters  model width 


ndx = 100 # steps in x direction 


ndy = 1 # steps in y direction 


\end{verbatim} 


The material constants and the temperature variables must also be defined. For the iron rod in the model they are defined as: 


\begin{verbatim} 


#PDE related 


q=200. * Celsius #Kelvin  our heat source temperature 


Tref = 0. * Celsius # Kelvin  starting temp of iron bar 


rho = 7874. *kg/m**3 #kg/m^{3} density of iron 


cp = 449.*J/(kg*K) #jules/Kg.K thermal capacity 


rhocp = rho*cp 


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


\end{verbatim} 


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: 


\begin{verbatim} 


#Script/Iteration Related 


t=0 #our start time, usually zero 


tend=5.*60. #seconds  time to end simulation 


outputs = 200 # number of time steps required. 


h=(tendt)/outputs #size of time step 


i=0 #loop counter 


#the folder to put our outputs in, leave blank "" for script path 


#note this folder path must exist to work 


save_path = "data/onedheatdiff001" 


\end{verbatim} 


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: 


\begin{verbatim} 


rod = Rectangle(l0=mx,l1=my,n0=ndx,n1=ndy) 


\end{verbatim} 


In this form \verb rod does not represent any discrete points, but rather an area of \verb ndx*ndy cells that fit into a rectangular space with opposing vertices at the origin and the point \verb [mx,my] . Our domain is contructed this way to allow the user to determine where the discrete points of each model will be located. Discretisation may be at the corners of each cell, the middle point of a cell or halfway along each side of the cell etc. Depending on the PDE or the model there may be advantages and disadvantages for each case. Fortunately \verb escript offers an easy way to extract finite points from the domain \verbrod using the domain property function \verbgetX() . This function sets the vertices of each cell as finite points to solve in the solution. If we let \verbx be these finite points, then; 


\begin{verbatim} 


x = rod.getX() 


\end{verbatim} 


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 escript. 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 can define by: 


\begin{verbatim} 


mypde=LinearSinglePDE(rod) 


\end{verbatim} 


In the next step we need to define the coefficients of the PDE. The linear 


PDEs in \ESCRIPT provide a general interface to do this. Here we will only discuss a simplified form that suits our heat diffusion problem and refer to the \ESCRIPT user's guide for the general case. This simpler form 


\footnote{In the form of the \ESCRIPT users guide which using the Einstein convention is written as 

39 
$(A\hackscore{jl} u\hackscore{,l})\hackscore{,j}+D u =Y$} 
$(A\hackscore{jl} u\hackscore{,l})\hackscore{,j}+D u =Y$} 
40 
is described by; 
and is described by; 
41 
\begin{equation}\label{eqn:commonform nabla} 
\begin{equation}\label{eqn:commonform nabla} 
42 
\nabla.(A.\nabla u) + Du = f 
\nabla.(A.\nabla u) + Du = f 
43 
\end{equation} 
\end{equation} 
44 
where $A$, $D$ and $f$ are known values. The symbol $\nabla$ which is called the \textit{Nabla operator} or \textit{del operator} represents 
where $A$, $D$ and $f$ are known values. The symbol $\nabla$ which is called the \textit{Nabla operator} or \textit{del operator} represents 
45 
the spatial derivative of its subject  in this case $u$. Lets assume for a moment that we deal with a onedimensional problem then 
the spatial derivative of its subject  in this case $u$. Lets assume for a moment that we deal with a onedimensional problem then ; 
46 
\begin{equation} 
\begin{equation} 
47 
\nabla = \frac{\partial}{\partial x} 
\nabla = \frac{\partial}{\partial x} 
48 
\end{equation} 
\end{equation} 
49 
and we can write equation \ref{eqn:commonform nabla} as 
and we can write \ref{eqn:commonform nabla} as; 
50 
\begin{equation}\label{eqn:commonform} 
\begin{equation}\label{eqn:commonform} 
51 
A\frac{\partial^{2}u}{\partial x^{2}} + Du = f 
A\frac{\partial^{2}u}{\partial x^{2}} + Du = f 
52 
\end{equation} 
\end{equation} 
53 
if $A$ is constant then equation \ref{eqn:commonform} is consistent with our heat diffusion problem in Equation \ref{eqn:hd}  except for $u$. When comparing equations \eqref{eqn:hd} and \eqref{eqn:commonform} we see that; 
if $A$ is constant then \ref{eqn:commonform} is consistent with our heat diffusion problem in \ref{eqn:hd}  except for $u$ and when comparing equations \ref{eqn:hd} and \ref{eqn:commonform} we see that; 
54 
\begin{equation} 
\begin{equation} 
55 
A = \kappa; D = \rho c \hackscore{p}; f = q \hackscore{H} 
A = \kappa; D = \rho c \hackscore{p}; f = q \hackscore{H} 
56 
\end{equation} 
\end{equation} 
57 


58 
We can write the partial $\frac{\partial T}{\partial t}$ in terms of $u$ by discretising the time of our solution. Many methods could be used here but we have decided on the Backwards Euler approximation which states; 
We can write the partial $\frac{\partial T}{\partial t}$ in terms of $u$ by discretising the time of our solution. The method we will use is the Backwards Euler approximation, which states; 
59 
\begin{equation} 
\begin{equation} 
60 
f'(x) \approx \frac{f(x+h)f(x)}{h} 
f'(x) \approx \frac{f(x+h)f(x)}{h} 
61 
\label{eqn:beuler} 
\label{eqn:beuler} 
62 
\end{equation} 
\end{equation} 
63 
where h is the the discrete step size $\Delta x$. 
where h is the the discrete step size $\Delta x$. 
64 
Now let $f(x) = T(t)$ and from equation \ref{eqn:beuler} we see that; 
Now let $f(x) = T(t)$ and from \ref{eqn:beuler} we see that; 
65 
\begin{equation} 
\begin{equation} 
66 
T'(t) \approx \frac{T(t+h)  T(t)}{h} 
T'(t) \approx \frac{T(t+h)  T(t)}{h} 
67 
\end{equation} 
\end{equation} 
70 
T\hackscore{,t}^{(n)} \approx \frac{T^{(n)}  T^{(n1)}}{h} 
T\hackscore{,t}^{(n)} \approx \frac{T^{(n)}  T^{(n1)}}{h} 
71 
\label{eqn:Tbeuler} 
\label{eqn:Tbeuler} 
72 
\end{equation} 
\end{equation} 
73 
where $n$ denotes the n\textsuperscript{th} time step. Substituting equation \ref{eqn:Tbeuler} into equation \ref{eqn:hd} we get; 
where $n$ denotes the n\textsuperscript{th} time step. Substituting \ref{eqn:Tbeuler} into \ref{eqn:hd} we get; 
74 
\begin{equation} 
\begin{equation} 
75 
\frac{\rho c\hackscore p}{h} (T^{(n)}  T^{(n1)})  \kappa \frac{\partial^{2} T}{\partial x^{2}} = q\hackscore H 
\frac{\rho c\hackscore p}{h} (T^{(n)}  T^{(n1)})  \kappa \frac{\partial^{2} T}{\partial x^{2}} = q\hackscore H 
76 
\label{eqn:hddisc} 
\label{eqn:hddisc} 
77 
\end{equation} 
\end{equation} 
78 
To fit our simplified general form we can rearrange so that; 
To fit our simplified general form we can rearrange \ref{eqn:hddisc} so that; 
79 
\begin{equation} 
\begin{equation} 
80 
\frac{\rho c\hackscore p}{h} T^{(n)}  (\kappa T^{(n)}\hackscore{,i})\hackscore{,i} = q\hackscore H + \frac{\rho c\hackscore p}{h} T^{(n1)} 
\frac{\rho c\hackscore p}{h} T^{(n)}  (\kappa T^{(n)}\hackscore{,i})\hackscore{,i} = q\hackscore H + \frac{\rho c\hackscore p}{h} T^{(n1)} 
81 
\label{eqn:hdgenf} 
\label{eqn:hdgenf} 
82 
\end{equation} 
\end{equation} 
83 
This is the form required for escript to solve our PDE across the domain for successive time nodes $t^{(n)}$ where 
This is the form required for \ESCRIPT to solve our PDE across the domain for successive time nodes $t^{(n)}$ where 
84 
$t^{(0)}=0$ and $t^{(n)}=t^{(n1)}+h$ where $h>0$ is the step size which is assumed to be constant. 
$t^{(0)}=0$ and $t^{(n)}=t^{(n1)}+h$ where $h>0$ is the step size and assumed to be constant. 
85 
In the following the upper index ${(n)}$ refers to a value at time $t^{(n)}$. When comparing equation \ref{eqn:hdgenf} with equation \ref{eqn:commonform} we see that; 
In the following the upper index ${(n)}$ refers to a value at time $t^{(n)}$. When comparing \ref{eqn:hdgenf} with \ref{eqn:commonform} we see that; 
86 
\begin{equation} 
\begin{equation} 
87 
A = \kappa; D = \frac{\rho c \hackscore{p}}{h}; f = q \hackscore{H} + \frac{\rho c\hackscore p}{h} T^{(n1)} 
A = \kappa; D = \frac{\rho c \hackscore{p}}{h}; f = q \hackscore{H} + \frac{\rho c\hackscore p}{h} T^{(n1)} 
88 
\end{equation} 
\end{equation} 
89 


90 
\TODO{DISCUSS BOUNDARY CONDITIONS} 
Now that we have established the general form we will submit to \ESCRIPT; 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 \ref{eqn:Tbeuler}. Are model stipulates a starting temperature in our iron bar of 0\textcelsius. Thus the temperature distribution is simply; 
91 
It is pointed out that the initial conditions satisfy the boundary condition defined by 
\begin{equation} 
92 

T(x,0) = T\hackscore{ref} = 0 
93 

\end{equation} 
94 

for all $x$ in the domain. 
95 


96 

\subsection{Boundary Conditions} 
97 

With our PDE sorted, consideration must now be given to the boundary conditions of our model. Typically there are two main types of boundary conditions known as Neumann and Dirichlet\footnote{More information on Boundary Conditions is available at wikipedia \url{http://en.wikipedia.org/wiki/Boundary_conditions}}. In this example, we have utilised both conditions. Dirichlet is conceptually simpler and is used to prescribe a known value to the model on its boundary. This is like holding a snake by the tail; we know where the tail will be as we hold it however, we have no control over the rest of the snake. Dirichlet boundary conditions exist where we have applied our heat source, As the heat source is a constant, we can simulate its presence on that boundary. This is done by continuously resetting the temperature of the boundary, so that is is the same as the heat source. 
98 


99 
Together with the natural boundary condition 
Neumann boundary conditions describe the radiation or flux normal to the surface. This aptly describes our insulation conditions as we do not want to exert a constant temperature as with the heat source, however, we do want to prevent any loss of energy from the system. These natural boundary conditions can be described by specifying a radiation condition which prescribes the normal component of the flux $\kappa T\hackscore{,i}$ to be proportional 
100 

to the difference of the current temperature to the surrounding temperature $T\hackscore{ref}$; 
101 
\begin{equation} 
\begin{equation} 
102 
\kappa T\hackscore{,i}^{(n)} n\hackscore i = \eta (T\hackscore{ref}T^{(n)}) 
\kappa T\hackscore{,i} n\hackscore i = \eta (T\hackscore{ref}T) 
103 
\label{DIFFUSION TEMP EQ 2222} 
\label{eqn:hdbc} 
104 
\end{equation} 
\end{equation} 
105 
taken from 
where $\eta$ is a given material coefficient depending on the material of the block and the surrounding medium and $n\hackscore i$ is the $i$th component of the outer normal field \index{outer normal field} at the surface of the domain. These two conditions form a boundary value problem that has to be solved for each time step. Note as $\eta = 0$ due to zero flux  no energy in or out  we do not need to worry about the Neumann terms for this example. 

this forms a boundary value problem that has to be solved for each time step. 


As a first step to implement a solver for the temperature diffusion problem we will 


first implement a solver for the boundary value problem that has to be solved at each time step. 

106 


107 
We need to revisit the general PDE equation ~\ref{eqn:commonform nabla} under the light of a two dimensional domain. \ESCRIPT is inherently designed to solve problems that are greater than one dimension and so \ref{eqn:commonform nabla} needs to be read as a higher dimensional problem. In the case of two spatial dimensions the Nabla operator has in fact two components $\nabla = (\frac{\partial}{\partial x}, \frac{\partial}{\partial y})$. In full the general equation ~\ref{eqn:commonform nabla} assuming a constant coefficient $A$ takes the form; 
\subsection{A \textit{1D} Clarification} 
108 

It is necessary for clarification that we revisit the general PDE from \refeq{eqn:commonform nabla} under the light of a two dimensional domain. \ESCRIPT is inherently designed to solve problems that are greater than one dimension and so \ref{eqn:commonform nabla} needs to be read as a higher dimensional problem. In the case of two spatial dimensions the Nabla operator has in fact two components $\nabla = (\frac{\partial}{\partial x}, \frac{\partial}{\partial y})$. In full \ref{eqn:commonform nabla} assuming a constant coefficient $A$ takes the form; 
109 
\begin{equation}\label{eqn:commonform2D} 
\begin{equation}\label{eqn:commonform2D} 
110 
A\hackscore{00}\frac{\partial^{2}u}{\partial x^{2}} 
A\hackscore{00}\frac{\partial^{2}u}{\partial x^{2}} 
111 
A\hackscore{01}\frac{\partial^{2}u}{\partial x\partial y} 
A\hackscore{01}\frac{\partial^{2}u}{\partial x\partial y} 
114 
+ Du = f 
+ Du = f 
115 
\end{equation} 
\end{equation} 
116 
We notice that for the higher dimensional case $A$ becomes a matrix. It is also 
We notice that for the higher dimensional case $A$ becomes a matrix. It is also 
117 
important to notice that the usage of the Nable operator creates 
important to notice that the usage of the Nabla operator creates 
118 
a compact formulation which is also independant from the spatial dimension. 
a compact formulation which is also independent from the spatial dimension. 
119 
So to make the general PDE~\ref{eqn:commonform2D} one dimensional as 
So to make the general PDE~\ref{eqn:commonform2D} one dimensional as 
120 
shown in~\ref{eqn:commonform} we need to set 
shown in~\ref{eqn:commonform} we need to set 
121 
\begin{equation}\label{eqn:commonform2D} 
\begin{equation}\label{eqn:commonform2D} 
122 
A\hackscore{00}=A; A\hackscore{01}=A\hackscore{10}=A\hackscore{11}=0 
A\hackscore{00}=A; A\hackscore{01}=A\hackscore{10}=A\hackscore{11}=0 
123 
\end{equation} 
\end{equation} 
124 


125 
Now that we have established the general form we will submit to escript 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. We have chosen the starting temperature of our iron bar to be 0$\deg C$ . The temperature becomes; 
\subsection{Developing a PDE Solution Script} 
126 
\begin{equation} 
To solve \ref{eqn:hd} we will write a simple python script which uses the \modescript, \modfinley and \modmpl modules. At this point we assume that you have some basic understanding of the python programming language. If not there are some pointers and links available in Section \ref{sec:escpybas} . 

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


\end{equation} 


for all $x$ in the domain. 

127 


128 
Because we have a symmetrical problem we will also need to set the symmetry on by: 
Our goal here is to develop a script for \ESCRIPT that will solve the heat equation at successive time steps for a predefined period using our general form \ref{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 sin and cos functions or more complicated like those from our \ESCRIPT library.} 
129 

that we will require. 
130 
\begin{verbatim} 
\begin{verbatim} 
131 
myPDE.setSymmetryOn() 
from esys.escript import * 
132 

from esys.escript.linearPDEs import SingleLinearPDE 
133 

from esys.finley import Rectangle 
134 

from esys.escript.unitsSI import * 
135 

import os 
136 
\end{verbatim} 
\end{verbatim} 
137 

It is generally a good idea to import all of the \modescript library, although if you know the packages you need you can specify them individually. The function \verbLinearPDE has been imported for ease of use later in the script. \verbRectangle is going to be our type of domain. The package \verb unitsSI is a module of \esc that provides support for units definitions with our variables; and the \verbos package is needed to handle file outputs once our PDE has been solved. 
138 


139 
Additionally we must also consider the boundary conditions of our PDE. They take the form: 
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 \ESCRIPT 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 domain upon which we wish to solve our problem needs to be defined. There are many different types of domains in \modescript which we will demonstrate in later tutorials but for our iron rod, we will simply use a rectangular domain. 

\begin{equation} 


\eta \hackscore{j} A\hackscore{jl} u\hackscore{,l} + du = y 


\end{equation} 

140 


141 

Using a rectangular domain simplifies our rod which would probably 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 domain, 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 domain 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 our \textit{1D} problem we will define our bar as being 1 metre long. An appropriate \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 so the temperature does not vary with $y$. Thus the domain 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. 
142 

\begin{verbatim} 
143 

#Domain related. 
144 

mx = 1*m #meters  model lenght 
145 

my = .1*m #meters  model width 
146 

ndx = 100 # steps in x direction 
147 

ndy = 1 # steps in y direction 
148 

\end{verbatim} 
149 

The material constants and the temperature variables must also be defined. For the iron rod in the model they are defined as: 
150 

\begin{verbatim} 
151 

#PDE related 
152 

q=200. * Celsius #Kelvin  our heat source temperature 
153 

Tref = 0. * Celsius # Kelvin  starting temp of iron bar 
154 

rho = 7874. *kg/m**3 #kg/m^{3} density of iron 
155 

cp = 449.*J/(kg*K) #jules/Kg.K thermal capacity 
156 

rhocp = rho*cp 
157 

kappa = 80.*W/m/K #watts/m.K thermal conductivity 
158 

\end{verbatim} 
159 

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: 
160 

\begin{verbatim} 
161 

#Script/Iteration Related 
162 

t=0 #our start time, usually zero 
163 

tend=5.*60. #seconds  time to end simulation 
164 

outputs = 200 # number of time steps required. 
165 

h=(tendt)/outputs #size of time step 
166 

i=0 #loop counter 
167 

#the folder to put our outputs in, leave blank "" for script path 
168 

#note this folder path must exist to work 
169 

save_path = "data/onedheatdiff001" 
170 

\end{verbatim} 
171 

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: 
172 

\begin{verbatim} 
173 

rod = Rectangle(l0=mx,l1=my,n0=ndx,n1=ndy) 
174 

\end{verbatim} 
175 

In this form \verb rod does not represent any discrete points, but rather an area of \verb ndx*ndy cells that fit into a rectangular space with opposing vertices at the origin and the point \verb [mx,my] . Our domain is constructed this way to allow the user to determine where the discrete points of each model will be located. Discretisation may be at the corners of each cell, the middle point of a cell or halfway along each side of the cell etc. Depending on the PDE or the model there may be advantages and disadvantages for each case. Fortunately \modescript offers an easy way to extract finite points from the domain \verbrod using the domain property function \verbgetX() . This function sets the vertices of each cell as finite points to solve in the solution. If we let \verbx be these finite points, then; 
176 

\begin{verbatim} 
177 

x = rod.getX() 
178 

\end{verbatim} 
179 

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 \ESCRIPT. 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. 
180 

\begin{verbatim} 
181 

mypde=LinearSinglePDE(rod) 
182 

mypde.setValue(A=kappa*kronecker(rod),D=rhocp/h) 
183 

\end{verbatim} 
184 


185 
NEED TO WORK ON THIS SECTION 
In a few special cases it may be possible to decrease the computational time of the solver if our PDE is symmetric. Symmetry of a PDE is defined by; 
186 

\begin{equation}\label{eqn:symm} 
187 

A\hackscore{jl}=A\hackscore{lj} 
188 

\end{equation} 
189 

Note that from the general form $D$ and $d$ may have any value. The right hand sides $Y$, $y$ as well as the constraints 
190 

are not relevant either. 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; 
191 

\begin{verbatim} 
192 

myPDE.setSymmetryOn() 
193 

\end{verbatim} 
194 


195 
We now need to specify Our boundary conditions and initial values. The initial values required to solve this PDE are temperatures for each discrete point in our domain that we wish to solve for. We will set our bar to: 
We now need to specify our boundary conditions and initial values. The initial values required to solve this PDE are temperatures for each discrete point in our domain. We will set our bar to: 
196 
\begin{verbatim} 
\begin{verbatim} 
197 
T = Tref 
T = Tref 
198 
\end{verbatim} 
\end{verbatim} 
199 
Boundary conditions are a little more difficult. Fortunately the escript solver will handle our insulated boundary conditions. However, we will need to apply our heat source $q_{H}$ to the end of the bar at $x=0$ . escript makes this easy by letting us define areas in our domain. To retrieve all the finite points in our domain we will use 
Boundary conditions are a little more difficult. Fortunately the escript solver will handle our insulated boundary conditions. However, we will need to apply our heat source $q_{H}$ to the end of the bar at $x=0$ . \ESCRIPT makes this easy by letting us define areas in our domain. To retrieve all the finite points in our domain we will use 
200 

\begin{verbatim} 
201 

# extract finite points 
202 

x=rod.getX() 
203 

# ... set heat source: .... 
204 

qH=q*whereZero(x[0]) 
205 

\end{verbatim} 
206 


207 
END WORK ON THIS SECTION 
Finally we will initialise an iteration loop to solve our PDE for all the time steps we specified in the variable section. As the RHS of the general form is dependent on the previous values for temperature \verb T across the bar this must be updated in the loop. 
208 

\begin{verbatim} 
209 

# ... start iteration: 
210 

while t<=tend: 
211 

i+=1 
212 

t+=h 
213 

mypde.setValue(Y=qH+rhocp/h*T) 
214 

T=mypde.getSolution() 
215 

#some plotting stuff here  waiting on matplotlib binary distribution 
216 

\end{verbatim} 
217 


218 
\section{Plot total heat} 
\subsection{Plot total heat} 
219 
\TODO{show the script} 
\TODO{show the script} 
220 


221 
\TODO{explain how to use matlibplot to visualize the total heat integral(rho*c*T) over time} 
\TODO{explain how to use matlibplot to visualize the total heat integral(rho*c*T) over time} 
222 


223 
\section{Plot Temperature Distribution} 
\subsection{Plot Temperature Distribution} 
224 
\TODO{explain how to use matlibplot to visualize T} 
\TODO{explain how to use matlibplot to visualize T} 
225 


226 

