11 
% 
% 
12 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 
13 



We will start by examining a simple one dimensional heat diffusion example. This problem will provide a good launch pad to build our knowledge of \esc and demonstrate how to solve simple partial differential equations (PDEs)\footnote{Wikipedia provides an excellent and comprehensive introduction to \textit{Partial Differential Equations} \url{http://en.wikipedia.org/wiki/Partial_differential_equation}, however their relevance to \esc and implementation should become a clearer as we develop our understanding further into the cookbook.} 





\section{One Dimensional Heat Diffusion in an Iron Rod} 





%\label{Sec:1DHDv0} 


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

14 
\begin{figure}[h!] 
\begin{figure}[h!] 
15 
\centerline{\includegraphics[width=4.in]{figures/onedheatdiff}} 
\centerline{\includegraphics[width=4.in]{figures/onedheatdiff001}} 
16 
\caption{One dimensional model of an Iron bar.} 
\caption{Temperature differential along a single interface between two granite blocks.} 
17 
\label{fig:onedhdmodel} 
\label{fig:onedgbmodel} 
18 
\end{figure} 
\end{figure} 
19 


20 

\section{One Dimensional Heat Diffusion in Granite} 
21 

\label{Sec:1DHDv00} 
22 


23 

The first model consists of two blocks of isotropic material, for instance granite, sitting next to each other. 
24 

Initially, \textit{Block 1} is of a temperature 
25 

\verbT1 and \textit{Block 2} is at a temperature \verbT2. 
26 

We assume that the system is insulated. 
27 

What would happen to the temperature distribution in each block over time? 
28 

Intuition tells us that heat will transported from the hotter block to the cooler until both 
29 

blocks have the same temperature. 
30 


31 
\subsection{1D Heat Diffusion Equation} 
\subsection{1D Heat Diffusion Equation} 
32 
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}}; 
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}}; 
33 
which is defined as: 
which is defined as: 
42 


43 
\subsection{PDEs and the General Form} 
\subsection{PDEs and the General Form} 
44 
Potentially, it is now possible to solve PDE \refEq{eqn:hd} analytically and this would produce an exact solution to our problem. However, it is not always possible or practical to solve a problem this way. Alternatively, computers can be used to solve these kinds of problems. To do this, a numerical approach is required to discretised 
Potentially, it is now possible to solve PDE \refEq{eqn:hd} analytically and this would produce an exact solution to our problem. However, it is not always possible or practical to solve a problem this way. Alternatively, computers can be used to solve these kinds of problems. To do this, a numerical approach is required to discretised 
45 
the PDE \refEq{eqn:hd} in time and space so finally we are left with a finite number of equations for a finite number of spatial and time steps in the model. While discretization introduces approximations and a degree of error, we find that a sufficiently sampled model is generally accurate enough for the requirements of the modeler. 
the PDE \refEq{eqn:hd} in time and space so finally we are left with a finite number of equations for a finite number of spatial and time steps in the model. While discretization introduces approximations and a degree of error, we find that a sufficiently sampled model is generally accurate enough for the requirements of the modeller. 
46 


47 
Firstly, we will discretise the PDE \refEq{eqn:hd} in the time direction which will 
Firstly, we will discretise the PDE \refEq{eqn:hd} in the time direction which will 
48 
leave as with a steady linear PDE which is involving spatial derivatives only and needs to be solved in each time 
leave as with a steady linear PDE which is involving spatial derivatives only and needs to be solved in each time 
76 
Notice that we evaluate the spatial derivative term at current time $t^{(n)}$  therefore the name \textbf{backward Euler} scheme. Alternatively, one can use evaluate the spatial derivative term at the previous time $t^{(n1)}$. This 
Notice that we evaluate the spatial derivative term at current time $t^{(n)}$  therefore the name \textbf{backward Euler} scheme. Alternatively, one can use evaluate the spatial derivative term at the previous time $t^{(n1)}$. This 
77 
approach is called the \textbf{forward Euler} scheme. This scheme can provide some computational advantages which 
approach is called the \textbf{forward Euler} scheme. This scheme can provide some computational advantages which 
78 
we are not discussed here but has the major disadvantage that depending on the 
we are not discussed here but has the major disadvantage that depending on the 
79 
material parameter as well as the discretiztion of the spatial derivative term the time step size $h$ needs to be chosen sufficiently small to achieve a stable temperature when progressing in time. The term \textit{stable} means 
material parameter as well as the discretization of the spatial derivative term the time step size $h$ needs to be chosen sufficiently small to achieve a stable temperature when progressing in time. The term \textit{stable} means 
80 
that the approximation of the temperature will not grow beyond its initial bounds and becomes unphysical. 
that the approximation of the temperature will not grow beyond its initial bounds and becomes nonphysical. 
81 
The backward Euler which we use here is unconditionally stable meaning that under the assumption of 
The backward Euler which we use here is unconditionally stable meaning that under the assumption of 
82 
physically correct problem setup the temperature approximation remains physical for all times. 
physically correct problem setup the temperature approximation remains physical for all times. 
83 
The user needs to keep in mind that the discretization error introduced by \refEq{eqn:beuler} 
The user needs to keep in mind that the discretization error introduced by \refEq{eqn:beuler} 
118 
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)} 
119 
\end{equation} 
\end{equation} 
120 



% Now that the general form has been established, it can be submitted to \esc. Note that it is necessary to establish the state of our system at time zero or $T^{(n=0)}$. This is due to the time derivative approximation we have used from \refEq{eqn:Tbeuler}. Our model stipulates a starting temperature in the iron bar of 0\textcelsius. Thus the temperature distribution is simply; 


% \begin{equation} 


% T(x,0) = \left 


% \end{equation} 


% for all $x$ in the domain. 




121 
\subsection{Boundary Conditions} 
\subsection{Boundary Conditions} 
122 
\label{SEC BOUNDARY COND} 
\label{SEC BOUNDARY COND} 
123 
With the PDE sufficiently modified, consideration must now be given to the boundary conditions of our model. Typically there are two main types of boundary conditions known as \textbf{Neumann} and \textbf{Dirichlet} boundary conditions\footnote{More information on Boundary Conditions is available at Wikipedia \url{http://en.wikipedia.org/wiki/Boundary_conditions}}, respectively. 
With the PDE sufficiently modified, consideration must now be given to the boundary conditions of our model. Typically there are two main types of boundary conditions known as \textbf{Neumann} and \textbf{Dirichlet} boundary conditions\footnote{More information on Boundary Conditions is available at Wikipedia \url{http://en.wikipedia.org/wiki/Boundary_conditions}}, respectively. 
124 
A \textbf{Dirichlet boundary condition} is conceptually simpler and is used to prescribe a known value to the unknown  in our example the temperature  on parts of or the entire boundary of the region of interest. 
A \textbf{Dirichlet boundary condition} is conceptually simpler and is used to prescribe a known value to the unknown  in our example the temperature  on parts of the boundary or on the entire boundary of the region of interest. 
125 
For our model problem we want to keep the initial temperature setting on the left side of the 
We discuss Dirichlet boundary condition in our second example presented in Section~\ref{Sec:1DHDv0}. 
126 
iron bar over time. This defines a Dirichlet boundary condition for the PDE \refEq{eqn:hddisc} to be solved at each time step. 

127 

We make the model assumption that the system is insulated so we need 
128 
On the other end of the iron rod we want to add an appropriate boundary condition to define insulation to prevent 
to add an appropriate boundary condition to prevent 
129 
any loss or inflow of energy at the right end of the rod. Mathematically this is expressed by prescribing 
any loss or inflow of energy at boundary of our domain. Mathematically this is expressed by prescribing 
130 
the heat flux $\kappa \frac{\partial T}{\partial x}$ to zero on the right end of the rod 
the heat flux $\kappa \frac{\partial T}{\partial x}$ to zero. In our simplified one dimensional model this is expressed 

In our simplified one dimensional model this is expressed 

131 
in the form; 
in the form; 
132 
\begin{equation} 
\begin{equation} 
133 
\kappa \frac{\partial T}{\partial x} = 0 
\kappa \frac{\partial T}{\partial x} = 0 
137 
\kappa \nabla T \cdot n = 0 
\kappa \nabla T \cdot n = 0 
138 
\end{equation} 
\end{equation} 
139 
where $n$ is the outer normal field \index{outer normal field} at the surface of the domain. 
where $n$ is the outer normal field \index{outer normal field} at the surface of the domain. 
140 
For the iron rod the outer normal field on the right hand side is the vector $(1,0)$. The $\cdot$ (dot) refers to the 
The $\cdot$ (dot) refers to the dot product of the vectors $\nabla T$ and $n$. In fact, the term $\nabla T \cdot n$ is the normal derivative of 

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

141 
the temperature $T$. Other notations which are used are\footnote{The \esc notation for the normal 
the temperature $T$. Other notations which are used are\footnote{The \esc notation for the normal 
142 
derivative is $T\hackscore{,i} n\hackscore i$.}; 
derivative is $T\hackscore{,i} n\hackscore i$.}; 
143 
\begin{equation} 
\begin{equation} 
145 
\end{equation} 
\end{equation} 
146 
A condition of the type \refEq{NEUMAN 1} defines a \textbf{Neuman boundary condition} for the PDE. 
A condition of the type \refEq{NEUMAN 1} defines a \textbf{Neuman boundary condition} for the PDE. 
147 


148 
The PDE \refEq{eqn:hdgenf} together with the Dirichlet boundary condition set on the left face of the rod 
The PDE \refEq{eqn:hdgenf} 
149 
and the Neuman boundary condition~\ref{eqn:hdgenf} define a \textbf{boundary value problem}. 
and the Neuman boundary condition~\ref{eqn:hdgenf} (potentially together with the Dirichlet boundary condition set) define a \textbf{boundary value problem}. 
150 
It is a nature of a boundary value problem that it allows to make statements on the solution in the 
It is a nature of a boundary value problem that it allows to make statements on the solution in the 
151 
interior of the domain from information known on the boundary only. In most cases 
interior of the domain from information known on the boundary only. In most cases 
152 
we use the term partial differential equation but in fact mean a boundary value problem. 
we use the term partial differential equation but in fact mean a boundary value problem. 
166 
\begin{equation}\label{NEUMAN 2b} 
\begin{equation}\label{NEUMAN 2b} 
167 
\kappa \frac{\partial T}{\partial x} = 0 
\kappa \frac{\partial T}{\partial x} = 0 
168 
\end{equation} 
\end{equation} 
169 
for the right hand side of the rod. This is identical to the Neuman boundary condition we want to set. \esc will take care of this condition for us. We will discuss the Dirichlet boundary condition later. 
for the boundary of the domain. This is identical to the Neuman boundary condition we want to set. \esc will take care of this condition for us. We will discuss the Dirichlet boundary condition later. 
170 


171 
\subsection{Outline of the Implementation} 
\subsection{Outline of the Implementation} 
172 
\label{sec:outline} 
\label{sec:outline} 
173 
To solve the heat diffusion equation (equation \refEq{eqn:hd}) we will write a simple \pyt script. At this point we assume that you have some basic understanding of the \pyt programming language. If not there are some pointers and links available in Section \ref{sec:escpybas}. The script we will discuss later in details will have four major steps. Firstly we need to define the domain where we want to 
To solve the heat diffusion equation (equation \refEq{eqn:hd}) we will write a simple \pyt script. At this point we assume that you have some basic understanding of the \pyt programming language. If not there are some pointers and links available in Section \ref{sec:escpybas}. The script we will discuss later in details will have four major steps. Firstly we need to define the domain where we want to 
174 
calculate the temperature. For our problem this is the iron rod which has a rectangular shape. Secondly we need to define the PDE 
calculate the temperature. For our problem this is the joint blocks of granite which has a rectangular shape. Secondly we need to define the PDE 
175 
we need to solve in each time step to get the updated temperature. Thirdly we need to define the the coefficients of the PDE and finally we need to solve the PDE. The last two steps need to be repeated until the final time marker has been reached. As a work flow this takes the form; 
we need to solve in each time step to get the updated temperature. Thirdly we need to define the the coefficients of the PDE and finally we need to solve the PDE. The last two steps need to be repeated until the final time marker has been reached. As a work flow this takes the form; 
176 
\begin{enumerate} 
\begin{enumerate} 
177 
\item create domain 
\item create domain 
185 
\item end of calculation 
\item end of calculation 
186 
\end{enumerate} 
\end{enumerate} 
187 
In the terminology of \pyt the domain and PDE are represented by \textbf{objects}. The nice feature of an object is that it defined by it usage and features 
In the terminology of \pyt the domain and PDE are represented by \textbf{objects}. The nice feature of an object is that it defined by it usage and features 
188 
rather than its actual representation. So we will create a domain object to describe the geometry of our iron rod. The main feature 
rather than its actual representation. So we will create a domain object to describe the geometry of the two 
189 

granite blocks. The main feature 
190 
of the object we will use is the fact that we can define PDEs and spatially distributed values such as the temperature 
of the object we will use is the fact that we can define PDEs and spatially distributed values such as the temperature 
191 
on a domain. In fact the domain object has many more features  most of them you will 
on a domain. In fact the domain object has many more features  most of them you will 
192 
never use and do not need to understand. Similar a PDE object is defined by the fact that we can define the coefficients of the PDE and solve the PDE. At a 
never use and do not need to understand. Similar a PDE object is defined by the fact that we can define the coefficients of the PDE and solve the PDE. At a 
207 


208 
There are various ways to construct domain objects. The simplest form is as rectangular shaped region with a length and height. There is 
There are various ways to construct domain objects. The simplest form is as rectangular shaped region with a length and height. There is 
209 
a ready to use function call for this. Besides the spatial dimensions the function call will require you to specify the number 
a ready to use function call for this. Besides the spatial dimensions the function call will require you to specify the number 
210 
elements or cells to be used along the length and height, see Figure~\ref{fig:fs}. Any specially distributed value 
elements or cells to be used along the length and height, see \reffig{fig:fs}. Any spatially distributed value 
211 
and the PDE is represented in discrete form using this element representation\footnote{We will use the finite element method (FEM), see \url{http://en.wikipedia.org/wiki/Finite_element_method}i for details.}. Therefore we will have access to an approximation of the true PDE solution only. 
and the PDE is represented in discrete form using this element representation\footnote{We will use the finite element method (FEM), see \url{http://en.wikipedia.org/wiki/Finite_element_method} for details.}. Therefore we will have access to an approximation of the true PDE solution only. 
212 
The quality of the approximation depends  besides other factors mainly on the number of elements being used. In fact, the 
The quality of the approximation depends  besides other factors mainly on the number of elements being used. In fact, the 
213 
approximation becomes better the more elements are used. However, computational costs and compute time grow with the number of 
approximation becomes better the more elements are used. However, computational costs and compute time grow with the number of 
214 
elements being used. It therefore important that you find the right balance between the demand in accuracy and acceptable resource usage. 
elements being used. It therefore important that you find the right balance between the demand in accuracy and acceptable resource usage. 
215 


216 
In general, one can thinks about a domain object as a composition of nodes and elements. 
In general, one can thinks about a domain object as a composition of nodes and elements. 
217 
As shown in Figure~\ref{fig:fs}, an element is defined by the nodes used to describe its vertices. 
As shown in \reffig{fig:fs}, an element is defined by the nodes used to describe its vertices. 
218 
To represent spatial distributed values the user can use 
To represent spatial distributed values the user can use 
219 
the values at the nodes, at the elements in the interior of the domain or at elements located at the surface of the domain. 
the values at the nodes, at the elements in the interior of the domain or at elements located at the surface of the domain. 
220 
The different approach used to represent values is called \textbf{function space} and is attached to all objects 
The different approach used to represent values is called \textbf{function space} and is attached to all objects 
229 
location of the socalled \textbf{sample points} used to represent values with the particular function space attached to it. So the 
location of the socalled \textbf{sample points} used to represent values with the particular function space attached to it. So the 
230 
call \verbContinuousFunction(domain).getX() will return the coordinates of the nodes used to describe the domain while 
call \verbContinuousFunction(domain).getX() will return the coordinates of the nodes used to describe the domain while 
231 
the \verbFunction(domain).getX() returns the coordinates of numerical integration points within elements, see 
the \verbFunction(domain).getX() returns the coordinates of numerical integration points within elements, see 
232 
Figure~\ref{fig:fs}. 
\reffig{fig:fs}. 
233 


234 
This distinction between different representations of spatial distributed values 
This distinction between different representations of spatial distributed values 
235 
is important in order to be able to vary the degrees of smoothness in a PDE problem. 
is important in order to be able to vary the degrees of smoothness in a PDE problem. 
270 
\label{sec:key} 
\label{sec:key} 
271 
\sslist{onedheatdiffbase.py} 
\sslist{onedheatdiffbase.py} 
272 
We will write a simple \pyt script which uses the \modescript, \modfinley and \modmpl modules. 
We will write a simple \pyt script which uses the \modescript, \modfinley and \modmpl modules. 
273 
By developing a script for \esc, the heat diffusion equation can be solved at successive time steps for a predefined period using our general form \refEq{eqn:hdgenf}. Firstly it is necessary to import all the libraries\footnote{The libraries contain predefined scripts that are required to solve certain problems, these can be simple like $sine$ and $cosine$ functions or more complicated like those from our \esc library.} 
By developing a script for \esc, the heat diffusion equation can be solved at successive time steps for a predefined period using our general form \refEq{eqn:hdgenf}. Firstly it is necessary to import all the libraries\footnote{The libraries contain predefined scripts that are required to solve certain problems, these can be simple like sine and cosine functions or more complicated like those from our \esc library.} 
274 
that we will require. 
that we will require. 
275 
\begin{python} 
\begin{python} 
276 
from esys.escript import * 
from esys.escript import * 
284 
\end{python} 
\end{python} 
285 
It is generally a good idea to import all of the \modescript library, although if the functions and classes required are known they can be specified individually. The function \verbLinearPDE has been imported explicitly for ease of use later in the script. \verbRectangle is going to be our type of model. The module \verb unitsSI provides support for SI unit definitions with our variables. 
It is generally a good idea to import all of the \modescript library, although if the functions and classes required are known they can be specified individually. The function \verbLinearPDE has been imported explicitly for ease of use later in the script. \verbRectangle is going to be our type of model. The module \verb unitsSI provides support for SI unit definitions with our variables. 
286 


287 
Once our library dependencies have been established, defining the problem specific variables is the next step. In general the number of variables needed will vary between problems. These variables belong to two categories. They are either directly related to the PDE and can be used as inputs into the \esc solver, or they are script variables used to control internal functions and iterations in our problem. For this PDE there are a number of constants which will need values. Firstly, the model upon which we wish to solve our problem needs to be defined. There are many different types of models in \modescript which we will demonstrate in later tutorials but for our iron rod, we will simply use a rectangular model. 
Once our library dependencies have been established, defining the problem specific variables is the next step. In general the number of variables needed will vary between problems. These variables belong to two categories. They are either directly related to the PDE and can be used as inputs into the \esc solver, or they are script variables used to control internal functions and iterations in our problem. For this PDE there are a number of constants which will need values. Firstly, the model upon which we wish to solve our problem needs to be defined. There are many different types of models in \modescript which we will demonstrate in later tutorials but for our granite blocks, we will simply use a rectangular model. 
288 


289 
Using a rectangular model simplifies our rod which would be a \textit{3D} object, into a single dimension. The iron rod will have a lengthways cross section that looks like a rectangle. As a result we do not need to model the volume of the rod because a cylinder is symmetrical about its centre. There are four arguments we must consider when we decide to create a rectangular model, the model \textit{length}, \textit{width} and \textit{step size} in each direction. When defining the size of our problem it will help us determine appropriate values for our model arguments. If we make our dimensions large but our step sizes very small we will to a point, increase the accuracy of our solution. Unfortunately we also increase the number of calculations that must be solved per time step. This means more computational time is required to produce a solution. In this \textit{1D} problem, the bar is defined as being 1 metre long. An appropriate step size \verbndx would be 1 to 10\% of the length. Our \verbndy need only be 1, this is because our problem stipulates no partial derivatives in the $y$ direction. Thus the temperature does not vary with $y$. Hence, the model parameters can be defined as follows; note we have used the \verb unitsSI convention to make sure all our input units are converted to SI. 
Using a rectangular model simplifies our granite blocks which would in reality be a \textit{3D} object, into a single dimension. The granite blocks will have a lengthways cross section that looks like a rectangle. As a result we do not need to model the volume of the block. There are four arguments we must consider when we decide to create a rectangular model, the model \textit{length}, \textit{width} and \textit{step size} in each direction. When defining the size of our problem it will help us determine appropriate values for our model arguments. If we make our dimensions large but our step sizes very small we will to a point, increase the accuracy of our solution. Unfortunately we also increase the number of calculations that must be solved per time step. This means more computational time is required to produce a solution. In this \textit{1D} problem, the bar is defined as being 1 metre long. An appropriate step size \verbndx would be 1 to 10\% of the length. Our \verbndy need only be 1, this is because our problem stipulates no partial derivatives in the $y$ direction. Thus the temperature does not vary with $y$. Hence, the model parameters can be defined as follows; note we have used the \verb unitsSI convention to make sure all our input units are converted to SI. 
290 
\begin{python} 
\begin{python} 
291 
#Domain related. 
mx = 500.*m #meters  model length 
292 
mx = 1*m #meters  model length 
my = 100.*m #meters  model width 
293 
my = .1*m #meters  model width 
ndx = 50 # mesh steps in x direction 
294 
ndx = 100 # mesh steps in x direction 
ndy = 1 # mesh steps in y direction 
295 
ndy = 1 # mesh steps in y direction  one dimension means one element 
boundloc = mx/2 # location of boundary between the two blocks 
296 
\end{python} 
\end{python} 
297 
The material constants and the temperature variables must also be defined. For the iron rod in the model they are defined as: 
The material constants and the temperature variables must also be defined. For the granite in the model they are defined as: 
298 
\begin{python} 
\begin{python} 
299 
#PDE related 
#PDE related 
300 
rho = 7874. *kg/m**3 #kg/m^{3} density of iron 
rho = 2750. *kg/m**3 #kg/m^{3} density of iron 
301 
cp = 449.*J/(kg*K) # J/Kg.K thermal capacity 
cp = 790.*J/(kg*K) # J/Kg.K thermal capacity 
302 
rhocp = rho*cp 
rhocp = rho*cp 
303 
kappa = 80.*W/m/K # watts/m.Kthermal conductivity 
kappa = 2.2*W/m/K # watts/m.Kthermal conductivity 
304 
qH=0 * J/(sec*m**3) # J/(sec.m^{3}) no heat source 
qH=0 * J/(sec*m**3) # J/(sec.m^{3}) no heat source 
305 
T0=100 * Celsius # initial temperature at left end of rod 
T1=20 * Celsius # initial temperature at Block 1 
306 
Tref=20 * Celsius # base temperature 
T2=2273. * Celsius # base temperature at Block 2 
307 
\end{python} 
\end{python} 
308 
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: 
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: 
309 
\begin{python} 
\begin{python} 
315 
print "Expected Number of time outputs is: ", (tendt)/h 
print "Expected Number of time outputs is: ", (tendt)/h 
316 
i=0 #loop counter 
i=0 #loop counter 
317 
\end{python} 
\end{python} 
318 
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: 
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 model as: 
319 
\begin{python} 
\begin{python} 
320 
#generate domain using rectangle 
#generate domain using rectangle 
321 
rod = Rectangle(l0=mx,l1=my,n0=ndx, n1=ndy) 
blocks = Rectangle(l0=mx,l1=my,n0=ndx, n1=ndy) 
322 
\end{python} 
\end{python} 
323 
\verb rod now describes a domain in the manner of Section \ref{ss:domcon}. There is an easy way to extract 
\verb blocks now describes a domain in the manner of Section \ref{ss:domcon}. T 

the coordinates of the nodes used to describe 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{python} 


#extract data points  the solution points 


x=rod.getX() 


\end{python} 


The data locations of specific function spaces can be returned in a similar manner by extracting the relevant function space from the domain followed by the \verb.getX() method. 

324 


325 
With a domain and all our required variables established, it is now possible to set up our PDE so that it can be solved by \esc. The first step is to define the type of PDE that we are trying to solve in each time step. In this example it is a single linear PDE\footnote{in comparison to a system of PDEs which will be discussed later.}. We also need to state the values of our general form variables. 
With a domain and all our required variables established, it is now possible to set up our PDE so that it can be solved by \esc. The first step is to define the type of PDE that we are trying to solve in each time step. In this example it is a single linear PDE\footnote{in comparison to a system of PDEs which will be discussed later.}. We also need to state the values of our general form variables. 
326 
\begin{python} 
\begin{python} 
327 
mypde=LinearSinglePDE(rod) 
mypde=LinearPDE(blocks) 
328 
A=zeros((2,2))) 
A=zeros((2,2))) 
329 
A[0,0]=kappa 
A[0,0]=kappa 
330 
q=whereZero(x[0]) 
mypde.setValue(A=A, D=rhocp/h) 

mypde.setValue(A=A, D=rhocp/h, q=q, r=T0) 

331 
\end{python} 
\end{python} 

The argument \verbq has not been discussed yet: In fact the arguments \verbq and \verbr are used to define 


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


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


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


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


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


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


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





In order to set \verbq we use 


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


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


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


zero elsewhere which is exactly what we need. 




332 
In a many cases it may be possible to decrease the computational time of the solver if the PDE is symmetric. 
In a many cases it may be possible to decrease the computational time of the solver if the PDE is symmetric. 
333 
Symmetry of a PDE is defined by; 
Symmetry of a PDE is defined by; 
334 
\begin{equation}\label{eqn:symm} 
\begin{equation}\label{eqn:symm} 
338 
\begin{python} 
\begin{python} 
339 
myPDE.setSymmetryOn() 
myPDE.setSymmetryOn() 
340 
\end{python} 
\end{python} 
341 
Next we need to establish the initial temperature distribution \verbT. We want to have this initial 
Next we need to establish the initial temperature distribution \verbT. We need to 
342 
value to be \verbTref except at the left end of the rod $x=0$ where we have the temperature \verbT0. We use; 
assign the value \verbT1 to all sample points left to the contact interface at $x\hackscore{0}=\frac{mx}{2}$ 
343 

and the value \verbT2 right to the contact interface. \esc 
344 

provides the \verbwhereNegative function to construct this. In fact, 
345 

\verbwhereNegative returns the value $1$ at those sample points where the argument 
346 

has a negative value. Otherwise zero is returned. If \verbx are the $x\hackscore{0}$ 
347 

coordinates of the sample points used to represent the temperature distribution 
348 

then \verbx[0]boundloc gives us a negative value for 
349 

all sample points left to the interface and nonnegative value to 
350 

the right of the interface. So with; 
351 
\begin{python} 
\begin{python} 
352 
# ... set initial temperature .... 
# ... set initial temperature .... 
353 
T = T0*whereZero(x[0])+Tref*(1whereZero(x[0])) 
T= T1*whereNegative(x[0]boundloc)+T2*(1whereNegative(x[0]boundloc)) 
354 
\end{python} 
\end{python} 
355 
Finally we will initialize an iteration loop to solve our PDE for all the time steps we specified in the variable section. As the right hand side of the general form is dependent on the previous values for temperature \verb T across the bar this must be updated in the loop. Our output at each time step is \verb T the heat distribution and \verb totT the total heat in the system. 
we get the desired temperature distribution. To get the actual sample points \verbx we use 
356 

the \verbgetX() method of the function space \verbSolution(blocks) 
357 

which is used to represent the solution of a PDE; 
358 

\begin{python} 
359 

x=Solution(blocks).getX() 
360 

\end{python} 
361 

As \verbx are the sample points for the function space \verbSolution(blocks) 
362 

the initial temperature \verbT is using these sample points for representation. 
363 

Although \esc is trying to be forgiving with the choice of sample points and to convert 
364 

where necessary the adjustment of the function space is not always possible. So it is 
365 

advisable to make a careful choice on the function space used. 
366 


367 

Finally we will initialise an iteration loop to solve our PDE for all the time steps we specified in the variable section. As the right hand side of the general form is dependent on the previous values for temperature \verb T across the bar this must be updated in the loop. Our output at each time step is \verb T the heat distribution and \verb totT the total heat in the system. 
368 
\begin{python} 
\begin{python} 
369 
while t < tend: 
while t < tend: 
370 
i+=1 #increment the counter 
i+=1 #increment the counter 
374 
totE = integrate(rhocp*T) #get the total heat (energy) in the system 
totE = integrate(rhocp*T) #get the total heat (energy) in the system 
375 
\end{python} 
\end{python} 
376 
The last statement in this script calculates the total energy in the system as volume integral 
The last statement in this script calculates the total energy in the system as volume integral 
377 
of $\rho \c_p T$ over the rod. 
of $\rho c\hackscore{p} T$ over the block. As the blocks are insulated no energy should be get lost or added. 
378 

The total energy should stay constant for the example discussed here. 
379 


380 

\subsection{Running the Script} 
381 

The script presented so for is available under 
382 

\verbonedheatdiffbase.py. You can edit this file with your favourite text editor. 
383 

On most operating systems\footnote{The you can use \texttt{escript} launcher is not supported under {\it MS Windows} yet.} you can use the \program{escript} command 
384 

to launch {\it escript} scripts. For the example script use; 
385 

\begin{verbatim} 
386 

escript onedheatdiffbase.py 
387 

\end{verbatim} 
388 

The program will print a progress report. Alternatively, you can use 
389 

the python interpreter directly; 
390 

\begin{verbatim} 
391 

python onedheatdiffbase.py 
392 

\end{verbatim} 
393 

if the system is configured correctly (Please talk to your system administrator). 
394 


395 

\begin{figure} 
396 

\begin{center} 
397 

\includegraphics[width=4in]{figures/ttblockspyplot150} 
398 

\caption{Total Energy in the Blocks over Time (in seconds).} 
399 

\label{fig:onedheatout1} 
400 

\end{center} 
401 

\end{figure} 
402 


403 
\subsection{Plotting the Total Energy} 
\subsection{Plotting the Total Energy} 
404 
\sslist{onedheatdiff001.py} 
\sslist{onedheatdiff001.py} 
405 


406 
\esc does not include its own plotting capabilities. However, it is possible to use a variety of free \pyt packages for visualization. 
\esc does not include its own plotting capabilities. However, it is possible to use a variety of free \pyt packages for visualisation. 
407 
Two types will be demonstrated in this cookbook; \mpl\footnote{\url{http://matplotlib.sourceforge.net/}} and \verb VTK \footnote{\url{http://www.vtk.org/}} visualisation. 
Two types will be demonstrated in this cookbook; \mpl\footnote{\url{http://matplotlib.sourceforge.net/}} and \verb VTK \footnote{\url{http://www.vtk.org/}} visualisation. 
408 
The \mpl package is a component of SciPy\footnote{\url{http://www.scipy.org}} and is good for basic graphs and plots. 
The \mpl package is a component of SciPy\footnote{\url{http://www.scipy.org}} and is good for basic graphs and plots. 
409 
For more complex visualisation tasks in particular when it comes to two and three dimensional problems it is recommended to us more advanced tools for instance \mayavi \footnote{\url{http://code.enthought.com/projects/mayavi/}} 
For more complex visualisation tasks in particular when it comes to two and three dimensional problems it is recommended to us more advanced tools for instance \mayavi \footnote{\url{http://code.enthought.com/projects/mayavi/}} 
410 
which bases the \verbVTK toolkit. We will discuss the usage of \verbVTK based 
which bases on the \verbVTK toolkit. We will discuss the usage of \verbVTK based 
411 
visualization in Chapter~\ref{Sec:2DHD} where will discuss a two dimensional PDE. 
visualization in Chapter~\ref{Sec:2DHD} where will discuss a two dimensional PDE. 
412 


413 
For our simple problem we have two plotting tasks: Firstly we are interested in showing the 
For our simple problem we have two plotting tasks: Firstly we are interested in showing the 
414 
behavior of the total energy over time and secondly in how the temperature distribution within the rod is 
behaviour of the total energy over time and secondly in how the temperature distribution within the block is 
415 
developing over time. Lets start with the first task. 
developing over time. Lets start with the first task. 
416 



\begin{figure} 


\begin{center} 


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


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


\label{fig:onedheatout1} 


\end{center} 


\end{figure} 




417 
The trick is to create a record of the time marks and the corresponding total energies observed. 
The trick is to create a record of the time marks and the corresponding total energies observed. 
418 
\pyt provides the concept of lists for this. Before 
\pyt provides the concept of lists for this. Before 
419 
the time loop is opened we create empty lists for the time marks \verbt_list and the total energies \verbE_list. 
the time loop is opened we create empty lists for the time marks \verbt_list and the total energies \verbE_list. 
435 
\begin{python} 
\begin{python} 
436 
import pylab as pl # plotting package. 
import pylab as pl # plotting package. 
437 
\end{python} 
\end{python} 
438 
Here we are not using the \verbfrom pylab import * in order to name clashes for function names 
Here we are not using the \verbfrom pylab import * in order to avoid name clashes for function names 
439 
with \esc. 
within \esc. 
440 


441 
The following statements are added to the script after the time loop has been completed; 
The following statements are added to the script after the time loop has been completed; 
442 
\begin{python} 
\begin{python} 
443 
pl.plot(t_list,E_list) 
pl.plot(t_list,E_list) 
444 
pl.title("Total Energy") 
pl.title("Total Energy") 
445 

pl.axis([0,max(t_list),0,max(E_list)*1.1]) 
446 
pl.savefig("totE.png") 
pl.savefig("totE.png") 
447 
\end{python} 
\end{python} 
448 
The first statement hands over the time marks and corresponding total energies to the plotter. 
The first statement hands over the time marks and corresponding total energies to the plotter. 
449 
The second statement is setting the title for the plot. The last statement renders the plot and writes the 
The second statment is setting the title for the plot. The third statement 
450 
result into the file \verbtotE.png which can be displayed by (almost) any image viewer. Your result should look 
sets the axis ranges. In most cases these are set appropriately by the plotter. 
451 
similar to Figure~\ref{fig:onedheatout1}. 
The last statement renders the plot and writes the 
452 

result into the file \verbtotE.png which can be displayed by (almost) any image viewer. 
453 

As expected the total energy is constant over time, see \reffig{fig:onedheatout1}. 
454 


455 
\subsection{Plotting the Temperature Distribution} 
\subsection{Plotting the Temperature Distribution} 
456 
\sslist{onedheatdiff001b.py} 
\sslist{onedheatdiff001b.py} 
480 
\begin{verbatim} 
\begin{verbatim} 
481 
save_path = os.path.join("data","onedheatdiff001") 
save_path = os.path.join("data","onedheatdiff001") 
482 
\end{verbatim} 
\end{verbatim} 
483 
The second argument of \verb join \xspace contains a string which is the filename or subdirectory name. We can use the operator \verb% to increment our file names with the value \verbi denoting a incrementing counter. The substring \verb %03d does this by defining the following parameters; 
The second argument of \verb join \xspace contains a string which is the file name or subdirectory name. We can use the operator \verb% to increment our file names with the value \verbi denoting a incrementing counter. The substring \verb %03d does this by defining the following parameters; 
484 
\begin{itemize} 
\begin{itemize} 
485 
\item \verb 0 becomes the padding number; 
\item \verb 0 becomes the padding number; 
486 
\item \verb 3 tells us the amount of padding numbers that are required; and 
\item \verb 3 tells us the amount of padding numbers that are required; and 
488 
\end{itemize} 
\end{itemize} 
489 
To increment the file name a \verb %i is required directly after the operation the string is involved in. When correctly implemented the output files from this command would be place in the directory defined by \verb save_path as; 
To increment the file name a \verb %i is required directly after the operation the string is involved in. When correctly implemented the output files from this command would be place in the directory defined by \verb save_path as; 
490 
\begin{verbatim} 
\begin{verbatim} 
491 
rodpyplot.png 
blockspyplot.png 
492 
rodpyplot.png 
blockspyplot.png 
493 
rodpyplot.png 
blockspyplot.png 
494 
... 
... 
495 
\end{verbatim} 
\end{verbatim} 
496 
and so on. 
and so on. 
502 
will check for the existence of \verb save_path and if missing, make the required directories. 
will check for the existence of \verb save_path and if missing, make the required directories. 
503 


504 
We start by modifying our solution script from before. 
We start by modifying our solution script from before. 
505 
Prior to the \verbwhile loop we will need to extract our finite solution points to a data object that is compatible with \mpl. First we create the node coordinates of the data points used to represent 
Prior to the \verbwhile loop we will need to extract our finite solution points to a data object that is compatible with \mpl. First we create the node coordinates of the sample points used to represent 
506 
the temperature as a \pyt list of tuples. As a solution of a PDE 
the temperature as a \pyt list of tuples or a \numpy array as requested by the plotting function. 
507 
the temperature has the \verbSolution(rod) function space attribute. We use 
We need to convert thearray \verbx previously set as \verbSolution(blocks).getX() into a \pyt list 
508 
the \verbgetX() method to get the coordinates of the data points as an \esc object 
and then to a \numpy array. The $x\hackscore{0}$ component is then extracted via an array slice to the variable \verbplx; 

which is then converted to a \numpy array. The $x$ component is then extracted via an array slice to the variable \verbplx; 

509 
\begin{python} 
\begin{python} 
510 
import numpy as np # array package. 
import numpy as np # array package. 
511 
#convert solution points for plotting 
#convert solution points for plotting 
512 
plx = Solution(rod).getX().toListOfTuples() 
plx = x.toListOfTuples() 
513 
plx = np.array(plx) # convert to tuple to numpy array 
plx = np.array(plx) # convert to tuple to numpy array 
514 
plx = plx[:,0] # extract x locations 
plx = plx[:,0] # extract x locations 
515 
\end{python} 
\end{python} 
516 


517 
\begin{figure} 
\begin{figure} 
518 
\begin{center} 
\begin{center} 
519 
\includegraphics[width=4in]{figures/rodpyplot001} 
\includegraphics[width=4in]{figures/blockspyplot001} 
520 
\includegraphics[width=4in]{figures/rodpyplot050} 
\includegraphics[width=4in]{figures/blockspyplot050} 
521 
\includegraphics[width=4in]{figures/rodpyplot200} 
\includegraphics[width=4in]{figures/blockspyplot200} 
522 
\caption{Temperature ($T$) distribution in rod at time steps $1$, $50$ and $200$.} 
\caption{Temperature ($T$) distribution in the blocks at time steps $1$, $50$ and $200$.} 
523 
\label{fig:onedheatout} 
\label{fig:onedheatout} 
524 
\end{center} 
\end{center} 
525 
\end{figure} 
\end{figure} 
538 
# set scale (Temperature should be between Tref and T0) 
# set scale (Temperature should be between Tref and T0) 
539 
pl.axis([0,mx,Tref*.9,T0*1.1]) 
pl.axis([0,mx,Tref*.9,T0*1.1]) 
540 
# add title 
# add title 
541 
pl.title("Temperature across rod at time %e minutes"%(t/minutes)) 
pl.title("Temperature across the blocks at time %e minutes"%(t/day)) 
542 
#save figure to file 
#save figure to file 
543 
pl.savefig(os.path.join(save_path,"tempT","rodpyplot%03d.png") %i) 
pl.savefig(os.path.join(save_path,"tempT","blockspyplot%03d.png") %i) 
544 
\end{python} 
\end{python} 
545 
Some results are shown in Figure~\ref{fig:onedheatout}. 
Some results are shown in \reffig{fig:onedheatout}. 
546 


547 
\subsection{Make a video} 
\subsection{Make a video} 
548 
Our saved plots from the previous section can be cast into a video using the following command appended to the end of the script. \verb mencoder is Linux only however, and other platform users will need to use an alternative video encoder. 
Our saved plots from the previous section can be cast into a video using the following command appended to the end of the script. \verb mencoder is Linux only however, and other platform users will need to use an alternative video encoder. 
549 
\begin{python} 
\begin{python} 
550 
# compile the *.png files to create a *.avi videos that show T change 
# compile the *.png files to create a *.avi videos that show T change 
551 
# with time. This operation uses Linux mencoder. For other operating 
# with time. This operation uses Linux mencoder. For other operating 
552 
# systems it is possible to use your favorite video compiler to 
# systems it is possible to use your favourite video compiler to 
553 
# convert image files to videos. 
# convert image files to videos. 
554 


555 
os.system("mencoder mf://"+save_path+"/tempT"+"/*.png mf type=png:\ 
os.system("mencoder mf://"+save_path+"/tempT"+"/*.png mf type=png:\ 