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% Copyright (c) 20032009 by University of Queensland 
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% Licensed under the Open Software License version 3.0 
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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.} 
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\section{One Dimensional Heat Diffusion in an Iron Rod} 
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%\label{Sec:1DHDv0} 
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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. 
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\begin{figure}[h!] 
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\centerline{\includegraphics[width=4.in]{figures/onedheatdiff}} 
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\caption{One dimensional model of an Iron bar.} 
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\label{fig:onedhdmodel} 
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\end{figure} 
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\subsection{1D Heat Diffusion Equation} 
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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}}; 
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which is defined as: 
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\begin{equation} 
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\rho c\hackscore p \frac{\partial T}{\partial t}  \kappa \frac{\partial^{2} T}{\partial x^{2}} = q\hackscore H 
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\label{eqn:hd} 
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\end{equation} 
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where $\rho$ is the material density, $c\hackscore p$ is the specific heat and $\kappa$ is the thermal 
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conductivity\footnote{A list of some common thermal conductivities is available from Wikipedia \url{http://en.wikipedia.org/wiki/List_of_thermal_conductivities}}. Here we assume that these material 
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parameters are \textbf{constant}. 
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The heat source is defined by the right hand side of \refEq{eqn:hd} as $q\hackscore{H}$; this can take the form of a constant or a function of time and space. For example $q\hackscore{H} = q\hackscore{0}e^{\gamma t}$ where we have the output of our heat source decaying with time. There are also two partial derivatives in \refEq{eqn:hd}; $\frac{\partial T}{\partial t}$ describes the change in temperature with time while $\frac{\partial ^2 T}{\partial x^2}$ is the spatial change of temperature. As there is only a single spatial dimension to our problem, our temperature solution $T$ is only dependent on the time $t$ and our position along the iron bar $x$. 
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\subsection{PDEs and the General Form} 
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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 
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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. 
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Firstly, we will discretise the PDE \refEq{eqn:hd} in the time direction which will 
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leave as with a steady linear PDE which is involving spatial derivatives only and needs to be solved in each time 
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step to progress in time  \esc can help us here. 
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For the discretization in time we will use is the Backwards Euler approximation scheme\footnote{see \url{http://en.wikipedia.org/wiki/Euler_method}}. It bases on the 
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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 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\hackscore p}{h} (T^{(n)}  T^{(n1)})  \kappa \frac{\partial^{2} T^{(n)}}{\partial x^{2}} = q\hackscore 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 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 
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approach is called the \textbf{forward Euler} scheme. This scheme can provide some computational advantages which 
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we are not discussed here but has the major disadvantage that depending on the 
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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 
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that the approximation of the temperature will not grow beyond its initial bounds and becomes unphysical. 
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The backward Euler which we use here is unconditionally stable meaning that under the assumption of 
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physically correct problem setup the temperature approximation remains physical for all times. 
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The user needs to keep in mind that the discretization error introduced by \refEq{eqn:beuler} 
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is sufficiently small so a good approximation of the true temperature is calculated. It is 
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therefore crucial that the user remains critical about his/her results and for instance compares 
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the results for different time and spatial step sizes. 
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To get the temperature $T^{(n)}$ at time $t^{(n)}$ we need to solve the linear 
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differential equation \refEq{eqn:hddisc} which is only including spatial derivatives. To solve this problem 
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we want to to use \esc. 
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\esc interfaces with any given PDE via a general form. For the purpose of this introduction we will illustrate a simpler version of the full linear PDE general form which is available in the \esc user's guide. A simplified form that suits our heat diffusion problem\footnote{In the form of the \esc users guide which using the Einstein convention is written as 
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$(A\hackscore{jl} u\hackscore{,l})\hackscore{,j}+D u =Y$} 
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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 symbol $\nabla$ which is called the \textit{Nabla operator} or \textit{del operator} represents 
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the spatial derivative of its subject  in this case $u$. Lets assume for a 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 \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\hackscore p}{h} T^{(n)}  \kappa \frac{\partial^2 T^{(n)}}{\partial x^2} = q\hackscore H + \frac{\rho c\hackscore p}{h} T^{(n1)} 
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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 required for \esc to solve our PDE. This can be done by generating a solution for successive increments in the time nodes $t^{(n)}$ where 
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$t^{(0)}=0$ and $t^{(n)}=t^{(n1)}+h$ where $h>0$ is the step size and assumed to be constant. 
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In the following the upper index ${(n)}$ refers to a value at time $t^{(n)}$. Finally, by comparing \refEq{eqn:hdgenf} with \refEq{eqn:commonform} it can be seen that; 
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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 \hackscore{p}}{h}; f = q \hackscore{H} + \frac{\rho c\hackscore p}{h} T^{(n1)} 
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\end{equation} 
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% 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; 
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% \begin{equation} 
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% T(x,0) = \left 
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% \end{equation} 
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% for all $x$ in the domain. 
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\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 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. 
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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. 
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For our model problem we want to keep the initial temperature setting on the left side of the 
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iron bar over time. This defines a Dirichlet boundary condition for the PDE \refEq{eqn:hddisc} to be solved at each time step. 
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On the other end of the iron rod we want to add an appropriate boundary condition to define insulation to prevent 
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any loss or inflow of energy at the right end of the rod. Mathematically this is expressed by prescribing 
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the heat flux $\kappa \frac{\partial T}{\partial x}$ to zero on the right end of the rod 
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In our simplified 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 of the domain. 
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For the iron rod the outer normal field on the right hand side is the vector $(1,0)$. The $\cdot$ (dot) refers to the 
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dot product of the vectors $\nabla T$ and $n$. In fact, the term $\nabla T \cdot n$ is the normal derivative of 
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the temperature $T$. Other notations which are used are\footnote{The \esc notation for the normal 
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derivative is $T\hackscore{,i} n\hackscore i$.}; 
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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{Neuman boundary condition} for the PDE. 
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The PDE \refEq{eqn:hdgenf} together with the Dirichlet boundary condition set on the left face of the rod 
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and the Neuman boundary condition~\ref{eqn:hdgenf} define a \textbf{boundary value problem}. 
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It is a nature of a boundary value problem that it allows to make statements on the solution in the 
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interior of the domain from information known on the boundary only. In most cases 
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we use the term partial differential equation but in fact mean a boundary value problem. 
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It is important to keep in mind that boundary conditions need to be complete and consistent in the sense that 
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at any point on the boundary either a Dirichlet or a Neuman boundary condition must be set. 
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Conveniently, \esc makes default assumption on the boundary conditions which the user may modify where appropriate. 
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For a problem of the form in~\refEq{eqn:commonform nabla} the default condition\footnote{In the form of the \esc users guide which is using the Einstein convention is written as 
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$n\hackscore{j}A\hackscore{jl} u\hackscore{,l}=0$.} 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 field. 
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Notice that the coefficient $A$ is the same as in the \esc PDE~\ref{eqn:commonform nabla}. 
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With the settings for the coefficients we have already identified in \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 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. 
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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 (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 
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calculate the temperature. For our problem this is the iron rod which has a rectangular shape. Secondly we need to define the PDE 
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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; 
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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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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 
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rather than its actual representation. So we will create a domain object to describe the geometry of our iron rod. The main feature 
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of the object we will use is the fact that we can define PDEs and spatially distributed values such as the temperature 
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on a domain. In fact the domain object has many more features  most of them you will 
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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 
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later stage you may use more advanced features of the PDE class but you need to worry about them only at the point when you use them. 
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\begin{figure}[t] 
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\centering 
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\includegraphics[width=6in]{figures/functionspace.pdf} 
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\label{fig:fs} 
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\caption{\esc domain construction overview} 
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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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It is helpful to have a better understanding how spatially distributed value such as the temperature or PDE coefficients are interpreted in \esc. Again 
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from the user's point of view the representation of these spatially distributed values is not relevant. 
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There are various ways to construct domain objects. The simplest form is as rectangular shaped region with a length and height. There is 
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a ready to use function call for this. Besides the spatial dimensions the function call will require you to specify the number 
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elements or cells to be used along the length and height, see Figure~\ref{fig:fs}. Any specially distributed value 
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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. 
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The quality of the approximation depends  besides other factors mainly on the number of elements being used. In fact, the 
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approximation becomes better the more elements are used. However, computational costs and compute time grow with the number of 
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elements being used. It therefore important that you find the right balance between the demand in accuracy and acceptable resource usage. 
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In general, one can thinks about a domain object as a composition of nodes and elements. 
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As shown in Figure~\ref{fig:fs}, an element is defined by the nodes used to describe its vertices. 
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To represent spatial 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 elements located at the surface of the domain. 
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The different approach used to represent values is called \textbf{function space} and is attached to all objects 
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in \esc representing a spatial distributed value such as the solution of a PDE. The three 
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function spaces we will 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) . 
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\end{enumerate} 
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A function space object such as \verbContinuousFunction(domain) has the method \verbgetX attached to it. This method returns the 
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location of the socalled \textbf{sample points} used to represent values with the particular function space attached to it. So the 
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call \verbContinuousFunction(domain).getX() will return the coordinates of the nodes used to describe the domain while 
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the \verbFunction(domain).getX() returns the coordinates of numerical integration points within elements, see 
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Figure~\ref{fig:fs}. 
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This distinction between different representations of spatial distributed values 
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is important in order to be able to vary the degrees of smoothness in a PDE problem. 
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The coefficients of a PDE need not be continuous thus this qualifies as a \verbFunction() type. 
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On the other hand a temperature distribution must be continuous and needs to be represented with a \verbContinuousFunction() function space. 
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An influx may only be defined at the boundary and is therefore a \verb FunctionOnBoundary() object. 
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\esc allows certain transformations of the function spaces. A \verb ContinuousFunction() can be transformed into a \verbFunctionOnBoundary() 
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or \verbFunction(). On the other hand there is not enough information in a \verb FunctionOnBoundary() to transform it to a \verb ContinuousFunction() . 
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These transformations, which are called \textbf{interpolation} are invoked automatically by \esc if needed. 
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Later in this introduction we will discuss how 
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to define specific areas of geometry with different materials which are represented by different material coefficients such the 
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thermal conductivities $kappa$. A very powerful technique to define these types of PDE 
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coefficients is tagging. Blocks of materials and boundaries can be named and values can be defined on subregions based on their names. 
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This is simplifying PDE coefficient and flux definitions. It makes for much easier scripting. We will discuss this technique in Section~\ref{STEADYSTATE HEAT REFRACTION}. 
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\subsection{A Clarification for the 1D Case} 
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It is necessary for clarification that we revisit the general PDE from \refeq{eqn:commonform nabla} under the light of a two dimensional domain. \esc is inherently designed to solve problems that are greater than one dimension and so \refEq{eqn:commonform nabla} needs to be read as a higher dimensional problem. In the case of two spatial dimensions the \textit{Nabla operator} has in fact two components $\nabla = (\frac{\partial}{\partial x}, \frac{\partial}{\partial y})$. In full, \refEq{eqn:commonform nabla} assuming a constant coefficient $A$, takes the form; 
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\begin{equation}\label{eqn:commonform2D} 
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A\hackscore{00}\frac{\partial^{2}u}{\partial x^{2}} 
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A\hackscore{01}\frac{\partial^{2}u}{\partial x\partial y} 
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A\hackscore{10}\frac{\partial^{2}u}{\partial y\partial x} 
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A\hackscore{11}\frac{\partial^{2}u}{\partial y^{2}} 
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+ Du = f 
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\end{equation} 
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Notice that for the higher dimensional case $A$ becomes a matrix. It is also 
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important to notice that the usage of the Nabla operator creates 
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a compact formulation which is also independent from the spatial dimension. 
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So to make the general PDE \refEq{eqn:commonform2D} one dimensional as 
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shown in \refEq{eqn:commonform} we need to set 
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\begin{equation} 
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A\hackscore{00}=A; A\hackscore{01}=A\hackscore{10}=A\hackscore{11}=0 
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\end{equation} 
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\subsection{Developing a PDE Solution Script} 
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\label{sec:key} 
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\sslist{onedheatdiffbase.py} 
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We will write a simple \pyt script which uses the \modescript, \modfinley and \modmpl modules. 
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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.} 
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that we will require. 
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\begin{python} 
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from esys.escript import * 
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# This defines the LinearPDE module as LinearPDE 
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from esys.escript.linearPDEs import LinearPDE 
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# This imports the rectangle domain function from finley. 
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from esys.finley import Rectangle 
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# A useful unit handling package which will make sure all our units 
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# match up in the equations under SI. 
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from esys.escript.unitsSI import * 
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\end{python} 
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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. 
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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. 
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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. 
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\begin{python} 
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#Domain related. 
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mx = 1*m #meters  model length 
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my = .1*m #meters  model width 
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ndx = 100 # mesh steps in x direction 
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ndy = 1 # mesh steps in y direction  one dimension means one element 
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\end{python} 
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The material constants and the temperature variables must also be defined. For the iron rod in the model they are defined as: 
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\begin{python} 
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#PDE related 
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rho = 7874. *kg/m**3 #kg/m^{3} density of iron 
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cp = 449.*J/(kg*K) # J/Kg.K thermal capacity 
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rhocp = rho*cp 
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kappa = 80.*W/m/K # watts/m.Kthermal conductivity 
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qH=0 * J/(sec*m**3) # J/(sec.m^{3}) no heat source 
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T0=100 * Celsius # initial temperature at left end of rod 
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Tref=20 * Celsius # base temperature 
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\end{python} 
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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: 
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\begin{python} 
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t=0 * day #our start time, usually zero 
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tend=1. * day #  time to end simulation 
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outputs = 200 # number of time steps required. 
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h=(tendt)/outputs #size of time step 
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#user warning statement 
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print "Expected Number of time outputs is: ", (tendt)/h 
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i=0 #loop counter 
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\end{python} 
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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: 
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\begin{python} 
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#generate domain using rectangle 
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rod = Rectangle(l0=mx,l1=my,n0=ndx, n1=ndy) 
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\end{python} 
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\verb rod now describes a domain in the manner of Section \ref{ss:domcon}. There is an easy way to extract 
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the coordinates of the nodes used to describe the domain \verbrod using the 
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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; 
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\begin{python} 
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#extract data points  the solution points 
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x=rod.getX() 
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\end{python} 
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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. 
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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. 
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\begin{python} 
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mypde=LinearSinglePDE(rod) 
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A=zeros((2,2))) 
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A[0,0]=kappa 
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q=whereZero(x[0]) 
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mypde.setValue(A=A, D=rhocp/h, q=q, r=T0) 
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\end{python} 
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The argument \verbq has not been discussed yet: In fact the arguments \verbq and \verbr are used to define 
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Dirichlet boundary condition as discussed in Section~\ref{SEC BOUNDARY COND}. In the \esc 
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PDE from the argument \verbq indicates by a positive value for which nodes we want to apply a 
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Dirichlet boundary condition, ie. where we want to prescribe the value of the PDE solution 
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rather then using the PDE. The actually value for the solution to be taken is set by the argument \verbr. 
346 
In our case we want to keep the initial temperature $T0$ on the left face of the rode for all times. Notice, 
347 
that as set to a constant value \verbr is assumed to have the same value 
348 
at all nodes, however only the value at those nodes marked by a positive value by \verbq are actually used. 
349 

350 
In order to set \verbq we use 
351 
\verbwhereZero function. The function returns the value (positive) one for those data points (=nodes) where the argument is equal to zero and otherwise returns (nonpositive) value zero. 
352 
As \verbx[0] given the $x$coordinates of the nodes for the domain, 
353 
\verbwhereZero(x[0]) gives the value $1$ for the nodes at the left end of the rod $x=x_0=0$ and 
354 
zero elsewhere which is exactly what we need. 
355 

356 
In a many cases it may be possible to decrease the computational time of the solver if the PDE is symmetric. 
357 
Symmetry of a PDE is defined by; 
358 
\begin{equation}\label{eqn:symm} 
359 
A\hackscore{jl}=A\hackscore{lj} 
360 
\end{equation} 
361 
Symmetry is only dependent on the $A$ coefficient in the general form and the other coefficients $D$ as well as the right hand side $Y$ may take any value. From the above definition we can see that our PDE is symmetric. The \verb LinearPDE class provides the method \method{checkSymmetry} to check if the given PDE is symmetric. As our PDE is symmetrical we will enable symmetry via; 
362 
\begin{python} 
363 
myPDE.setSymmetryOn() 
364 
\end{python} 
365 
Next we need to establish the initial temperature distribution \verbT. We want to have this initial 
366 
value to be \verbTref except at the left end of the rod $x=0$ where we have the temperature \verbT0. We use; 
367 
\begin{python} 
368 
# ... set initial temperature .... 
369 
T = T0*whereZero(x[0])+Tref*(1whereZero(x[0])) 
370 
\end{python} 
371 
Finally we will initialize an iteration loop to solve our PDE for all the time steps we specified in the variable section. As the right hand side of the general form is dependent on the previous values for temperature \verb T across the bar this must be updated in the loop. Our output at each time step is \verb T the heat distribution and \verb totT the total heat in the system. 
372 
\begin{python} 
373 
while t < tend: 
374 
i+=1 #increment the counter 
375 
t+=h #increment the current time 
376 
mypde.setValue(Y=qH+rhocp/h*T) # set variable PDE coefficients 
377 
T=mypde.getSolution() #get the PDE solution 
378 
totE = integrate(rhocp*T) #get the total heat (energy) in the system 
379 
\end{python} 
380 
The last statement in this script calculates the total energy in the system as volume integral 
381 
of $\rho \c_p T$ over the rod. 
382 

383 
\subsection{Plotting the Total Energy} 
384 
\sslist{onedheatdiff001.py} 
385 

386 
\esc does not include its own plotting capabilities. However, it is possible to use a variety of free \pyt packages for visualization. 
387 
Two types will be demonstrated in this cookbook; \mpl\footnote{\url{http://matplotlib.sourceforge.net/}} and \verb VTK \footnote{\url{http://www.vtk.org/}} visualisation. 
388 
The \mpl package is a component of SciPy\footnote{\url{http://www.scipy.org}} and is good for basic graphs and plots. 
389 
For more complex visualisation tasks in particular when it comes to two and three dimensional problems it is recommended to us more advanced tools for instance \mayavi \footnote{\url{http://code.enthought.com/projects/mayavi/}} 
390 
which bases the \verbVTK toolkit. We will discuss the usage of \verbVTK based 
391 
visualization in Chapter~\ref{Sec:2DHD} where will discuss a two dimensional PDE. 
392 

393 
For our simple problem we have two plotting tasks: Firstly we are interested in showing the 
394 
behavior of the total energy over time and secondly in how the temperature distribution within the rod is 
395 
developing over time. Lets start with the first task. 
396 

397 
\begin{figure} 
398 
\begin{center} 
399 
\includegraphics[width=4in]{figures/ttrodpyplot150} 
400 
\caption{Total Energy in Rod over Time (in seconds).} 
401 
\label{fig:onedheatout1} 
402 
\end{center} 
403 
\end{figure} 
404 

405 
The trick is to create a record of the time marks and the corresponding total energies observed. 
406 
\pyt provides the concept of lists for this. Before 
407 
the time loop is opened we create empty lists for the time marks \verbt_list and the total energies \verbE_list. 
408 
After the new temperature as been calculated by solving the PDE we append the new time marker and total energy 
409 
to the corresponding list using the \verbappend method. With these modifications the script looks as follows: 
410 
\begin{python} 
411 
t_list=[] 
412 
E_list=[] 
413 
# ... start iteration: 
414 
while t<tend: 
415 
t+=h 
416 
mypde.setValue(Y=qH+rhocp/h*T) # set variable PDE coefficients 
417 
T=mypde.getSolution() #get the PDE solution 
418 
totE=integrate(rhocp*T) 
419 
t_list.append(t) # add current time mark to record 
420 
E_list.append(totE) # add current total energy to record 
421 
\end{python} 
422 
To plot $t$ over $totE$ we use the \mpl a module contained within \pylab which needs to be loaded before used; 
423 
\begin{python} 
424 
import pylab as pl # plotting package. 
425 
\end{python} 
426 
Here we are not using the \verbfrom pylab import * in order to name clashes for function names 
427 
with \esc. 
428 

429 
The following statements are added to the script after the time loop has been completed; 
430 
\begin{python} 
431 
pl.plot(t_list,E_list) 
432 
pl.title("Total Energy") 
433 
pl.savefig("totE.png") 
434 
\end{python} 
435 
The first statement hands over the time marks and corresponding total energies to the plotter. 
436 
The second statement is setting the title for the plot. The last statement renders the plot and writes the 
437 
result into the file \verbtotE.png which can be displayed by (almost) any image viewer. Your result should look 
438 
similar to Figure~\ref{fig:onedheatout1}. 
439 

440 
\subsection{Plotting the Temperature Distribution} 
441 
\sslist{onedheatdiff001b.py} 
442 
For plotting the spatial distribution of the temperature we need to modify the strategy we have used 
443 
for the total energy. Instead of producing a final plot at the end we will generate a 
444 
picture at each time step which can be browsed as slide show or composed to a movie. 
445 
The first problem we encounter is that if we produce an image in each time step we need 
446 
to make sure that the images previously generated are not overwritten. 
447 

448 
To develop an incrementing file name we can use the following convention. It is convenient to 
449 
put all image file showing the same variable  in our case the temperature distribution  
450 
into a separate directory. As part of the \verbos module\footnote{The \texttt{os} module provides 
451 
a powerful interface to interact with the operating system, see \url{http://docs.python.org/library/os.html}.} \pyt 
452 
provides the \verbos.path.join command to build file and 
453 
directory names in a platform independent way. Assuming that 
454 
\verbsave_path is name of directory we want to put the results the command is; 
455 
\begin{python} 
456 
import os 
457 
os.path.join(save_path, "tempT%03d.png"%i ) 
458 
\end{python} 
459 
where \verbi is the time step counter. 
460 
There are two arguments to the \verb join command. The \verb save_path variable is a predefined string pointing to the directory we want to save our data in, for example a single subfolder called \verb data would be defined by; 
461 
\begin{verbatim} 
462 
save_path = "data" 
463 
\end{verbatim} 
464 
while a subfolder of \verb data called \verb onedheatdiff001 would be defined by; 
465 
\begin{verbatim} 
466 
save_path = os.path.join("data","onedheatdiff001") 
467 
\end{verbatim} 
468 
The second argument of \verb join \xspace contains a string which is the filename or subdirectory name. We can use the operator \verb% to increment our file names with the value \verbi denoting a incrementing counter. The substring \verb %03d does this by defining the following parameters; 
469 
\begin{itemize} 
470 
\item \verb 0 becomes the padding number; 
471 
\item \verb 3 tells us the amount of padding numbers that are required; and 
472 
\item \verb d indicates the end of the \verb % operator. 
473 
\end{itemize} 
474 
To increment the file name a \verb %i is required directly after the operation the string is involved in. When correctly implemented the output files from this command would be place in the directory defined by \verb save_path as; 
475 
\begin{verbatim} 
476 
rodpyplot.png 
477 
rodpyplot.png 
478 
rodpyplot.png 
479 
... 
480 
\end{verbatim} 
481 
and so on. 
482 

483 
A subfolder check/constructor is available in \esc. The command; 
484 
\begin{verbatim} 
485 
mkDir(save_path) 
486 
\end{verbatim} 
487 
will check for the existence of \verb save_path and if missing, make the required directories. 
488 

489 
We start by modifying our solution script from before. 
490 
Prior to the \verbwhile loop we will need to extract our finite solution points to a data object that is compatible with \mpl. First we create the node coordinates of the data points used to represent 
491 
the temperature as a \pyt list of tuples. As a solution of a PDE 
492 
the temperature has the \verbSolution(rod) function space attribute. We use 
493 
the \verbgetX() method to get the coordinates of the data points as an \esc object 
494 
which is then converted to a \numpy array. The $x$ component is then extracted via an array slice to the variable \verbplx; 
495 
\begin{python} 
496 
import numpy as np # array package. 
497 
#convert solution points for plotting 
498 
plx = Solution(rod).getX().toListOfTuples() 
499 
plx = np.array(plx) # convert to tuple to numpy array 
500 
plx = plx[:,0] # extract x locations 
501 
\end{python} 
502 

503 
\begin{figure} 
504 
\begin{center} 
505 
\includegraphics[width=4in]{figures/rodpyplot001} 
506 
\includegraphics[width=4in]{figures/rodpyplot050} 
507 
\includegraphics[width=4in]{figures/rodpyplot200} 
508 
\caption{Temperature ($T$) distribution in rod at time steps $1$, $50$ and $200$.} 
509 
\label{fig:onedheatout} 
510 
\end{center} 
511 
\end{figure} 
512 

513 
For each time step we will generate a plot of the temperature distribution and save each to a file. We use the same 
514 
techniques provided by \mpl as we have used to plot the total energy over time. 
515 
The following is appended to the end of the \verb while loop and creates one figure of the temperature distribution. We start by converting the solution to a tuple and then plotting this against our \textit{x coordinates} \verb plx we have generated before. We add a title to the diagram before it is rendered into a file. 
516 
Finally, the figure is saved to a \verb*.png file and cleared for the following iteration. 
517 
\begin{python} 
518 
# ... start iteration: 
519 
while t<tend: 
520 
.... 
521 
T=mypde.getSolution() #get the PDE solution 
522 
tempT = T.toListOfTuples() # convert to a tuple 
523 
pl.plot(plx,tempT) # plot solution 
524 
# set scale (Temperature should be between Tref and T0) 
525 
pl.axis([0,mx,Tref*.9,T0*1.1]) 
526 
# add title 
527 
pl.title("Temperature across rod at time %e minutes"%(t/minutes)) 
528 
#save figure to file 
529 
pl.savefig(os.path.join(save_path,"tempT","rodpyplot%03d.png") %i) 
530 
\end{python} 
531 
Some results are shown in Figure~\ref{fig:onedheatout}. 
532 

533 
\subsection{Make a video} 
534 
Our saved plots from the previous section can be cast into a video using the following command appended to the end of the script. \verb mencoder is Linux only however, and other platform users will need to use an alternative video encoder. 
535 
\begin{python} 
536 
# compile the *.png files to create a *.avi videos that show T change 
537 
# with time. This operation uses Linux mencoder. For other operating 
538 
# systems it is possible to use your favorite video compiler to 
539 
# convert image files to videos. 
540 

541 
os.system("mencoder mf://"+save_path+"/tempT"+"/*.png mf type=png:\ 
542 
w=800:h=600:fps=25 ovc lavc lavcopts vcodec=mpeg4 oac copy o \ 
543 
onedheatdiff001tempT.avi") 
544 
\end{python} 
545 
