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 1 ahallam 2401 2 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 3 % 4 jfenwick 2881 % Copyright (c) 2003-2010 by University of Queensland 5 ahallam 2401 % Earth Systems Science Computational Center (ESSCC) 6 7 % 8 % Primary Business: Queensland, Australia 9 % Licensed under the Open Software License version 3.0 10 11 % 12 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 13 14 gross 2905 \begin{figure}[h!] 15 \centerline{\includegraphics[width=4.in]{figures/onedheatdiff001}} 16 gross 2952 \caption{Example 1: Temperature differential along a single interface between two granite blocks.} 17 gross 2905 \label{fig:onedgbmodel} 18 \end{figure} 19 ahallam 2801 20 gross 2949 \section{Example 1: One Dimensional Heat Diffusion in Granite} 21 gross 2905 \label{Sec:1DHDv00} 22 gross 2878 23 gross 2905 The first model consists of two blocks of isotropic material, for instance granite, sitting next to each other. 24 artak 2957 Initial temperature in \textit{Block 1} is \verb|T1| and in \textit{Block 2} is \verb|T2|. 25 gross 2905 We assume that the system is insulated. 26 What would happen to the temperature distribution in each block over time? 27 artak 2958 Intuition tells us that heat will be transported from the hotter block to the cooler one until both 28 gross 2905 blocks have the same temperature. 29 30 ahallam 2495 \subsection{1D Heat Diffusion Equation} 31 artak 2957 We can model the heat distribution of this problem over time using 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}}; 32 ahallam 2494 which is defined as: 33 ahallam 2401 \begin{equation} 34 \rho c\hackscore p \frac{\partial T}{\partial t} - \kappa \frac{\partial^{2} T}{\partial x^{2}} = q\hackscore H 35 \label{eqn:hd} 36 \end{equation} 37 gross 2861 where $\rho$ is the material density, $c\hackscore p$ is the specific heat and $\kappa$ is the thermal 38 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 39 parameters are \textbf{constant}. 40 ahallam 2975 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 signed distance from the the block-block interface $x$. 41 ahallam 2401 42 gross 2861 \subsection{PDEs and the General Form} 43 ahallam 2975 It is possible to solve PDE \refEq{eqn:hd} analytically and obtain an exact solution to our problem. However, it is not always possible or practical to solve the problem this way. Alternatively, computers can be used to find the solution. To do this, a numerical approach is required to discretise 44 the PDE \refEq{eqn:hd} across time and space, this reduces the problem to a finite number of equations for a finite number of spatial points and time steps. These parameters together define the model. While discretisation introduces approximations and a degree of error, a sufficiently sampled model is generally accurate enough to satisfy the accuracy requirements for the final solution. 45 gross 2477 46 artak 2958 Firstly, we discretise the PDE \refEq{eqn:hd} in time. This leaves us with a steady linear PDE which involves spatial derivatives only and needs to be solved in each time step to progress in time. \esc can help us here. 47 gross 2861 48 artak 2958 For time discretization we use the Backwards Euler approximation scheme\footnote{see \url{http://en.wikipedia.org/wiki/Euler_method}}. It is based on the 49 gross 2861 approximation 50 \begin{equation} 51 \frac{\partial T(t)}{\partial t} \approx \frac{T(t)-T(t-h)}{h} 52 \label{eqn:beuler} 53 \end{equation} 54 for $\frac{\partial T}{\partial t}$ at time $t$ 55 where $h$ is the time step size. This can also be written as; 56 \begin{equation} 57 \frac{\partial T}{\partial t}(t^{(n)}) \approx \frac{T^{(n)} - T^{(n-1)}}{h} 58 \label{eqn:Tbeuler} 59 \end{equation} 60 where the upper index $n$ denotes the n\textsuperscript{th} time step. So one has 61 \begin{equation} 62 \begin{array}{rcl} 63 t^{(n)} & = & t^{(n-1)}+h \\ 64 T^{(n)} & = & T(t^{(n-1)}) \\ 65 \end{array} 66 \label{eqn:Neuler} 67 \end{equation} 68 Substituting \refEq{eqn:Tbeuler} into \refEq{eqn:hd} we get; 69 \begin{equation} 70 \frac{\rho c\hackscore p}{h} (T^{(n)} - T^{(n-1)}) - \kappa \frac{\partial^{2} T^{(n)}}{\partial x^{2}} = q\hackscore H 71 \label{eqn:hddisc} 72 \end{equation} 73 artak 2958 Notice that we evaluate the spatial derivative term at current time $t^{(n)}$ - therefore the name \textbf{backward Euler} scheme. Alternatively, one can evaluate the spatial derivative term at the previous time $t^{(n-1)}$. This 74 approach is called the \textbf{forward Euler} scheme. This scheme can provide some computational advantages, which 75 ahallam 2975 are not discussed here. However, the \textbf{forward Euler} scheme has a major disadvantage. Depending on the 76 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. Stabiliy is achieved if the temperature does not grow beyond its initial bounds and become non-physical. 77 artak 2958 The backward Euler scheme, which we use here, is unconditionally stable meaning that under the assumption of 78 physically correct problem set-up the temperature approximation remains physical for all time steps. 79 gross 2861 The user needs to keep in mind that the discretization error introduced by \refEq{eqn:beuler} 80 artak 2958 is sufficiently small, thus a good approximation of the true temperature is computed. It is 81 ahallam 2975 therefore crucial that the user remains sceptical about their results and for instance compares 82 the results for different time and spatial step sizes for correlation. 83 gross 2861 84 To get the temperature $T^{(n)}$ at time $t^{(n)}$ we need to solve the linear 85 artak 2958 differential equation \refEq{eqn:hddisc} which only includes spatial derivatives. To solve this problem 86 we want to use \esc. 87 gross 2861 88 ahallam 2975 In \esc any given PDE can be described by the general form. For the purpose of this introduction we illustrate a simpler version of the general form for full linear PDEs which is available in the \esc user's guide. A simplified form that suits our heat diffusion problem\footnote{The form in the \esc users guide which uses the Einstein convention is written as 89 gross 2477 $-(A\hackscore{jl} u\hackscore{,l})\hackscore{,j}+D u =Y$} 90 ahallam 2606 is described by; 91 gross 2477 \begin{equation}\label{eqn:commonform nabla} 92 jfenwick 2657 -\nabla\cdot(A\cdot\nabla u) + Du = f 93 ahallam 2411 \end{equation} 94 gross 2861 where $A$, $D$ and $f$ are known values and $u$ is the unknown solution. The symbol $\nabla$ which is called the \textit{Nabla operator} or \textit{del operator} represents 95 ahallam 2495 the spatial derivative of its subject - in this case $u$. Lets assume for a moment that we deal with a one-dimensional problem then ; 96 gross 2477 \begin{equation} 97 \nabla = \frac{\partial}{\partial x} 98 \end{equation} 99 gross 2861 and we can write \refEq{eqn:commonform nabla} as; 100 ahallam 2411 \begin{equation}\label{eqn:commonform} 101 gross 2477 -A\frac{\partial^{2}u}{\partial x^{2}} + Du = f 102 ahallam 2411 \end{equation} 103 gross 2861 if $A$ is constant. To match this simplified general form to our problem \refEq{eqn:hddisc} 104 we rearrange \refEq{eqn:hddisc}; 105 ahallam 2411 \begin{equation} 106 ahallam 2645 \frac{\rho c\hackscore p}{h} T^{(n)} - \kappa \frac{\partial^2 T^{(n)}}{\partial x^2} = q\hackscore H + \frac{\rho c\hackscore p}{h} T^{(n-1)} 107 ahallam 2494 \label{eqn:hdgenf} 108 \end{equation} 109 ahallam 2775 The PDE is now in a form that satisfies \refEq{eqn:commonform nabla} which is required for \esc to solve our PDE. This can be done by generating a solution for successive increments in the time nodes $t^{(n)}$ where 110 ahallam 2495 $t^{(0)}=0$ and $t^{(n)}=t^{(n-1)}+h$ where $h>0$ is the step size and assumed to be constant. 111 artak 2958 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} one can see that; 112 gross 2862 \begin{equation}\label{ESCRIPT SET} 113 gross 2861 u=T^{(n)}; 114 ahallam 2494 A = \kappa; D = \frac{\rho c \hackscore{p}}{h}; f = q \hackscore{H} + \frac{\rho c\hackscore p}{h} T^{(n-1)} 115 \end{equation} 116 117 ahallam 2495 \subsection{Boundary Conditions} 118 gross 2878 \label{SEC BOUNDARY COND} 119 gross 2862 With the PDE sufficiently modified, consideration must now be given to the boundary conditions of our model. Typically there are two main types of boundary conditions known as \textbf{Neumann} and \textbf{Dirichlet} boundary conditions\footnote{More information on Boundary Conditions is available at Wikipedia \url{http://en.wikipedia.org/wiki/Boundary_conditions}}, respectively. 120 artak 2958 A \textbf{Dirichlet boundary condition} is conceptually simpler and is used to prescribe a known value to the unknown solution (in our example the temperature) on parts of the boundary or on the entire boundary of the region of interest. 121 gross 2905 We discuss Dirichlet boundary condition in our second example presented in Section~\ref{Sec:1DHDv0}. 122 ahallam 2495 123 ahallam 2975 However, for this example we have made the model assumption that the system is insulated, so we need 124 gross 2905 to add an appropriate boundary condition to prevent 125 ahallam 2975 any loss or inflow of energy at the boundary of our domain. Mathematically this is expressed by prescribing 126 gross 2905 the heat flux $\kappa \frac{\partial T}{\partial x}$ to zero. In our simplified one dimensional model this is expressed 127 gross 2862 in the form; 128 ahallam 2494 \begin{equation} 129 gross 2862 \kappa \frac{\partial T}{\partial x} = 0 130 ahallam 2494 \end{equation} 131 gross 2862 or in a more general case as 132 \begin{equation}\label{NEUMAN 1} 133 \kappa \nabla T \cdot n = 0 134 \end{equation} 135 where $n$ is the outer normal field \index{outer normal field} at the surface of the domain. 136 gross 2905 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 137 artak 2958 the temperature $T$. Other notations used here are\footnote{The \esc notation for the normal 138 gross 2862 derivative is $T\hackscore{,i} n\hackscore i$.}; 139 ahallam 2645 \begin{equation} 140 gross 2862 \nabla T \cdot n = \frac{\partial T}{\partial n} \; . 141 ahallam 2645 \end{equation} 142 gross 2862 A condition of the type \refEq{NEUMAN 1} defines a \textbf{Neuman boundary condition} for the PDE. 143 ahallam 2494 144 gross 2905 The PDE \refEq{eqn:hdgenf} 145 artak 2958 and the Neuman boundary condition~\ref{eqn:hdgenf} (potentially together with the Dirichlet boundary conditions) define a \textbf{boundary value problem}. 146 It is the nature of a boundary value problem to allow making statements about the solution in the 147 gross 2862 interior of the domain from information known on the boundary only. In most cases 148 artak 2958 we use the term partial differential equation but in fact it is a boundary value problem. 149 gross 2862 It is important to keep in mind that boundary conditions need to be complete and consistent in the sense that 150 at any point on the boundary either a Dirichlet or a Neuman boundary condition must be set. 151 152 gross 2878 Conveniently, \esc makes default assumption on the boundary conditions which the user may modify where appropriate. 153 gross 2862 For a problem of the form in~\refEq{eqn:commonform nabla} the default condition\footnote{In the form of the \esc users guide which is using the Einstein convention is written as 154 $n\hackscore{j}A\hackscore{jl} u\hackscore{,l}=0$.} is; 155 \begin{equation}\label{NEUMAN 2} 156 gross 2948 -n\cdot A \cdot\nabla u = 0 157 gross 2862 \end{equation} 158 which is used everywhere on the boundary. Again $n$ denotes the outer normal field. 159 Notice that the coefficient $A$ is the same as in the \esc PDE~\ref{eqn:commonform nabla}. 160 With the settings for the coefficients we have already identified in \refEq{ESCRIPT SET} this 161 condition translates into 162 gross 2867 \begin{equation}\label{NEUMAN 2b} 163 gross 2862 \kappa \frac{\partial T}{\partial x} = 0 164 \end{equation} 165 artak 2958 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 discuss the Dirichlet boundary condition later. 166 gross 2862 167 gross 2870 \subsection{Outline of the Implementation} 168 \label{sec:outline} 169 ahallam 2975 To solve the heat diffusion equation (\refEq{eqn:hd}) we 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 (discussed in \refSec{sec:key}) has four major steps. Firstly, we need to define the domain where we want to 170 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 to solve in each time step to get the updated temperature. Thirdly, we need to define 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. The work flow described in \reffig{fig:wf}. 171 % \begin{enumerate} 172 % \item create domain 173 % \item create PDE 174 % \item while end time not reached: 175 % \begin{enumerate} 176 % \item set PDE coefficients 177 % \item solve PDE 178 % \item update time marker 179 % \end{enumerate} 180 % \item end of calculation 181 % \end{enumerate} 182 183 \begin{figure}[h!] 184 \centering 185 \includegraphics[width=1in]{figures/workflow.png} 186 \caption{Workflow for developing an \esc model and solution.} 187 \label{fig:wf} 188 \end{figure} 189 190 artak 2959 In the terminology of \pyt, the domain and PDE are represented by \textbf{objects}. The nice feature of an object is that it is defined by its usage and features 191 gross 2905 rather than its actual representation. So we will create a domain object to describe the geometry of the two 192 artak 2959 granite blocks. Then we define PDEs and spatially distributed values such as the temperature 193 on this domain. Similarly, to define a PDE object we use the fact that one needs only to define the coefficients of the PDE and solve the PDE. The PDE object has advanced features, but these are not required in simple cases. 194 gross 2870 195 196 \begin{figure}[t] 197 \centering 198 \includegraphics[width=6in]{figures/functionspace.pdf} 199 ahallam 2975 \caption{\esc domain construction overview} 200 gross 2870 \label{fig:fs} 201 \end{figure} 202 203 \subsection{The Domain Constructor in \esc} 204 \label{ss:domcon} 205 ahallam 2975 While it is not strictly relevant or necessary, it can be helpful to have a better understanding of how values are spatially distributed and how PDE coefficients are interpreted in \esc. The example in this case would be temperature. 206 gross 2870 207 artak 2963 There are various ways to construct domain objects. The simplest form is a rectangular shaped region with a length and height. There is 208 a ready to use function for this named \verb rectangle(). Besides the spatial dimensions this function requires to specify the number of 209 gross 2905 elements or cells to be used along the length and height, see \reffig{fig:fs}. Any spatially distributed value 210 artak 2963 and the PDE is represented in discrete form using this element representation\footnote{We 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. 211 gross 2870 The quality of the approximation depends - besides other factors- mainly on the number of elements being used. In fact, the 212 ahallam 2975 approximation becomes better when more elements are used. However, computational costs and time grow with the number of 213 artak 2963 elements being used. It is therefore important that you find the right balance between the demand in accuracy and acceptable resource usage. 214 gross 2870 215 artak 2963 In general, one can think about a domain object as a composition of nodes and elements. 216 As shown in \reffig{fig:fs}, an element is defined by the nodes that are used to describe its vertices. 217 gross 2870 To represent spatial distributed values the user can use 218 artak 2963 the values at the nodes, at the elements in the interior of the domain or at the elements located at the surface of the domain. 219 gross 2870 The different approach used to represent values is called \textbf{function space} and is attached to all objects 220 in \esc representing a spatial distributed value such as the solution of a PDE. The three 221 artak 2963 function spaces we use at the moment are; 222 gross 2870 \begin{enumerate} 223 \item the nodes, called by \verb|ContinuousFunction(domain)| ; 224 \item the elements/cells, called by \verb|Function(domain)| ; and 225 \item the boundary, called by \verb|FunctionOnBoundary(domain)| . 226 \end{enumerate} 227 A function space object such as \verb|ContinuousFunction(domain)| has the method \verb|getX| attached to it. This method returns the 228 artak 2963 location of the so-called \textbf{sample points} used to represent values of the particular function space. So the 229 gross 2870 call \verb|ContinuousFunction(domain).getX()| will return the coordinates of the nodes used to describe the domain while 230 the \verb|Function(domain).getX()| returns the coordinates of numerical integration points within elements, see 231 gross 2905 \reffig{fig:fs}. 232 gross 2870 233 artak 2963 This distinction between different representations of spatially distributed values 234 gross 2870 is important in order to be able to vary the degrees of smoothness in a PDE problem. 235 ahallam 2975 The coefficients of a PDE do not need to be continuous, thus this qualifies as a \verb|Function()| type. 236 gross 2870 On the other hand a temperature distribution must be continuous and needs to be represented with a \verb|ContinuousFunction()| function space. 237 An influx may only be defined at the boundary and is therefore a \verb FunctionOnBoundary() object. 238 \esc allows certain transformations of the function spaces. A \verb ContinuousFunction() can be transformed into a \verb|FunctionOnBoundary()| 239 or \verb|Function()|. On the other hand there is not enough information in a \verb FunctionOnBoundary() to transform it to a \verb ContinuousFunction() . 240 gross 2878 These transformations, which are called \textbf{interpolation} are invoked automatically by \esc if needed. 241 gross 2870 242 artak 2963 Later in this introduction we discuss how 243 gross 2870 to define specific areas of geometry with different materials which are represented by different material coefficients such the 244 ahallam 2975 thermal conductivities $\kappa$. A very powerful technique to define these types of PDE 245 gross 2870 coefficients is tagging. Blocks of materials and boundaries can be named and values can be defined on subregions based on their names. 246 ahallam 2975 This is a method for simplifying PDE coefficient and flux definitions. It makes scripting much easier and we will discuss this technique in Section~\ref{STEADY-STATE HEAT REFRACTION}. 247 gross 2870 248 249 \subsection{A Clarification for the 1D Case} 250 gross 2931 \label{SEC: 1D CLARIFICATION} 251 artak 2963 It is necessary for clarification that we revisit our general PDE from \refeq{eqn:commonform nabla} for 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 $\nabla = (\frac{\partial}{\partial x}, \frac{\partial}{\partial y})$. Assuming the coefficient $A$ is constant, the \refEq{eqn:commonform nabla} takes the following form; 252 gross 2477 \begin{equation}\label{eqn:commonform2D} 253 -A\hackscore{00}\frac{\partial^{2}u}{\partial x^{2}} 254 -A\hackscore{01}\frac{\partial^{2}u}{\partial x\partial y} 255 -A\hackscore{10}\frac{\partial^{2}u}{\partial y\partial x} 256 -A\hackscore{11}\frac{\partial^{2}u}{\partial y^{2}} 257 + Du = f 258 \end{equation} 259 ahallam 2606 Notice that for the higher dimensional case $A$ becomes a matrix. It is also 260 ahallam 2495 important to notice that the usage of the Nabla operator creates 261 a compact formulation which is also independent from the spatial dimension. 262 artak 2963 To make the general PDE \refEq{eqn:commonform2D} one dimensional as 263 gross 2861 shown in \refEq{eqn:commonform} we need to set 264 ahallam 2606 \begin{equation} 265 ahallam 2494 A\hackscore{00}=A; A\hackscore{01}=A\hackscore{10}=A\hackscore{11}=0 266 gross 2477 \end{equation} 267 268 gross 2867 269 ahallam 2495 \subsection{Developing a PDE Solution Script} 270 ahallam 2801 \label{sec:key} 271 gross 2949 \sslist{example01a.py} 272 artak 2963 We write a simple \pyt script which uses the \modescript, \modfinley and \modmpl modules. 273 gross 2905 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 ahallam 2495 that we will require. 275 ahallam 2775 \begin{python} 276 ahallam 2495 from esys.escript import * 277 ahallam 2606 # This defines the LinearPDE module as LinearPDE 278 from esys.escript.linearPDEs import LinearPDE 279 # This imports the rectangle domain function from finley. 280 from esys.finley import Rectangle 281 # A useful unit handling package which will make sure all our units 282 # match up in the equations under SI. 283 from esys.escript.unitsSI import * 284 ahallam 2775 \end{python} 285 artak 2963 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 \verb|LinearPDE| has been imported explicitly for ease of use later in the script. \verb|Rectangle| is going to be our type of domain. The module \verb unitsSI provides support for SI unit definitions with our variables. 286 gross 2477 287 artak 2963 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 need values. Firstly, the domain upon which we wish to solve our problem needs to be defined. There are different types of domains in \modescript which we demonstrate in later tutorials but for our granite blocks, we simply use a rectangular domain. 288 ahallam 2401 289 ahallam 2975 Using a rectangular domain 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 due to symmetry. There are four arguments we must consider when we decide to create a rectangular domain, the domain \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 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 \verb|ndx| would be 1 to 10\% of the length. Our \verb|ndy| 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 ahallam 2775 \begin{python} 291 gross 2905 mx = 500.*m #meters - model length 292 my = 100.*m #meters - model width 293 ndx = 50 # mesh steps in x direction 294 ndy = 1 # mesh steps in y direction 295 boundloc = mx/2 # location of boundary between the two blocks 296 ahallam 2775 \end{python} 297 gross 2905 The material constants and the temperature variables must also be defined. For the granite in the model they are defined as: 298 ahallam 2775 \begin{python} 299 ahallam 2495 #PDE related 300 gross 2905 rho = 2750. *kg/m**3 #kg/m^{3} density of iron 301 cp = 790.*J/(kg*K) # J/Kg.K thermal capacity 302 gross 2878 rhocp = rho*cp 303 gross 2905 kappa = 2.2*W/m/K # watts/m.Kthermal conductivity 304 gross 2878 qH=0 * J/(sec*m**3) # J/(sec.m^{3}) no heat source 305 gross 2905 T1=20 * Celsius # initial temperature at Block 1 306 T2=2273. * Celsius # base temperature at Block 2 307 ahallam 2775 \end{python} 308 ahallam 2495 Finally, to control our script we will have to specify our timing controls and where we would like to save the output from the solver. This is simple enough: 309 ahallam 2775 \begin{python} 310 gross 2878 t=0 * day #our start time, usually zero 311 tend=1. * day # - time to end simulation 312 ahallam 2495 outputs = 200 # number of time steps required. 313 h=(tend-t)/outputs #size of time step 314 ahallam 2606 #user warning statement 315 print "Expected Number of time outputs is: ", (tend-t)/h 316 i=0 #loop counter 317 ahallam 2775 \end{python} 318 gross 2905 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 ahallam 2775 \begin{python} 320 ahallam 2606 #generate domain using rectangle 321 gross 2905 blocks = Rectangle(l0=mx,l1=my,n0=ndx, n1=ndy) 322 ahallam 2775 \end{python} 323 artak 2963 \verb blocks now describes a domain in the manner of Section \ref{ss:domcon}. 324 ahallam 2658 325 artak 2963 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 we discuss later.}. We also need to state the values of our general form variables. 326 ahallam 2775 \begin{python} 327 gross 2905 mypde=LinearPDE(blocks) 328 gross 2878 A=zeros((2,2))) 329 A[0,0]=kappa 330 gross 2905 mypde.setValue(A=A, D=rhocp/h) 331 ahallam 2775 \end{python} 332 gross 2878 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; 334 ahallam 2495 \begin{equation}\label{eqn:symm} 335 A\hackscore{jl}=A\hackscore{lj} 336 \end{equation} 337 artak 2963 Symmetry is only dependent on the $A$ coefficient in the general form and the other coefficients $D$ as well as the right hand side $Y$. 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 enable symmetry via; 338 ahallam 2775 \begin{python} 339 ahallam 2495 myPDE.setSymmetryOn() 340 ahallam 2775 \end{python} 341 gross 2905 Next we need to establish the initial temperature distribution \verb|T|. We need to 342 assign the value \verb|T1| to all sample points left to the contact interface at $x\hackscore{0}=\frac{mx}{2}$ 343 and the value \verb|T2| right to the contact interface. \esc 344 provides the \verb|whereNegative| function to construct this. In fact, 345 \verb|whereNegative| returns the value $1$ at those sample points where the argument 346 has a negative value. Otherwise zero is returned. If \verb|x| are the $x\hackscore{0}$ 347 coordinates of the sample points used to represent the temperature distribution 348 then \verb|x-boundloc| gives us a negative value for 349 all sample points left to the interface and non-negative value to 350 the right of the interface. So with; 351 gross 2878 \begin{python} 352 # ... set initial temperature .... 353 gross 2905 T= T1*whereNegative(x-boundloc)+T2*(1-whereNegative(x-boundloc)) 354 gross 2878 \end{python} 355 gross 2905 we get the desired temperature distribution. To get the actual sample points \verb|x| we use 356 the \verb|getX()| method of the function space \verb|Solution(blocks)| 357 which is used to represent the solution of a PDE; 358 gross 2878 \begin{python} 359 gross 2905 x=Solution(blocks).getX() 360 \end{python} 361 As \verb|x| are the sample points for the function space \verb|Solution(blocks)| 362 the initial temperature \verb|T| 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 artak 2963 Finally we 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 gross 2905 \begin{python} 369 gross 2878 while t < tend: 370 i+=1 #increment the counter 371 t+=h #increment the current time 372 mypde.setValue(Y=qH+rhocp/h*T) # set variable PDE coefficients 373 T=mypde.getSolution() #get the PDE solution 374 totE = integrate(rhocp*T) #get the total heat (energy) in the system 375 \end{python} 376 The last statement in this script calculates the total energy in the system as volume integral 377 gross 2905 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 ahallam 2401 380 gross 2905 \subsection{Running the Script} 381 ahallam 2975 The script presented so far is available under 382 gross 2949 \verb|example01a.py|. You can edit this file with your favourite text editor. 383 jfenwick 2911 On most operating systems\footnote{The you can use \texttt{run-escript} launcher is not supported under {\it MS Windows} yet.} you can use the \program{run-escript} command 384 gross 2905 to launch {\it escript} scripts. For the example script use; 385 \begin{verbatim} 386 gross 2949 run-escript example01a.py 387 gross 2905 \end{verbatim} 388 The program will print a progress report. Alternatively, you can use 389 the python interpreter directly; 390 \begin{verbatim} 391 gross 2949 python example01a.py 392 gross 2905 \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 gross 2952 \caption{Example 1b: Total Energy in the Blocks over Time (in seconds).} 399 gross 2905 \label{fig:onedheatout1} 400 \end{center} 401 \end{figure} 402 403 gross 2878 \subsection{Plotting the Total Energy} 404 gross 2949 \sslist{example01b.py} 405 gross 2878 406 gross 2905 \esc does not include its own plotting capabilities. However, it is possible to use a variety of free \pyt packages for visualisation. 407 gross 2878 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. 409 ahallam 2975 For more complex visualisation tasks in particular, two and three dimensional problems we recommend the use of more advanced tools. For instance, \mayavi \footnote{\url{http://code.enthought.com/projects/mayavi/}} 410 which is based upon the \verb|VTK| toolkit. The usage of \verb|VTK| based 411 visualization is discussed in Chapter~\ref{Sec:2DHD} which focusses on a two dimensional PDE. 412 gross 2878 413 ahallam 2975 For our simple granite block problem, we have two plotting tasks. Firstly, we are interested in showing the 414 behaviour of the total energy over time and secondly, how the temperature distribution within the block is 415 gross 2878 developing over time. Lets start with the first task. 416 417 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 419 the time loop is opened we create empty lists for the time marks \verb|t_list| and the total energies \verb|E_list|. 420 ahallam 2975 After the new temperature has been calculated by solving the PDE we append the new time marker and the total energy value for that time 421 artak 2964 to the corresponding list using the \verb|append| method. With these modifications our script looks as follows: 422 ahallam 2775 \begin{python} 423 gross 2878 t_list=[] 424 E_list=[] 425 # ... start iteration: 426 while t