 # Contents of /trunk/doc/cookbook/example01.tex

Revision 2957 - (show annotations)
Tue Mar 2 01:28:19 2010 UTC (9 years, 6 months ago) by artak
File MIME type: application/x-tex
File size: 37823 byte(s)
more corrections

 1 2 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 3 % 4 % Copyright (c) 2003-2010 by University of Queensland 5 % 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 \begin{figure}[h!] 15 \centerline{\includegraphics[width=4.in]{figures/onedheatdiff001}} 16 \caption{Example 1: Temperature differential along a single interface between two granite blocks.} 17 \label{fig:onedgbmodel} 18 \end{figure} 19 20 \section{Example 1: 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 Initial temperature in \textit{Block 1} is \verb|T1| and in \textit{Block 2} is \verb|T2|. 25 We assume that the system is insulated. 26 What would happen to the temperature distribution in each block over time? 27 Intuition tells us that heat will be transported from the hotter block to the cooler until both 28 blocks have the same temperature. 29 30 \subsection{1D Heat Diffusion Equation} 31 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 which is defined as: 33 \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 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 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 $x$. 41 42 \subsection{PDEs and the General Form} 43 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 44 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. 45 46 Firstly, we will discretise the PDE \refEq{eqn:hd} in the time direction which will 47 leave as with a steady linear PDE which is involving spatial derivatives only and needs to be solved in each time 48 step to progress in time - \esc can help us here. 49 50 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 51 approximation 52 \begin{equation} 53 \frac{\partial T(t)}{\partial t} \approx \frac{T(t)-T(t-h)}{h} 54 \label{eqn:beuler} 55 \end{equation} 56 for $\frac{\partial T}{\partial t}$ at time $t$ 57 where $h$ is the time step size. This can also be written as; 58 \begin{equation} 59 \frac{\partial T}{\partial t}(t^{(n)}) \approx \frac{T^{(n)} - T^{(n-1)}}{h} 60 \label{eqn:Tbeuler} 61 \end{equation} 62 where the upper index $n$ denotes the n\textsuperscript{th} time step. So one has 63 \begin{equation} 64 \begin{array}{rcl} 65 t^{(n)} & = & t^{(n-1)}+h \\ 66 T^{(n)} & = & T(t^{(n-1)}) \\ 67 \end{array} 68 \label{eqn:Neuler} 69 \end{equation} 70 Substituting \refEq{eqn:Tbeuler} into \refEq{eqn:hd} we get; 71 \begin{equation} 72 \frac{\rho c\hackscore p}{h} (T^{(n)} - T^{(n-1)}) - \kappa \frac{\partial^{2} T^{(n)}}{\partial x^{2}} = q\hackscore H 73 \label{eqn:hddisc} 74 \end{equation} 75 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 $t^{(n-1)}$. This 76 approach is called the \textbf{forward Euler} scheme. This scheme can provide some computational advantages which 77 we are not discussed here but has the major disadvantage that depending on the 78 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 79 that the approximation of the temperature will not grow beyond its initial bounds and becomes non-physical. 80 The backward Euler which we use here is unconditionally stable meaning that under the assumption of 81 physically correct problem set-up the temperature approximation remains physical for all times. 82 The user needs to keep in mind that the discretization error introduced by \refEq{eqn:beuler} 83 is sufficiently small so a good approximation of the true temperature is calculated. It is 84 therefore crucial that the user remains critical about his/her results and for instance compares 85 the results for different time and spatial step sizes. 86 87 To get the temperature $T^{(n)}$ at time $t^{(n)}$ we need to solve the linear 88 differential equation \refEq{eqn:hddisc} which is only including spatial derivatives. To solve this problem 89 we want to to use \esc. 90 91 \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 92 $-(A\hackscore{jl} u\hackscore{,l})\hackscore{,j}+D u =Y$} 93 is described by; 94 \begin{equation}\label{eqn:commonform nabla} 95 -\nabla\cdot(A\cdot\nabla u) + Du = f 96 \end{equation} 97 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 98 the spatial derivative of its subject - in this case $u$. Lets assume for a moment that we deal with a one-dimensional problem then ; 99 \begin{equation} 100 \nabla = \frac{\partial}{\partial x} 101 \end{equation} 102 and we can write \refEq{eqn:commonform nabla} as; 103 \begin{equation}\label{eqn:commonform} 104 -A\frac{\partial^{2}u}{\partial x^{2}} + Du = f 105 \end{equation} 106 if $A$ is constant. To match this simplified general form to our problem \refEq{eqn:hddisc} 107 we rearrange \refEq{eqn:hddisc}; 108 \begin{equation} 109 \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)} 110 \label{eqn:hdgenf} 111 \end{equation} 112 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 113 $t^{(0)}=0$ and $t^{(n)}=t^{(n-1)}+h$ where $h>0$ is the step size and assumed to be constant. 114 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; 115 \begin{equation}\label{ESCRIPT SET} 116 u=T^{(n)}; 117 A = \kappa; D = \frac{\rho c \hackscore{p}}{h}; f = q \hackscore{H} + \frac{\rho c\hackscore p}{h} T^{(n-1)} 118 \end{equation} 119 120 \subsection{Boundary Conditions} 121 \label{SEC BOUNDARY COND} 122 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. 123 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. 124 We discuss Dirichlet boundary condition in our second example presented in Section~\ref{Sec:1DHDv0}. 125 126 We make the model assumption that the system is insulated so we need 127 to add an appropriate boundary condition to prevent 128 any loss or inflow of energy at boundary of our domain. Mathematically this is expressed by prescribing 129 the heat flux $\kappa \frac{\partial T}{\partial x}$ to zero. In our simplified one dimensional model this is expressed 130 in the form; 131 \begin{equation} 132 \kappa \frac{\partial T}{\partial x} = 0 133 \end{equation} 134 or in a more general case as 135 \begin{equation}\label{NEUMAN 1} 136 \kappa \nabla T \cdot n = 0 137 \end{equation} 138 where $n$ is the outer normal field \index{outer normal field} at the surface of the domain. 139 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 140 the temperature $T$. Other notations which are used are\footnote{The \esc notation for the normal 141 derivative is $T\hackscore{,i} n\hackscore i$.}; 142 \begin{equation} 143 \nabla T \cdot n = \frac{\partial T}{\partial n} \; . 144 \end{equation} 145 A condition of the type \refEq{NEUMAN 1} defines a \textbf{Neuman boundary condition} for the PDE. 146 147 The PDE \refEq{eqn:hdgenf} 148 and the Neuman boundary condition~\ref{eqn:hdgenf} (potentially together with the Dirichlet boundary condition set) define a \textbf{boundary value problem}. 149 It is a nature of a boundary value problem that it allows to make statements on the solution in the 150 interior of the domain from information known on the boundary only. In most cases 151 we use the term partial differential equation but in fact mean a boundary value problem. 152 It is important to keep in mind that boundary conditions need to be complete and consistent in the sense that 153 at any point on the boundary either a Dirichlet or a Neuman boundary condition must be set. 154 155 Conveniently, \esc makes default assumption on the boundary conditions which the user may modify where appropriate. 156 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 157 $n\hackscore{j}A\hackscore{jl} u\hackscore{,l}=0$.} is; 158 \begin{equation}\label{NEUMAN 2} 159 -n\cdot A \cdot\nabla u = 0 160 \end{equation} 161 which is used everywhere on the boundary. Again $n$ denotes the outer normal field. 162 Notice that the coefficient $A$ is the same as in the \esc PDE~\ref{eqn:commonform nabla}. 163 With the settings for the coefficients we have already identified in \refEq{ESCRIPT SET} this 164 condition translates into 165 \begin{equation}\label{NEUMAN 2b} 166 \kappa \frac{\partial T}{\partial x} = 0 167 \end{equation} 168 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. 169 170 \subsection{Outline of the Implementation} 171 \label{sec:outline} 172 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 173 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 174 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; 175 \begin{enumerate} 176 \item create domain 177 \item create PDE 178 \item while end time not reached: 179 \begin{enumerate} 180 \item set PDE coefficients 181 \item solve PDE 182 \item update time marker 183 \end{enumerate} 184 \item end of calculation 185 \end{enumerate} 186 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 187 rather than its actual representation. So we will create a domain object to describe the geometry of the two 188 granite blocks. The main feature 189 of the object we will use is the fact that we can define PDEs and spatially distributed values such as the temperature 190 on a domain. In fact the domain object has many more features - most of them you will 191 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 192 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. 193 194 195 \begin{figure}[t] 196 \centering 197 \includegraphics[width=6in]{figures/functionspace.pdf} 198 \label{fig:fs} 199 \caption{\esc domain construction overview} 200 \end{figure} 201 202 \subsection{The Domain Constructor in \esc} 203 \label{ss:domcon} 204 It is helpful to have a better understanding how spatially distributed value such as the temperature or PDE coefficients are interpreted in \esc. Again 205 from the user's point of view the representation of these spatially distributed values is not relevant. 206 207 There are various ways to construct domain objects. The simplest form is as rectangular shaped region with a length and height. There is 208 a ready to use function call for this. Besides the spatial dimensions the function call will require you to specify the number 209 elements or cells to be used along the length and height, see \reffig{fig:fs}. Any spatially distributed value 210 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. 211 The quality of the approximation depends - besides other factors- mainly on the number of elements being used. In fact, the 212 approximation becomes better the more elements are used. However, computational costs and compute time grow with the number of 213 elements being used. It therefore important that you find the right balance between the demand in accuracy and acceptable resource usage. 214 215 In general, one can thinks 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 used to describe its vertices. 217 To represent spatial distributed values the user can use 218 the values at the nodes, at the elements in the interior of the domain or at elements located at the surface of the domain. 219 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 function spaces we will use at the moment are; 222 \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 location of the so-called \textbf{sample points} used to represent values with the particular function space attached to it. So the 229 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 \reffig{fig:fs}. 232 233 This distinction between different representations of spatial distributed values 234 is important in order to be able to vary the degrees of smoothness in a PDE problem. 235 The coefficients of a PDE need not be continuous thus this qualifies as a \verb|Function()| type. 236 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 These transformations, which are called \textbf{interpolation} are invoked automatically by \esc if needed. 241 242 Later in this introduction we will discuss how 243 to define specific areas of geometry with different materials which are represented by different material coefficients such the 244 thermal conductivities $kappa$. A very powerful technique to define these types of PDE 245 coefficients is tagging. Blocks of materials and boundaries can be named and values can be defined on subregions based on their names. 246 This is simplifying PDE coefficient and flux definitions. It makes for much easier scripting. We will discuss this technique in Section~\ref{STEADY-STATE HEAT REFRACTION}. 247 248 249 \subsection{A Clarification for the 1D Case} 250 \label{SEC: 1D CLARIFICATION} 251 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; 252 \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 Notice that for the higher dimensional case $A$ becomes a matrix. It is also 260 important to notice that the usage of the Nabla operator creates 261 a compact formulation which is also independent from the spatial dimension. 262 So to make the general PDE \refEq{eqn:commonform2D} one dimensional as 263 shown in \refEq{eqn:commonform} we need to set 264 \begin{equation} 265 A\hackscore{00}=A; A\hackscore{01}=A\hackscore{10}=A\hackscore{11}=0 266 \end{equation} 267 268 269 \subsection{Developing a PDE Solution Script} 270 \label{sec:key} 271 \sslist{example01a.py} 272 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.} 274 that we will require. 275 \begin{python} 276 from esys.escript import * 277 # 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 \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 \verb|LinearPDE| has been imported explicitly for ease of use later in the script. \verb|Rectangle| 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 granite blocks, we will simply use a rectangular model. 288 289 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 \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 \begin{python} 291 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 \end{python} 297 The material constants and the temperature variables must also be defined. For the granite in the model they are defined as: 298 \begin{python} 299 #PDE related 300 rho = 2750. *kg/m**3 #kg/m^{3} density of iron 301 cp = 790.*J/(kg*K) # J/Kg.K thermal capacity 302 rhocp = rho*cp 303 kappa = 2.2*W/m/K # watts/m.Kthermal conductivity 304 qH=0 * J/(sec*m**3) # J/(sec.m^{3}) no heat source 305 T1=20 * Celsius # initial temperature at Block 1 306 T2=2273. * Celsius # base temperature at Block 2 307 \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: 309 \begin{python} 310 t=0 * day #our start time, usually zero 311 tend=1. * day # - time to end simulation 312 outputs = 200 # number of time steps required. 313 h=(tend-t)/outputs #size of time step 314 #user warning statement 315 print "Expected Number of time outputs is: ", (tend-t)/h 316 i=0 #loop counter 317 \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 model as: 319 \begin{python} 320 #generate domain using rectangle 321 blocks = Rectangle(l0=mx,l1=my,n0=ndx, n1=ndy) 322 \end{python} 323 \verb blocks now describes a domain in the manner of Section \ref{ss:domcon}. T 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. 326 \begin{python} 327 mypde=LinearPDE(blocks) 328 A=zeros((2,2))) 329 A[0,0]=kappa 330 mypde.setValue(A=A, D=rhocp/h) 331 \end{python} 332 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 \begin{equation}\label{eqn:symm} 335 A\hackscore{jl}=A\hackscore{lj} 336 \end{equation} 337 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; 338 \begin{python} 339 myPDE.setSymmetryOn() 340 \end{python} 341 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 \begin{python} 352 # ... set initial temperature .... 353 T= T1*whereNegative(x-boundloc)+T2*(1-whereNegative(x-boundloc)) 354 \end{python} 355 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 \begin{python} 359 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 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} 369 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 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 \verb|example01a.py|. You can edit this file with your favourite text editor. 383 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 to launch {\it escript} scripts. For the example script use; 385 \begin{verbatim} 386 run-escript example01a.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 example01a.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{Example 1b: 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} 404 \sslist{example01b.py} 405 406 \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. 408 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/}} 410 which bases on the \verb|VTK| toolkit. We will discuss the usage of \verb|VTK| based 411 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 414 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. 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 After the new temperature as been calculated by solving the PDE we append the new time marker and total energy 421 to the corresponding list using the \verb|append| method. With these modifications the script looks as follows: 422 \begin{python} 423 t_list=[] 424 E_list=[] 425 # ... start iteration: 426 while t