# Contents of /trunk/esys2/doc/user/diffusion.tex

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 1 % $Id$ 2 \chapter{How to Solve A Diffusion Problem} 3 \label{DIFFUSION CHAP} 4 5 \begin{figure} 6 \centerline{\includegraphics[width=\figwidth]{DiffusionDomain}} 7 \caption{Temperature Diffusion Problem with Circular Heat Source} 8 \label{DIFFUSION FIG 1} 9 \end{figure} 10 11 \section{\label{DIFFUSION OUT SEC}Outline} 12 In this chapter we will discuss how to solve the time dependent-temperature diffusion\index{diffusion equation} for 13 a block of material. Within the block there is a heat source which drives the temperature diffusion. 14 On the surface, energy can radiate into the surrounding environment. 15 \fig{DIFFUSION FIG 1} shows the configuration. 16 17 In the next \Sec{DIFFUSION TEMP SEC} we will present the relevant model. A 18 time integration scheme is introduced to calculate the temperature at given time nodes $t^{(n)}$. 19 We will see that at each time step a Helmholtz equation \index{Helmholtz equation} 20 must be solved. 21 The implementation of a Helmholtz equation solver will be discussed in \Sec{DIFFUSION HELM SEC}. 22 In Section~\ref{DIFFUSION TRANS SEC} the solver of the Helmholtz equation is used to build a 23 solver for the temperature diffusion problem. 24 25 \section{\label{DIFFUSION TEMP SEC}Temperature Diffusion} 26 27 The unknown temperature $T$ is a function of its location in the domain and time $t>0$. The governing equation 28 in the interior of the domain is given by 29 \begin{equation} 30 \rho c\hackscore p T\hackscore{,t} - (\kappa T\hackscore{,i})\hackscore{,i} = q 31 \label{DIFFUSION TEMP EQ 1} 32 \end{equation} 33 where $\rho c\hackscore p$ and $\kappa$ are given material constants. In case of a composite 34 material the parameters depend on their location in the domain. $q$ is 35 a heat source (or sink) within the domain. We are using Einstein summation convention \index{summation convention} 36 as introduced in \Chap{FirstSteps}. In our case we assume $q$ to be equal to a constant heat production rate 37 $q^{c}$ on a circle or sphere with center $x^c$ and radius $r$ and $0$ elsewhere: 38 \begin{equation} 39 q(x,t)= 40 \left\{ 41 \begin{array}{lcl} 42 q^c & & \|x-x^c\| \le r \\ 43 & \mbox{if} \\ 44 0 & & \mbox{else} \\ 45 \end{array} 46 \right. 47 \label{DIFFUSION TEMP EQ 1b} 48 \end{equation} 49 for all $x$ in the domain and all time $t>0$. 50 51 On the surface of the domain we are 52 specifying a radiation condition 53 which precribes the normal component of the flux $\kappa T\hackscore{,i}$ to be proportional 54 to the difference of the current temperature to the surrounding temperature $T\hackscore{ref}$: 55 \begin{equation} 56 \kappa T\hackscore{,i} n\hackscore i = \eta (T\hackscore{ref}-T) 57 \label{DIFFUSION TEMP EQ 2} 58 \end{equation} 59 $\eta$ is a given material coefficient depending on the material of the block and the surrounding medium. 60 As usual $n\hackscore i$ is the $i$-th component of the outer normal field \index{outer normal field} 61 at the surface of the domain. 62 63 To solve the time dependent \eqn{DIFFUSION TEMP EQ 1} the initial temperature at time 64 $t=0$ has to be given. Here we assume that the initial temperature is the surrounding temperature: 65 \begin{equation} 66 T(x,0)=T\hackscore{ref} 67 \label{DIFFUSION TEMP EQ 4} 68 \end{equation} 69 for all $x$ in the domain. It is pointed out that 70 the initial conditions satisfy the 71 boundary condition defined by \eqn{DIFFUSION TEMP EQ 2}. 72 73 The temperature is calculated at discrete time nodes $t^{(n)}$ where 74 $t^{(0)}=0$ and $t^{(n)}=t^{(n-1)}+h$ where $h>0$ is the step size which is assumed to be constant. 75 In the following the upper index ${(n)}$ refers to a value at time $t^{(n)}$. The simplest 76 and most robust scheme to approximate the time derivative of the the temperature is 77 the backward Euler 78 \index{backward Euler} scheme, see~\cite{XXX} for alternatives. The backward Euler 79 scheme is based 80 on the Taylor expansion of $T$ at time $t^{(n)}$: 81 \begin{equation} 82 T^{(n-1)}\approx T^{(n)}+T\hackscore{,t}^{(n)}(t^{(n-1)}-t^{(n)}) 83 =T^{(n-1)} - h \cdot T\hackscore{,t}^{(n)} 84 \label{DIFFUSION TEMP EQ 6} 85 \end{equation} 86 This is inserted into \eqn{DIFFUSION TEMP EQ 1}. By separating the terms at 87 $t^{(n)}$ and $t^{(n-1)}$ one gets for $n=1,2,3\ldots$ 88 \begin{equation} 89 \frac{\rho c\hackscore p}{h} T^{(n)} - (\kappa T^{(n)}\hackscore{,i})\hackscore{,i} = q + \frac{\rho c\hackscore p}{h} T^{(n-1)} 90 \label{DIFFUSION TEMP EQ 7} 91 \end{equation} 92 where $T^{(0)}=T\hackscore{ref}$ is taken form the initial condition given by \eqn{DIFFUSION TEMP EQ 4}. 93 Together with the natural boundary condition 94 \begin{equation} 95 \kappa T\hackscore{,i}^{(n)} n\hackscore i = \eta (T\hackscore{ref}-T^{(n)}) 96 \label{DIFFUSION TEMP EQ 2222} 97 \end{equation} 98 taken from \eqn{DIFFUSION TEMP EQ 2} 99 this forms a boundary value problem that has to be solved for each time step. 100 As a first step to implement a solver for the temperature diffusion problem we will 101 first implement a solver for the boundary value problem that has to be solved at each time step. 102 103 \section{\label{DIFFUSION HELM SEC}Helmholtz Problem} 104 The partial differential equation to be solved for $T^{(n)}$ has the form 105 \begin{equation} 106 \omega u - (\kappa u\hackscore{,i})\hackscore{,i} = f 107 \label{DIFFUSION HELM EQ 1} 108 \end{equation} 109 where $u$ plays the role of $T^{(n)}$ and we set 110 \begin{equation} 111 \omega=\frac{\rho c\hackscore p}{h} \mbox{ and } f=q+\frac{\rho c\hackscore p}{h}T^{(n-1)} \;. 112 \label{DIFFUSION HELM EQ 1b} 113 \end{equation} 114 With $g=\eta T\hackscore{ref}$ the radiation condition defined by \eqn{DIFFUSION TEMP EQ 2222} 115 takes the form 116 \begin{equation} 117 \kappa u\hackscore{,i} n\hackscore{i} = g - \eta u\mbox{ on } \Gamma 118 \label{DIFFUSION HELM EQ 2} 119 \end{equation} 120 The partial differential 121 \eqn{DIFFUSION HELM EQ 1} together with boundary conditions of \eqn{DIFFUSION HELM EQ 2} 122 is called the Helmholtz equation \index{Helmholtz equation}. 123 124 We want to use the \LinearPDE class provided by \escript to define and solve a general linear PDE such as the 125 Helmholtz equation. We have used a special case of the \LinearPDE class, namely the 126 \Poisson class already in \Chap{FirstSteps}. 127 Here we will write our own specialized sub-class of the \LinearPDE to define the Helmholtz equation 128 and use the \method{getSolution} method of parent class to actually solve the problem. 129 130 The form of a single PDE that can be handled by the \LinearPDE class is 131 \begin{equation}\label{EQU.FEM.1} 132 -(A\hackscore{jl} u\hackscore{,l})\hackscore{,j}+D u =Y \; . 133 \end{equation} 134 We show here the terms which are relevant for the Helmholtz problem. 135 The general form and systems is discussed in \Sec{SEC LinearPDE}. 136 $A$, $D$ and $Y$ are the known coeffecients of the PDE. \index{partial differential equation!coefficients} 137 Notice that $A$ is a matrix or tensor of order 2 and $D$ and $Y$ are scalar. 138 They may be constant or may depend on their 139 location in the domain but must not depend on the unknown solution $u$. 140 The following natural boundary conditions \index{boundary condition!natural} that 141 are used in the \LinearPDE class have the form 142 \begin{equation}\label{EQU.FEM.2} 143 n\hackscore{j}A\hackscore{jl} u\hackscore{,l}+du=y \;. 144 \end{equation} 145 where, as usual, $n$ denotes the outer normal field on the surface of the domain. Notice that 146 the coefficient $A$ is already used in the PDE in \eqn{EQU.FEM.1}. $d$ and $y$ are given scalar coefficients. 147 148 By inspecting the Helmholtz equation \index{Helmholtz equation} 149 we can easily assign values to the coefficients in the 150 general PDE of the \LinearPDE class: 151 \begin{equation}\label{DIFFUSION HELM EQ 3} 152 \begin{array}{llllll} 153 A\hackscore{ij}=\kappa \delta\hackscore{ij} & D=\omega & Y=f \\ 154 d=\eta & y= g & \\ 155 \end{array} 156 \end{equation} 157 $\delta\hackscore{ij}$ is the Kronecker symbol \index{Kronecker symbol} defined by $\delta\hackscore{ij}=1$ for 158 $i=j$ and $0$ otherwise. 159 160 We want to implement a 161 new class which we will call \class{Helmholtz} that provides the same methods as the \LinearPDE class but 162 is described using the coefficients $\kappa$, $\omega$, $f$, $\eta$, 163 $g$ rather than the general form given by \eqn{EQU.FEM.1}. 164 Python's mechanism of subclasses allows us to do this in a very simple way. 165 The \Poisson class of the \linearPDEsPack module, 166 which we have already used in \Chap{FirstSteps}, is in fact a subclass of the more general 167 \LinearPDE class. That means that all methods (such as the \method{getSolution}) 168 from the parent class \LinearPDE are available for any \Poisson object. However, new 169 methods can be added and methods of the parent class can be redefined. In fact, 170 the \Poisson class redefines the \method{setValue} of the \LinearPDE class which is called to assign 171 values to the coefficients of the PDE. This is exactly what we will do when we define 172 our new \class{Helmholtz} class: 173 \begin{python} 174 from esys.linearPDEs import LinearPDE 175 import numarray 176 class Helmholtz(LinearPDE): 177 def setValue(self,kappa=0,omega=1,f=0,eta=0,g=0) 178 ndim=self.getDim() 179 kronecker=numarray.identity(ndim) 180 self._setValue(A=kappa*kronecker,D=omega,Y=f,d=eta,y=g) 181 \end{python} 182 \code{class Helmholtz(linearPDE)} declares the new \class{Helmholtz} class as a subclass 183 of \LinearPDE which we have imported in the first line of the script. 184 We add the method \method{setValue} to the class which overwrites the 185 \method{setValue} method of the \LinearPDE class. The new method which has the 186 parameters of the Helmholtz \eqn{DIFFUSION HELM EQ 1} as arguments 187 maps the parameters of the coefficients of the general PDE defined 188 in \eqn{EQU.FEM.1}. We are actually using the \method{_setValue} of 189 the \LinearPDE class (notice the leeding underscoure). 190 The coefficient \var{A} involves the Kronecker symbol. We use the 191 \numarray function \function{identity} which returns a square matrix with ones on the 192 main diagonal and zeros off the main diagonal. The argument of \function{identity} gives the order of the matrix. 193 The \method{getDim} of the \LinearPDE class object \var{self} to get the spatial dimensions of the domain of the 194 PDE. As we will make use of the \class{Helmholtz} class several times, it is convenient to 195 put its definition into a file which we name \file{mytools.py} available in the \ExampleDirectory. 196 You can use your favourite editor to create and edit the file. 197 198 An object of the \class{Helmholtz} class is created through the statments: 199 \begin{python} 200 from mytools import Helmholtz 201 mypde = Helmholtz(mydomain) 202 mypde.setValue(kappa=10.,omega=0.1,f=12) 203 u = mypde.getSolution() 204 \end{python} 205 In the first statement we import all definition from the \file{mytools.py} \index{scripts!\file{mytools.py}}. Make sure 206 that \file{mytools.py} is in the directory from where you started Python. 207 \var{mydomain} is the \Domain of the PDE which we have undefined here. In the third statment values are 208 assigned to the PDE parameters. As no values for arguments \var{eta} and \var{g} are 209 specified, the default values $0$ are used. \footnote{It would be better to use the default value 210 \var{escript.Data()} rather then $0$ as then the coefficient would be defined as being not present and 211 would not be processed when the PDE is evaluated}. In the fourth statement the solution of the 212 PDE is returned. 213 214 To test our \class{Helmholtz} class on a rectangular domain 215 of length $l\hackscore{0}=5$ and height $l\hackscore{1}=1$, we choose a simple test solution. Actually, we 216 we take $u=x\hackscore{0}$ and then calculate the right hand side terms $f$ and $g$ such that 217 the test solution becomes the solution of the problem. If we assume $\kappa$ as being constant, 218 an easy calculation shows that we have to choose $f=\omega \cdot x\hackscore{0}$. On the boundary we get 219 $\kappa n\hackscore{i} u\hackscore{,i}=\kappa n\hackscore{0}$. 220 So we have to set $g=\kappa n\hackscore{0}+\eta x\hackscore{0}$. The following script \file{helmholtztest.py} 221 \index{scripts!\file{helmholtztest.py}} which is available in the \ExampleDirectory 222 implements this test problem using the \finley PDE solver: 223 \begin{python} 224 from mytools import Helmholtz 225 from esys.escript import Lsup 226 from esys.finley import Rectangle 227 #... set some parameters ... 228 kappa=1. 229 omega=0.1 230 eta=10. 231 #... generate domain ... 232 mydomain = Rectangle(l0=5.,l1=1.,n0=50, n1=10) 233 #... open PDE and set coefficients ... 234 mypde=Helmholtz(mydomain) 235 n=mydomain.getNormal() 236 x=mydomain.getX() 237 mypde.setValue(kappa,omega,omega*x[0],eta,kappa*n[0]+eta*x[0]) 238 #... calculate error of the PDE solution ... 239 u=mypde.getSolution() 240 print "error is ",Lsup(u-x[0]) 241 \end{python} 242 The script is similar to the script \file{mypoisson.py} dicussed in \Chap{FirstSteps}. 243 \code{mydomain.getNormal()} returns the outer normal field on the surface of the domain. The function \function{Lsup} 244 is imported by the \code{from esys.escript import Lsup} statement and returns the maximum absulute value of its argument. 245 The error shown by the print statement should be in the order of $10^{-7}$. As piecewise bi-linear interpolation is 246 used to approximate the solution and our solution is a linear function of the spatial coordinates one might 247 expect that the error would be zero or in the order of machine precision (typically $\approx 10^{-15}$). 248 However most PDE packages use an iterative solver which is terminated 249 when a given tolerance has been reached. The default tolerance is $10^{-8}$. This value can be altered by using the 250 \method{setTolerance} of the \LinearPDE class. 251 252 \section{The Transition Problem} 253 \label{DIFFUSION TRANS SEC} 254 Now we are ready to solve the original time dependent problem. The main 255 part of the script is the loop over time $t$ which takes the following form: 256 \begin{python} 257 t=0 258 T=Tref 259 mypde=Helmholtz(mydomain) 260 while t

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