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\chapter{How to Solve The Diffusion Equation} |
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\label{DIFFUSION CHAP} |
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\begin{figure} |
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\centerline{\includegraphics[width=\figwidth]{DiffusionDomain}} |
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\caption{Temperature Diffusion Problem with Circular Heat Source} |
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\label{DIFFUSION FIG 1} |
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\end{figure} |
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\begin{figure} |
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\centerline{\includegraphics[width=\figwidth]{DiffusionRes1}} |
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\centerline{\includegraphics[width=\figwidth]{DiffusionRes16}} |
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\centerline{\includegraphics[width=\figwidth]{DiffusionRes32}} |
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\centerline{\includegraphics[width=\figwidth]{DiffusionRes48}} |
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\caption{Results of the Temperture Diffusion Problem for Time Steps $1$ $16$, $32$ and $48$.} |
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\label{DIFFUSION FIG 2} |
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\end{figure} |
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\section{\label{DIFFUSION OUT SEC}Outline} |
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In this chapter we will discuss how to solve the time depeneded-temperature diffusion\index{diffusion equation} within |
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a block of material. Within the block there is a heat source which drives the temperature diffusion. |
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On the surface energy can radiate into the surrounding environment. |
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\fig{DIFFUSION FIG 1} shows the configuration. |
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In the next \Sec{DIFFUSION TEMP SEC} we will present the relevant model. A |
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time integration scheme is introduced to calculate the temperature at given time nodes $t^{(n)}$. |
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We will see that at time step a so-called Helmholtz equation \index{Helmholtz equation} has to be solved. |
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The implementation iof a Helmholtz equation solver will be discussed in \Sec{DIFFUSION HELM SEC}. |
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In Section~\ref{DIFFUSION TRANS SEC} the solver of the Helmholtz equation is used to build a |
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solver for the temperature diffusion problem. |
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\section{\label{DIFFUSION TEMP SEC}Temperature Diffusion} |
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The temperature $T$ is a function of its location in the domain and time $t>0$. The governing equation |
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in the interior of the domain is given by |
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\begin{equation} |
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\rho c\hackscore p T\hackscore{,t} - (\kappa T\hackscore{,i})\hackscore{,i} = q |
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\label{DIFFUSION TEMP EQ 1} |
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\end{equation} |
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where $\rho c\hackscore p$ and $\kappa$ are given material constants. In case of a composite |
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material the parameters are depending on the their location in the domain. $q$ is |
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a heat source (or sink) within the domain. We are using Einstein summation convention \index{summation convention} |
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as introduced in \Chap{FirstSteps}. In our case we assume $q$ to be equal to a constant $q^{c}$ |
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on a circle or sphere with center $x^c$ and radius $r$ and $0$ elsewhere: |
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\begin{equation} |
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q(x,t)= |
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\left\{ |
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\begin{array}{lcl} |
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q^c & & \|x-x^c\| \le r \\ |
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& \mbox{if} \\ |
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0 & & \mbox{else} \\ |
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\end{array} |
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\right. |
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\label{DIFFUSION TEMP EQ 1b} |
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\end{equation} |
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for all $x$ in the domain and all time $t>0$. |
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On the surface of the domain we are |
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are specifying a radiation condition |
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which precribes the normal component of the flux $J\hackscore i= \kappa T\hackscore{,i}$ to be proportional |
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to the difference of the current temperature to the surrounding temperature $T\hackscore{ref}$: |
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\begin{equation} |
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\kappa T\hackscore{,i} n\hackscore i = \eta (T\hackscore{ref}-T) |
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\label{DIFFUSION TEMP EQ 2} |
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\end{equation} |
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$\eta$ is a given material coefficient depending on the material and the surrounding medium. |
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As usual $n_i$ is the $i$-th component of the outer normal field \index{outer normal field} |
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at the surface of the domain. |
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To solve the the time depended \eqn{DIFFUSION TEMP EQ 1} the initial temperature at time |
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$t=0$ has to be given. Here we assume that the initial temperature is the surrounding temperature: |
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\begin{equation} |
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T(x,0)=T\hackscore{ref} |
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\label{DIFFUSION TEMP EQ 4} |
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\end{equation} |
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for all $x$ in the domain. It is pointed out that |
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the initial conditions is fullfilling the |
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boundary condition defined by \eqn{DIFFUSION TEMP EQ 2}. |
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The temperature is calculated discrete time nodes $t^{(n)}$ where |
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$t^{(0)}=0$ and $t^{(n)}=t^{(n-1)}+h$ where $h>0$ is the step size which is assumed to be constant. |
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In the following the upper index ${(n)}$ is refering to a value at time $t^{(n)}$. The simplest |
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and most robust scheme to approximate the time derivative of the the temperature is the |
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\index{backward Euler} scheme, see~\cite{XXX} for alternatives. The backward Euler scheme bases |
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on the Taylor expansion |
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\begin{equation} |
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T^{(n-1)}\approx T^{(n)}+T\hackscore{,t}^{(n)}(t^{(n-1)}-t^{(n)}) |
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=T^{(n-1)} + h \cdot T\hackscore{,t}^{(n)} |
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\label{DIFFUSION TEMP EQ 6} |
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\end{equation} |
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which is inserted into \eqn{DIFFUSION TEMP EQ 1}. By separating the terms at |
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$t^{(n)}$ and $t^{(n-1)}$ one gets for $n=1,2,3\ldots$ |
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\begin{equation} |
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\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)} |
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\label{DIFFUSION TEMP EQ 7} |
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\end{equation} |
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where $T^{(0)}=T\hackscore{ref}$ is taken form the initial condition given by \eqn{DIFFUSION TEMP EQ 4}. |
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Together with the natural boundary condition from \eqn{DIFFUSION TEMP EQ 2}. |
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this forms a boundary value problem that has to be solved for each time step. |
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As a first step to implement a solver for the temperature diffusion problem we will |
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first implement a solver for the boundary value problem that has to be solved at each time step. |
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\section{\label{DIFFUSION HELM SEC}Helmholtz Problem} |
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The partial differential equation to be solved for $T^{(n)}$ has the form |
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\begin{equation} |
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\omega u - (\kappa u\hackscore{,i})\hackscore{,i} = f |
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\label{DIFFUSION HELM EQ 1} |
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\end{equation} |
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where $u$ plays the role of $T^{(n)}$ and we set |
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\begin{equation} |
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\omega=\frac{\rho c\hackscore p}{h} \mbox{ and } f=q+\frac{\rho c\hackscore p}{h}T^{(n-1)} \;. |
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\label{DIFFUSION HELM EQ 1b} |
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\end{equation} |
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With $g=\eta T\hackscore{ref}$ the radiation condition defined by \eqn{DIFFUSION TEMP EQ 2} |
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takes the form |
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\begin{equation} |
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\kappa u\hackscore{,i} n\hackscore{i} = g - \eta u\mbox{ on } \Gamma |
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\label{DIFFUSION HELM EQ 2} |
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\end{equation} |
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The partial differential |
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\eqn{DIFFUSION HELM EQ 1} together with boundary conditions of \eqn{DIFFUSION HELM EQ 2} |
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is called a Helmholtz equation \index{Helmholtz equation}. |
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We want to use the \LinearPDE class provided by \escript to define and solve a general linear PDE such as the |
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Helmholtz equation. We have used a special case of the \LinearPDE class, namely the |
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\Poisson class already in \Chap{FirstSteps}. |
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Here we will write our own specialized class of the \LinearPDE to solve the Helmholtz equation. |
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The general form of a single PDE that can be handeled by the \LinearPDE class is |
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\begin{equation}\label{EQU.FEM.1} |
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-(A\hackscore{jl} u\hackscore{,l})\hackscore{,j}+D u =Y \; . |
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\end{equation} |
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The general form and systems is discussed in \Sec{SEC LinearPDE}. |
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$A$, $D$ and $Y$ are the known coeffecients of the PDE \index{partial differential equation!coefficients}. |
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Notice that $A$ is a matrix or tensor of order 2 and $D$ and $Y$ are scalar. |
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They may be constant or may depend on their |
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location in the domain but must not depend on the unknown solution $u$. |
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The following natural boundary conditions \index{boundary condition!natural} that |
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are used in the \LinearPDE class have the form |
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\begin{equation}\label{EQU.FEM.2} |
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n\hackscore{j}A\hackscore{jl} u\hackscore{,l}+du=y \;. |
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\end{equation} |
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where, as usual, $n$ denotes the outer normal field on the surface of the domain. Notice that |
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the coefficient $A$ is already used in the PDE in \eqn{EQU.FEM.1}. $d$ and $y$ are given scalar coefficients. |
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By inpecting the Helmholtz equation \index{Helmholtz equation} |
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we can easily assign values to the coefficients in the |
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general PDE of the \LinearPDE class: |
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\begin{equation}\label{DIFFUSION HELM EQ 3} |
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\begin{array}{llllll} |
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A\hackscore{ij}=\kappa \delta\hackscore{ij} & D=\omega & Y=f \\ |
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d=\eta & y= g & \\ |
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\end{array} |
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\end{equation} |
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$\delta\hackscore{ij}$ is the Kronecker symbol \index{Kronecker symbol} defined by $\delta=\hackscore{ij}=1$ for |
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$i=j$ and $0$ otherwise. |
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We want to implement a |
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new class which we will call \class{Helmholtz} that provides the same methods like the \LinearPDE class but |
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is defined through the coefficeints $\kappa$, $\omega$, $f$, $\eta$, |
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$g$ rather than the general form given by \eqn{EQU.FEM.1}. |
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Python's |
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mechanism of inhertence allows doing this in a very easy way. |
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The advantage is that our new \class{Helmholtz} can be used in any context |
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that works with a \LinearPDE class but with an easier interface to define the PDE. |
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This improves reuasablity as well as maintainability of program codes. |
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We want to implement a |
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new class which we will call \class{Helmholtz} that provides the same methods like the \LinearPDE class but |
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is defined through the coefficeints $\kappa$, $\omega$, $f$, $\alpha$, |
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$g$ rather than the general form given by \eqn{EQU.FEM.1}. |
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Python's mechanism of subclasses allows doing this in a very easy way. |
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The \Poisson class of the \linearPDEsPack module, |
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which we have already used in \Chap{FirstSteps}, is in fact a subclass of the |
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\LinearPDE class. That means that it all methods (such as the \method{getSolution}) |
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from the parent class \LinearPDE are defined for any \Poisson object. However, new |
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methods can be added and methods of the parent class can be redefined. In fact, |
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the \Poisson class redefines the \method{setValue} of the \LinearPDE class which is called to assign |
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values to the coefficients of the PDE. This is exactly what we will do when we define |
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our new \class{Helmholtz} class: |
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\begin{python} |
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from esys.linearPDEs import LinearPDE |
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import numarray |
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class Helmholtz(LinearPDE) |
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def setValue(self,kappa=0,omega=1,f=0,eta=0,g=0) |
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self._setValue(A=kappa*numarray.identity(self.getDim()),D=omega,Y=f,d=eta,y=g) |
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\end{python} |
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\code{class Helmholtz(linearPDE)} declares the new \class{Helmholtz} class as a subclass |
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of the \LinearPDE which have imported in the first line of the script. |
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We add the method \method{setValue} to the class which overwrites the |
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\method{setValue} method of the \LinearPDE class. The new methods which has the |
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parameters of the Helmholtz \eqn{DIFFUSION HELM EQ 1} as arguments |
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maps the parameters of the coefficients of the general PDE defined |
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in \eqn{EQU.FEM.1}. The coefficient \var{A} is defined by Kroneckers symbol. we use the |
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\numarray function \function{identity} which return a square matrix which has ones on the |
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main diagonal and zeros out side the main diagonal. The argument of \function{identity} gives the order of the matrix. |
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Here we use |
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the \method{getDim} of the \LinearPDE class object \var{self} to get the spatial dimension of the domain of the |
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PDE. As we will make use of the \class{Helmholtz} class several times, it is convient to |
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put is definition into a file which we name \file{mytools.py} available in the \ExampleDirectory. |
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You can use your favourite editor to create and edit the file. |
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An object of the \class{Helmholtz} class is created through the statments: |
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\begin{python} |
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from mytools import * |
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mypde=Helmholtz(mydomain) |
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mypde.setValue(kappa=10.,omega=0.1,f=12) |
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u=mypde.getSolution() |
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\end{python} |
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In the first statement we import all definition from the \file{mytools.py} \index{scripts!\file{mytools.py}}. Make sure |
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that \file{mytools.py} is in directory from where you have started Python. |
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\var{mydomain} is the \Domain of the PDE. In the third statment values are |
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assigned to the PDE parameters. As no values for arguments \var{eta} and \var{g} are |
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specified the default values $0$ are used \footnote{It would be better to use the default value |
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\var{escript.Data()} rather then $0$ as then the coefficient would be defined as being not present and |
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would not be processed when the PDE is evaluated.}. In the forth statement the solution of the |
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PDE is returned. |
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We want to test our \class{Helmholtz} class on a rectangular domain |
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of length $l$ and height $h$. We do this by choosing a simple test solution, |
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here we take $u=x\hackscore{0}$ and then calculate the right hand side terms $f$ and $g$ such that |
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the test solution becomes the solution of the problem. If we take $\kappa=1$ |
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an easy calculation shows that we have to choose $f=\omega \cdot x\hackscore{0}$. On the boundary we get |
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$\kappa n\hackscore{i} u\hackscore{,i}=n\hackscore{0}$. |
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So we have to set $g=n\hackscore{0}+\eta x\hackscore{0}$. The following script \file{helmholtztest.py} |
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\index{scripts!\file{helmholtztest.py}} which is available in the \ExampleDirectory |
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implements this test problem using the \finley PDE solver: |
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\begin{python} |
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from mytools import * |
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from esys.escript import * |
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import esys.finley |
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#... set some parameters ... |
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omega=0.1 |
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eta=10. |
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#... generate domain ... |
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mydomain = esys.finley.Rectangle(l0=5.,l1=1.,n0=50, n1=10) |
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#... open PDE and set coefficients ... |
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mypde=Helmholtz(mydomain) |
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n=mydomain.getNormal() |
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x=mydomain.getX() |
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mypde.setValue(1,omega,omega*x[0],eta,n[0]+eta*x[0]) |
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#... calculate error of the PDE solution ... |
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u=mypde.getSolution() |
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print "error is ",Lsup(u-x[0]) |
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\end{python} |
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The script is similar to the script \file{mypoisson.py} dicussed in \Chap{FirstSteps}. |
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\code{mydomain.getNormal()} returns the outer normal field on the surface of the domain. The function \function{Lsup} |
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is imported by the \code{from escript import *} statement and returns the maximum absulute value of it argument. To run |
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The error shown by the print statement should be in the order of $10^{-7}$. As piecewise linear interpolation is |
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used to approximate the solution and our solution is a linear function of the spatial coordinates one may |
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expect that the error is zero. However, as most PDE packages uses an iterative solver which is terminated |
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when a given tolerance has been reached. The default tolerance is $10^{-8}$. Thsi value can be altered by using the |
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\method{setTolerance} of the \LinearPDE class. |
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\section{The Transition Problem} |
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\label{DIFFUSION TRANS SEC} |
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Now we are ready to solve the original time dependent problem. The main |
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part of the script is the loop over the time $t$ which takes the following form: |
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\begin{python} |
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mypde=Helmholtz(mydomain) |
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while t<t_end: |
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mypde.setValue(kappa,rhocp/h,q+rhocp/h*T,eta,eta*Tref) |
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T=mypde.getSolution() |
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t+=h |
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\end{python} |
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\var{kappa}, \var{rhocp}, \var{eta} and \var{Tref} are input parameters of the model. \var{q} is the heat source |
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in the domain and \var{h} is the time step size which has to be chosen. Notice that the \class{Hemholtz} |
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is created before the loop over time is entered while in each time step only the coefficients |
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are reset in each time step. This way some information about the reperesentation of the PDE can be reused |
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\footnote{The efficience can be improved further by setting the coefficients in the operator |
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\var{kappa}, \var{omega} and \var{eta} before entering the \code{while}-loop and only update the coefficients |
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in the right hand side \var{f} and \var{g}. This needs a more careful implementation of the \method{setValue} |
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method but gives the advantage that the \LinearPDE class can save rebuilding the PDE operator}. The variable \var{T} |
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holds the current temperature. It is used to calculate the right hand side coefficient \var{f} in the |
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Helmholtz \eqn{DIFFUSION HELM EQ 1}. Statement \code{T=mypde.getSolution()} overwrites \var{T} with the |
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temperature of the new time step $\var{t}+\var{h}$. To get this iterative process going we need to sepcify the |
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initial temperature distribution, which equal to $T\hackscore{ref}$. |
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The heat source \var{q} which is defined in \eqn{DIFFUSION TEMP EQ 1b} shall be \var{q0} |
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at an area defined as a circle of radius \var{r} and center \var{xc} and zero outside this circle. |
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\var{q0} is a fixed constant. The following script defines \var{q} as desired: |
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\begin{python} |
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xc=[0.02,0.002] |
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r=0.001 |
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x=mydomain.getX() |
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q=q0*(length(x-xc)-r).whereNegative() |
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\end{python} |
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\var{x} is a \Data class object of |
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the \escript module defining the locations of points in the \Domain \var{mydomain}. |
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\code{length(x-xc)} calculates the distances in the Euclidean norm |
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of the locations \var{x} to the center of the circle \var{xc} where the heat source is acting. |
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Notice that the coordinates of \var{xc} are defined as a list of floating point numbers. It is independently |
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converted into a \Data class object before subtracted from \var{x}. The method \method{whereNegative} of |
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a \Data class object, in this case the result of the expression |
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\code{length(x-xc)-r}, returns a \Data class which is equal one where the object negative and |
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zero elsewhere. After multiplication with \var{q0} we get a function with the deired property. |
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|
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Now we can put the components together to the script \file{diffusion.py} which is available in the \ExampleDirectory: |
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\index{scripts!\file{diffusion.py}}: |
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\begin{python} |
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from mytools import * |
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from esys.escript import * |
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import esys.finley |
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#... set some parameters ... |
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x_c=[0.02,0.002] |
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r=0.001 |
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q0=50.e6 |
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Tref=0. |
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rhocp=2.6e6 |
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eta=75. |
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kappa=240. |
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t_end=5. |
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# ...time step size and counter ... |
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h=0.1 |
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i=0 |
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t=0 |
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#... generate domain ... |
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mydomain = esys.finley.Rectangle(l0=0.05,l1=0.01,n0=250, n1=50) |
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#... open PDE ... |
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mypde=Helmholtz(mydomain) |
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# ... set heat source: .... |
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x=mydomain.getX() |
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q=q0*(length(x-x_c)-r).whereNegative() |
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# ... set initial temperature .... |
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T=Tref |
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# ... start iteration: |
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while t<t_end: |
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i+=1 |
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t+=h |
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print "time step :",t |
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mypde.setValue(kappa=kappa,omega=rhocp/h,f=q+rhocp/h*T,eta=eta,g=eta*Tref) |
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T=mypde.getSolution() |
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T.saveDX("T%d.dx"%i) |
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\end{python} |
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The script will create the files \file{T.1.dx}, |
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\file{T.2.dx}, $\ldots$, \file{T.50.dx} in the directory where the script has been started. The files give the |
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temperature distributions at time steps $1$, $2$, $\ldots$, $50$ in the \OpenDX file format. |
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An easy way to visualize the results is the command |
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\begin{verbatim} |
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dx -edit diffusion.net |
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\end{verbatim} |
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where \file{diffusion.net} is an \OpenDX script available in the \ExampleDirectory. |
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\fig{DIFFUSION FIG 2} shows the result for some selected time steps. |
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