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% $Id$ |
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\section{The Diffusion Problem} |
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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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\subsection{\label{DIFFUSION OUT SEC}Outline} |
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In this chapter we will discuss how to solve the time dependent-temperature diffusion\index{diffusion equation} for |
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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 each time step a Helmholtz equation \index{Helmholtz equation} |
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must be solved. |
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The implementation of 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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\subsection{\label{DIFFUSION TEMP SEC}Temperature Diffusion} |
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The unknown 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 depend on 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 heat production rate |
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$q^{c}$ 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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specifying a radiation condition |
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which precribes the normal component of the flux $\kappa T\hackscore{,i}$ 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 of the block and the surrounding medium. |
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As usual $n\hackscore i$ 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 time dependent \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 satisfy the |
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boundary condition defined by \eqn{DIFFUSION TEMP EQ 2}. |
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The temperature is calculated at 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 ${(n)}$ refers 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 |
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the backward Euler |
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\index{backward Euler} scheme, see~\cite{XXX} for alternatives. The backward Euler |
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scheme is based |
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on the Taylor expansion of $T$ at time $t^{(n)}$: |
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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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This 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 |
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\begin{equation} |
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\kappa T\hackscore{,i}^{(n)} n\hackscore i = \eta (T\hackscore{ref}-T^{(n)}) |
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\label{DIFFUSION TEMP EQ 2222} |
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\end{equation} |
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taken 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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\subsection{\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 2222} |
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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 \eqn{DIFFUSION HELM EQ 1} together with boundary conditions of \eqn{DIFFUSION HELM EQ 2} |
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is called the Helmholtz equation \index{Helmholtz equation}. |
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|
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We want to use the \LinearPDE class provided by \escript to define and solve a general linear,steady, second order PDE such as the |
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Helmholtz equation. For a single PDE the \LinearPDE class supports the following form: |
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\begin{equation}\label{LINEARPDE.SINGLE.1} |
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-(A\hackscore{jl} u\hackscore{,l})\hackscore{,j}+(B\hackscore{j} u)\hackscore{,j}+C\hackscore{l} u\hackscore{,l}+D u =-X\hackscore{j,j}+Y \; . |
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\end{equation} |
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The coefficients $A$, $B$, $C$, $D$, $X$ and $Y$ have to be specified through \Data objects in the |
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\Function on the PDE or objects that can be converted into such \Data objects. |
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$A$ is a \RankTwo, $B$, $C$ and $X$ are \RankOne and $D$ and $Y$ are scalar. |
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The following natural |
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boundary conditions are considered \index{boundary condition!natural} on $\Gamma$: |
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\begin{equation}\label{LINEARPDE.SINGLE.2} |
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n\hackscore{j}(A\hackscore{jl} u\hackscore{,l}+B\hackscore{j} u)+d u=n\hackscore{j}X\hackscore{j} + y \;. |
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\end{equation} |
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Notice that the coefficients $A$, $B$ and $X$ are the same like in the PDE~\eqn{LINEARPDE.SINGLE.1}. The coefficients $d$ and $y$ are |
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each a \Scalar in the \FunctionOnBoundary. Constraints \index{constraint} for the solution prescribing the value of the |
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solution at certain locations in the domain. They have the form |
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\begin{equation}\label{LINEARPDE.SINGLE.3} |
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u=r \mbox{ where } q>0 |
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\end{equation} |
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$r$ and $q$ are each \Scalar where $q$ is the characteristic function |
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\index{characteristic function} defining where the constraint is applied. |
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The constraints defined by \eqn{LINEARPDE.SINGLE.3} override any other condition set by \eqn{LINEARPDE.SINGLE.1} |
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or \eqn{LINEARPDE.SINGLE.2}. |
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The \Poisson class of the \linearPDEs module, |
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which we have already used in \Chap{FirstSteps}, is in fact a subclass of the more general |
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\LinearPDE class. The \linearPDEs module provides a \Helmholtz class but |
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we will make direct use of the general \LinearPDE class. |
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By inspecting 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. Undefined coefficients are assumed to be not present\footnote{There is a difference |
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in \escript of being not present and set to zero. As not present coefficients are not processed, |
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it is more efficient to leave a coefficient undefined insted assigning zero to it.} |
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Defining and solving the Helmholtz equation is very easy now: |
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\begin{python} |
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from esys.escript import * |
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from linearPDEs import LinearPDE |
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mypde=LinearPDE(mydomain) |
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mypde.setValue(A=kappa*kronecker(mydomain),D=omega,Y=f,d=eta,y=g) |
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u=mypde.getSolution() |
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\end{python} |
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where we assume that \code{mydomain} is a \Domain object and |
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\code{kappa} \code{omega} \code{eta} and \code{g} are given scalar values |
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typically \code{float} or \Data objects. The \method{setValue} method |
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assigns values to the coefficients of the general PDE. The \method{getSolution} method solves |
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the PDE and returns the solution \code{u} of the PDE. \function{kronecker} is \escript function |
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returning the Kronecker symbol. |
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The coefficients can set by several calls of \method{setValue} where the order can be chosen arbitrarily. |
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If a value is assigned to a coefficint several times, the last assigned value is used when |
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the solution is calculated: |
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\begin{python} |
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mypde=LinearPDE(mydomain) |
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mypde.setValue(A=kappa*kronecker(mydomain),d=eta) |
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mypde.setValue(D=omega,Y=f,y=g) |
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mypde.setValue(d=2*eta) # overwrites d=eta |
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u=mypde.getSolution() |
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\end{python} |
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In some cases the solver of the PDE can make use of the fact that the PDE is symmetric\index{symmetric PDE} where the |
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PDE is called symmetric if |
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\begin{equation}\label{LINEARPDE.SINGLE.4} |
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A\hackscore{jl}=A\hackscore{lj} \mbox{ and } B\hackscore{j}=C\hackscore{j} \;. |
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\end{equation} |
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Note that $D$ and $d$ may have any value and the right hand side $X$, $Y$, $y$ as well as the constraints |
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are not relevant. The Helmholtz problem is symmetric. |
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The \LinearPDE class provides the method \method{checkSymmetry} method to check if the given PDE is symmetric. |
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\begin{python} |
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mypde=LinearPDE(mydomain) |
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mypde.setValue(A=kappa*kronecker(mydomain),d=eta) |
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print mypde.checkSymmetry() # returns True |
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mypde.setValue(B=kronecker(mydomain)[0]) |
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print mypde.checkSymmetry() # returns False |
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mypde.setValue(C=kronecker(mydomain)[0]) |
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print mypde.checkSymmetry() # returns True |
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\end{python} |
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Unfortunately, a \method{checkSymmetry} is very expensive and is recommendable to use for |
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testing and debugging purposes only. The \method{setSymmetryOn} method is used to |
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declare a PDE symmetric: |
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\begin{python} |
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mypde = LinearPDE(mydomain) |
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mypde.setValue(A=kappa*kronecker(mydomain)) |
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mypde.setSymmetryOn() |
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\end{python} |
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Now we want to see how we actually solve the Helmholtz equation. |
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on a rectangular domain |
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of length $l\hackscore{0}=5$ and height $l\hackscore{1}=1$. We choose a simple test solution such that we |
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can verify the returned solution against the exact answer. Actually, we |
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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 assume $\kappa$ as being constant, |
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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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$\kappa n\hackscore{i} u\hackscore{,i}=\kappa n\hackscore{0}$. |
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So we have to set $g=\kappa n\hackscore{0}+\eta x\hackscore{0}$. The following script \file{helmholtz.py} |
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\index{scripts!\file{helmholtz.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 esys.escript import * |
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from esys.escript.linearPDEs import LinearPDE |
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from esys.finley import Rectangle |
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#... set some parameters ... |
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kappa=1. |
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omega=0.1 |
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eta=10. |
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#... generate domain ... |
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mydomain = Rectangle(l0=5.,l1=1.,n0=50, n1=10) |
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#... open PDE and set coefficients ... |
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mypde=LinearPDE(mydomain) |
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mypde.setSymmetryOn() |
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n=mydomain.getNormal() |
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x=mydomain.getX() |
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mypde.setValue(A=kappa*kronecker(mydomain),D=omega,Y=omega*x[0], \ |
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d=eta,y=kappa*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{poisson.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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imported by the \code{from esys.escript import *} statement and returns the maximum absulute value of its argument. |
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The error shown by the print statement should be in the order of $10^{-7}$. As piecewise bi-linear interpolation is |
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used by \finley approximate the solution and our solution is a linear function of the spatial coordinates one might |
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expect that the error would be zero or in the order of machine precision (typically $\approx 10^{-15}$). |
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However most PDE packages use an iterative solver which is terminated |
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when a given tolerance has been reached. The default tolerance is $10^{-8}$. This value can be altered by using the |
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\method{setTolerance} of the \LinearPDE class. |
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|
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\subsection{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 time $t$ which takes the following form: |
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\begin{python} |
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t=0 |
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T=Tref |
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mypde=LinearPDE(mydomain) |
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mypde.setValue(A=kappa*kronecker(mydomain),D=rhocp/h,d=eta,y=eta*Tref) |
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while t<t_end: |
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mypde.setValue(Y=q+rhocp/h*T) |
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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. |
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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 equation in \eqn{DIFFUSION HELM EQ 1}. The 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 specify the |
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initial temperature distribution, which equal to $T\hackscore{ref}$. |
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The \LinearPDE class object \var{mypde} |
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and coefficients not changing over time are set up before the loop over time is entered. In each time step only the coefficient |
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$Y$ is reset as it depends on the temperature of the previous time step. This allows the PDE solver to reuse information |
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from previous time steps as much as possible. |
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|
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The heat source \var{q} which is defined in \eqn{DIFFUSION TEMP EQ 1b} is \var{qc} |
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in 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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from esys.escript import length |
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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*whereNegative(length(x-xc)-r) |
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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 locations in the \Domain \var{mydomain}. |
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The \function{length()} imported from the \escript module returns the |
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Euclidean norm: |
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\begin{equation}\label{DIFFUSION HELM EQ 3aba} |
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\|y\|=\sqrt{ |
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y\hackscore{i} |
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y\hackscore{i} |
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} = \function{esys.escript.length}(y) |
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\end{equation} |
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So \code{length(x-xc)} calculates the distances |
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of the location \var{x} to the center of the circle \var{xc} where the heat source is acting. |
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Note 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 being subtracted from \var{x}. The function \function{whereNegative} |
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applied to |
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\code{length(x-xc)-r}, returns a \Data class which is equal to one where the object is negative and |
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zero elsewhere. After multiplication with \var{qc} we get a function with the desired property. |
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|
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Now we can put the components together to create 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 esys.escript import * |
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from esys.escript.linearPDEs import LinearPDE |
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from esys.finley import Rectangle |
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#... set some parameters ... |
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xc=[0.02,0.002] |
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r=0.001 |
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qc=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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tend=5. |
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# ... time, time step size and counter ... |
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t=0 |
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h=0.1 |
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i=0 |
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#... generate domain ... |
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mydomain = Rectangle(l0=0.05,l1=0.01,n0=250, n1=50) |
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#... open PDE ... |
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mypde=LinearPDE(mydomain) |
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mypde.setSymmetryOn() |
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mypde.setValue(A=kappa*kronecker(mydomain),D=rhocp/h,d=eta,y=eta*Tref) |
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# ... set heat source: .... |
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x=mydomain.getX() |
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q=qc*whereNegative(length(x-xc)-r) |
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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<tend: |
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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(Y=q+rhocp/h*T) |
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T=mypde.getSolution() |
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saveVTK("T.%d.xml"%i,temp=T) |
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\end{python} |
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The script will create the files \file{T.1.xml}, |
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\file{T.2.xml}, $\ldots$, \file{T.50.xml} 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 \VTK file format. |
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|
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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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An easy way to visualize the results is the command |
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\begin{verbatim} |
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mayavi -d T.1.xml -m SurfaceMap & |
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\end{verbatim} |
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Use the \texttt{Configure Data} |
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to move forward and and backwards in time. |
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\fig{DIFFUSION FIG 2} shows the result for some selected time steps. |
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