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1 jgs 102 % $Id$
2 jgs 121 \section{The Diffusion Problem}
3 jgs 102 \label{DIFFUSION CHAP}
4    
5     \begin{figure}
6 gross 599 \centerline{\includegraphics[width=\figwidth]{figures/DiffusionDomain.eps}}
7 jgs 102 \caption{Temperature Diffusion Problem with Circular Heat Source}
8     \label{DIFFUSION FIG 1}
9     \end{figure}
10    
11 jgs 121 \subsection{\label{DIFFUSION OUT SEC}Outline}
12 jgs 107 In this chapter we will discuss how to solve the time dependent-temperature diffusion\index{diffusion equation} for
13 jgs 102 a block of material. Within the block there is a heat source which drives the temperature diffusion.
14 jgs 107 On the surface, energy can radiate into the surrounding environment.
15 jgs 102 \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 jgs 107 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 jgs 102 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 jgs 121 \subsection{\label{DIFFUSION TEMP SEC}Temperature Diffusion}
26 jgs 107 The unknown temperature $T$ is a function of its location in the domain and time $t>0$. The governing equation
27 jgs 102 in the interior of the domain is given by
28     \begin{equation}
29 lkettle 575 \rho c\hackscore p T\hackscore{,t} - (\kappa T\hackscore{,i})\hackscore{,i} = q\hackscore H
30 jgs 102 \label{DIFFUSION TEMP EQ 1}
31     \end{equation}
32     where $\rho c\hackscore p$ and $\kappa$ are given material constants. In case of a composite
33 lkettle 575 material the parameters depend on their location in the domain. $q\hackscore H$ is
34 jgs 102 a heat source (or sink) within the domain. We are using Einstein summation convention \index{summation convention}
35 lkettle 575 as introduced in \Chap{FirstSteps}. In our case we assume $q\hackscore H$ to be equal to a constant heat production rate
36 jgs 107 $q^{c}$ on a circle or sphere with center $x^c$ and radius $r$ and $0$ elsewhere:
37 jgs 102 \begin{equation}
38 lkettle 575 q\hackscore H(x,t)=
39 jgs 102 \left\{
40     \begin{array}{lcl}
41     q^c & & \|x-x^c\| \le r \\
42     & \mbox{if} \\
43     0 & & \mbox{else} \\
44     \end{array}
45     \right.
46     \label{DIFFUSION TEMP EQ 1b}
47     \end{equation}
48     for all $x$ in the domain and all time $t>0$.
49    
50     On the surface of the domain we are
51 jgs 107 specifying a radiation condition
52     which precribes the normal component of the flux $\kappa T\hackscore{,i}$ to be proportional
53 jgs 102 to the difference of the current temperature to the surrounding temperature $T\hackscore{ref}$:
54     \begin{equation}
55     \kappa T\hackscore{,i} n\hackscore i = \eta (T\hackscore{ref}-T)
56     \label{DIFFUSION TEMP EQ 2}
57     \end{equation}
58 jgs 107 $\eta$ is a given material coefficient depending on the material of the block and the surrounding medium.
59     As usual $n\hackscore i$ is the $i$-th component of the outer normal field \index{outer normal field}
60 jgs 102 at the surface of the domain.
61    
62 jgs 107 To solve the time dependent \eqn{DIFFUSION TEMP EQ 1} the initial temperature at time
63 jgs 102 $t=0$ has to be given. Here we assume that the initial temperature is the surrounding temperature:
64     \begin{equation}
65     T(x,0)=T\hackscore{ref}
66     \label{DIFFUSION TEMP EQ 4}
67     \end{equation}
68     for all $x$ in the domain. It is pointed out that
69 jgs 107 the initial conditions satisfy the
70 jgs 102 boundary condition defined by \eqn{DIFFUSION TEMP EQ 2}.
71    
72 jgs 107 The temperature is calculated at discrete time nodes $t^{(n)}$ where
73 jgs 102 $t^{(0)}=0$ and $t^{(n)}=t^{(n-1)}+h$ where $h>0$ is the step size which is assumed to be constant.
74 jgs 107 In the following the upper index ${(n)}$ refers to a value at time $t^{(n)}$. The simplest
75     and most robust scheme to approximate the time derivative of the the temperature is
76     the backward Euler
77     \index{backward Euler} scheme, see~\cite{XXX} for alternatives. The backward Euler
78     scheme is based
79     on the Taylor expansion of $T$ at time $t^{(n)}$:
80 jgs 102 \begin{equation}
81 lkettle 573 T^{(n)}\approx T^{(n-1)}+T\hackscore{,t}^{(n)}(t^{(n)}-t^{(n-1)})
82     =T^{(n-1)} + h \cdot T\hackscore{,t}^{(n)}
83 jgs 102 \label{DIFFUSION TEMP EQ 6}
84     \end{equation}
85 jgs 107 This is inserted into \eqn{DIFFUSION TEMP EQ 1}. By separating the terms at
86 jgs 102 $t^{(n)}$ and $t^{(n-1)}$ one gets for $n=1,2,3\ldots$
87     \begin{equation}
88 lkettle 575 \frac{\rho c\hackscore p}{h} T^{(n)} - (\kappa T^{(n)}\hackscore{,i})\hackscore{,i} = q\hackscore H + \frac{\rho c\hackscore p}{h} T^{(n-1)}
89 jgs 102 \label{DIFFUSION TEMP EQ 7}
90     \end{equation}
91     where $T^{(0)}=T\hackscore{ref}$ is taken form the initial condition given by \eqn{DIFFUSION TEMP EQ 4}.
92 jgs 107 Together with the natural boundary condition
93     \begin{equation}
94     \kappa T\hackscore{,i}^{(n)} n\hackscore i = \eta (T\hackscore{ref}-T^{(n)})
95     \label{DIFFUSION TEMP EQ 2222}
96     \end{equation}
97     taken from \eqn{DIFFUSION TEMP EQ 2}
98 jgs 102 this forms a boundary value problem that has to be solved for each time step.
99     As a first step to implement a solver for the temperature diffusion problem we will
100     first implement a solver for the boundary value problem that has to be solved at each time step.
101    
102 jgs 121 \subsection{\label{DIFFUSION HELM SEC}Helmholtz Problem}
103 jgs 102 The partial differential equation to be solved for $T^{(n)}$ has the form
104     \begin{equation}
105 lkettle 575 \omega T^{(n)} - (\kappa T^{(n)}\hackscore{,i})\hackscore{,i} = f
106 jgs 102 \label{DIFFUSION HELM EQ 1}
107     \end{equation}
108 lkettle 575 and we set
109 jgs 102 \begin{equation}
110 lkettle 575 \omega=\frac{\rho c\hackscore p}{h} \mbox{ and } f=q\hackscore H +\frac{\rho c\hackscore p}{h}T^{(n-1)} \;.
111 jgs 102 \label{DIFFUSION HELM EQ 1b}
112     \end{equation}
113 jgs 107 With $g=\eta T\hackscore{ref}$ the radiation condition defined by \eqn{DIFFUSION TEMP EQ 2222}
114 jgs 102 takes the form
115     \begin{equation}
116 lkettle 575 \kappa T^{(n)}\hackscore{,i} n\hackscore{i} = g - \eta T^{(n)}\mbox{ on } \Gamma
117 jgs 102 \label{DIFFUSION HELM EQ 2}
118     \end{equation}
119 gross 568 The partial differential \eqn{DIFFUSION HELM EQ 1} together with boundary conditions of \eqn{DIFFUSION HELM EQ 2}
120 jgs 107 is called the Helmholtz equation \index{Helmholtz equation}.
121 jgs 102
122 gross 568 We want to use the \LinearPDE class provided by \escript to define and solve a general linear,steady, second order PDE such as the
123     Helmholtz equation. For a single PDE the \LinearPDE class supports the following form:
124     \begin{equation}\label{LINEARPDE.SINGLE.1}
125     -(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 \; .
126 jgs 102 \end{equation}
127 gross 568 The coefficients $A$, $B$, $C$, $D$, $X$ and $Y$ have to be specified through \Data objects in the
128     \Function on the PDE or objects that can be converted into such \Data objects.
129     $A$ is a \RankTwo, $B$, $C$ and $X$ are \RankOne and $D$ and $Y$ are scalar.
130     The following natural
131     boundary conditions are considered \index{boundary condition!natural} on $\Gamma$:
132     \begin{equation}\label{LINEARPDE.SINGLE.2}
133     n\hackscore{j}(A\hackscore{jl} u\hackscore{,l}+B\hackscore{j} u)+d u=n\hackscore{j}X\hackscore{j} + y \;.
134 jgs 102 \end{equation}
135 gross 568 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
136     each a \Scalar in the \FunctionOnBoundary. Constraints \index{constraint} for the solution prescribing the value of the
137     solution at certain locations in the domain. They have the form
138     \begin{equation}\label{LINEARPDE.SINGLE.3}
139     u=r \mbox{ where } q>0
140     \end{equation}
141     $r$ and $q$ are each \Scalar where $q$ is the characteristic function
142     \index{characteristic function} defining where the constraint is applied.
143     The constraints defined by \eqn{LINEARPDE.SINGLE.3} override any other condition set by \eqn{LINEARPDE.SINGLE.1}
144     or \eqn{LINEARPDE.SINGLE.2}.
145     The \Poisson class of the \linearPDEs module,
146     which we have already used in \Chap{FirstSteps}, is in fact a subclass of the more general
147     \LinearPDE class. The \linearPDEs module provides a \Helmholtz class but
148     we will make direct use of the general \LinearPDE class.
149 jgs 102
150 jgs 107 By inspecting the Helmholtz equation \index{Helmholtz equation}
151 lkettle 575 (\ref{DIFFUSION HELM EQ 1}) and boundary condition (\ref{DIFFUSION HELM EQ 2}) and
152     substituting $u$ for $T^{(n)}$
153 jgs 102 we can easily assign values to the coefficients in the
154     general PDE of the \LinearPDE class:
155     \begin{equation}\label{DIFFUSION HELM EQ 3}
156     \begin{array}{llllll}
157     A\hackscore{ij}=\kappa \delta\hackscore{ij} & D=\omega & Y=f \\
158     d=\eta & y= g & \\
159     \end{array}
160     \end{equation}
161 jgs 107 $\delta\hackscore{ij}$ is the Kronecker symbol \index{Kronecker symbol} defined by $\delta\hackscore{ij}=1$ for
162 lkettle 575 $i=j$ and $0$ otherwise. Undefined coefficients are assumed to be not present.\footnote{There is a difference
163 gross 568 in \escript of being not present and set to zero. As not present coefficients are not processed,
164 lkettle 575 it is more efficient to leave a coefficient undefined instead of assigning zero to it.}
165     In this diffusion example we do not need to define a characteristic function $q$ because the
166     boundary conditions we consider in \eqn{DIFFUSION HELM EQ 2} are just the natural boundary
167     conditions which are already defined in the \LinearPDE class (shown in \eqn{LINEARPDE.SINGLE.2}).
168 jgs 102
169 gross 568 Defining and solving the Helmholtz equation is very easy now:
170     \begin{python}
171     from esys.escript import *
172     from linearPDEs import LinearPDE
173     mypde=LinearPDE(mydomain)
174     mypde.setValue(A=kappa*kronecker(mydomain),D=omega,Y=f,d=eta,y=g)
175     u=mypde.getSolution()
176     \end{python}
177     where we assume that \code{mydomain} is a \Domain object and
178     \code{kappa} \code{omega} \code{eta} and \code{g} are given scalar values
179     typically \code{float} or \Data objects. The \method{setValue} method
180     assigns values to the coefficients of the general PDE. The \method{getSolution} method solves
181 gross 569 the PDE and returns the solution \code{u} of the PDE. \function{kronecker} is \escript function
182 gross 568 returning the Kronecker symbol.
183    
184     The coefficients can set by several calls of \method{setValue} where the order can be chosen arbitrarily.
185 lkettle 573 If a value is assigned to a coefficient several times, the last assigned value is used when
186 gross 568 the solution is calculated:
187     \begin{python}
188     mypde=LinearPDE(mydomain)
189     mypde.setValue(A=kappa*kronecker(mydomain),d=eta)
190     mypde.setValue(D=omega,Y=f,y=g)
191     mypde.setValue(d=2*eta) # overwrites d=eta
192     u=mypde.getSolution()
193     \end{python}
194     In some cases the solver of the PDE can make use of the fact that the PDE is symmetric\index{symmetric PDE} where the
195     PDE is called symmetric if
196     \begin{equation}\label{LINEARPDE.SINGLE.4}
197     A\hackscore{jl}=A\hackscore{lj} \mbox{ and } B\hackscore{j}=C\hackscore{j} \;.
198     \end{equation}
199     Note that $D$ and $d$ may have any value and the right hand side $X$, $Y$, $y$ as well as the constraints
200     are not relevant. The Helmholtz problem is symmetric.
201 gross 569 The \LinearPDE class provides the method \method{checkSymmetry} method to check if the given PDE is symmetric.
202 gross 568 \begin{python}
203     mypde=LinearPDE(mydomain)
204     mypde.setValue(A=kappa*kronecker(mydomain),d=eta)
205     print mypde.checkSymmetry() # returns True
206     mypde.setValue(B=kronecker(mydomain)[0])
207     print mypde.checkSymmetry() # returns False
208     mypde.setValue(C=kronecker(mydomain)[0])
209     print mypde.checkSymmetry() # returns True
210     \end{python}
211 gross 569 Unfortunately, a \method{checkSymmetry} is very expensive and is recommendable to use for
212     testing and debugging purposes only. The \method{setSymmetryOn} method is used to
213 gross 568 declare a PDE symmetric:
214     \begin{python}
215     mypde = LinearPDE(mydomain)
216     mypde.setValue(A=kappa*kronecker(mydomain))
217     mypde.setSymmetryOn()
218     \end{python}
219 gross 569 Now we want to see how we actually solve the Helmholtz equation.
220     on a rectangular domain
221     of length $l\hackscore{0}=5$ and height $l\hackscore{1}=1$. We choose a simple test solution such that we
222     can verify the returned solution against the exact answer. Actually, we
223 lkettle 575 take $T=x\hackscore{0}$ (here $q\hackscore H = 0$) and then calculate the right hand side terms $f$ and $g$ such that
224 jgs 107 the test solution becomes the solution of the problem. If we assume $\kappa$ as being constant,
225 jgs 102 an easy calculation shows that we have to choose $f=\omega \cdot x\hackscore{0}$. On the boundary we get
226 jgs 107 $\kappa n\hackscore{i} u\hackscore{,i}=\kappa n\hackscore{0}$.
227 gross 569 So we have to set $g=\kappa n\hackscore{0}+\eta x\hackscore{0}$. The following script \file{helmholtz.py}
228 gross 568 \index{scripts!\file{helmholtz.py}} which is available in the \ExampleDirectory
229 jgs 102 implements this test problem using the \finley PDE solver:
230     \begin{python}
231 gross 568 from esys.escript import *
232 gross 569 from esys.escript.linearPDEs import LinearPDE
233 jgs 107 from esys.finley import Rectangle
234 jgs 102 #... set some parameters ...
235 jgs 107 kappa=1.
236 jgs 102 omega=0.1
237     eta=10.
238     #... generate domain ...
239 jgs 107 mydomain = Rectangle(l0=5.,l1=1.,n0=50, n1=10)
240 jgs 102 #... open PDE and set coefficients ...
241 gross 568 mypde=LinearPDE(mydomain)
242     mypde.setSymmetryOn()
243 jgs 102 n=mydomain.getNormal()
244     x=mydomain.getX()
245 gross 569 mypde.setValue(A=kappa*kronecker(mydomain),D=omega,Y=omega*x[0], \
246     d=eta,y=kappa*n[0]+eta*x[0])
247 jgs 102 #... calculate error of the PDE solution ...
248     u=mypde.getSolution()
249     print "error is ",Lsup(u-x[0])
250 lkettle 575 saveVTK("x0.xml",sol=u)
251 jgs 102 \end{python}
252 lkettle 575 To visualise the solution `x0.~xml' just use the command
253     \begin{python}
254     mayavi -d u.xml -m SurfaceMap &
255     \end{python}
256     and it is easy to see that the solution $T=x\hackscore{0}$ is calculated.
257    
258 gross 569 The script is similar to the script \file{poisson.py} dicussed in \Chap{FirstSteps}.
259 jgs 102 \code{mydomain.getNormal()} returns the outer normal field on the surface of the domain. The function \function{Lsup}
260 lkettle 573 imported by the \code{from esys.escript import *} statement and returns the maximum absolute value of its argument.
261 jgs 107 The error shown by the print statement should be in the order of $10^{-7}$. As piecewise bi-linear interpolation is
262 gross 569 used by \finley approximate the solution and our solution is a linear function of the spatial coordinates one might
263 jgs 107 expect that the error would be zero or in the order of machine precision (typically $\approx 10^{-15}$).
264     However most PDE packages use an iterative solver which is terminated
265     when a given tolerance has been reached. The default tolerance is $10^{-8}$. This value can be altered by using the
266 jgs 102 \method{setTolerance} of the \LinearPDE class.
267    
268 jgs 121 \subsection{The Transition Problem}
269 jgs 102 \label{DIFFUSION TRANS SEC}
270     Now we are ready to solve the original time dependent problem. The main
271 jgs 107 part of the script is the loop over time $t$ which takes the following form:
272 jgs 102 \begin{python}
273 jgs 107 t=0
274     T=Tref
275 gross 569 mypde=LinearPDE(mydomain)
276     mypde.setValue(A=kappa*kronecker(mydomain),D=rhocp/h,d=eta,y=eta*Tref)
277 jgs 102 while t<t_end:
278 gross 569 mypde.setValue(Y=q+rhocp/h*T)
279 jgs 102 T=mypde.getSolution()
280     t+=h
281     \end{python}
282     \var{kappa}, \var{rhocp}, \var{eta} and \var{Tref} are input parameters of the model. \var{q} is the heat source
283 gross 569 in the domain and \var{h} is the time step size.
284     The variable \var{T}
285 jgs 102 holds the current temperature. It is used to calculate the right hand side coefficient \var{f} in the
286 jgs 107 Helmholtz equation in \eqn{DIFFUSION HELM EQ 1}. The statement \code{T=mypde.getSolution()} overwrites \var{T} with the
287     temperature of the new time step $\var{t}+\var{h}$. To get this iterative process going we need to specify the
288 jgs 102 initial temperature distribution, which equal to $T\hackscore{ref}$.
289 gross 569 The \LinearPDE class object \var{mypde}
290     and coefficients not changing over time are set up before the loop over time is entered. In each time step only the coefficient
291     $Y$ is reset as it depends on the temperature of the previous time step. This allows the PDE solver to reuse information
292     from previous time steps as much as possible.
293 jgs 102
294 lkettle 575 The heat source $q\hackscore H$ which is defined in \eqn{DIFFUSION TEMP EQ 1b} is \var{qc}
295 jgs 107 in an area defined as a circle of radius \var{r} and center \var{xc} and zero outside this circle.
296 lkettle 575 \var{q0} is a fixed constant. The following script defines $q\hackscore H$ as desired:
297 jgs 102 \begin{python}
298 lkettle 573 from esys.escript import length,whereNegative
299 jgs 102 xc=[0.02,0.002]
300     r=0.001
301     x=mydomain.getX()
302 lkettle 575 qH=q0*whereNegative(length(x-xc)-r)
303 jgs 102 \end{python}
304     \var{x} is a \Data class object of
305 jgs 107 the \escript module defining locations in the \Domain \var{mydomain}.
306     The \function{length()} imported from the \escript module returns the
307     Euclidean norm:
308     \begin{equation}\label{DIFFUSION HELM EQ 3aba}
309     \|y\|=\sqrt{
310     y\hackscore{i}
311     y\hackscore{i}
312     } = \function{esys.escript.length}(y)
313     \end{equation}
314     So \code{length(x-xc)} calculates the distances
315     of the location \var{x} to the center of the circle \var{xc} where the heat source is acting.
316     Note that the coordinates of \var{xc} are defined as a list of floating point numbers. It is independently
317 gross 569 converted into a \Data class object before being subtracted from \var{x}. The function \function{whereNegative}
318     applied to
319 jgs 107 \code{length(x-xc)-r}, returns a \Data class which is equal to one where the object is negative and
320     zero elsewhere. After multiplication with \var{qc} we get a function with the desired property.
321 jgs 102
322 jgs 107 Now we can put the components together to create the script \file{diffusion.py} which is available in the \ExampleDirectory:
323 jgs 102 \index{scripts!\file{diffusion.py}}:
324     \begin{python}
325 gross 569 from esys.escript import *
326     from esys.escript.linearPDEs import LinearPDE
327 jgs 107 from esys.finley import Rectangle
328 jgs 102 #... set some parameters ...
329 jgs 107 xc=[0.02,0.002]
330 jgs 102 r=0.001
331 jgs 107 qc=50.e6
332 jgs 102 Tref=0.
333     rhocp=2.6e6
334     eta=75.
335     kappa=240.
336 jgs 107 tend=5.
337     # ... time, time step size and counter ...
338     t=0
339 jgs 102 h=0.1
340     i=0
341     #... generate domain ...
342 jgs 107 mydomain = Rectangle(l0=0.05,l1=0.01,n0=250, n1=50)
343 jgs 102 #... open PDE ...
344 gross 569 mypde=LinearPDE(mydomain)
345     mypde.setSymmetryOn()
346     mypde.setValue(A=kappa*kronecker(mydomain),D=rhocp/h,d=eta,y=eta*Tref)
347 jgs 102 # ... set heat source: ....
348     x=mydomain.getX()
349 lkettle 575 qH=qc*whereNegative(length(x-xc)-r)
350 jgs 102 # ... set initial temperature ....
351     T=Tref
352     # ... start iteration:
353 jgs 107 while t<tend:
354 jgs 102 i+=1
355     t+=h
356     print "time step :",t
357 lkettle 575 mypde.setValue(Y=qH+rhocp/h*T)
358 jgs 102 T=mypde.getSolution()
359 gross 569 saveVTK("T.%d.xml"%i,temp=T)
360 jgs 102 \end{python}
361 gross 569 The script will create the files \file{T.1.xml},
362     \file{T.2.xml}, $\ldots$, \file{T.50.xml} in the directory where the script has been started. The files give the
363     temperature distributions at time steps $1$, $2$, $\ldots$, $50$ in the \VTK file format.
364 jgs 107
365     \begin{figure}
366 gross 599 \centerline{\includegraphics[width=\figwidth]{figures/DiffusionRes1.eps}}
367     \centerline{\includegraphics[width=\figwidth]{figures/DiffusionRes16.eps}}
368     \centerline{\includegraphics[width=\figwidth]{figures/DiffusionRes32.eps}}
369     \centerline{\includegraphics[width=\figwidth]{figures/DiffusionRes48.eps}}
370 lkettle 581 \caption{Results of the Temperature Diffusion Problem for Time Steps $1$ $16$, $32$ and $48$.}
371 jgs 107 \label{DIFFUSION FIG 2}
372     \end{figure}
373 jgs 102 An easy way to visualize the results is the command
374     \begin{verbatim}
375 gross 569 mayavi -d T.1.xml -m SurfaceMap &
376 jgs 102 \end{verbatim}
377 gross 569 Use the \texttt{Configure Data}
378     to move forward and and backwards in time.
379 jgs 121 \fig{DIFFUSION FIG 2} shows the result for some selected time steps.

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