ViewVC logotype

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

Parent Directory Parent Directory | Revision Log Revision Log

Revision 575 - (show annotations)
Fri Mar 3 03:33:07 2006 UTC (16 years, 9 months ago) by lkettle
File MIME type: application/x-tex
File size: 18371 byte(s)
I have changed some of the documentation and added more explanations for
the online reference guide for esys13. I have modified two of the
example source codes to write out the results for Helmholtz problem and
changed one variable name in the diffusion.py code to avoid confusion.

1 % $Id$
2 \section{The Diffusion Problem}
5 \begin{figure}
6 \centerline{\includegraphics[width=\figwidth]{DiffusionDomain}}
7 \caption{Temperature Diffusion Problem with Circular Heat Source}
8 \label{DIFFUSION FIG 1}
9 \end{figure}
11 \subsection{\label{DIFFUSION OUT SEC}Outline}
12 In this chapter we will discuss how to solve the time dependent-temperature diffusion\index{diffusion equation} for
13 a block of material. Within the block there is a heat source which drives the temperature diffusion.
14 On the surface, energy can radiate into the surrounding environment.
15 \fig{DIFFUSION FIG 1} shows the configuration.
17 In the next \Sec{DIFFUSION TEMP SEC} we will present the relevant model. A
18 time integration scheme is introduced to calculate the temperature at given time nodes $t^{(n)}$.
19 We will see that at each time step a Helmholtz equation \index{Helmholtz equation}
20 must be solved.
21 The implementation of a Helmholtz equation solver will be discussed in \Sec{DIFFUSION HELM SEC}.
22 In Section~\ref{DIFFUSION TRANS SEC} the solver of the Helmholtz equation is used to build a
23 solver for the temperature diffusion problem.
25 \subsection{\label{DIFFUSION TEMP SEC}Temperature Diffusion}
26 The unknown temperature $T$ is a function of its location in the domain and time $t>0$. The governing equation
27 in the interior of the domain is given by
28 \begin{equation}
29 \rho c\hackscore p T\hackscore{,t} - (\kappa T\hackscore{,i})\hackscore{,i} = q\hackscore H
30 \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 material the parameters depend on their location in the domain. $q\hackscore H$ is
34 a heat source (or sink) within the domain. We are using Einstein summation convention \index{summation convention}
35 as introduced in \Chap{FirstSteps}. In our case we assume $q\hackscore H$ to be equal to a constant heat production rate
36 $q^{c}$ on a circle or sphere with center $x^c$ and radius $r$ and $0$ elsewhere:
37 \begin{equation}
38 q\hackscore H(x,t)=
39 \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 $t>0$.
50 On the surface of the domain we are
51 specifying a radiation condition
52 which precribes the normal component of the flux $\kappa T\hackscore{,i}$ to be proportional
53 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 $\eta$ 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 at the surface of the domain.
62 To solve the time dependent \eqn{DIFFUSION TEMP EQ 1} the initial temperature at time
63 $t=0$ 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 the initial conditions satisfy the
70 boundary condition defined by \eqn{DIFFUSION TEMP EQ 2}.
72 The temperature is calculated at discrete time nodes $t^{(n)}$ where
73 $t^{(0)}=0$ and $t^{(n)}=t^{(n-1)}+h$ where $h>0$ is the step size which is assumed to be constant.
74 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 \begin{equation}
81 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 \label{DIFFUSION TEMP EQ 6}
84 \end{equation}
85 This is inserted into \eqn{DIFFUSION TEMP EQ 1}. By separating the terms at
86 $t^{(n)}$ and $t^{(n-1)}$ one gets for $n=1,2,3\ldots$
87 \begin{equation}
88 \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 \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 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 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.
102 \subsection{\label{DIFFUSION HELM SEC}Helmholtz Problem}
103 The partial differential equation to be solved for $T^{(n)}$ has the form
104 \begin{equation}
105 \omega T^{(n)} - (\kappa T^{(n)}\hackscore{,i})\hackscore{,i} = f
106 \label{DIFFUSION HELM EQ 1}
107 \end{equation}
108 and we set
109 \begin{equation}
110 \omega=\frac{\rho c\hackscore p}{h} \mbox{ and } f=q\hackscore H +\frac{\rho c\hackscore p}{h}T^{(n-1)} \;.
111 \label{DIFFUSION HELM EQ 1b}
112 \end{equation}
113 With $g=\eta T\hackscore{ref}$ the radiation condition defined by \eqn{DIFFUSION TEMP EQ 2222}
114 takes the form
115 \begin{equation}
116 \kappa T^{(n)}\hackscore{,i} n\hackscore{i} = g - \eta T^{(n)}\mbox{ on } \Gamma
117 \label{DIFFUSION HELM EQ 2}
118 \end{equation}
119 The partial differential \eqn{DIFFUSION HELM EQ 1} together with boundary conditions of \eqn{DIFFUSION HELM EQ 2}
120 is called the Helmholtz equation \index{Helmholtz equation}.
122 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 \end{equation}
127 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 \end{equation}
135 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.
150 By inspecting the Helmholtz equation \index{Helmholtz equation}
151 (\ref{DIFFUSION HELM EQ 1}) and boundary condition (\ref{DIFFUSION HELM EQ 2}) and
152 substituting $u$ for $T^{(n)}$
153 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 $\delta\hackscore{ij}$ is the Kronecker symbol \index{Kronecker symbol} defined by $\delta\hackscore{ij}=1$ for
162 $i=j$ and $0$ otherwise. Undefined coefficients are assumed to be not present.\footnote{There is a difference
163 in \escript of being not present and set to zero. As not present coefficients are not processed,
164 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}).
169 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 the PDE and returns the solution \code{u} of the PDE. \function{kronecker} is \escript function
182 returning the Kronecker symbol.
184 The coefficients can set by several calls of \method{setValue} where the order can be chosen arbitrarily.
185 If a value is assigned to a coefficient several times, the last assigned value is used when
186 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 The \LinearPDE class provides the method \method{checkSymmetry} method to check if the given PDE is symmetric.
202 \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 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 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 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 take $T=x\hackscore{0}$ (here $q\hackscore H = 0$) and then calculate the right hand side terms $f$ and $g$ such that
224 the test solution becomes the solution of the problem. If we assume $\kappa$ as being constant,
225 an easy calculation shows that we have to choose $f=\omega \cdot x\hackscore{0}$. On the boundary we get
226 $\kappa n\hackscore{i} u\hackscore{,i}=\kappa n\hackscore{0}$.
227 So we have to set $g=\kappa n\hackscore{0}+\eta x\hackscore{0}$. The following script \file{helmholtz.py}
228 \index{scripts!\file{helmholtz.py}} which is available in the \ExampleDirectory
229 implements this test problem using the \finley PDE solver:
230 \begin{python}
231 from esys.escript import *
232 from esys.escript.linearPDEs import LinearPDE
233 from esys.finley import Rectangle
234 #... set some parameters ...
235 kappa=1.
236 omega=0.1
237 eta=10.
238 #... generate domain ...
239 mydomain = Rectangle(l0=5.,l1=1.,n0=50, n1=10)
240 #... open PDE and set coefficients ...
241 mypde=LinearPDE(mydomain)
242 mypde.setSymmetryOn()
243 n=mydomain.getNormal()
244 x=mydomain.getX()
245 mypde.setValue(A=kappa*kronecker(mydomain),D=omega,Y=omega*x[0], \
246 d=eta,y=kappa*n[0]+eta*x[0])
247 #... calculate error of the PDE solution ...
248 u=mypde.getSolution()
249 print "error is ",Lsup(u-x[0])
250 saveVTK("x0.xml",sol=u)
251 \end{python}
252 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.
258 The script is similar to the script \file{poisson.py} dicussed in \Chap{FirstSteps}.
259 \code{mydomain.getNormal()} returns the outer normal field on the surface of the domain. The function \function{Lsup}
260 imported by the \code{from esys.escript import *} statement and returns the maximum absolute value of its argument.
261 The error shown by the print statement should be in the order of $10^{-7}$. As piecewise bi-linear interpolation is
262 used by \finley approximate the solution and our solution is a linear function of the spatial coordinates one might
263 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 \method{setTolerance} of the \LinearPDE class.
268 \subsection{The Transition Problem}
270 Now we are ready to solve the original time dependent problem. The main
271 part of the script is the loop over time $t$ which takes the following form:
272 \begin{python}
273 t=0
274 T=Tref
275 mypde=LinearPDE(mydomain)
276 mypde.setValue(A=kappa*kronecker(mydomain),D=rhocp/h,d=eta,y=eta*Tref)
277 while t<t_end:
278 mypde.setValue(Y=q+rhocp/h*T)
279 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 in the domain and \var{h} is the time step size.
284 The variable \var{T}
285 holds the current temperature. It is used to calculate the right hand side coefficient \var{f} in the
286 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 initial temperature distribution, which equal to $T\hackscore{ref}$.
289 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.
294 The heat source $q\hackscore H$ which is defined in \eqn{DIFFUSION TEMP EQ 1b} is \var{qc}
295 in an area defined as a circle of radius \var{r} and center \var{xc} and zero outside this circle.
296 \var{q0} is a fixed constant. The following script defines $q\hackscore H$ as desired:
297 \begin{python}
298 from esys.escript import length,whereNegative
299 xc=[0.02,0.002]
300 r=0.001
301 x=mydomain.getX()
302 qH=q0*whereNegative(length(x-xc)-r)
303 \end{python}
304 \var{x} is a \Data class object of
305 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 converted into a \Data class object before being subtracted from \var{x}. The function \function{whereNegative}
318 applied to
319 \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.
322 Now we can put the components together to create the script \file{diffusion.py} which is available in the \ExampleDirectory:
323 \index{scripts!\file{diffusion.py}}:
324 \begin{python}
325 from esys.escript import *
326 from esys.escript.linearPDEs import LinearPDE
327 from esys.finley import Rectangle
328 #... set some parameters ...
329 xc=[0.02,0.002]
330 r=0.001
331 qc=50.e6
332 Tref=0.
333 rhocp=2.6e6
334 eta=75.
335 kappa=240.
336 tend=5.
337 # ... time, time step size and counter ...
338 t=0
339 h=0.1
340 i=0
341 #... generate domain ...
342 mydomain = Rectangle(l0=0.05,l1=0.01,n0=250, n1=50)
343 #... open PDE ...
344 mypde=LinearPDE(mydomain)
345 mypde.setSymmetryOn()
346 mypde.setValue(A=kappa*kronecker(mydomain),D=rhocp/h,d=eta,y=eta*Tref)
347 # ... set heat source: ....
348 x=mydomain.getX()
349 qH=qc*whereNegative(length(x-xc)-r)
350 # ... set initial temperature ....
351 T=Tref
352 # ... start iteration:
353 while t<tend:
354 i+=1
355 t+=h
356 print "time step :",t
357 mypde.setValue(Y=qH+rhocp/h*T)
358 T=mypde.getSolution()
359 saveVTK("T.%d.xml"%i,temp=T)
360 \end{python}
361 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.
365 \begin{figure}
366 \centerline{\includegraphics[width=\figwidth]{DiffusionRes1}}
367 \centerline{\includegraphics[width=\figwidth]{DiffusionRes16}}
368 \centerline{\includegraphics[width=\figwidth]{DiffusionRes32}}
369 \centerline{\includegraphics[width=\figwidth]{DiffusionRes48}}
370 \caption{Results of the Temperture Diffusion Problem for Time Steps $1$ $16$, $32$ and $48$.}
371 \label{DIFFUSION FIG 2}
372 \end{figure}
373 An easy way to visualize the results is the command
374 \begin{verbatim}
375 mayavi -d T.1.xml -m SurfaceMap &
376 \end{verbatim}
377 Use the \texttt{Configure Data}
378 to move forward and and backwards in time.
379 \fig{DIFFUSION FIG 2} shows the result for some selected time steps.


Name Value
svn:eol-style native
svn:keywords Author Date Id Revision

  ViewVC Help
Powered by ViewVC 1.1.26