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{SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT 256 13 "Heat Equation" }}{PARA 0 
"" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 187 "A derivation is \+
given in the notes (jpg-files in the folder). We follow David Betounes
: Partial Differential Equations for Computational Science, Springer V
erlag 1997, ISBN 0-387-98300-7." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 176 "We begin with a one-dimensional example.
(Example 2.2 in the book). A thin rod lies on the x-axis from 0 to 1 (
appropriate units are chosen to make the equation dimensionless). " }}
{PARA 0 "" 0 "" {TEXT -1 65 "At time zero the rod is at zero temperatu
re (inital condition).  " }{TEXT 19 10 "u(x,0) = 0" }{TEXT -1 1 "." }}
{PARA 0 "" 0 "" {TEXT -1 158 "On the left boundary the rod is kept at \+
zero temperature by a heat bath (make it zero Celsius to avoid trouble
 with physics, zero Kelvin are not reachable!). " }}{PARA 0 "" 0 "" 
{TEXT 19 10 "u(0,t) = 0" }{TEXT -1 6 "  for " }{TEXT 19 5 "t > 0" }}
{PARA 0 "" 0 "" {TEXT -1 101 "On the right boundary the rod is insulat
ed, i.e., no flow of heat across this boundary is permitted. " }{TEXT 
19 28 "diff(u(x,t),x)[@x=1,t>0] = 0" }{TEXT -1 28 " (this is not Maple
 syntax!)" }}{PARA 0 "" 0 "" {TEXT -1 172 "The heat equation contains \+
a source term, i.e., it is not homogeneous. The inhomogeneity is chose
n as a constant, i.e., heat is being generated in all locations of the
 rod " }{TEXT 19 9 "0 < x < 1" }{TEXT -1 9 " (not at " }{TEXT 19 5 "x \+
= 0" }{TEXT -1 2 ")." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 96 "The heat equation has a steady-state solution, i.e., a \+
solution that is being reached for large " }{TEXT 19 1 "t" }{TEXT -1 
38 ". For this to be possible, a solution " }{TEXT 19 6 "u(x,t)" }
{TEXT -1 127 " has to be found with vanishing time derivative in the l
imit of large time. This function will depend on the position variable
 " }{TEXT 19 1 "x" }{TEXT -1 6 " only." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 
"HEq:=diff(u(x,t),t) = diff(u(x,t),x$2) + 1;" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%$HEqG/-%%diffG6$-%\"uG6$%\"xG%\"tGF-,&-F'6$F)-%\"$G6$
F,\"\"#\"\"\"F5F5" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 46 "In order to \+
express the boundary condition at " }{TEXT 19 3 "x=1" }{TEXT -1 31 " w
e have to understand how the " }{TEXT 19 1 "D" }{TEXT -1 50 " operator
 works on functions of several variables:" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 28 "BC:=u(0,t)=0,D[1](u)(1,t)=0;" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%#BCG6$/-%\"uG6$\"\"!%\"tGF*/--&%\"DG6#\"\"\"6#F(6$F2F
+F*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "IC:=u(x,0)=0;" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#ICG/-%\"uG6$%\"xG\"\"!F*" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "sol:=pdsolve(\{HEq,BC,IC\},u
(x,t));" }{TEXT -1 0 "" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$solG<#/-%
\"uG6$%\"xG%\"tG,,**%$_C1G\"\"\"%$_C3GF/-%$expG6#*&&%#_cG6#F/F/F+F/F/-
F26#*&F5#F/\"\"#F*F/F/F/**F0F/F1F/%$_C2GF/F8!\"\"F/*&F<F?F*F<F?*(F.F?F
>F/F*F/F?*&F.F?F0F/F?" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 138 "It seem
s as if Maple has not found a unique solution. Four constants appear i
n the answer. Is it possible that many solutions truly exist?" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 141 "Let us e
xtract the steady state solution by solving the ordinary differential \+
equation that follows from setting the time derivative to zero." }}
{PARA 0 "" 0 "" {TEXT -1 38 "Let us call the steady-state solution " }
{TEXT 19 4 "w(x)" }{TEXT -1 1 ":" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 23 "DE:=diff(w(x),x$2)+1=0;" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#>%#DEG/,&-%%diffG6$-%\"wG6#%\"xG-%\"$G6$F-\"\"#\"\"\"F2F2\"\"!" 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "BCode:=w(0)=0,D(w)(1)=0;
" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&BCodeG6$/-%\"wG6#\"\"!F*/--%\"D
G6#F(6#\"\"\"F*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "solDE:=d
solve(\{DE,BCode\});" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&solDEG/-%\"
wG6#%\"xG,&*&#\"\"\"\"\"#F-*$)F)F.F-F-!\"\"F)F-" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 24 "plot(rhs(solDE),x=0..1);" }}{PARA 13 "" 1 "" 
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FB7$$\"3A++D\"oK0e*FB$\"3CM]UeB?\"*\\FB7$$\"3A++v=5s#y*FB$\"3I[`@#\\Rw
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LABELSG6$Q\"x6\"Q!Fa[l-%%VIEWG6$;F(Fbz%(DEFAULTG" 1 2 0 1 10 0 2 9 1 
4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" }}}}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 61 "The heat equation is supposed to describe the evolut
ion from " }{TEXT 19 8 "u(x,0)=0" }{TEXT -1 9 " towards " }{TEXT 19 
18 "u(x,infinity)=w(x)" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 
177 "Note how the boundary conditions are satisified (prescribed tempe
rature value on the left, and zero derivative on the right). These con
ditions have to be satisfied at all times." }}{PARA 0 "" 0 "" {TEXT 
-1 201 "The equation is usually solved by the separation of variables \+
method. Before we do that let us look at the solution generated by map
le. Without the boundary conditions the following answer is obtained:
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "sol:=pdsolve(HEq);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$solG-%'&whereG6$/-%\"uG6$%\"xG%\"tG
,&*&-%$_F1G6#F,\"\"\"-%$_F2G6#F-F3F3*&,(*(\"\"#!\"\"%$_C1GF3F,F:F3*&%$
_C2GF3F,F3F3%$_C3GF3F3F<F;F;7#<$/-%%diffG6$F4F-*&&%#_cG6#F3F3F4F3/-FD6
$F0-%\"$G6$F,F:*&FGF3F0F3" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 76 "Now \+
it becomes clearer what the  constants mean: pdsolve is telling us tha
t:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 152 "1)
 there is a time-independent, additive part to the solution, which is \+
characterized by three constants. From our asymptotic solution it is c
lear that " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "solHEq:=subs(
_C1=1,_C2=1,_C3=0,rhs(op(1,sol)));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#
>%'solHEqG,(*&-%$_F1G6#%\"xG\"\"\"-%$_F2G6#%\"tGF+F+*&\"\"#!\"\"F*F1F2
F*F2" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 128 "The heat equation admits
 a solution of the form where the transient part is expressed as a pro
duct of one-dimensional functions " }{TEXT 19 11 "F1(x)*F2(t)" }{TEXT 
-1 77 ". The two functions satisfy simple ordinary differential equati
ons listed in " }{TEXT 19 3 "sol" }{TEXT -1 1 "." }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 28 "eq1:=diff(F2(t),t)=c1*F2(t);" }}{PARA 11 "" 
1 "" {XPPMATH 20 "6#>%$eq1G/-%%diffG6$-%#F2G6#%\"tGF,*&%#c1G\"\"\"F)F/
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "eq2:=diff(F1(x),x$2)=c1
*F1(x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$eq2G/-%%diffG6$-%#F1G6#%
\"xG-%\"$G6$F,\"\"#*&%#c1G\"\"\"F)F3" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 46 "sol_x:=dsolve(\{eq2,F1(0)=0,D(F1)(1)=0\},F1(x));" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&sol_xG/-%#F1G6#%\"xG\"\"!" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 68 "We are not interested in the trivi
al solution, so what do we do now?" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 35 "sol_x:=dsolve(\{eq2,F1(0)=0\},F1(x));" }}{PARA 11 "" 
1 "" {XPPMATH 20 "6#>%&sol_xG/-%#F1G6#%\"xG,&*&%$_C2G\"\"\"-%$expG6#*&
%#c1G#F-\"\"#F)F-F-!\"\"*&F,F--F/6#,$F1F5F-F-" }}}{EXCHG {PARA 0 "" 0 
"" {TEXT -1 131 "Somehow we will have to find acceptable solutions (de
rivative to vanish on the right-hand boundary) by choosing c1 rather t
han _C2." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "sol_x1:=subs(_C
2=1,rhs(sol_x));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'sol_x1G,&-%$exp
G6#*&%#c1G#\"\"\"\"\"#%\"xGF,!\"\"-F'6#,$F)F/F," }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 40 "plot(subs(x=1,diff(sol_x1,x)),c1=0..50);" }}
{PARA 13 "" 1 "" {GLPLOT2D 846 231 231 {PLOTDATA 2 "6%-%'CURVESG6$7S7$
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{EXCHG {PARA 0 "" 0 "" {TEXT -1 35 "This shows that positive values of
 " }{TEXT 19 2 "c1" }{TEXT -1 88 " won't do the trick. So we try negat
ive values, which makes the solution complex-valued." }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 72 "[Re(subs(x=1,c1=-5,diff(sol_x1,x))),Im(su
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\"\"!,$*(\"\"#\"\"\"\"\"&#F(F'-%$cosG6#*$F)F*F(!\"\"" }}}{EXCHG {PARA 
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bs(x=1,diff(sol_x1,x)))],c1=-220..0,color=[red,blue],axes=boxed,numpoi
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0 0 "Curve 1" "Curve 2" }}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 185 "What \+
are we learning here? We took the condition for the right-hand boundar
y, and we showed that it can be satisfied by the solution to the spati
al ODE provided the separation constant " }{TEXT 19 2 "c1" }{TEXT -1 
132 " is chosen to be negative. The real part of the condition is alwa
ys zero (boring, but important). The derivative of the solution at " }
{TEXT 19 3 "x=1" }{TEXT -1 105 " does vanish when the imaginary part o
f the expression crosses zero. We can use the integration constant " }
{TEXT 19 3 "_C2" }{TEXT -1 59 " to make the solution real-valued (by m
aking it imaginary)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 117 "Therefore, the product part of the solution, i.e., the
 transient part, is not unique. There are different choices of " }
{TEXT 19 2 "c1" }{TEXT -1 152 " that work, and we have to take them al
l into account in order to find a unique solution that matches our bou
ndary conditions and the initial condition." }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 32 "Each spatial solution (function
 " }{TEXT 19 5 "F1(x)" }{TEXT -1 86 " that satisfies the ODE and the b
oundary conditions) has an appropriate time solution " }{TEXT 19 5 "F2
(t)" }{TEXT -1 112 " that goes with it. Before we find it, let us look
 at the spatial modes. First we determine a few of the lowest " }
{TEXT 19 2 "c1" }{TEXT -1 19 " values accurately." }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 82 "sol_c1:='sol_c1': sol_c1[1]:=fsolve(Im(subs(
x=1,diff(sol_x1,x))),c1=-10..-0.0001);" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#>&%'sol_c1G6#\"\"\"$!++6SnC!\"*" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 60 "sol_c1[2]:=fsolve(Im(subs(x=1,diff(sol_x1,x))),c1=-50
..-10);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>&%'sol_c1G6#\"\"#$!+!*4m?A
!\")" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 61 "sol_c1[3]:=fsolve(I
m(subs(x=1,diff(sol_x1,x))),c1=-100..-50);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>&%'sol_c1G6#\"\"$$!+^F]oh!\")" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 62 "sol_c1[4]:=fsolve(Im(subs(x=1,diff(sol_x1,x))),c
1=-150..-100);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>&%'sol_c1G6#\"\"%$!
+Rl-47!\"(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 62 "sol_c1[5]:=fs
olve(Im(subs(x=1,diff(sol_x1,x))),c1=-250..-150);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>&%'sol_c1G6#\"\"&$!+\"*[f)*>!\"(" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 203 "Of course, there are infinitely many solutions, a
nd it is not unreasonable to ask at this point whether we will get any
where at all... Nevertheless, we'll proceed by looking at some of the \+
spatial modes:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 90 "plot([seq
(subs(c1=sol_c1[i],I*sol_x1),i=1..5)],x=0..1,color=[red,blue,green,bla
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F(Fcz%(DEFAULTG" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 45.000000 
0 0 "Curve 1" "Curve 2" "Curve 3" "Curve 4" "Curve 5" }}}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 67 "These functions look familiar. So let us \+
understand why that is so." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 
"" 0 "" {TEXT -1 233 "The differential equation for the spatial modes \+
is asking: which functions are proportional to their second derivative
. This selects the class of exponentials, including complex exponentia
ls. Then we asked the functions to vanish at " }{TEXT 19 3 "x=0" }
{TEXT -1 165 ", and we wanted real-valued functions. This selected fro
m sines and cosines just sine functions. Next we asked only for those \+
functions which had zero derivative at " }{TEXT 19 3 "x=1" }{TEXT -1 
25 ". This selected from all " }{TEXT 19 15 "sin(sqrt(c1)*x)" }{TEXT 
-1 54 " those with an appropriate wavenumber, i.e., value of " }{TEXT 
19 2 "c1" }{TEXT -1 91 ". Of those we plotted the lowest four. Functio
ns which do not have too much variation with " }{TEXT 19 1 "x" }{TEXT 
-1 216 " should be representable with a few low-lying modes to a reaso
nable degree of accuracy. This is where our hope comes from: we procee
d even though we know that in principle there are infinitely many prod
uct solutions." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 51 "Let us now understand how the modes evolve in time." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 27 "sol_t:=dsolve(\{eq1\},F2(t));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6
#>%&sol_tG<#/-%#F2G6#%\"tG*&%$_C1G\"\"\"-%$expG6#*&%#c1GF-F*F-F-" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 43 "We observe that we want negative v
alues of " }{TEXT 19 2 "c1" }{TEXT -1 185 ", because this generates dy
ing solutions. Each of the spatial modes dies with time with a time co
nstant of its own! The higher modes (many oscillations) die faster tha
n the lower modes." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 113 "Now one might ask, why do we need to worry about dying m
odes, when the solution to the heat equation starts from " }{TEXT 19 
8 "u(x,0)=0" }{TEXT -1 286 " ? The answer is that we have the additive
 asymptotic piece. To be consistent with it, and with the stated initi
al condition we have to superimpose the modes in such a way that they \+
cancel the asymptotic piece. This we do by a projection method (formal
ly we construct a Fourier series)." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 6 "solDE;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"wG6#%\"
xG,&*&#\"\"\"\"\"#F+*$)F'F,F+F+!\"\"F'F+" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 54 "basis:=seq(evalc(subs(c1=sol_c1[i],I*sol_x1)),i=1..5)
;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&basisG6',$*&\"\"#\"\"\"-%$sinG
6#,$*&$\"+Fjzq:!\"*F)%\"xGF)F)F)F),$*&F(F)-F+6#,$*&$\"+!)*)Q7ZF1F)F2F)
F)F)F),$*&F(F)-F+6#,$*&$\"+M;)R&yF1F)F2F)F)F)F),$*&F(F)-F+6#,$*&$\"+Hu
b*4\"!\")F)F2F)F)F)F),$*&F(F)-F+6#,$*&$\"+%p;PT\"FKF)F2F)F)F)F)" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 130 "We claim that the inner product (
dot product) of two basis functions is just the integral of the produc
t of functions from 0 to 1." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
30 "int(basis[1]*basis[2],x=0..1);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#
$\"+$)=QEV!#>" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "int(basis[
1]*basis[3],x=0..1);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$!+pm+D#)!#@" 
}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 81 "To numerical accuracy the funct
ions are orthogonal! What about the normalization?" }}}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 30 "int(basis[1]*basis[1],x=0..1);" }}{PARA 11 
"" 1 "" {XPPMATH 20 "6#$\"+++++?!\"*" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 30 "int(basis[2]*basis[2],x=0..1);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#$\"+++++?!\"*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 353 "
OK, so the functions are normalized in a funny way, but that shouldn't
 deter us. Verify the orthogonality and normalization for some of the \+
other basis functions (or basis vectors in the Hilbert space). Let us \+
expand by calculating projections. To make up for the wrong normalizat
ion we re-define the basis functions by dividing by the square root of
 2." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "b[1]:=int(rhs(solDE)
*basis[1]/sqrt(2),x=0..1);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>&%\"bG6
#\"\"\"$\"+#fW)[O!#5" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "b[2
]:=int(rhs(solDE)*basis[2]/sqrt(2),x=0..1);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>&%\"bG6#\"\"#$\"+IRU^8!#6" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 46 "b[3]:=int(rhs(solDE)*basis[3]/sqrt(2),x=0..1);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>&%\"bG6#\"\"$$\"+wc2>H!#7" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "b[4]:=int(rhs(solDE)*basis[4]/sqrt(
2),x=0..1);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>&%\"bG6#\"\"%$\"+GH!Q1
\"!#7" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "b[5]:=int(rhs(solD
E)*basis[5]/sqrt(2),x=0..1);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>&%\"b
G6#\"\"&$\"+2KF0]!#8" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 168 "Good new
s: the projection of the function to be expanded onto higher basis sta
tes are getting smaller. We are ready to compare the expansion with th
e original function:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "FE:
=add(b[i]*basis[i]/sqrt(2),i=1..5);" }}{PARA 12 "" 1 "" {XPPMATH 20 "6
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}}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 202 "We see that the expansion is \+
better than 1% accurate (otherwise we would notice a difference in the
 graph). Repeat the graph with less than five basis states and observe
 how the convergence comes about!" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 218 "Now we are ready to put everything toget
her. The total solution is obtained by multiplying the basis functions
 with the appropriate time-factors, and by adding the transient and st
eady-state parts together as required:" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 6 "sol_t;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#<#/-%#F2G6#%
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he solution rapidly approaches the steady state. Explore the time beha
vior in detail (by choosing different time intervals and more steps " 
}{TEXT 19 1 "j" }{TEXT -1 2 ")." }}{PARA 0 "" 0 "" {TEXT -1 120 "Repea
t the observation for a more approximate solution by selecting a small
er expansion in the Fourier series, i.e., in " }{TEXT 19 3 "FEt" }
{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 165 "Here is a graph in three dimensions. It shows, how well \+
the Fourier expansion approximates the initial condition, and how quic
kly the asymptotic values are attained." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 37 "plot3d(FEt,x=0..1,t=0..2,axes=boxed);" }}{PARA 13 "" 
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-49.000000 0 0 "Curve 1" }}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 69 "Rotat
e the graph to look at the solution from different perspectives." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 106 "To remin
d ourselves that we do not have an exact solution we show the deviatio
n from the intial condition:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 27 "plot(subs(t=0,FEt),x=0..1);" }}{PARA 13 "" 1 "" {GLPLOT2D 651 
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$\"35+]P40O\"*)*Fiq$!3]56d(GHUS#F07$$\"31+voa-oX**Fiq$!3\"pu;#z70KCF07
$$\"\"\"F)$!3nT$*3P![8W#F0-%'COLOURG6&%$RGBG$\"#5!\"\"F(F(-%+AXESLABEL
SG6$Q\"x6\"Q!F[_m-%%VIEWG6$;F(F\\^m%(DEFAULTG" 1 2 0 1 10 0 2 9 1 4 2 
1.000000 45.000000 45.000000 0 0 "Curve 1" }}}}{EXCHG {PARA 0 "" 0 "" 
{TEXT 257 9 "Exercise:" }}{PARA 0 "" 0 "" {TEXT -1 103 "Increase the e
xpansion to 10 basis functions and observe how well the initial condit
ion can be matched." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}
}{MARK "62 0 0" 69 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 
1 2 33 1 1 }