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{SECT 0 {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 256 61 "Epilog: A Perspec
tive for the Separation of Variables Problem" }}{PARA 0 "" 0 "" {TEXT
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 859 " In the summer of 2004, I \+
taught the Fourier Series and Boundary Value Problem course for Georgi
a Tech one more time. At the end, a good graduate student in engineeri
ng told me that he was confident he could solve any of those Separatio
n-of-Variables problems which arise in the course, but he had not yet \+
organized a unified perspective for working the problems. I thought it
was a good idea for him to think about formulating a unified perspect
ive. I suggested that he take another Math class: one that introduced \+
Linear Algebra in an Infinite Dimensional Vector Space. The School of \+
Mathematics at Georgia Tech offers such a course: Math 6580: An Introd
uction to Hilbert Spaces. Most mathematics departments with graduate c
ourses will offer such a course. Before retiring, I taught that course
many times and have notes for it in Item 6 on my web page." }}{PARA
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 388 "Graduate stude
nts in the early part of their studies have restrictions on their time
. Such restrictions may make it hard for my student that Summer of 200
4 to put what he was learning into appropriate contexts. In the best o
f situations, students would be encouraged to do that, but there is so
much to tell them that giving them time for organizing should be incl
uded in the curriculum. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT -1 367 "For that reason, I add this Epilog. It is meant t
o help these harried students. I will try to re-work several examples \+
from the lectures contained in these notes, but each time, I will make
the telling of how to do the problem have the same structure as each \+
of the other problems. That structure will be the one which is indicat
ed in the first sections of the notes." }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT -1 577 "We start with a space of functions d
efined on an interval and on which there is defined an inner product, \+
or a dot product. For each problem, the space will be a space of funct
ions which is rich enough that all the sectionally smooth functions on
that interval are contained in the space. Also, the space will be ric
h enough that if we identify a sequence of functions in the space that
converges in the norm associated with the dot product, then there is \+
something in the space to which it converges. That is, limits exist fo
r sequences that converge in norm within the space." }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 381 "While the space is def
ined the same way for all the problems we encounter in this summary, t
he dot product is not. What the dot product is to be is neither the pr
erogative of the student nor of the professor. The partial differentia
l equation (PDE) and boundary conditions dictate what should be the in
terval over which the functions are defined and what should be the dot
product." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1
158 "We put these problems into a context by working several examples \+
from the previous lectures and by using exactly the same steps each ti
me. Here are the steps:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 ""
0 "" {TEXT 261 77 "Start with a PDE and boundary conditions such as pr
esented in these lectures." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0
"" 0 "" {TEXT 257 7 "Step 1." }{TEXT -1 1 " " }{TEXT 258 90 "Choose th
e appropriate dot product and maximal orthogonal family associated wit
h this PDE." }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA
0 "" 0 "" {TEXT 259 69 "Step 2. Give a typical function having as valu
es points in the space." }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0
"" }}{PARA 0 "" 0 "" {TEXT 260 136 "Step 3. Give the infinite system o
f ordinary differential equations that provide the coefficients for a \+
function that will solve the PDE" }{TEXT -1 2 ". " }}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 262 352 "Step 4. \+
Determine the initial values or boundary values that will be used to s
olve each of the equations in the infinite system of ordinary differen
tial equations.\n\nStep 5. Make an approximation for the solution for \+
the PDE by solving enough of the infinite number of ordinary different
ial equations to get a good approximation for the boundary values." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 256 "" 0 "" {TEXT -1 34 "Step 6.
Check the approximation. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT
1 {PARA 3 "" 0 "" {TEXT -1 30 "Example 1: A Diffusion Problem" }}
{PARA 0 "" 0 "" {TEXT -1 19 "Here is the PDE: " }{XPPEDIT 18 0 "diff
(u,t);" "6#-%%diffG6$%\"uG%\"tG" }{TEXT -1 5 " = " }{XPPEDIT 18 0 "d
iff(u,`$`(x,2));" "6#-%%diffG6$%\"uG-%\"$G6$%\"xG\"\"#" }{TEXT -1 4 " \+
- " }{XPPEDIT 18 0 "u(t,x);" "6#-%\"uG6$%\"tG%\"xG" }{TEXT -1 1 "." }
}{PARA 0 "" 0 "" {TEXT -1 46 "Here are the homogeneous boundary condit
ions: " }{XPPEDIT 18 0 "diff(u,x);" "6#-%%diffG6$%\"uG%\"xG" }{TEXT
-1 13 " (t,0) = 0 = " }{XPPEDIT 18 0 "diff(u,x);" "6#-%%diffG6$%\"uG%
\"xG" }{TEXT -1 6 "(t,3)." }}{PARA 0 "" 0 "" {TEXT -1 31 "Here is the \+
initial condition: " }{XPPEDIT 18 0 "u(0,x);" "6#-%\"uG6$\"\"!%\"xG" }
{TEXT -1 27 " = | x (2 x - 3) (x - 3) |." }}{PARA 0 "" 0 "" {TEXT -1
0 "" }}{PARA 0 "" 0 "" {TEXT 263 7 "Step 1:" }{TEXT -1 94 " Separation
of variables leads to two ordinary differential equations and boundar
y conditions:" }}{PARA 0 "" 0 "" {TEXT -1 14 " X '' = " }
{XPPEDIT 18 0 "mu;" "6#%#muG" }{TEXT -1 41 " X, with X '(0) = 0 = X '(
3), and T ' = (" }{XPPEDIT 18 0 "mu;" "6#%#muG" }{TEXT -1 7 " +1) T."
}}{PARA 0 "" 0 "" {TEXT -1 114 "We should recognize that the homogeneo
us boundary conditions with the X equation gives a Sturm-Liouville pro
blem. " }}{PARA 0 "" 0 "" {TEXT -1 68 "The eigenfunctions form the max
imal orthogonal family and are cos(n " }{XPPEDIT 18 0 "Pi;" "6#%#PiG"
}{TEXT -1 10 " x/3) and " }{XPPEDIT 18 0 "mu;" "6#%#muG" }{TEXT -1 3 "
= " }{XPPEDIT 18 0 "-n^2*Pi^2/9;" "6#,$*(%\"nG\"\"#%#PiGF&\"\"*!\"\"F
)" }{TEXT -1 32 ". The appropriate dotproduct is " }}{PARA 0 "" 0 ""
{TEXT -1 60 " < f(x) , g(x) >
= " }{XPPEDIT 18 0 "int(f(x)*g(x),x = 0 .. 3);" "6#-%$intG6$*&-%\"fG6
#%\"xG\"\"\"-%\"gG6#F*F+/F*;\"\"!\"\"$" }{TEXT -1 1 "." }}{PARA 0 ""
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 264 8 "Step 2: " }{TEXT -1
149 "A typical point in this space can be expressed as an infinite sum
of the eigenfunctions. Thus a typical function of t in the space can \+
be written as " }}{PARA 0 "" 0 "" {TEXT -1 23 " T(t) = \+
" }{XPPEDIT 18 0 "sum(T[n](t)*phi[n],n);" "6#-%$sumG6$*&-&%\"TG6#%\"nG
6#%\"tG\"\"\"&%$phiG6#F+F.F+" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT
-1 10 "where the " }{XPPEDIT 18 0 "phi[n];" "6#&%$phiG6#%\"nG" }{TEXT
-1 76 " 's are the eigenfunctions. Thus, the solution to the PDE will \+
have the form" }}{PARA 0 "" 0 "" {TEXT -1 24 " u(t,x) = \+
" }{XPPEDIT 18 0 "sum(T[n](t)*cos(n*Pi*x/3),n);" "6#-%$sumG6$*&-&%\"TG
6#%\"nG6#%\"tG\"\"\"-%$cosG6#**F+F.%#PiGF.%\"xGF.\"\"$!\"\"F.F+" }
{TEXT -1 2 " ." }}{PARA 0 "" 0 "" {TEXT 266 7 "Step 3:" }{TEXT -1 119
" To determine the infinite system of equations, we suppose that u sat
isfies the PDE and determine the implications for " }{XPPEDIT 18 0 "T[
n];" "6#&%\"TG6#%\"nG" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 5 " \+
" }{XPPEDIT 18 0 "diff(u,t);" "6#-%%diffG6$%\"uG%\"tG" }{TEXT -1
3 " = " }{XPPEDIT 18 0 "sum(diff(T[t],t)*cos(n*Pi*x/3),n);" "6#-%$sumG
6$*&-%%diffG6$&%\"TG6#%\"tGF-\"\"\"-%$cosG6#**%\"nGF.%#PiGF.%\"xGF.\"
\"$!\"\"F.F3" }}{PARA 0 "" 0 "" {TEXT -1 3 "and" }}{PARA 0 "" 0 ""
{TEXT -1 5 " " }{XPPEDIT 18 0 "diff(u,`$`(x,2));" "6#-%%diffG6$%\"
uG-%\"$G6$%\"xG\"\"#" }{TEXT -1 3 " - " }{TEXT 265 6 "u(t,x)" }{TEXT
-1 3 " = " }{XPPEDIT 18 0 "sum(-(1+n^2*Pi^2/9)*T[n](t)*cos(n*Pi*x/3),n
);" "6#-%$sumG6$,$*(,&\"\"\"F)*(%\"nG\"\"#%#PiGF,\"\"*!\"\"F)F)-&%\"TG
6#F+6#%\"tGF)-%$cosG6#**F+F)F-F)%\"xGF)\"\"$F/F)F/F+" }{TEXT -1 0 "" }
}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 71 "Because \+
we are supposing that u satisfies the PDE, then it must be that" }}
{PARA 0 "" 0 "" {TEXT -1 14 " " }{XPPEDIT 18 0 "diff(T[n]
,t);" "6#-%%diffG6$&%\"TG6#%\"nG%\"tG" }{TEXT -1 6 " = - (" }{XPPEDIT
18 0 "1+n^2*Pi^2/9;" "6#,&\"\"\"F$*(%\"nG\"\"#%#PiGF'\"\"*!\"\"F$" }
{TEXT -1 3 " ) " }{XPPEDIT 18 0 "T[n](t);" "6#-&%\"TG6#%\"nG6#%\"tG" }
{TEXT -1 21 ", for each integer n." }}{PARA 0 "" 0 "" {TEXT -1 112 "We
list a few of these differential equations. First, we choose how many
terms we will use in the approximation." }}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 14 "restart:\nN:=4;" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 76 "for n from 0 to N do\n ode[n]:=diff(T[n](t),t)=-(1+
n^2*Pi^2/9)*T[n](t);\nod;" }}}{PARA 0 "" 0 "" {TEXT 267 8 "Step 4: " }
{TEXT -1 266 "In order to solve these first order equations, we need t
he initial value appropriate for this problem. We find this from the i
nitial value of the PDE, but rewritten in term of the orthogonal seque
nce that is associated with this problem. We compute the coefficients.
" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "f:=x->abs(x*(2*x-3)*(x-3
));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 101 "for n from 0 to N d
o\n A[n]:=int(f(x)*cos(n*Pi*x/3),x=0..3)/int(cos(n*Pi*x/3)^2,x=0..3)
;\nod;\nn:='n':" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 272 7 "Step 5
." }{TEXT -1 66 " We now solve several of the ordinary differential eq
uations in t." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 102 "for n from
0 to N do\n dsolve(\{ode[n],T[n](0)=A[n]\},T[n](t)):\n T[n]:=unap
ply(rhs(%),t);\nod;\nn:='n':" }}}{PARA 0 "" 0 "" {TEXT -1 58 "We now m
ake an approximation for the solutions of the PDE." }}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 44 "u:=(t,x)->sum(T[n](t)*cos(n*Pi*x/3),n=0..N);
" }}}{PARA 0 "" 0 "" {TEXT 273 7 "Step 6." }{TEXT -1 73 " We check to \+
see that this u satisfied the partial differential equation." }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "simplify(diff(u(t,x),t)-diff
(u(t,x),x,x)+u(t,x));" }}}{PARA 0 "" 0 "" {TEXT -1 45 "We check the ho
mogeneous boundary conditions." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1
0 27 "D[2](u)(t,0); D[2](u)(t,3);" }}}{PARA 0 "" 0 "" {TEXT -1 72 "We \+
check to see if u(0,x) is a good approximation for the initial value.
" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 60 "plot([f(x),u(0,x)],x=0..
3,color=[blue,red],thickness=[5,8]);" }}}{PARA 0 "" 0 "" {TEXT -1 103
"The fit for the initial value is not so good. It can be improved by i
ncreasing the number of terms, N. " }}{PARA 0 "" 0 "" {TEXT -1 38 "Nex
t, we draw a graph of the solution." }}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 62 "plot3d(u(t,x),x=0..3,t=0..1,axes=NORMAL,orientation=[
-30,65]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 303 6 "EXTRA:" }{TEXT -1
232 " We add for this problem an error estimation. Here is the idea. S
uppose U(t,x) is the real solution. We have the approximate solutions \+
u(t,x) from above. Consider the function v defined by \n \+
v(t,x) = U(t,x) - u(t,x)." }}{PARA 0 "" 0 "" {TEXT -1 67 "Take the \+
derivative of v with respect to t. A little calculus gives" }}{PARA 0
"" 0 "" {TEXT -1 5 " " }{XPPEDIT 18 0 "diff(v(t,x),t);" "6#-%%diff
G6$-%\"vG6$%\"tG%\"xGF)" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "diff(v(t,x)
,`$`(x,2));" "6#-%%diffG6$-%\"vG6$%\"tG%\"xG-%\"$G6$F*\"\"#" }{TEXT
-1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 14 "Further, " }{XPPEDIT 18
0 "diff(v,x);" "6#-%%diffG6$%\"vG%\"xG" }{TEXT -1 13 " (t,0) = 0 = " }
{XPPEDIT 18 0 "diff(v,x);" "6#-%%diffG6$%\"vG%\"xG" }{TEXT -1 6 "(t,3)
." }}{PARA 0 "" 0 "" {TEXT -1 225 "The maximum value of v will occur o
n the boundary. Consequently, the real solution is approximated by the
cosine approximation with accuracy at least as good as the initial va
lue is approximated by its Fourier cosine series.." }}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 0 "" }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT
1 {PARA 3 "" 0 "" {TEXT -1 49 "Example 2: The Wave Equation in a Visco
us Medium." }}{PARA 0 "" 0 "" {TEXT -1 19 "Here is the PDE: " }
{XPPEDIT 18 0 "diff(u,`$`(t,2));" "6#-%%diffG6$%\"uG-%\"$G6$%\"tG\"\"#
" }{TEXT -1 4 " = " }{XPPEDIT 18 0 "1/9;" "6#*&\"\"\"F$\"\"*!\"\"" }
{TEXT -1 2 " " }{XPPEDIT 18 0 "diff(u,`$`(x,2));" "6#-%%diffG6$%\"uG-
%\"$G6$%\"xG\"\"#" }{TEXT -1 4 " - " }{XPPEDIT 18 0 "1/5;" "6#*&\"\"
\"F$\"\"&!\"\"" }{TEXT -1 1 " " }{XPPEDIT 18 0 "diff(u,t);" "6#-%%diff
G6$%\"uG%\"tG" }{TEXT -1 2 " ." }}{PARA 0 "" 0 "" {TEXT -1 46 "Here ar
e the homogeneous boundary conditions: " }{XPPEDIT 18 0 "u(t,0);" "6#-
%\"uG6$%\"tG\"\"!" }{TEXT -1 7 " = 0 = " }{XPPEDIT 18 0 "u(t,1);" "6#-
%\"uG6$%\"tG\"\"\"" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 31 "Her
e is the initial condition: " }{XPPEDIT 18 0 "u(0,x);" "6#-%\"uG6$\"\"
!%\"xG" }{TEXT -1 11 " = 5 sin(2 " }{XPPEDIT 18 0 "Pi;" "6#%#PiG" }
{TEXT -1 8 " x) and " }{XPPEDIT 18 0 "diff(u,t);" "6#-%%diffG6$%\"uG%
\"tG" }{TEXT -1 10 "(0,x) = 0." }}{PARA 0 "" 0 "" {TEXT -1 78 "With de
liberation, we follow exactly the same steps as in the previous proble
m" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 268 7 "Step
1:" }{TEXT -1 94 " Separation of variables leads to two ordinary diff
erential equations and boundary conditions:" }}{PARA 0 "" 0 "" {TEXT
-1 19 " X '' = 9 " }{XPPEDIT 18 0 "mu;" "6#%#muG" }{TEXT -1
37 " X with X(0) = 0 = X(1) and T '' + " }{XPPEDIT 18 0 "1/5;" "6#*&
\"\"\"F$\"\"&!\"\"" }{TEXT -1 7 " T ' = " }{XPPEDIT 18 0 "lambda^2;" "
6#*$%'lambdaG\"\"#" }{TEXT -1 2 "T." }}{PARA 0 "" 0 "" {TEXT -1 114 "W
e should recognize that the homogeneous boundary conditions with the X
equation gives a Sturm-Liouville problem. " }}{PARA 0 "" 0 "" {TEXT
-1 75 "Again, the eigenfunctions form the maximal orthogonal family an
d are sin(n " }{XPPEDIT 18 0 "Pi;" "6#%#PiG" }{TEXT -1 8 " x) and " }
{XPPEDIT 18 0 "mu;" "6#%#muG" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "-n^2*P
i^2/9;" "6#,$*(%\"nG\"\"#%#PiGF&\"\"*!\"\"F)" }{TEXT -1 32 ". The appr
opriate dotproduct is " }}{PARA 0 "" 0 "" {TEXT -1 55 " \+
< f(x) , g(x) > = " }{XPPEDIT 18 0 "int(f(x)*g(x
),x = 0 .. 1);" "6#-%$intG6$*&-%\"fG6#%\"xG\"\"\"-%\"gG6#F*F+/F*;\"\"!
F+" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT 269 8 "Step 2: " }{TEXT
-1 174 "As in the above problem, a typical point in this space can be \+
expressed as an infinite sum of the eigenfunctions. Thus a typical fun
ction of t in the space can be written as " }}{PARA 0 "" 0 "" {TEXT
-1 23 " T(t) = " }{XPPEDIT 18 0 "sum(T[n](t)*phi[n],n);
" "6#-%$sumG6$*&-&%\"TG6#%\"nG6#%\"tG\"\"\"&%$phiG6#F+F.F+" }{TEXT -1
0 "" }}{PARA 0 "" 0 "" {TEXT -1 10 "where the " }{XPPEDIT 18 0 "phi[n]
;" "6#&%$phiG6#%\"nG" }{TEXT -1 76 " 's are the eigenfunctions. Thus, \+
the solution to the PDE will have the form" }}{PARA 0 "" 0 "" {TEXT
-1 24 " u(t,x) = " }{XPPEDIT 18 0 "sum(T[n](t)*sin(n*Pi*
x),n);" "6#-%$sumG6$*&-&%\"TG6#%\"nG6#%\"tG\"\"\"-%$sinG6#*(F+F.%#PiGF
.%\"xGF.F.F+" }{TEXT -1 2 " ." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT 270 7 "Step 3:" }{TEXT -1 119 " To determine the
infinite system of equations, we suppose that u satisfies the PDE and
determine the implications for " }{XPPEDIT 18 0 "T[n];" "6#&%\"TG6#%
\"nG" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 5 " " }{XPPEDIT
18 0 "diff(u,`$`(t,2));" "6#-%%diffG6$%\"uG-%\"$G6$%\"tG\"\"#" }{TEXT
-1 3 " + " }{XPPEDIT 18 0 "1/5;" "6#*&\"\"\"F$\"\"&!\"\"" }{TEXT -1 1
" " }{XPPEDIT 18 0 "diff(T,t);" "6#-%%diffG6$%\"TG%\"tG" }{TEXT -1 3 "
= " }{XPPEDIT 18 0 "sum((diff(T[t],`$`(t,2))+diff(T(t),t)/5)*sin(n*Pi
*x),n);" "6#-%$sumG6$*&,&-%%diffG6$&%\"TG6#%\"tG-%\"$G6$F.\"\"#\"\"\"*
&-F)6$-F,6#F.F.F3\"\"&!\"\"F3F3-%$sinG6#*(%\"nGF3%#PiGF3%\"xGF3F3F?" }
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 3 "and" }}{PARA 0 "" 0 ""
{TEXT -1 3 " " }{XPPEDIT 18 0 "1/9;" "6#*&\"\"\"F$\"\"*!\"\"" }
{TEXT -1 2 " " }{XPPEDIT 18 0 "diff(u,`$`(x,2));" "6#-%%diffG6$%\"uG-
%\"$G6$%\"xG\"\"#" }{TEXT -1 4 " = " }{XPPEDIT 18 0 "sum(-n^2*Pi^2/9*
T[n](t)*sin(n*Pi*x),n);" "6#-%$sumG6$,$*,%\"nG\"\"#%#PiGF)\"\"*!\"\"-&
%\"TG6#F(6#%\"tG\"\"\"-%$sinG6#*(F(F3F*F3%\"xGF3F3F,F(" }{TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 71 "Becau
se we are supposing that u satisfies the PDE, then it must be that" }}
{PARA 0 "" 0 "" {TEXT -1 10 " " }{XPPEDIT 18 0 "diff(T[n],`$`
(t,2));" "6#-%%diffG6$&%\"TG6#%\"nG-%\"$G6$%\"tG\"\"#" }{TEXT -1 3 " +
" }{XPPEDIT 18 0 "1/5;" "6#*&\"\"\"F$\"\"&!\"\"" }{TEXT -1 2 " " }
{XPPEDIT 18 0 "diff(T[n],t);" "6#-%%diffG6$&%\"TG6#%\"nG%\"tG" }{TEXT
-1 6 " = - (" }{XPPEDIT 18 0 "n^2*Pi^2/9;" "6#*(%\"nG\"\"#%#PiGF%\"\"*
!\"\"" }{TEXT -1 3 " ) " }{XPPEDIT 18 0 "T[n](t);" "6#-&%\"TG6#%\"nG6#
%\"tG" }{TEXT -1 21 ", for each integer n." }}{PARA 0 "" 0 "" {TEXT
-1 112 "We list a few of these differential equations. First, we choos
e how many terms we will use in the approximation." }}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 5 "N:=4;" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 92 "for n from 1 to N do\n ode
[n]:=diff(T[n](t),t,t)+diff(T[n](t),t)/5=-n^2*Pi^2/9*T[n](t);\nod;" }}
}{PARA 0 "" 0 "" {TEXT 271 8 "Step 4: " }{TEXT -1 268 "In order to sol
ve these second order equations, we need the initial values appropriat
e for this problem. We find this from the initial value of the PDE, bu
t rewritten in term of the orthogonal sequence that is associated with
this problem. We compute the coefficients." }}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 29 "f:=x->5*sin(2*Pi*x);\ng:=x->0;" }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 162 "for n from 1 to N do\n A[n]:=int(f(x)*sin
(n*Pi*x),x=0..1)/int(sin(n*Pi*x)^2,x=0..1);\n B[n]:=int(g(x)*sin(n*P
i*x),x=0..1)/int(sin(n*Pi*x)^2,x=0..1);\nod;\nn:='n':" }}}{PARA 0 ""
0 "" {TEXT -1 0 "" }{TEXT 274 7 "Step 5." }{TEXT -1 66 " We now solve \+
several of the ordinary differential equations in t." }}{EXCHG {PARA
0 "> " 0 "" {MPLTEXT 1 0 118 "for n from 1 to N do\n dsolve(\{ode[n]
,T[n](0)=A[n],D(T[n])(0)=B[n]\},T[n](t)):\n T[n]:=unapply(rhs(%),t);
\nod;\nn:='n':" }}}{PARA 0 "" 0 "" {TEXT -1 57 "These provide an appro
ximation for a solution of the PDE." }}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 42 "u:=(t,x)->sum(T[n](t)*sin(n*Pi*x),n=0..N);" }}}{PARA
0 "" 0 "" {TEXT 275 6 "Step 6" }{TEXT -1 74 ". We check to see that th
is u satisfied the partial differential equation." }}{EXCHG {PARA 0 ">
" 0 "" {MPLTEXT 1 0 63 "simplify(diff(u(t,x),t,t)-diff(u(t,x),x,x)/9+
diff(u(t,x),t)/5);" }}}{PARA 0 "" 0 "" {TEXT -1 45 "We check the homog
eneous boundary conditions." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0
15 "u(t,0); u(t,1);" }}}{PARA 0 "" 0 "" {TEXT -1 72 "We check to see i
f u(0,x) is a good approximation for the initial value." }}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 64 "plot([f(x),u(0,x)],x=0..3,color=[re
d,black],style=[line,point]);" }}}{PARA 0 "" 0 "" {TEXT -1 42 "The fit
for the initial value is perfect. " }}{PARA 0 "" 0 "" {TEXT -1 38 "Ne
xt, we draw a graph of the solution." }}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 63 "plot3d(u(t,x),x=0..1,t=0..10,axes=NORMAL,orientation=
[-30,65]);" }}}{PARA 0 "" 0 "" {TEXT -1 37 "Here is an animation of th
e solution." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "plots[animate
](u(t,x),x=0..1,t=0..5);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "
restart;" }{TEXT -1 0 "" }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1
{PARA 3 "" 0 "" {TEXT -1 45 "Example 3: Laplace's Equation on a Rectan
gle." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 42 "H
ere is the partial differential equation:" }}{PARA 0 "" 0 "" {TEXT -1
14 " 0 = " }{XPPEDIT 18 0 "diff(u,`$`(x,2));" "6#-%%diffG6$%
\"uG-%\"$G6$%\"xG\"\"#" }{TEXT -1 3 " + " }{XPPEDIT 18 0 "diff(u,`$`(y
,2));" "6#-%%diffG6$%\"uG-%\"$G6$%\"yG\"\"#" }{TEXT -1 2 " ." }}{PARA
0 "" 0 "" {TEXT -1 61 "Here are the homogeneous boundary conditions: u
( 0, y) = 0 = " }{XPPEDIT 18 0 "diff(u,x);" "6#-%%diffG6$%\"uG%\"xG" }
{TEXT -1 2 "( " }{XPPEDIT 18 0 "Pi/2;" "6#*&%#PiG\"\"\"\"\"#!\"\"" }
{TEXT -1 6 " , y)." }}{PARA 0 "" 0 "" {TEXT -1 83 "Here are the bounda
ry conditions on the remaining sides: u(x,0) = sin(x) = u(x,1)." }}
{PARA 0 "" 0 "" {TEXT -1 68 "Again, we follow exactly the same steps a
s in the previous problems." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA
0 "" 0 "" {TEXT 276 7 "Step 1:" }{TEXT -1 94 " Separation of variables
leads to two ordinary differential equations and boundary conditions:
" }}{PARA 0 "" 0 "" {TEXT -1 15 " X '' = " }{XPPEDIT 18 0 "mu;
" "6#%#muG" }{TEXT -1 25 " X, with X(0) = 0 = X '( " }{XPPEDIT 18 0 "P
i/2;" "6#*&%#PiG\"\"\"\"\"#!\"\"" }{TEXT -1 15 " ) and Y '' = " }
{XPPEDIT 18 0 "-mu;" "6#,$%#muG!\"\"" }{TEXT -1 3 " Y." }}{PARA 0 ""
0 "" {TEXT -1 82 "From Section 3.3, we can see that the Sturm-Liouvill
e problem in X has eigenvalues" }}{PARA 0 "" 0 "" {TEXT -1 20 " \+
" }{XPPEDIT 18 0 "mu;" "6#%#muG" }{TEXT -1 33 " = 1 + 2 n
, for n = 0, 1, 2, ..." }}{PARA 0 "" 0 "" {TEXT -1 115 "and the eigen
functions are sin((1+2 n) x). These from the maximal orthogonal family
. The appropriate dotproduct is " }}{PARA 0 "" 0 "" {TEXT -1 40 " \+
< f(x) , g(x) > = " }{XPPEDIT 18 0 "int(f(x)*g(x),x =
0 .. Pi/2);" "6#-%$intG6$*&-%\"fG6#%\"xG\"\"\"-%\"gG6#F*F+/F*;\"\"!*&
%#PiGF+\"\"#!\"\"" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT 277 8 "Step
2: " }{TEXT -1 149 "A typical point in this space can be expressed as
an infinite sum of the eigenfunctions. Thus a typical function of y i
n the space can be written as " }}{PARA 0 "" 0 "" {TEXT -1 23 " \+
Y(y) = " }{XPPEDIT 18 0 "sum(Y[n](y)*phi[n],n);" "6#-%$sumG6$
*&-&%\"YG6#%\"nG6#%\"yG\"\"\"&%$phiG6#F+F.F+" }{TEXT -1 0 "" }}{PARA
0 "" 0 "" {TEXT -1 10 "where the " }{XPPEDIT 18 0 "phi[n];" "6#&%$phiG
6#%\"nG" }{TEXT -1 76 " 's are the eigenfunctions. Thus, the solution \+
to the PDE will have the form" }}{PARA 0 "" 0 "" {TEXT -1 24 " \+
u(x,y) = " }{XPPEDIT 18 0 "sum(Y[n](y)*sin((1+2*n)*x),n);" "6#-
%$sumG6$*&-&%\"YG6#%\"nG6#%\"yG\"\"\"-%$sinG6#*&,&F.F.*&\"\"#F.F+F.F.F
.%\"xGF.F.F+" }{TEXT -1 2 " ." }}{PARA 0 "" 0 "" {TEXT 278 7 "Step 3:
" }{TEXT -1 119 " To determine the infinite system of equations, we su
ppose that u satisfies the PDE and determine the implications for " }
{XPPEDIT 18 0 "T[n];" "6#&%\"TG6#%\"nG" }{TEXT -1 1 "." }}{PARA 0 ""
0 "" {TEXT -1 5 " " }{XPPEDIT 18 0 "diff(u,`$`(x,2));" "6#-%%diffG
6$%\"uG-%\"$G6$%\"xG\"\"#" }{TEXT -1 8 " = - " }{XPPEDIT 18 0 "sum(
(1+2*n)^2,n);" "6#-%$sumG6$*$,&\"\"\"F(*&\"\"#F(%\"nGF(F(F*F+" }
{XPPEDIT 18 0 "Y[n](y);" "6#-&%\"YG6#%\"nG6#%\"yG" }{TEXT -1 15 " sin(
(1+2 n) x)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT
-1 3 "and" }}{PARA 0 "" 0 "" {TEXT -1 5 " " }{XPPEDIT 18 0 "diff(u
,`$`(y,2));" "6#-%%diffG6$%\"uG-%\"$G6$%\"yG\"\"#" }{TEXT -1 4 " = "
}{XPPEDIT 18 0 "sum(diff(Y[n],`$`(y,2)),n);" "6#-%$sumG6$-%%diffG6$&%
\"YG6#%\"nG-%\"$G6$%\"yG\"\"#F," }{TEXT -1 15 " sin((1+2 n) x)" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 71 "Because w
e are supposing that u satisfies the PDE, then it must be that" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 6 " " }
{XPPEDIT 18 0 "diff(Y[n],`$`(y,2));" "6#-%%diffG6$&%\"YG6#%\"nG-%\"$G6
$%\"yG\"\"#" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "(1+2*n)^2;" "6#*$,&\"\"
\"F%*&\"\"#F%%\"nGF%F%F'" }{TEXT -1 1 " " }{XPPEDIT 18 0 "Y[n](y);" "6
#-&%\"YG6#%\"nG6#%\"yG" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1
112 "We list a few of these differential equations. First, we choose h
ow many terms we will use in the approximation." }}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 5 "N:=3;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1
0 72 "for n from 0 to N do\n ode[n]:=diff(Y[n](y),y,y)=(1+2*n)^2*Y[n
](y);\nod;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "
" 0 "" {TEXT 279 8 "Step 4: " }{TEXT -1 245 "In order to solve these s
econd order equations, we need what are boundary values appropriate fo
r this problem. We find this from the initial value of the PDE, but re
written in term of the orthogonal sequence that is associated with thi
s problem." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "f:=x->sin(x);
\ng:=x->sin(x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 186 "for n f
rom 0 to N do\n A[n]:=int(f(x)*sin((1+2*n)*x),x=0..Pi/2)/int(sin((1+
2*n)*x)^2,x=0..Pi/2);\n B[n]:=int(g(x)*sin((1+2*n)*x),x=0..Pi/2)/int
(sin((1+2*n)*x)^2,x=0..Pi/2);\nod;\nn:='n':" }}}{PARA 0 "" 0 "" {TEXT
-1 0 "" }{TEXT 280 7 "Step 5." }{TEXT -1 66 " We now solve several of \+
the ordinary differential equations in t." }}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 115 "for n from 0 to N do\n dsolve(\{ode[n],Y[n](0)=A[n
],Y[n](1)=B[n]\},Y[n](y)):\n Y[n]:=unapply(rhs(%),y);\nod;\nn:='n':
" }}}{PARA 0 "" 0 "" {TEXT -1 57 "These provide an approximation for a
solution of the PDE." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "u:=
(x,y)->sum(Y[n](y)*sin((1+2*n)*x),n=0..N);" }}}{PARA 0 "" 0 "" {TEXT
-1 43 "Sometimes, I just like to see the solution." }}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 7 "u(x,y);" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT 281 6 "Step 6" }{TEXT -1
74 ". We check to see that this u satisfied the partial differential e
quation." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "simplify(diff(u(
x,y),x,x)+diff(u(x,y),y,y));" }}}{PARA 0 "" 0 "" {TEXT -1 45 "We check
the homogeneous boundary conditions." }}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 24 "u(0,y); D[1](u)(Pi/2,y);" }}}{PARA 0 "" 0 "" {TEXT
-1 84 "We check to see if u(0,x) is a good approximation for the non-h
omogeneous boundary.." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 67 "plo
t([f(x),u(x,0)],x=0..Pi/2,color=[red,black],style=[line,point]);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 67 "plot([g(x),u(x,1)],x=0..Pi/2
,color=[red,black],style=[line,point]);" }}}{PARA 0 "" 0 "" {TEXT -1
42 "The fit for the initial value is perfect. " }}{PARA 0 "" 0 ""
{TEXT -1 38 "Next, we draw a graph of the solution." }}{EXCHG {PARA 0
"> " 0 "" {MPLTEXT 1 0 66 "plot3d(u(x,y),x=0..Pi/2,y=0..1,axes=NORMAL,
orientation=[-120,70]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "r
estart;" }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 3 "" 0 ""
{TEXT -1 42 "Example 4: Laplace's Equation on a Washer." }}{PARA 0 ""
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 42 "Here is the partial \+
differential equation:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 ""
0 "" {TEXT -1 10 " " }{XPPEDIT 18 0 "diff(r*diff(u(t,theta),r
),r)/r+diff(u(r,theta),`$`(theta,2))/(r^2);" "6#,&*&-%%diffG6$*&%\"rG
\"\"\"-F&6$-%\"uG6$%\"tG%&thetaGF)F*F)F*F)!\"\"F**&-F&6$-F.6$F)F1-%\"$
G6$F1\"\"#F**$F)F;F2F*" }{TEXT -1 6 " = 0." }}{PARA 0 "" 0 "" {TEXT
-1 67 "Recall that there is the implicit homogeneous boundary conditio
ns: " }}{PARA 0 "" 0 "" {TEXT -1 18 " " }{TEXT 283 1
"u" }{TEXT -1 4 "(r, " }{XPPEDIT 18 0 "Pi;" "6#%#PiG" }{TEXT -1 4 ") =
" }{TEXT 284 1 "u" }{TEXT -1 4 "(r, " }{XPPEDIT 18 0 "-Pi;" "6#,$%#Pi
G!\"\"" }{TEXT -1 6 ") and " }{XPPEDIT 18 0 "diff(u,theta);" "6#-%%dif
fG6$%\"uG%&thetaG" }{TEXT -1 5 " (r, " }{XPPEDIT 18 0 "theta;" "6#%&th
etaG" }{TEXT -1 4 ") = " }{XPPEDIT 18 0 "diff(u,theta);" "6#-%%diffG6$
%\"uG%&thetaG" }{TEXT -1 7 " (r, - " }{XPPEDIT 18 0 "theta;" "6#%&thet
aG" }{TEXT -1 3 "). " }}{PARA 0 "" 0 "" {TEXT -1 94 "We suppose that t
his washer has inner radius 1 and outer radius 2. The boundary conditi
ons are" }}{PARA 0 "" 0 "" {TEXT -1 37 " \+
u(1," }{XPPEDIT 18 0 "theta" "6#%&thetaG" }{TEXT -1 8 ") = u(2," }
{XPPEDIT 18 0 "theta" "6#%&thetaG" }{TEXT -1 8 ") = sin(" }{XPPEDIT
18 0 "theta" "6#%&thetaG" }{TEXT -1 9 ") for - " }{XPPEDIT 18 0 "Pi;
" "6#%#PiG" }{XPPEDIT 18 0 "` ` <= ` `;" "6#1%\"~GF$" }{XPPEDIT 18 0 "
theta;" "6#%&thetaG" }{XPPEDIT 18 0 "` ` <= ` `;" "6#1%\"~GF$" }
{XPPEDIT 18 0 "Pi;" "6#%#PiG" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 131 "Again, to illustrate that all \+
these problems have the same structure, we follow exactly the same ste
ps as in the previous problems." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT 282 7 "Step 1:" }{TEXT -1 94 " Separation of var
iables leads to two ordinary differential equations and boundary condi
tions:" }}{PARA 0 "" 0 "" {TEXT -1 5 " " }{XPPEDIT 18 0 "theta;" "
6#%&thetaG" }{TEXT -1 6 " '' + " }{XPPEDIT 18 0 "lambda^2*theta;" "6#*
&%'lambdaG\"\"#%&thetaG\"\"\"" }{TEXT -1 6 " = 0, " }{XPPEDIT 18 0 "th
eta;" "6#%&thetaG" }{TEXT -1 4 " (- " }{XPPEDIT 18 0 "pi;" "6#%#piG" }
{TEXT -1 4 ") = " }{XPPEDIT 18 0 "theta;" "6#%&thetaG" }{TEXT -1 3 " (
" }{XPPEDIT 18 0 "pi;" "6#%#piG" }{TEXT -1 7 "), and " }{XPPEDIT 18
0 "theta;" "6#%&thetaG" }{TEXT -1 5 " '(- " }{XPPEDIT 18 0 "pi;" "6#%#
piG" }{TEXT -1 4 ") = " }{XPPEDIT 18 0 "theta;" "6#%&thetaG" }{TEXT
-1 4 " '( " }{XPPEDIT 18 0 "pi;" "6#%#piG" }{TEXT -1 2 ")." }}{PARA 0
"" 0 "" {TEXT -1 3 "and" }}{PARA 0 "" 0 "" {TEXT -1 18 " r( rR ') \+
' - " }{XPPEDIT 18 0 "lambda^2;" "6#*$%'lambdaG\"\"#" }{TEXT -1 7 " R \+
= 0." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 136 "
The first equation has homogeneous, periodic boundary conditions. Reca
lling Section 3.3, we know that eigenvalues and eigenfunctions are" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 5 " " }
{XPPEDIT 18 0 "lambda;" "6#%'lambdaG" }{TEXT -1 9 " = 0, " }
{XPPEDIT 18 0 "theta;" "6#%&thetaG" }{TEXT -1 5 " = 1." }}{PARA 0 ""
0 "" {TEXT -1 5 " " }{XPPEDIT 18 0 "-lambda^2;" "6#,$*$%'lambdaG\"
\"#!\"\"" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "-n^2;" "6#,$*$%\"nG\"\"#!
\"\"" }{TEXT -1 5 ", " }{XPPEDIT 18 0 "Theta;" "6#%&ThetaG" }{TEXT
-1 1 "(" }{XPPEDIT 18 0 "theta;" "6#%&thetaG" }{TEXT -1 10 ") = cos(n \+
" }{XPPEDIT 18 0 "theta;" "6#%&thetaG" }{TEXT -1 13 ") or sin(n " }
{XPPEDIT 18 0 "theta;" "6#%&thetaG" }{TEXT -1 2 ")." }}{PARA 0 "" 0 "
" {TEXT -1 118 " These sines and cosines functions form the maximal or
thogonal family for this problem. The appropriate dotproduct is " }}
{PARA 0 "" 0 "" {TEXT -1 40 " < f(x) , g(x) > = \+
" }{XPPEDIT 18 0 "int(f(x)*g(x),x = -Pi .. Pi);" "6#-%$intG6$*&-%\"fG6
#%\"xG\"\"\"-%\"gG6#F*F+/F*;,$%#PiG!\"\"F2" }{TEXT -1 1 "." }}{PARA 0
"" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 285 8 "Step 2: " }{TEXT
-1 148 "A typical point in the space can be expressed as an infinite s
um of the eigenfunctions. Thus a typical function of y in the space ca
n be written as " }}{PARA 0 "" 0 "" {TEXT -1 23 " R(r) \+
= " }{XPPEDIT 18 0 "sum(R[n](r)*phi[n],n);" "6#-%$sumG6$*&-&%\"RG6#%\"
nG6#%\"rG\"\"\"&%$phiG6#F+F.F+" }{TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT -1 10 "where the " }{XPPEDIT 18 0 "phi[n];" "6#&%$phiG6#%\"nG" }
{TEXT -1 114 " 's are the eigenfunctions. A convenient ordering of the
terms presents the solution to the PDE as having the form" }}{PARA 0
"" 0 "" {TEXT -1 23 " u( r, " }{XPPEDIT 18 0 "theta;"
"6#%&thetaG" }{TEXT -1 5 " ) = " }{XPPEDIT 18 0 "sum(R[n,c](r)*cos(n*t
heta),n);" "6#-%$sumG6$*&-&%\"RG6$%\"nG%\"cG6#%\"rG\"\"\"-%$cosG6#*&F+
F/%&thetaGF/F/F+" }{TEXT -1 3 " + " }{XPPEDIT 18 0 "sum(R[n,s](r)*sin(
n*theta),n);" "6#-%$sumG6$*&-&%\"RG6$%\"nG%\"sG6#%\"rG\"\"\"-%$sinG6#*
&F+F/%&thetaGF/F/F+" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}{PARA 0 "" 0 "" {TEXT 286 7 "Step 3:" }{TEXT -1 160 " To determine th
e infinite system of equations, we suppose that u satisfies the PDE an
d determine the implications for R(r). First, we multiply the equation
by " }{XPPEDIT 18 0 "r^2;" "6#*$%\"rG\"\"#" }{TEXT -1 30 ". This chan
ges the equation to" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 10 " " }{XPPEDIT 18 0 "r*diff(r*diff(u(t,theta),r)
,r)+diff(u(r,theta),`$`(theta,2));" "6#,&*&%\"rG\"\"\"-%%diffG6$*&F%F&
-F(6$-%\"uG6$%\"tG%&thetaGF%F&F%F&F&-F(6$-F.6$F%F1-%\"$G6$F1\"\"#F&" }
{TEXT -1 6 " = 0." }}{PARA 0 "" 0 "" {TEXT -1 38 "We now compute the \+
requirements for R." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 5 " " }{XPPEDIT 18 0 "r*diff(r*diff(u,r),r);" "6#*&%\"r
G\"\"\"-%%diffG6$*&F$F%-F'6$%\"uGF$F%F$F%" }{TEXT -1 5 " = " }
{XPPEDIT 18 0 "sum(r^2*diff(R[n,c],`$`(r,2))+r*diff(R[n,c],r)*cos(n*th
eta),n);" "6#-%$sumG6$,&*&%\"rG\"\"#-%%diffG6$&%\"RG6$%\"nG%\"cG-%\"$G
6$F(F)\"\"\"F5*(F(F5-F+6$&F.6$F0F1F(F5-%$cosG6#*&F0F5%&thetaGF5F5F5F0
" }{TEXT -1 2 " +" }}{PARA 0 "" 0 "" {TEXT -1 60 " \+
" }{XPPEDIT 18 0 "sum(r^2*diff
(R[n,s],`$`(r,2))+r*diff(R[n,s],r)*sin(n*theta),n);" "6#-%$sumG6$,&*&%
\"rG\"\"#-%%diffG6$&%\"RG6$%\"nG%\"sG-%\"$G6$F(F)\"\"\"F5*(F(F5-F+6$&F
.6$F0F1F(F5-%$sinG6#*&F0F5%&thetaGF5F5F5F0" }{TEXT -1 1 " " }}{PARA 0
"" 0 "" {TEXT -1 3 "and" }}{PARA 0 "" 0 "" {TEXT -1 5 " " }
{XPPEDIT 18 0 "diff(u,`$`(theta,2));" "6#-%%diffG6$%\"uG-%\"$G6$%&thet
aG\"\"#" }{TEXT -1 5 " = -" }{XPPEDIT 18 0 "sum(n^2*R[n,c],n = 1 .. i
nfinity);" "6#-%$sumG6$*&%\"nG\"\"#&%\"RG6$F'%\"cG\"\"\"/F';F-%)infini
tyG" }{TEXT -1 5 " cos(" }{XPPEDIT 18 0 "n*theta;" "6#*&%\"nG\"\"\"%&t
hetaGF%" }{TEXT -1 6 ") - " }{XPPEDIT 18 0 "sum(n^2*R[n,s],n = 1 .. \+
infinity);" "6#-%$sumG6$*&%\"nG\"\"#&%\"RG6$F'%\"sG\"\"\"/F';F-%)infin
ityG" }{TEXT -1 5 " sin(" }{XPPEDIT 18 0 "n*theta;" "6#*&%\"nG\"\"\"%&
thetaGF%" }{TEXT -1 2 ")." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT -1 226 "Since the left side of the above equations add to
give zero, so do the right sides. This produces an infinite pair of e
quations. One infinite set goes with the cosine series, the other goes
with the sine series. We list a few." }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 5 "N:=3;" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 56 "ODE[0,c]:=r^2*diff(R[0,c](r),r,r)+r*diff(R[0,c](r),r)
=0;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 204 "for n from 1 to N d
o\n ODE[n,c]:=r^2*diff(R[n,c](r),r,r)+r*diff(R[n,c](r),r)-\n \+
n^2*R[n,c](r)=0;\n ODE[n,s]:=r^2*diff(R[n,s](r),r,r)+r*diff(R[
n,s](r),r)-\n n^2*R[n,s](r)=0;\nod;" }}}{PARA 0 "" 0 "
" {TEXT 287 8 "Step 4: " }{TEXT -1 242 "In order to solve these second
order equations, we need the boundary values appropriate for this pro
blem. We find this from the boundary values of the PDE, but rewritten \+
in term of the orthogonal sequence that is associated with this proble
m." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 58 "We \+
input the boundary condition at radius 1 and radius 2. " }}{PARA 0 ""
0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "fr1:=x-
>sin(x);\nfr2:=x->sin(x);" }}}{PARA 0 "" 0 "" {TEXT -1 57 "Here are th
e coefficients for the boundary conditions of " }{XPPEDIT 18 0 "R[n,c]
;" "6#&%\"RG6$%\"nG%\"cG" }{TEXT -1 24 " at r = 1 and at r = 2." }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 178 "for n from 0 to N do\n at
_r1[n,c]:=int(fr1(x)*cos(n*x),x=-Pi..Pi)/int(cos(n*x)^2,x=-Pi..Pi);\n \+
at_r2[n,c]:=int(fr2(x)*cos(n*x),x=-Pi..Pi)/int(cos(n*x)^2,x=-Pi..Pi)
;\nod;\nn:='n':" }}}{PARA 0 "" 0 "" {TEXT -1 57 "Here are the coeffici
ents for the boundary conditions of " }{XPPEDIT 18 0 "R[n,s];" "6#&%\"
RG6$%\"nG%\"sG" }{TEXT -1 24 " at r = 1 and at r = 2." }}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 178 "for n from 1 to N do\n at_r1[n,s
]:=int(fr1(x)*sin(n*x),x=-Pi..Pi)/int(sin(n*x)^2,x=-Pi..Pi);\n at_r2
[n,s]:=int(fr2(x)*sin(n*x),x=-Pi..Pi)/int(sin(n*x)^2,x=-Pi..Pi);\nod;
\nn:='n':" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 288 7 "Step 5." }
{TEXT -1 80 " We solve N of the ordinary differential equations that d
efine the cosine terms." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 137 "
for n from 0 to N do\n dsolve(\{ODE[n,c],R[n,c](1)=at_r1[n,c],R[n,c]
(2)=at_r2[n,c]\},R[n,c](r));\n R[n,c]:=unapply(rhs(%),r);\nod;\nn:='
n';" }}}{PARA 0 "" 0 "" {TEXT -1 77 "We solve N of the ordinary differ
ential equations that define the sine terms." }}{EXCHG {PARA 0 "> " 0
"" {MPLTEXT 1 0 137 "for n from 1 to N do\n dsolve(\{ODE[n,s],R[n,s]
(1)=at_r1[n,s],R[n,s](2)=at_r2[n,s]\},R[n,s](r));\n R[n,s]:=unapply(
rhs(%),r);\nod;\nn:='n';" }}}{PARA 0 "" 0 "" {TEXT -1 21 "We make the \+
solution." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 99 "u:=(r,theta)->s
um(R[n,c](r)*cos(n*theta),n=0..N)+\n sum(R[n,s](r)*sin(n*
theta),n=1..N);" }}}{PARA 0 "" 0 "" {TEXT -1 32 "Let's just look at th
e solution." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "u(r,theta);"
}}}{PARA 0 "" 0 "" {TEXT 289 7 "Step 6." }{TEXT -1 98 " We check to se
e the solution satisfies the partial differential equation and boundar
y conditions." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 77 "diff(r*diff
(u(r,theta),r),r)/r+diff(u(r,theta),theta,theta)/r^2;\nsimplify(%);" }
}}{PARA 0 "" 0 "" {TEXT -1 42 "We check the periodic boundary conditio
ns." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "u(r,Pi); u(r,-Pi);" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "D[2](u)(r,Pi); D[2](u)(r,-
Pi);" }}}{PARA 0 "" 0 "" {TEXT -1 38 "We check the edge boundary condi
tions." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "u(1,theta); u(2,th
eta);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 107 "plots[cylinderplo
t]([r,theta,u(r,theta)],r=1..2,theta=0..2*Pi,\n orientation=[
-20,70],axes=NORMAL);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }
}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 3 "" 0 "" {TEXT -1
38 "Example 5: The Heat Equation on a Disk" }}{PARA 0 "" 0 "" {TEXT
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 323 "We consider the diffusion equa
tion on a disk. The assumption is that the steady state has been deter
mined and we are in the steady state situation. For this illustration,
we take the initial value to be independent of theta. Assume the disk
to have radius a. Thus, the equation and boundary conditions are give
n as follows:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT -1 16 " " }{XPPEDIT 18 0 "diff(r*diff(u(r,theta),
r),r)/r;" "6#*&-%%diffG6$*&%\"rG\"\"\"-F%6$-%\"uG6$F(%&thetaGF(F)F(F)F
(!\"\"" }{TEXT -1 4 " = " }{XPPEDIT 18 0 "diff(u,t);" "6#-%%diffG6$%
\"uG%\"tG" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT -1 57 "Here are the homogeneous boundary conditions: u(t,
2) = 0." }}{PARA 0 "" 0 "" {TEXT -1 41 "Here is the initial value: u(0
, r) = 1 -" }{XPPEDIT 18 0 "r^2;" "6#*$%\"rG\"\"#" }{TEXT -1 2 ". " }
}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 290 7 "Step 1:
" }{TEXT -1 94 " Separation of variables leads to two ordinary differe
ntial equations and boundary conditions:" }}{PARA 0 "" 0 "" {TEXT -1
20 "(1) r ( r R ' ) ' + " }{XPPEDIT 18 0 "lambda^2;" "6#*$%'lambdaG\"
\"#" }{TEXT -1 1 " " }{XPPEDIT 18 0 "r^2;" "6#*$%\"rG\"\"#" }{TEXT -1
51 " R = 0, with R(a) = 0, and R bounded on [0, a), and" }}{PARA 0 ""
0 "" {TEXT -1 10 "(2) T ' + " }{XPPEDIT 18 0 "lambda^2;" "6#*$%'lambda
G\"\"#" }{TEXT -1 7 " T = 0." }}{PARA 0 "" 0 "" {TEXT -1 144 "Equation
(1) is a Sturm-Liouville problem. We ask Maple for solutions. This is
a second order differential equation, so we expect two solutions." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0
14 "restart;\na:=2;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 82 "dsol
ve(\{r^2*diff(R(r),r,r)+r*diff(R(r),r)+lambda^2*r^2*R(r)=0\},R(r),outp
ut=basis);" }}}{PARA 0 "" 0 "" {TEXT -1 78 "Because we ask that soluti
ons remain bounded on [0, a), we exclude BesselY(0, " }{XPPEDIT 18 0 "
lambda;" "6#%'lambdaG" }{TEXT -1 69 " r). The requirement that R(a) sh
ould be zero imposes a condition on " }{XPPEDIT 18 0 "lambda;" "6#%'la
mbdaG" }{TEXT -1 31 ". This requirement asks that " }{XPPEDIT 18 0 "
a*lambda;" "6#*&%\"aG\"\"\"%'lambdaGF%" }{TEXT -1 64 " should be a z
ero of BesselJ(0, r). We list several values of " }{XPPEDIT 18 0 "lamb
da;" "6#%'lambdaG" }{TEXT -1 6 " with " }{TEXT 291 1 "a" }{TEXT -1 5 "
= 2." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 82 "a:=2;\nfor n from 1 to 10 do\n lambda[n]:=evalf(Be
sselJZeros(0,n))/a;\nod;\nn:='n';" }}}{PARA 0 "" 0 "" {TEXT -1 325 "Kn
owing what is the Sturm Liouville problem and knowing what are its eig
envalues and eigenvectors, we have a maximal orthogonal family. Also, \+
we know the dot product and can verify that the eigenvalues are orthog
onal with that dot product. The dot product for functions f and g on [
0,a] in this problem is defined as follows:" }}{PARA 0 "" 0 "" {TEXT
-1 18 " < f, g> = " }{XPPEDIT 18 0 "int(f(r)*g(r)*r,r = 0 .. a)
;" "6#-%$intG6$*(-%\"fG6#%\"rG\"\"\"-%\"gG6#F*F+F*F+/F*;\"\"!%\"aG" }
{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 178 "In confirmation, we che
ck the orthogonality of two of the eigenfunctions. Since our eigenvalu
es are approximations, we expect the value of these next integrals to \+
be almost zero." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 121 "int(Bess
elJ(0,lambda[2]*r)*BesselJ(0,lambda[5]*r)*r,r=0..a);\nint(BesselJ(0,la
mbda[4]*r)*BesselJ(0,lambda[5]*r)*r,r=0..a);" }}}{PARA 0 "" 0 ""
{TEXT 292 8 "Step 2: " }{TEXT -1 149 "A typical point in this space ca
n be expressed as an infinite sum of the eigenfunctions. Thus a typica
l function of t in the space can be written as " }}{PARA 0 "" 0 ""
{TEXT -1 23 " T(t) = " }{XPPEDIT 18 0 "sum(T[n](t)*phi[
n],n);" "6#-%$sumG6$*&-&%\"TG6#%\"nG6#%\"tG\"\"\"&%$phiG6#F+F.F+" }
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 9 "where the" }}{PARA 0 "" 0
"" {TEXT -1 27 " " }{XPPEDIT 18 0 "phi[n];"
"6#&%$phiG6#%\"nG" }{TEXT -1 20 " (r) = BesselJ(0, " }{XPPEDIT 18 0
"lambda[n];" "6#&%'lambdaG6#%\"nG" }{TEXT -1 5 " r). " }}{PARA 0 "" 0
"" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 48 "Thus, the solution to \+
the PDE will have the form" }}{PARA 0 "" 0 "" {TEXT -1 25 " \+
u(t, r) = " }{XPPEDIT 18 0 "sum(T[n],n);" "6#-%$sumG6$&%\"TG6#%\"n
GF)" }{TEXT -1 1 " " }{TEXT 294 10 "BesselJ(0," }{TEXT -1 1 " " }
{XPPEDIT 18 0 "lambda[n];" "6#&%'lambdaG6#%\"nG" }{TEXT -1 2 " " }
{TEXT 293 4 "r )." }}{PARA 0 "" 0 "" {TEXT 295 7 "Step 3:" }{TEXT -1
119 " To determine the infinite system of equations, we suppose that u
satisfies the PDE and determine the implications for " }{XPPEDIT 18
0 "T[n];" "6#&%\"TG6#%\"nG" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT
-1 5 " " }{XPPEDIT 18 0 "diff(u,t);" "6#-%%diffG6$%\"uG%\"tG" }
{TEXT -1 3 " = " }{XPPEDIT 18 0 "sum(diff(T[t],t),n);" "6#-%$sumG6$-%%
diffG6$&%\"TG6#%\"tGF,%\"nG" }{TEXT -1 1 " " }{TEXT 297 10 "BesselJ(0,
" }{TEXT -1 1 " " }{XPPEDIT 18 0 "lambda[n];" "6#&%'lambdaG6#%\"nG" }
{TEXT -1 2 " " }{TEXT 296 4 "r )." }}{PARA 0 "" 0 "" {TEXT -1 66 "It \+
is easy to see what taking the derivative of u with respect to " }
{TEXT 298 1 "t" }{TEXT -1 49 " does. We use Maple to help us see how t
o compute" }}{PARA 0 "" 0 "" {TEXT -1 35 " \+
" }{XPPEDIT 18 0 "diff(r*diff(u(r,theta),r),r)/r;" "6#*&-%%diff
G6$*&%\"rG\"\"\"-F%6$-%\"uG6$F(%&thetaGF(F)F(F)F(!\"\"" }{TEXT -1 3 " \+
." }}{PARA 0 "" 0 "" {TEXT -1 105 "Here is the experiment, take u to \+
be made of the simple one term product of T(t) and the Bessel function
." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "u:=(t,r)->T(t)*BesselJ(
0,lambda*r);" }}}{PARA 0 "" 0 "" {TEXT -1 29 "Now, take the derivative
s of " }{TEXT 299 1 "u" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "expand(diff(r*diff(u(t,r
),r),r)/r);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0
"" 0 "" {TEXT -1 77 "This makes the complicated looking derivatives wi
th respect to r look simple:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA
0 "" 0 "" {TEXT -1 16 " " }{XPPEDIT 18 0 "diff(r*diff(u
(r,theta),r),r)/r;" "6#*&-%%diffG6$*&%\"rG\"\"\"-F%6$-%\"uG6$F(%&theta
GF(F)F(F)F(!\"\"" }{TEXT -1 8 " = - " }{XPPEDIT 18 0 "sum(T[n](t),n
);" "6#-%$sumG6$-&%\"TG6#%\"nG6#%\"tGF*" }{TEXT -1 1 " " }{XPPEDIT 18
0 "lambda[n]^2;" "6#*$&%'lambdaG6#%\"nG\"\"#" }{TEXT -1 1 " " }
{XPPEDIT 18 0 "BesselJ(0, lambda[n]*r);" "6#-%(BesselJG6$\"\"!*&&%'lam
bdaG6#%\"nG\"\"\"%\"rGF," }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1
0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 111 "B
ecause we are supposing that u satisfies the PDE, then the infinite sy
stem of equations that must be solved is" }}{PARA 0 "" 0 "" {TEXT -1
14 " " }{XPPEDIT 18 0 "diff(T[n],t);" "6#-%%diffG6$&%\"TG
6#%\"nG%\"tG" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "-lambda[n]^2;" "6#,$*$
&%'lambdaG6#%\"nG\"\"#!\"\"" }{TEXT -1 1 " " }{XPPEDIT 18 0 "T[n];" "6
#&%\"TG6#%\"nG" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 20 "for each
integer n. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT
300 7 "Step 4:" }{TEXT -1 29 " We know solutions for this: " }
{XPPEDIT 18 0 "T[n](t);" "6#-&%\"TG6#%\"nG6#%\"tG" }{TEXT -1 3 " = " }
{XPPEDIT 18 0 "c[n];" "6#&%\"cG6#%\"nG" }{TEXT -1 5 " exp(" }{XPPEDIT
18 0 "-lambda[n]^2;" "6#,$*$&%'lambdaG6#%\"nG\"\"#!\"\"" }{TEXT -1 27
" t). To find the constants " }{XPPEDIT 18 0 "c[n];" "6#&%\"cG6#%\"nG
" }{TEXT -1 68 ", we use the intial values. They come from having that
u(0,r) = 1 - " }{XPPEDIT 18 0 "r^2;" "6#*$%\"rG\"\"#" }{TEXT -1 9 ". \+
That is" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 7
" 1 - " }{XPPEDIT 18 0 "r^2;" "6#*$%\"rG\"\"#" }{TEXT -1 3 " = " }
{XPPEDIT 18 0 "sum(c[n],n);" "6#-%$sumG6$&%\"cG6#%\"nGF)" }{TEXT -1 1
" " }{XPPEDIT 18 0 "BesselJ(0,lambda[n]*r);" "6#-%(BesselJG6$\"\"!*&&%
'lambdaG6#%\"nG\"\"\"%\"rGF," }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT
-1 61 "These initial values can be computed as Fourier Coefficients."
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "a:=2;f:=r->1-(r/2)^2;" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 168 "for n from 1 to 10 do\n c
[n]:=evalf(Int(f(r)*BesselJ(0,BesselJZeros(0,n)/a*r)*r,r=0..a))/\n \+
evalf(Int(BesselJ(0,BesselJZeros(0,n)/a*r)^2*r,r=0..a));\nod;\nn:
='n';" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{PARA
0 "" 0 "" {TEXT 301 7 "Step 5:" }{TEXT -1 33 " We now approximate the \+
solution." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 71 "u:=(t,r)->sum(c
[n]*exp(-lambda[n]^2*t)*BesselJ(0,lambda[n]*r),n=1..10);" }}}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 302 7 "Step 6:" }{TEXT
-1 88 " Here is a check that this is correct. First we check the parti
al differential equation." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 52 "simplify(diff(u(t,r),t)-diff(r*diff
(u(t,r),r),r)/r);" }}}{PARA 0 "" 0 "" {TEXT -1 97 "Next, we check the \+
boundary condition at r = a. To make this check, we need to know the v
alue of " }{XPPEDIT 18 0 "lambda[n];" "6#&%'lambdaG6#%\"nG" }{TEXT -1
1 "." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 69 "for n from 1 to 10 d
o\n lambda[n]:=BesselJZeros(0,n)/a;\nod:\nn:='n';" }}}{EXCHG {PARA
0 "> " 0 "" {MPLTEXT 1 0 7 "u(t,a);" }}}{PARA 0 "" 0 "" {TEXT -1 36 "F
inally, we check the initial value." }}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 27 "plot([f(r),u(0,r)],r=0..a);" }}}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "T[n](t)
;" "6#-&%\"TG6#%\"nG6#%\"tG" }{TEXT -1 21 ", for each integer n." }}
{PARA 0 "" 0 "" {TEXT -1 112 "We list a few of these differential equa
tions. First, we choose how many terms we will use in the approximatio
n." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "plot(\{u(0,r),f(r)\},r
=0..a);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 76 "for n from 0 to \+
N do\n ode[n]:=diff(T[n](t),t)=-(1+n^2*Pi^2/9)*T[n](t);\nod;" }}}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 596 "Does this structure unify the problems? \+
Why did the student who asked me about creating an overview not see th
is himself? I think I know. The presentation that has come before has \+
elements which are anachronisms from the time before the power of comp
uter algebra systems. Back then, we chose solutions of boundary value \+
problems to make computations easy. For example, we used sinh(x) and s
inh(1 - x) because they are zero at zero and one, respectively. This m
ade computing coefficients easier for humans. The task of making the c
omputations easy is much less important with a tool such as Maple." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 332 "So, it s
eemed well to look at those linear problems and see that their solutio
ns fit into the standard structure of a Hilbert Space. With this basis
for the linear problems, the nest step should be to get a good unders
tanding for non-linear problems. The numerical methods presented in th
ese notes is a first step in that direction." }}{PARA 0 "" 0 "" {TEXT
-1 1 " " }}}{MARK "0 1" 61 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }
{PAGENUMBERS 0 1 2 33 1 1 }