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{SECT 0 {EXCHG {PARA 256 "" 0 "" {TEXT -1 12 "MTHSC 434/63" }}{PARA 
258 "" 0 "" {TEXT 256 26 "Second Summer Session 1998" }}{PARA 257 "" 
0 "" {TEXT -1 24 "Final Exam Answser Sheet" }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 128 "This answer sheet uses Maple t
o handle the integrations, plotting, and other tedious bookkeeping, wh
ich are items that were not " }{TEXT 257 7 "heavily" }{TEXT -1 8 " gra
ded." }}{PARA 0 "" 0 "" {TEXT -1 116 "The problem setup cannot be done
 by Maple or any other computer assistance and was counted much more i
n the grading." }}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 9 "Problem 1" }}
{PARA 0 "" 0 "" {TEXT -1 49 "Find the Fourier Cosine series for the fu
nction, " }{XPPEDIT 18 0 "f(x) = x*(Pi-x);" "6#/-%\"fG6#%\"xG*&F'\"\"
\",&%#PiGF)F'!\"\"F)" }{TEXT -1 17 " on the interval " }{XPPEDIT 18 0 
"0 <= x;" "6#1\"\"!%\"xG" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "x <= Pi;
" "6#1%\"xG%#PiG" }{TEXT -1 1 "." }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
38 "First, clear the workspace and define " }{XPPEDIT 18 0 "f(x);" "6#
-%\"fG6#%\"xG" }{TEXT -1 1 "." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 30 "restart: f := x -> x*(Pi - x);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 14 "Next, compute " }{XPPEDIT 18 0 "a[0] = 2*Int(f(x),x = 0 .. Pi)/
(2*Pi);" "6#/&%\"aG6#\"\"!*(\"\"#\"\"\"-%$IntG6$-%\"fG6#%\"xG/F1;F'%#P
iGF**&\"\"#F*F4F*!\"\"" }{TEXT -1 1 "." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 37 "a[0] := (2/(2*Pi))*int(f(x),x=0..Pi);" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 38 "Then compute the general coefficient, " }
{XPPEDIT 18 0 "a[k] = 2*Int(f(x)*cos(k*x),x = 0 .. Pi)/Pi;" "6#/&%\"aG
6#%\"kG*(\"\"#\"\"\"-%$IntG6$*&-%\"fG6#%\"xGF*-%$cosG6#*&F'F*F2F*F*/F2
;\"\"!%#PiGF*F:!\"\"" }{TEXT -1 1 "." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 42 "a[k] := (2/Pi)*int(f(x)*cos(k*x),x=0..Pi);" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 13 "Note that if " }{XPPEDIT 18 0 "k;
" "6#%\"kG" }{TEXT -1 40 " is an integer, then the term involving " }
{XPPEDIT 18 0 "sin(k*Pi);" "6#-%$sinG6#*&%\"kG\"\"\"%#PiGF(" }{TEXT 
-1 21 " will be zero. Also, " }{XPPEDIT 18 0 "cos(k*Pi) = (-1)^k;" "6#
/-%$cosG6#*&%\"kG\"\"\"%#PiGF)),$\"\"\"!\"\"F(" }{TEXT -1 2 ". " }}
{PARA 0 "" 0 "" {TEXT -1 83 "To use this coefficient expression in the
 series, we need to make it a function of " }{XPPEDIT 18 0 "k;" "6#%\"
kG" }{TEXT -1 1 "." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "coef \+
:= unapply(a[k],k);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 51 "It is easy
 to see the pattern for the coefficients." }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 96 "a[1] := coef(1);a[2] := coef(2);a[3] := coef(3);a[4
] := coef(4);a[5] := coef(5);a[6] := coef(6);" }}}{EXCHG {PARA 0 "" 0 
"" {TEXT -1 87 "The series - actually an arbitrary partial sum of the \+
series - can be simply defined by" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 49 "CF := (x,N) -> a[0]+sum(coef(k)*cos(k*x),k=1..N);" }}
}{EXCHG {PARA 0 "" 0 "" {TEXT -1 86 "We can verify the convergence of \+
the series by looking a the first few approximations." }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 53 "plot([f(x),CF(x,0),CF(x,2),CF(x,4),
CF(x,6)],x=0..Pi);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 79 "A more info
rmative plot for higher order approximations is a plot of the error." 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "plot((f(x)-CF(x,50)),x=0.
.Pi);" }}}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 9 "Problem 2" }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 47 "Find the Fourier S
ine series for the function, " }{XPPEDIT 18 0 "f(x) = x*(Pi-x);" "6#/-
%\"fG6#%\"xG*&F'\"\"\",&%#PiGF)F'!\"\"F)" }{TEXT -1 17 " on the interv
al " }{XPPEDIT 18 0 "0 <= x;" "6#1\"\"!%\"xG" }{TEXT -1 5 " and " }
{XPPEDIT 18 0 "x <= Pi;" "6#1%\"xG%#PiG" }{TEXT -1 1 "." }}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 38 "First, clear the workspace and define " }
{XPPEDIT 18 0 "f(x);" "6#-%\"fG6#%\"xG" }{TEXT -1 1 "." }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "restart: f := x -> x*(Pi - x);" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 38 "Next compute the general coefficie
nt, " }{XPPEDIT 18 0 "b[k] = 2*Int(f(x)*sin(k*x),x = 0 .. Pi)/Pi;" "6#
/&%\"bG6#%\"kG*(\"\"#\"\"\"-%$IntG6$*&-%\"fG6#%\"xGF*-%$sinG6#*&F'F*F2
F*F*/F2;\"\"!%#PiGF*F:!\"\"" }{TEXT -1 1 "." }}}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 42 "b[k] := (2/Pi)*int(f(x)*sin(k*x),x=0..Pi);" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 13 "Note that if " }{XPPEDIT 18 0 "k;
" "6#%\"kG" }{TEXT -1 40 " is an integer, then the term involving " }
{XPPEDIT 18 0 "sin(k*Pi);" "6#-%$sinG6#*&%\"kG\"\"\"%#PiGF(" }{TEXT 
-1 21 " will be zero. Also, " }{XPPEDIT 18 0 "cos(k*Pi) = (-1)^k;" "6#
/-%$cosG6#*&%\"kG\"\"\"%#PiGF)),$\"\"\"!\"\"F(" }{TEXT -1 2 ". " }}
{PARA 0 "" 0 "" {TEXT -1 83 "To use this coefficient expression in the
 series, we need to make it a function of " }{XPPEDIT 18 0 "k;" "6#%\"
kG" }{TEXT -1 1 "." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "coef \+
:= unapply(b[k],k);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 51 "It is easy
 to see the pattern for the coefficients." }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 96 "b[1] := coef(1);b[2] := coef(2);b[3] := coef(3);b[4
] := coef(4);b[5] := coef(5);b[6] := coef(6);" }}}{EXCHG {PARA 0 "" 0 
"" {TEXT -1 87 "The series - actually an arbitrary partial sum of the \+
series - can be simply defined by" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 44 "SF := (x,N) -> sum(coef(k)*sin(k*x),k=1..N);" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 86 "We can verify the convergence of t
he series by looking a the first few approximations." }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 45 "plot([f(x),SF(x,1),SF(x,3),SF(x,5)],x=0..
Pi);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 79 "A more informative plot f
or higher order approximations is a plot of the error." }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "plot((f(x)-SF(x,25)),x=0..Pi);" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 101 "One immediate observation is that
 the Sine series converges much more quickly than the Cosine series." 
}}{PARA 0 "" 0 "" {TEXT -1 108 "The maximum error is almost 100 times \+
smaller with only half as many terms. This is because the function and
" }}{PARA 0 "" 0 "" {TEXT -1 60 "the first derivative are continuous u
nder the odd extension." }}}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 9 "Probl
em 3" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 19 "F
ind the solution, " }{XPPEDIT 18 0 "u(x,t);" "6#-%\"uG6$%\"xG%\"tG" }
{TEXT -1 83 ", to the heat equation on a bar where one end is insulate
d, the temperature is held" }}{PARA 0 "" 0 "" {TEXT -1 108 "constant a
t the other end, and the initial temperature distribution is specified
.  More specifically, solve " }}{PARA 0 "" 0 "" {XPPEDIT 18 0 "Diff(u,
t) = Diff(u,`$`(x,2));" "6#/-%%DiffG6$%\"uG%\"tG-F%6$F'-%\"$G6$%\"xG\"
\"#" }{TEXT -1 11 "    with   " }{XPPEDIT 18 0 "Diff(u(0,t),x) = 0;" "
6#/-%%DiffG6$-%\"uG6$\"\"!%\"tG%\"xGF*" }{TEXT -1 4 ",   " }{XPPEDIT 
18 0 "u(Pi,t) = 0;" "6#/-%\"uG6$%#PiG%\"tG\"\"!" }{TEXT -1 10 ",   and
   " }{XPPEDIT 18 0 "u(x,0) = x*(Pi-x);" "6#/-%\"uG6$%\"xG\"\"!*&F'\"
\"\",&%#PiGF*F'!\"\"F*" }{TEXT -1 1 "," }}{PARA 0 "" 0 "" {TEXT -1 7 "
where  " }{XPPEDIT 18 0 "0 <= x;" "6#1\"\"!%\"xG" }{TEXT -1 3 ",  " }
{XPPEDIT 18 0 "x <= Pi;" "6#1%\"xG%#PiG" }{TEXT -1 8 ",  and  " }
{XPPEDIT 18 0 "0 <= t;" "6#1\"\"!%\"tG" }{TEXT -1 1 "." }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 38 "First, clear \+
the workspace and define " }{XPPEDIT 18 0 "f(x);" "6#-%\"fG6#%\"xG" }
{TEXT -1 1 "." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "restart: f
 := x -> x*(Pi - x);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 51 "Start by \+
assuming a separable solution of the form " }{XPPEDIT 18 0 "G(t)*H(t);
" "6#*&-%\"GG6#%\"tG\"\"\"-%\"HG6#F'F(" }{TEXT -1 43 ".  Then we can c
onclude that if there is a " }}{PARA 0 "" 0 "" {TEXT -1 28 "solution o
f this form, then " }{XPPEDIT 18 0 "G(t);" "6#-%\"GG6#%\"tG" }{TEXT 
-1 5 " and " }{XPPEDIT 18 0 "H(x);" "6#-%\"HG6#%\"xG" }{TEXT -1 61 " m
ust satisfy two ordinary differential equations of the form" }}{PARA 
0 "" 0 "" {XPPEDIT 18 0 "Diff(G,t) = lambda*G;" "6#/-%%DiffG6$%\"GG%\"
tG*&%'lambdaG\"\"\"F'F+" }{TEXT -1 9 "   and   " }{XPPEDIT 18 0 "Diff(
H,`$`(x,2)) = lambda*H;" "6#/-%%DiffG6$%\"HG-%\"$G6$%\"xG\"\"#*&%'lamb
daG\"\"\"F'F/" }{TEXT -1 35 ",   where the separation constant, " }
{XPPEDIT 18 0 "lambda;" "6#%'lambdaG" }{TEXT -1 33 ", is the same for \+
both equations." }}{PARA 0 "" 0 "" {TEXT -1 35 "We consider the possib
le values of " }{XPPEDIT 18 0 "lambda;" "6#%'lambdaG" }{TEXT -1 37 ", \+
and their role in the equation for " }{XPPEDIT 18 0 "H(x);" "6#-%\"HG6
#%\"xG" }{TEXT -1 24 ". Note that the boundary" }}{PARA 0 "" 0 "" 
{TEXT -1 23 "conditions imply that  " }{XPPEDIT 18 0 "Diff(H(0),x) = 0
;" "6#/-%%DiffG6$-%\"HG6#\"\"!%\"xGF*" }{TEXT -1 8 "  and   " }
{XPPEDIT 18 0 "H(Pi) = 0;" "6#/-%\"HG6#%#PiG\"\"!" }{TEXT -1 1 "." }}}
{SECT 1 {PARA 4 "" 0 "" {XPPEDIT 18 0 "lambda = 0;" "6#/%'lambdaG\"\"!
" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 22 "In this case, we have \+
" }{XPPEDIT 18 0 "Diff(H,`$`(x,2)) = 0;" "6#/-%%DiffG6$%\"HG-%\"$G6$%
\"xG\"\"#\"\"!" }{TEXT -1 42 ", and the general solution is of the for
m " }{XPPEDIT 18 0 "H(x) = A*x+B;" "6#/-%\"HG6#%\"xG,&*&%\"AG\"\"\"F'F
+F+%\"BGF+" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 14 "The conditi
on " }{XPPEDIT 18 0 "Diff(H(0),x) = 0;" "6#/-%%DiffG6$-%\"HG6#\"\"!%\"
xGF*" }{TEXT -1 14 " implies that " }{XPPEDIT 18 0 "A = 0;" "6#/%\"AG
\"\"!" }{TEXT -1 20 ", and the condition " }{XPPEDIT 18 0 "H(Pi) = 0;
" "6#/-%\"HG6#%#PiG\"\"!" }{TEXT -1 14 " implies that " }{XPPEDIT 18 
0 "B = 0;" "6#/%\"BG\"\"!" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 
58 "We can instruct Maple to verify this reasoning as follows:" }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "de0 := diff(H(x),x$2)=0;" }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "sol0 := dsolve(de0,H(x));" 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "H:=unapply(rhs(sol0),x);
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "solve(\{D(H)(0)=0,H(Pi)
=0\},\{_C1,_C2\});" }}}}{SECT 1 {PARA 4 "" 0 "" {XPPEDIT 18 0 "lambda \+
= omega^2;" "6#/%'lambdaG*$%&omegaG\"\"#" }{TEXT -1 0 "" }}{PARA 0 "" 
0 "" {TEXT -1 22 "In this case, we have " }{XPPEDIT 18 0 "Diff(H,`$`(x
,2)) = omega^2*H;" "6#/-%%DiffG6$%\"HG-%\"$G6$%\"xG\"\"#*&%&omegaG\"\"
#F'\"\"\"" }{TEXT -1 42 ", and the general solution is of the form " }
{XPPEDIT 18 0 "H(x) = A*cosh(omega*x)+B*sinh(omega*x);" "6#/-%\"HG6#%
\"xG,&*&%\"AG\"\"\"-%%coshG6#*&%&omegaGF+F'F+F+F+*&%\"BGF+-%%sinhG6#*&
F0F+F'F+F+F+" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 14 "The condi
tion " }{XPPEDIT 18 0 "Diff(H(0),x) = 0;" "6#/-%%DiffG6$-%\"HG6#\"\"!%
\"xGF*" }{TEXT -1 14 " implies that " }{XPPEDIT 18 0 "B = 0;" "6#/%\"B
G\"\"!" }{TEXT -1 20 ", and the condition " }{XPPEDIT 18 0 "H(Pi) = 0;
" "6#/-%\"HG6#%#PiG\"\"!" }{TEXT -1 14 " implies that " }{XPPEDIT 18 
0 "A = 0;" "6#/%\"AG\"\"!" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 
58 "We can instruct Maple to verify this reasoning as follows:" }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "H:='H';de1 := diff(H(x),x$2)
=omega^2*H(x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "sol1 := d
solve(de1,H(x));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "H:=unap
ply(rhs(sol1),x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "solve(
\{D(H)(0)=0,H(Pi)=0\},\{_C1,_C2\});" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" 
}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 0 "" }{XPPEDIT 18 0 "lambda = -ome
ga^2;" "6#/%'lambdaG,$*$%&omegaG\"\"#!\"\"" }{TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT -1 22 "In this case, we have " }{XPPEDIT 18 0 "Diff(H,`$`
(x,2)) = -omega^2*H;" "6#/-%%DiffG6$%\"HG-%\"$G6$%\"xG\"\"#,$*&%&omega
G\"\"#F'\"\"\"!\"\"" }{TEXT -1 42 ", and the general solution is of th
e form " }{XPPEDIT 18 0 "H(x) = A*cos(omega*x)+B*sin(omega*x);" "6#/-%
\"HG6#%\"xG,&*&%\"AG\"\"\"-%$cosG6#*&%&omegaGF+F'F+F+F+*&%\"BGF+-%$sin
G6#*&F0F+F'F+F+F+" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 14 "The \+
condition " }{XPPEDIT 18 0 "Diff(H(0),x) = 0;" "6#/-%%DiffG6$-%\"HG6#
\"\"!%\"xGF*" }{TEXT -1 14 " implies that " }{XPPEDIT 18 0 "B = 0;" "6
#/%\"BG\"\"!" }{TEXT -1 20 ", and the condition " }{XPPEDIT 18 0 "H(Pi
) = 0;" "6#/-%\"HG6#%#PiG\"\"!" }{TEXT -1 14 " implies that " }
{XPPEDIT 18 0 "A = 0;" "6#/%\"AG\"\"!" }{TEXT -1 4 " or " }{XPPEDIT 
18 0 "cos(omega*Pi) = 0;" "6#/-%$cosG6#*&%&omegaG\"\"\"%#PiGF)\"\"!" }
{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 43 "This last condition will
 be true whenever  " }{XPPEDIT 18 0 "omega = (2*k-1)/2;" "6#/%&omegaG*
&,&*&\"\"#\"\"\"%\"kGF)F)\"\"\"!\"\"F)\"\"#F," }{TEXT -1 6 "  for " }
{XPPEDIT 18 0 "k;" "6#%\"kG" }{TEXT -1 16 " = 1, 2, 3, ...." }}{PARA 
0 "" 0 "" {TEXT -1 58 "We can instruct Maple to verify this reasoning \+
as follows:" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "H:='H';de2 :=
 diff(H(x),x$2)=-omega^2*H(x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 25 "sol2 := dsolve(de2,H(x));" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 24 "H:=unapply(rhs(sol2),x);" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 39 "solve(\{D(H)(0)=0,H(Pi)=0\},\{omega,_C2\});" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 56 "Note that it is up to us to notice
 that other values of " }{XPPEDIT 18 0 "omega;" "6#%&omegaG" }{TEXT 
-1 11 " also work." }}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 21 "With these
 values of " }{XPPEDIT 18 0 "omega;" "6#%&omegaG" }{TEXT -1 41 " we se
e that there are solutions for the " }{XPPEDIT 18 0 "G(t);" "6#-%\"GG6
#%\"tG" }{TEXT -1 27 " equation that are given by" }}{PARA 0 "" 0 "" 
{TEXT -1 8 "        " }{XPPEDIT 18 0 "G(t) = C*exp(-((2*k-1)/2)^2*t);
" "6#/-%\"GG6#%\"tG*&%\"CG\"\"\"-%$expG6#,$*&*&,&*&\"\"#F*%\"kGF*F*\"
\"\"!\"\"F*\"\"#F6\"\"#F'F*F6F*" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 57 "Thus there are solutions to the heat equation of the form
" }}{PARA 0 "" 0 "" {TEXT -1 6 "      " }{XPPEDIT 18 0 "G(t)*H(x) = c[
k]*exp(-((2*k-1)/2)^2*t)*cos((2*k-1)*x/2);" "6#/*&-%\"GG6#%\"tG\"\"\"-
%\"HG6#%\"xGF)*(&%\"cG6#%\"kGF)-%$expG6#,$*&*&,&*&\"\"#F)F2F)F)\"\"\"!
\"\"F)\"\"#F=\"\"#F(F)F=F)-%$cosG6#*(,&*&\"\"#F)F2F)F)\"\"\"F=F)F-F)\"
\"#F=F)" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 48 "We can instruc
t Maple to verify this as follows:" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 61 "u[k] := (x,t) -> c[k]*exp(-((2*k-1)/2)^2*t)*cos((2*k-
1)*x/2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "D[2](u[k])(x,t)
 - D[1,1](u[k])(x,t);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 53 "We have \+
one remaining condition to satisfy, namely   " }{XPPEDIT 18 0 "u(x,0) \+
= x*(Pi-x);" "6#/-%\"uG6$%\"xG\"\"!*&F'\"\"\",&%#PiGF*F'!\"\"F*" }
{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 82 "Since we have an infini
te set of solutions, each with an unspecified coefficient, " }
{XPPEDIT 18 0 "c[k];" "6#&%\"cG6#%\"kG" }{TEXT -1 22 ", can we find a \+
linear" }}{PARA 0 "" 0 "" {TEXT -1 64 "combination, which yields the s
olution?   That is, can we write " }{XPPEDIT 18 0 "u(x,t) = sum(u[k](x
,t),k = 1 .. infinity);" "6#/-%\"uG6$%\"xG%\"tG-%$sumG6$-&F%6#%\"kG6$F
'F(/F/;\"\"\"%)infinityG" }{TEXT -1 3 " ? " }}{PARA 0 "" 0 "" {TEXT 
-1 14 "The answer is " }{TEXT 258 3 "yes" }{TEXT -1 36 ", provided we \+
can find coefficients " }{XPPEDIT 18 0 "c[k];" "6#&%\"cG6#%\"kG" }
{TEXT -1 11 ", such that" }}{PARA 0 "" 0 "" {XPPEDIT 18 0 "u(x,0) = su
m(c[k]*cos((2*k-1)*x/2),k = 1 .. infinity);" "6#/-%\"uG6$%\"xG\"\"!-%$
sumG6$*&&%\"cG6#%\"kG\"\"\"-%$cosG6#*(,&*&\"\"#F1F0F1F1\"\"\"!\"\"F1F'
F1\"\"#F:F1/F0;\"\"\"%)infinityG" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 41 "But this is just the answ
er to Problem 1 " }{TEXT 259 11 "- or is it?" }{TEXT -1 1 " " }}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 260 82 
"Unfortunately, the answer to Problem 1 does NOT give us the answer to
 this problem" }}{PARA 0 "" 0 "" {TEXT 261 49 "because the frequency t
erm in Problem 1 is simply" }{TEXT -1 1 " " }{XPPEDIT 18 0 "k;" "6#%\"
kG" }{TEXT -1 1 " " }{TEXT 262 32 "while the frequency term here is" }
{TEXT -1 1 " " }{XPPEDIT 18 0 "(2*k-1)/2;" "6#*&,&*&\"\"#\"\"\"%\"kGF'
F'\"\"\"!\"\"F'\"\"#F*" }{TEXT -1 1 "." }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 98 "The preceding work constitutes the answer I was looking f
or on the test. Curiosity demands that we" }}{PARA 0 "" 0 "" {TEXT -1 
97 "say something about the coefficients of the series that we just de
rived.  As the book mentions in" }}{PARA 0 "" 0 "" {TEXT -1 99 "Sectio
n 16.1 on page 791, the Sturm-Liouville theorem (Section 6.1) asserts \+
that this will work and" }}{PARA 0 "" 0 "" {TEXT -1 45 "that the coeff
icients can be calculated from " }}{PARA 0 "" 0 "" {TEXT -1 5 "     " 
}{XPPEDIT 18 0 "c[k] = Int(f(x)*cos((2*k-1)*x/2),x = 0 .. Pi)/Int(cos(
(2*k-1)*x/2)^2,x = 0 .. Pi);" "6#/&%\"cG6#%\"kG*&-%$IntG6$*&-%\"fG6#%
\"xG\"\"\"-%$cosG6#*(,&*&\"\"#F1F'F1F1\"\"\"!\"\"F1F0F1\"\"#F:F1/F0;\"
\"!%#PiGF1-F*6$*$-F36#*(,&*&\"\"#F1F'F1F1\"\"\"F:F1F0F1\"\"#F:\"\"#/F0
;F>F?F:" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 102 "Note that the \+
term in the denominator is exactly what we need if we want to reproduc
e a single term of" }}{PARA 0 "" 0 "" {TEXT -1 11 "the series." }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 101 "One immediate observation is that
 the Sine series converges much more quickly than the Cosine series." 
}}{PARA 0 "" 0 "" {TEXT -1 108 "The maximum error is almost 100 times \+
smaller with only half as many terms. This is because the function and
" }}{PARA 0 "" 0 "" {TEXT -1 60 "the first derivative are continuous u
nder the odd extension." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "
d[k] := int((cos((2*k-1)*x/2))^2,x=0..Pi);" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 52 "c[k] := (1/d[k])*int(f(x)*cos((2*k-1)*x/2),x=0..Pi)
;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "ccoef := unapply(c[k],
k);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 68 "c[1] := ccoef(1);c[2
] := ccoef(2);c[3] := ccoef(3);c[4] := ccoef(4);" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 53 "HF := (x,N) -> sum(ccoef(k)*cos((2*k-1)*x/2),
k=1..N);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 53 "plot([f(x),HF(x
,1),HF(x,2),HF(x,3),HF(x,4)],x=0..Pi);" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 62 "Notice that each of the partial sums has a zero derivativ
e at " }{XPPEDIT 18 0 "x = 0;" "6#/%\"xG\"\"!" }{TEXT -1 51 ", which i
s one of the requirements of the solution." }}{PARA 0 "" 0 "" {TEXT 
-1 15 "However, since " }{XPPEDIT 18 0 "f(x);" "6#-%\"fG6#%\"xG" }
{TEXT -1 21 " has a derivative of " }{XPPEDIT 18 0 "Pi;" "6#%#PiG" }
{TEXT -1 4 " at " }{XPPEDIT 18 0 "x = 0;" "6#/%\"xG\"\"!" }{TEXT -1 
33 ", the series converges slowly at " }{XPPEDIT 18 0 "x = 0;" "6#/%\"
xG\"\"!" }{TEXT -1 22 ". On the other hand it" }}{PARA 0 "" 0 "" 
{TEXT -1 93 "definitely seems to be converging.  This is further suppo
rted by the error plot for 25 terms." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 28 "plot(f(x)-HF(x,25),x=0..Pi);" }}}{EXCHG {PARA 0 "" 0 
"" {TEXT -1 56 "An approximation to the solultion using 25 terms is th
us" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 79 "u[25] := (x,t) -> sum
(ccoef(k)*exp(-((2*k-1)/2)^2*t)*cos((2*k-1)*x/2),k=1..25);" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 125 "We can get a plot of this approximate so
lution.  It shows the temperature rising at the insulated end with a z
ero derivative." }}{PARA 0 "" 0 "" {TEXT -1 123 "It also shows that th
e total heat in the rod is decreasing as the temperature in the rod de
cays to zero, the temperature at" }}{PARA 0 "" 0 "" {TEXT -1 12 "the f
ar end." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 57 "plot3d(u[25],0..
Pi,0..1,axes=BOXED,labels=['x','t','u']);" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 70 "We can also use Maple's plot package to show this solutio
n in a movie." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "with(plots
): animate(u[25](x,t), x=0..Pi, t=0..4);" }}}}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 9 "Problem 4" }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 19 "Find the solution, \+
" }{XPPEDIT 18 0 "u(x,t);" "6#-%\"uG6$%\"xG%\"tG" }{TEXT -1 79 ", to t
he wave equation for a vibrating string which is initially flat but ha
s a" }}{PARA 0 "" 0 "" {TEXT -1 54 "specified initial velocity.  More \+
specifically, solve " }}{PARA 0 "" 0 "" {XPPEDIT 18 0 "Diff(u,`$`(t,2)
) = Diff(u,`$`(x,2));" "6#/-%%DiffG6$%\"uG-%\"$G6$%\"tG\"\"#-F%6$F'-F)
6$%\"xG\"\"#" }{TEXT -1 11 "    with   " }{XPPEDIT 18 0 "Diff(u(0,t),x
) = 0;" "6#/-%%DiffG6$-%\"uG6$\"\"!%\"tG%\"xGF*" }{TEXT -1 4 ",   " }
{XPPEDIT 18 0 "u(Pi,t) = 0;" "6#/-%\"uG6$%#PiG%\"tG\"\"!" }{TEXT -1 4 
",   " }{XPPEDIT 18 0 "u(x,0) = 0;" "6#/-%\"uG6$%\"xG\"\"!F(" }{TEXT 
-1 11 ",   and    " }{XPPEDIT 18 0 "Diff(u(x,0),x) = x*(Pi-x);" "6#/-%
%DiffG6$-%\"uG6$%\"xG\"\"!F**&F*\"\"\",&%#PiGF-F*!\"\"F-" }{TEXT -1 1 
"," }}{PARA 0 "" 0 "" {TEXT -1 7 "where  " }{XPPEDIT 18 0 "0 <= x;" "6
#1\"\"!%\"xG" }{TEXT -1 3 ",  " }{XPPEDIT 18 0 "x <= Pi;" "6#1%\"xG%#P
iG" }{TEXT -1 8 ",  and  " }{XPPEDIT 18 0 "0 <= t;" "6#1\"\"!%\"tG" }
{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 38 "First c lear the workspace and define " }{XPPEDIT 18 0 
"f(x);" "6#-%\"fG6#%\"xG" }{TEXT -1 1 "." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 28 "restart: f := x -> x*(Pi-x);" }}}{EXCHG {PARA 0 "" 0 
"" {TEXT -1 110 "First  note that the specific conditions do not exact
ly match the verbal description. The derivative condition" }}{PARA 0 "
" 0 "" {TEXT -1 3 "at " }{XPPEDIT 18 0 "x = 0;" "6#/%\"xG\"\"!" }
{TEXT -1 103 " implies that we have a sliding connection at that end s
o that the string always has a zero derivative." }}{PARA 0 "" 0 "" 
{TEXT -1 107 "If we keep this condition, then the problem is essential
ly identical to Problem 3.  Let's consider the case" }}{PARA 0 "" 0 "
" {TEXT -1 6 "where " }{XPPEDIT 18 0 "u(0,t) = 0;" "6#/-%\"uG6$\"\"!%
\"tGF'" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 
0 "" {TEXT -1 51 "Start by assuming a separable solution of the form \+
" }{XPPEDIT 18 0 "G(t)*H(t);" "6#*&-%\"GG6#%\"tG\"\"\"-%\"HG6#F'F(" }
{TEXT -1 43 ".  Then we can conclude that if there is a " }}{PARA 0 "
" 0 "" {TEXT -1 28 "solution of this form, then " }{XPPEDIT 18 0 "G(t)
;" "6#-%\"GG6#%\"tG" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "H(x);" "6#-%
\"HG6#%\"xG" }{TEXT -1 61 " must satisfy two ordinary differential equ
ations of the form" }}{PARA 0 "" 0 "" {XPPEDIT 18 0 "Diff(G,`$`(t,2)) \+
= lambda*G;" "6#/-%%DiffG6$%\"GG-%\"$G6$%\"tG\"\"#*&%'lambdaG\"\"\"F'F
/" }{TEXT -1 9 "   and   " }{XPPEDIT 18 0 "Diff(H,`$`(x,2)) = lambda*H
;" "6#/-%%DiffG6$%\"HG-%\"$G6$%\"xG\"\"#*&%'lambdaG\"\"\"F'F/" }{TEXT 
-1 35 ",   where the separation constant, " }{XPPEDIT 18 0 "lambda;" "
6#%'lambdaG" }{TEXT -1 33 ", is the same for both equations." }}{PARA 
0 "" 0 "" {TEXT -1 35 "We consider the possible values of " }{XPPEDIT 
18 0 "lambda;" "6#%'lambdaG" }{TEXT -1 37 ", and their role in the equ
ation for " }{XPPEDIT 18 0 "H(x);" "6#-%\"HG6#%\"xG" }{TEXT -1 24 ". N
ote that the boundary" }}{PARA 0 "" 0 "" {TEXT -1 23 "conditions imply
 that  " }{XPPEDIT 18 0 "H(0) = 0;" "6#/-%\"HG6#\"\"!F'" }{TEXT -1 8 "
  and   " }{XPPEDIT 18 0 "H(Pi) = 0;" "6#/-%\"HG6#%#PiG\"\"!" }{TEXT 
-1 1 "." }}}{SECT 1 {PARA 4 "" 0 "" {XPPEDIT 18 0 "lambda = 0;" "6#/%'
lambdaG\"\"!" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 22 "In this ca
se, we have " }{XPPEDIT 18 0 "Diff(H,`$`(x,2)) = 0;" "6#/-%%DiffG6$%\"
HG-%\"$G6$%\"xG\"\"#\"\"!" }{TEXT -1 42 ", and the general solution is
 of the form " }{XPPEDIT 18 0 "H(x) = A*x+B;" "6#/-%\"HG6#%\"xG,&*&%\"
AG\"\"\"F'F+F+%\"BGF+" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 14 "
The condition " }{XPPEDIT 18 0 "H(x) = 0;" "6#/-%\"HG6#%\"xG\"\"!" }
{TEXT -1 14 " implies that " }{XPPEDIT 18 0 "B = 0;" "6#/%\"BG\"\"!" }
{TEXT -1 20 ", and the condition " }{XPPEDIT 18 0 "H(Pi) = 0;" "6#/-%
\"HG6#%#PiG\"\"!" }{TEXT -1 14 " implies that " }{XPPEDIT 18 0 "A = 0;
" "6#/%\"AG\"\"!" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 58 "We ca
n instruct Maple to verify this reasoning as follows:" }}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 24 "de0 := diff(H(x),x$2)=0;" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "sol0 := dsolve(de0,H(x));" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "H:=unapply(rhs(sol0),x);" }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "solve(\{H(0)=0,H(Pi)=0\},\{
_C1,_C2\});" }}}}{SECT 1 {PARA 4 "" 0 "" {XPPEDIT 18 0 "lambda = omega
^2;" "6#/%'lambdaG*$%&omegaG\"\"#" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 22 "In this case, we have " }{XPPEDIT 18 0 "Diff(H,`$`(x,2)) \+
= omega^2*H;" "6#/-%%DiffG6$%\"HG-%\"$G6$%\"xG\"\"#*&%&omegaG\"\"#F'\"
\"\"" }{TEXT -1 42 ", and the general solution is of the form " }
{XPPEDIT 18 0 "H(x) = A*cosh(omega*x)+B*sinh(omega*x);" "6#/-%\"HG6#%
\"xG,&*&%\"AG\"\"\"-%%coshG6#*&%&omegaGF+F'F+F+F+*&%\"BGF+-%%sinhG6#*&
F0F+F'F+F+F+" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 14 "The condi
tion " }{XPPEDIT 18 0 "H(0) = 0;" "6#/-%\"HG6#\"\"!F'" }{TEXT -1 14 " \+
implies that " }{XPPEDIT 18 0 "A = 0;" "6#/%\"AG\"\"!" }{TEXT -1 20 ",
 and the condition " }{XPPEDIT 18 0 "H(Pi) = 0;" "6#/-%\"HG6#%#PiG\"\"
!" }{TEXT -1 14 " implies that " }{XPPEDIT 18 0 "B = 0;" "6#/%\"BG\"\"
!" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 58 "We can instruct Mapl
e to verify this reasoning as follows:" }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 44 "H:='H':  de1 := diff(H(x),x$2)=omega^2*H(x);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "sol1 := dsolve(de1,H(x));" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "H:=unapply(rhs(sol1),x);" 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "solve(\{D(H)(0)=0,H(Pi)=0
\},\{_C1,_C2\});" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{SECT 1 {PARA 4 
"" 0 "" {TEXT -1 0 "" }{XPPEDIT 18 0 "lambda = -omega^2;" "6#/%'lambda
G,$*$%&omegaG\"\"#!\"\"" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 22 
"In this case, we have " }{XPPEDIT 18 0 "Diff(H,`$`(x,2)) = -omega^2*H
;" "6#/-%%DiffG6$%\"HG-%\"$G6$%\"xG\"\"#,$*&%&omegaG\"\"#F'\"\"\"!\"\"
" }{TEXT -1 42 ", and the general solution is of the form " }{XPPEDIT 
18 0 "H(x) = A*cos(omega*x)+B*sin(omega*x);" "6#/-%\"HG6#%\"xG,&*&%\"A
G\"\"\"-%$cosG6#*&%&omegaGF+F'F+F+F+*&%\"BGF+-%$sinG6#*&F0F+F'F+F+F+" 
}{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 14 "The condition " }
{XPPEDIT 18 0 "H(0) = 0;" "6#/-%\"HG6#\"\"!F'" }{TEXT -1 14 " implies \+
that " }{XPPEDIT 18 0 "A = 0;" "6#/%\"AG\"\"!" }{TEXT -1 20 ", and the
 condition " }{XPPEDIT 18 0 "H(Pi) = 0;" "6#/-%\"HG6#%#PiG\"\"!" }
{TEXT -1 14 " implies that " }{XPPEDIT 18 0 "B = 0;" "6#/%\"BG\"\"!" }
{TEXT -1 4 " or " }{XPPEDIT 18 0 "sin(omega*Pi) = 0;" "6#/-%$sinG6#*&%
&omegaG\"\"\"%#PiGF)\"\"!" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 
43 "This last condition will be true whenever  " }{XPPEDIT 18 0 "omega
 = k;" "6#/%&omegaG%\"kG" }{TEXT -1 6 "  for " }{XPPEDIT 18 0 "k;" "6#
%\"kG" }{TEXT -1 16 " = 1, 2, 3, ...." }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT -1 58 "We can instruct Maple to verify this \+
reasoning as follows:" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "H:=
'H':  de2 := diff(H(x),x$2)=-omega^2*H(x);" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 25 "sol2 := dsolve(de2,H(x));" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 24 "H:=unapply(rhs(sol2),x);" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 36 "solve(\{H(0)=0,H(Pi)=0\},\{omega,_C2\});" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 56 "Note that it is up to us to notice
 that other values of " }{XPPEDIT 18 0 "omega;" "6#%&omegaG" }{TEXT 
-1 11 " also work." }}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 21 "With these
 values of " }{XPPEDIT 18 0 "omega;" "6#%&omegaG" }{TEXT -1 42 " we se
e that the general solution for the " }{XPPEDIT 18 0 "G(t);" "6#-%\"GG
6#%\"tG" }{TEXT -1 26 " equation that is given by" }}{PARA 0 "" 0 "" 
{TEXT -1 8 "        " }{XPPEDIT 18 0 "G(t) = C*cos(k*t)+D*sin(k*t);" "
6#/-%\"GG6#%\"tG,&*&%\"CG\"\"\"-%$cosG6#*&%\"kGF+F'F+F+F+*&%\"DGF+-%$s
inG6#*&F0F+F'F+F+F+" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}{PARA 0 "" 0 "" {TEXT -1 57 "Thus there are solutions to the heat equ
ation of the form" }}{PARA 0 "" 0 "" {TEXT -1 6 "      " }{XPPEDIT 18 
0 "G(t)*H(x) = (c[k]*cos(k*t)+d[k]*sin(k*t))*sin(k*x);" "6#/*&-%\"GG6#
%\"tG\"\"\"-%\"HG6#%\"xGF)*&,&*&&%\"cG6#%\"kGF)-%$cosG6#*&F4F)F(F)F)F)
*&&%\"dG6#F4F)-%$sinG6#*&F4F)F(F)F)F)F)-F>6#*&F4F)F-F)F)" }{TEXT -1 0 
"" }}{PARA 0 "" 0 "" {TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 48 "We \+
can instruct Maple to verify this as follows:" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 56 "u[k] := (x,t) -> (c[k]*cos(k*t)+d[k]*sin(k*t))*s
in(k*x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "D[2,2](u[k])(x,
t) - D[1,1](u[k])(x,t);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "
simplify(%);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 53 "We have two remai
ning conditions to satisfy, namely  " }{XPPEDIT 18 0 "u(x,0) = 0;" "6#
/-%\"uG6$%\"xG\"\"!F(" }{TEXT -1 9 ",  and   " }{XPPEDIT 18 0 "Diff(u(
x,0),x) = x*(Pi-x);" "6#/-%%DiffG6$-%\"uG6$%\"xG\"\"!F**&F*\"\"\",&%#P
iGF-F*!\"\"F-" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 83 "Since we
 have an infinite set of solutions, each with an unspecified coefficie
nts, " }{XPPEDIT 18 0 "c[k];" "6#&%\"cG6#%\"kG" }{TEXT -1 6 ", and " }
{XPPEDIT 18 0 "d[k];" "6#&%\"dG6#%\"kG" }{TEXT -1 22 ", can we find a \+
linear" }}{PARA 0 "" 0 "" {TEXT -1 40 "combination, which yields the s
olution? " }}{PARA 0 "" 0 "" {TEXT -1 22 "That is, can we write " }
{XPPEDIT 18 0 "u(x,t) = sum(u[k](x,t),k = 1 .. infinity);" "6#/-%\"uG6
$%\"xG%\"tG-%$sumG6$-&F%6#%\"kG6$F'F(/F/;\"\"\"%)infinityG" }{TEXT -1 
3 " ? " }}{PARA 0 "" 0 "" {TEXT -1 11 "First, set " }{XPPEDIT 18 0 "t \+
= 0;" "6#/%\"tG\"\"!" }{TEXT -1 38 " and plug this into the series to \+
get " }}{PARA 0 "" 0 "" {TEXT -1 7 "       " }{XPPEDIT 18 0 "u(x,0) = \+
sum(c[k]*sin(k*x),k = 1 .. infinity);" "6#/-%\"uG6$%\"xG\"\"!-%$sumG6$
*&&%\"cG6#%\"kG\"\"\"-%$sinG6#*&F0F1F'F1F1/F0;\"\"\"%)infinityG" }
{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 34 "But this can only be true for all " }{XPPEDIT 18 0 "x;" "
6#%\"xG" }{TEXT -1 5 " if  " }{XPPEDIT 18 0 "c[k] = 0;" "6#/&%\"cG6#%
\"kG\"\"!" }{TEXT -1 28 ".  This leaves the condition" }}{PARA 0 "" 0 
"" {TEXT -1 5 "     " }{XPPEDIT 18 0 "Diff(u(x,0),x) = sum(d[k]*(-k)*s
in(k*x),k = a .. infinity);" "6#/-%%DiffG6$-%\"uG6$%\"xG\"\"!F*-%$sumG
6$*(&%\"dG6#%\"kG\"\"\",$F3!\"\"F4-%$sinG6#*&F3F4F*F4F4/F3;%\"aG%)infi
nityG" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 48 "This is the answe
r to Problem 2 provided we set " }{XPPEDIT 18 0 "d[k] = -b[k]/k;" "6#/
&%\"dG6#%\"kG,$*&&%\"bG6#F'\"\"\"F'!\"\"F." }{TEXT -1 1 "." }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "d[k] := (2/(k*Pi))*int(f(x)*sin(k*x
),x=0..Pi);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "dcoef := una
pply(d[k],k);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 85 "d[1] := dc
oef(1);d[2] := dcoef(2);d[3] := dcoef(3);d[4] := dcoef(4);d[5] := dcoe
f(5);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 56 "An approximation to the \+
solultion using 25 terms is thus" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 58 "u[25] := (x,t) -> sum(dcoef(k)*sin(k*t)*sin(k*x),k=1.
.25);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 118 "We can get a plot of th
is approximate solution.  It shows the string starting out flat and th
en vibrating periodically" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
58 "plot3d(u[25],0..Pi,0..10,axes=BOXED,labels=['x','t','u']);" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 70 "We can also use Maple's plot packa
ge to view this solution as a movie." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 52 "with(plots):  animate(u[25](x,t), x=0..Pi, t=0..10);
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}}{MARK "1" 0 }
{VIEWOPTS 1 1 0 1 1 1803 }