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{SECT 0 {PARA 263 "" 0 "" {TEXT -1 0 "" }}{PARA 18 "" 0 "" {TEXT -1
37 "Linear Methods of Applied Mathematics" }}{PARA 265 "" 0 "" {TEXT
-1 41 "Review of Ordinary Differential Equations" }}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{SECT 0 {PARA 3 "" 0 "" {TEXT -1 79 "Copyright 2000 by
Evans M. Harrell II and James V. Herod. All rights reserved." }}}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA
0 "" 0 "" {TEXT -1 184 " We will use information about elementary,
ordinary differential equations in the process of solving partial dif
ferential equations. It is well to remember some of the basic ideas."
}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 120 " S
imple, constant coefficient, initial value problems are classical prob
lems in elementary diffferential equations. " }}{PARA 0 "" 0 "" {TEXT
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 3 "" 0 ""
{TEXT -1 85 "General solutions for linear, constant-coefficient differ
ential equations and systems" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }
}{PARA 0 "" 0 "" {TEXT -1 108 "The methods for solving these are a par
t of the standard tools. Here is an example to recall the techniques.
" }}{PARA 0 "" 0 "" {TEXT -1 40 " Consider the differential equati
on," }}{PARA 0 "" 0 "" {TEXT -1 33 " y '' + 3 y ' + 2 y = 0.
" }}{PARA 0 "" 0 "" {TEXT -1 192 "Using the ideas that have come befor
e, the consideration of such a differential equation could be viewed a
s the consideration of the nullspace of the linear operator L(y) = y '
' + 3 y ' + 2 y." }}{PARA 0 "" 0 "" {TEXT -1 193 " Recall that the
re are two linearly independent solutions for this second order equati
on. Said in our context, the nullspace of this linear operation had a \+
basis consisting of two elements." }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 42 "deq:=diff(y(x),x,x)+3*diff(y(x),x)+2*y(x);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "dsolve(deq,y(x),output=basis
);" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 153 "T
he nullspace is two dimensional and any solution of the equation -- an
y element of the nullspace -- can be written as a linear combination o
f these two." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "dsolve(deq,y
(x));" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 4 "" 0 ""
{TEXT -1 51 "Linear, constant-coefficient initial-value problems" }}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 157 "I
f we want to know a particular solution, say the solution of an initia
l value problem, we must choose the constants so that the initial valu
es are attained." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "inits:=
y(0)=1,D(y)(0)=5;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "Part:=
dsolve(\{deq,inits\},y(x));" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA
0 "" 0 "" {TEXT -1 54 "We can now draw the graph of this particular so
lution." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "y1:=unapply(rhs(P
art),x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "plot(y1(x),x=0.
.1,y=0..3);" }}}{PARA 0 "" 0 "" {TEXT -1 174 "You might choose to draw
a family of solutions. For example, we draw a family of solutions for
the differential equation that begin at zero and have different initi
al slopes." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "inits:=y(0)=1,
D(y)(0)=b;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "Part:=dsolve(
\{deq,inits\},y(x));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "plo
t([seq(subs(b=n,rhs(Part)),n=-2..2)],x=0..1);" }}}}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 4 "" 0 "
" {TEXT -1 52 "Linear, constant-coefficient boundary-value problems" }
}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 243 "
Boundary value problems are a little more complicated in concept
. There are second order, constant coefficient differential equations \+
for which there is exactly one solution, or there is no solution, or t
here is an infinity of solutions." }}{PARA 0 "" 0 "" {TEXT -1 85 " \+
Suppose you want solutions for the above differential equations to st
art off at " }{TEXT 261 1 "a" }{TEXT -1 15 " and end up at " }{TEXT
262 1 "b" }{TEXT -1 14 ". Not so hard." }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 78 "Part2:=\n dsolve(\{diff(y(x),x,x)+3*diff(y(x),x)+2*y
(x)=0,y(0)=a,y(1)=b\},y(x));" }}}{PARA 0 "" 0 "" {TEXT -1 55 "For exam
ple, we draw the graph in case a = 1 and b = 1." }}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 47 "plot(subs(\{a=1,b=1\},rhs(Part2)),x=0..1,y=0..
2);" }}}{PARA 0 "" 0 "" {TEXT -1 95 "As an alternate, you might choose
the slope at either end, instead of where the solutions lies." }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 74 "dsolve(\{diff(y(x),x,x)+3*di
ff(y(x),x)+2*y(x)=0,D(y)(0)=a,D(y)(1)=b\},y(x));" }}}{PARA 0 "" 0 ""
{TEXT -1 130 "It is of value to note that we can make second order, co
nstant coefficient, differential equations for which there is no solut
ion." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 74 "dsolve(\{diff(y(x),x
,x)+3*diff(y(x),x)+0*y(x)=0,D(y)(0)=a,D(y)(1)=b\},y(x));" }}}{PARA 0 "
" 0 "" {TEXT -1 118 "And, there are second order, constant coefficient
, differential equations for which there are an infinity of solution.
" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 74 "dsolve(\{diff(y(x),x,x)+
0*diff(y(x),x)+0*y(x)=0,D(y)(0)=0,D(y)(1)=0\},y(x));" }}}}{PARA 0 ""
0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 40 "Linear syste
ms of differential equations" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }
}{PARA 0 "" 0 "" {TEXT -1 241 " Often, one has a coupled system of
equations instead of a single differential equation. Review almost an
y text on elementary differential equations, and you will find systems
of equations, with initial conditions, such as the following:" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1
0 69 "sys:=diff(x(t),t)=-2*x(t)-11*y(t), \n diff(y(t),t)=11*x(t)-2
*y(t);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "inits:=x(0)=1,y(0
)=2;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "dsolve(\{sys,inits
\},\{x(t),y(t)\});" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "assig
n(%);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "plot([x(t),y(t)],t
=0..1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}}
{PARA 268 "" 0 "" {TEXT -1 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{SECT 1 {PARA 3 "" 0 "" {TEXT -1 37 "Non-homogeneous equations and sys
tems" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT
-1 306 " We might have a differential equation for which the right h
and side is not zero. For example, think of the simplified Newton's La
w of Cooling. It says that a body cools at a rate proportioned to the \+
difference in its temperature and the temperature of the surrounding m
edium. We could write this idea as" }}{PARA 0 "" 0 "" {TEXT -1 33 " \+
T ' = K (A-T), T(0) = C" }}{PARA 0 "" 0 "" {TEXT -1 267 "where \+
A is the ambient temperature and C is the initial temperature of the o
bject. The number K is the constant of proportionality. We see that if
K is negative and the object is hotter than the ambient temperature, \+
it will cool. This equations could be rewritten as " }}{PARA 0 "" 0 "
" {TEXT -1 41 " T ' + K T = K A, with T(0) = C." }}{PARA 0 "
" 0 "" {TEXT -1 158 "This introduces a differential equation with righ
t hand side not zero. Such systems are common. Techniques for obtainin
g their solutions need to be recalled. " }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 44 "dsolve(\{diff(T(t),t)+2*T(t)=3,T(0)=5\},T(t));" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT
-1 47 "The right side may, itself, be a function of t." }}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "dsolve(\{diff(T(t),t)+2*T(t)=sin(t)
,T(0)=5\},T(t));" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT -1 54 "Indeed, the right hand side may not be differentiable." }
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 56 "dsolve(\{diff(T(t),t)+2*T(t
)=signum(1/2-t),T(0)=5\},T(t));" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 271 "" 0 "" {TEXT -1 46 "Numerical solutions for Differential Eq
uations" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1
48 "One might choose to obtain a numerical solution." }}{EXCHG {PARA
0 "> " 0 "" {MPLTEXT 1 0 96 "Cool:=dsolve(\{diff(T(t),t)+2*T(t)=signum
(1/2-t),T(0)=5\},T(t),type=numeric,output=listprocedure);" }}}{PARA 0
"" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "y2:=
subs(Cool,T(t));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "plot('y
2(t)',t=0..1);" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT -1 62 "Numerical solutions can be used for systems of equations,
too." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 159 "Drug1:=dsolve(\{di
ff(x(t),t)=sin(t)-2*x(t)-3*y(t),\n diff(y(t),t)=3*x(t)-2
*y(t),\n x(0)=1,y(0)=0\},\{x(t),y(t)\},type=numeric,output=listproc
edure);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "x3:=subs(Drug1,x
(t));\ny3:=subs(Drug1,y(t));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1
0 33 "plot(['x3(t)','y3(t)'],'t'=0..5);" }}}}{PARA 261 "" 0 "" {TEXT
-1 24 "Non-homogeneous Systems." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{SECT 1 {PARA 3 "" 0 "" {TEXT -1 8 "Exercise" }}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 38 "Draw one graph with five solutions for" }}{PARA 0 "" 0
"" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 65 " x '' + 2 x '
+ 2 x = 20 cos(t), x(0) = a, x '(0) = 0," }}{PARA 0 "" 0 "" {TEXT
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 259 "using five different values of
a. Looking at the graph, it seems that all solutions to dampen to the
same curve. Can you find that steady periodic solution for the equati
on? What is its initial value? (This problem is found on page 194 of t
he attractive text " }{TEXT 263 50 "Differential Equations and Boundar
y Value Problems" }{TEXT -1 55 ", by Edwards and Penny and published b
y Prentice Hall.)" }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA
3 "" 0 "" {TEXT -1 44 "Bounded solutions, eigenvalues, and all that" }
}{EXCHG {PARA 0 "" 0 "" {TEXT -1 51 " We continue our review by ask
ing six questions." }}{PARA 0 "" 0 "" {TEXT 264 11 "Question 1." }
{TEXT -1 9 " Suppose " }{XPPEDIT 18 0 "lambda;" "6#%'lambdaG" }{TEXT
-1 76 " > 0. Which of these is a pair of linearly independent solution
s for Y '' - " }{XPPEDIT 18 0 "lambda^2;" "6#*$%'lambdaG\"\"#" }{TEXT
-1 14 " Y = 0 on [0, " }{XPPEDIT 18 0 "pi;" "6#%#piG" }{TEXT -1 2 "]?
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 7 "a. exp
(" }{XPPEDIT 18 0 "lambda;" "6#%'lambdaG" }{TEXT -1 14 " x) and exp( -
" }{XPPEDIT 18 0 "lambda;" "6#%'lambdaG" }{TEXT -1 18 " x), c. si
nh(" }{XPPEDIT 18 0 "lambda;" "6#%'lambdaG" }{TEXT -1 14 " x ) and cos
h(" }{XPPEDIT 18 0 "lambda;" "6#%'lambdaG" }{TEXT -1 4 " x)," }}{PARA
0 "" 0 "" {TEXT -1 7 "b. sin(" }{XPPEDIT 18 0 "lambda;" "6#%'lambdaG"
}{TEXT -1 12 " x) and cos(" }{XPPEDIT 18 0 "lambda;" "6#%'lambdaG" }
{TEXT -1 21 " x ), d. sinh(" }{XPPEDIT 18 0 "lambda;" "6#%'lamb
daG" }{TEXT -1 15 " x ) and sinh( " }{XPPEDIT 18 0 "lambda;" "6#%'lamb
daG" }{TEXT -1 2 " (" }{XPPEDIT 18 0 "pi;" "6#%#piG" }{TEXT -1 7 " - x
))." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}{PARA 0 "" 0 "" {TEXT 265 11 "Question 2." }{TEXT -1 9 " Suppose " }
{XPPEDIT 18 0 "lambda;" "6#%'lambdaG" }{TEXT -1 76 " > 0. Which of the
se is a pair of linearly independent solutions for Y '' + " }{XPPEDIT
18 0 "lambda^2;" "6#*$%'lambdaG\"\"#" }{TEXT -1 14 " Y = 0 on [0, " }
{XPPEDIT 18 0 "pi;" "6#%#piG" }{TEXT -1 2 "]?" }}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 7 "a. exp(" }{XPPEDIT 18 0 "l
ambda;" "6#%'lambdaG" }{TEXT -1 14 " x) and exp( -" }{XPPEDIT 18 0 "la
mbda;" "6#%'lambdaG" }{TEXT -1 18 " x), c. sinh(" }{XPPEDIT 18 0
"lambda;" "6#%'lambdaG" }{TEXT -1 14 " x ) and cosh(" }{XPPEDIT 18 0 "
lambda;" "6#%'lambdaG" }{TEXT -1 4 " x)," }}{PARA 0 "" 0 "" {TEXT -1
7 "b. sin(" }{XPPEDIT 18 0 "lambda;" "6#%'lambdaG" }{TEXT -1 12 " x) a
nd cos(" }{XPPEDIT 18 0 "lambda;" "6#%'lambdaG" }{TEXT -1 21 " x ), \+
d. sinh(" }{XPPEDIT 18 0 "lambda;" "6#%'lambdaG" }{TEXT -1 15 " x
) and sinh( " }{XPPEDIT 18 0 "lambda;" "6#%'lambdaG" }{TEXT -1 2 " (
" }{XPPEDIT 18 0 "pi;" "6#%#piG" }{TEXT -1 7 " - x))." }}{PARA 0 "" 0
"" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 202 "Remark: If you have f
orgotten enough calculus that you do not know how to find the answer t
o the above two questions, you need to do two things: review and use M
aple in some manner such as the following" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 8 "restart:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0
20 "y:=x->exp(lambda*x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29
"diff(y(x),x,x)-lambda^2*y(x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT
1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1
0 "" }}{PARA 0 "" 0 "" {TEXT 266 11 "Question 3." }{TEXT -1 9 " Suppos
e " }{XPPEDIT 18 0 "lambda;" "6#%'lambdaG" }{TEXT -1 54 " > 0. Which o
f these is a bounded solution for Y '' - " }{XPPEDIT 18 0 "lambda^2;"
"6#*$%'lambdaG\"\"#" }{TEXT -1 14 " Y = 0 on [0, " }{XPPEDIT 18 0 "inf
inity;" "6#%)infinityG" }{TEXT -1 2 ")?" }}{PARA 0 "" 0 "" {TEXT -1 0
"" }}{PARA 0 "" 0 "" {TEXT -1 7 "a. exp(" }{XPPEDIT 18 0 "lambda;" "6#
%'lambdaG" }{TEXT -1 15 " x) b. exp(-" }{XPPEDIT 18 0 "lambda;" "6#
%'lambdaG" }{TEXT -1 16 " x) c. sinh(" }{XPPEDIT 18 0 "lambda;" "6
#%'lambdaG" }{TEXT -1 17 " x) d. cosh(" }{XPPEDIT 18 0 "lambda;"
"6#%'lambdaG" }{TEXT -1 3 " x)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 94 "There are two issues here: which is a sol
ution and which is bounded on the specified interval." }}{PARA 0 "" 0
"" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT 267 11 "Question 4." }{TEXT -1 89 " Which of these is a bounded \+
solution on the interval [0, 5] of the differential equation" }}{PARA
0 "" 0 "" {TEXT -1 10 " " }{XPPEDIT 18 0 "r^2;" "6#*$%\"rG\"
\"#" }{TEXT -1 33 " R ''(r) + r R '(r) - 9 R(r) = 0?" }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 21 "a. exp(3 r) b. "
}{XPPEDIT 18 0 "r^3;" "6#*$%\"rG\"\"$" }{TEXT -1 49 " c. sin(3 r
) d. exp(-3 r) e. 1/" }{XPPEDIT 18 0 "r^3;" "6#*$%\"rG\"\"
$" }{TEXT -1 19 " f. cosh(3 r)" }}{PARA 0 "" 0 "" {TEXT -1 0 ""
}}{PARA 0 "" 0 "" {TEXT 268 11 "Question 5." }{TEXT -1 14 " If u(x,y)
= " }{XPPEDIT 18 0 "sum(a[p]*sin(p*x)*sinh(p*y));" "6#-%$sumG6#*(&%\"
aG6#%\"pG\"\"\"-%$sinG6#*&F*F+%\"xGF+F+-%%sinhG6#*&F*F+%\"yGF+F+" }
{TEXT -1 3 " + " }{XPPEDIT 18 0 "sum(b[p]*sin(p*x)*sinh(p*(Pi-y)));" "
6#-%$sumG6#*(&%\"bG6#%\"pG\"\"\"-%$sinG6#*&F*F+%\"xGF+F+-%%sinhG6#*&F*
F+,&%#PiGF+%\"yG!\"\"F+F+" }{TEXT -1 6 " and " }}{PARA 0 "" 0 ""
{TEXT -1 43 " u(x,0) = 0, u(x," }{XPPEDIT
18 0 "pi;" "6#%#piG" }{TEXT -1 13 ") = sin(2 x) " }}{PARA 0 "" 0 ""
{TEXT -1 13 "what are the " }{XPPEDIT 18 0 "a[p];" "6#&%\"aG6#%\"pG" }
{TEXT -1 13 " 's and " }{XPPEDIT 18 0 "b[p];" "6#&%\"bG6#%\"pG" }
{TEXT -1 4 " 's?" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 116 "u:=(x,y
)->sum(a[p]*sin(p*x)*sinh(p*y),p=1..infinity)+ \n sum(b[p]
*sin(p*x)*sinh(p*(Pi-y)),p=1..infinity);" }}}{PARA 0 "" 0 "" {TEXT -1
34 "We have two pieces of information." }}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 9 "0=u(x,0);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0
17 "sin(2*x)=u(x,Pi);" }}}{PARA 0 "" 0 "" {TEXT -1 40 "The first of th
ese implies that all the " }{XPPEDIT 18 0 "b[p];" "6#&%\"bG6#%\"pG" }
{TEXT -1 49 "'s are zero, and the second implies that all the " }
{XPPEDIT 18 0 "a[p];" "6#&%\"aG6#%\"pG" }{TEXT -1 52 "'s is zero excep
t the second one, and it is 1/sinh(" }{XPPEDIT 18 0 "2*pi;" "6#*&\"\"
#\"\"\"%#piGF%" }{TEXT -1 30 "). Thus here is a graph for u." }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 69 "plot3d(1/sinh(2*Pi)*sin(2*x)
*sinh(2*y), x=0..Pi,y=0..Pi,axes=NORMAL);" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT 269 11 "Question 6." }{TEXT -1 9 " If u(r, " }{XPPEDIT 18 0 "t
heta;" "6#%&thetaG" }{TEXT -1 4 ") = " }{XPPEDIT 18 0 "sum(a[p]*sin(p*
theta)*r^p,p);" "6#-%$sumG6$*(&%\"aG6#%\"pG\"\"\"-%$sinG6#*&F*F+%&thet
aGF+F+)%\"rGF*F+F*" }{TEXT -1 3 " + " }{XPPEDIT 18 0 "sum(b[p]*cos(p*t
heta)*r^p,p);" "6#-%$sumG6$*(&%\"bG6#%\"pG\"\"\"-%$cosG6#*&F*F+%&theta
GF+F+)%\"rGF*F+F*" }{TEXT -1 6 " and " }}{PARA 0 "" 0 "" {TEXT -1 10
" u(1, " }{XPPEDIT 18 0 "theta;" "6#%&thetaG" }{TEXT -1 16 ") = 1 \+
+ 3 cos(3 " }{XPPEDIT 18 0 "theta;" "6#%&thetaG" }{TEXT -1 13 ") + 5 s
in( 2 " }{XPPEDIT 18 0 "theta;" "6#%&thetaG" }{TEXT -1 2 ") " }}{PARA
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 18 "then what is u(
r, " }{XPPEDIT 18 0 "theta;" "6#%&thetaG" }{TEXT -1 22 "), u(0,0), and
u(1/2, " }{XPPEDIT 18 0 "pi;" "6#%#piG" }{TEXT -1 4 "/4)?" }}{PARA 0
"" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 117 "u:=
(r,theta)->sum(a[p]*sin(p*theta)*r^p,p=1..infinity) + \n \+
sum(b[p]*cos(p*theta)*r^p,p=0..infinity);" }}}{PARA 0 "" 0 ""
{TEXT -1 18 "Our information is" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT
1 0 43 "1+3*cos(3*theta)+5*sin(2*theta)=u(1,theta);" }}}{PARA 0 "" 0 "
" {TEXT -1 22 "This implies that all " }{XPPEDIT 18 0 "a[p];" "6#&%\"a
G6#%\"pG" }{TEXT -1 7 "'s and " }{XPPEDIT 18 0 "b[p];" "6#&%\"bG6#%\"p
G" }{TEXT -1 19 "'s are zero except " }{XPPEDIT 18 0 "b[0];" "6#&%\"bG
6#\"\"!" }{TEXT -1 6 " = 1, " }{XPPEDIT 18 0 "b[3];" "6#&%\"bG6#\"\"$
" }{TEXT -1 10 " = 3, and " }{XPPEDIT 18 0 "a[2];" "6#&%\"aG6#\"\"#" }
{TEXT -1 5 " = 5." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 54 "u:=(r,t
heta)->5*sin(2*theta)*r^2+1+3*cos(3*theta)*r^3;" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 7 "u(0,0);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT
1 0 12 "u(1/2,Pi/4);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 144 "pl
ot3d([r,theta,u(r,theta)],r=0..1,theta=-Pi..Pi,\n coords=c
ylindrical,axes=normal,orientation=[20,40],\n numpoints=20
00);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0
"> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 5 " " }}
{PARA 0 "" 0 "" {TEXT 270 11 "Question 7." }{TEXT -1 73 " What are all
the eigenvalues of the selfadjoint, Sturm-Liouville Problem" }}{PARA
0 "" 0 "" {TEXT -1 17 " y '' = " }{XPPEDIT 18 0 "mu;" "6#%#mu
G" }{TEXT -1 25 " y, with y(0) = y(1) = 0?" }}{PARA 0 "" 0 "" {TEXT
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 57 "We break the problem into two c
ases. First, suppose that " }{XPPEDIT 18 0 "mu;" "6#%#muG" }{TEXT -1
41 " is positive. To remind us of this, take " }{XPPEDIT 18 0 "mu = la
mbda^2;" "6#/%#muG*$%'lambdaG\"\"#" }{TEXT -1 25 " . Thus, we seek num
bers " }{XPPEDIT 18 0 "lambda;" "6#%'lambdaG" }{TEXT -1 10 " such that
" }}{PARA 0 "" 0 "" {TEXT -1 17 " y '' = " }{XPPEDIT 18 0 "la
mbda^2;" "6#*$%'lambdaG\"\"#" }{TEXT -1 24 " y with y(0) = y(1) = 0."
}}{PARA 0 "" 0 "" {TEXT -1 186 "The differential equation is a second \+
order constant coefficient equation. We know how to solve it. Solution
s are of the form exp(r x). In this case, the general solutions is of \+
the form" }}{PARA 0 "" 0 "" {TEXT -1 29 " y(x) = A exp( \+
" }{XPPEDIT 18 0 "lambda;" "6#%'lambdaG" }{TEXT -1 13 " x) + B exp( "
}{XPPEDIT 18 0 "-lambda;" "6#,$%'lambdaG!\"\"" }{TEXT -1 4 " x)." }}
{PARA 0 "" 0 "" {TEXT -1 83 "To ask that y(0) = 0 asks that 0 = A + B.
To ask that y(1) = 0 asks that 0 = A exp(" }{XPPEDIT 18 0 "lambda;" "
6#%'lambdaG" }{TEXT -1 10 ") + B/exp(" }{XPPEDIT 18 0 "lambda;" "6#%'l
ambdaG" }{TEXT -1 100 ") . The only solution is A = 0 = B. This is the
trivial solutions and will not be of interest to us." }}{PARA 0 "" 0
"" {TEXT -1 29 " The second case is that " }{XPPEDIT 18 0 "mu;" "6
#%#muG" }{TEXT -1 41 " is negative. To remind us of this, take " }
{XPPEDIT 18 0 "mu = -lambda^2;" "6#/%#muG,$*$%'lambdaG\"\"#!\"\"" }
{TEXT -1 24 ". Thus, we seek numbers " }{XPPEDIT 18 0 "lambda;" "6#%'l
ambdaG" }{TEXT -1 11 " such that" }}{PARA 0 "" 0 "" {TEXT -1 17 " \+
y '' = " }{XPPEDIT 18 0 "-lambda^2;" "6#,$*$%'lambdaG\"\"#!\"\"
" }{TEXT -1 24 " y with y(0) = y(1) = 0." }}{PARA 0 "" 0 "" {TEXT -1
79 "We also know how to solve this differential equation. Solutions ar
e of the form" }}{PARA 0 "" 0 "" {TEXT -1 28 " y(x) = A \+
sin(" }{XPPEDIT 18 0 "lambda;" "6#%'lambdaG" }{TEXT -1 12 " x) + B cos
(" }{XPPEDIT 18 0 "lambda;" "6#%'lambdaG" }{TEXT -1 4 " x)." }}{PARA
0 "" 0 "" {TEXT -1 84 "To ask that y(0) = 0 asks that 0 = B. To also a
sk that y(1) = 0 asks that 0 = A sin(" }{XPPEDIT 18 0 "lambda;" "6#%'l
ambdaG" }{TEXT -1 82 "). We don't want to have the trivial solutions a
gain, so it must be that 0 = sin(" }{XPPEDIT 18 0 "lambda;" "6#%'lamb
daG" }{TEXT -1 81 ") . But we know all the places where the sine func
tion is zero. It must be that " }{XPPEDIT 18 0 "lambda = n*Pi;" "6#/%'
lambdaG*&%\"nG\"\"\"%#PiGF'" }{TEXT -1 6 ", and " }{XPPEDIT 18 0 "mu =
-n^2*Pi^2;" "6#/%#muG,$*&%\"nG\"\"#%#PiG\"\"#!\"\"" }{TEXT -1 1 "." }
}{PARA 0 "" 0 "" {TEXT -1 55 "The answer to the question is that the e
igenvalues are " }{XPPEDIT 18 0 "-n^2*Pi^2;" "6#,$*&%\"nG\"\"#%#PiG\"
\"#!\"\"" }{TEXT -1 32 " and the eigenfuntions are sin(" }{XPPEDIT
18 0 "n*Pi*x;" "6#*(%\"nG\"\"\"%#PiGF%%\"xGF%" }{TEXT -1 2 ")." }}
{PARA 0 "" 0 "" {TEXT -1 29 "Somehow, this is no surprise." }}}{SECT
1 {PARA 3 "" 0 "" {TEXT -1 8 "Exercise" }}{EXCHG {PARA 0 "" 0 ""
{TEXT -1 68 "Find all the eigenvalues of the selfadjoint, Sturm-Liouvi
lle Problem" }}{PARA 0 "" 0 "" {TEXT -1 17 " y '' = " }
{XPPEDIT 18 0 "mu;" "6#%#muG" }{TEXT -1 28 " y, with y '(0) = y '(1) =
0" }}}}}{MARK "4 0 0" 10 }{VIEWOPTS 1 1 0 1 1 1803 }