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{SECT 0 {PARA 18 "" 0 "" {TEXT -1 37 "Linear Methods of Applied Mathem
atics" }}{PARA 256 "" 0 "" {TEXT -1 52 "Solving the wave equation by s
eparation of variables" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 0
{PARA 3 "" 0 "" {TEXT -1 62 "Copyright 1998, 2000 by Evans M. Harrell \+
II and James V. Herod" }}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 52 "Solving
the wave equation by separation of variables" }}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 57 " We begin now a study of the classical wave equatio
n." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 48 " \+
The classical ,linearized wave equation is " }}{PARA 0 "" 0 ""
{TEXT -1 39 " " }{XPPEDIT 18 0 "
diff(u(t,x),`$`(t,2)) = c^2*diff(u(t,x),`$`(x,2));" "6#/-%%diffG6$-%\"
uG6$%\"tG%\"xG-%\"$G6$F*\"\"#*&%\"cG\"\"#-F%6$-F(6$F*F+-F-6$F+\"\"#\"
\"\"" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 108 " We classify this PDE as a special case of the mor
e general constant coefficient, second order equation:" }}{PARA 0 ""
0 "" {TEXT -1 9 " " }{XPPEDIT 18 0 "a*diff(u(t,x),`$`(t,2))+b*
diff(u(t,x),t,x)+c*diff(u(t,x),`$`(x,2))+d*diff(u(t,x),t)+e*diff(u(t,x
),x)+f;" "6#,.*&%\"aG\"\"\"-%%diffG6$-%\"uG6$%\"tG%\"xG-%\"$G6$F-\"\"#
F&F&*&%\"bGF&-F(6%-F+6$F-F.F-F.F&F&*&%\"cGF&-F(6$-F+6$F-F.-F06$F.\"\"#
F&F&*&%\"dGF&-F(6$-F+6$F-F.F-F&F&*&%\"eGF&-F(6$-F+6$F-F.F.F&F&%\"fGF&
" }{TEXT -1 5 " = 0." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT 269 12 "Second order" }{TEXT -1 319 " refers to the lack of de
rivatives of more than second order.We will solve the wave equation by
the method of separation of variables in this worksheet. We will find
there are alternate methods for this equation. We will also see that \+
slightly more complicated situations lead to the more general second o
rder equation." }}{PARA 0 "" 0 "" {TEXT -1 39 " The method takes a
dvantage of the " }{TEXT 270 22 "superposition property" }{TEXT -1 28
". That is, if two functions " }{XPPEDIT 18 0 "u[1]" "6#&%\"uG6#\"\"\"
" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "u[2]" "6#&%\"uG6#\"\"#" }{TEXT
-1 49 " are solutions then so is any linear combination:" }}{PARA 0 "
" 0 "" {TEXT -1 25 " " }{XPPEDIT 18 0 "alpha*u
[1]+beta*u[2]" "6#,&*&%&alphaG\"\"\"&%\"uG6#\"\"\"F&F&*&%%betaGF&&F(6#
\"\"#F&F&" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT -1 177 "The wave equation commonly comes with boundary co
nditions and initial conditions. At this start, we take homogeneous bo
undary conditions, assuming that x ranges between 0 and L." }}{PARA 0
"" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 51 "Boundary Conditio
ns: u(t, 0) = 0, and u(t, L) = 0." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 76 "Because the equation is second order in t
, we expect two initial conditions." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}{PARA 0 "" 0 "" {TEXT -1 40 "Initial Conditions: u(0, x) = f(x) and \+
" }{XPPEDIT 18 0 "diff(u,t);" "6#-%%diffG6$%\"uG%\"tG" }{TEXT -1 30 "(
0, x) = g(x) for x in [0, L]." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 303 "The usual physical realization made for \+
this model is that of a string, held fixed at two ends, displaced by a
n amount f(x) initially, and given an initial velocity g(x). This mode
l will guide our intuition and, with small displacements, gives an acc
urate impression for the vibrations of a taut string." }}{PARA 0 "" 0
"" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT -1 513 " We often set c = 1 for convenience. We do this in t
his worksheet, but remove the restriction later. It is of value to con
sider the difference in the left and right side of the wave equation. \+
That difference reminds us that we are solving linear operator equatio
ns and looking for the null space of these linear operators. This exam
ple is simply a linear equation which has an infinite number of soluti
ons. We seek one which satisfies the boundary and initial conditions. \+
To give understanding, we define the " }{TEXT 271 14 "wave operator."
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "waveop:=diff(u(t,x),t,t)-
diff(u(t,x),x,x);" }}}{PARA 0 "" 0 "" {TEXT -1 139 "(The restriction t
hat c = 1 is not a severe restriction. Equivalently if c is not one, w
e could choose a clock with time variable T = c*t.)" }}{PARA 0 "" 0 "
" {TEXT -1 113 " Let's test out the wave operator. We will eventua
lly try to get functions in the null space of the operator." }}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "u:=(t,x)->t*sin(x)^2; waveop;" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "u:=(t,x)->sin(t-x); waveop;
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "u:='u':" }}}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 273 " \+
We have had experience with the method of separation of variables \+
and know that this method for solving partial differential equations i
s precisely what it says: One assumes that solutions can be written as
products of separate functions of t and x. Thus, we make the " }
{TEXT 272 6 "ansatz" }{TEXT -1 37 " that u(t, x) is of the special for
m " }}{PARA 0 "" 0 "" {TEXT -1 26 " T(t) X(x), " }}
{PARA 0 "" 0 "" {TEXT -1 28 "known as a product solution." }}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "eq:=subs(u(t,x)=T(t)*X(x),waveop)=0
;" }}}{PARA 0 "" 0 "" {TEXT -1 55 " This simplifies if we divide t
hrough by T(t) X(x)." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "eq/X
(x)/T(t);\nexpand(%);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "se
p:=(%)+(diff(X(x),x,x)/X(x)=diff(X(x),x,x)/X(x));" }}}{PARA 0 "" 0 ""
{TEXT -1 35 " The left side of the equation " }{TEXT 275 3 "sep" }
{TEXT -1 17 " depends only on " }{TEXT 273 1 "t" }{TEXT -1 36 " and th
e right side depends only on " }{TEXT 274 1 "x" }{TEXT -1 142 ". Thus,
each side must be constant. We do not know the value of this constant
, yet. As in the heat equation, it will be negative; we call it -" }
{XPPEDIT 18 0 "mu^2" "6#*$%#muG\"\"#" }{TEXT -1 1 "." }}{EXCHG {PARA
0 "> " 0 "" {MPLTEXT 1 0 39 "dsolve(diff(X(x),x,x)=-mu^2*X(x),X(x));"
}}}{PARA 0 "" 0 "" {TEXT -1 65 "In order to satisfy the boundary condi
tion at x = 0, we must have" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0
16 "X:=x->sin(mu*x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "sol
ve(X(L)=0,L);" }}}{PARA 0 "" 0 "" {TEXT -1 243 "There are an infinite \+
number of solutions for this equation, they change with L and mu. Mapl
e did not pick out any solution except L = 0. We know, however where t
he sine function is zero: the sine function is zero at all integral mu
ltiples of " }{XPPEDIT 18 0 "pi;" "6#%#piG" }{TEXT -1 14 " . Thus, tak
e " }{XPPEDIT 18 0 "mu*L=n*Pi" "6#/*&%#muG\"\"\"%\"LGF&*&%\"nGF&%#PiGF
&" }{TEXT -1 15 " and solve for " }{XPPEDIT 18 0 "mu" "6#%#muG" }
{TEXT -1 1 "." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "solve(mu*L=
n*Pi,mu);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "mu:=%;" }}}
{PARA 0 "" 0 "" {TEXT -1 35 " We now have the function X(x)." }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 5 "X(x);" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 20 "diff(X(x),x,x)/X(x);" }}}{PARA 0 "" 0 ""
{TEXT -1 71 " We now look at the other part of the PDE which we \"
separated off\"." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "dsolve(d
iff(T(t),t,t)=-mu^2*T(t),T(t));" }}}{PARA 0 "" 0 "" {TEXT -1 229 " \+
Both the sine and cosine functions give possible solutions. We have n
o reason to eliminate either of these. The general solution we find is
a linear combination of the particular solutions we get by separating
the equation. " }}{PARA 0 "" 0 "" {TEXT -1 56 " The most general \+
linear combination is of the form." }}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 103 "u:=(t,x)->sum((A[n]*cos(Pi*n*t/L)+B[n]*sin(Pi*n*t/L)
)*sin(n*Pi*x/L),\n n=1..infinity);" }}}{PARA 0 "" 0
"" {TEXT -1 154 " The coefficients A and B have to be determined f
rom the initial conditions in a manner that is familiar. We use the in
itial conditions. Suppose that " }{XPPEDIT 18 0 "u(0,x)=f(x)" "6#/-%\"
uG6$\"\"!%\"xG-%\"fG6#F(" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "u[t](0,x
)=g(x)" "6#/-&%\"uG6#%\"tG6$\"\"!%\"xG-%\"gG6#F+" }{TEXT -1 1 "." }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "u(0,x)=f(x);\nD[1](u)(0,x)=g
(x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "
" {TEXT -1 110 "Doesn't this look like a job for Fourier Series? To sp
ell out what how to compute the coefficients recall that" }}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 55 "assume(n,integer):\nint(sin(n*Pi*x/
L)^2,x=0..L);\nn:='n':" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 ""
}}}{PARA 0 "" 0 "" {TEXT -1 17 " We have that" }}{PARA 0 "" 0 ""
{TEXT -1 25 " " }{XPPEDIT 18 0 "A[n];" "6#&%\"
AG6#%\"nG" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "2/L;" "6#*&\"\"#\"\"\"%\"
LG!\"\"" }{TEXT -1 1 " " }{XPPEDIT 18 0 "int(f(x)*sin(n*Pi*x/L),x = 0 \+
.. L);" "6#-%$intG6$*&-%\"fG6#%\"xG\"\"\"-%$sinG6#**%\"nGF+%#PiGF+F*F+
%\"LG!\"\"F+/F*;\"\"!F2" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 3 "
and" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 24 " \+
" }{XPPEDIT 18 0 "B[n]*n*Pi/L;" "6#**&%\"BG6#%\"
nG\"\"\"F'F(%#PiGF(%\"LG!\"\"" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "2/L;
" "6#*&\"\"#\"\"\"%\"LG!\"\"" }{TEXT -1 1 " " }{XPPEDIT 18 0 "int(g(x)
*sin(n*Pi*x/L),x = 0 .. L);" "6#-%$intG6$*&-%\"gG6#%\"xG\"\"\"-%$sinG6
#**%\"nGF+%#PiGF+F*F+%\"LG!\"\"F+/F*;\"\"!F2" }{TEXT -1 5 " , or" }}
{PARA 0 "" 0 "" {TEXT -1 23 " " }{XPPEDIT 18 0 "
B[n];" "6#&%\"BG6#%\"nG" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "2/(n*Pi);"
"6#*&\"\"#\"\"\"*&%\"nGF%%#PiGF%!\"\"" }{TEXT -1 1 " " }{XPPEDIT 18 0
"int(g(x)*sin(n*Pi*x/L),x = 0 .. L);" "6#-%$intG6$*&-%\"gG6#%\"xG\"\"
\"-%$sinG6#**%\"nGF+%#PiGF+F*F+%\"LG!\"\"F+/F*;\"\"!F2" }{TEXT -1 3 " \+
." }}{PARA 0 "" 0 "" {TEXT -1 371 "Here is an example. We can animate
and expect to see the vibrations of the string. The following two exa
mples may give considerable insight for how the wave equation models t
he motion of a string. In the first example, take f as below and g to \+
be zero. This will mean that all the B coefficients will be zero and A
coefficients will be determined by the Fourier quotient." }}{PARA 0 "
" 0 "" {TEXT -1 7 " " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0
96 "L:=4;\nf:=x->(x-1)*(Heaviside(x-1)-Heaviside(x-2))+\n (3-x)*
(Heaviside(x-2)-Heaviside(x-3));" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 18 "plot(f(x),x=0..L);" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 93 "for n from 1 to 5 do\n A[n]:=2/L*int(f(x)*sin(n*P
i*x/L),x=0..L);\n B[n]:=0;\nod;\nn:='n':" }}}{EXCHG {PARA 0 "> "
0 "" {MPLTEXT 1 0 96 "u:=(t,x)->sum((A[n]*cos(Pi*n*t/L)+B[n]*sin(Pi*n*
t/L))*sin(n*Pi*x/L),\n n=1..5);" }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 65 "plot3d(u(t,x),x=0..L,t=0..2*L,axes=NORMAL,or
ientation=[-135,45]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "wi
th(plots):\nanimate(u(t,x),x=0..L,t=0..2*L);" }}}{EXCHG {PARA 0 "> "
0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 286 "In the second \+
example, take f to be zero and g to be as below. The formula for f and
g suggests that the string is at rest and we give the middle an initi
al velocity. This will mean that all the B coefficients will be zero \+
and A coefficients will be determined by the Fourier quotient." }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "L:=4;\ng:=x->-(Heaviside(x-1
)-Heaviside(x-3));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "plot(
g(x),x=0..L,discont=true,color=RED);" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 106 "n:='n':\nfor n from 1 to 5 do\n A[n]:=0;\n B
[n]:=2/(n*Pi)*int(g(x)*sin(n*Pi*x/L),x=0..L);\nod;\nn:='n':" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 96 "u:=(t,x)->sum((A[n]*cos(Pi*n
*t/L)+B[n]*sin(Pi*n*t/L))*sin(n*Pi*x/L),\n n=1..5);
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 65 "plot3d(u(t,x),x=0..L,t=
0..2*L,axes=NORMAL,orientation=[-135,45]);" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 32 "animate(u(t,x),x=0..L,t=0..2*L);" }}}{SECT 1 {PARA
4 "" 0 "" {TEXT -1 8 "Exercise" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 81 "
Examine the vibrations of a string with these three different initial \+
conditions." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0
"" {MPLTEXT 1 0 31 "L:=2:\nf:=x->sin(Pi*x);\ng:=x->0;" }}}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "L:=2:\nf:=x->0;\ng:=x->sin(Pi*x);"
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "L:=Pi;\nf:=x->Pi/2-abs(x-
Pi/2);\ng:=x->0;" }}}}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 61 "Orthogonal
families of functions and Sturm-Liouville problems" }}{EXCHG {PARA 0
"" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 408 " We have fou
nd that having an orthogonal family in a vector space is useful for th
e purposes of representing other elements of the space. If one were lo
oking for a basis for the space, an orthogonal family would be the ide
al choice. We have one method for generating an orthogonal family: per
form the Gramm-Schmidt process. In this worksheet we introduce another
method for getting an orthogonal family. " }}{PARA 0 "" 0 "" {TEXT
-1 70 " We compute eigenvectors for self-adjoint, linear transform
ations." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 262
11 "Definition:" }{TEXT -1 35 " A function A on a vector space is " }
{TEXT 263 6 "linear" }{TEXT -1 3 " if" }}{PARA 0 "" 0 "" {TEXT -1 10 "
" }{XPPEDIT 18 0 "A(lambda*x+y) = lambda*A(x)+A(y);" "6#/-%
\"AG6#,&*&%'lambdaG\"\"\"%\"xGF*F*%\"yGF*,&*&F)F*-F%6#F+F*F*-F%6#F,F*
" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 16 "for all numbers " }
{XPPEDIT 18 0 "lambda;" "6#%'lambdaG" }{TEXT -1 25 " and all vectors x
and y." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 257 "" 0 "" {TEXT -1
9 "Examples." }}{PARA 0 "" 0 "" {TEXT -1 73 "1. Let A be a square matr
ix defined on a finite dimensional vector space." }}{PARA 0 "" 0 ""
{TEXT -1 29 "2. Let the space be C ''([0, " }{XPPEDIT 18 0 "infinity;
" "6#%)infinityG" }{TEXT -1 43 " )) and A(f) = f '' for any f in the s
pace." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 268 11
"Definition:" }{TEXT -1 116 " A linear function A is self-adjoint on t
he space \{ E, < *, * > \} if , for all x and y in the domain of the o
perator" }}{PARA 0 "" 0 "" {TEXT -1 10 " " }{XPPEDIT 18 0 "an
glebracket(Ax,y) = anglebracket(x,Ay);" "6#/-%-anglebracketG6$%#AxG%\"
yG-F%6$%\"xG%#AyG" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 258 "" 0 "" {TEXT -1 9 "Examples." }}{PARA 0 "" 0 "" {TEXT -1
191 "3. One class of examples consists of all matrices which have real
number entries and which are symmetric about the main diagonal. To be
specific, we illustrate with a two dimensional example." }}}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "restart:\nwith(linalg):" }}}{PARA
0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 98 "as
sume(alpha,real):\nassume(beta,real):\nassume(Gamma,real):\nA:=matrix(
[[alpha,beta],[beta,gamma]]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1
0 52 "is(dotprod(A&*[x,y],[u,v])=dotprod([x,y],A&*[u,v]));" }}}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}{PARA 0 "" 0 "" {TEXT -1 362 "4. As a second example, we take A to be
a differential operator on a subset of C ''([0, 1]) with the usual do
t product. The subset of C ''([0, 1]) consists of functions f such tha
t f(0) = f(1) = 0. The definition of A is A(f) = f ''. We establish th
at this transformaton is self adjoint. To see how this is done the rea
der needs to remember integration-by-parts:" }}{PARA 0 "" 0 "" {TEXT
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 10 " " }{XPPEDIT 18 0 "int
(diff(u,x)*v,x = a .. b) = u(b)*v(b)-u(a)*v(a)-int(u*diff(v,x),x = a .
. b);" "6#/-%$intG6$*&-%%diffG6$%\"uG%\"xG\"\"\"%\"vGF-/F,;%\"aG%\"bG,
(*&-F+6#F2F--F.6#F2F-F-*&-F+6#F1F--F.6#F1F-!\"\"-F%6$*&F+F--F)6$F.F,F-
/F,;F1F2F>" }{TEXT -1 2 " ." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA
0 "" 0 "" {TEXT -1 16 "We must consider" }}{PARA 0 "" 0 "" {TEXT -1 0
"" }}{PARA 0 "" 0 "" {TEXT -1 12 " " }{XPPEDIT 18 0 "int(di
ff(f(x),`$`(x,2))*g(x),x = 0 .. 1)-int(f(x)*diff(g(x),`$`(x,2)),x = 0 \+
.. 1);" "6#,&-%$intG6$*&-%%diffG6$-%\"fG6#%\"xG-%\"$G6$F.\"\"#\"\"\"-%
\"gG6#F.F3/F.;\"\"!\"\"\"F3-F%6$*&-F,6#F.F3-F)6$-F56#F.-F06$F.\"\"#F3/
F.;F9\"\"\"!\"\"" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 77 "Using the integration by parts with u = f
'' and v = g, the first integral is" }}{PARA 0 "" 0 "" {TEXT -1 0 ""
}}{PARA 0 "" 0 "" {TEXT -1 39 " f '(1) g(1) - f '(0) g(0) - \+
" }{XPPEDIT 18 0 "int(diff(f,x)*diff(g,x),x = a .. b);" "6#-%$intG6$*&
-%%diffG6$%\"fG%\"xG\"\"\"-F(6$%\"gGF+F,/F+;%\"aG%\"bG" }{TEXT -1 2 " \+
." }}{PARA 0 "" 0 "" {TEXT -1 105 "On the other hand, using integratio
n by parts with u = g '' and v = f we have that the second integral is
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 38 " \+
g '(1) f(1) - g '(0) f(0) - " }{XPPEDIT 18 0 "int(diff(f,x)*diff(
g,x),x = 0 .. 1);" "6#-%$intG6$*&-%%diffG6$%\"fG%\"xG\"\"\"-F(6$%\"gGF
+F,/F+;\"\"!\"\"\"" }{TEXT -1 2 " ." }}{PARA 0 "" 0 "" {TEXT -1 60 "Wh
en we subtract the second integral from the first, we have" }}{PARA 0
"" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 64 " f '(1) \+
g(1) - f '(0) g(0) - g '(1) f(1) + g '(0) f(0)." }}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 109 "This last line will be z
ero if f and g satisfy the boundary conditions to be in the domain of \+
A, that is, if " }}{PARA 0 "" 0 "" {TEXT -1 40 " f(1) = 0 = f(0) an
d g(1) = 0 = g(0)." }}{PARA 0 "" 0 "" {TEXT -1 258 "We have establishe
d that this A -- two derivatives and boundary conditions -- is self ad
joint. (The reader might observe in passing that there are other bound
ay conditions that would achieve this same condition. We will see some
of these in future problems.)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 23 "Definition: The number " }{XPPEDIT 18 0 "
lambda;" "6#%'lambdaG" }{TEXT -1 86 " is an eigenvalue and the vector \+
v is an eigenvector of the linear transformation A if" }}{PARA 0 "" 0
"" {TEXT -1 17 " A(v) = " }{XPPEDIT 18 0 "lambda;" "6#%'lambd
aG" }{TEXT -1 3 " v." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 8 "Example." }}{PARA 0 "" 0 "" {TEXT -1 74 "5. The numbers \+
-1 and -2 are eigenvalues corresponding to the eigenvectors" }}{PARA
0 "" 0 "" {TEXT -1 10 " " }{XPPEDIT 18 0 "matrix([[1], [1]]);
" "6#-%'matrixG6#7$7#\"\"\"7#\"\"\"" }{TEXT -1 6 " and " }{XPPEDIT
18 0 "matrix([[1], [-1]]);" "6#-%'matrixG6#7$7#\"\"\"7#,$\"\"\"!\"\""
}{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 25 "for the following matri
x." }}{PARA 0 "" 0 "" {TEXT -1 10 " " }}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "A:=matrix([[
-3/2,1/2],[1/2,-3/2]]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "
eigenvects(A);" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT -1 23 "6. Each of the numbers " }{XPPEDIT 18 0 "-pi^2;" "6#,$*$%
#piG\"\"#!\"\"" }{TEXT -1 2 ", " }{XPPEDIT 18 0 "-4*Pi^2;" "6#,$*&\"\"
%\"\"\"*$%#PiG\"\"#F&!\"\"" }{TEXT -1 3 " , " }{XPPEDIT 18 0 "-9*Pi^2;
" "6#,$*&\"\"*\"\"\"*$%#PiG\"\"#F&!\"\"" }{TEXT -1 7 ", ..., " }
{XPPEDIT 18 0 "-n^2*Pi^2;" "6#,$*&%\"nG\"\"#%#PiG\"\"#!\"\"" }{TEXT
-1 58 " ... is an eigenvalue corresponding to the eigenfunction " }}
{PARA 0 "" 0 "" {TEXT -1 10 " " }{XPPEDIT 18 0 "sin(n*Pi*x);
" "6#-%$sinG6#*(%\"nG\"\"\"%#PiGF(%\"xGF(" }{TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT -1 64 "for the linear operator of Example 4 above. We ver
ify this here." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 56 "assume(n,i
nteger):\nu:=x->sin(n*Pi*x);\nlambda:=-n^2*Pi^2;" }}}{EXCHG {PARA 0 ">
" 0 "" {MPLTEXT 1 0 55 "is(diff(u(x),x,x)=lambda*u(x));\nis(u(0)=0);
\nis(u(1)=0);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 99 "Remark: We have
presented the eigenvalues and eigenvectors for these two linear trans
formations as " }{TEXT 264 14 "faits accompli" }{TEXT -1 307 ". The pr
ocess for getting eigenvalues and eigenvectors for a matrix is the sub
ject of another course. The process of getting the eigenvalues and eig
enfunctions for this differential operator is a subject for this cours
e, and we will derive the results in time. For now, we observe what is
true in Example 6." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 49 "We do not forget the purpose of this development." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 266 7 "Fact 1:"
}{TEXT -1 98 " Eigenvalues corresponding to self-adjoint transformatio
ns are real numbers - not complex numbers." }}{PARA 0 "" 0 "" {TEXT
-1 0 "" }}{PARA 0 "" 0 "" {TEXT 267 7 "Fact 2:" }{TEXT -1 101 " Eigenv
ectors for self adjoint transformations corresponding to different eig
envalues are orthogonal." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT -1 188 "Revisit Examples 3 and 4. Note that the eigenvect
ors are orthogonal. In the matrix example, this is seen quickly. For t
he differential operator, recall that the dot product is an integral.
" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 55 "assume(m,integer):\nint(
sin(n*Pi*x)*sin(m*Pi*x),x=0..1);" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT -1 216 "Remark. Concerning Fact 2 above, suppose we have two eig
envectors corresponding to one eigenvalue. The eigenvectors found may \+
not be orthogonal. Not to worry! Perform the Gramm-Schmidt process to \+
these eigenvectors. " }}{PARA 0 "" 0 "" {TEXT -1 1 " " }}{PARA 0 "" 0
"" {TEXT 265 48 "Definition of a regular Sturm-Liouville problem:" }
{TEXT -1 47 " Suppose that s, s ', and q are continuous and " }
{XPPEDIT 18 0 "s(x) <> 0;" "6#0-%\"sG6#%\"xG\"\"!" }{TEXT -1 94 " for \+
x in [a,b]. In the presence of appropriate boundary conditions, the di
fferential operator" }}{PARA 0 "" 0 "" {TEXT -1 27 " L(f) = (s f '
) ' - q f" }}{PARA 0 "" 0 "" {TEXT -1 133 "is self adjoint in C ''([a,
b]). Example of appropriate boundary conditions are that functions f i
n the domain of L should satisfy are" }}{PARA 0 "" 0 "" {TEXT -1 23 "(
1) f(0) = f(1) = 0, or" }}{PARA 0 "" 0 "" {TEXT -1 27 "(2) f '(0) = f \+
'(1) = 0, or" }}{PARA 0 "" 0 "" {TEXT -1 11 "(3) f(0) = " }{XPPEDIT
18 0 "alpha;" "6#%&alphaG" }{TEXT -1 19 " f '(0) and f(1) = " }
{XPPEDIT 18 0 "alpha;" "6#%&alphaG" }{TEXT -1 11 " f '(1), or" }}
{PARA 0 "" 0 "" {TEXT -1 111 "(4) if s(0) = s(1), then f(0) = f(1) and
f '(0) = f '(1). (These last are called periodic boundary conditions.
)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 8 "Examp
le:" }}{PARA 0 "" 0 "" {TEXT -1 95 "We provide two eigenvectors (eigen
functions) for each eigenvalue of the Sturm-Liouville problem" }}
{PARA 0 "" 0 "" {TEXT -1 17 " f '' = " }{XPPEDIT 18 0 "lambda
;" "6#%'lambdaG" }{TEXT -1 39 " f, f(-1) = f(1), f '(-1) = f '(1).
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 148 "In t
his example, there are an infinite number of eigenvalues, and for each
one, there are two eigenfunctions. We will see that the eigenvalues a
re " }{XPPEDIT 18 0 "-n^2*pi^2;" "6#,$*&%\"nG\"\"#%#piG\"\"#!\"\"" }
{TEXT -1 22 " , for each integer n." }}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 18 "assume(n,integer);" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 39 "y1:=x->sin(n*Pi*x);\ny2:=x->cos(n*Pi*x);" }}}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "A:=y->diff(y(x),x,x)+n^2*Pi^2*y(x);
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "A(y1);\nA(y2);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "y1(-1); y1(1);\nD(y1)(-1);D(
y1)(1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "y2(-1); y2(1);\n
D(y2)(-1);D(y2)(1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 12 "Assignmen
t: " }}{PARA 0 "" 0 "" {TEXT -1 23 " Verify that sin(n " }
{XPPEDIT 18 0 "pi;" "6#%#piG" }{TEXT -1 54 " x) is an eigenvalue corre
sponding to the eigenvector " }{XPPEDIT 18 0 "-n^2*pi^2;" "6#,$*&%\"nG
\"\"#%#piG\"\"#!\"\"" }{TEXT -1 17 " for the operator" }}{PARA 0 "" 0
"" {TEXT -1 46 " A(f) = f ''." }}
{PARA 0 "" 0 "" {TEXT -1 34 "(Hint: show that if f(x) = sin(n " }
{XPPEDIT 18 0 "pi;" "6#%#piG" }{TEXT -1 8 " x) and " }{XPPEDIT 18 0 "l
ambda;" "6#%'lambdaG" }{TEXT -1 4 " = " }{XPPEDIT 18 0 "-n^2*pi^2;" "
6#,$*&%\"nG\"\"#%#piG\"\"#!\"\"" }{TEXT -1 14 ", then f '' = " }
{XPPEDIT 18 0 "lambda;" "6#%'lambdaG" }{TEXT -1 5 " f. )" }}}{PARA 0 "
" 0 "" {TEXT -1 3 " " }}}{MARK "4" 0 }{VIEWOPTS 1 1 0 1 1 1803 }