Hassani S. Mathematical Physics: A Modern Introduction to Its Foundations
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13.4 THEWRONSKIAN 363
Dividing (13.t7) by this w and substituting for w yields
u" +
S(x)u
= 0, where
W'
w"
12
II
Sex) = q +
p-
+ - = q - 4
P
- 'lP .
w w
oscillation
of
the
Bessel
function
of
order
zero
A
useful
special
case
of
the
comparison
theorem
is
given
as
the
following
corollary
whose
straightforward
but
instructive
proof
is
left
as a
problem.
13.4.13.
Corollary.
If
q(x)
::: 0 for all x E [a, bj, then no nontrivial solution
of
the differential equation v" +
q(x)v
= 0 can have more than one zero.
13.4.14.
Example.
It should be clear from the preceding discussion that the oscillations
of the solutions of
s"
+q(x)v = 0 are mostly determined by the sign and magnitude of
q(x).
For
q(x)
::s
0 there is no oscillation; that is, there is no solution that changes sign
more than once. Now suppose that
q(x)
::: k
2
> 0 for some real k. Then, by Theorem
13.4.11, any solution of v" +
q(x)v
= 0
must
have at least
one
zero between any two
successivezeros of the solutionsin
kx
of u
lf
+ k
2u
=
O.
This means that any solutionof
v" +
q(x)v
= ohas a zero in any interval oflength1l"jk
if
q(x)
2: k
2
>
O.
Let us apply this to the Bessel DE,
"
I,
( n
2
)
y +
-y
+
1-
- y =
O.
x x
2
We can eliminate the y' term by substituting v
j.JX
for y.
5
This transforms the Bessel DE
into.
(
4n2-1)
v" + 1 -
---
v =
O.
4x
2
We compare this, for n = 0, with u"+U = 0, whichhas a solutionu = sin x, and conclude
that
each
interval
of
length7r
of
the positivex-axis contains at leastone zero
of
any solution
of
orderzero (n = 0)
of
the Bessel equation. Thus, in particular, the zeroth Besselfunction,
denoted by
Jo(x), has a zero in
each
interval
of
length 7r
of
the x-axis.
On the other hand, for 4n
2
- I > 0, or n > !'we have I > [t - (4n
2
-
l)j4x
2
j.
This implies that sin x has at least one zerobetween any two successive zeros
of
the Bessel
functions
of
order greater than
~.
It
follows that such a Bessel function can have at most
one
zero between any two successive zeros
of
sin x (or in
each
interval
of
length 7r on the
positive x-axis). 11
13.4.15.
Example.
Let us apply Corollary 13.4.13 to v" - v = 0 in
whichq(x)
=
-I
<
O.According to the corollary, the
most
general solution,
cj
eX +
C2e-x,
can have at
most
one zero. Indeed,
x
-x
0 1I 1
c21
qe+c2e
=
;::}x=zn-
q
,
and this (real) x
(if
it exists) is the only possiblesolution, as predictedby the corollary. 11
SBecause
of
the square root in the denominator, the range
of
x will have to be restricted to positive values.
364 13.
SECOND-ORDER
LINEAR
DIFFERENTIAL
EQUATIONS
13.5 AdjointDifferentialOperators
We discussed adjointoperators in detail in the context of finite-dimensional vector
spaces in Chapter 2.
In
particular, the importance of self-adjoint, or hermitian,
operators was clearlyspelledout by the spectraldecompositiontheoremof Chapter
4. A consequence of that theorem is the completeness of the eigenvectors of a
hermitian operator, the fact that an arbitrary vector can be expressed as a linear
combination of the (orthonormal) eigenvectors
of
a hermitian operator.
Self-adjoint differential operators are equally importantbecause their "eigen-
functions" also form complete orthogonal sets, as we shall see later. This section
will generalize the concept
of
the adjoint to the case of a differential operator (of
second degree).
13.5.1. Definition. The HSOLVE
L[y]
'"
p2(X)y" +PI
(x)y'
+
po(x)y
= 0
exact
SOLDE
is said to be exact
if
(13.18)
(13.19)
integrating
factor
for for all f E e
2[a,
b]andfor some A, B E
ella,
bJ, An integratingfactorfor
L[y]
SOLDE
is afunction /L(x) such that /L(x
)L[y]
is exact.
If
an integrating factor exists, then Equation (13.18) reduces to
d,
,
d)A(x)y
+
B(x)y]
=0
=>
A(x)y
+
B(x)y
=C,
a FOLDE with a constantinhomogeneous term. Even the ISOLDE corresponding
to Equation (13.18) can be solved, because
d ,
/L(x)L[y] =/L(x)r(x)
=>
dx[A(x)y
+
B(x)y]
=/L(x)r(x)
=>
A(x)y'
+
B(x)y
=
LX
/L(t)r(t)dt,
which is a general FOLDE. Thus, the existence of an integrating factor completely
solves a SOLDE.
It
is therefore importantto know whether
ornot
a SOLDEadmits
an integrating factor. First let us give a criterion for the
exactness of a SOLDE.
13.5,2. Proposition.
The
SOWE
of
Equation (13.18) is exact
ifand
only
if
P~
-
pi
+Po =
o,
Proof
If
the SOLDEis exact, then Equation (13.19) holds for all
f.
implying that
P2 = A, PI = A' +B. and
PO
= B'.
It
follows that
P~
=
A",
pi
=
A"
+B', and
PO
= B'. which in
tum
give
P~
-
pi
+
PO
=
O.
13.5 ADJOINTDIFFERENTIAL
OPERATORS
365
Conversely if Pz - PI +
PO
= 0, then, substituting
PO
= -pz +PI in the
LHS of Equation (13.18), we obtain
and the DE is exact.
D
A general SOLDE is clearly not exact. Can we make it exact by multiplying
it by an integrating factor as we did with a FOLDE? The following proposition
contains
the
answer.
13.5.3. Proposition.
Afunetion
J.L
is an integratingfactor
of
the
SainE
ofEqua-
tion (13.18) ifand only ifit is a solution
of
the
HSOInE
M[J.L]
es
(P2J.L)"
- (PIJ.L)'+
PoJ.L
= O.
(13.20)
Proof
This is an immediate conseque,:!ce
of
Proposition 13.5.2.
We can expand Equation (13.20) to obtain the equivalent equation
P2J.L"
+
(2p~
-
PI)J.L'
+
(pz
- PI +
pO)J.L
=
O.
D
(13.21)
(13.22)
The operator Mgiven by
2
~
d r d " ,
M es P2dx
2
+(2p2 -
PI)
dx
+(P2 -
PI+
PO)
is called the
adjoint
of the operator L and denoted by Mes
Lt.
The reason for the
use of the word "adjoint" will be
made
clear below.
Proposition 13.5.3 confirms the existence
of
an integrating factor. However,
the latter can be obtained only by solving Equation (13.21), which is at least as
difficult as solving the original differential equation!
In contrast, the integrating
factor for a FOLDEcan be obtained by a mere integration [see Equation (13.8)].
Althoughintegratingfactors for SOLDEs are
not
as usefnl as their counterparts
for FOLDEs, they can facilitate the study
of
SOLDEs.
Let
us first note that the
adjoint
of
the adjoint of a differential operator is the original operator:
(Lt)t
= L
(see Problem 13.11). This suggests that if v is an integrating factor of L[u], then u
will be an integratingfactor ofM[v] sa
L
t[
v]. In particular, multiplyingthe first one
by v and the second one by u and subtracting the results, we obtain [see Equations
(13.18) and (13.20)] vL[u] - uM[v]
= (VP2)U" - U(P2V)" +(VPI)U' +U(PI v)',
which can be simplified to
adjoint
ofa
second-order
linear
differential
operator
d , ,
vL[u] - uM[v] =
-[p2VU
- (P2V) u +PIUV].
dx
(13.23)
(13.24)
366 13.
SECOND-ORDER
LINEAR
DIFFERENTIAL
EQUATIONS
Lagrange
identities
Integrating this from a to b yields
l
b
(vL[u]-
uM[v])dx =
[pzvu'
-
(pzv)'u
+
Pluv]I~.
Equations(13.23) and (13.24)are calledthe
Lagrange
identities.Equation(13.24)
embodies the reasonfor calling
Mthe adjoint
of
L:
If
we consideru and v as abstract
vectors
Iu) and Iv), L and M as operators in a Hilbert space with the inner product
(ul v) =
J:
u*(x)v(x)
dx,
then Equation (13.24)
can
be written as
(vi L lu) - (ul MIv) =(ul Lt [u)" - (ul M [u} =[pzvu' -
(pzv)'u
+
Pluv]I~.
If
the
RHS
is zero, then (ul
Lt
Iv)* = (ul M Iv) for all ]«) , [u), and since all these
operators and functions are real,
Lt = M.
As in the case
of
finite-dimensional vector spaces, a self-adjoint differential
operator merits special consideration.
For
M[v] es Lt [v] to be equal to L, we must
have [see Equations (13.18)
and(13.21)12p~
- PI = PI and
P~
-
pi
+
PO
= Po·
The
first equation gives
P~
= PI, which also solves the second equation.
If
this
conditionholds, then we
can
write Equation(13.18) as L[y] = PZy" +
p~y'
+
POY,
or
d [ d
Y]
(
L[y] =
dx
pz(x)
dx
+
Po(x)y
=
O.
Can we make all SOLDEs self-adjoint? Let us multiply both sides
of
Equation
(13.18) by a function
h(x),
to be detenntined later. We get the new DE
h(x)pz(x)y"
+
h(x)PI(x)y'
+
h(x)po(x)y
= 0,
which we desire to be self-adjoint. This will be accomplished
if
we choose
h(x)
such
thathpi
=(hpz)', or
pzh'
+h(p~
- PI) =0, which
can
be readilyintegrated
to give
h(x)
=
~
exp
[fX
PI
(t)
dt]
.
PZ
pz(t)
We have
just
proved the following:
all
SOLOEs
canbe 13.5.4.
Theorem.
The
SOInE
of
Equation (13.18) is self-adjoint if and only if
made
self-adjoint
P~
= PI, in which case the DE has the form
d [ d
Y]
dx
pz(x)
dx
+
po(x)y
=
O.
If
it is not self-adjoint, it can be made so by multiplying it through by
I
[fX
PI(t)
]
h(x)
=
-exp
--dt.
pz
pz(t)
13.6
POWER-SERIES
SOLUTIONS
OF
SOLOES
367
13.5.5. Example. (a)TheLegendreequationin normalform,
"
2x,
x
y
---y
+--y=O,
I-x
lI-x
2
isnotself-adjoint.
However,
we get-aself-adjoint versionif we multiply
through
byh(x) =
I-x
2:
(I - x
2
)y"
- 2xy' +Ay= 0,
or [(I - x
2
)y'1' +Ay= o.
(b)Similarly, thenormalformof the Besselequation
I ( n
2
)
y"+-y'+
1--
y=O
x x
2
isnot self-adjoint, butmultiplying throughby
h(x)
= x yields
~(XdY)
+
(x
_n
2
)
y = 0,
dx.
dx x
whichis clearlyself-adjoint.
13.6 Power-Series Solutions
of
SOLDEs
l1li
Analysis is one of the richest branches
of
mathematics, focusing ou the eudless
variety of objects we call functions. The simplest kind
of
functiou is a polyno-
mial, which is obtainedby performing the simple algebraic operations
of
addition
and multiplication on the independent variable
x. The next in complexity are the
trigonometric functions, which are obtained by taking ratios of geometric objects.
If
we demand a simplistic, intuitive approach to functions, the list ends there.
It
was only with the advent of derivatives, integrals, and differential equations that a
vastly richvariety of functions exploded into existence in the eighteenth and nine-
teenth
centuries.
For
instance,
e",nonexistentbeforethe
invention
of
calculus,
can
be thought of as the function that solves
dy
[dx
= y.
Although the definition of a function in terms
of
DEs and integrals seems a bit
artificial, for most applications it is the only way to define a function.
For
instance,
the
error
function,
usedin statistics, is.definedas
I
IX
2
erf(x)
==
r;;
e-
t
dt.
y'Jr
-00
Such a function carmot be expressed in terms of elementary functions. Similarly,
functions (of
x)
such as
1
00
sint
-dt,
x t
368 13.
SECOND-ORDER
LINEAR
DIFFERENTIAL
EQUATIONS
and soon are enconnteredfrequently in applications. None of these functionscan
be expressedin terms of other well-knownfunctions.
An effectiveway of studying such functions is to study the differentialequa-
tions they satisfy.
Infact, the majority of functions encountered in mathematical
physicsobey theHSOLDEof Equation (13.I8)in whichthe
Pi
(x)
areelementary
functions,mostlyratiosofpolynomials(ofdegree at most
2). Ofcourse, to specify
functionscompletely,appropriatebonndaryconditions
are necessary.Forinstance,
the errorfunctionmentionedabovesatisfiesthe HSOLDE
y" +
2xy'
= 0 withthe
boundaryconditions
yeO)
=!and y'(O) =l/..,m.
Thenatural tendencyto resist the idea of a function as a solutionof a SOLDE
is mostlydue to theabstractnature of differentialequations.Afterall,it iseasier to
imagineconstructingfnnctionsby simplemultiplicationsorwithsimplegeometric
figuresthat have been aronndfor centuries.The followingbeautiful example (see
[Birk78, pp.
85--S7])
should overcome this resistance and convince the skeptic
that differentialequations contain all the information about a function.
13.6.1. Example.
We
can
show
thatthe
solutions
toy" + Y =0
have
all the
properties
we expectof sinx
and
cosx. Letus
denote
thetwo
linearly
independent
solutions
of this
equationbyC(x) andS(x).
Tospecifythesefunctiooscompletety,
weset
C(O)
= S'(O) = t,
and
C'(O) = S(O)=
O.
We
claimthatthis
information
is
enoogh
to
ideotify
C(x)
andSex)
as
cos x and
sinx,
respectively.
First,
let us show
that
thesolutions exist andarewell-behaved
functions.
With
C(O)
and c'(0) given, the equation y" + y = 0 can generate all derivatives of C(x) at zero:
C"(O)
= -C(O) =
-I,
CII/(O) =
-C'(O)
= 0, C(4)(0) =
-C"(O)
=
+1,
and,
in
general,
{
o
ifn i odd
C(n)
(0) = 1 a is ,
(-I)k
if
a = 2k wherek= 0, 1,2, .. ,.
Thns,
the
Taylor
expansion
ofC(x) is
00
2k
C(x)
=
L(-I)k~,
k=O
e2k)!
Shnilarty,
00
2k+l
Sex) =
"e_l)k_X;-:-:c-;
6
(Zk+I)!'
(13.25)
e13.26)
Example
illustrates
that
all
intormation
about
sine
and
cosine
is
hidd,m
in
their
differential
equation
A simple
ratio
testontheseries
representation
of C
(x)
yields
.
ak+t
'
e-l)k+l
x2(k+I)/e2k+2)!,
_x
2
100
--=
too = 1hn
=0
k....oo ak k....oo
e-l)k
x2k/e2k)!
k....oo e2k+ 2)(2
k+
I) ,
whichshows
that
theseriesfor
C(x)
converges
forallvaluesof x.
Similarly,
theseriesfor
S(x) is also
convergent.
Thus,
we are
dealing
withwell-defined
finite-valued
functions.
Letusnow
enumerate
and
prove
some
properties
of
C(x)
andS(x).
(a) C'ex) =
-Sex),
We
prove
this
relation
by
differentiating
c"
(x)
+ C
(x)
= 0 and
writing
the
result
as
13.6 POWER-SERIES SOLUTIONS
OF
SOLOES
369
[C'
(x)1"+
C'
(x) = 0 to makeevidentthe fact
that
C'
(x) is also a solution. Since
C'
(0) = 0
and
[C'eO)]'
= C"(O) =
-1,
and since
-Sex)
satisfies the same initialconditions, the
uniqueness theorem implies
that
C'(x)
=
-S(x).
Similarly,
S'(x)
=
C(x).
(b) C
2(x)
+S2(x)
= I.
Sincethe
p(x)
term
is
absent
from
the
SOLDE,
Proposition 13.4.2implies
that
the
Wron-
skiau
of
C(x)
aud S(x) is constaut. On the
otherhaud,
W(C, S; x) =
C(x)S'
(x) - C'
(x)S(x)
= C
2(x)
+ S2(x)
= W(C, S; 0) = C
2(0)
+S2(O)=
I.
(c) S(a +x) =
S(a)C(x)
+
C(a)S(x).
Theuseoftbechainruleeasilyshows
that
Sea
+x)
is asolution
of
the
equation
y"
+y =
O.
Thus,it canbe
written
as a
linear
combination
of
C(x)
andSex) [which
are
linearly
independent
because
their
Wronskian
is nonzeroby (b)]:
S(a
+x) =
AS(x)
+
BC(x).
(\3.27)
This is a functional identity, which for x = 0 gives S(a) = BC(O) =
B.lfwe
differentiate
both sides
of
Equation (13.27), we get
C(a
+x)
=
AS'(x)
+
BC'(x)
=
AC(x)
-
BS(x),
which
for x = 0 gives C(a) = A. Substitoting the values
of
A aud B in Equatiou (13.27)
yieldsthe
desired
identity.
A
similar
argument
leadsto
C(a
+x) =
C(a)C(x)
- S(a)S(x).
(d) Periodicity
of
C(x)
aud
S(x).
Let xo be the smallest positive real
uumber
such
that
S(xo) = C(xo).
Then
property (b)
implies
that
C(xo) = S(xo) =
1/./2.
On the
otherhaud,
S(xo +x) = S(xO)C(x) +
C(xo)S(x)
= C(XO)C(x) +S(xO)S(x)
=
C(xo)C(x)
-
S(xo)S(-x)
=C(xO -
x).
The
third equality follows becaose by Equation (13.26), S(x) is au
odd
function
of
x. This
is
truefor
all x; in particular, for x = xo it yields S(2xo) = C(O) =
I,
aud
by property (b),
C(2xO) =
O.
Using property (c)
once
more, we
get
S(2xo +x) =S(2xo)C(x) +C(2xo)S(x) =
C(x),
C(2xo +x) =
C(2xo)C(x)
- S(2xO)S(x) =
-S(x).
Substitoting x = 2xo yields S(4xO) = C(2xo) = 0
aud
C(4xo) = -S(2xO) =
-I.
Continuing
inthis
manner,
we caneasily
obtain
S(8xo +x) =S(x),
C(8xo
+ x) =
C(x),
which
prove
the
periodicity
of Sex) and
C(x)
andshow
that
their
periodis
8xQ.
It is even
possibleto
determine xo.This
determination
is leftas a
problem,
buttheresultis
xo =
{l/.Ji
_d_t_.
10
JJ=t2
A
numerical
calculation
will showthatthisis
11:/4.
370 13.
SECOND-ORDER
LINEAR
DIFFERENTIAL
EQUATIONS
13.6.1 Frobenius Method of Undetermined Coefficients
A proper treatment of SOLDEs requires the medium of complex analysis and will
be undertakeu in the uext chapter. At this poiut, however, we are seekiug
aformal
iufiuite series solutiou to the SOLDE
y" +
p(x)y'
+
q(x)y
= 0,
where
p(x)
and
q(x)
are real and analytic. This means that
p(x)
and
q(x)
can he
represeuted by convergent power series in some interval
(a, b). [The interesting
case where
p(x)
and
q(x)
may have siugularities will be treated in the context of
complex solutions.]
The general procedure is to write the expansions''
00
p(x)
=
Lakxk,
k~O
00
q(x)
=
LbkXk,
k~O
(13.28)
for the coefficient functions
p and q and the solution y, substitute them in the
SOLDE, and equate the coefficient of each power of
x to zero.
For
this purpose,
we need expansions for derivatives of y:
00 00
y'
=
LkCkX
k-
1
=
L(k
+l)ck+I
Xk,
k=l
k=O
00 00
v"
=
L(k
+l)kck+IXk-1 =
L(k
+
2)(k
+
I)Ck+2
xk
k~l
k=O
Thus
00 00
p(x)y'
= L L amxm(k +l)Ck+I
Xk
=
L(k
+
l)a
mCk+lX
k+
m.
k=Om=O k,m
Letk+m
==
n andsumovern.Thenthe
other
sum,saym,
cannot
exceedn.Thus,
00
n
p(x)y'
=
LL(n-m+l)a
mc
n-
m+IX
n.
n=Om=O
Similarly,
q(x)y
=
I:~o
I:~~o
bmcn_mx
n.
Substituting these sums and the se-
ries for
y" in the SOLDE, we obtain
~
{(n +
l)(n
+2)C
n+2
+
1;
[en- m +
l)a
mC
n-
m+1
+
b~cn-m]
} x
n
=
O.
6Herewe are
expanding
about
the
origin.
If
suchan
expansion
is impossible or
inconvenient,
onecan
expand
about
another
point,sayxo. Onewouldthen
replace
allpowersof x in allexpressions belowwithpowersof x - xo.These
expansions
assume
that
p, q, andy haveno
singularity
atx =
O.
In general, this
assumption
isnotvalid,anda differentapproach, inwhichthewhole
seriesis
multiplied
bya(notnecessarily positive
integer)
powerof x, oughttobe
taken.
Detailsare
provided
in
Chapter
14".
13.6
POWER-SERIES
SOLUTIONS
OF
SOLOES
371
For this to be true for all x, the coefficient
of
each power of x mustvanish:
"
(n +
lien
+2)c"+2 = -
L[(n
- m +l)amc,,-m+1 +bmc,,_m]
m=O
or
for
n 2: 0,
,,-I
n(n
+I)C"+l = -
L[(n
- m)amc,,_m +bmc,,_m_tl
m=O
for n 2: 1.
(13.29)
the
SOLOE
existence
theorem
If
we know
Co
and CI (for instance from bonndary conditions), we
can
uniquely
determine
c"for n 2:2 from Eqnation (13.29). This, in tnm, gives a uniquepower-
series expansion for
y, and we have the following theorem.
13.6.2.
Theorem.
(the existence theorem) For any SOLVE
of
the form
s"
+
p(x)y' +q(x) y = 0 with analytic coefficient functions given by the first two
equations
of
(13.28), there exists a unique power series, given by the third equa-
tion of(13.28) that formally satisfies the
SOWEfor
each choice
of
coand
ct.
This theorem merely states the existence
of
a formal power series and says
nothing about its convergence. The following examplewill demonstrate that con-
vergence is not necessarily guaranteed.
13.6.3. Example. The
formal
power-series
solutionforx
2
s'- y+x = 0canbe
obtained
by
letting
y =
L~o
e-x",Theny' =
L~o(n
+ l)cn+lXn, and
substitution
in theDE
gives
L~O(n
+ l)Cn+lXn+2 -
E~o
cnx
n
+x = 0, or
00 00
L(n+
1)cn+lXn+2 -
Co
-qx
-
LCnxn
+x
=
O.
n=O
n=2
We see that
Co
= 0,
C1
= 1, and (n + l)C
n+l
= c
n+2
for n
~
O.
Thus,we have the
recursion
relation
nCn
= c
n+l
forn
:::
1whose
unique
solution
is en = (n -
1)1,
which
generates
thefollowing solution forthe
DE:
Y = x +x
z
+(2!)x
3
+(3!)x
4
+...+(n -
t)!x"
+....
Thisseriesis not
convergent
foranynonzerox.
III
As we shall see later, for normal SOLDEs, the power series of y in Equation
(13.28) converges to an analytic function. The SOLDE solved in the preceding
example is not normal.
13.6.4.
Example.
As an application of Theorem
13.6.2,
tet us cousiderthe Legendre
equation
inits
normal
form
"
2x!
A
Y
---ZY
+--zy=O.
I-x
t-x
372
13.
SECOND-ORDER
LINEAR
DIFFERENTIAL
EQUATIONS
For IxI < I both p and q are analytic, and
00 00
p(x)
=
-2x
L (x
2)m
= L
(_2)x
2m+1,
m=O m=O
00 00
q(x)
= ic L (x
2)m
= L !cx
2m
.
m=O m=O
Thus, the coefficients of Eqoation (13.28) are
{
o
if
m is even.
am =
-2
ifmisodd
and
b
m
=
{~
if m is even,
ifm
is odd.
Wewantto
substitute
for am andb
m
in
Equation
(13.29)to findC
n
+l.It is
convenient
to
consider
two
cases:
when n is odd andwhen n is even. For n = 2r + 1,
Equation
(13.29)-after
some
algebra-yields
r
(2r +
1)(2r
+2)C2'+2 = L (4r - 4m - ic)C2(,-m)'
m=O
With
r
---+
r + 1,this
becomes
,+1
(2r +
3)(2r
+4)C2'+4 = L
(4r
+4 - 4m - ic)C2(,+I-m)
m=O
(13.30)
forevenk.
(13.31)
,+1
= (4r +4 - ic)C2(,+I) + L (4r +4 - 4m - ic)C2(,+I_m)
m=l
r
=(4r +4 - ic)c2r+2 + L (4r - 4m - ic)C2(,_m)
m=O
= (4r +4 - ic)C2,+2+(2r +1)(2r +2)02,+2
=
[-ic
+(2r +
3)(2r
+2)]c2,+2,
where
in goingfromthesecond
equality
tothe
third
we
changed
the
dummy
index,
andin
going from the thirdequality to the fourth we used Eqoation (13.30). Now we
let2r+2
eek
to obtain (k +
I)(k
+2)Ck+2=
[k(k
+I) - iclck, or
k(k+
I)
-ic
ck+2 = (k +
I)(k
+2)
Ck
Itis not
difficult
toshow
that
starting
withn = 2r, thecaseof evenn, we
obtain
this
same
equation
foroddk.
Thus,
we can
write
n(n+I)-ic
c
n+2
= (n +
I)(n
+2)
Cn·
For
arbitrary
coandcj ,we
obtain
two
independent
solutions, oneofwhichhasonlyeven
powers
ofx andthe
other
onlyodd
powers.
Thegeneralizedratiotest(see[Hass99,
Chapter
5]) showsthatthe seriesis
divergent
for x = ±1 unless A =
1(1
+ 1) for some positive