Chapter 4 Power Series Solutions
4.1 Introduction
This chapter will focus on the discussion of linear equation
with nonconstant-coefficient equations
Solving the differential equation listed in the following
with the series method
dy
y0
dx
(1)
To solve by the series method, we seek a solution in the
form of a power series expansion about any desired point x=x0,

y ( x)   an ( x  x0 ) n
n 0
1
We choose x0=0 for simplicity, then

y ( x)   an x n  a0  a1 x  a2 x 2 
(2a)
n 0
and
dy d
 (a0  a1 x  a2 x 2 
dx dx
)  a1  2a2 x1  3a3 x 2 
(2b)
Putting Eq. (2a,b) into (1) gives
(a1  a0 )  (2a2  a1 ) x  (3a3  a2 ) x 
1
2
0
(4)
Because the right hand side of Eq.( 4) is 0, it implies that
a1  a0  0  a1  a0
2a2  a1  0  a2  a1 / 2  a0 / 2
3a3  a2  0  a3  a2 / 3  a0 / 6
2
Where a0 remains arbitrary. Thus, we have
1 2 1 3
y ( x)  a0 (1  x  x  x 
2
6
)  Ce
x
(6)
Eq. (6) us derived based on the following operations
d
d
n
(i)  an x   (an x n )
dx
dx
(ii) An   Bn   ( An  Bn )
(iii) if  An x   Bn x  An  Bn
n
n
In this chapter, we will discuss the following topics.
4.2 Power series solutions
4.3 The method of Frobenius
4.4 Legendre functions
4.5 Singular integrals
4.6 Bessel functions
3
4.2 Power series solutions
Finite sum of a series
N
a
k 1
 a1  a2 
n
 aN
(1)
Infinite sum of a series

a
k 1
k
 a1  a2  a3
(2)
The sequence of partial sums of the series (2) as
s1  a1 , s2  a1  a2 , s3  a1  a2  a3
(3)
and so on
n
sn   ak
(4)
k 1
4
If the limit of the sequence Sn exists, as n→∞, and equals
some number s, then we say that the series (2) is convergent,
and that it converges to s; otherwise it is divergent. That is the
limit of an infinite series is defined as the limit (if that limit exists)
of its sequence of partial sums.

a
k 1
k
n
 lim  ak  lim sn  s
n 
k 1
n 
sn  s , it means that to each number
By definition lim
n 
ε>0, no matter how small, there exists an integer N such
that∣s-sn∣<εfor all n>N.
Theorem 4.2.1 Cauchy Convergence Theorem
An infinite series is convergent if and only if its sequence
of partial sums sn is a Cauchy sequence-that is, if to each e >o
there corresponds an integer N(e) such that ∣sm-sn∣<e for all
5
m and n greater than N.
A series of the form

n
2
a
(
x

x
)

a

a
(
x

x
)

a
(
x

x
)

 n
0
0
1
0
2
0
,
(6)
n 0
Where the an’s are numbers called the coefficients of the
series, x is a variable, and x0 is a fixed point called the center
of the series. The series may converge at some points on the
x axis and diverge at others.
Theorem 4.2.2 Interval of Convergence of Power Series
The power series (6) converges at x=x0. If it converges at
other points as well, then those points necessarily comprise
an interval ∣x-x0∣<R centered at x0 and, possibly, one or
both endpoints of that interval (Fig. 1), where R can be
determined from either of the formulas if the limits in the
denominators exist and are nonzero.
6
Fig. 1 Interval of convergence
of power series
1
1
R
or R 
an 1
lim n an
lim
n 
n  a
n
(7a,b)
If the limits in Eq. (7a,b) are zero, then Eq. (6) converges for all
x (i.e. for every finite x, no matter how large), and we say that
“R = ∞.“ If the limits fail to exist by virtue of being infinite, then
R = 0 and Eq. (6) converges only at x0.
We call∣x-x0∣<R the interval of convergence, and R the
radius of convergence. If a power series converges to a
function f on some interval, we say that it represents f on that
interval, and we call f its sum function.
7

Example 1 Find R of the series
n
n
!
x

0

Example 2 Find R of the series
n
n
(

1)
[(
x

5)
/
2]

0
( x  1)n
Example 3 Find R of the series 
n
(
n

1)
4

8
Theorem 4.2.3 Manipulation of power series
(a) Termwise differentiation (or integration) permissible
A power series may be differentiated (or integrated)
termwise within its interval of convergence I. The series that
results has the same interval of convergence I and presents
the derivative (or integral) of the sum function of the original
series.

If f(x)= a n ( x  x0 )n within I , then
0


d 
d
f ( x)   an ( x - x0 ) n   [an ( x - x0 ) n ]   nan ( x - x0 ) n-1
dx 0
0 dx
1
(11)

b
a
n 1
n 1
(
b

x
)

(
a

x
)
0
0
f ( x)dx   an  ( x - x0 )n dx   an
a
n 1
0
0

b

(12)
9
(b) Termwise addition (or subtraction or multiplication)
permissible.
Two power series (about the same point xo) may be added
(or subtracted or multiplied) termwise within their common
interval of convergence I. The series that results has the
same interval of convergence I and presents the sum (or
difference or product) of their two sum functions.


0
0
If f ( x)   an ( x  x0 ) n and g ( x)   bn ( x  x0 ) n

f ( x)  g ( x)   (an  bn )( x - x0 ) n
(13)
0
Letting z = x – x0


0
0
f ( x) g ( x)  ( (an z n )( bn z n )

= (a 0 bn +a1bn-1 +
0
+a n b0 ) z n
(14)
10
(c) If two power series are equal, then their corresponding
coefficients must be equal. That is, for


n 0
n 0
n
n
a
(
x

x
)

b
(
x

x
)
 n
n
0
0
to hold in some common interval of convergence, it must
be true that an=bn for each n. In particular, if

n
a
(
x

x
)
0
 n
0
n 0
in some interval, then each an must be zero.
11
Taylor Series: Taylor series of a given function f(x) about a
chosen point x0, which we denote here as TS f
, is
x0
defined as the infinite series
f ( x0 )
f ( x0 )
TS f x0  f ( x0 ) 
( x  x0 ) 
( x  x0 ) 2 
1!
2!


0
f ( n ) ( x0 )
( x  x0 ) n
n!
The purpose of Taylor series is to represent the given function.

f ( x)  
0
f ( n ) ( x0 )
( x  x0 )n
n!
Three conditions need to be met.
1. f have a Taylor series about that point. That means f must
be infinitely differentiable at x0. So that all the coefficients
f(n)(xo)/n! in Eq. (15) exist.
2. The series converge in some interval∣x-x0∣<R, for R>0.
12
3. The sum of the Taylor series is equal to f in the interval.
If a function is represented in some nonzero interval
∣x-x0∣<R by its Taylor series (i.e.
and
TS f exists,
x0
converges to f(x)), then f is said to be analytic at x0. If a
function is not analytic at x0, then it is singular there and
X0 is called singular point of f.
4.2.2 Power series solution of differential equations
Theorem 4.2.4 Power series solution
If p and q are analytic at x0, then every solution of
y  p( x) y  q ( x) y  0
is too, and can be found in the form

y ( x)   an ( x  x0 ) n
n 0
13
Example 4
Solve
y  y  0 by the power series method.
14
Example 5
Solve the initial-value problem
( x  1) y  y  2( x  1) y  0
y (4)  5, y(4)  0
by the power series method on the interval 4≦x＜∞.
15
4.3 The Method of Frobenius
4.3.1 Singular points
y  p( x) y  q ( x) y  0
(1)
For the differential equation (1), we can find two LI solutions
as power series expansions about any point xo at which both
p and q are analytic. We call such a point xo an ordinary point
of the equation (1)
Definition 4.3.1 Regular and Irregular Singular Point of (1)
Let x0 be a singular point of p and/or q. Xo is a regular
singular point of Eq. (1) if (x-x0)p(x) and (x-x0)2q(x) are
analytic at x0. Otherwise, x0 is an irregular singular point
of Eq.(1) if it is not a regular singular point.
16
2
x
(
x

1)
y  3 y  5 y  0
Example 1
Example 2
y  x y  0
17
4.3.2 Method of Frobenius
Consider the equation
y  p( x) y  q ( x) y  0
(5)
to have a regular singular point at the origin. There is no loss
of generality in assuming it to be at the origin since, if it is at
x=x0≠0, we can always make a change of variableξ= x-x0
to move it to the origin in terms of the new variable ξ .
Multiplying Eq. (5) by x2 and rearranging terms as
x 2 y  x[ xp( x)] y  [ x 2 q( x)] y  0
(6)
Since x = 0 is a regular singular point, it follows that xp(x)
and x2q(x) can be expanded about the origin in convergent
Taylor series,
18
x 2 y  x( p0  p1 x 
) y  (q0  q1 x 
)y  0
(7)
Locally, in the neighborhood of x = 0, we can approximate (7)
as
x 2 y  p0 xy  q0 y  0
(8)
which is Cauchy-Euler equation. As such, Eq. (8) has at least
one solution in the form xr, for some constant r.
Returning to Eq. (7), it is reasonable to expect that equation,
likewise, to have at least one solution that behaves like xr
(for some value of r) in the neighborhood of x = 0. That is, we
expect it to have at least one solution of the form
y ( x)  x r (a0  a1 x  a2 x 2 
)
(9)
19
where the power series factor is needed to account for the
deviation of y(x), away from x= 0, from its asymptotic behavior
y(x)~a0xr as x → 0. That is, in place of the power series
expansion

y ( x )   an x n
(10)
n0
that is guaranteed to work when x = 0 is an ordinary point
of Eq.(5), it appears that we should seek y(x) in the more
general form


y ( x)  x r  an x n   an x n  r
n 0
(11)
n 0
If x=0 is a regular singular point.
20
Example 3
6 x 2 y  7 xy  (1  x 2 ) y  0
(0 < x <∞) (12)
21
Theorem 4.3.1 Regular Singular Point; Frobenius Solution
Let x = 0 be a singular point of the differential equation
y  p( x) y  q ( x) y  0
with xp( x)  p0  p1 x 
(x > 0)
(37)
and x 2 q( x)  q0  q1 x 
Having radii of convergence R1, R2, respectively. Let r1, r2 be
the roots of the indicial equation
r 2  ( p0  1)r  q0  0
(38)
where r1≧r2 if the roots are real.
Seeking y(x) in the form


y ( x)  x r  an x n   an x n  r
n 0
n 0
a0≠ 0
(39)
22
with r =r1 inevitably leads to a solution
y1 ( x)  x

r1
a x
n 0
n
n
(a 0  0)
(40)
where a1, a2, … are known multiples of a0, which remains
arbitrary. For definiteness, we choose a0=1 in Eq. (40).
The form of the second LI solution, y2(x), depends on r1
and r2 as follows:
(i) r1 and r2 distinct and not differing by an integer. Then
with r = r2, Eq. (39) yields
y2 ( x )  x

r2
b x
n
n
(b0  0)
(41a)
0
(ii) Repeated roots, r1 = r2 ≡r. Then y2(x) can be found in
the form

y2 ( x)  y1 ( x) ln x  x r  cn x n
1
(41b)
23
(iii) r1 - r2 equal to an integer. Then y2(x) can be found in
the form

y2 ( x)   y1 ( x) ln x  x r2  d n x n
(41c)
0
Example 4
x 2 y  ( x  x 2 ) y  y  0
(0 < x < )
(49)
24

r
n
y
(
x
)

y
(
x
)
ln
x

x
c
x
Derive 2
 n in case of r1 = r2 = r
1

Assume

1
y1 ( x)  x r  an x n   an x n  r
n 0
is known, let us
n 0
assume y2(x)=A(x)y1(x).
Substituting y2 into Eq.(37) gives
Ay1  A(2 y  py1 )  ( y1  py  qy1 )  0
(42)
The last term in Eq. (42) is zero, so Eq. (42) becomes
Ay1  A(2 y  py1 )  0
(43)
replacing A by dA/dx , multiplying through by dx, and
dividing by y1 and A  gives
dy1
dA
2
 pdx  0
A
y1
(44)
25
Integrating
and
ln A  2 ln y1   p( x)dx  ln c for c  0
ln
so
A y1
2
   p( x)dx
C
p0
 p1  p2 x L )dx
x
e
e
e p0 ln x e(  p1x L )
A( x)  C 2
C
 C 2r
2
y1 ( x)
x (1  2a1 x  L )
 x r (1  a1 x  L ) 
 (
 p ( x )dx
(45)
A( x)  C
1
2 r  p0
(1  k1 x  L )  A( x)  C (ln x  k1 x  L )
(47)
x
setting C = 1, we have
y2 ( x)  A( x) y1 ( x)  (ln x  k1 x  L ) y1 ( x)
 y1 ( x) ln x  (k1 x  L ) x r (1  a1 x  L )

 y1 ( x) ln x  x r  cn x n
(48)
26
Example 5
xy  y  0
(0 < x < )
27
4.4 Legendre Functions
4.4.1 Legendre Polynomials
The differential equation
(1  x 2 ) y  2 xy   y  0
(1)
where  is a constant, is known as Legendre’s equation
The x interval of interest is -1<x<1, and Eq. (1) has regular
singular points at each endpoints, x =1 and -1.
Find the power series solutions about the ordinary point x = 0.

y ( x )   ak x
k 0
k
(2)
putting Eq. (2) into Eq. (1) leads to the recursion formula
28
ak  2
k (k  1)  

ak (k  0,1, 2, )
(k  1)(k  2)
(3)

2
(6   )
a2   a0 , a3 
a1 , a4  
a0
2
6
24
The general solution is
  2 (6   ) 4 (20   )(6   ) 6
y ( x)  a0 1  x 
x 
x 
24
720
 2
2   3 (12   )(2   ) 5


 a1  x 
x 
x  
6
120


 a0 y1 ( x)  a1 y2 ( x)


(4)
29
Radii of convergence
1
1
R
=
1
k (k  1)  
an 1
lim
lim
k  ( k  1)( k  2)
k  a
n
So each series converges in -1<x<1. However, for arbitrary
values of the parameter  the functions y1(x) and y2(x) given
in Eq. (4) grow unboundly as x→±1, as illustrated in Fig. 1
for = 1.
Nonetheless, for certain specific
values of l one series of the other,
in Eq. (4), will terminate and thereby
be bounded on the interval since it
is then a finite degree polynomial!
Fig. 1 y1(x) and y2(x)
in Eq. 4 for  = 1.
30
If  happens to be such that
  n(n  1)
(6)
Then Eq. (4) becomes
 n(n  1) 2 n(n  2)(n  1)(n  3) 4

y ( x)  a0 1 
x 
x  
2!
4!


 (n  1)(n  2) 3 (n  1)(n  3)(n  2)(n  4) 5
 a1  x 
x 
x 
3!
5!

 a0 y1 ( x)  a1 y2 ( x)


(6’)
For any integer n = 0, 1, 2, …, then we can see from Eq. (3)
that one of the two series terminates at k = n: if n is odd, then
the second series contains only a finite number of terms; if n is
even, then the first series contains only a finite number of
terms. In either of these cases the series which reduces to a
finite sum is known as a Legendre polynomial.
31
Fig. 2 Graphs of the first
five Legendre polynomials
In fact, it can be shown that they are given explicitly by the
formula:
1 dn
2
n


Pn ( x)  n
(
x

1)
n 
2 n ! dx
(8)
which result is known as Rodrigues’s formula
32
4.5 Singular Integrals; Gamma Function
4.5.1 Singular integrals
An integral is said to be singular (or improper) if one or
both integration limits are infinite and/or if the integrand is
unbounded on the interval; otherwise, it is regular (or proper).
For example, if

I1   xe 2 x dx
3
dx
I3  
1 x  1
2
I2  
e x
dx
x
5
0
I4  
100
0
(1)
xe x dx

Considering the function of
Analogous to the definition
 f ( x)dx
a
a
n 0
n
N
 lim  an
N 
n 0
(2)
33
Define

X
I   f ( x)dx  lim  f ( x)dx
a
X  a
If the limit exists, we say that I is convergent; if not, it is divergent.


0
0
Comparison tests: If S1   an and S 2   bn are series of
finite positive terms, and an~Kbn as n→∞ for some finite
constant K, then S1 and S2 both converge or both diverge.

1
In calculus, the p-series  p converges if p > 1 and
1 n
diverges if p ≦ 1.
2n  3
2n  3 2
 3
Considering the series  4
, we observe that 4
n 5 n
1 n 5
as n→∞.

1
1 n3 is convergent because it is a p - series with p  3.
34

Analogous to the p-series, we study the horizontal p-integral
I 

a
1
dx (a>0)
p
x
(5)
where p is a constant.

X
lim
ln
x
 X 
a ( p  1)
 1
X 1

I   p dx  lim  p dx = 
(6)
a x
a x
X 
1-p aX
1
 lim
x ( p  1)
 X  1-p

lim ln X is infinite and hence does not exist, and similar for
X 
lim X 1- p if p  1, whereas the latter does exist if p  1.
X 
35
Theorem 4.5.1 Horizontal p-Integral
 1
The horizontal p-integral, a x p dx
converges if p > 1 and diverges if p ≦ 1.
Theorem 4.5.2 Comparison Tests
Let I1 


a

f ( x)dx and I2  g ( x)dx, where f(x) and g(x) are
a
positive and bounded on a  x  
(a) If there exist constants K and X such that f(x) ≦ Kg(x)
(for all x ≧ X), then the convergence of I2 implies the
convergence of I1, and divergence of I1 implies the
divergence of I2.
(b) If f(x) ～C g(x) as x→∞, for some finite constant C, then
I1 and I2 both converge or both diverge.
36
Theorem 4.5.3 Absolute Convergence




If
f ( x) dx converges, then so does a f ( x)dx ,
a
and we say that the latter converges absolutely.
Example 1 Consider I  

9
Example 2 Consider
I 
2x  3
dx
4
x 5

2
sin x
dx
2
3x  1
37
Example 3 Consider I 
Example 4 Consider


0
x100e0.01x dx
I 

3
dx
x ln x
38
All the aforementioned cases are that the upper limit is ∞,
let us consider the left endpoint x = a. If the intergrand f(x)
blows up as x→ a. We define
b
I   f ( x)dx  lim 
a
b
e 0 a  e
f ( x)dx
(10)
where ε→ 0 through positive values.
Considering vertical p-integral
I 
b
0
1
dx
p
x
(11)
According to Eq. (10)

b
lim
ln
x
 e 0
e ( p  1)
b 1
b 1

I   p dx  lim  p dx = 
0 x
e 0 e x
1-p eb
1
lim
x ( p  1)
 e 0 1-p

(12)
39
Theorem 4.5.4 Vertical p-Integral
b 1
The vertical p-integral, 0 p dx (b  0)
x
converges if p < 1 and diverges if p > 1.
Theorem 4.5.5 Comparison Test

b
If I  f ( x)dx , where 0 < b <∞. If f(x)~K/xp as x→ 0 for
0
some constant K and p, and f(x) is continuous on 0 < x ≦ b,
then I converges if p < 1 and diverges if p≧ 1.
Example 5 Test the integral
divergence.

4
sin 2x
0
x
3
2
dx for convergence/
40
4.5.2 Gamma function

( x)   t x1et dt
0
The integral is singular for two reasons: first, the upper
limit is ∞ and, second, the integrand blows up as t→0
if the exponent (x-1) is negative.


0
T
e
f (t )dt  lim
e 0
T 

T

f (t )dt  lim  f (t )dt   f (t )dt 
 e


e 0
T 

T


= lim  f (t )dt  lim  f (t )dt   f (t )dt   f (t )dt
e 0 e
T  
0
( x)  ( x  1)( x  1)


( x)  t e  ( x  1)  t x2et dt
x 1 t
0
(15)
(16)
41
(n  1)  n( x)  n(n  1)(n  1)
  n( n  1)( n  2)( n  3)
since
(1)(1)
(18)

(1)   et dt  1
(19)
0
Eq. (18) becomes
(n  1)  n !
and
(20)
1
( )  
2
(21)
For x < 0, let us define
Fig. Gamma function
( x  1)
( x) 
for x  0
x
(22)

2
3  t
Example 6 Evaluate ( x)   t e
0
dt
42
4.6 Bessel Functions
The differential equation
x 2 y  xy  ( x 2  2 ) y  0
(1)
where νis a nonnegative real number, is known as
Bessel’s equation of order ν. The equation appears in a wide
Range of application such as steady and unsteady diffusion in
cylindrical region.
Putting the equation into a standard form by dividing x2
through, we see
2
2
1
( x  )
p( x)  , and q( x) 
x
x2
There is one singular point, x = 0, and that it is a regular
singular point because xp(x)=1, and x2q(x) = x2 – ν2 are
analytic at x = 0.
43
4.6.1 ν≠integer
Seeking a Frobenius solution for the Bessel’s equation about
the regular singular point x = 0 by letting

y ( x)   an x k  r


k 0
k 0
(a 0  0)
(2)

 k  r 2  2  ak  ak  2 x k  r =0


(3)
where a0 ≠ 0 and a-2=a-1≡0. Equating to zero the coefficient
of each power of x in Eq. (3) gives
k  0 : (r 2  2 )a0  0
(4a)
k  1: (( r  1)  ) a1  0
(4b)
k  2 : ((r  k ) 2  2 )ak  ak  2  0
(4c)
2
2
44
Since a0 ≠ 0, Eq. (4a) gives the indicial equation r2-2 = 0.
(a) Considering r=+, then Eq. (4b) gives a1=0 and Eq. (4c)
gives the recursion relation
1
ak  
ak  2 (k  2)
k (k  2 )
(5)
From Eq. (5), together with the fact that a1 = 0, it follows
that a1 = a3 = a5 =…= 0 and that
(1) k
a2 k  2 k
a0
2 k !(  k )(  k  1) L (  1)
(1) k (  1)
= 2k
a0
2 k !(  k  1)
(6)
(7)
45
we have the solution
k
(

1)
x 2 k 

y( x)  a0 2 (  1) 
( )
K 0 k !(  k  1) 2

(8)
Dropping the a02νΓ(ν+1) scale factor, we call the resulting
solution the Bessel function of the first kind, of order :
x 
(1)k
x 2k
J ( x)  ( ) 
( )
2 K 0 k !(  k  1) 2
(9)
(b) Considering r=-. Changing  to – everywhere on the
right side of Eq. (9). Denoting the result as J(x), the
Bessel function of the first kind, of order –, we have
x  
(1)k
x 2k
J  ( x)  ( ) 
( )
2 K 0 k !(k   1) 2
46
The general solution of Eq. (1) is
y( x)  AJ ( x)  BJ  ( x)
Writing Eqs. (9) and (10)

1
1
2
J ( x)  x 

x 

 2
 (  1)2 (  2)2

(11)




1
1
2
J  ( x)  x 

x 

  2
(2  )2
 (1  )2

Fig. 1 J1/2 (x) and J-1/2(x)
(12)



(13)
2
J 1 ( x) 
sin x
x
2
(14a)
2
J 1 ( x) 
cos x

x
2
(14b)
47
4.6.2 ν= integer
If  is a positive integer n, then
(  k  1)  (n  k  1)  (n  k )!
And the solution expressed by Eq. (9) gives
(1)k
x 2k n
J n ( x)  
( )
k 0 k !(k  n)! 2

(15)
We need to be careful with Eq. (10) because if  = -n, then the
Γ(k-n+1) in Eq. (10) is undefined when its argument is zero
or a negative integer, namely, for k = 0, 1, 2, …, n-1, and it
equals 1/(k-n)! For k = n, n+1, n+2,…, in which case Eq. (10)
becomes
(1)k
x 2 k n
J  n ( x)  
( )
k  n k !(k  n)! 2

(17)
48
Since Γ(k-n+1) is undefined at k=0,1,2,…,(n-1), rather
than “∞”. Replacing the dummy summation index k by
m according to k-n=m,
(1)m n x 2 m n
J  n ( x)  
( )
m 0 (m  n)!m ! 2

If (-1)n is factored out, the series that remains is the same
as that given in Eq. (15), so that
J  n ( x)  (1) J n ( x)
n
(18)
JJ n ( x) and J  n ( x) are linearly dependent. Thus
whereas J ( x) and J  ( x) are LI if  is not an integer.
We have only one linearly independent solution thus far for
the case where  = n. namely, y1(x) = Jn(x).
49
Let us begin with n=0, i.e. y1(x)=J0(x), for repeated indicial roots,
r=±n=0, which corresponds to case (ii) of that theorem4.3.1, to
obtain a second LI solution y2(x) we rely on Theorem 4.3.1.
Accordingly, we seek y2(x) in the form

y2 ( x)  J 0 ( x) ln x   ck x k
(19)
1
and obtain
x 2
1 1 x 4
y2 ( x)  J 0 ( x) ln x  ( )  (1  )
( ) 
2
2
2 (2!) 2
(20)
which is called Y0(x), the Neumann function of order zero.
Thus, we have two LI solutions, y1(x) = J0(x), and y2(x)=Y0(x).
However, following Weber, it proves to be convenient and
standard to use, in place of Y0(x), a linear combination of
J0(x) and Y0(x), namely,
y2 ( x ) 
2

 Y0 ( x)  (  ln 2) J 0 ( x)  Y0 ( x)
(21)
50
where
 x
x2
1
x4 
(ln 2   ) J 0 ( x)  22  (1  2 ) 24 (2!) 2 
2

Y0 ( x) 


1 1
x6

 (1   ) 6

2
2 3 2 (3!)


(22)
is Weber’s Bessel function of the second kind, of order zero.
γ= 0.5772157 is known as Euler’s constant. The graphs of
J0(x) and Y0(x) are shown in Fig. 2.
J 0 ( x) ~ 1, Y0 ( x) ~
and
J 0 ( x) ~
2

ln x
2

cos( x  ),
x
4
2

Y0 ( x) ~
sin( x  )
x
4
as x→0
as x→∞
51
For n = 1, 2, 3,…the indicial roots r =± n differ by an integer,
which corresponds to case (iii) of Theorem 4.3.1. Using the
theorem, and the ideas given above for Y0 we obtain Weber’s
Bessel function of the second kind, of order n.
1 n 1 (n  k  1)! x 2 k n 
 x
( ) 
(ln 2   ) J n ( x)  2 
k!
2
2
k 0

Yn ( x) 
k

  1  (1)  (k )   (k  n)  x 2 k  n
(25)
( )
 

k !(k  n)!
2
 2 k 0

4.6.3 General solution of Bessel equation
The solution of Bessel equation
x y  xy  ( x  ) y  0
2
If
2
2
Eq.(9)
Eq.(10)
 is not an integer
y( x)  AJ ( x)  BJ  ( x)
 is an integer and
 = n = 0, 1, 2, …
y( x)  AJ n ( x)  BYn ( x)
Eq.(15)
Eq.(25)
52
If we define
(cos x) J ( x)  J  ( x)
Y ( x) 
sin x
(26)
for noninteger n, then the limit of Y(x) as →n (n = 0,1, 2,..)
gives the same result as Eq. (25). Furthermore, J(x) and Y(x)
are LI so we can express the general solution of Eq. (1) as
y( x)  AJ ( x)  BY ( x)
(27)
for all value of .
4.6.5 Modified Bessel equation
2
2


x y  xy  ( x  ) y  0
2
(33)
 is an integer
53
Let t = ix (or x = it), we can convert Eq. (33) to the Bessel
equation
t 2Y   tY   (t 2  n2 )Y  0
(34)
The general solution is
y( x)  AJ n (ix)  BYn (ix)
(35)
(1) k
ix 2 k  n
J n (ix)  
( )
k  0 k !( k  n)! 2


1
x 2k n
 i 
( )
k  0 k !( k  n)! 2
Define I n ( x)  i  n J n (ix )

1
x 2k n
I n ( x)  
( )
and
k  0 k !( k  n)! 2
n
(36)
(37)
known as the modified Bessel function of the first kind,
and order n.
54
In place of Yn(ix) it is standard to introduce, as a second
real-valued solution, the modified Bessel function of the
second kind, and order n.
K n ( x) 

2
i
n 1
 J n (ix)  iYn (ix)
(38)
and the graphs of these functions are plotted in Fig. 4.
As a general solution of Eq. (33) we have
y( x)  AI n ( x)  BK n ( x)
(40)
Whereas the Bessel functions are
oscillatory, the modified Bessel
functions are not.
Fig. 4 Graphs of I0(x) and K0(x).
55
Example 1
xy  y   2 xy  0
Solve
(43)
x 2 y  xy   2 x 2 y  0
More generally, the equation
d a dy
(x
)  bx c y  0
dx
dx
1
t   bx  and u  x
(46)



y
(47)
2
d
u
du
2
2
2
t

t

(
t


)u  0
2
dt
dt
u ( x)  AJ (t )  BY (t )  x



(48)
y  AJ (t )  BY (t )
56
If we choose
2
1 a

and  
ca2
ca2


(49)
1
y( x)  x Z ( b x  )
where
(50)
Z ( x)  AJ ( x)  BY ( x)
Example 2
y  3 x y  0
57
Problems for Chapter 4
Exercise 4.2
3.(b)、(h)、(k)
6.； 7.(b)、(f)
11.(b)、(c)、(e)
Exercise 4.3
1.(a)-(g)、(l)、(p)
2.;
3.(a)、(b)
6.(b)、(g)、(m)、(s)
10.
Exercise 4.4
1.; 5.
6.(a)； 7. (a)
10.(a)
11.(a)、(b)
Exercise 4.6
4.(a)； 5.(a)； 6.(a)； 7.(a)
8.(b)； 12.(a)、(e)、(i)
Exercise 4.5
1.; 2.
3.(a)、(c)、(d)、(f)、(g)、(h)
4.
6.(a)、(b)
7.(a)、(b)、(f)
9.
10.(a)、(b)、(d)
14.
16.(b)
17.
19.(a)、(b)、(f)、(g)
58