Chapter 6
Series Solutions of Linear Equations
1
NCCU
Wireless Comm. Lab.
Outline
 Using power series to solve a differential equation. First, we
should decide the point we choose to be the expanding point that
is ordinary or not.
 If the point is not an ordinary point, decide it a regular or irregular
singular point, then use the Frobenius’ series to solve the problem.
 Introduce the Bessel equation and the Legendre’s equation.
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NCCU
Wireless Comm. Lab.
Introduction
 In applications, higher order linear equations with variable
coefficients are just as important as, if not more important than,
differential equations with constant coefficient.
 Considering a equation y   xy  0, it does not possess elementary
solutions. But we can find two linear independent solutions of
y   xy  0 by using the series expansion.
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NCCU
Wireless Comm. Lab.
6.1 Solutions About Ordinary Points
 In section 4.7, without understanding that the most higher-order
ordinary equations with variable coefficients cannot be solved in
terms of elementary functions.
 The usual strategy for solving differential equations of this sort is
to assume a solution in the form of an infinite series and proceed
in a manner similar to the method of undetermined coefficients.
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NCCU
Wireless Comm. Lab.
6.1.1 Review of Power Series
 Definition
 A power series in x  a is an infinite series of the form

c 0  c1 ( x  a )  c 2 ( x  a ) 

2
c
(x  a)
n
n
n 0
Such a series is also said to be a power series centered at a.

 For example, the power series
c
n
( x  1)
n
is centered at a =1.
n0
 Convergence

 A power series  c n ( x  a ) n is convergent at a specified value of x if its
n0
sequence of partial sums  S N ( x ) converges- that is,
N
lim S  x   lim  c
N
N 
( x  a )  constant.
n
n
N   n0
 If the limit does not exist at x, the series is said to be divergent.
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NCCU
Wireless Comm. Lab.
 Interval of Convergence
 Every power series has an interval of convergence. The interval of
convergence is the set of all real number x for which the series converges.
 Note: We will use the ratio test to see the series is convergence or divergence for
 x x  the region of R  , w here R w ill be defined im m edia tely
 Radius of Convergence
and R is the radius of convergence .
 As we mentioned that the R is assigned to be an interval boundary to check
the series for its convergence property and R is also called the radius of
convergence.
 What’s the meaning for the value R?
 It means a distance from the point x to the nearest singular point.(see in Theorem 6.1)
 Bringing a concept, the singular point possess between convergent and
divergent region.
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NCCU
Wireless Comm. Lab.

 If R  0, then a power series  c
diverges for x  a  R .
n
(x  a)
n
converges for
xa  R
and
n0
 For example, if the series converges for x=a or for all x, then R is equal to
0 or  .
 Recall that
xa  R
is equivalent to
a  R  x  a  R.
 Note: A power series may or may not converge at the endpoints a-R or a+R of this
interval.
 Absolute Convergence
 Within its interval of convergence a power series converges absolutely.


cn ( x  a )
n
co n verg es , w h ere a  R  x  a  R .
n0
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Wireless Comm. Lab.
 Ratio test

 Convergence of a power series  c n ( x  a ) can often be determined by
n0
the ratio test.
n
 Suppose that
c n  0 ,  n and that
lim
n 
c n 1 ( x  a )
cn ( x  a )
n 1
n
 xa
lim
n 
c n 1
 L.
cn
 If L<1 the series converges absolutely, if L>1 the series diverges, and if L=1
the test is inconclusive.
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Wireless Comm. Lab.
 Example:

A power series 
n0
( x  5)
4 ( n  1)
n
( x  5)
lim
4
n 1
,
the ratio test gives
n 1
(n  2)
( x  5)
n 
n
n
 x 5
lim
n 
4 ( n  1)
( n  1)
4(n  2)

1
x 5 (
L );
4
n
 The series converges absolutely for
1
4
x5 1
(L<1), we get 1  x  9 .
 The series diverges for L>1, that is x  9 o r x  1 .
 Test for the convergence of the boundary for x=1 or 9.


n0
(4)
4 ( n  1)
n


n 1
(4)


n0
(  1)


n 1
n
( n  1)

n
4 ( n  1)
n

n
1
( n  1)
S o th e in t e r v a l o f
fo r x  1, it w ill c o n v e r g e .
fo r x  9 , it w ill d iv e r g e .
convergence of
9
th e s e r ie s is [1, 9 ).
NCCU
Wireless Comm. Lab.
 A Power Series Defines a Function

 A power series defines a function f ( x )   c ( x  a ) whose domain is the
interval of convergence of the series. If the radius of the convergence is
R>0, then f is continuous, differential, and integrable on the interval
( a  R , a  R ). Thus, f  ( x ) and  f ( x ) dx can be found by term-by-term
differentiation and integration.
n
n
n0

 If
y
c
n
x
n
is a power series in x, then the first two derivatives are
n0

S in ce y  c 0  c1 x  c 2 x 
 cn x 
2

n
c
n
x
n
n0
S o y   c1  2 c 2 x 
 ncn x
n 1



 nc
n
x
n 1
n 1
y   2 c 2  3  2 c 3 x 
 n  ( n  1) c n x
n2



 n ( n  1) c
n
x
n2
.
n2
 It will be useful to substitute
y , y , and y 
10
into the 2 n d differential equation.
NCCU
Wireless Comm. Lab.
 Identity Property

c
If
( x  a )  0, R  0 fo r a ll n u m b ers x in th e in terval o f co n verg en ce ,
n
n
n0
th en c n  0  n .
 Analytic at a Point
 A function f is analytic at a point a if it can be represented by a power
series in x-a with a positive or infinite radius of convergence.
 For example,
e
x
 1 x 
x
2


2!
co s x  1 
x
x

n

 (  1)
n
2!
sin x  x 
x
x


(  1)
n2
and integrable in the interval of convergence .
n!


x
F or analytic , the function is continuous , differ ential ,
n
2n
2n!
3
3!

n0
2


n!
x


n0
2 n 1
( 2 n  1) !
(  1) x



n0
n
2n
2n!
(  1)
n2
x
2 n 1
( 2 n  1) !
fo r
x  .
T h ese T a ylo r series cen tere d a t 0, ca lled M a cla u rin series ,
sh o w th a t e , co s x , a n d sin x a re a n a lytic a t x  0 .
x
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Wireless Comm. Lab.
 Arithmetic of Power Series
 Power series can be combined through the operations of addition,
multiplication, and division.
 Example:
U sin g th e series ex p an sio n to d escrib e th e sin x  co s x

sin x  co s x 

(  1)
 (x 
3


(  1)
 x
3
3

x
5
20

2 n 1
n2
x
x


m0
x
2m
2m !
2 n 1
( 2 n  1) !
3!
x
x
( 2 n  1) !
n0
x
n2
)  (1 
x
2

2!

x
2n

)
2n!
7

 x,
x  .
252
It is o b vio u s th a t th e p o w er series sin x  co s x co n verg es o n th e sa m e in terval.
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Wireless Comm. Lab.
 Shifting the Summation Index
 It is important to combine two or more summations with different index, so
it may need to shift the summation index. You may see the rule by the
following example.
 Example 1:Adding Two Power Series


W r ite

n ( n  1) c n x
n2


n2
c
n
x
n 1
n 0

( If y 

cn x
n
is a T a y lo r s e r ie s a t c e n te r e d p o in t 0 , th e s u b je c t
n0
is lik in g to c o m b in e th e y   x y .)
F r o m th e o r ig in a l s u b je c t ,
s ta r t fr o m x
s ta r t fr o m x
0
s ta r t fr o m x



n2
n ( n  1) c n x



cn x
s ta r t fr o m x
w ith n  3
1

n2
1
w ith n  0

n 1

 2 c2 x
n0
0


n3
13
n ( n  1) c n x
1

n2



cn x
n 1
n0
NCCU
Wireless Comm. Lab.

 n ( n  1) c
C onsidering the term of
n
x
n2
n3
 n ( n  1) c
n
x


n3
c
n
x
n 1

n0
 [( k  3)( k  2 )c
k 3
n
x
n 1
let k  n

 ( k  3)( k  2 ) c
k 3
x
k 1


k 0


c

let k  n - 3
n2

n0



 ck ] x
c
k
x
k 1
k 0
k 1
k 0

  n ( n  1) c n x
n2
n2


c
n
x
n 1

 2c2 
n0
 [( k  3)( k  2 )c
k 3
 ck ]x
k 1
k 0
N ote : F or the shifting sum m ation approach ,
St ep 1.Y ou m ay figure out the m inim um order of each com ponent , then take the term s
out from the sum m ation .
Step 2. Substitute a new index to the old one for prep aring to com bin e the sum m ation .
Step 3.C reate a new sum m ation for the original sub ject .
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Wireless Comm. Lab.
6.1.2 Power Series Solutions
 Definition 6.1 Ordinary and Singular Points
 A point x 0 is said to be an ordinary point of the differential equation
a ( x ) y   a ( x ) y   a ( x ) y  0, if both P(x) and Q(x) in the standard form
y   P ( x ) y   Q ( x ) y  0 are analytic at x 0 . A point that is not an ordinary
point is said to be a singular point of the equation.
2
1
0
 F or a n a lytic , it m e a n s th a t a 2 ( x )  0 in y  
 Considering the differential equation
–
a1 ( x )
y 
a2 ( x)
y   cos x  y   e y  0
x
a0 ( x )
y  0.
a2 ( x)
and
y   ln x  y   e y  0.
x
y   cos x  y   e y  0
y   ln x  y   e y  0
F or y   cos x  y   e y  0
F or y   ln x  y   e y  0
P ( x )  cos x , Q ( x )  e
P ( x )  ln x , Q ( x )  e
x
x
x
x
x
x
x  0 is the ordinary point,
x  0 is a singular point,
because P ( x ) and Q ( x ) are analytic .
because P ( x ) is not a nalyti c .
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Wireless Comm. Lab.
 Polynomial Coefficients
A polynom ial is analytic at any value x , and a rational function
is analytic except at points w here its denom inator is zero .
T hus if a 2 ( x ), a 1 ( x ), ,and a 0 ( x ) are polynom ials w ith no com m on factors,
then both rational function P (x)=
a1( x)
and Q (x)=
a2 ( x)
a 0 ( x)
are analytic except
a2 ( x)
w here a 2 ( x )  0.
F or x  x 0 is an ordinary point of a 2 ( x ) y   a 1 ( x ) y   a 0 ( x ) y  0, if a 2 ( x 0 )  0,
w hereas x  x 0 is a singular point of a 2 ( x ) y   a1 ( x ) y   a 0 ( x ) y  0, if a 2 ( x 0 )  0.
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NCCU
Wireless Comm. Lab.
E xam ple : If the equation is ( x  1)( x  3) y   ( x  1) y   ( x  1)( x  3) y  0
2
2
C onsider for the ordinary and singular points.
T he original equation can be rew ritten to y  
( x  1)
( x  1)( x  3)
2
2
y 
( x  1)( x  3)
( x  1)( x  3)
2
2
y  0.
1. F or a 2 ( x )  ( x  1)( x  3) , let a 2 ( x )  0 to calculate the possible point that m a ke the function
2
2
not analytic , so x   1, -3.
2. P ( x ) 
Q ( x) 
( x  1)
( x  1)( x  3)
2
2
( x  1)( x  3)
( x  1)( x  3)
2
2


1
( x  1)( x  3)
1
x 1
3.C om bining 1 and 2.
So x  1, -3 are the singular point for this function .
N ote : x  1, -3 are also called the regular singular point
in D efinition 6.2.
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Wireless Comm. Lab.
 Theorem 6.1 Existence of Power Series Solutions
 If
x  x0
is an ordinary point of the differential equation
a 2 ( x ) y   a1 ( x ) y   a 0 ( x ) y  0, we can always find two linearly independent
solutions in the form of a power series centered at x 0 -that is,

y 
c
(x  x0 ) .
n
n
A series solution converges at least on some interval
n 0
defined by x 
singular point.
x0  R ,
 A solution of the form
whereas R is the distance from x 0
to the closest

y
c
n
( x  x0 )
n
is said to be a solution about the ordinary point
n0
x0 .
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NCCU
Wireless Comm. Lab.
 Example 2. Power Series Solutions
 Solve
y   xy  0.
 T h ere a re n o fin ite sin gu lar p oin ts, T h eo rem 6 .1 g u a ra n tees
tw o p o w er series so lu tio n s cen tered a t 0, co n verg en t fo r x   .
S u b stitu tin g y 

 cn x
n
a n d y  
n0

 c n n ( n - 1) x
n -2
in to th e
n2
d ifferen tia l eq u a tio n .
y   xy 

 c n n ( n - 1) x
n -2

n2
let k  n -3
fo r th e sec o n d term

let k  n
and
fo r th e th ird term
 2 c2 
2 c2 x 
0

 cn x
n 1
n0
 2 c2 x 
0

 c k  3 ( k  3)( k  2 ) x
k 0

  ( k  3)( k  2 ) c k  3  c k  x
k 1

 c n n ( n - 1) x
n -2
n3
k 1


 ck x


 cn x
n 1
n0
k 1
k 0
0
k 0
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Wireless Comm. Lab.

 x,
x  
s a tis fie s 2 c 2 
 (k
 3)( k  2 ) c k  3  c k
 x k 1
 0.
k 0
W e s h o u ld le t 2 c 2  0 a n d ( k  3)( k  2 ) c k  3  c k  0 , w h e r e k  0 , 1, 2 ,
So w e get c2  0 and ck 3  
S u b s titu te th e v a lu e o f
k  0 , c3  
k  1, c 4  
k  2 , c5  
k  3, c 6  
k  4, c7  
k  5, c 8  
c0
23
c1
43
c2
5 4
c3
65
c4
76
c5
8 7
ck
( k  3)( k  2 )
, k  0 , 1, 2
.
.
k in to th e r e c u r s iv e te r m ,
c0
 
6
c1
 
12
 
0
 0
20
 
c 
c0

 0  
30
6 
180

 
c 
c1

 1  
42
12 
504

 
c 

 2   0
56
20 

1
1
1
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NCCU
Wireless Comm. Lab.

 (k
S o w e r e c o n s tr u c t th e te r m
 3)( k  2 ) c k  3  c k
 x k 1
k 0

x
x
c0  1 


6
180

3
to b e th e fo r m o f
6


x
x

c
x



1


12
504


4



7

 W e s u b s titu e it in to th e o r ig in s e r ie s y 
c
n
n
x .
n0

x
x
S o , w e g e t y  c0  1 


6
180

3


x
x


  c1  x 
12
504


6
4
7

.

A n o th e r d e s c r ip tio n o f y  c 0 y 1 ( x )  c 1 y 2 ( x ),

w e g e t y1 ( x )  1 

n 1
(  1)
2 3

y2 ( x )  x 

n 1
n
( 3 n  1) 3 n
(  1)
34 
x
3n 2
n
( 3 n )( 3 n  1)
21
x
3 n 1
NCCU
Wireless Comm. Lab.
 Example 3. Power Series Solution
 Solve
( x  1) y   xy   y  0.
2
T h ere sh o w s sin g u lar p o in ts a t x   i , a n d so th e p o w er series so lu tio n
cen tered a t 0 w ill co n verg e a t lea st fo r
x  1, w h ere 1 is th e d istan ce
in th e co m p lex p la n e fro m th e o rig in to x   i .
S u b stitu in g y 

 cn x
n0
y  

n
, y 
 nc
n
x
n 1
, and
n 1


c n n ( n -1) x n -2 in to th e d ifferen tia l eq u a tio n .
n2
22
NCCU
Wireless Comm. Lab.

( x  1) y   xy   y 
2

n2

c n n ( n -1) x   c n n ( n -1) x
n
n2

 (2 c 2  c 0 ) x  (6 c 3  c1  c1 ) x 
0
1
 c n n ( n -1) x

let k  n
for another sum m ations

2 c 2  c0  6 c3 x 
1

k 2
 2 c 2  c0  6 c3 x 

nc n x 


cn x n
n0
n 1
 c n n ( n -1) x


n -2
n4

 nc
x 
n
n

 cn x
n
n2
n2


c k k ( k -1) x   c k  2 ( k  2) ( k  1) x   kc k x   c k x k
k
k
k 2

1

n

n
n2
le t k  n - 2
for the fourth term


n -2
  k ( k - 1)  k  1 c
k
k 2
k 2
 ( k  2)( k  1) c k  2  x  0
k
k
k 2
F or the equality of the equation to 0  x , x  1.
W e have 2 c 2  c 0  0, 6 c 3  0, and
 c 0  2 c 2 , c 3  0, c k  2  -
 k ( k - 1)  k  1  c k  ( k  2)( k  1) c k  2
( k  1)( k  1)
( k  2)( k  1)
ck  -
k -1
k2
23
c k , k  2.3.4.
 0, k= 2,3,4,
.
.
NCCU
Wireless Comm. Lab.
B y itera tio n o f k , w e g et
k  2, c 4  -
1
k  3, c 5  -
2
k  4, c 6  -
4
5
3
6
c2  -
11
1

c

c0
0 

42
8

c3  0
c4  -
1 1 
1
c0
  c0  
2  8  16
k  5, c 7  0

S o w e rew ritten th e p o w er series y 
c
n
x
n
to y  c 0 (1 
n0
T h u s y1 ( x )  1 
1
2

x 
2
 (  1)
n2
n 1
13 5
 ( 2 n  3)
n
x
1
2
x 
2
1
x 
4
8
)  c 1 x  c 0 y 1 ( x )  c1 y 2 ( x ).
2n
2 n!
y2 ( x)  x
, w h ere x  1 .
24
NCCU
Wireless Comm. Lab.
 Example 4. Three-Term Recurrence Relation

 If we seek a power series solution
y


c n n ( n  1) x
n2


c n n ( n  1) x
n3

Let

 c


cn x
n

k 2
n2




n 1

cn x
n


cn x
n 1
n0

k  n -2
cn x
n0
n 1

 2 c2  c0 
for the differential equation
n
n0

 ( 2 c2  c0 ) x
x

n2
0
n
n0
y   (1  x ) y  0.
y   (1  x ) y 
c

k  n
k  n 1
( k  2 ) ( k  1)  c k  c k  1 x
k
 0
k 1
F o r th e e q u a lity  x ,
W e get c2 
1
2
x  .
c0 and ck 2 
25
c k  c k 1
( k  2 ) ( k  1)
k  1 .2 .3 .
.
NCCU
Wireless Comm. Lab.
Ite r a tin g th e v a lu e k in to th e r e c u r r e n c e r e la tio n
1
w ith th e c o n d itio n c 2 
c1  c 0
k  1, c 3 
32
c1


6
k  2, c4 
k  3, c 5 
4 3
c0
6
1
c 2  c1
c0 ,
2
 2
c 0  c1
12
c3  c 2
54
c1

c1


12

120
1
24
c0
c0
30

 y 

cn x
n
n0
 c 0 (1 
1
2
x
2

1
6
x
3

1
24
x
4

1
x
5

)  c1 ( x 
30
26
1
6
x
3

1
12
x
4

1
x
5

).
120
NCCU
Wireless Comm. Lab.
 Nonpolynomial Coefficients
 The next example illustrates how to find a power series solution about the
ordinary point x  0 of a differential equation when its coefficients are
not polynomials.
0
 Example 5. ODE with Nonpolynomial Coefficients
 Solve
y   (cos x ) y  0.
W e see th a t x  0 is a n o rd in a ry p o in t o f th e eq u a tio n ,
b eca u se w e kn o w co s x is a n a lytic a t x  0 .

S u b stitu tin g y 

n0

co s x 

n0
(  1) x
n

c n x , y  
n

n ( n - 1) c n x
n2
, and
n2
2n
in to th e eq u a tio n .
2n !
27
NCCU
Wireless Comm. Lab.

y   (cos x ) y 

n ( n - 1) c n x
n2


n2

(  1) x
k
2k !
k 0
 2 c 2  6 c 3 x  12 c 4 x  20 c 5 x 
2
 (1 
3
2k
1


1
2
n
n0
1
x 
2
2
 2 c 2  c 0  (6 c 3  c1 ) x  (12 c 4  c 2 
cn x
1
x 
4
24
x 
)( c 0  c1 x  c 2 x 
6
2
720
c 0 ) x  (20 c 5  c 3 
2
1
2
c1 ) x 
 0.
1
c1  0,
3
)
F or the equality  x , x   .
W e get 2 c 2  c 0  0, 6 c 3  c1  0, 12 c 4  c 2 
So c 2  
1
2
c0 , c3 
1
6

y

n0
c n x  c 0 (1 
n
c1 , c 4 
1
2
x 
2
1
12
1
12
c0 , c5 
x 
4
1
30
1
2
c 0  0, 20 c 5  c 3 
c1 ,
)  c1 ( x 
28
2
.
.
1
6
x 
3
1
x 
5
).
30
NCCU
Wireless Comm. Lab.
 Solution Curves

y
c
x
n
 The approximate graph of a power series solution
can be
n0
obtained in several ways. We can always resort to graphing the terms in the
sequence of partial sums of the series—in other words, the graphs of the
N
n
polynomials
S N  x    cn x .
n
n0
 For a large value of N,
By this,
SN  x 
lim
N is l arg e
SN
x 
N
lim
N is l arg e
c
n
x
n
y.
n0
N
c
n
x
n
will give us some information about the
n0
behavior of y(x) near the ordinary point.
29
NCCU
Wireless Comm. Lab.
 Remarks
 Even though we can generate as many terms as desired in
series solution y   c x either through the use of a
recurrence relation or, as in Example 4, by multiplication, it
may not be possible to deduce any general term for the
coefficients c n . We may have to settle, as we did in Example 4
and 5, for just writing out the first few terms of the series.

n
n
n0
30
NCCU
Wireless Comm. Lab.
6.2 Solutions About Singular Points
 The two differential equations y   xy  0 and x y   y  0 are similar
only in that they are both examples of simple linear second-order
equations with variable coefficients.
2
 We saw in the preceding section that since x=0 is an ordinary
point of the first equation, there is no problem in finding two
linear independent power series solutions centered at that point.
 In the contrast, because x=0 is a singular point(which is defined
in Definition 6.1) of the second ODE, finding two infinite series
solutions of the equation about that point becomes a more
difficult task.
31
NCCU
Wireless Comm. Lab.
 Regular and Irregular Singular Points
A sin gu la r p oin t a t x  x 0 o f a lin e a r d iffe re n tia l e q u a tio n
a 2 ( x ) y   a 1 ( x ) y   a 0 ( x ) y  0
is fu rth e r c la ssifie d a s e ith e r re g u la r o r irre g u la r . T h e c la ssific a tio n a g a in
d e p e n d s o n th e fu n c tio n s P a n d Q in th e sta n d a rd fo rm
y   P ( x ) y   Q ( x ) y  0 .
N o te : W h e re P ( x ) 
a1 ( x )
and Q ( x ) 
a2 ( x)
a0 ( x )
.
a2 ( x)
 Definition 6.2 Regular and Irregular Singular Points
A sin g u lar p o in t x 0 is sa id to b e a reg u la r sin g u lar p o in t o f th e d ifferen tia l eq u a tio n
a 2 ( x ) y   a1 ( x ) y   a 0 ( x ) y  0, if th e fu n ctio n p ( x )  P ( x )  ( x - x 0 ) a n d q ( x )  Q ( x )  ( x - x 0 )
2
a re b o th a n a lytic a t x 0 . A sin g u lar p o in t th a t is n o t reg u la r is sa id to b e a n
irreg u la r sin g u lar p o in t o f th e eq u a tio n .
32
NCCU
Wireless Comm. Lab.
If x  x 0 is a reg u la r sin g u lar p o in t o f th e d ifferen tia l eq u a tio n y   P ( x ) y   Q ( x )  0,
th en m u ltip lyin g ( x - x 0 )
2
in b o th sid e ,
( x - x 0 ) y   ( x - x 0 ) P ( x ) y   ( x - x 0 ) Q ( x ) y  0  ( x - x 0 ) y   ( x - x 0 ) p ( x ) y   q ( x ) y  0 .
2
2
2
2
F o r reg u la r sin g u lar p o in t p ( x )  P ( x )  ( x - x 0 ) a n d q ( x )  Q ( x )  ( x - x 0 )
2
a re b o th a n a lytic a t x 0 .
S o ( x - x 0 ) y   ( x - x 0 ) p ( x ) y   q ( x ) y  0, w h ere p a n d q a re a n a lytic a t th e p o in t x  x 0 .
2
 Example 1. Classification of Singular Points
It s h o u ld b e c le a r th a t x  2 a n d x  -2 a r e s in g u la r p o in ts o f
(x
2
- 4)
2
y   3( x  2 ) y   5 y  0 .
C o m p a r in g w ith th e g e n e r a l fo r m o f
it is c le a r ly th a t P ( x ) 
3( x  2 )
(x
2
- 4)
B y th e e q u iv a le n t fo r m ( x - x 0 )
2
2

y   P ( x ) y   Q ( x ) y  0 ,
3
( x  2) ( x  2)
2
5
and Q ( x) 
(x
2
- 4)
2
.
y   ( x - x 0 ) p ( x ) y   q ( x ) y  0 ,
w h ere p ( x )  P ( x )  ( x - x0 ) a n d q ( x )  Q ( x )  ( x - x0 ) .
2
33
NCCU
Wireless Comm. Lab.
C h eckin g x  2 a n d x  -2 is th e reg u la r o r irreg u la r sin g u lar p o in t.
F irst if x  2, p ( x )  ( x - 2 ) P ( x ) 
3
( x  2)
and q ( x)  ( x - 2) Q ( x) 
5
2
2
( x  2)
2
.
F ro m p ( x ) a n d q ( x ) a re a n a lytic a t x  2, so x  2 is a reg u la r sin g u lar p o in t.
S eco n d if x  -2, p ( x )  ( x  2 ) P ( x ) 
3
( x  2 )( x  2 )
and q ( x)  ( x  2) Q ( x) 
2
5
( x  2)
2
.
A lth o u g h q ( x ) is a n a lytic a t x  -2, p ( x ) is n o t a n a lytic a t x  -2 .
C o m b in in g th ese co n d itio n s , x  -2 is a n irreg u la r sin g u lar p o i n t.
34
NCCU
Wireless Comm. Lab.
 Theorem 6.2 Frobenius’ Theorem
 If
x  x0
is a regular singular point of the differential equation
a 2 ( x ) y   a1 ( x ) y   a 0 ( x ) y  0, then there exists at least one solution of the
form


y  ( x  x0 )
r
c
( x  x0 ) 
n
n
n0
c
n
( x  x0 )
nr
,
n0
where the number r is a constant to be determined. The series will
converge at least on some interval 0  x  x 0  R .
 If we consider a differential equation that has a regular singular point,

n  r into the DE like the approach
then we can substitute y 
c (x  x )

n
0
n0
we did before by using the power series to solve with the ordinary point.
35
NCCU
Wireless Comm. Lab.
 Example 2. Two series solutions
B eca u se x  0 is a reg u la r sin g u lar p o in t o f th e d ifferen tia l eq u a tio n
3 xy   y  - y  0,

w e try to fin d a so lu tio n o f th e fo rm y 
c
n
x
nr
.
n0
F ro m y  c 0 x  c1 x
r
y   c 0 rx
r 1
1 r

 cn x
 c1 (1  r ) x 
r
nr

,
 cn (n  r ) x
n  r 1



 (n  r )c
n
x
n  r 1
n0
y   c 0 r ( r  1) x
r2
 c1 (1  r ) rx
r 1

 c n ( n  r )( n  r  1) x
nr2



c
n
( n  r )( n  r  1) x
nr2
,
n0
n o w w e su b stitu te y , y , a n d y  in to th e eq u a tio n .

3 xy   y  - y 
 3c
n
( n  r )( n  r  1) x
n0
  3 r ( r  1)  r  c 0 x
r 1
n  r 1


 (n  r )c


 3c
n
( n  r )( n  r  1) x
n  r 1
n 1
let k  n
th
fo r 4 term
x
n0
let k  n -1
nd
rd
fo r 2
and 3
term

n
n  r 1
r (3 r - 2 ) c 0 x
r 1


c
x
 (n  r )c
n
x
n  r 1
n 1


  ( k  r  1)(3 k  3 r  1) c
nr
n0


n
k 1
 ck  x


c
n
x
nr
n0
kr
0
k 0
36
NCCU
Wireless Comm. Lab.
T his im plies r (3 r - 2) c 0  0 and ( k  r  1)(3 k  3 r  1) c k  1  c k  0, k  0,1, 2,
.
If w e let c 0  0, w e w ill see the all term s in recursive term that w ill equal to zero , it ' s a trivial solutio n .
So , w e choose r (3 r - 2)  0 a nd c k  1 
If r1  0, c k  1 
r2 
2
3
ck
( k  1)(3 k  1)
, c k 1 
( k  1)(3 k  5)
, k  0,1, 2,
.
(2)
F rom (2) w e get
c1 
c2 
c0
c2 
c3 
c0
cn 
c3 
168
c0
n ! 1  4
 (3 n  2)
.
(1)
c1  c 0
8
, k  0,1, 2,
( k  r  1)(3 k  3 r  1)
, k  0,1, 2,
ck
F rom (1) w e get
ck
cn 
c0
5
c0
80
c0
2640
c0
n ! 5  8  11 
37
 (3 n  2)
.
NCCU
Wireless Comm. Lab.
F rom each set contains the sam e coefficient c 0 , so w e om it the term .
So w e find tw o independent solutions on the entired x - axis .

y1 ( x )  x 1 

0

y 2 ( x )  x 1 

2
3


n 1


n 1

x 
 (3 n  2)

1
n ! 1  4
n
1
n ! 5  8  11 
n 
x .
 (3 n  2)

B y the ratio test it can be dem onstrated that both y1 ( x ) and y 2 ( x ) converge
for all finite values of x that is , x   .
H ence , by the superposition principle , y  c1 y 1 ( x )  c 2 y 2 ( x ) is another solution
of this differential equation . O n any interval not containing the origin , this linear
com binition represents t he general solution of the differential equation .
38
NCCU
Wireless Comm. Lab.
Indicial Equation
 Equation r (3 r  2)  0 is called the indicial equation of the previous example,
and the value r  0 and r  2 are called the indicial roots, or exponents, of
3
the singularity x=0.
1
2

 In general, after substituting y   c n x
into the given
n0
differential equation and simplifying, the indicial equation is a quadratic
equation in r that results from equating the total coefficient of the lowest
power of x to zero.
nr
 We solve for the two values of r and substitute these value
ck
into a recurrence relation such as c k 1 
( k  r  1)(3 k  3 r  1)
.
By Theorem 6.2, there is at least one solution of the assumed
series form that can be found.
39
NCCU
Wireless Comm. Lab.
 Example for Indicial Equation
2
F rom y   P ( x ) y   Q ( x ) y  0, if w e m ultiply x on both side ,
w e get x y   x  xP ( x )  y    x Q ( x )  y  0.
2
2
A s the previous concept , w e know that the p ( x )  xP ( x ) an d
q ( x )  x Q ( x ) are both analytic .
2
So , w e let p ( x )  a 0  a 1 x  a 2 x 
2

Substituting y 
c
n
x
nr
and q ( x )  b0  b1 x  b 2 x 
2
.
, p ( x ), and q ( x ) into the differential equ ation ,
n0

 ( n  r )( n  r  1) c n x
n0
nr


 ( n  r )  a 0  a 1 x  a 2 x 
2
n0
 cn x

nr



n0
 b 0  b1 x  b 2 x 

2
 cn x

nr
 0.
r
A s w e find the indicial equation , w e w ill take from th e low est order of n like x ,
so w e let n  0, w e get the indicial equation r ( r - 1)  ra 0  b 0  0.
40
NCCU
Wireless Comm. Lab.
Three Cases
 Case I:
If
r1 a n d r2 a r e r is tin c t a n d d o n o t d iffe r b y a n in te g e r ,
th e n th e r e e x is t tw o lin e a r ly in d e p e n d e n t s o lu tio n s y 1 ( x )
and y2 ( x) of
e q u a tio n

th e fo r m y 

cn x
nr
a 2 ( x ) y   a 1 ( x ) y   a 0 ( x ) y  0 o f
. T h is is th e c a s e o f
E x a m p le 2 .
n0
 Case II:
If r1  r2  N , w h e re N is a p o sitiv e in te g e r , th e n th e re e x ist
tw o l in e a rly
in d e p e n d e n t so lu tio n s o f e q u a tio n a 2 ( x ) y   a 1 ( x ) y   a 0 ( x ) y  0 o f th e fo rm

y1 ( x ) 
c
n
x
n  r1
, c0  0
n0

y 2 ( x )  C y 1 ( x ) ln x 
b
n
x
n  r2
, b 0  0,
n 0
w h e re C is a c o n sta n t th a t c o u ld b e ze ro .

N o te : W h y y 2 ( x ) is e q u a l to C y 1 ( x ) ln x 
b
n
x
n  r2
a n d h o w to o b ta in it ?
n0
T h e so lu tio n y 2 ( x ) c a n b e o b ta in e d b y y 2 ( x )  y 1 ( x ) 
41
exp(   P ( x )dx )
2
y1 ( x )
dx .
NCCU
Wireless Comm. Lab.
F r o m y   P ( x ) y   Q ( x ) y  0 , w e g e t
y1  P ( x ) y 1  Q ( x ) y 1  0  y 2  y 1  P ( x ) y 1  Q ( x ) y 1   0  (1)




y 2   P ( x ) y 2   Q ( x ) y 2  0  y 1  y 2   P ( x ) y 2   Q ( x ) y 2   0  ( 2 )




B y ( 2 ) - (1),
y y   y y   0
y  )   P ( x )  y y   y y    0  " F ir s t o r d e r lin e a r d iffe r e n tia l e q u a tio n "
y1 y 2   y 2 y 1  P ( x )
( y1 y 2   y 2
1
1
2
1
2
2
1
2
1
y1 y 2   y 2 y 1  c e x p (   P ( x ) d x )
y1 y 2   y 2 y 1
2

c exp(  P ( x )dx )
2
y1
d(
y2
) 
y1
c exp(  P ( x )dx )
2
y1
y2
y1

y1

c exp(  P ( x )dx )
s o y 2  y1 
2
, le t c  1,
y1
exp(  P ( x )dx )
2
y1
. S o y o u c a n c a lc u la te y 2 ( x ) b y u s in g th e fo r m u la .
42
NCCU
Wireless Comm. Lab.
 Case III: If r  r then there always exists two linearly
independent solutions of a ( x ) y   a ( x ) y   a ( x ) y  0 of the form
1
2
2

y1 ( x ) 
c
n
x
n  r1
1
0
, c0  0
n0

y 2 ( x )  C y 1 ( x ) ln x 
b
n
x
n  r2
,
n 1
T h e so lu tio n y 2 ( x ) c a n a lso b e o b ta in e d b y y 2 ( x )  y 1 ( x ) 
exp(   P ( x )dx )
2
dx .
y1 ( x )
N o te : W e w ill se e h o w th e fo rm u la o f y 2 ( x ) w o rk s b y th e fo llo w in g e x a m p le .
43
NCCU
Wireless Comm. Lab.
 Example 5.
 Find the general solution of
Y o u c a n u s e y1 ( x ) o f
y 2 ( x )  y1 ( x ) 
y1 ( x )  x 
1
x
2
xy   y  0.
E x a m p le 4 . a n d fin d y 2 ( x ) b y s u b s titu t in g y 1 ( x ) in to
exp(  P ( x )dx )
2
dx
.
y1 ( x )
2

1
x
3
12

1
x
4

144
N o te : T h e r e is w r o n g in E x a m p le 4 . fo r y 1 ( x ).
y 2 ( x )  y1 ( x ) 
 y1 ( x ) 
exp(  0 dx )

2
3
4
x 
x 
x 
x 

2
12
144

1
1
dx
5
7

3
4
5
x - x 
x 
x 

12
12

1



2
dx
1
7
19
 1
 y1 ( x )   2 


x 
x
12
72

 x


1
7
19

2
 y1 ( x )  
 ln x 
x 
x 
x
12
144


d x




1
7
19

2
 y 2 ( x )  y 1 ( x ) ln x  y 1  

x 
x 
x
12
144


.

O n th e in te rv a l ( 0 ,  ) , th e g e n e r a l s o lu tio n is y  c1 y 1 ( x )  c 2 y 2 ( x ).
44
NCCU
Wireless Comm. Lab.
Remark
 When the difference of indicial roots r  r is a positive integer
( r  r ), it sometimes pays to iterate the recurrence relation using
the smaller root r2 first.
1
1
2
2
 Since r is the root of a quadratic equation, it could be complex.
Here we do not concern this case.
 If x=0 is an irregular singular point, we may not be able to find

any solution of the form y   c n x n  r .
n0
45
NCCU
Wireless Comm. Lab.
6.3 Two Special Equations
 The two differential equations
x y   x y   ( x   ) y  0  (1)
2
and
2
2
(1 - x ) y  - 2 x y   n ( n  1) y  0  ( 2 )
2
occur frequently in advanced studies in applied mathematics,
physics, and engineering.
 They are called Bessel’s equation and Legendre’s equation,
respectively.
 In solving (1) we shall assume   0, whereas in (2) we shall
consider only the case when n is a nonnegative integer.
46
NCCU
Wireless Comm. Lab.
Solution of Bessel’s Equation

 Substituting
y 
c
n
x
into the Bessel’s equation,
nr
n 0

 ( n  r )( n  r  1) c
n
x
nr
n0
 c0 ( r

2
    k
k 0


 (n  r )c
n
x
nr


n0
r
2
 2  r   r  ck 2  ck

T a k in g r
2
2

2
x
k r2
n
x
n0
 r  r   ) x  c1  r ( r  1)  ( r  1)  
2
c
nr2
2
x



2
c
n
x
nr
n0
r 1

 0
2
 0, ( r1   , r2  - ) a n d   k  2  r   r  c k  2  c k  0 ,


2
w e g e t r  r1   , c1  0, a n d c k  2 
 ck
( k  2 )( k  2  2 )
If c1  0, it is e a sy to sa y c k  0, k  1, 3, 5, 7 ,
L e t k  2  2 n , n  1, 2 , 3,
, so c 2 n 
47
.
.
 cn 2
2 n(n   )
2
, k  0,1, 2 ,
.
NCCU
Wireless Comm. Lab.
T hus c2 
c4 
c6 
 c0
2  1  (1   )
2
 c2
2  2  (2   )
2
 c4
2  4  (4   )
2
 c0

2  1  2  (1   )( 2   )
4
 c0

2 1  2  3  (1   )( 2   )(3   )
6
(  1) c 0
n
c2 n 
2
2n
n !(1   )( 2   )
(n   )
, n  1, 2, 3,
.
F irst w e d efin e th e g a m m a fu n ctio n  (1   )    ( ),
 (1    1)  (1   )  (1   )  (1   )     ( )  (1   )    (  1)  (  1)

 (1   ) !
S o , w e m u ltip ly  (1   ) a t th e n u m era to r a n d d en o m in ato r ,
(  1) c 0  (1   )
n
c2 n 
2
2n
n !  (1   )(1   )( 2   )
F ro m  (1   )(1   )( 2   )
(n   )
(  1) c 0  (1   )
2
2n
n ! !(1   )( 2   )
.
( n   )   !(1   )( 2   )
n
c2 n 
, n  1, 2, 3,
(  1) c 0  (1   )
( n   )   (1  n   ),
n
(n   )

2
2n
n !  (1  n   )
48
.
NCCU
Wireless Comm. Lab.
S o , y1  x
 x



n0
r1
 c0  c2 x

2
  x


(  1) c 0  (1   )
2n
n !  (1  n   )

x
2n

c2n x

(  1) c 0  (1   ) 2
n

L e t a 1  c 0  (1   ) 2 , th e n y 1  a 1 

n0
W e d e n o te d th a t J  ( x ) 

n0

n !  (1  n   )
n0

2n
n0

n
2


(  1)
(  1)
n
 x 


 2 
 x 


n !  (1  n   )  2 
n
 x 


n !  (1  n   )  2 
2 n 
2 n 
 a 1 J  ( x ).
2 n 
.
If w e ta k e r2   in to th e e q u a tio n , w e w ill g e t

y2  a2 
n0
(  1)
n
 x 


n !  (1  n   )  2 
2 n 

W e d e n o te d th a t J   ( x ) 

n0
 a 2 J   ( x ).
(  1)
n
 x 


n !  (1  n   )  2 
2 n 
T h u s y  y1 ( x )  y 2 ( x )  a1 J  ( x )  a 2 J   ( x ).
N o te : J  ( x ) a n d J - ( x ) a r e c a lle d B e s s e l fu n c tio n s o f th e fir s t k in d
o f th e o r d e r  a n d
-  , r e p e c tiv e ly .
49
NCCU
Wireless Comm. Lab.
 Example 1. General Solution:  Not an Integer
P u t o u t a tte n e io n o n th is d iffe r e n tia l e q u a tio n
x y   x y   ( x
2
2

1
) y  0 o n ( 0 ,  ), w e c a n s e e th a t  
4
.
2
T h e n th e g e n e r a l s o lu tio n is y  c1 J
1
( x )  c2 J

1
( x ).
2
2

1
F irst w e ta lk a b o u t th e p ro p erties o f B essel fu n ctio n s
o f o rd er m , m  0,1, 2,
.
( i ) J  m ( x )  (  1) J m ( x )
( ii ) J m (  x )  (  1) J m ( x )
 0, m  0
( iii ) J m (0 )  
 1, m  1
( iv ) lim Y m ( x )   
m
m
x 0
N o te : J m ( x ) a n d J  m ( x ) a re li n ea r d ep en d en t , w h ere m  0,1, 2,
50
.
NCCU
Wireless Comm. Lab.
 Bessel Functions of the Second Kinds
If   in teg er , b y lin ea r m a p p in g Y ( x ) 
co s   J  ( x ) - J - ( x )
sin  
,
J  ( x ) a n d Y ( x ) ca n b e th e lin ea rly in d ep en d en t so lu ti o n s o f
x y   xy   ( x
2
2
-  ) y  0 , so y  c1 J  ( x )  c 2 J   ( x ) o r y  c 1 J  ( x )  c 2Y ( x ).
2
If   in teg er , a cco rd in g to L ' H o sp ita l ' s ru le th a t lim Y ( x ) exists .
m
B y lin ea r m a p p in g Y m ( x )  lim Y ( x ), Y m ( x ) a n d J m ( x ) a re th e lin ea rly
m
in d ep en d en t so lu tio n s o f x y   xy   ( x
2
2
- m ) y  0 , so y  c 1 J  ( x )  c 2Y ( x ).
2
N o te : W e d ro p th e so lu tio n o f y  c 1 J  ( x )  c 2 J   ( x ),
b eca u se J  ( x ) a n d J   ( x ) a re lin ea r d ep en d en t fro m th e p revio u s p a g e .
S o , fo r a n y va lu e o f  th e g en era l so lu tio n o f
x y   xy   ( x
2
2
 ) y  0
2
o n (0 ,  ) ca n b e m o d el ed b y y  c1 J  ( x )  c 2 Y ( x ).
N o te : Y ( x ) is ca lled th e B essel fu n ctio n o f th e seco n d kin d o f o rd er  .
51
NCCU
Wireless Comm. Lab.
Following plot will show us the first and second kind of Bessel
function.
52
NCCU
Wireless Comm. Lab.
 Example 2. General Solution:  an Integer
C o n sid e rin g x y   x y   ( x - 9 ) y  0 o n ( 0,  ),
2
w e se e 
2
2
 9 a n d   3, so th e g e n e ra l so lu tio n is
y  c1 J 3 ( x )  c 2Y 3 ( x ).
 Parametric Bessel Equation
F o r th e p a ra m etic eq u a tio n x y   xy   (  x -  ) y  0,
2
2
2
2
th e g en era l so lu tio n y  c1 J  (  x )  c 2 Y (  x ) o r
y  c1 J  (  x )  c 2 J   (  x ), w h ere   in teg er.
N o te : W e w ill p ro ve it a t th e n ext p a g e .
53
NCCU
Wireless Comm. Lab.
F ro m x y   x y   (  x -  ) y  0, le t z =  x ,
2
dy
th e n
2

dx
2

dz
dz
, y = 
dx
2
dy
,
dz
2
d  dy  dz
2 d y




.


2
dz  dz  dx
dz
d y
dx
dy
2
2
S u b stitu tin g in to th e D E , w e g e t
2
x 
2
2
d y
dz
2
 x
2
z
2
d y
dz
2
 z
dy
 ( x - ) y  0
2
2
2
dz
dy
 ( z - ) y  0
2
2
dz
S o , y  c1 J  ( z )  c 2Y ( z ) o r y  c1 J  ( z )  c 2 J  ( z ).
T h e n y  c1 J  (  x )  c 2Y (  x ) o r y  c1 J  (  x )  c 2 J  (  x ).
54
NCCU
Wireless Comm. Lab.
 Example 4. Derivation Using the Series Definition
D e r iv e th e fo r m u la x J   ( x )   J  ( x )  x J   1 ( x ) .

F ro m J  ( x ) 

n0
xJ ( x ) 


n0



n0
(  1)
n
 x 


n !  (1  n   )  2 
(  1) ( 2 n   )  x 


n !  (1  n   )  2 
n
(  1)
n
 x 


n !  (1  n   )  2 

  J ( x )  x 
n 1

  J ( x )  x 
k 0
2 n 
,
2 n 
2 n 

 2
n0
(  1) n
n
(  1) n
n
 x 


n !  (1  n   )  2 
 x 


( n  1) !  (1  n   )  2 
(  1) ( k  1)
k
 x 


k !  (2  k   )  2 
55
2 n 
2 n   1
, le t k  n - 1
2 k   1
  J  ( x )  x J  1 ( x ) .
NCCU
Wireless Comm. Lab.

F ro m th e e x a m p le 4 . x J   ( x )   J  ( x )  x J   1 ( x ), w e d iv id e
x a t b o th sid e ,
J ( x ) 
x
-

x
J ( x ) 

x
J  ( x )  J   1 ( x ).
J  ( x )   J   1 ( x ), m u ltip ly x
J ( x )   x
-  1
J ( x )   x
-
J  1 ( x ) 
-
a t b o th sid e .
d
x
dx 
-
J ( x )
  x
-
J   1 ( x ).
S o , J 0  ( x )   J 1 ( x ) a n d Y 0    Y 1 ( x ).
 Spherical Bessel Functions

W h e n th e o rd e r  is h a lf a n o d d in te g e r, th a t is ,
  
1
2
,
3
2
,
5
2
,
, th e B e sse l fu n c tio n o f first k in d J  ( x )
c a n b e e x p re sse d in te rm s o f th e e le m e n ta ry fu n c tio n s sin x ,
c o s x , a n d p o w e rs o f x . S u c h B e sse l fu n c tio n s a re c a lle d
S p h e ric a l B e sse l fu n c tio n s .
56
NCCU
Wireless Comm. Lab.
 Example 5. Spherical Bessel Function with
F in d
a n a lte r n a tiv e e x p r e s s io n fo r J
1
 
1
.
2
( x ).
2
W ith 

1

,
we get J
1
2
n0
2
B y  (1   )    (  ) a n d
n  0 :
 (1 
1
1
) 
2
n  1 :
3
 (1 
3
) 
1

2
1
 (1  n 

2
2
So,
J
1
( x) 
n!
2
s in x  x -
1
2 n 1
x
3

3!
so w e get J
1
2
(x) 
2n
1
2
.
 , w e fin d
) 
3!

2
2 n 1

n!
1

3
n!
(  1)
( 2 n - 1) !
2
F rom
 x 


1
2 
n !  (1  n 
) 
2
2
n

n0
2
1
( 2 n - 1) !
) 

(
n

2
2
n  n :

(x) 
(  1)
 .
 x 


 2 
 x
1
2

x
5


5!
2
2n

n0

2
 x
(  1)


n0
(  1)
n
( 2 n  1) !
x
2 n 1
.
n
( 2 n  1) !
x
2 n 1
,
s in x .
57
NCCU
Wireless Comm. Lab.
 Now we bring out an extra concept that is not mentioned in
textbook, that is, the Modified Bessel Equation.

x y   x y   ( x
2
 u ) y  0 is c a lle d th e M o d ifie d B e sse l E q u a tio n .
2
2
It c h a n g e s fro m x y   x y   ( i x
2
2
2
 u ) y  0 . W e g e t th e g e n e ra l
2
so lu tio n y  c 1 J u ( ix )  c 2Y u ( ix ), w h e re u  0,1, 2 , 3,
y  c1 J u ( ix )  c 2 J  u ( ix ), w h e re u  0,1, 2 ,

A n d J u ( ix ) 
(  1)

n0

 i
u

n0
(  1) i
n
 ix 


n !  (1  n  u )  2 
2n
 x 
 
n !  (1  n  u )  2 

W h e re I u ( x ) 
n

n0
2nu
.


, or

n0
(  1) i
n
2n u
 x 
 
n !  (1  n  u )  2 
i
2nu
2nu
(  1) i
n
 i I u ( x ).
u
2n
 x 
 
n !  (1  n  u )  2 
2nu
is c a lle d th e
" M o d ifie d B e sse l fu n c tio n o f 1 st k in d " .
F o llo w in g th e sim ila r a p p ro a c h , y o u c a n g e t J - u ( ix )  i
58
u
I  u ( x ).
NCCU
Wireless Comm. Lab.
 F ro m th e eq u a tio n x 2 y   xy   ( x 2  u 2 ) y  0, th e so lu tio n is
y  c1 J u ( ix )  c 2 J  u ( ix ), w h ere u  m , m  0,1, 2,
.
W e m o d ify it to y  A1 I u ( x )  A 2 I  u ( x ).
y  c1 J u ( ix )  c 2 Y u ( ix ), w h ere u  m , m  0,1, 2,
.
W e m o d ify it to y  A1 I m ( x )  A2 K m ( x ),
w h ere K u ( x ) 

2

I u ( x)  Iu ( x)
sin u 
, K m (x) 
lim K
u
( x ).
um
K u ( x ) is ca lled th e " M o d ified B essel fu n ctio n o f 2 n d k in d " .
59
NCCU
Wireless Comm. Lab.
 Example.
S o lv e x
2
y   x y   (  4 x
x y   x y   [ ( 2 i ) x
2
2
2
 (
2
 6) y  0.
6) ]y  0
2
S o , th e g e n e r a l s o lu tio n
y  c1 I
6
(2 x )  c2 K
6
( 2 x ).
 Solution of Legendre’s Equation
S in ce x  0 is th e o rd in a ry p oin t o f th e eq u a tio n (1 - x ) y  - 2 xy    (  1) y  0,
2

let y 
c
n
x
n
su b stitu te in to th e eq u a tio n a n d co m b in e th e su m m a tio n ,
n0
(1 - x ) y  - 2 xy    (  1) y   (  1) c 0  2 c 2    (  1)(  2 ) c1  6 c 3  x 
2

  ( n  2 )( n  1) c
 ( n   )( n    1) c n  x  0 .
n
n2
n2
60
NCCU
Wireless Comm. Lab.
F ro m th e p re v io u s e q u a tio n ,  (  1) c 0  2 c 2  0
(  1)(  2 ) c1  6 c 3  0
( n  2 )( n  1) c n  2  ( n   )( n    1) c n  0
or c2  
c3  
 (  1) c 0
2!
(  1)(  2 ) c1
cn 2  
3!
( n   )( n    1) c n
( n  2 )( n  1)
T a k e n  2 , 3, 4 ,
c4  
c5  
c6  
c7  
.
in to c n  2 , w e g e t
(  2 )(  3) c 2
4 3
(  3)(  4 ) c 3
54
(  4 )(  5) c 4
6 5
(  5)(  6 ) c 5
7 6
, n  2 , 3, 4 ,

(  2 ) (  1)(  3) c 0
4!

(  3)(  1)(  2 )(  4 ) c1
5!
 
(  4 )(  2 ) (  1)(  3)(  5) c 0
6!
 
(  5)(  3)(  1)(  2 )(  4 )(  6 ) c1
7!
61
NCCU
Wireless Comm. Lab.
 (  1) 2 (  2 ) (  1)(  3) 4

y  c 0 1 
x 
x 
2!
4!




(  1)(  2 ) 3
(  3)(  1)(  2 )(  4 ) 5

c1  x 
x 
x 
3!
5!




 c 0U  ( x )  c1V ( x ), w h ere
U (x)  1 
V ( x )  x 
 (  1)
x 
2
(  2 ) (  1)(  3)
2!
x 
4
4!
(  1)(  2 )
x 
3
(  3)(  1)(  2 )(  4 )
3!
( i ) If   n , n  0,1, 2, 3,
x 
5
.
5!
 " N o te : H ere n m ea n s a n o n n eg a tive in teg er."
U  ( x ) a n d V ( x ) w ill d iverg e a t x   1, it vio la te th e p h ysic p h en o m en o n .
S o ,   n , n  0,1, 2, 3,
, th e o rig in a l eq u a tio n b eco m es
(1 - x ) y  - 2 xy   n ( n  1) y  0, n  0,1, 2, 3,
2
, th e so lu tio n is
y  c 0U n ( x )  c1V n ( x )
62
NCCU
Wireless Comm. Lab.
( ii )W h en n  0, 2, 4,
n  1, 3, 5,
, U n ( x ) h a s a fin ite term s , b u t V n ( x ) h a s a n in fin it e term s .
, U n ( x ) h a s a n in fin ite term s , b u t V n ( x ) h a s a fin ite term s .
S o , w e reco m b in e th e U n ( x ) a n d V n ( x ), p u t th e sa m e term to g eth er .
N o te : S ep a ra te th e in fin ite term s a n d th e fin ite term s .
U n ( x)
 U (1) , n  0, 2, 4,
 n
D efin e Pn ( x )  
 V n ( x ) , n  1, 3, 5,

 V n (1)
,
w h ere Pn ( x ) is th e " first kin d o f n - th o rd er L eg en d re ' s e q u a tio n " .
V n ( x )  U n (1), n  0, 2, 4,
Qn ( x)  
  U n ( x )V n (1), n  1, 3, 5,
,
w h ere Q n ( x ) is th e "seco n d kin d o f n - th o rd er L eg en d re ' s eq u a tio n " .
63
NCCU
Wireless Comm. Lab.
W e g e t P0 ( x )  1, P1 ( x )  x , P2 ( x ) 
Q0 ( x)  x 
1
x 
3
3
Q1 ( x )  x ( x 
1
5
1
x 
3
1
5
3x  1
1
(x 
1
ln (
2
3
2

5
2
Q2 ( x) 
x 
3
x 
) 
5
x
2
x 
3
1
5
x 
5
1
2
( 3 x  1), a n d P3 ( x ) 
2
1 x
1 x
ln (
(5 x  3 x ).
3
2
)
1 x
1 x
)
1
3
)1
3x  1
2
x 
2
4
ln (
1 x
1 x
)
3
x.
2
T h e g e n e ra l so l u tio n c a n b e e x p re sse d b y y  A Pn ( x )  B Q n ( x ).
x   1, Q n ( x ) w ill d iv e rg e ,  B  0 .
 o d d fu n c tio n , n  1, 3, 5,
S o , th e so lu tio n is y  A Pn ( x ), w h e re Pn ( x )  
 e v e n fu n c tio n , n  0, 2 , 4 ,
.
W e w ill sh o w th e p lo t o f Pn ( x ) a t n e x t p a g e .
64
NCCU
Wireless Comm. Lab.
 Example from P305. Q19
F o r (1 - x ) y  - 2 x y   2 y  0,
2
x  1.
W e g e t n  1, so y  A P1 ( x )  B Q 1 ( x ), w h e re
P1 ( x )  x a n d Q 1 ( x ) 
 Properties of
x
ln (
2
1 x
1 x
)  1.
Pn ( x )
 ( i ) Pn (  x )  (  1) n Pn ( x )
( iii ) Pn (  1)  (  1)
( ii ) Pn (1)  1
( iv ) Pn ( 0 )  0, n 0 d d
n
( v ) Pn  ( 0 )  0, n e v e n .
65
NCCU
Wireless Comm. Lab.
 Conclusion
 Here we point out two type of these special function, it will be
categorized to represent with a special solution, that we will
find out in the mathematics handbook.
66
NCCU
Wireless Comm. Lab.