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## C HAPTER 5. S ERIES S OLUTIONS OF

ODE S .

S PECIAL F UNCTIONS

2019.5 서울대학교 조선해양공학과 서유택

※ 본 강의 자료는 이규열, 장범선, 노명일 교수님께서 만드신 자료를 바탕으로 일부 편집한 것입니다.

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## Introduction

 In the previous chapters, linear ODEs with constant coefficients (

상수계수

) can be solved by algebraic methods (

대수적인 방법

), and that their

solutions are elementary functions (

초등함수

) known from calculus (

미적분

).

 For ODEs with variable coefficients (

## 변수계수

) the situation is more complicated, and their solutions may be nonelementary functions.

 In this chapter, the three main topics are Legendre polynomials, Bessel functions, and hypergeometric functions (초기하함수).

 Legendre’s ODE and Legendre polynomials are obtained by the power series method (거듭제곱급수 또는 멱급수 해법).

 Bessel’s ODE and Bessel functions are obtained by the Frobenius method, an extension of the power series method.

1 x2

y'' 2 xy'n n

1

y 0

 

2 2 2

'' ' 0

x yxyx  y

* Elementary functions (초등함수): 다항 함수, 로그 함수, 지수 함수, 삼각 함수와 이들 함수의 합성 함수들을 총칭

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Power Series (거듭제곱):

Coefficients: a0, a1 , a2, …

Center: x0

Power Series in powers of x if x0 = 0:

Ex 1 Maclaurin series

0

0 1

0

2

0

2

0

m m

m

a x x a a x x a x x

 

2

0 1 2

0 m m m

a x a a x a x

 

 

   

   

2 0

2 3

0

2 2 4

0

2 1 3 5

0

1 1 1, geometric series

1

! 1 2! 3!

cos 1 1

2 ! 2! 4!

sin 1

2 1 ! 3! 5!

m m

m x

m

m m

m

m m

m

x x x x

x

x x x

e x

m

x x x

x m

x x x

x x

m

  

  

   

   

## 5.1 Power Series Method (거듭제곱급수해법, 멱급수해법)

 The power series method is the standard method for solving linear ODEs with variable coefficients.

등비급수

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5.1 Power Series Method

3 5

*

) sin ,

3! 5!

x x ex x   x  

x sin x

! 5

! 3

5

3

x

x  x 

! 3 x

3

x 

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Idea of the Power Series Method ODE

 We represent p(x) and q(x) by power series in powers of x.

 We assume a solution in the form of a power series with unknown coefficients.

 Series obtained by termwise differentiation

 Insert the series into the ODE.

   

'' ' 0

y p x y q x y

2 3

0 1 2 3

0

m m m

y a x a a x a x a x

    

## 5.1 Power Series Method

1 2

1 2 3

1

' m m 2 3

m

y ma x a a x a x

   





2 2 3 4 2

2

3 4 2

3 2

) 1

(m a x a a x a x

m

y m m

m

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Ex.2 Solve the following ODE by power series.

Insert the series into the ODE We collect like powers of x.

General solution:

' 0

y  y

2 3

0 1 2 3

0

m m m

y a x a a x a x a x

1 1 2 3 2

1

' m m 2 3

m

y ma x a a x a x

   

2 3

2 3

0 0

0 0 0 1 0

2! 3! 2! 3!

a a x x x

y a a x x x axa e

            

 

a1 2a x2 3a x3 2

 

a0 a x1 a x2 2

0

0 0 3 0

1 2

1 0, 2 , 3 , 4 ,

2 2! 3 3! 4 4!

a a a a

a a

a a a a a

       

5.1 Power Series Method

0 )

3 ( ) 2

( )

(a1a0a2a1 xa3a2 x2 

 , 0 )

3 ( , 0 ) 2

( , 0 )

(a1a0a2a1a3a2 x

Maclaurin series for ex

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Ex.3 A Special Legendre Equation

Insert the series into the ODE.

2 3

0 1 2 3

0 m m m

y a x a a x a x a x

1 1 2 3 2

1

' m m 2 3

m

y ma x a a x a x

 

## 5.1 Power Series Method

0 2

2 )

1

(  x2 y xy y



2 2 3 4 2

2

3 4 2

3 2

) 1

(m a x a a x a x

m

y m m

m

General solution:

5 ) 1 3

1 1

( 2 4 6

0

1     

a x a x x x y

 



 

4 4 3

3 2

2 1

0

4 4 3

3 2

2 1

4 4 3

3 2

2 2

4 6 3

5 2

4 3

2

2 2

2 2

2 2

8 6

4 2

2

12 6

2

30 20

12 6

2

x a x

a x

a x

a a

y

x a x

a x

a x

a y

x

x a x

a x

a y

x

x a x

a x

a x

a a

y

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Theory of the Power Series Method

nth Partial Sum:

Remainder:

For example,

0

0 1

0

 

2 0

2

0

m m

m

a x x a a x x a x x

      

 

0 1

0

 

2 0

2

0

n

n n

s x  a a x x a x x  a x x

 

1

0

1 2

0

2

n n

n n n

R xa x xa x x

5.1 Power Series Method

 

x x2 xn 1

2 3

0 0

2 3 4

1 1

2 3 4 5

2 2

1, ,

1 , ,

1 , ,

s R x x x

s x R x x x

s x x R x x x etc

    

     

      

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Convergent (

수렴

): If this sequence converges at x=x1, say,

→ the series called convergent at x=x1, s(x1): convergent value or sum

Divergent (

발산

): If this sequence diverges at x=x1, say,

5.1 Power Series Method

) ( )

(

lim s

n

x

1

s x

1

n

0

0 1

1

) ( )

(

m

m

m

x x

a x

s

n

n x x

s

x x s

x s

s

 1

, 1

, 1

, 1

2 2

1 0

 

0 1

0

 

2 0

2

0

n

n n

s x  a a x x a x x  a x x

### For example,

 

2 0

2 0

2

0

2 1

! 1 1

1

! 1 2

!

x x x

m

x x x x

x x m

e x

m

m m

m m

m x

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In the case of convergence,

→ for every n,

 If all sn(x1) with n > N lie between s(x1) ‒ ε and s(x1) + ε → s(x1) ≈ sn(x1),

 And at x=x0, s(x0)=a0 → the series converges at x0.

 If there are other values of x for which the series converges besides x0, there values form an interval → “convergence interval (

## 수렴구간

)”

 The series converges for all x in the interval:

The series diverges for all x:

5.1 Power Series Method

0

0 1

1) ( )

(

m

m

m x x

a x

s

) ( )

( )

( x

1

s x

1

R x

1

s 

n

n

N n

x s x

s x

R

n

(

1

)  (

1

) 

n

(

1

)   for all 

 

0 1

0

 

2 0

2

0

n

n n

s x  a a x x a x x  a x x

 

1

0

1 2

0

2

n n

n n n

R x a x x a x x

R x

x 

0

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Convergence Interval (수렴구간). Radius of Convergence (수렴반지름)

Case 1. (useless) The series always converges at the center at x = x0, s(x0) = a0.

Case 2. (usual) If there are further values of x for which the series converges, these values form an interval, called the convergence interval.

## 5.1 Power Series Method

 

0 1

0

 

2 0

2

0

n

n n

s x   a a x x   a x x    a x x 

a L x a

x x x a

x x a

m m m m

m

m m

m

  

1 0

0 1 0

1

lim

) (

) lim (

1 1 1

 L L

L

: converges : diverges : inconclusive

R

Radius of convergence of the series (R): | x- x0 | < R

m m m

m m m

a a x

a x x a

x

1 0

1 0

lim 1 1

lim

   

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1

lim

1

0

m

m

m

a

x a x

Case 3. (best) The convergence interval may sometimes be infinite, that is, the series converges for all x.

5.1 Power Series Method

### 0

 0

1 x

0

x

   R

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5.1 Power Series Method

 

 

xm x x aa mm m R

e

m m m

m

x 0,

1 1

! / 1

)!

1 /(

1

! 1 2

!

1 2

0

1 ,

1 1 1 1

1

1 2 1

0

 

x x x aa R

x m

m m

m

0 ,

! 1 )!

1 2 (

1

! 2 1

0

 

m x x x aa mm m R m

m m

m

1

1 1 / ( 1)! 1

lim lim lim 0, for all x, the series converges

1 / ! 1

m m

m m m m

m

a x m

x x R

a x m m

  

      

1

1 1

lim lim only for 1, the series converges

1

m m

m m m

m

a x

x x x R 1

a x

     

1

1 ( 1)!

lim lim lim ( 1) , for any x, the series diverge 0

!

m m

m m m m

m

a x m

x x m R

a x m

  

       

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## 5.1 Power Series Method

Example

m m

m m

k x

2 0

1

 ) (

m m

m

x2

0 3

2



 

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Theorem 1 Existence of Power Series Solutions

Ifp, q, and r in yʺ(x)+p(x)yʹ+q(x)y = r(x) are analytic at x = x0, (power series representations)

then every solution is analytic and can thus be represented by a power series in powers of x- x0 with radius of convergence R > 0.

5.1 Power Series Method

* 함수f가 한 점 x0에서 해석적이라는 것은 그 점 근방에서의 테일러 급수가 수렴하는 것과 같은 의미

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Legendre’s equation:

Substituting and its derivatives

The first expression as two separate series.

Set m‒2 = s in the first series

and simply write s instead of m in the other three series.

1x2

y'' 2 xy'n n

1

y 0

0 m m m

y a x

2

  

2 1

 

2 1 0

1 1 m m 2 m m 1 m m 0

m m m

x m m - a x x ma x n n a x

## 5.2 Legendre’s Equation. Legendre Polynomials P

n

(x)

0 )

1 ( 2

) 1 (

) 1 (

0 1

2 2

2      

  

m

m m m

m m m

m m m

m

mx m m a x ma x n n a x

a m

m

0 )

1 ( 2

) 1 ( )

1 )(

2 (

0 1

2 0

2      

  

s

s s s

s s s

s s s

s

s x s s a x sa x n n a x

a s

s

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 Recurrence relation (

점화관계

):

  

    

2

1 0 1

2 1

s s

n s n s

a a s , ,

s s

  

  

 

5.2 Legendre’s Equation. Legendre Polynomials P

n

(x)

 The sum of the coefficients of each power of x on the left must be zero.

0 )

1 ( 2

) 1 ( )

1 )(

2 (

0 1

2 0

2      

  

s

s s s

s s s

s s s

s

s x s s a x sa x n n a x

a s

s

0 2

0 0 0

2 0

! 2

) 1 0 (

) 1 ( 1

2

: n n a

a x

a n

n x a

x

1 3

1 3

1

! 3

) 2 )(

1 0 (

)]

1 ( 2 [ 2

3

: n n a

a a

n n a

x  

2 2

( )( 1)

: ( 2)( 1) [ ( 1) 2 ( 1)] 0

( 2)( 1)

s

s s s s

n s n s

x s s a s s s n n a a a

s s

  

           

 

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 

  

    

  

  

    

2 0 3 1

4 2 5 3

0 1

1 1 2

2! 3!

2 3 3 4

4 3 5 4

2 1 3 3 1 2 4

4! 5!

n n n n

a a a a

n n n n

a a a a

n n n n n n n n

a a

  

   

   

   

 

      

 

        

         

2 4

1

3 5

2

1 2 1 3

1 2! 4!

1 2 3 1 2 4

3! 5!

n n n n n n

y x x x

n n n n n n

y x x x x

   

   

 

0 1

 

1 2

 

y x a y x a y x

5.2 Legendre’s Equation. Legendre Polynomials P

n

(x)

 We express successively except for a0 and a1.

General solution:

0

m m m

y a x

converge for |x|<1

 y1 and y2 are independent

 x=±1 → no longer analytic

1x2

y'' 2 xy'n n

1

y 0





2

1 0 1

2 1

s s

n s n s

a a s , ,

s s

 

 

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Legendre Polynomials

The reduction of power series ( , m→ ∞) to polynomials (m is finite) is a great advantage

because then we have solutions for all x, without convergence restrictions.

## 5.2 Legendre’s Equation. Legendre Polynomials

0 m m m

y a x

        

         

2 4

1

3 5

2

1 2 1 3

1 2! 4!

1 2 3 1 2 4

3! 5!

n n n n n n

y x x x

n n n n n n

y x x x x

   

   

  

    

2

1 0 1

2 1

s s

n s n s

a a s , ,

s s

 

 

2 0, 4 0, 6 0,...

n n n

a a a

   

2 2 4 4 2 2 2 2

1 1 a x a x an xn anxn an xn

y

   

1 1 3 3 1 1 1 1 3 3

2

n n n

n n

n x a x a x

a x

a x a y If s=n

If n is odd If n is even

Polynomials (finite terms)

Polynomials (finite terms)

### “Legendre polynomials”

1x2

y'' 2 xy'n n

1

y 0
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Legendre Polynomials

The reduction of power series ( , m → ∞) to polynomials (m is finite) is a great advantage

because then we have solutions for all x, without convergence restrictions.

## 5.2 Legendre’s Equation. Legendre Polynomials

0 m m m

y a x

        

         

2 4

1

3 5

2

1 2 1 3

1 2! 4!

1 2 3 1 2 4

3! 5!

n n n n n n

y x x x

n n n n n n

y x x x x

   

   

  

    

2

1 0 1

2 1

s s

n s n s

a a s , ,

s s

 

 

2 0, 4 0, 6 0,...

n n n

a a a

   

2 2 4 4 2 2 2 2

1 1 a x a x an xn anxn an xn

y

   

1 1 3 3 1 1 1 1 3 3

2

n n n

n n

n x a x a x

a x

a x a y If s=n

If n is odd If n is even

Polynomials (finite terms)

Polynomials (finite terms)

### “Legendre polynomials”

1x2

y'' 2 xy'n n

1

y 0
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Legendre Polynomials

We choose the coefficient an of the highest power xn as

We solved for as in terms of as+2

  

  

2

 

2 1

2

s 1 s

s s

a a s n

n s n s

   

 

## 5.2 Legendre’s Equation. Legendre Polynomials

  

    

2

1 0 1

2 1

s s

n s n s

a a s , ,

s s

 

 

n pn(1)1 for every

)!

2 )(

1 ( )!

1 )(

( 2

)!

2 2 )(

1 2 )(

2 ( )

1 2 ( 2

) 1 ( )

! ( 2

)!

2 ( ) 1 2 ( 2

) 1 ( )

1 2 ( 2

) 1 (

2 2   

 

 

 

 

n n

n n

n

n n

n n

n n n

n n

n a n

n n

an n n n n

2

(2 2)!

2 ( 1)!( 2)!

n n

a n

n n

  

 

The choice of an makes

 

 

2

 

1 a positive integer 2 !

1 3 5 2 1

2 ! 0

!

n n

n n

a n

n n

n

  



for n = 0

for n ≠ 0, n is a positive integer

2 s  n

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Legendre Polynomials

     

   

2

0

2 2 ! 1

1 or

2 ! ! 2 ! 2 2

M m n m

n n

m

n m n n

P x x M

m n m n m

   

   

2

2 2 !

2 0, 1

2 ! ! 2 !

m

n m n

n m

n m a

m n m n m

    

 

5.2 Legendre’s Equation. Legendre Polynomials

)!

2 ( )!

1 ( 2

)!

2 2 (

2  

 

n n

an n n

)!

4 ( )!

2 (

! 2 2

)!

4 2

( )

3 2 ( 4

) 3 )(

2 (

2

4  

 

 

 

n n

a n n

n

an n n n

### When n-2m ≥ 0

  

  

2

 

2 1

2

s 1 s

s s

a a s n

n s n s

   

 

whichever is an integer

4 s  n

2 2

(2 )! (2 2)!

2 ( !) 2 1!( 1)!( 2)!

n n

n n

n n

x x

n n n

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Examples of the Legendre Polynomials

## 5.2 Legendre’s Equation. Legendre Polynomials

     

   

2

0

2 2 ! 1

1 or

2 ! ! 2 ! 2 2

M m n m

n n

m

n m n n

P x x M

m n m n m

   

     

1x2

y'' 2 xy'n n

1

y 0

 

0

0 0 2

0! 1

2 (0!)

P xx1

 

1 2 1

2!

2 (1!)

P xxx

2 2

(2 )! (2 2)!

2 ( !) 2 1!( 1)!( 2)!

n n

n n

n n

x x

n n n

  

 

Q: Derive P

2

(x) and prove it satisfies the ODE.

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Indicial Equation (결정방정식), Indicating the Form of Solutions

 We expand b(x) and c(x) in power series

 We differentiate term by term

 The equation corresponding to the power xr is .

 By assumption a0 ≠ 0, Indicial Equation:

   

2 '' ' 0

x y xb x y c x y

   

b x  b0 b x b x1 2 2 , c x  c0 c x c x1 2 2

 

1 0

0 1

 

0

 

0 1

 

0 1

0

r r r

x r r  a   bb xx ra   cc xx aa x 

 1 

0 0

### 0

r r   b r  c 

   

'' b x ' c x2 0

y y y

x x

  

0 1 2 2

0 0

r m m r r

m m

m m

y x x a x a x x a a x a x

1

0 0 0 0 r r  b rc a

 

 

5.3 Extended Power Series Method: Frobenius Method

Multiply x2

] )

1 ( [

) (

)

( 1

0

0 1

1    

 

x ra r a x

x a r m x

y

m

r r

m m

] )

1 ( )

1 ( [ )

1 )(

( )

( 1

0

0 2

2     

 

x r r a r ra x

x a r

m r m x

y

m

r r

m m

doesn’t change!

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Theorem 2 Frobenius Method. Basis of Solutions. Three Cases

Suppose that the ODE satisfies the assumptions in Theorem 1. Let r1 and r2 be the roots of the indicial equation. Then we have the following three cases.

Case 1. Distinct Roots Not Differing by an Integer. A basis is

Case 2. Double Root (r1 = r2 =r). A basis is

Case 3. Roots Differing by an Integer. A basis is

where the roots are so denoted that r1 - r2 > 0.

 

1

2

  

2

2

1 r 0 1 2 and 2 r 0 1 2

y xx aa x a x  y xx AA xA x

  

2

     

2

1 r 0 1 2 and 2 1 ln r 1 2

y xx aa x a x  y xy x x xA xA x

 

1

2

    

2

2

1 r 0 1 2 and 2 1 ln r 0 1 2

y xx aa x a x  y xky x x xAA xA x

## 5.3 Extended Power Series Method: Frobenius Method

(두 근의 차가 정수가 아닌 서로 다른 근들)

(두 근의 차가 정수인 서로 다른 근들)

(이중근)

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Ex.2 Illustration of Case 2 (Double Root)

Solve the ODE .

Inserting and its derivatives into the ODE.

First Solution. We insert r = 0.

1

''

3 1

' 0

x xyxy  y

 

1

: 1 0 0 0

xr r r  r a r

  

0 1 2 2

0

r m r

m m

y x x a x x a a x a x

     

   

1

0 0

1

0 0 0

1 1

3 0

m r m r

m m

m m

m r m r m r

m m m

m m m

m r m r a x m r m r a x

m r a x m r a x a x

 

 

   

참조

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