중앙대학교 건설환경플랜트공학과 교수

김 진 홍

- 9주차 강의 내용 -

The power series method is the standard method for solving liner ODEs

with variable coefficients. It gives solutions in the form of power series.

0

2

020100 )()()()1(m

m

m xxaxxaaxxa

where, x ; variable ; coefficients (constants)210 ,, aaa

0x ; center (constant)

- Power series is an infinite series of the form

0

3

3

2

210)2(m

m

m xaxaxaaxa

we obtain a power series in powers of,00 xIf x

5.1 Power Series Method

Maclaurin series

0

3211

1

m

m xxxxx

!3!21

!

32

0

xxx

m

xe

m

mx

!4!21

)!2(

)1(cos

42

0

2 xx

m

xx

m

mm

!5!3)!12(

)1(sin

53

0

12 xxx

m

xx

m

mm

← m does not include negative or fractional values.

,1x Geometric series

Ideas of the Power Series Methods

For a given ODE

0)()( yxqyxpy

and are represented by power series of)(xp )(xq x (or of )0xx or polynomials.

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

0

3

3

2

210)3(m

m

m xaxaxaaxay

0

2

321

1 32)4(m

m

m xaxaaxmaya

0

2

4

1

32

2 34232)1()4(m

m

m xaxaaxammyb

We collect like powers of and equate the coefficients of each powers of

to zero, starting with the constant terms.

xx

Ex. 1) Solve by power series.xyy 2

Let

3

3

2

2

0

10 xaxaxaaxaym

m

m

(3) and (4a) are inserted into the given eq. xyy 2

)(232 2

210

2

321 xaxaaxxaxaa

5

6

4

5

3

4

2

321 65432 xaxaxaxaxaa

5

4

4

3

3

2

2

10 22222 xaxaxaxaxa

,26,25,24,23,22,0 46352413021 aaaaaaaaaaa

Hence, ,0,0 53 aa and for the coefficients with even subscripts,

!33,

!22, 04

602

402

aaa

aaaaa

Thus,2

0

8642

0 )!4!3!2

1( xeaxxx

xay

More rapidly, (3) and (4) are used for ODE xyy 2

0

1

0

1

2

0

1 221m

m

m

m

m

m

m

m

m xaxaxxmaxa

To get the same general power on both sides, we make a shift of index

and into the above eq. ;2 sm sm

0

11

0

21 2)2(s

s

s

s

s

s xaxasa

⇒ 01 a and ,2)2( 2 ss aas

For ,)4/2()4/2(,0)3/2(,)2/2(,2,1,0 0241302 aaaaaaas

ss as

a2

22

or

2

0

8642

0 )!4!3!2

1( xeaxxx

xa

6

6

5

5

4

4

3

3

2

210 xaxaxaxaxaxaay

,)!4/1()8/2(,)!3/1()6/2(,0 0680465 aaaaaaa

Ex. 2) Solve 0 yy by power series.

Let

3

3

2

2

0

10 xaxaxaaxaym

m

m

0)1(0

2

2

m

m

m

m

m

m

xaxamm

To get the same general power on both sides, we make a shift of index

and into the above eq. ;2 sm sm

00

2)1)(2(s

s

s

s

s

s xaxass

⇒ ),1,0()2)(1(

2

sss

aa s

s

!545,

!434,

!323,

!212

135

024

113

002

aaa

aaa

aaa

aaa

and so on

xaxaxx

xaxx

ay sincos)!5!3

()!4!2

1( 10

53

1

42

0

5140312010

!5!4!3!2x

ax

ax

ax

axaay

Ex. 3) Solve 02 yyxy by power series.

Let

3

3

2

2

0

10 xaxaxaaxaym

m

m

02)1(01

2

2

m

m

m

m

m

m

m

m

m

xaxmaxamm

To get the same general power on both sides, we make a shift of index

and into the above eq. ;2 sm sm

0)12()1)(2(00

2

s

s

s

s

s

s xasxass

⇒ ),1,0()2)(1(

)12(2

sa

ss

sa ss

1350241

30

2!5

51

54

5,

!4

3

43

3,

32,

21aaaaaa

aa

aa

and so on

)!8

1173

!6

73

!4

3

!2

11( 8642

0

xxxxay

9753

1!9

13951

!7

951

!5

51

!3

1( xxxxxa

,!6

73

65!4

73

65

70046 aaaa

1157!7

951

76!5

951

76

9aaaa

5.2 Theory of Power Series Method

- Basic Concepts

0

2

020100 )()()()1(m

m

m xxaxxaaxxa

where, ; variablex ; coefficients (constants)210 ,, aaa

0x ; center (constant)

n

nn xxaxxaxxaaxs )()()()()2( 0

2

02010

2

02

1

01 )()()()3( n

n

n

nn xxaxxaxR

← remainder of (1) after term ofn

n xxa )( 0

If for some the sequence converges, say,1xx ),()(lim 11 xsxsnn

then the (1) is convergent at1xx , and

If that sequence diverges at

,)()( 0

0

11

m

m

m xxaxs

1xx , the series (1) is divergent at 1xx

series

)()()( 111 xRxsxs nn

- Convergence Interval. Radius of Convergence

With respect to the convergence of the power series (1) there are three cases,

the useless Case 1, the usual Case 2, the best Case 3, as follows.

Case 1 ; The series (1) converges always at 0xx because for 0xx all its

terms are zero, except for the first one, .0a Such a series is of no practical interest.

Case 2 ; The series (1) converges and these values form an interval, called the

convergence interval. If this interval is finite, it has the midpoint 0x

(6) Rxx 0

and the series (1) converges for all such thatx Rxx 0 and diverges for all x

such that .0 Rxx R is called the radius of convergence,

(7)m

m

m

mm

m a

aRbaRa 1lim/1)(lim/1)(

Case 3 ; The series converges for all .x It corresponds to R

Ex. 3) Case 3 ; Series

0

2

!21

!m

mx x

xm

xe

1

1lim/1

)!1(

!lim/1lim/1!/1 1

mm

m

a

aRma

mmm

m

mm

The series converges for all .x

Ex. 1) Case 1 ; Series

32

0

621! xxxxm m

m

0)1(lim/1!

)!1(lim/1lim/1! 1

m

m

m

a

aRma

mmm

m

mm

This series converges only at .0x Such a series is useless.

Ex. 2) Case 2 ; Geometric series )1(11

1

0

32

xxxxxx m

m

for all1ma m and .1R

Geometric series converges and represents )1/(1 x when 1x

Ex. 4) Case 3 ; Series

022

2

)!(2

)1(

mm

mm

m

x

22 )!(2/)1( ma mm

m

The series converges for all .x

2

22

2)1(2

1

1

)1(4

1lim/1

)1(

)!(2

])!1[(2

)1(lim/1lim/1

m

m

ma

aR

mm

m

m

m

mm

m

m

Ex. 5) Find the radius of convergence of the series

,1

)3(

ma

m

m

The series converges for 3/1x

0 1

)3(

m

mm

m

x

)/2(1

)/1(13

2

13

)3(

1

2

)3( 1

1

m

m

m

mm

ma

am

m

m

m

3/1)/2(1

)/1(1lim3/1lim/1 1

m

m

a

aR

mm

m

m

Ex. 6) Find the radius of convergence of the series

5126481

8

)1( 96

0

33 xxx

xm

m

m

m

8,8

1

8

8,

8

)1(1

1

Ra

aa

m

m

m

m

m

m

m

The series converges for ,83 x that is 2x

5.3 Legendre's Equation

Important equation in the field of physics is Legendre's equation,

(1)

Legendre's equation has power series solutions of the form

(2)

0)1(2)1( 2 ynnyxyx

0m

m

mxay

Substituting (2) and its derivatives into (1), and denoting ,)1( knn

To obtain the same general power

02 1

122 02)1()1(m

m

m

m m

m

m

m

m xakxmaxxammx

2 2 1 0

2 02)1()1(m m m m

m

m

m

m

m

m

m

m xakxmaxammxamm

sx in all four series, we set sm 2

in the first series and sm in the other three series.

02)1()1)(2(00 2 1

2

s

s

s

s s s

s

s

s

s

s

s xkaxsaxassxass

Term ;

Term ;

0x

1x

0)1(12 02 anna

0)]1(2[23 13 anna

Subsequently, 0)]1(2)1([)1)(2( 2 ss annsssass

),1,0(,)1)(2(

)1)((2

sa

ss

snsna ss

← recurrence relation, recursion formula

02!2

)1(a

nna

0

24

!4

)3)(1()2(

34

)3)(2(

annnn

ann

a

13!3

)2)(1(a

nna

1

35

!5

)4)(2)(1)(3(

45

)4)(3(

annnn

ann

a

By inserting these expressions for the coefficients into (2), we obtain

(5) )()()( 2110 xyaxyaxy

where

42

1!4

)3)(1()2(

!2

)1(1)( x

nnnnx

nnxy

53

2!5

)4)(2)(1)(3(

!3

)2)(1()( x

nnnnx

nnxxy

These series converges for .1x1y and 2y are not proportional and thus

linearly independent solutions.

.11 x

(5) is a general solutions of (1) on the interval

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