Series Solutions of Linear Differential Equations

CHAPTER 5

Ch5_2

Contents

5.1 Solutions about Ordinary Points5.2 Solution about Singular Points5.3 Special Functions

Ch5_3

5.1 Solutions about Ordinary Point

Review of Power SeriesRecall from that a power series in x – a has the form

Such a series is said to be a power series centered at a.

2210

0

)()()( axcaxccaxcn

nn

Ch5_4

Convergence exists.

Interval of ConvergenceThe set of all real numbers for which the series converges.

Radius of ConvergenceIf R is the radius of convergence, the power series converges for |x – a| < R and diverges for |x – a| > R.

N

nn

nNNN axCxS0

)(lim)(lim

Ch5_5

Absolute ConvergenceWithin its interval of convergence, a power series converges absolutely. That is, the following converges.

Ratio Test Suppose cn 0 for all n, and

If L < 1, this series converges absolutely, if L > 1, this series diverges, if L = 1, the test is inclusive.

0

|)(|n

nn axc

LC

Cax

axC

axC

n

n

nnn

nn

n

11

1 lim||)(

)(lim

Ch5_6

A Power Defines a FunctionSuppose then

Identity PropertyIf all cn = 0, then the series = 0.

(1) )1(",'0

20

1

n

nn

n xnnyxny

0n

nnxcy

Ch5_7

Analytic at a PointA function f is analytic at a point a, if it can be represented by a power series in x – a with a positive radius of convergence. For example:

(2)

!6!4!21cos

!5!3sin ,

!2!11

642

532

xxxx

xxxx

xxex

Ch5_8

Arithmetic of Power SeriesPower series can be combined through the operations of addition, multiplication and division.

303

241

121

1201

61

61

21

61

)1()1(

5040120624621

sin

532

5432

753432

xxxx

xxxxx

xxxx

xxxx

xex

Ch5_9

Example 1

Write as one power series.

SolutionSince

we let k = n – 2 for the first series and k = n + 1 for the second series,

01

22)1( n

nnn

nn xcxcnn

2 0 3 0

1202

12 )1(12)1(n n n n

nn

nn

nn

nn xcxcnnxcxcxcnn ．

Ch5_10

then we can get the right-hand side as

(3)We now obtain

(4)

1 1122 )1)(2(2

k k

kk

kk xcxckkc

1122

2 0

12

])1)(2[(2

)1(

k

kkk

n n

nn

nn

xcckkc

xcxcnn

Example 1 (2)

Ch5_11

Suppose the linear DE

(5)is put into

(6)

A Solution

0)()()( 012 yxayxayxa

0)()( yxQyxPy

A point x0 is said to be an ordinary point of (5) if both

P and Q in (6) are analytic at x0. A point that is not an ordinary point is said to be a singular point.

DEFINITION 5.1

Ch5_12

Since P and Q in (6) is a rational function, P = a1(x)/a2(x), Q = a0(x)/a2(x)

It follows that x = x0 is an ordinary point of (5) if a2(x0) 0.

Polynomial Coefficients

Ch5_13

A series solution converges at least of some interval defined by |x – x0| < R, where R is the distance from x0 to the nearest singular point.

If x = x0 is an ordinary point of (5), we can always findtwo linearly independent solutions in the form of powerseries centered at x0, that is,

THEOREM 5.1Criterion for an Extra Differential

0 0)(

nn

n xxcy

Ch5_14

Example 2

Solve

SolutionWe know there are no finite singular points. Now, and

then the DE gives

(7)

0" xyy

0n

nnxcy

22)1("

nn

nxcnny

0

1

2

2

2 0

2

)1(

)1(

n

nn

n

nn

n n

nn

nn

xcxnnc

xcxxnncxyy

Ch5_15

Example 2 (2)

From the result given in (4),

(8)

Since (8) is identically zero, it is necessary all the coefficients are zero, 2c2 = 0, and

(9)Now (9) is a recurrence relation, since (k + 1)(k + 2) 0, then from (9)

(10)

1122 0])2)(1[(2

k

kkk xcckkcxyy

,3,2,1,0)2)(1( 12 kcckk kk

,3,2,1,)2)(1(

12

kkk

cc k

k

Ch5_16

Example 2 (3)

Thus we obtain

,1k32

03 ．

cc

,2k43

14 ．

cc

,3k 054

25

．c

c

,4k 03

6 65321

65c

cc

．．．．

,5k 14

7 76431

76c

cc

．．．．

Ch5_17

and so on.

Example 2 (4)

,6k 087

58

．c

c

,7k 06

9 9865321

98c

cc

．．．．．．

,8k 17

10 10976431

109c

cc

．．．．．．

,9k 01110

811

．c

c

Ch5_18

Example 2 (5)

Then the power series solutions are y = c0y1 + c1y2

....07．6．4．3

6．5．3．20

4．33．20

71

60413010

xc

xc

xc

xc

xccy

Ch5_19

1

13

10742

)13)(3(43)1(

10976431

76431

431

1)(

k

kk

xkk

x

xxxxy

．

．．．．．．．．．

Example 2 (6)

1

3

9631

)3)(13(32)1(

1

9865321

65321

321

1)(

k

kk

xkk

xxxxy

．

．．．．．．．．．

Ch5_20

Example 3

Solve

SolutionSince x2 + 1 = 0, then x = i, −i are singular points. A power series solution centered at 0 will converge at least for |x| < 1. Using the power series form of y, y’ and y”, then

0'")1( 2 yxyyx

012

2

2

2 01

122

)1()1(

)1()1(

n

nn

n

nn

n

nn

n

nn

n n

nn

n

nn

nn

xcxncxcnnxcnn

xcxncxxcnnx

Ch5_21

nk

n

nn

nk

n

nn

nk

n

nn

nk

n

nn

xcxncxcnn

xcnnxcxcxcxcxc

22

2

4

2

2113

00

02

)1(

)1(62

Example 3 (2)

22302

22302

0])1)(2()1)(1[(62

])1)(2()1([62

k

kkk

k

kkkkk

xckkckkxccc

xckcckkckkxccc

Ch5_22

Example 3 (3)

From the above, we get 2c2 － c0 = 0, 6c3 = 0 , and

Thus c2 = c0/2, ck+2 = (1 – k)ck/(k + 2)Then

0)1)(2()1)(1( 2 kk ckkckk

02024 !22

142

141

cccc ．

352

35 cc

03046 !32

31642

363

cccc．

．．

074

57 cc

Ch5_23

Example 3 (4)

and so on.

04068 !42

5318642

5385

cccc．．

．．．．

096

79 cc

050810 !52

7531108642

753107

cccc．

．．．．．．．

．．

Ch5_24

Example 3 (5)

Therefore,

)()(

!52

7531

!42

531

!32

31

!22

121

1

2110

110

58

46

34

22

0

1010

99

88

77

66

55

44

33

2210

xycxyc

xcxxxxxc

xcxcxcxcxc

xcxcxcxcxccy

．．．．．．

1||,!2

)32(531)1(

21

1)( 2

2

121

xxn

nxxy n

nn

n ．．

xxy )(2

Ch5_25

Example 4

If we seek a power series solution y(x) for

we obtain c2 = c0/2 and the recurrence relation is

Examination of the formula shows c3, c4, c5, … are expresses in terms of both c1 and c2. However it is more complicated. To simplify it, we can first choose c0 0, c1 = 0. Then we have

,3,2,1,)2)(1(

12

k

kk

ccc kk

k

0)1( yxy

02 21

cc

Ch5_26

Example 4 (2)

and so on. Next, we choose c0 = 0, c1 0, then

0012

4 241

43243c

cccc

．．．

0023

5 301

21

61

5454c

cccc

．．

0001

3 61

3232c

cccc

．．

021

02 cc

Ch5_27

Example 4 (3)

and so on. Thus we have y = c0y1 + c1y2, where

1101

3 61

3232c

cccc

．．

1112

4 121

4343c

cccc

．．

1123

5 1201

65454c

cccc

．．．

54321 30

1241

61

21

1)( xxxxxy

5432 120

1121

61

)( xxxxxy

Ch5_28

Example 5

Solve

SolutionWe see x = 0 is an ordinary point of the equation. Using the Maclaurin series for cos x, and using , we find

0nn

nxcy

0)(cos" yxy

2 0

6422

!6!4!21)1(

)(cos

n n

nn

nn xc

xxxxcnn

yxy

0

21

2021

12)6(2 3135

20241302

xcccxcccxcccc

Ch5_29

Example 5 (2)

It follows that

and so on. This gives c2 = － 1/2c0, c3 = － 1/6c1, c4 = 1/12c0, c5 = 1/30c1,…. By grouping terms we get the general solution y = c0y1 + c1y2, where the convergence is |x| < , and

021

20,021

12,06,02 1350241302 cccccccccc

421 12

121

1)( xxxy

532 30

161

1)( xxxy

Ch5_30

5.2 Solutions about Singular Points

A DefinitionA singular point x0 of a linear DE

(1)is further classified as either regular or irregular. This classification depends on

(2)

0)()()( 012 yxayxayxa

0)()( yxQyxPy

Ch5_31

A singular point x0 is said to be a regular singular point of (1), if p(x) = (x – x0)P(x), q(x) = (x – x0)2Q(x) are both analytic at x0 . A singular point that is not regular is said to be irregular singular point.

DEFINITION 5.2Regular/Irregular Singular Points

Ch5_32

Polynomial Ciefficients

If x – x0 appears at most to the first power in the denominator of P(x) and at most to the second power in the denominator of Q(x), then x – x0 is a regular singular point.

If (2) is multiplied by (x – x0)2,

(3)

where p, q are analytic at x = x0

0)()()()( 02

0 yxqyxpxxyxx

Ch5_33

Example 1

It should be clear x = 2, x = – 2 are singular points of(x2 – 4)2y” + 3(x – 2)y’ + 5y = 0

According to (2), we have

2)2)(2(

3)(

xxxP

22 )2()2(

5)(

xxxQ

Ch5_34

Example 1 (2)

For x = 2, the power of (x – 2) in the denominator of P is 1, and the power of (x – 2) in the denominator of Q is 2. Thus x = 2 is a regular singular point.

For x = −2, the power of (x + 2) in the denominator of P and Q are both 2.Thus x = − 2 is a irregular singular point.

Ch5_35

If x = x0 is a regular singular point of (1), then there

exists one solution of the form

(4)

where the number r is a constant to be determined.

The series will converge at least on some interval

0 < x – x0 < R.

THEOREM 5.2Frobenius’ Theorem

00

000 )()()(

n

rnn

n

nn

r xxCxxCxxy

Ch5_36

Example 2: Frobenius’ Method

Because x = 0 is a regular singular point of

(5)we try to find a solution .Now,

03 yyyx

0n

rnnxcy

0

1)(n

rnnxcrny

0

2)1)((n

rnnxcrnrny

Ch5_37

Example 2 (2)

00

1

00

1

0

1

)233)((

)()1)((3

3

n

rnn

n

rnn

n

rnn

n

rnn

n

rnn

xcxcrnrn

xcxcrnxcrnrn

yyyx

0])133)(1[()23(

)233)(()23(

01

10

1

1

1

110

k

kkk

r

nk

n

nn

nk

n

nn

r

xccrkrkxcrrx

xcxcrnrnxcrrx

Ch5_38

Example 2 (3)

which implies r(3r – 2)c0 = 0 (k + r + 1)(3k + 3r + 1)ck+1 – ck = 0, k = 0, 1, 2, …Since nothing is gained by taking c0 = 0, then

r(3r – 2) = 0 (6)and

(7)

From (6), r = 0, 2/3, when substituted into (7),

,2,1,0,)133)(1(1

k

rkrkc

c kk

Ch5_39

Example 2 (4)

r1 = 2/3, k = 0,1,2,… (8)

r2 = 0, k = 0,1,2,… (9)

,)1)(53(1

kkc

c kk

,)13)(1(1

kkc

c kk

Ch5_40

Example 2 (5)

From (8) From(9)

Ch5_41

Example 2 (6)

These two series both contain the same multiple c0. Omitting this term, we have

(10)

(11)

1

3/21 )23(1185!

11)(

n

nxnn

xxy．．

1

02 )23(741!

11)(

n

nxnn

xxy．．

Ch5_42

Example 2 (7)

By the ratio test, both (10) and (11) converges for all finite value of x, that is, |x| < . Also, from the forms of (10) and (11), they are linearly independent. Thus the solution is

y(x) = C1y1(x) + C2y2(x), 0 < x <

Ch5_43

Indicial Equation

Equation (6) is called the indicial equation, where the values of r are called the indicial roots, or exponents.

If x = 0 is a regular singular point of (1), then p = xP and q = x2Q are analytic at x = 0.

Ch5_44

Thus the power series expansionsp(x) = xP(x) = a0 ＋ a1x ＋ a2x2 …＋q(x) = x2Q(x) = b0 ＋ b1x ＋ b2x2 …＋ (12)

are valid on intervals that have a positive radius of convergence. By multiplying (2) by x2, we have

(13)

After some substitutions, we find the indicial equation, r(r – 1) + a0r + b0 = 0 (14)

0)]([)]([ 22 yxQxyxxPxyx

Ch5_45

Example 3

Solve

SolutionLet , then

0nrn

nxcy

00

1

00

0

1

0

1

)1()122)((

)(

)()1)((2

)1(2

n

rnn

n

rnn

n

rnn

n

rnn

n

rnn

n

rnn

xcrnxcrnrn

xcxcrn

xcrnxcrnrn

yyxyx

0')1("2 yyxxy

Ch5_46

Example 3 (2)

which implies r(2r – 1) = 0 (15)

(16)

01

10

0

1

1

110

])1()122)(1[()12(

)1()122)(()12(

k

kkk

r

nk

n

nn

nk

n

nn

r

xcrkcrkrkxcrrx

xcrnxcrnrnxcrrx

,2,1,0,0)1()122)(1( 1 kcrkcrkrk kk

Ch5_47

Example 3 (3)

From (15), we have r1 = ½ , r2 = 0.Foe r1 = ½ , we divide by k + 3/2 in (16) to obtain

(17)

Foe r2 = 0 , (16) becomes

(18)

,2,1,0,)1(21

kk

cc k

k

,2,1,0,121

kk

cc k

k

Ch5_48

Example 3 (4)

From (17) From (18)

1

1．20

10

1c

cc

c

Ch5_49

Example 3 (5)

Thus for r1 = ½

for r2 = 0

and on (0, ), the solution is y(x) = C1y1 + C2y2.

0

2/1

1

2/11 !2

)1(

!2

)1(1)(

n

nn

n

n

nn

n

xn

xn

xxy

||,

)12(7531)1(

1)(1

2 xxn

xyn

nn

．．．

Ch5_50

Example 4

Solve

SolutionFrom xP = 0, x2Q = x, and the fact 0 and x are their own power series centered at 0, we conclude a0 = 0, b0 = 0. Then form (14) we have r(r – 1) = 0, r1 = 1, r2 = 0. In other words, there is only a single series solution

0" yxy

...122)!1(!

)1()(

321

01

xx

xxnn

xy n

n

n

Ch5_51

Three Cases

(1) If r1, r2 are distinct and do not differ by an integer, there exists two linearly independent solutions of the form:

0

20

121 )( and )(

n

rnn

n

rnn xbxyxcxy

Ch5_52

(2) If r1 – r2 = N, where N is a positive integer, there

exists two linearly independent solutions of the form:

(20) 0 ,ln)()(

(19) 0 ,)(

00

12

00

1

2

1

bxbxxCyxy

cxcxy

n

rnn

n

rnn

Ch5_53

(3) If r1 = r2, there exists two linearly independent solutions of the form:

(22) ln)()(

(21) 0 ,)(

012

00

1

2

1

n

rnn

n

rnn

xbxxyxy

cxcxy

Ch5_54

Finding a Second Solution

If we already have a known solution y1, then the second solution can be obtained by

(23) )( 212

1

dxy

eyxy

Pdx

Ch5_55

Example 5

Find the general solution of

SolutionFrom the known solution in Example 4,

we can use (23) to find y2(x). Here please use a CAS for the complicated operations.

0" yxy

4321 144

1121

21

)( xxxxxy

Ch5_56

Example 5 (2)

21

21

54321

2432

121

0

12

14419

127

ln1

)(

7219

12711

)(

127

125

)(

1441

121

21

)()]([

)()(

xxxx

xy

dxxxx

xy

xxxx

dxxy

xxxx

dxxydx

xy

exyxy

dx

2

112 14419

1271

)(ln)()( xxx

xyxxyxy

Ch5_57

5.3 Special Functions

Bessel’s Equation of order v (1)where v 0, and x = 0 is a regular singular point of (1). The solutions of (1) are called Bessel functions.

Lengender’s Equation of order n

(2)where n is a nonnegative integer, and x = 0 is an ordinary point of (2). The solutions of (2) are called Legender functions.

0)( 222 yvxyxyx

0)1(2)1( 2 ynnyxyx

Ch5_58

The Solution of Bessel’s Equation

Because x = 0 is a regular singular point, we know there exists at least one solution of the form . Then from (1),

(3)

0n

rnnxcy

0

2

1

22220

0

22

1

220

00

22

00

222

])[()(

])()1)([()(

)()1)((

)(

n

nn

r

n

nn

rr

n

nn

rn

nn

rr

n

rnn

n

rnn

n

rnn

n

rnn

xcxxvrncxxvrc

xcxxvrnrnrncxxvrrrc

xcvxcxrncxrnrnc

yvxyxyx

Ch5_59

From (3) we have the indicial equation r2 – v2 = 0, r1 = v, r2 = −v. When r1 = v, we have

(1 + 2v)c1 = 0(k + 2)(k + 2+ 2v)ck+2 + ck = 0

or (4)

The choice of c1 = 0 implies c3 = c5 = c7 = … = 0, so for k = 0, 2, 4, …., letting k + 2 = 2n, n = 1, 2, 3, …, we have

(5)

,2,1,0,)22)(2(2

k

vkkc

c kk

)(2222

2 vnn

cc n

n

Ch5_60

Thus

(6)

,3,2,1,)()2)(1(!2

)1(

)3)(2)(1(3212)3(32

)2)(1(212)2(22

)1(12

20

2

60

24

6

40

22

4

20

2

nvnvvn

cc

vvv

c

v

cc

vv

c

v

cc

v

cc

n

n

n

．．．．

．．．

．．

Ch5_61

We choose c0 to be a specific value

where (1 + v) is the gamma function. See Appendix II. There is an important relation:

(1 + ) = ()so we can reduce the denominator of (6):

)1(2

10 v

c v

)1()1)(2()2()2()21(

)1()1()11(

vvvvvv

vvv

Ch5_62

Hence we can write (6) as

,...2,1,0,)1(!2

)1(22

n

nvnc vn

n

n

Ch5_63

Bessel’s Functions of the First Kind

We define Jv(x) by

(7)

and

(8)

In other words, the general solution of (1) on (0, ) is y = c1Jv(x) + c2J-v(x), v integer (9)

See Fig 5.3

0

2

2)1(!)1(

)(n

vnn

vx

nvnxJ

0

2

2)1(!)1(

)(n

vnn

vx

nvnxJ

Ch5_64

Fig 5.3

Ch5_65

Example 1

Consider the DE

We find v = ½, and the general solution on (0, ) is

0)1/4('" 22 yxxyyx

)()( 1/221/21 xJcxJcy

Ch5_66

Bessel’s Functions of the Second Kind

If v integer, then

(10)

and the function Jv(x) are linearly independent. Another solution of (1) is y = c1Jv(x) + c2Yv(x).

As v m, m an integer, (10) has the form 0/0. From L’Hopital’s rule, the function

and Jv(x) are linearly independent solutions of

vxJxJv

xY vvv sin

)()(cos)(

)(lim)( xYxY vmv

m

0)('" 222 ymxxyyx

Ch5_67

Hence for any value of v, the general solution of (1) is

(11)

Yv(x) is called the Bessel function of the second kind of order v. Fig 5.4 shows y0(x) and y1(x).

)()( 21 xYcxJcy vv

Ch5_68

Fig 5.4

Ch5_69

Example 2

Consider the DE

We find v = 3, and from (11) the general solution on (0, ) is

0)9('" 22 yxxyyx

)()( 3231 xYcxJcy

Ch5_70

DEs Solvable in Terms of Bessel Function

Let t = x, > 0, in

(12)then by the Chain Rule,

0)( 2222 yvxyxyx

dtdy

dxdt

dtdy

dxdy

2

22

2

2

dt

yddxdt

dxdy

dtd

dx

yd

Ch5_71

Thus, (12) becomes

The solution of the above DE is y = c1Jv(t) + c2Yv(t)Let t = x, we have

y = c1Jv(x) + c2Yv(x)(13)

0

0

222

22

222

22

2

yvtdtdy

tdt

ydt

yvtdtdyt

dt

ydt

Ch5_72

Another equation is called the modified Bessel equation order v,

(14)This time we let t = ix, then (14) becomes

The solution will be Jv(ix) and Yv(ix). A real-valued solution, called the modified Bessel function of the first kind of order v is defined by

(15)

0)( 222 yvxyxyx

0)( 222

22 yt

dtdy

tdt

ydt

)()( ixJixI

Ch5_73

Analogous to (10), the modified Bessel function of the second kind of order v integer is defined by

(16)

and for any integer v = n,

Because Iv and Kv are linearly independent on (0, ), the general solution of (14) is

(17)

sin)()(

2)(

xIxIxK

)(lim)( xKxKn

n

)()( 21 xKcxIcy

Ch5_74

We consider another important DE:

(18)

The general solution of (18) is

(19)

We shall not supply the details here.

0 ,021

2

2222222

pyx

cpaxcby

xa

y c

)]()([ 21c

pc

pa bxYcbxJcxy

Ch5_75

Example 3

Find the general solution of on (0, )

SolutionWriting the DE as

according to (18)1 – 2a = 3, b2c2 = 9, 2c – 2 = −1, a2 – p2c2 = 0

then a = −1, c = ½ . In addition we take b= 6, p = 2.From (19) the solution is

093 yyyx

093 yx

yx

y

)]6()6([ 2/122

2/121

1 xYcxJcxy

Ch5_76

Example 4

Recall the model in Sec. 3.8

You should verify that by letting

we have

0 ,0 xkexm t

2/ 2 te

nk

s

022

22 xs

dsdx

sds

xds

Ch5_77

Example 4 (2)

The solution of the new equation is x = c1J0(s) + c2Y0(s),

If we resubstitute

we get the solution.

2/ 2 te

nk

s

2/

022/

0122

)( tt emk

Ycemk

Jctx

Ch5_78

Properties

(1)

(2)

(3)

(4)

)()1()( xJxJ mm

m

)()1()( xJxJ mm

m

0,1

0,0)0(

m

mJm

)(lim

0xYmx

Ch5_79

Example 5

Derive the formula

SolutionIt follows from (7)

1

1

12

0

2

0

2

0

2

2)1()!1()1(

)(

2)1(!)1(

22)1(!

)1(

2)1(!)2()1(

)(

nk

n

vnn

v

n

vnn

n

vnn

n

vnn

v

xnvn

xxvJ

xnvn

nxnvn

v

xnvnvn

xJx

)()()(' 1 xxJxvJxxJ vvv

Ch5_80

Example 5 (2)

)()(

2)2(!)1(

)(

1

0

12

xxJxvJ

xkvk

xxvJ

vv

k

vkk

v

Ch5_81

The result in example 5 can be written as

which is a linear DE in Jv(x). Multiplying both sides the integrating factor x-v, then

(20)

It can be shown(21)

When y = 0, it follows from (14) that(22)

)()()( 1 xJxJxv

xJ vvv

)()]([ 1 xJxxJxdxd

vv

vv

)()]([ 1 xJxxJxdxd

vv

vv

,)()( 10 xJxJ )()( 10 xYxY

Ch5_82

Spherical Bessel Functions

When the order v is half an odd number, that is, 1/2, 3/2, 5/2, …..

The Bessel function of the first kind Jv(x) can be expressed as spherical Bessel function:

Since (1 + ) = () and (1/2) = ½, then

0

2/12

2/1 2)2/11(!

)1()(

n

nn x

nnxJ

!2

)!12(

2

11

12 n

nn

n

Ch5_83

Hence

and

0

12

0

2/12

12

2/1 )!12()1(2

2!2

)!12(!

)1()(

n

nn

n

n

n

n

xnx

x

n

nn

xJ

(24) cos2

)(

(23) sin2

)(

2/1

2/1

xx

xJ

xx

xJ

Ch5_84

The Solution of Legender Equation

Since x = 0 is an ordinary point of (2), we use

After substitutions and simplifications, we obtain

or in the following forms:

0n

nnxcy

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

06)2)(1(

02)1(

2

31

20

jj cjnjncjj

ccnn

ccnn

Ch5_85

Using (25), at least |x| < 1, we obtain

6

4201

!6)5)(3)(1()2)(4(

!4)3)(1()2(

!2)1(

1)(

xnnnnnn

xnnnn

xnn

cxy

(25) ,4,3,2,)1)(2(

)1)((!3

)2)(1(!2

)1(

2

13

02

jcjj

jnjnc

cnn

c

cnn

c

jj

Ch5_86

Notices: If n is an even integer, the first series terminates, whereas y2 is an infinite series. If n is an odd integer, the series y2 terminates with xn.

(26) !7

)6)(4)(2)(1)(3)(5(

!5)4)(2)(1)(3(

!3)2)(1(

)(

7

5312

xnnnnnn

xnnnn

xnn

xcxy

Ch5_87

Legender Polynomials

The following are nth order Legender polynomials:

(27))157063(

81

)(),33035(81

)(

3)5(21

)(),13(21

)(

)(,1)(

355

24

33

22

10

xxxxPxxxP

xxxPxxP

xxPxP

Ch5_88

They are in turn the solutions of the DEs. See Fig 5.5

(28)

0122)1(:3

062)1(:2

022)1(:1

02)1(:0

2

2

2

2

yyxyxn

yyxyxn

yyxyxn

yxyxn

Ch5_89

Fig 5.5

Ch5_90

Properties

(1)

(2)

(3)

(4)

(5)

)()1()( xPxP nn

n

1)1( nP

nnP )1()1(

odd ,0)0( nPn

even ,0)0(' nP n

Ch5_91

Recurrence Relation

Without proof, we have

(29)which is valid for k = 1, 2, 3, …Another formula by differentiation to generate Legender polynomials is called the Rodrigues’ formula:

(30)

0)()()12()()1( 11 xkPxxPkxPk kkk

... ,2 ,1 ,0 ,)1(!2

1)( 2 nx

dx

d

nxP n

n

n

nn