L-6 de Series Solution

8/4/2019 L-6 de Series Solution

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Solution in Series


2/88

A linear second order homogeneous differential equation is of the

form

where are the functions of x (may be complexvalued) real or complex.

10210 yxyxyx x,x,x 210

The function is not identically zero, but may vanish at

certain points.

x0

The points where are called Ordinary Points 00 x

0 yxQyxPy

Dividing the equation (1) by we get x0

The points where are called Singular Points. 00 x

A singular pointx = a is called regular singular point or

regular singularity if xQaxxPax 2and have derivatives of all orders in the neighbourhood ofa.


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We note the following theorems for the solutions of the equation:

10210 yxyxyx Theorem 1. Whenx = a is an ordinary point of (1), its every

solution can be expressed in the form

22210 .......axaaxaay

Theorem 2. Whenx = a is a regularsingular point of (1), at

least one solution can be expressed in the form

32210 ... ........axaaxaaaxym

Theorem 3. The series (2) and (3) are uniformly convergent at

every point within the circle of convergence ata.

This series is known as Frobenius series.


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Solve the equation

10 yyxy

in the neighbourhood ofx = 0.Sincex = 0 is an ordinary point, we take

......xaxaay 2210

Differentiating (2), we get.......xaxaay 2321 32 31

0

1

n

nn xan

20

n

nnxa

and ......2432 34232 xaxaay

4120

2

n

nn xann

Substitute (2), (3) and (4) in (1)


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or

011200

1

0

2

n

nn

n

nn

n

nn xaxanxxann

011200

110

2

nnn

nnn

nnn xaxanxann

or 012010

2

n

nn

n

nn

n

nn xaxnaxann

or 011221

202

n

n

nn xanannaa

This holds for all values of x if, and only if

,........,,,nanannaa

nn 4321011202

2

02

or ,. .......,,,nan

aaa nn 4321for2

1and

2

1202


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06

04

02

246

1

24

1

2

1

aa

aa

aa

17

15

13

357

1

35

1

3

1

aa

aa

aa

and so on.

Using these values in (2), we get

........xaxaxaxaxaay

514

03

12

01035

1

24

1

3

1

2

1

..... ....xxxxa

..... ....xxxay

5531

6420

357

1

35

1

3

1

246

1

24

1

2

11

This the general solution of (1) with a0 and a1 as arbitrary

constants.


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To solve the differential equation

whenx = a is a regular singularity, that is

and

have derivatives of all orders, then at least one solution can

be in the form of the Frobenius series

0210 yxyxyx

00 a

xxax

x

xax,

0

22

0

1

002

210 aaxaaxaaaxym

,...........

To understand steps involved in this method (Frobenius

Method) we consider the Bessel differential equation oforder , where is any constant,

02222

2

yxxxdx

dy

dx

yd


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Comparing with we find that

102222

2

yxxxdx

dy

dx

yd We note thatx = 0 is a singular point and dividing byx2 we

get 0

2

22

2

21

y

x

x

dx

dy

xdx

yd 0 yxQyxPy

222

and1

xxQxxxPthereforex = 0 is a regular singularity of the DE and we take

...........xaxaaxy m 2210 20

n

nmnxa

The first and second derivatives of y are

,xanmyn

nmn

0

1

0

231

n

nmnxanmnmy

Substituting for in (1) we getyy,y and


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0

1

0

22

0

1

0

22

n

nmn

n

nmn

n

nmn

xaxxanmx

xanmnmx

01

111

22

2

11

20

2

n

nmnn

mm

xaanmnmnm

xammmxammm

or

010

2

0

2

n

nmn

n

nmn xaxanmnmnm or

012

2

0

2

n

nmn

n

nmn xaxanmnmnm or


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Equating the coefficients of powers of x, stating with the

least power, we get

401 02

ammm

5011 12 ammm

.................,,,, 5432

601 22

n

aanmnmnm nn

Sincea0 0, therefore from (4)

701 2 mmmThis equation is known as indicial equation. The indicial

equation for a second order DE is a quadratic and gives twovalues of index m. We expect a solution of DE

corresponding to each value of m. From (7) we have

022 m 21 and mm


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From (5)

011 12 amm 01 122 am or

Since first factor is not zero for any value of m , therefore

From (6), forn= 2, 3, 4, 5, ..

2

2

1

nmnmnm

aa nn

22

nmnm

an

)(82

nmnm

an

,01

aSince form (8) we find that

............... 7530 aaa

nmnm

aa nn

2forn= 2, 4, 6,

.a 01


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nmnm

aa

nn

22

122

)(or forn= 1, 2, 3,

From this relation we can find in terms of.............,,, 642 aaa .0a

It is expected that each value of m will provide one set of

values of these coefficients and hence two solutions of DE.

,forNow 1 mm 22212

2

nn

aa nn)(

,forand 2 mm

222

122

nn

aa

nn

)(

nna n2

12

2)(

nn

a n2

12

2

)(

Clearly if = 0 we will get identical values and if is an

integer some coefficients will become infinite after somen.


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Case I: When neither = 0 nor is an integer.

In this case we have two distinct solutions of indicial equation

which do not differ by an integer. Two linearly independent

solutions shall be obtained in this case.

1For mm

nn

aa

nn 2

122

2

)(

1220

2 aa

22222

4 aa

121224

0a

33224

6a

a

12312326

0a

1212

12

02

......! nnn

aa

n

nn

and so on.

For general value of n, (n = 1, 2, 3, ), we have


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2forsimilarlyProceeding mm

solutionget theweforTherefore 1 mm

.......!!

123321222121 6

6

4

4

2

2

01

xxx

xay

.......!!

123321222121 6

6

4

4

2

2

02

xxx

xay

2211issolutiongeneralThe ycycy


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Gamma Function

The Gamma function is defined as

.)( 0for10

dxxe xThe integral is known as Eulers integral of second kind.

dxe x0

1)(

0

xe 1

dxxe x 0

1)(

dxxeexxx 1

00

)( )()( 112 1

)()( 223 !212 )()( 334 !! 323 !)( nn 1 dxxe x 2

1

021


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)( 121

0 a

In

.......

!!

123321222121

6

6

4

4

2

2

01xxx

xay

..........)(!

)(!)()(

112332

1122211211

2

6

6

4

4

2

2

1

x

xxxy

0

2

121

1

k

kkx

kky

)(!

0

2

211

1

k

kkx

kk

)()(

take


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0

2

211

1

k

kkx

kkxJ

)()()(

Definition:

is known as Bessel function of the first kind of order .

Corresponding to the other solution y2 , we have

0

2

211

1

k

kkx

kkxJ

)()()(

This is known as Bessel function of the first kind of order -.

Hence the complete solution ofBessel equation can be

expressed as.)()( xJbxJay


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This gives only one solution instead of two. The second

solution is given by when = 0.m

y

1

....

log

4

2

2

2

422

2

222

4

2

2

0

11

mmmm

x

mm

xxa

xymy

m

Therefore for m = 0

....log211

422 224

22

010

12

xxaxymyy

m

....

222

6

22

4

2

2

01642422

1xxx

ay

The general solution is then 2211 ycycy


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....

222

6

22

4

2

2

01642422

1xxx

ay

Consider the solution

....

)()(26

6

24

4

2

2

032122122

1xxx

a

....)!()!( 26

6

24

4

2

2

0322221

xxx

a

From the definition of Bessel function

0

2

0

211

1

k

kkx

kk

xJ

)()(

)(

0

2

2

2

1

k

kkx

k )!(10 aThus with is Bessel function of first kind of order zero.1y

The other solution with is Bessel function of second

kind or Neumann function of order zero and is denoted by .

10 a2y)(xY0


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Case III When is an integer, we have seen some of the

coefficients become infinite.Now suppose that = n a positive integer, then we obtain the

solution

0

2

211

1

k

knk

nx

nkkxJ

)()()(

To find the other solution since one solution is known we take

the general solution as y = u(x) y1(x)

Submit the answer to the above question on the day of

Next Tutorial

Show that

and the complete solution is

constantsareandwhere

1

2badx

xJxbaxu

n

)(

)(

.

)(

)()( dx

xJx

xJbxJay

n

nn 2

1


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In general if the roots of indicial equation differ by an integer, we

can obtain only one solution. The second solution is given by

1

1mmm

y

The function is the Bessel

function of second kind of order n or Neumann function.

dx

xJxxJxY

n

nn 2

1

)()()(


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Recurrence relations forBessel functions

1

1

1 1

1 1

1 1

1

(1) [ ( )] ( )

(2) [ ( )] ( )

(3) 2 ( ) [ ( ) ( )]

(4) 2 ( ) ( ) ( )

(5) ( ) 2 ( ) ( )

(6) ( ) ( ) ( )

n n

n n

n n

n n

n n n

n n n

n n n

n n n

dx J x x J x

dx

d

x J x x J xdx

nJ x x J x J x

J x J x J x

xJ x nJ x xJ x

xJ x nJ x xJ x


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Show that )()]([ xJxxJxdx

dn

nn

n1

We havekn

k

kn

n

n x

knkxxJx

2

0 21

1

)(!

)()(

kn

kkn

k

xknk

22

02

12

1

)(!

)(

122

02

12

122

kn

kkn

k

nn xknk

knxJxdx

d

)(!

))(()]([

122

012

2

1

kn

kkn

k

xknk )(!

)(

122

02

12

12

kn

kkn

k

xknk

kn

)(!

))((

12

012

2

1

kn

kkn

kn x

knkx

)(!

)(

)(xJx nn 1


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Show that )()]([ xJxxJxdx

dn

nn

n1

We have )(xJxn

n

k

kkn

k

xknk

2

02

12

1

)(!

)(

kn

k

kn x

knkx

2

0 21

1

)(!)(

)]([ xJxdx

dnn 12

12

12

12

k

kkn

k

xknk

k

)(!

)(

12

112

112

1

k

kkn

k

xknk )()!(

)(

112

112

1

11112

1

)(

)( )()!(

)( kkn

kx

knk

0k

12

012

112

1

k

kkn

k

xknk )()!(

)(


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12

012

112

1

nk

kkn

kn x

knkx

)()!(

)(

)(xJx

n

n

1

)]([ xJxdx

d

n

n

Show that

)()()(

)]()([)(

xJxJxJ

xJxJxxnJ

nnn

nnn

11

11

2

2

We have

)()]([

)()]([

xJxxJxdx

d

xJxxJxdx

d

n

n

n

n

nn

nn

1

1


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)()]([)(

)()]([)(

xJxxJ

dx

dxxJnx

xJxxJdx

dxxJnx

nn

nn

nn

nn

nn

nn

11

11

Therefore

or

)()()]([)(

)()()]([)(

2

1

1

1

xxJxJdx

dxxnJ

xxJxJdx

dxxnJ

nnn

nnn

Subtracting (2) from (1) )()]()([)( 32 11 xJxJxxnJ nnn

Adding (1) and (2)1 12 ( ) ( ) ( ) (4)n n nJ x J x J x

1( ) ( ) ( ) (6)n n nxJ x nJ x xJ x From (2)

From (3) 1 1( ) 2 ( ) ( ) (5)n n nxJ x nJ x xJ x


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All the Bessel functionsJn(x) and their derivatives can be

expressed in terms ofJ0(x) andJ1(x) by using recurrence

relations.

Example: ExpressJ5(x) in terms ofJ0(x) andJ1(x).

)]()([)( xJxJxxnJ nnn 112 We have

)(xJn 1Or

)()()( xJxJxxJ 123

4

Therefore )()()( xJxJx

xJ 0122

)()()( xJxJxJxx 101

24

)()( xJx

xJx

012

41

8

)()( xJxJx

n

nn 1

2


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)()()( xJxJx

xJ 2346

)()()()( xJxJx

xJx

xJxx

01012 24186

)()( xJ

x

xJ

xx

0213

241

848

)()()( xJxJx

xJ 3458

)()()()( xJx

xJx

xJx

xJxxx 0120213

41

8241

8488

)()( xJxx

xJxx

03124

192121

72384


31/88

Example: Express in terms ofJ0(x) andJ1(x).)(xJ1

)()()( xJxJxJ nnn 112

We have

)()()( xJxJxJ nnn 112 Differentiating we get

)()()()()( xJxJxJxJxJ nnnnn 222

1

2

12

)()()()( xJxJxJxJ nnnn 224

1

2

1

4

1

)()(2)(4

122 xJxJxJ nnn

)()(2)(4

1)( 3111 xJxJxJxJ Putting n =1


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)]()([)(2 11 xJxJxxnJ nnn From the relationFor n = 0 )]()([0 11 xJxJx )]()( 11 xJxJ

)(4

)(18

)( 0123 xJxxJ

xxJ

Also from the previous example

)()(2)(41)( 3111 xJxJxJxJ Therefore substituting in

)(4)(18)(2)(4

1)( 012111 xJx

xJx

xJxJxJ

)(4)(484

1)( 0121 xJx

xJx

xJ )(1

)(12

012xJ

xxJ

x


33/88

We havekn

k

k

nx

knk

xJ

2

0 21

1

)(!

)()(

nrkeirkn ..Putting

rn

nr

nr

nx

rnrxJ

2

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

R l ti b t B l d


34/88

Relation between Bessel andtrigonometric functions

,cos2

)(21

xx

xJ

Show that

0

2

21

1

k

kkx

kkxJ

)(!)( We have

0

2

21

21

21 21

1

k

kkx

kkxJ

)(!)(

0

2

21 21

12

k

kkx

kkx )(!

.........

)(!)(!)()(

6

27

4

25

2

23

21 23

1

22

1

2

112 xxx

x


35/88

..........)(!

)(!)()()(

6

21

21

23

25

4

21

21

23

2

21

21

21

23

1

22

1

2

112

21

x

xx

xxJ

..........

)(

6

2

1

2

2

2

3

2

4

2

5

2

6

4

21

22

23

24

2

2

21

21

2

1

2

1

212

1

1

12

x

xx

x

..........

!!!

642

6

1

4

1

2

11

2xxx

x xx cos2


36/88

Bessel function is the coefficients of tn in the

expansion of)(xJn

)(t

txe

121

This function is known as generating function of Bessel

function of first kind.

nn

ntx xJte t )()( 1

21


37/88

Reduce the equation

to Bessel form

)(1022222

2

ynxxxdx

dy

dx

yd

02222

2

yxxxdx

dy

dx

yd Write down the solution of (1) in terms of Bessel function.


38/88

Orthogonality of Bessel Functions

where , are roots of Jn (x) = 0

,)]([

,)()( 2

12

1

1

0

0

n

nn

J

dxxJxxJ

The solutions of the equations

and

are respectively.)()( xJvxJu nn and

2

2

2 2 2 2 0 (2)d v dvdxdx

x x x n v

2

2

2 2 2 2

0 (1)

d u du

dxdxx x x n u

Multiply (1) with v, (2) with u and subtract

022222

2

2

2

uvxuvxuvxdxdv

dxdu

dx

vd

dx

ud )()(


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2 2

2 2

2 2or ( ) 0d u d v du dvdx dxdx dx

x v u v u xuv

dxdv

dxdu

dx

vd

dx

ud uvuvxxuv )(2

2

2

222or

Integrate with respect to x from 0 to 1

10

1

0

22)(

dxdv

dxdu uvxxuvdx )()]([ 31 xdx

dvdxdu uvx

Since ( )nu J x

[ ( ) ]du d ndx dxJ x ( )( )

[ ( ) ] d xd nd x dxJ x

( )nJ x

Similarly as )( xJv n ])( xJndxdv Substituting in (3)

)]([dxdv

dxdu

dxd uvx

dxdv

dxdu

dxdv

dxdu

dxd uvuvxxuv )(22or


40/88

22

1

0or

)()()()(

)()(nnnn

nn

JJJJ

dxxJxxJ

)()()()()()( nnnnnn JJJJdxxJxxJ 1

0

22

Since and are the roots of , therefore if 0)(xJn

0

1

0 dxxJxxJ nn )()(

If = , then

1

0

2dxxJx n )]([ 22

)()()()(lim nnnn

JJJJ

22

)()(lim nn

JJ])([ 0nJ


41/88

12

0

[ ( )]nx J x dx ( ) ( )

lim2

n nJ J

1( ) ( ) ( )n n nxJ x nJ x xJ x

21[ ( )]2

nJ

From the relation

1( ) ( )n nJ J

12

0

[ ( )]nx J x dx 2

11

[ ( )]2

nJ

Therefore

3


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3

1

n 1

Expand ( ) in the interval 0 3 in terms of ( ),

where are given by (3 ) 0.

n

n

f x x x J x

J

3 11

n n

n

x c J x

Let

Multiply both side by and integrate with respect to

x from 0 to 3.

1 mxJ x

3 3

4

1 1 1

10 0

(1)m n n mn

x J x dx c J x x J x dx

3

1 1

1 0

n n m

n

c xJ x J x dx

3

1 1

10

Right hand side of (1) n n mn

c J x x J x dx

1

1 1

1 0

R.H.S. 3 3 3 (3 )n n m

n

c tJ t J t d t

Putting x = 3t


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1

1 1

1 0

R.H.S. of (1) 3 3 3 3n n mn

c tJ t J t dt

1

2

1 11 0

3 3 3n n mn c tJ t J t dt

22

2

3

3 22 [ ( )] ( )m mc J

3 4 10

L.H.S. of (1) = mx J x dx

1We have [ ( )] ( )n n

n n

dx J x x J xdx

1

1

( ) ( ) (3) n nn nx J x dx x J x1

( ) ( ) ( ) ( ) ( )n nn nx J x x J x d x 1( ) ( ) n nn nx J x x J x dx

3 2 2 10

= [ ]m

x x J x dx

3 4 10

L.H.S. of (1) = mx J x dx

Integrate by parts using (3)

32 23


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32 232 2200

L.H.S. of (1) = 2

m mm m

x J x x J xx x dx

4 3

2 3

2

0

3 3 2=

m mm mJ x J x dx 34 32 30

3 3 2=

m m

m m m

J x J x[By using (3)]

4 32 32

3 3 2 3 3= 4( )

m mm m

J J x

22

2

33

2

[ ( )]m m

c J 4 32 3

2

3 3 2 3 3=

m mm m

J J x

2 32 22

63 3 2 3

3[ ]

[ ( )]m

m m m

m

c J JJ


45/88

Hence

3 2 3 12 21 2

63 3 2 3

3[ ]

[ ( )]m m m

m m m

x J J J xJ


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2 4 6 8 10x

-0.4

-0.2

0.2

0.4

0.6

0.8

1

Jn x

J1

J0

Bessel Fuctions


47/88

2 4 6 8 10

x

-0.5

0.5

1

1.5

2

Jn x

J 0.5

J0.5

Bessel Fns


48/88

Solution in Series

Legendre polynomials


49/88

The Legendre equation

arises in the problems with spherical symmetry. 21 2 1 0 1( ) ( )x y xy n n y

This equation has two regular singular pointsx = -1 and 1.

0

mmm

y a x

Substituting in (1), simplifying and equating the coefficients

of powers of x to zero we get

2 0

3 1

2

2 1 06 2 1 0

2 1 1 2 1 0

2 3

( )[ ( )]

( )( ) [ ( ) ( )]

, , ............

m m

a n n aa n n a

m m a m m m n n a

m

x = 0 is an ordinary point. Therefore in the neighbourhood of

0 we can have a series solution of the form.


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2 0 3 12 1 0 6 2 1 0( ) [ ( )]a n n a a n n a 2

1for 2 3

2 1

( )( ), , ............

( )( )

m mn m n m

a a m

m m

This is the recurrence relation or recursion formula.

Two independent solutions are

2 4

1

1 2 1 3

1 2 4

( ) ( ) ( )( )

..................! !

n n n n n n

y x x

3 52

1 2 3 1 2 4

3 5

( )( ) ( )( )( )( ).........

! !

n n n n n ny x x x

Both the series converge for |x| < 1.

If n is a non-negative integer, one of them terminates and one

solution then is a polynomial.

The general solution isy(x) = a0 y1 + a1y2


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We rewrite the recursion formula

21

for 2 32 1

( )( ), , ............

( )( )m m

n m n ma a m

m m

2

2 1as where 2

1

( )( )

( )( )m m

m ma a m n

n m n m

2na 12 2 1

( )( )

nn n an

Then using this relation we can express all the coefficients

in terms of2 4 6, , ..................n n na a a .na

2

2

2

( )!

( !)n n

na

nWe take

21 2

2 2 1 2( ) ( )!

( ) ( !)n

n n nn n

2

1 2 2 1 2 2

2 2 1 2

( ) ( )( )!

( ) ( !)n

n n n n n

n n

2 2

2 1 2

( )!

( )!( )!n

n

n n

The value of is so selected that resulting polynomial takes

the value 1 atx = 1.na

2 3( )( ) 2 3 2 2( )( ) ( )!


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4na 22 34 2 3

( )( )

( )n

n na

n

2 3 2 2

4 2 3 2 1 2

( )( ) ( )!

( ) ( )!( )!n

n n n

n n n

2 3 2 2 2 3 2 4

4 2 3 2 1 2

( )( ) ( )( ) ( )!( ) ( )!( )!n

n n n n nn n n

1 2 4

2 2 2 4

( )!

( )!( )!n

n

n n

2 4

2 2 2 4

( )!

! ( )!( )!n

n

n n

6na 2 63 2 3 6

( )!

! ( )!( )!n

n

n n

2n ka 2 212 2

( )!( )

! ( )!( )!

k

n

n k

k n k n k

In general when 2 0,n k


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The resulting polynomial solution of Legendre DE is

2 4

2

6

2 2 2 2 4

2 2 1 2 2 2 2 42 6

3 2 3 6

( )! ( )! ( )!

( !) ( )!( )! ! ( )!( )!( )!

.....................! ( )!( )!

n n n

n n n

n

n

n n nx x x

n n n n nn

xn n

20

12 2

2 2

1 12 2

where or whichever is an integer.

( )!

( )!( )!( )!

Mm n m

nm

n n

n m

xm n m n m

M

The polynomial (1) is known as Legendre Polynomial of

degree n andis denoted by ( ).nP x


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20

12 2

2 21

2 2where or whichever is an integer.

( )!( )

!( )!( )!

Mm n m

n n

mn n

n mP x x

m n m n mM

Since

Therefore

0 1( )P x

212

3 1( )x 2( )P x

4 218

35 30 3( )x x 4( )P x

x1( )P x

31

2

5 3( )x x3( )P x

5 318

63 70 15( )x x x5( )P x Observe that 1 1( )nP

h f d l i l 1


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Graph of Legendre PolynomialP0 = 1.

-1 -0.5 0.5 1

x

0.5

1

1.5

2

P0x

G h f L d P l i l


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Graphs of Legendre Polynomials

0 11( ) ( )P x P x x

-1 -0.5 0.5 1

x

-1

-0.5

0.5

1

Pn x

P1

P0

G h f L d P l i l


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-1 -0.5 0.5 1

x

-1

-0.5

0.5

1

Pn x

P2

P1

P0


210 1 2 2

1 3 1( ) ( ) ( )P x P x x P x

G h f L d P l i l


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210 1 2 2

313 2

1 3 1

5 3

( ) ( ) ( )

( )

P x P x x P x

P x x

-1 -0.5 0.5 1

x

-1

-0.5

0.5

1

Pn x

P3

P2

P1

P0


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-1 -0.5 0.5 1

x

-1

-0.5

0.5

1

Pn x

P5

P4

P3

P2

P1

P0


0 1( )P x 21

23 1( )x 2( )P x

4 218

35 30 3( )x x 4( )P x

x1( )P x 31

25 3( )x x3( )P x

5 318

63 70 15( )x x x5( )P x

3


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Question: Obtain the 7th derivative of

1 2

1 2

If and be two function of possesing derivatives

of nth order, then

( ) ( ) ( ) ( )

( ) ( ) ( )

( ) ( )

( ) ...................

n n n n n n

n n r r nr

u x v x x

uv u v C u v C u vC u v uv

Leibnitzs Theorem

3sinxe x

Question: Obtain the 7th derivative of2 2 31( ) xx e


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Rodrigues Formula

21 1

2

( ) [( ) ]

!

nn

n n n

dP x x

n dx

2 1( )nv x Let Then 2 12 1( )ndv nx x

dx

2 21 2 1( ) ( )

ndvx nx x

dx

2 1 2 0( )

dvx nxv

dx

Differentiate n+1 timesusing Leibnitzs theorem

2 2 1

1

11 1 2 2

2

2 1 0

( ) ( ) ( )

( ) ( )

( )( ) ( )

!

[ ( ) ]

n n n

n n

n nx v n x v v

n xv n v


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Express in terms of Legendre polynomials3 24 2 3 8


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Express in terms of Legendre polynomials.3 24 2 3 8x x x

Therefore3 24 2 3 8x x x

0 1( )P x 21

23 1( )x 2( )P x

x1( )P x 31

25 3( )x x3( )P x

31 2 35

[ ( ) ]P x x3x 3 11 2 35

[ ( ) ( )]P x P x2

12 1

3[ ( ) ]P x 2x

3 1 2 14 2

2 3 2 1 3 85 3

[ ( ) ( )] [ ( ) ] ( )P x P x P x P x 3 2 1 0

8 4 3 22

5 3 5 3( ) ( ) ( ) ( )P x P x P x P x

Since is a cubic polynomial, it can be

expressed in terms of

3 24 2 3 8x x x

E i Sh h122 n


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1 12 221 1

22 2

1 2 1 2

1

11 2 22 2

( )

( )

( ) [ ( )] ....!

xt t t x t

t x t t x t

Exercise: Show that 220

1 2 ( )n

nn

xt t t P x

Hint: Using Binomial theorem

Express right hand side in powers of t to obtain the result.

Generating function of Legendre Polynomials

1221 2xt t


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Recurrence formulae for Legendre polynomials

1 11 1 2 1. ( ) ( ) ( ) ( ) ( )n n nn P x n xP x n P x 12. ( ) ( ) ( )n n nnP x x P x P x

1 13 2 1. ( ) ( ) ( ) ( )n n nn P x P x P x 1 14. ( ) ( ) ( )n n nP x x P x n P x

215 1. ( ) ( ) [ ( ) ( )]n n nx P x n P x x P x

All these formulas can be proved using generating function for ( ).nP x

We have 1220

1 2 ( ) ( )n

nn

xt t t P x I

Differentiating with respect tot, we get 322 112

0

1 2 2 2( ) ( )n

nn

xt t x t n t P x

3


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122 2 101 2 1 2( ) ( )n nnx t xt t xt t n t P x 2 1

0 0

1 2( ) ( ) ( )n nn nn n

x t t P x xt t n t P x

Equating the coefficients oftn on both sides, we get

1 1 11 2 1( ) ( ) ( ) ( ) ( ) ( ) ( )n n n n nxP x P x n P x nxP x n P x 1 11 2 1 1( ) ( ) ( ) ( ) ( ) ( )n n nn P x n x P x n P x

322 10

1 2( ) ( ) ( )n

nn

x t xt t n t P x II


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Differentiating (I) with respect tox, we get

322120

1 2 2( ) ( )n

n

n

xt t t t P x

322

0

1 2 ( ) ( )n

nn

t xt t t P x III

322 10

1 2( ) ( ) ( )n

nn

x t xt t n t P x II

Also , we haveFrom (II) and (III)

1

0 0

( ) ( )n n

n nn n

n t P x t P x

x t t

1


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1

0 0

( ) ( ) ( )n n

n nn n

t n t P x x t t P x

Equating the coefficients oftn on both sides, we get

1 2( ) ( ) ( ) ( )n n nn P x xP x P x

1 11 2 1 1( ) ( ) ( ) ( ) ( ) ( )n n nn P x n x P x n P x Differentiating the relation

with respect tox, we get

1 11 2 1 2 1( ) ( ) ( ) ( ) ( ) ( ) ( )n n n nn P x n P x n xP x n P x Substituting for x from (2)( )nP x

1

1 1

12 1 2 1

( ) ( )( ) ( ) ( )[ ( ) ( )] ( )

n

n n n n

n P xn P x n nP x P x n P x

1 12 1 3( ) ( ) ( ) ( ) ( )n n nn P x P x P x

O th lit f L d l i l


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Orthogonality of Legendre polynomials

1

1

0 for

2 for2 1

( ) ( )n m

m n

P x P x dx m nn

Proof:

We know that are solution of the

Legendre equations of ordern andm respectively.

and( ) ( )n mP x P x

2

2

1 2 1 0 1

and 1 2 1 0 2

( ) ( )

( ) ( )

n n n

m m m

x P xP n n P

x P xP m m P

Therefore

Multiplying (1) by , (2) by and subtracting, we get( )mP x ( )nP x

21 2

1 1 0

( )( ) ( )

[ ( ) ( )]

n m n m n m n m

n m

x P P P P x P P P P

n n m m P P


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2or 1 1 0[( )( )] ( )( )n m n m n md

x P P P P m n m n P Pdx

Integrating with respect to x from -1 to 1, we get

112

11

1 1 0( )( ) ( )( )n m n m n mx P P P P m n m n P P dx

1

11 0 3( )( ) ( )n mm n m n P P dx

11

Therefore 0 when( ) ( )n mP x P x dx m n

When m = n, the relation (3) does not provide any information.

When m= n, it is proved by using the


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generating function

Squaring both sides and integrating w.r.t. xover [-1, 1],we obtain

From the left hand side, we get

n

n

ntxptxt )()21(

0

2

1

2

.])([

)21(

21

1 0

1

1

2dxtxp

txt

dx n

n

n

)]21ln()21[ln(2

1

]

2

)21ln([

2122

1

1

21

1

2

ttttt

t

txt

txt

dx

])1ln()1[ln(22

tt


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)1.....](12

.......53

1[2

....)]2

(......)2

[(1

)]1ln()1[ln(1

])1ln()1[ln(2

242

22

n

ttt

tt

tt

t

ttt

ttt

n

From the right hand side , we have11


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(2)

Comparing the coefficients of in equation (1) and (2),we get

Hence we have the result.

dxxpt

dxtxptxp

nn

n

n

nn

n

nn

)(

)(])([

1

1

2

0

2

2

0

1

1

221

1 0

nt2

12

2)(

1

1

2

ndxxpn

Ch b h P l i l


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Chebyshev Polynomials

( )nT xChebyshev Polynomials denoted by , are the solutions

of the differential equation2 21 0 1( ) ( )x y x y n y 1x Singularities of this equation are

0

( ) mmm

y x a x

0x We find a series solution about of the form

This series solution is convergent for Substituting in

(1), we get

1| | .x

2 2 1 2

2 1 0

1 1 0( ) ( ) m m mm m mm m m

x m m a x x ma x n a x

Solution

21 1( ) ( )m m
http://de%20series%20solution%20i.ppt/http://de%20series%20solution%20i.ppt/


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2

2 2

2

1 0

1 1

0

( ) ( )m mm mm m

m mm m

m m

m m a x m m a x

ma x n a x

20 2

2

1 0

2 1 1

0

( )( ) ( )m m

m mm m

m mm m

m m

m m a x m m a x

ma x n a x

22 3 1 0 1

22

2

2 6

2 1 1 0

( )

[( )( ) ( ) ]m

m m m mm

a a x a x n a a x

m m a m m a ma n a x

Equating the coefficients of various powers ofx to zero, we get

2 22 0 3 1 12 0 6 0a n a a a n a


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2 0 3 1 1

22

2 0 6 0

2 1 1 0 for 2( )( ) ( )m m m m

a n a a a n a

m m a m m a ma n a

2

2 02

na a 23 116

na a 2 2

2 0 for 22 1( )( )

m m

n m

a a

m m

2 2

4 22

4 3

na a

2 2 2

02

4 3 2

n na 2 2 2 0

12

4( )

!n n a

2 2

5 33

5 4

na a

2 2 2

13 1

5 4 6

n na

2 2 2

11 3

5

( )( )

!

n na

and so on.

Substituting in the assumed power series solution, we get

2

11

3!n a

2 2 2 2 2 40

1 11 2( ) ( )y x a n x n n x


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0

2 2 2 2 2 23 5

1

1 22 4

1 1 3

3 5

( ) ( ) .............! !

( )( )..........

! !

y x a n x n n x

n n na x x x

The two series

2 2 2 2 2 41

1 11 2

2 4( ) ( ) ............

! !y x n x n n x

2 2 2 2 2 23 5

21 1 3

and3 5

( )( )( ) ..........

! !

n n ny x x x x

are two linearly independent solution of Chebyshev differential

equation and converge for |x | 1.

Forn= 0, 2, 4, ..,y1(x) reduces to the polynomials

1 1( ) ,y x 21 1 2( ) ,y x x 2 41 1 8 8( ) , ..............y x x x

Forn= 1, 3, 5, ..,y2(x) reduces to the polynomials


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, , , , y2( ) p y

2 ( ) ,y x x

32

13 4

3( ) ( ),y x x x

3 52

15 20 16

5

( ) ( )y x x x x and so on.

These polynomials give rise to important polynomials called

Chebyshev Polynomials.

To define Chebyshev Polynomials in different form


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we again consider the Chebyshev differential equation

2 21 0( )x y x y n y cos ,x Put then dyy

dx dy d

d dx 1sin dyd

2

2

d yy

dx 1

sin

d dy d

d d dx

2

2 21 1 cos

sin sin sin

d y dy

dd

2

2 2 31 cos

sin sin

d y dydd

Substituting in the differential equation

22 2

2 2 31 1 0cossin cos

sinsin sin

d y dy dy n yd dd

2

2

20

d yn y

d

The general solution of this differential equation is


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The general solution of this differential equation is

( ) cos siny A n B n 1 1

( ) cos( cos ) sin( cos )y x A n x B n x Therefore the general solution of Chebyshev differential equation is1 1Hence and are linerly independent

soltuions.

cos( cos ) sin( cos )n x n x

When n is an integer, we define

1 1(x)=( ) cos( cos ) sin( cos )n nT x n x U n x

( )nT x is Chebyshev polynomial of first kind of degreen.

( )nT x 1.

1 1Show that 2 0( ) ( ) ( )n n nT x xT x T x


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1 where( ) cos( cos ) cos cos .nT x n x n x 1 1S ow t at 0( ) ( ) ( )n n nx x x x

We have

11

11

1 1

1 1

( ) cos[( )cos ] cos( )

( ) cos[( )cos ] cos( )

n

n

T x n x n

T x n x n

1

1

or ( ) cos( )cos sin( )sin

( ) cos( )cos sin( )sin

n

n

T x n n

T x n n

Adding

1 1 2( ) ( ) cos( )cosn nT x T x n 1 1 2( ) ( ) ( )n n nT x T x xT x 1 1or 2 0( ) ( ) ( )n n nT x xT x T x

Using the recurrence relation


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g

and the definition

we obtain

1 12 0( ) ( ) ( ) ,n n nT x xT x T x

1 where( ) cos( cos ) cos cos .nT x n x n x 0 ( )T x 1 1( )T x x

2 1 02( ) ( ) ( )T x xT x T x

1 1or 2( ) ( ) ( )n n nT x xT x T x

22 1x 3 2 12( ) ( ) ( )T x xT x T x 34 3x x4 3 22( ) ( ) ( )T x xT x T x 4 28 8 1x x 5 4 32( ) ( ) ( )T x xT x T x 5 316 20 5x x x

Graphs of Chebyshev Polynomials


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-1 -0.5 0.5 1

x

-1

-0.5

0.5

1

Tn x

T2

T1

T0

G ap s o C ebys ev o y o a s

20 1 21 2 1( ) ( ) ( )T x T x x T x x

Graphs of Chebyshev Polynomials


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-1 -0.5 0.5 1

x

-1

-0.5

0.5

1

Tn x

T5

T4

T3

T2

T1

T0

p y y

0 1

2 3

2 34 2 5 3

4 5

1

2 1 4 38 8 1 16 20 5

( ) ( )

( ) ( )( ) ( )

T x T x x

T x x T x x xT x x x T x x x x

Problem:


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Express the polynomial in terms of Chebyshev

polynomials of first kind

4 35 3x x

Solution:

We have 0 12 3

2 3

4 2 5 3

4 5

1

2 1 4 3

8 8 1 16 20 5

( ) ( )

( ) ( )

( ) ( )

T x T x x

T x x T x x x

T x x x T x x x x

Therefore1( )x T x

4 34 3 2 1 0

15 3 [5 2 20 18 15 ]

8x x T T T T T

22 2 0

1 11

2 2[ ( ) ] [ ( ) ( )]x T x T x T x

33 3 1

1 13 3

4 4

[ ( ) ] [ ( ) ( )]x T x x T x T x 4 2

4 4 2 0

1 18 1 4 5

8 8[ ( ) ] [ ( ) ( ) ( )]x T x x T x T x T x


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Exercise: Show that 120

1 1 2( ) ( )n

nn

xt xt t t T x

12Generating function of Chebyshev Polynomials

1 1 2( )xt xt t

Orthogonality of Chebyshev polynomials


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g y y p y

1

21

if 0

1

if 02

0 if

( ) ( ),

,

,

n mT x T xdx m n

x

m n

m n

Proof:Case I: m = n = 0

1 1

2 21 1

1

1 1

( ) ( )n mT x T x dx dx

x x


88/88

Case II: m = n 0

1 1 12 2 1

2 2 21 1 11 1 1

( ) ( ) [ ( )] cos ( cos )n m nT x T x T x n x

dx dx dxx x x

Substitute , then obtain1cos x t 1 22

12

1

[ ( )]nT x dx

x

Case III: m n

1 1 1 1

2 21 11 1

( ) ( ) cos( cos )cos( cos )n mT x T x n x m xdx dx

x x

Substitute , then obtain1cos x t 12

0( ) ( )n mT x T x dx

L-6 de Series Solution

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