Legendre functions

Legendre Functions

SOLO HERMELIN

Updated: 20.02.13

1

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Table of Content

SOLO

2

Legendre Functions

IntroductionLegendre Polynomials HistorySecond Order Linear Ordinary Differential Equation (ODE)Laplace’s Homogeneous Differential Equation

Legendre PolynomialsThe Generating Function of Legendre Polynomials

Rodrigues' Formula

Series Solutions – Frobenius’ MethodRecursive Relations for Legendre Polynomial Computation

Orthogonality of Legendre Polynomials

Expansion of Functions, Legendre Series

Schlaefli IntegralLaplace’s Integral Representation

Neumann Integral

Table of Content (continue)

SOLO

3

Legendre Functions

Associate Lagrange Differential EquationLaplace Differential Equation in Spherical Coordinates

Analysis of Associate Lagrange Differential Equation

Associate Lagrange Differential Equation (2nd Way)

Generating Function for Pnm (t)

Recurrence Relations for Pnm (t)

Orthogonality of Associated Legendre FunctionsRecurrence Relations for Θn

m FunctionsSpherical Harmonics

References

Boundary Value Problems and Sturm–Liouville Theory

SOLO

4

Legendre Polynomials

Adrien-Marie Legendre(1752 –1833(

In mathematics, Legendre functions are solutions to Legendre's differential equation:

011 2

xPnnxP

xd

dx

xd

dnn

They are named after Adrien-Marie Legendre. This ordinary differential equation is frequently encountered in physics and other technical fields. In particular, it occurs when solving Laplace's equation (and related partial differential equations) in spherical coordinates.

The Legendre polynomials were first introduced in 1785 by Adrien-Marie Legendre, in “Recherches sur l’attraction des sphéroides homogènes”, as the coefficients in the expansion of the Newtonian potential

'md

r

'r

0

222

cos'1

cos'

2'

1

11

cos'2'

1

'

1

nn

n

Pr

r

r

rr

rrrrrrrrr

Return to Table of Content

Second Order Linear Ordinary Differential Equation (ODE)

Consider a Second Order Linear Ordinary Differential Equation (ODE) define by the Operator

SOLO

xuxpxd

udxp

xd

udxpxu 212

2

0: L

defined in the interval a ≤ x ≤ b, with the coefficients p0 (x), p1 (x), p2 (x) real in this region and with the first 2 – i derivatives of pi (x) continuous . Also we require that p0 (x) is nonzero in the interval.Define the Quadratic Form of the Operator L as:

dxuupxd

udp

xd

udpdxxuxuxu

b

a

b

a

212

2

0: LLu,

b

a

b

a

b

a

b

a

b

a

b

a

b

a

b

a

b

a

b

a

b

a

b

a

b

a

b

a

xdupuxdupxd

duupudxup

xd

duup

xd

du

xd

udup

xdupuxdupxd

duupudx

xd

udup

xd

d

xd

udup

xdupxduxd

udpxdu

xd

udp

21102

2

00

21100

2212

2

0

5

SOLO

dxuupxd

udp

xd

udpdxxuxuxu

b

a

b

a

212

2

0: LLu,

b

a

b

a

b

a

b

a

b

a

xuxd

pdpxu

upuxd

udupu

xd

pdu

xd

udupupuup

xd

du

xd

udup

01

100

0100

Therefore:

b

a

b

a

dxupupxd

dup

xd

duxu

xd

pdpxu 2102

20

1

Define the Adjoint Operator:

upupxd

dup

xd

dxu 2102

2

: L

6


SOLO

dx

u

upxd

udp

xd

udpuxu

b

a

L

Lu, 212

2

0

The two Integrals are equal if:

b

a

b

a

dxupupxd

dup

xd

duxu

xd

pdpxu

uL

2102

20

1

or:

xuxqxd

xudxp

xd

dxuxu

LL

02 101

20

2

2102

2

212

2

0

xd

udp

xd

pduu

xd

pd

xd

pdu

upupxd

dup

xd

duup

xd

udp

xd

udpu

xd

xpdxp 0

1

If this condition is satisfied, then the terms at the boundary x = a and x = b also vanish, and we have by defining p (x): = p0 (x) and q (x) := p2 (x)

7


SOLO

xuxqxd

xudxp

xd

dxuxu

LL

This Operator is called Self-Adjoint

xuxp

xd

udxp

xd

udxptd

tp

tp

xpxutd

tp

tp

xp xx

212

2

00

1

00

1

0

expexp1

L1

xutdtp

tp

xp

xp

xd

udtd

tp

tp

xd

d

xq

x

xp

x

0

1

0

2

0

1 expexp This is clearly Self-Adjoint.

We can see that p0 (x) is in the denominator. This is the reason that we requested the p0 (x) to be nonzero in the interval a ≤ x ≤ b. p0 (x1) = 0 means that the Differential Equation is not Second Order at that point.

We can always transform the Non-Self-Adjoint Operator to a Self-Adjoint one

by multipling L by

x

tdtp

tp

xp 0

1

0

exp1

8


Consider a Second Order Linear Ordinary Differential Equation (ODE)

SOLO

2

2

210 :'':'0''':xd

udu

xd

uduxuxpxuxpxuxpxu L

If we know one Solution u1 (x) we can find a second u2 (x).

Proof

If we have two Solutions u1 (x) and u2 (x), then

0'''

0'''

222120

121110

upupup

upupup

Multiply first equation by u2 (x) and second by u1 (x) and subtract:

0'''''' 1221112210 uuuupuuuup

The Wronskian W is defined as:

122121

21 ''''

: uuuuuu

uuW

0' 10 WpWp

9

Theorem: “Method of Reduction of Order”


Consider a Second Order Linear Ordinary Differential Equation (ODE)

SOLO

2

2

210 :'':'0''':xd

udu

xd

uduxuxpxuxpxuxpxu L

Theorem: “Method of Reduction of Order”

If we know one Solution u1 (x) we can find a second u2 (x).

Proof (continue -1)

0' 10 WpWp xdp

p

W

Wd

0

1

x

x

dp

pxWxW

0 0

10 exp

From: xuxuxuxuxW 1221 ''

q.e.d.

Therefore:

x

x

du

Wxuxu

0

21

12

1

222

1

1

1

22

1

''

u

u

xd

dxu

xu

xu

xu

xu

xu

xWDivide by u1

2 (x)

x0 and W (x0) are given (or not).

10


If the Second Order Linear Ordinary Differential Equation (ODE) is in the Self-Adjoint Mode:

SOLO

Then:

The Wronskian is given by

0

xuxq

xd

xudxp

xd

dxuxu LL

0

1

0

2

2

xuxqxd

xud

xd

xpd

xd

xudxp

p

p

xpxp

xW

p

pdxWd

p

pxWxW

x

x

x

x

000

0

0

00

0

10

00

expexp

11



12

SOLO

The Laplace’s Homogeneous Differential Equation is:

We want to find the Potential Φ at the point F (field) due to all the sources (S) in the volume V, including its boundaries .

n

iiSS

1

iS

nS

n

iiSS

1

dV

dSn

1

V

Fr

Sr

F

0rSF rrr iS

nS

dV

dSn

1

V

Fr

Sr

F

0r SF rrr

F inside V F on the boundary of V

Therefore is the vector from S to F.SF rrr

zzyyxxr 111

zzyyxxr SSSS 111

zzyyxxr FFFF 111

Let define the operator

that acts only on the source coordinate.

zz

yy

xx SSS

S 111

Sr

Laplace’s Homogeneous Differential Equation

02 rPierre-Simon,

marquis de Laplace (1749 1827)

http://en.wikipedia.org/wiki/File:Pierre-Simon_Laplace.jpg

13

SOLO Laplace’s Homogeneous Differential Equation

Laplace Differential Equation in Spherical Coordinates

02 r

Spherical Coordinates:

cos

sinsin

cossin

rz

ry

rx

r

y

x

z

A Solution in Spherical Coordinates is: 01

rr

r

0111

22 r

r

r

rr

rrr

0033111

353332

r

rrr

rr

rr

rr

rrr

r

r

r

z

r

y

r

xzyx

zyxr zyxzyx

111111 222

3111111

zyxzyx

r zyxzyx

2222 zyxr

14




cos

sinsin

cossin

rz

ry

rx

r

y

x

z

cos

z

r

2222 zyxr

0cos11

22

rrz

r

rrzz

01cos3331

3

2

5

22

4332

2

rrr

rz

z

r

r

z

rr

z

zz

0cos9cos159153

526

34

23

7

23

6

22

55

22

3

3

rr

r

r

rzz

z

r

r

rz

r

zr

rz

r

rz

zz

We note that ∂nΦ/∂zn is a n-degree polynomial in cos θ divided by rn+1.

zyxzn

hzyxz

hzyxz

hzyxhzyxn

nnn ,,

!

1,,

!2

1,,,,,,

2

22

Using Taylor’s Series development we obtain

15




cos

sinsin

cossin

rz

ry

rx

r

y

x

z

2222 zyxr

rzn

hrz

hrz

hrhzyx

hzyxn

nnn 1

!

11

!2

1111,,

2

22

222

0

22

1

!

1

cos2

1

nn

nnn

rznh

hhrr or:

Let define:

rznrP

n

nnn

n

1

!

1:cos 1

rhPr

h

rhhrr nn

n

022

cos1

cos2

1

Therefore:

16




cos

sinsin

cossin

rz

ry

rx

r

y

x

z

2222 zyxr

Let define w:=Pn (cos θ) and substitute w/rn+1 in the Laplace Equations in Spherical Coordinates instead of Φ

Therefore, since w:=Pn (cos θ) (Partial Derivative becomes Total Derivative):

0sin

1sin

sin

11

0

12

2

221212

2

nnn r

w

rr

w

rr

w

rr

rr

0sinsin

1111

w

rr

n

rw

nnor:

0sinsin

11

d

wd

d

dwnn

Substitute t=cos θ (dt = - sinθ dθ):

td

wdt

td

d

td

wdt

td

d

d

td

td

wd

d

td

d

d

d

wd

d

d 2

sin

2

1sin

11sin

1sin

sin

1sin

sin

1

2

Finally we obtain: 1011 2

twnn

td

wdt

td

d


17

SOLO

This is the Legendre Differential Equation and Pn (t) the Legendre Polynomialsare one of the two solutions of the ODE.

1011 2

ttPnntP

td

dt

td

dnn

02

cos

cos21

1

nn

n

Pr

h

rh

rh

We found

1coscos21

1

02

uPuuu n

nn

For u:=h/r

The left-hand side is called “The Generating Function of Legendre Polynomials”


The Generating Function of Legendre Polynomials

18

SOLO

Let use Taylor expansion for the function:

nn

xn

fx

fx

ffxf

!

0

!2

0

1

00 2

21

1,21

1

02

tutPutuu n

nn

101 2

1

fxxf

2

101

2

1 12

31 fxxf

2

3

2

101

2

3

2

1 22

52 fxxf

!2

!2

!2

!2

2

1231

2

12

2

3

2

101

2

12

2

3

2

12

2

12

n

n

n

nnnfx

nxf

nn

n

nn

nn

Tacking x = 2 u t - u2 we obtain

12!2

!2

21

1

0

2222

uutun

n

utu n

n

n

Let prove that Pn (t) are indeed Polynomials.



19

SOLO

Using Taylor expansion we obtained:

1,21

1

02

tutPutuu n

nn

Take the binomial expansion of (2 u t - u2)n we obtain

12!2

!2

21

1

0

2222

uutun

n

utu n

n

n

12

!!!2

!212

!!

!1

!2

!2

21

1

0 02

0 0222

uutknkn

nut

knk

n

n

n

utu n

n

k

knkn

n

k

n

n

k

kknk

n

Change Variables in the second sum from n+k to n

evennifn

oddnifnnuut

knknk

kn

utu n

n

k

nkn

kn

k

2/1

2/

212

!2!!2

!221

21

1

0

2/

0

2

222

Equating in the two power series the un coefficients, we obtain

2/

0

2 1!2!!2

!221

n

k

knn

kn tt

knknk

kntP

Polynomialof order n in t

Return to Rodrtgues Formula

Return to Frobenius Series

We can see that for n odd the polynomial Pn (t) has only odd powers of t and for n even only even powers of t.



20

SOLO

Let use Taylor expansion for the function:

nn

xn

fx

fx

ffxf

!

0

!2

0

1

00 2

21

1,21

1

02

tutPutuu n

nn

Tacking x = u2-2 u t we obtain

101 2

1

fxxf

2

101

2

1 12

31 fxxf

2

3

2

101

2

3

2

1 22

52 fxxf

2

12

2

3

2

1101

2

12

2

3

2

11 2

12

n

fxn

xf nnn

nn

0

244

33

220

4232222

2

8

33035

2

35

2

131

2128

352

16

52

8

32

11

21

1

nn

n tPutt

utt

ut

utuu

tuutuutuutuuttuu



SOLO

21


The first few Legendre polynomials are:

256/63346530030900901093954618910

128/31546201801825740121559

128/35126069301201264358

16/353156934297

16/51053152316

8/1570635

8/330354

2/353

2/132

1

10

246810

3579

2468

357

246

35

24

3

2

xxxxx

xxxxx

xxxx

xxxx

xxx

xxx

xx

xx

x

x

xPn n

http://en.wikipedia.org/wiki/File:Legendrepolynomials6.svg

22

SOLO

1,21

1

02

tutPuutu n

nn

Substitute u by – u in this equation:

1121

1

002

utPutPuutu n

nn

nn

nn

which results in the following identity: tPtP nn

n 1

For t =1 we have 111

1

21

1

02

uPuuuu n

nn

But 11

1

0

uuu n

n

By equalizing the coefficients of un in the two sums, we obtain:

nPn 11

and nnn

n PP 1111



23

SOLO

1,21

1

02

tutPuutu n

nn

For t = 0 we obtain:


101

1

02

uPuu n

nn

022

2

0

1

2

0

2

22222/12

2

!2

!21

!2

222642

!2

125311

!2

125311

!

2/123/12/1

!2

3/12/1

2

111

1

1

nn

nn

nnn

nn

nn

nn

n

n

un

n

nn

n

un

n

un

tn

nttu

u

Therefore we have:

10!2

!21

1

1

0022

2

2

uPun

un

u nn

n

nn

nn

By equating coefficients of un on both sides we obtain:

00

!2

!210

12

222

n

n

nn

P

n

nP



Derivation of Legendre Polynomials via Rodrigues’ Formula

SOLO

24


Benjamin Olinde Rodrigues (1794-1851)

In mathematics, Rodrigues' Formula (formerly called the Ivory–Jacobi formula) is a formula for Legendre polynomials independently introduced by Olinde Rodrigues (1816), Sir James Ivory (1824) and Carl Gustav Jacobi (1827). The name "Rodrigues’ formula" was introduced by Heine in 1878, after Hermite pointed out in 1865 that Rodrigues was the first to discover it, and is also used for generalizations to other orthogonal polynomials. Askey (2005) describes the history of the Rodrigues’ Formula in detail.

Rodrigues stated his formula for Legendre polynomials Pn

Carl Gustav Jacob Jacobi (1804 –1851)

n

n

n

nn xxd

d

nxP 1

!2

1 2

SOLO

25


Olinde Rodrigues (1794-1851)

Start from the function: .12 constkxkyn

12 12:'

n

xxknxd

ydy

222122

2

11412:''

nn

xxnknxknxd

ydy

Let compute:

'12211412''112222 yxnynxxnknxknyx

nn

or: 02'12''12 ynyxnyx

Let differentiate the last equation n times with respect to x:

''1''2''1

00''13

''12

''11

''1''1

2

2

1

12

3

3

0

23

3

2

22

2

2

1

1222

yxd

dnny

xd

dxny

xd

dx

yxd

dx

xd

dny

xd

dx

xd

dny

xd

dx

xd

dny

xd

dxyx

xd

d

n

n

n

n

n

n

n

n

n

n

n

n

n

n

n

n

''12'1

'12'121

1

1

1

yxd

dny

xd

dxny

xd

dx

xd

dny

xd

dxnyx

xd

dn

n

n

n

n

n

n

n

n

n

n


SOLO

26



Start from the function: .12 constkxkyn

02'12''12 ynyxnyxDifferentiate n times with respect to x:

''1''2''12

2

1

12 y

xd

dnny

xd

dxny

xd

dx

n

n

n

n

n

n

02''121

1

yxd

dny

xd

dny

xd

dxn

n

n

n

n

n

n

Define: a Polynomial n

n

n

n

n

xxd

dk

xd

ydxw 1: 2

02'121'2''12 wnwnwxnwnnwxnwx

02121'2''12 wnnnnnwxnxxnwx

01'2''12 wnnwxwx

This is Legendre’s Differential Equation. We proved that one of the solutions are Polynomials. We can rewrite this equation in a Sturm-Liouville Form:

0112

wnnw

xd

dx

xd

d


SOLO

27



Let find k such that:

by using the fact that Pn (1) = 1

n

n

n

n

n

n xxd

dk

xd

ydxP 12

0

22 1!2111i

inn

v

n

u

n

n

nn

n

n

n xxaxnkxxxd

dkxk

xd

dxP

!2

1

nk

n

We recover the Rodrigues Formula: n

n

n

nn xxd

d

nxP 1

!2

1 2

Let use Leibnitz’s Rule (Binomial Expansion for the n Derivative of a Product - with u:=(x-1)n and v:=(x+1)n ):

udvudvdnvddunn

vddunvdu

vdudmnm

nvud

nnnnn

n

m

mnmn

1221

0

!2

1

!!

!

We have:

1!2

!2

11

1!20

12

0

21

00

nkudvudvdnvddunn

vddunvdukxP n

xn

nnnnnn

n

We can see from this Formula that Pn (x) is indeed a Polynomial of Order n in t.


SOLO

28



Let find an explicit expression for Pn (x) from Rodrigues’ Formula:

n

n

n

nn

n

n xxd

d

nxd

ydxP 1

!2

1 2

Start with:

n

m

mnmnx

mnm

nx

0

22 1!!

!1

oddnn

evennnp

xknknk

kn

xnm

m

mnm

xnmmmmnm

n

nx

xd

d

nxP

p

k

knk

n

knm

n

pm

nmmn

n

n

pm

nmmn

n

n

n

n

nn

2/12

2/

!2!!

!221

2

1

!2

!2

!!

11

2

1

12122!!

!1

!2

11

!2

1

0

2

2

22

2/0

2/121222

nm

nmnmmmx

xd

d mn

n

We recover the result obtained by the Generating Function of Legendre Polynomials

Return to Frobenius Series



SOLO

Series Solutions – Frobenius’ Method

Ferdinand Georg Frobenius

(1849 –1917)

.01212

222 constrealryrr

xd

ydyx

xd

ydx

Example: General Legendre Equation

0

kxay

Let check the Frobenius’s expansion:

We have:

0

22

2

0

1 1&

kk xkka

xd

ydxka

xd

yd

Substitute in the General Legendre Equation:

0111

121

00

2

000

2

kk

kkkk

xkkrraxkka

xrraxkaxxkka

29


SOLO

Example: Legendre Equation (continue – 1)

011100

2

kk xkkrraxkka

Denote, in the first sum λ = j +2 and in the second sum λ = j, to obtain:

01112

11

02

11

20

j

jkjj

kk

xjkjkrrajkjka

xkkaxkka

All the coefficients of xk+j must be zero, therefore

001 00 akka

011 kka

011122 jkjkrrajkjka jj

30

Ferdinand Georg Frobenius

1849 - 1917



SOLO


,2,1,0

12

1

12

112

jjkjk

jkrjkra

jkjk

jkjkrraa jjj

01 kk

011 kka

011122 jkjkrrajkjka jj

0&1.3

0&0.2

0&0.1

1

1

1

ak

ak

akThree possiblesolutions

The equation that, k (k+1) = 0, comes from the coefficient of the lowest power of x, and is called Indicial Equation. It has two solutions for k

k = 0 and k = 1

The equation

gives the recursive relation

31



SOLO


1001: korkkkEquationIndicial

1. Using k = 0 and a1 = 0 we obtain a series of even powers of x

,2,1,0

21

1

21

112

jjj

jrjra

jj

jjrraa jjj

42

0 !4

312

!2

11: x

rrrrx

rraxpxy reven

01231 naaa

The recurrence relation results in the following expression for the coefficients

,3,2,1

!2

1231242221 02

ma

m

mrrrrrmrmra m

m

32

,2,1,0

12

1

12

112

jjkjk

jkrjkra

jkjk

jkjkrraa jjj

0

22m

mmeven xaxy



SOLO

Example: Legendre Equation (continue – 4) 1001: korkkkEquationIndicial

3. Using k = 1 and a1 = 0 we obtain a series of odd powers of x

.2,1,0

32

21

32

2112

j

jj

jrjra

jj

jjrraa jjj

53

0 !4

4213

!3

21: x

rrrrx

rrxaxqxy rodd

01231 naaa


,3,2,1

!12

242132121 02

mam

mrrrrmrmra m

m

33Since pr (x) and qr (x) are two linearly independent solutions of the 2nd Order Linear Lagrange ODE, the final solution is y = c1 pn (x) + c2 qn (x)

,2,1,0

12

1

12

112

jjkjk

jkrjkra

jkjk

jkjkrraa jjj

0

122m

mmodd xaxy



SOLO


Using d’Alembert – Cauchy test for convergence of an Infinite Series, we have

Convergence Test:

divergex

convergexxx

jkjk

jkjkrr

xa

xajj

j

jj

j 11

11

12

11limlim 22

22

The even series stops at j = n. The expansion is a Polynomial of order n (even).

1. If n is even, using k = 0 and a1 = 0 we obtain a series of even powers of x .

,..1,0

21

112

jjj

jjnnaa jj

For the case that r = n, a positive integer:

,...2,1,,02 nnnja j

3. If n is odd, using k = 1 and a1 = 0 we obtain a series of odd powers of x .

,..2,1

32

2112

jjj

jjnnaa jj

,...2,1,,1,02 nnnnja j

The odd series stops at j = n-1. The expansion is a Polynomial of order n (odd).34



SOLO


1. If n is even, using k = 0 and a1 = 0 we obtain a series of even powers of x

42

0 !4

312

!2

11 x

nnnnx

nnaxyeven

3. If n is odd using k = 1 and a1 = 0 we obtain a series of odd powers of x

53

0 !4

4231

!3

21x

nnnnx

nnxaxyodd

420

420

20

20

0

6

70101

!4

3414244

!2

14414

31!2

12212

0

xxaxxayn

xaxayn

ayn

even

even

even

30

30

0

3

5

!3

23133

1

xxaxxayn

xayn

odd

odd

We obtain the Legendre Polynomials Solutions for a0 = 1. 35



SOLO

Example: Legendre Equation (continue – 7)1. The recurrence relation for even x powers results in the following expression for the coefficients

2/,,3,2,1

!2

1231242221 02 nma

m

mnnnnnmnmna m

m

!!

2121224222 112

mp

pppmpmpnnmnmn mm

pn

!!22

!!22

212

1

!2

!22

242

2421231

!

!1231

22

mpp

pmp

mpppp

mp

mnnn

mnnnmnnn

n

nmnnn

m

pn

m

pn

2/,,3,2,1

!!!22

!221

!!!2!2

!2

!22

1!!2!2

!!22

!

!

2

11

0

0

2

202 nmsnssn

sna

snssnn

nsn

ampmp

pmp

mp

pa

n

sthatsuchachoose

snmps

mm

2/

0

2 1!2!!2

!221

n

s

snn

seven xx

snsns

sny 36



SOLO

Example: Legendre Equation (continue – 8)3. The recurrence relation for odd x powers results in the following expression for the coefficients

2

1,,3,2,1

!12

242132121 02

n

mam

mnnnnmnmna m

m

!!

222242222213212 112

mp

pppmpmpnmnmn m

pn

!!122

!!122

212

2242221225232

!12

!12242

1

1

12

mpp

pmp

mppp

mpppmppp

p

pmnnn

m

m

pn

2

1,,3,2,1

!2!1!!

!2

1!122

1!12!!!12

!!1221 0

2

0

2

2

nma

snsnsn

nsn

ammpmpp

pmpa sp

mpsm

m

37

2

1,...,1,0

!2!!!2

!2

1!22

1!2!12!!

!2

1!12222

1 0

2

0

2

n

sasnsnsn

nsn

asnsnsnsn

nsnsn

spmps

spmps



Example: Legendre Equation (continue – 9)3. The recurrence relation for odd x powers results in the following expression for the coefficients

2

1,...,1,0

!2!!2

!221

!2!!

!22!

2

1

!2

11

0

0

2

2

1

2

n

ssnsns

sna

snsns

snn

na

n

sthatsucha

Choose

sn

m

1&

!2!!

!221

2

1 2/1

0

2

xoddnxsnsns

snxy

n

s

sns

nodd

We also found that the solution for k = 0 and a1 = 0 is

1&

!2!!2

!221

2/

0

2

xevennxsnsns

snxy

n

s

snn

seven

We recover the result obtained by the Generating Function of Legendre Polynomialsand by Rodrigues’ Formula 38

Therefore we can unify those two relations to obtain:

1

!2!!2

!221

2/

0

2

xxsnsns

snxP

n

s

snn

sn


SOLOLegendre Polynomials

SOLO


2. Using k = 0 and a1 ≠ 0 we obtain an infinite series and not a polynomial. This is the Second Solution of the Legendre Differential Equation. The solution is the sum of the two infinite series, one with even powers of x and the other with odd powers of x. The series solution, in this case,

diverges at x = ± 1.

Those are Legendre Functions of the Second Kind.The Polynomial solutions are Legendre Functions of the First Kind.

39

,2,1,0

21

1

21

112

jjj

jnjna

jj

jjnnaa jjj

In this case we have


,3,2,1

!2

1231242221 02

ma

m

mnnnnnmnmna m

m

,3,2,1

!12

1231242221 112

mam

mnnnnnmnmna m

m



SOLO


40

We want to find a series that converges for |x| > 1. Let return to the conditions to have a series solution for Legendre ODE

For k = 0 and a0 = 0

By substituting j = m – 2 we obtain

,2,1,0

1

212

jajnjn

jja jj

mm a

mnmn

mma

12

12

.01212

222 constrealryrr

xd

ydyx

xd

ydx

,2,1,0

21

1

21

112

jjj

jrjra

jj

jjrraa jjj

We can make am+2, am+4, am+6,…. To vanish for m = r = n or m = -n – 1. If we start for am ≠ 0 we can obtain the following recursive formula



SOLO


41

mm a

mnmn

mma

12

12

Tacking m = n we obtain

nnn

nn

ann

nnnna

n

nna

an

nna

321242

321

324

32

122

1

24

2

The first solution can be written as

innnn

n xinnni

ininnnx

nn

nnnnx

n

nnxaxy 242

1 1223212242

1221

321242

321

122

1


divergex

convergexxx

ini

inin

xa

xaiin

in

inin

j 11

11

1222

122limlim 22

2222

22



SOLO


42

mm a

mnmn

mma

12

12

Tacking m =- n - 1 we obtain

135

13

523242

4321

524

43

322

21

nnn

nn

ann

nnnna

n

nna

an

nna

The second solution can be written as

1253112 1225232242

21221

523242

4321

322

21 innnnn x

innni

ininnnx

nn

nnnnx

n

nnxaxy


divergex

convergexxx

ini

inin

xa

xaiin

in

inin

j 11

11

1222

212limlim 22

1212

1212



SOLO


43

!

1353212

n

nnan

By choosingwe have

1]

1223212242

12211

321242

321

122

1[

!

1353212

2

421

xxinnni

ininnn

xnn

nnnnx

n

nnx

n

nnxyxP

ini

nnnn


Finally 1

!2!!2

!221

0

2

xxinini

inxP

i

inn

in

!2!!2

!221

!2

!22

!2!2

11

!2!2

1351221

1223212242

12211

!

1353212

inini

in

in

in

iiniin

in

innni

ininnn

n

nn

n

i

ini

i

i

i

i



SOLO


44

1351212

!1 nn

naa nn

By choosingwe have

1]

1225232242

21221

523242

4321

322

21[

1351212

!

12

5312

xxinnni

ininnn

xnn

nnnnx

n

nnx

nn

nxyxQ

in

nnnn

!122!

!2!

!

!12

!

!2

!122

!12

!

!2

2

1

!122

2222!12

!

!2

!2

1

1225232242

21221

2

ini

inin

n

n

n

in

in

n

n

in

in

innn

n

in

iinnni

ininnn

i

i

i

1!122!

!2!2

0

12

xxini

ininxQ

i

innn

Go to Neumann Integral

!12

!2!

1351212

!

n

nn

nn

n n

Finally



SOLO


45

Summarize

1!122!

!2!2

0

12

xxini

ininxQ

i

innn


1

!2!!

!221

2

1

0

2

xxinini

inxP

i

ini

nn

1

!2!!

!221

2

1 2/

0

2

xxinini

inxP

n

i

ini

nn

Return toSimilar to Rodrigues Formula

Using Frobenius’ Method we found that solutions of Legendre ODE

are

integerpositive01212

222 nynn

xd

ydyx

xd

ydx



4,3,2,1

1

l

xxPl 4,3,2,1

1

l

xxQl

46

SOLO

121

1

02

utPuutu n

nn

For u=0 we obtain 10 tP

For t = 1 we obtain 1111

1

21

1

002

n

nn

n

n

n PPuuuuu

For t = -1 we obtain nnn

nn

n

nn PPuuuuu

11111

1

21

1

002

Let find a Recursive Relation for Legendre Polynomial computation

Start with:

11

1

1

11

21 cos

1

11

!

1

1

11

!1

1cosn

nn

nn

n

nn

nn

r

P

znrznnrznr

P

zd

Pd

rz

r

r

Pn

nr

P nnn

nn

n cos

cos

cos1cos1

1

1cos122

1

Recursive Relations for Legendre Polynomial ComputationFirst Recursive Relation


47

SOLO

Recursive Relations for Legendre Polynomial Computation

zd

Pd

rz

r

r

Pn

nr

P nnn

nn

n cos

cos

cos1cos1

1

1cos122

1

cosrz cos

z

r

zr

z

r

z

z

coscos1

cos

rz

2cos1cos

rd

Pd

rr

Pn

nr

P nnn

nn

n 2

1221 cos1

cos

cos1cos

cos1

1

1cos

cos

cos

1

cos1coscoscos

2

1 d

Pd

nPP n

nn

Substituting t = cos θ we obtain

td

tPd

n

ttPttP n

nn 1

1 2

1

First Recursive Relation (continue – 1)


48

SOLO


0,1

1

1 2

1

nttd

tPd

n

ttPttP n

nn

Use to start 01 0

0 td

tPdtP

8

157063

8

33035

2

35

2

13

35

5

24

4

3

3

2

2

1

ttttP

tttP

tttP

ttP

ttP

First Recursive Relation (continue – 2)


http://en.wikipedia.org/wiki/File:Legendrepolynomials6.svg

49

SOLO


Start from 121

1:,

02

utPuutu

tugn

nn

Let differentiate both sides with respect to u and rearranging

1

12

221

21 nn

n tPunutuutu

ut

u

g

1

12

0

21n

nn

nn

n tPunututPuut

1

11

0

1 2n

nnnn

nn

nn tPunutnuntPuut

2

110

11

10

121n

nn

nn

n

nn

n

nn

n

nn

n tPuntPutntPuntPutPut

2112

122

0

11

1201

10

ntPuntPuntPutn

ntPtutPutPutuPt

ntPtPt

nn

nn

nn

We can see that the last relation agrees also with the previousrelations, for n = 0 and n = 1.

Second Recursive Relation


50

SOLO


We find the Recursive Relation:

111

1211

ntPn

ntPt

n

ntP nnn

10 tPThis is called the Bonnet’s Recursion Relation. It starts with:

Examples:

2

31 01

2

tPtPttPn

2

1

2

3 22 ttP

tP

tP

ttttPn1

2

3

2

2

1

2

3

3

52 2

3

2

35 3

3

tttP

tPtP

tttttPn

23

2

1

2

3

4

3

2

3

2

5

4

73 23

4

8

33035 24

4

tttP

ttP 1

Second Recursive Relation (continue)


51

SOLO


Start from 121

1:,

02

utPuutu

tugn

nn

Let differentiate both sides with respect to t and rearranging

02/3221 n

nn

td

tPdu

utu

u

t

g

0

2

0

21n

nn

nn

n

td

tPduututPuuor

Equaling coefficients of each power of u gives

td

tPd

td

tPdt

td

tPdtP nnn

n11 2

tPntPntPtn nnn 11112 Differentiate the Second Recursive Relation

with respect to t and rearranging

td

tPd

n

n

td

tPd

n

n

td

tPdttP nnn

n11

1212

1

Third Recursive Relation


52

SOLO


We found td

tPdttP

td

tPd

td

tPd nn

nn 211

td

tPdttP

td

tPd

n

n

td

tPd

n

n nn

nn

11

1212

1

Third Recursive Relation (continue)

Let solve for and in terms of and tPn

td

tPd n td

tPd n 1 td

tPd n 1

td

tPdttPn

td

tPd nn

n 1

td

tPdttPn

td

tPd nn

n 11

Subtracting the first relation from the second gives the Third Recursive Relation

td

tPd

td

tPdtPn nn

n1112


53

SOLO


yPxPyPxPyx

nyPxPk

tPtntPntd

tPdt

tPntPtntd

tPdt

tPntd

tPdt

td

tPd

tPntd

tPd

td

tPdt

tPntd

tPd

td

tPd

tPntPtntPn

tPn

ntP

n

ntPt

tPintd

tPd

nnnn

n

kkk

nnn

nnn

nnn

nnn

nnn

nnn

nnn

l

iin

n

110

12

12

11

1

11

11

11

12

1

012

1129

1118

17

6

5

124

01213

1212

12

1421

Recursive Relation between Legendre Polynomials and their Derivatives)



54

SOLO


Define tPwtPv nm :&:

We use Legendre’s Differential Equations:

011 2

vmm

td

vdt

td

d

011 2

wnn

td

wdt

td

d

Multiply first equation by w and integrate from t = -1 to t = +1.

0111

1

1

1

2

dtwvmmdtw

td

vdt

td

d

Integrate the first integral by parts we get

01111

1

1

1

2

0

1

1

2

dtwvmmdttd

wd

td

vdtw

td

vdt

t

t

In the same way, multiply second equation by v and integrate from t = -1 to t = +1. 011

1

1

1

1

2

dtwvnndt

td

wd

td

vdt


55

SOLO


0111

1

1

1

2

dtwvmmdt

td

wd

td

vdt

Subtracting those two equations we obtain

0111

1

1

1

2

dtwvnndt

td

wd

td

vdt

011111

1

1

1

dttPtPnnmmdtwvnnmm nm

This gives the Orthogonality Condition for m ≠ n

nmdttPtP nm

0

1

1

To find let square the relation and integrate between t = -1 to t = +1. Due to orthogonality only the integrals of terms having Pn

2(t) survive on the right-hand side. So we get

1

1

2 dttPn

0

221

1

nn

n tPuutu

0

1

1

221

1 221

1

nn

n dttPudtutu


56

SOLO


0

1

1

221

1 221

1

nn

n dttPudtutu

11

1ln

1

1

1ln

2

121ln

2

1

21

12

21

1

21

1 2

uu

u

uu

u

utuu

udt

tuu

t

t

0

11

0

1

0

1

11

1

11

1

11

11ln

11ln

1

n

uu

un

u

un

u

uu

uu

u

nnn

nn

nn

0

2

0

12

0

0

121212

0

12122

12

2

12

12

121

1

121

1 nnnn

nnn

n unn

u

un

uu

un

uu

u

Let compute first

Therefore

0

1

1

22

0

21

1 2 12

2

21

1dttPuu

ndt

utu nnn

Comparing the coefficients of u2n we get 12

21

1

2

ndttPn


nmmn ndttPtP

12

21

1

Hence

57

SOLO

Using Rodrigues’ Formula let calculate

1

1

21

1

1

!2

1dt

td

tdtP

ndttPtP

n

nn

knnk


n

nn

nn td

td

ntP

1

!2

1 2

Orthogonality of Legendre Polynomials (Second Method)

Assume, without loss of generality, that n > k, and integrate by parts

1

1

21

1 1

21

0

1

1

1

21

1

1

21

1

1!2

11

1

!2

11

!2

1

1

!2

1

dttd

tPdt

ndt

td

td

td

tPd

ntd

tdtP

n

dttd

tdtP

ndttPtP

nk

nn

n

n

n

nnk

nn

nn

kn

n

nn

knnk

Since Pk (t) is a Polynomial of Order k and we assume that n > k, we have

0n

kn

td

tPd

Therefore nkdttPtP nk

0

1

1

58

SOLO

For k = n we have

Legendre PolynomialsOrthogonality of Legendre Polynomials (Second Method) (continue – 1)

1

1

21

1

2 1!2

11 dt

td

tPdt

ndttP

nn

nn

n

nn

But we found that Pn (t) is given by:

2/

0

2

!2!!2

!221

n

k

knn

kn t

knknk

kntP

Therefore to compute it is sufficient to consider only the highest power of tin the series, i.e. for k = 0, and we obtain

n

nn

td

tPd

!2

!2!

!2

!22 n

nn

n

n

td

tPdnnn

nn

1

1

222

1

1

21

1

2 1!2

!2

!2

!21

!2

11 dtt

n

ndt

n

nt

ndttP

n

nn

n

n

nn


Orthogonality of Legendre Polynomials (Second Method) (continue – 2)

1

1

222

1

1

21

1

2 1!2

!2

!2

!21

!2

11 dtt

n

ndt

n

nt

ndttP

n

nn

n

n

nn

0

120

12cos1

1

2 sinsin1 dddtt nntn

Therefore

!12

!2cos

2222

!2

11212

!2sin

31212

2222sin

12

2sin

212

2

0

1

00

12

0

12

n

n

nn

n

nn

nd

nn

nnd

n

nd

nnnnn

0

1212

0

212

0

0

2

0

2

0

12 sinsin2cossin2cossincossinsin dndndd nnnnnn

,2,1,0

12

2

!12

!2

!2

!21

!2

!2 212

22

1

1

222

1

1

2

nnn

n

n

ndtt

n

ndttP

n

n

n

nn


59



Using Sturm-Liouville Theory it can be seen that the LegendrePolynomials that are Solution of the Legendre ODE, form an orthogonal and “Complete” Set, meaning that we can expand any function f (t) ,Piecewise Continuous in the interval -1 ≤ t ≤+1. Therefore we can define a series of Legendre Polynomials that converges in the mean to the function f (t)

110

ttPatfn

nn

tf

t1 1

The coefficients an can be defined using theOrthogonality Property of Legendre Polynomials

mn

m

mnnm am

tdtPtPatdtPtf

mn

12

2

0

2

12

1

1

1

1

1

12

12tdtPtf

ma mm

112

12

0

1

1

ttPtdtPtfn

tfn

nn

60



tf

t1 1

0t

0tf

0tf

At any discontinuous point t0 ( f(t0- ) ≠ f(t0+) ) we have

112

12

2

10

00

1

1

00

ttPtdtPtf

ntftf

nnn

If f (t) is defined in the interval –a ≤ t ≤ +a then

ataatPatfn

nn

0

/

a

a

nn tdatPtfa

na /

2

12

61



If f (t) is an odd function ( f(-t ) =- f(t) ) we have

1

0

1

0

1

0

1

1

0

0

1

1

1

2

0

12

2

oddndttPtf

evenndttPtfdttPtf

dttPtfdttPtfdttPtfan

n

n

tP

n

tf

n

tt

nnn

nn

If f (t) is an even function ( f(-t ) = f(t) ) we have

oddn

evenndttPtfdttPtfdttPtf

dttPtfdttPtfdttPtfan

nn

tP

n

tf

n

tt

nnn

nn 0

2

12

2

1

01

0

1

0

1

1

0

0

1

1

1

62



Using Rodrigues’ Formula let calculate

n

nn

nn td

td

ntP

1

!2

1 2

1

1

21

11

21

0

1

1

1

21

1

1

21

1

1!2

11

1

!2

11

!2

1

1

!2

1

tdtd

tfdt

ntd

td

tfd

td

td

ntd

tdtf

n

tdtftd

td

ntdtftP

n

nn

n

n

n

nn

nn

nn

n

n

nn

nn

1

1

21

1

21

1

1!2

11

!2

11 td

td

tfdt

ntd

td

tfdt

ntdtPtf

n

nn

nn

nn

n

nn

or

63



Example f (t) = tk ,|t| < 1

nktdttnkkkn

n

nk

tdtd

tdt

n

ntdtPt

na nkn

nn

knn

nnk

n1

1

21

1

1

21

1

1111

!2

12

0

1!2

12

2

12

We have

1

1

2

1

1

22222

1

1

422222

1

1

222122

0

1

1

122122

12

1

1

1

2222

1!2

!22

!2

!2

12222122

12222322122

1222122

322122

1122

1221

122

11

122

222

tdtnm

nm

mn

n

tdttmnnn

nmnmnmnm

tdttnn

nmnm

tdttn

nmtt

ntdtt

mn

nmnm

nmnmmn

nmn

nmnnmntu

tdtdv

nmnnm

n

Take k = 2m and since t2m is even, they are only even coefficients nonzero sowe take 2 n instead of n, and 2n ≤ 2m

64



Example f (t) = tk , |t| < 1 (continue – 1)

nmtdtt

nm

m

n

n

nm

a nmn

n

n1

1

222212

2 1!22

!2

!22

14

0

We have for f (t) = t2m

!122

!2

!2

!22

!2

!2

1!2

!22

!2

!21

2122

1

1

21

1

2222

nm

mn

nm

nm

mn

n

tdtnm

nm

mn

ntdtt

nm

nmnm

mn

nmnmnmn

!122!

!!2142

!122

!2

!2

!22

!2

!2

!22

!2

!22

14 22122

122

nmnm

nmmn

nm

nm

nm

nm

nm

n

nm

m

n

na

nnm

nmnmnn

!12

!21

2121

1

2

n

ndtt

nnWhere we used the previous result

Therefore

m

nn

nm tP

nmnm

nmmnt

02

22

!122!

!!2142

65



nktdttnkkkn

n

nk

tdtd

tdt

n

ntdtPt

na nkn

nn

knn

nnk

n1

1

21

1

1

21

1

1111

!2

12

0

1!2

12

2

12

We have

1

1

122122

1

1

222212

1

1

422322

1

1

222222

0

1

1

122222

12

1

1

1

22122

1!122

!2

!2

!22

!12

!12

112322222

12222322122

1322222

322122

1222

1221

222

11

122

2122

tdtnm

mn

nm

nm

mn

n

tdttmnnn

nmnmnmnm

tdttnn

nmnm

tdttn

nmtt

ntdtt

mnnm

nmnm

nmnmmn

nmn

nmnnmntu

tdtdv

nmnnm

n

Take k = 2m+1 and since t2m+1 is odd, they are only odd coefficients nonzero sowe take 2 n+1 instead of n, and 2n+1 ≤ 2m+1

66




nmtdtt

nm

m

n

n

nm

a nmn

n

n1

1

2212222

12 1!22

!2

!122

34

0

We have for f (t) = t2m+1

!322

!12

!2

!22

!12

!12

1!122

!2

!2

!22

!12

!121

2322

1

1

1221221

1

22122

nm

mn

nm

nm

mn

n

tdtnm

mn

nm

nm

mn

ntdtt

nm

nmnm

mnnm

nmnmnmn

!322!

!1!12342

!322

!12

!2

!22

!12

!12

!22

!2

!122

34 122322

2212

nmnm

nmmn

nm

nm

nm

nm

nm

n

nm

m

n

na

nnm

nmnmnn

!12

!21

2121

1

2

n

ndtt

nnWhere we used the previous result

Therefore

m

nn

nm tP

nmnm

nmmnt

012

1212

!322!

!1!12342

Return to Neumann Integral

67




Neumann Integral

11

1

111

02

12

012

2

01

0

x

t

x

t

x

t

x

t

x

t

xxtxtx m

m

m

mm

m

mm

m

m

m

Start from

Use

m

nn

nm tP

nmnm

nmmnt

012

1212

!322!

!1!12342

m

nn

nm tP

nmnm

nmmnt

02

22

!122!

!!2142

0 012

212

0 02

2122

012

212

02

122

0 0

212

12

0

12

02

2

!322!

!12!122342

!124!

!2!22142

!322!

!1!12342

!122!

!!2142

!322!

!1!12342

!122!

!!21421

n in

mn

n in

inninm

n nmn

mn

n nmn

mnOrder

SummationChange

m

m

n

mn

n

m

mm

nn

n

tPxnmi

ininntPx

ini

ininn

tPxnmnm

nmmntPx

nmnm

nmmn

xtPnmnm

nmmnxtP

nmnm

nmmn

tx

m

n

m

n 0

nm

mn

0 0m

m

n

00 0 n nmm

m

n

0n nm

ChangeSummation

Order

68



Neumann Integral (continue – 1)

012

0

212

02

0

2122

!322!

!12!122342

!124!

!2!221421

nn

i

mn

nn

i

inn

tPxnmi

ininntPx

ini

ininn

tx

We found

1!122!

!2!2

0

12

xxini

ininxQ

i

innn

Use the Frobenius Series development of Legendre Functions of the Second Kind Qn (x)

We have

0

12120

22 34141

nnn

nnn tPxQntPxQn

tx

0

121

nnn xttPxQn

tx

Return to Qn

Frobenius Series

69



Neumann Integral (continue – 2)We found

0

121

nnn tPxQn

tx

Multiply both sides by Pm (t) and integrate between -1 to +1

xQtdtPtPxQntdtx

tPm

n

n

mnnm

nm

2120

12

2

1

1

1

1

We obtain

1

12

1td

tx

tPxQ n

n

Franz Neumann's Integral of 1848

Franz Ernst Neumann (1798 –1895)


70

http://en.wikipedia.org/wiki/File:Franz_Ernst_Neumann_by_Carl_Steffeck_1886.jpg

SOLO

71


Schlaefli Integral

Start with

Using Rodrigues's Formula we obtain n

n

n

nn xxd

d

nxP 1

!2

1 2

td

zt

tf

jzf

2

1Cauchy's Integral

with nzzf 12

td

zt

t

jz

nn 1

2

11

22

Differentiate n times this equation with respect to z and multiply by 1/ (2n n!)

tdzt

t

jz

zd

d

n n

nnn

n

n

n 1

22 1

2

21

!2

1

with the contour enclosing the point t = z.

Schlaefli Integral

tdzt

t

jzP n

nn

n 1

2 1

2

2


SOLO

72


Laplace’s Integral Representation

Start with

220

1

20

22

02

022

2/tan

022

2

02

20

121

2

1

1tan

1

2

11

1

11

1

2

11

2

11

2

2/tan12/tan1

2/tan1

2/tan12/tan1

1cos1

t

t

td

t

td

tt

tdddd t

0

21

2

22

1

1

cos11cos111cos11

1

cos1

11

1

cos1

1

2

n

nn

xu

xu

xxuxuxxuxuxuxu

xu

xuxu

Let write

where we used

0

11n

naa

2222

2

22

1

1

2 21

1

11

1

1

11

1

1

1

2

uxu

xu

xuxu

xu

xu

xu

xu

xu

22

0 0

2

0 0

2

0 21

1

1cos11cos11

cos1 uxu

xudxxuxudxxuxu

d

n

nn

n

nn

SOLO

73


Laplace’s Integral Representation (continue - 1)

0

20 0

2

21cos1

nn

n

n

nn xPu

uxudxxu

We obtained

Equating un coefficients we obtain :

0

2 cos11

dxxxPn

n

0112

ynny

xd

dx

xd

d

If we replace in the Legendre ODE n by –n – 1

the equation does not change. Therefore , and xPxP nn 1

0

12 cos1

1dxxxP

n

n

Substitute x = cosθ

0

cossincos1

cos djP nn

Laplace’s First Integral

Laplace’s Second Integral

SOLO

74



Use the Generating Function

0

2/1221

1

nn

n tPuuut

Substitute t = cosθ and u = ejφ

0

2/12cos

cos21

1

nn

njjj

Peee

2/12/12/12/12/12 coscos2cos2cos21 jjjjjj eeeeeeBut

Therefore

2/12/

2/12/

0

coscos2

1

coscos2

1

cos

j

j

nn

nj

e

ePe

Equating the real and imaginary parts, we obtain

2/1

2/1

0

coscos2

2/sin2

coscos2

2/cos2

coscosn

nPn

2/1

2/1

0

coscos2

2/cos2

coscos2

2/sin2

cossinn

nPn

SOLO

75



Let multiply first relation by cos (nφ) and the second by sin (nφ) and integrate over φ on (0,π), we obtain two integrals

2/1

2/1

0

coscos2

2/sin2

coscos2

2/cos2

coscosi

iPi

2/1

2/1

0

coscos2

2/cos2

coscos2

2/sin2

cossini

iPi

0

2/12/10

2

0 coscos2

cos2/sin2

coscos2

cos2/cos2cos

2coscoscos d

nd

nPPdni n

ii

in

0

2/12/10

2

0 coscos2

sin2/cos2

coscos2

sin2/sin2cos

2cossinsin d

nd

nPPdni n

ii

in

02/12/1 coscos

cos2/sin

coscos

cos2/cos2cos d

nd

nPn

02/12/1 coscos

sin2/cos

coscos

sin2/sin2cos d

nd

nPn

Dirichlet Integrals

Johann Peter Gustav Lejeune Dirichlet

(1805 –1859)

SOLO

76


Add and subtract those two equations

02/12/1 coscos

cos2/sin

coscos

cos2/cos2cos d

nd

nPn

02/12/1 coscos

sin2/cos

coscos

sin2/sin2cos d

nd

nPn

Dirichlet Integrals

02/12/1 coscos

2/1sin

coscos

2/1cos

2

1cos d

nd

nPn

02/12/1 coscos

2/1sin

coscos

2/1cos0 d

nd

n

Replace n by n + 1 in the last equation and substitute in the previous

02/12/1 coscos

2/1sin2

coscos

2/1cos2cos d

nd

nPn

Mehler Integrals

Gustav Ferdinand Mehler

(1835 - 1895)



SOLO

77


We found


Integrals in terms of sin(iθ) and cos(iθ)

mnnk ndttPtP

12

21

1

mnnk n

dPP

12

2sincoscos

0

cost

nm

nmnm

nmm

nm

an

tdtPt nnn

m

!122!

!!22

0

14

2 122

1

1

22

nmnmnm

nmm

nm

an

tdtPt nnn

m

!322!

!1!122

0

34

2 2212

1

1

1212

nm

nmnm

nmm

nm

dP nn

m

!122!

!!22

0

sincoscos 12

0

22

nm

nmnm

nmm

nm

dP nn

m

!322!

!1!122

0

sincoscos 22

0

1212

Ordinary Differential EquationsSOLO


78

Legendre Functions of the Second Kind Qn (x)

02

12/12

0

12

1cosh21

cosh1

n

nn

n

n

txQx

xtttx

dxxxQ

nn

n

nn xxd

d

nxP 1

!2

1 2

1 xxQBxPAy nn

xPxd

dxxP nm

mmm

n

2/21

xQxd

dxxQ nm

mmm

n

2/21

xWx

xxPxQ nnn 11

1ln

2

1

3

2

2

52

3

11

1ln

2

1

2033

022

011

0

xxQxPxQ

xxQxPxQ

xQxPxQx

xxQ

n

mmnmn xPxP

mxW

111

1

evennifn

oddnifnn

xPm

nm

mn

x

xxP

xPrnr

rn

x

xxPxQ

n

mmnn

mr

n

rrnnn

2/2

2/1

2

1

21122

1

1ln

2

1

12

142

1

1ln

2

1

1

12

2

1

012

http://en.wikipedia.org/wiki/Ordinary_differential_equation

Legendre Ordinary Differential EquationSOLO

79

Legendre Functions of the Second Kind Qn (x)

xPxuxy nnn

With Pn (x) being a solution of the Legendre Differential Equation

we look for the second solution having the form

01212

22 wnn

xd

wdx

xd

wdx

2

2

2

2

2

2

2xd

xPdxu

xd

xPd

xd

xudxP

xd

xud

xd

xyd

xd

xPdxuxP

xd

xud

xd

xyd

nn

nnn

nn

nnn

nn

012211212

222

2

22 xPxunn

xd

xPdxuxxP

xd

xudx

xd

xPdxxu

xd

xPd

xd

xudxxP

xd

xudx nn

nnn

nnn

nnn

n

Substituting in the Legendre ODE we obtain

01212121

0

2

222

2

22

xuxPnn

xd

xPdx

xd

xPdxxP

xd

xudx

xd

xPd

xd

xudxxP

xd

xudx nn

nnn

nnnn

n

or

SOLO

80

Legendre Functions of the Second Kind Qn (x) (continue – 1)

02121 22

22 xP

xd

xudx

xd

xPd

xd

xudxxP

xd

xudx n

nnnn

n

Equivalent to

0

1

2/2

/

/2

22

x

x

xP

xdxPd

xdxud

xdxud

n

n

n

n

01lnln2ln 2 xxd

dxP

xd

d

xd

xud

xd

dn

n

or

Integrating we obtain

.1lnlnln 22 constxxPxd

xudn

n

Therefore AconstxxPxd

xudn

n .1 22

22 1 xxP

xdAxu

n

n

This means that the second solution has the form

22 1 xxP

xdxPxuxPxQ

n

nnnn

Legendre Ordinary Differential Equation

SOLO

81

Legendre Functions of the Second Kind Qn (x) (continue – 2)We obtained

1

1 22

xxxP

xdxPxuxPxQ

n

nnnn

x

xxxxd

xxxxP

xdxPxQ

1

1ln

2

11ln1ln

2

1

1

1

1

1

2

1

1 2

1

201

00

Let calculate Q0 (x), Q1 (x)

11

1ln

2

1

1

2/1

1

2/1

1

1

1 222221

11

2

x

xxxd

xxxxxd

xxx

xxP

xdxPxQ

x

x


SOLO

82


3

2

2

5

1

1ln

2

1

3

2

2

5

1

1ln

4

35

2

3

1

1ln

2

1

2

3

1

1ln

4

43

11

1ln

2

11

1

1ln

2

1

1ln

2

1

2

3

23

3

2

2

2

11

0

x

x

xxP

x

x

xxxxQ

x

x

xxP

x

x

xxxQ

x

xxP

x

xxxQ

x

xxQ


SOLO

83

Legendre Functions of the Second Kind Qn (x) (continue – 3)To obtain a general formula for Qn (x), start from the Polynomial Pn (x) that has n zeros αi, i=1,2,…,n

nnn xxxkxP 21

n

i i

i

i

i

nnn

x

d

x

c

x

b

x

a

xxxxxkxxP

12

00

222

21

222

11

11

1

1

1

n

i i

i

i

innn

x

d

x

cxxPxPxbxPxa

12

2220

20 1111

If we put x=1 and x = -1, and remembering that Pn (1) = 1 and Pn(-1)=(-1)n, we obtain

2/100 ba

Let prove that

0

1

122

2

ixn

iixxP

xxd

dc


SOLO

84


Let prove that 0

1

122

2

ixn

iixxP

xxd

dc

Start with

i

x

iixi xatfinitexfprovidedxd

xfdxxfxxfx

xd

d

i

i

02 22

The only terms that are not finite in

at x = αi are the terms ci/(x-αi) and di/(x-αi)2, therefore

n

i i

i

i

i

n x

d

x

c

x

b

x

a

xxP 12

0022 111

1

i

x

iii

xi

i

i

ii

xn

i cdxcxd

d

x

d

x

cx

xd

d

xxPx

xd

d

iii

2

2

22

2

1

1

Therefore if we write Pn (x) = (x-αi) Li (x), we have

22

2

223

2

2322222

1

/12

1

/122

1

/2

1

2

1

1

iii

iiiiii

xi

ii

xi

i

ixi

i

L

xdxLdL

xxL

xdxLdxxLx

xxL

xdxLd

xxL

x

xxLxd

dc

iii


SOLO

85


We proved that

ixn

iixxP

xxd

dc

22

2

1

1

22

2

1

/12

iii

iiiiiii

L

xdxLdLc

Therefore if we write Pn (x) = (x-αi) Li (x), we have

Since Pn (x) = (x-αi) Li (x) satisfies the Legendre ODE, we have

01212

22 xLxnnxLx

xd

dxxLx

xd

dx iiiiii

Performing the calculation and substituting x = αi, we have

012212

22

ix

iiii

iii

i xLxnnxLxd

xLdxx

xd

xLd

xd

xLdxx

0212 2

iiiii

i Lxd

xLd

Substituting in ci equation we get

01

/12 22

2

iii

iiiiiii

L

xdxLdLc

Therefore

n

i i

i

n x

d

xxxxP 1222 1

1

1

1

2

1

1

1


SOLO

86


We can prove that but the exact value is not important, as we shall see

ixn

ii

xxP

xd

22

2

1

Therefore

n

i i

i

n x

d

xxxxP 1222 1

1

1

1

2

1

1

1

n

i i

in

i i

i

n x

d

x

xdx

x

ddx

xxxxP

dx

11222 1

1ln

2

1

1

1

1

1

2

1

1

n

i i

ninn x

xPd

x

xxPxQ

11

1ln

2

1

Since Pn (x)/(x-αi) is a polynomial of order (n-1) the sum above is also a polynomial of order (n-1), and we define it as

n

i i

nin x

xPdxW

11 :

so 1

1

1ln

2

11

xxWx

xxPxQ nnn


SOLO

87


To find Wn-1 (x) let use the fact that Qn (x) is a solution of Legendre’s ODE

or

01212

22 xQnn

xd

xQdx

xd

xQdx n

nn

xd

xWdxP

xx

x

xd

xPd

xd

xQd nn

nn 121

1

1

1ln

2

1

21

2

2222

2

2

2

1

2

1

2

1

1ln

2

1

xd

xWdxP

x

x

xd

xPd

xx

x

xd

xPd

xd

xQd nn

nnn

01211

22

1

2

1

1ln

2

1121

11

21

22

22

0

2

22

xWnnxd

xWdx

xd

xWdxxP

x

x

xd

xPdxP

x

x

x

xxPnn

xd

xPdx

xd

xPdx

nnn

nn

n

nnn

xd

xPdxWnn

xd

xWdx

xd

xWdx n

nnn 2121 1

121

22


11

1ln

2

11

xxWx

xxPxQ nnn

SOLO

88


Use the Recursive Formula

xd

xPdxWnn

xd

xWdx

xd

d nn

n 211 112

12

1

0121421

l

iin

n xPinxd

xPd

Since Wn-1(x) is a polynomial of order n – 1, let write

12

1

01231101

n

iininnn xPaxPaxPaxW

Substitute those two equation in the Wn-1(x) O.D.E. to obtain

12

1

012

12

1

012

12

1

012

2 142211

l

iin

n

iini

n

iini xPinxPannxPx

xd

da

But by Legendre O.D.E.: 02121 12122 xPininxPx

xd

dinin

12

1

012

12

1

012 14221212

l

iin

n

iini xPinxPnninina

Let solve


SOLO

89


The coefficients of the same polynomial in both sides must be equal

12

1

012

12

1

012 14221212

l

iin

n

iini xPinxPnninina

14221212 innnininai

12224244

1221212222

2

iininininniniinn

nnininnnininBut

which gives

12

142

iin

inai

n

m

n

mmnmn

min

iinn xP

mn

mn

mxP

mn

mn

mxP

iin

inxW

1 1

121

2

1

0121 12

12221

21

1221

12

142

and

n

mmnmn xPxP

mxW

111

1?????


SOLO

90

Legendre Functions of the Second Kind Qn (x) (continue -10)

3,2,1,0

10

n

xxQn


3

2

2

5

1

1ln

2

1

3

2

2

5

1

1ln

4

35

2

3

1

1ln

2

1

2

3

1

1ln

4

43

11

1ln

2

11

1

1ln

2

1

1ln

2

1

2

3

23

3

2

2

2

11

0

x

x

xxP

x

x

xxxxQ

x

x

xxP

x

x

xxxQ

x

xxP

x

xxxQ

x

xxQ

SOLO

91


Similar to Rodrigues Formula for Legendre Functions of the Second Kind Qn (x)

Start with

122212

11

1

1

n

nn xxx

1

32211

11,

,121,11,1:

ini

nnn

uinnuf

unnufunufuuf

1

!!

!

!!

!1

1

1 12

0

2

02212

xx

in

inx

in

in

xx

in

i

i

inn

1!

21

!

0

00

uui

innnu

i

fuf

i

i

i

ii

Taylor expansion around u = 0

Use u = x-2


SOLO

92


Similar to Rodrigues Formula for Legendre Functions of the Second Kind Qn (x)

Integrate relative to x

0

122

0

122

12

02

12

0212

122!!

!

122!!

!

!

!

!

!

1

i

in

i

xin

x

in

ix

in

ixn

xinni

in

inni

in

dn

ind

n

ind

Integrate this n more times

0

12

0

12

0

12212

1

!122!!

!2!

122222122!!

!

122!!

!

1

i

in

i

in

i x

nin

xn

n

xinni

inin

xininininni

in

dinni

ind


SOLO

93


Similar to Rodrigues Formula for Legendre Functions of the Second Kind Qn (x)(continue)

1!122!

!2!

!

1

1 0

1212

1

xxini

inin

n

d

i

in

xn

n

We found, using Frobenius Series

By comparing those two relations we obtain

11

!2 12

1

xd

nxQx

n

nn

n

Return to

Frobenius Series

1!122!

!2!2

0

12

xxini

ininxQ

i

innn


SOLO

94


Another Expression for Legendre Functions of the Second Kind Qn (x)

Start with the following Differential Equation

021212

22 un

xd

udxn

xd

udx

One of the Solutions is . Check: nxxu 121

2221221

2121 11412&12

nnnxxnnxn

xd

udxxn

xd

ud

01211411412

2121

21221222

11

21

22

nnnnxnxnxnxxnnxn

unxd

udxn

xd

udx

The Second Solution can be find using u1 (x) by finding a function v (x) that satisfies:

nxxvxu 122


SOLO

95


Another Expression for Legendre Functions of the Second Kind Qn (x)(continue – 1)

01221

21212121

112

2

112

0

11

21

22

22

22

22

xd

vduxnu

xd

vdu

xd

vd

xd

udx

unxd

udxn

xd

udxvun

xd

udxn

xd

udx

The Second Solution can be find using u1 (x) by finding a function v (x) that satisfies:

nxxvxu 122

21

21

12

2

22

21

12 2&

xd

udv

xd

ud

xd

vdu

xd

vd

xd

ud

xd

udvu

xd

vd

xd

ud

1121

22

112

21

12

/

/2

1

21

21

1

12

122

x

xn

ux

uxnx

uxn

u

xdud

xuxn

xdvd

xdvd


SOLO

96



Therefore

1

12

/

/2

22

x

xn

xdvd

xdvd

x

nn

dxv

xxd

vdxn

xd

vd1212

2

11

11ln1

x

n

n dxxuxvxu 12

212

11

Differentiate the Differential Equation 021212

22 un

xd

udxn

xd

udx

0122221

2121221

2

2

3

32

2

2

2

22

xd

udn

xd

udxn

xd

udx

xd

udn

xd

ud

xd

dxn

xd

udn

xd

udx

xd

ud

xd

dx


SOLO

97



Differentiate relative to x the Ordinary Differential Equation n times

0122221:2

2

3

32

xd

udn

xd

udxn

xd

udxODE

xd

d

0223321:2

2

4

42

2

2

xd

udn

xd

udxn

xd

udxODE

xd

d

02121:2

22 un

xd

udxn

xd

udxODE

0121:1

1

2

22

n

n

n

n

n

n

n

n

xd

udnn

xd

udx

xd

udxODE

xd

d

Derive 021121:1

1

2

22

i

i

i

i

i

i

i

i

xd

udini

xd

udxin

xd

udxODE

xd

d

0122221

21121221:

1

1

1

1

3

32

1

1

1

1

3

32

i

i

i

i

i

i

i

i

i

i

i

i

i

i

xd

udini

xd

udxin

xd

udx

xd

udiniin

xd

udxin

xd

udxODE

xd

d

xd

dProofbyInduction q.e.d.

i = n

i → i+1


SOLO

98


Another Expression for Legendre Functions of the Second Kind Qn (x) (continue – 4)

This is the Legendre Ordinary Differential Equation for dnu/dxn, having the solutionPn (x) and Qn (x). Thus, the solution Pn (x) and Qn (x) can be written in the following form:

0121:1

1

2

22

n

n

n

n

n

n

n

n

xd

udnn

xd

udx

xd

udxODE

xd

d

We obtain

n

n

n

nn

n

nn xxd

d

nxd

ud

nxP 1

!2

1

!2

1 21

1

11

!2

!21

!2

!2112

22

xd

xxd

d

n

n

xd

ud

n

nxQ

xn

n

n

nnn

n

nnn

n

Rodrigues Formula

An integral for Qn (x) valid in |x| < 1 can be obtained from the previous result

1

11

!2

!21

012

2

x

dx

xd

d

n

nxQ

x

n

n

n

nnn

n



99

SOLO


0sin

1sin

sin

112

2

2222

22

rrr

rrr

Let solve this equation by the method of Separation of Variables, by assuming a solution of the form :

,SrR


cos

sinsin

cossin

rz

ry

rx

In Spherical Coordinates the Laplace equation becomes:

Substituting in the Laplace Equation and dividing by Φ gives:

0sinsinsin

112

2

222

2

SS

Srrd

Rdr

rd

d

Rr

The first term is a function of r only, and the second of angular coordinates. For the sum to be zero each must be a constant, therefore:

2

2

2

2

sinsinsin

1

1

SS

S

rd

Rdr

rd

d

R

r

y

x

z

Associate Legendre Differential Equation

100

SOLO

2

2

2sinsin

sin

1 SS

S

,,, SrRr

We obtain:

Multiply this by S sin2θ and put to get: ,S

01

sinsinsin1

2

22

d

d

d

d

Again, the first term, in the square bracket, and the last term must be equal and opposite constants, which we write m2, -m2. Thus:

22

2

2

2

0sin

sinsin

1

md

d

m

d

d

d

d

The Φ (ϕ) must be periodical in ϕ (a period of 2 π) and because this we choose the constant m2, with m an integer. Thus:

mbma sincos


mjmj ee ,or

With m integer, we have the Orthogonality Condition

21

21,

2

0

2 mmmjmj dee



101

SOLO

rR

We obtain:

or:

2

2

2

2

sincot

m

d

d

d

d

0sin

sinsin

12

2

m

d

d

Analysis of Associate Lagrange Differential Equation


Change of variables: t = cos θ

dtd sin

td

d

d

d

sin

td

dt

td

dt

td

d

d

d

td

d

td

d

td

d

td

d

d

d

d

d

d

d

2

22

cossin/1

2

22

2

2

1sin

sinsinsinsin

2

2

2

2cos

2

2

2

2

1sincot0

t

m

td

dt

td

dt

td

dm

d

d

d

d t

We obtain:

1cos

01

22

2

2

2

tt

t

m

td

dt

td

d


102

SOLO

gfrR

We obtain:

or:ff

m

d

fd

d

fd

2

2

2

2

sincot

Let try to factorize the left-hand side, second order differential equation into two first-order operators:

0sin

sinsin

12

2

fm

d

fd

The two equations are identical if the coefficient are equal. This is obtained by choosing α and β as follows:

integer1

12

mm

ffd

fd

d

fd

ffd

fd

d

fd

fd

fd

d

df

d

d

d

d

22

2

1sin

1

222

2

sin

11cot

cotsin

cot

cotcotcotcot

2


Analysis of Associate Lagrange Differential Equation (continue – 1)


103

SOLO

gfrR

We obtain:

and:

ffm

d

fd

d

fd

2

2

2

2

sincot

We have two solutions for α and β as follows:1.β1 = m, α1 = 1-m2.β2 = -m, α2 = 1+mSince m is an integer α, β are also integers.

integer1

12

mm

fffd

fd

d

fd

'sin

11cot

22

2

1

integer22 mm

Let define the two operators:

cot1

cot

md

dM

md

dM

m

m




104

SOLO

gfrRWe obtain:

We have two solutions for α and β as follows:1.β1 = m, α1 = 1-m2.β2 = -m, α2 = 1+mSince m is an integer α, β are also integers.

cot1

cot

md

dM

md

dM

m

m

ffd

d

d

d

cotcot

11

11

11

11

1 mmmm fmmfMM

We have:

22

22

22

1 mmmm fmmfMM

fm(1) – the solution for α1, β1

fm(2) – the solution for α2, β2




105

SOLO

gfrR

We obtain:

1111 1 mmmm fmmfMM

22 1 mmmm fmmfMM

Let operate on first of those equations with 1mM

11

1111 1 mmmmmm fMmmfMMM

In the Second Equation replace m by m-1

21

2111 1

mmmm fmmfMM

Let operate on second of those equations with mM

1

2

22 1 mmmmmm fMmmfMMM

In the First Equation replace m by m+1 1

11

1 1 mmmm fmmfMM

211 mmmm fpfM

11

2

mmmm fqfM


pm is a constant

qm is a constant

mm 1

mm 1



106

SOLO

We obtained: 21

1 mmmm fpfM

11

2

mmmm fqfM


cot

cot1

md

dM

md

dM

m

m

where:

Theorem:

00

sinsin dgMfdfMg mm

where f and g are arbitrary bounded function of θ.Proof:

00

00

0

00

sinsinsin

cossin

cos1sinsin

cot1sinsin

dgMfdgfmgd

df

dgfmgd

dffg

dfmd

dgdfMg

m

m

Those are Recursive Relations in m.



107

SOLO


00

sinsin dgMfdfMg mm

0

11

0

11 sinsin dfMMfdfMfM mmmmmmmm

0

22

00

11

sin

sinsin

dfp

dfpfpdfMfM

mm

mmmmmmmm

Choose f := fm+1 and g := Mm(-) fm+1:

0

21

0

11

0

11

sin1

1sinsin

dfmm

dfmmfdfMMf

m

mmmmmm

Assuming: we have1sinsin0

21

0

2

dfdf mm

11 mmmmpm



108

SOLO


00

sinsin dgMfdfMg mm

Choose now f := Mm(+) fm and g := fm, to obtain as before

We obtained

11 mmmmqm

11 mmmmpm

We see that m, an integer, is bounded by λ, therefore we must choose λ as

0integer1 lll

In this case we have mMAX = l, mmin = -(l+1), or

integer0,integer1 mllml

Therefore

lmlmmllqp mm 111



109

SOLO


We obtained

lmlmmllqp mm 111

cot

cot1

md

dM

md

dM

m

m

211 mmmm fpfM

11

2

mmmm fqfM

lml

mmllfmd

d

fmmllfmd

d

m

mm

1

11cot

11cot1

2

211

Substituting m = - (l+1) in the First Equation and m = l in the Second we obtain:

0cot

0cot

2

1

l

l

fld

d

fld

d

sin

sin

sin

sin

2

2

1

1

dl

f

fd

dl

f

fd

l

l

l

l

llll Cff sin21

If we can find a Solution for a particular m, we can use the Recursive Relations above to find the others.



110

SOLO


We obtained llll Cff sin21

Cl must be chosen to normalize fi and f-l:

1sinsin0

122

0

2

dCdf lll

!12

!21 212

2

n

n

C

n

l

212 !2

!12

n

nC

nl

0

1212

0

212

0

0

2

0

2

0

12 sinsin2cossin2cossincossinsin dndndd nnnnnn

Therefore

!12

!2cos

2222

!2

11212

!2sin

31212

2222sin

12

2sin

212

2

0

1

00

12

0

12

n

n

nn

n

nn

nd

nn

nnd

n

nd

nnnnn



111

SOLO


The Normalized Solution for m = l is defined as

ln

lml

lml

n

nsin

!2

!12212

The Solutions for m < l can be found using the Recursive Relation:

1,,,0,,1,cot

11

111

1

1

llllmmd

d

mmllM

pm

lm

lmm

ml

211 mmmm fpfM

From which the Normalized Solutions for m < l are given by

Or we can find the Solutions for m >- l by using the Recursive Relation:

11

2

mmmm fqfM

llllmm

d

d

mmllM

qm

lm

lmm

ml ,1,,0,,,1cot

11

111



112

SOLO


ln

lml

lml

n

nsin

!2

!12212

1,,,0,,1,cot

11

111

1

1

llllmmd

d

mmllM

pm

lm

lmm

ml

Examples:

llllmm

d

d

mmllM

qm

lm

lmm

ml ,1,,0,,,1cot

11

111


2cos

11

cos0

1

2cos

11

00

12

3sin

2

3

2

3cos

2

3

12

3sin

2

31

2

10

x

x

xl

l

x

x

x

2cos

222

2cos

12

2cos

202

2cos

12

2cos

222

14

15sin

4

15

12

15cossin

2

15

1322

51cos3

22

5

12

15cossin

2

15

14

15sin

4

152

x

xx

x

xx

xl

x

x

x

x

x



SOLO

113

Associated Legendre Functions

Let Differentiate this equation m times with respect to t, and use Leibnitz Rule of Product Differentiation:

im

im

i

im

im

m

td

tgd

td

tsd

imi

mtgts

td

d

0 !!

!

Start with: 1011 2

ttwnntw

td

dt

td

dnn

or: 101212

22 ttwnntw

td

dttw

td

dt nnn

twmmtwtmtwttwtd

dt

td

d mn

mn

mnnm

m

1211 1222

22

twmtwttwtd

dt

td

d mn

mnnm

m

1

twnntwmtwttwmmtwtmtwt mn

mn

mn

mn

mn

mn 122121 1122

011121 122 twmmnntwtmtwt mn

mn

mn

2nd Way

SOLO

114


011121 122 twmmnntwtmtwt mn

mn

mn

Define: twty mn:

011121 122 tymmnntytmtyt

Now define: tyttum

221: Let compute:

122122 11 ytyttm

td

ud mm

1122222 111 ytyttm

td

udt

mm

21221221221

2222222 1121111 ytyttmyttmyttmytmtd

udt

td

d mmmmm

yt

tmmnnmmtymmnnytmytt

mm

2

22222

0

12222

111111211

We get: 01

112

22

u

t

mnn

td

udt

td

d

2nd Way

SOLO

115


Define: twtd

dttu nm

mm

221:

We get: 01

112

22

u

t

mnn

td

udt

td

d

Start with Legendre Differential Equation:

1011 2

ttwnntw

td

dt

td

dnn

Summarize

But this is the Differential Equation of f (θ) obtained by solving Laplace’s Equation

by Separation of Variables in Spherical Coordinates .

02 rRr ,,

The Solutions of this Differential Equation are called Associated Legendre Functions, because they are derived from the Legendre Polynomials, and Legendre Functions of the Second Kind Qn (x) and are denoted:

tPtd

dttP nm

mmm

n221:

2nd Way

tQtd

dttQ nm

mmm

n221:

SOLO

116


Let use Rodrigues Formula for Pn (t):

We see that we can define the Associated Legendre Function even for negative m (In the Differential equation m2 appears):

tPtd

dttP nm

mmm

n221:

n

n

n

nn ttd

d

ntP 1

!2

1 2

we obtain:


n

mn

mnm

nm

n ttd

dt

ntP 11

!2

1: 222

From this equation we obtain:

tPtP nn 0

2nd Way

SOLO

117


Using Rodrigues Formula for Pn (t) the Associated Legendre Functions are

n

mn

mnm

n

mn t

td

dt

ntP 11

!2

1: 222

Let Differentiate this equation n+m times with respect to t, and use Leibnitz Rule of Product Differentiation:

im

im

i

im

im

m

td

tgd

td

tsd

imi

mtgts

td

d

0 !!

!

nn

mn

mnn

mn

mn

tttd

dt

td

d1112

Since we differentiate (t-1)n and (t+1)n, all the differentials greater then n will vanish,therefore we get

mn

i

n

im

imn

in

in

n

n

nn

m

nmn

n

m

mn

n

mmnnn

mn

mn

ttd

dt

td

d

imin

mn

ttd

dt

td

d

mn

mn

ttd

dt

td

d

mn

mntt

td

d

0

11!!

!

0011!!

!

11!!

!0011

2nd Way

SOLO

118



n

mn

mnm

n

mn t

td

dt

ntP 11

!2

1: 222

mn

i

imni

mn

i

n

im

imn

in

innn

mn

mn

timlnntinnimin

mn

ttd

dt

td

d

imin

mntt

td

d

0

0

111111!!

!

11!!

!11

!11!11

1111

!!

1111

imnimnininiiimim

imnnninn

inim

imnnninn

!!

1111

imni

innnimnn

mn

i

minmi

n

m

n

mn

mnm

n

mn

tinnntimnnimni

mn

n

ttd

dt

ntP

0

2/2/2/

222

111111!!

!

!2

1

11!2

1:

2nd Way

SOLO

119



mn

i

minmi

n

m

n

mn

mnm

n

mn

tinnntimnnimni

mn

n

ttd

dt

ntP

0

2/2/2/

222

111111!!

!

!2

1

11!2

1:

Using this formula we can write the Associated Legendre Functions for – m as

n

mn

mnm

n

mn t

td

dt

ntP 11

!2

1: 222

Using Leibnitz Rule of Product Differentiation we obtain

mn

i

n

i

in

imn

imnnn

mn

mn

ttd

dt

td

d

imni

mntt

td

d

0

11!!

!11

mn

i

inim tinnntimnnimni

mn

0

111111!!

!

mn

i

minmi

n

m

n

mn

mnm

n

mn

tinnntimnnimni

mn

n

ttd

dt

ntP

0

2/2/2/

222

111111!!

!

!2

1

11!2

1:

2nd Way

SOLO

120


We obtained

mn

i

minmi

n

m

n

mn

mnm

n

mn

tinnntimnnimni

mn

n

ttd

dt

ntP

0

2/2/2/

222

111111!!

!

!2

1

11!2

1:

mn

i

minmi

n

m

n

mn

mnm

n

mn

tinnntimnnimni

mn

n

ttd

dt

ntP

0

2/2/2/

222

111111!!

!

!2

1

11!2

1:

tP

mn

mntP m

nmm

n !

!1

Therefore

2nd Way

SOLO

121


Examples

tPtd

dttP nm

mmm

n221:

10 00

0 tPtPn

sin1

cos:

sin111

cos2

121

11

1

cos

10

1

cos2

122

121

1

1

t

t

t

tP

ttPtP

ttPtP

tttd

dttPn

tP

mn

mntP m

nmm

n !

!1

2cos

222

222

cos2

1211

2

2cos2

20

2

cos2

12

2

2

121

2

2cos

22

2

2

2

222

2

sin8

11

8

1

!4

!01

2

cossin13

!3

11

2

1cos3

2

13

cossin3132

131

sin3132

1312

12

2

1

t

t

tP

t

t

tP

t

tP

ttPtP

tttP

ttPtP

ttt

td

dttP

tt

td

dttPn

10 tP 2

1

2

3 22 ttP ttP 1

2nd Way

SOLO

122


Examples

tPtd

dttP nm

mmm

n221:

tPmn

mntP m

nmm

n !

!1

2nd Way


http://en.wikipedia.org/wiki/File:Associated_Legendre_Poly.svg

SOLO

123



Start with 121

1:,

02

utPuutu

tugn

nn

Let derivate this function relative to t

m

m

mmm

m

m

m

um

mutuu

m

mmutu

t

tug

uutut

tug

uutut

tug

uutut

tug

!12

!1221

!12

2242123121

,

22

5

2

3

2

121

,

22

3

2

121

,

22

121

,

1

2/12

1

1

2/12

32/723

3

22/522

2

2/32

00

222/12

2/2

1

2/2 121

1

!12

!12,1:,

n

mn

n

nnm

mmn

m

mm

mm

mm

m tPutPtd

dtu

utu

ut

m

m

t

tugttug

tPtd

dttP nm

mmm

n221: Use

We obtain

SOLO

124


02/12

2/2

1

2/2

21

1

!12

!12,1,

n

mn

nm

mm

mm

mm

m tPuutu

ut

m

m

t

tugttug

The Generating Function for Pnm (t) is

The Generating Function for Pn (t) is

121

1:,

02

utPuutu

tugn

nn



SOLO

125



Start with

Because of the existence of two indices in Pnm (t), we have a wide variety of

recurrence relations.

0

2/12

2/21

21

11

!12

!12,

n

mn

nm

m

c

m

mm tPuutu

utm

mtug

m

or

02/1221

1

n

mn

mnmm tPu

utuc

Differentiate with respect to u

0

12

0

0

122/12

0

12/32

2112

2121

12

21

222/1

n

mn

n

n

mn

m

n

mn

nm

mm

n

mn

mnmm

tPumnututPuutm

tPumnutuutu

ucutm

tPumnutu

utmc

Arrange so all the terms, in the last relation, have the same power of u, by shifting indexes

n

mn

mn

mn

n

n

mn

mn

m tPmntPmnttPmnutPmttPmu 111 1211212

First Recurrence Relation

SOLO

126


Equating the terms in um, and rearranging gives the First Recurrence Equation:

n

mn

mn

mn

n

n

mn

mn

m tPmntPmnttPmnutPmttPmu 111 1211212

First Recurrence Relation (continue)

tPmntPmnttPn mn

mn

mn 11 112

Second Recurrence Relation

Start with

n

m

mmm

nn

m

mm

m tm

mctPu

utu

uctug

2/212/12

1!12

!12

21,

or 2/2

1

2/12 1!12

!1221

m

nmm

mn

nmmm t

m

mctPuutuuc

Differentiate with respect to u

n

mn

nmnmmm tPunutuuututumumc 12/122/121 2121222/1

n

mn

nnm

mm tPunutuutum

utu

umc 122/12

1

21222/121

ud

d


SOLO

127


Second Recurrence Relation (continue - 1)

or

1

2/1212/22

2/122/21

1121!22

!321

12

22121

!12

!12

m

m

m

m

mm ctmtm

mt

m

mmt

m

mc

n

mn

nnm

mm tPunutuutum

utu

ucmtm 12

2/12

112/12 21222/1

21112

RHS

n

mn

nn

LHS

n

mn

n tPunutuutumtPutmm

1212/12 21222/1112

Rearrange the RHS , by changing indexes, to have powers of un

n

nmn

mn

mn

n

nmn

mn

mn

mn

mn

n

mn

nn

utPntPtnmtPnm

utPntPtntPntPtmtPm

tPunutuutum

11

111

12

11222

1211212

21222/1

Equating the terms in un of LHS and RHS we obtain:

tPntPtnmtPnmtPtmm mn

mn

mn

mn 11

12/12 11222112


SOLO

128


Second Recurrence Relation (continue - 2)

Use the First Recurrence Formula in RHS2:

2

11

2

12/12 11222112RHS

mn

mn

mn

LHS

mn tPntPtnmtPnmtPtmm

tPmntPmnttPn mn

mn

mn 11 112

tPmntPmnn

nmtPntPnmRHS mn

mn

mn

mn 1111 1

12

1122122

tPmmtPmm

tPmnnmnntPmnnmnmnRHSnm

nm

n

mn

mn

11

11

1212

1122112122212212

Equating (2 n + 1) LHS2 with (2 n + 1) RHS2 we obtain the Second Recurrence Formula:

tPtPtPtn mn

mn

mn 11

12112


SOLO

129


tPmnmntPmnmnn

tPt

tPtPn

tPt

tPmntPmntPtn

tPmmnntPt

tmtP

mn

mn

mn

mn

mn

mn

mn

mn

mn

mn

mn

mn

11

11

2

11

11

2

11

12

1

211112

114

12

113

1122

0111

21



SOLO

130


Orthogonality of Associated Legendre Functions

n

mn

mnm

nm

n ttd

dt

ntP 11

!2

1: 222

Let Compute

1

1

2221

1

111!!2

1dtt

td

dt

td

dt

qpdttPtP

q

mq

mqp

mp

mpm

qpm

qm

p

Define X := x2 -1

1

1

1

1 !!2

1dtX

td

dX

td

dX

qpdttPtP q

mq

mqp

mp

mpm

qp

mm

qm

p

If p ≠ q, assume q > p and integrate by parts q + m times

mqidtXtd

dvdX

td

dX

td

du q

imq

imqp

mp

mpm

i

i

,,1,0

All the integrated parts will vanish at the boundaries t = ± 1 as long as there is a factorX = x2-1. We have, after integrating m + q times

1

1

1

1 !!2

11dtX

td

dX

td

dX

qpdttPtP p

mp

mpm

mq

mqq

qp

mqmm

qm

p

SOLO

131


1:!!2

11 21

1

1

1

xXdtXtd

dX

td

dX

qpdttPtP p

mp

mpm

mq

mqq

qp

mqmm

qm

p

Because the term Xm contains no power greater than x2m, we must haveq + m – i ≤ 2 m

or the derivative will vanish. Similarly,p + m + i ≤ 2 p

Adding both inequalities yieldsq ≤ p

which contradicts the assumption that q > p, therefore is no solution for i and the integral vanishes.

Let expand the integrand on the right-side using Leibniz’s formula

mq

i

pimp

impm

imq

imqqp

mp

mpm

mq

mqq X

td

dX

td

d

imqi

mqXX

td

dX

td

dX

0 !!

!

qpdttPtP mq

mp

01

1

This proves that the Associated Legendre Functions are Orthogonal (for the same m).


SOLO

132


1:

!!2

11 21

12

21

1

2

xXdtXtd

dX

td

dX

ppdttP p

mp

mpm

mp

mpp

p

mpm

p

Let expand the integrand on the right-side using Leibniz’s formula

mp

i

pimp

impm

imp

imppp

mp

mpm

mp

mpp X

td

dX

td

d

impi

mpXX

td

dX

td

dX

0 !!

!

For the case p = q we have

Because X = x2 – 1 the only non-zero term is for i = p - m

1:!2!!2

!1 21

1

!2

2

2

!2

2

2

22

1

1

2

xXdtXtd

dX

td

dX

mmpp

mpdttP

p

pp

p

m

mm

mp

p

pm

p

!

!

12

2

sin1!!2

!2!11

!!2

!2!1

!12

!21

0

1222

cos1

1

222

212

mp

mp

p

dmpp

pmpdtx

mpp

pmp

p

p

pp

p

ptp

X

p

p

pp


SOLO

133


qp

mq

mp

tm

qm

p mp

mp

pdPPdttPtP ,

0

cos1

1 !

!

12

2sincoscos

Therefore the Orthonormal Associated Legendre Functions is

mnmP

mn

mnnΘ m

nm

n

cos!

!

2

12cos


SOLO

134


Examples

10 00

0 tPtPn

sin1

cos

sin11

cos2

121

1

cos0

1

cos2

121

1

t

t

t

ttP

ttP

ttPn

2cos

222

cos2

121

2

2cos20

2

cos2

121

2

2cos

222

sin8

11

8

12

cossin1

2

12

1cos3

2

13

cossin313

sin3132

t

t

t

t

t

tP

tttP

ttP

tttP

ttPn

2nd Way

mnmP

mn

mnnΘ m

nm

n

cos!

!

2

12cos

2cos

11

cos0

1

2cos

11

00

12

3sin

2

3

2

3cos

2

3

12

3sin

2

31

2

10

t

t

tl

l

t

t

t

2cos

222

2cos

12

2cos

202

2cos

12

2cos

222

14

15sin

4

15

12

15cossin

2

15

1322

51cos3

22

5

12

15cossin

2

15

14

15sin

4

152

t

tt

t

tx

tl

t

t

t

t

t

We recovered the previous results

SOLO

135

Associated Legendre FunctionsRecurrence Relations for Θn

m Functions

2nd Way

Use the First Recurrence Formula for Pnm Functions:

mnmtΘ

mn

mn

ntP m

nm

n

!

!

12

2

tPmntPmnttPn mn

mn

mn 11 112

tΘ

mn

mn

nmntΘ

mn

mn

nmntΘ

mn

mn

ntn m

nm

nm

n 11 !1

!1

32

21

!1

!1

12

2

!

!

12

212

We have

to obtain

Cancellation of common terms leads to

tΘnn

mnmntΘ

nn

mnmntΘt m

nm

nm

n 11 12123212

11

First Recurrence Relations for Θnm Functions

SOLO

136

Associated Legendre Functions 2nd Way

Use the Second Recurrence Formula for Pnm Functions:

mnmtΘ

mn

mn

ntP m

nm

n

!

!

12

2

We have

to obtain

Cancellation of common terms leads to

Second Recurrence Relations for Θnm Functions

tPtPtPtn mn

mn

mn 11

12112

tΘ

mn

mn

ntΘ

mn

mn

ntΘ

mn

mn

ntn m

nm

nm

n 1112

!1

!1

12

2

!1

!1

32

2

!1

!1

12

2112

tΘnn

mnmntΘ

nn

mnmntΘt m

nm

nm

n 1112

1212

1

3212

11


Recurrence Relations for Θnm Functions

SOLO

137

Spherical HarmonicsThe Spherical Harmonics that forms an orthogonal system, were first introduced by Pierre Simon de Laplace in 1782

Pierre-Simon, marquis de Laplace

(1749-1827)

Spherical Harmonics were first investigated in connection with the Newtonian potential of Newton's law of universal gravitation in three dimensions. In 1782, Pierre-Simon de Laplace had, in his “Mécanique Céleste”, determined that the gravitational potential at a point x associated to a set of point masses mi located at points xi was given by

In 1867, William Thomson (Lord Kelvin) and Peter Guthrie Tait introduced the solid spherical harmonics in their “Treatise on Natural Philosophy”, and also first introduced the name of "spherical harmonics" for these functions. The solid harmonics were homogeneous solutions of Laplace's equation

William Thomson, 1st Baron Kelvin

(1824 –1907)

Peter Guthrie Tait(1831 –1901)

http://en.wikipedia.org/wiki/File:Pierre-Simon_Laplace.jpg

http://en.wikipedia.org/wiki/File:Lord_Kelvin_photograph.jpg

SOLO

138

Spherical Harmonics

mnmP

mn

mnnΘ m

nm

n

cos!

!

2

12cos

0sin

1sin

sin

112

2

2222

22

rrr

rrr

By Separation of Variables, we assume a solution of the form rRr ,,

Laplace Equation in Spherical Coordinates is:

f (θ) and g (φ) are described by the Differential Equations:

22

2

2

2

0sin

1sinsin

1

md

d

mnn

d

d

The Orthonormal Solutions, as functions of the parameter m are:

mjm eG

2

1

The functions Θnm are Orthonormal with respect to the angle θ and the

functions Gm are Orthonormal with respect to angle φ.

SOLO

139

Spherical Harmonics

mnmeP

mn

mnnGΘY mjm

nmmm

nmm

n

cos!

!

4

121cos1:,

For r = constant (Spherical Surfaces) we define the function:

We can see that:

The Functions Ynm are Orthonormal with respect to both angles θ and φ.

2121

21

2,1

212

2

1

1

21

2

2

1

1

,

02

22

1

11

2

0022

222

11

111

2

0 0

sincoscos!

!

2

12

!

!

2

12

2

1sincoscos1

!

!

2

12

!

!

2

12

sin,,

nnm

nm

n

mmm

mmjmn

mn

mm

mn

mn

dPPmn

mnn

mn

mnn

dedPPmn

mnn

mn

mnn

ddYY

mm

We can write 2121

2

2

1

1 ,,

4

,, mmnnm

nm

n dYY

Where d Ω = sin θ dθ dφ is the element of solid angle.

SOLO

140

Spherical Harmonics

One of the most important properties of Spherical Harmonics lies in the Completeness Property, a consequence of the Sturm-Liouville form of the Laplace Equation. This property means that any function f (θ,φ) (with sufficient continuity properties) can be extended in a uniformly Convergent Double Series of Spherical Harmonics

This expansion holds in the sense of mean-square convergence, which is to say that

0

,,n

n

nm

mn

mn Yff

0sin,,2

0 0

2

0

ddYffn

n

nm

mn

mn

The expansion coefficients are the analogs of Fourier coefficients, and can be obtained by multiplying the above equation by the complex conjugate of a spherical harmonic, integrating over the solid angle Ω, and utilizing the above orthogonality relationships. This is justified rigorously by basic Hilbert space theory. For the case of orthonormalized harmonics, this gives:

2

0 0

sin,,,, dYfddYff mn

mn

mn

SOLO

141

Spherical Harmonics

To prove the fnm relation let compute

0

,,n

n

nm

mn

mn Yff

0,,

00

1

1

1

1

11 ,,,,,,

l

mn

l

lmmmln

ml

l

l

lm

ml

mn

ml

l

l

lm

ml

ml

mn

mn

ff

dYYfdYfYdYf

Let choose f (θ,φ) = Pnm (θ,φ)

2

0 0

sin, ddYPf kl

mn

kl

klkeP

kl

klnY kjk

lkk

l

cos!

!

4

121:,

2

0

!

!

12

2

0

,

sin!

!

4

121 dedPP

ml

mlnf mj

mn

mn

n

ml

mn

mml

ln

qp

mq

mp mp

mp

pdPP ,

0 !

!

12

2sincoscos

SOLO

142

Spherical Harmonics

Proof

,1, mn

mmn YY

mjm

nmm

n ePmn

mnnY

cos!

!

4

121,

mnmeP

mn

mnnGΘY mjm

nmmm

nmm

n

cos!

!

4

121cos1:,

,1cos!

!

4

1211

cos!

!1

!

!

4

121

mn

mmjmn

mm

mjmn

mm

YePmn

mnn

ePmn

mn

mn

mnn

q.e.d.

SOLO

143

Spherical Harmonics

Visualization of Spherical HarmonicsThe Laplace spherical harmonics Yl

m can be visualized by considering their “nodal lines", that is, the set of points on the sphere where Re[Yl

m], or alternatively where Im[Ylm]. Nodal

lines of are composed of circles: some are latitudes and others are longitudes. One can determine the number of nodal lines of each type by counting the number of zeros of Yl

m in the latitudinal and longitudinal directions independently. For the latitudinal direction, the real and imaginary components of the associated Legendre polynomials each possess ℓ−|m| zeros, whereas for the longitudinal direction, the trigonometric sin and cos functions possess 2|m| zeros.

Schematic representation of Ylm on

the unit sphere and its nodal lines. Re[Yl

m] is equal to 0 along m great circles passing through the poles, and along n-m circles of equal latitude. The function changes sign each time it crosses one of these lines.

When the spherical harmonic order m is zero (upper-left in the figure), the spherical harmonic functions do not depend upon longitude, and are referred to as zonal. Such spherical harmonics are a special case of zonal spherical functions. When ℓ = |m| (bottom-right in the figure), there are no zero crossings in latitude, and the functions are referred to as sectoral. For the other cases, the functions checker the sphere, and they are referred to as tesseral.

http://en.wikipedia.org/wiki/File:Harmoniques_spheriques_positif_negatif.png

SOLO

144

Spherical Harmonics

j

j

j

j

j

j

eY

eY

Y

eY

eY

eY

Y

eY

Y

2222

12

202

12

2222

11

01

11

00

sin2

15

4

1,

cossin2

15

2

1,

1cos35

4

1,

cossin2

15

2

1,

sin2

15

4

1,

sin2

3

2

1,

cos3

2

1,

sin2

3

2

1,

1

2

1,


http://en.wikipedia.org/wiki/File:Spherical_harmonics.png

145

SOLO

We obtain:

This is a Sturm-Liouville Differential Equation:

22

0sinsin

sin2

ffm

d

fd

r

q

p

Laplace’s Homogeneous Differential Equation

Analysis of f (θ) Equation

bxaconstyxryxqxd

ydxp

xd

d

0


Ordinary Differential Equations


Jacques Charles François Sturm

(1855–1803)

Joseph Liouville (1809–1882),

bxaconstyxryxqxd

ydxp

xd

d

0

Many problems of mathematical physics lead to differential equations of the form

00'

00'2

22

121

22

2121

bbbybbyb

aaayaaya

with the Boundary Conditions B.C. [y] = 0:

When p (x), q (x), r (x) are continuous on [a,b] and p (x)>0 and r (x)>0 on [a,b], we refer to this system as a Regular Sturm-Liouville Boundary Value Problem.

The Sturm-Liouville Problem is a two-point boundary second-order linear differential equation problem, in which we want to define the parameter λ such that the equation has non trivial solutions y(t)≠0. Such problems are called Eigenvalues Problems and the corresponding number λ an Eigenvalue.

SOLO

146Return to Rodrigues’ Formula


http://en.wikipedia.org/wiki/File:Charles_Sturm.jpeg

http://en.wikipedia.org/wiki/File:Joseph_liouville.jpeg


Jacques Charles François Sturm

(1855–1803)

Joseph Liouville (1809–1882),

bxaconstyxryxqxd

ydxp

xd

d

0

Many problems of mathematical physics lead to differential equations of the form

with the Boundary Conditions B.C. [y] = 0:

When p (x), q (x), r (x) are continuous on (a,b) and p (x)>0 and r (x)>0 on (a,b). we refer to this system as a Singular Sturm-Liouville Boundary Value Problem if one of the following conditions occurs:

unboundedisbac

xrorxqorxporxrorxqorxpb

xporxpa

bxbx

axax

bxbx

axax

,

limlim

0lim0lim

SOLO

00'

00'2

22

121

22

2121

bbbybbyb

aaayaaya

147



http://en.wikipedia.org/wiki/File:Charles_Sturm.jpeg

http://en.wikipedia.org/wiki/File:Joseph_liouville.jpeg


Define the Operator: yxqxd

ydxp

xd

dxyL

:

The Sturm–Liouville Equation can be rewritten as: 0 xyxrxyL

Lagrange's Identity (Boundary Value Problem)

Proof:

Joseph-Louis Lagrange (1736 –1813),

vuWxpxd

duLvvLu ,

where is the Wronskian of u and v. xd

udv

xd

vduvuW :,

Theorem 1: Let u and v be functions having continuous second derivatives on the interval [a,b]. Then

uvxqxd

udxp

xd

dvvuxq

xd

vdxp

xd

duuLvvLu

xd

udxp

xd

dv

xd

udxp

xd

vd

xd

vdxp

xd

ud

xd

vdxp

xd

du

xd

udv

xd

vduxp

xd

d

xd

udxpv

xd

d

xd

vdxpu

xd

d

q.e.d.

SOLO

148




Proof:

0////

,,

21212121

avauaaavaaauapbvbubbbvbbbubp

avuWapbvuWbpxdxuLvvLub

a

q.e.d.

Green’s Formula (Boundary Value Problem)

When a2≠0 and b2≠0: bvbbbvavaaav

bubbbuauaaau

2121

2121

/',/'

/',/'

SOLO

149


Corollary 1: Let u and v be functions having continuous second derivatives on the interval [a,b]. Then

George Green1793-1841tomb stone

If u and v satisfies the Boundary Conditions (y=u,v):

then:

b

a

b

a

vuWxpxdxuLvvLu ,

0b

a

xdxuLvvLu

00'

00':..

2121

2121

bbnotaybayb

aanotayaayaCB


Is the Inner Product on the set of Continuous Real-valued functions on [a,b].

Green’s Formula (Boundary Value Problem)

b

a

xdxgxfgf :,

Takes a particularly useful and elegant form when expressed in form of a Inner Product:

Equation 0b

a

xdxuLvvLu

Using this notation we can rewrite the Green’s Formula:

vuLvLu ,,

A Linear Differential Operator that satisfies this relation for all u and v in its domain is called a Selfadjoint Operator. We have shown that the Sturm-Luville

Operator with the domain of functions that have continuous second derivatives on [a,b] ans satisfy the Boundary Conditions, then L is a Selfadjoint Operator. Selfadjoint Operators are like symmetric matrices in that that their Eigenvalues are Real-valued. We will prove this for Regular Sturm-Liouville Boundary Value Problems.

yxqxd

ydxp

xd

dxyL

:

SOLO

150



Proof:

Theorem 2: The eigenvalues λ for the Regular Sturm-Liouville Boundary Value Problem

are Real and have Real-valued Eigenvalues

0 xyxrxyL

Assume λ, possible a complex number, is an Eigenvalue for the Sturm-Liouville Problem with eigenfunction φ (x)

0 xxrxL and φ (x) satisfies the Boundary Conditions:

00'

00'

2121

2121

bbnotbbbb

aanotaaaa

Take the Complex Conjugate and use the fact that p, q and r are Real-valued:

0 xxrxLxxrxL Since a1, a2, b1, b2 are Real-valued, also satisfies the Boundary Conditions, hence is an Eigenvalue with the Eigenfunction .

b

a

b

a

xdxxxrxdxxL

SOLO

151



Proof (continue):

q.e.d.

b

a

b

a

xdxxxrxdxxL

Using Green’s Formula for Boundary Value:

0b

a

xdxuLvvLu

b

a

b

a

b

a

xdxxxrxdxLxxdxxL

Therefore: b

a

b

a

xdxxxrxdxxxr

0

0

0

b

a

xdxxxr

If φ (x) is not real-valued, since λ is real, we can see that Real {φ (x)} and Im {φ (x)} are also Eigenfunctions, and we can use Real {φ (x)} as the Real-valued Eigenfunction.

λ is Real.

SOLO

152


Theorem 2: The eigenvalues λ for the Regular Sturm-Liouville Boundary Value Problem

are Real and have Real-valued Eigenvalues

..&0 CBxyxrxyL


Proof :

q.e.d.

Theorem 3: All the eigenvalues λ of the Regular Sturm-Liouville Boundary Value Problem

are simple (for each Eigenvalue λ there is only one Linearly Independent Eigenfunction)

Assume φ (x) and ψ (x) are two Linearly Independent Eigenfunctions corresponding to the same Eigenvalue λ (thus φ (x) and ψ (x) ) satisfy the differential equation for the same λ and the Boundary Condition):

Assume a2 ≠ 0 (the same argument if a1≠ 0):

If the Wronskian of two solutions of a second-order linear homogeneous equation is zero at a point in the interval [a,b] then the solutions are linearly dependent, and this is a contradiction.

00'&0' 212121 aanotaaaaaaaa

aaaaaaaa 2121 /'&/'

Compute the Wronskian W [φ,ψ] at x = a

0//'', 2121 aaaaaaaaaaaaW

..&0 CBxyxrxyL

SOLO

153



Proof :

q.e.d.

Theorem 4: Eigenfunctions of the Regular Sturm-Liouville Boundary Value Problem

are Orthogonal with respect to the weight function r (x) on [a,b].

Let λ and μ be distinct (λ≠μ) and Real Eigenvalues with corresponding Real-valued Eigenfunctions φ (x) and ψ (x) respectively

Use Green’s Formula:

0&0 xxrxLxxrxL

b

a

b

a

b

a

dxxxxr

dxxxrxxxrxdxxLxxLx

0

0b

a

dxxxxr

..&0 CBxyxrxyL

SOLO

154



Uniform Convergence of Eigenfunction Extension

SOLO

155


Theorem 5: The Eigenvalues of the Regular Sturm-Liouville Boundary Value Problem

form a countable, increasing sequence λ1<λ2<λ3<… with lim n→∞ λn=+∞

..&0 CBxyxrxyL


Uniform Convergence of Eigenfunction Expension

Theorem 6: Let be an Orthonormal Sequence of Eigenfunctions for the Regular Sturm-Liouville Boundary Value Problem

Let f (x) be a continuous function on [a,b] with f’(x) piecewise continuous in [a,b]. If f satisfies the Boundary Conditions then

1nn nkkn ,

..&0 CBxyxrxyL

bxaxcxfn

nn

,1

where b

a

nn dxxfxxrc

The Eigenfunction expension converges uniformly in [a,b]. Since this is true for all continuous f (x) and piecewise continuous f’(x) this Functional Space is Completely Covered by the Eigenfunctions.

Eigenfunction Expension

The series of functions is uniformly convergent to f (x)

if for all ε>0 and for all x(a,b) we can find a number N (ε)

such that Nnxfxfn

n

iiin xcxf

1

SOLO

156



Pointwise Convergence of Eigenfunction Expension

SOLO

157


Theorem 7: Let be an Orthonormal Sequence of Eigenfunctions for the Regular Sturm-Liouville Boundary Value Problem

Let f (x) and f’(x) piecewise continuous in [a,b]. There for any x in (a,b)

where

..&0 CBxyxrxyL

1nn nkkn ,

b

a

nn dxxfxxrc

n

xxxx

n

xxxx

xfxfxfxfn

n

n

n

lim&lim

bxaxfxfxcn

nn

,

2

1

1


EquationRegular singularity

x =

Iregular singularity

x =

1. Hypergeometric x (x-1) y”+[(1+a+b) x –c] + aby = 0

0, 1, ∞-

2. Legendre (1-x2) y” – 2 x y’ + l (l+1) = 0

-1, 1, ∞-

3. Chebyshev (1-x2) y”–x y’ + n2y= 0

-1, 1, ∞-

4. Confluent Hypergeometric x y” +(c –x) y’ –a y = 0

0∞

5. Bessel x2 y”+x y’ +(x2- n2) y= 0

0∞

6. Laguerre x y” +(1 –x) y’ +a y = 0

0∞

7. Simple Harmonic Oscillator y” + ω2 y = 0

-∞

8. Hermite y”-2 x y’ 2 α y= 0

-∞ 158



Equationp (x)q (x)λr (x)

Legendre1-x20l (l+1)1

Shifted Legendrex (1-x)0l (l+1)1

Associated Legendre1-x2-m2/(1-x2)l (l+1)1

Chebyshev I(1-x2)1/20n2(1-x2)-1/2

Shifted Chebyshev I[x (1-x)] 1/20n2[x (1-x)] -1/2

Chebyshev II(1-x2)3/20n (n+2)(1-x2)1/2

Ultraspherical (Gegenbauer)

(1-x2)α+1/20n (n+2 α)(1-x2)α-1/2

Besselx-n2/xa2x

Laguerrex e-x0αe-x

Associated Laguerrexk+1 e-x0α-kxk e-x

Hermitee-x202 αe-x2

Simple Harmonic Oscillator

10n21

Self-Adjoint ODE

159



EquationIntervalr (x)Standard Normalization

Legendre-1 ≤ x ≤11

Shifted Legendre0 ≤ x ≤11

Chebyshev I-1 ≤ x ≤1(1-x2)-1/2

Shifted Chebyshev I0 ≤ x ≤1[x (1-x)] -1/2

Chebyshev II-1 ≤ x ≤1(1-x2)1/2

Laguerre0 ≤ x < ∞e-x

Associated Laguerre0 ≤ x < ∞xk e-x

Hermite-∞ ≤ x < ∞e-x2

Orthogonal Polynomials Generated by Gram-Schmidt Orthogonalization

12

21

1

2

ndxxPn

12

1*

1

0

2

ndxxPn

0

02/

1

1

12/12

2

n

ndx

x

xTn

0

02/

1

*1

02/12

2

n

ndx

x

xTn

2

11

1

2/122

dxxxU n

10

2

dxexL xn

!

!

0

2

n

kndxexxL xkk

n

!2 2/1

0

2 2

ndxexH nxn

160



Equationabr (x)

Legendre-111

Shifted Legendre011

Associated Legendre-111

Chebyshev I-11(1-x2)-1/2

Shifted Chebyshev I01[x (1-x)] -1/2

Chebyshev II-11(1-x2)1/2

Laguerre0∞e-x

Associated Laguerre0∞xk e-x

Hermite-∞∞e-x2

Simple Harmonic Oscillator

0-π

2ππ

11

The Weighting Function r (x) is established by putting the ODE in Self-Adjoint form.Return to Expansion of Functions, Legendre Series

Return to Table of Content Return to Spherical Harmonics

161


References

SOLO

162

Legendre Functions

http://en.wikipedia.org/wiki/

Nagle, Saff, Snider, “Fundamentals of Differential Equations and Boundary Value Problems“, 4th Ed., Pearson/Addison Wesley, 2004, Ch. 11, “Eigenvalue Problems and Sturm-Liouville Equations”

Byron Jr., F.W., Fuller, R.W., “Mathematics of Classical and Quantum Physics”, Dover, 1969, 1970

G.M. Wysin, “Associated Legendre Functions and Dipole Transition Matrix Elements”, http://www.phys.ksu.edu/personal/wysin

Wallace, P.R., “Mathematical Analysis of Physical Problems”, Dover 1972, 1984

Afken, Weber, “Mathematical Methods for Physicists”, Harcourt Academic Press, 5th Ed., 2001

W. W. Bell, “Special Functions for Scientists and Engineers”, D. van Nostrand Company ltd., 1968

S.I. Hayek, “Advanced Mathematical Methods in Science and Engineering”, Marcel Dekker Inc., 2001

L.C. Andrews, “Special Functions for Engineers and Applied Mathematics”, MacMillan Publishing Companie, 1985


163

SOLO

TechnionIsraeli Institute of Technology

1964 – 1968 BSc EE1968 – 1971 MSc EE

Israeli Air Force1970 – 1974

RAFAELIsraeli Armament Development Authority

1974 – 2013

Stanford University1983 – 1986 PhD AA