講者： 許永昌 老師 1. Contents Prefaces Generating function for integral order Integral representation Orthogonality Reference: .

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Ch11.1 & Ch11.2 (Ch12.1e) Bessel Functions of the 1st Kind &

Orthogonality講者： 許永昌 老師

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ContentsPrefacesGenerating function for integral orderIntegral representationOrthogonality

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

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Preface of Bessel function(N) Means only for nZ. n R.

(N) Generating function Jn (x)

Bessel’s ODE series (Ch9.5

~Ch9.6)

Contour integrals

(N) Integral representati

onRecurrence Relations of

Jn.

Hankel functions Hn

(1), Hn

(2).

1st kind of Bessel

function Jn.2nd kind of

Bessel function Nn (or Yn).

Orthogonality of Jn.

(N) Integral

representation of

N0.

Wronskian

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Preface of Modified Bessel functions

Recurrence Relations of In

and Kn.

Modified Bessel functions In and

Kn.

Modified Bessel’s ODE

Bessel functions Jn

and Hn(1).

Asymptotic expansion of J n , N n ,

I n , Kn.

P n and Qn: for

asymptotic

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Preface of Spherical Bessel functions

Recurrence Relations

(N) Spherical Bessel functions

jn and nn, hn(1)

and hn(2).

Helmholtz eq. Bessel’s ODE.

Bessel functions Jn , Nn , Hn

(1) and Hn

(2).

Orthogonality

Series forms

Limiting values:

x << 1

Asymptotic exp. as shown in P4.

1

2

n nx

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Generating function for integral order ( 請預讀 P675~P678)

(N) Generating function:

From this generating function, we can get:

J- n(x)=(-1)nJn(x)=Jn(-x). ------(2) Gotten from g(x,t)=g(-x,t-1)

Recurrence relations: From xg=(t-1/t)/2*g= S J ’n(x)tn 2J ’n=Jn-1-Jn+1 --------(3)

J ’0=-J1. From tg=x/2*g*(1+1/t2)= S Jn(x)ntn -1 ---(4) From g(0,t)=g(x,1)=1 J0(0)=1,Jn(0)=0 & 1=J0(x)+2SJ2n(x).

By Eqs. (3) & (4) we can get Bessel’s equation.

1

2, .xt

ntn

n

g x t e J x t

2

0

1, 1

! ! 2

s n s

ns

xJ x

n s s

1 1

2.n n n

nJ J J

x

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Generating function for integral order ( 請預讀 P679~P680)

From g(u+v,t)=g(u,t)g(v,t) --(5)

(N) Integral representation: From g(x,eiq)=exp(ixsinq)=SJn(x)exp(inq). Therefore,

.n l n ll

J u v J u J v

2 sin

0

0

1

21

cos sin . 6

ix innJ x e d

x n d

2 sin0 0

2 cos

0

1

21

. 72

ix

ix

J x e d

e d

exp cos

,

.

i k r

i

n inn

n

e ikr

g kr ie

i J kr e

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Example ( 請預讀 P680~P682) Fraunhofer Diffraction, Circular Aperture:

The net light wave will be

Based on Eq. (7), we get

0s

a

r

s

0

0

0

0

0

0 0

0

, exp

exp

ˆ~ exp

~ 2 exp sin cos

iks i t

iks i t

iks i t

iks i t

t Ce rdrd ik s s

Ce rdrd ik s r s

Ce rdrd iks r

Ce rdrd ikr

0

00

0 1 12 00

, ~ 2 sin

1where sin .

sin

ai ks t

a a

t Ce rdrJ kr

ardrJ r xJ x J ka

k

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Example (continue)

Therefore, when kasina~3.8317… , |y|0.We get:

3.8317 1.22sin ~ ~ .

2 a D

-5 0 5-5

0

5

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Orthogonality ( 請預讀 P694~P695)

anm: the mth zero of J n .

Derivation: Bessel Eq. :

x1J3(a3mx)

1

0?m n mnJ x J x xdx C

2 2

20. 0.

m

m m

d d d dx J x x J x x x J x

dx dx x dx dx x

L

1

0

1

2 2

0

2 2

0

' ' if 1 .

n m m m n n

x

m nm n mn n m

x

m n m n n mmn

n m

J x J x J x J x dx

dJ x dJ xC xJ x x J x

dx dx

J J J JC

L L

Why?

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Orthogonality (continue)

When mn, we get Cmn=0.When m=n, use L’Hospital role:

2 2

' '

m n

m n m n n m

mnn

n m

m

m n

J J J J

C

J

'' n nJ J 2' ' ' '

2 2

m n n n n

n

n

J J J J

J

2

1 21 .

2 2n

JJ

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Orthogonality (continue)

Therefore, we get

In x[0,1] and f(x=1)=0, fm(x)=Jn(anmx) will form a complete set, because Hn is a hermitian operator when these boundary condition are held.

2 21 1

0

'.

2 2n n

n m mn mn

J JJ x J x xdx

2

2m m m

d dx x x x

dx dx x

H

2

21 1m m m

d dor x x x x x

dx dx xx x

H

Q: However, fm(0)=0 when n0. Do we need to force the function space obey f(0)=0?

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Bessel Series (continue)

A function may be expanded in

1

1

,, where 0, 0 , and 1.

,

.

m m

m m m

m mm

J x J x f axf f a a

J x J x

c Ja

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Homework11.1.1 (12.1.1e)11.1.22 (12.1.14e)11.2.3

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NounsAddition theorem of 1st kind of Bessel

function:Bessel (Fourier-Bessel) Series: