View
666
Download
7
Category
Preview:
Citation preview
2
ContentsPrefacesGenerating function for integral orderIntegral representationOrthogonality
Reference: http://en.wikipedia.org/wiki/Bessel_function
3
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
4
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
5
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
6
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
7
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
8
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
9
Example (continue)
Therefore, when kasina~3.8317… , |y|0.We get:
3.8317 1.22sin ~ ~ .
2 a D
-5 0 5-5
0
5
10
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?
11
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
12
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?
13
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
Recommended