講者： 許永昌 老師 1

Contents Prefaces Generating function for integral order Integral representation Orthogonality Reference: 2

Preface of Bessel function 3 (N) Means only for n  Z.  R. (N) Means only for n  Z.  R. (N) Generating function  J n  (x) Bessel’s ODE  series (Ch9.5 ~Ch9.6) Contour integrals (N) Integral representation Recurrence Relations of J. 1 st kind of Bessel function J. 2 nd kind of Bessel function N (or Y ). Orthogonality of J. (N) Integral representati on of N 0. Wronskian

Preface of Modified Bessel functions 4 Recurrence Relations of I and K. Modified Bessel functions I and K. Modified Bessel’s ODE Bessel functions J and H . Asymptotic expansion of J , N , I , K. P  and Q : for asymptotic

Preface of Spherical Bessel functions 5 Recurrence Relations (N) Spherical Bessel functions j n and n n, h n (1) and h n (2). Helmholtz eq.  Bessel’s ODE. Bessel functions J, N, H  and H . Orthogonality Series forms Limiting values: x << 1 Asymptotic exp. as shown in P4.

Generating function for integral order ( 請預讀 P675~P678) (N) Generating function: From this generating function, we can get: J  n (x)=(-1) n J n (x)=J n (-x) (2) Gotten from g(x,t)=g(-x,t -1 ) Recurrence relations: From  x g=(t-1/t)/2*g=  J ’ n (x)t n  2J ’ n =J n-1  J n (3) J ’ 0 =  J 1. From  t g=x/2*g*(1+1/t 2 )=  J n (x)nt n  1  ---(4) From g(0,t)=g(x,1)=1  J 0 (0)=1,J n (0)=0 & 1=J 0 (x)+2  J 2n (x). By Eqs. (3) & (4) we can get Bessel’s equation. 6

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,e i  )=exp(ixsin  )=  J n (x)exp(in  ). Therefore, 7

Example ( 請預讀 P680~P682) Fraunhofer Diffraction, Circular Aperture: The net light wave will be Based on Eq. (7), we get 8 

Example (continue) Therefore, when kasin  ~3.8317…, |  |  0. We get: 9

Orthogonality ( 請預讀 P694~P695)  m : the m th zero of J . Derivation: Bessel Eq. : 10 x 1 J3(3mx)J3(3mx) Why?

Orthogonality (continue) When m  n, we get C mn =0. When m=n, use L’Hospital role: 11

Orthogonality (continue) Therefore, we get In x  [0,1] and f(x=1)=0,  m (x)=J (  m x) will form a complete set, because H is a hermitian operator when these boundary condition are held. 12 Q: However,  m (0)=0 when 0. Do we need to force the function space obey f(0)=0?

Bessel Series (continue) A function may be expanded in 13

Homework (12.1.1e) ( e)

Nouns 15