Contents Prefaces Generating function for integral order Integral representation Orthogonality Reference: http://en.wikipedia.org/wiki/Bessel_function http://en.wikipedia.org/wiki/Bessel_function 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+1 --------(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