投影片 1 - Ncku

14. Bessel Functions 1. Bessel Functions of the 1st Kind, J (x) 2. Orthogonality 3. Neumann Functions, Bessel Functions of the 2nd Kind 4. Hankel Functions, I (x) and K (x) 5. Asymptotic Expansions 6. Spherical Bessel Functions Defining Properties of Special Functions 1. Differential eq. 2. Series form / Generating function. 3. Recurrence relations. 4. Integral representation. Basic Properties : 1. Orthonormality. 2. Asymptotic form. Ref : 1.M.Abramowitz & I.A.Stegun, Handbook of Mathematical Functions, Dover Publ. (1970) http://people.math.sfu.ca/~cbm/aands/abramowitz_ and_stegun.pdf. 2.NIST Digital Library of Mathematical Functions: http://dlmf.nist.gov/ Usage of Bessel Functions Solutions to equations involving the Laplacian, 2 , in or circular cylindrical coordinates : Bessel / Modified Bessel functions spherical coordinates : Spherical Bessel functions 1. Bessel Functions of the 1st Kind, J (x) Bessel functions are Frobenius solutions of the Bessel ODE J x a0 2 ! j 0 j x j ! j ! 2 x 2 J xJ x 2 2 J 0

2 j for 1, 2, 3, (eq.7.48) a0 1st kind Jn (x) : n = 0, 1, 2, 3, regular at x = 0. Periodic with amp x 1/2 as x . Mathematica 1 2n n ! cf. gen. func. Generating Function for Integral Order x 1 n g x, t exp t J n x t 2 t n Generating function : g x, t e x t /2 x /2 t e tr x r r 0 r ! 2 s s t s! s 0 x 2 s r s r s s t x r ! s ! 2 r 0 s 0 n r s s n 2s n x

t n s max 0, n n s ! s ! 2 s n 2s x J n x s 0 n s ! s ! 2 For n 0 n = m < 0 : J m x s m Generalize: n s 0 s max 0, n s 0 J x s 0 s x m s ! s ! 2 s 1 in eq.7.48 ) n 2 n! t m m 2t x t 0 t ! t m ! 2 m 2s x s 1 s ! 2 ( a0 2s

m J m x J m x 1, 2, 3, g x, t e Recurrence 1 x x t g x , t g x , t 1 2 t 2 2 Jn x n t n 1 n n n Jn x t t Jn x t n n 2 x 2 n t n J n x J n 2 x n 1 1 1 x g x, t g x, t t t 2 2 J n 1 x n 1 t n 2 n 1 J n x J n 2 x J n 1 x

x e n xt /2 x /2 t J n x t n n n J n x J n x n Jn x t J n 1 x J n 1 x n 1 t n 1 1 2 t n J n 1 x J n 1 x 2 J n x J 1 x J 0 x n 2n Jn x x J n 1 x J n 1 x Ex. 14.1.4 d x n J n x x n J n 1 x dx d x n J n x x n J n 1 x dx J n x J n 1 x n 1 J n 1 x

x x 2 J x J x 2 2 J 0 Bessels Differential Equation J n 1 x J n 1 x Any set of functions Z (x) satisfying the 2n Jn x x J n 1 x J n 1 x 2 J n x recursions must also satisfy the ODE, J x though not necessarily the series expansion. s 0 s x s 1 s ! 2 2s Proof : 1 Z x Z 1 x Z 1 x 2 x 2 Z x Z 1 x Z 1 x 2 x xZ x Z 1 x Z 1 x 2 x2 x Z x Z 1 x Z1 x 2 2 x 2 Z x xZ x 2 Z x x2 1 Z 1 x Z1 x Z 1 x Z 1 x Z 1 x Z 1 x 2 x x x 2 Z x xZ x 2 Z x x2 1 Z 1 x Z1 x Z 1 x Z 1 x Z 1 x Z 1 x 2 x x x2 1

Z 1 x Z 1 x 2 x J n x J n 1 x 1 Z1 x Z 1 x Z x x x 2 Z x xZ x 2 Z x x 2 Z x x k n 1 J n 1 x x 1 Z x Z x 1 1 x QED d2 d 2 2 2 Z k Z k k Z k 0 2 d d 2 Integral Representation : Integral Order

e x /2 t 1/t Jm x C d t t n 1 m x exp t 2 1 n J x t n t n C encloses t = 0. n = integers m n 1 d t t 2 i J n x C C = unit circle centered at origin : t e dt i n i e d n 1 t i 2 i n i x sin i d e d cos x sin n i sin x sin n 0 Re : 1 2 1 e i e i 2i sin t 2 2 i J n x i 1 Jn x 2

t 0 2 d cos x sin n 0 2 Im : d sin x sin n 0 0 d cos x sin n 1 J n x d cos x sin n 0 n = integers 2 J n x d cos x sin n d cos x sin n 0 d cos x sin n n 0 n d cos x sin n n = integers 0 n d cos x sin n J n x J n x 0 1 J0 x 2 2 d e i x sin

0 1 d cos x sin 0 1 J0 x 2 1 2 3 /2 i x cos d e /2 2 i x cos d e 0 Zeros of Bessel Functions nk : kth zero of Jn(x) nk : kth zero of Jn(x) kth zero of J0(x) = kth zero of J1(x) kth zero of Jn(x) ~ kth zero of Jn-1(x) Mathematica Example 14.1.1. Fraunhofer Diffraction, Circular Aperture Kirchhoff's diffraction formula (scalar amplitude of field) : 1 r 4 e i k r r ei k r r r S r r r r r d S k Fraunhofer diffraction (far field) for incident plane wave, circular aperture : r r sin , 0, cos r r r a cos , sin , 0 2 r sin a cos a 2 sin 2 r 2 cos2 r a sin cos a 2

~ d r r d ei b r cos 0 b k sin 0 Mathematica 2 a 2 ~ d r r 0 1 J0 x 2 i b r cos d e b k sin 0 2 d e a i x cos 0 Intensity: 2 2 a J 1 ka sin k sin 1st min: 2 k a sin 11 3.8317 14 ~ 2 d r r J 0 b r 0 d x n J n x x n J n 1 x dx 2 a 2 ~ J 1 ba b

Primes on variables dropped for clarity. Mathematica 2 ~ 2 b ba d x 0 d 2 a x J 1 x J 1 ba dx b Example 14.1.2. Cylindrical Resonant Cavity 1 2 2 2 2 F 0 c t Wave equation in vacuum : F e it 2 2 c2 F 0 F E or B 0 E 0 // E// Circular cylindrical cavity, axis along z-axis : TM mode : Bz 0, E// S 0 1 1 2 2 2 2 z2 2 // means tangent to wall S

Ez z S 0 0 z 0, h E z , , z R Z z 1 d d R 1 d 2 d 2 Z 2 2 2 0 2 2 R d d d Z d z c 1 d d R 1 d 2 d 2 Z 2 2 k 0 2 2 R d d d Z d z d2 2 l 2 Z 0 dz Ez z Ez 0 c d2 2 m d 2 0 d d R 2 2 2 2 k l m R 0

d d 0 p Z p z cos h m A sin m B cos m z 0, h Ez k 2 z l p h p 0,1, 2, 3, m 0, 1, 2, 3, d2 d 2 2 2 Z k Z k k Z k 0 d2 d 2 Ez a 0 R m j p J m m j a with

p m j k2 m j p c h a 2 2 mj = jth zero of Jm(x) . 2 mn p a h 2 resonant frequency Bessel Functions of Nonintegral Order Formally, x 1 exp t J n x t n 2 t n J n x j 0 j x n j ! j ! 2 gives only Jn of integral order. n2 j with n J n x J n x However, the series expansion can be extended to J of nonintegral order : J x j 0 Caution: j x j ! j ! 2 J x & J x 2 j for 1, 2, 3, are linearly independent.

Schlaefli Integral 1 Jn x 2 i For nonintegral , e x /2 t 1/t C d t t n 1 1 t 1 is multi-valued. C encloses t = 0. n = integers e x /2 t 1/t lim 0 1 Re t t Possible candidate for J x is 1 F x 2 i e x /2 t 1/t dt t 1 C1 Strategy for proving F x J x 1. Show F satisfies Bessel eq. for J . 2. Show F x for J x x 0. Mathematica Consider any open contour C that doesnt cross the branch cut 1 e x /2 t 1/t F x dt 2 i C t 1 x 2 J x J x 2 2 J 0 1 e x /2 t 1/ t 1 1 F x dt t 1 2 i C t 4 t 1 e x /2 t 1/t 1 1 F x dt t 1 2 i C t 2 t

x /2 t 1/ t 1 e 2 2 2 x F xF x F dt 2 i C t 1 d e x /2 t 1/t e x /2 t 1/t dt t t 1 d e x /2 t 1/ t d t t x 1 t 2 t x /2 t 1/ t x 1 e 2 t t t 1 x 2 F xF x 2 2 F x2 1 2 x 1 2 t t t 2 t 4 1 e 2 i 2 x x 2 1 2 t t 2 4 t

x /2 t 1/ t t x 1 t 2 t 1 t tend tstart 2 x 2 F xF x 2 2 F For C1 : 1 e RHS 2 i For t x 1 t 2 t t x 1 t 2 t tend tstart t i 0 0 t i 0 this F is a solution of the Bessel eq. 1 x 0

1 F x 2 i Set x /2 t 1/ t 1 e 2 i x /2 t 1/ t u e x /2 t 1/t 1 d t 1 t 2 i C 1 xt 2 : e xt /2 d t 1 t C C = spatial inversion of C , ( same as that for ; B.cut. on +axis ) . 1 x i F x e 2 i 2 1 Mathematica u 1 d u e u C 1 x i 2 i 1 1 e e 2 i 2 1 x sin

1 1 2 x 2 J x 1 z 1 z QED sin z 2. Orthogonality 2 Z k Z k k 2 2 2 Z k 0 2 L Z k k Z k d2 1 d 2 L 2 2 d d where i.e., Z (k) is the eigenfunction, with eigenvalue k2 , of the operator L . ( Sturm-Liouville eigenvalue probem ) Helmholtz eq. 2 2 0 P Z z P J m k in cylindrical coordinates with P a 0 k 2 2 lz2 d2 1 d

m2 2 d 2 d 2 P k P mn k a mn = nth root of Jm d2 1 d 2 L 2 2 d d w is not self-adjoint 1 p 1 exp d 1 exp d p0 p0 d d 2 L L d d L J k k 2 J k is self-adjoint L J k k 2 J k L is Hermitian, i.e., L L J is an eigenfunction of with eigenvalue k2. , if the inner product is defined as d * 0

d d L L d d * * * d * d d d d d L L d d * * * d d * d d d d * * d d d d a d 0 J k L J k L J k J k L L d J k k J k d a J k a k J k a k J k a J k a 1 L J k k 2 J k a k 2 k 2 d J k J k 0 a

2 2 a k J k a J k a J k a k J k a k k d J k J k 0 J i J Orthogonal Sets j 0 a a k J k a J ka J k a k J ka k 2 k 2 d J k J k 0 Let k a i k a j 2 j 2 i 0 a2 a d J J i j a a 0 a Orthogonality : d J J i j i j a a

0 J ; j 0, 1, 2, j a orthogonal set Let k a i k a j a Orthogonality : 2j 2i 0 a2 a d J J i j a a 0 d J J i j i j a a 0

J ; j 0, 1, 2, j a orthogonal set 1 d J J i j i j 0 1 d J J i 0 Mathematica j i j Normalization a a k J k a J ka J k a k J ka k 2 k 2 d J k J k 0 a k J k a J ka J k a k J ka d J k lim a 2 2 k k k k 0 2 y J y J x J y x J x lim a y x x2 y2 2 J x J y xJ x y J y

lim a 2 y x 2y 2 x J x J x xJ x 2 a 2x a 2 2 1 2 d J i a J i a 2 0 x ka y k a 2 a 2 1 2 d J i a J i a 2 0 J n x J n 1 x n 1 J n 1 x x J n 1 x J n x n Jn x x J 1 i J i 2 a 2

1 2 d J i a J 1 i a 2 0 Similarly : (see Ex.14.2.2) a 2 1 2 d J a i a 2 0 Mathematica 2 2 1 2 J i i Bessel Series : J ( i / a ) a 2 1 2 d J i J j a J 1 i i j a a 2 0 For any well-behaved function f () with f (a) 0 : f c j J j a j 1 with c j 2 a for any > 1

d J j f 2 2 a a J 1 j 0 Bessel Series J ( i / a ) 2 2 1 2 2 d J a 1 J i 2 i a 2 i 0 a For any well-behaved function f () with f (a) 0 : f d j J j a j 1 with d j 2 2 2 a 1 2 J 1 j j 2

for any > 1 a d J j f a 0 Example 14.2.1. Electrostatic Potential: Hollow Cylinder Hollow cylinder : axis // z-axis, ends at z = 0, h ; radius = a. Potentials at boundaries : V z 0 V a 0 Electrostatics, no charges : V f , 2 V 0 Eigenstates with cylindrical symmetry : z h , , z P Z z 1 d d P 1 d 2 d 2 Z 2 0 2 2 P d d d Z d z d 2 2 m 2 d m e i m d d P m2 2 2 l P 0 d d d 2Z

2 l Z 2 dz Z 0 0 Z sinh l z m e i m Z sinh l z P J m l J m m j a m j l a d d P m2 2 2 l P 0 d d V , , z a 0 z m j J m m j ei m sinh m j a a c mj m j m j 1 a 2 1 2 d J i J j a J 1 i i j a

a 2 0 2 d e i m m 2 m m 0 cm j 1 2 h a J m m j sinh m j a 2 2 d e 0 a i m d J m m j f , a 0

Recently Viewed Presentations

  • Inside the Earth

    Inside the Earth

    The Core. The core is split into two parts inner and outer. The outer core is made of melted iron.It is totally liquid and the inner core floats in it. The inner core is a solid sphere.It is the center...
  • Vernier Science: The Greenhouse effect

    Vernier Science: The Greenhouse effect

    The probe tips should be 3 cm from the ruler ends and the tape should not cover the probe tips. Connect the temperature probes-as shown on the next slide. Start the Vernierdata-collection program and open the file "03 The Greenhouse...
  • The Prayer of Intercession

    The Prayer of Intercession

    Our enemies Matthew 5:44 44 "But I say to you, love your enemies, bless those who curse you, do good to those who hate you, and pray for those who spitefully use you and persecute you, 45 that you may...
  • Arvutikäsitusõpetus MTAT.03.010 2 ap, arvestus www.ut.ee ...

    Arvutikäsitusõpetus MTAT.03.010 2 ap, arvestus www.ut.ee ...

    Terminaal - süsteem, mis koosneb klaviatuurist ja monitorist. Kõik arvutused tehakse keskarvutis, millega terminaal on võrgu abil ühendatud. Terminaaliks võib olla ka personaalarvuti või tööjaam. Kohtvõrk (LAN = Local Area Network) - arvutivõrk piiratud geograafilisel alal, näiteks ühes ruumis.
  • Credit and Inventory Management

    Credit and Inventory Management

    4% semi-annual cupon payment is due just before futures contract expires. Costaguanan pulgas (currency) 9300 pulgas=$1. 6900 pulgas =$1. Costaguanan interest rate is 95% per year. Establishment Industries common stock. Establishment pays dividends of $2 per quarter. Next dividend is...
  • 2 CHAPTER Strategic Planning and the Marketing Process

    2 CHAPTER Strategic Planning and the Marketing Process

    SWOT ANALYSIS • SWOT analysis Analysis that helps planners compare internal organizational strengths and weaknesses with external opportunities and threats. THE STRATEGIC WINDOW • Strategic window Limited periods during which the key requirements of a market and the particular competencies...
  • Properties of Exponents Test-Out Question (1)33  32 4

    Properties of Exponents Test-Out Question (1)33 32 4

    ANSWERS Test-Out Question 35 or 243 cannot combine c-4 d24 Common Core #3 Know and apply the properties of exponents HOW MANY ARE THERE? THERE ARE 3 RULES FOR EXPONENTS Multiplying Powers With the Same Base 23 • 24 =...
  • Chapter 3 Flow of Control - Computer Science &amp; E

    Chapter 3 Flow of Control - Computer Science & E

    Input Validation You should check your input to ensure that it is within a valid or reasonable range. For example, consider a program that converts feet to inches. You might write the following: What if: The user types a negative...