1 Solutions in cylindrical coordinates: Bessel functions

Transcription

1 Bessels 19 1 Solutions in cylinricl coorintes: Bessel functions 1.1 Bessel functions Lplce s eqution in cylinricl coorintes is: 1 Φ Φ + Φ φ φ z Seprte vribles: Let Φ R () W (φ) Z (z). Ten we fin: 1 R + 1 W R W φ + 1 Z Z z Te lst term is function of z only, wile te sum of te first two terms is function of n φ only. Tus we tke ec prt to be constnt clle k. Ten Z z k Z n te solutions re Z e ±kz Tis is te pproprite solution outsie of crge istribution, sy bove plne, (Φ s z ± ), or insie cyliner wit groune wlls n non-zero potentil on one en. Te remining eqution is: 1 R Now multiply troug by : R R R + 1 W W φ + k + k + 1 W W φ Here te lst term is function of φ only n te first two terms re functions of only. Agin we often wnt solution tt is perioic wit perio π, so we coose negtive seprtion constnt: W φ m W W e ±imφ Finlly we ve te eqution for te function of : R + k R m R 1

2 To see tt tis eqution is of Sturm-Liouville form, ivie troug by : R + k R m R (1) Now we ve Sturm-Liouville eqution wit f (), g() m /, eigenvlue λ k n weigting function w (). Eqution (1) is Bessel s eqution. Te solutions re ortogonl functions. Since f (), we o not nee to specify ny bounry conition t if our rnge is, s is frequently te cse. (We o specify tt R remin finite.) We o nee bounry conition t. It is simpler n more elegnt to solve Bessel s eqution if we cnge to te imensionless vrible x k. Ten: k k k R k x + k R k m k R x R + xr m x x R Te eqution s singulr point t x. So we look for series solution of te Frobenius type (cf Le Cpter ): R x p n x n x R x x R n (n + p) n x n+p 1 n (n + p) n x n+p 1 n Ten te eqution becomes: (n + p) n x n+p 1 + n x n+p+1 m n n n Te inicil eqution is given by te coefficient of x p 1 : p m p ±m n x n+p 1 Tus one of te solutions (wit p m) is nlytic t x, n one (wit p m) isnot. Tofin te recursion reltion, look t te k + p 1 power of x: (k + p) k + k m k

3 n so k k (k + p) m k k +kp + p m k k k +kp k (k ± m) Let s look first t te solution wit p +m. Wecnstepowntofin ec k. If we strt te series wit, ten k will lwys be even, k n, n n 1 1 n (n +m) (n ) (n +m) n 4 ( 1) 3 3 n (n 1) (n ) 3 (n + m)(n + m 1) (n + m ) n 6 ( 1) n n n! Te usul convention is to tke Ten 1 n (n + m)(n + m 1) (m +1) 1 m Γ (m +1) 1 ( 1) n 1 n m Γ (m +1) n n! n (n + m)(n + m 1) (m +1) ( 1) n n!γ (n + m +1) n+m (3) n te solution is te Bessel function: J m (x) n () ( 1) n x m+n (4) n!γ (n + m +1) Te function J m (x) s only even powers if m is n even integer n only o powers if m is n o integer. Te series converges for ll vlues of x. Let s see wt te secon solution looks like. Wit p m te recursion reltion is: k k (5) k (k m) were gin k n. Now if m is n integer we will not be ble to etermine m becuse te recursion reltion blows up. One solution to tis ilemm is to strt te series wit m. Ten we cn fin te succeeing n : (n+m) (n 1)+m (n + m) n (n )+m 4 (n + m)(n + m 1) n (n 1) ( 1)n Γ (m +1) n!γ (n + m +1) n m wic is te sme recursion reltion we before. (Compre te eqution bove wit eqution (3). Tus we o not get linerly inepenent solution 3

4 tis wy 1. (Tis ilemm oes not rise if te seprtion constnt is tken to be ν wit ν non-integer. In tt cse te secon recursion reltion provies seriesj ν (x) tt is linerly inepenent of te first.) Inee we fin: nifwecoose J m (x) n n ( 1) n Γ (m +1) n!γ (n + m +1) n mx (n+m) m ( 1) n Γ (m +1) m x m+n m n!γ (n + m +1) ( 1) m m Γ (m +1) m ten J m (x) ( 1) m J m (x) (6) Wit tis coice J ν (x) is continuous function of ν. (Noticettwecnlso express te series using eqution (3) for te coefficients, wit m mn n k + m,n were n for n<m.) We still ve to etermine te secon, linerly inepenent solution of te Bessel eqution. We cn fin it by tking te limit s ν m of liner combintion of J ν n J ν known s te Neumnn function N ν (x) : N m (x) lim J ν (x)cosνπ J ν (x) ν (x) lim ν m ν m sin νπ J m+ε (x)cos(m + ε) π J (m+ε) (x) lim ε sin (m + ε) π J m+ε (x)(cosmπ cos επ sin mπ sin επ) J (m+ε) (x) lim ε sin mπ cos επ +cosmπsin επ J m+ε (x)( 1) m cos επ J (m+ε) (x) lim ε ( 1) m sin επ Now we expn te functions to first orer in ε. We use Tylor series for te Bessel functions. Note tt ε ppers in te inex, not te rgument, so we ve to ifferentite wit respect to ν.. J m+ε (x)( 1) m J (m+ε) (x) N m (x) lim ε ( 1) m επ ( 1) m lim ( 1) m J m + ε J ν ε επ ν J m + ε J ν νm ν νm Using reltion (6), we ve: N m (x) 1 π Jν ν ( 1) m J ν νm ν νm 1 Tis ppens becuse te two roots of te inicil eqution iffer by n integer: m. 4

5 Te erivtive s logritmic term: J ν x ν ( 1) n x n ν ν n!γ (n + ν +1) xν ν n n ( 1) n x n + x ν n!γ (n + ν +1) ν n ( 1) n x n n!γ (n + ν +1) n x ν ν ν eν ln x lnxe ν ln x x ν ln x n so J ν /x s term contining J ν ln x. Tis term iverges s x provie tt J v () is not zero, i.e. for ν. Te function N v (x) lso iverges s x for ν, becuse it contins negtive powers of x. (Te series for J ν strts wit term x ν.) N ν is finite s x becuse J ν goes to zero sufficiently fst. Two itionl functions clle Hnkel functions re efine s liner combintions of J n N : H (1) m (x) J m (x)+in n (x) (7) n H m () (x) J m (x) in n (x) (8) Compre te reltion between sine, cosine, n exponentil: e ±ix cosx ± i sin x 1. Properties of te functions Te Bessel functions (Js) re well beve bot t te origin n s x. Tey ve infinitely mny zeroes. All of tem, except for J, re zero t x. Te first few functions re sown in te figure. J Te first tree Bessel functions. J,J 1 (re) n J x 5

6 For smll vlues of te rgument, we my pproximte te function wit te first term in te series: 1 x m J m (x) for x 1 (9) Γ (m +1) Te Neumnn functions re not well beve t x.n s logritmic singulrity, n for m>, N m iverges s n inverse power of x : N (x) ln x for x 1 π m (m 1)! N m (x) for x 1, m > (1) π x For lrge vlues of te rgument, bot J n N oscillte: tey re like mpe cosine or sine functions: J m (x) πx cos x mπ π for x 1,m (11) 4 N m (x) πx sin x mπ π for x 1,m (1) 4 n tus te Hnkel functions re like complex exponentils: H m (1,) πx exp ±i x mπ π for x 1,m (13) 4 Notice tt if m>1, te lrge rgument expnsions pply for x m rter tn te usul x Reltions between te functions As we foun wit te Legenre functions, we cn etermine set of recursion reltions tt relte successive J m (x). For exmple (Le 8.4.3) Jm (x) x x m J m+1 (x) x m (14) wic is vli for m. In prticulr, wit m we obtin: J 1 (x) J (x) (15) x (xm J m (x)) x m J m 1 (x) (16) From (14) n (16) we my obtin J m+1 + J m 1 m x J m (17) n similrly J m+1 J m 1 J m x Te sme reltions ol for te Ns ntehs. (18) 6

7 1.4 Ortogonlity of te J m Since te Bessel eqution is of Sturm-Liouville form, te Bessel functions re ortogonl if we emn tt tey stisfy bounry conitions of te form (SL review notes eqn ). In prticulr, suppose te region of interest is to, n te bounry conitions re J m (k). We o not nee bounry conition t becuse te function f () is zero tere. Ten te eigenvlues re k mn α mn were α mn is te nt zero of J m. (Te zeros re tbulte in stnr references suc s Abrmowitz n Stegun. Also progrms suc s Mtemtic n Mple cn compute tem.) Ten J m (k mn ) J m (k mn ) [J m (k mn )] δ nn (19) 1.5 Solving potentil problem. Exmple. A cyliner of rius n eigt s its curve surfce n its bottom groune. Te top surfce s potentil V. Wt is te potentil insie te cyliner? Te potentil s no epenence on φ n so only eigenfunctions wit m contribute. Te potentil is zero t,sotesolutionweneeisj (k)wit eigenvlues cosen to mke J (k). Tus te eigenvlues re given by k n α n, were α n re te zeros of te function J. Te remining function of z must be zero t z, sowecooseteyperbolicsine. Tustepotentil is: Φ (,z) n J (k n )sin(k n z) Now we evlute tis t z : n1 V Φ (,) n J (k n )sin(k n ) n1 Next we mke use of te ortogonlity of te Bessel functions. Multiply bot sies by J (k r )n integrte from to. (Note ere tt te weigt function w (). Tis is te firsttimeweveseenweigtfunctionttisnot1.) Only one term in te sum, wit n r, survives te integrtion. V J (k r ) J (k r ) n J (k n )sin(k n ) n1 r [J (k r )] sin (k r ) r [J (k r )] sin (k r ) 7

8 To evlute te left n sie, we use eqution (16) wit m 1: J (k) 1 k k (kj 1 (k)) 1 k J 1 (k) k J 1 (k) So r V J 1 (k r ) k r [J (k r)] sin (k r ) V k r J 1 (k r )sin(k r ) wereweuseteresultfromeqution(15)ttj J 1. Finlly our solution is: J (α n /) sin (α n z/) Φ V α n J 1 (α n ) sin (α n /) n1 Te first two zeros of J re: α 1.448, α 5.51, n tus te first two terms in te potentil re: J.448 sin.448 z Φ V.448J 1 (.448) sin J 5.51 sin 5.51 z 5.51J 1 (5.51) sin Moifie Bessel functions Suppose we cnge te potentil problem so tt te top n bottom of te cyliner re groune but te outer wll t s potentil V (φ,z). Ten we woul nee to coose negtive seprtion constnt so tt te solutions of te z-eqution re trigonometric functions: Z z k Z Z sin kz + b cos kz At z,z(z), so we nee te sine, n terefore set b. We lso nee Z (), so we coose te eigenvlue k nπ/. Tis cnge in sign of te seprtion constnt lso ffects te eqution for te function R ()becuse te sign of te k term cnges. R or, cnging vribles to x k : x x R x k R m R xr m R () x 8

9 wic is clle te moifie Bessel eqution. Te solutions to tis eqution re J m (ik). It is usul to efine te moifie Bessel function I m (x) by te reltion: I m (x) 1 i m J m (ix) (1) so tt te function I m is lwys rel (weter or not m is n integer). Using eqution (4) we cn write series expnsion for I m : I m (x) 1 i m n n ( 1) n n!γ (n + m +1) m+n ix 1 x m+n () n!γ (n + m +1) As wit te Js, if m is n integer, I m is not inepenent of I m : in fct: I m (x) i m J m (ix) i m ( 1) m J m (ix) ( 1) m i m I m (x) I m (x) (3) Te secon inepenent solution is usully cosen to be: Ten tese functions ve te limiting forms: n I m (x) K m (x) π im+1 H (1) m (ix) (4) 1 x m for x 1 (5) Γ (m +1) K (x).577 ln x for x 1 (6) K m (x) Γ (m) m for m>, x 1 (7) x At lrge x, x 1,m, te symptotic forms re: I m (x) 1 πx e x (8) n π K m (x) x e x (9) (cf Le Cpter 3 Exmple 3.9) Tese functions, like te rel exponentils, o not ve multiple zeros n re not ortogonl functions. Note tt te Is re well beve t te origin but iverge t infinity. For te Ks, te reverse is true. Tey iverge t te origin but re well beve t infinity (See figure below). 9

10 I,K Moifie Bessel functions. K soli I se Blck m,blue 1, Re x Te recursion reltions stisife by te moifie Bessel functions re similr to, but not ienticl to, te reltions stisfie by te Js. For te Is, gin we cn strt wit te series: x Now let k n 1: x Im Im x m x m 1 m 1 m 1 m 1 x m x n n1 k k Im x m 1 x n n!γ (n + m +1) n x n 1 n!γ (n + m +1) 1 x k+1 k!γ (k + m +) 1 x k+m+1 k!γ (k + m +) I m+1 (x) x m (3) n similrly x (xm I m )x m I m 1 (31) Expning out n combining, we get: Im I m+1 + I m 1 m x I m I m 1 I m+1 (3) 1

11 For te Ks, te reltions re: n consequently: x (xm K m ) x m K m 1 ; x Km (x) x m K m+1 (x) x m 1.7 Combining functions K m 1 K m+1 m x K m K m 1 + K m+1 K m (33) Wen solving pysics problem, we strt wit prtil ifferentil eqution n set of bounry conitions. Seprtion of vribles prouces set of couple orinry ifferentil equtions in te vrious coorintes. Te stnr solution meto ("Ortogonl" notes ) requires tt we coose te seprtion constnts by fitting te zero bounry conitions first. In stnr 3-imensionl problem, once we ve cosen te two seprtion constnts we ve no more freeom n te tir function is etermine. Wen solving Lplce s eqution in cylinricl coorintes, te functions couple s follows: Zero bounry conitions in : J m α mn Te eigenfunctions re of te form: e ±imφ A mn sin α mn z + B mn cos α mn z Te set of functions J m αmn e ±imφ form complete ortogonl set on te surfces z constnt tt boun te region. Zero bounry conitions in z : Te eigenfunctions re of te form: nπ nπ nπz A mn I m + B mn K m sin e ±imφ Te set of functions sin nπz e ±imφ form complete ortogonl set on te bounry surfce constnt. Tus in solutions of Lplce s eqution, Jsin lwys couple wit te yperbolic sines n yperbolic cosines (or rel exponentils) in z, wile te Is n Ks in lwys couple wit te sines n cosines (or complex exponentils) in z. Exmple: Suppose te potentil on te curve wll of te cyliner is V (, φ,z)v sin φ Te top n bottom re groune Te solution is of te form Φ (, φ,z) + sin m n1 nπz nπ e imφ A mn I m 11 nπ + B mn K m

12 Te solution must be finite on xis t, so B mn. Now we evlute te potentil t Φ (, φ,z) + sin m n1 nπz nπ e imφ A mn I m V sin φ We mke use of ortogonlity by mutltiplying by sin pπz/ n integrting from to. + m n1 sin pπz nπz nπ sin ze imφ A mn I m + m pπ eimφ A mp I m Te result is non-zero only for o p. Next we mke use of te ortogonlity of te e imφ + m π pπ e imφ e imφ φa mp I m π V pπ p even V sin φ V sin φ pπ sin pπz z cos pπz V pπ sin φ [( 1)p 1] sin φe imφ φ To o te integrl on te rigt n sie, express te sine in terms of exponentils. π π sin φe imφ e iφ e iφ φ e imφ φ i π i (δ m 1 δ m, 1) Only one term wit m m survives te integrtion on te left n sie. Tus Φ (, φ,z) 4V π πa m pi m n1, o But I 1 I 1, (eqn 3), so pπ V π pπ i (δ m 1 δ m, 1) 1 (δ m 1 δ m, 1) i A m p 4V pπ sin Φ (, φ,z) 4V π nπz 1 in n1, o sin 1 e iφ I 1 nπz I m nπ I 1 nπ sin φ n pπ e iφ I 1 I nπ 1 I 1 nπ nπ I 1 nπ p o

13 Te imensions re correct, n te 1/n gives us resonble convergence. Te rtio I nπ 1 I nπ 1 for. 1 Te potentil t is zero, s expecte. Te m 1"source" (te potentil on te surfce) genertes n m 1response. Agin we cn look t te first few terms (ere 1). Let s tke / n φ π/4 (re) n Φ π/ (blck) pi/v ro/ 1.8 Continuous set of eigenvlues: te Fourier Bessel Trnsform In Le Cpter 7 we pproce te Fourier trnsform by letting te lengt of te omin in Fourier series problem become infinite. Te ortogonlity reltion for te exponentil functions: 1 L exp L L i nπx L exp i mπx L x δ mn becomes 1 e ikx e ikx x δ (k k ) π Tt is, te Kronecker elt becomes elt function, n te countble set of eigenvlues nπ/l becomes continuous set of vlues <k<. Te sme ting ppens wit Bessel functions. Wit finite omin in, sy, we cn etermine countble set of eigenvlues from te set of 13

14 zeros of te Bessel functions J m. If our omin in becomes infinite, ten we cnnot etermine te eigenvlues, n instewevecontinuousset. Te ortogonlity reltion: J m α mn J m α mk [J m (α mn )] δ nk becomes: J m (k) J m (k ) δ (k k ) (34) k (Te proof of tis reltion is in Le Appenix 7.) Ten te solution to te pysics problem is etermine s n integrl over k. For exmple, solution of Lplce s eqution my be written s: Φ (, φ,z) e imφ A (k) f (kz) J m (k) k m were f (kz) epens on te bounry conitions in z. It will be combintion of te exponentils e kz n e +kz. Exmple: Suppose te potentil is V () V sin on plne t z, n we wnt to fin te potentil for z>. Ten te pproprite function of z is e kz, cosen so tt Φ s z ( long wy from te plne). We lso nee te J m, wic remin finite t. Ten te solution is of te form: Φ (, φ,z) e imφ A m (k) e kz J m (k) k m Evluting Φ on te plne t z, we get: V sin Φ (, φ, ) e imφ A m (k) J m (k) k m Now we cn mke use of te ortogonlity of te e imφ. Multiply bot sies by e imφ n integrte over te rnge to π. On te LHS, only te term wit m survives te integrtion, n on te RHS only te term wit m m survives. π V sin π e imφ φ e imφ e imφ φ A m (k) J m (k) k m πv sin δ m π A m (k) J m (k) k Fourier Bessel trnsform. Next multiply bot sies by J m (k ), integrte from to in, n use eqution (34) to get: V sin J m (k ) δ m A m (k) J m (k) J m (k ) k A m (k) δ (k k ) k A m (k ) k k We lso rop te primes on te m for convenience. 14

15 wic etermines te coefficient A m (k ) in terms of te known potentil on te plne. Only A is non-zero, s expecte from te zimutl symmetry. On te RHS, let / x. We get (ropping te primes on k) V Tis integrl is GR 6.671#7. A (k) V k sin xj (xk) x A (k) k So if < 1 <k 1 1 (k) if <k<1 So finlly we ve Φ (,z) V 1/ kj (k) 1 (k) e kz k 1 xj x V 1 x e xz/ x For z we ve 1 xj x Φ (, ) V x V 1 1 x / sin wic gives us bck te potentil on te plne. Te crge ensity on te plne is given by E ẑ σ z ε GR 6.554# So we nee Φ z V z V V 1 1 x J x x 1 x x 1 x ( 1) n n [n!] ( 1) n n [n!] x n 1 n x x (n+1) 1 x x Let x sinθ, x cosθθ Ten te integrl is I n Tus π/ sin (n+1) θ cos θ cos θθ π/ sin (n+1) θ θ (n +1)!! π (n +)!! GR 3.61#3, Le C P9 15

16 Φ z V π z V π 4 ( 1) n [n!] ( 1) n n n n so te crge ensity on te plne is σ ε E z ε V π 4 (n!) ( 1) n n (n!) Teplotsowsteseriesupton 4. 8 n (n +1)!! n+1 (n +1)! n (n +1)!! n (n +1)! n (n +1)!! (n +1)! sigm /epsv ro/ Convince yourself tt tis grees wit te fiel line igrm. (Te se line is te potentil.) 16