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# Classical Electrodynamics I PHYS 685

Mason

GPA 3.66

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This 61 page Class Notes was uploaded by Sonny Breitenberg on Monday September 28, 2015. The Class Notes belongs to PHYS 685 at George Mason University taught by Staff in Fall. Since its upload, it has received 56 views. For similar materials see /class/215193/phys-685-george-mason-university in Physics 2 at George Mason University.

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Date Created: 09/28/15

BoundaryValue Problems in Electrostatics 11 Reading Jackson 31 through 33 35 through 310 Legendre Polynomials These functions appear in the solution of Laplace39s eqn in cases with azimuthal symmetry Consider a point charge q located at x y z 0 0 a d d2a27 272arc056 r Potential 95 at point r 6 1 7 2 l q q a dz 1 if 7 l 7 27 9 i l sud lfrsor r COS r2 Take t ar and r E cost 1 lt1 7 r r q g t i y 7 2 2 12 n 47mm 1 Where 90v 1 NH Is the generatmg functionquot for Legendre polynomials If ltz 2 Itl lt i then gt x can be expanded in a binomial series To derive the binomial expansion no 1 u start With Maclaurin s Thm 2 712 0 For fu1u l2 fl1u 71u32 fWu 12lu 52 3 7 7 3135 772 f u i 1 lu etc W ilquot 2714 1 2714 gt f Uun Zn 271 246 270 71 271 7 l u 239171 00 So 1 1912 2 110 2n 2TH 1 21 eg n 5 1098765432 2n 2n 2 719 53 n 39 1086i2 gum4394 2mm 71quot 2n So 1u L2 2 2230 u Vith u 72Iff2 12If2 1 2n5 1 1 1 9t712ow1 1 2177 u r r n 7 1 71 n Blnomlal Thm 1 i b i anik hi i 70511 I 22nn2 g knik co n 2n 1 n EEEIVWQI kt H We would like to express this as a power series in t with coeffs that depend on x 9amp1 2 PW t n0 To this end the following theorem is useful Do 11 00 112 n n2 n even 2 me Z Z Manat Wlth 7 710 k 110 k 2 n 7 1 2 71 add co co Proof Start with S Z Z akm 10 k0 no 11 Set 171217k20 571 2 5Zak k 110k0 VI2 so Alternativelyset m 71 7 21c gt k 3 712 i S Z Z aka172k n0 k0 4 7 71V 271 7 Lquot 10 2r k t k2k H 22n7kkgt n 7 L k k n 7 1 71mm 7 2k k 21quot 2 t 2quot2 i akn72k 22n72kkgt n 7 2k k k n 7 2k 71 6 271 7 2k 2I 2k t 22391 2 c k n 7 k n 7 2k 00 722 Emmi 2k ll be n7 in 2 P in 2 90 27239 A4014 717210 21 go I 112 k 7 1 2quot 2 l 3721 j P 271k 71 7k n 7 2k 1 A few examples Pom 1 PAI 7 JAN 2 3 2 1 13235 2242727 5 gm 17 2It rw 7 2 mm 110 89 l 7 t 7 7 7 2 mm W at l72xtt232 0 LHs7r 7ta 7I 7tiptn 172zt229 quotI 172Itt2 M quotI so so I7t Emmiquot172xttznPnzt 1 110 n0 00 gt 2m Pnz 1 Pm W1 7 2 nIPnz t 7 anx t nPnz rm 0 720 Since this vanishes for all values of I each power of I must vanish separately 2 71 7 lP1r Pn7lI 7 271 1131 7 IP1I 71 1 Pn1l 0 n PH1 7 2n 1 11212 n l Pn1r 0 recurrence relation This can be used to find higher order Legendre polynomials For example given P0x l and P1 x x we find with n 1 P0017 31P1L 2P21r 0 71 3 2 3 1 2 EM ff all 5 as we found With the explicit formula With repeated Pom 1 application we find P105 x P200 7 61371 Note that P3 3513 7 31 P x 1 1 f PM g 35254 7 3002 3 1351 63x5 7 7013 l5z PM 23116 7 315254 10532 7 5 Pm i 429357 7 6932c5 31513 7 35m Psm 1738 6435328 7 HOUR 693014 7 l26012 35 8g t 7 7 7 2 PW 9x 17212t23Z HA co co 7 t Zan t 1 7 2Itt2 Z Rztquot 120 710 co ZiR M W i 210137200 W P430 1 PHI tquot 0 110 1351 205 QIPLJI Pn71I PHI 0 With 71 7gt n 1 P3911r P1I 21131 PHI 1 Differentiating recurrence relation wrt x 71P11 7271 l1 Pr 7 271 1Pn1 11 1 P711 0 n 1R 1znPL1r 271 1 IPLM 2n 1Pnr 2 Subtract eqns 2 2 7 271 1 1 2111P391I 2nP 11I 22711IPI 22711Pn1 QTIJVUPQKN Zn f39UPJHO 72 271 1 1131 271 1 Pnr A PAHU 7Plr 2111Pnr 3 g l 3 gt 1 139 P31 n l PquotJ with 71 a 11 1 PAH l P V l 71Pquot112 1 gm 3 gt PHI rain 7 n PM 5 4 1 5 3 1341 1 Plr IP 1J 71 P 41 1 2 131 n IP41 1 12 n Pnlr 71 I Pn1 G 16 11 5 L l 12 PHI 7 2IPQ1 71P1I11P1r7 HTPIHP11 7 I P1 7n2 Pn1 l 7 12 Pr Kr 7 2m P405 7101 1 Pnx 0 quotLegendre39s differential eqnquot Legendre polynomials satisfy Legendre39s eqn hence the name Legendre39s eqn may also be written as i 17 I ELM 71 n 1 Pnr 0 7 Emmi 17 I 1221 7101 lmnomxr 7 0 Pm 1 7 r2 12111 m m 1 Pmm PM 7 0 interchanging m and n Subtract eqns and integrate from x l to l 1 mm lt1 7 Human 7 PM lt1 7 rolenun dx 7 mltm17nn11 1 am am dr Integrate l 11 Pmx 1 7 n2Pn d by parts awnammtl 7 517 12mm dr 0 12 l 1 7 ms 7 lt1 7 H mm PM dz 7 lt1 7 r2 Puss Rim dz 7 0 L 7i 1 mm l 7 nn 1 ll Pmx Pnx dz U l Form n Pmr Pnr d1 0 71 gt Legendre polynomials are onhogonal on interval 1 1 What about n m 7 1x 2 t 1 l 7 2 It t2 i L an in 0 1 dm 1 co co P P tmnd so t2 1 P2 d In n05 1 no as cross terms vanish because of onhogonality 1 dz 7 1 i 2duillu1t 112Itt272t 172 u it 177 co 2111 lnlt1t 2 t 14 lit Hzn co t2n co 2 1 2 Th2 7 t39 P39d us 72nl 7 ii quotw T 21P2139d 7 2 71 I I72nl Legendre polynomials are also complete on 1 1 Legendre series 1 fr Zn Pnx with an 272 fzPnz dI 110 71 Next we39ll derive a useful alternative formula for computing Legendre polynomials called Rodrigues39 formula 1 d 77 2717 2quotndzquotltx P1100 112 15 7 1 2n 7 2r 7172 Pnx 7 Es 71 J as derived earlier 7121 v 1 n 2172 quot1 1 7 1 d 1quot 39 2H 7 grwn7ryl 2 l d n 1 lt3 1 Differentiating x2 n times yields r 721v 7 271 27 1172139 2n7272n72r71n72rlI 7W1 Tl 47 2 2quot 71721 v dquot 271721 W I 1th drill i replace 2 The sum is extended from fl2 to n In each of these tenns the exponent of x is lt n gt n differentiations yields zero So we39re just adding zero n 7l 3 Binomial Thm 1 7 lquot E if n VIZ PZ VB 7 n 7 7 Laplace eqn in 1 82 EFF 1 8 681 1 821 7 spherical coords r 8r2 T r2 sine 86 SM 8639 r2 51112 9 8432 7 Separation of variables lt1 Um P6 6207 7 dZU l d dP 1 d2 5 r2 sin2 639 i lt Q 1 if 7 mi 770 Udr2 Przsin d 5 d6 Qd 2 If the region under consideration includes the full range of azimuth 0 S I lt 2n then Q must be periodic with period 2n 1 d2Q gt const must be negative 7 im a Q olt s Q do ii in o with m an integer leU 1 d IF 2 26 if 7 39 97 2 Thus r 5111 U drz Prz si119d9 gm d9 m 1 am 1 d dP 17 2 2 7 i i 2 T 5 9 U 172 Pr siu d SM 19 m 9 c Zll if 7 T A 7 5111 7 U W sin a P 51110 M cons r2 d2 U m2 1 d d6 1 d 13 m2 gt milt5um gt711 rlimP0 dQU 7 11 1 1172 72 70 5 UAr39lBrquot sin2 6 1P sin 6 d6 lz1iiP0 sin2 9 1 d Eqn for P6 mg 39ith J GOSH II 7 sin 0419 d 7 1P 2 a 1 7 I E W l 7 1152 P 0 quotgenerallzed Legendre eqnquot When m 0 as for cases with azimuthal symmetry this reduces to the 18 Legendre eqn We know that Legendre polynomials satisfy Legendre39s eqn for non negative integer values of f Other values of f are excluded on physical grounds since in these cases the soln diverges atx 1 or 1 9 0 or Tr The same is true of the 2nd linearly independent soln Legendre39s eqn is a 2nd order diff eqn for cases with f 0 1 2 For example the 2nd soln for f 0 is lntan 92 So the general soln of the Laplace eqn in spherical coords for cases with azimuthal symmetry is NT 639 gm Tl Bl T7ll Pcos 639 0 Now we return to the generalized Legendre eqn which is required in cases that lack azimuthal symmetry d 2rlP m2 El71Elll7liz2P70 T712 l 7 I2 0r 17m2P 72rP lll7 JP0 Solutions are obtained by differentiating Legendre39s eqn Recall Leibnitz39s formula for differentiating products dn 11 n dais d5 5W Ban 7 2 Am M 50 n 7 s s dzn s BUB Legendre39s eqn 17 z2P M 1 R 0 d Differentiate quot1 m 1 dmti s d51 de Z I li dui 1P1zi o m 1111682 50 771 l 7 s s drm dIs davm d117 Pt 20 dJcm 7 0 mL m 1 d1nl7s 2 dsi 7 l 7 7P l l l g 771 17 9 s drquot11 5 I dz51 l d2 d3 E 1 139120 13 2 if 3171 2r dra 1719 72 So only the terms with s m1 m m 1 are non zero dm2 m 1 M m 1 439quot ima 3 172712772 P772 P 11 1 I lll m2 I m I am l 2 7n 71 111quot 1 dl39quot 1717131 Define u E 7 liftquot 1 7 I2 11 7 27n 1 1 u W l 7 7n7n Lu 0 Replace u1t with l 7 I2quot 2 111 j u l 7 IZ quot2 U u l 7 rm2 U39 lm I I I217I2 lixzy 2mIv 7711 mm2 121 H if 2 7m2 H 7 u z I 1 2 2 17122vquot2mrvl mv er 1 21 7 27n1rv lll77n7nLv 7 0 7 2 n7 1 7 m2 7 l I U 21U lll 170 1712 This is the generalized Legendre eqn or associated Legendre eqn dmp So solns are 1 7 I2mz l m The quotassociated Legendre functionsquot are defined as m 31111 7mm 1 7 I21n2 d for m 2 0 Note Not all texts include the factor 1 Since the highest power ofx in Plx is x1 we must have m S 1 otherwise P r E 0 21 We can use Rodrigues39 formula to define associated Legendre fcns with negative m dl 1 Recall Rodrlgues39 formula Pr mz 7 ll d111 1 dl I m 7 1n 2 39m 2 2 l 77120 PI77l l7zm r7l Um 1 2 quot12 de 2 1 l 1 lrdf tr 7 211 r I WWW a s0 va 1 or nega we m if Iml S Z Next we39ll show that PF 139 olt P r Apply Leibnitz39s formula to I2 7 1 r l I 7 1l dH m I dH m 2 l l l ma 71 7mm 71 7 1m dH39H T l d5 7 50 I m 7 s 3 dz 39quot 5 I dzs x7 w 22 zm lm I lrn7s s 23 d r2 L Z MLzilddx71 r dIH 50 1 771 7 s s dim Both 3 and lm s must be S 1 otherwise one of the derivatives is zero lm s 1 gt 52771 Thus for m 2 0 the sum runs from s m to l mlt0 s0tolmlm dquotri1 1 T i L I d 39 2 l 7 l m 1 57m 1 is 80 dac quot1r 7 72lm7ss S7771I1 173 I71 dzm 1 1 m 1 ls7vnI 7 75 F 32 1 l l 2 0 7 dr quot l z 1n7ss s77nl7s 7711 7177 s39m 7539 For 7 m L 12 7 1 I2 Z I my I 1 I 1 drl m 50 17m7s s s ml7s d 39 Hquot I 7n I 1 n 1 1I s 24 F397 7271112 m m dmquotquot 1 J J 5201 771 s s 5 m l s s s m L1H 121a2 2 li7nl 1 177 1W drl m Smliss7mslmis I 7 1y 7 1 PFquot air2rm2 diligently szh satisfy the generalized Legendre eqn for integer values of l m 25 with l 2 0 and l S m S I As with azimuthally symmetric Laplace eqn and Legendre polynomials other values of l and m as well as other solns are unphysical P111 7l7zzl2 7si116 P21 z 7325 l 7 r212 73 cos 39sin P2 2 3 3 75 5x2 7 l l 7 I212 7E 5 c0526 7 l sin P I 15x l 7 1 2 15 059 sinz 39 PM 7151 7 m232 715 51139 I 3l7x23si1126 5 5 7E 713 7 3x 1 7 zzl2 7 7 00539 7 3 0059 sin 15 i5 P z g 7x2 7 l l 7 12 g 7 c052 639 71 sinz 639 1331 7105241 7 3232 7105 c059 51139 105l 7 22 105 sin4 6 For fixed m and 2 Iml P I are orthogonal over 1 S x S 1 26 i m my 7 2 lm i P 1 d1 I 7 m 0 The proof is straightforward using Rodrigues formula and integrating by parts but cumbersome so we39ll omit it Summary of angular functions When the region of interest includes 0 6 0 n 2ll 17m 2 1m P 9 PZWCDS 9 orthonormal When the region of interest includes 4 6 0 211 20 0 eii39m with m an integer and m 2 0 Qua 0c emquot with m any integer Recall the orthogonal functions in a Fourier series can be expressed as 27 swim 1 for r 6 0 a With a 21T the functions are 6 So Q are orthogonal r 2 21 Normalization quotquot4 equotquot do dd 2w 0 CI 1 imc m 5 orthonormal cm Define quotspherical harmonicsquot 211 l i m an 171094 139 24 4W 1 m Pl c059 e quot These fcns are orthonormal and complete over the unit sphere OSQSTI39 and OS lt21T 28 131H1COS 9 Eiimw 211 1m 4T 17m 2111m l m 7 if 1 117131 6 1 mm 417 17m 1m l COS 5 1171109 45 21 1 l 7 m 417 l m UmW 9 0 Lm 1 m m COS 9 57mm 2W Tr Orthonormality dq siu deijfmf mewb6y U 0 2w 1 0r j rm 1 muse YFmGytp we 6l3916m39m m I Completeness Z Y6 o Ylmw 1 1966 6 64 0 SH 0 m l 6c056 7 cos 9 54 05 g 39 1 H m o 7 sin 9 5quot cos 9 W H 7 Ln 1 A J M wm Err M7 10 51 27f 7r 0 w sin2 9 a sin 9 cos 9 e N m q 1 w H 5 4 01 2m id 1 c052 9 7 7 2 sin3 9 53 39112 9 c059 52 sin 9 5 c052 9 7 1 517 3 7 0053 9 7 Enos 9 2 29 9 Note for m 0 1709411 JET WCDSH 30 Expansion of a function in spherical harmonics 1 994 Z Z AszImWMI I with A4 10 quot14 1191mm gm 2n 1 dzp dcos i 6mg6q6 O 1 As a special case consider 0 0 cos 0 1 PI quotI X 17 raw2 for m gt 0 and 13 1 X P I gt le 036 1 0 when m 0 w Zl1 9014 Azuxu907 Z4m 4 BCDS 1 10 0 quotT w 21 Z Am since P1 1 47f 1 General soln of Laplace39s eqn when region of interest 31 includes OS lt2Tr and OSQSTr co 1 137 Y 6 Z Z Aim Tl Bhn F HD 3113 0quot 10 quot171 Consider the Green function Gff WI 1 i 17 3339 n 9 1 I T 79 1q5 DO I From sample problem 33 G Z 4 Pcosw where rlt r is the smaller larger of r r and y is the angle btwn if and f Since G is a soln of the Laplace eqn everywhere except f iquot G M8 l 2 Aim Tl Bl111r7l1 Yl11197 except at f iquot 7 l n 7 l H 0 Al and BIquot depend on 7quot 6 q 1 Since we require G gt 0 as r gt so and G remains finite as r gt 0 32 00 I G Z Z Azmr m9o rltr 1 0 177 co 1 Z 2 Elm I M mum yr gt 7quot IU quot17 G is continuous at r r gt A1 r B1r 1 1 V G 476f f 1 q 67 7quot609 6 60 0 rsrul9 39 2 7 3 23 radial part of V G r2 3r 7 37 4 4 i fdrVZGz 726 T Br 69 7 6 60 0 l H2 r 1quot2 sin 9 BC 87 7 m 2 Z 1A1mr gtquot 310w I 0 quot271 7 Z 2 1 1 Blm 1quot 2 2m90 a f i 01BMr3939zAzltr39 iimltao4w 1 6079396lt 7a 0 a quot14 SLL1 9 With Ann 094sz and Blm 7quotIsz 1 is automatically satisfied and 2 becomes 1 quot 1 Z Z 1mam1C1ylm9zp4n a quot14 51119 59 i 9 607 W Using the completeness relation l 0 I 2 21 manage M Z X 122194 33114940 0 m l ux l M8 F o 4 2 am 134626 2I1 M l 47f G 2 EM mw39w39wlmam rltr39 0 1nil J M l 4ft IL y 9 I 7l1ymg gt r g1glr21l lm1 r z 14 T T 33 l i GfflEW x l l l 7 4W 21721 1 54w 0 mm m 0 1117 Tgt Comparing this with the expansion in Legendre polynomials yields mom i we w m6 gs my quotAddition thm for spherical harmonicsquot Recall Topic 2 notes p 19 603 case cos 6 sin a sine cosd 7 4 Next let39s find an analogous expression for 113511739 for the region outside a sphere with radius a From the method of images Topic 2 p 16 GDff 5 2 a if jquot Mad and fquot dive74 r 35 Since r lt a rlt r and rgt r in this case n a2 l am My l l M l a T gt mwmmwm w gt G 5 4T 7 7 7 7 D 211 r91 r39 rm l 1 TL am1 4quot Z fwd 5zm9 Ym9 0 1114 Dirichlet Green function for the region between 2 spherical shells with radii a and b a lt b Aim Tl BlinT H39Lu K111091115 7 T lt T M 71 m T gt7 M8 EME Q1117 l Dim PM mm 4 M it H o m Alma Bl111u7li 0 CmblDlnb lgt0 Dim Am 7 am FM me M m fl M8 EMB cm 7 by FM we a Mx H o m it 7 2li i Blm 7 Azm 70ml b211 G is continuous at r gt Am r 7 112 WWW sz T i 52 T39HHW l 7 arlzz1 Cam Arm 1 As before 28 r x 1 vs 7mm g An 2 Z maymum r 20 quot17 3639 l I 11 v 221 I rzz 37 ET Am 1 r z 1 a r 114614 lG m 8G l r 171 211 I 42 x E 2 2 Clquot 1m l1b 7 1 MM l0m4 gt i i Aim 1 W lt2 1 r quot 7 01m 1 W l 1 172 W l Wv W 0 m it 00 l 4WZ Z xlwzqs nimwm lt2 0 1727 Soln of 1 and 2 hr 1 VJb Ylfnw 39 1 71quot12 i 7 4K 71121H sz SWWVMQ 211 34nlt9 agt39ymlt9w 38 gt G 7 4 Eng W hn Wlth 1 r 271 2H1 41 I 13027 Trl1 17 Z T7u T TltT 2li 2z1 I 7 T17 b211 FWD 1 T gt Tr a 1 TI a2l1 1 J flth TII1 7 b211 7 7 r11 I T lt7 G2 I 1 T1 1 I Gun1 7 b211 7 W r gt T ZH L 1 HM Tllt quot11 Iii 7 271 Tlt 7 gt b 2zi z a 1 r f 7 l gt M gt 00 5 9 a2ll 1 Tl a at 7 In 7 m 1 z gt W Z n1 Ha11W Tquot rg bm 717l Bessel functions These appear in the soln 0f Laplace39s eqn in cylindrical coords First recall the Gamma function F E 10 e tquot1d1 3 gt 0 no ri E H l G m F 1 0 tildt 40 0 Integrate by pans Fz 1 z eittzil dt 0 1 Z 172 quotrecursion formulaquot used to define Hz for z lt 0 1F11 2 r221 3F3321 etc NH L 71 for 71 0 L2 Bessel39s equation I2 y r y 12 7 12 y 0 42 Try 9 Is I with 00 i 0 fully general if 00 were 0 just increment s m 00 y n 5Cn1ns l y 201 3 3 71cnIn572 110 710 to Z n 5 391 5 1 671 In 71 s an 111 on NH 12 cu 1 0 110 Is i n s2 7 V2 quotIn CHIHH 0 quot0 Must be true for any x gt term in each power of x must vanish IO 32 12 co 0 gt s b since no 75 0 I11 S l2 712c1 0 IfsI Es7 higher powers of x an 1 I ll12 l2 10 212lcl0 gt I 3 0139 310 1 712 712c391 U i 614 n32 7 12 ni 12 712 43 1 L727 10 511 or 10 71 s2 712c 322 0 71 l l 2 712 i 21m 77 n i 21 7 71 7 71 c072i2174i1 04142 71 71 C0762 0744i21 41i17481iI2iI 13 2524 71 l 71 etc c0 2 66i2lz 39431iu2iu 48121i12iI3j1 For the special case v i 12 62 1 Co iii 9 sv when 11112 2 1 4 c04812 6 em 1 2 LA gt even terms y1eld ca 12 lt1 7 i i a 12 cost 2 4 For the odd terms 911 551513 171 i I Q s2 3 0176361 5 3 2 5 ec39 5 1 3 I Gilaz clriiZ 51111 44 7 V 45 gt solnwhenvi12 ls ycor 12 COSIClIU251nI Check l w 4 1 an 7w 1 3J2 y to I 39smriar 3051 61 1 CDSI SI SHIJ 1 1 12 751 yI cl cosr cu 51111 v 4 1 x y 76h Ly39 1 2y 1 31 sum ca 051 7 5 I 3 2 cl c051 to squ L L 392 32 72 32 73 51 yr cl cosric031ur7r y 714 51 39 cl 051760 51111 3 2 az 11 y y r ricosrfcosmr 1 5 Izy ry IZ 1y z 1 12 2 1 1 cus139 to 1111 3yr c1 cusri co sl1111 31 15 0 3 2 u yiryir39 4 For the general case that v i i 12 cl 0 gt only the even terms remain 4 16 2 Eli I y 6 4lizx48li12i1 48121iu2iz3i1 Take v 2 0 fully general since we have solns with iv If v i integer then we have 2 linearly independent solns to our 2nd order differential eqn If v is an integer then the terms in the v series with j 2 v have a term v v 0 in the denominator gt the series diverges We need to find a 2nd soln in this case more later For the v soln 1j 1 l 1j 1quot TNquot 1 y 6014 22139j rawyaw to 2mm F1jl Absorbing the 2quot Fvl into the constant we find solns of the form no 1 j n2j Jx Bessel function of the first kind 46 For v 5 integer 47 Do 1 J I 7u2j 39 J 5 is also a soln of Bessel39s eqn It can be shown that Jyx and 1406 are linearly independent When v integer define 1406 using the above eqn NZ Him for 501 17 21u gt terms withj0 to jv l vanish 7 00 71 I 42 M W 39 i 7 co inkw p2k Substitute k j v J JI g kufk L 2 klrklk39k1quot39kl39FC1kFIk1 1V I 339 J71I771 il 111 So J Vx and 1406 are NOT linearly independent for v integer Define the Neurnann function or Bessel fcn of the 2nd kind by Jr CDSI 7 JI sin 1m NM J Vx and Nyx are linearly independent for v i integer since JV and JVquot are For v integer Nyx 00 But for n an integer 1311 KAI is defined and linearly independent of Jnx Note that some authors denote Neurnann fcns as Yvx rather than Nyx Note that 100 1 and JVO 0 v gt 0 easily obtained from the series The Neurnann fcns diverge at x 0 50 Define the Hankel functions Hf r 51 or Bessel fcns of the 3rd kind by H mm JUL 7 L39Nr H 5 3 j These are more convenient for some problems and are also linearly independent J M Hf Hf all satisfy the following recursion relations as can be verified from the series representation mm 1mm 11 J m 7 1mm 2mm Asymptotic forms For x ltlt 1 Mr E 9quot V 2 I I wherey w 057721566 is the s i 39x U I w n 2 Euler Mascheroni constant n Nx a frogy I0 For x gtgt 1 2 VTF 7r Jc 4 7005I7i77gt WI 2 4 2 17r 7r N 7 39 7 7 7 7 V00 a Tm 5111 I 2 4 The transition between the limiting forms occurs when x v The asymptotic forms reveal that each Bessel fcn has an infinite number of roots Denoting the n root of Jquot by x 1 0 IDquot 7 240483 552008 865373 1179153 1 1 11 383171 701559 1017347 1332369 772 29 7 513562 841724 1161984 1470595 1 3 3 6380l6 976102 1301520 1022347 1 7r For 71 Z 4 rm m mr 1 7 5 E accurate to at least 3 figures Orthogonality 53 Consider fixed v 2 0 and roots x n l 2 a 2 0 JV JV rdr 3 Hangmann proof on pp 114 115 of Jackson The set of Bessel fcns Jyxmra is also complete gt an arbitrary fcn f r can be expanded in a Bessel series also called a Fourier Bessel or Bessel Fourier series m 2 AW o s r s a n1 a with Am L uafr J rdr 2 1 I39Al 1 m works for all v 2 l 2 Return to Bessel39s eqn 2d y dy 54 I EIEI2I2y0 Replace x in u ix Solns are Jyx J in Nyiu etc dyd d2y d dy dui dzy dzy 11 dudr du Ei dz dmiil 71 73 dzy rly Bessel39s eqn becomes rug iz w 71 i u2 7 12 y 0 du du zdzy fly 2 2 ll iu77u Vy0 1112 u Solns are Jyiu Nyiu etc By ry 7 r2 12y 0 is called the modified Bessel eqn The linearly independent solns are taken as 1J1 1quot JUGS These are real valued for real x and v and KAI g F H39EW 13 are called modified Bessel fcns Iyx diverges as x a so and Kvx diverges as x a 0 55 0f0rvgt0 x only has a zero atx v I x and Kvx have no zeros I 0 Laplace eqn in cylindrical coords r I z la 83 63 82370 6 18d law 3 0 T37quot 723w 622 gt mt Ha z92 Separation of variables ltIgtr p 3 Rr QM Z FR ldR RZJZQ 22 Z 7 if if R 0 Q er 7quot dr 7 2 dri2 d22 1 rl 92 U2 R d7 2 39r dr 39er 11 2 Z dz2 7 2 2 gig coust ik2 gt kzz0 I 1 FR ldR llsz 2 l El 6472ik 0 56 r2 d2 1 dB 1 dZQ 57 R 7 if ikZ 2 770 dr2r tr T qu 12 2 l caust 712 sign must be negative if full range of azimuth Q Q is allowed so b 0 2n yield the same Q dZQ 2 mp2 I Q 0 2 r2 1 1 dB 2 2 2 7170 Ma 1 dB 2 12 WE ii 7 R70 3 dZR dR Replacmg x kr 3 becomes I2 F z T inf 7 12 R 0 1 139 This is the Bessel eqn sign or the modified Bessel eqn sign Assuming the full range of azimuth is allowed the solns are 20 A slumqb B cos 111 v m integer SO that Q2Tr Soln for k2 22 05 Dequot RU FL11 TGNmkT Soln for kZ 22 Csink D cask R FImUW GKmUW So the general soln is r 152 I Aklm 5111th Bklm cosmqb Chm 6quot Dklme k X 3117 Ft1quot Jnxklr G39th Nmk17 l Z Akzym sinmd Bkwn cosmq Chm sin kzz Dkzvm cos kzz X 1 m Pkgan Ime 0162111 Kmk2rl Boundary conditions will restrict the values of k1 and k2 and in typical applications will force many of the coeffs to vanish Example cylinder with length L and radius a zaxis is the symmetry axis 15 O for z 0 bottom face 15 VLr 4 for z L top face and 15 V Z for r a curved face Find 15 everywhere inside the cylinder Consider 2 separate problems 1 I 0 for z 0 and z L I Vq39z for 7quot a 50111 is 104152 2 I 0 for 2 0 and r a I VLr 139 for z L 50111 is 12rzg39z 1 bi 4 2 solves the Laplace eqn and satisfies the full set of boundary conditions In both cases Nm klr and Km kzr do not contribute since they diverge at r O 1lt150forz0andzL 60 k 7k Terms 111 e 1 and e 1 do not contribute s1nce no linear combination of these can vanish at 2 different points Also cos kzz cannot contribute i 0 at z 0 sinCZL0 e 162 n12 39x M TUH 7m 2 k171rgb Z ZL mu sinmo an cosmo Im sin m0 v11 Do be 39ITTLH W71 At r a V40 2 7 E 2 AW 511177112 Bm cos 7711 In L 5111 L 1110 This is a 2D Fourier series a standard Fourier series for qb and a Fourier sine series for z 2 1 2 L nz 61 Amn d39 dv39 39 39 TFL Im muL a q a 7 422 suimd 5111 L Trnz 2 1 2w L Empii ddvg m39 TFL annam O 7 O z 7 z cosm 5111 L except m 0 l l 2739 L TH LZ 14010 and Bun mO dQL1dzlazpzsln L 2 950 for z0 and ra Since Imk2r has no zeros besides r 0 these terms cannot contribute This leaves only Jmk1r among the radial fcns Jmkla 0 gt kia mm 71 12 xm is the n root of Jmx k1 rm a

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