Transcription of 1 Solutions in cylindrical coordinates: Bessel functions
1 Bessels 20201 Solutions in cylindrical coordinates: Bessel functionsLaplace s equation in cylindrical coordinates is:1 +1 1 + 2 z2=0 Separate variables: Let =R( )W( )Z(z).Then wefind:1R R +1W 2 2W 2+1Z 2Z z2=0 The last term is a function ofzonly, while the sum of thefirst two terms is afunction of and only. Thus we take each part to be a constant 2Z z2=k2 Zand the Solutions areZ=e kzThis is the appropriate solution outside of a charge distribution, say above aplane, ( 0asz ),or inside a cylinder with grounded walls andnon-zero potential on one remaining equation is:1R R +1W 2 2W 2+k2=0 Now multiply through by 2: R R +k2 2+1W 2W 2=0 Here the last term is a function of only and thefirst two terms are functionsof only.
2 Again we often want a solution that is periodic with period2 ,sowe choose a negative separation constant: 2W 2= m2W W=e im Finally we have the equation for the function of : R +k2 2R m2R=01To see that this equation is of Sturm- liouville form, divide through by : R +k2 R m2 R=0(1)Now we have a Sturm- liouville equation (slreview notes eqn. 1) withf( )= ,g( )=m2/ ,eigenvalue =k2and weighting functionw( )= .Equation (1)is Bessel s equation. The Solutions are orthogonal functions . Sincef(0) = 0,we do not need to specify any boundary condition at =0if our range is0 a,as is frequently the case. (We do specify thatRremainfinite.) Wedo need a boundary condition at = is simpler and more elegant to solve Bessel s equation if we change to thedimensionless variablex=k.
3 Then:k k k R k +k2 R km2k R=0ddx xdRdx +xR m2xR=0 The equation has a singular point atx= we look for a series solution ofthe Frobenius type (cf Lea Chapter 3 ):R=xp [n=0anxnR3= [n=0(n+p)anxn+p 1ddx xdRdx = [n=0(n+p)2anxn+p 1 Then the equation becomes: [n=0(n+p)2anxn+p 1+ [n=0anxn+p+1 m2 [n=0anxn+p 1=0 The indicial equation is given by the coefficient ofxp 1:p2 m2=0 p= mThus one of the Solutions (withp=m)is analytic atx=0,and one (withp= m)isnot. Tofind the recursion relation, look at thek+p 1power ofx:(k+p)2ak+ak 2 m2ak=0and soak= ak 2(k+p)2 m2= ak 2k2+2kp+p2 m2= ak 2k2+2kp= ak 2k(k 2m)2 Let s lookfirst at the solution withp=+ we start the series witha0,thenkwill always be even,k=2n,anda2n= 12n(2n+2m) 1(2n 2) (2n 2+2m)a2n 4=( 1)323n(n 1) (n 2) 23(n+m)(n+m 1) (n+m 2)a2n 6=a0( 1)n2nn!]]]]]]
4 12n(n+m)(n+m 1) (m+1)The usual convention is to takea0=12m (m+1)(2)Thena2n=12m (m+1)( 1)n2nn!12n(n+m)(n+m 1) (m+1)=( 1)nn! (n+m+1)22n+m(3)and the solution is the Bessel function:Jm(x)= [n=0( 1)nn! (n+m+1) x2 m+2n(4)The functionJm(x)has only even powers ifmis an even integer and only oddpowers ifmis an odd integer. The series converges for all values s see what the second solution looks like. Withp= mthe recursionrelation is:ak=ak 2k(k 2m)(5)where againk= ifmis an integer we will not be able to determinea2mbecause the recursion relation blows up. One solution to this dilemma isto start the series we canfind the succeedinga2n:a2(n+m)= a2(n 1)+2m22(n+m)n= a2(n 2)+2m24(n+m)(n+m 1)n(n 1)=( 1)n (m+1)n!]
5 (n+m+1)22na2mwhich is the same recursion relation we had before. (Compare the equationabove with equation (3). Thus we do not get a linearly independent solutionthis way1. (This dilemma does not arise if the separation constant is taken tobe 2with non-integer. In that case the second recursion relation provides1 This happens because the two roots of the indicial equation differ by an (x)that is linearly independent of thefirst.) Indeed wefind:J m(x)= [n=0( 1)n (m+1)n! (n+m+1)22na2mx2(n+m) m= [n=0( 1)n (m+1)2mn! (n+m+1)a2m x2 m+2nandifwechoosea2m=( 1)m (m+1)2mthenJ m(x)=( 1)mJm(x)(6)With this choiceJ (x)is a continuous function of .(Noticethatwecanalsoexpress the series using equation (3) for the coefficients, withm mandn k+m,and wherea2n 0forn<m.))]]
6 We still have to determine the second, linearly independent solution of theBessel equation. We canfind it by taking the limit as mof a linearcombination ofJ andJ known as the Neumann functionN (x):Nm(x) = lim mN (x) = lim mJ (x)cos J (x)sin = lim 0Jm+ (x)cos(m+ ) J (m+ )(x)sin (m+ ) = lim 0Jm+ (x)(cosm cos sinm sin ) J (m+ )(x)sinm cos +cosm sin = lim 0Jm+ (x)( 1)mcos J (m+ )(x)( 1)msin Now we expand the functions tofirst order in .We use a Taylor series for theBessel functions . Note that appears in the index, not the argument, so wehave to differentiate with respect to ..Nm(x) = lim 0Jm+ (x)( 1)m J (m+ )(x)( 1)m = lim 0( 1)m ( 1)m Jm+ dJ d =m J m+ dJ d =m Using relation (6), we have:Nm(x)=1 dJ d =m ( 1)mdJ d =m 4 The derivative has a logarithmic term:dJ d =dd %x [n=0( 1)nn!]
7 (n+ +1) x2 2n&=dx d [n=0( 1)nn! (n+ +1) x2 2n+x dd [n=0( 1)nn! (n+ +1) x2 2nanddx d =dd e lnx=lnxe lnx=x lnxand sodJ /dxhas a term containingJ term diverges asx 0provided thatJv(0)is not zero, for = functionNv(x)alsodiverges asx 0for 9=0,because it contains negative powers ofx.(Theseries forJ starts with a termx .)N isfinite asx becauseJ goesto zero sufficiently additional functions called Hankel functions are defined as linear com-binations ofJandN:H(1)m(x)=Jm(x)+iNn(x)(7)andH(2)m (x)=Jm(x) iNn(x)(8)Compare the relation between sine, cosine, and exponential:e ix=cosx Properties of the functionsThe Bessel functions (Js)are well behaved both at the origin and asx.]]
8 They have infinitely many zeroes. All of them, except forJ0,are zero atx= few functions are shown in three Bessel ,J1(red) andJ25 For small values of the argument, we may approximate the function with thefirst term in the series:Jm(x) 1 (m+1) x2 mforx 1(9)The Neumann functions are not well behaved atx= a logarithmicsingularity, and form>0,Nmdiverges as an inverse power ofx:N0(x) 2 lnxforx 1Nm(x) (m 1)! 2x mforx 1,m>0(10)For large values of the argument, bothJandNoscillate: they are like dampedcosine or sine functions :Jm(x) u2 xcos x m 2 4 forx 1,m(11)Nm(x) u2 xsin x m 2 4 forx 1,m(12)and thus the Hankel functions are like complex exponentials:H(1,2)m u2 xexpk i x m 2 4 lforx 1,m(13)Notice that ifm>1,the large argument expansions apply forx mratherthan the usualx Relations between the functionsAs we found with the Legendre functions , we can determine a set of recursionrelations that relate successiveJm(x).
9 For example (Lea )ddx Jm(x)xm = Jm+1(x)xm(14)which is valid form particular, withm=0we obtain:J1(x)= J30(x)(15)ddx(xmJm(x)) =xmJm 1(x)(16)From (14) and (16) we may obtainJm+1+Jm 1=2mxJm(17)and similarlyJm+1 Jm 1= 2dJmdx(18)The same relations hold for Orthogonality of theJmSince the Bessel equation is of Sturm- liouville form, the Bessel functions areorthogonal if we demand that they satisfy boundary conditions of the form(slreview notes eqn 2). In particular, suppose the region of interest is =0to =a,and the boundary conditions areJm(ka)= do not need aboundary condition at =0because the functionf( )= is zero there. Thenthe eigenvalues arekmn= mnawhere mnis thenth zero ofJm.
10 (The zeros are tabulated in standard referencessuch as Abramowitz and Stegun. Also programs such as Mathematica andMaple can compute them.) Then]a0 Jm(kmn )Jm(kmn3 )d =a22[J3m(kmna)]2 nn3(19) Solving a potential cylinder of radiusaand heighthhas its curved surface and itsbottom grounded. The top surface has is the potential insidethe cylinder?The potential has no dependence on and so only eigenfunctions withm=0contribute. The potential is zero at =a,sothesolutionweneedisJ0(k )witheigenvalues chosen to makeJ0(ka)= the eigenvalues are given byk0na= 0n,where 0nare the zeros of the remaining functionofzmust be zero atz=0,sowechoosethehyperbolicsine. Thusthepotentialis: ( ,z)= [n=1anJ0(k0n )sinh(k0nz)Now we evaluate this atz=h:V= ( ,h)= [n=1anJ0(k0n )sinh(k0nh)Next we make use of the orthogonality of the Bessel functions .]
