1 Solutions in cylindrical coordinates: Bessel functions

Transcription

1 1 Solutions in cylindrical coordinates: Bessel functions 1.1 Bessel functions Bessel functions arise as solutions of potential problems in cylindrical coordinates. Laplace s equation in cylindrical coordinates is: 1 Φ Φ + Φ Separate variables: Let Φ () () () Then we find: The last term is a function of only, while the sum of the first two terms is a function of and only. Thus we take each part to be a constant called Then and the solutions are ± This is the appropriate solution outside of a charge distribution, say above a plane, (Φ as ± ) or inside a cylinder with grounded walls and non-zero potential on one end. The remaining equation is: 1 Now multiply through by : Here the last term is a function of only and the first two terms are functions of only. Again we often want a solution that is periodic with period so we choose a negative separation constant: ± Finally we have the equation for the function of : + 1

2 To see that this equation is of Sturm-Liouville form, divide through by : + (1) Now we have a Sturm-Liouville equation with () () eigenvalue and weighting function () Equation (1) is Bessel s equation. The solutions are orthogonal functions. Since () we do not need to specify any boundary condition at if our range is as is frequently the case. We do need a boundary condition at It is simpler and more elegant to solve Bessel s equation if we change to the dimensionless variable Then: + + The equation has a singular point at So we look for a series solution of the Frobenius type (cf Chapter ): X ( + ) + 1 ( + ) + 1 Then the equation becomes: ( + ) X The indicial equation is given by the coefficient of 1 : ± + 1 Thus one of the solutions (with ) is analytic at and one (with ) is not. To find the recursion relation, look at the + 1 power of : ( + ) + and so ( + ) ( ± )

3 Let s look first at the solution with + Wecanstepdowntofind each If we start the series with then will always be even, and 1 1 ( +) ( ) ( +) 4 ( 1) 3 3 ( 1) ( ) 3 ( + )( + 1) ( + ) 6 ( 1)! The usual convention is to take Then 1 ( + )( + 1) ( +1) 1 Γ ( +1) 1 ( 1) 1 Γ ( +1)! ( + )( + 1) ( +1) ( 1)!Γ ( + +1) + (3) and the solution is the Bessel function: () () ( 1) + (4)!Γ ( + +1)³ The function () has only even powers if is an even integer and only odd powers if is an odd integer. The series converges for all values of Let s see what the second solution looks like. With the recursion relation is: (5) ( ) where again Now if is an integer we will not be able to determine because the recursion relation blows up. One solution to this dilemma is to start the series with Then we can find the succeeding : (+) ( 1)+ ( + ) ( )+ 4 ( + )( + 1) ( 1) ( 1) Γ ( +1)!Γ ( + +1) which is the same recursion relation we had before. (Compare the equation above with equation (3). Thus we do not get a linearly independent solution this way. (This dilemma does not arise if the separation constant is taken to be with non-integer. In that case the second recursion relation provides aseries () that is linearly independent of the first.) Indeed we find: ( 1) Γ ( +1) ()!Γ ( + +1) (+) ( 1) Γ ( +1) ³ +!Γ ( + +1) 3

4 and if we choose ( 1) Γ ( +1) then () ( 1) () (6) With this choice () is a continuous function of (Notice that we can also express the series using equation (3) for the coefficients, with and + and where for ) We still have to determine the second, linearly independent solution of the Bessel equation. We can find it by taking the limit as of a linear combination of and knownastheneumannfunction () : () ()cos () lim () lim sin + ()cos( + ) (+) () lim sin ( + ) + ()(cos cos sin sin ) (+) () lim sin cos +cos sin + ()( 1) cos (+) () lim ( 1) sin Now we expand the functions to first order in We use a Taylor series for the Bessel functions. Note that appears in the index, not the argument, so we have to differentiate with respect to. + ()( 1) (+) () () lim ( 1) ( 1) lim ( 1) + + Using relation (6), we have: () 1 ( 1) The derivative has a logarithmic term: Ã! X ( 1)!Γ ( + +1)³ and ( 1) +!Γ ( + +1)³ ln ln ln ln 4 ( 1)!Γ ( + +1)³

5 and so has a term containing ln This term diverges as provided that () is not zero, i.e. for The function () also diverges as for 6 because it contains negative powers of (The series for starts with a term ) is finite as because goes to zero sufficiently fast. Two additional functions called Hankel functions are defined as linear combinations of and : (1) () ()+ () (7) and () () () () (8) Compare the relation between sine, cosine, and exponential: ± cos ± sin 1. Properties of the functions The Bessel functions (s) are well behaved both at the origin and as They have infinitely many zeroes. All of them, except for are zero at The first few functions are shown in the figure. J x The first three Bessel functions. 1 (red) and For small values of the argument, we may approximate the function with the first term in the series: 1 () for 1 (9) Γ ( +1)³ The Neumann functions are not well behaved at has a logarithmic singularity, and for diverges as an inverse power of : () ln for 1 ( 1)! () for 1 (1) 5

6 For large values of the argument, both and oscillate: they are like damped cosine or sine functions: r ³ () cos for À 1 (11) 4 r ³ () sin for À 1 (1) 4 and thus the Hankel functions are like complex exponentials: r h ³ (1) exp ± i for À 1 (13) 4 Notice that if 1 the large argument expansions apply for À rather than the usual À Relations between the functions As we found with the Legendre functions, we can determine a set of recursion relations that relate successive () For example: () +1 () (14) which is valid for In particular, with we obtain: 1 () () (15) ( ()) 1 () (16) (17) and similarly, by subtracting instead of adding, we can find +1 1 Thesamerelationsholdforthes andthes. (18) 1.4 Orthogonality of the Since the Bessel equation is of Sturm-Liouville form, the Bessel functions are orthogonal if we demand that they satisfy boundary conditions of the form (??). In particular, suppose the region of interest is to and the boundary conditions are () We do not need a boundary condition at because the function () is zero there. Then the eigenvalues are 6

7 where is the th zero of (The zeros are tabulated in standard references such as Abramowitz and Stegun. Also programs such as Mathematica and Maple can compute them.) Then Z 1.5 Solving a potential problem. [ ( )] [ ( )] (19) Example. A cylinder of radius and height has its curved surface and its bottom grounded. The top surface has potential What is the potential inside the cylinder? The potential has no dependence on and so only eigenfunctions with contribute. The potential is zero at sothesolutionweneedis ()with eigenvalues chosen to make () Thus the eigenvalues are given by where are the zeros of the function The remaining function of must be zero at so we choose the hyperbolic sine. Thus the potential is: Φ ( ) ( )sinh( ) Nowweevaluatethisat : 1 Φ ( ) ( )sinh( ) 1 Next we make use of the orthogonality of the Bessel functions. Multiply both sides by ( )and integrate from to (Note here that the weight function () This is the first time we have a weight function that is not 1.) Only one term in the sum, with survives the integration. Z ( ) Z ( ) ( )sinh( ) 1 Z [ ( )] sinh ( ) [ ( )] sinh ( ) To evaluate the left hand side, we use equation (16) with : Z Z () 1 ( 1 ()) 1 1 () 1 () 7

8 So 1 ( ) [ ( )] sinh ( ) 1 ( )sinh( ) where we used the result from equation (15) that 1 Finally our solution is: ( ) sinh ( ) Φ 1 ( ) sinh ( ) 1 The first two zeros of are: 1.448, 5.51, and thus the first two terms in the potential are: Ã 448 sinh 448 Φ (448) sinh sinh! (551) sinh Modified Bessel functions Suppose we change the potential problem so that the top and bottom of the cylinder are grounded but the outer wall at has potential ( ) Then we would need to choose a negative separation constant so that the solutions of the -equation are trigonometric functions: sin + cos At () so we need the sine, and therefore set We also need () so we choose the eigenvalue. This change in sign of the separation constant also affects the equation for the function ()because the sign of the term changes. or, changing variables to : () which is called the modified Bessel equation. Thesolutionstothisequation are () It is usual to define the modified Bessel function () by the relation: () 1 () (1) 8

9 so that the function is always real (whether or not is an integer). Using equation (4) we can write a series expansion for : () 1 ( 1)!Γ ( + +1) ()!Γ ( + +1)³ As with the s, if is an integer, is not independent of : in fact: () () ( 1) () ( 1) () () (3) The second independent solution is usually chosen to be: Then these functions have the limiting forms: and () () +1 (1) () (4) 1 for 1 (5) Γ ( +1)³ () 577 ln for 1 (6) () Γ () for 1 (7) At large À 1, the asymptotic forms are: () 1 (8) and r () (9) (cf Lea Chapter 3 Example 3.9) These functions, like the real exponentials, do not have multiple zeros and are not orthogonal functions. Note that the s are well behaved at the origin but diverge at infinity. For the s the reverse is true. They diverge at the origin but are well behaved at infinity (See Figure below). 9

10 I,K Modified Bessel functions. solid dashed Black Blue 1 Red x The recursion relations satisifed by the modified Bessel functions are similar to, but not identical to, the relations satisfied by the s. For the s, again we can start with the series: Now let 1: !Γ ( + +1)³ 1!Γ ( + +1)³ 1 +1!Γ ( + +)³ 1 ++1!Γ ( + +)³ +1 () (3) and similarly ( ) 1 (31) Expanding out and combining, we get: (3) For the s, the relations are: ( ) 1 ; 1 () +1 ()

11 and consequently: 1.7 Combining functions (33) When solving a physics problem, we start with a partial differential equation and a set of boundary conditions. Separation of variables produces a set of coupled ordinary differential equations in the various coordinates. The standard solution method ("Orthogonal" notes ) requires that we choose the separation constants by fitting the zero boundary conditions first. In a standard 3-dimensional problem, once we have chosen the two separation constants we have no more freedom and the third function is determined. When solving Laplace s equation in cylindrical coordinates, the functions couple as follows: Zero boundary conditions in : ³ The eigenfunctions are of the form: ± ³ sinh + cosh The set of functions ± form a complete orthogonal set on the surfaces constant that bound the region. Zero boundary conditions in : The eigenfunctions are of the form: ³ ³ ³ ³ + sin ± The set of functions sin ± form a complete orthogonal set on the boundary surface constant. Thus in solutions of Laplace s equation, s in always couple with the hyperbolic sines and hyperbolic cosines (or real exponentials) in while the s and s in always couple with the sines and cosines (or complex exponentials) in Example: Suppose the potential on the curved wall of the cylinder is ( ) sin The top and bottom are grounded The solution is of the form + ³ ³ ³ ³ Φ ( ) sin + 1 The solution must be finite on axis at so Now we evaluate the potential at Φ ( ) + sin 1 ³ 11 ³ sin

12 We make use of orthogonality by mutltiplying by sin and integrating from to + 1 Z sin ³ ³ sin + ³ The result is non-zero only for odd Next we make use of the orthogonality of the + Z ³ Z even sin Z sin sin cos sin [( 1) 1] sin To do the integral on the RHS, express the sine in terms of exponentials. Z Z sin ( 1 1) Only one term with survives the integration on the LHS. Thus Φ ( ) 4 1 odd But 1 1 (eqn 3), so ³ ( 1 1) 1 ( 1 1) 4 sin Φ ( ) 4 ³ 1 " 1 odd sin 1 ³ 1 sin # odd The dimensions are correct, and the 1 gives us reasonable convergence. The ratio 1 1 for 1 The potential at is zero, as expected. The 1"source" (the potenial on the surface) generates an 1response. Again we can look at the first few 1

13 terms (here 1). Let s take and 4 (red) and Φ (black) phi/v rho/a 1.8 Continuous set of eigenvalues: the Fourier Bessel Transform In Chapter 7 we approached the Fourier transform by letting the length of the domain in a Fourier series problem become infinite. The orthogonality relation for the exponential functions: Z 1 ³ exp ³ exp becomes Z 1 ( ) That is, the Kronecker delta becomes a delta function, and the countable set of eigenvalues becomes a continuous set of values. The same thing happens with Bessel functions. With a finite domain in say we can determine a countable set of eigenvalues from the set of zeros of the Bessel functions If our domain in becomes infinite, then we cannot determine the eigenvalues, and instead we have a continuous set. The orthogonality relation: Z ³ ³ [ ( )] 13

14 becomes: Z () ( ) ( ) (34) (The proof of this relation is in Lea Appendix 7.) Then the solution to the physics problem is determined as an integral over For example, a solution of Laplace s equation may be written as: Φ ( ) X Z () () () where () depends on the boundary conditions in It will be a combination of the exponentials and + ³ Example: Suppose the potential is () sin on a plane at and we want to find the potential for Then the appropriate function of is, chosen so that Φ as ( a long way from the plane). We also need the which remain finite at Then the solution is of the form: Φ ( ) X Z () () Evaluating Φ on the plane at we get: sin Φ ( ) X Z () () Now we can make use of the orthogonality of the Multiply both sides by and integrate over the range to On the LHS, only the term with survives the integration, and on the RHS only the term with survives. Z sin Z X Z () () sin Z () () a Fourier Bessel transform. Next 1 multiply both sides by ( ) integrate from to in and use equation (34) to get: Z sin Z Z ( ) () () ( ) Z () ( ) ( ) which determines the coefficient ( ) in terms of the known potential on the plane. Only is non-zero, as expected from the azimuthal symmetry. 1 We also drop the primes on the for convenience. 14

15 On the RHS, let We get (dropping the primes on ) Z sin () () This integral is GR 6.671#7. So ( 1 () () So finally we have Φ ( ) Z 1 () q1 () Z 1 1 For we have Z 1 Φ ( ) 1 1 sin which gives us back the potential on the plane. Thechargedensityontheplaneisgivenby ˆ GR 6.554# So we need Φ Z 1 Z ( 1) ( 1) [!] Let sin cos Then the integral is Thus Z sin (+1) cos cos Z [!] ³ ³ Z 1 sin (+1) (+1) 1 ( +1)!! ( +)!! GR 3.61#3, LeaChP9d Φ 4 ( 1) [!] ( 1) (!) 15 ³ ( +1)!! +1 ( +1)! ³ ( +1)!! ( +1)!

16 and so the charge density on the plane is 4 ( 1) (!) Theplotshowstheseriesupto 4 8 ( +1)!! ( +1)! sigma a/epsv rho/a Convince yourself that this agrees with the field line diagram. (The dashed line is the potential.) 16