Bessels 2018

1 Solutions in cylindrical coordinates: Bessel

functions

1.1 Bessel functions

Bessel functions arise as solutions of potential problems in cylindrical coordi-

nates. Laplace’s equation in cylindrical coordinates is:

1

µΦ

¶+1

µ1

Φ

¶+

2Φ

2= 0

Separate variables: Let Φ = () () () Then we find:

1

µ

¶+

1

22

2+1

2

2= 0

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 2

Then2

2= 2

and the solutions are

= ±

This is the appropriate solution outside of a charge distribution, say above a

plane, (Φ → 0 as → ±∞) or inside a cylinder with grounded walls andnon-zero potential on one end.

The remaining equation is:

1

µ

¶+

1

22

2+ 2 = 0

Now multiply through by 2 :

µ

¶+ 22 +

1

2

2= 0

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 2 so

we choose a negative separation constant:

2

2= −2 ⇒ = ±

Finally we have the equation for the function of :

µ

¶+ 22−2 = 0

1

To see that this equation is of Sturm-Liouville form, divide through by :

µ

¶+ 2− 2

= 0 (1)

Now we have a Sturm-Liouville equation with () = () = 2 eigen-

value = 2 and weighting function () = Equation (1) is Bessel’s equation.

The solutions are orthogonal functions. Since (0) = 0 we do not need to spec-

ify any boundary condition at = 0 if our range is 0 ≤ ≤ 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:

µ

¶+ 2−

2

= 0

µ

¶+ − 2

= 0

The equation has a singular point at = 0 So we look for a series solution of

the Frobenius type (cf Lea Chapter 3 §3.3.2):

= ∞X=0

0 =

∞X=0

(+ ) +−1

µ

¶=

∞X=0

(+ )2

+−1

Then the equation becomes:

∞X=0

(+ )2

+−1 +∞X=0

++1 −2

∞X=0

+−1 = 0

The indicial equation is given by the coefficient of −1:

2 −2 = 0⇒ = ±Thus one of the solutions (with = ) is analytic at = 0 and one (with

= −) is not. To find the recursion relation, look at the +−1 power of :( + )

2 + −2 −2 = 0

and so

= − −2( + )

2 −2= − −2

2 + 2+ 2 −2

= − −22 + 2

= − −2 ( ± 2)

2

Let’s look first at the solution with = + We can step down to find each

If we start the series with 0then will always be even, = 2 and

2 =−1

2 (2+ 2)

−1(2− 2) (2− 2 + 2)2−4

=(−1)3

23 (− 1) (− 2) 23 (+) (+− 1) (+− 2)2−6

= 0(−1)2!

1

2 (+) (+− 1) · · · (+ 1)

The usual convention is to take

0 =1

2Γ (+ 1)(2)

Then

2 =1

2Γ (+ 1)

(−1)2!

1

2 (+) (+− 1) · · · (+ 1)

=(−1)

!Γ (++ 1) 22+(3)

and the solution is the Bessel function:

() =

∞X=0

(−1)!Γ (++ 1)

³2

´+2(4)

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:

=−2

( − 2) (5)

where again = 2 Now if is an integer we will not be able to determine

2 because the recursion relation blows up. One solution to this dilemma is

to start the series with 2 Then we can find the succeeding 2 :

2(+) =−2(−1)+222 (+)

=−2(−2)+2

24 (+) (+− 1) (− 1) =(−1) Γ (+ 1)

!Γ (++ 1) 222

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 −2 with non-integer. In that case the second recursion relation provides

a series − () that is linearly independent of the first.) Indeed we find:

− () =

∞X=0

(−1) Γ (+ 1)

!Γ (++ 1) 222

2(+)−

=

∞X=0

(−1) Γ (+ 1) 2

!Γ (++ 1)2

³2

´+23

and if we choose

2 =(−1)

Γ (+ 1) 2

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 2 ≡ 0 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 − known as the Neumann function () :

() = lim→

() = lim→

() cos − − ()sin

= lim→0

+ () cos (+ ) − −(+) ()sin (+ )

= lim→0

+ () (cos cos − sin sin )− −(+) ()sin cos + cos sin

= lim→0

+ () (−1) cos − −(+) ()(−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 .

() = lim→0

+ () (−1) − −(+) ()(−1)

= lim→0

(−1)

∙(−1)

µ +

¯̄̄̄=

¶−µ− +

−

¯̄̄̄=

¶¸Using relation (6), we have:

() =1

∙

¯̄̄̄=

− (−1) −

¯̄̄̄=

¸The derivative has a logarithmic term:

=

̰X=0

(−1)!Γ (+ + 1)

³2

´2!

=

∞X=0

(−1)!Γ (+ + 1)

³2

´2+

∞X=0

(−1)!Γ (+ + 1)

³2

´2and

=

ln = ln ln = ln

4

and so has a term containing ln This term diverges as → 0

provided that (0) is not zero, i.e. for = 0 The function () also

diverges as → 0 for 6= 0 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 com-

binations of and :

(1) () = () + () (7)

and

(2) () = ()− () (8)

Compare the relation between sine, cosine, and exponential:

± = cos± sin

1.2 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 0 are zero at = 0

The first few functions are shown in the figure.

2 4 6 8 10 12 14

-0.4

-0.2

0.0

0.2

0.4

0.6

0.8

1.0

x

J

The first three Bessel functions. 0 1(red) and 2

For small values of the argument, we may approximate the function with the

first term in the series:

() ≈ 1

Γ (+ 1)

³2

´for ¿ 1 (9)

The Neumann functions are not well behaved at = 0 0 has a logarithmic

singularity, and for 0 diverges as an inverse power of :

0 () ≈ 2

ln for ¿ 1

() ≈ −(− 1)!

µ2

¶for ¿ 1 (10)

5

For large values of the argument, both and oscillate: they are like damped

cosine or sine functions:

() ≈r2

cos³−

2−

4

´for À 1 (11)

() ≈r2

sin³−

2−

4

´for À 1 (12)

and thus the Hankel functions are like complex exponentials:

(12) ≈

r2

exp

h±³−

2−

4

´ifor À 1 (13)

Notice that if 1 the large argument expansions apply for À rather

than the usual À 1

1.3 Relations between the functions

As we found with the Legendre functions, we can determine a set of recursion

relations that relate successive () For example:

µ ()

¶= −+1 ()

(14)

which is valid for ≥ 0 In particular, with = 0 we obtain:

1 () = − 00 () (15)

Another relation is:

( ()) = −1 () (16)

Expanding out the derivatives, and combining the two relations,we obtain:

+1 + −1 =2

(17)

and similarly,

+1 − −1 = −2

(18)

The same relations hold for the s and the s.

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 (SL

review notes eqn 2). In particular, suppose the region of interest is = 0

to = and the boundary conditions are () = 0 We do not need a

6

boundary condition at = 0 because the function () = is zero there. Then

the eigenvalues are

=

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.) ThenZ

0

[ ()]2 =

2

2[ 0 ()]

2(19)

1.5 Solving a potential problem.

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 = 0

contribute. The potential is zero at = so the solution we need is 0 ()with

eigenvalues chosen to make 0 () = 0 Thus the eigenvalues are given by

0 = 0 where 0 are the zeros of the function 0 The remaining function

of must be zero at = 0 so we choose the hyperbolic sine. Thus the potential

is:

Φ ( ) =

∞X=1

0 (0) sinh (0)

Now we evaluate this at = :

= Φ ( ) =

∞X=1

0 (0) sinh (0)

Next we make use of the orthogonality of the Bessel functions. Multiply both

sides by 0 (0)and integrate from 0 to (Note here that the weight function

() = This is the first time we have seen a weight function that is not 1.)

Only one term in the sum, with = survives the integration.

Z

0

0 (0) =

Z

0

0 (0)

∞X=1

0 (0) sinh (0)

=

Z

0

[0 (0)]2 sinh (0)

= 2

2[ 00 (0)]

2sinh (0)

To evaluate the left hand side, we use equation (16) with = 1 :Z

0

0 () =1

Z

0

(1 ())

=1

1 ()|0 =

1 ()

7

So

=

01 (0)

2

2 [ 00 (0)]2sinh (0)

=

0

2

1 (0) sinh (0)

where we used the result from equation (15) that 00 = −1 Finally our solutionis:

Φ = 2

∞X=1

0 (0)

01 (0)

sinh (0)

sinh (0)

The first two zeros of 0 are: 01 = 2.4048, 02 = 5.5201, and thus the first

two terms in the potential are:

Φ = 2

Ã0¡24048

¢240481 (24048)

sinh¡24048

¢sinh

¡24048

¢ + 0¡55201

¢552011 (55201)

sinh¡55201

¢sinh

¡55201

¢ + · · ·!

1.6 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:

2

2= −2 ⇒ = sin + cos

At = 0 () = 0 so we need the sine, and therefore set = 0 We also need

() = 0 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 2 term changes.

µ

¶− 2− 2

= 0

or, changing variables to = :

µ

¶− − 2

= 0 (20)

which is called the modified Bessel equation. The solutions to this equation

are () It is usual to define the modified Bessel function () by the

relation:

() =1

() (21)

8

so that the function is always real (whether or not is an integer). Using

equation (4) we can write a series expansion for :

() =1

∞X=0

(−1)!Γ (++ 1)

µ

2

¶+2=

∞X=0

1

!Γ (++ 1)

³2

´+2(22)

As with the s, if is an integer, − is not independent of : in fact:

− () = − () = (−1) () = (−1) 2 () = () (23)

The second independent solution is usually chosen to be:

() =

2+1(1)

() (24)

Then these functions have the limiting forms:

() ≈ 1

Γ (+ 1)

³2

´for ¿ 1 (25)

and

0 () ≈ −05772− ln 2for ¿ 1 (26)

() ≈ Γ ()

2

µ2

¶ 0 for ¿ 1 (27)

At large À 1, the asymptotic forms are:

() ≈ 1√2

(28)

and

() ≈r

2− (29)

(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

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.00

2

4

6

8

10

x

I,K

Modified Bessel functions. solid dashed Black = 0Blue 1 Red 2

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:

µ

¶=

1

2

∞X=0

1

!Γ (++ 1)

³2

´2=

1

2

∞X=1

!Γ (++ 1)

³2

´2−1Now let = − 1 :

µ

¶=

1

2

∞X=0

1

!Γ ( ++ 2)

³2

´2+1=

1

∞X=0

1

!Γ ( ++ 2)

³2

´2++1

µ

¶=

+1 ()

(30)

and similarly

() = −1 (31)

Expanding out and combining, we get:

2 0 = +1 + −12

= −1 − +1 (32)

For the s, the relations are:

() = −−1;

µ ()

¶= −+1 ()

10

and consequently:

−1 −+1 = −2

−1 ++1 = −20 (33)

1.7 Combining functions

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 cou-

pled ordinary differential equations in the various coordinates. The standard

solution method ("Orthogonal" notes §2) requires that we choose the separa-

tion 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 : The eigenfunctions are of the form:

³

´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

´´sin³

´±

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 hy-

perbolic 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

Φ ( ) =

+∞X=−∞

∞X=1

sin³

´

³

³

´+

³

´´The solution must be finite on axis at = 0 so = 0 Now we evaluate the

potential at =

Φ ( ) =

+∞X=−∞

∞X=1

sin³

´

³

´= sin

11

We make use of orthogonality by mutltiplying by sin and integrating from

0 to

+∞X=−∞

∞X=1

Z

0

sin

sin³

´

³

´= sin

Z

0

sin

+∞X=−∞

2

³

´= − sin

cos

¯̄̄0

=

sin [1− (−1)]

The result is non-zero only for odd

Next we make use of the orthogonality of the

+∞X=−∞

2

Z 2

0

−0

³

´= 2

Z 2

0

sin−0 odd

= 0 even

To do the integral on the RHS, express the sine in terms of exponentials.Z 2

0

sin−0 =

Z 2

0

− −

2−

0

=2

2(01 − 0−1)

Only one term with = 0 survives the integration on the LHS. Thus

2200

³

´= 2

2

2(01 − 0−1)

0 =4

1

2

(01 − 0−1)0

¡

¢Φ ( ) =

4

∞X=1 odd

sin³

´ 1

2

"

1¡

¢1¡

¢ − −−1

¡

¢−1

¡

¢#But 1 = −1 (eqn 23), so

Φ ( ) =4

∞X=1 odd

sin³

´ sin

1¡

¢1¡

¢The dimensions are correct, and the 1 gives us reasonable convergence. The

ratio1¡

¢1¡

¢ ≤ 1 for ≤

The potential at = 0 is zero, as expected. The = 1 "source" (the potential

on the surface) generates an = 1 response. Again we can look at the first few

12

terms (here 10) . Let’s take = 2 and = 4 (red) and Φ = 2 (black)

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.00.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

rho/a

phi/V

1.8 Continuous set of eigenvalues: the Fourier Bessel Trans-

form

In Lea 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:

1

2

Z

−exp

³

´exp

³−

´ =

becomes1

2

Z ∞−∞

−0 = ( − 0)

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 0 ≤ ≤ 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

0

³

´

³

´ =

2

2[ 0 ()]

2

13

becomes: Z ∞0

() (0) =

( − 0)

(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 ∞0

() () ()

where () depends on the boundary conditions in It will be a combination

of the exponentials − and +

Example: Suppose the potential is () = 0

³

´sin

on a plane at = 0

and we want to find the potential for 0 Then the appropriate function of

is −, chosen so that Φ→ 0 as →∞ ( a long way from the plane). We also

need the which remain finite at = 0 Then the solution is of the form:

Φ ( ) =X

Z ∞0

() − ()

Evaluating Φ on the plane at = 0 we get:

0

µ

¶sin

= Φ ( 0) =

X

Z ∞0

() ()

Now we can make use of the orthogonality of the Multiply both sides by

−0 and integrate over the range 0 to 2 On the LHS, only the term with

= 0 survives the integration, and on the RHS only the term with = 0

survives.Z 2

0

0

µ

¶sin

−

0 =

Z 2

0

X

−0

Z ∞0

0 ()0 ()

20

µ

¶sin

00 = 2

Z ∞0

0 ()0 ()

—a Fourier Bessel transform. Next1 multiply both sides by (0) integrate

from 0 to ∞ in and use equation (34) to get:

0

Z ∞0

µ

¶sin

(

0) 00 =

Z ∞0

Z ∞0

() () (0)

=

Z ∞0

() ( − 0)

=

(0)

0

which determines the coefficient (0) in terms of the known potential on the

plane. Only 0 is non-zero, as expected from the azimuthal symmetry.

1We also drop the primes on the 0 for convenience.

14

On the RHS, let = We get (dropping the primes on )

02

Z ∞0

sin0 () =0 ()

This integral is GR 6.671#7. So

() = 02

(0 0 1 1√

1−()2 0 1

So finally we have

Φ ( ) = 02

Z 1

0

0 ()q1− ()2

−

= 0

Z 1

0

0¡

¢√1− 2

−

For = 0 we have

Φ ( 0) = 0

Z 1

0

0¡

¢√1− 2

= 01

sin

GR 6.554#2

which gives us back the potential on the plane.

The charge density on the plane is given by

· ̂¯̄̄=0

=

0

So we need

Φ

¯̄̄̄=0

= −0

Z 1

0

20¡

¢√1− 2

= −0

Z 1

0

2√1− 2

∞X=0

(−1)[!]

2

³2

´2

= −0

∞X=0

(−1)[!]

2

³

2

´2 Z 1

0

2(+1)√1− 2

Let = sin = cos Then the integral is

=

Z 2

0

sin2(+1)

cos cos =

Z 2

0

sin2(+1) =(2+ 1)!!

(2+ 2)!!

2GR 3.621#3, Lea Ch2 P29d

Thus

Φ

¯̄̄̄=0

= −0

2

∞X=0

(−1)[!]

2

³

2

´2 (2+ 1)!!

2+1(+ 1)!

= −0

4

∞X=0

(−1)(!)

2

³

2

´2 (2+ 1)!!2(+ 1)!

15

and so the charge density on the plane is

= 0 = 00

4

∞X=0

(−1)(!)

2

µ2

82

¶(2+ 1)!!

(+ 1)!

The plot shows the series up to = 40

1 2 3 4 5 6 7 8 9 10

-0.2

0.0

0.2

0.4

0.6

0.8

1.0

rho/a

sigma a/epsV

Convince yourself that this agrees with the field line diagram. (The dashed line

is the potential.)

16