Chapter 7

Bessel Functions

In general, Bessel functions are those which satisfy the di↵erential equation

x2d2Jndx2

+ xdJndx

+�x2

� n2�Jn(x) = 0. (7.1)

We know from our discussions of power series that the series solution to this di↵er-ential equation can be written as

Jn(x) =1X

s=0

(�1)s

s!(n+ s)!

⇣x2

⌘n+2s

. (7.2)

7.1 Recurrence relations: a first go

By manipulating the series (7.2) we can prove the following recurrence relations:

Jn�1(x) + Jn+1(x) =2n

xJn(x) (7.3a)

Jn�1(x) =n

xJn(x)� J 0

n(x). (7.3b)

Typical notation here is that n is an integer, although in fact the recurrence relationsas long as the index is any real number ⌫.

Proof of first recurrence relation The left-hand side of Eq. (7.3a) is equal to:

1X

s=0

(�1)s

s!(n+ s� 1)!

⇣x2

⌘n+2s�1

+(�1)s

s!(n+ s+ 1)!

⇣x2

⌘n+2s+1�. (7.4)

There are some nice common factors here. But before combining coe�cients we wantto arrange things so that the powers match. To do that we pull o↵ the first time inthe first sum and re-index the second sum so we have:

1

(n� 1)!

⇣x2

⌘n�1

+1X

s=1

(�1)s

s!(n+ s)!

⇣x2

⌘n+2s�1

+(�1)s�1

(s� 1)!(n+ s)!

⇣x2

⌘n+2s�1�.

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Now we can combine the coe�cients and when we do we have

1

(n� 1)!

⇣x2

⌘n�1

+1X

s=1

(�1)s

(s� 1)!(n+ s� 1)!

✓1

s�

1

n+ s

◆⇣x2

⌘n+2s�1

=1

(n� 1)!

⇣x2

⌘n�1

+ n1X

s=1

(�1)s

(s)!(n+ s)!

⇣x2

⌘n+2s�1

. (7.5)

Meanwhile, the right-hand side of Eq. (7.3a) is:

n1X

s=0

(�1)s

s!(n+ s)!

⇣x2

⌘n+2s�1

. (7.6)

Comparing this against Eq. (7.5) we see that the coe�cients are equal and so, by theuniqueness of power series, the left-hand side and right-hand side must be the samefunction.

7.2 Neumann function

In fact we can derive a series for the Bessel function and prove recurrence relationsfor all real numbers, then we usually say e.g. xZ 0

⌫(x) = xZ⌫�1(x) � ⌫Z⌫(x) as therecurrence relation for the Bessel function of order ⌫. In other words, the J ’s arereserved for the Bessel functions of integer order.

Now for most real numbers, Z⌫(x) and Z�⌫(x) are linearly independent. In thatcase we can take linear combinations of Z⌫ and Z�⌫ to obtain the general solution ofBessel’s di↵erential equation (7.1 with n ! ⌫. However, it is more typical to insteadtake the second solution to be the Neumann function associated with ⌫. This can bedefined as:

N⌫(x) =cos (⌫⇡)Z⌫ � Z�⌫(x)

sin (n⇡)

⌫!n�!

(�1)nJn(x)� J�n(x)

sin (n⇡)!

0

0.

In other words, for integer ⌫ this linear combination is an indeterminate form. If wewish we can compute the indeterminate form with l’Hopital’s rule. This yields

Nn(x) = (�n� 1)!

✓2

x

◆n 1

⇡+ · · ·+

2

⇡

⇣x2

⌘n✓

1

n!

◆ln⇣x2

⌘.

This expression diverges as 1

xn as x ! 0 (ln(x) if n = 0). But if n is not an integerthen the N⌫(x)s obeys the same recurrence relations as the J⌫(x)s. Then, by limits,Nn(x) also obeys the same recurrence relation as Jn(x)0s.

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7.3 Generating Function

In the homework, you will be asked to derive the generating function for Besselfunctions. You should begin by constructing the power series,

g(x, t) =1X

n=�1Jn(x)t

n. (7.7)

Using the recurrence relation proven above (7.3a) you can arrive at

✓t+

1

t

◆g =

2t

x

@g

@t.

Solving this PDE for g and imposing the condition that the coe�cient of t0 in theexpansion of g in powers of t is J0(x), we have:

g(x, t) = e(x2 )(t�

1t ). (7.8)

That the series in (7.7) is equal to this function of two variables can also be provenby inserting the series for the Bessel function (7.2) in Eq. (7.7) and rearranging thedouble sum that results. See Arfken, Weber, and Harris for this proof.

We quickly note that g has the property g(x, 1t ) = g(�x, t).

1X

n=�1Jn

✓1

t

◆n

=1X

n=�1Jn(�x)tn =

1X

n=�1J�m(x)t

m

by allowing dummy variables n = m, we have the relationship Jn(�x) = J�n(x).However, we also note that the generating function has the additional propertyg(�x, t) = g(x,�t). Satisfying both of these observations, we write

1X

n=�1Jn(�x)tn =

1X

n=�1Jn(x)(�t)n =

1X

n=�1J�n(x)t

n

... (�1)nJn(x) = J�n(x).

Note that we did establish this previously, but the proof via the generating functionis more elegant and faster.

7.4 Recurrence Relations Redux

In addition to the afore mentioned recurrence relations which can be relatively easilyshown using the series solution for the Bessel di↵erential equation, the next recurrencerelation we will look at will come from a partial derivative of the generating function.

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First, observe that the following identity holds for g(x, t):

@g

@x=

1

2

✓t�

1

t

◆e

x2 (t�

1t ) =

1

2

✓t�

1

t

◆g(x, t).

Inserting the series for g(x, t), Eq. (7.7) this becomes:

1X

n=�1J 0n(x)t

n =1

2

✓t�

1

t

◆ 1X

n=�1Jn(x)t

n

1X

n=�1J 0n(x)t

n =1

2

" 1X

n=�1Jn(x)t

n+1�

1X

n=�1Jn(x)t

n�1

#

Because limits of the summation are infinite, a shift in their placement does notchange their meaning and thus,

2J 0n(x) = Jn�1(x)� Jn+1(x) (7.9)

Now,d

dx

Jn(x)

xn

�=

J 0n(x)x

n� nJn(x)xn�1

x2n=

J 0n(x)

xn�

nJn(x)

xn+1

=1

2[Jn�1(x)� Jn+1(x)]

xn�

nJn(x)

xn+1=

1

2[Jn�1(x)� Jn+1(x)]

xn�

1

2[Jn�1(x) + Jn+1(x)]

xn.

Where the last substitution is a use of (7.3a). The resulting equality is

d

dx

Jn(x)

xn

�=

�Jn+1(x)

xn

This relationship of Bessel functions is particularly useful for solving integrals.We note also that combining Eqs. (7.3a) and (7.3b) we can obtain

d

dx[xnJn(x)] = xnJn�1(x). (7.10)

7.5 Integral Relations

If we choose to set t to be ei✓ then we have, for part of the exponent in the generatingfunction:

t�1

t= ei✓ � e�i✓ = 2i sin ✓

... g�x, ei✓

�= eix sin ✓ =

1X

n=�1Jn(x)

�ei✓

�n.

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This is a Fourier series in ✓ in which the coe�cients Jn(x)0s are Bessel functions. Weknow that if we compute the average of the Fourier series over one period we get1/2a0. That = J0(x) here, so applying that logic to this Fourier series yields:

Z2⇡

0

eix sin ✓d✓ = 2⇡J0(x) ! J0(x) =1

2⇡

Z2⇡

0

eix sin ✓d✓.

This happens because when we integrate over ✓ from 0 to 2⇡ we have, for all naturalnumbers n: Z

2⇡

0

ein✓d✓ = 0

Z2⇡

0

e�in✓d✓ = 0.

This is why J0(x) is the only term in the series that survives the integration over ✓.Furthermore, J0(x) is real, so we can take the real part of the right-hand side, whichimplies that

J0(x) =1

2⇡

Z2⇡

0

cos (x sin ✓)d✓.

More generally, we can apply this kind of Fourier series logic to equate themth Fouriercoe�cient in the Fourier series (in ✓) of eix sin ✓ to Jn(x). This says that:

Jm(x) =1

2⇡

Z2⇡

0

cos (x sin ✓ �m✓) d✓.

7.6 Example: Di↵raction Patternof Circular Aperture

We know

sin ✓ = 1.22�

d,

where d is the diameter of the aperture and � is the wavelength. We will now showthat the value 1.22 comes from the value of the first zero of the Bessel function.

We begin by parameterizing the aperture

~r = r cos ✓x+ r sin ✓y

r 2 [0, a];

✓ 2 [0, 2⇡].

A wave propagating from r to a point on the screen that is located at an angle ↵away from the perpendicular line from the center of the aperture to the screen has awave vector ~k = 2⇡

� sin↵y + 2⇡� cos↵z. This means we have taken the positive z-axis

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to be from the aperture to the screen and the y-axis as the direction along the screen.Then we have defined the aperture using polar co-ordinates in the x� y plane. If ~x0

is the vector from the center of the aperture to the point of interest on the screen,then the phase di↵erence of a wave from an arbitrary point on the aperture to thatpoint on the screen is compared to the wave coming from the center is ~k · ~r. But inthis co-ordinate system we have:

~k · ~r =2⇡

�r sin↵ sin ✓.

To calculate the amplitude of the wave on the screen at that point, �(~x0) we addup the amplitudes of waves coming from all di↵erent points on the aperture:

�(~x0) =

Z

aperture

ei~k·~r d2r,

where ~k for the center of the aperture is the vector defined above. Now we assumethat the distance from the aperture to the screen is large compared to the size of theaperture itself. Therefore we do not care about the variation in the direction of thewave vector ~k over the aperture. In fact, to write down this equation we have alreadyassumed that there is not a significant di↵erence in the distance that waves fromdi↵erent points on the aperture have to travel to ~x0, and so the only thing we haveto keep track of across the aperture is the phase di↵erence. Under this assumption,and using the polar co-ordinate parameterization of the aperture:

�(~x0) =

Z a

0

Z2⇡

0

ei~k·~r r d✓dr

We now define b = 2⇡� sin↵. b then depends on ~x0 through the variable sin↵, and

� is purely a function of b and a:

�(b) =

Z a

0

Z2⇡

0

eibr sin ✓d✓

�rdr

) � =

Z a

0

2⇡J0(br)rdr.

where we have used the integral identity derived in the previous subsection. We canmake the upper integration bound dimensionless by writing

�(b) =1

b2

Z ab

0

J0(x)xdx

To do this integral we remember

xJ0(x) =d

dx[xJ1(x)]

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and so

�(b) =1

b2

Z ab

0

d

dx[xJ1(x)] dx =

1

b2baJ1(ba)

=�a

2⇡ sin↵J1

✓2⇡a

�sin↵

◆.

Since J1(x) / x for small x this goes to a constant as ↵ ! 0, which indicatesthe existence of a bright central maximum. There is a minimum if J1

�2⇡a� sin↵

�= 0.

This means that the zeros of the bessel function are minimums of the di↵ractionpattern

2⇡a

�sin↵ = ↵11,↵12,↵13, · · · = 3.8317, ... )

d

�sin↵1 =

3.8317

⇡

) sin↵1 =1.22�

d,

where ↵1 is the angle at which the first di↵raction minimum occurs.

7.7 Orthogonality

Before continuing, we will make a few modifications to the structure of our Besselfunction. First we define

x = k⇢

wherek has units of inverse length and ⇢ has units of length. For the present purposes,it will be the case that k is fixed and ⇢ is varied. Bessel’s equation (7.1) will now bewritten as

⇢d2Z⌫(k⇢)

d⇢2+

dZ⌫(k⇢)

d⇢+

✓k2⇢�

⌫2

⇢

◆Z⌫(k⇢) = 0, (7.11)

where we are adopting the Z⌫ notation for the Bessel function because we want tomake it clear that ⌫ could be not-an-integer and we also want the proof we’re doingto encompass Neumann functions of non-integer order.

This equation can now be written in a form that makes it clear that the di↵erentialoperator here is “self-adjoint”. If you want to know more about what this means Iencourage you to go and read Arfken, Weber, and Harris on the subject. If you don’tcare, that’s fine. :) Anyway, regardless of what you call it, the new form is:

d

d⇢

⇢d

d⇢Z⌫(k⇢)

�+

✓k2⇢�

⌫2

⇢

◆Z⌫(k⇢) = 0.

We multiply this equation from the left by Z⌫(k0⇢) and integrate from 0 to a. (Youshould think of a as the radius of the solution domain for our partial-di↵erential-equation problem.) That gives:

Z a

0

Z⌫(k0⇢)

d

d⇢

⇢d

d⇢Z⌫(k⇢)

�d⇢+

Z a

0

Z⌫(k0⇢)

✓k2⇢�

⌫2

⇢

◆Z⌫(k⇢) d⇢ = 0

75

) Z⌫(k0⇢)⇢

d

d⇢Z⌫(k⇢)

����a

0

�

Z a

0

dZ⌫(k0⇢)

d⇢⇢dZ⌫(k⇢)

d⇢d⇢

+

Z a

0

Z⌫(k0⇢)

✓k2⇢�

⌫2

⇢

◆Z⌫(k⇢) d⇢ = 0 (7.12)

Z⌫(k0a)a

d

d⇢Z⌫(k⇢)

����⇢=a

�

Z a

0

dZ⌫(k0⇢)

d⇢⇢dZ⌫(k⇢)

d⇢d⇢

+

Z a

0

Z⌫(k0⇢)

✓k2⇢�

⌫2

⇢

◆Z⌫(k⇢) d⇢, (7.13)

where we have used the fact that the functions we are considering are regular as⇢ ! 0.

Now write down (7.13) again, but with the roles of k and k0 switched. Take thedi↵erence of the two equations and you get:

Z⌫(k0a)a

d

d⇢Z⌫(k⇢)

����⇢=a

� Z⌫(ka)ad

d⇢Z⌫(k

0⇢)

����⇢=a

+

Z a

0

Z⌫(k⇢)�k2

� k02� ⇢Z⌫(k0⇢) d⇢ = 0. (7.14)

So far what we’ve done is true for any ⌫. But we now specialize to integer ⌫, so wereplace ⌫ by n and Z⌫ by Jn (because the proof wouldn’t work if it was Nn; why?).Suppose now we pick specific wave numbers—which are actually the wave numberswe’d have if we had to satisfy a zero boundary condition at a, then the first two termsvanish. I.e., we pick k and k0 such that Jn(ka) = Jn(k0a) = 0. Those wave numbersare:

ka = ↵np p = 1, 2, 3...k0a = ↵nq q = 1, 2, 3...

For these wave numbers we have:

�k2

� k02�Z a

0

Jn(k⇢)⇢Jn(k0⇢) d⇢ = 0

Therefore, provided that k 6= k0, i.e., p 6= q, we haveZ a

0

Jn⇣↵np

a⇢⌘⇢Jn

⇣↵nq

a⇢⌘d⇢ = 0.

This is the statement of orthogonality for Bessel functions. If we remember, thenormal modes found for the drum were

Unp(⇢, ✓) = Jn⇣↵np

a⇢⌘(An cos (n✓) + Bn sin (n✓))

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This orthogonality means that the normal modes are orthogonal in both p and nwhen we integrate over the circle, i.e. integrate Unp against Umq, i.e.

Z a

0

Z2⇡

0

Unp(⇢, ✓)Umq(⇢, ✓)⇢d⇢d✓ = N �nm�pq, (7.15)

where N is a normalization integral that we are not computing yet. What we aresaying is that the ✓ integral forces m and n to be equal, while the ⇢ integral forcesp and q to be equal. It is important to note that this way of thinking about theorthogonality tells you two things it’s good to remember:

• The Bessel functions are orthogonal for the di↵erent discrete wave numbers. Itis not an orthogonality on the index n.

• There is a factor of ⇢ in the Bessel function orthogonality relation.

This doesn’t tell you how to normalize the Bessel functions though. We’ll do thatfor homework, by extending the above arguments. The answer is:

Z a

0

hJn

⇣↵np

a⇢⌘i2

⇢d⇢ =a2

2[Jn+1(↵np)]

2 . (7.16)

7.7.1 Analogy:

One dimenson Two dimensionsin

�p⇡xa

�, cos

�p⇡xa

�Jn

�↵np

a ⇢�, Nn

�↵np

a ⇢�

Argument of sin function scaled s.t. Argument of Jn scaled s.t.sine function =0 when x = a Bessel function =0 when ⇢ = a

Obey the x-part of 1D wave eq. Obey radial part of 2D wave eq.

y00 +�p⇡a

�2y = 0 ⇢ d

d⇢

⇣⇢dyd⇢

⌘+⇣

↵2np⇢

2

a2 � n2

⌘y = 0

R a

0sin

�p⇡xa

�sin

�q⇡xa

�dx = 0

R a

0Jn

�↵np

a ⇢�⇢Jn

�↵nq

a ⇢�d⇢ = 0

So you should think of the Bessel function zeroes as the 2D analogs of the rootsof the sine function p⇡. It is just that they are not evenly spaced. Although theybecome so for large p. Another key di↵erence is that there are infinitely many sets ofBessel function zeroes: there’s one for each value of n.

7.8 Bessel Series (plural! n = 0, 1, 2, 3, ...)

Now we have a set of orthogonal functions. What do we like to do with those? Expandother functions in them! Except here we have the opportunity to do infinitely manyorthogonal-function expansions (yay!) because there is one set of orthogonal functions

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for each n. Once I pick an n I make an orthogonal set of functions by forming the set{Jn

�↵np

a ⇢�: p = 1, 2, . . .}. The function f(⇢) is then expanded as:

f(⇢) =1X

p=1

CpnJn⇣↵np

a⇢⌘.

Here the sum is over p, but the coe�cients also depend on which Bessel function Jnwe chose before we formed our set of orthogonal functions. Those coe�cients arefound by multiplying both sides by ⇢ and by Jn(

�↵nq

a ⇢�and then integrating from 0

to a.

Cpn =2

a2 [Jn+1(↵np)]2

Z a

0

f(⇢)Jn⇣↵np

a⇢⌘⇢d⇢.

7.9 Example: Electrostatic Potentialin a Hollow Cylinder

Consider Laplace’s equation in cylindrical polar coordinates

r2 = 0.

The normal modes for this problem are:

kn(⇢,', z) = Jn(k⇢) (an cos (n') + bn sin (n'))⇥cke

kz + dke�kz

⇤,

where we have imposed the condition that the coe�cient of Nn(k⇢) = 0 to force afinite solution. We now consider the problem of the electrostatic potential inside acylinder that is held at zero potential on the left end-cap and on its surface. Hencewe impose the boundary conditions

(a, ⇢, z) = 0, (7.17a)

(⇢,', 0) = 0, (7.17b)

for '[0, 2⇡] and z[0, L] (7.17c)

By writing the boundary conditions in terms of a separation-of-variables solutionwe find that:

R(a)�(')Z(z) = 0 ! Jn(ka) = 0 ! k =↵np

aR(⇢)�(')Z(0) = 0 ! ck = �dk

With these 2 boundary conditions imposed the normal modes become discrete, andare indexed by n, which denotes the number of nodes in the angular variable, and p,which denotes the number of nodes for 0 < ⇢ a (including the one at a):

np(⇢,', z) = Jn⇣↵np⇢

a

⌘[anp cos (n') + bnp sin (n')] sinh

⇣↵npz

a

⌘.

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Here the coe�cient and factor of 2 in the z part has been absorbed into the coe�cientsof the ' part, and so now they are also indexed by p. As we know, the general solutionis a sum of normal modes, so

(⇢,', z) =1X

n=0

1X

p=1

Jn⇣↵np⇢

a

⌘[anp cos (n') + bnp sin (n')] sinh

⇣↵npz

a

⌘.

If I then specify a functional form for the other end of the cylinder,

kn(⇢,', L) = f(⇢,')

then we must have

f(⇢,') =1X

n=0

1X

p=1

Jn⇣↵np⇢

a

⌘[anp cos (n') + bnp sin (n')] sinh

✓↵npL

a

◆.

We are left with a Fourier series in n and a Bessel series in p. We use orthogonalityof the trig functions and the Bessel functions to pick out a particular coefificent, e.g.,

b65,32 /

Z a

0

Z2⇡

0

f(⇢,') sin (65')J65⇣↵65,32⇢

a

⌘⇢d'd⇢

More precisely, i.e., keeping track of the factors associated with the normalization ofour set of (doubly orthogonal) functions:

⇢anpbnp

�= 2

⇡a2 sinh⇣

↵npLa

⌘J2n+1(↵np)

R a

0

R2⇡

0f(⇢,')

⇢cos (n')sin (n')

�Jn

�↵np⇢a

�⇢d'd⇢.

7.10 Spherical Bessel Functions

The Schrodinger eequation in spherical polar coordinates can be written as

r2d2R

dr2+ 2r

dR

dr+⇥k2r2 � `(`+ 1)

⇤R = 0 (7.18)

we try

R(r) =Z(r)pkr

!dR

dr=

1pkr

dZ(r)

dr�

1

2r

Z(r)pkr

!d2Z(r)

dr2=

1pkr

d2Z(r)

dr2�

1

rprk

dZ(r)

dr+

3

4r2Z(r)pkr

Plugging this substitution back into (7.18),

d2Z(r)

dr2+

1

r

dZ

dr+

"k2r2 �

✓`+

1

2

◆2#Z(r) = 0

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where ` is an integer 0,1,2,.. and thus `+ 1

2is a half integer. With our knowledge of

the solutions of a di↵erential equation of this form we quickly spot that

R(r) =J`+ 1

2(kr)

pkr

.

The ubiquity of this kind of solution of the three-dimension Schrodinger equationleads us to define new “spherical” Bessel functions:

jL(x) =

r⇡

2xJL+ 1

2(x)

and corresponding “spherical Neumann functions”:

nL(x) =

r⇡

2xNL+ 1

2(x).

The series form of jL(x) is:

jL(x) =1X

s=0

(�1)s(s+ L)!

(2s+ 2(`+ 1))!s!x2s

�(2x)L.

So it turns out that:

j0(x) =1X

s=0

(�1)s

(2s+ 1)!x2s = 1�

x2

3!+

x4

5!� · · · =

sin x

x,

while

n0(x) =� cos x

x.

7.11 Hankel functions and Modified Bessel func-tions

We can also define Hankel functions, essentially the travelling-wave counterparts tothe standing-wave Bessel functions:

H(1)

⌫ (x) = J⌫(x) + iN⌫(x)

H(2)

⌫ (x) = J⌫(x)� iN⌫(x)

Lastly, there are solutions to the modified Bessel equation

x2d2Indx2

+ xdIndx

��x2 + n2

�In = 0, (7.19)

they can be thought of as Bessel functions with an imaginary argument. As such theyare a generalization of the hyperbolic sine and cosine functions with an associatedgenerating function of

e(x2 )(t+

1t ) =

1X

n=0

In(x)tn. (7.20)

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