Kreh

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Bessel Functions

Project for the Penn State - Gttingen Summer School on Number Theory

Martin Kreh

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Content i

Content

1 The Bessel differential equation 1

1.1 Bessel functions of first and second kind . . . . . . . . . . . . . . . . . . . 1

1.2 Modified Bessel functions of first and second kind . . . . . . . . . . . . . . 3

2 Properties of the Bessel functions 5

2.1 Generating function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

2.2 Special values . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

2.3 Integral representations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

2.4 Defining Bessel functions through the generating function . . . . . . . . . 15

2.5 Asymptotics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162.6 Graphs of Bessel functions . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

2.7 Integral Transforms. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

References 21

Martin Kreh The Bessel Functions

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1 The Bessel differential equation

In the first chapter, we introduce the (modified) Bessel differential equation and deduce

from it the (modified) Bessel functions of first and second kind. This will be done via a

power series approach. We decided to start from a differential equation, since this seems

kind of naturally. We will point out later that we could have defined the Bessel function

in an other, equivalent, way.

1.1 Bessel functions of first and second kind

In this section we study the the so called Bessel equation

x2y+xy+ (x2 n2)y = 0

for n N, which can for example be obtained by the separation of the wave equation incylindric or polar coordinates. Since this is a second order differential equation, we will

have two linearly independent solutions. We will now start to construct a first solution.

We take as approach for a solution y a power series. It will be convenient to write

y(x) =xnk=0

bkxk.

Then we have

xy(x) =nxnk=0

bkxk +xn

k=0

kbkxk

and

x2y(x) =n(n 1)xnk=0

bkxk + 2nxn

k=0

kbkxk +xn

k=0

k(k 1)xk.

We further have

x2y(x) =xnk=2

bk2xk

when settingb2:= b1:= 0. Plugging in the differential equation yields

0 =n(n 1)xnk=0

bkxk + 2nxn +nxn

k=0

bkxk

+xnk=0

kbkxk +xn

k=2

bk2xk n2xnk=0

bkxk


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=xnk=0

(n(n 1)bk+ 2nkbk+k(k 1)bk+nbk+kbk+bk2 n2bk)xk

=xn

k=0

(2nkbk+k2bk+bk2)xk.

Comparing coefficients gives

k(k+ 2n)bk+bk2= 0.

Since b1 = 0 we get b1 = b11+2n = 0 and recursively b2k1 = 0 for all k as long as2n is not a negative whole number. But in this case the recursion will also be fullfilled if

b2k1 = 0 for all k. So it remains to consider even k. Here we have no condition on b0.We can choose it freely and will do so later. Now ifn / N we get

b2k = b2k2

4k(n+k)

and this yields inductively

b2k = (1)k b04kk!(n+k)(n+k 1) (n+ 1) .

It is convenient to choose b0= 12nn! . So together we get

y(x) =x

2n

k=0

(

1)k

k!(n+k)!x

22k

.

This is convergent due to the quotient criterion, therefore we have indeed constructed a

solution for the Bessel equation.

We also want to construct a solution for complex order . Here we need to exchane

the factorial with the Gamma function. Then we get

J(x) =x

2

k=0

(1)k(k+ 1)(+k+ 1)

x2

2k

This is valid unlessn N. But in this case we can just write

Jn(x) :=x

2

n k=n

(1)k(k+ 1)(n+k+ 1)

x2

2k

=x

2

n k=0

(1)n+k(n+k+ 1)(k+ 1)

x2

2n+2k


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= (1)nJn(x)

and of course this solves the Bessel equation.

Now we want to determine a linearly independent solution. First, we describe how J

behaves ifx 0. From the power series expansion, we immediately get

limx0

J(x) =

0, ()> 01, = 0

, ()< 0, / Z.

This means that J and Jare two linearly independent solutions if / Z. If Z,they are linearly dependent. Since (1)n = cos n, the function J(z)cos J(z)isa solution of the Bessel equation, which vanishes ifn

N0. Therefore, we define

Y(x) =cos()J(x) J(x)

sin() .

for / Z. Then for n Zthe limit

Yn(x) := limnY(x)

exists, as can be shown by LHospitals rule, and J andYare two linearly independent

solutions of the Bessel equation for all C. This can e.g. be shown be computing the

Wronskian determinant.

1.2 Modified Bessel functions of first and second kind

The modified Bessel equation is given by

x2y+xy (x2 +2)y= 0.

Analog to the previous section we can compute a solution of this differential equation

using the power series approach. This gives

I(x) =x

2

k=0

1

(k+ 1)(+ 1 +k)

x2

2k


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As before, we search for a second, linearly independent solution. This will, for / Z, begiven by

K(x) =

2

I(x) I(x)sin()

and its limit for n Zexists.


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2 Properties of the Bessel functions

In this chapter we will prove some properties of Bessel functions. There are of course

more interesting facts, in particular, there are connections between Bessel functions and,

e.g. Legendre polynomials, hypergeometric functions, the usual trigonometric functions

and much more. These can be found in[Wat22]and [GR00]

2.1 Generating function

Many facts about Bessel functions can be proved by using its generating function. Here

we want to determine the generating function.

Theorem 2.1 We have

ex2(zz1) =

n=

Jn(x)zn (2.1)

i.e.ex2(zz1) is the generating function ofJn(x).

Proof: We have

ex2ze

x2

1z =

m=0

(x2 )m

m! zm

k=0

(1)k(x2 )kk!

zk

=

n=

mk=nm,k0

(

1)k(x2 )

m+k

m!k!

zn

=

n=

k=0

(1)k(n+k)!k!

x2

2k x2

nzn

=

n=Jn(x)z

n.

q.e.d.

We can use this generating function to prove some standard results.

Lemma 2.2We have

cos x= J0(x) + 2n=1

(1)nJ2n(x),

sin(x) = 2n=0

(1)nJ2n+1(x),


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1 =J0(x) + 2n=1

J2n(x).

Proof: Directly from (2.1) withz = ei, i sin = 12(z

1z )we get

cos(x sin )+i sin(x sin ) =eix sin =

n=Jn(x)e

in =

n=Jn(x)(cos(n)+i sin(n))

so

cos(x sin ) =J0(x) + 2n=1

J2n(x) cos(2n)

and

sin(x sin ) = 2

n=0J2n+1(x) sin((2n+ 1)).

This gives the results with = 2 and = 0, respectively. q.e.d.

Lemma 2.3We have

Jn(x) =Jn(x) = (1)nJn(x)

for all n Z.

Proof: To show the first equality, we make the change of variables x x, z z1 inthe generating function. Then the function does not change and we get

n=

Jn(x)zn =

n=Jn(x)zn =

n=

Jn(x)zn

and comparing coefficients gives the result. The second equality follows analog with the

change of variable z z1. q.e.d.

Lemma 2.4We have for any n Z

2Jn(x) =Jn1(x) Jn+1(x),2n

xJn(x) =Jn+1(x) +Jn1(x),

ddx

(xnJn(x)) =xnJn1(x). (2.2)

Proof: The first statement follows, when we differentiate equation (2.1) with respect to

x:


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n=Jn(x)z

n = 1

2

z 1

z

ex2(z 1

z) =

1

2z

n=Jn(x)z

n 12z

n=Jn(x)z

n

=

n=

1

2(Jn1(x) Jn+1(x))zn.

Comparing coeffivients yields the result.

The second statement follows completely analog when considering the derivative with

respect to z .

Adding the two equations and multiplying with xn/2 gives the third result. q.e.d.

With the last two lemmata we can describe some integrals with Bessel functions via a

different Bessel function.

Lemma 2.5For any n Zwe have xn+1Jn(x) dx= x

n+1Jn+1(x) and

xn+1Jn(x) dx= xn+1Jn+1(x).

Proof: The first equation follows directly from (2.2). For the second one, we use (2.2)

and lemma2.3 to get

xn+1Jn(x) dx=

xn+1(1)nJn(x) dx= (1)nxn+1Jn+1(x) = xn+1Jn+1.

q.e.d.

We have already seen, that J20 (x) + 2

n=1J2n(x) = 1. With a bit more work, we can

also deduce a result for

Jn+m(x)Jn(x) withm = 0.Lemma 2.6We have for any m = 0

J20 (x) + 2n=1

J2n(x) = 1 and

n=Jn+m(x)Jn(x) = 0.

Proof:

We multiply equation (2.1) with the equation where z z1

to get

1 =

k=Jk(x)z

k

n=Jn(x)z

n

=

m=

n=

Jn+m(x)Jn(x)

zm


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=

n=J2n(x) +

mZm=0

n=

Jn+m(x)Jn(x)

zn.

Comparing coefficients gives

1 =

n=J2n(x) =J

20 (x) + 2

n=1

J2n(x)

and

0 =

n=Jn+m(x)Jn(x)

for all m = 0. q.e.d.

Lemma 2.7We have nZ

Jn(x) = 1.

Proof: This follow by setting z = 1 in (2.1). q.e.d.

Lemma 2.8We have

Jn(x+y) =kZ

Jk(x)Jnk(y)

for all n Z.

Proof: This follow through

nZ

Jn(x+y) = exp

1

2(x+y)(t t1)

= exp

1

2x(t t1)

exp

1

2y(t t1)

=kZ

Jk(x)tkmZ

Jm(y)tm

=nZ

kZ

Jk(x)Jnk(y)

tn

and comparing coefficients. q.e.d.

Most of the results hold (in a similar form) also for the other three Bessel functions.

Since the proofs are anlog, we will only state the results.

Theorem 2.9 (Results for Jn(x), Yn(x)and Kn(x))


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The generating function for the In(x)is ex2(z+z1), i.e.

ex2(z+z1) =

n=In(x)z

n.

We have for all n Z

Yn(x) = Yn(x) = (1)nYn(x),In(x) = (1)nIn(x) =In(x),

Kn(x) = (1)nKn(x) =Kn(x).

For all n Zwe have

Yn1(x) +Yn+1(x) =2nx

Yn(x), Yn1(x) Yn+1(x) = 2Yn(x),

In1(x) In+1(x) =2nx

In(x), In1(x) +In+1(x) = 2In(x),

Kn+1(x)Kn1(x) =2nx

Kn(x), Kn1(x) +Kn+1(x) = 2Kn(x).

Indeed this and the next two sets of formulae, as well as those for Jn(x), hold also

for complex orders, as can be shown through the series representation.

We have for all n Z

ddx

(xnYn(x)) =xnYn1(x),

d

dx(xnIn(x)) =x

nIn1(x),

d

dx(xnKn(x)) = xnKn1(x).

For all n Zthere holds xn+1Yn(x) dx= x

n+1Yn+1(x),

xn+1Yn(x) = xn+1Yn+1(x),

xn+1In(x) dx= xn+1In+1(x),

xn+1In(x) = xn+1In+1(x),

xn+1Kn(x) dx= xn+1Kn+1(x),

xn+1Kn(x) =xn+1Kn+1(x).


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We have for all m Z, m = 0

I20(x) + 2

n=1(1)nI2n(x) = 1 and

n=(1)nIn+m(x)In(x) = 0.

We have n=

Ik(x) =ex.

It holdsIn(x+y) =

kZ

Ik(x)Ink(x).

2.2 Special values

In this section we want to examine how the Bessel functions look like when we plug in

some special values.

Lemma 2.10 If 12 + Z, the Bessel functions are elementary functions.

1. We have

J12

(x) =Y 12

(x) =

2

xsin(x),

J

1

2

(x) =

Y12

(x) = 2

x

cos(x),

I12

(x) =

2x sinh(x),

I 12

(x) =

2

xcosh(x),

K12

(x) =K 12

(x) =

2xex.

2. There are polynomialsPn(x), Qn(x)with

deg Pn= deg Qn= n, Pn(

x) = (

1)nPn(x), Qn(

x) = (

1)nQn(x)

such that for any k N0

Jk+ 12

(x) =

2

x

Pk

1

x

sin(x)Qk1

1

x

cos(x)

,

Jk 12

(x) = (1)k

2

x

Pk

1

x

cos(x) +Qk1

1

x

sin(x)

,


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Yk+ 12

(x) = (1)k1Jk 12

(x)

Yk12

= (1)kJk+ 12

(x),

Ik+1

2 (x) = 2

x

i

k

Pk i

x

sinh(x) +i

k

1

Qk1 i

x

cosh(x)

,

Ik 12

(x) =

2

x

Pk

1

x

cosh(x)Qk1

1

x

sinh(x)

,

Kk+ 12

(x) =Kk 12

(x) =

2x

ikPk

1

ix

sin(x) +ik1Qk1

1

ix

ex

.

Proof: The first formulae follow directly from the series representations of J and I,

the definitions ofY and Kand the properties of the -function. For example, we have

J12 (x) =x

2

k=0

(

1)k

(k+ 1)(k+ 32)x

22k

=

x

2

k=0

(1)kk! (2k+1)!

k!22k+1

x2

2k

=

2

x

k=0

(1)k(2k+ 1)!

x2k+1

=

2

xsin x

and

Y 12

(x) = cos(2 )J 12 (x) J12 (x)

sin(2 ) =J1

2

(x).

The rest follows from the recurrence formulas of the last section. q.e.d.

Directly from the definitions, we also get

In(x) =inJn(ix) and Kn(x) =

2in1(Jn(ix) +iYn(ix)).

2.3 Integral representations

In this section we will give an integral representation for each of the four Bessel functions

that we have introduced. There are much more representations as can be shown here, see

e.g. [Wat22] and[GR00].


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Theorem 2.11 For all n, x C we have

J(x) = 1

0

cos(x sin t t) dtsin()

0

ex sinh(t)t dt,

Y(x) = 1

0

sin(x sin t t) dt 1

0

ex sinh(t)(et + cos()et) dt,

I(x) = 1

0

ex cos t cos(t) dt sin()

0

ex cosh(t)t dt,

K(x) =

0

ex cosh(t) cosh(t) dt.

Proof: We use the representation 1(z) = 12itzet dt where is some contour from

, turning around 0 in positive direction and going back to. Then we get

J(x) =(x/2)

2i

k=0

(1)k(x/2)2ktk1k!

et dt=(x/2)

2i

t1etx2/(4t) dt.

Let t = x2u. Then we get

J(x) = 1

2i

u1e(x/2)(u1u) du

for some contour of the same kind. Now let u = ew. Then the contour , given as

rectangular with vertices i,i,i,+i without the right edge is the image ofa suitable(see figure2.1). So we get

Figure 2.1: The contours for the integral.


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J(x) = 1

2i

ewex sinh(w) dw= 1

2i(I1+I2 I3)

with

I1=

eitex sinh(it)i dt,

I2=

0

e(i+t)ex sinh(i+t) dt,

I3=

0

e(i+t)ex sinh(i+t) dt.

We have

1

2iI1=

1

2i

eitex sinh(it)i dt

= 1

2

0

ei(x sin tt) dt+0

ei(x sin tt) dt

= 1

2

0

ei(x sin tt) +ei(x sin tt) dt

= 1

0

cos(x sin t t) dt

which gives the first part. Further,

1

2i(I2 I3) = 1

2i

0

e(i+t)ex sinh(i+t) dt0

e(i+t)ex sinh(i+t) dt

=

1

2i

0

ex sinh t

t

(ei

ei

) dt

= sin()

0

ex sinh tt dt

which gives the second part. This proves the first result.


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For Y, we note that

sin()Y(x) = cos()J(x) J(x) = I1sin()

I2

with

I1= cos()

0

cos(x sin(t) (t)) dt

0

cos(x sin(t) +t) dt,

I2=

0

ex sinh(t)(cos()et +et) dt.

Since

cos() cos(x sin(t) t) = cos(x sin(t) +( t)) + sin() sin(x sin(t) t)

and

0

cos(x sin(t) +( t)) dt=

0

cos(x sin(t) +t) dt,

we have I1 = sin()0 sin(x sin(t) t) dt and this proves the formula for / Z.

Because of the continuity, we therefore get it for all .

The formula for Iis proven completely analog to that for Jand that for K follows

then from the definition, similarly to those for Y. q.e.d.

With this theorem we can again follow that Yn and Kn exist for n Z.With the generating function, these formulae can be proven even simplier, at least for

n Z. We will show this as an example for the functions Jn.We use

eint dt= 20n

to get

eizsin

ein

d=kZ

Jk(x)

eik

ein

d = 2Jn(x).

So

Jn(x) = 1

2

ei(x sin n) = 1

0

cos(x sin n) d.


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2.4 Defining Bessel functions through the generating function

As we mentioned before, we wanted to start with the defining differential equation. We

could have also defined the Bessel function Jn(x) (at least for integer n) through its

generating function. On this way, we also could have developed the series and integral

representation:

Theorem 2.12 Let the functions yn(x)be defined by

ex2(zz1) =

n=

yn(x)zn

Then we have

yn(x) =x

2n

k=0

(

1)k

(n+k)!k!x

22k

and yn(x) = 1

0

cos(x sin n) d.

Proof: First, the Cauchy integral formula gives us

yn(x) = 1

2i

ex2(tt1)

tn+1 dt

for any simply closed contour around 0.

First we use the substitution t = 2ux and the series representation ofex to get

yn(x) = 1

2i

e4u2x2

4u

( 2x)n+1un+1

2

xdu

= 1

2i

x2

n

euex2

4u un1 du

=x

2

n m=0

k=0

(1)km!k!

x2

2k 12i

umkn1 du

=x

2n

l=0

(

1)k

(n+k)!k!x

22k

.

Choosing the contourt = ei we get

yn= 1

2

20

ex2(eiei)

eni d


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= 1

2

20

exp(xi sin ni) d

=

1

0

cos(x sin n) d.

q.e.d.

From this integral representation we can then easily, using partial integration, see that

this function solves the Bessel differential equation.

2.5 Asymptotics

Now we want to use the proven integral representations to get asymptotic formulae.Theorem 2.13 Forx R, x we have

J(x)

2

xcos(x

4

2 ),

Y(x)

2

xsin(x

4

2 ),

I(x)

1

2xex,

K

(x)

2xex.

Proof: J and Y:The second integrals in both of the representations go to 0 exponentially, therefore

we have

J(x) +iY(x) = 1

0

ei(x sin tt) dt+O(xA)

for all A. Substitution gives

J(x) +iY(x) =2ei/2

(I1(x) +I2(x)) + O(xA)

with

I1(x) =

/30

eix cos t cosh(t) dt, I2(x) =

/2/3

eix cos t cosh(t) dt.


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InI2 we substitutecos(t) =u to get

I2(x) =

1/2

0

eixu(u) du

with(u) = cosh(cos1(u))

1u2 . Integrating by parts give

I2(x) = eixu

ix (u)

1/2

0

1ix

1/20

eixu(u) du= O(x1),

sinceu stays away from any singularites of and at 1.

InI1, we setu =

2x sin(t/2)to get

I1(x) =

2eix

x

x/2

0

eiu2

cosh

2sin1

u

2x

du

1 u22x

and so

I1(x)

2xeix0

eit2

dt.

Using 0 e

it2 dt=2 e

i/4 we have

J(x) +iY(x)

2

xei(xpi/4/2),

which gives the result.

I:Again we have

0 exp(x cosh tt) dt 0so we only have to consider the other

integral. We split this in two integrals, on from0 to /2and the other from /2to

. Since cos(t) 0 in the second interval, the second integral is bounded. For thefirst one we setu = 2x sin(t/2)to get

I(x) = ex

x

2x

0

eu2

2 cos

2sin1

u

2

x

du

1 u24x+O(1).

If now x , the integral goes to 0 eu22 = 2 .


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2.7 Integral Transforms

In the last section, we want to compute three of the most used integral transforms. Here

we will restrict to the Bessel function J0(x).

Theorem 2.14 (Laplace Transform) We have

L[J0](s) = 11 +s2

.

Proof: The Laplace transform of the equation x(y + y) + y= 0, which is equivalent tothe Bessel equation of order 0, gives

0 = ddsL[y+y] + L[y]

= d

ds (s2

L[y](s) + L[y](s) sy(0) y(0)) +sL[y](s) y(0)= (1 +s2)L[y](s) sL[y](s),

soL[J0](s) = c1+s2 with a constant c. We can compute c through

0 = limxL[J

0](x) = limx(xL[J0](x) J0(0)) =c 1.

This means we have

L[J0](s) = 11 +s2

.

q.e.d.

Theorem 2.15 (Fourier Transform) For s R we have

F[J0](s) =

21s2 , |s|

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= lim0

1

1 + ( is)2 + 1

1 + (+is)2

= 21s2 , |s|

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References 21

References

[Coh07] Cohen, Henri:Number Theory Volume II: Analytic and Modern Tools. Springer,

2007

[Dav01] Davies, Brian: Integral Transforms and Their Applications. Third Edition.

Springer, 2001

[GR00] Gradshteyn, I.S. ; Ryzhik, I.M.: Table of Integrals, Series and Products.

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[Wat22] Watson, G.N.: A Treatise on the Theory of Bessel Functions. Cambridge

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