3 bessel's functions

Bessel Functions
The differential equation
where p is a non negative constant, is called Bessels
equation, and its solutions are known as Bessel
functions.
These functions first arose in Daniel Bernoulli’s
investigation of the oscillations of a hanging chain
appeared again in Euler’s theory of the vibration of a
circular membrane and Besell’s studies of planetary
motion.
,0)( 222
=−+′+′′ ypxyxyx

Bessel Functions continued…
Friedrich Wilhelm Bessel (1784 – 1846)
studied disturbances in planetary motion,
which led him in 1824 to make the first
systematic analysis of solutions of this
equation.
The solutions became known as Bessel
functions.

Bessel’s Equation of order p:
The origin that is x = 0 is a regular singular point.
The indicial equation is m2
-p2
=0, and the exponents are
m1=p and m2 = -p.
It follows from Theorem 30-A that BE has a solution of
the form
where and the power series converges
for all x.
( ) .0222
=−+′+′′ ypxyxyx
∑∑
∞
=
+
∞
=
==
00
,
n
pn
n
n
n
n
p
xaxaxy
00 ≠a ∑ n
n xa

Hence we have
The terms of the equation becomes
( )
( )( )∑
∑ ∑
∞
=
−+
∞
=
∞
=
−++
−++=′′
+=′=
0
2
0 0
1
.1)(
,)(,)(
n
pn
n
n n
pn
n
pn
n
xpnpnaxy
xpnaxyxaxy
( ) ( )( )∑∑
∑ ∑
∞
=
+
∞
=
+
∞
=
∞
=
+
−
+
−++=′′+=′
=−=−
0
2
0
0 0
2
222
.1,
,,
n
pn
n
n
pn
n
n n
pn
n
pn
n
xpnpnayxxpnayx
xayxxapyp

Inserting into the equation and equating the coefficients
of xn+p
to zero, we get the following recursion formula
We know that a0 is arbitrary and a-1=0 tells us that a1=0;
and repeated application of this recursion formula yields
the fact that an=0 for every odd subscript n.
.
n)pn(
a
a
aanpn
n
n
nn
+
−=
=++
−
−
2
or
0)2(
2
2

The nonzero coefficients of our solution are therefore
And the solution is
,
)62)(42)(22(642)62(6
,
)42)(22(42)42(4
,
)22(2
,
04
6
02
4
0
20
+++⋅⋅
−=
+
−=
++⋅
=
+
−=
+
−=
ppp
a
p
a
a
pp
a
p
a
a
p
a
aa












+
+++
−
++
+
+
−
=

)3)(2)(1(!32
)2)(1(!22)1(2
1
6
6
4
4
2
2
0
ppp
x
pp
x
p
x
xay p

Or we have
This solution y is known as Bessel function of the first
kind of order p.
This is denoted by Jp(x) and is defined by replacing the
arbitrary constant a0 by . So we have
.
)()2)(1(!2
)1(
0
2
2
0 ∑
∞
= +++
−=
n
n
n
np
npppn
x
xay

)!2/(1 pp
∑
∑
∞
=
+
∞
=
+
−=
+++
−=
0
2
0
2
2
.
)!(!
)2/(
)1(
)()2)(1(!2
)1(
!2
)(
n
pn
n
n
n
n
n
p
p
p
npn
x
npppn
x
p
x
xJ


The most useful Bessel functions are those of order 0
and 1 which are
And
+
⋅⋅
−
⋅
+−=
−= ∑
∞
=
222
6
22
4
2
2
0
2
0
642422
1
!!
)2/(
)1()(
xxx
nn
x
xJ
n
n
n
−





+





−=






+
−=
+∞
=
∑
53
12
0
1
2!3!2
1
2!2!1
1
2
2)!1(!
1
)1()(
xxx
x
nn
xJ
n
n
n

Graphs of J0(x) and J1(x), and J2(x).

These graphs displays several interesting properties of
the functions J0(x) and J1(x); each has a damped
oscillating behavior producing an infinite number of
positive zeros; and these zeros occur alternatively, in a
manner suggesting the functions cos x and sin x.
(positive zeros- the positive real numbers for which the
function Jp(x) vanishes).
We have seen that in the denominator of Jp(x) there is a
term (p+n)! , but it is meaning less if p is not a positive
integer.
Now our next turn is to overcome this difficulty.

The Gamma Function:
The gamma function is defined by
The factor so rapidly as that this
integral converges at the upper limit regardless of the
value of p. However at the lower limit we have
and the factor whenever p<1.
The restriction that p must be positive is necessary in
order to guarantee convergence at the lower limit.
∫
∞
−−
>=Γ
0
1
.0,)( pdtetp tp
0→−t
e ∞→t
1→−t
e
∞→−1p
t

We can see that
For integration by parts yields
Since as .
);()1( ppp Γ=+Γ
).(lim
lim
lim)1(
0
1
0
1
0
0
ppdtetp
dtetpet
dtetp
b
tp
b
b
tpbtp
b
b
tp
b
Γ=



=




 +−=
=+Γ
∫
∫
∫
−−
∞→
−−−
∞→
−
∞→
0/ →bp
eb ∞→b

If we use the fact that
Then the formula yields
And in general for any integer greater
than equal to zero.
Its extension to other values of p.
Now with the help of the formula
,1)1(
0
==Γ ∫
∞
−
dte t
);()1( ppp Γ=+Γ
,!3123)3(3)4(,!212)2(2)3(,1)1(1)2( =⋅⋅=Γ=Γ=⋅=Γ=Γ=Γ=Γ
!)1( nn =+Γ
.
)1(
)(
p
p
p
+Γ
=Γ

We extend it to other values of p (it is necessary for the
applications).
If –1<p<0, then 0<p+1<1, so the right side of the
equation G1 has a value and the left side of G1 is
defined to have the value given by right side.
The next step is that when –2<p<-1, then –1<p+1<0, so
we can use the formula G1 to define on the
interval –2<p<-1 in terms of the values of .
We can continue this process indefinitely.
.
)1(
)(
p
p
p
+Γ
=Γ G1
)( pΓ
)1( +Γ p

Also we can see that
Accordingly as from the right or left. This
function behaves in a similar way near all negative
integers. Its graph is shown in the next slide.
Gamma function gives us
Since never vanishes the function will
be defined and well behaved for all values of p, if we
agree that .
±∞=
+Γ
=Γ
→→ p
p
p
pp
)1(
lim)(lim
00
0→p
)( pΓ
.
2
1
π=





Γ
)( pΓ )(/1 pΓ
,2,1,0for0)(/1 −−==Γ pp

These ideas enable us to define p! by
for all values of p except negative integers.
Also
is defined for all values of p and has the value 0 when
ever p is negative integer.
Now the Bessel function Jp(x) defined by the formula
has a meaning for all values of p greater than equal to 0.
)1(! +Γ= pp
)1(
1
!
1
+Γ
=
pp
∑
∞
=
+
+
−=
0
2
.
)!(!
)2/(
)1()(
n
pn
n
p
npn
x
xJ

Graph of the gamma function.

The general Solution of Bessel’s Equation:
We have found one particular solution for the exponent
m1=p, namely Jp(x).
To get the general solution we must have to construct a
second linearly independent solution, that is one which
is not a constant multiple of Jp(x).
Any such solution is called a Bessel function of
second kind.
For the second LI solution the procedure is to try the
other exponent m2=-p, but in doing this we expect to
encounter difficulties whenever the difference m1-m2
=2p is zero or a positive integer.

The difficulties are serious if the difference is zero. So
we assume p is not an integer. We replace p by –p in our
earlier treatment.
We see the change in
and if it happens that p=1/2, then by letting n=1 we see
that there is no compulsion to choose a1=0.
But since we want a particular solution it is permissible
to put a1=0. The same problem arise when p=3/2 and
n=3, and so on; and we solve it by putting a1=a3=… =0.
;0)2( 2 =++− −nn aanpn

Other calculations goes as before. Hence we obtain the
second solution as
the first term of the series is
so J-p(x) is unbounded near x=0. Since Jp(x) is bounded
near x=0, these two solutions are independent and
is the general solution of Bessel’s equation.
,
)!(!
)2/(
)1()(
2
0 npn
x
xJ
pn
n
n
p
+−
−=
−∞
=
− ∑
,
2)!(
1
p
x
p
−






−
integer,annot),()( 21 pxJcxJcy pp −+=
B2
B3

The solution is entirely different when p is an integer
Formula B2 becomes
since the factors 1/(-m+n)! are zero when n=0,1,…, m-1.
On replacing the dummy variable n by n+m and
compensating by beginning the summation at n=0.
.0≠m
)!(!
)2/(
)1(
)!(!
)2/(
)1()(
2
2
0
nmn
x
nmn
x
xJ
mn
mn
n
mn
n
n
m
+−
−=
+−
−=
−∞
=
−∞
=
−
∑
∑

We obtain
This shows that J-m(x) is not independent of Jm(x) so in
this case
is not the general solution.
).()1(
)!(!
)2/(
)1()1(
!)!(
)2/(
)1()(
2
0
)(2
0
xJ
nmn
x
nmn
x
xJ
m
m
mn
n
nm
mmn
n
mn
m
−=
+
−−=
+
−=
+∞
=
−+∞
=
+
−
∑
∑
),()( 21 xJcxJcy mm −+=

One possible approach is by Section 16, which yield
As a second solution independent of Jm(x).
When p is not an integer any function of the form B3
with is a Bessel function of the second kind
including J-p(x) it self.
The standard Bessel function of second kind is defined
by
∫ 2
)(
)(
xxJ
dx
xJ
m
m
02 ≠c
.
sin
)(cos)(
)(
π
π
p
xJpxJ
xY
pp
p
−−
=

This choice of Yp is made for two reasons:
(We have a problem when p is an integer but we avoid it
by taking the limit which is explained in the first reason)
First reason: (for p is an integer we take the limit)
We can see that (detail analysis is omitted) the function
exists and is Bessel functions of the second kind and it
follows that
is the general solution of Bessel's equation in all cases,
whether p is an integer or not.
)(lim)( xYxY p
mp
m
→
=
)()( 21 xYcxJcy pp +=

Second reason:
By introducing a new dependent variable
we transform the Bessel equation into
When x is very large, the above equation closely
approximates the familiar differential equation
Which has independent solutions
We therefore expect that for large values of x, any
Bessel function y(x) will behave like some linear
combination of
),()( xyxxu =
.0
4
41
1 2
2
=




 −
++′′ u
x
p
u
.0=+′′ uu
.sin)(andcos)( 21 xxuxxu ==

And this expectation is supported by the fact
and
where r1(x) and r2(x) are bounded as
.sin
1
cos
1
x
x
andx
x
2/3
1 )(
24
cos
2
)(
x
xrp
x
x
xJ p +





−−=
ππ
π
,
)(
24
sin
2
)( 2/3
2
x
xrp
x
x
xYp +





−−=
ππ
π
.∞→x

Graph of Y (x), Y (x) and Y (x)

Properties of Bessel Functions:
The Bessel function Jp(x) is defined for any real number
p by
Identities and the function Jm+1/2(x).
We prove the following identity
∑
∞
=
+
+
−=
0
2
.
)!(!
)2/(
)1()(
n
pn
n
p
npn
x
xJ
[ ] )()( 1 xJxxJx
dx
d
p
p
p
p
−= I1

Proof:
[ ]
)(
)!1(!
)2/()1(
)!1(!2
)1(
)!(!2
)1(
)(
1
0
12
0
12
122
0
2
22
xJx
npn
x
x
npn
x
npn
x
dx
d
xJx
dx
d
p
p
n
pnn
p
n
pn
pnn
n
pn
pnn
p
p
−
∞
=
−+
∞
=
−+
−+
∞
=
+
+
=
−+
−
=
−+
−
=
+
−
=
∑
∑
∑

And the second identity is
Proof:
[ ] ).()( 1 xJxxJx
dx
d
p
p
p
p
+
−−
−= I2
[ ]
).(
)!1(!
)2/()1(
)!()!1(2
)1(
)!(!2
)1(
)(
1
0
12
1
12
12
0
2
2
xJx
npn
x
x
npn
x
npn
x
dx
d
xJx
dx
d
p
p
n
pnn
p
n
pn
nn
n
pn
nn
p
p
+
−
∞
=
++
−
∞
=
−+
−
∞
=
+
−
−=
++
−
−=
+−
−
=
+
−
=
∑
∑
∑

If the differentiation is carried out and the results are
divided by , then the formula becomes
If I3 and I4 are added we get
p
x±
)()()(
and
)()()(
1
1
xJxJ
x
p
xJ
xJxJ
x
p
xJ
ppp
ppp
+
−
−=−′
=+′ I3
I4
),()()(2 11 xJxJxJ ppp +− −=′ I5

and if subtracted, we get
or
These formulas enables us to write Bessel functions and
their derivatives in terms of other Bessel functions.
An interesting applications begins with identity 6 and
with the formulas (already established in the previous
exercise)
).()()(
2
11 xJxJxJ
x
p
ppp +− += I6
,cos
2
)(sin
2
)( 2/12/1 x
x
xJandx
x
xJ
ππ
== −
).(
2
)()( 11 xJ
x
p
xJxJ ppp +−= −+

From I6 it follows that
and
also
and






−=−= − x
x
x
x
xJxJ
x
xJ cos
sin2
)()(
1
)( 2/12/12/3
π
.sin
cos3sin32
)()(
3
)( 22/12/32/5 





−−=−= x
x
x
x
x
x
xJxJ
x
xJ
π
.sin
cos2
)()(
1
)( 2/12/12/3 





−−=−−= −− x
x
x
x
xJxJ
x
xJ
π
.cos
cos3cos32
)()(
3
)( 22/12/32/5 





−+=−−= −−− x
x
x
x
x
x
xJxJ
x
xJ
π

It can be continued indefinitely, and therefore every
Bessel function Jm+1/2(x) (where m is an integer) is
elementary.
It has been proved by Liouville that these functions are
the only cases in which Jp(x) is elementary.
The I6 yields Lamberts continued fraction
for tan x, and this continued fraction led to the
first proof of the fact that is not a rational
number.
π

When the differentiation formulas I1 and I2 are written
in the form
Then they serve for the integration of many simple
expressions containing Bessel functions. For example,
when p=1, we get
∫
∫
+−=
+=
−
+
−
−
,)()(
and
,)()(
1
1
cxJxdxxJx
cxJxdxxJx
p
p
p
p
p
p
p
p
∫ += .)()( 10 cxxJdxxxJ

Zeros and Bessel Series:
For every value of p, the function Jp(x) has an infinite
number of positive zeros.
For J0(x) the first five positive zeros are
2.4048, 5.5201, 8.6537, 11.7915, and 14.9309.
For J1(x) the first five positive zeros are
3.8317, 7.0156, 10.1735, 13.3237, and 16.4706.
The purpose of concern of these zeros of Jp(x) is:
It is often necessary in mathematical physics to expand
a given function in terms of Bessel functions.

The simplest and most useful expansion of this kind are
the series of the form
Where f(x) is defined on the interval and the
are the positive zeros of some fixed Bessel function
Jp(x) with p greater than equal to zero.
As in the previous experience in Legendre series here
also we need to determine the coefficients of the
expansion, which depend on certain integral properties
of the function .
+++== ∑
∞
=
)()()()()( 3322
1
11 xJaxJaxJaxJaxf pp
n
pnpn λλλλ
10 ≤≤ x nλ
)( xJ np λ

Here we need the fact
The function are said to be orthogonal with
respect to the weight function x on the interval
If the expansion is assumed to be possible then
multiplying through , and integrating term by
term from 0 to 1, and using the above fact we get
∫




=
≠
=
+
1
0
2
1 .)(
2
1
,0
)()(
nmifJ
nmif
dxxJxxJ
np
npmp
λ
λλ
)( xJ np λ
.10 ≤≤ x
)( xxJ mp λ
∫ +=
1
0
2
1 ;)(
2
)()( mp
m
mp J
a
dxxJxxf λλ

and on replacing m by n we obtain the formula for the
coefficients an as:
The series
With the coefficients given by the above formula is
know as Bessel series (or sometimes the Fourier-Bessel
series) of the function f(x).
∫+
=
1
02
1
.)()(
)(
2
dxxJxxf
J
a np
np
n λ
λ
+++== ∑
∞
=
)()()()()( 3322
1
11 xJaxJaxJaxJaxf pp
n
pnpn λλλλ

This theorem is given without proof which tells about
the conditions under which the series actually converges.
Theorem A. (Bessel expansion theorem).
Assume that f(x) and have at most finite
number of jump discontinuities on the interval
. If 0<x<1, then the Bessel series
converges to f(x) when x is a point of continuity
of this function, and converges to
when x is a point of discontinuity.
)(xf ′
10 ≤≤ x
[ ])()(
2
1
++− xfxf

At the end point x=1 the series converges to zero
regardless of the nature of the function because every
is zero. The series also converges at x=0, to zero if p>1
and to f(0+) if p=0.
Example 1.
Bessel series of the function in terms
of the functions .
Solution: We use this formula for the coefficients
)( npJ λ
,10,1)( ≤≤= xxf
)(0 xJ nλ
∫+
=
1
02
1
.)()(
)(
2
dxxJxxf
J
a np
np
n λ
λ

But
By using the formula
so we get
Now it follows that
is the desired Bessel series.
,
)(
)(
1
)()()(
1
1
0
1
1
0
1
0
00
n
n
n
n
nn
J
xxJ
dxxxJdxxJxxf
λ
λ
λ
λ
λλ
=





=
=∫ ∫
∑
∞
=
≤≤==
1
0
1
)10()(
)(
2
1)(
n
n
nn
xxJ
J
xf λ
λλ
∫ +=− ,)()(1 cxJxdxxJx p
p
p
p
)(
2)(
)(
2
)()(
)(
2
1
1
2
1
1
02
1
nnn
n
n
np
np
n
J
J
J
dxxJxxf
J
a
λλλ
λ
λ
λ
λ
==
= ∫+

Proofs of Orthogonality Properties
To establish
We begin with the fact that y=Jp(x) is a solution of
If a and b are two distinct positive constants, it follows
that the functions u(x)=Jp(ax) and v(x)=Jp(bx) satisfy the
equations
∫




=
≠
=
+
1
0
2
1 .)(
2
1
,0
)()(
nmifJ
nmif
dxxJxxJ
np
npmp
λ
λλ
.01
1
2
2
=





−+′+′′ y
x
p
y
x
y

and
We now multiply these equations by v and u and
subtract the results, to obtain
and after multiplying by x, this becomes
0
1
2
2
2
=





−+′+′′ u
x
p
au
x
u
.0
1
2
2
2
=





−+′+′′ v
x
p
bv
x
v
;)()(
1
)( 22
uvabuvvu
x
uvvu
dx
d
−=′−′+′−′
.)()]([ 22
xuvabuvvux
dx
d
−=′−′

When this expression is integrated from x=0 to x=1, we get
The expression in brackets vanishes at x=0 and at the other
end of the interval we have u(1)=Jp(a) and v(1) = Jp(b).
So when x=1 the right side becomes
It follows that the integral on the left is zero if a and b are
distinct positive zeros and of Jp(x); that is we have
which is the first part of the proof.
.)]([)( 1
0
1
0
22
uvvuxdxxuvab ′−′=− ∫
∫ =
1
0
,0)()( dxxJxxJ npmp λλ
mλ nλ
).()()()( aubvbvau ′−′

Now we have to evaluate the integral when m=n. If
is multiplied by , it becomes
or
so on integrating form x=0 to x=1,
0
1
2
2
2
=





−+′+′′ u
x
p
au
x
u
ux ′2
2
02222 22222
=′−′+′+′′′ uupuuxauxuux
,0)(2)()( 222222222
=−−+′ up
dx
d
xuauxa
dx
d
ux
dx
d

we obtain
when x=0. The expression in brackets vanishes; and
since we get
Now putting we get
since Jp(a)=0 and
and the proof of the second part is complete.
,])([2 1
0
222222
1
0
22
upxauxdxxua −+′=∫
),()1( aJau p
′=′
.)(1
2
1
)(
2
1
)( 2
2
2
2
1
0
2
aJ
a
p
aJdxaxxJ ppp 





−+′=∫
na λ=
,)(
2
1
)(
2
1
)( 2
1
2
1
0
2
npnpnp JJdxxxJ λλλ +=′=∫
)()()()()( 11 npnpppp JJxJxJ
x
p
xJ λλ ++ =
′
⇒=−
′

Exercise 1: Show that
And between any two zeros of J0(x) there is a zero of
J1(x), and also between any two zeros of J1(x) there is a
zero of J0(x).
Solution: Taking the derivative of J0(x) we get
).()]([and);()( 0110 xxJxxJ
dx
d
xJxJ
dx
d
=−=
[ ]
).(
)!1(!
)2/()1(
)!1(!
)2/()1(
!!2
2)1(
!!2
)1(
)(
1
0
121
1
12
1
2
12
0
2
2
0
xJ
nn
x
nn
x
nn
nx
nn
x
dx
d
xJ
dx
d
n
nn
n
nn
n
n
nn
n
n
nn
−=
+
−
=
−
−
=
−
=
−
=
∑∑
∑∑
∞
=
++∞
=
−
∞
=
−∞
=

Suppose x1 and x2 are any two positive zeros of J0(x), then
J0(x1)=0=J0(x2) and J0(x) is differentiable for all positive
values of x.
But by Rolle’s theorem if f(x) is continuous on [a,b] and
differentiable on (a,b) and f(a)=f(b)=0 then there exists at
least one number c in (a,b) such that
Hence applying this theorem there exists at least one
number x3 in (x1, x2) such that
Hence in between any two positive zeros of J0(x) there is
a zero for J1(x). Other part is left to you.
.0)( =′ cf
.0)(.0)()(or0)( 31313030 ==−=
′
=
′
xJeixJxJxJ

Exercise 2: Express J2(x), J3(x) in terms of J0(x) and
J1(x).
Solution:Using the formula
We get
Similarly others.
).(
2
)()( 11 xJ
x
p
xJxJ ppp +−= −+
).()(
2
)(
12
)()()(
01
10211
xJxJ
x
xJ
x
xJxJxJ
−=
⋅
+−==+

Exercise 3: If f(x)=xp
, for the interval show that
its Bessel series in the functions where the
are the positive zeros of Jp(x) is
Solution: The coefficients are given by the formula
Hence replacing f(x) by xp
and calculating the
coefficients we get
10 ≤≤ x
)( xJp nλ nλ
∑
∞
= +
=
1 1
).(
)(
2
n
np
npn
p
xJ
J
x λ
λλ
∫+
=
1
02
1
.)()(
)(
2
dxxJxxf
J
a np
np
n λ
λ

The coefficient an as
Putting the coefficient in the series we get
Which completes the proof.
.
)(
2)(
)(
2
)(
)(
2
)(
)(
2
1
1
2
1
1
01
1
2
1
1
0
1
2
1
npnn
np
np
np
n
p
np
np
p
np
n
J
J
J
xJ
x
J
dxxJx
J
a
λλλ
λ
λ
λ
λλ
λ
λ
+
+
+
+
+
+
+
+
==
=
= ∫
∑∑
∞
= +
∞
=
===
1 11
)(
)(
2
)()(
n
np
npnn
npn
p
xJ
J
xJaxxf λ
λλ
λ

Exercise 4: Prove that
Solution:
[ ] [ ].)()()()(
2
1
2
1 xJxJxxJxxJ
dx
d
pppp ++ −=
[ ] [ ]
[ ] [ ]
[ ]
[ ].)()(
)()()()(
)()()()(
)()()()(
2
1
2
11
11
1
1
1
1
1
1
1
xJxJx
xJxxJxxJxxJx
xJx
dx
d
xJxxJx
dx
d
xJx
xJxxJx
dx
d
xJxxJ
dx
d
pp
p
p
p
p
p
p
p
p
p
p
p
p
p
p
p
p
p
p
p
p
pp
+
+
−
+
++−
−
+
+
+
+−
+
+−
+
−=
−+=
+=
=

Special Functions
Chapter 8
THANK YOU
THE END

3 bessel's functions

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