Upload
others
View
3
Download
0
Embed Size (px)
Citation preview
MATHEMATICAL PHYSICS
UNIT โ 2
BESSELโS EQUATION
DR. RAJESH MATHPAL
ACADEMIC CONSULTANT
SCHOOL OF SCIENCES
UTTARAKHAND OPEN UNIVERSITY
TEENPANI, HALDWANI
UTTRAKHAND
MOB:9758417736,7983713112
Email: [email protected]
STRUCTURE OF UNIT
โข 2.1 INTRODUCTION
โข 2.2 BESSELโS EQUATION
โข 2.4 BESSELโS FUNCTIONS, Jn (x)
โข 2.5 Besselโs function of the second kind of order n
โข 2.6 RECURRENCE FORMULAE
โข 2.7 ORTHOGONALITY OF BESSEL FUNCTION
โข 2.8 A GENERATING FUNCTION FOR Jn (x)
โข 2.9 SOME EXAMPLES
2.1 INTRODUCTION
We find the Besselโs equation while solving Laplace equation in polar coordinates by the needed of separation of variables. This equation has a number of applications in engineering.
Besselโs function are involved in
โข The Oscillatory motion of a hanging chain
โข Eulerโs theory of a circular membrane
โข The studies of planetary motion
โข The propagation of waves
โข The Elasticity
โข The fluid motion
โข The potential theory
โข Cylindrical and spherical waves
โข Theory of plane waves
โข Besselโs function are also known as cylindrical and spherical function.
2.2 BESSELโS EQUATION
The differential equation
๐ฅ2 ๐2๐ฆ
๐๐ฅ2 + ๐ฅ๐๐ฆ
๐๐ฅ+ ๐ฅ2 โ ๐ฅ๐ ๐ฆ = 0
is called the Besselโs differential equation, and particular solutions of this equation are
called Besselโs fraction of order n.
2.3 SOLUTION OF BESSELโS EQUATION
๐ฅ2 ๐2๐ฆโฒ
๐๐ฅ2 + ๐ฅ๐๐ฆ
๐๐ฅ+ ๐ฅ2 โ ๐ฅ๐ ๐ฆ = 0. โฆ(1)
Let ฯ๐=0โ ๐๐๐ฅ
๐+๐ ๐๐ ๐ฆ = ๐0๐ฅ๐ + ๐1๐ฅ
๐+1 + ๐2๐ฅ๐+2 + โฏ โฆ(2)
So that๐๐ฆ
๐๐ฅ= ฯ๐=0
โ ๐๐ ๐ + ๐ ๐ฅ๐+๐โ1
and ๐2๐ฆ
๐๐ฅ2 = ฯ๐=0โ ๐๐ ๐ + ๐ (๐ + ๐ โ1)๐ฅ๐+๐โ2
Substituting these values in (1), we get
๐ฅ2
๐=0
โ
๐๐ ๐+๐ (๐+๐ โ1)๐ฅ๐+๐โ2 +๐ฅ
๐=0
โ
๐๐ ๐ + ๐ ๐ฅ๐+๐โ1 + (
)
๐ฅ2
โ ๐2
๐=0
โ
๐๐๐ฅ๐+๐ = 0
2 2
0 0 0 0
2 2
0 0
2 2 2
0 0
( )( 1) ( ) 0
[( )( 1) ( ) ] 0
[( ) ] 0.
m r m r m r m r
r r r r
r r r r
m r m r
r r
r r
m r m r
r r
r r
a m r m r x a m r x a x n a x
a m r m r m r n x a x
a m r n x a x
+ + + + +
= = = =
+ + +
= =
+ + +
= =
+ + โ + + + โ =
+ + โ + + โ + =
+ โ + =
Equating the coefficient of lowest degree term of xm in the identity (3) to zero,
by putting r = 0 in the first summation we get the indicial equation.
a0[m+0)2 โ n2] = 0. (r = 0)
โ m2 = n2 i.e. m = n, m = - n a0 โ 0
Equating the coefficient of the next lowest degree term xm+1 in the identity (3), we put r = 1 in the first summation
a1 [m + 1)2 โ n2] = 0 i.e. a1 = 0, since m + 1)2 โ n2 โ 0
Equating the coefficient of xm + r + 2 in (3) to zero, to find relation in successive coefficients, we get
ar +2[(m+r+2)2 โ n2] +ar = 0
โ ๐๐+2 = โ1
๐+๐+2 2โ๐2 . ar
Therefore, a3 = a5 = a1 = โฆ. = 0, since a1 = 0
If r = 0, ๐2 = โ1
๐+2 2โ๐2 . a0
If r = 2, ๐4 = โ1
๐+4 2โ๐2 a2 =1
๐+2 2โ๐2 [(๐+4)2โ๐2 a0 and so on.
On substituting the values of the coefficients a1, a2, a3, a4 โฆโฆ.. in (2), we have
y = a0xm = โ
๐0
๐+2 2โ๐2 ๐ฅ๐+2 +๐0
๐+2 2โ๐2 [ ๐+4)2โ๐2 ๐ฅ๐+4 + โฏ
y = a0xm = 1 โ
1
๐+2 2โ๐2 ๐ฅ2 +1
๐+2 2โ๐2 [ ๐+4)2โ๐2 ๐ฅ4 โ โฏ
For m = n
y = a0xn 1 โ
1
4 ๐+1๐ฅ2 +
1
42.2! ๐+1 ๐+2๐ฅ4 โ โฏ
where a0 is an arbitrary constant.
For m = โ n
y = a0x-n 1 โ
1
4 โ๐+1๐ฅ2 +
1
42.2! โ๐+1 โ๐+2๐ฅ4 โ โฏ
2.4 BESSELโS FUNCTIONS, Jn (x)
The Besselโs equation is ๐ฅ2 ๐2๐ฆ
๐๐ฅ 2+ ๐ฅ
๐๐ฆ
๐๐ฅ+ (๐ฅ2 โ ๐ฅ๐)๐ฆ = 0. โฆ(1)
Solution of (1) is
y = a0x-n 1 โ๐ฅ2
2.2(๐+1)+
๐ฅ4
2.4.22(๐+1)(๐+2)โ โฏ + (โ1)๐ ๐ฅ2๐
(2๐๐!).2๐(๐+1)(๐+2)โฆ(๐+๐)+ โฏ
2
0 20
( 1)2 . !( 1)( 2)...( )
rn r
rr
xa x
r n n n r
=
= โ+ + +
where a0 is an arbitrary constant.
If a0 = 1
2๐ (๐+1)
The above solution is called Besselโs function denoted by Jn (x).
Thus ๐ฝ๐(๐ฅ) = 1
2๐ (๐+1)ฯ(โ1)๐ ๐ฅ๐+2๐
22๐ .๐!(๐+1)(๐+2)โฆ(๐+๐) ๐ + 1 = ๐!
โ ๐ฝ๐(๐ฅ) = x
2
n
1
(๐+1)โ
1
1! (๐+2)
x
2
2
+1
2! (๐+3)
x
2
4
โ1
3! (๐+4)
x
2
6
+ โฏ
โ ๐ฝ๐(๐ฅ) =๐ฅ๐
2๐ ๐+1 1 โ
๐ฅ2
2.(2๐+2)+
๐ฅ4
2.4.(2๐+2)(2๐+4)+ โฏ โฆ(2)
2 2
0 0
( 1) ( 1) ( ) ( )
2 2! ( 1) ! ( )!
n r n rr r
n n
r r
x xJ x J x
r n r r n r
+ +
= =
โ โ = =
+ + +
If n = 0, J0 (x) = ฯ(โ1)๐
(๐!)2 ๐ฅ
2
2๐
โ J0 (x) = 1 โ๐ฅ2
22+
๐ฅ4
22 .42โ
๐ฅ6
22 .42 .62+ โฏ
If n = 1, J1 (x) = ๐ฅ
2โ
๐ฅ3
22.4+
๐ฅ5
22.42.6โ โฏ
We draw the length of these two functions. Both the functions are oscillatory with a varying period and a decreasing amplitude.
Replacing n by โ n in (2), we get J-n (x) = ฯ๐=0โ โ1 ๐
๐! โ๐+๐+1
๐ฅ
2
โ๐+2๐
Case I. If n is not integer or zero, then complete solution of (1) is
Case II. If n = 0, then y1 = y2 and complete solution of (1) is the Besselโs function of order zero.
Case III. If n is positive integer, then y2 is not solution of (1). And y1 fails to give a solution for negative values of n. Let us find out the general solution when n is an integer.
2.5 Besselโs function of the second kind of
order n
๐ฅ2 ๐2๐ฆ
๐๐ฅ2 + ๐ฅ๐๐ฆ
๐๐ฅ+ ๐ฅ2 โ ๐2 ๐ฆ = 0 โฆ(1)
Let y = u(x) Jn (x) be the second of the Besselโs equation when n integer.
๐๐ฆ
๐๐ฅ= uโ Jn + u Jโn
๐2๐ฆ
๐๐ฅ2 = uโโ Jn + 2uโ Jโn + u Jโโn
Substituting these values of y, yโ, yn in (1), we get
x2 (uโโ Jn + 2uโ Jโn + u Jโโn) + x(uโ Jn + u.Jโn) + (x2 โ n2) u Jn = 0
โ u [x2 Jnn + x Jโn + (x2 โ n2) Jn] + x2 uโโ Jn + 2x2 uโ Jn + x uโ Jn = 0 โฆ(2)
โ x2 Jโโn + x Jโn + (x2 โ n2) Jn = 0 [Since Jn is a solution of (1)]
(2) becomes x2 uโโ Jn + 2x2 uโ Jโn + xuโ Jn = 0 โฆ(3)
Dividing (3) by x2 uโ Jn, we have
๐ข๐
๐ขโฒ + 2๐ฝ๐โฒ
๐ฝ๐+
1
๐ฅ= 0
(4) Can also be written as โฆ(4)
๐
๐๐ฅ[log๐ขโฒ + 2
๐
๐๐ฅ[log ๐ฝ๐] +
๐
๐๐ฅ(log ๐ฅ) = 0
โ๐
๐๐ฅ[log๐ขโฒ + 2 log ๐ฝ๐ + log ๐ฅ] = 0
โ๐
๐๐ฅlog(๐ขโฒ. ๐ฝ๐
2 ๐ฅ = 0 โฆ(5)
Integrating (5), we get
log ๐ขโฒ. ๐ฝ๐2 . ๐ฅ = log ๐ถ1
โ ๐ขโฒ. ๐ฝ๐2. ๐ฅ =๐ถ1 โ ๐ขโฒ =
๐ถ1
๐ฝ๐2.๐ฅ
โฆ(6)
On integrating (6), we obtain
๐ข = ๐ถ1
๐ฝ๐2 .๐ฅ
๐๐ฅ +๐ถ2
Putting the value of ๐ข in the assumed solution y = u (x). ๐ฝ๐2(๐ฅ), we get
2.6 RECURRENCE FORMULAE
These formulae are very useful in solving the questions. So, they are to be
committed to memory.
1. x ๐ฝ๐โฒ = ๐๐ฝ๐ โ ๐ฅ ๐ฝ๐+1
2. x ๐ฝ๐โฒ = โ๐๐ฝ๐ + ๐ฅ ๐ฝ๐โ1
3. 2 ๐ฝ๐โฒ = ๐ฝ๐โ1 โ ๐ฝ๐+1
4. 2๐ ๐ฝ๐ = ๐ฅ ๐ฝ๐โ1 + ๐ฝ๐+1
5. ๐
๐๐ฅ๐ฅโ๐๐ฝ๐ = โ๐ฅโ๐ ๐ฝ๐+1
6. ๐
๐๐ฅ๐ฅโ๐๐ฝ๐ = ๐ฅ๐ ๐ฝ๐โ1
Formula I. x ๐ฝ๐โฒ = ๐๐ฝ๐ โ ๐ฅ๐ฝ๐+1
Proof. We know that
๐ฝ๐ = ฯ๐=0โ โ1 ๐
๐! ๐+๐+1
๐ฅ
2
๐+2๐
Differentiating with respect to x, we get
๐ฝ๐โฒ = ฯ
โ1 ๐ ๐+2๐
๐! ๐+๐+1
๐ฅ
2
๐+2๐โ1 1
2
โ ๐ฅ๐ฝ๐โฒ = ๐ฯ
โ1 ๐
๐! ๐+๐+1
๐ฅ
2
๐+2๐+ ๐ฅ ฯ
โ1 ๐.2๐
2.๐! ๐+๐+1
๐ฅ
2
๐+2๐โ1
= ๐ฅ๐ฝ๐ + ๐ฅ ฯ๐=0โ โ1 ๐
๐โ1 ! ๐+๐+1
๐ฅ
2
๐+2๐โ1
= ๐๐ฝ๐ + ๐ฅ ฯ๐ =0โ โ1 ๐ +1
๐ ! ๐+๐ +2
๐ฅ
2
๐+2๐ โ1[Putting r โ 1 = s]
= ๐๐ฝ๐ โ ๐ฅ ฯ๐ =0โ โ1 ๐
๐ ! ๐+1 +๐ +1
๐ฅ
2
๐+1 +2๐
๐ฅ๐ฝ๐โฒ = ๐๐ฝ๐ โ ๐ฅ๐ฝ๐+1 Proved.
Formula II. ๐ฅ๐ฝ๐โฒ = โ๐๐ฝ๐ + ๐ฅ๐ฝ๐โ1
Proof. We know that ๐ฝ๐ = ฯ๐=0โ โ1 ๐
๐! ๐+๐+2
๐ฅ
2
๐+2๐
Differentiating w.r.t. โxโ, we get ๐ฝ๐โฒ = ฯ๐=0
โ โ1 ๐ ๐+2๐
๐! ๐+๐+1
๐ฅ
2
๐+2๐โ1 1
2
๐ฝ๐โฒ = ฯ๐=0
โ โ1 ๐ ๐+2๐
๐! ๐+๐+1
๐ฅ
2
๐+2๐= ฯ๐=0
โ โ1 ๐ 2๐+2๐ โ๐
๐! ๐+๐+1
๐ฅ
2
๐+2๐
= ฯ๐=0โ โ1 ๐ 2๐+2๐
๐! ๐+๐+1
๐ฅ
2
๐+2๐โ ๐ ฯ๐=0
โ โ1 ๐
๐! ๐+๐+1
๐ฅ
2
๐+2๐
= ฯ๐=0โ โ1 ๐2
๐! ๐+๐
๐ฅ
2
๐+2๐โ ๐๐ฝ๐
= ๐ฅ
๐=0
โโ1 ๐
๐! ๐ โ 1 + ๐ + 1
๐ฅ
2
๐โ1+2๐
โ ๐๐ฝ๐
โ ๐๐ฑ๐โฒ = ๐๐ฑ๐โ๐ โ ๐๐ฑ๐
Formula III. ๐๐ฑ๐โฒ = ๐ฑ๐โ๐ โ ๐ฑ๐+๐
Proof.
We know that
๐ฅ๐ฝ๐โฒ = ๐ฝ๐ โ ๐ฅ๐ฝ๐+1 โฆ(1) (Recurrence formula I)
๐ฅ๐ฝ๐โฒ = โ๐๐ฝ๐ + ๐ฅ๐ฝ๐โ1 โฆ(2) (Recurrence formula II)
Adding (1) and (2), we get
2๐ฅ๐ฝ๐โฒ = โ๐ฅ๐ฝ๐+1 + ๐ฅ๐ฝ๐โ1 โ 2๐ฝ๐
โฒ = ๐ฝ๐โ1 โ ๐ฝ๐+1
Formula IV. 2๐๐ฝ๐ = ๐ฅ (๐ฝ๐โ1 + ๐ฝ๐+1)
Proof.
We know that
๐ฅ๐ฝ๐โฒ = ๐๐ฝ๐ โ ๐ฅ๐ฝ๐+1 โฆ(1) (Recurrence formula I)
๐ฅ๐ฝ๐โฒ = โ๐๐ฝ๐ + ๐ฅ๐ฝ๐โ1 โฆ(2) (Recurrence formula II)
subtracting (2) from (1), we get
0 = 2 ๐ ๐ฝ๐ โ๐ฅ๐ฝ๐+1 โ๐ฅ๐ฝ๐โ1
โ 2 ๐ ๐ฝ๐ = ๐ฅ (๐ฝ๐โ1 +๐ฝ๐+1) โฆ(3)
Formula V.๐
๐ ๐(x-n. Jn) = -x-n Jn+1
Proof. We know that ๐ฅ๐ฝ๐โฒ = ๐๐ฝ๐ โ ๐ฅ๐ฝ๐+1
(Recurrence formula I)
Multiplying by x-n-1, we obtain x-n ๐ฝ๐โฒ = nx-n-1 Jn โ x-n Jn+1
i.e., x-n ๐ฝ๐โฒ = nx-n-1 Jn =โ x-n Jn+1
โ๐
๐๐ฅ(x-n Jn) = - x-n Jn + 1
Formula VI.๐
๐ ๐(xn Jn) = xn Jn-1
Proof.
We know that x-n ๐ฝ๐โฒ = -nJn + x Jn-1 (Recurrence formula II)
Multiplying by xn+1, we have
xn ๐ฝ๐โฒ = -nxn-1 Jn + xn Jn-1 i.e., xn ๐ฝ๐
โฒ +nxn-1 Jn = xn Jn-1
โ๐
๐ ๐(xn Jn) = xn Jn-1
2.7 ORTHOGONALITY OF BESSEL
FUNCTION
Proof. We know that
๐ฅ2 ๐2๐ฆ
๐๐ฅ2 + ๐ฅ๐๐ฆ
๐๐ฅ+ ๐ผ2๐ฅ2 โ ๐2 ๐ฆ = 0 โฆ(1)
โ ๐ฅ2 ๐2๐ง
๐๐ฅ2 + ๐ฅ๐๐ง
๐๐๐ฅ+ ๐ฝ2๐ฅ2 โ ๐2 ๐ง = 0 โฆ(2)
Solution of (1) and (2) are y = Jn (๐ผ ๐ฅ), z = Jn (๐ฝ ๐ฅ) respectively.
Multiplying (1) by ๐ง
๐ฅand (2) by โ
๐ฆ
๐ฅand adding, we get
๐ฅ ๐ง๐2๐ฆ
๐๐ฅ2 โ ๐ฆ๐2๐ง
๐๐ฅ2 + ๐ง๐๐ฆ
๐๐ฅโ ๐ฆ
๐๐ง
๐๐ฅ+ ๐ผ2 โ๐ฝ2 ๐ฅ๐ฆ๐ง = 0.
โ๐
๐๐ฅ๐ฅ ๐ง
๐๐ฆ
๐๐ฅโ ๐ฆ
๐๐ง
๐๐ฅ+ ๐ผ2 โ๐ฝ2 ๐ฅ๐ฆ๐ง = 0 โฆ(3)
Integrating (3) w.r.t. โxโ between the limits 0 and 1, we get
๐
๐๐ฅ๐ฅ ๐ง
๐๐ฆ
๐๐ฅโ ๐ฆ
๐๐ง
๐๐ฅ 0
1+ ๐ผ2 โ๐ฝ2 0
1๐ฅ ๐ฆ ๐ง ๐๐ฅ = 0
โ ๐ฝ2 โ๐ผ2 01๐ฅ ๐ฆ ๐ง ๐๐ฅ = ๐ฅ ๐ง
๐๐ฆ
๐๐ฅโ ๐ฆ
๐๐ง
๐๐ฅ 0
1= ๐ง
๐๐ฆ
๐๐ฅโ ๐ฆ
๐๐ง
๐๐ฅ ๐ฅ=1โฆ(4)
Putting the values of y = Jn (๐ผ ๐ฅ), ๐๐ฆ
๐๐ฅ= ๐ผ ๐ฝ๐
โฒ ๐ผ๐ฅ , ๐ง = ๐ฝ๐ ๐ฝ๐ฅ ,๐๐ง
๐๐ฅ= ๐ฝ, ๐ฝ๐
โฒ ๐ฝ๐ฅ in (4), we get
๐ฝ2 โ ๐ผ2 01๐ฅ๐ฝ๐ ๐ผ๐ฅ . ๐ฝ๐ ๐ฝ๐ฅ ๐๐ฅ = ๐ผ๐ฝ๐
โฒ ๐ผ๐ฅ ๐ฝ๐ ๐ฝ๐ฅ = ๐ฝ๐ฝ๐โฒ ๐ฝ๐ฅ ๐ฝ๐ ๐ผ๐ฅ ๐ฅ=1
= ๐ผ๐ฝ๐โฒ ๐ผ ๐ฝ๐ ๐ฝ โ ๐ฝ๐ฝ๐
โฒ ๐ฝ ๐ฝ๐ ๐ผ โฆ(5)
Since ๐ผ, ๐ฝ are the roots of Jn (x) = 0, so Jn ๐ผ = Jn ๐ฝ = 0
Putting the values of Jn (๐ผ) = Jn (๐ฝ) = 0 in (5), we get
(๐ผ2โ๐ฝ2) ๐ฅ1
0 Jn (๐ผ๐ฅ). Jn (๐ฝ๐ฅ) dx = 0
โ ๐ฅ1
0 Jn (๐ผ๐ฅ). Jn (๐ฝ๐ฅ) dx = 0 Proved.
We also know that Jn (๐ผ) = 0. Let ๐ฝ be a neighboring value of ๐ผ, which tends to ๐ผ.
Then
1 '
2 2
0
0 ( ). ( )lim ( ). ( ) lim n n
n n
J JxJ x J x dx
โ โ
+=
โ
As the limit is of the form 0
0, we apply Lโ Hopitalโs rule
1 ' '2
2 '
0
0 ( ). ( ) 1( ) lim ( )
2 2
n nn n
J JxJ x dx J
โ
+ = = =
Proved.
2.8 A GENERATING FUNCTION FOR
Jn (x)
Prove that Jn (x) is the coefficient of zn in the expansion of ๐๐ฅ
2๐งโ
1
๐ง
Proof. We know that et = 1 + t + ๐ก2
2!+
๐ก3
3!+ โฏ
๐๐ฅ๐ง
2 = 1 +๐ฅ๐ง
2+
1
2!
๐ฅ
2๐ง
2โ
1
3!
๐ฅ
2๐ง
3+ โฏ โฆ(1)
๐๐ฅ
2๐ง = 1 โ๐ฅ
2๐ง+
1
2!
๐ฅ
2๐ง
2โ
1
3!
๐ฅ
2๐ง
3+ โฏ โฆ(2)
On multiplying (1) and (2), we get
๐๐ฅ
2๐ง1
๐ง = 1 +๐ฅ๐ง
2+
1
2!
๐ฅ๐ง
2
2+
1
3!
๐ฅ๐ง
2
3+ โฏ ร 1 โ
๐ฅ
2๐ง+
1
2!
๐ฅ
2๐ง
2โ
1
3!
๐ฅ
2๐ง
3+ โฏ โฆ(3)
The coefficient of zn in the product of (3), we get
=1
๐ ! ๐ฅ
2 ๐
โ1
(๐+1)! ๐ฅ
2 ๐+2
+1
2!(๐+2)! ๐ฅ
2 ๐+4
โ โฏ = Jn (x)
Similarly, coefficient of z-n in the product of (3) = J-n(x)
โด ๐๐ฅ
2 ๐ง
1
๐ง = J0 + z J1 + z2 J2 + z3 J3 + โฆ + z-1 J-1 + z-2 J-2 + z-3 J-3 + โฆ
1
2 ( )
xz
nz
n
n
e z J x
=โ
=
For this reason ๐๐ฅ
2 ๐ง
1
๐ง is known as the generating function of Besselโs functions.
Proved.
2.9 SOME EXAMPLES
Example 1. Show that Besselโs Function Jn(x) is an even function when n is even and is
odd function when n is odd.
Solution. We know that
2
0
( 1)( ) ...(1)
2! 1
n rr
n
r
xJ x
r n r
+
=
โ =
+ +
Replacing x by โ x in (1), we get
2
0
( 1)( ) ...(2)
2! 1
n rr
n
r
xJ x
r n r
+
=
โ โ โ =
+ +
Case I. If n is even, then n + 2r is even โ โ๐ฅ
2 ๐+2๐
= ๐ฅ
2 ๐+2๐
Thus (2), becomes
2
0
( 1)( )
2! 1
n rr
n
r
xJ x
r n r
+
=
โ โ =
+ +
= Jn (x) ๐น๐๐ ๐๐ฃ๐๐ ๐๐ข๐๐๐ก๐๐๐
๐(โ๐ฅ) = ๐(๐ฅ)
Hence, Jn (x) is even function
Case II. If n is odd, then n + 2r is odd โ โ๐ฅ
2 ๐+2๐
= โ ๐ฅ
2 ๐+2๐
Thus (2). Becomes
2
0
( 1)( )
2! 1
n rr
n
r
xJ x
r n r
+
=
โ โ = โ
+ +
= โ Jn (x) ๐น๐๐ ๐๐๐ ๐๐ข๐๐๐ก๐๐๐
๐(โ๐ฅ) = โ๐(๐ฅ)
Proved.
Hence, Jn (x) os odd function.
Example 2. Prove that:
x 0
( ) 1lim ;( 1).
2 1
n
n n
J xn
x nโ= โ
+
Solution. From the equation (2) of Article 29.3 on page 798, we know that
Jn (x) = ๐ฅ๐
2๐ ๐+1 1 โ
๐ฅ2
2.(2๐+2)+
๐ฅ4
2.4(2๐+2)(2๐+4)โ โฏ
On taking limit on both sides when x โ 0, we get
2 4
x 0 x 0
( ) 1lim lim 1 ...
2.(2 2) 2.4.(2 2)(2 4)2 1
n
n n
J x x x
x n n nnโ โ
= โ + โ
+ + ++
= 1
2๐ ๐+1
Example 3. Find the value of J-1 (x) + J1 (x).
Solution. By using Recurrence relation IV for Jn (x) is
2n Jn = x (Jn โ 1 + Jn + 1)
Jn โ 1 (x) + Jn + 1 (x) = 2๐
๐ฅJn (x)
Put n = 0
J-1(X) + J1(x) = 0
Example 4.Prove that
Formula V.๐
๐ ๐(x-n. Jn) = -x-n Jn+1
Proof. We know that ๐ฅ๐ฝ๐โฒ = ๐๐ฝ๐ โ ๐ฅ๐ฝ๐+1 (Recurrence formula I)
Multiplying by x-n-1, we obtain x-n ๐ฝ๐โฒ = nx-n-1 Jn โ x-n Jn+1
i.e., x-n ๐ฝ๐โฒ = nx-n-1 Jn =โ x-n Jn+1
โ๐
๐๐ฅ(x-n Jn) = - x-n Jn + 1
THANKS