Download Bessel`s Equation

Boyce/DiPrima 9th ed, Ch 5.7:
Bessel’s Equation
Elementary Differential Equations and Boundary Value Problems, 9 th edition, by William E. Boyce and Richard C. DiPrima, ©2009 by John Wiley & Sons, Inc.
Bessel Equation of order :


x 2 y  xy  x 2  2 y  0
Note that x = 0 is a regular singular point.
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.
In this section, we study the following cases:
Bessel Equations of order zero:  = 0
Bessel Equations of order one-half:  = ½
Bessel Equations of order one:  = 1
Bessel Equation of Order Zero
(1 of 12)
The Bessel Equation of order zero is
x 2 y  xy  x 2 y  0
We assume solutions have the form

y ( x)   r , x    an x r  n , for a0  0, x  0
n 0
Taking derivatives,

y ( x )   an x
n 0
r n

, y( x)   an r  n x r  n 1 ,
n 0

y( x)   an r  n r  n  1x r  n  2
n 0
Substituting these into the differential equation, we obtain

 a r  nr  n  1x
n 0
n
r n

  an r  n x
n 0
r n

  an x r  n  2  0
n 0
Indicial Equation
(2 of 12)
From the previous slide,

 an r  nr  n  1x
r n
n 0

  an r  n x
r n
n 0

  an x r  n  2  0
n 0
Rewriting,
a0 r (r  1)  r  x r  a1 (r  1)r  (r  1) x r 1

  an r  n r  n  1  r  n   an 2  x r  n  0
n2
or
a0 r x  a1 (r  1) x
2
r
2
r 1



  an r  n   an  2 x r  n  0
n2
2
The indicial equation is r2 = 0, and hence r1 = r2 = 0.
Recurrence Relation
(3 of 12)
From the previous slide,
a0 r x  a1 (r  1) x
2
r
2
r 1



  an r  n   an  2 x r  n  0
n2
2
Note that a1 = 0; the recurrence relation is
an  
an  2
, n  2, 3, 
2
r  n 
We conclude a1 = a3 = a5 = … = 0, and since r = 0,
a2 m
a2 m  2

, m  1, 2,
2
( 2 m)
Note: Recall dependence of an on r, which is indicated by
an(r). Thus we may write a2m(0) here instead of a2m.
First Solution
(4 of 12)
From the previous slide,
a2 m
a2 m  2

, m  1, 2,
2
( 2 m)
Thus
a2  
a0
a0
a0
a0
a2
,
a




,
a


,
4
6
2
2
2
2
2 2
4
6
2
4
42
2 2 1
2 3  2 1
and in general,
a2 m
Thus
(1) m a0
 2m
, m  1, 2,
2
2 m !


(1) m x 2 m 
y1 ( x)  a0 1   2 m
, x0
2 
 m1 2 m! 
Bessel Function of First Kind,
Order Zero (5 of 12)
Our first solution of Bessel’s Equation of order zero is


(1) m x 2 m 
y1 ( x)  a0 1   2 m
, x0
2 
 m1 2 m! 
The series converges for all x, and is called the Bessel
function of the first kind of order zero, denoted by
(1) m x 2 m
J 0 ( x)   2 m
, x0
2
m!
m 0 2

The graphs of J0 and several
partial sum approximations
are given here.
Second Solution: Odd Coefficients
(6 of 12)
Since indicial equation has repeated roots, recall from Section
5.7 that the coefficients in second solution can be found using
an (r ) r 0
Now
a0 (r )r x  a1 (r )( r  1) x
2
r
Thus
Also,
2
r 1


n2
a 1 (r )  0  a1 (0)  0
an  2 ( r )
an ( r )  
, n  2, 3, 
2
r  n 
and hence

  an (r )r  n   an  2 (r ) x r  n  0
a2 m1 (0)  0, m  1, 2,
2
Second Solution: Even Coefficients
(7 of 12)
Thus we need only compute derivatives of the even
coefficients, given by
a2 m  2 ( r )
(1) m a0
a2 m ( r )  
 a2 m ( r ) 
, m 1
2
2
2
r  2m
r  2 r  2m
It can be shown that
a2 m (r )
1
1 
 1
 2 


a2 m (r )
r  2m 
r  2 r  4
and hence
1 
1 1
a2 m (0)  2     
a2 m (0)

2m 
2 4
Second Solution: Series Representation
Thus
(1) m a0
a2 m (0)   H m 2 m
, m  1, 2,
2
2 m!
where
Hm 
1 1
1
 
2 4
2m
Taking a0 = 1 and using results of Section 5.7,
(1) m1 H m 2 m
y2 ( x)  J 0 ( x) ln x   2 m
x , x0
2
m!
m 1 2

(8 of 12)
Bessel Function of Second Kind,
Order Zero (9 of 12)
Instead of using y2, the second solution is often taken to be a
linear combination Y0 of J0 and y2, known as the Bessel
function of second kind of order zero. Here, we take
Y0 ( x) 
2

 y2 ( x)    ln 2J 0 ( x)
The constant  is the Euler-Mascheroni constant, defined by
  lim H n  ln n   0.5772
n 
Substituting the expression for y2 from previous slide into
equation for Y0 above, we obtain
2 
Y0 ( x)     ln
 

(1) m1 H m 2 m 
x
x , x  0
 J 0 ( x)   2 m
2
2
m!
m 1 2

General Solution of Bessel’s Equation,
Order Zero (10 of 12)
The general solution of Bessel’s equation of order zero, x > 0,
is given by
y( x)  c1 J 0 ( x)  c2Y0 ( x)
where
(1) m x 2 m
J 0 ( x)   2 m
,
2
m !
m 0 2


(1) m 1 H m 2 m 
2 
x
Y0 ( x)     ln  J 0 ( x)   2 m
x 
2
 
2
m !
m 1 2

Note that J0  0 as x  0 while Y0 has a logarithmic
singularity at x = 0. If a solution which is bounded at the
origin is desired, then Y0 must be discarded.
Graphs of Bessel Functions,
Order Zero (11 of 12)
The graphs of J0 and Y0 are given below.
Note that the behavior of J0 and Y0 appear to be similar to sin x
and cos x for large x, except that oscillations of J0 and Y0 decay
to zero.
Approximation of Bessel Functions,
Order Zero (12 of 12)
The fact that J0 and Y0 appear similar to sin x and cos x for
large x may not be surprising, since ODE can be rewritten as
 v2 
1
x y  xy  x  v y  0  y  y  1  2  y  0
x
 x 

2
2
2

Thus, for large x, our equation can be approximated by
y  y  0,
whose solns are sin x and cos x. Indeed, it can be shown that
1/ 2


cos x  , as x  
4

1/ 2


sin  x  , as x  
4

 2 

J 0 ( x)  
 x 
 2 

Y0 ( x)  
 x 
Bessel Equation of Order One-Half
(1 of 8)
The Bessel Equation of order one-half is
 2 1



x y  xy   x   y  0
4

We assume solutions have the form
2

y ( x)   r , x    an x r  n , for a0  0, x  0
n 0
Substituting these into the differential equation, we obtain

 an r  n r  n  1x
n 0

  an x
n 0
r n2
r n

  an r  n x r  n
n 0
1 
  an x r  n  0
4 n 0
Recurrence Relation
(2 of 8)
Using the results of the previous slide, we obtain


1

r n
r  n2





r

n
r

n

1

r

n

a
x

a
x
0


n
n


4
n 0
n 0
or
 r n
1  r 1  
1
 2 1

2
r
2
 r   a0 x  (r  1)   a1 x   r  n    an  an 2  x  0
4
4
4


n  2 

The roots of the indicial equation are r1 = ½, r2 = - ½ , and
note that they differ by a positive integer.
The recurrence relation is
an  2 ( r )
an ( r )  
, n  2, 3,
2
r  n   1 / 4
First Solution: Coefficients
(3 of 8)
Consider first the case r1 = ½. From the previous slide,
 r n
1  r 1   
1

2
2
r  1 / 4 a0 x  (r  1)   a1 x    r  n    an  an 2  x  0
4
4

n2  


2

r
Since r1 = ½, a1 = 0, and hence from the recurrence relation,
a1 = a3 = a5 = … = 0. For the even coefficients, we have
a2 m
a2 m  2
a2 m  2


, m  1, 2, 
2
1 / 2  2m  1 / 4 2m2m  1
It follows that
and
a2  
a2 m
a0
a
a
, a4   2  0 , 
3!
5  4 5!
(1) m a0

, m  1, 2,
(2m  1) !
Bessel Function of First Kind,
Order One-Half (4 of 8)
It follows that the first solution of our equation is, for a0 = 1,
y1 ( x)  x
1/ 2


(1) m 2 m 
x , x  0
1  
 m 1 (2m  1) !

  (1) m 2 m 1 
 x 
x
, x  0
 m 0 (2m  1) !

 x 1/ 2 sin x, x  0
1 / 2
The Bessel function of the first kind of order one-half, J½,
is defined as
1/ 2
2
J1/ 2 ( x)   
 
1/ 2
 2 

y1 ( x)  
 x 
sin x, x  0
Second Solution: Even Coefficients
(5 of 8)
Now consider the case r2 = - ½. We know that
 r n
1  r 1   
1

2
2
r  1 / 4 a0 x  (r  1)   a1 x    r  n    an  an 2  x  0
4
4

n2  



2
r
Since r2 = - ½ , a1 = arbitrary. For the even coefficients,
a2 m
a2 m  2
a2 m  2


, m  1, 2, 
2
 1 / 2  2m  1 / 4 2m2m  1
It follows that
a0
a0
a2
a2   , a4  
 ,
2!
4  3 4!
and
a2 m
(1) m a0

, m  1, 2,
(2m) !
Second Solution: Odd Coefficients (6 of 8)
For the odd coefficients,
a2 m 1  
a2 m 1
a2 m 1


, m  1, 2, 
2
 1 / 2  2m  1  1 / 4 2m2m  1
It follows that
a3  
a1
3!
, a5  
a3
5 4

a1
5!
,
and
(1) m a1
a2 m1 
, m  1, 2,
(2m  1) !
Second Solution
(7 of 8)
Therefore
m 2m
m 2 m 1




(

1
)
x
(

1
)
x
1 / 2
y 2 ( x )  x  a0 
 a1 
, x  0
m  0 ( 2m  1) ! 
 m  0 ( 2m) !
 x 1/ 2 a0 cos x  a1 sin x , x  0
The second solution is usually taken to be the function
1/ 2
 2 

J 1/ 2 ( x)  
 x 
cos x, x  0
where a0 = (2/)½ and a1 = 0.
The general solution of Bessel’s equation of order one-half is
y( x)  c1 J1/ 2 ( x)  c2 J 1/ 2 ( x)
Graphs of Bessel Functions,
Order One-Half (8 of 8)
Graphs of J½ , J-½ are given below. Note behavior of J½ , J-½
similar to J0 , Y0 for large x, with phase shift of /4.
1/ 2
 2 

J 1/ 2 ( x)  
 x 
1/ 2
 2 

J 0 ( x)  
 x 
1/ 2
 2 

J1/ 2 ( x)  
 x 
cos x,


cos x  ,
4

sin x
1/ 2
 2 

Y0 ( x)  
 x 


sin  x  , as x  
4

Bessel Equation of Order One
(1 of 6)
The Bessel Equation of order one is
x 2 y  xy  x 2  1y  0
We assume solutions have the form

y ( x)   r , x    an x r  n , for a0  0, x  0
n 0
Substituting these into the differential equation, we obtain

 a r  n r  n  1x
n 0
n

  an x
n 0
r n2

r n

  an r  n x r  n
n 0
  an x r  n  0
n 0
Recurrence Relation
(2 of 6)
Using the results of the previous slide, we obtain

 r  n r  n  1  r  n   1a x
n
n 0
or
r
2



 1 a0 x  (r  1)  1 a1 x
r
2
r 1


n2
r n

  an x r  n  2  0
n 0
r  n  1a
2

r n

a
x
0
n
n2
The roots of indicial equation are r1 = 1, r2 = - 1, and note
that they differ by a positive integer.
The recurrence relation is
an  2 ( r )
an ( r )  
, n  2, 3, 
2
r  n   1
First Solution: Coefficients
(3 of 6)
Consider first the case r1 = 1. From previous slide,
r



 1 a0 x  (r  1)  1 a1 x
2
r
2
r 1


n2
r  n  1 a
2
n

 an  2 x r  n  0
Since r1 = 1, a1 = 0, and hence from the recurrence relation,
a1 = a3 = a5 = … = 0. For the even coefficients, we have
a2 m
a2 m  2
a2 m  2

 2
, m  1, 2,
2
1  2m  1 2 m  1 m
It follows that
a2  
and
a2 m
a0
a0
a2
,
a



,
4
2
2
4
2  2 1
2  3  2 2 3!2!
(1) m a0
 2m
, m  1, 2,
2 (m  1) !m !
Bessel Function of First Kind,
Order One (4 of 6)
It follows that the first solution of our differential equation is


(1) m
2m 
y1 ( x)  a0 x 1   2 m
x , x  0
 m1 2 (m  1) !m!

Taking a0 = ½, the Bessel function of the first kind of order
one, J1, is defined as
x
(1) m
2m 
J1 ( x)    2 m
x , x  0
2 m0 2 (m  1) !m!

The series converges for all x and hence J1 is analytic
everywhere.
Second Solution
(5 of 6)
For the case r1 = -1, a solution of the form

1 
2n 
y2 ( x)  a J1 ( x) ln x  x 1   cn x , x  0
 n 1

is guaranteed by Theorem 5.7.1.
The coefficients cn are determined by substituting y2 into the
ODE and obtaining a recurrence relation, etc. The result is:
m



(

1
)
H m  H m1  2 n 
1
y2 ( x)   J1 ( x) ln x  x 1  
x , x  0
2m
 m1 2 m!(m  1) !

where Hk is as defined previously. See text for more details.
Note that J1  0 as x  0 and is analytic at x = 0, while y2 is
unbounded at x = 0 in the same manner as 1/x.
Bessel Function of Second Kind,
Order One (6 of 6)
The second solution, the Bessel function of the second kind
of order one, is usually taken to be the function
Y1 ( x) 
2
 y2 ( x)    ln 2 J1 ( x),
x0

where  is the Euler-Mascheroni constant.
The general solution of Bessel’s equation of order one is
y( x)  c1J1 ( x)  c2Y1 ( x), x  0
Note that J1 , Y1 have same
behavior at x = 0 as observed
on previous slide for J1 and y2.