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The solutions and properties of Bessel's Equation, specifically focusing on the Bessel functions of the first kind of orders zero and one-half, and the Bessel function of the second kind of order one. the indicial equation, recurrence relations, and the computation of coefficients. It also discusses the behavior of the functions for large x and provides asymptotic approximations.
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y ′′^ +
α xs^ y ′^ +
β xt^ y = 0 , (i)
where α = 0 and β = 0 are real numbers, and s and t are positive integers that for the moment are arbitrary. (a) Show that if s > 1 or t > 2, then the point x = 0 is an irregular singular point. (b) Try to find a solution of Eq. (i) of the form
y =
∑^ ∞
n = 0
an xr^ + n^ , x > 0. (ii)
Show that if s = 2 and t = 2, then there is only one possible value of r for which there is a formal solution of Eq. (i) of the form (ii). (c) Show that if s = 1 and t = 3, then there are no solutions of Eq. (i) of the form (ii). (d) Show that the maximum values of s and t for which the indicial equation is quadratic in r [and hence we can hope to find two solutions of the form (ii)] are s = 1 and t = 2. These are precisely the conditions that distinguish a “weak singularity,” or a regular singular point, from an irregular singular point, as we defined them in Section 5.4.
As a note of caution we should point out that while it is sometimes possible to obtain a formal series solution of the form (ii) at an irregular singular point, the series may not have a positive radius of convergence. See Problem 20 for an example.
In this section we consider three special cases of Bessel’s^12 equation,
x^2 y ′′^ + x y ′^ + ( x^2 − ν^2 ) y = 0 , (1)
where ν is a constant, which illustrate the theory discussed in Section 5.7. It is easy to show that x = 0 is a regular singular point. For simplicity we consider only the case x > 0.
Bessel Equation of Order Zero. This example illustrates the situation in which the roots of the indicial equation are equal. Setting ν = 0 in Eq. (1) gives
L [ y ] = x^2 y ′′^ + x y ′^ + x^2 y = 0. (2)
Substituting
y = φ( r , x ) = a 0 xr^ +
n = 1
an xr^ + n^ , (3)
(^12) Friedrich Wilhelm Bessel (1784 –1846) embarked on a career in business as a youth, but soon became interested in astronomy and mathematics. He was appointed director of the observatory at K¨onigsberg in 1810 and held this position until his death. His study of planetary perturbations led him in 1824 to make the first systematic analysis of the solutions, known as Bessel functions, of Eq. (1). He is also famous for making the first accurate determination (1838) of the distance from the earth to a star.
we obtain
L [φ]( r , x ) =
n = 0
an [( r + n )( r + n − 1 ) + ( r + n )] xr^ + n^ +
n = 0
an xr^ + n +^2
= a 0 [ r ( r − 1 ) + r ] xr^ + a 1 [( r + 1 ) r + ( r + 1 )] xr^ +^1
n = 2
{ an [( r + n )( r + n − 1 ) + ( r + n )] + an − 2 } xr^ + n^ = 0. (4)
The roots of the indicial equation F ( r ) = r ( r − 1 ) + r = 0 are r 1 = 0 and r 2 = 0; hence we have the case of equal roots. The recurrence relation is
an ( r ) = −
an − 2 ( r ) ( r + n )( r + n − 1 ) + ( r + n )
an − 2 ( r ) ( r + n )^2
, n ≥ 2. (5)
To determine y 1 ( x ) we set r equal to 0. Then from Eq. (4) it follows that for the coefficient of xr^ +^1 to be zero we must choose a 1 = 0. Hence from Eq. (5), a 3 = a 5 = a 7 = · · · = 0. Further,
an ( 0 ) = − an − 2 ( 0 )/ n^2 , n = 2 , 4 , 6 , 8 ,... ,
or letting n = 2 m , we obtain
a 2 m ( 0 ) = − a 2 m − 2 ( 0 )/( 2 m )^2 , m = 1 , 2 , 3 ,....
Thus
a 2 ( 0 ) = −
a 0 22
, a 4 ( 0 ) =
a 0 2422
, a 6 ( 0 ) = −
a 0 26 ( 3 · 2 )^2
and, in general,
a 2 m ( 0 ) =
(− 1 ) m^ a 0 22 m^ ( m !)^2
, m = 1 , 2 , 3 ,.... (6)
Hence
y 1 ( x ) = a 0
m = 1
(− 1 ) m^ x^2 m 22 m^ ( m !)^2
, x > 0. (7)
The function in brackets is known as the Bessel function of the first kind of order zero and is denoted by J 0 ( x ). It follows from Theorem 5.7.1 that the series converges for all x , and that J 0 is analytic at x = 0. Some of the important properties of J 0 are discussed in the problems. Figure 5.8.1 shows the graphs of y = J 0 ( x ) and some of the partial sums of the series (7). To determine y 2 ( x ) we will calculate a n ′ ( 0 ). The alternative procedure in which we simply substitute the form (23) of Section 5.7 in Eq. (2) and then determine the bn is discussed in Problem 10. First we note from the coefficient of xr^ +^1 in Eq. (4) that ( r + 1 )^2 a 1 ( r ) = 0. It follows that not only does a 1 ( 0 ) = 0 but also a ′ 1 ( 0 ) = 0. It is easy to deduce from the recurrence relation (5) that a ′ 3 ( 0 ) = a ′ 5 ( 0 ) = · · · = a ′ 2 n + 1 ( 0 ) = · · · = 0; hence we need only compute a 2 ′ m ( 0 ), m = 1 , 2 , 3 ,.... From Eq. (5) we have
a 2 m ( r ) = − a 2 m − 2 ( r )/( r + 2 m )^2 , m = 1 , 2 , 3 ,....
The second solution of the Bessel equation of order zero is found by setting a 0 = 1 and substituting for y 1 ( x ) and a 2 ′ m ( 0 ) = b 2 m ( 0 ) in Eq. (23) of Section 5.7. We obtain
y 2 ( x ) = J 0 ( x ) ln x +
m = 1
(− 1 ) m +^1 Hm 22 m^ ( m !)^2
x^2 m^ , x > 0. (10)
Instead of y 2 , the second solution is usually taken to be a certain linear combination of J 0 and y 2. It is known as the Bessel function of the second kind of order zero and is denoted by Y 0. Following Copson (Chapter 12), we define^13
Y 0 ( x ) =
π
[ y 2 ( x ) + (γ − ln 2) J 0 ( x )]. (11)
Here γ is a constant, known as the Euler–Ma´scheroni (1750–1800) constant; it is defined by the equation γ = lim n →∞ ( Hn − ln n ) ∼= 0. 5772. (12)
Substituting for y 2 ( x ) in Eq. (11), we obtain
Y 0 ( x ) =
π
γ + ln
x 2
J 0 ( x ) +
m = 1
(− 1 ) m +^1 Hm 22 m^ ( m !)^2
x^2 m
, x > 0. (13)
The general solution of the Bessel equation of order zero for x > 0 is y = c 1 J 0 ( x ) + c 2 Y 0 ( x ). Note that J 0 ( x ) → 1 as x → 0 and that Y 0 ( x ) has a logarithmic singularity at x = 0; that is, Y 0 ( x ) behaves as ( 2 /π) ln x when x → 0 through positive values. Thus if we are interested in solutions of Bessel’s equation of order zero that are finite at the origin, which is often the case, we must discard Y 0. The graphs of the functions J 0 and Y 0 are shown in Figure 5.8.2. It is interesting to note from Figure 5.8.2 that for x large both J 0 ( x ) and Y 0 ( x ) are oscillatory. Such a behavior might be anticipated from the original equation; indeed it
–0.
2 4 6 8 10 12 14
1
y
x
y = Y 0 ( x )
y = J 0 ( x )
FIGURE 5.8.2 The Bessel functions of order zero. (^13) Other authors use other definitions for Y
is true for the solutions of the Bessel equation of order ν. If we divide Eq. (1) by x^2 , we obtain
y ′′^ +
x
y ′^ +
ν^2 x^2
y = 0.
For x very large it is reasonable to suspect that the terms ( 1 / x ) y ′^ and (ν^2 / x^2 ) y are small and hence can be neglected. If this is true, then the Bessel equation of order ν can be approximated by
y ′′^ + y = 0.
The solutions of this equation are sin x and cos x ; thus we might anticipate that the solutions of Bessel’s equation for large x are similar to linear combinations of sin x and cos x. This is correct insofar as the Bessel functions are oscillatory; however, it is only partly correct. For x large the functions J 0 and Y 0 also decay as x increases; thus the equation y ′′^ + y = 0 does not provide an adequate approximation to the Bessel equation for large x , and a more delicate analysis is required. In fact, it is possible to show that
J 0 ( x ) ∼=
π x
cos
x −
π 4
as x → ∞, (14)
and that
Y 0 ( x ) ∼=
π x
sin
x −
π 4
as x → ∞. (15)
These asymptotic approximations, as x → ∞, are actually very good. For example, Figure 5.8.3 shows that the asymptotic approximation (14) to J 0 ( x ) is reasonably accurate for all x ≥ 1. Thus to approximate J 0 ( x ) over the entire range from zero to infinity, one can use two or three terms of the series (7) for x ≤ 1 and the asymptotic approximation (14) for x ≥ 1.
y = J 0 ( x )
y
x
2
2 4 6 8 10
1
Asymptotic approximation: y = (2/ π x )1/2^ cos( x – π/4)
FIGURE 5.8.3 Asymptotic approximation to J 0 ( x ).
Corresponding to the root r 2 = − 12 it is possible that we may have difficulty in computing a 1 since N = r 1 − r 2 = 1. However, from Eq. (17) for r = − 12 the coef- ficients of xr^ and xr^ +^1 are both zero regardless of the choice of a 0 and a 1. Hence a 0 and a 1 can be chosen arbitrarily. From the recurrence relation (18) we obtain a set of even-numbered coefficients corresponding to a 0 and a set of odd-numbered coefficients corresponding to a 1. Thus no logarithmic term is needed to obtain a second solution in this case. It is left as an exercise to show that, for r = − 12 ,
a 2 n =
(− 1 ) n^ a 0 ( 2 n )!
, a 2 n + 1 =
(− 1 ) n^ a 1 ( 2 n + 1 )!
, n = 1 , 2 ,....
Hence
y 2 ( x ) = x −^1 /^2
a 0
n = 0
(− 1 ) n^ x^2 n ( 2 n )!
n = 0
(− 1 ) n^ x^2 n +^1 ( 2 n + 1 )!
= a 0
cos x x^1 /^2
sin x x^1 /^2
, x > 0. (21)
The constant a 1 simply introduces a multiple of y 1 ( x ). The second linearly independent solution of the Bessel equation of order one-half is usually taken to be the solution for which a 0 = ( 2 /π)^1 /^2 and a 1 = 0. It is denoted by J − 1 / 2. Then
J − 1 / 2 ( x ) =
π x
cos x , x > 0. (22)
The general solution of Eq. (16) is y = c 1 J 1 / 2 ( x ) + c 2 J − 1 / 2 ( x ). By comparing Eqs. (20) and (22) with Eqs. (14) and (15) we see that, except for a phase shift of π/4, the functions J − 1 / 2 and J 1 / 2 resemble J 0 and Y 0 , respectively, for large x. The graphs of J 1 / 2 and J − 1 / 2 are shown in Figure 5.8.4.
–0.
2 4 14
1
y
6 12^ x
J –1/2( x )
J 1/2( x )
8 10
FIGURE 5.8.4 The Bessel functions J 1 / 2 and J − 1 / 2.
Bessel Equation of Order One. This example illustrates the situation in which the roots of the indicial equation differ by a positive integer and the second solution involves a logarithmic term. Setting ν = 1 in Eq. (1) gives
L [ y ] = x^2 y ′′^ + x y ′^ + ( x^2 − 1 ) y = 0. (23) If we substitute the series (3) for y = φ( r , x ) and collect terms as in the preceding cases, we obtain
L [φ]( r , x ) = a 0 ( r^2 − 1 ) xr^ + a 1 [( r + 1 )^2 − 1] xr^ +^1
n = 2
{[( r + n )^2 − 1] an + an − 2 } xr^ + n^ = 0. (24)
The roots of the indicial equation are r 1 = 1 and r 2 = −1. The recurrence relation is
[( r + n )^2 − 1] an ( r ) = − an − 2 ( r ), n ≥ 2. (25)
Corresponding to the larger root r = 1 the recurrence relation becomes
an = −
an − 2 ( n + 2 ) n
, n = 2 , 3 , 4 ,....
We also find from the coefficient of xr^ +^1 in Eq. (24) that a 1 = 0; hence from the recurrence relation a 3 = a 5 = · · · = 0. For even values of n , let n = 2 m ; then
a 2 m = −
a 2 m − 2 ( 2 m + 2 )( 2 m )
a 2 m − 2 22 ( m + 1 ) m
, m = 1 , 2 , 3 ,....
By solving this recurrence relation we obtain
a 2 m =
(− 1 ) m^ a 0 22 m^ ( m + 1 )! m!
, m = 1 , 2 , 3 ,.... (26)
The Bessel function of the first kind of order one, denoted by J 1 , is obtained by choosing a 0 = 1 /2. Hence
J 1 ( x ) =
x 2
m = 0
(− 1 ) m^ x^2 m 22 m^ ( m + 1 )! m!
The series converges absolutely for all x , so the function J 1 is analytic everywhere. In determining a second solution of Bessel’s equation of order one, we illustrate the method of direct substitution. The calculation of the general term in Eq. (28) below is rather complicated, but the first few coefficients can be found fairly easily. According to Theorem 5.7.1 we assume that
y 2 ( x ) = a J 1 ( x ) ln x + x −^1
n = 1
cn xn
, x > 0. (28)
Computing y 2 ′( x ), y 2 ′′ ( x ), substituting in Eq. (23), and making use of the fact that J 1 is a solution of Eq. (23) give
2 ax J (^) 1 ′( x ) +
n = 0
[( n − 1 )( n − 2 ) cn + ( n − 1 ) cn − cn ] xn −^1 +
n = 0
cn xn +^1 = 0 , (29)
The second solution of Eq. (23), the Bessel function of the second kind of order one, Y 1 , is usually taken to be a certain linear combination of J 1 and y 2. Following Copson (Chapter 12), Y 1 is defined as
Y 1 ( x ) =
π
[− y 2 ( x ) + (γ − ln 2) J 1 ( x )], (33)
where γ is defined in Eq. (12). The general solution of Eq. (23) for x > 0 is
y = c 1 J 1 ( x ) + c 2 Y 1 ( x ).
Notice that while J 1 is analytic at x = 0, the second solution Y 1 becomes unbounded in the same manner as 1/ x as x → 0. The graphs of J 1 and Y 1 are shown in Figure 5.8.5.
–0.
2 4 8 10 14
1
y
6 12^ x
y = J 1 ( x ) y = Y 1 ( x )
FIGURE 5.8.5 The Bessel functions J 1 and Y 1.
PROBLEMS In each of Problems 1 through 4 show that the given differential equation has a regular singular point at x = 0, and determine two linearly independent solutions for x > 0.
x^2 y ′′^ + x y ′^ + ( x^2 − 94 ) y = 0 , x > 0.