# Elementary Differential Equations and Boundary Value Problems - Boyce W.E.

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13,

Other authors use other definitions for YQ. The present choice for YQ is also known as the Weber (184-1913) function.

284

Chapter 5. Series Solutions of Second Order Linear Equations

is true for the solutions of the Bessel equation of order v. If we divide Eq. (1) by x2, we obtain

Σ

For x very large it is reasonable to suspect that the terms (1/x)y' and (v2/x2)y are small and hence can be neglected. If this is true, then the Bessel equation of order v 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 Bessels 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 J0 and Y0 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

J0(x) =

( 2

\n x

) cos (x - 4)

and that

/2 N 1/2

r«(x) -(«) s,n (x - 4)

as

as

00,

.

(14)

(15)

These asymptotic approximations, as x ^ro, are actually very good. For example, Figure 5.8.3 shows that the asymptotic approximation (14) to J0(x) is reasonably accurate for all x > 1. Thus to approximate J0(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.

x

x

FIGURE 5.8.3 Asymptotic approximation to J0(x).

5.8 Bessels Equation

285

Bessel Equation of Order One-Half. This example illustrates the situation in which the roots of the indicial equation differ by a positive integer, but there is no logarithmic term in the second solution. Setting v 2 in Eq. (1) gives

L [y] x2y + xy' + (x2 - 2) y 0.

If we substitute the series (3) for y τ(γ, x), we obtain

CO

L[τ](γ, x) ^ [(r + n)(r + n - 1) + (r + n) - j] anXr +n + ^2

(16)

rT +n+2

n0

n0

= (r2 - 1 )a0xr + [(r + 1)2 - 2] a1 x

r + 1

+ E1 [(r + n)2 - j] an + a,-2} xr+n = 0

(17)

n2

The roots of the indicial equation are r1 = 1, r2 integer. The recurrence relation is

- 2; hence the roots differ by an

[(r + n)2 - j] an = -c

² '"Χ *

n-2

n > 2.

(18)

Corresponding to the larger root r1 = 2 we find from the coefficient of xr+1 in Eq. (17) that a1 = 0. Hence, from Eq. (18), a3 = a5 = ¦ ¦ ¦ = a2n+1 = ¦ ¦ ¦ = 0. Further, for

r

n2

n(n + 1)

n = 2, 4, 6.

or letting n = 2m, we obtain

2m-2

2m 2m (2m + 1)

By solving this recurrence relation we find that

m = 1, 2, 3,...

and, in general,

3!

(-1)ma0

(2m + 1)!

Hence, taking a0 = 1, we obtain

5!

m = 1, 2, 3,

y1(x) = x

- x 1/2

1+

m 2m

(-1)mx

(2m + 1)!

= x-1/2E

m0

(-1)mx 2m+1 - (2m + 1)! !

x > 0. (19)

The power series in Eq. (19) is precisely the Taylor series for sin x; hence one solution of the Bessel equation of order one-half is x-1/2 sin x. The Bessel function of the first kind of order one-half, J1/2, is defined as (2/n)1/2y1. Thus

2 \1/2 J1/2(x) = ( ) Sln x

x > 0.

(20)

2

a

a

0

0

a4

a2m =

286

Chapter 5. Series Solutions of Second Order Linear Equations

Corresponding to the root r = 1 it is possible that we may have difficulty in computing a1 since N = r1 r = 1. However, from Eq. (17) for r = the coefficients of xr and xr+1 are both zero regardless of the choice of a0 and a1. Hence a0 and a1 can be chosen arbitrarily. From the recurrence relation (18) we obtain a set of even-numbered coefficients corresponding to a0 and a set of odd-numbered coefficients corresponding to a1. 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 = ,

(1)4 (n)!

an + 1 =

(1)4

(n + 1)!

n = 1,,

Hence

Σ(x) = x

_ x1/

n0

a

cos x

0 x^

+ a

sin x

1 x1/

a, .

*=0 (n + 1)!

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