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

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n = 1

generates a multiple of J1, and y2 is only determined up to an additive multiple of J1. In accord with the usual practice we choose c2 = 1/22. Then we obtain

-1

24 -2 (-1)

+1

2

-1

242!

1 + |i +1

(H2 + H)

24 2!

It is possible to show that the solution of the recurrence relation (31) is

(-1)m + 1( Hm + Hm_ 1)

2m 2 m !(m - 1)!

with the understanding that H0 = 0. Thus

m 1 , 2,

y2(x) = -J1(x ) ln X +-------

^ Ζ (-1)m (Hm + Hm-1) X2m

22mm!(m - 1)!

x > 0. (32)

The calculation of y2(x) using the alternative procedure [see Eqs. (19) and (20) of Section 5.7] in which we determine the cn(r2) is slightly easier. In particular the latter procedure yields the general formula for c2m without the necessity of solving a recurrence relation of the form (31) (see Problem 11). In this regard the reader may also wish to compare the calculations of the second solution of Bessels equation of order zero in the text and in Problem 10.

c4 =

5.8 Bessel's Equation

289

The second solution of Eq. (3), the Bessel function of the second kind of order one, Y1, is usually taken to be a certain linear combination of J1 and y. Following Copson (Chapter 1), Y1 is defined as

Y1(x) = -[y(x) + (Y ln)J1(x)], (33)

where y is defined in Eq. (1). The general solution of Eq. (3) for x > 0 is

y = c1 J1(x) + cY1(x).

Notice that while J1 is analytic at x = 0, the second solution Y1 becomes unbounded in the same manner as 1/x as x ^ 0. The graphs of J1 and Y1 are shown in Figure 5.8.5.

FIGURE 5.8.5 The Bessel functions J1 and Y1.

PROBLEMS In each of Problems 1 through 4 show that the given differential equation has a regular singular

³ point at x = 0, and determine two linearly independent solutions for x > 0.

1. x yw + xy' + xy = 0 . x yw + 3xy; + (1 + x)y = 0

3. x y" + xy' + xy = 0 4. x yw + 4xy; + ( + x)y = 0

5. Find two linearly independent solutions of the Bessel equation of order ,

x2y" + xy' + (x 9)y = 0, x > 0.

6. Show that the Bessel equation of order one-half,

x2y" + xy' + (x ;1 )y = 0, x > 0,

can be reduced to the equation

v" + v = 0

by the change of dependent variable y = x1/v(x). From this conclude that y1(x) = x1/ cos x and y(x) = x1/ sin x are solutions of the Bessel equation of order one-half.

7. Show directly that the series for J0(x), Eq. (7), converges absolutely for all x.

8. Show directly that the series for J1(x), Eq. (7), converges absolutely for all x and that

J0 (x) = J1(x).

290

Chapter 5. Series Solutions of Second Order Linear Equations

9. Consider the Bessel equation of order v,

x2y" + xy' + (x2 v2) = 0, x > 0.

Take v real and greater than zero.

(a) Show that x = 0 is a regular singular point, and that the roots of the indicial equation are v and v.

(b) Corresponding to the larger root v, show that one solution is

Σ³ (x) = xv

1 +jr---------------------^2

1 ς! (1 + v) (2 + v) (m 1 + v)(m + v) \ 2 /

(c) If 2v is not an integer, show that a second solution is

y2(x) = x-

1 + E----------------------------------------------------------(x )2

m! (1 v) (2 v) (m 1 v)(m v) \ 2 /

Note that y1(x) ^ 0 as x ^ 0, and that y2(x) is unbounded as x ^ 0.

(d) Verify by direct methods that the power series in the expressions for y1 (x) and y2(x) converge absolutely for all x. Also verify that y2 is a solution provided only that v is not an integer.

10. In this section we showed that one solution of Bessels equation of order zero,

L [y] = x2 y11 + xy1 + x2 y = 0,

is J0, where J0(x) is given by Eq. (7) with a0 = 1. According to Theorem 5.7.1 a second solution has the form (x > 0)

y2(x) = J0(x) lnx + ^ bnxn.

n=1

(a) Show that

TO

L[y2](x) = ^2n(n 1)bnxn + ^2nbnxn + ^2 bnxn+2 + 2xJ0(x). (i)

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