# Elementary differential equations 7th edition - Boyce W.E

ISBN 0-471-31999-6

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y2(x ) = aJ1(x ) ln x + x 1

? + ?

c,xn

n=1

x > 0. (28)

Computing y2(x), y2" (x), substituting in Eq. (23), and making use of the fact that J1 is a solution of Eq. (23) give

TO

2axJ((x) + Y [(n - 1)(n - 2)cn + (n - 1)cn - cn]xn-1 + ^cnxn+1 = 0, (29)

n=0 n=0

288

Chapter 5. Series Solutions of Second Order Linear Equations

where c0 = 1. Substituting for J1(x) from Eq. (27), shifting the indices of summation in the two series, and carrying out several steps of algebra give

c1 + [0 ? c2 + c0]x + 2_^ [(n l)c+i + cn-\]x"

n=2

= a

x+

m = 1

(1)m (2m + 1)x2m+1

22m(m + 1)! m!

(30)

From Eq. (30) we observe first that c1 = 0, and a = c0 = 1. Further, since there are only odd powers of x on the right, the coefficient of each even power of x on the left must be zero. Thus, since c1 = 0, we have c3 = c5 = = 0. Corresponding to the odd powers of x we obtain the recurrence relation [let n = 2m + 1 in the series on the left side of Eq. (30)]

[(2m + 1)2 1]c2m+2 + c2m =

(1)m (2m + 1)

22m(m + 1)! m!

m = 1, 2, 3,

(31)

When we set m = 1 in Eq. (31), we obtain

(32 - 1)c4 + C2 = ( 1 )3/(22 ? 2!).

Notice that c2 can be selected arbitrarily, and then this equation determines c4. Also notice that in the equation for the coefficient of x, c2 appeared multiplied by 0, and that equation was used to determine a. That c2 is arbitrary is not surprising, since c2

TO

is the coefficient of x in the expression x1[1 + cnxn]. Consequently, c2 simply

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 + H1).

24 ? 2!

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

(1)m + 1( Hm + H_ 1)

2m 22mm !(m 1)!

with the understanding that H0 = 0. Thus

m 1 , 2,

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

^ ^ (1)m (Hm + Hm 1) x2m

^1 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 Bessels Equation

289

PROBLEMS

The second solution of Eq. (23), the Bessel function of the second kind of order one, Y1, is usually taken to be a certain linear combination of J1 and y2. Following Copson (Chapter 12), Y1 is defined as

2

Yi(x) = -[-y2(x) + (Y - ln2)Ji(x)], (33)

n

where y is defined in Eq. (12). The general solution of Eq. (23) for x > 0 is

y = c1 J1(x ) + c2Y1(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.

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.

1. x2y" + 2xy' + xy = 0 2. x2yw + 3xy; + (1 + x)y = 0

3. x2y" + xy' + 2xy = 0 4. x2yw + 4xy; + (2 + x)y = 0

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

x2y" + xy + (x2 9)y = 0, x > 0.

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

x2y" + xy + (x2 1 )y = 0, x > 0,

can be reduced to the equation

v" + v = 0

by the change of dependent variable y = x1/2v(x). From this conclude that y:(x) = x1/2 cos x and y2(x) = x1/2 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 Jx(x), Eq. (27), converges absolutely for all x and that J0 (x) = Jj(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

y1(x ) = xv

i + jr----------------------()m----------------------2

1 m ! (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

i + 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)

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