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

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x > 0.

(1)

The constant a1 simply introduces a multiple of y1 (x). The second linearly independent solution of the Bessel equation of order one-half is usually taken to be the solution for which a0 = (/ο)1/ and a1 = 0. It is denoted by J1/. Then

J1/(x) =

1/

cos x,

x > 0.

()

The general solution of Eq. (16) is y = c1 J1/(x) + cJ1(x).

By comparing Eqs. (0) and () with Eqs. (14) and (15) we see that, except for a phase shift of n/4, the functions J1/ and J1/ resemble J0 and Y0, respectively, for large x. The graphs of J1/ and J1/ are shown in Figure 5.8.4.

an =

a

FIGURE 5.8.4 The Bessel functions J1/ and J1/.

5.8 Bessels Equation

287

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 v = 1 in Eq. (1) gives

L [y] = x2y" + xy' + (x2 - 1)y = 0. (23)

If we substitute the series (3) for y = τ(γ, x) and collect terms as in the preceding

cases, we obtain

L[τ](γ, x) = a0(r2 1)xr + a1[(r + 1)2 1]xr+1

TO

+ ?{[(r + n)2 1]an + an2}xr+n = 0. (24)

n=2

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

[(r + n)2 1]an (r) = -an2(r), n > 2. (25)

Corresponding to the larger root r = 1 the recurrence relation becomes

a

a =------------n^-, n = 2, 3, 4,....

n (n + 2)n

We also find from the coefficient of xr+1 in Eq. (24) that a1 = 0; hence from the

n, let n = 2m

m = 1, 2, 3,

recurrence relation a3 = a5 = = 0. For even values of n, let n = 2m; then

2m (2m + 2)(2m) 22(m + 1)m

By solving this recurrence relation we obtain

(1)ma0

a, =^----------- 0-----, m = 1, 2, 3,.... (26)

2m 2 (m + 1)!m!

The Bessel function of the first kind of order one, denoted by J1, is obtained by choosing a0 = 1/2. Hence

V TO ( I'mv 2m

J1(x ) = -Y ------------)--------. (27)

^ 2 m=0 22m(m + 1)!m! V '

The series converges absolutely for all x, so the function J1 is analytic everywhere.

In determining a second solution of Bessels 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

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

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 ofEq. (23) give

TO

2axJ[(x) + Y [(n 1)(n 2)cn + (n 1)cn cn]xn1 + Ycnxn+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 + [° ¦ C2 + C0]X + 2_^ [(n - 1)cn + 1 + ρο-1]υΟ 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

Ζ

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

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