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n=2 n=1 n=1
(b) Substituting the series representation for J0(x) in Eq. (i), show that
2 2 ^ 2 n ^ (1)n2nx2n
b1x + 2 b2x +J2 (nbn + bn2)x =~2? 22n ( !)2 . (ii)
^ 3 ^ 1 2 (n!)
(c) Note that only even powers of x appear on the right side of Eq. (ii). Show that b1 = b3 = b5 = = 0, b2 = 1 /22 (1! )2, and that
(2n)2b2n + b2n_2 = 2(1)n(2n)/2ln(n!)2, n = 2, 3, 4,..
1 ( 1 \ 1 ( 11 b4 =------2Γ-2 I 1 + I and be = 2 2 2 I 1 + ~
4 224^ V 2/ 6 22426^ V 2 3
The general solution of the recurrence relation is b2n = (1)n+1 Hn /22n (n!)2. Substituting for bn in the expression for y2(x) we obtain the solution given in Eq. (10).
11. Find a second solution of Bessels equation of order one by computing the cn (r2) and a ofEq. (24) of Section 5.7 according to the formulas (19) and (20) of that section. Some guidelines along the way of this calculation are the following. First, use Eq. (24) of this
5.8 Bessels Equation
section to show that a1 (1) and a\ (1) are 0. Then show that c1(1) = 0 and, from the recurrence relation, that cn (1) = 0 for n = 3, 5,.... Finally, use Eq. (5) to show that
( ) (1)ma0
am (Γ ) =
2m (r + 1)(r + 3) (r + 2m 1)2(r + 2m + 1)
for m = 1, 2, 3,..., and calculate
cm (1) = (1)m+1(Hm + Hm-1)/mm !(m 1)!.
1. By a suitable change of variables it is sometimes possible to transform another differential equation into a Bessel equation. For example, show that a solution of
x2y" + (ΰ2β2xe + 1 v2e )y = 0, x > 0
is given by y = x1/2 f (axβ) where f ($) is a solution of the Bessel equation of order v.
13. Using the result of Problem 12 show that the general solution of the Airy equation
y" xy = 0, x > 0
is y = x 1/2[c1 f1(ixi/2) + cf2(3ixi/2)] where f1 ($) and f2($) are linearly independent solutions of the Bessel equation of order one-third.
14. It can be shown that J0 has infinitely many zeros for x > 0. In particular, the first three zeros are approximately'2.405, 5.520, and 8.653 (see Figure 5.8.1). Let kj, j = 1, 2, 3,..., denote the zeros of J0; it follows that
) =| 1 x = L
Verify thaty = J0(kjx) satisfies the differential equation
' 1 2,
y + y + kjy = 0, x > 0.
Hence show that
xJ0(kix)J^kx) dx = 0 if ki = kj.
This important property of J0(kix), known as the orthogonality property, is useful in solving boundary value problems.
Hint: Write the differential equation for J0(kix). Multiply it by xJ0(kjx) and subtract it from xJ0(kix) times the differential equation for J0(kjx). Then integrate from 0 to 1.
Coddington, E. A., An Introduction to Ordinary Differential Equations (Englewood Cliffs, NJ: Prentice Hall, 1961; New York: Dover, 1989).
Copson, E. T., An Introduction to the Theory of Functions of a Complex Variable (Oxford: Oxford University, 1935).
Proofs of Theorems 5.3.1 and 5.7.1 can be found in intermediate or advanced books; for example, see Chapters 3 and 4 of Coddington, or Chapters 3 and 4 of:
Rainville, E. D., Intermediate Differential Equations (nd ed.) (New York: Macmillan, 1964).
Also see these texts for a discussion of the point at infinity, which was mentioned in Problem 1 of Section 5.4. The behavior of solutions near an irregular singular point is an even more advanced topic; a brief discussion can be found in Chapter 5 of:
Chapter 5. Series Solutions ofSecond Order Linear Equations
Coddington, E. A., and Levinson, N., Theory of Ordinary Differential Equations (New York: McGraw-Hill, 1955).