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

ISBN 0-471-31999-6

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Hint: Write 1 + x = 2 + (x — 1) and x = 1 + (x — 1). Alternatively, make the change of variable x — 1 = t and determine a series solution in powers of t.

12. The Chebyshev equation is

(1 — x2) y" — xy + a2y = 0, where a is a constant; see Problem 10 of Section 5.3.

(a) Show that x = 1 and x = — 1 are regular singular points, and find the exponents at each of these singularities.

(b) Find two linearly independent solutions about x = 1.

272

Chapter 5. Series Solutions ofSecond Order Linear Equations

13. The Laguerre11 differential equation is

xy" + (1 — x )y' + Xy — 0.

Show that x = 0 is a regular singular point. Determine the indicial equation, its roots, the recurrence relation, and one solution (x > 0). Show that if X — m, a positive integer, this solution reduces to a polynomial. When properly normalized this polynomial is known as the Laguerre polynomial, Lm (x).

14. The Bessel equation of order zero is

x 2y" + xy1 + x 2y = 0.

Show that x = 0 is a regular singular point; that the roots of the indicial equation are r1 = r2 = 0; and that one solution for x > 0 is

^ (—1)nx2n

Jr(x ) — 1 + ^ ^ it .

0W ^ 2 (n\)2

Show that the series converges for all x . The function J0 is known as the Bessel function of the first kind of order zero.

15. Referring to Problem 14, use the method of reduction of order to show that the second solution of the Bessel equation of order zero contains a logarithmic term.

Hint: Ify2(x) — J0(x)v(x),then

f dx y2(x) = J0(x) 2 .

J x [ J0(x )]

Find the first term in the series expansion of 1 /x[ J0(x)]2.

16. The Bessel equation of order one is

x 2y" + xy' + (x2 — 1)y — 0.

(a) Show that x — 0 is a regular singular point; that the roots of the indicial equation are r1 — 1 and r2 — — 1; and that one solution for x > 0 is

x ( 1)nx 2n

J1(x) — ly (—1)x 2 .

1 2 n—0 (n + 1)! n! 2

Show that the series converges for all x . The function J1 is known as the Bessel function of the first kind of order one.

(b) Show that it is impossible to determine a second solution of the form

TO

x—1 b xn, x > 0.

n

n—0

5.7 Series Solutions near a Regular Singular Point, Part II

Now let us consider the general problem of determining a solution of the equation

L [y] — x2 y" + x [xp(x )]y' + [x 2q (x )]y — 0, (1)

11Edmond Nicolas Laguerre (1834-1886), a French geometer and analyst, studied the polynomials named for him about 1879.

5.7 Series Solutions near a Regular Singular Point, Part II

273

where

xp(x) = E PnX", x 2q (x) = E VnXn, (2)

nn n=0 n=0

and both series converge in an interval |x | < p for some p > 0. The point x = 0 is a regular singular point, and the corresponding Euler equation is

x 2 y" + P0 xZ + ^0 y = 0 (3)

We seek a solution of Eq. (1) for x > 0 and assume that it has the form

CO CO

y = 0(r, x) = x*J2 anxn = E anxr+n, (4)

n=0 n=0

where a0 = 0, and we have written y = 0(r, x) to emphasize that 0 depends on r as well as x .It follows that

OO

y' = E (r + n)anxr +n-1, y'' = J2 (r + n)(r + n - 1)anxr +n-2. (5)

n=0 n=0

Then, substituting from Eqs. (2), (4), and (5) in Eq. (1) gives

a0r(r - 1)xr + a1(r + 1)rxr+1 + ••• + an(r + n)(r + n - 1)xr+n + •••

+ (P0 + P\x + ••• + Pnxn + •••)

x [a0rxr + a1 (r + 1)xr+1------+-----+ an (r + n)xr+n--+-]

+ (q0 + q1 x + ••• + qnx + •••) x (a0xr + a1 xr+1 +---+ anxr+n-------+------) = 0.

Multiplying the infinite series together and then collecting terms, we obtain

a0 F (r )xr + [a1 F (r + 1) + a0(p1r + q1)]xr+1

+ a F (r + 2) + a0(P2r + q2) + a1^1(r + 1) + qJK+2

+ ••• + {anF (r + n) + a0(Pnr + qn) + a1[Pn-1(r + 1) + qn-1]

+ ••• + an-1L?1(r + n — 1) + q1]}xr+n + ••• = o,

or in a more compact form,

L[0](r, x) = a0F(r)xr

O n-1

+ J2 \F(r + n)an + E ak[(r +kk)Pn-k + qn-kHxr+n =0, (6)

n = 1 yk=0 J

where

F(r) = r (r - 1) + P0r + q0. (7)

For Eq. (6) to be satisfied identically the coefficient of each power of x must be zero.

Since a0 = 0, the term involving xr yields the equation F(r) = 0. This equation is called the indicial equation; note that it is exactly the equation we would obtain in looking for solutions y = xr of the Euler equation (3). Let us denote the roots of the indicial equation by r1 and r2 with r1 > r2 if the roots are real. If the roots are

complex, the designation of the roots is immaterial. Only for these values of r can we expect to find solutions of Eq. (1) of the form (4). The roots r1 and r2 are called the

274

Chapter 5. Series Solutions of Second Order Linear Equations

exponents at the singularity; they determine the qualitative nature of the solution in the neighborhood of the singular point.

Setting the coefficient of xr+n in Eq. (6) equal to zero gives the recurrence relation

F(r + n)an + ^ ak[(r + k)Pn-k + %-k] = °> n > L (8)

k=0

Equation (8) shows that, in general, an depends on the value of r and all the preceding coefficients a°, ax,, an_j. It also shows that we can successively compute av a2,..., an,... in terms of a° and the coefficients in the series for xp(x) and x2q (x) provided that F(r + 1), F(r + 2),..., F(r + n),... are not zero. The only values of r for which F(r) = 0 are r = r1 and r = r2; since r1 it follows that r1 + n is not equal to r1 or r2 for n > 1. Consequently, F(r1 + n) = 0 for n > 1. Hence we can always determine one solution of Eq. (1) in the form (4), namely,

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