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

**Download**(direct link)

**:**

**447**> 448 449 450 451 452 453 .. 609 >> Next

/ë2 /'~i2 /-,2 /-.2^. 2

a1 = a0, &2 = a1/2 = a0/2 , ^3 = a2/3 = a0/3 2 ,

*1 - &0, »2 - a1' ^ “ a0' ^ ' a3 “ a2' J “ Cl0/

2222 2

a4 = a3/4 = a0/4 -3-2 ,... and an = a0/(n!) . Thus one

solution (on setting a0 = 1) is y = xn/(n!) 2.

n=0

If we make the change of variable t = x-1 and let y = u(t), then the Legendre equation transforms to 2

(t + 2t)u (t) + 2(t + 1)u (t) - a(a+1)u(t) = 0. Since x = 1 is a regular singular point of the original equation, we know that t = 0 is a regular singular point

of the transformed equation. Substituting u =

n=0

in the transformed equation and shifting indices, we obtain

1 (n+r)(n+r-1)antn+r + 2 (n+r+1)(n+r)an+1tn+r

, n+r

ant

n=0 n=-1

+ 2 (n+r)antn+r + 2 (n+r+1)an+!tn+r

n+1

n=0 n=-1

a(a+1) antn+r = 0, or

n

n=0

[2r(r-1) + 2r]a0 tr1 + I {2(n+r+1) 2an+1

n=0

+ [(n+r)(n+r+1) - a(a+1)]an}tn+r = 0.

2

The indicial equation is 2r = 0 so r = 0 is a double root. Thus there will be only one series solution of the

oo

oo

Section 5.6

91

14.

form y = antn+r. The recurrence relation is

n=0

2

2(n+1) an+1 = [a(a+1) - n(n+1)]an,n = 0,1,2,... . We have a1 = [a (a+1)]a0/2-12, a2 = [a(a+1)][a(a+1) - 1-2]a0/22 22-12,

3222

a3 = [a(a+1)][a(a+1) - 1-2][a(a+1) - 2-3]a0/23-32-22-12,..., and an = [a(a+1)][a(a+1)-1-2]..[a(a+1)-(n-1)n]a0/2n(n!) 2.

Reverting to the variable x it follows that one solution of the Legendre equation in powers of x-1 is

y1(x) = X [a(a+1)][a(a+1) - 1-2] ...

n=0

[a(a+1) - (n-1)n](x-1) n/2n(n!)2 where we have set a0 = 1,

which is equivalent to the answer in the text if a (-1) is taken out of each square bracket.

The standard form is y" + p(x)y' + q(x)y = 0, with

p(x) = 1/x and q(x) = 1. Thus x = 0 is a singular point;

2

and since xp(x) ^ 1 and xq(x) ^ 0 as x ^ 0, x = 0 is a regular singular point. Substituting y = X a^^ into

n=0

22

x y" + xy + x y = 0 and shifting indices appropriately, we obtain

n+r n+r n+r

X (n+r)(n+r-1)anx + X (n+r)a^x + X an-2x = 0,

n=0 n=0 n=2

or

[r(r-1)+r]a0xr + [(1+r)r+1+r]a1xr+1

+X, [(n+r) 2an + an-2] xn+r = 0. The indicial equation

2

2n

n=2

2

is r = 0 so r = 0 is a double root. It is necessary to

r+1

take a1 = 0 in order that the coefficient of x be zero.

2

The recurrence relation in n an = -an-2, n = 2,3,... . Since a1 = 0 it follows that a3 = a5 = a7 = ... = 0. For

the even coefficients we let n = 2m, m = 1,2,... . Then

2 2 2. 2 2. 2. 2. 2

a2m = ^^^2 m so a2 = -a0/2 -1 , a4 = a0/2 -2 -1 -2 ,... ,

and a2m = (-1)ma0/22m(m!) 2. Thus one solution of the Bessel

92

Section 5.6

equation of order zero is J0(x) = 1 + (-1) mx2m/22m(m!) 2

m=1

where we have set a0 = 1. Using the ratio test it can be shown that the series converges for all x. Also note that J0(x) — 1 as x — 0.

15. In order to determine the form of the integral for x near zero we must study the integrand for x small. Using the above series for J0, we have

1 1 1

x[J0(x)] 2 x[1 - x2/2 +...] 2 x[1 - x2 +...]

12

—[1 + x + ...] for x small. Thus x

dx 1

= J0 (x) I [— + x + ...]dx

dx

y2(x) = J0(x) I------ = J0(x) I

x[J0(x)] 2 J x

2

x

= J0(x)[lnx + — + ...], and it is clear that y2(x)

0 x 2 will contain a logarithmic term.

16a. Putting the D.E. in the standard form

**447**> 448 449 450 451 452 453 .. 609 >> Next