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

Boyce W.E. Elementary Differential Equations and Boundary Value Problems - John Wiley & Sons, 2001. - 1310 p.
Download (direct link): elementarydifferentialequations2001.pdf
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/2 /'~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
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