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Elementary differential equations 7th edition - Boyce W.E

Boyce W.E Elementary differential equations 7th edition - Wiley publishing , 2001. - 1310 p.
ISBN 0-471-31999-6
Download (direct link): elementarydifferentialequat2001.pdf
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JJx ) =
( 2
\n x
) cos (x - 4)
and that
/2 N 1/2
^ > = ( „) sin (x - 4)
as
as
00,
00.
(14)
(15)
These asymptotic approximations, as x ^ro, are actually very good. For example, Figure 5.8.3 shows that the asymptotic approximation (14) to J0(x) is reasonably accurate for all x > 1. Thus to approximate J0(x) over the entire range from zero to infinity, one can use two or three terms of the series (7) for x < 1 and the asymptotic approximation (14) for x > 1.
x
x
FIGURE 5.8.3 Asymptotic approximation to J0(x).
5.8 Bessel’s Equation
285
Bessel Equation of Order One-Half. This example illustrates the situation in which the roots of the indicial equation differ by a positive integer, but there is no logarithmic term in the second solution. Setting v = 2 in Eq. (1) gives
L [y] = x2y" + xy' + (x2 - 2) y = 0.
If we substitute the series (3) for y = fi(r, x), we obtain
L[0](r, x) = Y2 [(r + n)(r + n - 1) + (r + n) - 2] anxr +n + ^
(16)
rr +n+2
n=0
n=0
= (r2 - 1 )a0xr + [(r + 1)2 - 2] a1 x
r + 1
+ ?{ [(r + n)2 - j] an + a,-} x'+n = 0
(17)
n2
The roots of the indicial equation are rl = 2, r2 integer. The recurrence relation is
- 2 ; hence the roots differ by an
[(r + n)2 - 2] an = -c
n-2’
n > 2.
(18)
Corresponding to the larger root r1 = 2 we find from the coefficient of xr+1 in Eq. (17) that a1 = 0. Hence, from Eq. (18), a3 = a5 = ??? = a2n+1 = ••• = 0. Further, for
r
n2
n(n + 1)
n = 2, 4, 6.
or letting n = 2m, we obtain
2m-2
2m 2m (2m + 1)
By solving this recurrence relation we find that
m = 1, 2, 3,...
and, in general,
3!
(-1)ma0
(2m + 1)!
Hence, taking a0 = 1, we obtain
5!
m = 1, 2, 3,
T1(x ) = x
- x 1/2
1+
m 2m
(-1)mx
^ (2m + 1)!
= x-1/2?
m0
(-1)mx 2m+1 - (2m + 1)! !
x > 0. (19)
The power series in Eq. (19) is precisely the Taylor series for sin x; hence one solution of the Bessel equation of order one-half is x-1/2 sin x. The Bessel function of the first kind of order one-half, J1/2, is defined as (2/n)1/2y1. Thus
2 \1/2 J1/2(x ) = ( —) Sln x ’
x > 0.
(20)
2
a
a
0
0
a2 =
a2m =
286
Chapter 5. Series Solutions ofSecond Order Linear Equations
Corresponding to the root r2 = — 1 it is possible that we may have difficulty in computing a1 since N = r1 — r2 = 1. However, from Eq. (17) for r = —2 the coefficients of xr and xr+1 are both zero regardless of the choice of a0 and a1. Hence a0 and a1 can be chosen arbitrarily. From the recurrence relation (18) we obtain a set of even-numbered coefficients corresponding to a0 and a set of odd-numbered coefficients corresponding to a1. Thus no logarithmic term is needed to obtain a second solution in this case. It is left as an exercise to show that, for r = — 2,
(— 1)nao (2n)!
a2n + 1 _
(—1)na1
(2n + 1)!
n _ 1, 2,
Hence
y2(x) _ x
- x — 1/2
“ (—1)nx2n ? (—1)nx2n+1
^ (2n)! +a1 ^
n= 0
a
cos x
0 ^
+ a
sin x
1x/2 ’
a, .
n^0 (2n + 1)!
x > 0.
(21)
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 = (2In f12 and a1 = 0. It is denoted by J—1/2. Then
J—1/2(x) _
1/2
cos x,
x > 0.
(22)
The general solution of Eq. (16) is y = c1 J1/2(x) + c2 J—1/2(x).
By comparing Eqs. (20) and (22) with Eqs. (14) and (15) we see that, except for a phase shift of n/4, the functions J—1/2 and J1/2 resemble J0 and F0, respectively, for large x. The graphs of J1/2 and J—1/2 are shown in Figure 5.8.4.
a2n _
a
2
FIGURE 5.8.4 The Bessel functions Jj/2 and J—j/2.
5.8 Bessel’s 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 = $(r, x) and collect terms as in the preceding
cases, we obtain
L [0](r, x) = a0(r2 — 1)xr + a1[(r + 1)2 — 1]xr+1
TO
+ ?{[(r + n)2 — 1]an + an—2}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) = -an—2(r), n > 2. (25)
Corresponding to the larger root r = 1 the recurrence relation becomes
a
a =-----------------, 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
a2 =^------------—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 (— ^\my. 2m
J (x) = - Y ^m—)---------------. (27)
1 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 Bessel’s 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
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