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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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The roots of the indicial equation F(r) = r (r - 1) + r = 0 are r1 = 0 and r2 = 0; hence we have the case of equal roots. The recurrence relation is
an(r) = -
a-2 (r )
(r + n)(r + n - 1) + (r + n)
an-2(r )
(r + n)2
n 2.
(5)
To determine >'1(x) we set r equal to 0. Then from Eq. (4) it follows that for the coefficient of xr+1 to be zero we must choose a1 = 0. Hence from Eq. (5), a3 = a5 = a7 = = 0. Further,
an(0) = -an-2(0)/n2, n = 2, 4, 6, 8,...,
or letting n = 2m, we obtain
a2m(0) = -a2m-2(0)/(2m)2, m = 1, 2, 3,... .
Thus
2422
26(3 2)2
and, in general,
Hence
(-1)ma0 a2m (0) = 2 2,
2m 22m (m !)2
m = 1, 2, 3,....
^1(x) = a0
m 2m
m1
(-1)mx 22m (m !)2
x > 0.
(6)
(7)
The function in brackets is known as the Bessel function of the first kind of order zero and is denoted by J0(x). It follows from Theorem 5.7.1 that the series converges for all x, and that J0 is analytic at x = 0. Some of the important properties of J0 are discussed in the problems. Figure 5.8.1 shows the graphs of y = J0(x) and some of the partial sums of the series (7).
To determine y2(x) we will calculate a'n (0). The alternative procedure in which we simply substitute the form (23) of Section 5.7 in Eq. (2) and then determine the bn is discussed in Problem 10. First we note from the coefficient of xr+1 in Eq. (4) that (r + 1)2a1(r) = 0. It follows that not only does a1(0) = 0 but also a1 (0) = 0. It is easy to deduce from the recurrence relation (5) that a'3 (0) = a'5 (0) = = a2n+1 (0) =
= 0; hence we need only compute a2m(0), m = 1, 2, 3,.... From Eq. (5) we have
a2m (r ) = a2m-2(r )/(r + 2m)
m = 1, 2, 3,....
282
Chapter 5. Series Solutions of Second Order Linear Equations

n = 4 n = 8 n = 12 n = 16 n= 20
2
1 1 1 1 1 1 1 1 1 1 1 >
2 4 6 8 10 x
J = J0
-1 -
n = 2 n = 6 n = 10 n = 14 n = 18
FIGURE 5.8.1 Polynomial approximations to J0(x). The value of n is the degree of the approximating polynomial.
By solving this recurrence relation we obtain
(1)ma0
a2m (r) =------------------------------------------------------------555-t > m = 1, 2, 3,.... (8)
(r + 2) (r + 4) (r + 2m 2) (r + 2m)
The computation of a2m (r) can be carried out most conveniently by noting that if
f (x) = (x 1) (x 2)2 (x 3) (x an), and if x is not equal to a1,a2,... ,an, then
f'(x) = 1 + _12_ + + _A
f (x) x a1 x a2 Applying this result to a2m (r) from Eq. (8) we find that
a2 m (r) J 1,1, , 1
= -2 ----- + --------7 + - +
a2m (r) V + 2 r + 4 r + 2m
and, setting r equal to 0, we obtain
a2 m (0) = -2
Substituting for a2m (0) from Eq. (6), and letting
a2m (0)-
we obtain, finally,
aL. (0) = H
2 (m!)
(1)ma0
a2m(0) = ~H^^k, m = 1> 2, 3,
5.8 Bessels Equation
283
The second solution of the Bessel equation of order zero is found by setting a0 = 1 and substituting for y1(x) and am (0) = bm (0) in Eq. (3) of Section 5.7. We obtain
ӗ (x) = J0(x) lnx + ^2
(-1)m+1 H
m (m!)
m1
m x m
x > 0.
(10)
Instead of y, the second solution is usually taken to be a certain linear combination of J0 and y. It is known as the Bessel function of the second kind of order zero and is denoted by Y0. Following Copson (Chapter 1), we define13

Y0(x) = -[y(x) + (Y - ln)J0(x)].
(11)
Here y is a constant, known as the Euler-Mascheroni (1750-1800) constant; it is defined by the equation
Y = lim (Hn - lnn) = 0.577.
Substituting for y(x) in Eq. (11), we obtain
Y0(x) =

(y + ln ) J0(x) + J2
, , (-1)m + 1 Hm x m
+ ?^ m (m!) x
x > 0.
(1)
(13)
The general solution of the Bessel equation of order zero for x > 0 is
= c1 J0(x) + c Y0(x).
Note that J0(x) ^ 1 as x ^ 0 and that Y0 (x) has a logarithmic singularity at x = 0; that is, Y0(x) behaves as (/n) ln x when x ^ 0 through positive values. Thus if we are interested in solutions of Bessels equation of order zero that are finite at the origin, which is often the case, we must discard Y0. The graphs of the functions J0 and Y0 are shown in Figure 5.8..
It is interesting to note from Figure 5.8. that for x large both J0(x) and Y0(x) are oscillatory. Such a behavior might be anticipated from the original equation; indeed it
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