7.1 Introduction
5.6Series Solutions near a Regular Singular Point, Part I
1
2
5.6 Series Solutions near a Regular Singular Point, Part I
We now consider the question of solving the general second order linear equation
ODE
P(x)y + Q(x)y + R(x)y = 0 (1)
in the neighborhood of a regular singular point x = x . For convenience we assume
0
that x = 0. If x = 0, the equation can be transformed into one for which the regular
0
0
singular point is at the origin by letting x − x equal t.
0
The fact that x = 0 is a regular singular point of Eq. (1) means that x Q(x)/P(x) = x p(x) and x2 R(x)/P(x) = x2q(x) have finite limits as x → 0, and are analytic at x = 0. Thus they have convergent power series expansions of the form
∞
∞
x p(x) = p xn,x2q(x) = q xn,
(2)
n
n
n=0
n=0
on some interval |x| < ρ about the origin, where ρ > 0. To make the quantities x p(x)
and x2q(x) appear in Eq. (1), it is convenient to divide Eq. (1) by P(x) and then to multiply by x2, obtaining x2 y + x[x p(x)]y + [x2q(x)]y = 0, (3) or x2 y + x( p + p x + · · · + p xn + · · ·)y
0
1
n
+ (q + q x + · · · + q xn + · · ·)y = 0.
(4)
0
1
n
If all of the coefficients p and q are zero except possibly
n
nx Q(x)x2 R(x)p = lim = lim ,
(5)
0
x→0 P (x ) and
q0
x→0
P(x) then Eq. (4) reduces to the Euler equation x2 y + p x y + q y = 0, (6)
0
0
which was discussed in the preceding section. In general, of course, some of the pn and q , n ≥ 1, are not zero. However, the essential character of solutions of Eq. (4) is
n
identical to that of solutions of the Euler equation (6). The presence of the terms p x +
1
· · · + p xn + · · · and q x + · · · + q xn + · · · merely complicates the calculations.
n
1
n
We restrict our discussion primarily to the interval x > 0. The interval x < 0 can be treated, just as for the Euler equation, by making the change of variable x = −ξ and then solving the resulting equation for ξ > 0.
Chapter 5. Series Solutions of Second Order Linear Equations Since the coefficients in Eq. (4) are “Euler coefficients” times power series, it is natural to seek solutions in the form of “Euler solutions” times power series. Thus we assume that
∞
∞
y = xr (a + a x + · · · + a xn + · · ·) = xra xn = a xr+n,
(7)
0
1
n
n
n
n=0
n=0
where a = 0. In other words, r is the exponent of the first term in the series and a is
0
0
its coefficient. As part of the solution we have to determine:
1.
The values of r for which Eq. (1) has a solution of the form (7).
2.
The recurrence relation for the coefficients a .
n
∞
3.
The radius of convergence of the series a xn.
n
n=0
The general theory is due to Frobenius10 and is fairly complicated. Rather than trying to present this theory we simply assume in this and the next two sections that there does exist a solution of the stated form. In particular, we assume that any power series in an expression for a solution has a nonzero radius of convergence, and concentrate on showing how to determine the coefficients in such a series. To illustrate the method of Frobenius we first consider an example.
Solve the differential equation E X A M P L E
1
2x2 y − x y + (1 + x)y = 0.
(8)
It is easy to show that x = 0 is a regular singular point of Eq. (8). Further, x p(x) = −1/2 and x2q(x) = (1 + x)/2. Thus p = −1/2, q = 1/2, q = 1/2, and all other
0
0
1
p’s and q’s are zero. Then, from Eq. (6), the Euler equation corresponding to Eq. (8) is 2x2 y − x y + y = 0.
(9) To solve Eq. (8) we assume that there is a solution of the form (7). Then y and y are given by
∞
y = a (r + n)xr+n−1
(10)
n
n=0
and
∞
y = a (r + n)(r + n − 1)xr+n−2.
(11)
n
n=0
By substituting the expressions for y, y, and y in Eq. (8) we obtain
∞
2x2 y − x y + (1 + x)y = 2a (r + n)(r + n − 1)xr+n
n
n=0
∞
∞
∞
−
a (r + n)xr+n + a xr+n + a xr+n+1.
(12)
n
n
n
n=0
n=0
n=0 10Ferdinand Georg Frobenius (1849 –1917) was (like Fuchs) a student and eventually a professor at the University of Berlin. He showed how to construct series solutions about regular singular points in 1874. His most distinguished work, however, was in algebra where he was one of the foremost early developers of group theory.
5.6Series Solutions near a Regular Singular Point, Part I
∞
The last term in Eq. (12) can be written as axr+n, so by combining the terms in
n−1
n=1 Eq. (12) we obtain 2x2 y − x y + (1 + x)y = a [2r(r − 1) − r + 1]xr
0
∞
+
{[2(r + n)(r + n − 1) − (r + n) + 1]a + a }xr+n = 0. (13)
n
n−1
n=1
If Eq. (13) is to be satisfied for all x, the coefficient of each power of x in Eq. (13) must be zero. From the coefficient of xr we obtain, since a = 0, 0
2r (r − 1) − r + 1 = 2r2 − 3r + 1 = (r − 1)(2r − 1) = 0.
(14)
Equation (14) is called the indicial equation for Eq. (8). Note that it is exactly the polynomial equation we would obtain for the Euler equation (9) associated with Eq. (8).
The roots of the indicial equation are r = 1,r = 1/2.
(15)
1
2
These values of r are called the exponents at the singularity for the regular singular point x = 0. They determine the qualitative behavior of the solution (7) in the neighborhood of the singular point.
Now we return to Eq. (13) and set the coefficient of xr+n equal to zero. This gives the relation [2(r + n)(r + n − 1) − (r + n) + 1]a + a = 0
(16)
n
n−1
or
aa = −
n−1
n
2(r + n)2 − 3(r + n) + 1 a
= −
n−1
,n ≥ 1.
(17)
[(r + n) − 1][2(r + n) − 1]
For each root r and r of the indicial equation we use the recurrence relation (17) to
1
2
determine a set of coefficients a , a , . . . . For r = r = 1, Eq. (17) becomes
1
2
1
aa = −
n−1
,n ≥ 1.
n(2n + 1)n
Thus
aa = − 0 , 1
3 · 1 a
aa = − 1 =
0
,
2
5 · 2 (3 · 5)(1 · 2) and
a
aa = − 2 = −
0 .
3
7 · 3 (3 · 5 · 7)(1 · 2 · 3)
In general we have (−1)na = a ,n ≥ 1.
(18)
n [3 · 5 · 7 · · · (2n + 1)]n! 0
Chapter 5. Series Solutions of Second Order Linear Equations
Hence, if we omit the constant multiplier a , one solution of Eq. (8) is
0
∞
(−1)nxny (x) = x 1 + ,
x > 0.
(19)
1 [3 · 5 · 7 · · · (2n + 1)]n!
n=1
To determine the radius of convergence of the series in Eq. (19) we use the ratio test:
a
xn+1
|x| lim n+1
= 0 n→∞
a xn = lim n→∞ (2n + 3)(n + 1)
n
for all x. Thus the series converges for all x.
Corresponding to the second root r = r = 1 , we proceed similarly. From Eq. (17)
2
2
we have a
aa = −
n−1
= −
n−1
,n ≥ 1.
n
2n(n − 1 )n(2n − 1)
2
Hence aa = − 0 , 1
1 · 1 a
aa = − 1 =
0
,
2
2 · 3 (1 · 2)(1 · 3)a
aa = − 2 = −
0
,
3
3 · 5 (1 · 2 · 3)(1 · 3 · 5) and in general (−1)na = a ,n ≥ 1.
(20)
n
n![1 · 3 · 5 · · · (2n − 1)] 0
Again omitting the constant multiplier a , we obtain the second solution
0
∞
(−1)nxny (x) = x1/2 1 + ,
x > 0.
(21)
2
n![1 · 3 · 5 · · · (2n − 1)]
n=1
As before, we can show that the series in Eq. (21) converges for all x. Since the leading terms in the series solutions y and y are x and x1/2, respectively, it follows that the
1
2
solutions are linearly independent. Hence the general solution of Eq. (8) is y = c y (x) + c y (x),x > 0.
1 1
2 2
The preceding example illustrates that if x = 0 is a regular singular point, then sometimes there are two solutions of the form (7) in the neighborhood of this point.
Similarly, if there is a regular singular point at x = x , then there may be two solutions
0
of the form
∞
y = (x − x )ra (x − x )n
(22)
0
n
0
n=0
that are valid near x = x . However, just as an Euler equation may not have two
0
solutions of the form y = xr , so a more general equation with a regular singular point may not have two solutions of the form (7) or (22). In particular, we show in the next section that if the roots r and r of the indicial equation are equal, or differ by an
1
2
integer, then the second solution normally has a more complicated structure. In all cases, though, it is possible to find at least one solution of the form (7) or (22); if r1 and r differ by an integer, this solution corresponds to the larger value of r . If there is
2
only one such solution, then the second solution involves a logarithmic term, just as for the Euler equation when the roots of the characteristic equation are equal. The method of reduction of order or some other procedure can be invoked to determine the second solution in such cases. This is discussed in Sections 5.7 and 5.8.
If the roots of the indicial equation are complex, then they cannot be equal or differ by an integer, so there are always two solutions of the form (7) or (22). Of course, these solutions are complex-valued functions of x. However, as for the Euler equation, it is possible to obtain real-valued solutions by taking the real and imaginary parts of the complex solutions.
Finally, we mention a practical point. If P, Q, and R are polynomials, it is often much better to work directly with Eq. (1) than with Eq. (3). This avoids the necessity of expressing x Q(x)/P(x) and x2 R(x)/P(x) as power series. For example, it is more convenient to consider the equation x(1 + x)y + 2y + x y = 0 than to write it in the form x2 y + 2x y + x2 y = 0, 1 + x 1 + x which would entail expanding 2x/(1 + x) and x2/(1 + x) in power series.
PROBLEMS
In each of Problems 1 through 10 show that the given differential equation has a regular singular point at x = 0. Determine the indicial equation, the recurrence relation, and the roots of the indicial equation. Find the series solution (x > 0) corresponding to the larger root. If the roots are unequal and do not differ by an integer, find the series solution corresponding to the smaller root also.
9
4
12. The Chebyshev equation is
Chapter 5. Series Solutions of Second Order Linear Equations 13. The Laguerre11 differential equation is
m
14. The Bessel equation of order zero is
1
2
∞
.
0
0
of the first kind of order zero.
15. Referring to Problem 14, use the method of reduction of order to show that the second
2
0
d x
.
2
0
0
0
16. The Bessel equation of order one is
1
2
∞
.
1
2
1
∞
b xn,
n
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[x p(x)]y + [x2q(x)]y = 0, (1)
11Edmond Nicolas Laguerre (1834 –1886), a French geometer and analyst, studied the polynomials named for him about 1879.
5.7Series Solutions near a Regular Singular Point, Part II where
∞
∞
x p(x) = p xn,x2q(x) = q xn,
(2)
n
n
n=0
n=0
and both series converge in an interval |x| < ρ for some ρ > 0. The point x = 0 is a regular singular point, and the corresponding Euler equation is x2 y + p x y + q y = 0.
(3)
0
0
We seek a solution of Eq. (1) for x > 0 and assume that it has the form
∞
∞
y = φ(r, x) = xra xn = a xr+n,
(4)
n
n
n=0
n=0
where a = 0, and we have written y = φ(r, x) to emphasize that φ depends on r as
0
well as x. It follows that
∞
∞
y = (r + n)a xr+n−1,y = (r + n)(r + n − 1)a xr+n−2.
(5)
n
n
n=0
n=0
Then, substituting from Eqs. (2), (4), and (5) in Eq. (1) gives a r (r − 1)xr + a (r + 1)r xr+1 + · · · + a (r + n)(r + n − 1)xr+n + · · ·
0
1
n
+ (p + p x + · · · + p xn + · · ·)
0
1
n
× [a rxr + a (r + 1)xr+1 + · · · + a (r + n)xr+n + · · ·]
0
1
n
+ (q + q x + · · · + q xn + · · ·)
0
1
n
× (a xr + a xr+1 + · · · + a xr+n + · · ·) = 0.
0
1
n
Multiplying the infinite series together and then collecting terms, we obtain a F (r)xr + [a F(r + 1) + a ( p r + q )]xr+1
0
1
0
1
1
+ {a F(r + 2) + a (p r + q ) + a [p (r + 1) + q ]}xr+2
2
0
2
2
1
1
1
+ · · · + {a F(r + n) + a (p r + q ) + a [p(r + 1) + q
]
n
0
n
n
1
n−1
n−1
+ · · · + a [ p (r + n − 1) + q ]}xr+n + · · · = 0,
n−1
1
1
or in a more compact form, L[φ](r, x) = a F(r)xr
0
∞
n−1
+
F (r + n)a + a [(r + k) p + q ] xr+n = 0, (6)
n
k
n−k
n−k
n=1
k=0
where F (r) = r(r − 1) + p r + q .
(7)
0
0
For Eq. (6) to be satisfied identically the coefficient of each power of x must be zero.
Since a = 0, the term involving xr yields the equation F(r) = 0. This equation
0
is called the 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 r and r with r ≥ r if the roots are real. If the roots are
1
2
1
2
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 r and r are called the
1
2
Chapter 5. Series Solutions of Second Order Linear Equations 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
n−1
F (r + n)a + a [(r + k) p + q ] = 0,n ≥ 1.
(8)
n
k
n−k
n−k
k=0
Equation (8) shows that, in general, a depends on the value of r and all the pre n
ceding coefficients a , a , . . . , a . It also shows that we can successively compute
0
1
n−1
a , a , . . . , a , . . . in terms of a and the coefficients in the series for x p(x) and x2q(x)
1
2
n
0
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 = r and r = r ; since r ≥ r , it follows that r + n is
1
2
1
2
1
not equal to r or r for n ≥ 1. Consequently, F(r + n) = 0 for n ≥ 1. Hence we can
1
2
1
always determine one solution of Eq. (1) in the form (4), namely,
∞
y (x) = xr1 1 + a (r )xn ,x > 0.
(9)
1
n
1
n=1
Here we have introduced the notation a (r ) to indicate that a has been determined
n
1
n from Eq. (8) with r = r . To specify the arbitrary constant in the solution we have
1
taken a to be 1.
0
If r is not equal to r , and r − r is not a positive integer, then r + n is not equal
2
1
1
2
2
to r for any value of n ≥ 1; hence F(r + n) = 0, and we can also obtain a second
1
2
solution
∞
y (x) = xr2 1 + a (r )xn ,x > 0.
(10)
2
n
2
n=1
Just as for the series solutions about ordinary points discussed in Section 5.3, the series in Eqs. (9) and (10) converge at least in the interval |x| < ρ where the series for both x p(x) and x2q(x) converge. Within their radii of convergence, the power series
∞
∞
1 + a (r )xn and 1 + a (r )xn define functions that are analytic at x = 0. Thus
n
1
n
2
n=1
n=1
the singular behavior, if there is any, of the solutions y and y is due to the factors xr1
1
2
and xr2 that multiply these two analytic functions. Next, to obtain real-valued solutions for x < 0, we can make the substitution x = −ξ with ξ > 0. As we might expect from our discussion of the Euler equation, it turns out that we need only replace xr1 in Eq. (9) and xr2 in Eq. (10) by |x|r1 and |x|r2, respectively. Finally, note that if r and r
1
2
are complex numbers, then they are necessarily complex conjugates and r = r + N .
2
1
Thus, in this case we can always find two series solutions of the form (4); however, they are complex-valued functions of x. Real-valued solutions can be obtained by taking the real and imaginary parts of the complex-valued solutions. The exceptional cases in which r = r or r − r = N , where N is a positive integer, require more discussion
1
2
1
2
and will be considered later in this section.
It is important to realize that r and r , the exponents at the singular point, are easy
1
2
to find and that they determine the qualitative behavior of the solutions. To calculate r1 and r it is only necessary to solve the quadratic indicial equation
2
r (r − 1) + p r + q = 0, (11)
0
0
whose coefficients are given by p = lim x p(x),q = lim x2q(x).
(12)
0
0
x→0
x→0
Note that these are exactly the limits that must be evaluated in order to classify the singularity as a regular singular point; thus they have usually been determined at an earlier stage of the investigation.
Further, if x = 0 is a regular singular point of the equation P(x)y + Q(x)y + R(x)y = 0, (13)
where the functions P, Q, and R are polynomials, then x p(x) = x Q(x)/P(x) and x2q(x) = x2 R(x)/P(x). Thus Q(x)
R(x)p = lim x,q = lim x2 .
(14)
0
0
x→0
P(x)
x→0
P(x) Finally, the radii of convergence for the series in Eqs. (9) and (10) are at least equal to the distance from the origin to the nearest zero of P other than x = 0 itself.
Discuss the nature of the solutions of the equation E X A M P L E
1
2x(1 + x)y + (3 + x)y − x y = 0 near the singular points.
This equation is of the form (13) with P(x) = 2x(1 + x), Q(x) = 3 + x, and R(x) = −x. The points x = 0 and x = −1 are the only singular points. The point x = 0 is a regular singular point, since Q(x) 3 + x lim x = lim x = 3,
x→0
P(x)
x→0
2x(1 + x)
2
R(x)
−x
lim x2 = lim x2 = 0.
x→0
P(x)
x→0
2x(1 + x) Further, from Eq. (14), p = 3 and q = 0. Thus the indicial equation is r(r − 1) +
0
2
0
3 r = 0, and the roots are r = 0, r = −1. Since these roots are not equal and do not
2
1
2
2
differ by an integer, there are two linearly independent solutions of the form
∞
∞
y (x) = 1 + a (0)xn and y (x) = |x|−1/2 1 + a (− 1 )xn
1
n
2
n
2
n=1
n=1
for 0 < |x| < ρ. A lower bound for the radius of convergence of each series is 1, the distance from x = 0 to x = −1, the other zero of P(x). Note that the solution y is bounded as x → 0, indeed is analytic there, and that the second solution y is
1
2
unbounded as x → 0.
The point x = −1 is also a regular singular point, since (x + 1)(3 + x) lim (x + 1) Q(x) = lim
= −1,x→−1
P(x)
x→−1
2x(1 + x)R(x)(x + 1)2(−x) lim (x + 1)2 = lim = 0.
x→−1
P(x)
x→−1
2x(1 + x)
Chapter 5. Series Solutions of Second Order Linear Equations
In this case p = −1, q = 0, so the indicial equation is r(r − 1) − r = 0. The roots
0
0
of the indicial equation are r = 2 and r = 0. Corresponding to the larger root there
1
2
is a solution of the form
∞
y (x) = (x + 1)2 1 + a (2)(x + 1)n .
1
n
n=1
The series converges at least for |x + 1| < 1 and y is an analytic function there. Since
1
the two roots differ by a positive integer, there may or may not be a second solution of the form
∞
y (x) = 1 + a (0)(x + 1)n.
2
n
n=1
We cannot say more without further analysis.
Observe that no complicated calculations were required to discover the information about the solutions presented in this example. All that was needed was to evaluate a few limits and solve two quadratic equations.
We now consider the cases in which the roots of the indicial equation are equal, or differ by a positive integer, r − r = N . As we have shown earlier, there is always one
1
2
solution of the form (9) corresponding to the larger root r of the indicial equation. By
1
analogy with the Euler equation, we might expect that if r = r , then the second solu 1
2
tion contains a logarithmic term. This may also be true if the roots differ by an integer.
Equal Roots.
The method of finding the second solution is essentially the same as the one we used in finding the second solution of the Euler equation (see when the roots of the indicial equation were equal. We consider r to be a continuous variable and determine a as a function of r by solving the recurrence relation (8). For
n this choice of a (r) for n ≥ 1, Eq. (6) reduces to
n
L[φ](r, x) = a F(r)xr = a (r − r )2xr , (15)
0
0
1
since r is a repeated root of F (r). Setting r = r in Eq. (15), we find that L[φ](r , x) =
1
1
1
0; hence, as we already know, y (x) given by Eq. (9) is one solution of Eq. (1). But
1
more important, it also follows from Eq. (15), just as for the Euler equation, that
∂φ
∂
L
( ,)2
∂
r
x) = a [xr (r − r
]
r
1
0 ∂r
1
r =r1
=
a [(r − r )2xr ln x + 2(r − r )xr ] = 0.
(16)
0
1
1
r =r1 Hence, a second solution of Eq. (1) is
∂φ(
r, x)
∂
∞
y (x) =
=
xra + a (r)xn
2
∂r
0
n
r =r
∂r
1
n=1 r =r1
∞
∞
= (xr1 ln x) a + a (r )xn + xr1 a (r )xn
0
n
1
n
1
n=1
n=1
∞
= y (x) ln x + xr1 a (r )xn,x > 0,
(17)
1
n
1
n=1
5.7Series Solutions near a Regular Singular Point, Part II where a (r ) denotes da /dr evaluated at r = r .
n
1
n
1
It may turn out that it is difficult to determine a (r) as a function of r from the n
recurrence relation (8) and then to differentiate the resulting expression with respect to r . An alternative is simply to assume that y has the form of Eq. (17), that is,
∞
y = y (x) ln x + xr1 b xn,x > 0,
(18)
1
n
n=1
where y (x) has already been found. The coefficients b are calculated, as usual, by
1
n
substituting into the differential equation, collecting terms, and setting the coefficient of each power of x equal to zero. A third possibility is to use the method of reduction of order to find y (x) once y (x) is known.
2
1
Roots Differing by an Integer.
For this case the derivation of the second solution is considerably more complicated and will not be given here. The form of this solution is stated in Eq. (24) in the following theorem. The coefficients c (r ) in Eq. (24) are
n
2
given by c (r ) = d [(r − r )a (r)] ,n = 1, 2, . . . ,
(19)
n
2
dr
2
n
r =r2 where a (r) is determined from the recurrence relation (8) with a = 1. Further, the n
0
coefficient a in Eq. (24) is a = lim (r − r )a (r).
(20)
r →r
2
N
2
If a (r ) is finite, then a = 0 and there is no logarithmic term in y . A full derivation
N
2
2
of the formulas (19), (20) may be found in Coddington (Chapter 4).
In practice the best way to determine whether a is zero in the second solution is simply to try to compute the a corresponding to the root r and to see whether it is
n
2
possible to determine a (r ). If so, there is no further problem. If not, we must use the
N
2
form (24) with a = 0.
When r − r = N , there are again three ways to find a second solution. First, we
1
2
can calculate a and c (r ) directly by substituting the expression (24) for y in Eq. (1).
n
2
Second, we can calculate c (r ) and a of Eq. (24) using the formulas (19) and (20). If
n
2
this is the planned procedure, then in calculating the solution corresponding to r = r be
1
sure to obtain the general formula for a (r) rather than just a (r ). The third alternative
n
n
1
is to use the method of reduction of order.
Theorem 5.7.1
Consider the differential equation (1), x2 y + x[x p(x)]y + [x2q(x)]y = 0, where x = 0 is a regular singular point. Then x p(x) and x2q(x) are analytic at x = 0 with convergent power series expansions
∞
∞
x p(x) = p xn,x2q(x) = q xn
n
n
n=0
n=0
Chapter 5. Series Solutions of Second Order Linear Equations for |x| < ρ, where ρ > 0 is the minimum of the radii of convergence of the power series for x p(x) and x2q(x). Let r and r be the roots of the indicial equation
1
2
F (r) = r(r − 1) + p r + q = 0, 0
0
with r ≥ r if r and r are real. Then in either of the intervals −ρ < x < 0 or
1
2
1
2
0 < x < ρ, there exists a solution of the form
∞
y (x) = |x|r1 1 + a (r )xn , (21)
1
n
1
n=1
where the a (r ) are given by the recurrence relation (8) with a = 1 and r = r .
n
1
0
1
If r − r is not zero or a positive integer, then in either of the intervals −ρ < x < 0
1
2
or 0 < x < ρ, there exists a second linearly independent solution of the form
∞
y (x) = |x|r2 1 + a (r )xn .
(22)
2
n
2
n=1
The a (r ) are also determined by the recurrence relation (8) with a = 1 and r = r .
n
2
0
2
The power series in Eqs. (21) and (22) converge at least for |x| < ρ.
If r = r , then the second solution is
1
2
∞
y (x) = y (x) ln |x| + |x|r1 b (r )xn.
(23)
2
1
n
1
n=1
If r − r = N , a positive integer, then
1
2
∞
y (x) = ay (x) ln |x| + |x|r2 1 + c (r )xn .
(24)
2
1
n
2
n=1
The coefficients a (r ), b (r ), c (r ), and the constant a can be determined by
n
1
n
1
n
2
substituting the form of the series solutions for y in Eq. (1). The constant a may turn out to be zero, in which case there is no logarithmic term in the solution (24). Each of the series in Eqs. (23) and (24) converges at least for |x| < ρ and defines a function that is analytic in some neighborhood of x = 0.
PROBLEMS
In each of Problems 1 through 12 find all the regular singular points of the given differential equation. Determine the indicial equation and the exponents at the singularity for each regular singular point.
2
In each of Problems 13 through 17:
5.7Series Solutions near a Regular Singular Point, Part II (c) Find the first three nonzero terms in each of two linearly independent solutions about
see Problem 1
see Problem 3
2
∞
Determine the first three nonzero terms in the series n
19. In several problems in mathematical physics (for example, the Schr¨odinger equation for a hydrogen atom) it is necessary to study the differential equation (i)
one solution of (i) is
αβ
1
x
2
+
.
(
(e) Show that the point at infinity is a regular singular point, and that the roots of the 20. Consider the differential equation x3 y + αx y + βy = 0, where α and β are real constants and α = 0.
∞
(b) By attempting to determine a solution of the form n
Chapter 5. Series Solutions of Second Order Linear Equations 21. Consider the differential equation α
β
y +
y + y = 0, (i)
xs
xt
where α = 0 and β = 0 are real numbers, and s and t are positive integers that for the moment are arbitrary.
(a) Show that if s > 1 or t > 2, then the point x = 0 is an irregular singular point.
(b) Try to find a solution of Eq. (i) of the form
∞
(ii)
n
(c) Show that if s = 1 and t = 3, then there are no solutions of Eq. (i) of the form (ii).
As a note of caution we should point out that while it is sometimes possible to obtain a formal series solution of the form (ii) at an irregular singular point, the series may not have a positive radius of convergence. See Problem 20 for an example.
5.8 Bessel’s Equation
In this section we consider three special cases of Bessel’s12 equation, x2 y
ODE
+ xy + (x2 − ν2)y = 0, (1)
where ν is a constant, which illustrate the theory discussed in Section 5.7. It is easy to show that x = 0 is a regular singular point. For simplicity we consider only the case x > 0.
Bessel Equation of Order Zero.
This example illustrates the situation in which the roots of the indicial equation are equal. Setting ν = 0 in Eq. (1) gives L[y] = x2 y + x y + x2 y = 0.
(2)
Substituting
∞
y = φ(r, x) = a xr + a xr+n,
(3)
0
n
n=1 12Friedrich Wilhelm Bessel (1784 –1846) embarked on a career in business as a youth, but soon became interested in astronomy and mathematics. He was appointed director of the observatory at K¨onigsberg in 1810 and held this position until his death. His study of planetary perturbations led him in 1824 to make the first systematic analysis of the solutions, known as Bessel functions, of Eq. (1). He is also famous for making the first accurate determination (1838) of the distance from the earth to a star.
5.8Bessel’s Equation we obtain
∞
∞
L[φ](r, x) = a [(r + n)(r + n − 1) + (r + n)]xr+n + a xr+n+2 n
n
n=0
n=0
= a [r(r − 1) + r]xr + a [(r + 1)r + (r + 1)]xr+1
0
1
∞
+
{a [(r + n)(r + n − 1) + (r + n)] + a }xr+n = 0.
(4)
n
n−2
n=2
The roots of the indicial equation F (r) = r(r − 1) + r = 0 are r = 0 and r = 0;
1
2
hence we have the case of equal roots. The recurrence relation is a
(r)a
(r)a (r) = −
n−2
= − n−2 ,n ≥ 2.
(5)
n(r + n)(r + n − 1) + (r + n)(r + n)2
To determine y (x) we set r equal to 0. Then from Eq. (4) it follows that for the
1
coefficient of xr+1 to be zero we must choose a = 0. Hence from Eq. (5), a = a =
1
3
5
a = · · · = 0. Further,
7
a (0) = −a(0)/n2,n = 2, 4, 6, 8, . . . ,
n
n−2
or letting n = 2m, we obtain a(0) = −a(0)/(2m)2,m = 1, 2, 3, . . . .
2m
2m−2
Thus
a
a
aa (0) = − 0 ,a (0) = 0 , a (0) = −
0
,
2
22
4
2422
6
26(3 · 2)2 and, in general, (−1)ma
a(0) = 0 ,m = 1, 2, 3, . . . .
(6)
2m
22m (m!)2
Hence
∞ (−1)mx2my (x) = a 1 + ,
x > 0.
(7)
1
0
22m (m!)2
m=1
The function in brackets is known as the Bessel function of the first kind of order zero and is denoted by J (x). It follows from Theorem 5.7.1 that the series converges
0
for all x, and that J is analytic at x = 0. Some of the important properties of J are
0
0
discussed in the problems. Figure 5.8.1 shows the graphs of y = J (x) and some of
0
the partial sums of the series (7).
To determine y (x) we will calculate a (0). The alternative procedure in which we
2
n
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)2a (r) = 0. It follows that not only does a (0) = 0 but also a (0) = 0. It is
1
1
1
easy to deduce from the recurrence relation (5) that a (0) = a (0) = · · · = a(0) =
3
5
2n+1
· · · = 0; hence we need only compute a (0), m = 1, 2, 3, . . . . From Eq. (5) we have
2m
a(r) = −a(r)/(r + 2m)2,m = 1, 2, 3, . . . .
2m
2m−2
Chapter 5. Series Solutions of Second Order Linear Equations
y
n = 4 n = 8 n = 12 n = 16 n = 20
2
1
2
4
6
8
10
x
y = J0(x)
–1
n = 2 n = 6 n = 10 n = 14 n = 18
FIGURE 5.8.1 Polynomial approximations to J (x). The value of n is the degree of the
0
approximating polynomial.
By solving this recurrence relation we obtain (−1)ma
a(r) =
0
,m = 1, 2, 3, . . . .
(8)
2m(r + 2)2(r + 4)2 · · · (r + 2m − 2)2(r + 2m)2
The computation of a (r) can be carried out most conveniently by noting that if
2m
f (x) = (x − α )β1(x − α )β2(x − α )β3 · · · (x − α )βn , 1
2
3
n
and if x is not equal to α , α , . . . , α , then
1
2
n
f (x)β
β
β
=
1
+
2
+ · · · +
n
.
f (x)x − αx − αx − α
1
2
n
Applying this result to a(r) from Eq. (8) we find that
2m
a (r)
1
2m
= −2 + 1 + · · · +
1
,
a
(r)r + 2 r + 4 r + 2m
2m
and, setting r equal to 0, we obtain
1
a (0) = −2 + 1 + · · · + 1 a (0).
2m
2
4
2m
2m
Substituting for a(0) from Eq. (6), and letting
2mH = 1 + 1 + 1 + · · · + 1 , (9)
m
2
3
m
we obtain, finally, (−1)maa (0) = −H 0 ,m = 1, 2, 3, . . . .
2m
m 22m(m!)2 and substituting for y (x) and a (0) = b (0) in Eq. (23) of Section 5.7. We obtain
1
2m
2m
∞ (−1)m+1Hy (x) = J (x) ln x + m x2m,x > 0.
(10)
2
0
22m (m!)2
m=1
Instead of y , the second solution is usually taken to be a certain linear combination
2
of J and y . It is known as the Bessel function of the second kind of order zero and is
0
2
denoted by Y . Following Copson (Chapter 12), we define13
0
Y (x) = 2 (x) + (γ − ln 2)J (x)].
(11)
0
π [y2
0
Here γ is a constant, known as the Euler–Ma´scheroni (1750–1800) constant; it is defined by the equation γ = lim (H − ln n) ∼ = 0.5772.
(12)
n→∞
n
Substituting for y (x) in Eq. (11), we obtain
2
x
∞ (−1)m+1HY (x) = 2 γ + ln J (x) + m x2m ,x > 0.
(13)
0
π
2
0
22m(m!)2
m=1
The general solution of the Bessel equation of order zero for x > 0 is y = c J (x) + c Y (x).
1 0
2 0
Note that J (x) → 1 as x → 0 and that Y (x) has a logarithmic singularity at x = 0;
0
0
that is, Y (x) behaves as (2/π) ln x when x → 0 through positive values. Thus if we
0
are interested in solutions of Bessel’s equation of order zero that are finite at the origin, which is often the case, we must discard Y . The graphs of the functions J and Y are
0
0
0
shown in Figure 5.8.2.
It is interesting to note from Figure 5.8.2 that for x large both J (x) and Y (x) are
0
0
oscillatory. Such a behavior might be anticipated from the original equation; indeed it y
1
y = J0(x)
0.5
y = Y0(x)
2
4
6
8
10
12
14
x
–0.5
FIGURE 5.8.2 The Bessel functions of order zero.
13Other authors use other definitions for Y . The present choice for Y is also known as the Weber (1842–1913)
0
0
function.
Chapter 5. Series Solutions of Second Order Linear Equations is true for the solutions of the Bessel equation of order ν. If we divide Eq. (1) by x2, we obtain
ν2
y + 1 y + 1 − y = 0.
x
x2
For x very large it is reasonable to suspect that the terms (1/x)y and (ν2/x2)y are small and hence can be neglected. If this is true, then the Bessel equation of order ν
can be approximated by y + y = 0.
The solutions of this equation are sin x and cos x; thus we might anticipate that the solutions of Bessel’s equation for large x are similar to linear combinations of sin x
and cos x. This is correct insofar as the Bessel functions are oscillatory; however, it is only partly correct. For x large the functions J and Y also decay as x increases; thus
0
0
the equation y + y = 0 does not provide an adequate approximation to the Bessel equation for large x, and a more delicate analysis is required. In fact, it is possible to show that
2
1/2
π
J (x) ∼
=
cos x − as
x → ∞,
(14)
0
πx
4
and that
2
1/2
π
Y (x) ∼
=
sin x − as x → ∞.
(15)
0
πx
4
These asymptotic approximations, as x → ∞, are actually very good. For example, Figure 5.8.3 shows that the asymptotic approximation (14) to J (x) is reasonably
0
accurate for all x ≥ 1. Thus to approximate J (x) over the entire range from zero to
0
infinity, one can use two or three terms of the series (7) for x ≤ 1 and the asymptotic approximation (14) for x ≥ 1.
y
2
Asymptotic approximation: y = (2/ πx)1/2 cos(x – /4)
π
1
y = J0(x)
2
4
6
8
10
x
–1
FIGURE 5.8.3
Asymptotic approximation to J (x).
0
5.8Bessel’s EquationBessel 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 ν = 1 in Eq. (1) gives
2
L[y] = x2 y + x y + x2 − 1 y = 0.
(16)
4
If we substitute the series (3) for y = φ(r, x), we obtain
∞
∞
L[φ](r, x) = (r + n)(r + n − 1) + (r + n) − 1 a xr+n + a xr+n+2 4
n
n
n=0
n=0
= (r2 − 1)a xr + (r + 1)2 − 1 a xr+1
4
0
4
1
∞
+
(r + n)2 − 1 a + axr+n = 0.
(17)
4
n
n−2
n=2
The roots of the indicial equation are r = 1 , r = − 1 ; hence the roots differ by an
1
2
2
2
integer. The recurrence relation is
(r + n)2 − 1 a = −a,n ≥ 2.
(18)
4
n
n−2
Corresponding to the larger root r = 1 we find from the coefficient of xr+1 in Eq. (17)
1
2
that a = 0. Hence, from Eq. (18), a = a = · · · = a = · · · = 0. Further, for
1
3
5
2n+1
r = 1 , 2
aa = −
n−2
,n = 2, 4, 6 . . . ,
n
n(n + 1) or letting n = 2m, we obtain a
a
= −
2m−2
,m = 1, 2, 3, . . . .
2m
2m(2m + 1)
By solving this recurrence relation we find that a
aa = − 0 ,a = 0 , . . .
2
3!
4
5!
and, in general, (−1)ma
a
=
0 ,m = 1, 2, 3, . . . .
2m(2m + 1)!
Hence, taking a = 1, we obtain
0
∞ (−1)mx2m
∞ (−1)mx2m+1 y (x) = x1/2 1 + = x−1/2 ,
x > 0.
(19)
1 (2m + 1)!
(2m + 1)! m=1
m=0
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, J , is defined as (2/π)1/2 y . Thus
1/2
1
2
1/2
J(x) = sin x,x > 0.
(20)
1/2
πx
Chapter 5. Series Solutions of Second Order Linear Equations
Corresponding to the root r = − 1 it is possible that we may have difficulty in
2
2
computing a since N = r − r = 1. However, from Eq. (17) for r = − 1 the coef 1
1
2
2
ficients of xr and xr+1 are both zero regardless of the choice of a and a . Hence a
0
1
0
and a can be chosen arbitrarily. From the recurrence relation (18) we obtain a set of
1
even-numbered coefficients corresponding to a and a set of odd-numbered coefficients
0
corresponding to a . Thus no logarithmic term is needed to obtain a second solution in
1
this case. It is left as an exercise to show that, for r = − 1 , 2
(−1)na(−1)na
a
=
0 ,a
=
1 ,n = 1, 2, . . . .
2n(2n)!
2n+1 (2n + 1)!
Hence
∞ (−1)nx2n
∞ (−1)nx2n+1 y (x) = x−1/2 a + a
2
0
(2n)!
1 (2n + 1)! n=0
n=0
= cos x sin xa + a,
x > 0.
(21)
0 x1/2
1 x1/2
The constant a simply introduces a multiple of y (x). The second linearly independent
1
1
solution of the Bessel equation of order one-half is usually taken to be the solution for which a = (2/π)1/2 and a = 0. It is denoted by J . Then
0
1
−1/2
2
1/2 J− (x) = cos x,x > 0.
(22)
1/2
πx
The general solution of Eq. (16) is y = c J(x) + c J(x).
1 1/2
2 −1/2 By comparing Eqs. (20) and (22) with Eqs. (14) and (15) we see that, except for a phase shift of π/4, the functions J− and J resemble J and Y , respectively, for
1/2
1/2
0
0
large x. The graphs of J and J are shown in Figure 5.8.4.
1/2
−1/2
y
1
J (x)
–1/2
0.5
J (x)
1/2
2
4
6
8
10
12
14
x
–0.5
FIGURE 5.8.4 The Bessel functions J and J .
1/2
−1/2
5.8Bessel’s EquationBessel 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 ν = 1 in Eq. (1) gives L[y] = x2 y + x y + (x2 − 1)y = 0.
(23)
If we substitute the series (3) for y = φ(r, x) and collect terms as in the preceding cases, we obtain L[φ](r, x) = a (r2 − 1)xr + a [(r + 1)2 − 1]xr+1
0
1
∞
+
{[(r + n)2 − 1]a + a }xr+n = 0.
(24)
n
n−2
n=2
The roots of the indicial equation are r = 1 and r = −1. The recurrence relation is
1
2
[(r + n)2 − 1]a (r) = −a(r),n ≥ 2.
(25)
n
n−2
Corresponding to the larger root r = 1 the recurrence relation becomes aa = −
n−2
,n = 2, 3, 4, . . . .
n
(n + 2)n
We also find from the coefficient of xr+1 in Eq. (24) that a = 0; hence from the
1
recurrence relation a = a = · · · = 0. For even values of n, let n = 2m; then
3
5
a
a
a
= −
2m−2
= −
2m−2
,m = 1, 2, 3, . . . .
2m(2m + 2)(2m) 22(m + 1)m
By solving this recurrence relation we obtain (−1)ma
a
=
0
,m = 1, 2, 3, . . . .
(26)
2m
22m(m + 1)!m!
The Bessel function of the first kind of order one, denoted by J , is obtained by choosing
1
a = 1/2. Hence
0
∞
(−1)mx2mJ (x) = x.
(27)
1
2
22m(m + 1)!m!
m=0
The series converges absolutely for all x, so the function J is analytic everywhere.
1
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
∞
y (x) = a J (x) ln x + x−1 1 + c xn ,x > 0.
(28)
2
1
n
n=1
Computing y (x), y(x), substituting in Eq. (23), and making use of the fact that J is
2
2
1
a solution of Eq. (23) give
∞
∞
2ax J (x) + [(n − 1)(n − 2)c + (n − 1)c − c ]xn−1 + c xn+1 = 0, (29)
1
n
n
n
n
n=0
n=0
Chapter 5. Series Solutions of Second Order Linear Equations where c = 1. Substituting for J (x) from Eq. (27), shifting the indices of summation
0
1
in the two series, and carrying out several steps of algebra give
∞
−c + [0 · c + c ]x + [(n2 − 1)c + c ]xn
1
2
0
n+1
n−1
n=2
∞
(−
= − 1)m(2m + 1)x2m+1 a x + .
(30) 22m (m + 1)! m!
m=1
From Eq. (30) we observe first that c = 0, and a = −c = −1. Further, since there
1
0
are only odd powers of x on the right, the coefficient of each even power of x on the left must be zero. Thus, since c = 0, we have c = c = · · · = 0. Corresponding to
1
3
5
the odd powers of x we obtain the recurrence relation [let n = 2m + 1 in the series on the left side of Eq. (30)] (−1)m(2m + 1) [(2m + 1)2 − 1]c + c = ,m = 1, 2, 3, . . . .
(31)
2m+2
2m
22m (m + 1)! m!
When we set m = 1 in Eq. (31), we obtain (32 − 1)c + c = (−1)3/(22 · 2!).
4
2
Notice that c can be selected arbitrarily, and then this equation determines c . Also
2
4
notice that in the equation for the coefficient of x, c appeared multiplied by 0, and
2
that equation was used to determine a. That c is arbitrary is not surprising, since c
2
2
∞
is the coefficient of x in the expression x−1[1 + c xn]. Consequently, c simply
n
2
n=1
generates a multiple of J , and y is only determined up to an additive multiple of J .
1
2
1
In accord with the usual practice we choose c = 1/22. Then we obtain
2
−1
3
−1 c = + 1 = 1 + 1 + 1
4
24 · 2 2 242!
2
(−
= 1) (H + H ).
24 · 2!
2
1
It is possible to show that the solution of the recurrence relation (31) is (−1)m+1(H + H)
c
=
m
m−1 ,m = 1, 2, . . .
2m
22m m!(m − 1)!
with the understanding that H = 0. Thus
0
∞
(−1)m(H + H)
y (x) = −J (x) ln x + 1 1 −
m
m−1
x2m ,x > 0.
(32)
2
1
x
22mm!(m − 1)! m=1
The calculation of y (x) using the alternative procedure [see Eqs. (19) and (20)
2
of Section 5.7] in which we determine the c (r ) is slightly easier. In particular the n
2
latter procedure yields the general formula for c without the necessity of solving a recurrence relation of the form (31) (see Problem 11). In this regard the reader may also wish to compare the calculations of the second solution of Bessel’s equation of order zero in the text and in Problem 10.
5.8Bessel’s Equation The second solution of Eq. (23), the Bessel function of the second kind of order one, Y , is usually taken to be a certain linear combination of J and y . Following
1
1
2
Copson (Chapter 12), Y is defined as
1
Y (x) = 2 (x) + (γ − ln 2)J (x)], (33)
1
π [−y2
1
where γ is defined in Eq. (12). The general solution of Eq. (23) for x > 0 is y = c J (x) + c Y (x).
1 1
2 1
Notice that while J is analytic at x = 0, the second solution Y becomes unbounded in
1
1
the same manner as 1/x as x → 0. The graphs of J and Y are shown in Figure 5.8.5.
1
1
y
1
y = J (x)
1
0.5
y = Y (x)
1
2
4
6
8
10
12
14
x
–0.5
FIGURE 5.8.5 The Bessel functions J and Y .
1
1
PROBLEMS
In each of Problems 1 through 4 show that the given differential equation has a regular singular point at x = 0, and determine two linearly independent solutions for x > 0.
5. Find two linearly independent solutions of the Bessel equation of order 3 ,
2
4
6. Show that the Bessel equation of order one-half, x2 y + x y + (x2 − 1 )y = 0,x > 0, 4
can be reduced to the equation v + v = 0 by the change of dependent variable y = x−1/2v(x). From this conclude that y (x) =
1
x−1/2 cos x and y (x) = x−1/2 sin x are solutions of the Bessel equation of order one-half.
2
0
8. Show directly that the series for J (x), Eq. (27), converges absolutely for all x and that
1
J (x) = −J (x).
0
1
Chapter 5. Series Solutions of Second Order Linear Equations 9. Consider the Bessel equation of order ν, x2 y + x y + (x2 − ν2) = 0,x > 0.
Take ν real and greater than zero.
(a) Show that x = 0 is a regular singular point, and that the roots of the indicial equation are ν and −ν.
(b) Corresponding to the larger root ν, show that one solution is
∞
(−
1)mx 2my (x) = xν 1 + .
1
m!(1 + ν)(2 + ν) · · · (m − 1 + ν)(m + ν) 2
m=1
(c) If 2ν is not an integer, show that a second solution is
∞
(−
1)mx 2my (x) = x−ν 1 + .
2
m!(1 − ν)(2 − ν) · · · (m − 1 − ν)(m − ν) 2
m=1
Note that y (x) → 0 as x → 0, and that y (x) is unbounded as x → 0.
1
2
(d) Verify by direct methods that the power series in the expressions for y (x) and y (x)
1
2
converge absolutely for all x. Also verify that y is a solution provided only that ν is not
2
an integer.
10. In this section we showed that one solution of Bessel’s equation of order zero, L[y] = x2 y + x y + x2 y = 0, is J , where J (x) is given by Eq. (7) with a = 1. According to Theorem 5.7.1 a second
0
0
0
solution has the form (x > 0)
∞
y (x) = J (x) ln x + b xn.
2
0
n
n=1
(a) Show that
∞
∞
∞
L[y ](x) = n(n − 1)b xn + nb xn + b xn+2 + 2x J (x).
(i)
2
n
n
n
0
n=2
n=1
n=1
(b) Substituting the series representation for J (x) in Eq. (i), show that
0
∞
∞ (−1)n2nx2nb x + 22b x2 + (n2b + b)xn = −2 .
(ii)
1
2
n
n−2
22n(n!)2
n=3
n=1
(c) Note that only even powers of x appear on the right side of Eq. (ii). Show that b = b = b = · · · = 0, b = 1/22(1!)2, and that
1
3
5
2
(2n)2b + b = −2(−1)n(2n)/22n(n!)2,n = 2, 3, 4, . . . .
2n
2n−2
Deduce that
b = − 1 1 + 1
and
b =
1
1 + 1 + 1 .
4
2242
2
6
224262
2
3
The general solution of the recurrence relation is b = (−1)n+1H /22n(n!)2. Substituting
2n
n for b in the expression for y (x) we obtain the solution given in Eq. (10).
n
2
11. Find a second solution of Bessel’s equation of order one by computing the c (r ) and an
2
of Eq. (24) of Section 5.7 according to the formulas (19) and (20) of that section. Some guidelines along the way of this calculation are the following. First, use Eq. (24) of this section to show that a (−1) and a (−1) are 0. Then show that c (−1) = 0 and, from the
1
1
1
recurrence relation, that c (−1) = 0 for n = 3, 5, . . . . Finally, use Eq. (25) to show that
n(−1)ma
a(r) =
0
,
2m
(r + 1)(r + 3)2 · · · (r + 2m − 1)2(r + 2m + 1) for m = 1, 2, 3, . . . , and calculate c(−1) = (−1)m+1(H + H)/22mm!(m − 1)!.
2m
m
m−1
12. By a suitable change of variables it is sometimes possible to transform another differential equation into a Bessel equation. For example, show that a solution of
4
13. Using the result of Problem 12 show that the general solution of the Airy equation 1 1 3 2 2 3
1
2
solutions of the Bessel equation of order one-third.
0
j
0
0
j
0
j
x
j
Hence show that
1
if
0
i
0
j
i
j
0
0
i
0
i
0
j
0
i
0
j
REFERENCES Coddington, E. A., An Introduction to Ordinary Differential Equations (Englewood Cliffs, NJ: Pren tice Hall, 1961; New York: Dover, 1989).
Copson, E. T., An Introduction to the Theory of Functions of a Complex Variable (Oxford: Oxford University, 1935).
Proofs of Theorems 5.3.1 and 5.7.1 can be found in intermediate or advanced books; for example, see Chapters 3 and 4 of Coddington, or Chapters 3 and 4 of: Rainville, E. D., Intermediate Differential Equations (2nd ed.) (New York: Macmillan, 1964).
Also see these texts for a discussion of the point at infinity, which was mentioned in Problem 21 of Section 5.4. The behavior of solutions near an irregular singular point is an even more advanced topic; a brief discussion can be found in Chapter 5 of:
Chapter 5. Series Solutions of Second Order Linear Equations Coddington, E. A., and Levinson, N., Theory of Ordinary Differential Equations (New York: McGraw Hill, 1955).
More complete discussions of the Bessel equation, the Legendre equation, and many of the other named equations can be found in advanced books on differential equations, methods of applied mathematics, and special functions. A text dealing with special functions such as the Legendre polynomials and the Bessel functions is: Hochstadt, H., Special Functions of Mathematical Physics (New York: Holt, 1961).
An excellent compilation of formulas, graphs, and tables of Bessel functions, Legendre functions, and other special functions of mathematical physics may be found in: Abramowitz, M., and Stegun, I. A. (eds.), Handbook of Mathematical Functions (New York: Dover, 1965); originally published by the National Bureau of Standards, Washington, DC, 1964.