Jim Lambers
MAT 415/515
Fall Semester 2013-14
Lecture 6 and 7 Notes
These notes correspond to Section 7.5 in the text.
Series Solutions–Frobenius’ Method
We now turn our attention to the solution of a linear, second-order, homogeneous ODE of the form
y 00 + P (x)y 0 + Q(x)y = 0.
Such an ODE has two linearly independent solutions, y1 (x) and y2 (x). If they can be found, then
every solution of the ODE can be expressed in the form
y(x) = c1 y1 (x) + c2 y2 (x),
where c1 and c2 are constants. This form of the solution is called the general solution.
Our goal is to find at least one series solution, which is a solution expressed as a power series
∞
X
y(x) = aj (x − x0 )xj+s ,
j=0
where x0 is the center of the power series and the {aj } are the coefficients. We will use examples
to describe how series solutions can be found.
Example: Linear Oscillator
To demonstrate how a series solution of an ODE can be obtained, we consider the linear oscillator
equation
y 00 + ω 2 y = 0.
Substituting the series
∞
X
y(x) = aj xj+s
j=0
into the equation yields
∞
X
aj [(j + s)(j + s − 1)xj+s−2 + ω 2 xj+s ] = 0.
j=0
Examining the lowest-degree terms, we have
∞
X ∞
X
s−2 s−1 j+s−2 2
a0 s(s − 1)x + a1 s(s + 1)x + aj (j + s)(j + s − 1)x +ω aj xj+s = 0.
j=2 j=0
Shifting the indices in the first summation yields
∞
X
a0 s(s − 1)xs−2 + a1 s(s + 1)xs−1 + [aj+2 (j + s + 2)(j + s + 1) + aj ω 2 ]xj+s = 0.
j=0
1
Because this equation must be satisfied for all x, all of the coefficients in this power series must
equal zero. However, we stipulate that a0 6= 0, because the lowest power of x in the solution has
yet to be determined. It follows that s must satisfy the indicial equation
s(s − 1) = 0.
Therefore, we must have s = 0 or s = 1.
Examining the second term, we see that if s = 0, the coefficient of this term, a0 s(s+1), vanishes
regardless of a1 . However, if s = 1, then we must have a1 = 0. We therefore set a1 = 0 regardless
of s. Subsequent coefficients are determined by the recurrence relation
aj ω 2
aj+2 = − , j ≥ 0.
(j + s + 2)(j + s + 1)
That is, we only have even-numbered coefficients in the series.
We first consider the case of s = 0. We then have, for j = 1, 2, 3,
a0 ω 2
a2 = − ,
2
a2 ω 2 a0 ω 4
a4 = − = ,
12 24
a4 ω 2 a0 ω 6
a6 = − =− .
30 720
From these values, we obtain the formula
a0 ω 2p
a2p = (−1)p , p ≥ 0,
(2p)!
which yields the solution
∞
X (−1)j ω 2j
y(x) = a0 = a0 cos ωx.
(2j)!
j=0
Then, we consider the case of s = 1. We then have, for j = 0, 2, 4,
a0 ω 2
a2 = − ,
6
a2 ω 2 a0 ω 4
a4 = − = ,
20 120
a4 ω 2 a0 ω 6
a6 = − =− .
42 5040
From these values, we obtain the formula
a0 ω 2p
a2p = (−1)p ,
(2p + 1)!
which yields the solution
∞
X (−1)j ω 2j a0
y(x) = a0 = sin ωx.
(2j + 1)! ω
j=0
This approach to obtaining a series solution is known as Frobenius’ method. Once the series
solution is obtained, it should be substituted into the differential equation to confirm that it really is
a solution. Also, it should be verified that the series actually converges for any x-values of interest.
2
Symmetry of Solutions
In the previous example, we obtained series solutions that included only odd powers of x, or even
powers of x. Such solutions are either odd functions (f (−x) = −f (x)) in the case of odd powers
of x, or even functions (f (−x) = f (x)) in the case of even powers of x. Whether this occurs for a
given ODE can be determined without computing the solutions explicitly.
For an ODE of the form
L(x)y(x) = 0,
where L(x) is a linear differential operator, solutions that have symmetry (even or odd) can be
found if the operator L is even; that is,
L(−x) = L(−x).
If this is the case, it follows that if y(x) is a solution of ODE, then y(−x) is also a solution.
If y(x) and y(−x) are linearly dependent, then they are proportional to one another, and
therefore are even (if the constant of proportionality is positive) or odd (if it is negative). If they
are independent, then the linear combinations
y(x) + y(−x), y(x) − y(−x)
are also solutions that are even and odd, respectively. Therefore, if L is even, then it is always
possible to find solutions that are even or odd functions.
Example: Bessel’s Equation
As a second example, we consider Bessel’s equation,
x2 y 00 + xy 0 + (x2 − n2 )y = 0.
Substituting the series
∞
X
y(x) = aj xs+j
j=0
into the equation yields
∞
X
aj [(s + j)(s + j − 1) + (s + j) − n2 ]xj + aj xs+j+2 = 0,
j=0
which can be rewritten as
∞
X ∞
X
2 2 2 2 2 2 s+j
a0 (s − n ) + a1 [(s + 1) − n ] + aj [(s + j) − n ]x + aj xs+j+2 = 0,
j=2 j=0
or, after shifting indices in the first summation,
∞
X
a0 (s2 − n2 ) + a1 [(s + 1)2 − n2 ]xs+1 + aj+2 [(s + j + 2)2 − n2 ] + aj xs+j = 0.
j=0
Requiring that a0 6= 0 yields the indicial equation
s2 − n2 = 0,
3
which has solutions s = ±n. Next, we consider the term with s1 . With s = ±n, we have the
requirement that
a1 (2n + 1) = 0 or a1 (1 − 2n) = 0,
so unless n = ±1/2, we must have a1 = 0. Subsequent coefficients are determined by the recurrence
relation
aj
aj+2 = − , j ≥ 0.
(s + j + 2)2 − n2
In the case of s = n, we obtain
aj
aj+2 = − .
(j + 2)(j + 2 + 2n)
Because a1 = 0, all of the odd-numbered coefficients vanish as well. Substituting even numbers for
j yields
a0
a2 = − ,
4(n + 1)
a2 a0
a4 = − =
8(n + 2) 32(n + 1)(n + 2)
a4 a0
a6 = − =− .
12(n + 3) 384(n + 1)(n + 2)(n + 3)
From these values, we can obtain the formula
a0 n!
a2p = (−1)p 2p
.
2 p!(n + p)!
We conclude that the solution is
∞
X (−1)j n!
y(x) = a0 xn+2j .
22j j!(n + j)!
j=0
Setting a0 = 1/2n n! yields the Bessel function
∞
X (−1)j x n+2j
Jn (x) = .
j!(n + j)! 2
j=0
As Bessel’s equation has symmetry, Jn (x) is an even function if n is even, and an odd function if
n is odd.
When s = −n and n is not an integer, we obtain a second solution, which we denote by J−n (x).
However, if n is an integer in this situation, a division by zero occurs in a2n−2 , so Frobenius’ method
fails.
Regular and Irregular Singularities
As seen in the preceding example, there are situations in which it is not possible to use Frobenius’
method to obtain a series solution. In particular, this can happen if the coefficients P (x) and Q(x)
in the ODE
y 00 + P (x)y 0 + Q(x)y = 0
fail to be defined at a point x0 . We classify a point x0 in the domain of the ODE as follows:
4
• If P (x0 ) and Q(x0 ) are both finite, then x0 is said to be an ordinary point of the ODE.
• If either P or Q becomes infinite as x → x0 , but (x − x0 )P (x) and (x − x0 )2 Q(x) are both
still finite at x0 , then x0 is a regular singular point of the ODE.
• If either (x − x0 )P (x) or (x − x0 )Q(x) becomes infinite as x → x0 , then x0 is an irregular or
essential singular point of the ODE.
For example, Bessel’s equation, rewritten in the form
n2
00 1 0
y + y + 1 − 2 y = 0,
x x
has a singular point at x = 0, as P (x) = 1/x and Q(x) = 1 − n2 /x2 both have a vertical asymptote
at x = 0. However, both xP (x) = 1 and x2 Q(x) = x2 − n2 are finite at x = 0, so the singular point
is regular. The linear oscillator equation, on the other hand, does not have any singular points.
A result known as Fuchs’ Theorem states that if x0 is not an essential singularity point of an
ODE, then it is always possible to obtain at least one series solution of the ODE using Frobenius’
method. However, it should be noted that the series may diverge at a point x that is equidistant
from x0 as a singularity.
