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Jim Lambers MAT 415/ Fall Semester 2013- 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′′^ + P (x)y′^ + Q(x)y = 0.
Such an ODE has two linearly independent solutions, y 1 (x) and y 2 (x). If they can be found, then every solution of the ODE can be expressed in the form
y(x) = c 1 y 1 (x) + c 2 y 2 (x),
where c 1 and c 2 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
y(x) =
∑^ ∞
j=
aj (x − x 0 )xj+s,
where x 0 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′′^ + ω^2 y = 0.
Substituting the series
y(x) =
∑^ ∞
j=
aj xj+s
into the equation yields
∑^ ∞
j=
aj [(j + s)(j + s − 1)xj+s−^2 + ω^2 xj+s] = 0.
Examining the lowest-degree terms, we have
a 0 s(s − 1)xs−^2 + a 1 s(s + 1)xs−^1 +
∑^ ∞
j=
aj (j + s)(j + s − 1)xj+s−^2 + ω^2
∑^ ∞
j=
aj xj+s^ = 0.
Shifting the indices in the first summation yields
a 0 s(s − 1)xs−^2 + a 1 s(s + 1)xs−^1 +
∑^ ∞
j=
[aj+2(j + s + 2)(j + s + 1) + aj ω^2 ]xj+s^ = 0.
Because this equation must be satisfied for all x, all of the coefficients in this power series must equal zero. However, we stipulate that a 0 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, a 0 s(s+1), vanishes regardless of a 1. However, if s = 1, then we must have a 1 = 0. We therefore set a 1 = 0 regardless of s. Subsequent coefficients are determined by the recurrence relation
aj+2 = −
aj ω^2 (j + s + 2)(j + s + 1) , j ≥ 0.
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,
a 2 = −
a 0 ω^2 2
a 4 = −
a 2 ω^2 12
a 0 ω^4 24
a 6 = −
a 4 ω^2 30
a 0 ω^6 720
From these values, we obtain the formula
a 2 p = (−1)p^ a 0 ω^2 p (2p)!
, p ≥ 0 ,
which yields the solution
y(x) = a 0
∑^ ∞
j=
(−1)j^ ω^2 j (2j)! = a 0 cos ωx.
Then, we consider the case of s = 1. We then have, for j = 0, 2 , 4,
a 2 = − a 0 ω^2 6
a 4 = − a 2 ω^2 20
a 0 ω^4 120
a 6 = − a 4 ω^2 42
a 0 ω^6 5040
From these values, we obtain the formula
a 2 p = (−1)p^
a 0 ω^2 p (2p + 1)!
which yields the solution
y(x) = a 0
∑^ ∞
j=
(−1)j^ ω^2 j (2j + 1)!
a 0 ω
sin ωx.
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.
which has solutions s = ±n. Next, we consider the term with s 1. With s = ±n, we have the requirement that a 1 (2n + 1) = 0 or a 1 (1 − 2 n) = 0,
so unless n = ± 1 /2, we must have a 1 = 0. Subsequent coefficients are determined by the recurrence relation aj+2 = − aj (s + j + 2)^2 − n^2
, j ≥ 0.
In the case of s = n, we obtain
aj+2 = −
aj (j + 2)(j + 2 + 2n)
Because a 1 = 0, all of the odd-numbered coefficients vanish as well. Substituting even numbers for j yields
a 2 = − a 0 4(n + 1)
a 4 = − a 2 8(n + 2)
a 0 32(n + 1)(n + 2) a 6 = −
a 4 12(n + 3)
a 0 384(n + 1)(n + 2)(n + 3)
From these values, we can obtain the formula
a 2 p = (−1)p^ a 0 n! 22 pp!(n + p)!
We conclude that the solution is
y(x) = a 0
∑^ ∞
j=
(−1)j^ n! 22 j^ j!(n + j)!
xn+2j^.
Setting a 0 = 1/ 2 nn! yields the Bessel function
Jn(x) =
∑^ ∞
j=
(−1)j j!(n + j)!
( (^) x 2
)n+2j .
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 a 2 n− 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′′^ + P (x)y′^ + Q(x)y = 0
fail to be defined at a point x 0. We classify a point x 0 in the domain of the ODE as follows:
- If P (x 0 ) and Q(x 0 ) are both finite, then x 0 is said to be an ordinary point of the ODE.
- If either P or Q becomes infinite as x → x 0 , but (x − x 0 )P (x) and (x − x 0 )^2 Q(x) are both still finite at x 0 , then x 0 is a regular singular point of the ODE.
- If either (x − x 0 )P (x) or (x − x 0 )Q(x) becomes infinite as x → x 0 , then x 0 is an irregular or essential singular point of the ODE.
For example, Bessel’s equation, rewritten in the form
y′′^ +
x
y′^ +
n^2 x^2
y = 0,
has a singular point at x = 0, as P (x) = 1/x and Q(x) = 1 − n^2 /x^2 both have a vertical asymptote at x = 0. However, both xP (x) = 1 and x^2 Q(x) = x^2 − n^2 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 x 0 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 x 0 as a singularity.
