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FROBENIUS SERIES SOLUTIONS
TSOGTGEREL GANTUMUR
Abstract. We introduce the Frobenius series method to solve second order linear equations, and illustrate it by concrete examples.
Contents
- Regular singular points 1
- Formulation of the method 2
- Examples 3
- Regular singular points Consider the differential equation a(x)y′′^ + b(x)y′^ + c(x)y = 0, (1)
where a, b, and c are polynomials, or equivalently,
y′′^ + p(x)y′^ + q(x)y = 0, (2)
where
p(x) =
b(x) a(x)
, and q(x) =
c(x) a(x)
Recall that a point x = α is called a singular point of (2) if
|p(x)| → ∞, or |q(x)| → ∞, as x → α. (4)
Recall also that an ordinary point of (2) is a point at which the functions p(x) and q(x) are continuous. We have seen that one can solve the equation in terms of a power series centred at an ordinary point. In these notes, we will generalize the power series method so that we can solve the equation (2) at least near some singular points. The method is called the Frobenius method, named after the mathematician Ferdinand Georg Frobenius.
Example 1. In fact, we have already encountered an equation with a singular point, and we have solved it near its singular point. The Cauchy-Euler equation
x^2 y′′^ + P xy′^ + Qy = 0, (5)
has a singular point at x = 0, and we know that a solution for x > 0 is given by
y(x) = xr^ = er^ log^ x, (6)
where r is a root of the characteristic (or auxiliary) equation
r^2 + (P − 1)r + Q = 0. (7)
This example shows that at least some singular points can be “tamed”.
Comparing (1) with a Cauchy-Euler equation leads us to the following definition.
Date: April 4, 2014. 1
2 TSOGTGEREL GANTUMUR
Definition 2. A singular point x = α is called a regular singular point of the equation (1) if (1) can be written as
(x − α)^2 ˜a(x)y′′^ + (x − α)˜b(x)y′^ + ˜c(x)y = 0, (8)
where ˜a(x), ˜b(x), and ˜c(x) are polynomials, with ˜a(α) 6 = 0.
A singular point that is not a regular singular point is called a irregular singular point.
Example 3. Consider the equation
(5 − x^2 )y′′^ +
1 + x x
y′^ −
x^2
y = 0. (9)
It is easy to see that the singular points are at x = 0 and x = ±5. Let us try to show that the point x = 0 is a regular singular point. We need Definition 2 with α = 0. According to the definition, all coefficients of the equation must be polynomials. We multiply (9) by x^2 , to get
x^2 (5 − x^2 )y′′^ + x(1 + x)y′^ − (3 + x^3 )y = 0, (10)
which is of the form (8), with
˜a(x) = 5 − x^2 , ˜b(x) = 1 + x, ˜c(x) = −(3 + x^3 ). (11)
The only remaining condition to check is the condition ˜a(α) 6 = 0. Since ˜a(0) = 5 6 = 0, this condition is true, and hence x = 0 is a regular singular point of (9).
Remark 4. An alternative, perhaps more direct way to identify the coefficients ˜a(x), ˜b(x), and ˜c(x) in (9) would be to write it as
˜a(x)y′′^ +
˜b(x) x
y′^ +
˜c(x) x^2
y = 0, (12)
that is,
(5 − x^2 )y′′^ +
1 + x x
y′^ −
3 + x^3 x^2
y = 0. (13)
Intuitively, the point x = 0 is a regular singular point if the coefficient of y′′^ is nonzero at x = 0, and as x → 0, the coefficients of y′^ and y do grow faster than x−^1 and x−^2 , respectively.
- Formulation of the method Recall for convenience that a singular point x = α is called a regular singular point of (1) if (1) can be written as
(x − α)^2 ˜a(x)y′′^ + (x − α)˜b(x)y′^ + ˜c(x)y = 0, (14)
where ˜a(x), ˜b(x), and ˜c(x) are polynomials, with ˜a(α) 6 = 0. Now, the central idea of the method we are about to see is the expectation that for x ≈ α, any solution y(x) of (14) must approximately solve the Cauchy-Euler equation
(x − α)^2 ˜a(α)y′′^ + (x − α)˜b(α)y′^ + ˜c(α)y = 0. (15)
Note carefully that the coefficients ˜a(x), ˜b(x), and ˜c(x) are evaluated at x = α, so that ˜a(α), ˜b(α), and ˜c(α) are numbers, not functions. Comparing (15) with (5), we identify the coefficients
P =
˜b(α) ˜a(α)
, Q =
˜c(α) ˜a(α)
and so a solution of (15) for x > α is given by
y˜(x) = (x − α)r, (17)
4 TSOGTGEREL GANTUMUR
Substituting these expressions into (22), we get
∑^ ∞
k=
5(k + r)(k + r − 1)akxk+r^ +
∑^ ∞
k=
(k + r)ak(xk+r^ + xk+r+1) −
∑^ ∞
k=
akxk+r^ = 0, (28)
which, upon some rearranging, becomes
∑^ ∞
k=
[5(k + r)(k + r − 1) + (k + r) − 1]akxk+r^ +
∑^ ∞
k=
(k + r)akxk+r+1^ = 0. (29)
The coefficient of each power of x appearing in the left hand side must vanish. We observe that the first term (k = 0) of the first sum is a term with xr, while the first term (k = 0) of the second sum is a term with xr+1. In other words, the coefficient of xr^ in the entire left hand side of (29) is
[5(0 + r)(0 + r − 1) + (0 + r) − 1]a 0 = [5r(r − 1) + r − 1]a 0 , (30)
which can be read off from the k = 0 term of the first sum in (29). Now the twist is that 5 r(r − 1) + r − 1 = 0, because r satisfies the indicial equation (24), and 5r(r − 1) + r − 1 = 0 is just (24) in disguise. This basically means that the constant a 0 can have arbitrary value, although we still have to check if there would be any constraint on a 0 induced by the conditions on the coefficients of the other powers of x in the left hand side of (29). To find those conditions, we write the second sum in (29) as ∑^ ∞
k=
(k + r)akxk+r+1^ =
∑^ ∞
n=
(n − 1 + r)an− 1 xn+r, (31)
where we have introduced the new variable n = k + 1. The first sum in (29) can be written as
∑^ ∞
k=
[5(k + r)(k + r − 1) + (k + r) − 1]akxk+r^ =
∑^ ∞
k=
[5(k + r)(k + r − 1) + (k + r) − 1]akxk+r, (32)
since the term k = 0 vanishes because of the indicial equation. So the left hand side of (29) is
∑^ ∞
k=
[5(k + r)(k + r − 1) + (k + r) − 1]akxk+r^ +
∑^ ∞
k=
(k + r)akxk+r+
∑^ ∞
k=
[5(k + r)(k + r − 1) + (k + r) − 1]akxk+r^ +
∑^ ∞
n=
(n − 1 + r)an− 1 xn+r
∑^ ∞
n=
[5(n + r)(n + r − 1) + (n + r) − 1]anxn+r^ +
∑^ ∞
n=
(n − 1 + r)an− 1 xn+r
∑^ ∞
n=
[5(n + r)(n + r − 1) + (n + r) − 1]an + (n − 1 + r)an− 1
xn+r,
where in the second step we have renamed the summation variable in the first sum in order to conveniently combine the terms of the two sums. By (29) the coefficient of each xn+r^ must vanish, meaning that
[5(n + r)(n + r − 1) + n + r − 1]an + (n − 1 + r)an− 1 = 0, n = 1, 2 ,... , (33)
which are the conditions we have been looking for. We can simplify it a bit, by dividing it by the common factor n − 1 + r, giving
[5(n + r) + 1]an + an− 1 = 0, n = 1, 2 ,... , (34)
FROBENIUS SERIES SOLUTIONS 5
or an = −
an− 1 5 n + 5r + 1
, n = 1, 2 ,.... (35)
Finally, we can use the concrete values r = 1 and r = − 15. For the case r = 1, we have
an = −
an− 1 5 n + 6
= (−1)na 0
∏^ n
k=
(5j + 1)−^1 , n = 1, 2 ,... , (36)
and for r = 15 , we have
an = −
an− 1 5 n
(−1)n 5 nn!
a 0 , n = 1, 2 ,.... (37)
In the latter case, the solution y(x) has a closed form expression
y(x) = x−^
15 ∑^ ∞
n=
(−1)n 5 nn!
a 0 xn^ = a 0 x−^
15 ∑^ ∞
n=
n!
x 5
)n = a 0 x−^
(^15) e−^
x 5
. (38)
Let us check if this expression indeed satisfies the equation (19). For simplicity, putting a 0 = 1, we compute
y′(x) = −
x−^
6 (^5) e−^
x (^5) − 1 5
x−^
1 (^5) e−^
x (^5) ,
y′′(x) =
x−^
11 (^5) e−^
x (^5) +^2 25
x−^
6 (^5) e−^
x (^5) +^1 25
x−^
1 (^5) e−^
x (^5).
Now writing all the relevant terms side by side
5 x^2 y′′(x) =
x−^
(^15) e−^
x 5
x
(^45) e−^
x 5
x
(^95) e−^
x 5 ,
xy′(x) = −
x−^
1 (^5) e−^
x (^5) − 1 5
x
4 (^5) e−^
x (^5) ,
x^2 y′(x) = −
x
4 (^5) e−^
x (^5) − 1 5
x
9 (^5) e−^
x (^5) ,
−y(x) = −x−^
(^15) e−^
x 5 ,
makes it clear that y(x) = x−^
1 (^5) e−^
x (^5) is a solution to (19).
Exercise 7. Find a solution of
3 x^2 y′′^ + (2x^2 − x)y′^ + y = 0, (39)
in terms of the Frobenius series
y(x) = xr
∑^ ∞
k=
akxk. (40)
