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Frobenius method∗
pahio†
2013-03-22 0:12:51
Let us consider the linear homogeneous differential equation
n
X
kν (x)y (n−ν) (x) = 0
ν=0
of order n. If the coefficient functions kν (x) are continuous and the coefficient
k0 (x) of the highest order derivative does not vanish on a certain interval (resp.
a domain in C), then all solutions y(x) are continuous on this interval (resp.
domain). If all coefficients have the continuous derivatives up to a certain order,
the same concerns the solutions.
If, instead, k0 (x) vanishes in a point x0 , this point is in general a singular
point. After dividing the differential equation by k0 (x) and then getting the
form
n
X
y (n) (x) +
cν (x)y (n−ν) (x) = 0,
ν=1
some new coefficients cν (x) are discontinuous in the singular point. However,
if the discontinuity is restricted so, that the products
(x − x0 )c1 (x),
(x − x0 )2 c2 (x),
...,
(x − x0 )n cn (x)
are continuous, and even analytic in x0 , the point x0 is a regular singular point
of the differential equation.
We introduce the so-called Frobenius method for finding solution functions
in a neighbourhood of the regular singular point x0 , confining us to the case of
a second order differential equation. When we use the quotient forms
(x − x0 )c1 (x) :=
p(x)
,
r(x)
(x − x0 )2 c2 (x) : =
q(x)
,
r(x)
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where r(x), p(x) and q(x) are analytic in a neighbourhood of x0 and r(x) 6= 0,
our differential equation reads
(x − x0 )2 r(x)y 00 (x) + (x − x0 )p(x)y 0 (x) + q(x)y(x) = 0.
(1)
Since a simple change x−x0 7→ x of variable brings to the case that the singular
point is the origin, we may suppose such a starting situation. Thus we can
study the equation
x2 r(x)y 00 (x) + xp(x)y 0 (x) + q(x)y(x) = 0,
(2)
where the coefficients have the converging power series expansions
r(x) =
∞
X
rn xn ,
n=0
p(x) =
∞
X
p n xn ,
q(x) =
n=0
∞
X
q n xn
(3)
n=0
and
r0 6= 0.
In the Frobenius method one examines whether the equation (2) allows a series
solution of the form
y(x) = xs
∞
X
an xn = a0 xs + a1 xs+1 + a2 xs+2 + . . . ,
(4)
n=0
where s is a constant and a0 6= 0.
Substituting (3) and (4) to the differential equation (2) converts the left
hand side to
[r0 s(s−1)+p0 s+q0 ]a0 xs +
[[r0 (s+1)s+p0 (s+1)+q0 ]a1 +[r1 s(s−1)+p1 s+q1 ]a0 ]xs+1 +
[[r0 (s+2)(s+1)+p0 (s+2)+q0 ]a2 +[r1 (s+1)s+p1 (s+1)+q1 ]a1 +[r2 s(s−1)+p2 s+q2 ]a0 ]xs+2 + . . .
Our equation seems clearer when using the notations fν (s) := rν s(s−1) + pν s +
qn u:
f0 (s)a0 xs + [f0 (s+1)a1 + f1 (s)a0 ]xs+1 + [f0 (s+2)a2 + f1 (s+1)a1 + f2 (s)a0 ]xs+2 + . . . = 0
(5)
Thus the condition of satisfying the differential equation by (4) is the infinite
system of equations


f0 (s)a0 = 0

f (s+1)a + f (s)a = 0
0
1
1
0
(6)
f0 (s+2)a2 + f1 (s+1)a1 + f2 (s)a0 = 0



···
···
···
2
In the first place, since a0 6= 0, the indicial equation
f0 (s) ≡ r0 s2 + (p0 − r0 )s + q0 = 0
(7)
must be satisfied. Because r0 6= 0, this quadratic equation determines for s
two values, which in special case may coincide.
The first of the equations (6) leaves a0 (6= 0) arbitrary. The next linear
equations in an allow to solve successively the constants a1 , a2 , . . . provided
that the first coefficients f0 (s+1), f0 (s+2), . . . do not vanish; this is evidently
the case when the roots of the indicial equation don’t differ by an integer (e.g.
when the roots are complex conjugates or when s is the root having greater
real part). In any case, one obtains at least for one of the roots of the indicial
equation the definite values of the coefficients an in the series (4). It is not hard
to show that then this series converges in a neighbourhood of the origin.
For obtaining the complete solution of the differential equation (2) it suffices
to have only one solution y1 (x) of the form (4), because another solution y2 (x),
linearly independent on y1 (x), is gotten via mere integrations; then it is possible
in the cases s1 −s2 ∈ Z that y2 (x) has no expansion of the form (4).
References
[1] Pentti Laasonen: Matemaattisia erikoisfunktioita. Handout No. 261.
Teknillisen Korkeakoulun Ylioppilaskunta; Otaniemi, Finland (1969).
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