Method of Frobenius: Equal Roots to the Indicial Equation
We solve the equation x y'' + 2 y' + y = 0 using a power series centered at the regular singular point x = 0. (You should check that
zero is really a regular singular point.)
n+r
Let y = Ú¥
. Then, inserting this series into the differential equation results in
n=0 an x
rHr + 1L a0 xr-1 + â Han Hn + r + 1L Hn + rL - an-1 L xn+r-1 ‡ 0
¥
n=1
The indicial equation is r Hr + 1L = 0 and we have roots r1 = 0 and r2 = -1. We also have the recurrence relation
an =
an-1
Hn + rL Hn + r + 1L
valid for n ³ 1.
For r1 = 0 we obtain
an =
an-1
nHn + 1L
and so we get (with a0 = 1L
an =
1
n! Hn + 1L!
The first solution is therefore
y1 = â
¥
n=0
xn
n! Hn + 1L!
To find the second solution, we solve the general recurrence relation to obtain an as a function of r. Doing this results in
an H rL =
a0
Hr + 1L Hr + 2L2 Hr + 3L2 º Hn + rL2 Hn + r + 1L
Note that this recurrence relation is undefined at the smaller zero r2 = -1. We need to calculate
d
dr
HHr + 1L an rL Èr=-1
We have, clearly,
Hr + 1L an HrL =
a0
Hr + 2L Hr + 3L º Hn + rL2 Hn + r + 1L
2
2
and so, to calculate the derivative, we apply the natural log to both sides and obtain
lnHHr + 1L an HrLL = ln
Therefore we have, for the derivative,
a0
Hr + 2L2 Hr + 3L2 º Hn + rL2 Hn + r + 1L
= lnIa0 M - 2 HlnHr + 2L + º + lnHr + nLL - lnHn + r + 1L
2
Roots_Differing_By_an_Integer.nb
d
dr
HHr + 1L an HrLLM
Hr + 1L an HrL
1
= -2
1
+
r+2
r+3
+º+
1
1
-
r+n
r+n+1
and so
d
dr
HHr + 1L an HrLL = Hr + 1L an HrL -2
a0
Hr + 2L Hr + 3L º Hn + rL Hn + r + 1L
2
2
1
r+2
-2
1
+
r+3
1
1
+
r+2
2
+º+
r+3
1
1
-
r+n
+º+
=
r+n+1
1
-
r+n
1
r+n+1
Thus, evaluating at r2 = -1, we have
d
dr
HHr + 1L an HrLL
=
a0
r=-1
1
H1L H2L º Hn - 1L HnL
2
2
-2 1 +
2
2
1
+º+
1
-
n-1
=
n
This forumla is valid for n ³ 2. The second solution is
â
¥
n=0
xn
n! Hn + 1L!
lnHxL -
1
x
1 - x -â
¥
n=2
HHn-1 + Hn L xn
n! Hn - 1L!
-a0 IHn-1 + Hn M
n! Hn - 1L!