6Series solutions
IA Differential Equations
6 Series solutions
Often, it is difficult to solve a differential equation directly. However, we can
attempt to find a Taylor series for the solution.
We will consider equations of the form
p(x)y
00
+ q(x)y
0
+ r(x)y = 0.
Definition (Ordinary and singular points). The point
x
=
x
0
is an ordinary
point of the differential equation if
q
p
and
r
p
have Taylor series about
x
0
(i.e. are
“analytic”, cf. Complex Analysis). Otherwise, x
0
is a singular point.
If x
0
is a singular point but the equation can be written as
P (x)(x − x
0
)
2
y
00
+ Q(x)(x − x
0
)y
0
+ R(x)y = 0,
where
Q
P
and
R
P
have Taylor series about
x
0
, then
x
0
is a regular singular point.
Example.
(i)
(1
− x
2
)
y
00
−
2
cy
0
+ 2
y
= 0.
x
= 0 is an ordinary point. However,
x
=
±
1
are (regular) singular points since p(±1) = 0.
(ii) sin xy
00
+
cos xy
0
+ 2
y
= 0.
x
=
nπ
are regular singular points while all
others are ordinary.
(iii)
(1 +
√
x
)
y
00
−
2
xy
0
+ 2
y
= 0.
x
= 0 is an irregular singular point because
√
x is not differentiable at x = 0.
It is possible to show that if
x
0
is an ordinary point, then the equation is
guaranteed to have two linearly independent solutions of the form
y =
∞
X
n=0
a
n
(x − x
0
)
n
,
i.e. Taylor series about
x
0
. The solution must be convergent in some neighbour-
hood of x
0
.
If
x
0
is a regular singular point, then there is at least one solution of the form
y =
∞
X
n=0
a
n
(x − x
0
)
n+σ
with
a
0
6
= 0 (to ensure
σ
is unique). The index
σ
can be any complex number.
This is called a Frobenius series.
Alternatively, it can be nice to think of the Frobenius series as
y = (x − x
0
)
σ
∞
X
n=0
a
n
(x − x
0
)
n
= (x − x
0
)
σ
f(x)
where f(x) is analytic and has a Taylor series.
We will not prove these results, but merely apply them.
Ordinary points
Example. Consider (1
− x
2
)
y
00
−
2
xy
0
+ 2
y
= 0. Find a series solution about
x = 0 (which is an ordinary point).
We try
y
=
P
∞
n=0
a
n
x
n
. First, we write the equation in the form of an
equidimensional equation with polynomial coefficients by multiplying both sides
by
x
2
. This little trick will make subsequent calculations slightly nicer. We
obtain
(1 − x)
2
(x
2
y
00
) − 2x
2
(xy
0
) + 2x
2
y = 0
X
a
n
[(1 − x
2
)n(n − 1) − 2x
2
n + 2x
2
]x
n
= 0
X
a
n
[n(n − 1) + (−n
2
− n + 2)x
2
]x
n
= 0
We look at the coefficient of
x
n
and obtain the following general recurrence
relation:
n(n − 1)a
n
+ [−(n − 2)
2
− (n − 2) + 2]a
n−2
= 0
n(n − 1)a
n
= (n
2
− 3n)a
n−2
Here we do not divide by anything since they might be zero.
First consider the case
n
= 0. The left hand side gives 0
· a
0
= 0 (the right
hand side is 0 since
a
n−2
= 0). So any value of
a
0
satisfies the recurrence
relationship, and it can take any arbitrary value. This corresponds to a constant
of integration. Similarly, by considering n = 1, a
1
is arbitrary.
For n > 1, n and n − 1 are non-zero. So we have
a
n
=
n − 3
n − 1
a
n−2
In this case (but generally not), we can further simplify it to obtain:
a
n
=
n − 3
n − 1
n − 5
n − 3
a
n−4
=
n − 5
n − 1
a
n−4
.
.
.
So
a
2k
=
−1
2k − 1
a
0
,
a
2k+1
= 0.
So we obtain
y = a
0
[1 −
x
2
1
−
x
4
3
−
x
6
5
− ···] + a
1
x
= a
0
1 −
x
2
ln
1 + x
1 − x
+ a
1
x
Notice the logarithmic behaviour near
x
=
±
1 which are regular singular points.
Regular singular points
Example. Consider 4
xy
00
+ 2(1
− x
2
)
y
0
− xy
= 0. Note that
x
= 0 is a singular
point. However, if we multiply throughout by
x
to obtain an equidimensional
equation, we obtain
4(x
2
y
00
) + 2(1 − x
2
)xy
0
− x
2
y = 0.
Since
Q
P
=
1−x
2
2
and
R
P
=
−
x
2
4
both have Taylor series,
x
= 0 is a regular singular
point. Try
y =
∞
X
n=0
a
n
x
n+σ
with a
0
6= 0.
Substituting in, we have
X
a
n
x
n+σ
[4(n + σ)(n + σ − 1) + 2(1 − x
2
)(n + σ) − x
2
]
By considering the coefficient of
x
n+σ
, we obtain the general recurrence relation
[4(n + σ)(n + σ − 1) + 2(n + σ)]a
n
− [2(n − 2 + σ) + 1]a
n−2
= 0.
Simplifying the equation gives
2(n + σ)(2n + 2σ − 1)a
n
= (2n + 2σ − 3)a
n−2
.
The n = 0 case gives the indicial equation for the index σ:
2σ(2σ − 1)a
0
= 0.
Since
a
0
6
= 0, we must have
σ
= 0 or
1
2
. The
σ
= 0 solution corresponds to an
analytic (“Taylor series”) solution, while
σ
=
1
2
corresponds to a non-analytic
one.
When σ = 0, the recurrence relation becomes
2n(2n − 1)a
n
= (2n − 3)a
n−2
.
When
n
= 0, this gives 0
· a
0
= 0. So
a
0
is arbitrary. For
n >
0, we can divide
and obtain
a
n
=
2n − 3
2n(2n − 1)
a
n−2
.
We can see that a
1
= 0 and so are subsequent odd terms.
If n = 2k, i.e. n is even, then
a
2k
=
4k − 3
4k(4k − 1)
a
2k−2
y = a
0
1 +
1
4 · 3
x
2
+
5
8 · 7 · 4 · 3
x
4
+ ···
Note that we have only found one solution in this case.
Now when σ =
1
2
, we obtain
(2n + 1)(2n)a
n
= (2n − 2)a
n−2
When
n
= 0, we obtain 0
· a
0
= 0, so
a
0
is arbitrary. To avoid confusion with
the a
0
above, call it b
0
instead.
When
n
= 1, we obtain 6
a
1
= 0 and
a
1
= 0 and so are subsequent odd terms.
For even n,
a
n
=
n − 1
n(2n + 1)
a
n−2
So
y = b
0
x
1/2
1 +
1
2 · 5
x
2
+
3
2 · 5 · 4 · 9
x
4
+ ···
Resonance of solutions
Note that the indicial equation has two roots
σ
1
, σ
2
. Consider the two different
cases:
(i)
If
σ
2
− σ
1
is not an integer, then there are two linearly independent
Frobenius solutions
y =
"
(x − x
0
)
σ
1
∞
X
n=0
a
n
(x − x
0
)
n
#
+
"
(x − x
0
)
σ
2
∞
X
n=0
b
n
(x − x
0
)
n
#
.
As x → x
0
, y ∼ (x − x
0
)
σ
1
, where Re(σ
1
) ≤ Re(σ
2
)
(ii)
If
σ
2
−σ
1
is an integer (including when they are equal), there is one solution
of the form
y
1
= (x − x
0
)
σ
2
∞
X
n=0
a
n
(x − x
0
)
n
with σ
2
≥ σ
1
.
In this case,
σ
=
σ
1
will not give a valid solution, as we will later see.
Instead, the other solution is (usually) in the form
y
2
= ln(x − x
0
)y
1
+
∞
X
n=0
b
n
(x − x
0
)
n+σ
1
.
This form arises from resonance between the two solutions. But if the
resonance somehow avoids itself, we can possibly end up with two regular
Frobenius series solutions.
We can substitute this form of solution into the differential equation to
determine b
n
.
Example. Consider
x
2
y
00
− xy
= 0.
x
= 0 is a regular singular point. It is
already in equidimensional form (x
2
y
00
) − x(y) = 0. Try
y =
∞
X
n=0
a
n
x
n+σ
with a
0
6= 0. We obtain
X
a
n
x
n+σ
[(n + σ)(n + σ − 1) − x] = 0.
The general recurrence relation is
(n + σ)(n + σ − 1)a
n
= a
n−1
.
n = 0 gives the indicial equation
σ(σ − 1) = 0.
Then
σ
= 0
,
1. We are guaranteed to have a solution in the form
σ
= 1. When
σ = 1, the recurrence relation becomes
(n + 1)na
n
= a
n−1
.
When n = 0, 0 · a
0
= 0 so a
0
is arbitrary. When n > 0, we obtain
a
n
=
1
n(n + 1)
a
n−1
=
1
(n + 1)(n!)
2
a
0
.
So
y
1
= a
0
x
1 +
x
2
+
x
2
12
+
x
3
144
+ ···
.
When σ = 0, we obtain
n(n − 1)a
n
= a
n−1
.
When
n
= 0, 0
· a
0
= 0 and
a
0
is arbitrary. When
n
= 1, 0
· a
1
=
a
0
. However,
a
0
6
= 0 by our initial constraint. Contradiction. So there is no solution in this
form (If we ignore the constraint that
a
0
6
= 0, we know that
a
0
is arbitrary. But
this gives exactly the same solution we found previously with σ = 1)
The other solution is thus in the form
y
2
= y
1
ln x +
∞
X
n=0
b
n
x
n
.