Series solutions of second order linear differential equations
We start with
Definition 1. A function f of a complex variable z is called analytic at z = z0 if there
exists a convergent Taylor series for f at z = z0 with positive radius of convergence:
f (z) =
∞
X
an (z − z0 )n
with
an =
n=0
f (n) (z0 )
,
n!
n = 0, 1, 2, . . .
for |z − z0 | < R with R > 0.
Now we consider second order linear differential equations of the form
y 00 + p(z)y 0 + q(z)y = 0
d2 y
dy
+ p(z) + q(z)y = 0.
2
dz
dz
or
(1)
Then we have the following definition:
Definition 2. A point z0 ∈ C is called a regular point of the differential equation (1) if both
p(z) and q(z) are analytic in z0 . Otherwise z0 is called a singular point of (1).
For singular points we distinguish between two cases:
Definition 3. If z0 ∈ C is a singular point of (1) and both (z − z0 )p(z) and (z − z0 )2 q(z) are
analytic in z0 , then z0 is called a regular singular point of (1). Otherwise z0 is called an
irregular singular point of (1).
For regular points we have:
Theorem 1. If z0 ∈ C is a regular point of the differential equation (1), then p(z) and q(z)
are both analytic at z = z0 . This implies that there exist positive constants R1 and R2 such
that
p(z) =
∞
X
n
pn (z − z0 ) ,
|z − z0 | < R1
and
q(z) =
n=0
∞
X
qn (z − z0 )n ,
|z − z0 | < R2 .
n=0
Then there exist solutions y of (1) which are analytic at z = z0 . In fact, if R = min(R1 , R2 ),
then the solution y can be written as
y(z) =
∞
X
cn (z − z0 )n
for
|z − z0 | < R.
n=0
Moreover, c0 = y(z0 ) and c1 = y 0 (z0 ) can be chosen arbitrary. Then
y(z) = c0 y1 (z) + c1 y2 (z)
with
{y1 (z), y2 (z)}
For regular singular points we have Frobenius’ method:
1
linearly independent.
Theorem 2. If z0 ∈ C is a regular singular point of the differential equation (1), then
(z − z0 )p(z) and (z − z0 )2 q(z) are both analytic at z = z0 . This implies that there exist
positive constants R1 and R2 such that
(z − z0 )p(z) =
∞
X
pn (z − z0 )n ,
|z − z0 | < R1
(2)
qn (z − z0 )n ,
|z − z0 | < R2 .
(3)
n=0
and
(z − z0 )2 q(z) =
∞
X
n=0
Then there exist solutions y of (1) which can be written in the form
y(z) = (z − z0 )r
∞
X
cn (z − z0 )n ,
c0 6= 0.
(4)
n=0
In fact, if R = min(R1 , R2 ), then we have
r
y(z) = (z − z0 )
∞
X
cn (z − z0 )n
for
0 < |z − z0 | < R.
n=0
Here r ∈ C denotes a solution of the indicial equation
r(r − 1) + p0 r + q0 = 0.
(5)
Moreover, if r1 , r2 ∈ C denote the solutions of the indicial equation (5) with Re r1 ≥ Re r2 ,
then we have:
1. If r1 − r2 ∈
/ {0, 1, 2, . . .}, then we have
r1
y1 (z) = (z − z0 )
∞
X
an (z − z0 )n ,
a0 6= 0
bn (z − z0 )n ,
b0 6= 0
n=0
and
y2 (z) = (z − z0 )r2
∞
X
n=0
are two linearly independent solutions of the differential equation (1).
2. If r1 − r2 = 0, then we have
y1 (z) = (z − z0 )r1
∞
X
an (z − z0 )n ,
a0 6= 0
n=0
and
y2 (z) = y1 (z) ln(z − z0 ) + (z − z0 )r2
∞
X
bn (z − z0 )n
n=0
are two linearly independent solutions of the differential equation (1).
2
3. If r1 − r2 ∈ {1, 2, 3, . . .}, then we have
y1 (z) = (z − z0 )
r1
∞
X
an (z − z0 )n ,
a0 6= 0
n=0
and
y2 (z) = Ay1 (z) ln(z − z0 ) + (z − z0 )r2
∞
X
bn (z − z0 )n ,
b0 6= 0
n=0
are two linearly independent solutions of the differential equation (1), where A is a
constant that might be zero or not.
Some remarks:
• A series of the form (4) is called a generalized power series. Only for r ∈ {0, 1, 2, . . .}
this generalized power series reduces to a normal power series.
• The solutions r1 and r2 of the indicial equation (5) are called indices.
• In the second case it is always possible to choose b0 = 0 since r1 = r2 . In that case the
second solution always has a logarithmic singularity at z = z0 .
• In the third case the second solution might have a logarithmic singularity at z = z0 (in
the case that A 6= 0) or not (in the case that A = 0).
The proof of this theorem due to Frobenius is constructive. We multiply the differential
equation (1) by (z − z0 )2 to obtain
(z − z0 )2 y 00 + (z − z0 )2 p(z)y 0 + (z − z0 )2 q(z)y = 0.
Since (z − z0 )p(z) and (z − z0 )2 q(z) are analytic at z = z0 we use (2) and (3) to find
!
!
∞
∞
X
X
2 00
0
n
n
(z − z0 ) y + (z − z0 )
pn (z − z0 ) y +
qn (z − z0 ) y = 0.
n=0
n=0
Now we set
y(z) = (z − z0 )r
∞
X
cn (z − z0 )n =
n=0
cn (z − z0 )n+r ,
c0 6= 0.
n=0
Then we have
0
∞
X
y (z) =
∞
X
(n + r)cn (z − z0 )n+r−1
n=0
and
y 00 (z) =
∞
X
(n + r)(n + r − 1)cn (z − z0 )n+r−2 .
n=0
Hence we have
∞
X
(n + r)(n + r − 1)cn (z − z0 )n+r +
n=0
∞
X
!
pn (z − z0 )n
n=0
+
∞
X
n=0
!
qn (z − z0 )n
∞
X
n=0
3
!
(n + r)cn (z − z0 )n+r
n=0
!
cn (z − z0 )n+r
∞
X
= 0.
This can only be true if the coefficients of all powers of z −z0 are equal to zero. The coefficient
of the lowest power of z − z0 , id est (z − z0 )r , is
{r(r − 1) + p0 r + q0 } c0 .
Since c0 6= 0 this leads to the indicial equation (5). More details are left out here.
Frobenius’ method is based on the theory of Euler’s differential equation:
x2 y 00 + αxy 0 + βy = 0,
α, β ∈ R.
(6)
By using the change of variables x = et or t = ln x we find that
dy
dy dt
1 dy
=
·
= ·
dx
dt dx
x dt
and
d
d2 y
=
2
dx
dx
1 dy
·
x dt
=
1 d2 y dt
1 dy
1 d2 y
1 dy
· 2 ·
− 2·
= 2· 2 − 2· .
x dt dx x dt
x dt
x dt
This implies that
dy
d2 y
d2 y dy
dy
=
and x2 · 2 = 2 − .
dx
dt
dx
dt
dt
Hence, with y(x) = Y (t) Euler’s differential equation (6) turns into
x·
Y 00 − Y 0 + αY 0 + βY = 0
⇐⇒
Y 00 + (α − 1)Y 0 + βY = 0,
which is a second order linear differential equation with constant coefficients. The characteristic equation for this differential equation is
r2 + (α − 1)r + β = 0
⇐⇒
r(r − 1) + αr + β = 0.
(7)
If r1 and r2 denote the roots of this characteristic equation, then we have
1. If r1 , r2 ∈ R and r1 6= r2 , then Y1 (t) = er1 t and Y2 (t) = er2 t are two linearly independent
solutions. This leads to y1 (x) = xr1 and y2 (x) = xr2 being two linearly independent
solutions of Euler’s differential equation (6).
2. If r1 , r2 ∈ R and r1 = r2 , then Y1 (t) = er1 t and Y2 (t) = ter1 t are two linearly independent
solutions. This leads to y1 (x) = xr1 and y2 (x) = xr2 ln x being two linearly independent
solutions of Euler’s differential equation (6).
3. If r1 , r2 ∈
/ R, say r1,2 = λ ± µi with λ, µ ∈ R and µ 6= 0, then Y1 (t) = eλt cos µt
and Y2 (t) = eλt sin µt are two linearly independent solutions. This leads to y1 (x) =
xλ cos(µ ln x) and y2 (x) = xλ sin(µ ln x) being two linearly independent solutions of
Euler’s differential equation (6).
This can also be obtained by looking for solutions of the form y(x) = xr of Euler’s differential
equation (6) directly. Then we have y 0 (x) = rxr−1 and y 00 (x) = r(r − 1)xr−2 which leads to
r(r − 1)xr + αrxr + βxr = 0
⇐⇒
This leads to the characteristic equation (7).
4
{r(r − 1) + αr + β} xr = 0.
Examples
1. The hypergeometric differential equation is given by
z(1 − z)y 00 + {c − (a + b + 1)z} y 0 − aby = 0,
a, b, c ∈ C.
(8)
It is clear that z = 0 and z = 1 are the only singular points of this differential equation.
Since
c − (a + b + 1)z
abz
zp(z) =
and z 2 q(z) = −
1−z
1−z
are both analytic at z = 0, we conclude that z = 0 is a regular singular point of (8).
Note that p0 = c and q0 = 0 which leads to the indicial equation
r(r − 1) + cr = 0
⇐⇒
r(r − 1 + c) = 0.
We also have that
(z − 1)p(z) = −
c − (a + b + 1)z
z
(z − 1)2 q(z) =
and
ab(z − 1)
z
are both analytic at z = 1, which implies that z = 1 is a regular singular point of (8)
too. In that case we have p0 = a + b − c + 1 and q0 = 0 which leads to the indicial
equation
r(r − 1) + (a + b − c + 1)r = 0
⇐⇒
r(r + a + b − c) = 0.
This implies that there exists solutions of the form
y1 (z) =
∞
X
an z n
and y2 (z) = z 1−c
n=0
∞
X
bn z n
n=0
for the hypergeometric differential equation (8). These two solutions are linearly independent at least for c ∈
/ Z. We also have solutions of the form
y3 (z) =
∞
X
n
cn (z − 1)
c−a−b
and y4 (z) = (z − 1)
n=0
∞
X
dn (z − 1)n
n=0
for the hypergeometric differential equation (8). These two solutions are linearly independent at least for c − a − b ∈
/ Z.
Note that we have
y(z) =
∞
X
an z
n
=⇒
0
y (z) =
n=0
∞
X
nan z
n−1
00
and y (z) =
n=0
∞
X
n(n − 1)an z n−2 .
n=0
Substitution into the differential equation (8) leads to
∞
X
n(n − 1)an z n−1 −
n=0
∞
X
n(n − 1)an z n
n=0
+c
∞
X
nan z n−1 − (a + b + 1)
n=0
∞
X
n=0
5
nan z n − ab
∞
X
n=0
an z n = 0.
Hence
∞
X
[{n(n + 1) + c(n + 1)} an+1 − {n(n − 1) + (a + b + 1)n + ab} an ] z n = 0.
n=0
This leads to the recurrence relation
(n + 1)(n + c)an+1 = (n + a)(n + b)an ,
n = 0, 1, 2, . . .
for the coefficients {an }∞
n=0 . The solution can be written as
an =
Hence
y(z) =
∞
X
(a)n (b)n
a0 ,
(c)n n!
an z n = a0
n=0
∞
X
(a)n (b)n z n
(c)n
n=0
Similarly, for
y(z) = z
n = 0, 1, 2, . . . .
1−c
∞
X
n
bn z =
n=0
n!
= a0 2 F1
∞
X
a, b
;z .
c
bn z n+1−c
n=0
we have
y 0 (z) =
∞
X
(n + 1 − c)bn z n−c
and y 00 (z) =
n=0
∞
X
(n + 1 − c)(n − c)bn z n−1−c .
n=0
Substitution into the differential equation (8) leads to
∞
X
(n + 1 − c)(n − c)bn z n−c −
n=0
∞
X
(n + 1 − c)(n − c)bn z n+1−c
n=0
+c
∞
X
(n + 1 − c)bn z n−c − (a + b + 1)
n=0
− ab
∞
X
(n + 1 − c)bn z n+1−c
n=0
∞
X
bn z n+1−c = 0.
n=0
Hence
∞
X
[{(n + 2 − c)(n + 1 − c) + c(n + 2 − c)} bn+1
n=0
− {(n + 1 − c)(n − c) + (a + b + 1)(n + 1 − c) + ab} bn ] z n+1−c = 0.
This leads to the recurrence relation
(n + 1)(n + 2 − c)bn+1 = (n + a + 1 − c)(n + b + 1 − c)bn ,
n = 0, 1, 2, . . .
for the coefficients {bn }∞
n=0 . The solution can be written as
bn =
(a + 1 − c)n (b + 1 − c)n
b0 ,
(2 − c)n n!
6
n = 0, 1, 2, . . . .
Hence
y(z) = z 1−c
∞
X
bn z n = b0 z 1−c
n=0
∞
X
(a + 1 − c)n (b + 1 − c)n z n
(2 − c)n
a + 1 − c, b + 1 − c
;z .
2−c
n!
n=0
= b0 z 1−c 2 F1
2. The confluent hypergeometric differential equation is given by
zy 00 + (c − z)y 0 − ay = 0,
a, c ∈ C.
(9)
Note that z = 0 is the only singular point of this differential equation. Further we have
in this case
zp(z) = c − z and z 2 q(z) = −az,
which are both analytic at z = 0. Note that p0 = c and q0 = 0, which implies that
z = 0 is a regular singular point of the differential equation (9) with indicial equation
r(r − 1) + cr = 0
⇐⇒
r(r − 1 + c) = 0.
Hence, the indices are r1 = 0 and r2 = 1 − c. This implies that for c ∈
/ Z we have two
linearly independent solutions of the form
y1 (z) =
∞
X
an z n
and y2 (z) = z 1−c
n=0
∞
X
bn z n .
n=0
If we substitute
y(z) =
∞
X
an z n ,
y 0 (z) =
n=0
∞
X
nan z n−1
and y 00 (z) =
n=0
∞
X
n(n − 1)an z n−2
n=0
into the differential equation (9) we obtain
∞
X
n(n − 1)an z n−1 + c
n=0
∞
X
nan z n−1 −
n=0
∞
X
nan z n − a
n=0
∞
X
an z n = 0
n=0
or equivalently
∞
X
[{(n + 1)n + c(n + 1)} an+1 − (n + a)an ] z n = 0.
n=0
This leads to the recurrence relation
(n + 1)(n + c)an+1 = (n + a)an ,
with solution
an =
(a)n
a0 ,
(c)n n!
7
n = 0, 1, 2, . . .
n = 0, 1, 2, . . . .
Hence we have a solution of the form
y(z) =
∞
X
an z n = a0
∞
X
(a)n z n
n=0
n=0
If we substitute
∞
X
y(z) = z 1−c
(c)n n!
bn z n =
n=0
y 0 (z) =
∞
X
(n + 1 − c)bn z n−c
= a0 1 F1
∞
X
a
c
;z .
bn z n+1−c ,
n=0
and y 00 (z) =
n=0
∞
X
(n + 1 − c)(n − c)bn z n−1−c
n=0
into the differential equation (9) we obtain
∞
X
(n + 1 − c)(n − c)bn z
n−c
+c
n=0
∞
X
(n + 1 − c)bn z n−c
n=0
−
∞
X
(n + 1 − c)bn z
n+1−c
−a
n=0
∞
X
bn z n+1−c = 0
n=0
or equivalently
∞
X
[{(n + 2 − c)(n + 1 − c) + c(n + 2 − c)} bn+1 − (n + a + 1 − c)bn ] z n+1−c = 0.
n=0
This leads to the recurrence relation
(n + 2 − c)(n + 1)bn+1 = (n + a + 1 − c)bn ,
with solution
bn =
(a + 1 − c)n
b0 ,
(2 − c)n n!
n = 0, 1, 2, . . .
n = 0, 1, 2, . . . .
Hence we have a solution of the form
y(z) = z
1−c
∞
X
n=0
n
bn z = b0 z
1−c
∞
X
(a + 1 − c)n z n
n=0
(2 − c)n
n!
= b0 z
1−c
1 F1
a+1−c
;z .
2−c
3. The Bessel differential equation is given by
z 2 y 00 + zy 0 + (z 2 − ν 2 )y = 0,
ν ≥ 0.
(10)
In this case z = 0 is the only singular point of the differential equation. Note that we
have
zp(z) = 1 and z 2 q(z) = z 2 − ν 2 ,
which are both analytic at z = 0. This implies, since p0 = 1 and q0 = −ν 2 , that z = 0
is a regular singular point of the Bessel differential equation (10) with indicial equation
r(r − 1) + r − ν 2
8
⇐⇒
r2 − ν 2 = 0.
This implies that the indices are r1 = ν and r2 = −ν. Hence, for 2ν ∈
/ {0, 1, 2, . . .} we
have two linearly independent solutions of the form
y1 (z) = z ν
∞
X
an z n
and y2 (z) = z −ν
n=0
∞
X
bn z n .
n=0
For ν = 0 we have two linearly independent solutions of the form
y1 (z) =
∞
X
an z n
and y2 (z) = y1 (z) ln z +
n=0
∞
X
bn z n .
n=0
Finally, for 2ν ∈ {1, 2, 3, . . .} we have two linearly independent solutions of the form
y1 (z) = z ν
∞
X
an z n
and y2 (z) = Ay1 (z) ln z + z −ν
n=0
∞
X
bn z n ,
n=0
where the constant A might be zero or not.
For instance, in the case ν 2 = 1/4 the indices are r1 = 1/2 and r2 = −1/2. In that case
we have two linearly independent solutions
y1 (z) = z 1/2
∞
X
(−1)n z 2n
n=0
sin z
= √
(2n + 1)!
z
and y2 (z) = z −1/2
∞
X
(−1)n z 2n
n=0
(2n)!
cos z
= √ .
z
This can easily be seen by setting y(z) = z −1/2 u(z) into the differential equation (10)
with ν 2 = 1/4. Then we have
1
y 0 (z) = z −1/2 u0 (z) − z −3/2 u(z)
2
3
and y 00 (z) = z −1/2 u00 (z) − z −3/2 u0 (z) + z −5/2 u(z),
4
which leads to
3
1
1
z 3/2 u00 (z) − z 1/2 u0 (z) + z −1/2 u(z) + z 1/2 u0 (z) − z −1/2 u(z) + z 3/2 u(z) − z −1/2 u(z) = 0
4
2
4
or
z 3/2 u00 (z) + u(z) = 0 =⇒ u00 (z) + u(z) = 0.
This leads to the linearly independent solutions u1 (z) = sin z and u2 (z) = cos z.
In the case ν = 0 we substitute
y(z) =
∞
X
an z n ,
n=0
y 0 (z) =
∞
X
nan z n
and y 00 (z) =
n=0
∞
X
n(n − 1)an z n−2
n=0
into the differential equation (10) with ν = 0 to find
∞
X
n(n − 1)an z n +
n=0
∞
X
n=0
nan z n +
∞
X
an z n+2 = 0,
n=0
which leads to
a1 z +
∞
X
[{(n + 2)(n + 1) + (n + 2)} an+2 + an ] z n+2 = 0.
n=0
9
This implies that
a1 = 0
(n + 2)2 an+2 = −an ,
and
n = 0, 1, 2, . . . .
The solution can be written as
a2k+1 = 0
and a2k =
(−1)k a0
,
22k (k!)2
k = 0, 1, 2, . . . .
This leads to the solution
y(z) =
∞
X
n
an z = a0
n=0
∞
X
(−1)k z 2k
22k (k!)2
k=0
= a0 0 F1
−
z2
;−
1
4
= a0 J0 (z),
where J0 (z) denotes the Bessel function of the first kind of order zero.
In the general case with ν > 0 we substitute
y(z) = z ν
∞
X
an z n =
n=0
∞
X
an z n+ν ,
y 0 (z) =
n=0
and
y 00 (z) =
∞
X
∞
X
(n + ν)an z n+ν−1
n=0
(n + ν)(n + ν − 1)an z n+ν−2
n=0
into the differential equation (10) to obtain
ν(ν − 1) + ν − ν 2 a0 z ν + (ν + 1)ν + (ν + 1) − ν 2 a1 z ν+1
∞
X
+
(n + ν + 2)(n + ν + 1) + (n + ν + 2) − ν 2 an+2 + an z n+ν+2 = 0.
n=0
This leads to the recurrence relation
(2ν + 1)a1 = 0 and
(n + ν + 2)2 − ν 2 an+2 = −an ,
n = 0, 1, 2, . . .
or equivalently, since ν > 0,
a1 = 0
(n + 2)(n + 2ν + 2)an+2 = −an ,
and
n = 0, 1, 2, . . . .
The solution can be written as
a2k+1 = 0
and a2k =
(−1)k a0
,
22k (ν + 1)k k!
k = 0, 1, 2, . . . .
This leads to the solution
y(z) = z
ν
∞
X
n=0
n
an z = a0 z
ν
∞
X
k=0
(−1)k z 2k
= a0 z ν 0 F1
22k (ν + 1)k k!
−
z2
;−
ν+1
4
The Bessel function Jν (z) of the first kind of order ν is defined by
(z/2)ν
−
z2
Jν (z) =
F
;
−
.
0 1
Γ(ν + 1)
ν+1
4
10
.