LECTURE 6: SINGULAR POWER SERIES NEAR A REGULAR SINGULAR POINT AND
BESSEL’S EQUATION
MINGFENG ZHAO
July 15, 2015
Series solutions near a regular singular points
Let x = a be a regular singular point of the following second order linear differential equation:
P (x)y 00 + Q(x)y 0 + R(x)y = 0.
(1)
R(x)
Q(x)
and q(x) =
, then y 00 + p(x)y 0 + q(x)y = 0. Since x = a is a regular singular point, then there
P (x)
P (x)
exists some R > 0 such that
Let p(x) =
(x − a)p(x) =
∞
X
pn (x − a)n ,
and
(x − a)2 q(x) =
∞
X
qn (x − a)n ,
for all |x − a| < R.
n=0
n=0
Then we get (x − a)2 y 00 + (x − a)2 p(x)y 0 + (x − a)2 q(x)y = 0, that is,
!
!
∞
∞
X
X
2 00
n
0
n
(2)
(x − a) y + (x − a)
pn (x − a)
y +
qn (x − a)
y = 0.
n=0
n=0
If pn = qn = 0 for all n ≥ 1, then we get (x − a)2 y 00 + (x − a)p0 y 0 + q0 y = 0, which is just the Cauchy-Euler equation.
In general, we seek a solution to (2) for R > x − a > 0 by using Frobenius’ method. Let
ψ(x, r) = (x − a)r
∞
X
an (r)(x − a)n ,
∀0 < x − a < R,
n=0
where {an (r)}∞
n=0 is a sequence of functions of r which will be defined later.
Then
(x − a)2 ψxx (x, r) + (x − a)2 p(x)ψx (x, r) + (x − a)2 q(x)ψ(x, r)
"
#
∞
n−1
X
X
r
= a0 (r)F (r)(x − a) +
an (r)F (n + r) +
ak (r)[(k + r)pn−k + qn−k ] (x − a)n+r ,
n=1
k=0
where F (r) = r(r − 1) + p0 r + q0 . The equation F (r) = 0 is called the indicial equation. Then F (r) = 0 has two roots
r1 and r2 , assume that Re r1 ≥ Re r2 . Roots r1 and r2 are called the exponents at the singularity.
1
2
MINGFENG ZHAO
Define {an (r)}∞
n=1 be the recurrence relation:
an (r)[(n + r)(n + r − 1) + (n + r)p0 + q0 ] +
n−1
X
ak (r)[(k + r)pn−k + qn−k ] = 0,
∀n ≥ 1.
k=0
which is called the recurrence relation of (1). That is, we have
Pn−1
an (r) = −
k=0
ak (r)[(k + r)pn−k + qn−k ]
,
F (n + r)
∀n ≥ 1.
So we get
(x − a)2 ψxx (x, r) + (x − a)2 p(x)ψx (x, r) + (x − a)2 q(x)ψ(x, r) = a0 (r)F (r)(x − a)r .
First solution: Let a0 (r) = 1, then
φ1 (x) = ψ(x, r1 ) = (x − a)r1
∞
X
an (r1 )(x − a)n ,
∀0 < x − a < R,
n=0
will be a non-zero solution.
Second solution: We have the following three cases:
• Case I: r1 − r2 is not a non-negative integer. Let a0 (r) = 1, then
φ2 (x)
= ψ(x, r2 )
=
r2
(x − a)
∞
X
an (r2 )(x − a)n ,
∀0 < x − a < R,
n=0
will be another non-zero solution which is linearly independent from φ1 (x).
• Case II: r1 = r2 = r0 . Let a0 (r) = 1, then
φ2 (x)
=
ψr (x, r0 )
=
r0
(x − a) ln(x − a)
∞
X
n
r0
an (r0 )(x − a) + (x − a)
n=0
∞
X
a0n (r0 )(x − a)n ,
∀0 < x − a < R,
n=0
will be another non-zero solution which is linearly independent from φ1 (x).
• Case III: r1 − r2 = N for some positive integer N . Let a0 (r) = r − r0 , then
φ2 (x)
=
ψr (x, r2 )
=
(x − a)r2 ln(x − a)
∞
X
n=0
an (r2 )(x − a)n + (x − a)r2
∞
X
a0n (r2 )(x − a)n ,
n=0
will be another non-zero solution which is linearly independent from φ1 (x).
∀0 < x − a < R,
LECTURE 6: SINGULAR POWER SERIES NEAR A REGULAR SINGULAR POINT AND BESSEL’S EQUATION
3
Example 1. For the differential equation 2x2 y 00 − xy 0 + (1 + x)y = 0, find all the regular singular points and determine
the indicial equation and the exponents at the singularity for each regular singular point.
Since 2x2 y 00 − xy 0 + (1 + x)y = 0, then
y 00 −
1 0 x+1
y = 0.
y +
2x
2x2
1
1
x+1
x+1
Let p(x) = −
, then x = 0 is a singular point. Notice that xp(x) = − and x2 q(x) =
and q(x) =
=
2
2x
2x
2
2
1 x
1
1
+ , both are analytic at x = 0, then x = 0 is a regular singular point. So p0 = − and q0 = , which implies that
2
2
2
2
the indicial equation is:
1
3
1
2r2 − 3r + 1
1
= 0.
0 = r(r − 1) + p0 r + q0 = r(r − 1) − r + = r2 − r + =
2
2
2
2
2
1
So the exponents at x = 0 are r1 = 1 and r2 = .
2
Example 2. Find the first two non-zero terms of two linearly independent solutions to 2x2 y 00 − xy 0 + (1 + x)y = 0
about x = 0.
1
By Example 1, x = 0 is a regular singular point, and the exponents at x = 0 are r1 = 1 and r2 = . Let
2
∞
∞
X
X
r
k
k+r
ψ(x, r) = x
ak x :=
ak x , then
k=0
k=0
ψx (x, r)
=
∞
X
(k + r)ak (R)xk+r−1
k=0
ψxx (x, r)
=
∞
X
(k + r)(k + r − 1)ak (R)xk+r−2 .
k=0
So we have
2x2 ψxx (x, r) − xψx (x, r) + (1 + x)ψ(x, r)
=
2x2
∞
X
(k + r)(k + r − 1)ak (r)xk+r−2 − x
k=0
=
2
∞
X
=
(k + r)(k + r − 1)ak (r)xk+r −
∞
X
(k + r)ak (r)xk+r +
[2(k + r)(k + r − 1) − (k + r) + 1]ak (r)xk+r +
∞
X
∞
X
ak (r)xk+r +
k=0
∞
X
∞
X
ak−1 (r)xk+r
k=1
[2r(r − 1) − r + 1]a0 (r)xr +
(2r2 − 3r + 1)a0 (r)xr +
ak (r)xk+r+1
k=0
∞
X
[2(k + r)(k + r − 1) − (k + r) + 1]ak (r)xk+r +
k=1
=
ak (r)xk+r
k=0
k=0
k=0
=
(k + r)ak (r)xk+r−1 + (1 + x)
k=0
k=0
∞
X
∞
X
∞
X
{[(k + r)(2k + 2r − 3) + 1]ak (r) + ak−1 (r)}xk+r .
k=1
∞
X
k=1
ak−1 (r)xk+r
4
MINGFENG ZHAO
Now let a0 (r) = 1 and define the sequence {ak (r)}∞
k=1 be the recurrence relation:
[(k + r)(2k + 2r − 3) + 1]ak (r) + ak−1 (r) = 0,
∀k ≥ 1.
That is,
−1
· ak−1 (r),
(k + r)(2k + 2r − 3) + 1
ak (r) =
∀k ≥ 1.
Then
ak (r)
=
=
=
=
−1
(k + r)(2k + 2r − 3) + 1
−1
(k + r)(2k + 2r − 3) + 1
−1
(k + r)(2k + 2r − 3) + 1
(−1)k
· ak−1 (r)
−1
· ak−2
(k − 1 + r)(2(k − 1) + 2r − 3) + 1
−1
−1
·
· ··· ·
· a0 (r)
(k − 1 + r)(2(k − 1) + 2r − 3) + 1
(1 + r)(2 · 1 + 2r − 3) + 1
·
k
Y
1
,
(i + r)(2i + 2r − 3) + 1
i=1
∀k ≥ 1.
Then we get
2x2 ψxx (x, r) − xψx (x, r) + (1 + x)ψ(x, r) = (2r2 − 3r + 1)xr = (2r − 1)(r − 1)xr .
∞
X
∞
X
1
xk , then φ1 and φ2 are two linearly independent solutions to
2
k=0
k=0
1
2 00
0
2x y − xy + (1 + x)y = 0. Since a0 (r) = 1 and a1 (r) = −
, then
(1 + r)(2 + 2r − 3) + 1
1
=1
a0 (1) = a0
2
Let φ1 (x) = x
1
ak (1)xk and φ2 (x) = x 2
ak
1
a1 (1) = −
3
1
= −1.
a1
2
Example 3. For the differential equation xy 00 + y 0 − y = 0, find all the regular singular points and determine the indicial
equation and the exponents at the singularity for each regular singular point.
Since xy 00 + y 0 − y = 0, then
y 00 +
1 0 1
y − y = 0.
x
x
1
1
and q(x) = − , then x = 0 is a singular point. Notice that xp(x) = 1 and x2 q(x) = −x, both are
x
x
analytic at x = 0, then x = 0 is a regular singular point. So p0 = 1 and q0 = 0, which implies that the indicial equation
Let p(x) =
LECTURE 6: SINGULAR POWER SERIES NEAR A REGULAR SINGULAR POINT AND BESSEL’S EQUATION
5
is:
0 = r(r − 1) + p0 r + q0 = r(r − 1) + r = r2 .
So the exponents at x = 0 are r1 = r2 = 0.
Example 4. Find the first two non-zero terms of two linearly independent solutions to xy 00 + y 0 − y = 0 about x = 0.
By Example 3, x = 0 is a regular singular point, and the exponents at x = 0 are r1 = r2 = 0. Let ψ(x, r) :=
∞
∞
X
X
xr
ak (r)xk =
ak (r)xk+r for some differentiable functions ak (r), for all x > 0, then
k=0
k=0
ψx (x, r)
∞
X
=
(k + r)ak (r)xk+r−1
k=0
ψxx (x, r)
∞
X
=
(k + r)(k + r − 1)ak (r)xk+r−2 .
k=0
So we get
xψxx (x, r) + ψx (x, r) − ψ(x, r)
=
x
∞
X
(k + r)(k + r − 1)ak (r)xk+r−2 +
k=0
=
∞
X
(k + r)ak (r)xk+r−1 −
k=0
(k + r)(k + r − 1)ak (r)xk+r−1 +
k=0
=
∞
X
∞
X
r(r − 1)a0 (r)xr−1 +
∞
X
ak (r)xk+r
k=0
(k + r)(k + r − 1)ak (r)xk+r−1 + ra0 (r)xr−1 +
[r(r − 1) + r]a0 (r)xr−1 +
∞
X
∞
X
[(k + r)(k + r − 1) + k + r]ak (r)xk+r−1 −
=
r2 a0 (r)xr−1 +
∞
X
ak−1 (r)xk+r−1
{[(k + r)(k + r − 1) + k + r]ak (r) − ak−1 (r)} xk+r−1
k=1
=
r2 a0 (r)xr−1 +
∞
X
[(k + r)2 ak (r) − ak−1 (r)]xk+r−1 .
k=1
Let a0 (r) = 1, for |r| < 1, define the sequence {ak (r)}∞
k=1 by the recurrence relation:
(k + r)ak (r) − ak1 (r) = 0,
for all k ≥ 1.
1
· ak−1 (r),
(k + r)2
for all k ≥ 1.
That is,
ak (r) =
∞
X
k=0
k=1
k=1
∞
X
(k + r)ak (r)xk+r−1 −
k=1
k=1
=
ak (r)xk+r
k=0
(k + r)ak (r)xk+r−1 −
k=0
∞
X
∞
X
ak (r)xk+r
6
MINGFENG ZHAO
Then we get
ak (r)
=
=
=
=
=
1
(k + r)2
1
(k + r)2
1
(k + r)2
1
(k + r)2
· ak−1 (r)
1
· ak−2 (r)
(k − 1 + r)2
1
1
·
·
· ak−3 (r)
2
(k − 1 + r) (k − 2 + r)2
1
1
·
· ··· ·
· a0 (r)
2
(k − 1 + r)
(1 + r)2
·
k
Y
1
,
(i
+
r)2
i=1
for all k ≥ 1.
Then we get
xψxx (x, r) + ψx (x, r) − ψ(x, r) = r2 xr−1 .
(3)
Then
• For φ1 (x): Let φ1 (x) =
∞
X
ak (0)xk =
k=0
∞
X
xk
, by (3), then φ1 (x) is a solution to xy 00 + y 0 − y = 0. Notice that
(k!)2
k=0
a0 (0) = a0 (0) = 1.
• For φ2 (x): Since a0 (r) = 1 and ak (r) =
k
Y
1
, then all ak (r)’s are differentiable for all |r| < 1. By (3),
(i + r)2
i=1
then
xψxxr (x, r) + ψxr (x, r) − ψr (x, r) = 2rxr−1 + r2 xr−1 ln x.
Let φ2 (x) = ψr (x, 0), then
xφ002 (x) + φ02 (x) − φ2 (x) = 0,
Since ψ(x, r) =
∞
X
∀x > 0.
ak (r)xk+r , then
k=0
ψr (x, r) =
∞
X
a0k (r)xk+r + ln x
k=0
So we get φ2 (x) =
∞
X
a0k (0)xk + ln x ·
k=1
a1 (r) =
∞
X
ak (r)xk+r .
k=0
∞
X
ak (0)xk is a solution to xy 00 + y 0 − y = 0. Since a0 (r) = 1,
k=0
1
1
1
and a2 =
= 2
, then
(r + 1)2
(r + 1)2 (r + 2)2
(r + 3r + 2)2
a0 (0) = 1 = a1 (0) = 1,
a01 (0) = −2,
3
and a02 (0) = − .
4
LECTURE 6: SINGULAR POWER SERIES NEAR A REGULAR SINGULAR POINT AND BESSEL’S EQUATION
7
Bessel’s equation
For any constant ν ≥ 0, the Bessel’s equation of order ν has the form:
x2 y 00 + xy 0 + (x2 − ν 2 )y = 0.
(4)
Since x2 y 00 + xy 0 + (x2 − ν 2 )y = 0, then
y 00 +
1 0 x2 − ν 2
y = 0.
y +
x
x2
1
x2 − ν 2
, then x = 0 is a singular point. Notice that xp(x) = 1 and x2 q(x) = x2 − ν 2 , both
and q(x) =
x
x2
are analytic at x = 0, then x = 0 is a regular singular point. So p0 = 1 and q0 = −ν 2 , which implies that the indicial
Let p(x) =
equation is:
0 = r(r − 1) + p0 r + q0 = r(r − 1) + r − ν 2 = r2 − ν 2 .
r
So the exponents at x = 0 are r1 = ν and r2 = −ν. Let ψ(x, r) := x
∞
X
k
ak (r)x =
k=0
∞
X
ak (r)xk+r for some
k=0
differentiable functions ak (r), for all x > 0, then
ψx (x, r) =
∞
X
(k + r)ak (r)xk+r−1 ,
and ψxx (x, r) =
∞
X
(k + r)(k + r − 1)ak (r)xk+r−2 .
k=0
k=0
So we get
x2 ψxx (x, r) + xψx (x, r) + (x2 − ν 2 )ψ(x, r)
= x2
∞
X
(k + r)(k + r − 1)ak (r)xk+r−2 + x
=
(k + r)(k + r − 1)ak (r)xk+r +
k=0
=
∞
X
(k + r)ak (r)xk+r−1 + (x2 − ν 2 )
∞
X
(k + r)ak (r)xk+r +
k=0
=
[(k + r)(k + r − 1) + k + r − ν 2 ]ak (r)xk+r +
=
ak (r)xk+r
ak (r)xk+r+2 − ν 2
∞
X
∞
X
ak (r)xk+r
k=0
ak (r)xk+r+2
k=0
[(k + r)2 − ν 2 ]ak (r)xk+r +
k=0
∞
X
k=0
k=0
∞
X
∞
X
k=0
k=0
k=0
∞
X
∞
X
∞
X
ak−2 (r)xk+r
k=2
(r2 − ν 2 )a0 (r)xr + [(1 + r)2 − ν 2 ]a1 (r)x1+r +
∞
X
{[(k + r)2 − ν 2 ]ak (r) + ak−2 (r)}xk+r .
k=2
Let a1 (r) = 0, define the sequence {ak (r)}∞
k=1 by the recurrence relation:
[(k + r)2 − ν 2 ]ak (r) + ak−2 (r) = 0,
for all k ≥ 2.
8
MINGFENG ZHAO
That is,
ak (r) =
−1
· ak−2 (r),
(k + r)2 − ν 2
for all k ≥ 1.
Then
and a2k (r) = (−1)k a0 (r)
a2k+1 (r) = 0,
k
Y
1
,
(2i + r)2 − ν 2
i=1
for all k ≥ 0.
So we get
ψ(x, r) = x
(5)
r
∞
X
a2k (r)x
2k
r
= a0 (r)x + a0 (r)x
r
k=0
(6)
2
2
∞
X
k
(−1)
k=1
2
x ψxx (x, r) + xψx (x, r) + (x − ν )ψ(x, r)
=
2
2
k
Y
1
(2i + r)2 − ν 2
i=1
!
x2k
r
(r − ν )a0 (r)x .
For any 1 ≤ i ≤ k, then (2i !
+ ν)2 − ν 2 = 4i2 + 4iν > 0, which implies that we can define φ1 (x) =
∞
k
X
Y
1
x2k . By (6), then φ1 (x) is a solution to (4). Notice that
xν + xν
(−1)k
2 − ν2
(2i
+
ν)
i=1
k=1
"
φ1 (x) = x
ν
1+
∞
X
(−1)
k
k=1
k
Y
1
(2i) · (2i + 2ν)
i=1
!
#
x
2k
"
=x
ν
∞
X
#
(−1)k x2k
1+
.
Qk
2k
i=1 (i + ν)
k=1 2 k! ·
k
Y
1
is differentiable
(2i
+
r)2
i=1
near r = 0. Let a0 (r) = 1, then all a2k (r)’s are differentiable near r = 0. Take the partial derivative with respect to r
Let’s find another solution φ2 (x) to (4) for ν = 0, then r1 = r2 = 0, which implies that
on the both sides of (6), then
x2 ψxxr (x, r) + xψxr (x, r) + x2 ψr (x, r) = 2ra0 (r)xr + r2 a00 (r)xr + r2 a0 (r)xr · ln x.
Let φ2 (x) = ψr (x, 0), then φ2 (x) is a non-zero solution to (4). Notice that ψr (x, r) = ln x · xr
xr
∞
X
∞
X
a2k (r)x2k +
k=0
a02k (r)x2k , then
k=0
φ1 (x)
=
1+
φ2 (x)
=
ln x
∞
X
∞
X (−1)k x2k
(−1)k x2k
=
Q
k
2k
22k (k!)2
i=1 i
k=1 2 k! ·
k=0
∞
X
a2k (0)x2k +
k=0
∞
X
k=0
a02k (0)x2k = ln x + ln x
∞
X
k=1
a2k (0)x2k +
∞
X
a02k (0)x2k ,
Since a0 (r) = 1.
k=1
Department of Mathematics, The University of British Columbia, Room 121, 1984 Mathematics Road, Vancouver, B.C.
Canada V6T 1Z2
E-mail address: mingfeng@math.ubc.ca