LECTURE 5: SERIES SOLUTIONS NEAR REGULAR SINGULAR POINTS
MINGFENG ZHAO
July 14, 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)
Q(x)
R(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 ,
(x − a)2 q(x) =
and
n=0
∞
X
qn (x − a)n ,
for all |x − a| < R.
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)
(x − a)2 y 00 + (x − a)
pn (x − a)n y 0 +
qn (x − a)n 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
r
ψ(x, r) = (x − a)
∞
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 we have
ψ(x, r)
=
∞
X
ak (r)(x − a)k+r
k=0
ψx (x, r)
=
∞
X
ak (r)(k + r)(x − a)k+r−1
k=0
ψxx (x, r)
=
∞
X
ak (r)(k + r)(k + r − 1)(x − a)k+r−2 .
k=0
By (2), we have
(x − a)2 ψxx (x, r) + (x − a)2 p(x)ψx (x, r) + (x − a)2 q(x)ψ(x, r)
1
2
MINGFENG ZHAO
=
(x − a)2
∞
X
ak (r)(k + r)(k + r − 1)(x − a)k+r−2
k=0
+(x − a)
∞
X
∞
X
!
k
pk (x − a)
·
k=0
∞
X
+
=
ak (r)(k + r)(x − a)
k=0
∞
X
!
qk (x − a)k
·
k=0
∞
X
!
k+r−1
!
ak (r)(x − a)k+r
k=0
ak (r)(k + r)(k + r − 1)(x − a)k+r
k=0
∞
X
+
∞
X
!
k
pk (x − a)
·
k=0
∞
X
+
=
ak (r)(k + r)(x − a)
k=0
∞
X
!
qk (x − a)k
·
k=0
∞
X
!
k+r
!
ak (r)(x − a)k+r
k=0
an (r)(n + r)(n + r − 1)(x − a)n+r
n=0
+
+
=
∞
n
X
X
n=0
k=0
∞
X
n
X
n=0
k=0
∞
X
!
pn−k ak (r)(k + r) (x − a)n+r
!
qn−k ak (r) (x − a)n+r
"
an (r)(n + r)(n + r − 1) +
n=0
=
∞
X
n
X
#
ak [(k + r)pn−k + qn−k ] (x − a)n+r
k=0
"
an (r)(n + r)(n + r − 1) + an (r)(n + r)p0 + q0 +
n=0
=
∞
X
n−1
X
#
ak (r)[(k + r)pn−k + qn−k ] (x − a)n+r
k=0
"
an (r)[(n + r)(n + r − 1) + (n + r)p0 + q0 ] +
n=0
n−1
X
#
ak (r)[(k + r)pn−k + qn−k ] (x − a)n+r
k=0
r
= a0 (r)[r(r − 1) + rp0 + q0 ](x − a)
"
#
∞
n−1
X
X
+
an (r)[(n + r)(n + r − 1) + (n + r)p0 + q0 ] +
ak (r)[(k + r)pn−k + qn−k ] (x − a)n+r .
n=1
k=0
Let F (r) = r(r − 1) + rp0 + q0 , then F (r) = 0 is called the indicial equation of (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)
#
"
∞
n−1
X
X
= a0 (r)F (r)(x − a)r +
an (r)F (n + r) +
ak (r)[(k + r)pn−k + qn−k ] (x − a)n+r .
n=1
k=0
LECTURE 5: SERIES SOLUTIONS NEAR REGULAR SINGULAR POINTS
3
Define {an (r)}∞
n=1 be the recurrence relation:
an (r)[(n + r)(n + r − 1) + (n + r)p0 + q0 ] +
(3)
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) = −
(4)
k=0
ak (r)[(k + r)pn−k + qn−k ]
,
F (n + r)
∀n ≥ 1.
So we get
(5)
(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 .
Since F (r) is a quadratic polynomial, 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. Since Re r1 ≥ Re r2 , then F (n + r1 ) 6= 0 for all n ≥ 1, which implies
that for given a0 (r1 ), we can define an (r1 ) for all n ≥ 1. Let φ1 (x) = ψ(x, r1 ) in which a0 (r) = 1, that is,
φ1 (x) = (x − a)r1
∞
X
bn (x − a)n ,
∀0 < x − a < R,
n=0
where b0 = 1 and {bn }∞
n=1 is defined by the recurrence relation:
Pn−1
bn = −
k=0
bk [(k + r1 )pn−k + qn−k ]
,
F (n + r1 )
∀n ≥ 1.
By (5), since b0 = 1, then φ1 (x) is a non-zero solution to (1). Now we want to seek another solution φ2 to (1) which
is linearly independent from φ1 , we have the following three cases:
Case I: If r1 − r2 is neither 0 nor a positive integer, then F (n + r2 ) 6= 0 for all n ≥ 1. Let φ2 (x) = ψ(x, r2 ) in which
a0 (r) = 1, that is,
φ2 (x) = (x − a)r2
∞
X
cn (x − a)n ,
∀0 < x − a < R,
n=0
where c0 = 1 and {cn }∞
n=1 is defined by the recurrence relation:
Pn−1
cn = −
k=0
ck [(k + r2 )pn−k + qn−k ]
,
F (n + r2 )
∀n ≥ 1.
By (5), since c0 = 1, then φ2 (x) is a non-zero solution to (1). Since c0 = 1 and r1 6= r2 , then it’s clear that
φ2 (x) is linear independent from φ1 (x).
Case II: If r1 = r2 = r0 , that is, F (r) = 0 has only one real root, then F (r) = r(r − 1) + rp0 + q0 = (r − r0 )2 . Since
r0 is the only solution to F (r) = 0, then there exists some > 0 such that F (n + r) 6= 0 for all n ≥ 1 and all
4
MINGFENG ZHAO
|r − r0 | < . Let a0 (r) = 1, by (4), then an (r) is differentiable for all |r − r0 | < and all n ≥ 1. Take the partial
derivative with respect to r on the both sides of (5), then
(x − a)2 ψxxr (x, r) + (x − a)2 p(x)ψxr (x, r) + (x − a)2 q(x)ψr (x, r)
2(r − r0 )(x − a)r + (r − r0 )2 (x − a)r ln(x − a).
=
Let φ2 (x) = φr (x, r0 ), then φ2 (x) is a solution to (1). Notice that
φ2 (x)
= ψr (x, r0 )
(x − a)r0 ln(x − a)
=
∞
X
an (r0 )(x − a)n + (x − a)r0
n=0
a0n (r0 )(x − a)n
n=0
ln(x − a)φ1 (x) + (x − a)r0
=
∞
X
∞
X
a0n (r0 )(x − a)n .
n=0
Claim I: φ2 (x) is linearly independent from φ1 (x).
If not, since φ1 (x) 6= 0, then there exists some constant C such that φ2 (x) = Cφ1 (x), that is,
ln(x − a)φ1 (x) + (x − a)r0
∞
X
a0n (r0 )(x − a)n = Cφ1 (x).
n=0
So we get
ln(x − a)(x − a)r0
∞
X
bn (x − a)n + (x − a)r0
∞
X
a0n (r0 )(x − a)n = C(x − a)r0
bn (x − a)n .
n=0
n=0
n=0
∞
X
That is,
ln(x − a)
∞
X
n=0
bn (x − a)n +
∞
X
a0n (r0 )(x − a)n = C
∞
X
bn (x − a)n ,
∀0 < x − a < R.
n=0
n=0
By taking x & a, since a0 (r0 ) = 1 and lim tα ln t = 0 for all α > 0, then we will get −∞ + a00 (r0 ) = C,
t&a
contradiction.
Case III: If r1 − r2 = N for some positive integer, since r1 , r2 are two different roots of the quadratic equation F (r) = 0,
then both r1 and r2 should be real numbers and F (r) = (r −r1 )(r −r2 ) = (r −r2 −N )(r −r2 ). Let a0 (r) = r −r2 ,
then we get ψ(x, r1 ) = (r1 − r2 )φ1 (x) and
(6)
(x − a)2 ψxx (x, r) + (x − a)2 p(x)ψx (x, r) + (x − a)2 q(x)ψ(x, r) = (r − r2 )F (r)(x − a)r
Since r1 − r2 = N , then F (r2 + k) 6= 0 for all 1 ≤ k ≤ N − 1, which implies that ak (r) is differentiable near
r = r2 for all 1 ≤ k ≤ N − 1. Since a0 (r2 ) = 0, by (4), then
a0 (r2 ) = a1 (r2 ) = · · · = aN −1 (r2 ) = 0.
LECTURE 5: SERIES SOLUTIONS NEAR REGULAR SINGULAR POINTS
5
Since a0 (r) = r − r2 , by (4), then ak (r) is a rational function of r, which implies that for each 1 ≤ k ≤ N − 1,
there exist some polynomials bk (r) and ck (r) such that ck (r2 ) 6= 0 and
ak (r) = (r − r2 ) ·
bk (r)
.
ck (r)
Hence we get
Pn−1
aN (r)
k=0
= −
ak (r)[(k + r)pn−k + qn−k ]
F (n + r)
Pn−1
k=0 (r
= −
− r2 ) ·
bk (r)
ck (r) [(k
+ r)pn−k + qn−k ]
(r − r2 − N )(r − r2 )
Pn−1
bk (r)
k=0 ck (r) [(k
= −
+ r)pn−k + qn−k ]
r − r2 − N
which implies that aN (r) is differentiable near r = r2 . Since F (r2 + k) 6= 0 for all k > N , and {ak (r)}N
k=0
are differentiable near r = r2 , by (4), then an (r) is differentiable near r = r0 for all n ≥ 0, that is, ψ(x, r) is
differentiable with respect to r near r = r0 . By taking the partial derivative with respect to r on the both sides
of (6), then
(x − a)2 ψxxr (x, r) + (x − a)2 p(x)ψxr (x, r) + (x − a)2 q(x)ψr (x, r)
= F (r)(x − a)r + (r − r2 )F 0 (r)(x − a)r + (r − r2 )F (r)(x − a)r ln(x − a)
Let φ2 (x) = φr (x, r2 ), then φ2 (x) is a solution to (1). Notice that
φ2 (x)
= ψr (x, r2 )
=
(x − a)r2 ln(x − a)
∞
X
an (r2 )(x − a)n + (x − a)r2
n=0
=
(x − a)r2 ln(x − a)
∞
X
a0n (r2 )(x − a)n
n=0
∞
X
an (r2 )(x − a)n + (x − a)r2
∞
X
a0n (r2 )(x − a)n
n=0
n=N
Since a0 (r2 ) = a1 (r2 ) = · · · = aN −1 (r2 ) = 0
=
∞
X
(x − a)r2 +N ln(x − a)
an (r2 )(x − a)n−N + (x − a)r2
=
(x − a)r1 ln(x − a)
ak+N (r2 )(x − a)k + (x − a)r2
ln(x − a) · (x − a)r1
∞
X
∞
X
a0n (r2 )(x − a)n
n=0
k=0
=
a0n (r2 )(x − a)n
n=0
n=N
∞
X
∞
X
dk (x − a)k + (x − a)r2
k=0
Let dk = ak+N (r2 ) for all k ≥ 0.
∞
X
n=0
a0n (r2 )(x − a)n
6
MINGFENG ZHAO
Since a0 (r2 ) = a1 (r2 ) = · · · = aN −1 (r2 ) = 0, by (4), then
dn
= an+N (r2 )
Pn+N −1
ak (r2 )[(k + r2 )pn+N −k + qn+N −k ]
= − k=0
F (n + N + r2 )
Pn+N −1
ak (r2 )[(k + r2 )pn+N −k + qn+N −k ]
= − k=N
F (n + N + r2 )
Pn−1
ai+N (r2 )[(i + N + r2 )pn+N −i−N + qn+N −i−N ]
= − i=0
F (n + N + r2 )
Pn−1
ai+N (r2 )[(i + r1 )pn−i + qn−i ]
= − i=0
Since r1 = r2 + N
F (n + r1 )
Pn−1
di (r2 )[(i + r1 )pn−i + qn−i ]
.
= − i=0
F (n + r1 )
That is, {dn }∞
n=1 satisfies the recurrence relation (4) with r = r2 , which implies that
for all n ≥ 1.
dn = d0 bn ,
So we get
(x − a)r1
∞
X
dk (x − a)k = d0 (x − a)r1
∞
X
bk (x − a)k = d0 φ1 (x) = aN (r2 )φ1 (x).
k=0
k=0
So we get
φ2 (x) = aN (r2 ) ln(x − a)φ1 (x) + (x − a)r2
∞
X
a0n (r2 )(x − a)n .
n=0
Claim II: φ2 (x) is linearly independent from φ1 (x).
If not, since φ1 (x) 6= 0, then there exists some constant C such that φ2 (x) = Cφ1 (x), that is,
aN (r2 ) ln(x − a)φ1 (x) + (x − a)r2
∞
X
a0n (r2 )(x − a)n = Cφ1 (x).
n=0
So we get
aN (r2 ) ln(x − a)(x − a)r1
∞
X
bn (x − a)n + (x − a)r2
n=0
∞
X
a0n (r2 )(x − a)n = C(x − a)r1
n=0
∞
X
bn (x − a)n .
n=0
That is, for all 0 < x − a < R, we have
aN (r2 ) ln(x − a)(x − a)r1 −r2
∞
X
n=0
bn (x − a)n +
∞
X
a0n (r2 )(x − a)n = C(x − a)r1 −r2
n=0
∞
X
bn (x − a)n .
n=0
α
By taking x & a, since r1 − r2 = N > 0, a00 (r2 ) = 1 and and lim t ln t = 0 for all α > 0, then we will get
t&a
1 = a00 (r2 ) = 0, contradiction.
LECTURE 5: SERIES SOLUTIONS NEAR REGULAR SINGULAR POINTS
7
Example 1. For the differential equation x2 y 00 + xy 0 + (x − 1)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 x2 y 00 + xy 0 + (x − 1)y = 0, then
y 00 +
1 0 x−1
y = 0.
y −
x
x2
1
x−1
and q(x) = − 2 , then x = 0 is a singular point. Notice that xp(x) = 1 and x2 q(x) = x − 1, both
x
x
are analytic at x = 0, then x = 0 is a regular singular point. So p0 = 1 and q0 = −1, which implies that the indicial
Let p(x) =
equation is:
0 = r(r − 1) + p0 r + q0 = r(r − 1) + r − 1 = r2 − 1.
So the exponents at x = 0 are r1 = 1 and r2 = −1.
Example 2. Find the first two non-zero terms of two linearly independent solutions to x2 y 00 + xy 0 + (x − 1)y = 0 about
x = 0.
By Example 1, x = 0 is a regular singular point, and the exponents at x = 0 are r1 = 1 and r2 = −1. Let
∞
∞
X
X
ψ(x, r) := 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
x2 ψxx (x, r) + xψx (x, r) + (x − 1)ψ(x, r)
=
x2
∞
X
(k + r)(k + r − 1)ak (r)xk+r−2 + x
k=0
=
∞
X
=
(k + r)ak (r)xk+r−1 + (x − 1)
k=0
(k + r)(k + r − 1)ak (r)xk+r +
k=0
∞
X
∞
X
∞
X
k=0
k=0
=
∞
X
[(k + r)2 − 1]ak (r)xk+r +
k=0
=
(r2 − 1)a0 (r)xr +
ak−1 (r)xk+r
{[(k + r)2 − 1]ak − ak−1 }xk+r .
k=1
ak (r)xk+r+1 −
ak (r)xk+r+1
k=1
∞
X
∞
X
k=0
∞
X
k=0
∞
X
ak (r)xk+r
k=0
(k + r)ak (r)xk+r +
[(k + r)(k + r − 1) + k + r − 1]ak (r)xk+r +
∞
X
∞
X
k=0
ak (r)xk+r
8
MINGFENG ZHAO
Define the sequence {ak (r)}∞
k=1 by the recurrence relation:
[(k + r)2 − 1]ak − ak−1 = 0,
for all k ≥ 1.
That is,
ak (r) =
1
· ak−1 (r),
(k + r)2 − 1
for all k ≥ 1.
Then we get
ak (r)
=
=
=
=
1
· ak−1 (r)
(k + r)2
1
1
·
(k + r)2 − 1 (k − 1 + r)2 − 1
1
1
·
(k + r)2 − 1 (k − 1 + r)2 − 1
1
1
·
(k + r)2 − 1 (k − 1 + r)2 − 1
= a0 (r)
k
Y
1
,
(i + r)2 − 1
i=1
· ak−2 (r)
1
· ak−3 (r)
(k − 2 + r)2 − 1
1
· ··· ·
· a0 (r)
(1 + r)2 − 1
·
for all k ≥ 1.
Then we get
x2 ψxx (x, r) + xψx (x, r) + (x − 1)ψ(x, r) = (r2 − 1)a0 (r)xr .
(7)
Then
• For φ1 (x): Let b0 = 1, bk =
∞
k
Y
X
1
2
=
,
and
φ
(x)
=
x
bk xk , by (7), then φ1 (x) is a
1
2−1
(i
+
1)
k!
·
(k
+
2)!
i=1
solution to x2 y 00 + xy 0 + (x − 1)y = 0. Notice that
b0
=
b1
=
b2
=
1
2
1
2
=
=
1! · (1 + 2)!
3!
3
2
2
1
=
=
.
2! · (2 + 2)!
2 · 4!
24
• For φ2 (x): Let a0 (r) = r + 1, then
a1 (r)
= a0 (r) ·
a2 (r)
=
=
1
r+1
=
(1 + r)2 − 1
r(r + 2)
1
· a1 (r)
(2 + r)2 − 1
1
r+1
·
(r + 1)(r + 3) r(r + 2)
k=0
LECTURE 5: SERIES SOLUTIONS NEAR REGULAR SINGULAR POINTS
=
ak (r)
1
r(r + 2)(r + 3)
= a2 (r)
=
9
k
Y
1
(i + r)2 − 1
i=3
k
Y
1
1
,
r(r + 2)(r + 3) i=3 (i + r)2 − 1
for all k ≥ 3.
So we know that all ak (r)’s are differentiable near r = −1. By taking the partial derivative with respect to r
on the both sides of (7), then
x2 ψxxr (x, r) + xψxr (x, r) + (x − 1)ψr (x, r)
=
2ra0 (r)xr + (r2 − 1)a00 (r)xr + (r2 − 1)a0 (r)xr ln x
=
2r(r + 1)xr + (r2 − 1)xr + (r2 − 1)(r + 1)xr ln x.
Let φ2 (x) = ψr (x, −1), then
x2 φ002 (x) + xφ02 (x) + (x − 1)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 (−1)xk−1 + ln x ·
∞
X
ak (r)xk+r .
k=0
∞
X
ak (−1)xk−1 =
k=0
k=0
k=1
∞
X
solution to x2 y 00 + xy 0 + (x − 1)y = 0. Since a0 (r) = r + 1, a1 (r) =
a3 (r) =
a0k+1 (0)xk + ln x ·
∞
X
ak (−1)xk−1 is a
k=0
r+1
1
, a2 (r) =
and
r(r + 2)
r(r + 2)(r + 3)
1
, then
r(r + 2)(r + 3)[(3 + r)2 − 1]
a0 (−1)
=
a1 (−1) = 0
a2 (−1)
=
−
a3 (−1)
=
a01 (−1)
=
−1
a02 (−1)
=
1
.
4
1
2
1
−
6
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