Electronic Journal of Differential Equations, Vol. 2009(2009), No. 27, pp. 1–8.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
REGULARIZED TRACE OF THE STURM-LIOUVILLE
OPERATOR WITH IRREGULAR BOUNDARY CONDITIONS
ALEXANDER MAKIN
Abstract. We consider the spectral problem for the Sturm-Liouville equation
with a complex-valued potential q(x) and with irregular boundary conditions
on the interval (0, π). We establish a formula for the first regularized trace of
the this operator.
1. Introduction and main result
This paper deals with the eigenvalue problem for the Sturm-Liouville equation
u00 − q(x)u + λu = 0
(1.1)
on the interval (0, π) with the boundary conditions
u0 (0) + (−1)θ u0 (π) + bu(π) = 0,
u(0) + (−1)θ+1 u(π) = 0,
(1.2)
where b is a complex number, b 6= 0, θ = 0, 1. The goal of this article is to calculate
the first-order regularized trace for (1.1)-(1.2).
The theory of regularized traces of ordinary differential operators has a long
history. First, the trace formulas for the Sturm-Liouville operator with the Dirichlet
boundary conditions and sufficiently smooth potential q(x) were established in [1, 2].
Afterwards these investigations were continued in many directions, for instance,
the trace formulas for the Sturm-Liouville operator with periodic or antiperiodic
boundary conditions were obtained in [3, 11], and for regular but not strongly
regular ones [9] similar formulas were found in [7]. A method for calculating trace
formulas for general problems involving ordinary differential equations on a finite
interval was proposed in [5]. The bibliography on the subject is very extensive and
we refer to the list of the works in [4, 10].
The trace formulas can be used for approximate calculation of the first eigenvalues of an operator [10], and in order to establish necessary and sufficient conditions
for a set of complex numbers to be the spectrum of an operator [11].
We will prove the following statement.
Theorem 1.1. Let q(x) be an arbitrary complex-valued function in W11 (0, π) and
denote by λn (n = 1, 2, . . . ) the eigenvalues of (1.1)-(1.2). Then we have the trace
2000 Mathematics Subject Classification. 34L05, 34B24.
Key words and phrases. Sturm-Liouville operator; eigenvalue problem; spectrum.
c
2009
Texas State University - San Marcos.
Submitted January 8, 2009. Published February 3, 2009.
1
2
A. MAKIN
EJDE-2009/27
formula
∞
X
n=1
Rπ
(−1)θ (q(π) − q(0) − 0 q 0 (t) cos(2nt)dt) λn − n − hqi −
πb
2
(1.3)
q(π) + q(0)
hqi (q(π) − q(0))2
+
−
+
= 0,
2
2
2b
4
Rπ
where hqi = π −1 0 q(t)dt.
This article is organized as follows. Section 2 is devoted to the analysis of the
characteristic determinant. In section 3, we obtain a formula of the regularized
trace in explicit form.
2. Analysis of the characteristic determinant
Denote by c(x, µ), s(x, µ) (µ2 = λ) the fundamental system of solutions to (1.1)
with the initial conditions c(0, µ) = s0 (0, µ) = 1, c0 (0, µ) = s(0, µ) = 0. Simple
calculations show that the characteristic equation of (1.1)-(1.2) can be reduced to
the form ∆(µ) = 0, where
∆(µ) = c(π, µ) − s0 (π, µ) − (−1)θ bs(π, µ).
(2.1)
Denote by ϕ(x, µ), ψ(x, µ) the system of solutions to (1.1) with the initial conditions
ϕ(0, µ) = ψ(0, µ) = 1, ϕ0x (0, µ) = iµ, ψx0 (0, µ) = −iµ. It is readily seen that
ϕ(x, µ) = ψ(x, −µ),
c(x, µ) = (ϕ(x, µ) + ψ(x, µ))/2,
s(x, µ) = (ϕ(x, µ) − ψ(x, µ))/(2iµ),
(2.2)
s0 (x, µ) = (ϕ0 (x, µ) − ψ 0 (x, µ))/(2iµ).
Rx
For convenience, we introduce I(x) = 0 q(t)dt. Asymptotic formulas for the functions ϕ(x, µ) and ϕ0 (x, µ) were established in [7]:
ϕ(x, µ) = eiµx [1 +
−
e−iµx
2iµ
1
1
I(x) − 2 I 2 (x)]
2iµ
8µ
Z x
e2iµt q(t)dt + A(x, µ) + R1 (x, µ),
(2.3)
0
and
1
ϕ0x (x, µ) = iµeiµx [1 +
I(x) − (8µ2 )−1 I 2 (x)]
2iµ
Z
e−iµx x 2iµt
+
e
q(t)dt + B(x, µ) + R3 (x, µ) + R4 (x, µ) + R5 (x, µ),
2
0
(2.4)
where
Z x
Z t
2 −1 −iµx
2iµt
q(s)ds
A(x, µ) = (4µ ) (e
e
q(t)dt
0
0
Z x
Z t
Z x
Z t
iµx
−2iµt
2iµs
−iµx
e2iµs q(s)ds),
+e
e
q(t)dt
e
q(s)ds − e
q(t)dt
0
0
0
0
EJDE-2009/27
REGULARIZED TRACE
3 −1
3
x
Z
R1 (x, µ) = −(4µ )
(eiµ(x−t) − e−iµ(x−t) )q(t)dt
0
(2.5)
Z s
Z t
iµ(t−s)
−iµ(t−s)
sin µ(s − y)q(y)ϕ(y, µ)dy,
(e
−e
)q(s)ds
×
0
0
−1
Z
x
−2iµt
t
Z
e2iµs q(s)ds
Z x
Z x
Z t
Z t
−iµx
2iµt
−iµx
−e
e
q(t)dt
e2iµs q(s)ds],
q(t)dt
q(s)ds + e
B(x, µ) = −(4iµ)
iµx
e
[e
q(t)dt
0
0
0
0
0
0
Z
1 iµx x
R2 (x, µ) = −
e
(1 + e2iµ(t−x) )q(t)I 2 (t)dt,
16µ2
0
Z
1 x iµ(x−t)
R3 (x, µ) =
(e
+ e−iµ(x−t) )q(t)R2 (t, µ)dt,
2 0
Z
1 x iµ(x−t)
(e
+ e−iµ(x−t) )q(t)A(t, µ)dt.
R4 (x, µ) =
2 0
(2.6)
(2.7)
(2.8)
We need more precise asymptotic formulas for the functions ϕ(x, µ) and ϕ0 (x, µ).
Substituting known the expression [8] for the function ϕ(y, µ) = eiµy +O(1/µ)e|Imµ|y
into (2.5), we obtain
Z x
3 −1
R1 (x, µ) = −(8iµ )
(eiµ(x−t) − e−iµ(x−t) )q(t)dt
0
Z t
Z s
iµ(t−s)
−iµ(t−s)
×
(e
−e
)q(s)ds
(eiµ(s−y) − e−iµ(s−y) )q(y)
0
0
× [eiµy + O(1/µ)e|Imµ|y ]dy
Z
Z t
eiµx x
−2iµ(x−t)
(1
−
e
)q(t)dt
(1 − e−2iµ(t−s) )q(s)ds
=−
8iµ3 0
0
Z s
−2iµ(s−y)
×
(1 − e
)q(y)dy + O(1/µ4 )e|Imµ|x
0
Z
Z t
Z s
eiµx x
=−
q(t)dt
q(s)ds
q(y)
8iµ3 0
0
0
× [1 − e−2iµ(x−t) − e−2iµ(t−s) − e−2iµ(s−y) + e−2iµ(x−s)
+ e−2iµ(x−t+s−y) + e−2iµ(t−y) − e−2iµ(x−y) ]dy + O(1/µ4 )e|Imµ|x
Z
Z t
Z s
1
eiµx x
q(t)dt
q(s)ds
q(y)dy −
K(x, µ) + O(1/µ4 )e|Imµ|x ,
=−
3
8iµ 0
8iµ3
0
0
where
K(x, µ) = eiµx
Z
x
Z
q(t)dt
Z
s
q(s)ds
0
0
−2iµ(x−t)
× [−e
t
0
−2iµ(t−s)
−e
q(y)
− e−2iµ(s−y) + e−2iµ(x−s)
+ e−2iµ(x−t+s−y) + e−2iµ(t−y) − e−2iµ(x−y) ]dy.
4
A. MAKIN
EJDE-2009/27
Arguing as in [6], we see that
Z x
Z t
Z
q(t)dt
q(s)ds
0
0
s
q(y)dy =
0
1 3
I (x).
6
The obvious inequality 0 ≤ y ≤ s ≤ t ≤ x, together with the inequality |x − 2z| ≤ x
valid for 0 ≤ z ≤ x and the Riemann lemma [8], implies
K(x, µ) = o(1)e|Imµ|x ,
R1 (x, µ) = −
1
I 3 (x) + o(1/µ3 )e|Imµ|x .
48iµ3
(2.9)
Continuing this line of reasoning, we reduce (2.6) and (2.8) to the form
R2 (x, µ) = −
1 iµx 3
e I (x) + o(1/µ2 )e|Imµ|x ,
48µ2
R4 (x, µ) = o(1/µ2 )e|Imµ|x . (2.10)
It follows from (2.7) and (2.9) that
R3 (x, µ) = O(1/µ3 )e|Imµ|x .
(2.11)
Combining (2.3) and (2.9), we obtain
1
1
I(x) − 2 I 2 (x)]
2iµ
8µ
Z
e−iµx x 2iµt
eiµx 3
−
e
q(t)dt + A(x, µ) −
I (x) + o(1/µ3 )e|Imµ|x ,
2iµ 0
48iµ3
(2.12)
It follows from (2.4), (2.10) and (2.11) that
ϕ(x, µ) = eiµx [1 +
1
I(x) − (8µ2 )−1 I 2 (x)]
ϕ0x (x, µ) = iµeiµx [1 +
2iµ
Z
e−iµx x 2iµt
1 iµx 3
+
e
q(t)dt + B(x, µ) −
e I (x) + o(1/µ2 )e|Imµ|x .
2
48µ2
0
(2.13)
Substituting into (2.1) the expressions for c(π, µ), s(π, µ), s0 (π, µ) of (2.2), (2.12),
(2.13), respectively, we obtain
∆(µ)
1 n iµπ
1
1
=
e [1 +
I(π) − 2 I 2 (π)]
2
2iµ
8µ
Z
e−iµπ π 2iµt
eiπµ 3
−
e
q(t)dt + A(π, µ) −
I (π) + o(1/µ3 )e|Imµ|π
2iµ 0
48iµ3
1
1
I(π) − 2 I 2 (π)]
+ e−iµπ [1 −
2iµ
8µ
Z
o
eiµπ π −2iµt
e−iπµ 3
3 |Imµ|π
+
e
q(t)dt + A(π, −µ) +
I
(π)
+
o(1/µ
)e
2iµ 0
48iµ3
n
1
1
−
iµeiµπ [1 +
I(π) − (8µ2 )−1 I 2 (π)]
2iµ
2iµ
Z
e−iµπ π 2iµt
1 iµπ 3
+
e
q(t)dt + B(π, µ) −
e I (π) + o(1/µ2 )e|Imµ|π
2
48µ2
0
1
− [−iµe−iµπ [1 −
I(π) − (8µ2 )−1 I 2 (π)]
2iµ
EJDE-2009/27
+
eiµπ
2
REGULARIZED TRACE
Z
π
e−2iµt q(t)dt + B(π, −µ) −
0
5
o
1 −iµπ 3
2 |Imµ|π
e
I
(π)
+
o(1/µ
)e
]
48µ2
b n iµπ
1
1
e [1 +
I(π) − 2 I 2 (π)]
2iµ
2iµ
8µ
Z
e−iµπ π 2iµt
e
q(t)dt + o(1/µ2 )e|Imµ|π
−
2iµ 0
Z
o
1
1 2
eiµπ π −2iµt
−iµπ
− [e
[1 −
e
I(π) − 2 I (π)] −
q(t)dt + o(1/µ2 )e|Imµ|π ] .
2iµ
8µ
2iµ 0
θ
− (−1)
Define ∆0 (µ) =
sion, gives
−(−1)θ b iπµ
(e
2iµ
− e−iπµ ). Combining like terms in the above expres-
Z π
Z
1 −iπµ π 2iµt
e−2iµt q(t)dt]
e
q(t)dt − eiπµ
[e
2iµ
0
0
Z π
Z t
Z π
Z t
1
+ 2 [eiπµ
e−2iµt q(t)dt
q(s) ds + e−iπµ
e2iµt q(t)dt
q(s) ds
4µ
0
0
0
0
Z π
Z t
Z π
Z t
− eiπµ
q(t)dt
e−2iµs q(s)ds − e−iπµ
q(t)dt
e2iµs q(s)ds]
∆(µ) = ∆0 (µ) −
0
0
0
0
b eiπµ + e−iπµ
eiπµ − e−iπµ 2
− (−1)
{
I(π) −
I (π)
2iµ
2iµ
8µ2
Z
Z π
1 −iπµ π 2iµt
−
(e
e
q(t)dt + eiπµ
e−2iµt q(t)dt)} + o(1/µ3 )e|Imµ|π .
2iµ
0
0
(2.14)
θ
3. Calculation of the regularized trace
First, consider the case hqi = 0. Formula (2.14) is then noticeably simplified:
∆(µ) = ∆0 (µ)(1 + r(µ)),
(3.1)
where
r(µ)
Z
Z π
1 n
1 −iπµ π 2iµt
−
[e
e
q(t)dt − eiπµ
e−2iµt q(t)dt]
∆0 (µ)
2iµ
0
0
Z
Z t
Z π
Z t
1 iπµ π −2iµt
−iπµ
2iµt
e
q(t)dt
+ 2 e
q(s) ds + e
e
q(t)dt
q(s) ds
4µ
0
0
0
0
Z π
Z t
Z π
Z t
− eiπµ
q(t)dt
e−2iµs q(s)ds − e−iπµ
q(t)dt
e2iµs q(s)ds
0
0
0
0
Z π
Z π
o
−iπµ
2iµt
iπµ
−2iµt
θ+1 b
(e
e
q(t)dt + e
e
q(t)dt) + o(1/µ3 )e|Imµ|π .
+ (−1)
2
4µ
0
0
(3.2)
Integrating by parts the terms on the right-hand side of (3.2), we have
=
r(µ) =
(−1)θ+1
{(eiπµ + e−iπµ )(q(π) − q(0))
4bµ sin πµ
Z π
Z π
q(π) + q(0)
iπµ
−2iµt 0
−iπµ
+ o(1/µ2 ).
− [e
e
q (t)dt + e
e2iµt q 0 (t)dt]} +
4µ2
0
0
(3.3)
6
A. MAKIN
EJDE-2009/27
Denote by ΓN the circle of radius N +1/2 centered at the origin. It is well known [8]
that the eigenvalues of (1.1)-(1.2) form a sequence λn = µ2n , where µn = n + o(1),
n = 1, 2, . . . . This asymptotic relation for the eigenvalues implies that, for all
sufficiently large N , the numbers µn with n ≤ N are inside ΓN , and the numbers
µn with n > N are outside ΓN . It follows that
I
N
X
1
∆0 (µ)
2
µ2
dµ;
2
µn =
2πi ΓN
∆(µ)
n=1
see [8]. Obviously, if µ ∈ ΓN , then |∆0 (µ)| ≥ c1 e|Imµ|π /|µ| (c1 > 0). This inequality
and the Riemann lemma [8] imply that maxµ∈ΓN |r(µ)| → 0 as N → ∞. Combining
this with (3.1) yields
I
I
1
1
r0 (µ) ∆0 (µ)
∆0 (µ)
µ2
dµ =
µ2 0
+
dµ
2πi ΓN
∆(µ)
2πi ΓN
∆0 (µ) 1 + r((µ)
I
N
X
1
=2
n2 +
µ2 d ln(1 + r(µ))
(3.4)
2πi
Γ
N
n=1
I
N
X
1
2
=2
n −
2µ ln(1 + r(µ))dµ.
2πi ΓN
n=1
Expanding ln(1 + r(µ)) by the Maclaurin formula and applying (3.3) and the Riemann lemma [8], we find that
ln(1 + r(µ)) = r(µ) −
(q(π) − q(0))2
o(1)
cot2 πµ + 2
2b2 µ2
µ
(3.5)
on ΓN . Evidently,
lim
|Imµ|→∞
(cot2 πµ + 1) = 0.
(3.6)
It follows from (3.5) and (3.6) that
I
1
2µ ln(1 + r(µ))dµ
2πi ΓN
I
(−1)θ+1
1
=
{(eiπµ + e−iπµ )(q(π) − q(0))
2πi ΓN 2b sin πµ
Z π
Z π
− (eiπµ
e−2iµt q 0 (t)dt + e−iπµ
e2iµt q 0 (t)dt)}
0
0
(q(π) − q(0))2 2
q(π) + q(0) o(1) −
cot πµ +
+
dµ
2
b µ
2µ
µ
Z
N
h
X
iπn
θ+1
−iπn
iπn
=
(−1)
(e
+e
)(q(π) − q(0)) − (e
0
n=−N
−int
Z
π
Z
+e
i
e2int q 0 (t)dt) (2bπ cos πn)
0
I
1
+
2πi
[
ΓN
θ+1
=
2(−1)
πb
(q(π) − q(0))2
q(π) + q(0)
+
]dµ + o(1)
b2 µ
2µ
N
X
n=1
(q(π) − q(0)
π
e−2int q 0 (t)dt
EJDE-2009/27
REGULARIZED TRACE
Z
−
π
q 0 (t) cos(2nt)dt) +
0
7
q(π) + q(0)
(q(π) − q(0))2
+
+ o(1).
b2
2
Combining this and (3.4), we obtain
2
N
X
µ2n = 2
n=1
N
X
n=1
n2 +
Z π
N
2(−1)θ X
(q(π) − q(0) −
q 0 (t) cos(2nt)dt)
πb n=1
0
(3.7)
q(π) + q(0)
(q(π) − q(0))2
−
−
+ o(1).
2
b
2
Passing to the limit as N → ∞ in (3.7), we have
Rπ
∞
X
(−1)θ (q(π) − q(0) − 0 q 0 (t) cos(2nt)dt) 2
λn − n −
πb
n=1
(3.8)
(q(π) − q(0))2
q(π) + q(0)
+
+
= 0.
2b2
4
Now consider the case hqi =
6 0. Let q̃(x) = q(x) − hqi. Then hq̃i = 0. Suppose
that (1.1)-(1.2) with potential q̃ has eigenvalues λ̃n . Then λ̃n = λn −hqi. According
to (3.8), we have
Rπ
∞
X
(−1)θ (q̃(π) − q̃(0) − 0 q̃ 0 (t) cos(2nt)dt) λ̃n − n2 −
πb
n=1
(q̃(π) − q̃(0))2
q̃(π) + q̃(0)
+
= 0.
2b2
4
Substituting the expressions for λ̃n and q̃(x) into this equality, we obtain formula
(1.3).
+
Acknowledgments. This work was supported by grant 07-01-00158 from the Russian Foundation for Basic Research.
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8
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Alexander Makin
Moscow State University of Instrument-Making and Informatics
Stromynka 20, Moscow, 107996, Russia
E-mail address: alexmakin@yandex.ru