Spectral problem for discontinuous integro-differential operator
A representation of solutions of a discontinuous integro-differential operator is obtained. The asymptotic behavior of the eigenvalues and eigenfunctions of this operator is described.
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| Дата: | 2012 |
|---|---|
| Автори: | , , , |
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| Опубліковано: |
Institute of Mathematics, NAS of Ukraine
2012
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Ukrains’kyi Matematychnyi Zhurnal| _version_ | 1860508493875773440 |
|---|---|
| author | Keskin, B. Ozkan, S. A. Кескін, Б. Озкан, С. А. |
| author_facet | Keskin, B. Ozkan, S. A. Кескін, Б. Озкан, С. А. |
| author_sort | Keskin, B. |
| baseUrl_str | https://umj.imath.kiev.ua/index.php/umj/oai |
| collection | OJS |
| datestamp_date | 2020-03-18T19:29:46Z |
| description | A representation of solutions of a discontinuous integro-differential operator is obtained. The asymptotic behavior of the eigenvalues and eigenfunctions of this operator is described. |
| first_indexed | 2026-03-24T02:26:05Z |
| format | Article |
| fulltext |
UDC 517.9
B. Keskin, A. S. Ozkan (Cumhuriyet Univ., Sivas, Turkey)
SPECTRAL PROBLEM FOR DISCONTINUOUS
INTEGRO-DIFFERENTIAL OPERATOR
СПЕКТРАЛЬНА ЗАДАЧА ДЛЯ РОЗРИВНОГО
IНТЕГРО-ДИФЕРЕНЦIАЛЬНОГО ОПЕРАТОРА
A representation of solutions of a discontinuous integro-differential operator is obtained. The asymptotic behavior of the
eigenvalues and eigenfunctions of this operator is described.
Отримано зображення розв’язкiв розривного iнтегро-диференцiального оператора. Описано асимптотичну поведiн-
ку власних чисел та власних функцiй цього оператора.
1. Introduction. Let us consider the boundary-value problem L :
`y : = −y′′ + q(x)y +
x∫
0
M(x− t)y(t)dt = λy, x ∈ (0, d) ∪ (d, π), (1)
U(y) : = y(0) = 0, (2)
V (y) : = y(π) = 0, (3)y(d+ 0) = αy(d− 0),
y′(d+ 0) = α−1y′(d− 0),
(4)
where α > 0, α 6= 1, d ∈ (0, π) and q(x) is a real valued function in L2(0, π), λ is the spec-
tral parameter. Moreover, the functions (π − x)M(x) and
∫ x
0
M(t)dt are real valued functions in
L2(0, π).
In [1] perturbation of a Sturm – Liouville operator by a Volterra integral operator is considered.
The presence of an ”aftereffect”in a mathematical model produces qualitative changes in the study
of the inverse problem.
Boundary-value problems with discontinuities inside the interval often appear in mathematics,
mechanics, physics, geophysics and other branches of natural properties. The inverse problem of
reconstructing the material properties of a medium from data collected outside of the medium is of
central importance in disciplins ranging from engineering to the geosciences.
For example, discontinuous inverse problems appear in electronics for constructing parameters of
heterogeneous electronic lines with desirable technical characteristics [4, 6]. Spectral information can
be used to reconstruct the permittivity and conductivity profiles of a one-dimentional discontinuous
medium [3, 5]. Further, it is known that spectral problems play an important role for investigating
some nonlinear evolution equations of mathematical physics.
Consider the operator
c© B. KESKIN, A. S. OZKAN, 2012
ISSN 1027-3190. Укр. мат. журн., 2012, т. 64, № 2 277
278 B. KESKIN, A. S. OZKAN
T : = − d2
dx2
+ q(x) +M,
with the domain
D(T ) = {y : y(x) and y′(x) are absolutely continuous in [0, d)∪ (d, π], `y ∈ L2(0, π), y(0) = 0,
y(π) = 0, y(d+ 0) = αy(d− 0), y′(d+ 0) = α−1y′(d− 0)}, such thatMy=
∫ x
0
M(x− t)y(t)dt,
whereM is an integral operator.
Let the function ϕ(x, λ) be the solution of equation (1) satisfying the initial conditions ϕ(0, λ) =
= 0, ϕ′(0, λ) = 1 and the jump conditions (4).
It is shown in [1] that, the solution ϕ(x, λ) has a representation as follows:
ϕ(x, λ) =
sin
√
λx√
λ
+
x∫
0
K(x, t)
sin
√
λt√
λ
dt,
where K(x, t) is a continuous function and K(x, 0) = 0.
Firstly, let us try to get a similar representation for the solution ϕ(x, λ) as follows:
ϕ(x, λ) =
sin
√
λx√
λ
+
x∫
−x
K(x, t)
sin
√
λt√
λ
dt, x < d,
α+ sin
√
λx√
λ
+ α−
sin
√
λ(2d− x)√
λ
+
x∫
−x
K(x, t)
sin
√
λt√
λ
dt, x > d.
(5)
It is clearly shown that the integral equation for the solution ϕ(x, k) is of the following type:
for x < d,
ϕ(x, k) =
sin kx
k
+
x∫
0
sin k(x− t)
k
q(t)ϕ(t, k)dt+
+
x∫
0
t∫
0
sin k(x− t)
k
M(t− τ)ϕ(τ, k)dτdt,
and, for x > d,
ϕ(x, k) = α+ sin kx
k
+ α−
sin k(2d− x)
k
+
1
k
d∫
0
(α+ sin k(x− t)+
+α− sin k(2d− x− t))q(t)ϕ(t, k)dt+
+
1
k
d∫
0
t∫
0
(α+ sin k(x− t) + α− sin k(2d− x− t))M(t− τ)ϕ(τ, k)dτdt+
ISSN 1027-3190. Укр. мат. журн., 2012, т. 64, № 2
SPECTRAL PROBLEM FOR DISCONTINUOUS . . . 279
+
1
k
x∫
d
sin k(x− t)q(t)ϕ(t, k)dt+
1
k
x∫
d
t∫
0
sin k(x− t)M(t− τ)ϕ(τ, k)dτdt,
where α± =
1
2
(
α+
1
α
)
and
√
λ = k.
In order to be solution of above equations of the function which has representation (5), the
equality
x∫
−x
K(x, t)
sin kt
k
dt =
1
k
d∫
0
(α+ sin k(x− t) + α− sin k(2d− x− t))q(t)sin kt
k
dt+
+
1
k
d∫
0
(α+ sin k(x− t) + α− sin k(2d− x− t))q(t)
t∫
−t
K(t, τ)
sin kτ
k
dτ
dt+
+
1
k
d∫
0
t∫
0
(α+ sin k(x− t) + α− sin k(2d− x− t))M(t− τ)
sin kτ
k
dτdt+
+
1
k
d∫
0
t∫
0
(α+ sin k(x− t) + α− sin k(2d− x− t))M(t− τ)
τ∫
−τ
K(τ, ξ)
sin kξ
k
dξ
dτdt+
+
1
k
x∫
d
sin k(x− t)q(t)
{
α+ sin kt
k
+ α−
sin k(2d− t)
k
}
dt+
+
1
k
x∫
d
sin k(x− t)q(t)
t∫
−t
K(t, τ)
sin kτ
k
dτ
dt+
+
1
k
x∫
d
t∫
0
sin k(x− t)M(t− τ)
{
α+ sin kτ
k
+ α−
sin k(2d− τ)
k
}
dτdt+
+
1
k
x∫
d
t∫
0
sin k(x− t)M(t− τ)
τ∫
−τ
K(τ, ξ)
sin kξ
k
dξ
dτdt
must be hold. For d < x < 2d, −x < t < x − 2d < 2d − x, it is easy to get the following integral
equation:
K(x, t) =
α+
2
x∫
(x−t)/2
q(s)ds+
α+
2
d∫
d−(x+t)/2
q(s)ds+
α+
2
x∫
0
t+x−s∫
t+s−x
M(s− τ)dτds+
ISSN 1027-3190. Укр. мат. журн., 2012, т. 64, № 2
280 B. KESKIN, A. S. OZKAN
+
α−
2
d∫
0
t+2d−x−s∫
t+s+x−2d
M(s− τ)dτds− α−
2
x∫
d
t−s+2d+x∫
t+s+2d−x
M(s− τ)dτds+
+
α+
2
d∫
0
q(s)
t+2d−x−s∫
t+s+x−2d
K(s, τ)dτds+
α−
2
d∫
0
q(s)
t+x−s∫
t−x+s
K(s, τ)dτds+
+
α+
2
d∫
0
s∫
0
t+x−s∫
t+s−x
M(s− τ)K(τ, ξ)dξdτds+
α−
2
d∫
0
s∫
0
t+2d−x−s∫
t+s+x−2d
M(s− τ)K(τ, ξ)dξdτds+
+
1
2
x∫
d
q(s)
t+x−s∫
t−x+s
K(s, τ)dτds+
1
2
d∫
0
s∫
0
t+x−s∫
t−x+s
M(s− τ)K(τ, ξ)dξdτds.
The same integral equations can be obtained for (i) 2d < x,−x < t < 2d− x; (ii) d < x < 2d,
x − 2d < t < 2d − x; (iii) 2d < x, 2d − x < t < x − 2d; (iv) 2d < x, 2d − x < t < x and
(v) d < x < 2d, x− 2d < t < x.
For solving this integral equation:
put
K0(x, t) =
α+
2
x∫
(x−t)/2
q(s)ds+
α+
2
d∫
d−(x+t)/2
q(s)ds+
α+
2
x∫
0
t+x−s∫
t+s−x
M(s− τ)dτds+
+
α−
2
d∫
0
t+2d−x−s∫
t+s+x−2d
M(s− τ)dτds− α−
2
x∫
d
t−s+2d+x∫
t+s+2d−x
M(s− τ)dτds
and
Kn+1(x, t) =
α+
2
d∫
0
q(s)
t+2d−x−s∫
t+s+x−2d
Kn(s, τ)dτds+
α−
2
d∫
0
q(s)
t+x−s∫
t−x+s
Kn(s, τ)dτds+
+
α+
2
d∫
0
s∫
0
t+x−s∫
t+s−x
M(s− τ)Kn(τ, ξ)dξdτds+
α−
2
d∫
0
s∫
0
t+2d−x−s∫
t+s+x−2d
M(s− τ)Kn(τ, ξ)dξdτd+
+
1
2
x∫
d
q(s)
t+x−s∫
t−x+s
Kn(s, τ)dτds+
1
2
d∫
0
s∫
0
t+x−s∫
t−x+s
M(s− τ)Kn(τ, ξ)dξdτds.
It is shown by the successive approximation method, the following theorem is true.
ISSN 1027-3190. Укр. мат. журн., 2012, т. 64, № 2
SPECTRAL PROBLEM FOR DISCONTINUOUS . . . 281
Theorem 1. For the solution of (1) which satisfies the initial conditions ϕ(0) = 0 ϕ′(0) = 1
and the jump condition (4) has the form,
ϕ(x, k) =
sin kx
k
+
x∫
−x
K(x, t)
sin kt
k
dt, x < d,
α+ sin kx
k
+ α−
sin k(2d− x)
k
+
x∫
−x
K(x, t)
sin kt
k
dt, x > d,
and also
x∫
−x
|K(x, t)| dt ≤ eCσ(x) − 1,
where
σ(x) =
x∫
0
(x− t)
|q(t)|+ t∫
0
|M(t− τ)| dτ
dt
and C = α+ + |α−|+ 1.
Denote ∆(λ) = ϕ(π, λ).The eigenvalues {λn}n≥1 of the boundary-value problem L coincide
with the zeros of the function ∆(λ).
Theorem 2. The eigenvalues λn and eigenfunctions ϕ(x, kn) of the problem (1) – (4) satisfy
the following asymptotic estimates for sufficiently large n :√
λn = kn = k0n + o
(
1
k0n
)
, (6)
ϕ(x, kn) =
sin k0nx
k0n
+ o
(
1
k0n
)
, x < d, (7)
ϕ(x, kn) = α+ sin k0nx
k0n
+ α−
sin k0n(2d− x)
k0n
+ o
(
1
k0n
)
, x > d, (8)
where k0n are the roots of ∆0(k) : = α+ sin kπ
k
+ α−
sin k(π − 2d)
k
and k0n = n+ hn, hn ∈ `∞.
Proof. From (5)
∆(k) = α+ sin kπ
k
+ α−
sin k(2d− π)
k
+
π∫
−π
K(π, t)
sin kt
k
dt.
Denote Γn : = {λ : λ = k2, |k| = k0n + δ}, n = 0, 1, . . . . Since ∆(k)−∆0(k) = o
(
e| Im kπ|
|k|
)
and |∆0(k)| ≥ C
e| Im kπ|
|k|
for all λ ∈ Γn, we establish by the Rouche’s theorem that kn =
ISSN 1027-3190. Укр. мат. журн., 2012, т. 64, № 2
282 B. KESKIN, A. S. OZKAN
= k0n + εn, where εn = o (1) . Moreover, εn = o
(
1
k0n
)
is obtained from the equality ∆(kn) =
=
(
∆′0(k
0
n) + o(1)
)
εn + o
(
1
k0n
)
= 0. This completes the proof of (6).
From (5) and (6), one can easily prove that the asymptotic formulae (7) and (8) are true.
Theorem 2 is proved.
1. Freiling G., Yurko V. A. Inverse Sturm – Liouville problems and their applications. – Huntington, NY: Nova Sci.,
2001.
2. Levitan B. M., Sargsyan I. S. Sturm – Liouville and Dirac Operators. – Moscow: Nauka, 1988 (in Russian).
3. Levitan B. M. Inverse Sturm – Liouville problems. – Moscow: Nauka, 1984 (Eglish transl.: Utrecht: VNU Sci. Press,
1987).
4. Litvinenko O. N., Soshnikov V. I. The theory of heteregeneous lines and their applications in radio engineering. –
Moscow: Radio, 1964 (in Russian).
5. Marchenko V. A. Sturm – Liouville operators and their applications. – Kiev: Naukova Dumka, 1977 (English transl.:
Basel: Birkhäuser, 1986).
6. Meschanov V. P.,Feldstein A. L. Automatic design of directional couplers. – Moscow: Sviaz, 1980.
Received 14.12.09
ISSN 1027-3190. Укр. мат. журн., 2012, т. 64, № 2
|
| id | umjimathkievua-article-2575 |
| institution | Ukrains’kyi Matematychnyi Zhurnal |
| keywords_txt_mv | keywords |
| language | Ukrainian English |
| last_indexed | 2026-03-24T02:26:05Z |
| publishDate | 2012 |
| publisher | Institute of Mathematics, NAS of Ukraine |
| record_format | ojs |
| resource_txt_mv | umjimathkievua/f2/1f0f087d20d05c47df02f1d10ecd43f2.pdf |
| spelling | umjimathkievua-article-25752020-03-18T19:29:46Z Spectral problem for discontinuous integro-differential operator Спектральна задача для розривного iнтегро-диференцiального оператора Keskin, B. Ozkan, S. A. Кескін, Б. Озкан, С. А. A representation of solutions of a discontinuous integro-differential operator is obtained. The asymptotic behavior of the eigenvalues and eigenfunctions of this operator is described. Отримано зображення розв’язкiв розривного iнтегро-диференцiального оператора. Описано асимптотичну поведiнку власних чисел та власних функцiй цього оператора. Institute of Mathematics, NAS of Ukraine 2012-02-25 Article Article application/pdf https://umj.imath.kiev.ua/index.php/umj/article/view/2575 Ukrains’kyi Matematychnyi Zhurnal; Vol. 64 No. 2 (2012); 277-282 Український математичний журнал; Том 64 № 2 (2012); 277-282 1027-3190 uk en https://umj.imath.kiev.ua/index.php/umj/article/view/2575/1909 https://umj.imath.kiev.ua/index.php/umj/article/view/2575/1910 Copyright (c) 2012 Keskin B.; Ozkan S. A. |
| spellingShingle | Keskin, B. Ozkan, S. A. Кескін, Б. Озкан, С. А. Spectral problem for discontinuous integro-differential operator |
| title | Spectral problem for discontinuous integro-differential operator |
| title_alt | Спектральна задача для розривного iнтегро-диференцiального оператора |
| title_full | Spectral problem for discontinuous integro-differential operator |
| title_fullStr | Spectral problem for discontinuous integro-differential operator |
| title_full_unstemmed | Spectral problem for discontinuous integro-differential operator |
| title_short | Spectral problem for discontinuous integro-differential operator |
| title_sort | spectral problem for discontinuous integro-differential operator |
| url | https://umj.imath.kiev.ua/index.php/umj/article/view/2575 |
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