Electronic Journal of Differential Equations, Vol. 2013 (2013), No. 237,... ISSN: 1072-6691. URL: or

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Electronic Journal of Differential Equations, Vol. 2013 (2013), No. 237, pp. 1–9.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
INVERSE SCATTERING PROBLEMS FOR
ENERGY-DEPENDENT STURM-LIOUVILLE EQUATIONS WITH
POINT δ-INTERACTION AND
EIGENPARAMETER-DEPENDENT BOUNDARY CONDITION
MANAF DZH. MANAFOV, ABDULLAH KABLAN
In memory of M. G. Gasymov, one of the pioneers of this subject
Abstract. We consider an inverse problem of the scattering theory for energydependent Sturm-Liouville equations on the half line [0, +∞) with point δinteraction and eigenparameter-dependent boundary condition. We define the
scattering data of the problem first, then consider the basic equation and study
an algorithm for finding the potentials with the given scattering data.
1. Introduction
We consider inverse scattering problem for the equation
− y 00 + q(x)y = λ2 y,
x ∈ (0, a) ∪ (a, +∞)
(1.1)
with the boundary condition
U (y) := λ2 (y 0 (0) − hy(0)) − (h1 y 0 (0) − h2 y(0)) = 0
and conditions at the point x = a,
(
y(a + 0) = y(a − 0) = y(a),
I(y) :=
y 0 (a + 0) − y 0 (a − 0) = 2iαλy(a),
(1.2)
(1.3)
where λ is a spectral parameter, q(x) is real-valued function satisfying the condition
Z ∞
(1 + x)|q(x)|dx < ∞,
0
and α < 0, h, h1 , h2 are real numbers such that
δ := hh1 − h2 > 0.
Notice that, we can understand problem (1.1) and (1.3) as one of the study of the
equation
y 00 + (λ2 − 2iλp(x) − q(x))y = 0, x ∈ (0, +∞),
(1.4)
2000 Mathematics Subject Classification. 34A55, 34B24, 47E05.
Key words and phrases. Inverse scattering; scattering data; point delta-interaction;
eigenvalue-dependent boundary condition.
c
2013
Texas State University - San Marcos.
Submitted July 29, 2103. Published October 24, 2013.
1
2
M. DZH. MANAFOV, A. KABLAN
EJDE-2013/237
when p(x) = αδ(x − a), where δ(x) is the Dirac function, (see [4]).
Sturm-Liouville spectral problems with potentials depending on the spectral parameter arise in various models of quantum and classical mechanics. For instance,
to this form can be reduced the corresponding evolution equations (such as the
Klein-Gordon equation [12], [21]) that are used to model interactions between colliding relativistic spinless particles. Then λ2 is related to the energy of the system,
this explaining the term “energy-dependent” in (1.4).
Problems of the form (1.4) have also appeared in the physical literature in the
context of scattering of waves and particles. In particular, in [9]-[11] the inverse
scattering problems for energy-dependent Schrödinger operators on the line are
studied, see also [2, 8, 13, 15, 19, 20].
The non-linear dependence of equation (1.4) on the spectral parameter λ should
be regarded as a spectral problem for a quadratic operator pencil. The problem
with p(x) ∈ W21 (0, 1) and q(x) ∈ L2 (0, 1) and with Robin boundary conditions
was discussed in [5]. Such problems for separated and nonseparated boundary
conditions were considered (see [1, 6, 22, 23] and the references therein).
The inverse scattering problem for equation (1.1) (in the case α = 0) with
boundary condition y 0 (0) − hy(0) = 0 was completely solved in [16] (in the case
h = ∞), in [14] (for arbitrary real number h). Similar problems were dealt with in
[3] and references in these works. The inverse scattering problem for equation (1.4)
(in the case α = 0) with spectral parameter in the boundary condition on the half
line were examined in [17], [18].
In this paper we consider the inverse problem scattering theory on the half line
[0, ∞) for the (1.4), (1.2) boundary-value problem.
2. Integral representation for the Jost solution and scattering data
In this section, we will find an integral representation for the Jost solution and
study some properties of scattering data of the problem.
Let us denote by e0 (x, λ) the solution of (1.1), when q(x) ≡ 0, satisfying conditions (1.3) and the condition at infinity
lim e0 (x, λ)e−iλx = 1 .
x→∞
This function is called the Jost solution.
It is obvious that the function e0 (x, λ) can be written as
(
(1 − α)eiλx + αeiλ(2a−x) , 0 < x < a
e0 (x, λ) =
eiλx ,
x>a
Analogously to [7] the following theorem can be proved.
R∞
Theorem 2.1. Let 0 (1 + x)|q(x)|dx < ∞. Then the Jost solution of (1.1) has
the form
Z
∞
K(x, t)eiλt dt
e(x, λ) = e0 (x, λ) +
(Im λ ≥ 0),
x
where for each fixed x 6= a the kernel K(x, ·) belong to the L1 (x, ∞) and
Z ∞
|K(x, t)|dt ≤ ecσ1 (x) − 1,
x
where σ1 (x) =
R∞
0
t|q(t)|dt, c = 1 − α2 .
(2.1)
EJDE-2013/237
INVERSE SCATTERING PROBLEMS
3
If the function q(x) is differentiable then the kernel K(x, t) satisfies the following
properties:
Kxx (x, t) − q(x)K(x, t) = Ktt (x, t), 0 < x < ∞, t > x,
( R∞
1
q(s)ds,
x>a
R∞
K(x, x) = 21 x
(1
−
α)
q(s)ds,
x<a
2
x
(2.2)
K(x, 2a − x + 0) − K(x, 2a − x − 0)
Z
Z x
α ∞
q(s)ds −
q(s)ds , x < a.
=
2 a
a
Now we learn some properties of scattering data of the problem. Since the
function q(x) is real-valued, it follows that for real λ together with e(x, λ) the
solution of (1.1) are also e(x, λ). Since the Wronskian of the two solutions y1 (x)
and y2 (x) of equation (1.1),
W {y1 (x), y2 (x)} = y10 (x)y2 (x) − y1 (x)y20 (x),
is independent of x, it coincides with its own value as x → ∞. Therefore
W {e(x, λ), e(x, λ)} = lim [e0 (x, λ)e(x, λ) − e(x, λ)e0 (x, λ)] = 2iλ.
x→+∞
(2.3)
Consequently for 0 6= λ ∈ R the pair {e(x, λ), e(x, λ)} is a fundamental system of
solutions of (1.1).
Let ϕ(x, λ) be the solution of (1.1) satisfying the initial condition
ϕ(0, λ) = λ2 − h1 ,
ϕ0 (0, λ) = λ2 h − h2 .
(2.4)
Lemma 2.2. The equality
(λ2
− h1
2iλϕ(x, λ)
= e(x, λ) − S(λ)e(x, λ)
− (λ2 h − h2 )e(0, λ)
)e0 (0, λ)
(2.5)
holds for all real λ 6= 0, where
S(λ) =
(λ2 − h1 )e0 (0, λ) − (λ2 h − h2 )e(0, λ)
(λ2 − h1 )e0 (0, λ) − (λ2 h − h2 )e(0, λ)
(2.6)
with
S(λ) = S(−λ) = [S(−λ)]−1
Proof. The functions e(x, λ) and e(x, λ) form a fundamental system of solutions of
(1.1) for all real λ 6= 0, then we can write
ϕ(x, λ) = c1 (λ)e(x, λ) + c2 (λ)e(x, λ),
(2.7)
where
(λ2 − h1 )e0 (0, λ) − (hλ2 − h2 )e(0, λ)
,
2iλ
(λ2 − h1 )e0 (0, λ) − (hλ2 − h2 )e(0, λ)
c2 (λ) =
.
2iλ
To prove that ψ(λ) ≡ (λ2 − h1 )e0 (0, λ) − (λ2 h − h2 )e(0, λ) 6= 0 for all real λ 6= 0,
we assume that λ0 ∈ (−∞, +∞)\{0} such that
c1 (λ) = −
(λ20 − h1 )e0 (0, λ0 ) − (λ20 h − h2 )e(0, λ0 ) = 0
4
M. DZH. MANAFOV, A. KABLAN
or
EJDE-2013/237
(λ20 h − h2 )
e(0, λ0 )
λ20 − h1
e0 (0, λ0 ) =
From (2.3), we obtained the equation
|e(0, λ0 )|2
|λ20
0
= 2iλ0 ;
− h1 |2
i.e., we have a contradiction. Hence ψ(λ) 6= 0 for all real λ 6= 0. Dividing both
sides of (2.7) by ψ(λ)/2iλ we obtain (2.5) where S(λ) is defined with (2.6). Since
ψ(λ) = ψ(−λ) = [ψ(−λ)]−1
we have
S(λ) = S(−λ) = [S(−λ)]−1 .
The function S(λ) is called the scattering function of the problem (1.1)-(1.2)(1.3).
Lemma 2.3. The function ψ(λ) may have only a finite number of zeros in the
half-plane Im λ > 0. Moreover, all these zeros are simple and lie in the imaginary
axis.
Proof. Since for real values of λ 6= 0 the inequality ψ(λ) 6= 0 holds, the only possible
real zero of the function ψ(λ) is λ = 0. Since the function ψ(λ) is analytic on the
upper half-plane it follows that its zeros form a (finite or countable) set.
Let us show that this set is bounded. Assume the converse and suppose that
there exist λk such that |λk | → ∞, with Im λk > 0 and
(λ2k − h1 )e0 (0, λk ) = (λ2k h − h2 )e(0, λk ).
Then as |λk | → ∞ yields limk→∞ e(0, λk ) = 0. On the other hand, it follows from
(2.3) that limk→∞ e(0, λk ) = 1. The resulting contradiction shows that the set
{λk } is bounded. Thus, the zeros of the function ψ(λ) form a bounded finite or
countable set whose unique limiting point can only be zero.
Now let us show that all the zeros of the function ψ(λ) lie on the imaginary axis.
Suppose that λ1 and λ2 are some zeros of the function ψ(λ). Since
−e00 (x, λ1 ) + q(x)e(x, λ1 ) = λ21 e(x, λ1 ),
−e00 (x, λ2 ) + q(x)e(x, λ2 ) = λ22 e(x, λ2 ).
If the first equation is multiplied by e(x, λ2 ), second equation is multiplied by
e(x, λ1 ) and subtracting them side by side and finally integrating over the interval
[0, +∞), the equality
Z ∞
2
2
(λ1 − λ2 )
e(x, λ1 )e(x, λ2 )dx + W {e(x, λ1 ), e(x, λ2 )}(|a0 + |+∞
a )=0
0
is attained.
If conditions (1.3) and
W {e(x, λ1 ), e(x, λ2 )}x=a−0 = W {e(x, λ1 ), e(x, λ2 )}x=a+0
are considered, then
Z
2
2
(λ1 − λ2 )
0
∞
e(x, λ1 )e(x, λ2 )dx + W {e(x, λ1 ), e(x, λ2 )}x=0 = 0
(2.8)
EJDE-2013/237
INVERSE SCATTERING PROBLEMS
5
is obtained. On the other hand, according to the definition of the function ψ(λ),
the following relation holds
ψ(λj ) = (λ2j − h1 )e0 (0, λj ) − (λ2j h − h0 )e(0, λj ) = 0,
j = 1, 2.
Also, by (1.2) we can write
[e0 (0, λ1 ) − he(0, λ1 )][e0 (0, λ2 ) − he(0, λ2 )] −2
(λ2 − λ21 ).
δ
(2.9)
Therefore, using (2.8) and (2.9) we have
Z ∞
1
(λ21 −λ−2
){
e(x, λ1 )e(x, λ2 )dx+ [e0 (0, λ1 )−he(0, λ1 )][e0 (0, λ2 )−he(0, λ2 )]} = 0.
2
δ
0
(2.10)
In particular, the choice λ2 = λ1 at (2.10) implies that λ1 = ik1 , where k1 ≥ 0.
Therefore, zeros of the function ψ(λ) can lie only on the imaginary axis. Now, let us
prove that function ψ(λ) has zeros in finite numbers. Since we can give an estimate
for the distance between the neighboring zeros of the function ψ(λ), it follows that
the number of zeros is finite, (see [16, p. 186]).
W {e(x, λ1 ), e(x, λ2 )}x=0 =
Let
m−2
p
Z
≡
0
∞
1
|e(x, ikp )|2 dx + |e0 (0, ikp ) − he(0, ikp )|2 ,
δ
p = 1, 2, . . . , n.
(2.11)
These numbers are called the norming constants for the boundary problem (1.4),
(1.2).
The collections {S(λ), (−∞ < λ < +∞); kp , mp (p = 1, 2, . . . , n)} are called the
scattering data of the boundary value problem (1.1)-(1.2)-(1.3) or (1.4)-(1.2).
3. Basic equation of the inverse scattering problem
To derive the basic equation for the kernel K(x, t) of the solution (2.1), we use
equality (2.5). Substituting expression (2.1) for e(x, λ) into this equality, we get
2iλϕ(x, λ)
− e0 (x, λ) + S0 (λ)e0 (x, λ)
ψ(λ)
Z ∞
Z ∞
=
K(x, t)e−iλt dt + [S0 (λ) − S(λ)]e0 (x, λ) − S0 (λ)
K(x, t)eiλt dt
x
x
Z ∞
iλt
+ (S0 (λ) − S(λ))
K(x, t)e dt.
(3.1)
x
Multiplying this equality by (1/2π)eiλr and integrating over λ from −∞ to +∞,
for r > x, at the right-hand side we obtain
Z ∞
1
K(x, r) +
[S0 (λ) − S(λ)]e0 (x, r)eiλr dλ
2π −∞
Z ∞
n 1 Z ∞
o
−
K(x, t)
S0 (λ)eiλ(t+r) dλ dt
(3.2)
2π −∞
x
Z ∞
Z
∞
n 1
o
+
K(x, t)
[S0 (λ) − S(λ)]eiλ(t+r) dλ dt.
2π −∞
x
6
M. DZH. MANAFOV, A. KABLAN
Now we will compute the integral (1/2π)
forms we obtain
R∞
−∞
EJDE-2013/237
S0 (λ)eiλ(t+r) dλ. By elementary trans∞
S0 (λ) =
X
τ2 − 1
+ τ e−2iλa = (τ 2 − 1)
(−1)k τ k e2iλak + τ e−2iλa
2iλa
1 + τe
k=0
where τ = α/(α − 1). Thus we have
Z ∞
∞
X
1
S0 (λ)eiλ(t+r) dλ = (τ 2 − 1) (−1)k τ k δ(t + r + 2ak) + τ δ(t + r − 2a).
2π −∞
k=0
Consequently (3.2) can be written as
Z ∞
K(x, r) + FS (x, r) +
K(x, t)F0S (t + r)dt − τ K(x, 2a − r)
x
− (τ 2 − 1)
∞
X
(−1)k τ k K(x, −2ak − r),
k=0
where
Z ∞
1
[S0 (λ) − S(λ)]eiλx dλ,
F0S (x) ≡
2π −∞
FS (x, r) ≡ (1 − α)F0S (x + r) + αF0S (2a − x + r).
We note that K(x, r) = 0 for r < x (see [7]). Therefore, for r > x, (2.4) takes the
form
Z ∞
K(x, r) + FS (x, r) +
K(x, r)F0S (t + r)dt − τ K(x, 2a − r).
(3.3)
x
On the left-hand side of (3.1) with help of Jordan’s lemma and the residue
theorem and by taking Lemma 2.3 into account for r > x, we obtain
−
n
X
2ikp ϕ(x, ikp )
p=1
e−kp r .
ψ 0 (ikp )
(3.4)
From the definition of norming constants mp (p = 1, 2, . . . , n) in (2.11) we have
−
n
X
2ikp ϕ(x, ikp )
p=1
=−
=−
ψ 0 (ikp )
e−kp r
n
X
2ikp e−kp r e(x, ikp )
[e0 (0, ikp ) − he(0, ikp )]ψ 0 (ikp )
p=1
n
X
(3.5)
m2p e(x, ikp )e−kp r
p=1
=−
n
X
p=1
m2p {e(x, ikp )e−kp (x+r) +
Z
∞
K(x, t)e−kp (x+r) dt}.
x
Thus, for x < min(r, 2a − r), by taking (3.3) and (3.5) into account, from (3.2) we
derive the relation
Z ∞
α
K(x, 2a − r) = 0
(3.6)
K(x, r) + F (x, r) +
K(x, t)F0 (t + r)dt +
1
−
α
x
EJDE-2013/237
INVERSE SCATTERING PROBLEMS
where
F0 (x) = F0S (x) +
n
X
m2p e−kp r ,
p=1
F (x, r) = FS (x, r) +
n
X
7
(3.7)
m2p e0 (x, ikp )e−kp (x+r) .
p=1
Equation (3.6) is called the basic equation of the inverse problem of the scattering
theory for the boundary problem (1.1)-(1.2)-(1.3) or (1.4)-(1.2). The basic equation
is different from the classical equation of Marchenko (see [16]). Thus, we have
proved the following theorem.
Theorem 3.1. For each x ≥ 0, the kernel K(x, r) of the solution (2.1) satisfies
the basic equation (3.6).
Obviously, to form the basic equation (3.6), it suffices to know the functions
F0 (x) and F (x, r). In turn, in order to find the functions F0 (x), F (x, r), it suffices
to know only the scattering data {S(λ) (−∞ < λ < ∞); kp , mp (p = 1, 2, . . . , n)}.
Given the scattering data, we can use formulas (3.7) to construct the functions
F0 (x), F (x, r) and write down the basic equation (3.6) for the unknown solution
K(x, r), the function q(x) may be found from formula (2.2).
Theorem 3.2. Equation (3.6) has a unique solution K(x, ·) ∈ L1 (x, +∞) for each
fixed x ≥ 0.
Proof. It is easy to show that for each fixed x ≥ 0 the operator
(
f (r),
x>a
(Mx f )(r) =
α
f (r) + 1−α f (2a − r), x < a,
acting in the space L1 (x, +∞) (and also in L2 (x, +∞)) is invertible. Therefore,
basic equation (3.6) is equivalent to
K(x, r) + (Mx )−1 F (x, r) + (Mx )−1 (F0 K(x, ·))(r) = 0;
i.e. to the equation with a compact operator (Mx )−1 F (for the compactness of
F , see [16, Lemma 3.3.1]). To prove the theorem, it is sufficient to show that the
homogeneous equation
Z ∞
α
fx (2a − r) +
fx (r) +
fx (t)F0 (t + r)dt = 0, x < min(r, 2a − r), (3.8)
1−α
x
has only the trivial solution fx (r) ∈ L1 (x, +∞). We can show that (see [16]) the
function F0 (r) belongs to the space L2 [0, +∞), is absolutely continuous on all the
intervals not containing the point 2a; and for all β ≥ 0
Z ∞
Z ∞
|F0 (r)|dr < +∞,
(1 + r)|F 0 (r)|dr < +∞.
β
β
Therefore, the function F (r) and the solution fx (r) are together bounded on the
semi-axis x ≤ y < +∞. Consequently, fx (r) ∈ L2 (x, +∞).
Now let us multiply (3.8) by fx (r) and integrate with respect to y over the
interval (x, +∞). Using (3.6), (3.7) and Parseval’s identity
Z ∞
Z ∞
1
|fx (r)|2 dr =
|fex (λ)|2 dλ,
2π
x
−∞
8
M. DZH. MANAFOV, A. KABLAN
α
1−α
Z
∞
fx (2a − r)fx (r)dr =
x
where
Z
1
2π
Z
EJDE-2013/237
∞
S0 (λ)fex (λ)fex (−λ)dλ,
−∞
∞
fex (λ) =
fx (t)e−iλt dt,
x
we obtain
Z ∞
Z ∞
n
X
1
1
|fex (λ)|2 dλ +
S(λ)fex (−λ)fex (λ)dλ = 0.
mp |fex (−ikp )|2 +
2π −∞
2π
−∞
p=1
Since |S(λ)| = |S(−λ)|, we obtain the estimate
Z ∞
Z ∞
1
1
|fex (λ)|2 dλ ≤
|S(λ)||fex (−λ)||fex (λ)|dλ
2π −∞
2π −∞
Z ∞
1
|fex (−λ)|2 + |fex (λ)|2
|S(λ)|
≤
dλ
2π −∞
2
Z ∞
1
=
|S(λ)||fex (λ)|2 dλ
2π −∞
or
Z ∞
1
{1 − |S(λ)|}|fex (λ)|2 dλ ≤ 0.
2π −∞
It follows from the above that fex (λ) ≡ 0 since 1 − |S(λ)| > 0 for all λ 6= 0. Thus,
the basic equation (3.6) is uniquely solvable.
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Manaf Dzh. Manafov
Faculty of Arts and Sciences, Department of Mathematics, Adiyaman University, Adiyaman, 02040, Turkey
E-mail address: mmanafov@adiyaman.edu.tr
Abdullah Kablan
Faculty of Arts and Sciences, Department of Mathematics, Gaziantep University,
Gaziantep, 27310, Turkey
E-mail address: kablan@gantep.edu.tr
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