\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2017 (2017), No. 12, pp. 1--8.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2017 Texas State University.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2017/12\hfil An inverse spectral problem]
{An inverse spectral problem for Sturm-Liouville operator with integral delay}

\author[Manaf Dzh. Manafov \hfil EJDE-2017/12\hfilneg]
{Manaf Dzh. Manafov}

\dedicatory{In memory of M. G. Gasymov (1939-2008)}

\address{Manaf Dzh. Manafov \newline
Ad{\i}yaman University,
Faculty of Science and Arts,
Department of Mathematics, 02040,
Ad{\i}yaman, Turkey}
\email{mmanafov@adiyaman.edu.tr}


\thanks{Submitted October 20, 2016. Published January 12, 2017.}
\subjclass[2010]{34A55, 34L05, 47G20}
\keywords{Sturm-Liouville Operator; inverse spectral problem; integral delay}

\begin{abstract}
 In this article, we study an inverse spectral problem for Sturm-Liouville
 operator with integral delay. We prove that the standard spectral
 asymptotic conditions are necessary and sufficient for unique solvability of 
 the inverse problem. 
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\allowdisplaybreaks

\section{Introduction}

We consider inverse problem for the boundary-value problem (BVP) 
generated by the integro-differential equation
\begin{equation}
l_y:=-y''+q(x)y+\int_0^{x}M(x-t)y(t)dt=\lambda ^{2}y,\quad
 x\in (0,a) \cup (a,\pi)  \label{1.1}
\end{equation}
with the Dirichlet boundary conditions
\begin{equation}
U(y):=y(0)=0,\quad V(y):=y(\pi ) =0,  \label{1.2}
\end{equation}
and the conditions at the point $x=a:$
\begin{equation}
I(y):=\begin{cases}
y(a+0)=y(a-0)\equiv y(a), \\
y'(a+0)-y'(a-0)=2\alpha \lambda y(a),
\end{cases}  \label{1.3}
\end{equation}
$q(x)$ and $M(x)$ are complex-valued functions, 
$q(x)\in L_2(0,\pi) $ and $(\pi -x) M(x)\in L_2(0,\pi )$,
$\alpha \in \mathbb{C}$, $a\in (\frac{\pi }{2},\pi ) $ and 
$\lambda $ is a spectral parameter.

Sturm-Liouville spectral problems with potentials depending on the spectral
parameter (in case $K(x)\equiv 0$) arise in various models of quantum and
classical mechanics. For instance, the evolution equations that are used to
model interactions between colliding relativistic spineless particles can be
reduced to the form \eqref{1.1}. Then $\lambda ^{2}$ is associated with the
energy of the system (see \cite{12,13}).

Spectral problems of differential operators are studied in two main
branches, namely, direct and inverse problems. Direct problems of spectral
analysis consist in investigating the spectral properties of an operator. On
the other hand, inverse problems aim at recovering operators from their
spectral characteristics. Such problems often appear in mathematics,
mechanics, physics, electronics, geophysics, meteorology and other branches
of naturel sciences and engineering. Direct and inverse problems for the
classical Sturm-Liouville operators have been extensively studied 
(see \cite{5,7,11} and the references therein).

For integro-differential and other classes of nonlocal operators inverse
problems are more difficult for investigation, and the classical methods
either are not applicable to them or require essential modifications (see
\cite{2,3,4,5,6,14,15}). In this aspect, various inverse spectral problems
for the \eqref{1.1}, \eqref{1.3} BVP (special case $M(x)\equiv 0)$ have
been investigated in \cite{8,9,10}).

In this article we establish uniqueness result for inverse spectral problem
for Sturm-Liouville operator with integral delay.

\section{Integral representations for solutions}

In this section, we construct an integral representation of the solution 
$y(x,\lambda )$ of  \eqref{1.1}, \eqref{1.3}, satisfying the initial
conditions
\begin{equation}
y(0,\lambda ) =1,\quad y'(0,\lambda )=i\lambda\,.   \label{2.1}
\end{equation}
Also we study some properties of the solutions.
Using the standard successive approximation methods (see \cite{11}), 
we can prove the following theorem.

\begin{theorem} \label{thm2.1}
The solution $y(x,\lambda )$ has the form
\begin{equation}
y(x,\lambda )=y_0(x,\lambda )+\int_{-x}^{x}A(x,t) e^{i\lambda t}dt,  \label{2.2}
\end{equation}
where
\[
y_0(x,\lambda )=\begin{cases}
e^{ix\lambda }, & x<a \\
(1-i\alpha ) e^{ix\lambda }+i\alpha e^{i\lambda }, & x>a
\end{cases}
\]
and the function $A(x,t) $ satisfies 
\begin{equation}
\int_{-x}^{x}| A(x,t) | dt\leq e^{C\sigma
_0(x)}-1  \label{2.3}
\end{equation}
with 
\[
\sigma _0(x)=\int_0^{x}(x-t) [ |q(t)| +\int_0^{t}| M(t-\tau ) |d\tau ] dt,
\]
and  $C=1+2|\alpha|$.
\end{theorem}

\begin{proof}
It is clear that when $\alpha =0$, if we consider the equation \eqref{1.1}
separately on the intervals $(0,a) $ and $(a,\pi ) $,
we can write the solutions as
\begin{gather}
e_0(x,\lambda )=e^{ix\lambda }+\int_{-x}^{x}K_0(x,t)
e^{i\lambda t}dt,\quad 0\leq x<a,  \label{2.4} \\
e_a(x,\lambda )=e^{i\lambda (x-a)}+\int_{-x+2a}^{x}K_a(x,t)
 e^{i\lambda (t-a)}dt,\quad x>a, \label{2.5}
\end{gather}
respectively. For the solutions of the above equations to solve the equation
that has representation \eqref{2.5}, the following equality must be satisfied:
\begin{align*}
&\int_{-x+2a}^{x}K_a(x,t) e^{i\lambda (t-a)}dt \\
&=\frac{1}{\lambda }\int_a^{x}\sin \lambda (x-t)
\Big\{ q(t)\Big[e^{i\lambda (t-a)}+\int_{-t+2a}^{t}K_a(t,\tau ) e^{i\lambda
(\tau -a) }d\tau \Big]   \\
&\quad + \int_0^{t}M(t-\tau ) \Big[ e^{i\lambda (\tau
-a)}+\int_{-\tau +2a}^{\tau }K_a(\tau ,s) e^{i\lambda (
s-a) }ds\Big] d\tau \Big\} dt\,.
\end{align*}
It is easy to obtain the  integral equation
\begin{equation} \label{2.6}
\begin{aligned}
K_a(x,t)
&= \frac{1}{2}\int_a^{\frac{x+t}{2}}q(u)du
 +\frac{1 }{2}\int_a^{x}q(u)\int_{t-(x-u) }^{t+(x-u)}K_a(u,v) \,dv\,du   \\
&\quad +\frac{1}{2}\int_a^{x}\int_{t-(x-u) }^{t+(x-u)
}M(u-v) \,dx\,du \\
&\quad +\frac{1}{2}\int_a^{x}\int_0^{u}\int_{t-(x-u) }^{t+(x-u) }M(u-v)
K_a(v,\xi ) d\xi \,dx\,du.
\end{aligned}
\end{equation}
Since $e_a(x,-\lambda ) $ is also the solution of \eqref{1.1},
\eqref{1.3}  on the interval $0<x\leq \pi $, the solution $y(x,\lambda ) $
has the  form
\begin{equation}
y(x,\lambda ) =\begin{cases}
e_0(x,\lambda ), & 0\leq x<a, \\
c_{1}e_a(x,\lambda ) +c_2e_a(x,-\lambda ),
& a<x\leq \pi ,
\end{cases}   \label{2.7}
\end{equation}
where the constants $c_{1},c_2$ are defined from  conditions \eqref{1.3}.
Hence, we have
\begin{equation}
y(x,\lambda ) =\begin{cases}
e_0(x,\lambda ), & 0\leq x<a, \\[4pt]
e_0(a,\lambda ) \frac{(1-2i\alpha ) e_a(
x,\lambda ) +(1+2i\alpha ) e_a(x,-\lambda )}{2} \\
+e_0'(a,\lambda ) \frac{e_a(x,\lambda) -e_a(x,-\lambda ) }{2i\lambda },
& 0<x\leq \pi .
\end{cases}   \label{2.8}
\end{equation}
Using \eqref{2.4}, \eqref{2.5} and \eqref{2.8}, after some simple
computations, we find the following expression for $y(x,\lambda
) $ $(a<x\leq \pi )$,
\begin{equation}
y(x,\lambda ) =e(x,\lambda )+\int_{-x+2a}^{x}K_a(x,t) e(t,\lambda ) dt,
\label{2.9}
\end{equation}
where
\begin{equation} \label{2.10}
\begin{aligned}
e(x,\lambda )
&= e_0(a,\lambda ) [ \cos\lambda (x-a) +2\alpha \sin \lambda (x-a)]
+e_0'(a,\lambda ) \frac{\sin \lambda (x-a) }{\lambda }   \\
&= (1-i\alpha ) e^{i\lambda x}+i\alpha e^{i\lambda
(2a-x)}+\int_{-x}^{x}A_{1}(x,t) e^{i\lambda t}dt,
\end{aligned}
\end{equation}
\begin{gather*}
A_{1}(x,t) =A_0+\frac{1}{2}K_0(a,t+2a-x) +\frac{1
}{2}K_0(a,t+x) +\frac{1}{2}\int_{t+2a-x}^{t+x}H(s)ds,\quad | t| <x, \\
A_0=\begin{cases}
\frac{1}{2}\int_0^{a}q(t)dt+\frac{1}{4}\int_{-a}^{a}q(\frac{a+t}{2}) dt,
& 2a-x<t<x, \\
0, & -x<t<2a-x,
\end{cases}
\end{gather*}
\begin{equation}
H(s)
=\frac{1}{2}\int_{\frac{a-s}{2}}^{a}K_0(\sigma ,s-a+\sigma) q(\sigma ) d\sigma
+\frac{1}{2}\int_{\frac{a+s}{2}}^{a}K_0(\sigma ,s+a-\sigma ) q(\sigma ) d\sigma .
\label{2.11}
\end{equation}
Here, we assume that $K_0(a,t)\equiv 0$, $H(t)\equiv 0$, for
$|t| >a$ and $A_{1}(x,t) =0$ for $|t| >x$. Now using the expression \eqref{2.10} in \eqref{2.9}, we
have for $a<x\leq \pi $ $(| t| <x)$
\begin{equation}
y(x,\lambda ) =(1-i\alpha ) e^{i\lambda x}+i\alpha
e^{i\lambda (2a-x)}+\int_{-x}^{x}A_2(x,t)e^{i\lambda t}dt,  \label{2.12}
\end{equation}
where
\begin{equation}
\begin{aligned}
A_2(x,t)&=A_{1}(x,t)+(1-i\alpha ) K_a(x,t)-i\alpha K_a(x,2a-t) \\
&\quad +\int_{2a-x}^{x}K_a(x,s) A_{1}(s,t)ds.
\end{aligned} \label{2.13}
\end{equation}
From \eqref{2.4} and \eqref{2.12}, we can write the formula \eqref{2.2} for
the solution $y(x,\lambda ) $, where
\begin{equation}
A(x,t) =\begin{cases}
K_0(x,t), & \text{if }0\leq x\leq a, \; |t| <x \\
A_2(x,t), & \text{if }a<x\leq \pi , \; | t| <x.
\end{cases}   \label{2.14}
\end{equation}
From \eqref{2.6} it is easy to obtain
\begin{equation}
\int_{2a-x}^{x}| K_a(x,t)| dt\leq e^{C_a\sigma_a(x)}-1,  \label{2.15}
\end{equation}
where $C_a>0$ is a constant and
\[
\sigma _a(x)=\int_a^{x}(x-t) \big[ q(t)+\int_0^{t}| M(t-\tau )
| d\tau \big] dt.
\]
Using \eqref{2.15}, from \eqref{2.11} and \eqref{2.13}, we have the
estimate
\begin{equation}
\int_{-x}^{x}| A_2(x,t)| dt\leq e^{C\sigma _0(x)}-1
\label{2.16}
\end{equation}
for some constant $C>0$. Hence, from \eqref{2.14} and \eqref{2.16}, we
arrive at \eqref{2.3}.
\end{proof}

Let $s(x,\lambda )$ be a solution of \eqref{1.1} with initial
conditions
\[
s(0,\lambda )=0,\quad s'(0,\lambda )=1.
\]
Because $y(x,\lambda )$ and $y(x,-\lambda )$ are two linearly independent
solutions of \eqref{1.1}, \eqref{1.3}, then
\[
s(x,\lambda )=\frac{y(x,\lambda )-y(x,-\lambda )}{2i}.
\]

Using integral representation \eqref{2.2}, we easily obtain
\begin{equation}
s(x,\lambda )=s_0(x,\lambda )+\int_0^{x}G(x,t)\frac{\sin \lambda t}{
\lambda }dt,  \label{2.17}
\end{equation}
where
\[
s_0(x,\lambda )=\begin{cases}
\frac{\sin \lambda x}{\lambda }, & x<a \\[4pt]
(1-i\alpha ) \frac{\sin \lambda x}{\lambda }+i\alpha \frac{\sin
\lambda (2a-x)}{\lambda }, & x>a,
\end{cases}
\]
$G(x,t)=A(x,t)-A(x,-t)$ is a continuous function, and $G(x,0)=0$.

\section{Properties of the spectral characteristics}

In the section, we study  properties of eigenvalues and eigenfunctions
of \eqref{1.1}.
Let $y(x)$ and $z(x)$ be continuously differentiable functions on 
$(0,a) $ and $(a,\pi ) $. Denote 
$\langle y,z\rangle :=yz'-y'z$. If $y(x)$ and $z(x)$
satisfy the matching conditions \eqref{1.3}, then
\begin{equation}
\langle y,z\rangle _{x=a-0}=\langle y,z\rangle_{x=a+0},  \label{3.1}
\end{equation}
i.e. the function $\langle y,z\rangle $ is continuous on $(0,\pi )$.

Denote $\Delta (\lambda )=s(\pi ,\lambda ) $. The eigenvalues 
$\{ \lambda _n^{2}\} _{n\geq 1}$ of the BVP \eqref{1.1} coincide with
the zeros of the function $\Delta (\lambda )$.

\begin{theorem} \label{thm3.1}
The eigenvalues $\lambda _n^{2}$ and eigenfunctions $s(x,\lambda _n) $
of the BVP \eqref{1.1} satisfy the following
asymptotic estimates for sufficiently large $n$,
\begin{gather}
\lambda _n=\lambda _n^{0}+o\big(\frac{1}{\lambda _n^{0}}\big) , \label{3.2} \\
s(x,\lambda _n)
= o\big(\frac{1}{\lambda ^{0}}\big)
+\begin{cases}
\frac{\sin \lambda _n^{0}x}{\lambda _n^{0}} , & x<a \\[4pt]
(1-i\alpha ) \frac{\sin \lambda _n^{0}x}{\lambda _n^{0}}
+i\alpha \frac{\sin \lambda _n^{0}(2a-x)}{\lambda _n^{0}}, & x>a,
\end{cases}   \label{3.3}
\end{gather}
where $\lambda _n^{0}$ are the roots of
 $\Delta _0(\lambda ):=( 1-i\alpha ) 
\frac{\sin \lambda \pi }{\lambda }+i\alpha \frac{\sin \lambda (2a-\pi )}{\lambda }$ 
and $\lambda _n^{0}=n+h_n$, $h_n\in l_{\infty }$.
\end{theorem}

\begin{proof}
From \eqref{2.17}, we have
\begin{equation}
\Delta (\lambda )=(1-i\alpha ) \frac{\sin \lambda \pi }{\lambda }
+i\alpha \frac{\sin \lambda (2a-\pi )}{\lambda }+\int_0^{\pi }G(\pi
,t) \frac{\sin \lambda t}{\lambda }dt.  \label{3.3.}
\end{equation}
Denote $\Gamma _n:=\{ \lambda :| \lambda |
=\lambda _n^{0}+\delta \} $, 
$n=0,1,\dots,(\delta >0)$.
 Since $\Delta (\lambda )-\Delta _0(\lambda )=o(\frac{e^{| \operatorname{Im}
\lambda | \pi }}{| \lambda | }) $ and
 $ | \Delta _0(\lambda )| \geq C_{\delta }
\frac{e^{| \operatorname{Im}\lambda | \pi }}{| \lambda| }$ for all 
$\lambda \in \Gamma _n$, we establish by the
Rouche's Theorem (see \cite[p. 125]{4c}) that
$\lambda _n=\lambda _n^{0}+\varepsilon _n$, where
 $\varepsilon _n=o(1)$. Moreover, 
$\varepsilon _n=o(\frac{1}{\lambda _n^{0}}) $ is obtained
from the equality 
$o=\Delta (\lambda _n)=(\Delta _0'(\lambda _n^{0})+o(1)) \varepsilon _n
+o(\frac{1}{\lambda _n^{0}}) $.
This completes the proof of \eqref{3.2}.

From \eqref{2.17} and \eqref{3.2}, one can easily prove that the asymptotic
formula \eqref{3.3} is true.
\end{proof}

\begin{theorem} \label{thm3.2} 
The specification of the spectrum $\{ \lambda_n^{2}\} _{n\geq 1}$ 
uniquely determines the characteristic function $\Delta (\lambda ) $ by the formula
\begin{equation}
\Delta (\lambda )=[ (1-i\alpha ) \pi +i\alpha (2a-\pi )] 
\prod_{n=1}^{\infty }\frac{\lambda _n^{2}-\lambda ^{2}}{(\lambda _n^{0}) ^{2}}. 
 \label{3.4}
\end{equation}
\end{theorem}

\begin{proof}
It follows from \eqref{3.3.} and consequently by Hadamard's factorization
theorem (see \cite[p. 289]{4c}), $\Delta (\lambda )$ is uniquely determined
up to a multiplicative constant by its zeros:
\begin{equation}
\Delta (\lambda )=C\prod_{n=1}^{\infty }(1-\frac{\lambda ^{2}}{
\lambda _n^{2}}) .  \label{3.6}
\end{equation}
Consider the function
\begin{align*}
\Delta _0(\lambda )
&:=(1-i\alpha ) \frac{\sin \lambda \pi }{
\lambda }+i\alpha \frac{\sin \lambda (2a-\pi )}{\lambda } \\
&=[ (1-i\alpha ) \pi +i\alpha (2a-\pi )] 
 \prod_{n=1}^{\infty}(1-\frac{\lambda ^{2}}{(\lambda _n^{0}) ^{2}}) .
\end{align*}
Then
\[
\frac{\Delta (\lambda )}{\Delta _0(\lambda )}
=C\frac{1}{[ (1-i\alpha ) \pi +i\alpha (2a-\pi )] }
\prod_{n=1}^{\infty}\frac{(\lambda _n^{0}) ^{2}}{\lambda ^{2}}
\prod_{n=1}^{\infty }\Big(1+\frac{\lambda _n^{2}-(\lambda_n^{0}) ^{2}}
{(\lambda _n^{0}) ^{2}-\lambda ^{2}}\Big) .
\]
Taking \eqref{3.2} and \eqref{3.3.} into account we calculate
\[
\lim_{\lambda \to -\infty }\frac{\Delta (\lambda )}{\Delta_0(\lambda )}=1,\quad
\lim_{\lambda \to -\infty }\prod_{n=1}^{\infty }\Big(1+\frac{\lambda _n^{2}-(\lambda
_n^{0}) ^{2}}{(\lambda _n^{0}) ^{2}-\lambda ^{2}} \Big) =1
\]
and hence
\[
C=[ (1-i\alpha ) \pi +i\alpha (2a-\pi )]
\prod_{n=1}^{\infty }\frac{\lambda _n^{2}}{(\lambda_n^{0}) ^{2}}.
\]
Substituting this into account \eqref{3.6} we arrive at \eqref{3.4}.
\end{proof}

\section{Formulation of the inverse problem uniqueness theorem}

In this section, we study inverse problem of recovering $M(x)$ from the given
spectral characteristics. We denote the BVP \eqref{1.1}-\eqref{1.3} by $
L=L(M)$. Together with $L=L(M)$ we consider a BVP 
$\widetilde{L}=L(\widetilde{M})$ of the same form, but with different kernel 
$\widetilde{M}$.
\smallskip

\noindent\textbf{Inverse Problem:}
Given a function $q(x)$, numbers $\alpha,a$, and
the spectrum $\{ \lambda _n\} _{n\geq 1}$, construct the
function $M(x)$.

Let us prove the uniqueness theorem for the solution of the Inverse
Problem. Everywhere below if a certain symbol $e$ denotes an object to $L$,
then the corresponding symbol $\widetilde{e}$ denotes the analogous object
related to $\widetilde{L}$ and $\widehat{e}=e-\widetilde{e}$.

\begin{theorem}
\label{teo 4.1}Fix $b\in (0,a) $. Let $\Lambda \subset \mathbb{N}$ 
be a subset of nonnegative integer numbers, and let 
$\Omega :=\{\lambda _n^{2}\} _{n\in \Lambda }$ be a part of the spectrum of 
$L$ such that the system of functions 
$\{ \cos \lambda _nx\} _{n\in \Lambda }$ is complete in $L_2(0,\pi ) $. 
Let $M(x)=\widetilde{M}(x)$ almost everywhere (a.e.) on $(b,\pi ) $, and 
$\Omega = \widetilde{\Omega }$. Then $M(x)=\widetilde{M}(x)$ a.e. on 
$(0,\pi) $.
\end{theorem}

\begin{proof}
Let $\chi (x,\lambda ) $ be the solution of the equation
\begin{equation}
l_{z}^{\ast }:=-z''+q(x)z+\int_{x}^{\pi }M(t-x)
z(t)dt=\lambda ^{2}z,\quad x\in (0,a) \cup (a,\pi )
\label{4.1}
\end{equation}
under the conditions $\chi (\pi ,\lambda ) =0$, 
$\chi '(\pi ,\lambda ) =-1$ and the conditions at the point 
$x=a:\chi(a+0,\lambda ) =\chi (a-0,\lambda ) \equiv \chi(a,\lambda ) $, 
$\chi '(a+0,\lambda ) -\chi '(a-0,\lambda ) =2\alpha \lambda \chi (a,\lambda) $.
Denote $\Delta ^{\ast }(\lambda )=\chi (0,\lambda ) $.
Then by \eqref{3.1} we have
\begin{align*}
&\int_0^{\pi }\chi (x,\lambda ) \int_0^{x}\widehat{M}(x-t) \widetilde{s}(t,\lambda ) 
\,dt\,dx  \\
&=\int_0^{\pi }\chi (x,\lambda ) l\widetilde{s}(x,\lambda )dx
-\int_0^{\pi }\chi (x,\lambda ) \widetilde{l}\widetilde{s} (x,\lambda ) dx \\
&=\int_0^{\pi }l^{\ast }\chi (x,\lambda ) \widetilde{s}(x,\lambda ) dx
 -\int_0^{\pi }\chi (x,\lambda ) \widetilde{l}\widetilde{s}(x,\lambda ) dx \\
&\quad +[ \widetilde{s}(x,\lambda ) \chi '(
x,\lambda ) -\widetilde{s}'(x,\lambda ) \chi(x,\lambda ) ] (| _0^{a}
 +|_a^{\pi }) \\
&=\Delta ^{\ast }(\lambda )-\widetilde{\Delta }(\lambda ) .
\end{align*}
For $\widetilde{l}=l$ we have 
$\Delta ^{\ast }(\lambda )\equiv \Delta (\lambda )$, and consequently
\begin{equation}
\int_0^{\pi }\chi (x,\lambda ) \int_0^{x}\widehat{M}(x-t) \widetilde{s}(t,\lambda ) 
\,dt\,dx=\widehat{\Delta }(\lambda ) .  \label{4.2}
\end{equation}
We transform \eqref{4.2} into
\begin{equation}
\int_0^{\pi }\widehat{M}(x) \Big(\int_{x}^{\pi }\chi (
t,\lambda ) \widetilde{s}(t-x,\lambda ) dt\Big) dx
= \widehat{\Delta }(\lambda ) .  \label{4.3}
\end{equation}
Denote $w(x,\lambda ) =\chi (\pi -x,\lambda ) $, 
$N(x)=M(\pi -x)$,
\begin{equation}
\varphi (x,\lambda ) =\int_0^{x}w(t,\lambda ) \widetilde{s}(x-t,\lambda ) dt. 
 \label{4.4}
\end{equation}
Then \eqref{4.2} takes the form
\begin{equation}
\int_0^{\pi }\widehat{N}(x)\varphi (x,\lambda ) dx
=\widehat{\Delta }(\lambda ) .  \label{4.5}
\end{equation}
For $x\leq a$ the following representation holds \cite{14},
\begin{equation}
\varphi (x,\lambda ) =\frac{1}{2\lambda ^{2}}
\Big(-x\cos \lambda x+\int_0^{x}V(x,t) \cos \lambda t\,dt\Big) ,
\label{4.6}
\end{equation}
where $V(x,t) $ is a continuous function which does not depend
on $\lambda $. Since $\Omega =\widetilde{\Omega }$, we have by 
Theorem \ref{thm3.2}
\[
\Delta (\lambda )\equiv \widetilde{\Delta }(\lambda )
\Longrightarrow \widehat{\Delta }(\lambda ) \equiv 0.
\]
Then, substituting \eqref{4.6} into \eqref{4.5}, we obtain
\[
\int_0^{b}\Big(-x\widehat{N}(x) +\int_{x}^{b}V(
t,x) \widehat{N}(t) \Big) \cos \lambda x\,dx\equiv 0,
\]
and consequently,
\[
-x\widehat{N}(x) +\int_{x}^{b}V(t,x) \widehat{N}
(t) dt=0\quad \text{a.e. on }(0,b).
\]
Since this homogeneous Volterra integral equation has only the trivial
solution it follows that $\widehat{N}(x) =0$ a.e. on $(0,b) $, i.e.
$M(x)=\widetilde{M}(x)$ a.e. on $(0,\pi ) $.
\end{proof}

\subsection*{Acknowledgements}
The author wants to thank the anonymous referees for their valuable suggestions 
that improving this article.
This work was supported by Grant No FEFMAP/2016-0002 from Adiyaman University 
of Research Project Coordination (ADYUBAP), Turkey. 



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