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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2017 (2017), No. 104, pp. 1--12.\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-2017/104\hfil Trace formula for a delay differential operator]
{A regularized trace formula for a discontinuous Sturm-Liouville operator with
delayed argument}

\author[M. Bayramoglu, A. Bayramov, E. \c{S}en \hfil EJDE-2017/104\hfilneg]
{Mehmet Bayramoglu, Azad Bayramov, Erdo\u{g}an \c{S}en}

\address{Mehmet Bayramoglu \newline
Institute of Mathematics and Mechanics of NAS
of Azerbaijan, 1141,
Baku, Azerbaijan}
\email{mamed.bayramoglu@yahoo.com}

\address{Azad Bayramov \newline
Azerbaijan State Pedagogical University,
1000, Baku, Azerbaijan}
\email{azadbay@gmail.com}

\address{Erdo\u{g}an \c{S}en \newline
Department of Mathematics,
Nam\i k Kemal University, 59030,
Tekirda\u{g}, Turkey}
\email{erdogan.math@gmail.com}

\thanks{Submitted June 16, 2016. Published April 18, 2017.}
\subjclass[2010]{47A10, 47A55}
\keywords{Delay differential equation; transmission conditions; regularized trace}

\begin{abstract}
 In this study, we obtain a formula for the regularized sums of 
 eigenvalues  for a Sturm-Liouville problem with delayed argument 
 at two points of discontinuity.
\end{abstract}

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

\section{Introduction}

We consider the boundary value problem for the differential equation
\begin{equation}
y''(x)+q(x)y(x-\Delta (x))+\lambda ^2y(x)=0,  \label{e1}
\end{equation}
on $[0,h_1) \cup (h_1,h_2) \cup (h_2,\pi ] $, with boundary conditions
\begin{gather}
y(0)\cos \alpha +y'(0) \sin \alpha =0,  \label{e2} \\
y(\pi)\cos \beta +y'(\pi) \sin \beta =0  \label{e3}
\end{gather}
and transmission conditions
\begin{gather}
y(h_1-0) =\delta y(h_1+0),  \label{e4} \\
y'(h_1-0) =\delta y'(h_1+0),  \label{e5} \\
y(h_2-0) =\gamma y(h_2+0),  \label{e6} \\
y'(h_2-0) =\gamma y'(h_2+0)  \label{e7}
\end{gather}
where the real-valued function $q(x) $ is continuous in
$[0,h_1) \cup (h_1,h_2) \cup (h_2,\pi ] $
and has finite limits
\begin{equation*}
q(h_1\pm 0) =\lim_{x\to h_1\pm 0}q(x) ,\quad
q(h_2\pm 0) =\lim_{x\to h_2\pm 0}q(x) ,
\end{equation*}
the real-valued function $\Delta (x)\geq 0$ is continuous in
$[0,h_1) \cup (h_1,h_2) \cup (h_2,\pi ] $
has finite limits
\begin{equation*}
\Delta (h_1\pm 0) =\lim_{x\to h_1\pm 0}\Delta (x) ,\quad
\Delta (h_2\pm 0) =\lim_{x\to h_2\pm 0}\Delta (x) ,
\end{equation*}
if $ x\in [0,h_1)$, then $x-\Delta (x)\geq 0$;
if $x\in (h_1,h_2)$, then $x-\Delta (x)\geq h_1$;
if $x\in (h_2,\pi ]$, then $x-\Delta (x)\geq h_2$;
where $\lambda $ is a spectral parameter;
$h_1,h_2,\alpha ,\beta ,\delta \neq 0$, $\gamma \neq 0$ are arbitrary real
 numbers such that $0<h_1<h_2<\pi $
and $\sin \alpha \sin \beta \neq 0$.

The goal of this article is to calculate the regularized trace for the
problem \eqref{e1}--\eqref{e7}.

The theory of regularized trace of ordinary differential operators has long
history. Gelfand and Levitan \cite{g1} first obtained the trace formula for the
Sturm-Liouville differential equation. After this study several
mathematicians were interested in developing trace formulas for different
differential operators. The current situation of this subject and studies
related to it are presented in the comprehensive survey paper \cite{s1}. The
regularized trace formulas for differential equations with the operator
coefficients are found in \cite{b1,b2,k1,m1,m2,s3}.
 However, there are a small number
of works on the regularized trace for the differential equations with
retarded argument; see \cite{p1,y1}.

Pikula \cite{p1} investigated the problem of second order:
\begin{gather*}
-y''+q(x)y(x-\tau)=\lambda y, \\
y'(0)-hy(0)=y'(\pi) +Hy(\pi) =0, \\
y(x-\tau)=y(0)\varphi (x-\tau),\quad x\leq \tau ,\; \varphi(0)=1
\end{gather*}
and obtained trace formula of first order:
if $\tau \geq \pi $, then
\[
\sum_{n=1}^{\infty }(\lambda _{n}(\tau) -n^2)
=q(0) \varphi (-\tau) -\frac{h^2+H^2}{2}
\]
and if $\tau \leq \pi $, then 
\begin{align*}
&\sum_{n=1}^{\infty }\Big[\lambda _{n}(\tau) -n^2
 -\frac{2}{\pi }\Big(h+H+\frac{\cos n\tau }{2}\int_{\tau }^{\pi }
 q(t) dt\big)) \Big]  \\
&= q(0) \varphi (-\tau) -\frac{h^2+H^2}{2}
 +\frac{h+H}{2\pi }\int_{\tau }^{\pi }q(t) dt
 +\Big[\frac{ q(\tau) +q(\pi) }{4}-q(\tau) \Big]
 \frac{\pi -\tau }{\pi } \\
&\quad +\Big[\frac{b}{4}-\Big(\frac{1}{\sqrt{8}}
 \int_{\tau }^{\pi}q(t) dt\Big) ^2\Big] \frac{\pi -2\tau }{\pi }.
\end{align*}
Here the definition of $b$ can be found in \cite{p1}.  Yang \cite{y1} obtained
formulas of the first regularized trace for a discontinuous boundary value
problem with retarded argument and with transmission conditions at the one
point of discontinuity. As in this problem, if the retardation function $
\Delta \equiv 0$ in \eqref{e1} and $\delta =1,\gamma =1$ we have the formula of the
first regularized trace for the classical Sturm-Liouville operator
(see \cite{g1})
which is called Gelfand-Levitan formula \cite{s1}.

Note that, the trace formulas can be used in the inverse problems of
spectral analysis of differential equations \cite{s1} and have applications in
the approximate calculation of the eigenvalues of the related operator
\cite{d1,s1}.


\section{A formula for the regularized trace}


First we give the asymptotic behavior of solutions for large values of the
spectral parameter. Let $\omega _1(x,\lambda)$ be a solution of \eqref{e1}
on $[0,h_1] $, satisfying the initial conditions
\begin{equation}
\omega _1(0,\lambda) =\sin \alpha \quad \text{and}\quad
\omega_1'(0,\lambda) =-\cos \alpha .  \label{e8}
\end{equation}
These  conditions  define a unique solution of \eqref{e1} on
 $[0,h_1] $  \cite[p. 12]{n1}.

After defining the above solution, then we will define the solution
$\omega_2(x,\lambda) $ of \eqref{e1} on $[h_1,h_2] $ by
means of the solution $\omega _1(x,\lambda) $ using the
initial conditions
\begin{equation}
\omega _2(h_1,\lambda) =\delta ^{-1}\omega _1(h_1,\lambda)
\quad \text{and} \quad
\omega _2'(h_1,\>\lambda)=\delta ^{-1}\omega _1'(h_1,\lambda). \label{e9}
\end{equation}
These  conditions define a unique solution of \eqref{e1} on $[h_1,h_2]$.

After describing the above solution, then we will give the solution
$\omega_3(x,\lambda) $ of \eqref{e1} on $[h_2,\pi ] $ by
means of the solution $\omega _2(x,\lambda) $ using the initial conditions
\begin{equation}
\omega _3(h_2,\lambda) =\gamma ^{-1}\omega _2(h_2,\lambda)
\quad\text{and}\quad \omega _3'(h_2,\lambda)=\gamma ^{-1}
\omega _2'(h_2,\lambda).
\label{e10}
\end{equation}
These conditions define a unique solution of \eqref{e1} on $[h_2,\pi]$.

Consequently, the function $\omega (x,\lambda) $ is defined on
$[0,h_1) \cup (h_1,h_2) \cup (h_2,\pi] $ by
\begin{equation}
\omega (x,\lambda)
=\begin{cases}
\omega _1(x,\lambda), & x\in \lbrack 0,h_1), \\
\omega _2(x,\lambda), & x\in (h_1,h_2) , \\
\omega _3(x,\lambda), & x\in (h_2,\pi ]
\end{cases}  \label{e11}
\end{equation}
is a solution of \eqref{e1} on $[0,h_1) \cup (h_1,h_2) \cup (h_2,\pi ] $,
 which satisfies one of the boundary conditions and the transmission conditions
\eqref{e4}--\eqref{e7} Then the
following integral equations hold:
\begin{gather}
\begin{aligned}
\omega _1(x,\lambda)
& =\sin \alpha \cos \lambda x-\frac{\cos \alpha }{
\lambda }\sin \lambda x   \\
&\quad -\frac{1}{\lambda }\int_0^{{x}}q(\tau) \sin \lambda
(x-\tau) \omega _1(\tau -\Delta (\tau),\lambda) d\tau ,
\end{aligned} \label{e12}\\
\begin{aligned}
\omega _2(x,\lambda)
& =\frac{1}{\delta }\omega _1(h_1,\lambda
) \cos \lambda (x-h_1) +\frac{\omega _1'(h_1,\lambda) }{\lambda \delta }
\sin \lambda ( x-h_1)    \\
&\quad  -\frac{1}{\lambda }\int_{h_1}^{{x}}q(\tau) \sin
\lambda (x-\tau) \omega _2(\tau -\Delta (\tau
) ,\lambda) d\tau ,  \label{e13}
\end{aligned} \\
\begin{aligned}
\omega _3(x,\lambda)
& =\frac{1}{\gamma }\omega _2(h_2,\lambda ) \cos \lambda (x-h_2)
+\frac{\omega _2'(h_2,\lambda) }{\lambda \gamma }\sin \lambda (x-h_2)    \\
&\quad  -\frac{1}{\lambda }\int_{h_2}^{{x}}q(\tau) \sin
\lambda (x-\tau) \omega _3(\tau -\Delta (\tau) ,\lambda) d\tau .
\end{aligned}  \label{e14}
\end{gather}
Differentiating \eqref{e12} and \eqref{e13} with respect to $x$, we obtain
\begin{gather}
\omega _1'(x,\lambda)=-\lambda \sin \alpha \sin \lambda x-\cos
\alpha \cos \lambda x-\int_0^{x}q(\tau)\cos \lambda (x-\tau
) \omega _1(\tau -\Delta (\tau) ,\lambda)d\tau ,
\label{e15} \\
\begin{aligned}
\omega _2'(x,\lambda)
&=-\frac{\lambda }{\delta }\omega _1(h_1,\lambda) \sin \lambda (x-h_1)
 +\frac{\omega_1'(h_1,\lambda) }{\delta }\cos \lambda (x-h_1)    \\
&\quad -\int_{h_1}^{{x}}q(\tau)\cos \lambda (x-\tau) \omega_2(\tau -\Delta (\tau) ,
 \lambda)d\tau ,  \label{e16}
\end{aligned} \\
\begin{aligned}
\omega _3'(x,\lambda)
&=-\frac{\lambda }{\gamma }\omega _2(
h_2,\lambda) \sin \lambda (x-h_2) +\frac{\omega
_2'(h_2,\lambda) }{\gamma }\cos \lambda (x-h_2)    \\
&\quad -\int_{h_2}^{{x}}q(\tau)\cos \lambda (x-\tau) \omega
_3(\tau -\Delta (\tau) ,\lambda)d\tau ,
\end{aligned}  \label{e17}
\end{gather}
Moreover
\[
| \omega _1(x,\lambda) | < \text{const.},\quad x\in [0,h_1] ,
\]
see [11].  Then from \eqref{e12} we have
\begin{equation}
\begin{aligned}
&\omega _1(x,\lambda)\\
&=\sin \alpha \cos \lambda x-\frac{\cos
\alpha }{\lambda }\sin \lambda x-\frac{\sin \alpha }{\lambda }
\int_0^{{x}}q(\tau) \sin \lambda (x-\tau)
\cos \lambda (\tau -\Delta (\tau))d\tau \\
&\quad +\frac{\cos \alpha }{\lambda ^2}\int_0^{{x}}q(\tau)
\sin \lambda (x-\tau) \sin \lambda (\tau -\Delta (\tau))d\tau
+O(\frac{1}{\lambda ^3}) \\
&=\sin \alpha \cos \lambda x-\frac{\cos \alpha }{\lambda }\sin \lambda x-
\frac{\sin \alpha }{2\lambda }\int_0^{{x}}q(\tau) \sin
\lambda (x-\Delta (\tau)) d\tau \\
&\quad -\frac{\sin \alpha }{2\lambda }\int_0^{{x}}q(\tau)
\sin \lambda (x-2\tau +\Delta (\tau)) d\tau -\frac{\cos \alpha
}{2\lambda ^2}\int_0^{{x}}q(\tau) \cos \lambda
(x-\Delta (\tau)) d\tau \\
&\quad +\frac{\cos \alpha }{2\lambda ^2}\int_0^{{x}}q(\tau)
\cos \lambda (x-2\tau +\Delta (\tau)) d\tau +O(\frac{1}{
\lambda ^3}) .
\end{aligned} \label{e18}
\end{equation}
Therefore,
\begin{equation}
\begin{aligned}
&\omega _2(x,\lambda)\\
&=\frac{\sin \alpha }{\delta }\cos
\lambda x-\frac{\cos \alpha }{\lambda \delta }\sin \lambda x-\frac{\sin
\alpha }{2\lambda \delta }\int_0^{{x}}q(\tau) \sin
\lambda (x-\Delta (\tau)) d\tau \\
&\quad -\frac{\sin \alpha }{2\lambda \delta }\int_0^{{x}}q(\tau
) \sin \lambda (x-2\tau +\Delta (\tau)) d\tau -\frac{\cos
\alpha }{2\lambda ^2\delta }\int_0^{{x}}q(\tau) \cos
\lambda (x-\Delta (\tau)) d\tau \\
&\quad  +\frac{\cos \alpha }{2\lambda ^2\delta }\int_0^{{x}}q(\tau
) \cos \lambda (x-2\tau +\Delta (\tau)) d\tau +O(
\frac{1}{\lambda ^3}) .
\end{aligned} \label{e19}
\end{equation}
and
\begin{equation}
\begin{aligned}
&\omega _3(x,\lambda)\\
& =\frac{\sin \alpha }{\delta \gamma }\cos
\lambda x-\frac{\cos \alpha }{\lambda \delta \gamma }\sin \lambda x-\frac{
\sin \alpha }{2\lambda \delta \gamma }\int_0^{{x}}q(\tau
) \sin \lambda (x-\Delta (\tau)) d\tau \\
&\quad -\frac{\sin \alpha }{2\lambda \delta \gamma }\int_0^{{x}}q(
\tau) \sin \lambda (x-2\tau +\Delta (\tau)) d\tau -\frac{
\cos \alpha }{2\lambda ^2\delta \gamma }\int_0^{{x}}q(\tau
) \cos \lambda (x-\Delta (\tau)) d\tau \\
&\quad  +\frac{\cos \alpha }{2\lambda ^2\delta \gamma }\int_0^{{x}
}q(\tau) \cos \lambda (x-2\tau +\Delta (\tau))
d\tau +O(\frac{1}{\lambda ^3}) .
\end{aligned}  \label{e20}
\end{equation}
The solution $\omega (x,\lambda)$ defined above is a nontrivial solution of
\eqref{e1} satisfying conditions \eqref{e2} and \eqref{e4}--\eqref{e7}.
Putting $\omega (x,\lambda)$ in \eqref{e3}, we obtain the characteristic equation
\begin{equation}
W(\lambda)\equiv \omega (\pi ,\lambda)\cos \beta +\omega '(\pi,\lambda)
\sin \beta =0.  \label{e21}
\end{equation}
The set of eigenvalues of boundary value problem \eqref{e1}--\eqref{e7}
 coincides with the set of the squares of roots of \eqref{e21}, and
eigenvalues are simple. From
\eqref{e12}-\eqref{e17} and \eqref{e21} we obtain
\begin{align*}
&W(\lambda) \\
&\equiv \Big\{ \frac{1}{\gamma }\Big[\frac{{1}}{\delta }(
\sin \alpha \cos {\lambda }h_1-\frac{{\cos \alpha }}{\lambda }\sin {
\lambda }h_1 \\
&\quad -{\frac{1}{\lambda }}\int_0^{h_1}q(\tau)\sin
\lambda (h_1-\tau)\omega _1(\tau -\Delta (\tau),\lambda)d\tau)
 \cos {\lambda }(h_2-h_1) \\
&\quad -{\frac{1}{\lambda \delta }}\Big(\lambda \sin \alpha \sin {\lambda }
h_1+\cos \alpha \cos \lambda h_1 \\
&\quad +\int_0^{h_1}q(\tau)\cos \lambda (h_1-\tau)
 \omega _1(\tau -\Delta (\tau),\lambda)d\tau\Big)
 \sin {\lambda }(h_2-h_1)  \\
&\quad -\frac{1}{\lambda }
\int_{h_1}^{h_2}q(\tau)\sin \lambda (h_2-\tau)\omega
_2(\tau -\Delta (\tau),\lambda)d\tau \Big] \cos {\lambda }(\pi
-h_2)  \\
&\quad + \frac{1}{\lambda \gamma }\Big[-\frac{\lambda }{\delta }\Big(\sin
\alpha \cos {\lambda }h_1-\frac{{\cos \alpha }}{\lambda }\sin {\lambda } h_1 \\
&\quad -{\frac{1}{\lambda }}\int_0^{h_1}q(\tau)\sin \lambda
(h_1-\tau)\omega _1(\tau -\Delta (\tau),\lambda)d\tau\Big)
 \sin {\lambda }(h_2-h_1) \\
&\quad -\frac{1}{\delta }(\lambda \sin \alpha \sin \lambda h_1+\cos \alpha
\cos \lambda h_1 \\
&\quad +\int_0^{h_1}q(\tau)\cos \lambda (h_1-\tau
)\omega _1(\tau -\Delta (\tau),\lambda)d\tau)
 \cos \lambda (h_2-h_1)  \\
&\quad -\int _{h_1}^{h_2}q(\tau)\cos \lambda (h_2-\tau)\omega _2(\tau
-\Delta (\tau),\lambda)d\tau \Big] \sin {\lambda }(\pi -h_2) \\
&\quad  -\frac{1}{\lambda }\int_{h_2}^{\pi }q(\tau)\sin \lambda ({
\pi }-\tau)\omega _3(\tau -\Delta (\tau),\lambda)d\tau \Big\}
\cos \beta \\
&\quad +\Big\{ -\frac{\lambda }{\gamma }\Big[\frac{{1}}{\delta }\Big(\sin
\alpha \cos {\lambda }h_1-\frac{{\cos \alpha }}{\lambda }\sin {\lambda }h_1\\
&\quad -{\frac{1}{\lambda }}\int_0^{h_1}q(\tau)\sin \lambda
(h_1-\tau)\omega _1(\tau -\Delta (\tau),\lambda)d\tau\Big)
 \cos {\lambda }(h_2-h_1)  \\
&\quad -{\frac{1}{\lambda \delta }}\Big(\lambda \sin \alpha \sin {\lambda }
h_1+\cos \alpha \cos {\lambda h_1+}\int_0^{h_1}q(\tau)\cos
\lambda (h_1-\tau)\omega _1(\tau -\Delta (\tau),\lambda)d\tau\Big) \\
&\quad \times \sin {\lambda }(h_2-h_1) -\frac{1}{\lambda }
\int_{h_1}^{h_2}q(\tau)\sin \lambda (h_2-\tau)\omega
_2(\tau -\Delta (\tau),\lambda)d\tau \Big] \sin {\lambda }(\pi -h_2) \\
&\quad +\frac{1}{\gamma }\Big[-\frac{\lambda }{\delta }\Big(\sin \alpha \cos {
\lambda }h_1-\frac{{\cos \alpha }}{\lambda }\sin {\lambda }h_1 \\
&\quad -{\frac{1}{\lambda }}\int_0^{h_1}q(\tau)\sin \lambda ({h}_1-\tau)\omega
_1(\tau -\Delta (\tau),\lambda)d\tau\Big)
\sin {\lambda }(h_2-h_1) \\
&\quad -\frac{1}{\delta }\Big(\lambda \sin \alpha \sin \lambda h_1+\cos \alpha
\cos \lambda h_1 \\
&\quad +\int_0^{h_1}q(\tau)\cos \lambda (h_1-\tau
)\omega _1(\tau -\Delta (\tau),\lambda)d\tau\Big) \cos {\lambda }(h_2-h_1)\\
&\quad -\int_{h_1}^{h_2}q(\tau)\cos \lambda (h_2-\tau)\omega
_2(\tau -\Delta (\tau),\lambda)d\tau \Big] \cos {\lambda }(\pi -h_2) \\
&\quad  -\int_{h_2}^{\pi }q(\tau)\cos \lambda ({\pi }-\tau)\omega
_3(\tau -\Delta (\tau),\lambda)d\tau \Big\} \sin \beta ,
\end{align*}
which implies
\begin{equation}
\begin{aligned} 
W(\lambda)
&=-\frac{\sin \alpha \sin \beta }{\delta \gamma }
\lambda \sin \lambda \pi +\frac{\sin (\alpha -\beta) }{\delta
\gamma }\cos \lambda \pi -{\frac{\cos \alpha \cos \beta }{\lambda \delta }
\sin \lambda \pi } \\
&\quad -\frac{\sin \beta }{\delta \gamma }\int_0^{h_1}q(\tau)\cos
\lambda (\pi -\tau)\omega _1(\tau -\Delta (\tau),\lambda)d\tau  \\
&\quad -\frac{\sin \beta }{\gamma }\int_{h_1}^{h_2}q(\tau)\cos
\lambda (\pi -\tau)\omega _2(\tau -\Delta (\tau),\lambda)d\tau \\
&\quad -\sin \beta \int_{h_2}^{\pi }q(\tau)\cos \lambda (\pi -\tau
)\omega _3(\tau -\Delta (\tau),\lambda)d\tau    \\
&\quad -\frac{\cos \beta }{\lambda \delta \gamma }\int_0^{h_1}q(\tau
)\sin \lambda (\pi -\tau)\omega _1(\tau -\Delta (\tau),\lambda)d\tau
 \\
&\quad -\frac{\cos \beta }{\lambda \gamma }\int_{h_1}^{h_2}q(\tau
)\sin \lambda (\pi -\tau)\omega _2(\tau -\Delta (\tau),\lambda)d\tau
\\
&\quad
-\frac{\cos \beta }{\lambda }\int_{h_2}^{\pi }q(\tau)\sin \lambda
(\pi -\tau)\omega _3(\tau -\Delta (\tau),\lambda)d\tau . 
\end{aligned} \label{e22}
\end{equation}
From \eqref{e18}-\eqref{e20}, we obtain
\begin{gather} 
\begin{aligned}
&\omega _1(\tau -\Delta (\tau),\lambda)  \\
&=\sin \alpha \cos \lambda (\tau -\Delta (\tau)) 
-\frac{\cos \alpha }{\lambda }\sin \lambda (\tau -\Delta (\tau)) \\
&\quad -\frac{\sin \alpha }{2\lambda }
 \int_0^{\tau -\Delta (\tau) }q(t_1)\sin \lambda 
 (\tau -\Delta (\tau)-\Delta (t_1)) dt_1 \\
&\quad -\frac{\sin \alpha }{2\lambda }
 \int_0^{\tau -\Delta (\tau) }q(t_1)\sin \lambda 
 (\tau -\Delta (\tau)-2t_1+\Delta (t_1)) dt_1+O(\frac{1}{\lambda^3}) , 
\end{aligned}  \label{e23}  \\
\begin{aligned}
\omega _2(\tau -\Delta (\tau),\lambda) 
&=\frac{\sin \alpha }{\delta }\cos \lambda (\tau -\Delta (\tau)) 
-\frac{\cos \alpha }{\lambda \delta }\sin \lambda (\tau -\Delta (\tau))
 \\
&\quad -\frac{\sin \alpha }{2\lambda \delta }\int_0^{\tau -\Delta (
\tau) }q(t_1)\sin \lambda (\tau -\Delta (\tau)
-\Delta (t_1)) dt_1, 
\end{aligned} \\
\begin{aligned} 
-\frac{\sin \alpha }{2\lambda \delta }
 \int_0^{\tau -\Delta (\tau) }q(t_1)\sin \lambda (\tau -\Delta (\tau)
 -2t_1+\Delta (t_1)) dt_1+O(\frac{1}{\lambda^3}), 
\end{aligned} \label{e24} \\
\begin{aligned}
&\omega _3(\tau -\Delta (\tau),\lambda) \\
&=\frac{\sin \alpha }{\delta \gamma }\cos \lambda (\tau -\Delta (\tau)) -
\frac{\cos \alpha }{\lambda \delta \gamma }\sin \lambda (\tau -\Delta(\tau))\\
&\quad -\frac{\sin \alpha }{2\lambda \delta \gamma }\int_0^{\tau -\Delta
(\tau) }q(t_1)\sin \lambda (\tau -\Delta (\tau) -\Delta (t_1)) dt_1 \\
&\quad -\frac{\sin \alpha }{2\lambda \delta \gamma }\int_0^{\tau -\Delta
(\tau) }q(t_1)\sin \lambda (\tau -\Delta (\tau) 
 -2t_1+\Delta (t_1)) dt_1+O(\frac{1}{\lambda ^3}) ,
\end{aligned} \label{e25} 
\end{gather}
Using \eqref{e22}--\eqref{e25} we obtain
\begin{equation}
\begin{aligned}
&W(\lambda) \\
&\equiv -\frac{\sin \alpha \sin \beta }{\delta \gamma
}\lambda \sin \lambda \pi +\frac{\sin (\alpha -\beta) }{\delta
\gamma }\cos \lambda \pi -{\frac{\cos \alpha \cos \beta }{\lambda \delta
\gamma }\sin \lambda \pi } \\
&\quad -\frac{\sin \alpha \sin \beta }{2\delta \gamma }\int_0^{\pi
}q(\tau)[\cos \lambda (\pi -\Delta (\tau))+\cos \lambda
(\pi -2\tau +\Delta (\tau))] d\tau  \\
&\quad -\frac{\sin \alpha \cos \beta }{2\lambda \delta \gamma }
\int_0^{\pi }q(\tau)[\sin \lambda (\pi -\Delta (\tau
))+\sin \lambda (\pi -2\tau +\Delta (\tau))] d\tau \\
&\quad +\frac{\cos \alpha \sin \beta }{2\lambda \delta \gamma }\int_0^{\pi
}q(\tau)[\sin \lambda (\pi -\Delta (\tau))-\sin \lambda
(\pi -2\tau +\Delta (\tau))] d\tau \\
&\quad +\frac{\sin \alpha \sin \beta }{4\lambda \delta \gamma }
 \int_0^{\pi}q(\tau)\Big[\int_0^{\tau -\Delta (\tau)}q(t_1)
 \sin \lambda (\pi -\Delta (\tau) -\Delta (t_1)) dt_1\Big] d\tau \\
&\quad -\frac{\sin \alpha \sin \beta }{4\lambda \delta \gamma }
 \int_0^{\pi}q(\tau)[\int_0^{\tau -\Delta (\tau)}q(t_1)
 \sin \lambda (\pi -2\tau +\Delta (\tau) +\Delta
(t_1)) dt_1] d\tau \\
&\quad +\frac{\sin \alpha \sin \beta }{4\lambda \delta \gamma }
 \int_0^{\pi}q(\tau)\Big[\int_0^{\tau -\Delta (\tau)}q(t_1)
 \sin \lambda (\pi -\Delta (\tau) -2t_1+\Delta (t_1)) dt_1\Big] d\tau \\
&\quad -\frac{\sin \alpha \sin \beta }{4\lambda \delta \gamma }
 \int_0^{\pi}q(\tau)\Big[\int_0^{\tau -\Delta (\tau)}q(t_1)
 \sin \lambda (\pi -2\tau +\Delta (\tau) \\
&\quad +2t_1-\Delta (t_1)) dt_1\Big] d\tau 
 +O(\frac{1}{\lambda ^3}).
\end{aligned} \label{e26}
\end{equation}
Denote
\begin{gather*}
A_1(\lambda) =\frac{1}{2}\int_0^{\pi }q(\tau)\cos
\lambda \Delta (\tau) d\tau ,\quad
A_2(\lambda) =\frac{1}{2}\int_0^{\pi }q(\tau)\sin \lambda
\Delta (\tau) d\tau , \\
B_1(\lambda) =\frac{1}{4}\int_0^{\pi }q(\tau)
\Big[\int_0^{\tau -\Delta (\tau) }q(t_1)\cos \lambda
(2\tau -\Delta (\tau) -\Delta (t_1)) dt_1\Big] d\tau , \\
B_2(\lambda) =\frac{1}{4}\int_0^{\pi }q(\tau)
\Big[ \int_0^{\tau -\Delta (\tau) }q(t_1)\sin \lambda
(2\tau -\Delta (\tau) -\Delta (t_1)) dt_1\Big] d\tau , \\
C_1(\lambda) =\frac{1}{4}\int_0^{\pi }q(\tau)
\Big[\int_0^{\tau -\Delta (\tau) }q(t_1)\cos \lambda
(\Delta (\tau) +\Delta (t_1)) dt_1 \Big] d\tau , \\
C_2(\lambda) =\frac{1}{4}\int_0^{\pi }q(\tau)
\Big[\int_0^{\tau -\Delta (\tau) }q(t_1)\sin \lambda
(\Delta (\tau) +\Delta (t_1)) dt_1 \Big] d\tau , \\
D_1(\lambda) =\frac{1}{2}\int_0^{\pi }q(\tau)\cos
 \lambda (2\tau -\Delta (\tau)) d\tau ,\\
D_2(\lambda) =\frac{1}{2}\int_0^{\pi }q(\tau)\sin
 \lambda (2\tau -\Delta (\tau)) d\tau , \\
E_1(\lambda) =\frac{1}{4}\int_0^{\pi }q(\tau)
\Big[\int_0^{\tau -\Delta (\tau) }q(t_1)\cos \lambda
(\Delta (\tau) +2t_1-\Delta (t_1)) dt_1\Big] d\tau , \\
E_2(\lambda) =\frac{1}{4}\int_0^{\pi }q(\tau)
\Big[\int_0^{\tau -\Delta (\tau) }q(t_1)\sin \lambda
(\Delta (\tau) +2t_1-\Delta (t_1)) dt_1\Big] d\tau , \\
H_1(\lambda) =\frac{1}{4}\int_0^{\pi }q(\tau)
\Big[\int_0^{\tau -\Delta (\tau) }q(t_1)\cos \lambda
(2\tau -\Delta (\tau) -2t_1+\Delta (t_1) ) dt_1\Big] d\tau , \\
H_2(\lambda) =\frac{1}{4}\int_0^{\pi }q(\tau)
\Big[\int_0^{\tau -\Delta (\tau) }q(t_1)\sin \lambda
(2\tau -\Delta (\tau) -2t_1+\Delta (t_1)) dt_1\Big] d\tau .
\end{gather*}
Then from \eqref{e26}, we obtain
\begin{equation}
\begin{aligned}
&W(\lambda) \\
&\equiv -\frac{\sin \alpha \sin \beta }{\delta \gamma
}\lambda \sin \lambda \pi +\frac{\sin (\alpha -\beta) }{\delta
\gamma }\cos \lambda \pi -{\frac{\cos \alpha \cos \beta }{\lambda \delta
\gamma }\sin \lambda \pi } \\
&\quad -\frac{\sin \alpha \sin \beta }{\delta \gamma }[A_1(\lambda
) \cos \lambda \pi +A_2(\lambda) \sin \lambda \pi
+D_1(\lambda) \cos \lambda \pi +D_2(\lambda)
\sin \lambda \pi ] \\
&\quad -\frac{\sin \alpha \cos \beta }{\lambda \delta \gamma }[A_1(
\lambda) \sin \lambda \pi -A_2(\lambda) \cos \lambda
\pi +D_1(\lambda) \sin \lambda \pi -D_2(\lambda
) \cos \lambda \pi ] \\
&\quad +\frac{\cos \alpha \sin \beta }{\lambda \delta \gamma }[A_1(
\lambda) \sin \lambda \pi -A_2(\lambda) \cos \lambda
\pi -D_1(\lambda) \sin \lambda \pi +D_2(\lambda
) \cos \lambda \pi ] \\
&\quad +\frac{\sin \alpha \sin \beta }{\lambda \delta \gamma }[C_1(
\lambda) \sin \lambda \pi -C_2(\lambda) \cos \lambda \pi ] \\
&\quad  -\frac{\sin \alpha \sin \beta }{\lambda \delta \gamma }
 [B_1(\lambda) \sin \lambda \pi -B_2(\lambda) \cos \lambda \pi ]\\
&\quad +\frac{\sin \alpha \sin \beta }{\lambda \delta \gamma }
 [E_1(\lambda) \sin \lambda \pi -E_2(\lambda) \cos \lambda \pi ]\\
&\quad -\frac{\sin \alpha \sin \beta }{\lambda \delta \gamma }
 [H_1(\lambda) \sin \lambda \pi -H_2(\lambda) \cos \lambda
\pi ] +O(\frac{1}{\lambda ^3}) .
\end{aligned} \label{e27}
\end{equation}
Define
\begin{equation}
W_0(\lambda) \equiv -\frac{\sin \alpha \sin \beta }{\delta
\gamma }\lambda \sin \lambda \pi .  \label{e28}
\end{equation}
Denote by $\lambda _{n}^{0},n\in
\mathbb{Z}$, zeros of the function $W_0(\lambda) $, then we have
$ \lambda _{n}^{0}=n,n\in \mathbb{Z}$, and it is simple algebraically
except for $\lambda _0^{0}$.

Denote by $C_{n}$ the circle of radius, $0<\varepsilon <\frac{1}{2}$,
centered at the origin $\lambda _{n}^{0}=n,n\in \mathbb{Z}$, and 
by $\Gamma _{N_0}$ the counterclockwise square contours with four
vertices
\begin{gather*}
A =N_0+\varepsilon +N_0i,\quad B=-N_0-\varepsilon +N_0i, \\
C =-N_0-\varepsilon -N_0i,\quad D=N_0+\varepsilon -N_0i,
\end{gather*}
where $i=\sqrt{-1}$ and $N_0$ is a natural number. Obviously, if 
$\lambda \in C_{n}$ or $\lambda \in \Gamma _{N_0}$, then 
$| W_0(\lambda) | \geq M| \lambda |e^{| \operatorname{Im}\lambda | \pi }$
 $(M>0) $ by
using a similar method in \cite{y2}. Thus, on $\lambda \in C_{n}$ or 
$\lambda \in \Gamma _{N_0}$, from \eqref{e27} and \eqref{e28}, we have
\begin{equation}
\begin{aligned}
&\frac{W(\lambda) }{W_0(\lambda) }\\
&=1+\frac{\cot
\alpha -\cot \beta +A_1(\lambda) +D_1(\lambda)
}{\lambda }\cot \lambda \pi +\frac{A_2(\lambda) +D_2(
\lambda) }{\lambda } \\
&\quad +\Big((\cot \alpha -\cot \beta) A_2(\lambda)
-(\cot \alpha +\cot \beta) D_2(\lambda)
+C_2(\lambda) -B_2(\lambda)  \\
&\quad +E_2( \lambda) -H_2(\lambda) \Big),\cot (\lambda \pi) /\lambda ^2\\
&\quad +\frac{\cot \alpha \cot \beta +(\cot \beta -\cot \alpha)
A_1(\lambda) +(\cot \alpha +\cot \beta)
D_1(\lambda) }{\lambda ^2} \\
&\quad +\frac{-C_1(\lambda) +B_1(\lambda)
-E_1(\lambda) +H_1(\lambda) }{\lambda ^2}
+O(\frac{1}{\lambda ^3}) .
\end{aligned} \label{e29}
\end{equation}
Expanding $\ln \frac{W(\lambda) }{W_0(\lambda) }$
by the Maclaurin formula, we find that
\begin{equation}
\begin{aligned}
&\ln \frac{W(\lambda) }{W_0(\lambda) }\\
&=\frac{\cot \alpha -\cot \beta +A_1(\lambda) +D_1(\lambda)
}{\lambda }\cot \lambda \pi +\frac{A_2(\lambda) +D_2(
\lambda) }{\lambda } \\
&\quad +\Big((\cot \alpha -\cot \beta) A_2(\lambda)
-(\cot \alpha +\cot \beta) D_2(\lambda)
+C_2(\lambda) -B_2(\lambda) \\
&\quad +E_2( \lambda) -H_2(\lambda) \Big) \cot(\lambda \pi)/ \lambda ^2 \\
&\quad +\frac{\cot \alpha \cot \beta +(\cot \beta -\cot \alpha)
A_1(\lambda) +(\cot \alpha +\cot \beta) D_1(\lambda) }{\lambda ^2} \\
&\quad +\frac{-C_1(\lambda) +B_1(\lambda) -E_1(\lambda) 
 +H_1(\lambda) }{\lambda ^2} \\
&\quad -\frac{1}{2}\Big[\frac{(\cot \alpha -\cot \beta +A_1(\lambda) 
 +D_1(\lambda)) ^2}{\lambda ^2}\cot^2\lambda \pi 
 +\frac{(A_2(\lambda) +D_2(\lambda)) ^2}{\lambda ^2}\\
&\quad +2\frac{(\cot \alpha -\cot \beta +A_1(\lambda)
+D_1(\lambda)) (A_2(\lambda)
+D_2(\lambda)) }{\lambda ^2}\cot \lambda \pi \Big]
+O(\frac{1}{\lambda ^3}) .
\end{aligned}
\end{equation}
It is well known (see \cite{s2}) that the spectrum of problem 
\eqref{e1}--\eqref{e7} is discrete and
\begin{equation}
\lambda _{n}\sim n+O(1) \quad \text{as }| n| \to \infty .  \label{e30}
\end{equation}
Next we present the more exact asymptotic distribution of the spectrum.
Using the residue theorem we have
\begin{align*}
\lambda _{n}-n
&=-\frac{1}{2\pi i}\oint_{C_{n}}\ln \frac{W(\lambda) }{W_0(\lambda) }d\lambda \\
&=-\frac{1}{2\pi i}\oint_{C_{n}}\frac{(\cot \alpha -\cot \beta
+A_1(\lambda) +D_1(\lambda)) \cot \lambda \pi }{\lambda }d\lambda  \\
&\quad -\frac{1}{2\pi i}\oint_{C_{n}}\frac{A_2(\lambda) 
 +D_2(\lambda) }{\lambda }d\lambda \\
&\quad -\frac{1}{2\pi i}\oint_{C_{n}}\Big[(\cot \alpha 
 -\cot\beta) A_2(\lambda) 
 -(\cot \alpha +\cot \beta) D_2(\lambda) \\
&\quad +C_2(\lambda)
-B_2(\lambda) +E_2(\lambda) -H_2(\lambda) \Big]
\frac{\cot \lambda \pi }{\lambda ^2}d\lambda \\
&\quad -\frac{1}{2\pi i}\oint_{C_{n}}\Big[\cot \alpha \cot \beta
+(\cot \beta -\cot \alpha) A_1(\lambda) +(
\cot \alpha +\cot \beta) D_1(\lambda)   \\
&\quad -C_1(\lambda) +B_1(\lambda) -E_1(\lambda) +H_1(\lambda) \Big] 
 \frac{d\lambda }{\lambda ^2}\\
&\quad +\frac{1}{2\pi i}\oint_{C_{n}}\frac{(\cot \alpha -\cot \beta
+A_1(\lambda) +D_1(\lambda)) ^2\cot
^2\lambda \pi }{2\lambda ^2}d\lambda \\
&\quad +\frac{1}{2\pi i}\oint_{C_{n}}\frac{(A_2(\lambda) +D_2(
\lambda)) ^2}{2\lambda ^2}d\lambda \\
&\quad +\frac{1}{2\pi i}\oint_{C_{n}}\frac{(\cot \alpha -\cot \beta
+A_1(\lambda) +D_1(\lambda)) (
A_2(\lambda) +D_2(\lambda)) \cot
\lambda \pi }{\lambda ^2}d\lambda \\
&\quad +O(\frac{1}{n^3}) .
\end{align*}
which implies, using residue calculation, that
\begin{equation}
\begin{aligned}
\lambda _{n}
&=n+\frac{\cot \beta -\cot \alpha -A_1(n)-D_1(n)}{n\pi } \\
&\quad +\Big(2D_2(n)\cot \alpha +(A_1(n) +D_1(n)) (A_2(n) 
 +D_2(n))-C_2(n)\\
&\quad +B_2(n)  -E_2(n)+H_2(n) \Big)/(n^2\pi)  \\
&\quad -\frac{(\cot \alpha -\cot \beta +A_1(n)+D_1(n)) ^2}{n^3}
+O(\frac{1}{n^3}) . 
\end{aligned} \label{e31}
\end{equation}
Thus we have proven the following theorem.

\begin{theorem} \label{thm2.1}
The spectrum of  problem \eqref{e1}--\eqref{e7} has the \eqref{e31} 
asymptotic distribution for sufficiently large $| n|$.
\end{theorem}

Finally, we will get regularized trace formula for
 problem \eqref{e1}--\eqref{e7}.
The asymptotic formula \eqref{e30} for the eigenvalues implies that for all
sufficiently large $N_0$, the numbers $\lambda _{n}$ with 
$|n| \leq N_0$ are inside $\Gamma _{N_0}$, and the numbers 
$\lambda _{n}$ with $| n| >N_0$ are outside $\Gamma_{N_0}$. It follows that
\begin{align*}
&\lambda _{-0}^2+\lambda _0^2+\sum_{0\neq n=-N_0}^{N_0}(\lambda _{n}^2-n^2) \\
&=-\frac{1}{2\pi i}\oint_{\Gamma_{n}}2\lambda \ln \frac{W(\lambda) }{W_0(\lambda
) }d\lambda \\
&=-\frac{1}{2\pi i}\oint_{\Gamma _{n}}2(\cot \alpha -\cot \beta
+A_1(\lambda) +D_1(\lambda)) \cot
\lambda \pi d\lambda  \\
&\quad -\frac{1}{2\pi i}\oint_{\Gamma _{n}}2(A_2(\lambda) +D_2(\lambda)) d\lambda\\
&\quad -\frac{1}{2\pi i}\oint_{\Gamma _{n}}2
 \Big[-2D_2(\lambda) \cot \alpha -(A_1(\lambda) +D_1(\lambda)) (A_2(\lambda) 
 +D_2(\lambda)) \\
&\quad  +C_2(\lambda) -B_2(\lambda)
+E_2(\lambda) -H_2(\lambda) \Big] 
\frac{\cot \lambda \pi }{\lambda }d\lambda \\
&\quad -\frac{1}{2\pi i}\oint_{\Gamma _{n}}2\Big[\cot \alpha \cot \beta
+(\cot \beta -\cot \alpha) A_1(\lambda) +(
\cot \alpha +\cot \beta) D_1(\lambda)  \\
&\quad -C_1(\lambda) +B_1(\lambda)-E_1(\lambda) +H_1(\lambda) \Big] 
 \frac{d\lambda }{\lambda } \\
&\quad +\frac{1}{2\pi i}\oint_{\Gamma _{n}}\frac{(\cot \alpha -\cot
\beta +A_1(\lambda) +D_1(\lambda))
^2\cot ^2\lambda \pi }{\lambda }d\lambda \\
&\quad +\frac{1}{2\pi i}\oint_{\Gamma _{n}}\frac{(A_2(\lambda) 
+D_2(\lambda)) ^2}{\lambda }d\lambda +O(\frac{1}{N_0})
\end{align*}
by calculations, which implies 
\begin{equation}
\begin{aligned}
&\lambda _{-0}^2+\lambda _0^2+\sum_{0\neq n=-N_0}^{N_0}(\lambda _{n}^2-n^2) \\
&=-\frac{2}{\pi }\sum_{0\neq n=-N_0}^{N_0}(\cot \alpha -\cot \beta
+A_1(n) +D_1(n)) \\
&\quad -\frac{2}{\pi }(\cot \alpha -\cot \beta +A_1(0) +D_1(0)) \\
&\quad -\sum_{0\neq n=-N_0}^{N_0}2\big[-2D_2(n) \cot \alpha
-(A_1(n) +D_1(n)) (A_2(n) +D_2(n))  \\
&\quad  +C_2(n) -B_2(n) +E_2(n)-H_2(n) \Big] \frac{1}{n\pi }+T\\
&\quad -2\Big[\cot \alpha \cot \beta +(\cot \beta -\cot \alpha)
A_1(0) +(\cot \alpha +\cot \beta) D_1(0)  \\
&\quad  -C_1(0) +B_1(0) -E_1(0)+H_1(0) \Big] \\
&\quad -(\cot \alpha -\cot \beta +A_1(0) +D_1(0)) ^2+(A_2(0) +D_2(0))^2
+O(\frac{1}{N_0}) , 
\end{aligned} \label{e32}
\end{equation}
where
\begin{align*}
T&=\operatorname{Res}_{\lambda =0}\Big\{ 2\Big[2D_2(\lambda) \cot \alpha
+(A_1(\lambda) +D_1(\lambda)) (A_2(\lambda) +D_2(\lambda)) \\
&\quad -C_2(\lambda) +B_2(\lambda) -E_2(\lambda) +H_2(\lambda) \Big] \frac{\cot
\lambda \pi }{\lambda }\Big\} .
\end{align*}
Passing to the limit as $N_0\to \infty $ in (32), we have
\begin{equation}
\begin{aligned}
&\lambda _{-0}^2+\lambda _0^2+\sum_{0\neq n=-\infty }^{\infty }
\Big\{\lambda _{n}^2-n^2-\frac{2}{\pi }[\cot \beta -\cot \alpha
-A_1(n) -D_1(n) ] \\
& +\Big[-2D_2(n) \cot \alpha +(A_1(n)+D_1(n)) (A_2(n) +D_2(n)) \\
& +C_2(n) -B_2(n) +E_2(n) +H_2(n) \Big] \frac{1}{n\pi }\Big\} \\
&=\frac{2}{\pi }(\cot \beta -\cot \alpha -A_1(0)
-D_1(0)) -\cot ^2\alpha -\cot ^2\beta \\
&\quad -2(\cot \alpha -\cot \beta) (A_1(0) +D_1(0)) 
-(A_1(0)  +D_1(0)) ^2 \\
&\quad +(A_2(0) +D_2(0)) ^2 
 -2\Big[\cot \alpha \cot \beta +(\cot \beta -\cot \alpha)A_1(0) \\
&\quad +(\cot \alpha +\cot \beta) D_1(0) 
 -C_1(0) +B_1(0) -E_1(0)+H_1(0) \Big] +T. 
\end{aligned} \label{e33}
\end{equation}
The series on the left side of this equality is called the regularized trace
of the problem \eqref{e1}--\eqref{e7}. The main result of this article 
is given by the following theorem.

\begin{theorem} \label{thm2.2}
Formula \eqref{e33} holds for the first regularized trace of the problem
\eqref{e1}--\eqref{e7}.
\end{theorem}

\subsection*{Acknowledgements}
The authors want to thank the anonymous referee for
the valuable comments.


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\end{document}