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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2017 (2017), No. 212, pp. 1--10.\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-2017/212\hfil Sturm-Liouville operator with parameter]
{Sturm-Liouville operator with parameter-dependent
boundary conditions \\ on time scales}

\author[A. S. Ozkan \hfil EJDE-2017/212\hfilneg]
{A. Sinan Ozkan}

\address{A. Sinan Ozkan \newline
Department of Mathematics Faculty of Art and Sci.
Cumhuriyet University 58140 \\
Sivas, Turkey}
\email{sozkan@cumhuriyet.edu.tr}

\dedicatory{Communicated by Jerome A. Goldstein}

\thanks{Submitted May 18, 2017. Published September 11, 2017.}
\subjclass[2010]{34B05, 34L05, 39A05}
\keywords{Time scale; Sturm-Liouville operator; 
\hfill\break\indent parameter-dependent boundary conditions}

\begin{abstract}
 In this study, we consider a boundary-value problem generated by a
 Sturm-Liouville dynamic equation on a time scale with boundary conditions
 depending on a spectral parameter. We introduce the operator formulation
 of the problem and give some properties of eigenvalues and eigenfunctions.
 We also formulate the number of eigenvalues when the time scale is finite.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{example}[theorem]{Example}
\allowdisplaybreaks


\section{Introduction and Preliminaries}

Time scale theory was introduced by Hilger in 1988. He gave a new derivation
in order to unify continuous and discrete analysis \cite{Hilger}. From then
on this approach has received a lot of attention and has applied quickly to
various area in mathematics. Sturm-Liouville theory on time scales was
studied first by Erbe and Hilger \cite{Erbe} in 1993. Some important results
on the properties of eigenvalues and eigenfunctions of the classical
Sturm-Liouville problem on time scales were given in various publications
(see e.g. \cite{Agarwal,Amster,Amster2,Davidson,Davidson2,Davidson3,Erbe2,
Guseinov,Guseinov2,Simon, Huseinov,Huseinov2,Kong,Rynne,Sun} and the
references therein).

In classical analysis, Sturm-Liouville problems with boundary conditions
which depend on the parameter were studied extensively. These kinds of
problems appear in physics, mechanics and engineering. There is a vast
literature about this subject. Particularly, an operator-theoretic
formulation of the problem with the spectral parameter contained linearly in
the boundary conditions has been given in \cite{F,F1,W}.
Oscillation and comparison results and some other properties of the
eigenvalues have been obtained in \cite{P1,P2,K}. Basis
properties and eigenfunction expansions have been considered in 
\cite{K2,K3,K4}. For Sturm-Liouville problem with
eigenparameter-dependent boundary conditions on arbitrary time scale we
refer to the study \cite{Bp} and the references therein.

In this article, Sturm-Liouville dynamic equation with boundary
conditions depending on the spectral parameter on a time scale is studied.
We define an operator which is appropriate to this boundary-value problem.
We prove that all eigenvalues are real, algebraically simple and two
eigenfunctions, corresponding to the different eigenvalues are orthogonal.
We also obtain a formula to give the number of eigenvalues of the problem
which is constructed on a finite time scale.

Before presenting our main results, we recall the some important concepts of
the time scale theory. For further knowledge, the reader is referred to
\cite{Atkinson,Bohner,Bohner2}.

If $\mathbb{T}$ is a closed subset of $\mathbb{R}$ it is called as a time scale.
The jump operators $\sigma $, $\rho $ and
graininess operator $\mu $ on $\mathbb{T}$ are defined as follows:
\begin{gather*}
 \sigma :\mathbb{T}\to \mathbb{T},\quad \sigma (
t) =\inf \{ s\in \mathbb{T}:s>t\} \quad \text{if }t\neq \sup
\mathbb{T},  \\
 \rho :\mathbb{T}\to \mathbb{T},\text{ }\rho ( t)
=\sup \{ s\in \mathbb{T}:s<t\} \quad \text{if }t\neq \inf \mathbb{T},  \\
 \sigma ( \sup \mathbb{T}) =\sup \mathbb{T},\quad
\rho ( \inf \mathbb{T}) =\inf \mathbb{T},  \\
\mu :\mathbb{T}\to [ 0,\infty )\quad
\mu (t) =\sigma ( t) -t.
\end{gather*}

A point of $\mathbb{T}$ is called as left-dense, left-scattered,
right-dense, right-scattered and isolated if $\rho (t)=t$, $\rho (t)<t$,
 $\sigma (t)=t$, $\sigma (t)>t$ and $\rho (t)<t<\sigma (t)$, respectively.

A function $f:\mathbb{T}\to \mathbb{R}$ is called rd-continuous on
$\mathbb{T}$ if it is continuous at all
right-dense points and has left-sided limits at all left-dense points in
$\mathbb{T}$. The set of rd-continuous functions on $\mathbb{T}$ is denoted
by $C_{rd}(\mathbb{T)}$ or $C_{rd}$. If $f$ is continuous on $\mathbb{T}$ it
is also rd-continuous.
Put
\[
\mathbb{T}^{k}:=\begin{cases}
\mathbb{T}-\{\sup \mathbb{T}\},  & \sup \mathbb{T<\infty }\text{ is
left-scattered} \\
\mathbb{T},  & \text{otherwise},
\end{cases}
\]
 $\mathbb{T}^{k^{2}}:=( \mathbb{T}^{k}) ^{k}$.

Let $t\in \mathbb{T}^{k}$. Suppose that for given any $\varepsilon >0$,
there exists a neighborhood
$U=( t-\delta ,t+\delta ) \cap \mathbb{T}$ such that
\begin{equation*}
| [ f( \sigma ( t) ) -f( s)
] -f^{\Delta }( t) [ \sigma ( t) -s]
| \leq \varepsilon | \sigma ( t) -s|
\end{equation*}
for all $s\in U$. Then $f$ is called $\Delta $-differentiable at
$t\in \mathbb{T}^{k}$. We call $f^{\Delta }( t) $ the $\Delta $-derivative
of $f$ at $t$. A function $F:\mathbb{T}\to \mathbb{R}$ defined as
 $F^{\Delta }(t)=f(t)$ for all $t\in \mathbb{T}^{k}$ is called
an antiderivative of $f$ on $\mathbb{T}$. In this case, the Cauchy integral
of $f$ is defined by
\begin{equation*}
\int_{a}^{b}f( t) \triangle t=F( b) -F(a) \text{ for }a,b\in \mathbb{T}.
\end{equation*}
We collect some necessary relations in the following lemma. Their proofs can
be found in \cite[Chapter 1]{Bohner}.

\begin{lemma} \label{lem1}
Let $f:\mathbb{T}\to \mathbb{R}$, $g:\mathbb{T}\to\mathbb{R}$
be two functions and $t\in \mathbb{T}^{k}$.
\begin{itemize}
\item[(i)] if $f^{\Delta }( t) $ exists, then $f$ is continuous at $t$;

\item[(ii)] if $t$ is right-scattered and $f$ is continuous at $t$, then $f$ is $
\Delta $-differentiable at $t$ and
$f^{\Delta }( t) =\frac{f^{\sigma }( t) -f( t) }{\sigma ( t) -t}$,
where $f^{\sigma }( t) =f( \sigma ( t) )$;

\item[(iii)] if $f^{\Delta }( t) $ exists, then $f^{\sigma }(
t) =f( t) +\mu (t)f^{\Delta }( t) $;
 if $f^{\Delta }\equiv 0$, then $f$ is constant;

\item[(iv)] if $f^{\Delta }( t) $ and $g^{\Delta }( t) $
exist, then $(f\pm g)^{\Delta }(t)=f^{\Delta }(t)\pm g^{\Delta }(t)$,
$(fg)^{\Delta }(t)=( f^{\Delta }g+f^{\sigma }g^{\Delta }) (t)$ and
if $( gg^{\sigma }) (t)\neq 0$, then
$( \frac{f}{g})^{\Delta }(t)=( \frac{f^{\Delta }g-fg^{\Delta }}{gg^{\sigma }})
(t)$;

\item[(v)] if $f\in C_{rd}(\mathbb{T)}$, then it has an antiderivative on
$\mathbb{T}$.
\end{itemize}
\end{lemma}

Now, let us recall the spaces $L_{p}$ and $H_1$. For
detailed knowledge related to Lebesgue measure, Lebesgue integration and
generalized derivation on the time scale we refer to \cite{Davidson}.

$L_{p}(\mathbb{T})=$\{$f:| f| ^{p}$ is integrable on $
\mathbb{T}$ in Lebesgue sense\} is a Banach space with the norm
$\|f\| _{L_{p}}=\big( \int_{a}^{b}| f( t)
| ^{p}\triangle t\big) ^{1/p}$. Moreover, the set
$C_{rd}^{1}(\mathbb{T)}=\{f:f \Delta $-differentiable on $\mathbb{T}$ and $
f^{\Delta }\in C_{rd}(\mathbb{T}^{k}\mathbb{)}$\} is a normed space with the
norm
$\| f\| _1:=\| f\|_{L_2}+\| f^{\Delta }\| _{L_2}$.
Finally, the Sobolev
space $H_1$ is defined to be the completion of $C_{rd}^{1}(\mathbb{T)}$
with respect to the norm $\|\cdot\| _1$. It is proven in
\cite{Davidson} that if $f\in H_1$, then there exists the generalized
derivative $f^{\Delta _{g}}$ of $f$ in $L_2(\mathbb{T})$ and the following
properties are valid.
\begin{itemize}
\item[(i)] If $f\in H_1$, then the function $f^{\Delta _{g}}$ is unique in
Lebesgue sense.

\item[(ii)] If $f\in C_{rd}^{1}(\mathbb{T)}$, then $f^{\Delta _{g}}=f^{\Delta }$.

\item[(iii)] item (ii)-(iv) in Lemma \ref{lem1} are valid with $f^{\Delta _{g}}$ instead of
$f^{\Delta }$.

\item[(iv)] $\int_{a}^{b}f^{\Delta _{g}}( t) \triangle t=f(b) -f( a) $
for $a,b\in \mathbb{T}$.
\end{itemize}

\section{Main Results}

Throughout this paper we assume that $\mathbb{T}$ is a bounded time scale.
Let us consider the boundary-value problem
\begin{gather}
 \ell y:=-y^{\Delta \Delta }(t)+q(t)y^{\sigma }(t)=\lambda
y^{\sigma }(t),\quad t\in \mathbb{T}^{k^{2}}  \label{e1}\\
 U(y):=( a_1\lambda +a_{0}) y^{\Delta }(\alpha
)-( b_1\lambda +b_{0}) y(\alpha )=0  \label{e2} \\
 V(y):=( c_1\lambda +c_{0}) y^{\Delta }(\beta
)-( d_1\lambda +d_{0}) y(\beta )=0, \label{e3}
\end{gather}
where $y^{\Delta \Delta }=( y^{\Delta }) ^{\Delta _{g}}$, $q(t)$
is real valued continuous function on $\mathbb{T}$,
$a_{i},b_{i},c_{i},d_{i} \in \mathbb{R}$, $i=0,1$, $\alpha =\inf \mathbb{T}$,
$\beta =\rho (\sup \mathbb{T)}$ and $\lambda $ is a spectral parameter.
 Additionally, we assume that $\alpha \neq \rho (\beta )$,
$K_1:=a_1b_{0}-a_{0}b_1>0$ and $ K_2:=c_{0}d_1-c_1d_{0}>0$.

\begin{definition} \label{def1} \rm
The values of the parameter for which equation \eqref{e1} has nonzero solutions
satisfying \eqref{e2} and \eqref{e3}, are called eigenvalues and the corresponding
nontrivial solutions are called eigenfunctions.
\end{definition}

Let the inner product in the Hilbert Space $H=L_2(\mathbb{T}^{k})\oplus
\mathbb{C}^{2}$ be defined by
\[
\langle Y,Z\rangle :=\int_{\alpha }^{\beta }y(t)\overline{
z(t)}\Delta t+\frac{1}{K_1}Y_{\alpha }\overline{Z_{\alpha }}+\frac{1}{
K_2}Y_{\beta }\overline{Z_{\beta }}
\]
for
\[
Y=( y(t),Y_{\alpha },Y_{\beta }) , \quad
Z=( z(t),Z_{\alpha},Z_{\beta }) \in H.
\]
Define an operator $L$ with $\ $the domain
$D(L)=\{Y\in H:y(t)\in C_{rd}^{1}(\mathbb{T})$,
$y^{\Delta }(t)\in H_1(\mathbb{T}^{k})$,
$Y_{\alpha}=a_1y^{\Delta }(\alpha )-b_1y(\alpha )$,
$Y_{\beta }=c_1y^{\Delta }(\beta )-d_1y(\beta )\}$ such that
\begin{gather*}
L(Y) =(z(t),Z_{\alpha },Z_{\beta }), \\
z(t) =-y^{\Delta \Delta }(t)+q(t)y^{\sigma }(t), \\
Z_{\alpha } = b_{0}y(\alpha )-a_{0}y^{\Delta }(\alpha ), \\
Z_{\beta }  = d_{0}y(\beta )-c_{0}y^{\Delta }(\beta )
\end{gather*}
It is obvious that the eigenvalue problem $L(Y)=\lambda Y^{\sigma }$
coincide with the problem \eqref{e1}-\eqref{e3}.

\begin{theorem} \label{thm1}
The relation $\langle LY,Z^{\sigma }\rangle =\langle
Y^{\sigma },LZ\rangle $ holds for all $Y,Z\in D(L)$.
\end{theorem}

\begin{proof}
Let $Y,Z\in D(L)$ be given as
\begin{gather*}
Y=( y(t),a_1y^{\Delta }(\alpha )-b_1y(\alpha ),c_1y^{\Delta }(\beta )
-d_1y(\beta )) , \\
Z=( z(t),a_1z^{\Delta }(\alpha )-b_1z(\alpha ),c_1z^{\Delta
}(\beta )-d_1z(\beta )) .
\end{gather*}
 It can be calculated by two partial
integrations that
\begin{align*}
\langle LY,Z^{\sigma }\rangle -\langle Y^{\sigma },LZ\rangle 
&=\int_{\alpha }^{\beta }y^{\sigma }(t)
 \overline{z}^{\Delta \Delta }(t)\Delta t-\int_{\alpha }^{\beta
 }y^{\Delta \Delta }(t)\overline{z^{\sigma }}(t)\Delta t \\
&\quad +\frac{1}{K_1}[ b_{0}y(\alpha )-a_{0}y^{\Delta }(\alpha )] [ a_1
\overline{z}^{\Delta }(\alpha )-b_1\overline{z}(\alpha )] \\
&\quad -\frac{1}{K_1}[ a_1y^{\Delta }(\alpha )-b_1y(\alpha )] [ b_{0}
 \overline{z}(\alpha )-a_{0}\overline{z}^{\Delta }(\alpha )] \\
&\quad +\frac{1}{K_2}[ d_{0}y(\beta )-c_{0}y^{\Delta }(\beta )] [ c_1
\overline{z}^{\Delta }(\beta )-d_1\overline{z}(\beta )] \\
&\quad -\frac{1}{ K_2}[ c_1y^{\Delta }(\beta )-d_1y(\beta )] [ d_{0}
\overline{z}(\beta )-c_{0}\overline{z}^{\Delta }(\beta )] =0.
\end{align*}
\end{proof}

\begin{corollary} \label{coro1}
All eigenvalues of  problem \eqref{e1}-\eqref{e3} are real numbers and two
eigenfunctions $y(t)$, $z(t)$ corresponding to different eigenvalues
$\lambda _1$, $\lambda _2$ are orthogonal, i.e.,
\begin{align*}
&\int_{\alpha }^{\beta }y(t)\overline{z(t)}\Delta t+\frac{1}{
K_1}[ a_1y^{\Delta }(\alpha )-b_1y(\alpha )]
[a_1z^{\Delta }(\alpha )-b_1z(\alpha )]  \\
&\quad +\frac{1}{K_2}[ c_1y^{\Delta }(\beta )-d_1y(\beta )]
 [ c_1z^{\Delta }(\beta )-d_1z(\beta )] =0
\end{align*}
\end{corollary}

Let $s(t,\lambda )$, $c(t,\lambda )$, $\varphi (t,\lambda )$ and
$\psi(t,\lambda )$ be the solutions of \eqref{e1} under the initial conditions
\begin{gather}
s(\alpha ,\lambda ) =0,\quad s^{\Delta }(\alpha ,\lambda )=1, \label{e4} \\
c(\alpha ,\lambda ) =1,\quad c^{\Delta }(\alpha ,\lambda )=0, \label{e5} \\
\varphi (\alpha ,\lambda ) =a_1\lambda +a_{0},\quad \varphi ^{\Delta
}(\alpha ,\lambda )=b_1\lambda +b_{0}, \label{e6} \\
\psi (\beta ,\lambda ) = c_1\lambda +c_{0},\quad \psi ^{\Delta }(\beta
,\lambda )=d_1\lambda +d_{0}, \label{e7}
\end{gather}
respectively. It is clear that the Wronskian 
\[
W[c(t,\lambda ),s(t,\lambda
)]=c(t,\lambda )s^{\Delta }(t,\lambda )-c^{\Delta }(t,\lambda )s(t,\lambda )
\]
does not depend on $t$. From \eqref{e4} and \eqref{e5} we have
$W[c(t,\lambda),s(t,\lambda )]=1$ and so $s(t,\lambda )$ and $c(t,\lambda )$
are linearly independent. Moreover, it can be seen that the relation
$\varphi(t,\lambda _n)=\chi _n\psi (t,\lambda _n)$ is valid for each
eigenvalue $\lambda _n$, where
\[
\chi _n=\frac{[ d_1\varphi
(\beta ,\lambda _n)-c_1\varphi ^{\Delta }(\beta ,\lambda _n)] }{K_2}.
\]

\begin{theorem} \label{thm2}
$s(t,\lambda )$, $c(t,\lambda )$, $\varphi (t,\lambda )$, $\psi (t,\lambda )$
and their $\Delta $-derivatives are entire functions of $\lambda $ for each
fixed $t$.
\end{theorem}

\begin{proof}
Let us prove that $\varphi (t,\lambda )$ is analytic for arbitrary $\lambda
_{0}\in\mathbb{C}$ and fixed $t\in \mathbb{T}^{k^{2}}$.
 Analyticity of the other functions
can be proven similarly. Put $\phi (t,\lambda ,\lambda _{0})=\varphi
(t,\lambda )-\varphi (t,\lambda _{0})$. From \eqref{e1} and \eqref{e6},
the function $\phi (t,\lambda ,\lambda _{0})$ is the solution of the
initial-value problem
\begin{equation} \label{e8}
\begin{gathered}
-\phi ^{\Delta \Delta }(t)+q(t)\phi ^{\sigma }(t)=\lambda _{0}\phi ^{\sigma
}(t)+( \lambda -\lambda _{0}) \varphi ^{\sigma }( t,\lambda)  \\
\phi (\alpha )=a_1( \lambda -\lambda _{0})  \\
\phi ^{\Delta }(\alpha )=b_1( \lambda -\lambda _{0}) .
\end{gathered}
\end{equation}
From the variation of parameters formula, we have
\begin{align*}
\phi (t,\lambda ,\lambda _{0})
&= Ac(t,\lambda _{0})+Bs(t,\lambda _{0})\\
&\quad +\int_{\alpha }^{t}[ c^{\sigma }(\xi ,\lambda _{0})s(t,\lambda
_{0})-s^{\sigma }(\xi ,\lambda _{0})c(t,\lambda _{0})] ( \lambda
-\lambda _{0}) \varphi ^{\sigma }( \xi ,\lambda ) \Delta t
\end{align*}
where $A=a_1( \lambda -\lambda _{0}) $ and
$B=( \lambda-\lambda _{0}) [ b_1+\mu (\alpha )\varphi ^{\sigma }(
\alpha ,\lambda ) ] $. Therefore,
\begin{align*}
\frac{\varphi (t,\lambda )-\varphi (t,\lambda _{0})}{\lambda -\lambda _{0}}
&= a_1c(t,\lambda _{0})+[ b_1+\mu (\alpha )\varphi ^{\sigma }(
\alpha ,\lambda ) ] s(t,\lambda _{0})+ \\
&\quad +\int_{\alpha }^{t}[ c^{\sigma }(\xi ,\lambda _{0})s(t,\lambda
_{0})-s^{\sigma }(\xi ,\lambda _{0})c(t,\lambda _{0})] \varphi
^{\sigma }( \xi ,\lambda ) \Delta t.
\end{align*}
Since the solution $\varphi (t,\lambda )$ is continuous, according to
 $\lambda $, it follows that $\varphi (t,\lambda )$ is also analytic on $\lambda $.
\end{proof}

Put
\begin{equation} \label{e9}
\Delta (\lambda )=W[\psi ,\varphi ]=( c_1\lambda +c_{0})
\varphi ^{\Delta }(\beta ,\lambda )-( d_1\lambda +d_{0})
\varphi (\beta ,\lambda ).
\end{equation}
From Theorem \ref{thm2}, $\Delta (\lambda )$ is an entire function.

\begin{theorem} \label{thm3}
The zeros of the function $\Delta (\lambda )$ coincide with the eigenvalues
of the problem \eqref{e1}-\eqref{e3}.
\end{theorem}

\begin{proof}
It is clear that, if $\Delta (\lambda _{0})=0$ for a number $\lambda _{0}$,
then $\lambda _{0}$ is also an eigenvalue of \eqref{e1}-\eqref{e3}.

On the other hand, if $\lambda _{0}$ is an eigenvalue and $y(t,\lambda
_{0})=C_1s(t,\lambda _{0})+C_2c(t,\lambda _{0})\neq 0$ is the
corresponding eigenfunction, then $y(t,\lambda _{0})$ satisfies \eqref{e2}
 and \eqref{e3}.
Therefore,
\begin{gather*}
 ( a_1\lambda _{0}+a_{0}) C_1-(
b_1\lambda _{0}+b_{0}) C_2=0,  \\
\begin{aligned}
&[ ( c_1\lambda _{0}+c_{0}) s^{\Delta
}(\beta )-( d_1\lambda _{0}+d_{0}) s(\beta )] C_1\\
&+[( c_1\lambda _{0}+c_{0}) c^{\Delta }(\beta )-(
d_1\lambda _{0}+d_{0}) c(\beta )] C_2=0
\end{aligned}
\end{gather*}
and so
\[
\det \begin{pmatrix}
( a_1\lambda _{0}+a_{0})  & -( b_1\lambda
_{0}+b_{0})  \\
( c_1\lambda _{0}+c_{0}) s^{\Delta }(\beta )-(
d_1\lambda _{0}+d_{0}) s(\beta ) & ( c_1\lambda
_{0}+c_{0}) c^{\Delta }(\beta )-( d_1\lambda _{0}+d_{0})
\end{pmatrix}
 =0.
\]
Since $\varphi (t,\lambda _{0})=( a_1\lambda _{0}+a_{0})
c(t,\lambda _{0})+( b_1\lambda _{0}+b_{0}) s(t,\lambda _{0})$,
we obtain $\Delta (\lambda _{0})=0$.
\end{proof}

Since $\Delta (\lambda )$ is entire function,
 problem \eqref{e1}-\eqref{e3} has a discrete spectrum.

\begin{theorem} \label{thm4}
The eigenvalues of problem \eqref{e1}-\eqref{e3} are algebraically simple.
\end{theorem}

\begin{proof}
Let $\lambda _n$ be an eigenvalue of \eqref{e1}-\eqref{e3}.
Obviously,
\[
\frac{d\Delta (\lambda )}{d\lambda }=( c_1\lambda +c_{0}) \varphi _{\lambda
}^{\Delta }(\beta ,\lambda )-( d_1\lambda +d_{0}) \varphi
_{\lambda }(\beta ,\lambda )+c_1\varphi ^{\Delta }(\beta ,\lambda
)-d_1\varphi (\beta ,\lambda ),
\]
 where
$\varphi _{\lambda }=\frac{\partial \varphi }{\partial \lambda }$ and
$\varphi _{\lambda }^{\Delta }=\frac{\partial \varphi ^{\Delta }}{\partial \lambda }$. It will be
sufficient to see $ \frac{d\Delta (\lambda )}{d\lambda }|
_{\lambda =\lambda _n}\neq 0$.

We write equation \eqref{e1} for $\varphi (t,\lambda )$ as
\begin{equation} \label{e10}
-\varphi ^{\Delta \Delta }(t,\lambda )+q(t)\varphi ^{\sigma }(t,\lambda
)=\lambda \varphi ^{\sigma }(t,\lambda ).
\end{equation}
By derivation according to $\lambda $, we obtain
\begin{equation} \label{e11}
-\varphi _{\lambda }^{\Delta \Delta }(t,\lambda )+q(t)\varphi _{\lambda
}^{\sigma }(t,\lambda )=\lambda \varphi _{\lambda }^{\sigma }(t,\lambda
)+\varphi ^{\sigma }(t,\lambda ).
\end{equation}
Now by the standard procedure of multiplying \eqref{e10} by
$\varphi _{\lambda}^{\sigma }$, \eqref{e11} by $\varphi ^{\sigma }$,
subtracting and integrating, we obtain
\begin{equation} \label{e12}
[ \varphi (t,\lambda )\varphi _{\lambda }^{\Delta }(t,\lambda )-\varphi
^{\Delta }(t,\lambda )\varphi _{\lambda }(t,\lambda )] _{\alpha
}^{\beta }=-\int_{\alpha }^{\beta }[ \varphi ^{\sigma
}(t,\lambda )] ^{2}\Delta t.
\end{equation}
Taking into account the initial conditions \eqref{e6} we obtain
\begin{equation*}
\varphi ^{\Delta }(\alpha ,\lambda )\varphi _{\lambda }(\alpha ,\lambda
)-\varphi (\alpha ,\lambda )\varphi _{\lambda }^{\Delta }(\alpha ,\lambda
)=K_1.
\end{equation*}
Therefore,
\begin{equation} \label{e13}
-\int_{\alpha }^{\beta }[ \varphi ^{\sigma }(t,\lambda )]
^{2}\Delta t=K_1+\varphi (\beta ,\lambda )\varphi _{\lambda }^{\Delta
}(\beta ,\lambda )-\varphi ^{\Delta }(\beta ,\lambda )\varphi _{\lambda
}(\beta ,\lambda ).
\end{equation}
Putting $\lambda _n$ instead of $\lambda $ in \eqref{e13} and doing the necessary
calculations we obtain
\begin{align*}
& -\int_{\alpha }^{\beta }[ \varphi ^{\sigma }(t,\lambda
_n)] ^{2}\Delta t=K_1+\varphi (\beta ,\lambda _n)\varphi
_{\lambda }^{\Delta }(\beta ,\lambda _n)-\varphi ^{\Delta }(\beta ,\lambda
_n)\varphi _{\lambda }(\beta ,\lambda _n) \\
&= K_1+\chi _n[ ( c_1\lambda _n+c_{0}) \varphi
_{\lambda }^{\Delta }(\beta ,\lambda _n)-( d_1\lambda
_n+d_{0}) \varphi _{\lambda }(\beta ,\lambda _n)]  \\
&= K_1+\chi _n[ ( c_1\lambda _n+c_{0}) \varphi
_{\lambda }^{\Delta }(\beta ,\lambda _n)-( d_1\lambda
_n+d_{0}) \varphi _{\lambda }(\beta ,\lambda _n)] + \\
&\quad +\chi _n[ c_1\varphi ^{\Delta }(\beta ,\lambda _n)-d_1\varphi
(\beta ,\lambda _n)] -\chi _n[ c_1\varphi ^{\Delta }(\beta
,\lambda _n)-d_1\varphi (\beta ,\lambda _n)] \\
&= K_1+\chi _n^{2}K_2+ \chi _n\frac{d\Delta (\lambda )}{
d\lambda }\big| _{\lambda =\lambda _n}
\end{align*}
Hence, we obtain
\begin{equation*}
-\frac{d\Delta (\lambda )}{d\lambda }\big| _{\lambda =\lambda_n}
=\frac{1}{\chi _n}\Big\{ \int_{\alpha }^{\beta }[
\varphi ^{\sigma }(t,\lambda _n)] ^{2}\Delta t+K_1+\chi
_n^{2}K_2\Big\} \neq 0\,.
\end{equation*}
This completes the proof.
\end{proof}

The next theorem gives the number of eigenvalues of problem \eqref{e1}-\eqref{e3}
on a finite time scale.

\begin{theorem}\label{thm5}
If $\mathbb{T}$ is finite, then the number of eigenvalues of \eqref{e1}-\eqref{e3} is
\begin{equation} \label{e14}
n-\operatorname{def}(c_1)-\operatorname{def}( a_1+\mu (\alpha )b_1)
\end{equation}
where $n$ is number of elements of $\mathbb{T}$ and
 $\operatorname{def}(u)=\begin{cases}
0, &u\neq 0 \\
1, &u=0
\end{cases}$.
\end{theorem}

\begin{proof}
Since $\mathbb{T}$ consists of finitely many elements, all points of
$\mathbb{T}$ are isolated. Let
$\mathbb{T}=\{ \alpha ,\sigma (\alpha),\sigma ^{2}(\alpha ),\ldots ,
\sigma ^{n-2}(\alpha ),\sigma ^{n-1}(\alpha )\} $,
$\sigma ^{n-2}(\alpha )=\beta $, where
$\sigma ^{j}=\sigma ^{j-1}\circ \sigma $, for $j\geq 2$.
It can be calculated from \eqref{e1} and \eqref{e6} that
\begin{gather} 
\varphi ^{\sigma }( \alpha )
=( \mu (\alpha)b_1+a_1) \lambda +\mu (\alpha )b_{0}+a_{0} \label{e15} \\
\varphi ^{\sigma ^{2}}( t) = [ f(t)\lambda +g(t)]
\varphi ^{\sigma }( t) +h(t)\varphi ( t) ,\quad t\in \mathbb{T}^{k^{2}} \label{e16}
\end{gather}
where $f(t)=-\mu (t)\mu ^{\sigma }(t)$,
$g(t)=\mu (t)\mu ^{\sigma }(t)q(t)+\frac{\mu ^{\sigma }(t)}{\mu (t)}+1$ and
$h(t)=-\frac{\mu ^{\sigma }(t)}{\mu (t)}$. It can be obtained that the
following equality is valid for $r\geq 3$,
\begin{equation} \label{e17}
\frac{\varphi ^{\sigma ^{r}}( \alpha ) }{\prod
_{j=0}^{r-2}f(\sigma ^{j}(\alpha ))}
=\begin{cases}
A_1\lambda ^{r}+p_{r-1}(\lambda ), & A_1\neq 0 \\
A_2\lambda ^{r-1}+p_{r-2}(\lambda ), & A_1=0,
\end{cases}
\end{equation}
where $A_1=\mu (\alpha )b_1+a_1$,
$A_2=\mu (\alpha )b_{0}+a_{0}+\frac{1}{( \mu (\alpha )) ^{2}}a_1$,
 $p_{r-1}(\lambda )$ and $p_{r-2}(\lambda )$ are polynomials with degrees
$r-1$ and $r-2$, respectively. We must note that,
if $A_1=0$, then $A_2\neq 0$.
Indeed, $A_1=0$ yields $\mu (\alpha )b_1=-a_1$. Since $K_1>0$, we obtain
\begin{equation*}
0<K_1\mu (\alpha )=a_1( b_{0}\mu (\alpha )+a_{0}) .
\end{equation*}
Thus, the signs of $a_1$ and $b_{0}\mu (\alpha )+a_{0}$ are the same and
so $A_2\neq 0$. It is obvious from \eqref{e9} that
\begin{equation} \label{e18}
\Delta (\lambda )=\frac{1}{\mu (\beta )}\{ ( c_1\lambda
+c_{0}) \varphi ^{\sigma }(\beta ,\lambda )-[ (
c_1+d_1\mu (\beta )) \lambda +( c_{0}+d_{0}\mu (\beta
)) ] \varphi (\beta ,\lambda )\} .
\end{equation}
Taking into account \eqref{e17} and \eqref{e18}, it is obtained that
\begin{equation} \label{e19}
\deg \Delta (\lambda )
=\begin{cases}
n, & c_1\neq 0\text{ and }  \mu (\alpha )b_1+a_1\neq 0 \\
n-1, & c_1\neq 0\text{ and } \mu (\alpha )b_1+a_1=0 \\
n-1, & c_1=0\text{ and } \mu (\alpha )b_1+a_1\neq 0 \\
n-2, & c_1=0\text{ and } \mu (\alpha )b_1+a_1=0\,.
\end{cases}
\end{equation}
Hence, we conclude that equality \eqref{e14} holds.
\end{proof}

\begin{corollary} \label{coro2}
If the number of  points in $\mathbb{T}$ is not less than three, then the
boundary-value problem \eqref{e1}-\eqref{e3} has at least one eigenvalue.
\end{corollary}

\begin{example} \label{examp1} \rm
Let us consider the following boundary-value problems on $\mathbb{T}
=\{0,1,2,\dots, n-1\}$:
\begin{gather*}
-y^{\Delta \Delta }(t)=\lambda y^{\sigma }(t)  \\
\lambda y^{\Delta }(0)+( \lambda -1) y(0)=0 \\
y^{\Delta }(n-2)-\lambda y(n-2)=0,
\end{gather*}
\begin{gather*}
-y^{\Delta \Delta }(t)=\lambda y^{\sigma }(t)  \\
\lambda y^{\Delta }(0)-y(0)=0 \\
y^{\Delta }(n-2)-\lambda y(n-2)=0
\end{gather*}
and
\begin{gather*}
-y^{\Delta \Delta }(t)=\lambda y^{\sigma }(t)  \\
\lambda y^{\Delta }(0)-y(0)=0 \\
\lambda y^{\Delta }(n-2)+y(n-2)=0\,.
\end{gather*}
According to Theorem \ref{thm5}, the numbers of the eigenvalues of these problems are
$n-2$, $n-1$ and $n$, respectively.
\end{example}

\begin{theorem} \label{thm6}
If $\lambda $ is not an eigenvalue, then the operator
$( L-\lambda I^{\sigma }) :D(L)\to L_2(\mathbb{T}^{k})\oplus \{0\}^{2}$
is bijective and its inverse operator is
\[
( L-\lambda I^{\sigma }) ^{-1}\begin{pmatrix}
f(t) \\
0 \\
0
\end{pmatrix}
=
\begin{pmatrix}
y(t) \\
Y_{\alpha } \\
Y_{\beta }
\end{pmatrix}
\]
 such that $y(t)=\int_{\alpha }^{\beta }G(t,\xi ,\lambda )f(\xi )\Delta \xi $,
\begin{gather*}
Y_{\alpha }=\int_{\alpha }^{\beta }[ a_1G^{\Delta }(\alpha
,\xi ,\lambda )-b_1G(\alpha ,\xi ,\lambda )] f(\xi )\Delta \xi , \\
Y_{\beta }=\int_{\alpha }^{\beta }[ c_1G^{\Delta }(\beta ,\xi
,\lambda )-d_1G(\beta ,\xi ,\lambda )] f(\xi )\Delta \xi ,
\end{gather*}
where 
\[
G(t,\xi ,\lambda )=\frac{1}{\Delta (\lambda )}
\begin{cases}
\varphi ^{\sigma }(\xi ,\lambda )\psi (t,\lambda ), &\alpha \leq \xi <t \\
\varphi (t,\lambda )\psi ^{\sigma }(\xi ,\lambda ),& t<\xi \leq \beta\,.
\end{cases}
 \]
\end{theorem}

\begin{proof}
If $\lambda $ is not an eigenvalue, then $\Delta (\lambda )\neq 0$ so
$L-\lambda I^{\sigma }$ is injective on $D(L)$. On the other hand, from the
method of variation of parameters, the boundary-value problem
\begin{gather*}
 -y^{\Delta \Delta }(t)+\{ q(t)-\lambda \}
y^{\sigma }(t)=f(t), \quad t\in \mathbb{T}^{k^{2}},   \\
( a_1\lambda +a_{0}) y^{\Delta }(\alpha
)-( b_1\lambda +b_{0}) y(\alpha )=0,   \\
 ( c_1\lambda +c_{0}) y^{\Delta }(\beta
)-( d_1\lambda +d_{0}) y(\beta )=0
\end{gather*}
has a solution for each $f(t)$ in $L_2(\mathbb{T}^{k})$ such that
 $y(t)=\int_{\alpha }^{\beta }G(t,\xi ,\lambda )f(\xi )\Delta \xi $ and
\[
G(t,\xi ,\lambda )=\frac{1}{\Delta (\lambda )}
\begin{cases}
\varphi ^{\sigma }(\xi ,\lambda )\psi (t,\lambda ),& \alpha\leq \xi <t \\
\varphi (t,\lambda )\psi ^{\sigma }(\xi ,\lambda ),& t<\xi \leq \beta\,.
\end{cases}
\]
 Therefore, $L-\lambda I^{\sigma }$ is surjective operator and its
inverse operator is given as
\[
( L-\lambda I^{\sigma })^{-1}\begin{pmatrix}
f(t) \\
0 \\
0
\end{pmatrix}
=
\begin{pmatrix}
y(t) \\
Y_{\alpha } \\
Y_{\beta }
\end{pmatrix}.
\]
Finally, since $Y=\begin{pmatrix}
y(t) \\
Y_{\alpha } \\
Y_{\beta }
\end{pmatrix} \in D(L)$, it follows that
\begin{gather*}
Y_{\alpha }=\int_{\alpha }^{\beta }[
a_1G^{\Delta }(\alpha ,\xi ,\lambda )-b_1G(\alpha ,\xi ,\lambda )]
f(\xi )\Delta \xi, \\
Y_{\beta }=\int_{\alpha }^{\beta }[
c_1G^{\Delta }(\beta ,\xi ,\lambda )-d_1G(\beta ,\xi ,\lambda )]
f(\xi )\Delta \xi .
\end{gather*}
\end{proof}

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\end{document}