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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2017 (2017), No. 307, pp. 1--11.\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-2017/307\hfil Fourth-order eigenvalue problem]
{Spectral properties of a fourth-order eigenvalue problem with
spectral parameter in the boundary conditions}

\author[Z. S. Aliyev, F. M. Namazov \hfil EJDE-2017/307\hfilneg]
{Ziyatkhan S. Aliyev, Faiq M. Namazov}

\address{Ziyatkhan S. Aliyev \newline
Baku State University,
Baku AZ1148, Azerbaijan.\newline
Institute of Mathematics and Mechanics NAS of Azerbaijan,
Baku AZ1141, Azerbaijan}
\email{z\_aliyev@mail.ru}

\address{Faiq M. Namazov \newline
Baku State University,
Baku AZ1148, Azerbaijan}
\email{faig-namazov@mail.ru}

\thanks{Submitted August 29, 2017. Published December 14, 2017.}
\subjclass[2010]{34B05, 34B09, 34B24}
\keywords{Bending vibrations of a homogeneous rod;
 fourth order ODE;
\hfill\break\indent  oscillation properties of eigenfunctions;
 basis properties of eigenfunctions}

\begin{abstract}
 In this article we consider eigenvalue problems for fourth-order
 ordinary differential equation with spectral parameter in boundary
 conditions. We study the location of eigenvalues on the real axis,
 find the multiplicities of eigenvalues, investigate the oscillation
 properties of eigenfunctions, and the basis properties in the space
 $L_p$, $1 < p < \infty$, of the subsystems of eigenfunctions of this
 problem.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{corollary}[theorem]{Corollary}
\allowdisplaybreaks

\section{Introduction}

We consider the  eigenvalue problem
\begin{gather}\label{e1.1}
y^{(4)} (x) - (q(x)y'(x))' = \lambda y(x),\quad 0 < x < 1, \\
\label{e1.2}
y''(0) = y'' (1) = 0, \\
\label{e1.3}
Ty(0) - a \lambda y (0) = 0, \\
\label{e1.4}
Ty(1) - c \lambda y(1) = 0,
\end{gather}
where $\lambda \in \mathbb{C}$ is a spectral parameter,
 $Ty \equiv y''' - qy'$, $q(x)$ is positive and absolutely continuous 
function on $[0,1]$, $a,c$ are real constants such that $a > 0$, $c < 0$.

Problem \eqref{e1.1}-\eqref{e1.4} arises in the dynamical boundary-value 
problem describing free bending vibrations of a homogeneous rod of constant
rigidity, in cross sections of which the longitudinal force acts, both ends 
of which are fixed elastically and on these ends the servocontrol forces 
in acting. For more details on the physical meaning of this problem, see 
\cite{b6,r1}.

The study of boundary-value problems for ordinary differential operators
with spectral parameter in boundary conditions has a long history. 
In his Memoire \cite{p1} Poisson solved the problem of the motion of a body 
suspended by the end of an inextensible thread. 
Krylov \cite{k6} and Timoshenko \cite{t2} considered the problem of the longitudinal 
vibrations of the rod, which is one of the most interesting exactly solvable
models. Additional information on specific physical problems leading
to the boundary-value problems for ordinary differential operators with
spectral parameter in boundary conditions can be found in the books 
\cite{b6,k6,r1,t1,t2}  and in the papers \cite{b4,b5,f1,k2,m2,s2,w1}. 
Spectral problems for ordinary differential operators with spectral 
parameter in the boundary conditions have been considered in various
formulations by many authors 
\cite{a1,a2,a3,a4,a5,a6,a7,b4,b5,f1,k1,k2,k3,k4,k5,m1,m2,m3,m4,p1,r2,s1,s2,w1}. 
In \cite{a3,a5,f1,k1,k2,k5,m1,m4,r2,s1,w1}
the authors studied the basis property in various function spaces of 
the root functions systems of the Sturm-Liouville problem with spectral 
parameter in the boundary conditions.
The basis properties of subsystems of root functions in the space 
$L_p$, $1 < p < \infty$, of the boundary-value problems for fourth order
 ordinary differential equations with spectral parameter in one of the 
boundary conditions are studied in \cite{a1,a2,a6,k4}.

In the recent paper \cite{a4} the basis properties of eigenfunctions of a 
fourth-order eigenvalue problem with spectral parameter entering in 
two of the boundary conditions at the point $x = 1$ are studied.
In this paper, are found sufficient conditions for the subsystems of 
eigenfunctions of this problem to form a basis in the space 
$L_p (0, 1),\, 1 < p < \infty$.

The purpose of the present paper is to study the basis property of the 
subsystems of eigenfunctions of boundary-value problem
\eqref{e1.1}-\eqref{e1.4} in $L_p (0, 1)$, $1 < p < \infty$.

This article has the following structure. Some statements necessary in 
the sequel are given in Section 2. In Section 3 we investigate the main 
properties of solution of problem \eqref{e1.1}-\eqref{e1.3} which play
an essential role in the study of the oscillatory properties of eigenfunctions 
of \eqref{e1.1}-\eqref{e1.4}. In Section 4 we give an operator
interpretation of boundary-value problem \eqref{e1.1}-\eqref{e1.4},
where we associate with the problem a self-adjoint operator in the 
Hilbert space $L_2(0, 1) \oplus \mathbb{C}^2$, and provide some spectral 
properties of the corresponding operator. Here we study the structure 
of root subspaces, the location of eigenvalues on the real axis and 
the oscillation properties of eigenfunctions of problem \eqref{e1.1}-\eqref{e1.4}.
We show that the eigenvalues of boundary-value problem \eqref{e1.1}-\eqref{e1.4}
are nonnegative, simple and they form an infinitely increasing sequence. 
In Section 5 we obtain sufficient conditions for the subsystems of 
eigenfunctions of \eqref{e1.1}-\eqref{e1.4} to form a basis in the space
$L_p (0, 1)$, $1 < p < \infty$. More precisely, we prove that the system 
of eigenfunctions of this problem after removing two eigenfunctions corresponding 
to eigenvalues with numbers of different parity forms a basis in the space
 $L_p(0, 1)$, $1 < p < \infty$, which is an unconditional basis for $p = 2$.


\section{Preliminaries}

To study the spectral properties of problem \eqref{e1.1}-\eqref{e1.4},
we will need the following statements.

\begin{lemma}[{\cite[Lemma 2.1]{b1}}]  \label{lem2.1}
Let  $y(x, \lambda)$ be a nontrivial solution of equation \eqref{e1.1}
for $\lambda > 0$. If $y,\, y',\, y'',\, Ty$ are nonnegative and not all 
equal zero at $x_0 \in (0,1)$, then they are positive for $x \in  (x_0, 1]$. 
If  $y, -y', y'', -Ty$ are nonnegative and not all equal zero at 
$x_0 \in (0,1)$, then they are positive for $x \in [0, x_0)$.
\end{lemma}

\begin{lemma}[{\cite[Lemma 2.2]{b1}}] \label{lem2.2}
 Let $y(x, \lambda)$ be a nontrivial solution of problem \eqref{e1.1}, \eqref{e1.2}
for $\lambda > 0$. If  $y(x_{0},\lambda) = 0$ or  $y''(x_{0},\lambda) = 0$,
then $y'(x,\lambda) Ty (x,\lambda) < 0$ in a some neighborhood of
$x_{0} \in (0,1)$; if  $y'(x_{0},\lambda) = 0\,$ or  $\,Ty(x_{0},\lambda) = 0$,
then $y(x,\lambda) y''(x,\lambda) < 0$ in a some neighborhood of $x_{0} \in (0,1)$.
\end{lemma}

Consider the boundary condition
\begin{gather}\label{e2.1}
y (0)\cos \beta + Ty (0)\sin \beta  = 0, \\
\label{e2.2}
y (1)\cos \delta - Ty (1)\sin \delta  = 0,
\end{gather}
where $\beta, \delta \in [0, \frac{\pi }{2}]$.

Alongside boundary-value problem \eqref{e1.1}-\eqref{e1.4} we shall
consider the spectral problem \eqref{e1.1}, \eqref{e1.2},
\eqref{e2.1}, \eqref{e2.2} and \eqref{e1.1}, \eqref{e1.2},
\eqref{e2.1}, \eqref{e1.4}. The spectral properties of  problem
\eqref{e1.1}, \eqref{e1.2}, \eqref{e2.1}, \eqref{e2.2} have been
investigated in \cite{b1}, and of 
 problem \eqref{e1.1}, \eqref{e1.2}, \eqref{e2.1}, \eqref{e1.4}
have been investigated in \cite{a2,a6,k4}.

\begin{theorem}[{\cite[Thms. 5.4 and 5.5]{b1}}] \label{thm2.1}
 The eigenvalues of boundary-value problem \eqref{e1.1}, \eqref{e1.2}, \eqref{e2.1}, 
\eqref{e2.2} are real, simple and form an infinitely increasing sequence 
$\{ \lambda_k (\beta, \delta)\}_{k = 1}^\infty$ such that 
$\lambda_1 (\beta, \delta) > 0$ for $\beta + \delta < \pi$ and 
$\lambda_1 \left( {\frac{\pi }{2},\frac{\pi }{2}} \right) = 0$.
 Moreover, the eigenfunction $u_k^{(\beta,\,\delta)} (x)$ corresponding 
to the eigenvalue $\lambda_k (\beta,\,\delta)$ has $k-1$ simple zeros in 
the interval $(0, 1)$.
\end{theorem}

\begin{theorem}[{\cite[Thm. 2.2]{k4}}] \label{thm2.2}
The eigenvalues of boundary-value problem \eqref{e1.1}, \eqref{e1.2}, \eqref{e2.1},
\eqref{e1.4} 
 are real, simple and form an infinitely increasing sequence \\
$\{ \tilde \lambda_k (\beta)\}_{k = 1}^\infty$ such that 
$\tilde \lambda_1 (\beta) > 0$ for $\beta < \pi/2$ and 
$\tilde \lambda_1 (\pi/2) = 0$. Moreover, 
the eigenfunction $\tilde u_k^{(\beta)} (x)$ corresponding to the eigenvalue 
$\tilde \lambda_k (\beta)$ has $k-1$ simple zeros in the interval $(0, 1)$.
\end{theorem}

\begin{remark} \label{rmk2.1} \rm 
By making the change of variables $x' = 1 - x$ and applying the conclusion of 
the Theorem \ref{thm2.2} we have: the eigenvalues of problem 
\eqref{e1.1}-\eqref{e1.3}, 
\eqref{e2.2}, are real, simple and form an infinitely increasing sequence 
$\{ \lambda_k (\delta)\}_{k = 1}^\infty$ such that  $\lambda_1 (\delta) > 0$ 
for $\delta \in [0, \pi/2)$ and 
$\lambda_1 ( \pi/2) = 0$; moreover, the eigenfunction 
$u_k^{(\delta)} (x)$ corresponding to the eigenvalue $\lambda_k (\delta)$ 
has $k-1$ simple zeros in the interval $(0, 1)$.
\end{remark}

\section{Properties of solution to \eqref{e1.1}-\eqref{e1.3}}

\begin{theorem} \label{thm3.1}
For each fixed $\lambda \in \mathbb{C}$ there exists a nontrivial solutions 
$y (x,\lambda)$  of problem  \eqref{e1.1}-\eqref{e1.3}, which is unique
up to a  constant coefficient. Moreover, the function $y (x,\lambda)$ 
for each fixed $x \in [0, 1]$  is an entire function of $\lambda$.
\end{theorem}

The proof of this theorem is similar to that of \cite[Lemma 2]{k3} 
with the use of Lemma \ref{lem2.1}.

\begin{remark} \label{rmk3.1}\rm 
Since any solution $y (x,\lambda)$ of problem \eqref{e1.1}-\eqref{e1.3}
has a representation $y (x,\lambda) = u(x,\lambda) + i v(x,\lambda)$, 
where the functions $u(x,\lambda)$ and  $v(x,\lambda)$ are real valued, 
and the coefficients $q(x)$ and $a$ are real, it follows that the functions 
$u(x,\lambda)$ and $v(x,\lambda)$ are solutions of problem 
\eqref{e1.1}-\eqref{e1.3} for $\lambda \in \mathbb{R}$. If $u(x,\lambda)$ 
is a nontrivial solution of \eqref{e1.1}-\eqref{e1.3},
then $u(0,\lambda) \ne 0$ for $\lambda > 0$ and $u(1,\lambda) \ne 0$ for
 $\lambda \le 0$. Indeed, if $u(0,\lambda) = 0$ for $\lambda > 0$, 
then it follows from \eqref{e1.2}-\eqref{e1.3} that
$u(0, \lambda) = u'' (0,\lambda) = Tu (0,\lambda) = 0$ for $\lambda > 0$. 
Hence $u'(0, \lambda) \ne 0$ for $\lambda > 0$ and it follows by continuity 
that $u'(x,\lambda) \ne 0$ in an open interval $(0,a)$ for some $a \in (0,1)$. 
We can assume without loss of generality that $u'(x, \lambda) > 0$ for 
$x \in(0, a)$. Then $u(x,\lambda) > 0$ for $x \in (0,a)$. Since $\lambda > 0$ 
it follows by \eqref{e1.1} that $\left( {Tu (x,\lambda )} \right)' > 0$  
for $x \in (0,a)$, so that $Tu (x, \lambda) > 0$ in $(0,a)$.
 In view of the equality $Tu (x,\lambda) = u'''(x,\lambda)- q(x)\, u'(x,\lambda)$ 
we obtain that $u'''(x,\lambda) > q(x)\,u'(x,\lambda) > 0$ for $x \in (0,a)$.
 Hence $u''(x,\lambda) > 0$ in $(0,a)$. Then the first statement of Lemma \ref{lem2.1} 
implies that $u''(1,\lambda) > 0$. But the boundary condition \eqref{e1.2}
 implies that $u''(1,\lambda)= 0$, a contradiction. If $u (1,\lambda) = 0$ for 
$\lambda \le 0$, then the function $u (x, \lambda)$ solves the problem 
\eqref{e1.1}-\eqref{e1.3}, \eqref{e2.2} for $\delta = 0$ which contradicts
the condition $\lambda \le 0$ in view of Remark \ref{rmk2.1}.

Now let $y(x,\lambda)$ be a solution of problem \eqref{e1.1}-\eqref{e1.3},
normalized for example by the condition
\begin{equation}\label{e3.1}
y(0,\lambda) = 1,
\end{equation}
if $\lambda > 0$, and by
\begin{equation}\label{e3.2}
y (1,\lambda)  = 1,
\end{equation}
if $\lambda \le 0$. If $\lambda > 0$ ($\lambda \le 0$), then it follows 
from representation $w(x,\lambda) = u(x,\lambda) + i v(x,\lambda)$ 
 that $u (0,\lambda) = 1$ and $v(0,\lambda) = 0$ ($u (1,\lambda) = 1$ and 
$v(1,\lambda) = 0$). Hence the above reasoning we see that $v (x,\lambda) \equiv 0$ 
for $\lambda \in \mathbb{R}$, i.e. $y(x,\lambda) = u(x,\lambda)$ for
 $\lambda \in \mathbb{R}$. Therefore, the solution $y(x,\lambda)$ of 
\eqref{e1.1}-\eqref{e1.3}, \eqref{e3.1} for $\lambda > 0$ and of
\eqref{e1.1}-\eqref{e1.3}, \eqref{e3.2} for $\lambda \le 0$ is a real
valued for $\lambda \in \mathbb{R}$.
\end{remark}

In the sequel we assume that the function 
$y(x,\lambda), x \in [0,1], \lambda \in \mathbb{C}$, is a solution of 
problem \eqref{e1.1}-\eqref{e1.3}, \eqref{e3.1} for $\lambda > 0$ and
of problem \eqref{e1.1}-\eqref{e1.3}, \eqref{e3.2} for $\lambda \le 0$.
Consider the equation
\[
y(x,\lambda) = 0,
\]
for $x \in [0, 1]$ and $\lambda \in \mathbb{R}$. 
The zeros of this equation are functions of $\lambda$.

\begin{lemma}\label{lem3.1}
Let $\lambda \in \mathbb{R}$. Then every zero $x(\lambda) \in (0, 1]$ of 
the function $y (x,\lambda)$ is simple and is a $C^1$ function of $\lambda$.
\end{lemma}

\begin{proof} 
Let $x_0 \in  (0, 1]$ and $\lambda_0 > 0$ such that 
$y (x_0,\lambda_0) = y' (x_0,\lambda_0) = 0$. If $x_0 \in (0,1)$ and
 $y ''(x_0,\lambda_0)$ $Ty (x_0,\lambda_0) \ge 0$, then the first statement 
of Lemma \ref{lem2.1} implies that $y'' (1, \lambda_0) > 0$. 
This is in contradiction 
with the condition $y'' (1, \lambda_0) = 0$. If $x_0 \in (0,1)$ and 
$y''(x_0,\lambda_0) T y(x_0,\lambda_0) < 0$, then the second part of the 
same lemma yield a contradiction with the boundary condition 
$y'' (0, \lambda_0) = 0$. If $x_0 = 1$, then by  \eqref{e1.2} we have
 $y (1,\lambda_0) = y' (1,\lambda_0) = y''(1,\lambda_0) = 0$. 
Let $b \in (0,1)$ be the fixed point such that $y''(x,\lambda_0) \ne 0$ 
for $x \in (b,1)$. We can assume without loss of generality that 
$y''(x, \lambda_0) > 0$ for $x \in (b, 1)$. Then $y'(x, \lambda_0) < 0$, 
$y(x, \lambda_0) > 0$ for $x \in (b, 1)$ and 
$Ty(1,\lambda_0) = y'''(1,\lambda_0) < 0$. Since $\lambda_0 > 0$ it follows
 by \eqref{e1.1} that $\left( {Ty(x,\lambda _0 )} \right)'   > 0$ for
$x \in (b, 1)$, so that $Ty(x,\lambda_0) < 0$ for $x \in (b, 1)$. 
Hence the second statement of Lemma \ref{lem2.1} implies that $y''(0,\lambda_0) > 0$ 
which contradicts the condition $y''(0,\lambda_0) = 0$.

Now let $x_0 \in  (0, 1]$ and $\lambda_0 \le 0$ such that 
$y (x_0,\lambda_0) = y' (x_0,\lambda_0) = 0$. Then $\lambda_0$ is a 
nonpositive eigenvalue of the problem defined on $[0, x_0]$ and determined by 
equation \eqref{e1.1} with the boundary conditions $y'' (0) = 0$, \eqref{e1.3}
 and $y (x_0) = y' (x_0) = 0$. By Remark \ref{rmk2.1} all the eigenvalues
of this problem are positive, contradiction.

The smoothness of the function $x(\lambda)$ follows from the implicit function 
theorem. The proof of this lemma is complete.
\end{proof}

From the continuity of the zeros of $y(x,\lambda)$ as functions of $\lambda$, 
together with Remark \ref{rmk3.1} (see \eqref{e3.1}, \eqref{e3.2}), it follows an 
important corollary.

\begin{corollary}  \label{coro3.1}
As $\lambda > 0$ $(\lambda \le 0)$ varies the function $y (x,\lambda)$ can 
lose or gain zeros only by these zeros leaving or entering the interval $[0, 1]$ 
through its endpoint $x = 1$ $(x = 0)$.
\end{corollary}

We consider the function
\[
F (x,\lambda) = \frac{{y (x,\lambda )}}{{Ty (x,\lambda )}}.
\]
It follows by Lemmas \ref{lem2.2}, \ref{lem3.1} and Remark \ref{rmk2.1} that the function $H(x,\lambda)$ 
is a finite order meromorphic function of $\lambda$ for all finite $\lambda$ 
and fixed $x \in (0,1]$.

Let $M_k  = \left( {\lambda _{k - 1} (0),\lambda_k (0)} \right)$,
$k \in \mathbb{N}$, where $\lambda_0 (0) =  - \infty$.

Obviously, the eigenvalues $\lambda_k (0)$ and 
$\lambda_k(\pi/2)$
$k \in \mathbb{N}$, of problem \eqref{e1.1}-\eqref{e1.3}, \eqref{e2.2} 
for $\delta = 0$ and 
$\delta = \pi/2$ are zeros of the entire functions 
$y (1,\lambda)$ and $Ty (1,\lambda)$, respectively. We note that the function
\[
G (\lambda ) = \frac {1}{F (1,\lambda)} = \frac {Ty (1,\lambda)}{y(1, \lambda)}
\]
is defined for
\[
\lambda  \in M \equiv \big( {\cup_{k = 1}^\infty  {M_k } } \big)\cup
 ( \mathbb{C}\backslash \mathbb{R}),
\]
and is a meromorphic function of finite order, 
$\lambda_k (\pi/2)$ and $\lambda_k(0), \, k \in \mathbb{N}$,
 are the zeros and poles of this function, respectively.

\begin{lemma} \label{lem3.2}
For each $\lambda \in M$ the following relation holds
\begin{equation}\label{e3.3}
\frac{{dG (\lambda )}}{{d\lambda }} 
= \frac{1}{{y^2 (1,\lambda )}}
\Big\{ {\int_0^1  y^2 (x,\lambda )dx + a y^2 (0,\lambda )} \Big\}.
\end{equation}
\end{lemma}

\begin{proof} By \eqref{e1.1} we have
\[
\left( {Ty(x,\mu )} \right)'  y(x,\lambda ) 
- \left( {Ty(x,\lambda )} \right)'  y(x,\mu )
 = \left( {\mu  - \lambda } \right)y(x,\mu ) y(x,\lambda ).
\]
Integrating this relation from $0\,$ to $1$, using the formula for the 
integration by parts and taking into account boundary conditions \eqref{e1.2} 
and \eqref{e1.3} we obtain
\begin{equation}\label{e3.4}
\begin{aligned}
& y(1,\lambda )\,Ty(1,\mu ) - y(1,\mu )\,Ty(1,\lambda )\\
& =  \left( {\mu  - \lambda } \right) 
\Big\{ {\int_0^1 {y(x,\mu ) y(x,\lambda )\,dx} 
+ a y(0,\mu ) y(0,\lambda )} \Big\}
\end{aligned}
\end{equation}

In view of \eqref{e3.4} for $\lambda, \mu \in M_k,\, k \in \mathbb{N}$, we have
\begin{equation}\label{e3.5}
\frac {Ty(1,\mu )} {y(1,\mu )} - \frac {Ty(1,\lambda )}{y(1,\lambda )}  
= (\mu  - \lambda )\,\frac{{\int_0^1 {y(x,\mu ) y(x,\lambda )dx}  
+ ay(0,\mu )y(0,\lambda )}}{{y(1,\mu )y(1,\lambda )}}
\end{equation}
Dividing both sides of relation \eqref{e3.5} by $\mu - \lambda$
 $(\mu \ne \lambda)$ and passing to the limit as $\mu \to \lambda$ 
we obtain \eqref{e3.3}. The proof of this lemma is complete.
\end{proof}

It follows from \eqref{e3.3} that
\begin{equation}\label{e3.6}
\frac{{\partial F(1,\lambda )}}{{\partial \lambda }}
 = - \frac{{\int_0^1 {y^2 (x,\lambda )dx}  + a y^2 (0,\lambda )}}
{( {Ty(1,\lambda )})^2 } < 0.
\end{equation}

\begin{lemma}  \label{lem3.3}
It holds
\begin{equation}\label{e3.7}
\lim_{\lambda  \to  - \infty } G (\lambda ) =  - \infty .
\end{equation}
\end{lemma}

The proof of this Lemma is similar to that of \cite[Lemma 3.4]{a4}; we omit it.

\begin{remark} \label{rmk3.2} \rm 
It follows from the relation 
$Ty( 1,\lambda_1 (\pi/2)) = 0$ that $G (0) = 0$. 
\end{remark}

\begin{remark} \label{rmk3.3}\rm 
By Remarks \ref{rmk2.1} and \ref{rmk3.2}, and Lemmas \ref{lem3.2} and
 \ref{lem3.3}, we have
\begin{equation}\label{e3.8}
0 = \lambda _1 \left( {\frac{\pi }{2}} \right) 
< \lambda_1 (0) < \lambda_2 \left( {\frac{\pi }{2}} \right) 
< \lambda_2 (0) < \dots .
\end{equation}
\end{remark}

\begin{lemma} \label{lem3.4}
Let $x_0 \in (0, 1]$ and $\lambda_0 > 0$ such that $y (x_0, \lambda_0) = 0$. Then
\[
\frac {\partial F (x_0, \lambda_0)}{\partial x} < 0.
\]
\end{lemma}

\begin{proof} 
Let $x_0 \in (0, 1]$ and $\lambda_0 > 0$ such that $y (x_0, \lambda_0 ) = 0$. 
If $x_0 \in (0, 1)$, then it follows from Lemma \ref{lem2.2} that 
$Ty (x_0,\lambda_0) \ne 0$. The same relation $Ty (x_0,\lambda_0) \ne 0$ 
follows from \eqref{e3.8} in the case $x_0 = 1$.  Hence we obtain
\[
\frac{{\partial F(x_0,\lambda_)}}{{\partial x}} 
= \frac{{y'(x_0,\lambda_0) Ty(x_0,\lambda_0) 
- y(x_0,\lambda_0) (Ty)' (x_0,\lambda_0)  }}
{{\left( {Ty\,(x_0,\lambda_0)} \right)^2 }} 
= \frac{{y'(x_0,\lambda_0)}}{{Ty\,(x_0,\lambda_0)}} < 0.
\]
The proof of Lemma \ref{lem3.4} is complete.
\end{proof}

\begin{lemma} \label{lem3.5}
Let $\,0 < \mu < \nu\,$ and $y (x, \mu)$ has $m$ zeros in the interval $(0, 1)$. 
Then $y (x, \nu)$ has at least $m$ zeros in $(0,1)$.
\end{lemma}

The proof of the above lemma is similar to that of \cite[Lemma 3.6]{a4}, 
using formula \eqref{e3.6} and Lemma \ref{lem3.4}.


Let $m (\lambda)$ be the number of zeros of the function $y (x, \lambda)$ 
in the interval $(0, 1)$. Then it follows from Remark \ref{rmk2.1} that
\begin{equation}\label{e3.9}
m (\lambda_k (0)) = k - 1,\, k \in \mathbb{N}.
\end{equation}
As an immediate consequence of Lemmas \ref{lem3.1}, \ref{lem3.5}, 
Corollary \ref{coro3.1} and relations \eqref{e3.8}, \eqref{e3.9}, we 
obtain the following result.

\begin{lemma} \label{lem3.6}
If $\lambda  \in (\lambda _{k - 1} (0),\lambda_k (0)] \cap (0, +\infty)$, 
then $m (\lambda) = k - 1$.
\end{lemma}

\section{Oscillatory properties of eigenfunctions of \eqref{e1.1}-\eqref{e1.4}}

Problem \eqref{e1.1}-\eqref{e1.4} can be reduced to the eigenvalue problem
for the linear operator $L$ in the Hilbert space  
$H = L_{2} (0, 1) \oplus \mathbb{C}^2$  with inner product
\begin{equation} \label{e4.1}
(\hat u,\hat v) = (\{ y,m,n\} ,\{ v,s,t\} ) 
= \int_0^1 {y(x)\overline {v(x)} } \,dx 
+ |a |^{ - 1} m \bar s + |c |^{ - 1} n \bar t,
\end{equation}
where
\[
L \hat y = L \{ y,m,n\} = \{\left( {Ty(x)} \right)',\,Ty (0),\,Ty(1)\}
\]
with the domain
\begin{align*}
 D(L) = \Big\{& \{ y (x),\,m,\,n\}: y \in W_2^4 (0,1), \left( {Ty (x)} \right)' 
\in L_{2} (0,1), \\ 
& y''(0) = y''(1) = 0,\; m = a y (0),\; n = c y(1) \Big\}
 \end{align*}
dense everywhere in $H$ \cite{s1}. It is obvious that the operator $L$ 
is well defined in $H$ and the eigenvalue problem \eqref{e1.1}-\eqref{e1.4}
becomes
\begin{equation} \label{e4.2}
L\hat y = \lambda \hat y,\quad \hat y \in D(L),
\end{equation}
i.e., the eigenvalues $\lambda_k, k \in \mathbb{N}$, of problem
\eqref{e1.1}-\eqref{e1.4} and those of the operator  $L$ coincide; moreover,
between the eigenfunctions, there is a one-to-one correspondence
\[
y_k (x) \leftrightarrow \{ y_k (x),m_k,\,n_k \} ,m_k
 = a y_k (0),\; n_k  = c y_k(1).
\]
Problem \eqref{e1.1}-\eqref{e1.4} is strongly regular in the sense of 
\cite{s1};
in particular, this problem has discrete spectrum.

\begin{theorem}  \label{thm4.1}
$L$ is a self-adjoint operator in $H$. The system of eigenvectors 
$\{ y_k (x),m_k ,n_k \}$ of the operator $L$ forms
a unconditional basis (Riesz basis after normalization) in the space $H$.
\end{theorem}

The proof of this theorem is similar to that of \cite[Theorem 5.1]{a4};
we omit it.


\begin{remark} \label{rmk4.1} \rm 
It follows  from Theorem \ref{thm4.1} that the eigenvalues of problem
 \eqref{e1.1}-\eqref{e1.4} are real. Moreover, by \eqref{e4.2} we have
\[
\left( {L \hat y (\lambda),\hat y (\lambda)} \right) 
= \lambda (\hat y(\lambda),\hat y(\lambda)),
\]
where $\hat y (\lambda) = \{y(x,\lambda), ay(0,\lambda),c y(1,\lambda)\}$,
 which implies by \eqref{e4.1} that
\begin{equation} \label{e4.3}
\begin{aligned}
&\int_0^1 \{ {y''^2 (x,\lambda ) + q(x)y'^2 (x,\lambda )} \} \,dx \\
&= \lambda \Big\{ {\int_0^1 {y^2 (x,\lambda )dx}  + ay^2 (0,\lambda ) 
- cy^2 (1,\lambda )} \Big\}.
\end{aligned}
\end{equation}
Hence all eigenvalues of problem \eqref{e1.1}-\eqref{e1.4} are nonnegative.
\end{remark}

We note that the eigenvalues (with regard to multiplicities) of the
 problem \eqref{e1.1}-\eqref {e1.4} are the roots of the equation
\begin{equation} \label{e4.4}
Ty (1,\lambda ) - c \lambda y (1,\lambda ) = 0.
\end{equation}

\begin{remark} \label{rmk4.2} \rm 
Let $\lambda$ be an eigenvalue of problem \eqref{e1.1}-\eqref{e1.4}. 
Hence by Remark \ref{rmk4.1} we have $\lambda \ge 0$. If $\lambda = 0$, then it 
follows from \eqref{e4.3} that $y(x,\lambda) \equiv {\rm const} \ne 0$, 
which implies by \eqref{e3.2} that $y(0,0) = 1$. If $\lambda > 0$, then 
$y (1,\lambda) \ne 0$ by virtue of \eqref{e3.8}. 
\end{remark}

In turn, by Remark \ref{rmk4.2} each root (with regard of multiplicities) of equation 
\eqref{e4.4}  is also a root of the equation
\begin{equation} \label{e4.5}
G (\lambda) = c \lambda\,.
\end{equation}

\begin{lemma} \label{lem4.1}
The eigenvalues of boundary-value problem \eqref{e1.1}-\eqref{e1.4} 
are simple and form a countable set without finite limit point.
\end{lemma}

\begin{proof} 
The entire function occurring on the left-hand side of \eqref{e4.4} 
does not vanish for nonreal $\lambda$ in view of Remark \ref{rmk4.1}. 
Hence it is distinct from the identically zero function and its zeros 
form an at most countable set without finite limit point.

Now we claim that \eqref{e4.5} has only simple roots. 
In fact, if $\lambda$ is a multiple root of equation \eqref{e4.5}, then
$G (\lambda) = c\lambda$ and $G' (\lambda) = c$.
Hence by Remarks \ref{rmk4.1} and \ref{rmk4.2} it follows from \eqref{e3.3} that
\[
\int_0^1 {y^2 (x,\lambda )dx}  + ay^2 (0,\lambda ) - cy^2 (1,\lambda ) = 0,
\]
which is impossible in view of conditions $a > 0$ and $c < 0$. 
The proof  is complete.
\end{proof}

\begin{theorem} \label{thm4.2}
Boundary-value problem \eqref{e1.1}-\eqref{e1.4} has a sequence 
of eigenvalues
\[
0 = \lambda_1 < \lambda_2 < \dots < \lambda_k \to +\infty.
\]
The corresponding eigenfunctions $y_k (x)$, $k \in \mathbb{N}$, have $k - 1$ 
simple zeros in $(0, 1)$.
\end{theorem}

\begin{proof}
 By  relations \eqref{e3.3}, \eqref{e3.7} and \eqref{e3.8},
 we have
\[
\lim_{\lambda  \to \lambda _k (0) - 0} G(\lambda ) 
=  + \infty ,\quad 
\lim_{\lambda  \to \lambda _{k - 1} (0) + 0} G(\lambda ) 
=  - \infty,\quad k \in \mathbb{N}.
\]
Hence the function $G (\lambda)$ takes each value in $(-\,\infty,+\,\infty)$ 
at a unique point in the interval $(\lambda_{k-1} (0), \lambda_k (0))$, 
$k \in \mathbb{N}$. For the function $H (\lambda) = c \,\lambda$ we have 
$H' (\lambda) = c$. Since $c < 0$ it follows that this function is strictly 
decreasing in the interval $(-\,\infty,+\,\infty)$.

It follows from the preceding considerations that in the interval 
$(\lambda_{k-1} (0), \lambda_k (0)),\,k \in \mathbb{N}$, 
there exists a unique $\lambda = \lambda_k$ such that
\[
G (\lambda) = H(\lambda)
\]
i.e., condition \eqref{e1.4} is satisfied. Therefore, $\lambda_k$ is an 
eigenvalue of boundary-value problem \eqref{e1.1}-\eqref{e1.4} and 
$y_k (x) = y (x, \lambda_k)$ is the corresponding eigenfunction.

By Remark \ref{rmk3.2} it follows from the preceding considerations that 
$\lambda_1 = \lambda _1 \left( {\frac{\pi }{2}} \right) = 0$ and 
$\lambda_k > \lambda _1 \left( {\frac{\pi }{2}} \right)$ for $k \ge 2$. 
Consequently, by Remark \ref{rmk2.1}, we have $\lambda_k > 0$ for $k > 2$. 
Hence it follows by Lemma \ref{lem3.6} and Remark \ref{rmk4.2} that $m (\lambda_k) = k - 1$. 
The proof is complete.
\end{proof}

It follows from \cite[(3.1)--(3.4)]{k4} that
\begin{gather} \label{e4.6}
\sqrt[4]{{\lambda _k (0,0)}} = k\pi  + O\big( {\frac{1}{k}} \big), \\
 \label{e4.7}
u_k^{(0,0)} (x) = \sin k\pi x + O\big( {\frac{1}{k}} \big), \\
 \label{e4.8}
\sqrt[4]{{\lambda _k (0)}} = (k - 1)\pi  + O\big( {\frac{1}{k}} \big), \\
 \label{e4.9}
u_k^{(0)} (x) = \sin (k - 1)\pi x + O\big( {\frac{1}{k}} \big),
\end{gather}
where relations \eqref{e4.7} and \eqref{e4.9} hold uniformly for $x \in [0, 1]$.

\begin{theorem} \label{thm4.3}
The following asymptotic formulas hold:
\begin{gather} \label{e4.10}
\sqrt[4]{{\lambda_k }} = (k-2) \pi  + O(1/k), \\
\label{e4.11}
 y_k (x) = \sin (k - 2)\pi x  + O(1/k)
\end{gather}
where relation \eqref{e4.11} holds uniformly for $x \in [0,1]$.
\end{theorem}

 The proof of the above theorem is similar to that of 
\cite[Theorem 3.1]{k4}  using Theorem \ref{thm4.2}. We omit the proof here.

\section{Basis property  in $L_p (0,1)$ , $1 < p < \infty$, of the 
eigenfunctions of \eqref{e1.1}-\eqref{e1.4}}

Let 
\begin{equation} \label{e5.1}
\delta_k  = \left( {\hat y_k ,\hat y_k } \right).
\end{equation}
Then by  conditions $a > 0,\,c < 0$ and \eqref{e4.1} it follows 
from \eqref{e5.1} that
\begin{equation} \label{e5.2}
\delta_k  = \|y_k \|_{L_{2} }^2  + a^{ - 1} m_k^2  - c^{- 1} n_k^2  > 0.
\end{equation}
Hence, the system of eigenvectors $\{\hat v_k \}_{k = 1}^\infty$,
$\hat v_k = \delta_k^{-1/2} \hat y_k$, of operator $L$ forms an orthonormal
 basis (i.e. Riesz basis) in $H$.

Let $r$ and $l$ ($r \ne l$) be arbitrary fixed natural numbers and
\begin{gather} \label{e5.3}
\tilde \Delta_{r,\,l} 
 =  \begin{vmatrix} a \delta_{r}^{- 1/2} y_{r} (0) 
& a \delta_1^{ -1/2} y_1 (0) \\
c \delta_{r}^{ -1/2} y_{r} (1)
& c \delta_1^{- 1/2}  y_1 (1) 
 \end{vmatrix}
= a c \delta_{r}^{- 1} \delta_1^{ - 1} 
 \begin{vmatrix} y_{r} (0) & y_1 (0) \\
 y_{r} (1) & y_1 (1) 
\end{vmatrix}, \\
\label{e5.4}
\Delta _{r,\,l}  
=  \begin{vmatrix} y_{r} (0) & y_1 (0)\\ 
y_{r} (1) & y_1 (1)
 \end{vmatrix}.
\end{gather}
By \eqref{e5.2} it follows from \eqref{e5.3} and \eqref{e5.4}  that
\begin{equation} \label{e5.5}
\tilde \Delta _{r,\,l}  \ne 0 \;\Leftrightarrow\; \Delta _{r,\,l}  \ne 0.
\end{equation}

\begin{theorem} \label{thm5.1}
Let $r$ and $l$ ($r \ne l$) be arbitrary fixed natural numbers. 
If $\Delta _{r,\,l}  \ne 0$, then the system of eigenfunctions 
$\{ y_k (x)\} _{k = 1,\,k \ne r,\,l}^\infty$ of problem \eqref{e1.1}-\eqref{e1.4} 
forms a basis in the space $L_{p} (0, 1), 1 < p < \infty$, which is an 
unconditional basis for $p = 2$; if $\Delta _{r,\,l} = 0$, then this system
 is incomplete and nonminimal in the space $L_{2} (0, 1)$.
\end{theorem}

The proof of the above theorem in the case $p = 2$ is similar to that of 
\cite[Theorem 4.1]{a5} using Theorem \ref{thm4.1} and relation \eqref{e5.5}.
In the case $p \in (1, + \infty )\backslash \{ 2\}$ is similar to that of
\cite[Theorem 5.1]{k4} using  asymptotic formulas \eqref{e4.6}-\eqref{e4.11}.

Using Theorem \ref{thm5.1}, we can obtain sufficient conditions for the subsystem of 
eigenfunctions $\{ y_k (x)\} _{k = 1,\,k \ne r,l}^\infty$ of 
problem \eqref{e1.1}-\eqref{e1.4} to form a basis in 
$L_{p} (0, 1), 1 < p < \infty$.

\begin{corollary} \label{coro5.1}
Let $r$ and $l$ ($r \ne l$) be arbitrary fixed natural numbers having 
different parity. Then the system of eigenfunctions 
$\{ y_k (x)\} _{k = 1,k \ne r,l}^\infty$ of problem \eqref{e1.1}-\eqref{e1.4} 
forms a basis in the space $L_{p} (0, 1), 1 < p < \infty$, which is an 
unconditional basis for $p=2$.
\end{corollary}

\begin{proof} 
By \eqref{e3.1}  from \eqref{e5.4} it follows that
\begin{equation} \label{e5.6}
\Delta _{r,\,l}  = y _l (1) - y _r (1).
\end{equation}
In view of \eqref{e3.1} and Theorem \ref{thm4.2} we have
\begin{equation} \label{e5.7}
\operatorname{sgn} y_k (1) = (-1)^{k + 1},\quad k \in \mathbb{N}.
\end{equation}
Taking this equality into account, from \eqref{e5.6} we obtain
\begin{equation} \label{e5.8}
\Delta_{r,\,l} = (-1)^{l+1} \{ {( - 1)^{r + l} | {y_l (1)} | 
- | {y_r (1)} |} \}.
\end{equation}
Now the statement of this corollary follows from Theorem \ref{thm5.1} in view of 
 relation \eqref{e5.8}. The proof is complete.
\end{proof}

\subsection*{Acknowledgements} 
The authors is deeply indebted to the reviewer, who made remarks and wishes 
which contributed to significant improvements in the text and in the
 transparency of the  obtained results.

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\end{document}