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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2015 (2015), No. 54, pp. 1--20.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2015 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2015/54\hfil Riesz basis and exponential stability]
{Riesz basis and exponential stability for Euler-bernoulli beams with
variable coefficients and indefinite damping under a force control in
position and velocity}

\author[K. A. Tour\'e, A. Coulibaly, A. A. H. Kouassi \hfil EJDE-2015/54\hfilneg]
{K. Augustin Tour\'e, Adama Coulibaly, Ayo A. Hermith Kouassi}

\address{K. Augustin Tour\'e \newline
Institut National Polytechnique Houphou\"et-Boigny de Yamoussoukro, BP
2444 Yamoussoukro, C\^ote d'Ivoire}
\email{latourci@yahoo.fr}

\address{Adama Coulibaly \newline
Universit\'e Felix Houphou\"et-Boigny and UFR Math\'ematiques Appliqu\'ees
et Informatique, C\^ote d'Ivoire}
\email{couliba@yahoo.fr}

\address{Ayo A. Hermith Kouassi \newline
Universit\'e Felix Houphou\"{e}t-Boigny \and UFR Math\'ematiques
Appliqu\'ees et Informatique, C\^ote d'Ivoire}
\email{hermithkouassi@gmail.com}

\thanks{Submitted January 23, 2015. Published February 26, 2015.}
\subjclass[2000]{93C20, 93D15, 35B35, 35P10}
\keywords{Beam equation; asymptotic analysis; Riesz basis; exponential stability}

\begin{abstract}
 This article concerns the Riesz basis property and the stability of
 a damped Euler-Bernoulli beam with nonuniform thickness or density,
 that is clamped at one end and is free at the other.
 To stabilize the system, we apply a linear boundary control
 force in position and velocity at the free end of the beam.
 We first put some basic properties for the closed-loop system and then
 analyze the spectrum of the system. Using the modern spectral analysis
 approach for two-points parameterized ordinary differential operators,
 we obtain the Riesz basis property. The spectrum-determined growth condition
 and the exponential stability are also concluded.
\end{abstract}

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


\section{Introduction}

We study the Riesz basis property and the stability of a flexible beam with
nonuniform thickness or density, that is clamped at one end and is submitted
to a linear boundary control force in position and velocity at the free end.
The equations of motion of the system are given by
\begin{gather}
m(x) u_{tt}(x,t) +(EI(x)u_{xx}(x,t) ) _{xx}+\gamma (x) u_{t}(
x) =0, \quad 0<x<1,\; t>0,  \label{1a} \\
u(0,t) =u_{x}(0,t) =u_{xx}(1,t) =0,
\quad t>0,  \label{1b} \\
(EI(\cdot) u_{xx}(.,t) ) _{x}(1)
=\alpha u(1,t) +\beta u_{t}(1,t) , \quad t>0,  \label{1c}\\
u(x,0) =u_0(x) ,\quad u_{t}(x,0)=u_1(x) , \quad 0<x<1,  \label{1d}
\end{gather}
where $\alpha ,\beta $ are two given positive constants, $u(x,t)$
stands for a transversal deviation of the beam at position $x$ and time $t$;
a subscript letter denotes the partial derivation with respect that
variable. The length of the beam is chosen to be unity, $EI(x) $
is the stiffness of the beam, $m(x) $ is the mass density and $
\gamma (x) $ is a continuous coefficient function of
feedback damping.  If $\gamma \geq 0$, it can be proven that the
energy of the system decays exponential (see theorem \ref{theo8}).

A question was raised in Wang and al. \cite{20}: Due to the nonuniform
physical thickness and / or density of the Euler-Bernoulli beam with the
variable coefficient damping $\gamma (x) $ in equation \eqref{1a}
, what kind of conditions on $\gamma $, can ensure that the system
remains exponentially stable?  In \cite{20}, the question is treated
without boundary conditions. Our work is a continuation study of
\cite{20} and follows the same arguments.
 In this paper, we shall always assume that:
\begin{equation}
m(x) , EI(x) \in\mathcal{C}^{4}(0,1) \quad\text{and}\quad m(x), EI(x) >0. \label{1}
\end{equation}

With the assumption \eqref{1}, we shall prove that system
\eqref{1a}--\eqref{1d} is a Riesz spectral system in the sense that the
generalized eigenfunctions of the system form a Riesz basis on the suitable
Hilbert space (see \cite{4}). The Riesz basis property, meaning that the
generalized eigenvectors of the system form an unconditional basis for the
state Hilbert space, is one of the fundamental properties of a linear
vibrating system. For this kind of system, the stability is usually
determined by the spectrum of the associated operator and one can also use
the theory of nonharmonic Fourier series to obtain important properties such
as the optimal decay rate of the energy.

There are two steps usually found in the study of linear systems with
variable coefficients. The first is to transform the "dominant term" of the
system under study into a uniform "dominant equation" by space scaling and
state transformation where no variable coefficient is involved any longer.
The second step is to approximate the eigenfunctions of the system by those
of uniform "dominant equation". This fundamental idea comes essentially from
Birkhoff's works \cite{1} and \cite{2} and  Naimark \cite{12} to
estimate the eigenvalues. This approach has been used in dealing with the
beam equations of variable coefficients (see Guo \cite{7,8},
 Wang \cite{19} or \cite{20} and the references therein).

Moreover, one of the methods on the verification of Riesz basis property
well developed recently and applied successfully, is the classical Bari's
theorem \cite{3}. When $\alpha =0 $, the undamped case ($\gamma =0$) has
been studied in \cite{7}, where the author used a corollary of Bari's
theorem on the Riesz basis property in \cite{3}. Our work shall make use a
result due to  Wang \cite{20}, which deals with the eigenvalue
problem of beams in the form of an ordinary differential equation $L(
f) =\lambda f$ with $\lambda $-polynomial boundary conditions
(Shkalikov \cite{14},  Tretter \cite{16},  Wang \cite{17}, \cite{8}
and the references therein). We establish conditions on the both positive
feedback parameters $\alpha $ and $\beta $ in order to get the Riesz basis
property and the exponential stability for system \eqref{1a}--\eqref{1d}.

The content of this article is as follows:
 in the next section, we convert system \eqref{1a}--\eqref{1d} into an abstract
Cauchy problem in Hilbert
state space, and discuss some basic properties of system. We show that
system \eqref{1a}--\eqref{1d} can be associated to a $\mathcal{C}_0$
-semigroup, and the generator $A_{\gamma }$ of $C_0$-semigroup
has compact resolvents. Furthermore, we obtain an asymptotic expression for
eigenvalues. In section 3, we discuss the Riesz basis property of the
eigenfunctions as well as the exponential stability of the system. Through a
bounded invertible transform $\mathcal{L}$, we establish the relationship
between $A_{\gamma }$ and $\mathbb{A}$ defined in
\eqref{01} and obtain the Riesz basis property from the strong regularity of
boundary conditions that has been verified in section 2. Incidentally, we
also obtain conditions for the exponential stability of the system for
indefinite damping.

\section{Basic properties of the problem \eqref{1a}--\eqref{1d}}

Let us introduce the following spaces:
\begin{gather}
H_{E}^2(0,1) =\{ u(x)\in H^2(
0,1) |u(0) =u_{x}(0) =0\} ,  \label{10} \\
H=H_{E}^2(0,1) \times L^2(0,1),  \label{11}
\end{gather}

The superscript $T$ stands for the transpose and the spaces
$L^2(0,1) $ and $H^{k}(0,1) $ are defined as
\begin{gather}
L^2(0,1) =\big\{ u:[ 0,1] \to \mathbb{C}
:\int_0^{1}| u| ^2dx<\infty \big\},  \label{12} \\
H^{k}(0,1) =\{ u:[ 0,1] \to \mathbb{C}
: u,u^{(1)},\ldots ,u^{(k)}\in L^2(0,1) \} .
\label{13}
\end{gather}

In the space $H$, we define the inner-product
\begin{equation}
\langle u,v\rangle _{H}=\int_0^{1}(
m(x)f_2(x) \overline{g_2(x) }+EI(x)
f_1''(x) \overline{g_1''(x) }) dx
+\alpha f_1(1) \overline{g_1(1) },  \label{14}
\end{equation}
where $u=(f_1,f_2) ^{T}\in H$ and
$v=(g_1,g_2) ^{T}\in H$ and we denote by $\|\cdot\| _{H}$
the associated norm. Next, we define an
unbounded linear operator
$A_{\gamma }:D(A_{\gamma}) \subset H\to H$ as follows:
\begin{equation}
A_{\gamma }(f,g) =\Big(g(x) ,-\frac{1}{m(x) }((EI(x) f''(
x) ) ''+\gamma (x) g(x)
)\Big) ^{T},  \label{15}
\end{equation}
where $D(A_{\gamma }) $, the domain of operator $A_{\gamma }$ is
\begin{equation}
\begin{aligned}
D(A_{\gamma }) =\Big\{&(f,g) ^{T}\in (H^{4}(0,1) \cap H_{E}^2(0,1)
)\times H_{E}^2(0,1) : \\
&f''(1) =0, \; (EI(\cdot)f''(\cdot) ) ''(1)
=\alpha f(1)+\beta g(1)\big\}.
\end{aligned}  \label{16}
\end{equation}
With this notation, the set of equations \eqref{1a}--\eqref{1d} can be
formally written as
\begin{equation}
\begin{gathered}
\frac{dY(t) }{dt}=A_{\gamma }Y(t) \\
Y(0) =Y_0\in H,
\end{gathered}  \label{17}
\end{equation}
where $Y(t) =(u(.,t) ,u_{t}(.,t)) ^{T}$, $Y(0) =(u_0,u_1) ^{T}$.
Here, it is clear that $A_0$ denotes the undamped case
 $\gamma (x) =0$ which was studied in \cite{23} and that
\[
\Gamma _{\gamma }(f,g)=A_{\gamma }-A_0=\Big(0,-\frac{
\gamma (x) g(x)}{m(x) }\Big)
\]
is a boundary linear operator on $H$. Therefore the following
result follows immediately from the theory of operator semigroups (see
Pazy \cite[theorem 1.1]{13}).

\begin{theorem} \label{theo1}
Let operators $A_{\gamma }$ and $A_0$ be defined as
before.Then $A_0$ is a m-dissipative operator and generates a
$C_0$-group on $H$, and hence $A_{\gamma }$ generates a
$C_0$-group $e^{A_{\gamma }t}$ on $H$.
\end{theorem}

\begin{proof}
In \cite{23} we applied the Lumer-Phillips theorem,
 (see, e.g., \cite[p.14]{13}) to prove that operator $A_0$ is m-dissipative.
 Then using
Hille-Yosida-Phillips theorem, we also obtained that operator
 $A_0 $ is infinitesimal generator of a $C_0$-semigroup
$S(t)=e^{A_0t}$on $H$, satisfying
\[
\| S(t) \| \leq Me^{\omega t}.
\]
Moreover we obtain $A_{\gamma }=\Gamma _{\gamma }+A_0$
where $\Gamma _{\gamma }$ is a boundary linear operator on $H$.
Then using the perturbation by bounded linear operator, we deduce that
 $A_{\gamma }=\Gamma _{\gamma }+A_0$ is infinitesimal
generator of a $C_0$-semigroup $T(t) =e^{A_{\gamma}t}$, satisfying
\[
\| T(t) \| \leq Me^{(\omega +M\| \Gamma _{\gamma }\| ) t}
\]
(see A. Pazy \cite[Theorem 1.1]{13}).
\end{proof}

\begin{theorem} \label{theo2}
$A_{\gamma }$ has compact resolvents and $0\in \rho (A_{\gamma }) $.
Therefore, the spectrum $\sigma (A_{\gamma }) $ consists entirely of isolated
eigenvalues.
\end{theorem}

\begin{proof}
Clearly, we only need to prove that $0\in \rho (A_{\gamma}) $ and
$A_{\gamma }^{-1}$ is compact on $H$. For
any $G=(g_1,\text{ }g_2) \in \mathit{H,}$ we need to find a
unique $F=(f_1, f_2) \in D(A_{\gamma}) $ such that
\[
A_{\gamma }F=G.
\]
In other words such that the following equations are satisfied:
\begin{gather}
f_2(x) = g_1(x) ,\quad g_1\in H _{E}^2(0,1)  \label{2a} \\
-\frac{1}{m(x) }((EI(x) f_1''(x) ) ''+\gamma (x)
f_2(x) ) = g_2(x) ,\quad g_2\in L^2(0,1)  \label{2b} \\
f_1(0) = f_1'(0) =f_1''(1) =0  \label{2c} \\
(EI(\cdot) f_1''(\cdot) )'(1) =\alpha f_1(1) +\beta f_2(1) . \label{2d}
\end{gather}
Using  \eqref{2b} we obtain
\[
(EI(x) f_1''(x) )''=-m(x) g_2(x) -\gamma (x) g_1(x)
\]
By integrating we obtain
\begin{gather*}
\int_{x}^{1}(EI(r) f_1''(r)) ''dr=-\int_{x}^{1}m(r) g_2(
r) +\gamma (r) g_1(r) \,dr\,, \\
(EI(\cdot) f_1''(\cdot) )'(1) -(EI(x) f_1''(x) ) '=-\int_{x}^{1}m(r)
g_2(r) +\gamma (r) g_1(r) \,dr\,.
\end{gather*}
Using the boundary condition \eqref{2d} we obtain
\begin{gather*}
\alpha f_1(1) +\beta g_1(1) -(EI(
x) f_1''(x) ) '=-\int_{x}^{1}m(r) g_2(r) +\gamma (r)g_1(r) \,dr\,,
\\
(EI(x) f_1''(x) )
'-\alpha f_1(1) =\int_{x}^{1}m(r)
g_2(r) +\gamma (r) g_1(r) \,dr+\beta g_1(1)\,.
\end{gather*}
By integrating again we obtain
\begin{align*}
&\int_{x}^{1}(EI(\eta ) f_1''(\eta
) ) 'd\eta -\alpha f_1(1)\int_{x}^{1}d\eta \\
&=\int_{x}^{1}\int_{\eta }^{1}m(r)
g_2(r) +\gamma (r) g_1(r) \,dr\,d\eta
+\beta g_1(1) \int_{x}^{1}d\eta\,,
\end{align*}
\begin{align*}
&EI(1) f_1''(1) -EI(x)
f_1''(x) -\alpha (1-x) f_1(1) \\
&=\int_{x}^{1}\int_{\eta }^{1}m(r) g_2(r) +\gamma (r) g_1(r) \,dr\,d\eta
+\beta (1-x) g_1(1)\,.
\end{align*}
Since $f_1''(1) =0$, we obtain
\begin{align*}
&f_1''(x) +\alpha \frac{(1-x) }{EI(x) }f_1(1) \\
&=-\frac{1}{EI(x) }
\int_{x}^{1}\int_{\eta }^{1}m(r) g_2(r) +\gamma
(r) g_1(r) \,dr\,d\eta -\beta \frac{(1-x)
}{EI(x) }g_1(1)\,,
\end{align*}
\begin{align*}
&\int_0^{x}f_1''(\xi ) d\xi +\alpha
f_1(1) \int_0^{x}\frac{(1-\xi ) }{EI(\xi) }d\xi \\
&=-\beta g_1(1) \int_0^{x}\frac{(1-\xi ) }{EI(\xi ) }d\xi 
-\int_0^{x}\frac{1}{EI(\xi ) }\int_{\xi }^{1}\int_{\eta
}^{1}m(r) g_2(r) +\gamma (r)
g_1(r) \,dr\,d\eta d\xi\,,
\end{align*}
\begin{align*}
&f_1'(x) +\alpha f_1(1) \int_0^{x}
\frac{(1-\xi ) }{EI(\xi ) }d\xi \\
&= \int_0^{x}-\frac{1}{EI(\xi ) }\int_{\xi }^{1}\int_{\eta
}^{1}m(r) g_2(r) +\gamma (r)
g_1(r) \,dr\,d\eta d\xi
-\beta g_1(1) \int_0^{x}\frac{(1-\xi ) }{
EI(\xi ) }d\xi\,,
\end{align*}
\begin{align*}
&\int_0^{x}f_1'(s) ds+\alpha f_1(1)
\int_0^{x}\int_0^{s}\frac{(1-\xi ) }{EI(\xi ) }\,d\xi\,ds \\
&=-\int_0^{x}\int_0^{s}\frac{1}{EI(\xi ) }\int_{\xi
}^{1}\int_{\eta }^{1}m(r) g_2(r) +\gamma (
r) g_1(r) \,dr\,d\eta\,d\xi\,ds\\
&\quad -\beta g_1(1) \int_0^{x}\int_0^{s}\frac{(1-\xi
) }{EI(\xi ) }\,d\xi\,ds\,.
\end{align*}
Using the boundary condition \eqref{2c} we have
\begin{align*}
&f_1(x) +\alpha f_1(1) \int_0^{x}\int_0^{s}
\frac{(1-\xi ) }{EI(\xi ) }\,d\xi\,ds\\
&=-\int_0^{x}\int_0^{s}\frac{1}{EI(\xi ) }\int_{\xi
}^{1}\int_{\eta }^{1}m(r) g_2(r) +\gamma (
r) g_1(r) \,dr\,d\eta\,d\xi\,ds\\
&\quad -\beta g_1(1) \int_0^{x}\int_0^{s}\frac{(1-\xi
) }{EI(\xi ) }\,d\xi\,ds\,.
\end{align*}
Next we determine $f(1)$. We obtain
\begin{align*}
&f_1(1) +\alpha f_1(1) \int_0^{1}\int_0^{s}
\frac{(1-\xi ) }{EI(\xi ) }\,d\xi\,ds \\
&=-\int_0^{1}\int_0^{s}\frac{1}{EI(\xi ) }\int_{\xi
}^{1}\int_{\eta }^{1}m(r) g_2(r) +\gamma (
r) g_1(r) \,dr\,d\eta\,d\xi\,ds\\
&\quad -\beta g_1(1) \int_0^{1}\int_0^{s}\frac{(1-\xi
) }{EI(\xi ) }\,d\xi\,ds\,,
\end{align*}
\begin{align*}
&f_1(1) (1+\alpha \int_0^{1}\int_0^{s}\frac{(1-\xi
) }{EI(\xi ) }\,d\xi\,ds) \\
&=-\int_0^{1}\int_0^{s}\frac{1}{EI(\xi ) }\int_{\xi
}^{1}\int_{\eta }^{1}m(r) g_2(r) +\gamma (
r) g_1(r) \,dr\,d\eta\,d\xi\,ds\\
&\quad -\beta g_1(1) \int_0^{1}\int_0^{s}\frac{(1-\xi
) }{EI(\xi ) }\,d\xi\,ds\,,
\end{align*}
\begin{align*}
f_1(1)
&=\Big(-\int_0^{1}\int_0^{s}\frac{1}{EI(\xi
) }\int_{\xi }^{1}\int_{\eta }^{1}m(r) g_2(
r) +\gamma (r) g_1(r) \,dr\,d\eta\,d\xi\,ds\\
&\quad -\beta g_1(1) \int_0^{1}\int_0^{s}\frac{(1-\xi ) }{
EI(\xi ) }\,d\xi\,ds\Big)\Big/
\Big(1+\alpha \int_0^{1}\int_0^{s}\frac{(
1-\xi ) }{EI(\xi ) }\,d\xi\,ds\Big)\,,
\end{align*}
then
\begin{align*}
f_1(x)
&=-K\int_0^{x}\int_0^{s}\frac{(1-\xi )
}{EI(\xi ) }\,d\xi\,ds-\beta g_1(1)
\int_0^{1}\int_0^{s}\frac{(1-\xi ) }{EI(\xi ) }
\,d\xi\,ds \\
&\quad -\int_0^{1}\int_0^{s}\frac{1}{EI(\xi ) }\int_{\xi
}^{1}\int_{\eta }^{1}m(r) g_2(r) +\gamma (
r) g_1(r) \,dr\,d\eta\,d\xi\,ds\,,
\end{align*}
with
\begin{align*}
K&=\alpha \Big(-\int_0^{1}\int_0^{s}\frac{1}{EI(\xi ) }
\int_{\xi }^{1}\int_{\eta }^{1}m(r) g_2(r) +\gamma
(r) g_1(r) \,dr\,d\eta\,d\xi\,ds\\
&\quad -\beta g_1(
1) \int_0^{1}\int_0^{s}\frac{(1-\xi ) }{EI(\xi
) }\,d\xi\,ds\Big)\Big/\Big(1+\alpha \int_0^{1}\int_0^{s}\frac{(1-\xi
) }{EI(\xi ) }\,d\xi\,ds\Big).
\end{align*}
Obviously, $(f_1,\text{ }f_2) \in D(A_{\gamma
})$, therefore
\[
F=(f_1,\text{ }f_2) =A_{\gamma }^{-1}G=(
B(x) ,\text{ }g_1)
\]
where
\begin{align*}
B(x) &=-K\int_0^{x}\int_0^{s}\frac{(1-\xi ) }{
EI(\xi ) }\,d\xi\,ds-\beta g_1(1)
\int_0^{1}\int_0^{s}\frac{(1-\xi ) }{EI(\xi ) }\,d\xi\,ds \\
&\quad -\int_0^{1}\int_0^{s}\frac{1}{EI(\xi ) }\int_{\xi
}^{1}\int_{\eta }^{1}m(r) g_2(r) +\gamma (
r) g_1(r) \,dr\,d\eta\,d\xi\,ds.
\end{align*}
Finally we obtain that $0\in \rho (A_{\gamma }) $ and
Sobolev's embedding theorem implies that $A_{\gamma }^{-1}$ is a
compact operator on $H$. Therefore, the spectrum
$\sigma (A_{\gamma }) $ consists entirely of isolated eigenvalues.
\end{proof}

\section{Spectral analysis and the Riesz basis property}

\subsection{Spectral analysis of operator $A_{\gamma }$}

In this section, we study the basic properties of system
\eqref{1a}--\eqref{1d}. Our work shall make use of the following
 result from \cite{20},
which deals with the eigenvalue problem of beams in the form of an ordinary
differential equation $L(f)=\lambda f$ with $\lambda $-polynomial boundary
conditions (see  Shkalikov \cite{14};  Tretter \cite{16}). To begin,
we recall some notations and definitions. Let $L(f) $ be an
ordinary differential operator of order $n=2m\in \mathbb{N}$,
\begin{equation}
L(f) =f^{(n) }(x) +\sum_{\nu
=1}^{n}f_{\nu }(x) f^{(n-\nu ) }(x) ,
\quad 0<x<1,  \label{3}
\end{equation}
and let the boundary conditions defined at the two points $x=0$, and $x=1$ be
\begin{equation}
B_j(f) =\sum_{\nu =0}^{k_j}(\alpha _{j_{\nu
}}f^{(k_j-\nu ) }(0) +\beta _{j_{\nu }}f^{(
k_j-\nu ) }(1) ) ,\quad 1\leq j\leq n,  \label{4}
\end{equation}
where $k_j\in \mathbb{N}$, $1\leq k_j\leq n-1$ and $\alpha _{j_{\nu }}$,
$\beta _{j_{\nu }}\in \mathbb{C}$,
$| \alpha _{j_0}| +| \beta _{j_0}| >0$. Suppose that the coefficient
functions $f_{\nu }(x) $ $(1\leq \nu \leq n) $ in \eqref{3}
 are sufficiently smooth in $(0,1) $, and
that the boundary conditions are normalized in the sense that
$\kappa =\sum_{j=1}^{n}k_j$ is minimal with respect to all equivalent
boundary conditions (see  Naimark \cite{12}).

Let $f_{k}(x,\rho ) $ $(k=1,2,\ldots ,n) $ be the
fundamental solutions for the equation:
\begin{equation}
L(f) +\rho ^{n}f+\rho ^{m}\mu (x) f(x)=0,\quad \rho \in \mathbb{C}  \label{5}
\end{equation}
where $\mu (x) $ being continuous in $[ 0,1] $, and
let $\omega _{k}$ $(k=1,2,\ldots ,n) $ be the n-th roots of
$\omega ^{n}+1=0$. If we denote by $\Delta (\rho ) $ the
characteristic determinant of \eqref{5} with respect to \eqref{4}
\[
\Delta (\rho ) =\det \big[B_j(f_{k}(.,\rho )
) \big]_{j,k=1,2,\ldots ,n}\,,
\]
then $\Delta (\rho ) $ can be expressed asymptotically in the
form, for $(r\geq 1)$,
\begin{equation}
\Delta (\rho ) =\rho ^{k}\sum_{\mathbb{K} _{k}}e^{\rho \mu \mathbb{K}
_{k}} [ F^{\mathbb{K} _{k}}] _{r}\,,  \label{6}
\end{equation}
whenever $\rho $ is large enough (see  Shkalikov \cite{14} and
Naimark \cite{12}). Here, $\mathbb{K}_{k}$ is a $k$-elements subset of
$\{1,2,\ldots ,n\} $, $\mu _{\mathbb{K} _{k}}=\sum_{j\in \mathbb{K}_{k}}\omega _j$,
\[
[ F^{\mathbb{K} _{k}}] _{r}=F_0^{\mathbb{K} _{k}}+\rho ^{-1}F_1^{\mathbb{K}
_{k}}+\ldots +\rho ^{-r+1}F_{r-1}^{\mathbb{K} _{k}}+\mathcal{O}(\rho
^{-r}) ,
\]
and the sum runs over all possible selections of $\mathbb{K}_{k}$. Here and
henceforth, $\mathcal{O}(\rho ^{-r}) $ means that
 $|\rho ^{r}\times \mathcal{O}(\rho ^{-r}) | $ is bounded
as $| \rho | \to \infty$.

\begin{definition}[{\cite[p. 461]{20}}] \label{definition1}  \rm
The boundary problem \eqref{5} with \eqref{4} is said to be regular
if the coefficients $F_0^{\mathbb{K} _{k}}$ in \eqref{6} are nonzero.
Furthermore, the regular boundary problem  \eqref{5}  with \eqref{4}
is said to be strongly regular if the zeros of $\Delta (\rho
) $ are asymptotically simple and isolated one from another.
\end{definition}

Let $W_2^{m}(0,1) $ be the usual Sobolev space of
order $m$ and let
\[
V_{E}^{m}(0,1)=\{ f(x)\in W_2^{m}(0,1) |B_j(f) =0,
\quad k_j<m\} .
\]
Define a Hilbert space
\[
\mathbb{H}=V_{E}^{m}(0,1) \times L^2(0,1),
\]
with the norm
\[
\| (f,g) \| _{\mathbb{H}}^2=\|f\| _{W_2^{m}}^2+\| g\| _2^2
\]
and define the operator $\mathbb{A}$ in $\mathbb{H}$ by
\begin{equation}
\begin{gathered}
\mathbb{A}(f,g) =(g,-L(f) -\mu (x)g) \\
D(\mathbb{A}) =\{(f,g) \in \mathbb{H}|\mathbb{A}
(f,g) \in \mathbb{H},\; B_j(f) =0,\; k_j\geq m\}.
\end{gathered}  \label{01}
\end{equation}
The following result used in \cite{20} was presented in \cite{17}. The
reader can  also be referred to \cite[chapter 3]{19}.

\begin{theorem}[{\cite[p. 461]{20}}] \label{theo6}
If the ordinary differential system with parameter $\lambda =\rho ^{m}$
\begin{equation}
\begin{gathered}
L(f,\lambda ) =L(f) +\lambda ^2f+\lambda \mu(x) f \\
B_j(f) =0,\quad 1\leq j\leq 2m
\end{gathered}  \label{7}
\end{equation}
has strongly regular boundary conditions, then the generalized
eigenfunction system of $\mathbb{A}$ form a Riesz basis in the
Hilbert space $\mathbb{H}$.
\end{theorem}

Now we are ready to study the eigenvalue problem of $A_{\gamma }$.
Let $\lambda \in \sigma (A_{\gamma }) $ and $\Phi
=(\phi ,\Psi ) $ be an eigenfunction of $A_{\gamma }$
corresponding to $\lambda $. Then  $\Psi =\lambda \phi $ and $\phi $
satisfy the following equations:
\begin{equation}
\begin{gathered}
\lambda ^2m(x)\phi (x) +(EI(x) \phi ''(x) ) ''+\lambda \gamma (
x) \phi (x) =0,\quad 0<x<1, \\
\phi (0) =\phi '(0) =\phi ''(1) =0 \\
\phi '''(1) =\frac{1}{EI(1) }
(\alpha +\beta \lambda ) \phi (1) .
\end{gathered}  \label{28}
\end{equation}
Expanding \eqref{28} yields
\begin{equation}
\begin{gathered}
\begin{aligned}
&\phi ^{(4) }(x) +\frac{2EI'(x)}{EI(x) }\phi '''(x) 
+\frac{EI''(x) }{EI(x) }\phi ''(x) \\
&+\frac{\lambda ^2m(x) }{EI(x) }\phi (x)
+ \frac{\lambda \gamma (x) }{EI(x) }\phi (x) =0,\quad 0<x<1,
\end{aligned} \\
\phi (0) =\phi '(0) =\phi ''(1) =0 \\
\phi '''(1) =\frac{1}{EI(1) }(\alpha +\beta \lambda ) \phi (1) .
\end{gathered}  \label{29}
\end{equation}
Two basic transformations are essential.

First, the ``dominant term'', $\phi ^{(4) }(x) +\frac{
\lambda ^2m(x) }{EI(x) }\phi (x) $ of \eqref{29}, is transformed
to become a uniform form by space scaling. In fact, set:
\begin{equation}
f(z) =\phi (x) ,\quad 
z=z(x) =\frac{1}{h}\int_0^{x}(\frac{m(\zeta ) }{EI(\zeta ) }
) ^{1/4}d\zeta  \label{30}
\end{equation}
where
\begin{equation}
h=\int_0^{1}(\frac{m(\zeta ) }{EI(\zeta ) }) ^{1/4}d\zeta
\end{equation}
Then, \eqref{29} together with its boundary conditions can be transformed
into
\begin{equation}
\begin{gathered}
\begin{aligned}
&f^{(4) }(z) +a(z) f'''(z) +b(z) f''(z)+c(z) f'(z) \\
&+\lambda ^2h^{4}f(z) + \frac{\lambda h^{4}\gamma (x) }{m(x) }f(x) =0,
\quad 0<z<1,
\end{aligned} \\
f(0) =f'(0) =0, \\
z_{x}^2(1) f''(1) +z_{xx}(1) f'(1) =0, \\
f'''(1) +\frac{3z_{xx}(1) }{
z_{x}^2(1) }f''(1) +\frac{z_{xxx}(1) }{z_{x}^{3}(1) }f'(1)
-\frac{(\alpha +\lambda \beta ) }{z_{x}^{3}(1) EI(1) }f(1) =0,
\end{gathered}  \label{32}
\end{equation}
with
\begin{gather}
a(z) =\frac{6z_{xx}}{z_{x}^2}+\frac{2EI'(x) }{z_{x}EI(x) } ,  \label{33} \\
b(z) =\frac{3z_{xx}^2}{z_{x}^{4}}
+\frac{6z_{xx}EI'(x) }{z_{x}^{3}EI(x) }
+\frac{EI''(x) }{z_{x}^2EI(x) }+\frac{4z_{xxx}}{z_{x}^{3}}, \label{34} \\
c(z) =\frac{z_{xxxx}}{z_{x}^{4}}+\frac{2z_{xxx}EI'(x) }{z_{x}^{4}EI(x) }
+\frac{z_{xx}EI''(x) }{z_{x}^{4}EI(x) } ,  \label{35} \\
z_{x}=\frac{1}{h}\big(\frac{m(x) }{EI(x) }
\big) ^{1/4},\quad z_{x}^{4}=\frac{1}{h^{4}}\frac{m(x) }{EI(x) },  \label{36} \\
z_{xx}=\frac{1}{4h}\big(\frac{m(x) }{EI(x) }
\big) ^{-3/4}\frac{d}{dx}\big(\frac{m(x) }{
EI(x) }\big) ^{1/4}.  \label{37}
\end{gather}
If we definite
\begin{equation}
d(x) =\frac{\gamma (x) }{m(x) },
\end{equation}
then equation in \eqref{32} is
\begin{equation}
\begin{aligned}
&f^{(4) }(z) +a(z) f'''(z) +b(z) f''(z) +c(z) f'(z)   \\
&+\lambda ^2h^{4}f(z) +\lambda h^{4}d(z) f(
z) =0,\quad 0<z<1.
\end{aligned}
\end{equation}

Second,  to cancel the term $a(z) f'''(z) $ in \eqref{32} as was done 
in Naimark \cite{12}, we make the invertible state transformation
\[
g(z) =\exp \Big(\frac{1}{4}\int_0^{z}a(\zeta )
d\zeta \Big) f(z) ,\quad 0<z<1,
\]
and we arrive at the following eigenvalue problem that is equivalent to the
original one:
\begin{equation}
\begin{gathered}
\begin{aligned}
&g^{(4) }(z) +b_1(z) g''(z) +c_1(z) g'(z)
+d_1(z) g(z) \\
&+ \lambda h^{4}d(z) g(z) +\lambda ^2h^{4}g(z) =0,\quad 0<z<1,
\end{aligned} \\
g(0) =g'(0) =0 \\
g''(1) +b_{11}g'(1) +b_{12}g(1) =0 \\
g'''(1) +b_{21}g''(1) +b_{22}g'(1) +b_{23}g(1) =0,
\end{gathered}  \label{388}
\end{equation}
where
\begin{gather}
b_1(z) =-\frac{3}{2}a'(z)-\frac{3}{8}a^2(z)+b(z), \label{40}\\
c_1(z)=\frac{1}{8}a^{3}(z)-\frac{1}{2}a(z)b(z)-a''(z)+c(z), \label{41} \\
\begin{aligned}
d_1(z) &=\frac{3}{16}a^{\prime 2}(z)-\frac{1}{4}a'''(z)
 +\frac{3}{32}a'(z)a^2(z)-\frac{3}{256}a^{4}(z)\\
&+ b(z)(\frac{1}{16}a^2(z)-\frac{1}{4}a'(z))
-\frac{a(z)c(z)}{4},
\end{aligned} \\
b_{11}=-\frac{1}{2}a(1)+\frac{z_{xx}(1)}{z_{x}^2(1)} , \label{43} \\
b_{12}=\frac{\frac{1}{16}z_{x}^2(1)a^2(1)-\frac{1}{4}
z_{x}^2(1)a'(1)-\frac{1}{4}z_{xx}(1)a(1)}{z_{x}^2(1)},
\label{44} \\
b_{21}=-\frac{3}{4}a(1)+\frac{3z_{xx}(1)}{z_{x}^2(1)} , \label{45} \\
b_{22}=-\frac{3}{4}a'(1)+\frac{3}{16}a^2(1)-\frac{
3z_{xx}(1)a(1)}{2z_{x}^2(1)}+\frac{z_{xxx}(1)}{z_{x}^{3}(1)},  \label{46} \\
\begin{aligned}
b_{23} &=-\frac{1}{4}a''(1)+\frac{3}{16}a'(1)a(1)-
\frac{1}{64}a^{3}(1)-\frac{3z_{xx}(1)a'(1)}{4z_{x}^2(1)}\\
&+\frac{3z_{xx}(1)a^2(1)}{16z_{x}^2(1)}
 -\frac{z_{xxx}(1)a(1)}{4z_{x}^{3}(1)}
 -\frac{(\alpha +\lambda \beta )}{z_{x}^{3}(1)EI(1)}.
\end{aligned}
\end{gather}
To  solve the eigenvalue problem \eqref{388}, we follow the procedure
in  Birkhoff \cite{1,2} and  Naimark \cite{12}, and
divide the complex plane into eight distinct sectors,
\begin{equation}
S_{k}=\big\{ z\in \mathbb{C}:\frac{k\pi }{4}\leq \arg z\leq \frac{(
k+1) \pi }{4}\big\} , \quad k=0,1,2,\ldots ,7  \label{48}
\end{equation}
and let $\omega _1$, $\omega _2$, $\omega _3$, $\omega _4$ be the
roots of equation $\theta ^{4}+1=0$ that are arranged so that
\begin{equation}
\operatorname{Re}(\rho \omega _1)
\leq \operatorname{Re}(\rho \omega_2)
\leq \operatorname{Re}(\rho \omega _3)
\leq \operatorname{Re}(\rho \omega _4) ,\quad \forall \rho \in S_{k}.  \label{49}
\end{equation}
Obviously, in sector $S_1$, we can choose
\begin{gather*}
\omega _1=\exp \big(i\frac{3}{4}\pi \big)
=- \frac{\sqrt{2}}{2}+ \frac{\sqrt {2}}{2}i,\quad
\omega _2=\exp \big(i\frac{1}{4}\pi\big)
= \frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i,\\
\omega _3=\exp \big(i\frac{5}{4}\pi \big)
=- \frac{\sqrt{2}}{2}-\frac{\sqrt {2}}{2}i,\quad
\omega _4=\exp \big(i\frac{7}{4}\pi\big)
= \frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i,
\end{gather*}
which satisfy the inequalities in \eqref{49} and choices can also
be made for other sectors. In the rest of this section, we shall derive the
asymptotic behavior of the eigenvalue of the sectors $S_1$ and $S_2$
because the same will hold for the other sectors with similar proofs.

Setting $\lambda =\rho ^2/h^2$, in each sector $S_{k}$, we have
the following result about the asymptotic fundamental solutions of system
\eqref{388}.

\begin{lemma}\label{lemma2}
For $\rho \in S_{k}$ with $\rho $ large enough, the equation
\begin{align*}
&g^{(4) }(z) +b_1(z) g''(z) +c_1(z) g'(z)
+d_1(z) g(z)  \\
&+\rho ^2h^2d(z) g(z) +\rho ^{4}g(z)  
&=&0,\quad 0<z<1,
\end{align*}
has four linearly independent asymptotic fundamental solutions,
\begin{equation}
\Phi _{s}(z,\rho ) =e^{\rho \omega _{s}z}\big(1+\frac{\Phi
_{s,1}(z) }{\rho }+\mathcal{O}(\rho ^{-2}) \big),\quad s=1,2,3,4  \label{50}
\end{equation}
and hence their derivatives for $s=1,2,3,4$ and $j=1,2,3$ are given by
\begin{equation}
\frac{d^{j}}{dz^{j}}\Phi _{s}(z,\rho ) =(\rho \omega
_{s}) ^{j}e^{\rho \omega _{s}z}\Big(1+\frac{\Phi _{s,1}(
z) }{\rho }+\mathcal{O}(\rho ^{-2}) \Big)   \label{51}
\end{equation}
where
\begin{equation}
\Phi _{s,1}(z) =-\frac{1}{4\omega _{s}}\int_0^{z}b_1(
\zeta ) d\zeta -\frac{h^2}{4\omega _{s}^{3}}\int_0^{z}d(
\zeta ) d\zeta ,\quad \Phi _{s,1}(0) =0,
 \label{52}
\end{equation}
for $s=1,2,3,4$, 
and
\begin{equation}
\Phi _{s,1}(z) =\frac{\omega _{s}^2\mu _1+\mu _2}{\omega
_{s}^{3}},\quad \text{with }\mu _1=-\frac{1}{4}\int_0^{1}b_1(\zeta ) d\zeta ,\quad 
\mu _2=-\frac{h^2}{4} \int_0^{1}d(\zeta ) d\zeta .  \label{53}
\end{equation}
\end{lemma}

\begin{proof}
The proof is a direct result in Birkhoff \cite{1}, \cite{2} and 
 Naimark \cite{12}. Here we briefly present a simple calculation to find
the asymptotic expansions of fundamental solutions in sector $S_{k}$. Let
\[
\widetilde{\Phi }_{s}(z,\rho ) =e^{\rho \omega _{s}z}\Big(\Phi
_{s,0}(z) +\frac{\Phi _{s,1}(z) }{\rho }+\mathcal{O}
(\rho ^{-2}) \Big) ,\quad s=1,2,3,4
\]
and
\begin{align*}
D(g) &=g^{(4) }(z) +b_1(z) g''(z) +c_1(z) g'(z) +d_1(z) g(z) \\
&+ \rho ^2h^2d(z) g(z) +\rho ^{4}g(z),\quad 0 <z<1.
\end{align*}
Then, substituting $\widetilde{\Phi }_{s}(z,\rho ) $ in the
expression of $e^{-\rho \omega _{s}z}D(g) $, for $
s=1,2,3,4$, it yields
\begin{align*}
&e^{-\rho \omega _{s}z}D(\widetilde{\Phi }_{s}(z,\rho )) \\
&=(\rho \omega _{s}) ^{4}\Big(\Phi _{s,0}(z)
+\frac{\Phi _{s,1}(z) }{\rho }\Big)
+4(\rho \omega _{s}) ^{3}\Big(\Phi _{s,0}'(
z) +\frac{\Phi _{s,1}'(z) }{\rho }\Big)\\
&\quad +6(\rho \omega _{s}) ^2\Big(\Phi _{s,0}''(z)
 +\frac{\Phi _{s,1}''(z) }{\rho }\Big)
 + 4\rho \omega _{s}\Big(\Phi _{s,0}'''(z)
 +\frac{\Phi _{s,1}'''(z) }{\rho }\Big)
 +\Phi _{s,0}^{(4)}(z) +\frac{\Phi _{s,1}^{(4)}(z) }{\rho }\\
&\quad + b_1(z) (\rho \omega _{s}) ^2\Big(\Phi_{s,0}(z)
 +\frac{\Phi _{s,1}(z) }{\rho }\Big)
 + 2b_1(z) \rho \omega _{s}\Big(\Phi _{s,0}'(z)
 +\frac{\Phi _{s,1}'(z) }{\rho }\Big) \\
&\quad + b_1(z) \Big(\Phi _{s,0}''(z) +\frac{\Phi _{s,1}''(z) }{\rho }\Big)
 +c_1(z) \rho \omega _{s}\Big(\Phi _{s,0}(z) +\frac{\Phi _{s,1}(z) }{\rho }\Big) \\
&\quad + c_1(z) \Big(\Phi _{s,0}'(z) +\frac{\Phi _{s,1}'(z) }{\rho }\Big)
 + (\rho ^{4}+d_1(z) +\rho ^2h^2d(z)) \Big(\Phi _{s,0}(z)
 +\frac{\Phi _{s,1}(z) }{\rho }\Big) \\
&=\rho ^{4}[ \Phi _{s,0}(z) -\Phi _{s,0}(z)]
 +\rho ^{3}[ -\Phi _{s,1}(z) +4\omega _{s}^{3}\Phi_{s,0}'(z)
 +\Phi _{s,1}(z) ] \\
&\quad + \rho ^2[4\omega _{s}^{3}\Phi _{s,1}'(z)
 +6\omega_{s}^2\Phi _{s,0}''(z) +b_1(z)\omega _{s}^2\Phi _{s,0}(z) \\
&\quad + h^2d(z) \Phi _{s,0}(z) ]+\rho R_{s}(z,\rho) ,
\end{align*}
where $R_{s}(z,\rho ) $ denote the remaining terms in the above
equation and satisfy the following estimates for some positive constant $M$:
\begin{equation}
R_{s}(z,\rho ) \leq M,\quad 0<z<1.  \label{54}
\end{equation}
Thus, by setting the coefficients of $\rho ^{3}$ and $\rho ^2$ to zero
respectively, we obtain
\begin{gather*}
\Phi _{s,0}'(z) =0,\\
4\omega _{s}^{3}\Phi _{s,1}'(z) +6\omega _{s}^2\Phi
_{s,0}''(z) +b_1(z) \omega
_{s}^2\Phi _{s,0}(z) +h^2d(z) \Phi _{s,0}(z) =0,
\end{gather*}
which yield that $\Phi _{s,0}(z) =1$ and
$\Phi _{s,1}( z) $ given in \eqref{53} are linear independent solutions.
Thus, as in the theorem in (Birkhoff \cite{1}, pp.225-226), we obtain the linearly
independent fundamental solutions of
\begin{align*}
&g^{(4) }(z) +b_1(z) g''(z) +c_1(z) g'(z)+d_1(z) g(z) \\
&+ \rho ^{4}g(z) +\rho ^2h^2d(z) g(z)
=0,\quad 0<z<1,
\end{align*}
given by $(s=1,2,3,4) $
\[
\Phi _{s}(z,\rho ) =\widetilde{\Phi }_{s}(z,\rho )
+e^{\rho \omega _{s}z}\mathcal{O}(\rho ^{-2}) ,
\]
from which we deduce the required results \eqref{50} and \eqref{51}
\end{proof}

For convenience, we introduce the notation $[ a] _2=a+\mathcal{O}(\rho ^{-2})$.

\begin{lemma}\label{lemme3}
For $\rho \in S_1$, if we set $\delta =\sin (\pi/4)=\sqrt{2}/2$,
then we have inequalities
\[
\operatorname{Re}(\rho \omega _1) \leq -| \rho| \delta ,\quad
 \operatorname{Re}(\rho \omega _4) \geq | \rho | \delta, \quad
e^{\rho \omega _1}=\mathcal{O}(\rho ^{-2})
\]
as $| \rho |\to \infty$.
\end{lemma}

Furthermore, substituting \eqref{50} and \eqref{51} into the boundary
conditions \eqref{388}, we obtain asymptotic expressions for the boundary
conditions for large enough $| \rho |$:
\begin{gather}
U_4(\Phi _{s},\rho ) =\Phi _{s}(0,\rho ) =1+
\mathcal{O}(\rho ^{-2}) =[ 1] _2,\quad s=1,2,3,4, \label{55}
\\
U_3(\Phi _{s},\rho ) =\Phi _{s}'(0,\rho)
 =\rho \omega _{s}(1+\mathcal{O}(\rho ^{-2})), 
\\
U_3(\Phi _{s},\rho ) =\rho \omega _{s}[ 1]_2,\quad s=1,2,3,4,\\
\begin{aligned}
&U_2(\Phi _{s},\rho ) \\
&=\Phi _{s}''(1,\rho ) +b_{11}\Phi _{s}'(1,\rho ) +b_{12}\Phi _{s}(1,\rho ), 
\quad\text{for }s=1,2,3,4,\\
&=(\rho \omega _{s}) ^2e^{\rho \omega _{s}}(1+(
\omega _{s}^2\mu _1+\mu _2) \rho ^{-1}\omega
_{s}^{-3}+b_{11}\rho ^{-1}\omega _{s}^{-1}+\mathcal{O}(\rho
^{-2}) ), 
\end{aligned} \\
U_2(\Phi _{s},\rho ) =(\rho \omega _{s})
^2e^{\rho \omega _{s}}[ 1+(\omega _{s}^2(\mu _1+b_{11})+\mu
_2) \omega _{s}^{-3}\rho ^{-1}] _2,  \label{57}
\\
\begin{aligned}
& U_1(\Phi _{s},\rho ) \\
&=\Phi _{s}'''(1,\rho ) +b_{21}\Phi_{s}''(1,\rho ) +b_{22}\Phi _{s}'(
1,\rho ) +b_{23}\Phi _{s}(1,\rho ) \\
&=(\rho \omega _{s}) ^{3}e^{\rho \omega _{s}}(1+(\omega
_{s}^2\mu _1+\mu _2) \rho ^{-1}\omega _{s}^{-3}+b_{21}\rho
^{-1}\omega _{s}^{-1}\\
&\quad +b_{23}(\rho \omega _{s}) ^{-3}+\mathcal{O}(\rho
^{-2}) ) \\
&=(\rho \omega _{s}) ^{3}e^{\rho \omega _{s}}(1+(\omega
_{s}^2(\mu _1+b_{21})+\mu _2) \omega _{s}^{-3}\rho ^{-1}\\
&\quad -\frac{\beta \omega _{s}^{-3}\rho ^{-1}}{z_{x}^{3}(1) EI(
1) h^2}+\mathcal{O}(\rho ^{-2}) )
\end{aligned} \\
U_1(\Phi _{s},\rho ) 
=(\rho \omega _{s}) ^{3}e^{\rho \omega _{s}}
[ 1+(\omega _{s}^2(\mu _1+b_{21})+\mu
_2) \omega _{s}^{-3}\rho ^{-1}-\chi ] _2,  \label{58}
\end{gather}
where 
\[
\chi =\frac{\beta \omega _{s}^{-3}\rho ^{-1}}{z_{x}^{3}(
1) EI(1) h^2}, \quad s=1,2,3,4,.
\]
Note that $\lambda =\rho ^2/h^2 \neq 0$ is the eigenvalue in
\eqref{388} if and only if $\rho $ satisfies the characteristic equation
\begin{equation}
\Delta (\rho ) =
\begin{vmatrix}
U_4(\Phi _1,\rho ) & U_4(\Phi _2,\rho ) &
U_4(\Phi _3,\rho ) & U_4(\Phi _4,\rho ) \\
U_3(\Phi _1,\rho ) & U_3(\Phi _2,\rho ) &
U_3(\Phi _3,\rho ) & U_3(\Phi _4,\rho ) \\
U_2(\Phi _1,\rho ) & U_2(\Phi _2,\rho ) &
U_2(\Phi _3,\rho ) & U_2(\Phi _4,\rho ) \\
U_1(\Phi _1,\rho ) & U_1(\Phi _2,\rho ) &
U_1(\Phi _3,\rho ) & U_1(\Phi _4,\rho )
\end{vmatrix}
 =0,  \label{60}
\end{equation}
so substituting \eqref{55}--\eqref{58} in \eqref{60}and using 
Lemma \ref{lemme3}, we obtain that
$\Delta (\rho )$ is the determinant of the matrix whose four columns are:
\begin{gather*}
\begin{pmatrix}[ 1] _2 \\ \rho \omega _1[ 1] _2 \\ 0 \\ 0\end{pmatrix},
\begin{pmatrix}
 [ 1] _2 \\
 \rho \omega _2[ 1] _2 \\
 (\rho \omega _2) ^2e^{\rho \omega _2}[ 1+(
\omega _2^2(\mu _1+b_{11})+\mu _2) \omega _2^{-3}\rho ^{-1}] _2 \\
 (\rho \omega _2) ^{3}e^{\rho \omega _2}[ 1+(
\omega _2^2(\mu _1+b_{21})+\mu _2) \omega _2^{-3}\rho
^{-1}-\chi ] _2
\end{pmatrix}, 
\\
\begin{pmatrix}
[ 1] _2 \\
\rho \omega _3[ 1] _2 \\
(\rho \omega _3) ^2e^{\rho \omega _3}[ 1+(
\omega _3^2(\mu _1+b_{11})+\mu _2) \omega _3^{-3}\rho ^{-1}] _2 \\
(\rho \omega _3) ^{3}e^{\rho \omega _3}[ 1+(
\omega _3^2(\mu _1+b_{21})+\mu _2) \omega _3^{-3}\rho
^{-1}-\chi ] _2
\end{pmatrix},
\\
\begin{pmatrix}
0 \\
0 \\
(\rho \omega _4) ^2e^{\rho \omega _4}[ 1+(
\omega _4^2(\mu _1+b_{11})+\mu _2) \omega _4^{-3}\rho ^{-1}] _2 \\
(\rho \omega _4) ^{3}e^{\rho \omega _4}[ 1+(
\omega _4^2(\mu _1+b_{21})+\mu _2) \omega _4^{-3}\rho
^{-1}-\chi ] _2
\end{pmatrix}.
\end{gather*}
Expanding the above determinant, we obtain the following expression:
\begin{align*}
&\Delta (\rho ) \\
&=\rho ^{6}e^{\rho \omega _4}\Big\{(\omega
_2-\omega _1) \omega _3^2\omega _4^2e^{\rho \omega
_3}[\omega _4-\omega _3+(\omega _3^{-1}\omega _4-\omega
_3\omega _4^{-1}) (\mu _1+b_{11}) \rho ^{-1}
\\
&\quad +(\omega _4^{-2}-\omega _3^{-2}) \mu _2\rho ^{-1}+(
\omega _4\omega _3^{-3}-\omega _3\omega _4^{-3}) \mu _2\rho
^{-1}+\frac{\beta \rho ^{-1}}{z_{x}^{3}(1) EI(1)
h^2}(\omega _3^{-2}-\omega _4^{-2}) ]\\
&\quad +
(\omega _1-\omega _3) \omega _2^2\omega _4^2e^{\rho
\omega _2}[\omega _4-\omega _2+(\omega _2^{-1}\omega
_4-\omega _2\omega _4^{-1}) (\mu _1+b_{11}) \rho
^{-1}\\
&\quad +(\omega _4^{-2}-\omega _2^{-2}) \mu _2\rho ^{-1}+(
\omega _4\omega _2^{-3}-\omega _2\omega _4^{-3}) \mu _2\rho
^{-1}+\frac{\beta \rho ^{-1}}{z_{x}^{3}(1) EI(1)
h^2}(\omega _2^{-2}-\omega _4^{-2}) ]\\
&\quad +\mathcal{O}(\rho ^{-2}) \}
\end{align*}

In sector $S_1$, the choices are such that:
$\omega _1^2=-i$,  $\omega _2^2=i$,  $\omega _3^2=i$,
$\omega _4^2=-i$,  $\omega _3^{-1}\omega _4=i$,
$\omega_2^{-1}\omega _4=-i$,  $\omega _3=-\omega _2$,
$\omega _4-\omega _3=\sqrt{2}$,  $\omega _1-\omega _3=\sqrt{2}i$,
$\omega _2-\omega _1=\sqrt{2}$,  $\omega _4-\omega _2=-i\sqrt{2}$,
$\omega _2^{-2}-\omega _4^{-2}=-2i$,
$\omega_3^{-2}-\omega _4^{-2}=-2i$,
$\omega _3^2\omega_4^2=1$, $\omega _2^2\omega _4^2=1$.
A straightforward simplification will arrive at the following result, which
is also true on all other sectors $S_{k}$ (see Naimark, \cite{12}).

\begin{theorem}\label{theo4}
Let $\Delta (\rho ) $ be the characteristic determinant of the
eigenvalue problem \eqref{388}. In sector $S_1$, an asymptotic expression
of $\Delta (\rho ) $ is given by:
\begin{equation}
\Delta (\rho ) =\rho ^{6}e^{\rho \omega _4}\{2e^{\rho \omega
_2}+2e^{-\rho \omega _2}+2\mu _3\rho ^{-1}e^{\rho \omega _2}+2i\mu
_4\rho ^{-1}e^{-\rho \omega _2}+\mathcal{O}(\rho ^{-2}) \},
\label{61}
\end{equation}
where
\begin{equation}
\begin{gathered}
\mu _3=\sqrt{2}(\mu _1+b_{11}) -\sqrt{2}\mu _2+\sqrt{2}
\frac{\beta }{z_{x}^{3}(1) EI(1) h^2} \\
\mu _4=\sqrt{2}(\mu _1+b_{11}) +\sqrt{2}\mu _2-\sqrt{2}
\frac{\beta }{z_{x}^{3}(1) EI(1) h^2}
\end{gathered}   \label{62}
\end{equation}
Thus, the boundary eigenvalue problem \eqref{388} is strongly regular.
\end{theorem}

Using \eqref{61}, we can deduce an asymptotic expression for the
eigenvalues of problem \eqref{388}. The equation 
$\Delta (\rho )=0$ and \eqref{61} imply that
\[
2e^{\rho \omega _2}+2e^{-\rho \omega _2}+2\mu _3\rho ^{-1}e^{\rho
\omega _2}+2i\mu _4\rho ^{-1}e^{-\rho \omega _2}+\mathcal{O}(
\rho ^{-2}) =0
\]
which is equivalent to
\begin{equation}
e^{\rho \omega _2}+e^{-\rho \omega _2}+\mu _3\rho ^{-1}e^{\rho \omega
_2}+i\mu _4\rho ^{-1}e^{-\rho \omega _2}+\mathcal{O}(\rho
^{-2}) =0  \label{63}
\end{equation}
and can be rewritten as
\begin{equation}
e^{\rho \omega _2}+e^{-\rho \omega _2}+\mathcal{O}(\rho
^{-1}) =0.  \label{64}
\end{equation}
Note that the  equation
$e^{\rho \omega _2}+e^{-\rho \omega _2}=0$
has solutions
\begin{equation}
\rho _{n}=\big(n+\frac{1}{2}\big) \frac{\pi i}{\omega _2},\quad
n=1,2,\ldots  \label{65}
\end{equation}

Let $\widetilde{\rho _{n}}$ be the solutions of \eqref{64}. Applying
Rouche's theorem see (Naimark \cite[p.70]{12}) to \eqref{64}, we obtain 
the  expression
\begin{equation}
\widetilde{\rho _{n}}=\rho _{n}+\alpha _{n}=(n+\frac{1}{2})
\frac{\pi i}{\omega _2}+\alpha _{n},\quad \alpha _{n}=\mathcal{O}(
n^{-1}) ,\text{ }n=N,N+1,\ldots ,  \label{66}
\end{equation}
where $N$ is a large positive integer. Substituting $\widetilde{\rho _{n}}$
into \eqref{63}, and using the fact that $e^{\rho \omega _2}=-e^{-\rho
\omega _2}$, we obtain
\[
e^{\alpha _{n}\omega _2}-e^{-\alpha _{n}\omega _2}+\mu _3\widetilde{
\rho _{n}}^{-1}e^{\alpha _{n}\omega _2}-i\mu _4\widetilde{\rho _{n}}
^{-1}e^{-\alpha _{n}\omega _2}+\mathcal{O}(\widetilde{\rho _{n}}
^{-2}) =0.
\]
Expanding the exponential function according to its Taylor series, we obtain
\[
\alpha _{n}=-\frac{\mu _3}{2\omega _2\rho _{n}}+\frac{\mu _4}{
2\omega _2\rho _{n}}i+\mathcal{O}(n^{-2}) ,\text{ }
n=N,N+1,\ldots
\]
Therefore,
\[
\widetilde{\rho _{n}}=(n+\frac{1}{2}) \frac{\pi i}{\omega _2}
+\frac{\mu _3}{2(n+\frac{1}{2}) \pi }i+\frac{\mu _4}{
2(n+\frac{1}{2}) \pi }+\mathcal{O}(n^{-2}) ,
\]
for $ n=N,N+1,\ldots$.
Note that $\lambda _{n}=\frac{\rho _{n}^2}{h^2}\neq 0$,
$\omega_2=e^{i\frac{\pi }{4}}$ and $\omega _2^2=i$. So we have
\begin{equation}
\lambda _{n}=\frac{\sqrt{2}}{2h^2}(\mu _4-\mu _3) +
\frac{1}{h^2}\big[ \frac{\sqrt{2}}{2}(\mu _4+\mu _3)
+(n+\frac{1}{2}) ^2\pi ^2\big] i+\mathcal{O}(
n^{-1}) ,  \label{67}
\end{equation}
where $n=N,N+1,\ldots $ with $N$ large enough.

The same proof can be applied to sector $S_2$ because the eigenvalues of
the problem \eqref{388} can be obtained by a similar calculation with the
choices
\begin{gather*}
\omega _1=\exp (i\frac{1}{4}\pi )=\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i, \quad 
\omega _2=\exp (i\frac{3}{4}\pi )=-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i,\\
\omega _3=\exp (i\frac{7}{4}\pi )=\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i, \quad
\omega _4=\exp (i\frac{5}{4}\pi )=-\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i,
\end{gather*}
so that \eqref{49} holds:
\[
\operatorname{Re}(\rho \omega _1) \leq \operatorname{Re}(\rho \omega
_2) \leq \operatorname{Re}(\rho \omega _3) \leq \operatorname{Re}
(\rho \omega _4) ,\quad \forall \rho \in S_2.
\]
Hence, in sector $S_2$, we have the following asymptotic expression of
$\Delta (\rho )$:
\[
\Delta (\rho ) =\rho ^{6}e^{\rho \omega _4}\{2e^{\rho \omega
_2}+2e^{-\rho \omega _2}-2\mu _3\rho ^{-1}e^{\rho \omega _2}+2i\mu
_4\rho ^{-1}e^{-\rho \omega _2}+\mathcal{O}(\rho ^{-2}) \}.
\]
By a direct calculation, we have
\begin{equation}
\widetilde{\rho _{n}}=(n+\frac{1}{2}) \frac{\pi i}{\omega _2}
-\frac{\mu _3}{2(n+\frac{1}{2}) \pi }i+\frac{\mu _4}{
2(n+\frac{1}{2}) \pi }+\mathcal{O}(n^{-2}) ,
 \label{68}
\end{equation}
for $n=N,N+1,\ldots$, with $N$ large enough.
Again, using $\lambda _{n}=\frac{\rho _{n}^2}{h^2}\neq 0$,
$\omega _2=e^{i\frac{3\pi }{4}}$ and $\omega _2^2=-i$.
\begin{equation}
\lambda _{n}=\frac{\sqrt{2}}{2h^2}(\mu _4-\mu _3)
-\frac{1}{h^2}\big[ \frac{\sqrt{2}}{2}(\mu _4+\mu _3) +(n+
\frac{1}{2}) ^2\pi ^2\big] i+\mathcal{O}(n^{-1}) ,
\label{69}
\end{equation}
where $n=N,N+1,\ldots $ with $N$ large enough.

Here we should point out that the eigenvalues generated from the other
sectors $S_{k}$ coincide with those from $S_1$ and $S_2$. The detailed
argument can be found in  Naimark \cite{12}. Combining with \eqref{67}
and \eqref{69}, we obtain the following result on the eigenvalues.

\begin{theorem}\label{theo5}
Let $A_{\gamma }$ be defined by $\eqref{16} $ and $
\eqref{17} $, then an asymptotic expression of the eigenvalues
of  problem \eqref{388} is given by
\begin{equation}
\lambda _{n}=\frac{\sqrt{2}}{2h^2}(\mu _4-\mu _3) \pm
\frac{1}{h^2}\big[ \frac{\sqrt{2}}{2}(\mu _4+\mu _3)
+(n+\frac{1}{2}) ^2\pi ^2\big] i+\mathcal{O}(
n^{-1}) ,  \label{70}
\end{equation}
where $n=N,N+1,\ldots $ with $N$ large enough, and
\begin{gather}
\mu _4-\mu _3=2\sqrt{2}\mu _2-2\sqrt{2}\frac{\beta }{z_{x}^{3}(
1) EI(1) h^2}=2\sqrt{2}\mu _2-\frac{2\sqrt{2}\beta h}{
EI(1) }(\frac{m(1) }{EI(1) }
) ^{-3/4},  \label{71}
\\
d(z) =\frac{\gamma (x) }{m(x) }, \quad 
z_{x}=\frac{dz}{dx}=\frac{1}{h}(\frac{m(x) }{EI(x) }) ^{1/4}, \nonumber
\end{gather}
so,
\[
\mu _2 
=-\frac{h^2}{4}\int_0^{1}\frac{\gamma (x) }{
m(x) }\frac{1}{h}(\frac{m(x) }{EI(x) }) ^{1/4}dx 
= -\frac{h}{4}\int_0^{1}\frac{\gamma (x) }{m(x)
}(\frac{m(x) }{EI(x) }) ^{1/4}dx.
\]
Moreover, $\lambda _{n}$ $(n=N,N+1,\ldots ) $ with sufficiently
large modulus are simple and distinct except for finitely many of them, and
satisfy
\begin{equation}
\lim_{n\to +\infty } \operatorname{Re}\lambda _{n}=-\frac{1}{2h}
\int_0^{1}\frac{\gamma (x) }{m(x) }(\frac{m(x) }{EI(x) }) ^{1/4}dx
-\frac{2\beta}{hEI(1) }(\frac{m(1) }{EI(1) }) ^{-3/4}.  \label{73}
\end{equation}
\end{theorem}

\subsection{Riesz basis property of eigenfunctions of $A_{\gamma }$}

In this subsection, we discuss the Riesz basis property of the
eigenfunctions of operator $A_{\gamma }$ of the system \eqref{17}.
We follow an idea due to Wang (see \cite{20} pp. 473-475). We begin with
showing that the generalized eigenfunctions of $A_{\gamma }$ form
an unconditional basis in Hilbert state space $H$. For this aim, we
introduce a transformation $\mathcal{L}$ via
\[
\mathcal{L}(f,g) =(\phi ,\psi )
\]
where
\begin{equation}
\phi (x) =f(z) ,\quad
\psi (x)=g(z) ,\quad
z=\frac{1}{h}\int_0^{x}\Big(\frac{m(
\zeta ) }{EI(\zeta ) }\Big) ^{1/4}d\zeta ,
\end{equation}
with
\begin{equation}
h=\int_0^{1}\Big(\frac{m(\zeta ) }{EI(\zeta ) }\Big) ^{1/4}d\zeta .
\end{equation}
It is easily seen that $\mathcal{L}$ is a bounded invertible operator on
$\mathbb{H}$.

Now we define the  ordinary differential operator:
\begin{equation}
\begin{gathered}
L(f) =f^{(4) }(z) +a(z)
f'''(z) +b(z) f''(z) +c(z) f'(z) , \\
\mu (z) =h^2d(z) =h^2\frac{\gamma (x) }{m(x) }, \\
B_1(f) =f(0) =0,\quad 
B_2(f) =f'(0) =0, \\
B_3(f) =z_{x}^2(1) f''(
1) +z_{xx}(1) f'(1) =0 \\
B_4(f) =f'''(1) +\frac{3z_{xx}(1) }{z_{x}^2(1) }f''(
1) +\frac{z_{xxx}(1) }{z_{x}^{3}(1) }f'(1) 
-\frac{(\alpha +\lambda \beta ) }{z_{x}^{3}(1)
EI(1) }f(1) =0,
\end{gathered}
\end{equation}
where the coefficients are given by \eqref{33} --\eqref{37}. 
Let $\mathbb{A}$ be defined as in $\eqref{01} $, 
$\eta \in \sigma (\mathbb{A}) $ be an eigenvalue of $\mathbb{A}$ 
and $(f,g) $ be an eigenfunction corresponding to $\eta $, 
then we have $g=\eta f$ and $f$ will satisfy the equation
\[
f^{(4) }(z) +a(z) f'''(z) +b(z) f''(z)
+c(z) f'(z) +\eta \mu (z)f(z) +\eta ^2f(z) =0,
\]
with boundary conditions $B_j(f) =0$, $j=1,2,3,4$. Now by
taking $\lambda =\frac{\eta }{h^2}$ and
\[
\mathcal{L}(f,g) =(\phi (x) ,\psi (x) )
\]
we see that $\psi =\lambda \phi $ and $\phi $ satisfies the equation
\begin{equation}
\begin{gathered}
\begin{aligned}
&\phi ^{(4) }(x) +\frac{2EI'(x)}{EI(x) }\phi '''(x)
+\frac{EI''(x) }{EI(x) }\phi ''(x) \\
&+\lambda \frac{\gamma (x) }{EI(x) }\phi (
x) +\frac{\lambda ^2m(x) }{EI(x) }\phi
(x) =0,\quad 0<x<1,
\end{aligned} \\
\phi (0) =\phi '(0) =\phi ''(1) =0 \\
\phi '''(1) =\frac{1}{EI(1) }(\alpha +\beta \lambda ) \phi (1) .
\end{gathered}
\end{equation}
Hence we have that
$\eta \in \sigma (\mathbb{A}) \Leftrightarrow \lambda \in \sigma (A_{\gamma}) $.

\begin{theorem}\label{theo7}
Let operator $A_{\gamma }$ of the system \eqref{17}. Then the
eigenvalues of operator $A_{\gamma }$ are all simple except for
finitely many of them, and the generalized eigenfunctions of
operator $A_{\gamma }$ form a Riesz basis for the Hilbert state space
$H$.
\end{theorem}

\begin{proof}
From the previous subsection, we have shown that the boundary problem
\eqref{388} is strongly regular because the coefficients of
 $F_0^{\mathbb{K}_{k}}$ are nonzero in $\Delta (\rho ) $
(see Definition \ref{definition1}). Therefore the eigenvalues are 
separated and simple except for finitely many of them. 
Thus the first assertion is true. Next,
according to Theorem \ref{theo6}, the strongly regular boundary
conditions ensure that the generalized eigenfunction sequence
 $F_{n}=(f_{n},\eta _{n}f_{n}) $ of operator $\mathbb{A}$ forms a Riesz basis
for $\mathbb{H}$. Since $\mathcal{L}$ is bounded and invertible on 
$\mathbb{H}$, it follows that $\Psi _{n}=(\phi _{n},\lambda _{n}\phi _{n})
=\mathcal{L}F_{n}$ also forms a Riesz basis on $H$.
\end{proof}

We are now in a position to investigate the exponential stability of 
system \eqref{17}. Since the Riesz basis property implies the
spectrum-determined growth condition (see Curtain and Zwart \cite{4}) and 
\eqref{73} describes the asymptote of $\sigma (A_{\gamma }) $, for any small 
$\varepsilon >0$ there are only finitely
many eigenvalues of $A_{\gamma }$ in the following half-plane:
\begin{equation}
\Sigma :\operatorname{Re}\lambda
 >-\frac{1}{2h}\int_0^{1}\frac{\gamma (
x) }{m(x) }\Big(\frac{m(x) }{EI(
x) }\Big) ^{1/4}dx-\frac{2\beta }{hEI(1) }
\Big(\frac{m(1) }{EI(1) }\Big) ^{-3/4}+\varepsilon .
\end{equation}
The following are two stability results that describe how stability depend
on the damping function $\gamma $.

\section{Exponential stability}

Following the idea in \cite[Theorem 2.4]{7}, all the properties
of operator $A_{\gamma }$ found above, allow us to claim
that for the semigroup $e^{A_{\gamma }t}$ generated by
$A_{\gamma }$
the spectrum-determined growth condition holds:
\[
\omega (A_{\gamma }) =s(A_{\gamma
}) ,
\]
where
\[
\omega (A_{\gamma }) =\underset{t\to +\infty }{
\lim }\frac{1}{t}\| e^{A_{\gamma }t}\| _{H}
\]
is the growth order of $e^{A_{\gamma }t}$ and
\[
s(A_{\gamma }) =\sup \big\{ \operatorname{Re}\lambda :\lambda
\in \sigma (A_{\gamma }) \big\}
\]
is the spectral bound of $A_{\gamma }$.

The Theorem \ref{theo7} is one of the fundamental properties of the
evolutive system \eqref{1a}--\eqref{1d}. Many other important properties of
this system can be concluded from Theorem \ref{theo7}. The
exponential stability stated below is one of such important property.

\begin{theorem}\label{theo8}
If $\gamma $ is continuous and nonnegative, the system
\eqref{1a}--\eqref{1d} is exponential stable for any $\beta >0$ and
$\alpha \geq 0$. That is, there are nonnegative constants $M$, $\omega $
such that the energy
$E(t) =\frac{1}{2}\| (u,u_{t})^{T}\| _{H}^2$ 
of system \eqref{1a}--\eqref{1d}
satisfies
\[
E(t) \leq ME(0) e^{-\omega t},\quad \forall t\geq 0,
\]
for any initial condition $(u(x,0) ,u_{t}(x,0)) \in H$.
\end{theorem}

\begin{proof}
We have $\gamma (x) \geq 0$, and for any $(f,g) \in D(A_{\gamma }) $,
\begin{align*}
&\langle A_{\gamma }(f,g) ,(f,g)\rangle \\
&=\big\langle (g(x) ,-\frac{1}{m(
x) }((EI(x) f''(x) ) ''+\gamma (x) g(x) )) ,(f,\text{ }g) \big\rangle \\
&=\int_0^{1}[ EI(x) g''(x)
\overline{f''(x) }-(EI(x)
f''(x) ) ''\overline{g(x) }-\gamma (x) | g(x) |^2] dx 
+\alpha g(1) \overline{f(1) } \\
&= \int_0^{1}EI(x) [ g''(x)\overline{f''(x) }-f''(
x) \overline{g(x) }] dx
+ \alpha \big(g(1) \overline{f(1) }-f(
1) \overline{g(1) }\Big) \\
&\quad -\beta | g(1) | ^2-\int_0^{1}\gamma (x) |g(x) | ^2dx;
\end{align*}
further
\[
\operatorname{Re}\langle A_{\gamma }(f,g) ,(f,g)\rangle
=-\beta | g(1) |^2-\int_0^{1}\gamma (x) | g(x)| ^2dx\leq 0.
\]
Thus $A_{\gamma }$ is dissipative and $e^{A_{\gamma }t}$
is a contraction semigroup on $H$. Moreover, the spectrum of $\mathit{
A}_{\gamma \text{ }}$ has an asymptote
\[
\operatorname{Re}\lambda \sim-\frac{1}{2h}\int_0^{1}\frac{\gamma (
x) }{m(x) }(\frac{m(x)}{EI(x) }) ^{1/4}dx
-\frac{2\beta }{hEI(1) }(\frac{m(1) }{EI(1) }) ^{-3/4}.
\]

If we can show that there is no eigenvalue on the imaginary axis,
then the exponential stability holds. Let $\lambda =ir$ with $r\in
\mathbb{R}^{\ast }$ be an eigenvalue of operator 
$A_{\gamma}$ on the imaginary axis and $\Psi =(\phi ,\psi ) ^{T}$
be the corresponding eigenfunction, then $\psi =\lambda \phi $. We have
\begin{gather*}
\operatorname{Re}(\langle A_{\gamma }\Psi ,\Psi \rangle _{
H})=-\beta | \psi (1) |
^2-\int_0^{1}\gamma (x) | \psi (x) | ^2dx, \\
0=\| \Psi \| _{H}^2\operatorname{Re}(\lambda) 
=\operatorname{Re}(\langle A_{\gamma }\Psi ,\Psi \rangle_{H})
=-\beta | \psi (1) |^2-\int_0^{1}\gamma (x) | \psi (x)| ^2dx,
\end{gather*}
since $\beta >0$, $\gamma (x) \geq 0$ and $\psi (x) $
are continuous, we obtain
\begin{equation}
\psi (1) =0\quad\text{and}\quad 
\gamma (x) | \psi (x) | ^2=0,\quad \forall x\in [0,1]. \label{009}
\end{equation}
Then $\phi (1) =0$ because $\psi =\lambda \phi$.
 We have the following equation satisfied by $\phi $,
\begin{equation}
\begin{gathered}
\lambda ^2m(x)\phi (x) +(EI(x) \phi ''(x) ) ''+\lambda \gamma (
x) \phi (x) =0,\quad 0<x<1, \\
\phi (0) =\phi '(0) =\phi ''(1) =\phi '''(1) =\phi (1) =0
\end{gathered} \label{010}
\end{equation}
The proof will be complete if we can show that there is only zero solution to 
\ref{010}.  For this aim we follow a method used in
\cite[p.1917]{7}.  First, we claim that there is at least one zero
of $\phi $ in $(0,1)$. In fact, by $\phi (0) =\phi (1) =0$, it follows from 
Rolle's theorem that there is a $\xi _1\in (0,1)$ such that $\phi '(\xi _1) =0$,
 which, together with $\phi '(0) =0$, claims that $(EI\phi'') (\xi _2) =0$ 
for some $\xi_2\in (0,\xi _1) $, and so $(EI\phi '') '(\xi _3) =0$ for some
$\xi_3\in (\xi _2,1) $ by condition $(EI\phi'') (1) =0$. Hence there is a 
$\xi_4\in (\xi _3,1) $ such that $(EI\phi '') ''(\xi _4) =0$ by the
condition $(EI\phi '') '(1) =0$.
 However, $\lambda ^2m(\xi _4)\phi (\xi _4) +(EI(\xi _4) \phi '') ''(\xi _4) 
+\lambda \gamma (\xi _4) \phi (\xi _4) =0$. We
have
\[
\lambda ^2m(\xi _4)\phi (\xi _4) +\lambda \gamma (
\xi _4) \phi (\xi _4) =0
\]
because $(EI(\xi _4) \phi '') ''(\xi _4) =0$. Then we obtain
$\lambda m(\xi _4)\lambda \phi (\xi _4) +\gamma
(\xi _4) \lambda \phi (\xi _4) =0$.
Since $\psi=\lambda\varphi$ we have
\[
\lambda m(\xi _4)\psi (\xi _4) +\gamma (\xi_4) \psi (\xi _4) =0.
\]
Multiplying the conjugate of $\psi(\xi_4)$ on both sides of the
above equation we obtain
\[
\lambda m(\xi _4)\psi (\xi _4) \overline{\psi (\xi _4) }
+\gamma (\xi _4) \psi (\xi_4) \overline{\psi (\xi _4) }=0.
\]
Then we obtain
\[
\lambda m(\xi _4)| \psi (\xi _4)| ^2+\gamma (\xi _4) | \psi(\xi _4) | ^2=0.
\]
Using \eqref{009} we have $\lambda m(\xi _4)| \psi (\xi _4) | ^2=0$, so
$\psi (\xi _4)=0$. Finally we conclude that $\phi (\xi _4) =0$ since
$\psi =\lambda \phi $ and $\lambda$ is nonzero.   Next, we
show that if there are $n$ different zeros of $\phi $ in $(0,1) $,
then there are at least $n+1$ number of different zeros of $
\phi $ in $(0,1) $.
Indeed, suppose that
\[
0<\xi _1<\xi _2<\dots <\xi _{n}<1,\quad
\phi (\xi _i) =0,\quad i=1,2,3,\dots ,n.
\]
Since $\phi (0) =\phi (1) =0$, it follows from
Rolle's theorem that there are $\eta _i$,  $i=1,2,3,\dots ,n+1$,
\[
0<\eta _1<\xi _1<\eta _2<\xi _2<\eta _3<\xi _3<\dots <\xi _{n}<\eta_{n+1}<1,
\]
such that $\phi '(\eta _i) =0$. Noting that
$\phi'(0) =0$, there are $\alpha _i$,  $i=1,2,3,\dots ,n+1$,
\[
0<\alpha _1<\eta _1<\alpha _2<\eta _2<\alpha _3<\eta
_3<\dots <\alpha _{n+1}<\eta _{n+1}<1,
\]
such that $(EI\phi '') (\alpha _i)=0$. Since $(EI\phi '') (1) =0$,
using Rolle's theorem, we have $\beta _i$,  $i=1,2,3,\dots ,n+1$,
\[
0<\alpha _1<\beta _1<\alpha _2<\beta _2<\alpha _3<\beta
_3<\dots <\alpha _{n+1}<\beta _{n+1}<1,
\]
such that $(EI\phi '') '(\beta_i) =0$. Finally, by the condition
$(EI\phi '') '(1) =0$, we have $\theta _i$,  $i=1,2,3,\dots ,n+1$,
\[
0<\beta _1<\theta _1<\beta _2<\theta _2<\beta _3<\theta
_3<\dots <\beta _{n+1}<\theta _{n+1}<1,
\]
such that $(EI\phi '') ''(\theta _i) =0$. Therefore
\[
\phi (\theta _i) =0,\quad i=1,2,3,\dots ,n+1
\]
Third, by mathematical induction, there is an infinite number of different
zeros $\{ x_i\} _1^{\infty }$ of $\phi $ in $(0,1) $. Let $x_0\in [ 0,1] $
be an accumulation point of $\{ x_i\} _1^{\infty }$. It is obvious that
\[
\phi ^{(i) }(x_0) =0,\quad i=0,1,2,3.
\]
Note that $\phi $ satisfies the linear differential equation
\eqref{010}.  Therefore, $\phi =0$ by uniqueness of the
solution of linear ordinary differential equations.  However
$\Psi =(\phi ,\psi ) ^{T}=(\phi ,\lambda \phi) ^{T}=0$ contradicts
$\Psi $ being an eigenfunction and so
there is no eigenvalue on the imaginary axis and we obtain
 $\operatorname{Re}(\lambda ) <0$.
From Theorem \ref{theo7} and the spectrum-determined growth
condition, the system is exponentially stable.
\end{proof}

Now we are ready to consider the case that $\gamma (x) $
is continuous and indefinite in $[ 0,1] $. We follow an
idea due to Wang \cite{20}.  Let $\gamma(x) =\gamma ^{+}(x) -\gamma ^{-}(x) $ 
for all $x\in [ 0,1] $ with
\begin{gather*}
\gamma ^{+}(x)  
=\max \{ \gamma (x),0\} 
=\begin{cases}
\gamma (x) & \text{if }\gamma (x) >0 \\
0&\text{otherwise},
\end{cases}
 \\
\gamma ^{-}(x)  =\max \{ -\gamma (x),0\} 
=\begin{cases}
-\gamma (x) , &\text{if }\gamma (x) <0 \\
0&\text{otherwise},
\end{cases}
\end{gather*}
and let
\[
A_{\gamma ^{+}}(f,g) =\Big(g(x) ,-\frac{1}{m(x) }((EI(x) f''(
x) ) ''+\gamma ^{+}(x) g(
x) )\Big) ^{T},
\]
for all $(f,g) \in D(A _{\gamma ^{+}}) =D(A_{\gamma })$,
and
\[
\Gamma _{-}(f,g) =(0,\frac{\gamma ^{-}(x) }{
m(x) }g(x) ) ^{T},\quad  \forall (f,g)\in H.
\]
Then $A_{\gamma }$ can be written as
$A_{\gamma }=A_{\gamma ^{+}}+\Gamma _{-}$.

\begin{theorem}
Let $s(A_{\gamma ^{+}}) =\sup \{ \operatorname{Re}\lambda
:\lambda \in \sigma (A_{\gamma ^{+}}) \} $. If
\[
\underset{x\in [ 0,1] }{\max }\big\{ \frac{\gamma ^{-}(
x) }{m(x) }\big\} <| s(A_{\gamma ^{+}}) | ,
\]
then system \eqref{17} is exponentially stable.
\end{theorem}

\begin{proof}
It is easy to verify that $\Gamma _{-}$ is self-adjoint operator and
\begin{equation}
\| \Gamma _{-}\| =\max_{x\in [ 0,1] }
\big\{ \frac{\gamma ^{-}(x) }{m(x) }\big\} .
\end{equation}
By Theorem \ref{theo8} and definition of operator $A_{\gamma ^{+}}$, 
$e^{A_{\gamma ^{+}}t}$ is a contraction
semigroup and $s(A_{\gamma ^{+}}) <0$. Applying
the perturbation theory of linear operators semigroup (see  Pazy
\cite[Theorem 1.1 page 76]{13}), we have $\lambda \in \rho
(A_{\gamma }) $ whenever $\operatorname{Re}\lambda >s(A_{\gamma ^{+}}) 
+\| \Gamma _{-}\| $. Again, Theorem \ref{theo7} gives
\[
\omega (A_{\gamma }) =s(A_{\gamma
}) \leq s(A_{\gamma ^{+}}) +\| \Gamma_{-}\| <0,
\]
where $\omega (A_{\gamma }) $ denotes the growth bound
of semigroup $e^{A_{\gamma }t}$. Therefore, system  \eqref{17} is
exponentially stable.
\end{proof}

\subsection*{Acknowledgments}
We want to thank the anonymous referee for the throughout reading of 
the manuscript and several suggestions that help us to improve the 
presentation of this article.

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