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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2015 (2015), No. 174, pp. 1--17.\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-2015/174\hfil A mathematical model for suspension bridges]
{A mathematical model for suspension bridges with energy dependent
 boundary conditions}

\author[Y. Wang \hfil EJDE-2015/174\hfilneg]
{Yongda Wang}

\address{Yongda Wang \newline
Dipartimento di Matematica, Politecnico di Milano,
Piazza Leonardo da Vinci 32, 20133 Milano, Italy}
\email{yongda.wang@polimi.it}

\thanks{Submitted November 25, 2014. Published June 23, 2015.}
\subjclass[2010]{35C10, 35L35, 74B20, 74K20}
\keywords{Higher order equation; suspension bridge;
\hfill\break\indent energy dependent boundary conditions}

\begin{abstract}
 We suggest a new mathematical model for dynamical suspension bridge
 with energy dependent boundary conditions. The roadway of the bridge is
 viewed as a long-narrow thin rectangular plate. After reducing the evolution
 problem corresponding to the model to a variational problem, we show that
 the original evolution problem admits a unique solution. Moreover,
 the unique solution is explicitly represented.
\end{abstract}

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

\section{Introduction}

 In a recent paper, Ferrero-Gazzola \cite{f1} suggested a
rectangular plate model for the description of dynamical suspension
bridges. The two short edges of the plate are hinged and the
remaining two edges are assumed to be free with no physical
constraints. Under these boundary conditions, the  problem they
derived has a global variational structure. Then according to
variational principles, they showed that there exists a unique
solution to the evolution problem with suitable initial data.
Subsequently, the same problem but not coercive was investigated by
Wang in \cite{w1}, where global existence and finite time blow-up of
solutions for different initial data are obtained.

Several decades ago, Pugsley in his monograph \cite{p1} analyzed the
theory on the oscillations of suspension bridges under lateral
winds. He stated that
\begin{quote}
in a complete bridge the wind forces are resisted partly by the
elastic flexure of the deck structure in a horizontal plane and
partly by gravity action induced by the cables.
\end{quote}
In order to understand the nature and origin of various oscillations
like purely vertical type and purely torsional type, Pugsley gave an
explanation by regarding the deck as a flat plate and by considering
the action of a lateral airstream upon a section of its length, see
\cite[pp. 123-125]{p1}. As the lateral wind blows harder in speed, the
magnitude and frequency of the fluctuating air force on the deck
will increase, and the bridge deck will gradually start to oscillate
in vertical type. The behavior is more sensitive to the wind speed,
and oscillations tend to change at some critical wind speed, which
is usually called in literature \emph{flutter speed}.

About 10 years later than Pugsley \cite{p1} and Scanlan \cite{s1} pointed out that
\begin{quote}
With such bridges natural structural vibration modes tend no longer
to be simply into ``bending'' and ``torsion'' but to be fully
three-dimensional in character, with components of vertical,
torsional and lateral sway motion $\ldots$
\end{quote}
The term ``lateral sway motion'' implies that the unsteady lateral
force acting on the plate has a crucial effect on the oscillation of
the bridge. For this reason, Gazzola \cite{g1} recently suggested to study
models with dynamical boundary conditions in order to display the
oscillations appearing in the actual suspension bridges. And indeed,
it appears necessary to combine variational methods arising from
energy balance with aerodynamics effects due to the lateral wind.

For a different model of suspension bridges,
Bleich-McCullough-Rosecrans-Vin\-cent \cite{b1} referred to the energies
involved in the structure. They made a careful quantitative analysis
of the energies, such as the kinetic energy, the potential energy
and the elastic energy. Concerning the qualitative analysis on the
energies, Gazzola \cite{g1} claims that the \emph{flutter speed} should be
seen as a \emph{critical energy threshold} which, if exceeded, gives
rise to uncontrolled phenomenon. Moreover, Arioli-Gazzola \cite{a1}
recently pointed out that
\begin{quote}
if the total energy increases over the critical energy threshold,
then tiny torsional oscillations may suddenly become, without
intermediate stages, wider oscillations.
\end{quote}
This implies that ``the total energy of the system'' and ``the
critical energy threshold'' have a direct influence on the
oscillations of the plate. Therefore, it is necessary to assume that
the total energy of the system and the critical energy threshold
appear explicitly in the model.

Motivated by these remarks, in this paper we suggest a new
mathematical model for dynamical suspension bridges. In this model,
the differential equation is deduced by applying a variational
principle to the energy functional $\mathscr{A}(u)$ given in \cite{f1}.
However, since we mainly focus on the effects of the unsteady
lateral wind acting on the edges of the bridge, the boundary
conditions we established are dynamical and depend on the energy of
the system and on the critical energy threshold. This paper could be
considered as a first step in order to obtain more interesting and
challenging results.

This article is organized as follows. In Section 2 a new mathematical
model is suggested for dynamical suspension bridges. The boundary
conditions we established depends on the energy and then we derive
the corresponding evolution problem to be solved. An auxiliary
problem is introduced in Section 3. The auxiliary problem, which
admits a unique solution, plays a crucial role in solving our
original problem. In Section 4, we first transfer the original
problem to the type of the auxiliary problem. Then we show that
there exists a unique solution to the original problem, see 
Theorem \ref{thm1}. Moreover, the unique solution is explicitly represented, 
see Theorem \ref{thm2}. Finally, we list several possible future developments
about the model in the last section.


\section{ The new mathematical model}


Assume that an open rectangular domain
$\Omega=(0,\pi)\times(-l,l)\subset\mathbb{R}^2$ represents the
roadway of a suspension bridge, where $\pi$ is the length of the
roadway and $2l$ is its width. A realistic assumption is that $2l\ll
\pi$. We recall the energy functional $\mathscr{A}(u)$ given in \cite{f1}
(see also \cite{g2,k1}):
\[
\mathscr{A}(u)=\int_0^{\infty}{\int_{\Omega}
{\Big(\frac{1}{2}u_t^2-\Big(\frac{1}{2}(\Delta
u)^2+(1-\sigma)(u_{xy}^2-u_{xx}u_{yy})+H-\varphi
u\Big)\Big)}\,dx\,dy}\,dt,
\]
where $u=u(x,y,t)$ denotes the displacement of the plate in the
vertical direction at the point $(x,y)$ and at time $t>0$,
$\sigma\in (0, 1/2)$ is the Poisson ratio that depends on the
material of the roadway of the bridges, $\int_{\Omega}{H}\,dx\,dy$ is a
potential energy due to the restoring force $h$ produced by the
hangers of the suspension bridge, $\varphi=\varphi(x,y,t)$
represents an external source, such as the wind and the weight of
the plate.

By applying variational principles to the energy functional
$\mathscr{A}(u)$ in the function space $C_c^{\infty}(\Omega)$, we
obtain the following fourth order equation by adding a damping term
$\mu u_t(\mu\geq 0)$ representing the positive structural damping of
the structure, including internal frictions
\begin{equation} \label{e2.1}
u_{tt}+ \Delta ^2 u + \mu u_t+h = \varphi,\quad \text{in }\Omega\times(0,+\infty).
\end{equation}

Since the plate is very narrow (compared to its length), we may
assume that the restoring elastic force due to the hangers acts on
every point of the plate and has the linear form $h=ku$ with an
elasticity constant $k>0$. Then we have an initial value problem
\begin{equation}  \label{e2.2}
\begin{gathered}
   u_{tt}+ \Delta ^2 u + \mu u_t+ku  = \varphi,  \quad (x,y) \in \Omega, t>0,   \\
   u(x,y,0) =u_0, \quad (x,y) \in \Omega , \\
   u_t(x,y,0) = u_1,  \quad (x,y) \in \Omega,
\end{gathered}
\end{equation}
which is to be complemented with some suitable boundary conditions.
Here, $u_0=u_0(x,y)$ is the initial position of the plate,
$u_1=u_1(x,y)$ is the initial vertical velocity of the plate.

As well-known, the external force $\varphi$ acting on the bridge
inserts an energy into the structure. We denote it by
\[
\mathcal {E}(t)=\int_\Omega{\varphi(x,y,t)^2}\,dx\,dy.
\]
Assume that $\overline{E}_{\mu}>0$ is given and represents the
critical energy threshold (see Arioli-Gazzola \cite{a1}) above which the
bridge displays self-excited oscillations and it increasingly
depends on the damping parameter $\mu$: for instance,
$\overline{E}_{\mu}=\overline{E}_0+c\mu$ for some $c>0$ and
$\overline{E}_0>0$ being the threshold of the undamped problem.

Now, we turn to set up the boundary conditions to be associated to
the initial value problem \eqref{e2.2}. Due to several physical constraints
being present on the sides of the plate, we seek the boundary
conditions which describe the physical situation appearing in the
actual suspension bridges. On the two short edges $x=0$ and $x=\pi$,
which are connected with the ground, we assume that they are hinged
and then
\begin{equation} \label{e2.3}
u(x,y,t) =u_{xx}(x,y,t) =0,\quad \text{for }(x,y) \in
\{0,\pi\}\times(-l,l),\; t>0,
\end{equation}
which is the same as any other model we met. While on the other two
sides $y=\pm l$, due to the continuous impact from the external
forces occurring on them, it is quite delicate to choose a kind of
``stationary'' boundary conditions. Hence, in the present paper, we
try to provide several suitable dynamical boundary conditions which
describes the situation more closely.

At first, since the plate is very narrow, it is natural to assume
that the cross section of the plate tends to remain straight. But
due to the appearance of the torsional oscillation, the cross
section cannot always be in a horizontal position. Therefore, we
assume that these two boundaries satisfy
\begin{equation} \label{e2.4}
\begin{gathered}
   u_y(x,-l,t)-u_y(x,l,t)=0, \quad x\in (0,\pi), t>0, \\
   u_y(x,-l,t)+u_y(x,l,t)=\alpha, \quad x\in (0,\pi), \; t>0,
\end{gathered}
\end{equation}
where $\alpha=\alpha(x,t)$ is a real function, which depends on
$\varphi$ and $u_0$ for $y=\pm l$. Furthermore, for any cross
section of the bridge, the vertical displacements at the two
endpoints also depend on the initial position and the external
forces acting on these two points. Thus, it is reasonable to assume
that the sum of the two displacements fulfills
\begin{equation} \label{e2.5}
   u(x,-l,t)+u(x,l,t)=\beta, \quad \text{ for } x\in (0,\pi),t>0,
\end{equation}
where $\beta=\beta(x,t)$ is a real function, which depends on
$\overline{\varphi}(x,t)$ and $\overline{u}_0(x)$ with
\[
\overline{\varphi}(x,t)=\frac{\varphi(x,l,t)+\varphi(x,-l,t)}{2},\quad
 \overline{u}_0(x)=\frac{u_0(x,l)+u_0(x,-l)}{2}.
\]
Moreover, the vibration of the cross section will switch to some
different kind of oscillations, such as the torsional oscillation,
when the inserted energy $\mathcal {E}(t)$ exceeds
$\overline{E}_{\mu}$. Therefore, as long as
$\mathcal {E}(t) \leq \overline{E}_{\mu}$, we assume that there is
only vertical vibration
appearing in the motion, then the difference between the two
displacements should be zero, that is,
\begin{equation} \label{e2.6}
u(x,-l,t)-u(x,l,t)=0, \quad\text{for } x\in (0,\pi).
\end{equation}
While if $\mathcal {E}(t) > \overline{E}_{\mu}$, the torsional
oscillation begins to arise and its amplitude increases as the
energy $\mathcal {E}(t)\to \infty$, which means that the
difference between the two displacements is related to the energy
$\mathcal {E}(t)$. Concretely, if $\mathcal {E}(t)\downarrow
\overline{E}_{\mu}$, then the motion tends to vertical-type so that
\[
u(x,-l,t)-u(x,l,t)\to 0,
\]
while if $\mathcal {E}(t)\to \infty$, then the difference
also increases. But it cannot increase to infinity because the
bridge will collapse earlier. Therefore, as long as
$\mathcal {E}(t) > \overline{E}_{\mu}$, the boundaries are assumed to satisfy the
 equation
\begin{equation} \label{e2.7}
\frac{d}{dt}\Big\{[u(x,-l,t)-u(x,l,t)]\frac{\mathcal
{E}(0)-\overline{E}_\mu}{\mathcal
{E}(t)-\overline{E}_\mu}\exp\Big(\frac{\mathcal
{E}(0)}{\overline{E}_\mu}-\frac{\mathcal
{E}(t)}{\overline{E}_\mu}\Big)\Big\}=\gamma, \quad\text{for }
x\in (0,\pi),
\end{equation}
here we assume $\mathcal {E}(0)>\overline{E}_{\mu}$ and
$\gamma=\gamma(x,t)$ is a real function satisfying
\[
\gamma \text{ has the same sign as } (u_0(x,-l)-u_0(x,l)).
\]

In this case, the conditions \eqref{e2.6}-\eqref{e2.7} can be combined into
\begin{equation} \label{e2.8}
   u_t(x,-l,t)-u_t(x,l,t)-\eta(t)(u(x,-l,t)-u(x,l,t))=\theta(t)\gamma, \quad\text{ for } x\in
   (0,\pi), t>0,
\end{equation}
where
\begin{gather} \label{e2.9}
\theta(t)=\frac{(\mathcal {E}(t)-\overline{E}_\mu)^+}{\mathcal
{E}(0)-\overline{E}_\mu}\exp\Big(\frac{\mathcal
{E}(t)}{\overline{E}_\mu}-\frac{\mathcal
{E}(0)}{\overline{E}_\mu}\Big), \\
 \label{e2.10}
\eta(t)=\mathcal
{E}'(t)E(t)\Big(\frac{1}{\overline{E}_\mu}+\frac{1}{(\mathcal
{E}(t)-\overline{E}_\mu)^+}\Big)
\end{gather}
with $(\mathcal {E}(t)-\overline{E}_\mu)^+
=\max\{\mathcal {E}(t)-\overline{E}_\mu,0\}$,
 $\mathcal {E}'(t)=\frac{d}{dt}\mathcal{E}(t)$ and
\begin{equation} \label{e2.11}
E(t)=\begin{cases}
   -1, & \mathcal {E}(t)\leq \overline{E}_{\mu},\\
   1, & \mathcal {E}(t)> \overline{E}_{\mu}.
\end{cases}
\end{equation}

Summarizing, the boundary conditions for a rectangular plate
$\Omega$ modeling dynamical suspension bridges are \eqref{e2.3}-\eqref{e2.5} and
\eqref{e2.8}, which together with \eqref{e2.2} yield an evolution problem
\begin{equation} \label{e2.12}
\begin{gathered}
   u_{tt}+ \Delta ^2 u + \mu u_t+ku  = \varphi,  \quad (x,y) \in \Omega, t>0,   \\
   u(x,y,0) =u_0, \quad (x,y) \in \Omega , \\
   u_t(x,y,0) = u_1,  \quad (x,y) \in \Omega ,
\end{gathered}
\end{equation}
with the boundary conditions: for every $t>0 $,
\begin{equation} \label{e2.13}
\begin{gathered}
   u(x,y,t) =u_{xx}(x,y,t) =0, \quad (x,y) \in \{0,\pi\}\times(-l,l), \\
   u_y(x,-l,t)-u_y(x,l,t)=0, \quad x\in (0,\pi), \\
   u_y(x,-l,t)+u_y(x,l,t)=\alpha, \quad x\in (0,\pi), \\
   u(x,-l,t)+u(x,l,t)=\beta, \quad x\in (0,\pi),\\
   u_t(x,-l,t)-u_t(x,l,t)-\eta(t)(u(x,-l,t)-u(x,l,t))=\theta(t)\gamma, \; x\in (0,\pi).
\end{gathered}
\end{equation}
The dynamical boundary conditions \eqref{e2.13} are energy dependent
boundary conditions since the energy $\mathcal {E}(t)$ appears
explicitly.

\section{An Auxiliary Problem}

In this section we give an auxiliary problem which plays a crucial
role in solving the original problem \eqref{e2.12}-\eqref{e2.13}. We first
introduce a subspace of $H^2(\Omega)$ denoted by
\[
V:=\{u\in H^2(\Omega):u=0\text{ on }\partial \Omega \text{ and
}u_y=0\text{ on }(0,\pi)\times\{-l,l\}\}.
\]
Clearly, $H_0^2(\Omega)\subset V\subset H_0^1(\Omega)\cap
H^2(\Omega)$. Hence, one may define a scalar product on the space
$V$ by
\[
(u,v)_V  = \int_\Omega  {\Delta u\Delta v  }\,dx\,dy, \quad
\text{for any }u,v \in V,
\]
which induces the norm
\[
\|u\|_V  = \Big(\int_\Omega  {|\Delta u|^2 }\,dx\,dy\Big)^{1/2},
\quad\text{for all }u\in V.
\]
Now we consider the  nonhomogeneous linear problem
\begin{equation} \label{e3.1}
\begin{gathered}
   v_{tt}+ \Delta ^2 v + \mu v_t+kv  = f,  \quad (x,y) \in \Omega, t>0,   \\
   v(x,y,t) =v_{xx}(x,y,t) =0, \quad (x,y) \in \{0,\pi\}\times(-l,l), t>0,\\
   v(x,y,t)=v_y(x,y,t)=0, \quad (x,y)\in (0,\pi)\times\{-l,l\}, t>0, \\
   v(x,y,0) =v_0, \quad (x,y) \in \Omega , \\
   v_t(x,y,0) = v_1, \quad (x,y) \in \Omega ,
\end{gathered}
\end{equation}
where $f=f(x,y,t)\in C^0([0,\infty);L^2(\Omega))$,
$v_0=v_0(x,y)\in V$ and $v_1=v_1(x,y)\in L^2(\Omega)$.
\smallskip

\noindent\textbf{Claim.} Problem \eqref{e3.1} admits
a unique weak solution 
\[
v \in C^0([0,\infty);V)\cap
C^1([0,\infty);L^2(\Omega)). 
\]
\smallskip

Actually, the variational problem \eqref{e3.1} is very analogous to 
\cite[problem (22)]{f1} if we take the nonlinear term $h=kv$. The only
difference lies in the boundary conditions on
$(0,\pi)\times\{-l,l\}$, but this has no influence when we solve it
by following the procedure in \cite{f1}. Hence, we only list the key
steps, for details see \cite[Section 8]{f1}. 
\smallskip

\noindent\textbf{Step 1.} 
Consider the approximated problems for $m\geq 1$
\begin{equation} \label{e3.2}
\begin{gathered}
   v''_m+ Lv_m + \mu v'_m+P_m(kv_m)  = P_m(f),  \quad t\in [0,\tau_m)   \\
   v(0) =v_0^m,\quad v'(0) = v_1^m 
\end{gathered}
\end{equation}
where $L$ is defined by $\langle Lu,v\rangle:=(u,v)_V$ for any
$u,v\in V$, $P_m:V\to W_m$ is the orthogonal projection onto $W_m$,
which is spanned by the eigenfunctions $\{w_m\}_{m\geq 1}$ of the
problem
\begin{gather*}
   \Delta ^2 w   = \lambda w,  \quad (x,y) \in \Omega,    \\
   w(x,y) =w_{xx}(x,y) =0, \quad (x,y) \in \{0,\pi\}\times(-l,l),\\
   w(x,y)=w_y(x,y)=0, \quad (x,y)\in (0,\pi)\times\{-l,l\}.
\end{gather*}

By Galerkin-type procedure, we obtain that problem \eqref{e3.2} has a
unique local solution $v_m\in C^2([0,\tau_m);V)$, where $[0,\tau_m)$
is the maximal interval of the continuation of $v_m$.
\smallskip

\noindent\textbf{Step 2.} The solution sequence $\{v_m\}$
is uniform bounded.

Indeed, testing \eqref{e3.2} with $v'_m$ and integrating over $(0,t)$, we
have by several estimates
\[
\|v_m\|_V^2+\|v'_m\|_{L^2}^2\leq C\quad \text{for any } t\in [0,\tau_m) 
\text{ and } m\geq 1,
\]
where $C$ is independent of $m$ and $t$. Hence, the solution $v_m$
is globally defined in $[0,\infty)$ and the sequence $\{v_m\}$ is
bounded in $C^0([0,\infty);V)\cap C^1([0,\infty);L^2(\Omega))$.
\smallskip

\noindent\textbf{Step 3.} $\{v_m\}$ admits a strongly
convergent subsequence in the set  $C^0([0,\infty);V)\cap
C^1([0,\infty);L^2(\Omega))$.

By Ascoli-Arzel\`{a} Theorem, we deduce that, up to subsequences,
there exists a function $v\in C^0([0,\infty);L^2(\Omega))$ such that
$v_m\to v$ strongly in $C^0([0,\infty);L^2(\Omega))$.

On the other hand, the sequence $\{v_m\}$ is a Cauchy sequence in
the space $C^0([0,\infty);V)\cap C^1([0,\infty);L^2(\Omega))$.
Hence, up to subsequences,
\[
v_m\to v\quad \text{in }C^0([0,\infty);V)\cap C^1([0,\infty);L^2(\Omega)) 
\quad \text{as } m\to \infty.
\]
\smallskip

\noindent\textbf{Step 4.}  Take the limit in \eqref{e3.2} and then
we prove the existence of solution to \eqref{e3.1}.
\smallskip

\noindent\textbf{Step 5.} The solution to \eqref{e3.1} is unique.

Assume that $v_1,v_2$ are two solutions of \eqref{e3.1}, denote
$v=v_1-v_2$. Then using $v'$ as a test function, we obtain after
integration over $(0,t)$
\[
\|v'\|_{L^2}^2+\|v\|_V^2=-2\mu \int_0^t{\|v'(s)\|_{L^2}^2}ds\leq 0,
\]
from which it immediately follows that $v=0$. Therefore, the problem
\eqref{e3.1} admits a unique weak solution $v\in C^0([0,\infty);V)\cap
C^1([0,\infty);L^2(\Omega))$.


\section{Main Results}

Due to the dynamical boundary conditions \eqref{e2.13}, it is not easy to
solve the original problem \eqref{e2.12}-\eqref{e2.13} directly. Hence, we first
transfer it to a simpler case, which uses the auxiliary problem.
Recalling the boundary conditions \eqref{e2.4}, we have
\[
u_y(x,l,t)=u_y(x,-l,t)=\alpha/2,\quad \text{ for any }x\in (0,\pi),~t>0.\]
Moreover, let $\beta$ be a $C^1$ function in $t$, then we obtain by
\eqref{e2.5} and \eqref{e2.8} for any $x\in (0,\pi)$
\begin{gather*}
u_t(x,l,t)-\eta(t)u(x,l,t)=g_1(x,t),\quad t>0,\\
u(x,l,0)=u_0(x,l),
\end{gather*}
and
\begin{gather*}
u_t(x,-l,t)-\eta(t)u(x,-l,t)=g_2(x,t),\quad t>0,\\
u(x,-l,0)=u_0(x,-l),
 \end{gather*}
 where
$g_1(x,t)=\frac{1}{2}(\beta_t(x,t)-\eta(t)\beta(x,t))
-\frac{1}{2}\theta(t)\gamma(x,t)$
and
$g_2(x,t)=\frac{1}{2}(\beta_t(x,t)-\eta(t)\beta(x,t))
+\frac{1}{2}\theta(t)\gamma(x,t)$.
For any fixed $x\in (0,\pi)$, they are first order ordinary
differential problems with respect to $t$, and then one gets the
explicit representation of the values of $u$ on
$(0,\pi)\times\{-l,l\}$, which we denote by $(h_1,h_2)=(u(x,l,t),
u(x,-l,t))$. Then the problem \eqref{e2.12}-\eqref{e2.13} reduces 
to an evolution linear problem
\begin{equation} \label{e4.1}
\begin{gathered}
   u_{tt}+ \Delta ^2 u + \mu u_t+ku  = \varphi,  \quad (x,y) \in \Omega,\; t>0,   \\
   u(x,y,0) =u_0, \quad (x,y) \in \Omega , \\
   u_t(x,y,0) = u_1,  \quad (x,y) \in \Omega
\end{gathered}
\end{equation}
with nonhomogeneous boundary conditions
\begin{equation} \label{e4.2}
\begin{gathered}
   u(x,y,t) =u_{xx}(x,y,t) =0, \quad (x,y) \in \{0,\pi\}\times(-l,l),\; t>0,\\
   u(x,y,t)=h_1, \quad u_y(x,y,t)=\alpha/2, \quad (x,y)\in (0,\pi)\times\{l\},\;
   t>0,\\
   u(x,y,t)=h_2,\quad u_y(x,y,t)=\alpha/2, \quad (x,y)\in (0,\pi)\times\{-l\},\;
   t>0.
\end{gathered}
\end{equation}

Now we want to reduce the boundary conditions \eqref{e4.2} to the
homogeneous case. To this end, it is necessary to construct a
suitable inverse trace operator. Note that $\Omega\subset
\mathbb{R}^2$ is a rectangular domain. Let us define the space
\[
H_{*}^4(\Omega):=\{u\in
H^4(\Omega):u=u_{xx}=0\text{ on }\{0,\pi\}\times(-l,l)\}
\] 
and a continuous map
\begin{align*}
T:H_{*}^4(\Omega) &\to \prod_{i=1}^2 \left(H^{7/2}(\Sigma_i)\times
H^{5/2}(\Sigma_i)\times H^{3/2}(\Sigma_i)\times
H^{1/2}(\Sigma_i)\right)\\
\phi &\mapsto \prod_{i=1}^2
\left(\phi_{|\Sigma_i},(\phi_y)_{|\Sigma_i},(\phi_{yy})_{|\Sigma_i},(\phi_{yyy})_{|\Sigma_i}
\right)\end{align*}
where $\Sigma_1:=(0,\pi)\times \{l\}$,
$\Sigma_2:=(0,\pi)\times \{-l\}$. For more details, see 
\cite[Section 2.5]{n1} or \cite[Chapter 1]{g3}.

Denote the range of $T$ by $R(T):=T(H_{*}^4(\Omega))$ and define the
following norm on $R(T)$,
\[
\|v\|_{R(T)}:=\inf\{\|w\|_{H^4(\Omega)}: w\in H_{*}^4(\Omega), T(w)=v\}\quad 
\text{for any }v\in R(T),
\]
then $R(T)$ is a Banach space with the norm
$\|\cdot\|_{R(T)}$. Therefore, the restriction of the previous map
$T$ to $(\ker (T))^{\perp}$, i.e, 
$T_{|(\ker (T))^{\perp}}: (\ker (T))^{\perp}\to R(T)$ is an isometric isomorphism.

Since $R(T)\subset \prod_{i=1}^2 \left(H^{7/2}(\Sigma_i)\times
H^{5/2}(\Sigma_i)\times H^{3/2}(\Sigma_i)\times
H^{1/2}(\Sigma_i)\right)$, one may represent $T$ as 
$(T_1, T_2, T_3, T_4)$ with
\begin{gather*}
T_1:H_{*}^4(\Omega)\to \prod_{i=1}^2 H^{7/2}(\Sigma_i),\quad  
T_2:H_{*}^4(\Omega)\to \prod_{i=1}^2 H^{5/2}(\Sigma_i),\\
T_3:H_{*}^4(\Omega)\to \prod_{i=1}^2 H^{3/2}(\Sigma_i),\quad 
 T_4:H_{*}^4(\Omega)\to \prod_{i=1}^2 H^{1/2}(\Sigma_i),
\end{gather*} 
 which are continuous maps from
$H_{*}^4(\Omega)$ to the respective spaces each of them endowed with
its normal norm. Then one may define the four subspaces
\begin{gather*}
V_1:=\{v\in R(T):v=(T_1(w),0,0,0), w\in H_{*}^4(\Omega)\},\\
V_2:=\{v\in R(T):v=(0,T_2(w),0,0), w\in H_{*}^4(\Omega)\},\\
V_3:=\{v\in R(T):v=(0,0,T_3(w),0), w\in H_{*}^4(\Omega)\},\\
V_4:=\{v\in R(T):v=(0,0,0,T_4(w)), w\in H_{*}^4(\Omega)\}.
\end{gather*} 
Concerning the fact that
$T_{|(\ker (T))^{\perp}}$ is an isometric isomorphism and the
continuity of the map $T$, one can show that $V_i$  $(i=1,2,3,4)$ are
closed in $R(T)$ endowed with $\|\cdot\|_{R(T)}$.

Now, let $\alpha\in C^2([0,\infty);V_2)$, $\beta\in
C^1([0,\infty);V_1)$, $\gamma\in C^0([0,\infty);V_1)$ and
$h_1,h_2$ be in $C^2([0,\infty);V_1)$. Then one may define the map
\[
w:=T_{|(\ker T)^{\perp}}^{-1}\left((h_1, \alpha/2,0,0)
\times(h_2, \alpha/2,0,0)\right)
\] 
with $w=h_i$, $w_y=\alpha/2$, $w_{yy}=0$ and $w_{yyy}=0$ on the boundary
$\Sigma_i$. In this way we have that $w\in C^2([0,\infty);H_*^{4}(
\Omega))$. Then putting $u=v+w$ ($v$ to be fixed) into the problem
\eqref{e4.1}-\eqref{e4.2}, we obtain the following variational problem
\begin{equation} \label{e4.3}
\begin{gathered}
   v_{tt}+ \Delta ^2 v + \mu v_t+kv  = \widetilde{\varphi},  \quad
 (x,y) \in \Omega,\; t>0,   \\
   v(x,y,t) =v_{xx}(x,y,t) =0, \quad (x,y) \in \{0,\pi\}\times(-l,l),\; t>0,\\
   v(x,y,t)=v_y(x,y,t)=0, \quad (x,y)\in (0,\pi)\times\{-l,l\},\; t>0, \\
   v(x,y,0) =v_0, \quad (x,y) \in \Omega , \\
   v_t(x,y,0) = v_1,  \quad (x,y) \in \Omega ,
\end{gathered}
\end{equation}
where $\widetilde{\varphi}=\varphi-w_{tt}-\Delta^2 w-\mu w_t-kw$,
$v_0=u_0(x,y)-w(x,y,0)$ and $v_1=u_1(x,y)-w_t(x,y,0)$.

Recalling the function space 
\[ 
H_{*}^2(\Omega):=\{u\in H^2(\Omega):u=0\text{ on }\{0,\pi\}\times(-l,l)\},
\] 
which is defined in \cite{f1}, we have the uniqueness result.


\begin{theorem} \label{thm1} 
Assume that 
$\varphi\in C^0([0,\infty);L^2(\Omega))$, $u_0\in H_*^2(\Omega)$ and
$u_1\in L^2(\Omega)$. Then there exists a unique solution to the
problem \eqref{e2.12}-\eqref{e2.13}.
\end{theorem} 

\begin{proof}
It is easy to see that the functions $\widetilde{\varphi}$, $v_0$ and
$v_1$ satisfy 
\[
\widetilde{\varphi}\in C^0([0,\infty);L^2(\Omega)),\quad 
v_0\in V,\quad v_1\in L^2(\Omega).
\] 
Hence, we obtain from Section 3 that the problem
\eqref{e4.3} has a unique solution 
\[
v\in C^0([0,\infty);V)\cap C^1([0,\infty);L^2(\Omega)).
\] 
Then, according to the arguments above, 
\[
u=v+w\in C^0([0,\infty);H_{*}^{2}( \Omega))\cap
C^1([0,\infty);L^2(\Omega))
\] 
is the unique solution of the original
problem \eqref{e2.12}-\eqref{e2.13} with the initial conditions
$u(x,y,0)=u_0,~u_t(x,y,0)=u_1$ and the boundary conditions \eqref{e2.3},
\eqref{e2.4}, \eqref{e2.5}, \eqref{e2.8}.
\end{proof}

Since \eqref{e2.12}-\eqref{e2.13} is a linear problem, it is possible to find an
explicit form of the unique solution. We first consider an initial
value problem for $S=S(t)$
\begin{equation} \label{e4.4}
\begin{gathered}
   S''+\mu S'+aS  = f(t),  \quad t>0,   \\
   S(0) =A,  \quad
   S'(0) =B  
\end{gathered}
\end{equation}
where $f(t)$ is a given function, $\mu$ and $a$ are positive constants, 
$A, B$ are constants.

We know that \eqref{e4.4} is a second-order ordinary differential problem.
 According to the eigenvalue method, we have the following three cases 
if we denote $\delta=\mu^2-4a$:
\smallskip

\noindent\textbf{Case 1.} 
If $\delta<0$, then the two eigenvalues are
$\lambda_1=-\frac{\mu}{2}+\frac{\sqrt{-\delta}}{2}i$ and
$\lambda_2=-\frac{\mu}{2}-\frac{\sqrt{-\delta}}{2}i$. Hence, we have
\begin{align*}
S(t)&=\exp\big(-\frac{\mu}{2}t\big)
 \cos\Big(\frac{\sqrt{-\delta}}{2}t\Big)\Big(A+\frac{2}{\sqrt{-\delta}}
\int_{0}^t{\exp\big(\frac{\mu}{2}s\big)\cos
\Big(\frac{\sqrt{-\delta}}{2}s\Big)f(s)}ds\Big)\\
&\quad -\frac{2}{\sqrt{-\delta}}\exp\big(-\frac{\mu}{2}t\big)
\sin\Big(\frac{\sqrt{-\delta}}{2}t\Big)
\Big(B+\frac{\mu}{2}A\\
&\quad +\int_{0}^t{\exp\big(\frac{\mu}{2}s\big)
\sin\Big(\frac{\sqrt{-\delta}}{2}s\Big)f(s)}ds\Big);
\end{align*}
\smallskip

\noindent\textbf{Case 2.} 
If $\delta=0$, then the two eigenvalues are
$\lambda_1=\lambda_2=-\frac{\mu}{2}$. Therefore,
\[
S(t)=\Big(A+\Big(B+\frac{\mu}{2}A\Big)t\Big)
\exp\big(-\frac{\mu}{2}t\big)-\exp\big(-\frac{\mu}{2}t\big)
\int_{0}^t{(s-t)\exp\big(\frac{\mu}{2}s\big)f(s)}ds;
\]
\smallskip

\noindent\textbf{Case 3.} If $\delta>0$, then the two eigenvalues are
$\lambda_1=-\frac{\mu}{2}+\frac{\sqrt{\delta}}{2}$ and
$\lambda_2=-\frac{\mu}{2}-\frac{\sqrt{\delta}}{2}$. Therefore,
\begin{align*}
S(t)&=\exp\Big(\Big(-\frac{\mu}{2}+\frac{\sqrt{\delta}}{2}\Big)t\Big)
\Big(\frac{\mu+\sqrt{\delta}}{2\sqrt{\delta}}A+\frac{1}{\sqrt{\delta}}B\\
&\quad +\frac{1}{\sqrt{\delta}}
\int_{0}^t{\exp\Big(-\Big(-\frac{\mu}{2}+\frac{\sqrt{\delta}}{2}\Big)s\Big)f(s)}ds
\Big)
\\
&\quad+\exp\Big(\Big(-\frac{\mu}{2}-\frac{\sqrt{\delta}}{2}\Big)t\Big)
\Big(\frac{\sqrt{\delta}-\mu}{2\sqrt{\delta}}A\\
&\quad -\frac{1}{\sqrt{\delta}}B-\frac{1}{\sqrt{\delta}}
\int_{0}^t{\exp\Big(-\Big(-\frac{\mu}{2}-\frac{\sqrt{\delta}}{2}\Big)s\Big)f(s)}
ds\Big).
\end{align*}

To obtain a Fourier series, we introduce the following
\begin{equation} \label{e4.5}
\begin{gathered}
u_{0m}(y)=\frac{2}{\pi}\int_0^{\pi}{u_0(x,y)\sin(mx)}dx,\quad
u_{1m}(y)=\frac{2}{\pi}\int_0^{\pi}{u_1(x,y)\sin(mx)}dx,\\
\alpha_m(t)= \frac{2}{\pi}\int_0^{\pi}{\alpha(x,t)\sin(mx)}dx,
\quad \beta_m(t)=\frac{2}{\pi}\int_0^{\pi}{\beta(x,t)\sin(mx)}dx\\
\gamma_m(t)=\frac{2}{\pi}\int_0^{\pi}{\gamma(x,t)\sin(mx)}dx,
\quad
\varphi_m(y,t)=\frac{2}{\pi}\int_0^{\pi}{\varphi(x,y,t)\sin(mx)}dx
\end{gathered}
\end{equation}
and define a function by
\begin{equation} \label{e4.6}
R_m(t)= \theta(t)\Big(\frac{u_{0m}(l)-u_{0m}(-l)}{2l}
-\int_0^t{\frac{\gamma_m(s)}{2l}}ds\Big), \quad t> 0,
\end{equation} 
here $\theta(t)$ is as in \eqref{e2.9}.

Then the unique solution to the original problem \eqref{e2.12}-\eqref{e2.13} can
be explicitly represented. 

\begin{theorem} \label{thm2} 
Assume that the functions $\varphi\in C^0([0,\infty);L^2(\Omega))$, 
$u_0\in H_*^2(\Omega)$ and $u_1\in L^2(\Omega)$. Then the unique solution to
\eqref{e2.12}-\eqref{e2.13} is given by 
\begin{equation} \label{e4.7} 
u(x,y,t)= \sum_{m=1}^{\infty}U_m(y,t)\sin(mx)
\end{equation}
with
\begin{equation} \label{e4.8}
U_m(y,t)=\sum_{n=1}^{\infty} (T_m)_n(t)\sin
\left(\frac{n\pi}{l}y\right)+\sum_{n=1}^{\infty}
(S_m)_n(t)\cos \left(\frac{n\pi}{l}y\right)+C_m(t)y+D_m(t),
\end{equation}
where $(T_m)_n(t)$, $(S_m)_n(t)$ and $D_m(t)$ are in the form of
$S(t)$, the solution of \eqref{e4.4}, $C_m(t)$ is as in \eqref{e4.6}.
\end{theorem}

\begin{proof} 
According to the boundary conditions for
$x=0, \pi$, we seek the solution $u$ of the problem \eqref{e2.12}-\eqref{e2.13} in
the form
\begin{equation} \label{e4.9}
u(x,y,t)= \sum_{m=1}^{\infty} U_m(y,t)\sin(mx).
\end{equation}
Inserting \eqref{e4.9} in \eqref{e2.12}-\eqref{e2.13} and recalling 
\eqref{e4.5},  for every $m\geq 1$, we have
\begin{equation} \label{e4.10}
\begin{gathered}
\begin{aligned}
&(U_m)_{tt}+ (U_m)_{yyyy} -2m^2(U_m)_{yy}+(m^4+k) U_m+ \mu (U_m)_t\\ 
&= \varphi_m(y,t),  \quad  y \in (-l,l), \; t>0,
\end{aligned}   \\
   U_m(y,0) =u_{0m}(y), \quad y \in (-l,l), \\
   (U_m)_t(y,0) = u_{1m}(y),  \quad  y \in (-l,l),
\end{gathered}
\end{equation}
with the boundary conditions
\begin{equation} \label{e4.11}
\begin{gathered}
   (U_m)_{y}(-l,t)-(U_m)_{y}(l,t)=0, \quad t>0, \\
   (U_m)_y(-l,t)+(U_m)_y(l,t)=\alpha_m(t), \quad t>0,\\
   U_m(-l,t)+U_m(l,t)=\beta_m(t), \quad t>0,\\
   (U_m)_t(-l,t)-(U_m)_t(l,t)-\eta(t)(U_m(-l,t)-U_m(l,t))=\gamma_m(t)\theta(t), 
\quad t>0.
\end{gathered}
\end{equation}

Now, we look for the solution to \eqref{e4.10}-\eqref{e4.11} in the form 
of \eqref{e4.8}; that is,
\begin{align*}
U_m(y,t)=\sum_{n=1}^{\infty} (T_m)_n(t)\sin
\left(\frac{n\pi}{l}y\right)+\sum_{n=1}^{\infty}
(S_m)_n(t)\cos \left(\frac{n\pi}{l}y\right)+C_m(t)y+D_m(t).
\end{align*}
Putting it into the equation in \eqref{e4.10}, we obtain
\begin{align*}
&\sum_{n=1}^{\infty}  \Big((T_m)_n''(t)+\mu
(T_m)_n'(t)+\Big(\Big(m^2+\left(\frac{n\pi}{l}\right)^2\Big)^2
+k\Big)(T_m)_n(t)\Big)
\sin\big(\frac{n\pi}{l}y\big)\\
&+\sum_{n=1}^{\infty} \Big((S_m)_n''(t)+\mu
(S_m)_n'(t)+\Big(\Big(m^2+\big(\frac{n\pi}{l}\big)^2\Big)^2
+k\Big)(S_m)_n(t)\Big)
\cos\big(\frac{n\pi}{l}y\big)\\
&+C_m''(t)y+\mu C_m'(t)y+(m^4+k)C_m(t)y+D_m''(t)+\mu D_m'(t)+(m^4+k)D_m(t)\\
&=\varphi_m(y,t).
\end{align*}
For $C_m(t)$ being as in the form in \eqref{e4.6}, we define the function 
$$
\phi_{m}(y,t)=C_m''(t)y+(m^4+k+\mu \eta(t))C_m(t)y-\frac{\mu
\gamma_m(t)\theta(t)}{2l}y
$$
and let $\varphi_m(y,t)=\varphi_m(y,t)-\phi_{m}(y,t)+\phi_{m}(y,t)$, 
then it follows that
\begin{equation} \label{e4.12}
C_m'(t)-\eta(t) C_m(t)=-\frac{
\gamma_m(t)\theta(t)}{2l},
\end{equation}
and
\begin{equation} \label{e4.13}
\begin{aligned}
&\sum_{n=1}^{\infty}  \Big((T_m)_n''(t)+\mu
(T_m)_n'(t)+\Big(\Big(m^2+\big(\frac{n\pi}{l}\big)^2\Big)^2+k\Big)
(T_m)_n(t)\Big) \sin\big(\frac{n\pi}{l}y\big)
\\
&+\sum_{n=1}^{\infty} \left((S_m)_n''(t)+\mu
(S_m)_n'(t)+\left(\left(m^2+\left(\frac{n\pi}{l}\right)^2\right)^2+k\right)(S_m)_n(t)\right)
\cos\left(\frac{n\pi}{l}y\right)\\
&\quad +D_m''(t)+\mu D_m'(t)+(m^4+k)D_m(t)
\\
&=\sum_{n=1}^{\infty}
\left(\varphi_{1m}\right)_n(t)\sin\left(\frac{n\pi}{l}y\right)+\sum_{n=1}^{\infty}
\left(\varphi_{2m}\right)_n(t)\cos\left(\frac{n\pi}{l}y\right)+\varphi_{m}(t),
\end{aligned}
\end{equation}
where
\begin{gather*}
\left(\varphi_{1m}\right)_n(t)
 =\frac{1}{l}\int_{-l}^l{\left(\varphi_{m}(y,t)-\phi_{m}(y,t)\right)
 \sin\left(\frac{n\pi}{l}y\right)}dy,\\
\left(\varphi_{2m}\right)_n(t)
 =\frac{1}{l}\int_{-l}^l{\left(\varphi_{m}(y,t)-\phi_{m}(y,t)\right)
 \cos\left(\frac{n\pi}{l}y\right)}dy,\\
\varphi_{m}(t)=\frac{1}{2l}\int_{-l}^l{\varphi_{m}(y,t)}dy.
\end{gather*}

The initial conditions yield 
\begin{equation} \label{e4.14}
\begin{gathered}
\begin{aligned}
&\sum_{n=1}^{\infty} (T_m)_n(0)\sin
\left(\frac{n\pi}{l}y\right)+\sum_{n=1}^{\infty}
(S_m)_n(0)\cos \left(\frac{n\pi}{l}y\right)+D_m(0)\\
&=\sum_{n=1}^{\infty} (A_{1m})_n\sin
\left(\frac{n\pi}{l}y\right)+\sum_{n=1}^{\infty}
(A_{2m})_n\cos \left(\frac{n\pi}{l}y\right)+A_m, \quad y \in (-l,l),
\end{aligned} \\
\begin{aligned}
&\sum_{n=1}^{\infty} (T_m)'_n(0)\sin
\left(\frac{n\pi}{l}y\right)+\sum_{n=1}^{\infty}
(S_m)'_n(0)\cos \left(\frac{n\pi}{l}y\right)+D'_m(0)\\
&=\sum_{n=1}^{\infty} (B_{1m})_n\sin
\left(\frac{n\pi}{l}y\right)+\sum_{n=1}^{\infty}
(B_{2m})_n\cos \left(\frac{n\pi}{l}y\right)+B_m,  \quad y \in (-l,l),
\end{aligned}
\end{gathered}
\end{equation}
where
\begin{gather*}
(A_{1m})_n=\frac{1}{l}\int_{-l}^{l}{\left(u_{0m}(y)-C_m(0)y\right)
\sin\left(\frac{n\pi}{l}y\right)}dy,\\
(A_{2m})_n=\frac{1}{l}\int_{-l}^{l}{\left(u_{0m}(y)\right)
 \cos\left(\frac{n\pi}{l}y\right)}dy,\\
(B_{1m})_n =\frac{1}{l}\int_{-l}^{l}{\left(u_{1m}(y)
 -C_m'(0)y\right)\sin\left(\frac{n\pi}{l}y\right)}dy,\\
(B_{2m})_n =\frac{1}{l}\int_{-l}^{l}{\left(u_{1m}(y)\right)
 \cos\left(\frac{n\pi}{l}y\right)}dy,\\
A_m =\frac{1}{2l}\int_{-l}^{l}{u_{0m}(y)}dy,\quad
B_m=\frac{1}{2l}\int_{-l}^{l}{u_{1m}(y)}dy.
\end{gather*}
Assume that the functions $\alpha_m(t)$, $\beta_m(t)$ and
$\gamma_m(t)$ are given by
\begin{equation} \label{e4.15}
\begin{gathered}
   \alpha_m(t)=2\sum_{n=1}^{\infty} \frac{n\pi}{l}(T_m)_n(t)\cos(n\pi), \quad t>0,\\
   \beta_m(t)=2\sum_{n=1}^{\infty}(S_m)_n(t)\cos(n\pi)+2D_m(t), \quad t>0,\\
   \gamma_m(t)=-2l(C'_m(t)-\eta(t)C_m(t))/\theta(t), \quad t>0,
\end{gathered}
\end{equation}
then the boundary conditions are satisfied. Moreover, we are led to
the several ordinary differential problems
\begin{equation} \label{e4.16}
\begin{gathered}
C_m'(t)-\eta(t)C_m(t)=-\frac{\gamma_m(t)\theta(t)}{2l}, \quad t>0,\\
C_m(t)=\frac{u_{0m}(l)-u_{0m}(-l)}{2l}, \quad t=0\,;
\end{gathered}
\end{equation}

\begin{equation} \label{e4.17}
\begin{gathered}
D_m''(t)+\mu D_m'(t)+(m^4+k)D_m(t)=\varphi_{m}(t), \quad t>0,\\
D_m(t)=A_m, \quad t=0,\\
D_m'(t)=B_m, \quad t=0\,;
\end{gathered}
\end{equation}

\begin{equation} \label{e4.18}
\begin{gathered}
(T_m)_n''(t)+\mu
(T_m)_n'(t)+\Big((m^2+\left(\frac{n\pi}{l}\right)^2)^2+k\Big)(T_m)_n(t)
=\left(\varphi_{1m}\right)_n(t), \quad  t>0,\\
(T_m)_n(t)=(A_{1m})_n, \quad t=0,\\
(T_m)_n'(t)=(B_{1m})_n, \quad t=0\,;
\end{gathered}
\end{equation}

\begin{equation} \label{e4.19}
\begin{gathered}
(S_m)_n''(t)+\mu
(S_m)_n'(t)+\Big((m^2+\left(\frac{n\pi}{l}\right)^2)^2+k\Big)(S_m)_n(t)
=\left(\varphi_{2m}\right)_n(t), \quad  t>0,\\
(S_m)_n(t)=(A_{2m})_n, \quad t=0,\\
(S_m)_n'(t)=(B_{2m})_n, \quad t=0.
\end{gathered}
\end{equation}

To complete the proof, we need to check that  problem \eqref{e4.16}
admits a solution in the form $R_m(t)$ and that the problems
\eqref{e4.17}-\eqref{e4.19} have solutions in the form of $S(t)$. Since the last
three problems depend on $C_m(t)$, we first consider the problem
\eqref{e4.16}. In fact, this is a first order ordinary differential problem
and we know that the solution is
\begin{align*}
C_m(t)&=C_m(0)\exp\Big(\int_{0}^t{\eta(s)}ds\Big)\\
&\quad -\exp\Big(\int_{0}^t{\eta(s)}ds\Big)
\int_0^t{\frac{\gamma_m(s)\theta(s)}{2l}\exp\left(-\int_{0}^s{\eta(\tau)}d\tau\right)}ds.
\end{align*}

Since $\eta(t)$ is not a continuous function, see \eqref{e2.10},  to compute 
conveniently, we denote $E(t)$ given in \eqref{e2.11} here by
\begin{equation} \label{e4.20}
E(t)=\begin{cases}
   -1, & \mathcal {E}(t)\leq \overline{E}_{\mu}-\varepsilon,\\
   \frac{1}{\varepsilon}\left(\mathcal {E}(t)-\overline{E}_{\mu}\right), 
& \overline{E}_{\mu}-\varepsilon<\mathcal {E}(t)\leq \overline{E}_{\mu}+\varepsilon,\\
   1, & \mathcal {E}(t)> \overline{E}_{\mu}+\varepsilon.
\end{cases}
\end{equation}
Then we consider the exponential function 
$\exp\big(\int_{0}^t{\eta(s)}ds\big)$. There are three cases:
\begin{itemize}
\item[(1)]  If $\mathcal {E}(t)>\overline{E}_\mu+\varepsilon$, then
\[
\exp\Big(\int_{0}^t{\eta(s)}ds\Big)
=\exp\Big(\frac{\mathcal
{E}(t)}{\overline{E}_\mu}-\frac{\mathcal
{E}(0)}{\overline{E}_\mu}+\ln{\frac{\mathcal
{E}(t)-\overline{E}_\mu}{\mathcal {E}(0)-\overline{E}_\mu}}\Big);
\]

\item[(2)] If $\overline{E}_\mu<\mathcal {E}(t)\leq
\overline{E}_\mu+\varepsilon$, then
\begin{align*}
&\exp\Big(\int_{0}^t{\eta(s)}ds\Big)\\
&=\exp\Big(\frac{(\mathcal
{E}(t)-\overline{E}_\mu)^2}{2\varepsilon\overline{E}_\mu}+\frac{\varepsilon-2\mathcal
{E}(0)}{2\overline{E}_\mu}+\ln{\frac{\varepsilon}{\mathcal
{E}(0)-\overline{E}_\mu}}+\frac{1}{\varepsilon}(\mathcal
{E}(t)-\overline{E}_\mu)\Big);
\end{align*}

\item[(3)] If $\mathcal {E}(t)\leq \overline{E}_\mu$, then
\[
\exp\Big(\int_{0}^t{\eta(s)}ds\Big)
=\exp\Big(\frac{\varepsilon-2\mathcal
{E}(0)}{2\overline{E}_\mu}+\ln{\frac{\varepsilon}{\mathcal
{E}(0)-\overline{E}_\mu}}+\int_{t_\mu}^t{\eta(s)}ds\Big),
\]
where $t_\mu$ satisfies $\mathcal {E}(t_\mu)=\overline{E}_\mu$.
\end{itemize}
Let $\varepsilon\to0$, we have
\[
\exp\Big(\int_{0}^t{\eta(s)}ds\Big)
= \begin{cases}
   0, & \mathcal {E}(t)\leq \overline{E}_{\mu},\\
 \frac{\mathcal {E}(t)-\overline{E}_\mu}{\mathcal
{E}(0)-\overline{E}_\mu}\exp\left(\frac{\mathcal
{E}(t)}{\overline{E}_\mu}-\frac{\mathcal
{E}(0)}{\overline{E}_\mu}\right), & \mathcal {E}(t)>
\overline{E}_{\mu}.
\end{cases}
\]
Therefore,
\[
C_m(t)= \begin{cases}
   0, & \mathcal {E}(t)\leq \overline{E}_{\mu},\\
   \frac{\mathcal {E}(t)-\overline{E}_\mu}{\mathcal
{E}(0)-\overline{E}_\mu}\exp\left(\frac{\mathcal
{E}(t)}{\overline{E}_\mu}-\frac{\mathcal
{E}(0)}{\overline{E}_\mu}\right)\left(\frac{u_{0m}(l)-u_{0m}(-l)}{2l}-\int_0^t{\frac{\gamma_m(s)}{2l}}ds\right)
, & \mathcal {E}(t)> \overline{E}_{\mu},
\end{cases}
\]
which is as in \eqref{e4.6} if one recalls that \eqref{e2.9}.

Next, for every $n\geq 1$ and $m\geq 1$, we solve the other three
ordinary differential problems \eqref{e4.17}, \eqref{e4.18} and \eqref{e4.19}. 
Recalling the initial value problem \eqref{e4.4} and let $a=m^4+k$ for \eqref{e4.17},
$a=\left(m^2+\left(\frac{n\pi}{l}\right)^2\right)^2+k$ for \eqref{e4.18}
and \eqref{e4.19}, we obtain that there exists a unique solution for
\eqref{e4.17}, \eqref{e4.18} and \eqref{e4.19} separately in the form of $S(t)$ 
with the constants $A, B$ replaced by
$A_m, B_m, (A_{1m})_n, (B_{1m})_n, (A_{2m})_n, (B_{2m})_n$. 
That is, if we denote $\delta=\mu^2-4(m^4+k)$, then:
\smallskip

\noindent\textbf{Case 1.} If $\delta<0$, then we have
\begin{align*}
D_m(t)&=\exp\big(-\frac{\mu}{2}t\big)\cos\Big(\frac{\sqrt{-\delta}}{2}t\Big)
\Big(A_m+\frac{2}{\sqrt{-\delta}}
\int_{0}^t{\exp\big(\frac{\mu}{2}s\big)
\cos\Big(\frac{\sqrt{-\delta}}{2}s\Big)\varphi_m(s)}ds\Big)\\
&\quad -\frac{2}{\sqrt{-\delta}}\exp\big(-\frac{\mu}{2}t\big)
\sin\Big(\frac{\sqrt{-\delta}}{2}t\Big)\Big(\frac{B_m+\mu A_m}{2}\\
&\quad +\int_{0}^t{\exp\big(\frac{\mu}{2}s\big)
 \sin\Big(\frac{\sqrt{-\delta}}{2}s\Big)\varphi_m(s)}ds\Big);
\end{align*}
\smallskip

\noindent\textbf{Case 2.} If $\delta=0$, then
\[
D_m(t)=\exp\big(-\frac{\mu}{2}t\big)\Big(A_m+
\left(B_m+\frac{\mu}{2}A_m\right)t
-\int_{0}^t{(s-t)\exp\big(\frac{\mu}{2}s\big)\varphi_m(s)}ds\Big);
\]
\smallskip

\noindent\textbf{Case 3.} 
If $\delta>0$, the two eigenvalues are
$\lambda_1=-\frac{\mu}{2}+\frac{\sqrt{\delta}}{2}$ and
$\lambda_2=-\frac{\mu}{2}-\frac{\sqrt{\delta}}{2}$, then
\begin{align*}
D_m(t)
&=\exp\left(\lambda_1t\right)\Big(\frac{ B_m-\lambda_2
A_m}{\sqrt{\delta}}+\frac{1}{\sqrt{\delta}}
\int_{0}^t{\exp\left(-\lambda_1s\right)\varphi_m(s)}ds\Big)\\
&\quad +\exp(\lambda_2t)\Big(\frac{\lambda_1A_m-
B_m}{\sqrt{\delta}}-\frac{1}{\sqrt{\delta}}
\int_{0}^t{\exp\left(-\lambda_2s\right)\varphi_m(s)}ds\Big).
\end{align*}

For \eqref{e4.18} and \eqref{e4.19}, denote 
$\gamma=\mu^2-4\big(\big(m^2+\big(\frac{n\pi}{l}\big)^2\big)^2+k\big)$, 
then we have:
\smallskip

\noindent\textbf{Case 1.} If $\gamma<0$, then
\begin{align*}
&(T_m)_n(t)\\
&=(A_{1m})_n\exp\big(-\frac{\mu}{2}t\big)
\cos\Big(\frac{\sqrt{-\gamma}}{2}t\Big)\\
&\quad -\frac{2(B_{1m})_n+\mu (A_{1m})_n}{\sqrt{-\gamma}}\exp\big(-\frac{\mu}{2}t\big)
\sin\Big(\frac{\sqrt{-\gamma}}{2}t\Big)
\\
&\quad +\frac{2}{\sqrt{-\gamma}}\exp\big(-\frac{\mu}{2}t\big)
 \cos\Big(\frac{\sqrt{-\gamma}}{2}t\Big)
\int_{0}^t{\exp\big(\frac{\mu}{2}s\big)
 \cos\Big(\frac{\sqrt{-\gamma}}{2}s\Big)\left(\varphi_{1m}\right)_n(s)}ds
\\
&\quad -\frac{2}{\sqrt{-\gamma}}\exp\big(-\frac{\mu}{2}t\big)
 \sin\Big(\frac{\sqrt{-\gamma}}{2}t\Big)
\int_{0}^t{\exp\big(\frac{\mu}{2}s\big)
 \sin\Big(\frac{\sqrt{-\gamma}}{2}s\Big)\left(\varphi_{1m}\right)_n(s)}ds;
\end{align*}
\begin{align*}
&(S_m)_n(t)\\
&=(A_{2m})_n\exp\big(-\frac{\mu}{2}t\big)
 \cos\Big(\frac{\sqrt{-\gamma}}{2}t\Big)\\
&\quad -\frac{2(B_{2m})_n
 +\mu (A_{2m})_n}{\sqrt{-\gamma}}\exp\big(-\frac{\mu}{2}t\big)
 \sin\Big(\frac{\sqrt{-\gamma}}{2}t\Big)\\
&\quad +\frac{2}{\sqrt{-\gamma}}\exp\big(-\frac{\mu}{2}t\big)
 \cos\Big(\frac{\sqrt{-\gamma}}{2}t\Big)
\int_{0}^t{\exp\big(\frac{\mu}{2}s\big)\cos\Big(\frac{\sqrt{-\gamma}}{2}s\Big)
 \left(\varphi_{2m}\right)_n(s)}ds\\
&-\frac{2}{\sqrt{-\gamma}}\exp\big(-\frac{\mu}{2}t\big)
 \sin\Big(\frac{\sqrt{-\gamma}}{2}t\Big)
\int_{0}^t{\exp\big(\frac{\mu}{2}s\big)
 \sin\Big(\frac{\sqrt{-\gamma}}{2}s\Big)\left(\varphi_{2m}\right)_n(s)}ds;
\end{align*}
\smallskip

\noindent\textbf{Case 2.} If $\gamma=0$, then
\begin{align*}
&(T_m)_n(t)\\
&=\left((A_{1m})_n+
\left((B_{1m})_n+\frac{\mu}{2}(A_{1m})_n\right)t\right)
 \exp\big(-\frac{\mu}{2}t\big)\\
&\quad -\exp\big(-\frac{\mu}{2}t\big)\Big(
\int_{0}^t{s\exp\big(\frac{\mu}{2}s\big)\left(\varphi_{1m}\right)_n(s)}ds
-t \int_{0}^t{\exp\big(\frac{\mu}{2}s\big)\left(\varphi_{1m}\right)_n(s)}ds\Big);
\end{align*}
\begin{align*}
&(S_m)_n(t)\\
&=\left((A_{2m})_n+ \left((B_{2m})_n+\frac{\mu}{2}(A_{2m})_n\right)t\right)
 \exp\big(-\frac{\mu}{2}t\big)\\
&\quad -\exp\big(-\frac{\mu}{2}t\big)\Big(
\int_{0}^t{s\exp\big(\frac{\mu}{2}s\big)\left(\varphi_{2m}\right)_n(s)}ds
-t \int_{0}^t{\exp\big(\frac{\mu}{2}s\big)\left(\varphi_{2m}\right)_n(s)}ds\Big);
\end{align*}
\smallskip

\noindent\textbf{Case 3.} If $\gamma>0$, the two eigenvalues are
$\lambda_1=-\frac{\mu}{2}+\frac{\sqrt{\gamma}}{2}$ and
$\lambda_2=-\frac{\mu}{2}-\frac{\sqrt{\gamma}}{2}$, then
\begin{align*}
&(T_m)_n(t)\\
&=\exp\left(\lambda_1t\right)\Big(\frac{(B_{1m})_n
 -\lambda_2(A_{1m})_n}{\sqrt{\gamma}}+\frac{1}{\sqrt{\gamma}}
\int_{0}^t{\exp\left(-\lambda_1s\right)\left(\varphi_{1m}\right)_n(s)}ds\Big)\\
&\quad+\exp\left(\lambda_2t\right)\left(\frac{\lambda_1(A_{1m})_n
 -(B_{1m})_n}{\sqrt{\gamma}}-\frac{1}{\sqrt{\gamma}}
\int_{0}^t{\exp\left(-\lambda_2s\right)\left(\varphi_{1m}\right)_n(s)}ds\right);
\end{align*}
\begin{align*}
&(S_m)_n(t)\\
&=\exp\left(\lambda_1t\right)\Big(\frac{(B_{2m})_n
 -\lambda_2(A_{2m})_n}{\sqrt{\gamma}}+\frac{1}{\sqrt{\gamma}}
\int_{0}^t{\exp\left(-\lambda_1s\right)\left(\varphi_{2m}\right)_n(s)}ds\Big)\\
&\quad +\exp\left(\lambda_2t\right)\Big(\frac{\lambda_1(A_{2m})_n
 -(B_{2m})_n}{\sqrt{\gamma}}-\frac{1}{\sqrt{\gamma}}
\int_{0}^t{\exp\left(-\lambda_2s\right)\left(\varphi_{2m}\right)_n(s)}ds\Big).
\end{align*}
Therefore, the solution to \eqref{e2.12}-\eqref{e2.13} has the
form of \eqref{e4.7} with $U_m(y,t)$ being given in \eqref{e4.8}. And then the
proof is finished.
\end{proof} 

Finally, by using the explicit form of the  solution to
\eqref{e2.12}-\eqref{e2.13}, we are able to analyze the amplitude of the torsional
oscillation appearing in suspension bridges.

\begin{corollary} \label{coro3} 
Assume that $u$ is the unique solution to the original problem 
\eqref{e2.12}-\eqref{e2.13}. Then the amplitude of the torsional 
oscillation on the two sides $y=\pm l$ reads
\[
|u(x,-l,t)-u(x,l,t)|=
\theta(t)\big|u_0(x,-l)-u_0(x,l)+2l\int_0^t{\gamma(x,s)}ds\big|,
\]
where $\theta(t)$ is given in \eqref{e2.9}.
\end{corollary}

 By \eqref{e2.9}, if $\mathcal {E}(t)\leq \overline{E}_\mu$,
then $\theta (t)=0$, which yields that $|u(x,l,t)-u(x,-l,t)|=0$ by
Corollary \ref{coro3}. That is, when $\mathcal {E}(t)\leq \overline{E}_\mu$,
there is no torsional oscillation appearing in the bridge structure.
However, once the energy $\mathcal {E}(t)$ exceeds
$\overline{E}_\mu$, then $\theta (t)>0$ and
$|u(x,l,t)-u(x,-l,t)|\neq 0$, which show that the torsional
oscillation appears. Since $\theta (t)$ is an increasing function
with respect to $\mathcal {E}(t)$ and by Corollary \ref{coro3}, if the energy
$\mathcal {E}(t)$ ($\mathcal {E}(t)>\overline{E}_\mu$) increases,
then the amplitude of the torsional oscillation will go up till the
bridges collapse.



\section{Future developments}

In this article, a new mathematical model for dynamical suspension
bridges is suggested. We solve the evolution problem corresponding
to the model and obtain a unique explicit solution which is to
display the behavior appearing in actual suspension bridges. In this
section, we indicate two possible future developments.

\subsection{Quantitative results}

In this article, we suggest a dynamical model for suspension bridges
and obtain an explicit solution from a theoretical point of view. In
this model, $\varphi(x,y,t)$, $u_0(x,y)$, $u_1(x,y)$, $\alpha(x,t)$,
$\beta(x,t)$ and $\gamma(x,t)$ are assumed to be real functions, but
we gave no hint on their explicit forms. Therefore, one can carry
out several numerical experiments to estimate them approximately.
Once they are given explicitly, an exact solution is to be expected
and one can even calculate the blow-up time of the solution, namely,
the time when the bridge collapses.

\subsection{Nonlinear restoring force}

The model we suggested in this paper is a linear model. According to
the opinion of McKenna \cite{m1}
\begin{quote}
We doubt that a bridge oscillating up and down by about 10 meters
every 4 seconds obeys Hooke's law.
\end{quote}
it may be not suitable to assume the restoring force $h$ due to the
hangers to be in a linear case, see also \cite{g1}. Therefore, it could be
better to consider the model with nonlinear restoring force
$h=h(x,y,u)$
\begin{align*}
u_{tt}+ \Delta ^2 u + \mu u_t+h = \varphi,\quad \text{in }\Omega\times(0,T).
\end{align*}
However, this problem appears significantly more difficult.



\subsection*{Acknowledgments}
The author is grateful to the anonymous referee for his useful suggestions on
the proper definition of the weak solution to the original problem.


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\end{document}