\documentclass[reqno]{amsart}

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2004(2004), No. 103, pp. 1--21.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu  (login: ftp)}
\thanks{\copyright 2004 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2004/103\hfil A nonlinear wave equation]
{A nonlinear wave equation with a nonlinear integral equation 
involving the boundary value}

\author[T. L. Nguyen, T. D. Bui\hfil EJDE-2004/103\hfilneg]
{Thanh Long Nguyen, Tien Dung Bui}

\address{Thanh Long Nguyen \hfill\break 
Department of Mathematics and Computer Science\\
University of Natural Science\\
Vietnam National University HoChiMinh City\\
227 Nguyen Van Cu Str., Dist.5, HoChiMinh City, Vietnam}
\email{longnt@hcmc.netnam.vn}

\address{Tien Dung Bui \hfill\break
Department of Mathematics\\
University of Architecture of HoChiMinh City\\
196 Pasteur Str., Dist. 3, HoChiMinh City, Vietnam}

\date{}
\thanks{Submitted January 14, 2004. Published September 3, 2004.}
\subjclass[2000]{35B30, 35L70, 35Q72}
\keywords{Galerkin method; system of integrodifferential
equations;\hfill\break\indent 
Schauder fixed point theorem; weak solutions;
stability of the solutions}

\begin{abstract}
 We consider the initial-boundary value problem for the 
 nonlinear wave equation 
 \begin{gather*}
 u_{tt}-u_{xx}+f(u,u_{t})=0,\quad x\in \Omega =(0,1),\; 0<t<T, \\
 u_{x}(0,t)=P(t),\quad u(1,t)=0, \\
 u(x,0)=u_0(x),\quad u_{t}(x,0)=u_1(x),
 \end{gather*}
 where $u_0,u_1,f$ are given functions, the unknown function 
 $u(x,t)$  and the unknown boundary value $P(t)$ satisfy the
 nonlinear integral equation 
 \begin{equation*}
 P(t)=g(t)+H(u(0,t))-\int_0^t K(t-s,u(0,s))ds,
 \end{equation*}
 where $g$, $K$, $H$ are given functions. We prove the existence 
 and uniqueness of weak solutions to this problem, and discuss 
 the stability of the solution with respect to the functions 
 $g$, $H$ and $K$. For the proof, we use the Galerkin method.
\end{abstract}

\maketitle

\numberwithin{equation}{section} 
\newtheorem{theorem}{Theorem}[section] 
\newtheorem{lemma}[theorem]{Lemma} 
\newtheorem{remark}[theorem]{Remark}

\section{Introduction}

In this paper we consider the problem of finding a pair of functions $(u,P)$
that satisfy 
\begin{gather}
u_{tt}-u_{xx}+f(u,u_{t})=0,\quad x\in \Omega =(0,1),\;0<t<T,  \label{e1.1} \\
u_{x}(0,t)=P(t)  \label{e1.2} \\
u(1,t)=0,  \label{e1.3} \\
u(x,0)=u_{0}(x),\quad u_{t}(x,0)=u_{1}(x),  \label{e1.4}
\end{gather}%
where $u_{0}$, $u_{1}$, $f$ are given functions satisfying conditions to be
specified later and the unknown function $u(x,t)$ and the unknown boundary
value $P(t)$ satisfy the nonlinear integral equation 
\begin{equation}
P(t)=g(t)+H(u(0,t))-\int_{0}^{t}K(t-s,u(0,s))ds,  \label{e1.5}
\end{equation}
where $g$, $H$, $K$ are given functions. Ang and Dinh \cite{a2} established
the existence of a unique global solution for the initial and boundary value
problem \eqref{e1.1}-\eqref{e1.4} with $u_{0}$, $u_{1}$, $P$ given functions
and $f(u,u_{t})=|u_{t}|^{\alpha }\mathop{\rm sign}(u_{t})$, $(0<\alpha <1)$.
As a generalization of the results in \cite{a2}, Long and Dinh \ \cite%
{d3,l2,l3} have considered problem \eqref{e1.1}, \eqref{e1.3}, \eqref{e1.4}
associated with the following nonhomogeneous boundary condition at $x=0$, 
\begin{equation}
u_{x}(0,t)=g(t)+H(u(0,t))-\int_{0}^{t}K(t-s,u(0,s))ds.  \label{e1.6}
\end{equation}
We have considered it with $K\equiv 0$, $H(s)=hs$, where $h>0$ \cite{l2}; 
$K\equiv 0$ \cite{d3}, $H(s)=hs$, $K(t,u)=k(t)u$, where $h>0$, $k\in
H^{1}(0,T)$, for all $T>0$ \cite{l3}. In the case of $H(s)=hs$, $%
K(t,u)=h\omega (\sin \omega t)u$, where $h>0$, $\omega >0$ are given
constants, the problem \eqref{e1.1}-\eqref{e1.5} is formed from the problem 
\eqref{e1.1}-\eqref{e1.4} wherein, the unknown function $u(x,t)$ and the
unknown boundary value $P(t)$ satisfy the following Cauchy problem 
\begin{gather}
P''(t)+\omega ^{2}P(t)=hu_{tt}(0,t),\quad 0<t<T,  \label{e1.7}
\\
P(0)=P_{0},\quad P'(0)=P_{1},  \label{e1.8}
\end{gather}%
where $\omega >0$, $h\geq 0$, $P_{0}$, $P_{1}$ are given constants \cite{l3}. 
An and Trieu \cite{a1}, studied a special case of problem 
\eqref{e1.1}-\eqref{e1.4}, \eqref{e1.7}, \eqref{e1.8} with $u_{0}=u_{1}=P_{0}=0$ 
and with 
$f(u,u_{t})$ linear, i.e., $f(u,u_{t})=Ku+\lambda u_{t}$ where $K$, $\lambda 
$ are given constants. In the later case the problem \eqref{e1.1}-\eqref{e1.4}, 
\eqref{e1.7}, and \eqref{e1.8} is a mathematical model
describing the shock of a rigid body and a linear visoelastic bar resting on
a rigid base \cite{a1}. Our problem is thus a nonlinear analogue of the
problem considered in \cite{a1}. In the case where 
$f(u,u_{t})=|u_{t}|^{\alpha }\mathop{\rm sign}(u_{t})$ the problem 
\eqref{e1.1}-\eqref{e1.4}, \eqref{e1.7}, and \eqref{e1.8} describes the shock 
between a solid body and
a linear viscoelastic bar with nonlinear elastic constraints at the side,
and constraints associated with a viscous frictional resistance. From 
\eqref{e1.7}, \eqref{e1.8} we represent $P(t)$ in terms of $P_{0}$, $P_{1}$, 
$\omega $, $h$, $u_{tt}(0,t)$ and then by integrating by parts, we have 
\begin{equation}
P(t)=g(t)+hu(0,t)-\int_{0}^{t}k(t-s)u(0,s)ds,  \label{e1.9}
\end{equation}
where 
\begin{gather}
g(t)=(P_{0}-hu_{0}(0))\cos \omega t+(P_{1}-hu_{1}(0))\frac{\sin \omega t}{\omega }, 
 \label{e1.10} \\
k(t)=h\omega (\sin \omega t).  \label{e1.11}
\end{gather}%
By eliminating an unknown function $P(t)$, we replace the boundary condition 
\eqref{e1.2} by 
\begin{equation}
u_{x}(0,t)=g(t)+hu(0,t)-\int_{0}^{t}k(t-s)u(0,s)ds.  \label{e1.12}
\end{equation}
Then, we reduce problem \eqref{e1.1}-\eqref{e1.4}, \eqref{e1.7}, \eqref{e1.8}
to \eqref{e1.1}-\eqref{e1.4}, \eqref{e1.9}-\eqref{e1.11} or to \eqref{e1.1}, 
\eqref{e1.3}, \eqref{e1.4}, \eqref{e1.10}-\eqref{e1.12}.

In this paper, we consider two main parts. In Part 1, we prove a theorem of
global existence and uniqueness of a weak solution of problem 
\eqref{e1.1}-\eqref{e1.5}. The proof is based on a Galerkin method 
associated to a priori
estimates, weak-convergence and compactness techniques. We remark that the
linearization method in \cite{d2,l4,o1} cannot be used for the problems in 
\cite{a2,b1,d1,d3,l2,l3}. In Part 2 we prove that the solution $(u,P)$ of
this problem is stable with respect to the functions $g,H$ and $K$. The
results obtained here generalize the ones in \cite{a1,a2,b1,d3,l2,l3}.

\section{The existence and uniqueness theorem}

We first set notations $\Omega =(0,1)$, $Q_{T}=\Omega \times (0,T)$, $T>0$, 
$L^{p}=L^{p}(\Omega )$, $H^{1}=H^{1}(\Omega )$, $H^{2}=H^{2}(\Omega )$, where 
$H^{1}$, $H^{2}$ are the usual Sobolev spaces on $\Omega $.

The norm in $L^{2}$ is denoted by $\|\cdot \|$. We also denote by $%
\langle \cdot ,\cdot \rangle $ the scalar product in $L^{2}$ or pair of dual
scalar product of continuous linear functional with an element of a function
space. We denote by $\|\cdot \|_{X}$ the norm of a Banach space $X$
and by $X'$ the dual space of $X$. We denote by $L^{p}(0,T;X)$, 
$1\leq p\leq \infty $ for the Banach space of the real functions 
$u:(0,T)\rightarrow X$ measurable, such that 
\begin{equation*}
\|u\|_{L^{p}(0,T;X)}=\Big(\int_{0}^{T}\|u(t)\|_{X}^{p}dt\Big)%
^{1/p}\quad \text{for }1\leq p<\infty ,
\end{equation*}
and
$$
\|u\|_{L^{\infty }(0,T;X)}=\mathop{\rm ess sup}_{0<t<T}\Vert
u(t)\|_{X}\quad\text{for }p=\infty .
$$
We put 
\begin{equation*}
V=\{v\in H^{1}:v(1)=0\},\quad a(u,v)=\langle \frac{\partial u}{\partial x},
\frac{\partial v}{\partial x}\rangle =\int_{0}^{1}\frac{\partial u}{\partial
x}\frac{\partial v}{\partial x}dx.
\end{equation*}
Here $V$ is a closed subspace of $H^{1}$and on $V$, $\|v\|_{H^{1}}$
and $\|v\|_{V}=\sqrt{a(v,v)}$ are two equivalent norms.

\begin{lemma} \label{lem1}
The imbedding $V\hookrightarrow C^{0}(\overline{\Omega})$ is compact and
\begin{equation}
\|v\|_{C^{0}(\overline{\Omega })}\leq \|v\|_{V} \label{e2.1}
\end{equation}
for all $v\in V$.
\end{lemma}The proof is straightforward and we omit it. We make the
following assumptions:

\begin{itemize}
\item[(A)] $u_0\in H^{1}$ and $u_1\in L^2$

\item[(G)] $g\in H^{1}(0,T)$ for all $T>0$

\item[(H)] $H\in C^{1}(\mathbb{R)}$, $H(0)=0$ and there exists a constant 
$h_{0}>0$ such that 
\begin{equation*}
\widehat{H}(y)=\int_{0}^{y}H(s)ds\geq -h_{0}
\end{equation*}

\item[(K1)] $K$ and $\frac{\partial K}{\partial t}$ are in 
$C^{0}(\mathbb{R}_{+}\times\mathbb{R};\mathbb{R})$

\item[(K2)] There exist the nonnegative functions $k_1\in L^2(0,T)$, 
$k_{2}\in L^{1}(0,T)$, $k_{3}\in L^2(0,T)$, and $k_{4}\in L^{1}(0,T)$, such
that\newline
(i) $|K(t,u)|\leq k_1(t)|u|+k_{2}(t)$, \newline
(ii) $|\frac{\partial K}{\partial t}(t,u)|\leq k_{3}(t)|u|+k_{4}(t)$.
\end{itemize}

The function $f:\mathbb{R}^2 \to \mathbb{R}$ satisfies $f(0,0)=0$ and the
following conditions:

\begin{itemize}
\item[(F1)] $$ 
(f(u,v)-f(u,\widetilde{v}))(v-\widetilde{v})\geq 0\quad\mbox{for all }
u,v,\widetilde{v}\in \mathbb{R}
$$

\item[(F2)] There is a constant $\alpha$ in $(0,1]$ and a function $B_1: 
\mathbb{R}_{+}\to \mathbb{R}_{+}$ continuous and satisfying 
\begin{equation*}
|f(u,v)-f(u,\widetilde{v})|\leq B_1(| u|)|v-\widetilde{v}|^{\alpha }\quad 
\text{for all }u,v,\widetilde{v}\in \mathbb{R}
\end{equation*}

\item[(F3)] There is a constant $\beta $ in $(0,1]$ and a function 
$B_{2}:\mathbb{R}_{+}\rightarrow \mathbb{R}_{+}$ continuous and satisfying 
\begin{equation*}
|f(u,v)-f(\widetilde{u},v)|\leq B_{2}(|v|)|u-\widetilde{u}|^{\beta }\quad 
\text{for all }u,\widetilde{u},v\in \mathbb{R}
\end{equation*}
\end{itemize}

We will use the notation $u^{\prime}=u_{t}=\partial u/\partial t$, 
$u''=u_{tt}=\partial ^2u/\partial t^2$. Then we have the
following theorem.

\begin{theorem} \label{thm1}
 Let (A),(G),(H),(K1),(K2), (F1), (F3) hold. Then, for every
$T>0$,  there exists a weak solution $(u,P)$ to  problem \eqref{e1.1}-\eqref{e1.5}
such that
\begin{gather}
u\in L^{\infty }(0,T;V),\quad u_{t}\in L^{\infty }(0,T;L^2),\quad
u(0,\cdot )\in H^{1}(0,T), \label{e2.2}\\
P\in H^{1}(0,T).\label{e2.3}
\end{gather}
Furthermore, if $\beta =1$ in (F3)  and the
functions $H$, $K$, $f$  satisfying, in addition
\begin{itemize}
\item[(H1)] $H\in C^2(\mathbb{R})$, $H'(s)>-1$ for all $s\in \mathbb{R}$
\item[(K3)] For all $M$ positive and all $T$ positive, there exists
$p_{M,T}$, $q_{M,T}$ in $L^2(0,T)$, $p_{M,T}(t)\geq 0$, $q_{M,T}(t)\geq 0$
such that
\begin{itemize}
\item[(i)] $|K(t,u)-K(t,v)|\leq p_{M,T}(t)|u-v|$
for all $u,v$ in $\mathbb{R}$, $|u|,|v|\leq M$,
\item[(ii)] $|\frac{\partial K}{\partial t}(t,u)
-\frac{\partial K}{\partial t}(t,v)|\leq q_{M,T}(t)|u-v|$
for all $u,v$ in $\mathbb{R}$, $|u|,|v|\leq M$.
\end{itemize}
\item[(F4)]
$B_{2}(|v|)\in L^2(Q_{T})$ for all $v\in L^2(Q_{T})$
for all $T>0$.
\end{itemize}
Then the solution is unique
\end{theorem}

\begin{remark} \label{rmk1} \rm
This result is stronger than that in \cite{l2}. Indeed,
corresponding to the same problem \eqref{e1.1}-\eqref{e1.5} with $K(t,u)\equiv 0$ 
and $H(s)=hs$, $h>0$ the following assumptions made in \cite{l2} are not
needed here:
$0<\alpha <1$, $B_1(|u|)\in L^{2/(1-\alpha)}(Q_{T})$
for all $u\in L^{\infty }(0,T;V)$ and all $T>0$;  %\label{e2.4}
$B_1$, $B_{2}$ are  nondecreasing functions. %\label{e2.5}
\end{remark}

\begin{proof}[Proof of Theorem \protect\ref{thm1}]
It is done in several steps.

\noindent \textit{Step 1. The Galerkin approximation.} Consider the
orthonormal basis on $V$ consisting of eigenvectors of the Laplacian, 
$-\partial ^{2}/\partial x^{2}$, 
\begin{equation*}
w_{j}(x)=\sqrt{2/(1+\lambda _{j}^{2})}\cos (\lambda _{j}x),\quad \lambda
_{j}=(2j-1)\frac{\pi }{2},\quad j=1,2,\dots .
\end{equation*}
Put 
\begin{equation*}
u_{m}(t)=\sum_{j=1}^{m}c_{mj}(t)w_{j},
\end{equation*}
where $c_{mj}(t)$ satisfy the system of nonlinear differential equations 
\begin{gather}
\langle u_{m}''(t),w_{j}\rangle
+a(u_{m}(t),w_{j})+P_{m}(t)w_{j}(0)+\langle f(u_{m}(t),u_{m}'(t)),w_{j}\rangle =0, 
\label{e2.6} \\
P_{m}(t)=g(t)+H(u_{m}(0,t))-\int_{0}^{t}K(t-s,u_{m}(0,s))ds,  \label{e2.7}
\end{gather}
with 
\begin{equation}
\begin{gathered} u_{m}(0)=u_{0m}=\sum_{j=1}^m \alpha _{mj}w_{j}\to u_0
\quad\text{strongly in }H^{1}, \\ u_{m}'(0)=u_{1m}=\sum_{j=1}^m \beta
_{mj}w_{j}\to u_1 \quad \text{strongly in }L^2, \end{gathered}  \label{e2.8}
\end{equation}
This system of equations is rewritten in form 
\begin{gather*}
c_{mj}''(t)+\lambda _{j}^{2}c_{mj}(t)=\frac{-1}{\Vert
w_{j}\|^{2}}(P_{m}(t)w_{j}(0)+\langle f(u_{m}(t),u_{m}'(t)),w_{j}\rangle ), \\
P_{m}(t)=g(t)+H(u_{m}(0,t))-\int_{0}^{t}K(t-s,u_{m}(0,s))ds, \\
c_{mj}(0)=\alpha _{mj},\quad c_{mj}'(0)=\beta _{mj},\quad 1\leq
j\leq m.
\end{gather*} % \eqref{e2.9}
This system is equivalent to the system of integrodifferential equations 
\begin{equation}
\begin{aligned} &c_{mj}(t)\\ 
&=G_{mj}(t)-\frac{1}{\|w_{j}\|^2} \int_0^t
N_{j}(t-\tau )( H(u_{m}(0,\tau ))w_{j}(0) +\langle f(u_{m}(\tau
),u_{m}'(\tau )),w_{j}\rangle ) d\tau \\ 
&\quad
+\frac{w_{j}(0)}{\|w_{j}\|^2} \int_0^t N_{j}(t-\tau )d\tau \int_0^\tau
K(\tau-s,u_{m}(0,s))ds, \quad 1\leq j\leq m, \end{aligned}  \label{e2.10}
\end{equation}
where $N_{j}(t)=\sin (\lambda _{j}t)/\lambda _{j}$ and 
\begin{equation} \label{e2.11}
G_{mj}(t)=\alpha _{mj}N_{j}'(t)+\beta _{mj}N_{j}(t)
-\frac{w_{j}(0)}{\|w_{j}\|^{2}}\int_{0}^{t}N_{j}(t-\tau )g(\tau )d\tau .
\end{equation}
We then have the following lemma.

\begin{lemma} \label{lem2}
 Let (A), (G), (H), (K1), (K2), (F1),(F3) hold.
For fixed $T>0$, the system \eqref{e1.10}-\eqref{e1.11} has solution
$c_{m}=(c_{m1},c_{m2},\dots ,c_{mm})$  on
an interval $[0,T_{m}]\subset [ 0,T)$.
\end{lemma}

\begin{proof}
Omitting the index $m$, system \eqref{e2.10}, \eqref{e2.11} is rewritten in
the form 
\begin{equation*}
c=Uc,
\end{equation*}
where $c=(c_{1},c_{2},\dots ,c_{m})$, $Uc=((Uc)_{1},(Uc)_{2},\dots ,(Uc)_{m})
$, 
\begin{gather}
(Uc)_{j}(t)=G_{j}(t)+\int_{0}^{t}N_{j}(t-\tau )(Vc)_{j}(\tau )d\tau ,
\label{e2.12} \\
(Vc)_{j}(t)=f_{1j}(c(t),c'(t))+\int_{0}^{t}f_{2j}(t-s,c(s))ds,
\label{e2.13} \\
G_{j}(t)=\alpha _{mj}N_{j}'(t)+\beta _{mj}N_{j}(t)
-\frac{w_{j}(0)}{\|w_{j}\|^{2}}\int_{0}^{t}N_{j}(t-\tau )g(\tau )d\tau ,
\label{e2.14}
\end{gather}
the functions $f_{1j}:\mathbb{R}^{2m}\rightarrow \mathbb{R}$ 
$f_{2j}:[0,T_{m}]\times \mathbb{R}^{m}\rightarrow \mathbb{R}$ satisfy 
\begin{gather}
f_{1j}(c,d)=\frac{-1}{\|w_{j}\|^{2}}\big[H(
\sum_{i=1}^{m}c_{i}w_{i}(0))w_{j}(0)+\langle
f(\sum_{i=1}^{m}c_{i}w_{i},\sum_{i=1}^{m}d_{i}w_{i}),w_{j}\rangle \big],
\label{e2.15} \\
f_{2j}(t,c)=\frac{w_{j}(0)}{\|w_{j}\|^{2}}K(t,
\sum_{i=1}^{m}c_{i}w_{i}(0)),\quad 1\leq j\leq m.  \label{e2.16}
\end{gather}%
For every $T_{m}>0$, $M>0$ we put 
\begin{gather*}
S=\{c\in C^{1}([0,T_{m}];\mathbb{R}^{m}):\|c\|_{1}\leq M\},\quad
\|c\|_{1}=\|c\|_{0}+\|c'\|_{0}, \\
\|c\|_{0}=\sup_{0\leq t\leq T_{m}}|c(t)|_{1},\quad
|c(t)|_{1}=\sum_{i=1}^{m}|c_{i}(t)|.
\end{gather*}%
Clearly $S$ is a closed convex and bounded subset of $Y=C^{1}([0,T_{m}];
\mathbb{R}^{m})$. Using the Schauder fixed point theorem we shall show that
the operator $U:S\rightarrow Y$ defined by \eqref{e2.12}-\eqref{e2.16} has a
fixed point. This fixed point is the solution of \eqref{e2.10}.

\noindent (a) First we show that $U$ maps $S$ into itself. Note that 
$(Vc)_{j}\in C^{0}([0,T_{m}];\mathbb{R})$ for all $c\in C^{1}([0,T_{m}];
\mathbb{R}^{m})$, hence it follows from \eqref{e2.12}, and the equality 
\begin{equation}
(Uc)_{j}'(t)=G_{j}'(t)+\int_{0}^{t}N_{j}'(t-\tau
)(Vc)_{j}(\tau )d\tau ,  \label{e2.17}
\end{equation}
that $U:Y\rightarrow Y$. Let $c\in S$, we deduce from \eqref{e2.11}, 
\eqref{e2.16} that 
\begin{gather}
|(Uc)(t)|_{1}\leq |G(t)|_{1}+\frac{1}{\lambda _{1}}T_{m}\|Vc\|_{0},
\label{e2.18} \\
|(Uc)'(t)|_{1}\leq |G'(t)|_{1}+T_{m}\|Vc\|_{0}.
\label{e2.19}
\end{gather}
On the other hand, it follows from (H), (K1), (K2),(F2),(F3), \eqref{e2.13}, 
\eqref{e2.15}, \eqref{e2.16} that 
\begin{equation}
\|Vc\|_{0}\leq
\sum_{j=1}^{m}[N_{1}(f_{1j},M)+TN_{2}(f_{2j},M,T)]\equiv \beta (M,T)\quad 
\text{for all }c\in S,  \label{e2.20}
\end{equation}
where 
\begin{equation}
\begin{gathered} N_1(f_{1j},M)=\sup \{|f_{1j}(y,z)|:\|y\|_{\mathbb{R}^{m}}
\leq M,\quad \|z\|_{\mathbb{R}^{m}}\leq M\}, \\ N_{2}(f_{2j},M,T)=\sup
\{|f_{2j}(t,y)|:0\leq t\leq T,\quad \|y\|_{\mathbb{R}^{m}}\leq M\}.
\end{gathered}  \label{e2.21}
\end{equation}
Hence, from \eqref{e2.18}-\eqref{e2.21} we obtain 
\begin{equation*}
\|Uc\|_{1}\leq \|G\|_{1T}+(1+\frac{1}{\lambda _{1}}%
)T_{m}\beta (M,T),
\end{equation*}
where 
\begin{equation*}
\|G\|_{1T}=\|G\|_{0T}+\|G'\|_{0T}
=\sup_{0\leq t\leq T}|G(t)|_{1}+\sup_{0\leq t\leq T}|G'(t)|_{1}.
\end{equation*}
Choosing $M$ and $T_{m}>0$ such that 
\begin{equation*}
M>2\|G\|_{1T}\quad \text{and}\quad (1+\frac{1}{\lambda _{1}}
)T_{m}\beta (M,T)\leq M/2.
\end{equation*}
Hence, $\|Uc\|_{1}\leq M$ for all $c\in S$, that is, the operator $U$
maps $S$ the set into itself.

\noindent (b) Now we show that the operator $U$ is continuous on $S$. Let 
$c,d\in S$, we have 
\begin{equation*}
(Uc)_{j}(t)-(Ud)_{j}(t)=\int_{0}^{t}N_{j}(t-\tau )[(Vc)_{j}(\tau
)-(Vd)_{j}(\tau )]d\tau .
\end{equation*}
Hence 
\begin{equation}
\|Uc-Ud\|_{0}\leq \frac{1}{\lambda _{1}}T_{m}\|Vc-Vd\|_{0}.
\label{e2.22}
\end{equation}
Similarly, we obtain from the equality 
\begin{equation*}
(Uc)_{j}'(t)-(Ud)_{j}'(t)=\int_{0}^{t}N_{j}'(t-\tau )((Vc)_{j}(\tau )-(Vd)_{j}(\tau ))d\tau ,
\end{equation*}
which implies 
\begin{equation}
\|(Uc)'-(Ud)'\|_{0}\leq T_{m}\|Vc-Vd\|_{0}.
\label{e2.23}
\end{equation}
By estimates \eqref{e2.22}, \eqref{e2.23}, we only have to prove that the
operator $V:Y\rightarrow C^{0}([0,T_{m}];\mathbb{R}^{m})$ is continuous on 
$S$. We have 
\begin{equation}
\begin{aligned}
(Vc)_{j}(t)-(Vd)_{j}(t)&=f_{1j}(c(t),c'(t))-f_{1j}(d(t),d'(t)) \\ 
&\quad +
\int_0^t (f_{2j}(t-s,c(s))-f_{2j}(t-s,d(s))) ds. \end{aligned}  \label{e2.24}
\end{equation}
 From the assumptions (H),(F2) and (F3), it follows that there exists a
constant $K_{M}>0$ such that 
\begin{equation}
\sup_{0\leq t\leq T_{m}}\sum_{j=1}^{m}|f_{1j}(c(t),c'(t))-f_{1j}(d(t),d'(t))|
\leq K_{M}(\|c-d\|_{0}+\Vert
c-d\|_{0}^{\beta }+\|c'-d'\|_{0}^{\alpha }),
\label{e2.25}
\end{equation}
for all $c,d\in S$. Then we have the following lemma.

\begin{lemma} \label{lem3}
 Let $f_{2j}:[0,T_{m}]\times\mathbb{R}^{m}\to R$ be continuous, and let
\begin{equation} \label{e2.26}
(W_{j}c)(t)=\int_0^t f_{2j}(t-s,c(s))ds,c\in C^{0}([0,T_{m}];\mathbb{R}^{m}).
\end{equation}
Then, the operator
$W_{j}:C^{0}([0,T_{m}];\mathbb{R}^{m})\to C^{0}([0,T_{m}];\mathbb{R})$
is continuous on $S$.
\end{lemma}

The proof of this lemma follows easily from $f_{2j}$ being uniformly
continuous on $[0,T_{m}]\times [-M,M]^{m}$. We omit the proof.

 From \eqref{e2.24}, \eqref{e2.25}, \eqref{e2.26}, we deduce that 
\begin{equation}
\begin{aligned} \|Vc-Vd\|_0 &=\sup_{0\leq \tau \leq T_{m}} \sum_{j=1}^m
|(Vc)_{j}(\tau )-(Vd)_{j}(\tau)|\\ &\leq K_{M}\big(
\|c-d\|_0+\|c-d\|_0^{\beta }+\|c'-d'\|_0^{\alpha }\big)\\ &\quad
+\sup_{0\leq t\leq T_{m}}\sum_{j=1}^m |(W_{j}c)(t)-(W_{j}d)(t)| , \quad
\forall c,d\in S\,. \end{aligned}  \label{e2.27}
\end{equation}
Thus, Lemma \ref{lem3} and inequality \eqref{e2.27} show that 
$V:S\rightarrow C^{0}([0,T_{m}];\mathbb{R}^{m})$ is continuous.

\noindent (c) Now, we shall show that the set $\overline{US}$ is a compact
subset of $Y$. Let $c\in S,t,t'\in [ 0,T_{m}]$. From 
\eqref{e2.12}, we rewrite 
\begin{equation}
\begin{aligned} &(Uc)_{j}(t)-(Uc)_{j}(t')\\ &=G_{j}(t)-G_{j}(t')+\int_0^t
N_{j}(t-\tau )(Vc)_{j}(\tau )d\tau -\int_0^{t'} N_{j}(t'-\tau )(Vc)_{j}(\tau
)d\tau \\ &=G_{j}(t)-G_{j}(t')+\int_0^t ( N_{j}(t-\tau)-N_{j}(t'-\tau ))
(Vc)_{j}(\tau )d\tau\\ &\quad -\int_t^{t'} N_{j}(t'-\tau )(Vc)_{j}(\tau
)d\tau \,. \end{aligned}  \label{e2.28}
\end{equation}
 From the inequality %\label{e2.29}
$|N_{j}(t)-N_{j}(s)|\leq |t-s|$ for all $t,s\in [ 0,T_{m}]$ and 
\eqref{e2.20}, we obtain 
\begin{equation}
\begin{aligned} |(Uc)(t)-(Uc)(t')|_1 
&=\sum_{j=1}^m |(Uc)_{j}(t)-(Uc)_{j}(t')|\\ 
&\leq |G(t)-G(t')|_1+(T_{m}+\frac{1}{\lambda
_1})|t-t'|\|Vc\|_0 \\ 
&\leq |G(t)-G(t')|_1+\beta
(M,T)(T_{m}+\frac{1}{\lambda _1})|t-t'|. \end{aligned}  \label{e2.30}
\end{equation}
Similarly, from \eqref{e2.17} and \eqref{e2.20}, we also obtain 
\begin{equation}
|(Uc)'(t)-(Uc)'(t')|_{1}\leq |G'(t)-G'(t')|_{1}+\beta (M,T)(\lambda
_{m}T_{m}+1)|t-t'|.  \label{e2.31}
\end{equation}
Since $US\subset S$, from estimates \eqref{e2.30}, \eqref{e2.31} we deduce
that the family of functions $US=\{Uc,c\in S\}$, are bounded and
equicontinuous with respect to the norm $\|\cdot \|_{1}$ of the
space $Y.$ Applying Arzela-Ascoli's theorem to the space $Y$, we deduce that 
$\overline{US}$  is compact in $Y$. By the Schauder fixed-point theorem, $U$
has a fixed point $c\in S$, which satisfies \eqref{e2.10}. The proof of
Lemma \ref{lem2} is complete.
\end{proof}

Using Lemma \ref{lem2}, for $T>0$, fixed, system \eqref{e2.6} - \eqref{e2.8}
has solution $(u_{m}(t),P_{m}(t))$ on an interval $[0,T_{m}]$. The following
estimates allow one to take $T_{m}=T$ for all $m$.

\noindent \textit{Step 2. A priori estimates}. Substituting \eqref{e2.7}
into \eqref{e2.6}, then multiplying the $j^{th}$ equation of \eqref{e2.6} by 
$c_{mj}'(t)$ and summing up with respect to $j$, integrating by
parts with respect to the time variable from $0$ to $t$, by (G) and (F1), we
have 
\begin{equation}
\begin{aligned} S_{m}(t)&\leq -2\widehat{H}(u_{m}(0,t))+2\widehat{H}
(u_{0m}(0))+S_{m}(0)+2g(0)u_{0m}(0) \\ 
&\quad -2g(t)u_{m}(0,t)+2 \int_0^t
g'(s)u_{m}(0,s)ds -2 \int_0^t \langle f(u_{m}(s),0),u_{m}'(s)\rangle ds \\
&\quad +2 \int_0^t u_{m}'(0,s)ds \int_0^s K(s-\tau ,u_{m}(0,\tau ))d\tau ,
\end{aligned}  \label{e2.32}
\end{equation}
where 
\begin{equation}
S_{m}(t)=\|u_{m}'(t)\|^{2}+\|u_{m}(t)\|_{V}^{2}\,.
\label{e2.33}
\end{equation}
Then, using \eqref{e2.8}, \eqref{e2.33}, (H), and Lemma \ref{lem1}, we have 
\begin{equation}
\begin{aligned}
&-2\widehat{H}(u_{m}(0,t))+2\widehat{H}(u_{0m}(0))+S_{m}(0)+2|
g(0)u_{0m}(0)| \\ 
&\leq 2h_0+2\widehat{H} (u_{0m}(0))+S_{m}(0)+2|g(0)u_{0m}(0)|\\ 
&\leq \frac{1}{4}C_1,\quad \text{for
all $m$ and all $t$}, \end{aligned}  \label{e2.34}
\end{equation}
where $C_{1}$ is a constant depending only on $u_{0}$, $u_{1}$, $h_{0}$, $H$, 
and $g$.

Again using Lemma \ref{lem1} and the inequality %\label{e2.35}
$2ab\leq 4a^{2}+\frac{1}{4}b^{2}$, we obtain 
\begin{equation}
\begin{aligned} &|-2g(t)u_{m}(0,t)+2\int_0^t g'(s)u_{m}(0,s)ds| \\ 
&\leq
4g^2(t)+4 \int_0^t |g'(s)|^2ds+\frac{1}{4}S_{m}(t) +\frac{1}{4} \int_0^t
S_{m}(s)ds. \end{aligned}  \label{e2.36}
\end{equation}
Using Lemma \ref{lem1}, from (F3) it follows that 
\begin{align*}
\big|-2\int_{0}^{t}\langle f(u_{m}(s),0),u_{m}'(s)\rangle ds\big|&
\leq 2B_{2}(0)\int_{0}^{t}S_{m}(s)^{(1+\beta )/2}ds \\
& \leq (1+\beta )B_{2}(0)\int_{0}^{t}S_{m}(s)ds+(1-\beta )B_{2}(0)t.
\end{align*}
Note that the last integral in \eqref{e2.32}, after integrating by parts,
gives 
\begin{align*}
I& =2\int_{0}^{t}u_{m}'(0,s)ds\int_{0}^{s}K(s-\tau ,u_{m}(0,\tau
))d\tau \\
& =2u_{m}(0,t)\int_{0}^{t}K(t-\tau ,u_{m}(0,\tau ))d\tau \\
& \quad -2\int_{0}^{t}u_{m}(0,s)ds\big[K(0,u_{m}(0,s))+\int_{0}^{s}
\frac{\partial K}{\partial t}(s-\tau ,u_{m}(0,\tau ))d\tau \big].
\end{align*}
Hence 
\begin{equation}
\begin{aligned} |I|&\leq 2\sqrt{S_{m}(t)} \int_0^t ( k_1(t-\tau
)\sqrt{S_{m}(\tau )}+k_{2}(t-\tau )) d\tau \\ &\quad +2 \int_0^t
\sqrt{S_{m}(s)}ds\big[ k_1(0)\sqrt{S_{m}(s)}+k_{2}(0)\\ &\quad +\int_0^s
(k_{3}(s-\tau )\sqrt{S_{m}(\tau )} +k_{4}(s-\tau )) d\tau \big] \\
&=2\sqrt{S_{m}(t)} \int_0^t k_1(t-\tau )\sqrt{S_{m}(\tau )}d\tau
+2\sqrt{S_{m}(t)} \int_0^t k_{2}(\tau )d\tau \\ &\quad +2k_1(0) \int_0^t
S_{m}(s)ds+2k_{2}(0) \int_0^t \sqrt{S_{m}(s)}ds \\ &\quad +2 \int_0^t
\sqrt{S_{m}(s)}ds \int_0^s k_{3}(s-\tau )\sqrt{S_{m}(\tau )}d\tau +2
\int_0^t \sqrt{S_{m}(s)}ds \int_0^s k_{4}(\tau)d\tau \\ &\equiv
I_1+I_{2}+2k_1(0) \int_0^t S_{m}(s)ds+I_{4}+I_{5}+I_{6}. \end{aligned}
\label{e2.37}
\end{equation}
By the inequality $2ab\leq 4a^{2}+\frac{1}{4}b^{2}$ and the Cauchy- Schwarz
inequality we estimate without difficulty the following integrals in the
right-hand side of the above expression as follows 
\begin{gather*}
I_{1}=2\sqrt{S_{m}(t)}\int_{0}^{t}k_{1}(t-\tau )\sqrt{S_{m}(\tau )}d\tau
\leq \frac{1}{4}S_{m}(t)+4\int_{0}^{t}k_{1}^{2}(\tau )d\tau
.\int_{0}^{t}S_{m}(\tau )d\tau , \\
I_{2}=2\sqrt{S_{m}(t)}\int_{0}^{t}k_{2}(\tau )\leq \frac{1}{4}S_{m}(t)+4\Big(
\int_{0}^{t}k_{2}(\tau )d\tau \Big)^{2}, \\
I_{4}=2k_{2}(0)\int_{0}^{t}\sqrt{S_{m}(s)}ds\leq 4k_{2}^{2}(0)+\frac{1}{4}
t\int_{0}^{t}S_{m}(s)ds, \\
I_{5}=2\int_{0}^{t}\sqrt{S_{m}(s)}ds\int_{0}^{s}k_{3}(s-\tau )\sqrt{
S_{m}(\tau )}d\tau \leq 2\sqrt{t}\Big(\int_{0}^{t}k_{3}^{2}(\tau )d\tau \Big)
^{1/2}\int_{0}^{t}S_{m}(s)ds, \\
I_{6}=2\int_{0}^{t}\sqrt{S_{m}(s)}ds\int_{0}^{s}k_{4}(\tau )d\tau \leq \frac{
1}{4}\int_{0}^{t}S_{m}(s)ds+4t\Big(\int_{0}^{t}k_{4}(\tau )d\tau \Big)^{2}.
\end{gather*}
It follows from the estimates for $I_{1},I_{2},I_{4},I_{5},I_{6}$ that 
\begin{equation}
\begin{aligned} |I|&\leq 4\Big( \int_0^t k_{2}(\tau )d\tau \Big) ^2
+4k_{2}^2(0)+4t\Big( \int_0^t k_{4}(\tau )d\tau \Big) ^2
+\frac{1}{2}S_{m}(t) \\ &\quad +\frac{1}{4}\Big[ 1+t +16 \int_0^t k_1^2(\tau
)d\tau +8k_1(0) +8\sqrt{t} \Big( \int_0^t k_{3}^2(\tau )d\tau
\Big)^{1/2}\Big] \int_0^t S_{m}(s)ds. \end{aligned}  \label{e2.38}
\end{equation}
It follows from \eqref{e2.32}-\eqref{e2.34}, \eqref{e2.36}-\eqref{e2.37},
and \eqref{e2.38} that 
\begin{equation}
S_{m}(t)\leq D_{1}(t)+D_{2}(t)\int_{0}^{t}S_{m}(\tau )d\tau ,  \label{e2.39}
\end{equation}
where 
\begin{equation}
\begin{aligned} D_1(t)&=C_1+16k_{2}^2(0)+4(1-\beta )B_{2}(0)t+16g^2(t) \\
&\quad +16 \int_0^t |g'(s)|^2ds+16\Big( \int_0^t k_{2}(\tau )d\tau \Big) ^2
+16t\Big( \int_0^t k_{4}(\tau )d\tau \Big) ^2, \end{aligned}  \label{e2.40}
\end{equation}
\begin{align*}
D_{2}(t)& =2+4(1+\beta )B_{2}(0)+8k_{1}(0)+t+\int_{0}^{t}k_{1}^{2}(\tau
)d\tau +8\sqrt{t}\Big(\int_{0}^{t}k_{3}^{2}(\tau )d\tau \Big)^{1/2} \\
& \leq 2+4(1+\beta )B_{2}(0)+8k_{1}(0)+T+\|k_{1}\|_{L^{2}(0,T)}^{2}+8
\sqrt{T}\|k_{3}\|_{L^{2}(0,T)}\equiv C_{T}^{(2)}.
\end{align*}
Since $H^{1}(0,T)\hookrightarrow C^{0}([0,T])$, from the assumptions (G),
(K2), we deduce that 
\begin{equation}
|D_{1}(t)|\leq C_{T}^{(1)},\quad \text{a.e. in }[0,T],  \label{e2.42}
\end{equation}
where $C_{T}^{(1)}$, is a constant depending only on $T$. By Gronwall's
lemma, from \eqref{e2.39}- \eqref{e2.42} we obtain that 
\begin{equation}
S_{m}(t)\leq C_{T}^{(1)}\exp (tC_{T}^{(2)})\leq C_{T}\quad \forall t\in
[ 0,T],\;\forall T>0.  \label{e2.43}
\end{equation}
Now we need an estimate on the integral $\int_{0}^{t}|u_{m}'(0,s)|^{2}ds$. Put 
\begin{equation}
K_{m}(t)=\sum_{j=1}^{m}\frac{\sin (\lambda _{j}t)}{\lambda _{j}},
\label{e2.44}
\end{equation}

\begin{align*}
\gamma _{m}(t)& =\sum_{j=1}^{m}w_{j}(0)[\alpha _{mj}\cos (\lambda
_{j}t)+\beta _{mj}\frac{\sin (\lambda _{j}t)}{\lambda _{j}}] \\
& \quad -\sqrt{2}\sum_{j=1}^{m}\int_{0}^{t}\frac{\sin [\lambda _{j}(t-\tau )]%
}{\lambda _{j}}\langle f(u_{m}(\tau ),u_{m}'(\tau )),\frac{w_{j}}{\|w_{j}\|}\rangle
 d\tau .
\end{align*}
Then $u_{m}(0,t)$ can be rewritten as 
\begin{equation}
u_{m}(0,t)=\gamma _{m}(t)-2\int_{0}^{t}K_{m}(t-\tau )P_{m}(\tau )d\tau .
\label{e2.45}
\end{equation}
We shall require the following lemma which proof can be found in \cite{a2}.

\begin{lemma} \label{lem4}
There exist a constant $C_{2}>0$  and a positive
continuous function $D(t)$ independent of $m$  such that
\[ %\label{e2.46}
\int_0^t |\gamma _{m}'(\tau )|
^2d\tau \leq C_{2}+D(t)\int_0^t \|
f(u_{m}(\tau ),u_{m}'(\tau ))\|^2d\tau \quad \forall
t\in [ 0,T],\forall T>0.
\]
\end{lemma}

\begin{lemma} \label{lem5}
There exist two positive constants $C_{T}^{(3)}$
and $C_{T}^{(4)}$ depending only on $T$ such that
\begin{equation} \label{e2.47}
\int_0^t ds|\int_0^s
K_{m}'(s-\tau )P_{m}(\tau )d\tau |^2\leq
C_{T}^{(3)}+C_{T}^{(4)}\int_0^t ds\int_0^s |u_{m}'(0,\tau )|^2d\tau ,
\end{equation}
for all $t\in [ 0,T]$ and all $T>0$.
\end{lemma}

\begin{proof}
Integrating by parts, we have 
\begin{equation*}
\int_{0}^{s}K_{m}'(s-\tau )P_{m}(\tau )d\tau
=K_{m}(s)P_{m}(0)+\int_{0}^{t}K_{m}(s-\tau )P_{m}'(\tau )d\tau ,
\end{equation*}
then 
\begin{equation}
\begin{aligned} &\int_0^t ds|\int_0^s K_{m}'(s-\tau )P_{m}(\tau )d\tau |^2\\
&\leq 2P_{m}^2(0) \int_0^t K_{m}^2(s)ds + 2 \int_0^t ds\int_0^s
K_{m}^2(r)dr\int_0^s |P_{m}'(\tau)|^2d\tau \\ &\leq 2\int_0^t K_{m}^2(s)ds
\big[ P_{m}^2(0) +\int_0^t ds\int_0^s |P_{m}'(\tau )|^2d\tau \big] .
\end{aligned}  \label{e2.48}
\end{equation}
 From \eqref{e2.7}, we have 
\begin{equation}
P_{m}(0)=g(0)+H(u_{0m}(0)),  \label{e2.49}
\end{equation}
\begin{equation}
P_{m}'(\tau )=g'(\tau )+H'(u_{m}(0,\tau
))u_{m}'(0,\tau )-K(0,u_{m}(0,\tau ))-\int_{0}^{\tau }
\frac{\partial K}{\partial t}(\tau -s,u_{m}(0,s))ds.  \label{e2.50}
\end{equation}
Using the inequality $(a+b+c+d)^{2}\leq 4(a^{2}+b^{2}+c^{2}+d^{2})$, 
for all $a,b,c,d\in \mathbb{R}$, we deduce from \eqref{e2.43}, \eqref{e2.50},
and (G),(H),(K2) that 
\begin{equation}
\begin{aligned} &\int_0^s |P_{m}'(\tau )| ^2d\tau \\ 
&\leq 4 \int_0^s
|g'(\tau)|^2d\tau +4\max_{|s|\leq \sqrt{C_{T}}} |H'(s)|^2\int_0^s
|u_{m}'(0,\tau )|^2d\tau \\ &\quad +4 \int_0^s |K(0,u_{m}(0,\tau ))|^2d\tau
+4 \int_0^s d\tau | \int_0^\tau \frac{\partial K}{\partial t}(\tau
-s,u_{m}(0,s))ds|^2 \\ &\leq 4 \int_0^s |g'(\tau )|^2d\tau +4\max_{|s|\leq
\sqrt{C_{T}}} |H'(s)|^2 \int_0^s |u_{m}'(0,\tau )|^2d\tau \\ 
&\quad
+8k_1^2(0)\int_0^s |u_{m}(0,\tau)|^2d\tau +8k_{2}^2(0)s \\ 
&\quad +8
\int_0^s d\tau \int_0^\tau k_{3}^2(s)ds\int_0^\tau u_{m}^2(0,s)ds+8 \int_0^s
d\tau ( \int_0^\tau k_{4}(s)ds) ^2 \\ &\leq 4 \int_0^s |g'(\tau )|^2d\tau
+8[k_1^2(0)C_{T}+k_{2}^2(0)]s +4C_{T} s^2 \int_0^s k_{3}^2(\tau )d\tau \\
&\quad +8s( \int_0^s k_{4}(\tau )d\tau ) ^2+4 \max_{|s|\leq \sqrt{C_{T}}}
|H'(s)|^2\int_0^s | u_{m}'(0,\tau )|^2d\tau . \end{aligned}  \label{e2.51}
\end{equation}
Hence 
\begin{align*}
\int_{0}^{t}ds\int_{0}^{s}|P_{m}'(\tau )|^{2}d\tau & \leq
4t\int_{0}^{t}|g'(\tau )|^{2}d\tau
+4[k_{1}^{2}(0)C_{T}+k_{2}^{2}(0)]t^{2}\quad \\
& \quad +\frac{4}{3}C_{T}t^{3}\int_{0}^{t}k_{3}^{2}(\tau )d\tau +4t^{2}\Big(%
\int_{0}^{t}k_{4}(\tau )d\tau \Big)^{2} \\
& \quad +4\max_{|s|\leq \sqrt{C_{T}}}|H'(s)|^{2}\int_{0}^{t}\,ds\int_{0}^{s}|u_{m}'(0,\tau )|^{2}d\tau .
\end{align*}%
 From this inequality, \eqref{e2.48}, and \eqref{e2.49}, it follows that 
\begin{equation}
\begin{aligned} &\int_0^t ds|\int_0^s K_{m}'(s-\tau )P_{m}(\tau )d\tau |^2\\ 
&\leq 2 \int_0^t K_{m}^2(s)ds \Big[ ( g(0)+H(u_{0m}(0)))^2 +4t \int_0^t
|g'(\tau )|^2d\tau +4[k_1^2(0)C_{T}+k_{2}^2(0)]t^2 \\ 
&\quad
+\frac{4}{3}C_{T} t^{3} \int_0^t k_{3}^2(\tau )d\tau +4t^2\Big( \int_0^t
k_{4}(\tau )d\tau \Big) ^2\\ &\quad + 4\max_{|s|\leq \sqrt{C_{T}}}
|H'(s)|^2\int_0^t \,ds\int_0^s | u_{m}'(0,\tau )|^2d\tau \Big] .
\end{aligned}  \label{e2.52}
\end{equation}
Note that for every $T>0$, $K_{m}\rightarrow \widetilde{K}$, strongly in 
$L^{2}(0,T)$ as $m\rightarrow +\infty $. Using the assumptions (G), (H),(K2)
and the results \eqref{e2.8} and \eqref{e2.52}, we obtain \eqref{e2.47}. The
proof of Lemma \ref{lem5} is complete.
\end{proof}

\begin{lemma} \label{lem6}
There exist two positive constants $C_{T}^{(5)}$
and $C_{T}^{(6)}$  depending only on $T$  such that
\begin{gather} \label{e2.53}
\int_0^t |u_{m}'(0,\tau )|
^2d\tau \leq C_{T}^{(5)}\quad \forall t\in [ 0,T],\forall T>0. \\
\label{e2.54}
\int_0^t |P_{m}'(\tau )|^2d\tau \leq C_{T}^{(6)}\quad \forall t\in [0,T],
\forall T>0.
\end{gather}
\end{lemma}

\begin{proof}
Since \eqref{e2.54} is a consequence of \eqref{e2.51} and \eqref{e2.53}, we
only have to prove \eqref{e2.53}. From \eqref{e2.45}, using Lemmas \ref{lem4}
and \ref{lem5}, we obtain 
\begin{equation}
\begin{aligned} \int_0^t |u_{m}'(0,s)|^2ds 
&\leq 2 \int_0^t
|\gamma_{m}'(s)|^2ds +8 \int_0^t ds|\int_0^s K_{m}'(s-\tau )P_{m}(\tau)d\tau
|^2 \\ 
&\leq 2C_{2}+2D(t)\int_0^t \|f(u_{m}(\tau),u_{m}'(\tau ))\|d\tau \\
&\quad +8C_{T}^{(3)}+8C_{T}^{(4)}\int_0^t ds \int_0^s |u_{m}'(0,\tau
)|^2d\tau . \end{aligned}  \label{e2.55}
\end{equation}
On the other hand, from the assumptions (F2),(F3), we obtain 
\begin{equation}
\|f(u_{m}(t),u_{m}'(t))\|^{2}\leq 2(\max_{|s|
\leq \sqrt{C_{T}}}B_{1}^{2}(s))\|u_{m}'(t)\|^{2\alpha
}+2B_{2}^{2}(0)\|u_{m}(t)\|_{V}^{2\beta },  \label{e2.56}
\end{equation}
since $0<\alpha \leq 1$ we have $\|\cdot \|\leq \|\cdot \Vert
_{L^{2\alpha }}$. Hence, using \eqref{e2.43} and \eqref{e2.56} we have 
\begin{equation}
\|f(u_{m}(t),u_{m}'(t))\|\leq C_{T}^{(7)}.  \label{e2.57}
\end{equation}
At last from this inequality and \eqref{e2.55} we obtain the inequality 
\begin{equation*}
\int_{0}^{t}|u_{m}'(0,s)|^{2}ds\leq
C_{T}^{(8)}+8C_{T}^{(4)}\int_{0}^{t}ds\int_{0}^{s}|u_{m}'(0,\tau
)|^{2}d\tau ,
\end{equation*}
which implies \eqref{e2.53}, by Gronwall's lemma. Therefore, Lemma \ref{lem6}
is proved.
\end{proof}

\noindent \textit{Step 3. Passing to limit}. From \eqref{e2.7}, \eqref{e2.33}, 
\eqref{e2.43}, \eqref{e2.53}, \eqref{e2.54}, and \eqref{e2.57}, we deduce
that, there exists a subsequence of sequence $\{(u_{m},P_{m})\}$, still
denoted by $\{(u_{m},P_{m})\}$, such that 
\begin{gather}
u_{m}\rightarrow u\quad \text{in }L^{\infty }(0,T;V)\text{ weak}\ast ,
\label{e2.58} \\
u_{m}'\rightarrow u'\quad \text{in }L^{\infty }(0,T;L^{2})%
\text{ weak}\ast ,  \label{e2.59} \\
u_{m}(0,t)\rightarrow u(0,t)\quad \text{in }L^{\infty }(0,T)\text{ weak}\ast
,  \label{e2.60} \\
u_{m}'(0,t)\rightarrow u'(0,t)\quad \text{in }L^{2}(0,T)
\text{ weak},  \label{e2.61} \\
f(u_{m},u_{m}')\rightarrow \chi \quad \text{in }L^{\infty
}(0,T;L^{2})\text{ weak}\ast ,  \label{e2.62} \\
P_{m}\rightarrow \widehat{P}\quad \text{in }H^{1}(0,T)\text{ weak},
\label{e2.63}
\end{gather}%
By the compactness lemma of Lions (see \cite{l2}), we can deduce from 
\eqref{e2.58}-\eqref{e2.61} that there exists a subsequence still denoted by 
$\{u_{m}\}$ such that 
\begin{gather}
u_{m}(0,t)\rightarrow u(0,t)\quad \text{strongly in }C^{0}([0,T]),
\label{e2.64} \\
u_{m}\rightarrow u\quad \text{strongly in }L^{2}(Q_{T})\text{ and a.e. }
(x.t)\in Q_{T}.  \label{e2.65}
\end{gather}%
By (H),(K) and using \eqref{e2.7}, \eqref{e2.64} we obtain 
\begin{equation}
P_{m}(t)\rightarrow g(t)+H(u(0,t))-\int_{0}^{t}K(t-s,u(0,s))ds\equiv
P(t)\quad \text{strongly in }C^{0}([0,T]).  \label{e2.66}
\end{equation}
 From \eqref{e2.63} and \eqref{e2.66} we have 
\begin{equation}
P\equiv \widehat{P}\text{ \thinspace a.e. in }Q_{T}.  \label{e2.67}
\end{equation}
Passing to the limit in \eqref{e2.6} by \eqref{e2.58}, \eqref{e2.59}, 
\eqref{e2.66}, and \eqref{e2.67} we have 
\begin{equation*}
\frac{d}{dt}\langle u'(t),v\rangle +a(u(t),v)+P(t)v(0)+\langle \chi
,v\rangle =0\quad \forall v\in V.
\end{equation*}
As in \cite{l2}, we can prove that 
\begin{equation*}
u(0)=u_{0},\quad u'(0)=u_{1}.
\end{equation*}
To prove the existence of solution $u$, we have to show that $\chi
=f(u,u')$. We need the following lemma which proof can be found in 
\cite{a2}.

\begin{lemma} \label{lem7}
Let $u$  be the solution of the problem
\begin{gather*}
u_{tt}-u_{xx}+\chi =0,\quad 0<x<1,\quad 0<t<T, \\
u_{x}(0,t)=P(t),\quad u(1,t)=0, \\
u(x,0)=u_0(x),\quad u_{t}(x,0)=u_1(x),  \\
u\in L^{\infty }(0,T;V),\quad u'\in L^{\infty }(0,T;L^2)\\
u(0,\cdot )\in H^{1}(0,T).
\end{gather*}
Then
\[ % \label{e2.68}
\frac{1}{2}\|u'(t)\|^2+ \frac{1}{2}\|u(t)\|_{V}^2+ \int_0^t
P(s)u'(0,s)ds+ \int_0^t \langle \chi(s),u'(s)\rangle ds
\geq \frac{1}{2}\|u_1\|^2+ \frac{1}{2}\|u_0\|_{V}^2\,,
\]
a.e. $t\in [ 0,T]$.
Furthermore, if $ u_0=u_1=0$ there is equality in the above
expression.
\end{lemma}

Now, from \eqref{e2.6}-\eqref{e2.8} we have 
\begin{equation}
\begin{aligned}
 &\int_0^t \langle f(u_{m}(s),u_{m}'(s)),u_{m}'(s)\rangle ds\\ 
&=\frac{1}{2}\|u_{1m}\|^2+ \frac{1}{2}\|u_{0m}\|_{V}^2
-\frac{1}{2}\|u_{m}'(t)\|^2- \frac{1}{2}\|u_{m}(t)\|_{V}^2 -\int_0^t
P_{m}(s)u_{m}'(0,s))ds. \end{aligned}  \label{e2.69}
\end{equation}
By Lemma \ref{lem7}, it follows from \eqref{e2.8}, \eqref{e2.58}, 
\eqref{e2.59}, \eqref{e2.61}, \eqref{e2.66} and \eqref{e2.69}, that 
\begin{align*}
& \limsup_{m\rightarrow +\infty }\int_{0}^{t}\langle
f(u_{m}(s),u_{m}'(s)),u_{m}'(s)\rangle ds \\
& \leq \frac{1}{2}\|u_{1}\|^{2}+\frac{1}{2}\|u_{0}\|_{V}^{2}-
\frac{1}{2}\|u'(t)\|^{2}-\frac{1}{2}\|u(t)\Vert
_{V}^{2}-\int_{0}^{t}P(s)u'(0,s))ds \\
& \leq \int_{0}^{t}\langle \chi (s),u'(s)\rangle ds,\quad \text{ a.e. }t\in [ 0,T].
\end{align*}%
Using the same arguments as in \cite{l2}, we can show that 
$\chi=f(u,u')$ a.e. in $Q_{T}$. The existence of the solution is proved.

\noindent \textit{Step 4. Uniqueness of the solution}. Assume now that 
$\beta =1$ in (F3), and that $H$, $K$, $f$ satisfy (H1),(K3), and (F4). 
Let $(u_{1},P_{1})$, $(u_{2},P_{2})$ be two weak solutions of the problem 
\eqref{e1.1}-\eqref{e1.5}. Then $u=u_{1}-u_{2}$, $P=P_{1}-P_{2}$ satisfy the
problem 
\begin{gather*}
u''-u_{xx}+\chi =0,\quad 0<x<1,\quad 0<t<T, \\
u_{x}(0,t)=P(t),\quad u(1,t)=0, \\
u(x,0)=u'(x,0)=0, \\
\chi =f(u_{1},u_{1}')-f(u_{2},u_{2}'), \\
\begin{aligned} P(t)&=P_1(t)-P_{2}(t)\\ &=H(u_1(0,t))-H(u_{2}(0,t)) \\
&\quad -\int_0^t (K(t-s,u_1(0,s))-K(t-s,u_{2}(0,s))) ds, \end{aligned} \\
u_{i}\in L^{\infty }(0,T;V),\quad u_{i}'\in L^{\infty
}(0,T;L^{2}),\quad u_{i}(0,\cdot )\in H^{1}(0,T), \\
P_{i}\in H^{1}(0,T),\quad i=1,2.
\end{gather*}
Using Lemma \ref{lem7} with $u_{0}=u_{1}=0$, we obtain 
\begin{equation}
\frac{1}{2}\|u'(t)\|^{2}+\frac{1}{2}\|u(t)\|_{V}^{2}
+\int_{0}^{t}P(s)u'(0,s)ds+\int_{0}^{t}\langle \chi
(s),u'(s)\rangle ds=0\,,  \label{e2.70}
\end{equation}
a.e. $t\in [ 0,T]$. Put 
\begin{gather*}
\sigma (t)=\|u'(t)\|^{2}+\frac{1}{2}\|u(t)\Vert
_{V}^{2}, \\
\widetilde{H}_{1}(t)=H(u_{1}(0,t))-H(u_{2}(0,t)), \\
\widetilde{K}_{1}(t,s)=K(t-s,u_{1}(0,s))-K(t-s,u_{2}(0,s)).
\end{gather*}%
Substituting $P(t)$, $\chi $ into \eqref{e2.70} and using that $f$ is
nondecreasing with respect to the second variable, we have 
\begin{equation}
\begin{aligned} &\sigma (t)+2 \int_0^t \widetilde{H}_1(s)u'(0,s)ds \\ 
&\leq 2 \int_0^t \|f(u_1(s),u_{2}'(s))-f(u_{2}(s),u_{2}'(s))\|\, \|u'(s)\|ds\\
&\quad +2 \int_0^t u'(0,s)ds \int_0^s \widetilde{K}_1(s,r)dr. \end{aligned}
\label{e2.71}
\end{equation}
Using assumption (F3), 
\begin{equation*}
\|f(u_{1}(s),u_{2}'(s))-f(u_{2}(s),u_{2}'(s))\Vert
\leq \|B_{2}(|u_{2}'(s)|)\|\|u(s)\|_{V}\,.
\end{equation*}
Using integration by parts in the last integral of \eqref{e2.71}, we get 
\begin{equation}
\begin{aligned} J
&=2 \int_0^t u'(0,s)ds \int_0^s \widetilde{K}_1(s,r)dr\\
&=2u(0,t) \int_0^t \widetilde{K}_1(t,r)dr-2 \int_0^t u(0,s)ds
\big[\widetilde{K}_1(s,s)+ \int_0^s \frac{\partial \widetilde{K}_1}{\partial
s} (s,r)dr\big]. \end{aligned}  \label{e2.72}
\end{equation}
 From assumption (K3), we have 
\begin{equation}
\begin{gathered} |\widetilde{K}_1(s,r)|\leq p_{M,T}(t-r)| u(0,r)|\leq
p_{M,T}(t-r)\sqrt{\sigma (r)}, \\ 
|\widetilde{K}_1(s,s)|\leq p_{M,T}(0)|
u(0,s)|\leq p_{M,T}(0)\sqrt{\sigma (s)},\\ 
|\frac{\partial
\widetilde{K}_1}{\partial s}(s,r)| \leq q_{M,T}(t-r)|u(0,r)|\leq
q_{M,T}(t-r)\sqrt{\sigma (r)}\,,
 \end{gathered}  \label{e2.73}
\end{equation}
where $M=\max_{i=1,2}\|u_{i}\|_{L^{\infty }(0,T;V)}$. It follows
from \eqref{e2.72} and \eqref{e2.73} that 
\begin{equation}
\begin{aligned} |J|&\leq 2\sqrt{\sigma (t)} \int_0^t
p_{M,T}(t-r)\sqrt{\sigma (r)}dr +2p_{M,T}(0) \int_0^ \sigma(s)\,ds \\ 
&\quad
+2 \int_0^t \sqrt{\sigma (s)}ds \int_0^s q_{M,T}(s-r)\sqrt{\sigma (r)}dr \\
&\leq \beta _1\sigma (t)+\frac{1}{\beta _1} \int_0^t p_{M,T}^2(r)dr \int_0^t
\sigma (r)dr\\ &\quad +2p_{M,T}(0) \int_0^t \sigma (s)ds 2\sqrt{t}\Big(
\int_0^t q_{M,T}^2(r)dr\Big)^{1/2} \int_0^t \sigma (s)ds \\ 
&=\beta _1\sigma
(t) +\Big[ 2p_{M,T}(0)+\frac{1}{\beta _1}\int_0^t p_{M,T}^2(r)dr\\ &\quad
+2\sqrt{t}\Big( \int_0^t q_{M,T}^2(r)dr\Big) ^{1/2}\Big] \int_0^t \sigma
(s)ds, \end{aligned}  \label{e2.74}
\end{equation}
for all $\beta _{1}>0$. Put 
\begin{equation}
m_{1}=\min_{|s|\leq M}H'(s),\quad m_{2}=\max_{|s|\leq M}{\max }%
|H''(s)|.  \label{e2.75}
\end{equation}
 From assumption (H1) we have 
\begin{equation}
m_{1}>-1\,.  \label{e2.76}
\end{equation}
On the other hand, using integration by parts and \eqref{e2.75} it follows
that 
\begin{align*}
& 2\int_{0}^{t}\widetilde{H}_{1}(s)u'(0,s)ds \\
& =2\int_{0}^{t}\Big[\int_{0}^{1}\frac{d}{d\theta }H(u_{2}(0,s)+\theta
u(0,s))d\theta \Big]u'(0,s)ds \\
& =u^{2}(0,t)\int_{0}^{1}H'(u_{2}(0,s)+\theta u(0,s))d\theta \\
& \quad -\int_{0}^{t}u^{2}(0,s)ds\int_{0}^{1}H^{\prime \prime
}(u_{2}(0,s)+\theta u(0,s))(u_{2}'(0,s)+\theta u'(0,s))d\theta \\
& \geq m_{1}u^{2}(0,t)-m_{2}\int_{0}^{t}u^{2}(0,s)(|u_{1}'(0,s)|+|u_{2}'(0,s)|)ds \\
& \geq m_{1}u^{2}(0,t)-m_{2}\int_{0}^{t}\sigma (s)(|u_{1}'(0,s)|+|u_{2}'(0,s)|)ds.
\end{align*}%
 From the above inequality, \eqref{e2.71}-\eqref{e2.72} and \eqref{e2.74}, we
obtain 
\begin{equation}
\begin{aligned} \sigma (t)+m_1u^2(0,t) 
&\leq m_{2}\int_0^t \sigma (s)(|u_1'(0,s)|+|u_{2}'(0,s)|) ds\\ 
&\quad +\int_0^t
\|B_{2}(|u_{2}'(s)|)\|\sigma (s)ds+|J|\equiv \eta (t). \end{aligned}
\label{e2.77}
\end{equation}
 From \eqref{e2.1}, \eqref{e2.76}, and \eqref{e2.77}, we have 
\begin{equation}
(1+m_{1})u^{2}(0,t)\leq \sigma (t)+m_{1}u^{2}(0,t)\leq \eta (t).
\label{e2.78}
\end{equation}
It follows from \eqref{e2.77} and \eqref{e2.78} that 
\begin{equation}
\begin{aligned} &\sigma (t)+[m_1+\beta _{2}(1+m_1)]u^2(0,t)\\ 
&\leq (1+\beta _{2})\eta (t) \\ &\leq (1+\beta _{2})\int_0^t [
m_{2}(|u_1'(0,s)|+|u_{2}'(0,s)| ) +\|B_{2}(|u_{2}'(s)|)\| ] \sigma (s)ds \\
&\quad +(1+\beta _{2})\beta _1\sigma (t)+(1+\beta _{2}) \Big[
2p_{M,T}(0)+\frac{1}{\beta _1}\int_0^t p_{M,T}^2(r)dr\\ 
&\quad
+2\sqrt{t}\Big( \int_0^t q_{M,T}^2(r)dr\Big) ^{1/2}\Big] \int_0^t \sigma
(s)ds, \end{aligned}  \label{e2.79}
\end{equation}
for all $\beta _{1}>0$, $\beta _{2}>0$. Choose $\beta _{1}>0$, $\beta _{2}>0$
such that $m_{1}+\beta _{2}(1+m_{1})\geq 1/2$, 
$(1+\beta _{2})\beta _{1}\leq1/2$ and denote 
\begin{equation}
\begin{aligned} R_1(t)&=2(1+\beta _{2})[ m_{2}( |u_1'(0,s)|+|u_{2}'(0,s)|)
+\|B_{2}(|u_{2}'(s)|)\| \\ 
&\quad +\frac{1}{\beta
_1}\|p_{M,T}\|_{L^2(0,T)}^2+2p_{M,T}(0) +2\sqrt{T}\|q_{M,T}\|_{L^2(0,T)}] .
\end{aligned}  \label{e2.80}
\end{equation}
Then from \eqref{e2.79} and \eqref{e2.80} we have 
\begin{equation}
\sigma (t)+u^{2}(0,t)\leq \int_{0}^{t}R_{1}(s)[\sigma (s)+u^{2}(0,s)]ds;
\label{e2.81}
\end{equation}
i.e. $\sigma (t)+u^{2}(0,t)\equiv 0$ by Gronwall's lemma. Then Theorem 
\ref{thm1} is proved.
\end{proof}

In the special cases 
\begin{gather*}
H(s)=hs,\quad h>0; \\
K(t,u)=k(t)u,\quad k\in H^{1}(0,T),\quad \forall T>0, k(0)=0,
\end{gather*}
the following theorem is a consequence of Theorem \ref{thm1}.

\begin{theorem} \label{thm2}
 Let $(A),(G)$ and $(F_1)-(F_{3})$ hold.
Then, for every $T>0$,  problem \eqref{e1.1}- \eqref{e1.4} and \eqref{e1.9}
has at least a weak solution $(u,P)$ satisfying \eqref{e2.2}, \eqref{e2.3}.

 Furthermore, if $\beta =1$ in (F3) and  $B_{2}$ satisfies (F4),
then this solution is unique.
\end{theorem}

%remark 2
We remark that Theorem \ref{thm2} gives the same result as in \cite{l3}, but
we do not need the assumption ``$B_{1}$ is nondecreasing" used there.

In the special case with $K(t,u)\equiv 0$, the following result is the
consequence of Theorem \ref{thm1}.

\begin{theorem} \label{thm3}
Let (A), (G), (H), (F1)--(F3) hold.
Then, for every $T>0$,  the problem \eqref{e1.1}-\eqref{e1.4}
corresponding to $P=g$  has at least a weak solution $u$
satisfying \eqref{e2.2}.

Furthermore, if $\beta =1$  in (F3)  and  the functions
 $H$, $B_{2}$  satisfy the assumptions (H1), (F4),  then this solution is unique.
\end{theorem}

We remark that Theorem gives same result in \cite{d3} but without using the
assumption ``$B_1$ is nondecreasing'' used there.

\section{Stability of the solutions}

In this section, we assume that $\beta =1$ in (F3) and that the functions 
$H$, $B_{2}$ satisfying (H), (H1), (F4), respectively. By Theorem \ref{thm1}
problem \eqref{e1.1}-\eqref{e1.5} admits a unique solution $(u,P)$ depending
on $g$, $H$, $K$: 
\begin{equation*}
u=u(g,H,K),\quad P=P(g,H,K),
\end{equation*}
where $g$, $H$, $K$ satisfy the assumptions (G), (H),(H1),(K1)-(K3), and 
$u_{0}$, $u_{1}$, $f$ are fixed functions satisfying (A), (F1)-(F4).

Let $h_0>0$ be a given constant and $H_0:\mathbb{R}_{+}\to\mathbb{R}_{+}$ be
a given function. We put 
\begin{align*}
\Im (h_0,H_0)&=\big\{H\in C^2(\mathbb{R}):H(0)=0,\; \int_0^x H(s)ds\geq
-h_0,\; \forall x\in\mathbb{R}, \\
&\quad H^{\prime}(s)>-1,\; \forall s\in\mathbb{R}, \sup_{|s|\leq M}(
|H(s)|+|H^{\prime}(s)|) \leq H_0(M),\;\forall M>0\big\}.
\end{align*}
Given $t\geq 0$, $M>0$, and $K\in C^{0}(\mathbb{R}_{+}\times\mathbb{R};%
\mathbb{R})$, we put 
\begin{equation*}
N_{h}(M,K,t)=\sup_{|u|,|v|\leq M,\;u\neq v} |\frac{K(t,u)-K(t,v)}{u-v}|.
\end{equation*}
Given the family $\{p_{M,T}\}$, $M>0$, $T>0$ which consists of nonnegative
functions $p_{M,T}(t)=p(M,T,t)$, $M>0$, $T>0$ such that $p_{M,T}\in L^2(0,T)$, 
for all $M, T>0$.

Let $k_1\in L^2(0,T)$, $k_{2}\in L^{1}(0,T)$, for all $T>0$. We put 
\begin{align*}
&\Gamma (k_1,k_{2},\{p_{M,T}\}) \\
&=\big\{ K\in C^{0}(\mathbb{R}_{+}\times\mathbb{R}):\partial K/\partial t
\in C^{0}(\mathbb{R}_{+}\times\mathbb{R}), \\
&\quad N_{h}(M,K,t)+N_{h}(M,\partial K/\partial t,t)\leq p_{M,T}(t),\;
\forall t\in [ 0,T],\;\forall M,T>0, \\
&\quad |K(t,u)|+|\partial K/\partial t(t,u)|\leq k_1(t)|u|+k_{2}(t),\;
\forall u\in\mathbb{R},\; \forall t\in [ 0,T],\; \forall T>0\big\}.
\end{align*}
Then we have the following theorem.

\begin{theorem} \label{thm4}
 Let $\beta =1$  and (A),(F1)--(F4) hold.
Then, for every $T>0$, the solutions of \eqref{e1.1}-\eqref{e1.5}
 are stable with respect to the data $g$, $H$, $K$; i.e.,
if $(g,H,K)$, $(g_{j},H_{j},K_{j})\in H^{1}(0,T)\times \Im
(h_0,H_0)\times \Gamma (k_1,k_{2},\{p_{M,T}\})$, are such that
\begin{equation} \label{e3.1}
(g_{j},H_{j})\to (g,H)\quad \text{in }H^{1}(0,T)\times C^{1}([-M,M])
\end{equation}
strongly, and
\begin{equation} \label{e3.2}
(K_{j},\partial K_{j}/\partial t)\to (K,\partial K/\partial t)\text{
in }[ C^{0}([0,T]\times [ -M,M])] ^2
\end{equation}
strongly, as $j\to +\infty $, for all $M,T>0$.
 Then
\[ %\label{e3.3}
(u_{j},u_{j}',u_{j}(0,t),P_{j})\to (u,u',u(0,t),P)
\]
in $L^{\infty }(0,T;V)\times L^{\infty }(0,T;L^2)\times C^{0}([0,T])\times
C^{0}([0,T])$
strongly, as $j\to +\infty $,  for all $M,T>0$,
where $u_{j}=u(g_{j},H_{j},K_{j})$, $P_{j}=P(g_{j},H_{j},K_{j})$.
\end{theorem}

\begin{proof}
First, we note that if the data $(g,H,K)$ satisfy 
\begin{equation}
\|g\|_{H^{1}(0,T)}\leq G_{0},\quad H\in \Im (h_{0},H_{0}),\quad K\in
\Gamma (k_{1},k_{2},\{p_{M,T}\}),  \label{e3.4}
\end{equation}
then, the a priori estimates of the sequences $\{u_{m}\}$ and $\{P_{m}\}$ in
the proof of the Theorem \ref{thm1} satisfy 
\begin{gather}
\|u_{m}'(t)\|^{2}+\|u_{m}(t)\|_{V}^{2}\leq
C_{T}^{2}\quad \forall t\in [ 0,T],\;\forall T>0,  \label{e3.5} \\
\int_{0}^{t}|u_{m}'(0,s)|^{2}ds\leq C_{T}^{2}\quad \forall t\in
[ 0,T],\;\forall T>0,  \label{e3.6} \\
\int_{0}^{t}|P_{m}'(s)|^{2}ds\leq C_{T}^{2}\quad \forall t\in
[ 0,T],\;\forall T>0,  \label{e3.7}
\end{gather}%
where $C_{T}$ is a constant depending only on $T$, $u_{0}$, $u_{1}$, $f$, 
$G_{0}$, $h_{0}$, $H_{0}$, $k_{1}$, $k_{2}$, $\{p_{M,T}\}$ (independent of 
$g,H,K$). Hence, the limit $(u,P)$ in suitable function spaces of the
sequence $\{(u_{m},P_{m})\}$ is defined by \eqref{e2.6}-\eqref{e2.8}, which
is a solution of \eqref{e1.1}-\eqref{e1.5} satisfying the a priori estimates 
\eqref{e3.5}-\eqref{e3.7}.

Now, by \eqref{e3.1}, \eqref{e3.2} we can assume that there exists constant 
$G_{0}>0$ such that the data $(g_{j},H_{j},K_{j})$ satisfy \eqref{e3.4} with 
$(g,H,K)=(g_{j},H_{j},K_{j})$. Then, by the above remark, we have that the
solutions $(u_{j},P_{j})$ of problem \eqref{e1.1}-\eqref{e1.5} corresponding
to $(g,H,K)=(g_{j},H_{j},K_{j})$ satisfy 
\begin{gather}
\|u_{j}'(t)\|^{2}+\|u_{j}(t)\|_{V}^{2}\leq
C_{T}^{2}\quad \forall t\in [ 0,T],\;\forall T>0,  \label{e3.8} \\
\int_{0}^{t}|u_{j}'(0,s)|^{2}ds\leq C_{T}^{2}\quad \forall t\in
[ 0,T],\;\forall T>0,  \label{e3.9} \\
\int_{0}^{t}|P_{j}'(s)|^{2}ds\leq C_{T}^{2}\quad \forall t\in
[ 0,T],\quad \forall T>0,  \label{e3.10}
\end{gather}%
Put %\label{e3.11}
$\widetilde{g}_{j}=g_{j}-g$, $\widetilde{H}_{j}=H_{j}-H$, $\widetilde{K}%
_{j}=K_{j}-K$. Then, $v_{j}=u_{j}-u$ and $\ Q_{j}=P_{j}-P$ satisfy the
problem 
\begin{gather*}
v_{j}''-v_{jxx}+\chi _{j}=0,\quad 0<x<1,\;0<t<T, \\
v_{jx}(0,t)=Q_{j}(t),\quad v_{j}(1,t)=0, \\
v_{j}(x,0)=v_{j}'(x,0)=0,
\end{gather*}%
where 
\begin{gather}
\begin{aligned} \chi _{j}&=f(u_{j},u_{j}')-f(u,u'), \\
Q_{j}(t)&=\widehat{g}_{j}(t)+H(u_{j}(0,t))-H(u(0,t))\\ 
&\quad -\int_0^t
[K(t-s,u_{j}(0,s))-K(t-s,u(0,s))] ds, \end{aligned}  \label{e3.12} \\
\widehat{g}_{j}(t)=\widetilde{g}_{j}(t)+\widetilde{H}_{j}(u_{j}(0,t))-%
\int_{0}^{t}\widetilde{K}_{j}(t-s,u_{j}(0,s))ds.  \label{e3.13}
\end{gather}
Applying Lemma \ref{lem7} with $u_{0}=u_{1}=0$, $\chi =\chi _{j}$, $P=Q_{j}$, 
we have 
\begin{equation*}
\|v_{j}'(t)\|^{2}+\|v_{j}(t)\Vert
_{V}^{2}+2\int_{0}^{t}Q_{j}(s)v_{j}'(0,s))ds+2\int_{0}^{t}\langle
\chi _{j}(s),v_{j}'(s)\rangle ds=0.
\end{equation*}
Let 
\begin{gather*}
S_{j}(t)=\|v_{j}'(t)\|^{2}+\|v_{j}(t)\Vert
_{V}^{2}+v_{j}^{2}(0,t), \\
M=C_{T},\quad m_{1}=\min_{|s|\leq M}H'(s)>-1,\quad
m_{2}=\max_{|s|\leq M}|H''(s)|.
\end{gather*}%
Then, we can prove the following inequality in a similar manner 
\begin{equation}
\begin{aligned} &\|v_{j}'(t)\|^2+ \|v_{j}(t)\|_{V}^2+m_1v_{j}^2(0,t)\\ 
&\leq \int_0^t \|B_{2}(|u'(s)|)\| S_{j}(s)ds+ m_{2} \int_0^t (
|u'(0,s)|+|u_{j}'(0,s)|)S_{j}(s)ds \\ 
&\quad +2\varepsilon
S_{j}(t)+\varepsilon \int_0^t S_{j}(s)ds+\frac{1}{\varepsilon }(
\widehat{g}_{j}^2(t) +\int_0^t |\widehat{g}_{j}'(s)|^2ds) \\ 
&\quad +(
\frac{1}{\varepsilon }\|p_{M,T}\|_{L^2(0,T)}^2
+2\sqrt{T}\|p_{M,T}\|_{L^2(0,T)}) \int_0^t S_{j}(s)ds\\ 
&=2\varepsilon S_{j}(t)+\frac{1}{\varepsilon }( \widehat{g}_{j}^2(t)+ \int_0^t
|\widehat{g}_{j}'(s)|^2ds) \\ 
&\quad+ \int_0^t [ \|B_{2}(|u'(s)|)\|+ m_{2}(
|u'(0,s)|+|u_{j}'(0,s)|) ] S_{j}(s)ds \\ 
&\quad +( \varepsilon
+\frac{1}{\varepsilon }\|p_{M,T}\|_{L^2(0,T)}^2
+2\sqrt{T}\|p_{M,T}\|_{L^2(0,T)}) \int_0^t S_{j}(s)ds\equiv y_{j}(t),
\end{aligned}  \label{e3.14}
\end{equation}
for all $\varepsilon >0$ and $t\in [ 0,T]$.

We remark that $v_{j}^{2}(0,t)\leq \|v_{j}(t)\|_{V}^{2}$,
consequently 
\begin{equation}
(1+m_{1})v_{j}^{2}(0,t)\leq \|v_{j}'(t)\|^{2}+\Vert
v_{j}(t)\|_{V}^{2}+m_{1}v_{j}^{2}(0,t)\leq y_{j}(t).  \label{e3.15}
\end{equation}
Multiplying the two members of \eqref{e3.15} by a number $\beta _{1}>0$ and
adding to \eqref{e3.14}, we have 
\begin{equation}
\begin{aligned} 
&\|v_{j}'(t)\|^2+ \|v_{j}(t)\|_{V}^2+[(1+m_1)\beta
_1+m_1]v_{j}^2(0,t)\\ 
&\leq (1+\beta _1)y_{j}(t) \\ 
&\leq (1+\beta _1)[2\varepsilon S_{j}(t)
+\frac{1}{ \varepsilon }( \widehat{g}_{j}^2(t)+\int_0^t
|\widehat{g}_{j}'(s)|^2ds) ] \\ 
&\quad + \int_0^t
\widetilde{R}_{j}(\varepsilon ,T,s)S_{j}(s)ds, \quad \forall \varepsilon
>0,\; \beta _1>0,\, t\in [ 0,T]. \end{aligned}  \label{e3.16}
\end{equation}
where 
\begin{equation}
\begin{aligned} \widetilde{R}_{j}(\varepsilon ,T,s) 
&=(1+\beta _1)\Big[
\varepsilon +\frac{1 }{\varepsilon }\|p_{M,T}\|_{L^2(0,T)}^2+2\sqrt{T}
\|p_{M,T}\|_{L^2(0,T)}\\ 
&\quad +\|B_{2}(|u'(s)|)\|+ m_{2}(
|u'(0,s)|+|u_{j}'(0,s)|) \Big] . \end{aligned}  \label{e3.17}
\end{equation}
Choose $\beta _{1}>0$ and $\varepsilon >0$ such that $(1+m_{1})\beta
_{1}+m_{1}\geq 1$, $2\varepsilon (1+\beta _{1})\leq 1/2$. From 
$H^{1}(0,T)\hookrightarrow C^{0}([0,T])$, and \eqref{e3.16} we have 
\begin{equation}
S_{j}(t)\leq 2(1+\beta _{1})\frac{1}{\varepsilon }C_{T}^{(9)}
\|\widehat{g}_{j}\|_{H^{1}(0,T)}^{2}+2\int_{0}^{t}\widetilde{R}_{j}
(\varepsilon,T,s)S_{j}(s)ds,  \label{e3.18}
\end{equation}
where $C_{T}^{(9)}$ is a constant depending only on $T$. By Gronwall's
lemma, we obtain from \eqref{e3.18} that 
\begin{equation}
S_{j}(t)\leq 2(1+\beta _{1})\frac{1}{\varepsilon }C_{T}^{(9)}\|\widehat{g}_{j}
\|_{H^{1}(0,T)}^{2}\exp \big(2\int_{0}^{T}\widetilde{R}_{j}
(\varepsilon ,T,s)S_{j}(s)ds\big),  \label{e3.19}
\end{equation}
for all $t\in [ 0,T]$. On the other hand, we from \eqref{e3.5}, 
\eqref{e3.12}, \eqref{e3.13}, \eqref{e3.17}, and \eqref{e3.19} obtain 
\begin{gather}
S_{j}(t)\leq C_{T}^{(10)}\|\widehat{g}_{j}\|_{H^{1}(0,T)}^{2}\quad
\forall t\in [ 0,T],  \label{e3.20} \\
|Q_{j}(t)|\leq |\widehat{g}_{j}(t)|+\max_{|s|\leq M}|H'(s)|\sqrt{%
S_{j}(t)}+\|p_{M,T}\|_{L^{2}(0,T)}\Big(\int_{0}^{t}S_{j}(s)ds\Big)%
^{1/2}.  \label{e3.21}
\end{gather}
We again use the embedding $H^{1}(0,T)\hookrightarrow C^{0}([0,T])$. Then,
it follows from \eqref{e3.20} and \eqref{e3.21} that 
\begin{equation*}
\|Q_{j}\|_{C^{0}([0,T])}\leq C_{T}^{(11)}\|\widehat{g}_{j}\Vert
_{H^{1}(0,T)}^{2}.
\end{equation*}
As a final step, we prove 
\begin{equation*}
\lim_{j\rightarrow +\infty }\|\widehat{g}_{j}\|_{H^{1}(0,T)}^{2}=0.
\end{equation*}
Indeed, from \eqref{e3.13} combined with \eqref{e3.9}, we deduce the
following inequality 
\begin{align*}
\|\widehat{g}_{j}\|_{H^{1}(0,T)}& \leq \|\widetilde{g}_{j}\Vert
_{H^{1}(0,T)}+\sqrt{T+M^{2}}\|\widetilde{H}_{j}\|_{C^{1}([-M,M])} \\
& \quad +\sqrt{2T(1+T^{2})}(\|\widetilde{K}_{j}\|_{C^{0}([0,T]\times
[ -M,M])}+\|\partial \widetilde{K}_{j}/\partial t\Vert
_{C^{0}([0,T]\times [ -M,M])}).
\end{align*}
Then the proof is complete.

\subsection*{Acknowledgment} The authors thank Professor Dung L. and the
referees for their valuable suggestions and help.
\end{proof}

\begin{thebibliography}{99}

\bibitem{a1} 1. An N. T., Trieu N. D., \textit{Shock between absolutely solid
body and elastic bar with the elastic viscous frictional resistance at the
side}, J. Mech. NCSR. Vietnam, \textbf{XIII} (2) (1991), 1-7.

\bibitem{a2} 2. Ang D. D., Dinh A. P. N., \textit{Mixed problem for some
semilinear wave equation with a nonhomogeneous condition}, Nonlinear Anal. 
\textbf{12} (1988), 581-592.

\bibitem{a3} 3. Adams R. A., \textit{Sobolev Spaces}, Academic Press,
NewYork, 1975.

\bibitem{b1} 4. Bergounioux M., Long N. T., Dinh A. P. N., \textit{Mathematical
model for a shock problem involving a linear viscoelastic bar}, Nonlinear
Anal. \textbf{43} (2001), 547-561.

\bibitem{d1} 5. Dinh A. P. N., \textit{Sur un probl\`{e}me hyperbolique
faiblement nonlin\'{e}aire en dimension} 1, Demonstratio Math. \textbf{16}
(1983), 269-289.

\bibitem{d2} 6. Dinh A. P. N., Long N. T., \textit{Linear approximation and
asymptotic expansion associated to the nonlinear wave equation in one
dimension}, Demonstratio Math. \textbf{19} (1986), 45-63.

\bibitem{d3} 7. Dinh A. P. N., Long N. T., \textit{The semilinear wave equation
associated with a nonlinear boundary}, Demonstratio Math. \textbf{30}
(1997), 557-572.

\bibitem{l1} 8. Lions J. L., \textit{Quelques m\'{e}thodes de r\'{e}solution
des probl\`{e} mes aux limites non-lin\'{e}aires}, Dunod- Gauthier- Villars,
Paris, 1969.

\bibitem{l2} 9. Long N. T., Dinh A. P. N., \textit{On the quasilinear wave
equation} $u_{tt}-\Delta u+f(u,u_{t})=0$ \textit{associated with a mixed
nonhomogeneous condition}, Nonlinear Anal.\textbf{\ 19} (1992), 613-623.

\bibitem{l3} 10. Long N. T., Dinh A. P. N., \textit{A semilinear wave equation
associated with a linear differential equation with Cauchy data}, Nonlinear
Anal.\textbf{\ 24} (1995), 1261-1279.

\bibitem{l4} 11. Long N. T., Diem T. N., \textit{On the nonlinear wave equation
} $u_{tt}-u_{xx}=f(x,t,u,u_{x},u_{t})$ \textit{associated with the mixed
homogeneous conditions}, Nonlinear Anal. \textbf{29} (1997), 1217-1230.

\bibitem{l5} 12. Long N. T., Thuyet T. M., \textit{A semilinear wave equation
associated with a nonlinear integral equation}, Demonstratio Math. 
\textbf{36} (2003), No.4, 915-938.

\bibitem{o1} 13. Ortiz E. L., Dinh A. P. N., \textit{Linear recursive schemes
associated with some nonlinear partial differential equations in one
dimension and the Tau method}, SIAM J. Math. Anal. \textbf{18}
(1987), 452-464.
\end{thebibliography}

\end{document}