\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2011 (2011), No. 78, pp. 1--11.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2011 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2011/78\hfil Existence and uniqueness of solutions]
{Existence and uniqueness of solutions for a second-order
 nonlinear hyperbolic system}

\author[R. Luca \hfil EJDE-2011/78\hfilneg]
{Rodica Luca}

\address{Rodica Luca \newline
Department of Mathematics, Gh. Asachi Technical University,
11 Blvd. Carol I, Iasi 700506, Romania}
\email{rluca@math.tuiasi.ro}

\thanks{Submitted May 20, 2010. Published June 20, 2011.}
\subjclass[2000]{35L55, 47J35, 47H05}
\keywords{Hyperbolic system; boundary conditions;
 Cauchy problem; \hfill\break\indent
monotone operator; strong solution; weak solution}

\begin{abstract}
 We study the existence, uniqueness and asymptotic behaviour
 of solutions to a second-order nonlinear hyperbolic system
 of equations. The spatial variable is in the positive
 half-axis.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{remark}[theorem]{Remark}
\allowdisplaybreaks


\section{Introduction}

We study the second-order nonlinear hyperbolic system
\begin{equation} \label{eS}
\begin{gathered}
\frac{\partial u}{\partial t}(t,x)
+\frac{\partial^2 v}{\partial x^2}(t,x)+\alpha(x,u)=f(t,x),\\
\frac{\partial v}{\partial t}(t,x)
-\frac{\partial^2 u}{\partial x^2}(t,x)+\beta(x,v)
=g(t,x),\quad t>0,\; x>0,
\end{gathered}
\end{equation}
with the boundary condition
\begin{equation} \label{BC}
\begin{pmatrix}
\operatorname{col}\big(-\frac{\partial u}{\partial x}(t,0),u(t,0)\big)\\
w'(t)\end{pmatrix}
\in -G\begin{pmatrix}
\operatorname{col}\big(v(t,0),\frac{\partial v}{\partial x}(t,0)\big)\\
w(t)
\end{pmatrix})+B(t),\quad
t>0,
\end{equation}
and the initial condition
\begin{equation} \label{IC}
\begin{gathered}
u(0,x)=u_0(x),\quad v(0,x)=v_0(x),\quad x>0,\\
w(0)=w_0.
\end{gathered}
\end{equation}

The unknowns $u$, $v$ and  $f$ and $g$ are the vectorial
functions depending on $(t,x)\in \mathbb{R}_+\times \mathbb{R}_+$
with values in $\mathbb{R}^n$, the unknown
$w$ is a vectorial function depending on $t\in\mathbb{R}_+$
with values in $\mathbb{R}^m$. In the system \eqref{eS},
the functions $\alpha$ and $\beta$ are of the form
$\alpha(x,u)=\operatorname{col}(\alpha_1(x,u_1),
\ldots,\alpha_n(x,u_n))$,
$\beta(x,v)=\operatorname{col}(\beta_1(x,v_1),\ldots,\beta_n(x,v_n))$,
$G$ is an operator in $\mathbb{R}^{2n+m}$ and
$B(t)=\operatorname{col}(b_1(t),\ldots,b_{2n+m}(t))
\in\mathbb{R}^{2n+m}$, for all $t>0$.

This problem is a generalization of the case studied by Luca
\cite{L3}, where $B(t)\equiv 0$. The methods we shall use to prove
the main results in this article are different from that used in
Luca \cite{L3}. We also mention the articles \cite{L1,L2,L4},
where we have investigated a $n$-order hyperbolic system,  for
spatial variable $x\in (0,1)$ and $t>0$, subject to some nonlinear
boundary conditions.

In the present work, we shall prove the existence, uniqueness,
some regularity properties and the asymptotic behaviour of the
strong and weak solutions for the problem
\eqref{eS}, \eqref{BC}, \eqref{IC}.
For the basic notation, concepts and results in the theory
of monotone operators and nonlinear evolution equations
of monotone type in Hilbert spaces we refer the reader
to Barbu \cite{Ba}, Brezis \cite{Br}, Lakshmikantham et al \cite{LL}.

Now we introduce the assumptions to be used in this article.
\begin{itemize}
\item[(A1)] (a) The functions $x\to \alpha_k(x,p)$ and
$x\to \beta_k(x,p)$ are measurable on $\mathbb{R}_+$,
for any fixed $p\in\mathbb{R}$. Besides, the
functions $p\to \alpha_k(x,p)$ and $p\to\beta_k(x,p)$ are
continuous and nondecreasing from $\mathbb{R}$ into $\mathbb{R}$,
for a.a. $x\in\mathbb{R}_+$, $k=\overline{1,n}$.

(b) There exist $a_k>0$, $b_k>0$, $k=\overline{1,n}$ and the
functions $c_k^1,c_k^2\in L^2(\mathbb{R}_+)$ such that
\[
|\alpha_k(x,p)|\le a_k|p|+c_k^1(x),\quad
|\beta_k(x,p)|\le b_k|p|+c_k^2(x),
\]
for a.a. $x\in\mathbb{R}_+$, for all $p\in\mathbb{R}$,
$k=\overline{1,n}$.

(c) There exist $\chi_1>0$, $\chi_2>0$ such that
\begin{gather*}
(\alpha_k(x,p_1)-\alpha_k(x,p_2))(p_1-p_2)\ge \chi_1(p_1-p_2)^2,\\
(\beta_k(x,p_1)-\beta_k(x,p_2))(p_1-p_2)\ge \chi_2(p_1-p_2)^2,
\end{gather*}
for a.a. $x\in \mathbb{R}_+$, for all $p_1,\,p_2\in\mathbb{R}$,
$k=\overline{1,n}$.

\item[(A2)] (a) $G:D(G)\subset\mathbb{R}^{2n+m}\to\mathbb{R}^{2n+m}$
is a maximal monotone operator (possibly multivalued). Moreover,
$G=\begin{pmatrix}
G_{11}&G_{12}\\
G_{21}&G_{22}\end{pmatrix}$
with
\begin{gather*}
G_{11}:D(G_{11})\subset\mathbb{R}^{2n}\to\mathbb{R}^{2n},\quad
G_{12}:D(G_{12})\subset\mathbb{R}^{m}\to\mathbb{R}^{2n},\\
G_{21}:D(G_{21})\subset\mathbb{R}^{2n}\to\mathbb{R}^{m},\quad
G_{22}:D(G_{22})\subset\mathbb{R}^{m}\to\mathbb{R}^{m},
\end{gather*}
where
\[
G((u_1,\ldots,u_{2n+m})^T)
=\begin{pmatrix}G_{11}((u_1,\ldots,u_{2n})^T)
+G_{12}((u_{2n+1},\ldots,u_{2n+m})^T)\\
G_{21}((u_1,\ldots,u_{2n})^T)
+G_{22}((u_{2n+1},\ldots,u_{2n+m})^T)
\end{pmatrix}.
\]

(b) There exists $\zeta_1>0$ such that for all $x,y\in D(G)$,
$x=\operatorname{col}(x^a,x^b)$,
$y=\operatorname{col}(y^a,y^b)\in\mathbb{R}^{2n}\times \mathbb{R}^m$
 and for all $w_1\in G(x)$, $w_2\in G(y)$ we have
\[
\langle w_1-w_2,x-y\rangle_{\mathbb{R}^{2n+m}}
\ge \zeta_1\|x^b-y^b\|^2_{\mathbb{R}^m}.
\]

(c) There exists $\zeta_2>0$ such that for all $x,y\in D(G)$ and
for all $w_1\in G(x)$,
$w_2\in G(y)$ we have
\[
\langle w_1-w_2,x-y\rangle_{\mathbb{R}^{2n+m}}
\ge \zeta_2\|x-y\|^2_{\mathbb{R}^{2n+m}}.
\]
The operator $G$ is a generalization of the matrix case.
It also covers some general boundary conditions for \eqref{eS}.
For example, if $G_{12}=0$ and $G_{21}=0$,
then the boundary condition \eqref{BC} becomes
\begin{gather}
\label{BC1}
\operatorname{col}(-u_x(t,0),u(t,0))\in
-G_{11}(\operatorname{col}(v(t,0),v_x(t,0)))+B_1(t),
\\
\label{BC2} w'(t)\in -G_{22}(w(t))+B_2(t),
\end{gather}
where $B(t)=\operatorname{col}(B_1(t),B_2(t))$.
\end{itemize}

The condition \eqref{BC2} with \eqref{IC} give us, by integration,
the function $w$.
For \eqref{BC1}, by making suitable choices of $G_{11}$, we
deduce many classical boundary conditions.
Here are some examples in the case $G_{11}=\partial l$,
the subdifferential of $l:\mathbb{R}^{2n}\to (-\infty,+\infty]$:
\begin{itemize}
\item[(a)] If
\[
l\begin{pmatrix}u\\v\end{pmatrix}
=\begin{cases}0,&\text{if } u=a,\; v=b\\
 +\infty, & \text{otherwise}.
\end{cases}
\]
then \eqref{BC1} becomes $v(t,0)=a$, $v_x(t,0)=b$.

\item[(b)] For $n=1$ and $l\begin{pmatrix}u\\v\end{pmatrix}
=\begin{cases} bv,&\text{if } u=a\\
\infty,&\text{otherwise},
\end{cases}$ we obtain $u(t,0)=B_2(t)-b$, $v(t,0)=a$.

\item[(c)] For $n=1$ and $l\begin{pmatrix}u\\v\end{pmatrix}
=\begin{cases} au,&\text{if }v=b\\
\infty,&\text{otherwise},
\end{cases}$ we obtain $u_x(t,0)=-B_1(t)+a$,
$v_x(t,0)=b$.

\item[(d)] For $n=1$ and $l\begin{pmatrix}u\\v\end{pmatrix}
=au+bv$, then \eqref{BC1} becomes $u_x(t,0)=-B_1(t)+a$,
$u(t,0)=B_2(t)-b$.
\end{itemize}

\section{Preliminary results}

Firstly we consider the case $B(t)\equiv b_0$ (a constant vector).
In this situation, we can replace $G$ by $\widetilde{G}$, defined
by $\widetilde{G}w=Gw-b_0$, which is under the assumption (A2)a,
a maximal monotone operator. So, we suppose without loss
of generality that $B(t)\equiv 0$. In what follows,
we shall recall some results from Luca \cite{L3}
relating to the existence and uniqueness of the solutions
of our problem \eqref{eS}, \eqref{BC}, \eqref{IC}.
We consider the spaces $X=(L^2(\mathbb{R}_+;\mathbb{R}^n))^2$,
$\mathbb{R}^m$
and $Y=X\times \mathbb{R}^m$ with the corresponding scalar products
\begin{gather*}
\langle f,g\rangle_X=\langle f_1,g_1\rangle_{L^2(\mathbb{R}_+;
\mathbb{R}^n)}+\langle f_2,g_2\rangle
_{L^2(\mathbb{R}_+;\mathbb{R}^n)},\quad
f=\begin{pmatrix}f_1\\f_2\end{pmatrix},\quad
g=\begin{pmatrix} g_1\\g_2\end{pmatrix}\in X,
\\
\langle x,y\rangle_{\mathbb{R}^m}
=\sum_{i=1}^m x_iy_i,\quad x,y\in\mathbb{R}^m,
\\
\big\langle \begin{pmatrix}f\\x\end{pmatrix},
\begin{pmatrix}g\\y\end{pmatrix}\big\rangle_Y
=\langle f,g\rangle_X+\langle x,y\rangle_{\mathbb{R}^m},
\quad \begin{pmatrix}f\\x\end{pmatrix},\quad
\begin{pmatrix}g\\y\end{pmatrix}\in Y.
\end{gather*}

We define the operator $\mathcal{A}:D(\mathcal{A})\subset Y\to Y$,
\begin{align*}
D(\mathcal{A})=\{&y\in Y, y=\operatorname{col}(u,v,w);\,
u,v\in H^2(\mathbb{R}_+;\mathbb{R}^n),\,\,\,w\in\mathbb{R}^m,\\
&\operatorname{col}(\gamma_0v,w)\in D(G),\,
\gamma_1 u\in -G_{11}(\gamma_0v)-G_{12}(w)\},
\end{align*}
where $\gamma_0v=\operatorname{col}(v(0),v'(0))$,
$\gamma_1 u=\operatorname{col}(-u'(0),u(0))$,
$$
\mathcal{A}\begin{pmatrix}u\\v\\w\end{pmatrix}
=\begin{pmatrix}v''\\ -u''\\ G_{21}(\gamma_0v)+G_{22}(w)\end{pmatrix},
\quad \begin{pmatrix}u\\v\\w\end{pmatrix}\in D(\mathcal{A}).
$$
We also define the operator $\mathcal{B}:D(\mathcal{B})\subset Y\to Y$,
$$
\mathcal{B}(y)=\operatorname{col}(\alpha(\cdot,u),\beta(\cdot,v),0),\;
D(\mathcal{B})
=\{y\in Y,\,y=\operatorname{col}(u,v,w);\,\mathcal{B}(y)\in Y\}.
$$
Under assumption (A2)a, we have $D(\mathcal{A})\neq \emptyset$,
$\overline{D(\mathcal{A})}=X\times \overline{D(G_{12})\cap D(G_{22})}$
and, under the assumptions (A1)ab, we obtain $D(\mathcal{B})=Y$.

\begin{lemma}\label{lemma:1}
 If {\rm (A2)a} holds, then
the operator $\mathcal{A}$ is maximal monotone.
\end{lemma}

\begin{lemma}\label{lemma:2}
If {\rm (A1)ab} hold, then the operator $\mathcal{B}$ is
maximal monotone.
\end{lemma}

\begin{remark}\label{remark:1}\rm
 By Lemma \ref{lemma:1}, Lemma \ref{lemma:2} and Rockafellar's
theorem (see Barbu \cite[Theorem 1.7, Chapter II]{Ba}),
it follows that, under the assumptions
(A1)ab and (A2)a, the operator
$\mathcal{A}+\mathcal{B}:D(\mathcal{A})\subset Y\to Y$
is maximal monotone in the space $Y$.
\end{remark}

Using the operators $\mathcal{A}$ and $\mathcal{B}$, our
problem \eqref{eS}, \eqref{BC}, \eqref{IC} can be equivalently
expressed as the following Cauchy problem in the space $Y$
\begin{equation} \label{eP}
\begin{gathered}
\frac{dy}{dt}(t)+\mathcal{A}(y(t))+\mathcal{B}((y(t))
\ni F(t,\cdot),\quad ,t>0,\\
y(0)=y_0,
\end{gathered}
\end{equation}
where $y(t)=\operatorname{col}(u(t),v(t),w(t))$,
 $F(t,\cdot)=\operatorname{col}(f(t,\cdot),g(t,\cdot),0)$,
$y_0=\operatorname{col}(u_0,v_0$, $w_0)$.

\begin{lemma}\label{lemma:3}
Assume that {\rm (A1)ab, (A2)a} hold. If
$f,g\in W^{1,1}(0,T;L^2(\mathbb{R}_+;\mathbb{R}^n))$
(with $T>0$ fixed), $u_0,v_0\in H^2(\mathbb{R}_+;\mathbb{R}^n)$,
$\operatorname{col}(\gamma_0v_0,w_0)\in D(G)$,
$\gamma_1u_0$ belongs to $-G_{11}(\gamma_0v_0)-G_{12}(w_0)$, then
problem \eqref{eS}, \eqref{BC}, \eqref{IC} has a unique strong
solution $\operatorname{col}(u,v,w)\in W^{1,\infty}(0,T;Y)$.
Moreover $u,v\in L^{\infty}(0,T;H^2(\mathbb{R}_+;\mathbb{R}^n))$.
\end{lemma}

\begin{remark}\label{remark:2}\rm
 For all $t\in [0,T)$, the above functions $u(t,\cdot),v(t,\cdot)$
satisfy the system \eqref{eS} for a.a. $x\in\mathbb{R}_+$
(with $\partial^+u/\partial t$, $\partial^+v/\partial t$ instead of
$\partial u/\partial t$, $\partial v/\partial t$), and
together with $w(t)$ verify the boundary condition \eqref{BC}
(with $d^+w/dt$ instead of $dw/dt$) and the initial data \eqref{IC}.
\end{remark}

\begin{lemma}\label{lemma:4}
Assume that {\rm (A1)ab, (A2)a} hold. If
$f,g\in L^1(0,T;L^2(\mathbb{R}_+;\mathbb{R}^n))$
(with $T>0$ fixed), $u_0,v_0\in L^2(\mathbb{R}_+;\mathbb{R}^n)$,
$w_0\in \overline{D(G_{12})\cap D(G_{22})}$ then
the problem \eqref{eS}, \eqref{BC}, \eqref{IC} has a unique
weak solution $\operatorname{col}(u,v,w)\in C([0,T];Y)$.
\end{lemma}

For the proofs of Lemmas \ref{lemma:1}--\ref{lemma:4}
see Luca \cite{L3}.

In what follows we shall present an existence result for
the stationary problem associated to \eqref{eP}.

\begin{lemma}\label{lemma:5}
If {\rm (A1)abc, (A2)ab} hold, then the stationary problem
\begin{equation}\label{eq:1}
\mathcal{A}(y)+\mathcal{B}(y)\ni 0
\end{equation}
has a unique solution $y=\operatorname{col}(u,v,w)\in D(\mathcal{A})$.
\end{lemma}

\begin{proof}
 By Remark \ref{remark:1}, the operator $\mathcal{A}+\mathcal{B}$
is maximal monotone in $Y$. In addition, it is strongly monotone.
Indeed, for all $y=\operatorname{col}(u,v,w)$,
$\widetilde{y}=\operatorname{col}
(\widetilde{u},\widetilde{v},\widetilde{w})\in D(\mathcal{A})$,
$h\in (\mathcal{A}+\mathcal{B})(y)$,
$\widetilde h\in (\mathcal{A}+\mathcal{B})(\widetilde y)$ we have
\begin{align*}
&\langle h-{\widetilde h},y-{\widetilde y}\rangle_Y\\
&=\langle g-{\widetilde g},z-{\widetilde z}\rangle_{\mathbb{R}^{2n+m}}
 +\sum_{k=1}^n\int_0^{\infty}[\alpha_k(x,u_k(x))
 -\alpha_k(x,\widetilde{u}_k(x))][u_k(x)-\widetilde{u}_k(x)]dx\\
&\quad +\sum_{k=1}^n\int_0^{\infty}[\beta_k(x,v_k(x))
 -\beta_k(x,\widetilde{v}_k(x))][v_k(x)-\widetilde{v}_k(x)]dx\\
&\ge \zeta_1\|w-\widetilde{w}\|^2_{\mathbb{R}^m}
 +\sum_{k=1}^n\chi_1\|u_k-\widetilde{u}_k\|^2_{L^2(\mathbb{R}_+)}
 +\sum_{k=1}^n\chi_2\|v_k-\widetilde{v}_k\|^2_{L^2(\mathbb{R}_+)}\\
&\ge \chi_0\|y-\widetilde{y}\|^2_Y,
\end{align*}
where $z=\operatorname{col}(\gamma_0v,w)$,
$\widetilde z=\operatorname{col}(\gamma_0\widetilde v,\widetilde w)$,
$g\in G(z)$, $\widetilde g\in G(\widetilde z)$ and
\[
\chi_0=\min\{\chi_1,\chi_2,\zeta_1/s_i,, i=\overline{1,m}\}.
\]
Therefore, this operator is coercive and then
$R(\mathcal{A}+\mathcal{B})=Y$. So we deduce that equation
\eqref{eq:1} has a unique solution
$y=\operatorname{col}(u,v,w)\in D(\mathcal{A})$.
\end{proof}

Now using Remark \ref{remark:1}, Lemma \ref{lemma:4},
Lemma \ref{lemma:5} and  Brezis \cite[Theorem 3.9]{Br}, we deduce
the following result.

\begin{lemma}\label{lemma:6}
Assume that {\rm (A1)abc, (A2)ab} hold,
$f,g\in L^1_{\rm loc}(\mathbb{R}_+;L^2(\mathbb{R}_+;\mathbb{R}^n))$
verify the conditions $\lim_{t\to\infty}f(t)=f^0$,
$\lim_{t\to\infty}g(t)=g^0$, strongly in
$L^2(\mathbb{R}_+;\mathbb{R}^n)$, and
$\delta=\operatorname{col}(p,q,r)$ is the unique solution of
equation \eqref{eq:1}.
Then $\lim_{t\to\infty}y(t)=\delta$, strongly in $Y$,
where $y(t)=\operatorname{col}(u(t),v(t),w(t))$, $t\ge 0$ is
an arbitrary weak solution of equation \eqref{eP}$_1$. More precisely
$$
\|y(t)-\delta\|_Y\le e^{-\chi_0t}\|y(0)
-\delta\|_Y+\int_0^te^{\chi_0(s-t)}\|F(s)-F^0\|_Yds,\quad t\ge 0,
$$
where $F^0=\operatorname{col}(f^0,g^0,0)$.

If $\frac{dF}{dt}\in L^1(\mathbb{R}_+;Y)$ and
$y(0)\in D(\mathcal{A})$ then
$\lim_{t\to\infty}\|\frac{d^+y}{dt}(t)\|_Y=0$ strongly in $Y$ and
$$
\int_0^{\infty}\|\frac{d^+y}{dt}(t)\|_Ydt
\le \frac{1}{\chi_0}\|((\mathcal{A}+\mathcal{B})(y(0))-F(0))^0\|_Y
+\frac{1}{\chi_0}\int_0^{\infty}\|\frac{dF}{dt}(t)\|_Ydt.
$$
\end{lemma}

\section{Existence, uniqueness and asymptotic behaviour of solutions}

In the general case $B(t)$ is not constant, we make a change
of variables $u_k=\widetilde{u}_k+\widetilde{\widetilde{u}}_k$,
where
$$
\widetilde{\widetilde{u}}_k(t,x)=(1+x)e^{-x}b_{n+k}(t)-xe^{-x}b_k(t),
\quad k=\overline{1,n}.
$$
Our problem \eqref{eS}, \eqref{BC}, \eqref{IC} can be written as
\begin{equation} \label{eStil}
\begin{gathered}
\frac{\partial\widetilde{u}}{\partial t}(t,x)
+\frac{\partial^2 v}{\partial x^2}(t,x)+\alpha(x,\widetilde{u}
+\widetilde{\widetilde{u}}(t,x))=\widetilde{f}(t,x),
\\
\frac{\partial v}{\partial t}(t,x)
-\frac{\partial^2\widetilde{u}}{\partial x^2}(t,x)
+\beta(x,v)=\widetilde{g}(t,x),\quad t>0,\;x>0,
\end{gathered}
\end{equation}
with the boundary condition
\begin{equation} \label{BCtil}
\begin{pmatrix}-\frac{\partial \widetilde{u}}{\partial x}(t,0)\\
\widetilde{u}(t,0)\\
w'(t)\end{pmatrix}
\in -G \begin{pmatrix}
v(t,0)\\ \frac{\partial v}{\partial x}(t,0)\\
w(t)\end{pmatrix}
+\begin{pmatrix} 0\\0\\B_2(t)\end{pmatrix},\quad
t>0,
\end{equation}
and the initial data
\begin{equation} \label{ICtil}
\begin{gathered}
\widetilde{u}(0,x)=\widetilde{u}_0(x),\quad
v(0,x)=v_0(x),\quad x>0,\\
w(0)=w_0,
\end{gathered}
\end{equation}
where
\begin{gather*}
\widetilde{f}_k(t,x)=f_k(t,x)
-\frac{\partial \widetilde{\widetilde{u}}_k}{\partial t}(t,x)
=f_k(t,x)-(1+x)e^{-x}b_{n+k}'(t)+xe^{-x}b_k'(t),
\\
\widetilde{g}_k(t,x)=g_k(t,x)
+\frac{\partial^2 \widetilde{\widetilde{u}}_k}{\partial x^2}(t,x)
=g_k(t,x)+(x-1)e^{-x}b_{n+k}(t)-(x-2)e^{-x}b_k(t),
\\
x>0, \quad t>0,\quad k=\overline{1,n},
\\
\widetilde{u}_{k0}(x)=u_{k0}(x)-(1+x)e^{-x}b_{n+k}(0)+xe^{-x}b_k(0),
\quad x>0,\quad k=\overline{1,n},
\\
B_2(t)=\operatorname{col}(b_{2n+1}(t),\ldots,b_{2n+m}(t)).
\end{gather*}

Using the operators $\mathcal{A}$ and $\mathcal{B}$, the problem
\eqref{eStil}, \eqref{BCtil}, \eqref{ICtil} can
be equivalently formulated as a time dependent Cauchy problem
in the space $Y$,
\begin{equation} \label{ePtil}
\begin{gathered}
\frac{d}{dt}\begin{pmatrix}\widetilde{u}\\v\\w\end{pmatrix}
+\mathcal{A}\begin{pmatrix}\widetilde{u}\\v\\w\end{pmatrix}
+\mathcal{B}\begin{pmatrix}\widetilde{u}+\widetilde{\widetilde{u}}(t)\\
v\\w\end{pmatrix}\ni\begin{pmatrix}
\widetilde{f}(t,\cdot)\\ \widetilde{g}(t,\cdot)\\B_2(t)\end{pmatrix}
\\
\begin{pmatrix}\widetilde{u}(0)\\v(0)\\w(0)\end{pmatrix}
=\begin{pmatrix}\widetilde{u}_0\\v_0\\w_0\end{pmatrix},
\end{gathered}
\end{equation}
where $\widetilde{f}=\operatorname{col}(\widetilde{f}_1,\ldots,
\widetilde{f}_n)$, $\widetilde{g}=\operatorname{col}
(\widetilde{g}_1,\ldots,\widetilde{g}_n)$,
$\widetilde{u}_0=\operatorname{col}(\widetilde{u}_{10},
\ldots,\widetilde{u}_{n0})$.

\begin{theorem}\label{theorem:5}
 Assume that {\rm (A1)ab, (A2)ac} hold,
$f,g\in W^{1,1}(0,T;L^2(\mathbb{R}_+$; $\mathbb{R}^n))$
 ($T>0$ fixed), $b_k\in W^{1,2}(0,T)$, $k=\overline{1,2n+m}$,
$u_0,v_0\in H^2(\mathbb{R}_+;\mathbb{R}^n)$,
$w_0\in\mathbb{R}^m$, $\operatorname{col}(\gamma_0v_0,w_0)\in D(G)$ and
$B_1(0)\in \gamma_1u_0+G_{11}(\gamma_0v_0)+G_{12}(w_0)$.
Then  problem \eqref{ePtil}
(equivalently problem \eqref{eStil}, \eqref{BCtil}, \eqref{ICtil})
has a unique strong solution
$\operatorname{col}(u,v,w)\in W^{1,\infty}(0,T;Y)$.
Moreover
$u,v\in L^{\infty}(0,T; H^2(\mathbb{R}_+;\mathbb{R}^n))$,
$(B_1(t)=\operatorname{col}(b_1(t),\ldots,b_{2n}(t)))$.
\end{theorem}

\begin{proof}
We shall use some similar techniques as those used in Luca \cite{L4}.
We assume in a first stage that
$f,g\in W^{1,\infty}(0,T;L^2(\mathbb{R}_+;\mathbb{R}^n))$,
$b_k\in W^{2,\infty}(0,T)$, $k=\overline{1,2n}$,
$b_j\in W^{1,\infty}(0,T)$, $j=\overline{2n+1,2n+m}$, and
the functions $\alpha_k(x,\cdot)$, $k=\overline{1,n}$
are Lipschitz continuous with Lipschitz constant $L$
independent of $x$. We consider the operators $\mathcal{C}(t)$,
$t\in[0,T]$, defined by
$D(\mathcal{C}(t))=D(\mathcal{A})$ and
$$
\mathcal{C}(t)\begin{pmatrix}\widetilde{u}\\v\\w\end{pmatrix}
=\mathcal{A}\begin{pmatrix}\widetilde{u}\\v\\w\end{pmatrix}
 +\mathcal{B}\begin{pmatrix}\widetilde{u}+\widetilde{\widetilde{u}}(t)\\
 v\\w\end{pmatrix}
 -\begin{pmatrix}\widetilde{f}(t,\cdot)\\ \widetilde{g}(t,\cdot)\\
 B_2(t)\end{pmatrix},\quad
\begin{pmatrix}\widetilde{u}\\v\\w\end{pmatrix}\in D(\mathcal{A}).
$$

Using Remark \ref{remark:1}, we deduce that the operators
$\mathcal{C}(t)$, $t\in[0,T]$ are maximal monotone in $Y$.
By the above assumption on the functions $\alpha_k$,
$k=\overline{1,n}$, we have
\begin{align*}
&|\alpha_k(x,\widetilde{u}_k+\widetilde{\widetilde{u}}_k(t,x))
 -\alpha_k(x,\widetilde{u}_k+\widetilde{\widetilde{u}}_k(s,x))|\\
&\le L|\widetilde{\widetilde{u}}_k(t,x)
 -\widetilde{\widetilde{u}}_k(s,x)|\\
&\le L[(1+x)e^{-x}|b_{n+k}(t)-b_{n+k}(s)|+xe^{-x}|b_k(t)-b_k(s)|],
\end{align*}
for all $t,\,s\in [0,T]$ for almost all $x>0$,
$k=\overline{1,n}$.
Therefore, we deduce
\begin{equation}\label{eq:2}
\begin{split}
&\|\alpha_k(\cdot,\widetilde{u}_k+\widetilde{\widetilde{u}}_k(t,\cdot))
-\alpha_k(\cdot,\widetilde{u}_k+\widetilde{\widetilde{u}}_k(s,
 \cdot))\|_{L^2(\mathbb{R}_+)}^2 \\
&\le 2L^2|b_{n+k}(t)-b_{n+k}(s)|^2\int_0^{\infty}(1+x)^2e^{-2x}dx
 +2L^2|b_k(t)-b_k(s)|^2\int_0^{\infty}x^2e^{-2x}dx\\
&=\frac{5L^2}{2}|b_{n+k}(t)-b_{n+k}(s)|^2+\frac{L^2}{2}|b_k(t)
 -b_k(s)|^2,\quad \forall t,s\in [0,T],\; k=\overline{1,n}.
\end{split}
\end{equation}

On the other hand for the functions $\widetilde f_k$,
$k=\overline{1,n}$ we obtain the inequality
\begin{align*}
|\widetilde{f}_k(t,x)-\widetilde{f}_k(s,x)|
&\le  |f_k(t,x)-f_k(s,x)|+(1+x)e^{-x}|b_{n+k}'(t)-b_{n+k}'(s)|\\
&\quad + xe^{-x}|b_k'(t)-b_k'(s)|,\quad
\forall t,s\in [0,T],\;x>0,\;k=\overline{1,n},
\end{align*}
and so
\begin{equation}\label{eq:3}
\begin{split}
\|\widetilde{f}_k(t,\cdot)-\widetilde{f}_k(s,\cdot)\|^2_{L^2
 (\mathbb{R}_+)}
&\le 3\|f_k(t,\cdot) -f_k(s,\cdot)\|^2_{L^2(\mathbb{R}_+)}\\
&\quad +\frac{15}{4}|b_{n+k}'(t)-b_{n+k}'(s)|^2
+\frac{3}{4}|b_k'(t)-b_k'(s)|^2,
\end{split}
\end{equation}
for all $t,s\in [0,T]$, $k=\overline{1,n}$.
For the functions $\widetilde g_k$, $k=\overline{1,n}$ we deduce
\begin{align*}
|\widetilde{g}_k(t,x)-\widetilde{g}_k(s,x)|
& \le  |g_k(t,x)-g_k(s,x)|+|x-1|e^{-x}|b_{n+k}(t)-b_{n+k}(s)|\\
&\quad +|x-2|e^{-x}|b_k(t)-b_k(s)|,\quad \forall t,\;s\in [0,T],\;
x>0,\;k=\overline{1,n},
\end{align*}
and so
\begin{equation}\label{eq:4}
\begin{split}
\|\widetilde{g}_k(t,\cdot)-\widetilde{g}_k(s,\cdot)
\|^2_{L^2(\mathbb{R}_+)}
&\le 3\|g_k(t,\cdot)-g_k(s,\cdot)\|^2_{L^2(\mathbb{R}_+)}
 +\frac{3}{4}|b_{n+k}(t)-b_{n+k}(s)|^2\\
&\quad +\frac{15}{4}|b_k(t)-b_k(s)|^2,\quad \forall t,s\in [0,T],\;
k=\overline{1,n}.
\end{split}
\end{equation}
Therefore, we obtain the following inequality for the operators
$\mathcal{C}(t)$,
\begin{align*}
&\|h_t-h_s\|_Y\\
&\le \|\alpha(\cdot,\widetilde{u}+\widetilde{\widetilde{u}}(t,\cdot))
 -\alpha(\cdot,\widetilde{u}+\widetilde{\widetilde{u}}(s,\cdot))
 \|_{L^2(\mathbb{R}_+;\mathbb{R}^n)}
+\|\widetilde{f}(t,\cdot)-\widetilde{f}(s,\cdot)\|_{L^2(\mathbb{R}_+;
 \mathbb{R}^n)}\\
&\quad +\|\widetilde{g}(t,\cdot)-\widetilde{g}(s,\cdot)
 \|_{L^2(\mathbb{R}_+;\mathbb{R}^n)}
 +\|B_2(t)-B_2(s)\|_{\mathbb{R}^m},
\end{align*}
for all $t,s\in [0,T]$, all
$\widetilde{y}=\operatorname{col}(\widetilde{u},v,w)
\in D(\mathcal{A})$, all $h_t\in \mathcal{C}(t)(\widetilde y)$,
all $h_s\in\mathcal{C}(s)(\widetilde y)$.


Using now the relations \eqref{eq:2}--\eqref{eq:4} and the
assumptions on the functions $f,g,b_k$, $k=\overline{1,2n+m}$,
from the last inequality we deduce that there exists $L_1>0$
such that
$$
\|h_t-h_s\|_Y\le L_1|t-s|,\quad \forall t,s\in [0,T],\;
\forall \widetilde{y}\in D(\mathcal{A}),\;
\forall h_t\in\mathcal{C}(t)(\widetilde y),\;
h_s\in\mathcal{C}(s)(\widetilde y).
$$
Therefore, the operator family
$\{\mathcal{C}(t); t\in [0,T]\}$ verifies the conditions
of Kato's Theorem (see Kato \cite{K}).
By the assumptions of our theorem, we deduce that
$\widetilde{y}_0=\operatorname{col}(\widetilde{u}_0,v_0,w_0)
\in D(\mathcal{A})$.
It follows that the problem \eqref{ePtil} has a unique strong
solution
$\widetilde{y}=\operatorname{col}
(\widetilde{u},v,w)\in W^{1,\infty}(0,T;Y)$,
$\operatorname{col}(\widetilde{u}(t),v(t),w(t))\in D(\mathcal{A})$,
for all $t\in [0,T]$. Moreover $\widetilde{y}$ is everywhere
differentiable from right on $[0,T)$ and
\begin{gather*}
\frac{d^+}{dt}\begin{pmatrix}\widetilde{u}(t)\\v(t)\\w(t)\end{pmatrix}
+\mathcal{A}\begin{pmatrix}\widetilde{u}(t)\\v(t)\\w(t)\end{pmatrix}
+\mathcal{B}\begin{pmatrix}\widetilde{u}(t)+\widetilde{\widetilde{u}}(t)
\\v(t)\\w(t)\end{pmatrix}\ni \begin{pmatrix}\widetilde{f}(t,\cdot)\\
\widetilde{g}(t,\cdot)\\B_2(t)\end{pmatrix}\\
\begin{pmatrix}\widetilde{u}(0)\\v(0)\\w(0)\end{pmatrix}
=\begin{pmatrix}\widetilde{u}_0\\v_0\\w_0\end{pmatrix}.
\end{gather*}
Hence $y(t)=\operatorname{col}(u(t),v(t),w(t))$ solves the problem
\begin{gather*}
\frac{d^+y}{dt}(t)+\mathcal{A}(y(t))+\mathcal{B}(y(t))
\ni F_1(t,\cdot),\quad 0\le t<T,\quad \text{in }Y\\
\gamma_1u(t)\in -G_{11}(\gamma_0v(t))-G_{12}(w(t))+B_1(t),\quad
0\le t<T\\
y(0)=y_0,
\end{gather*}
where $F_1(t,\cdot)=\operatorname{col}(f(t,\cdot),g(t,\cdot),B_2(t))$.
We deduce that $y=\operatorname{col}(u,v,w)$ is a solution of the
problem \eqref{eS}, \eqref{BC}, \eqref{IC}.

In a second stage, we suppose that $\alpha_k(x,\cdot)$,
$k=\overline{1,n}$ are not Lipschitz continuous and we replace
the functions
$\alpha_k(x,\cdot)$ by the Yosida approximations
$\alpha_k^{\lambda}(x,\cdot)$, $k=\overline{1,n}$, $\lambda>0$.
Using the above reasoning, we deduce that the problem \eqref{ePtil}
with $\alpha_k^{\lambda}$ instead of $\alpha_k$ has a unique strong
solution $\operatorname{col}(\widetilde{u}^{\lambda},v^{\lambda},
w^{\lambda})\in W^{1,\infty}(0,T;Y)$. Then
$y^{\lambda}=\operatorname{col}(u^{\lambda},v^{\lambda},w^{\lambda})$
solves the  problem
\begin{equation}\label{eq:5}
\begin{gathered}
\frac{d^+y^{\lambda}}{dt}(t)+\mathcal{A}(y^{\lambda}(t))
+\mathcal{B}_{\lambda}(y^{\lambda}(t))\ni F_1(t,\cdot),\quad
0\le t<T,\quad \text{in }Y\\
\gamma_1u^{\lambda}(t)\in -G_{11}(\gamma_0v^{\lambda}(t))
-G_{12}(w^{\lambda}(t))+B_1(t), \quad 0\le t<T\\
y^{\lambda}(0)=y_0,
\end{gathered}
\end{equation}
with
$$
\mathcal{B}_{\lambda} \begin{pmatrix}u\\v\\w\end{pmatrix}
=\begin{pmatrix}\operatorname{col}(\alpha_1^{\lambda}(\cdot,u_1),
 \ldots,\alpha_n^{\lambda}(\cdot,u_n))\\
\operatorname{col}(\beta_1(\cdot,v_1),\ldots,
\beta_n(\cdot,v_n))\\0\end{pmatrix},
\quad \lambda>0.
$$
We write the first relation in \eqref{eq:5} for $t+h$ and $t$,
 we subtract the relations and we multiply the obtained relation
by $y^{\lambda}(t+h)-y^{\lambda}(t)$ in the space $Y$.
We obtain after some computations
\begin{align*}
&\frac{1}{2}\frac{d^+}{dt}\|y^{\lambda}(t+h)-y^{\lambda}(t)\|_Y^2
+\langle g_{t+h}-g_t,z_{t+h}-z_t\rangle_{\mathbb{R}^{2n+m}}\\
&-\langle B_1(t+h)-B_1(t),\gamma_0v^{\lambda}(t+h)
 -\gamma_0v^{\lambda}(t)\rangle _{\mathbb{R}^n}\\
&\le \langle f(t+h,\cdot)-f(t,\cdot),u^{\lambda}(t+h)
-u^{\lambda}(t)\rangle _{L^2(\mathbb{R}_+;\mathbb{R}^n)}\\
&\quad +\langle g(t+h,\cdot)-g(t,\cdot),v^{\lambda}(t+h)
 -v^{\lambda}(t)\rangle _{L^2(\mathbb{R}_+;\mathbb{R}^n)}\\
&\quad +\langle B_2(t+h)-B_2(t),w^{\lambda}(t+h)
 -w^{\lambda}(t)\rangle_{\mathbb{R}^m},
\end{align*}
where $z_t=col(\gamma_0v^{\lambda}(t),w^{\lambda}(t))$,
$z_{t+h}=col(\gamma_0v^{\lambda}(t+h),w^{\lambda}(t+h))$,
$g_t\in G(z_t)$, $g_{t+h}\in G(z_{t+h})$.

Using (A2)c, the above inequality, we obtain
\begin{align*}
&\frac{1}{2}\frac{d^+}{dt}\|y^{\lambda}(t+h)
 -y^{\lambda}(t)\|_Y^2+\zeta_2\|\gamma_0v^{\lambda}(t+h)
 -\gamma_0v^{\lambda}(t)\|^2_{\mathbb{R}^{2n}}\\
&+\zeta_2\|w^{\lambda}(t+h)-w^{\lambda}(t)\|^2_{\mathbb{R}^m}\\
&\le\frac{1}{\zeta_0}\|B_1(t+h)-B_1(t)\|^2_{\mathbb{R}^{2n}}
 +\zeta_0\|\gamma_0v^{\lambda}(t+h)
 -\gamma_0v^{\lambda}(t)\|^2_{\mathbb{R}^n}\\
&\quad +\frac{1}{\zeta_0}\|B_2(t+h)-B_2(t)\|^2_{\mathbb{R}^m}
 +\zeta_0\|w^{\lambda}(t+h)-w^{\lambda}(t)\|^2_{\mathbb{R}^m}\\
&\quad +\|F_0(t+h,\cdot)-F_0(t,\cdot)\|_X\cdot\|y^{\lambda}(t+h)
-y^{\lambda}(t)\|_Y,
\end{align*}
for $0\le t<t+h<T$, $\lambda>0$,
where $F_0(t,\cdot)=\operatorname{col}(f(t,\cdot),g(t,\cdot))$.

When we choose $0<\zeta_0<\zeta_2$, we obtain
\begin{align*}
&\frac{1}{2}\frac{d^+}{dt}\|y^{\lambda}(t+h)
 -y^{\lambda}(t)\|^2_Y\\
&\le\frac{1}{\zeta_0}\|B(t+h)
 -B(t)\|^2_{\mathbb{R}^{2n+m}}
+\|F_0(t+h,\cdot)-F_0(t,\cdot)\|_X
\cdot\|y^{\lambda}(t+h)-y^{\lambda}(t)\|_Y,
\end{align*}
for $0\le t<t+h<T$, $\lambda>0$.
We integrate the above inequality over $[0,t]$ and we deduce that
\begin{align*}
&\frac{1}{2}\|y^{\lambda}(t+h)-y^{\lambda}(t)\|^2_Y\\
&\le\frac{1}{2}\Big(\|y^{\lambda}(h)-y^{\lambda}(0)\|^2_Y
 +\frac{2}{\zeta_0}\int_0^T\|B(s+h)-B(s)\|^2_{\mathbb{R}^{2n+m}}ds\Big)
\\
&\quad +\int_0^t\|F_0(s+h,\cdot)-F_0(s,\cdot)\|_X\cdot\|
y^{\lambda}(s+h)-y^{\lambda}(s)\|_Y ds,
\end{align*}
for $0\le t<t+h<T$, $\lambda>0$.
Using a variant of Gronwal's lemma, we obtain
\begin{align*}
&\|y^{\lambda}(t+h)-y^{\lambda}(t)\|_Y\\
&\le \|y^{\lambda}(h)-y^{\lambda}(0)\|_Y+\sqrt{\frac{2}{\zeta_0}}
\Big(\int_0^T\|B(s+h)-B(s)\|^2_{\mathbb{R}^{2n+m}}ds\Big)^{1/2}\\
&\quad +\int_0^t\|F_0(s+h,\cdot)-F_0(s,\cdot)\|_Xds,\quad
0\le t<t+h<T,\; \lambda>0.
\end{align*}
We deduce from the above inequality that
\begin{equation}\label{eq:6}
\begin{aligned}
\|\frac{d^+y^{\lambda}}{dt}(t)\|_Y
&\le \|\frac{d^+y^{\lambda}}{dt}(0)\|_Y+\sqrt{\frac{2}{\zeta_0}}
\Big(\int_0^T\|\frac{dB}{ds}(s)\|^2_{\mathbb{R}^{2n+m}}ds\Big)^{1/2}\\
&\quad +\int_0^T\|\frac{df}{ds}(s,\cdot)\|_{L^2(\mathbb{R}_+;
\mathbb{R}^n)}ds+\int_0^T\|\frac{dg}{ds}(s,\cdot)\|_{L^2(\mathbb{R}_+;
\mathbb{R}^n)}ds,
\end{aligned}
\end{equation}
for $0\le t<T$, $\lambda>0$.
Because
$\sup\big\{\|\frac{d^+y^{\lambda}}{dt}(0)\|_Y;\lambda>0\big\}$
is a positive constant independent of $\lambda$, using the
assumptions of the theorem, the inequality \eqref{eq:6} gives us
$$
\sup\big\{\|\frac{dy^{\lambda}}{dt}(t)\|_Y;\lambda>0,\,0<t<T\big\}
\le \text{const.}
$$
and then $\sup \{\|y^{\lambda}(t)\|_Y;\lambda>0,\,0<t<T\}\le
 \text{const}$.
Hence
\begin{gather*}
\{u^{\lambda};\lambda>0\},\; \{v^{\lambda};\lambda>0\}
 \text{ are bounded in }L^{\infty}(0,T;L^2(\mathbb{R}_+;\mathbb{R}^n)),
\\
\{w^{\lambda};\lambda>0\}\text{ is bounded in }
L^{\infty}(0,T;\mathbb{R}^m),
\end{gather*}
and, using the assumption (A1)b, we deduce that
\begin{equation}\label{eq:7}
\{\mathcal{B}_{\lambda}(y^{\lambda}(t));\lambda>0\}
\text{ is bounded in } L^{\infty}(0,T;Y).
\end{equation}
By \eqref{eq:5}, we have
\begin{equation}\label{eq:8}
\frac{1}{2}\frac{d}{dt}\|y^{\lambda}(t)
-y^{\mu}(t)\|^2_Y\le -\langle \mathcal{B}_{\lambda}(y^{\lambda}(t))
-\mathcal{B}_{\mu}(y^{\mu}(t)),y^{\lambda}(t)-y^{\mu}(t)\rangle _Y,
\end{equation}
for $0<t<T$, $\lambda>0$.
Using now the relations \eqref{eq:7} and \eqref{eq:8}, we obtain
$$
\|y^{\lambda}(t)-y^{\mu}(t)\|_Y\le \text{const.}(\lambda+\mu)^{1/2},
\quad 0\le t\le T,\; \lambda,\; \mu>0.
$$
Therefore, the sequence $\{y^{\lambda};\lambda>0\}$ converges
to some function $y = \operatorname{col}(u,v,w)$ in $C([0,T];Y)$
as $\lambda\to 0$.
Using Lebesgue's Dominated Convergence Theorem, we obtain
$\mathcal{B}_{\lambda}(y^{\lambda})\to\mathcal{B}(y)$,
as $\lambda\to 0$, strongly in $L^2(0,T;Y)$.
By letting $\lambda\to 0$ in \eqref{eq:5}, $\mathcal{A}$ and
$G$ being demi-closed operators, we obtain that $y$ is a
strong solution of the problem \eqref{eS}, \eqref{BC}, \eqref{IC}.

In the third stage (general case), we approximate
$f,g\in W^{1,1}(0,T;L^2(\mathbb{R}_+;\mathbb{R}^n))$
by  $\{f^j\}_{j\ge 1}$,
$\{g^j\}_{j\ge 1}\subset W^{1,\infty}(0,T;L^2(\mathbb{R}_+;
\mathbb{R}^n))$ in
$W^{1,1}(0,T;L^2(\mathbb{R}_+;\mathbb{R}^n))$, and
$b_k\in W^{1,2}(0,T)$ by $\{b_k^j\}_{j\ge 1}\subset W^{2,\infty}(0,T)$,
$k=\overline{1,2n}$, $b_i\in W^{1,2}(0,T)$ by
$\{b_i^j\}_{j\ge 1}\subset W^{1,\infty}(0,T)$,
$i=\overline{2n+1,2n+m}$, in  $W^{1,2}(0,T)$.

Fixing $y_0=\operatorname{col}(u_0,v_0,w_0)\in Y$ with
$\widetilde{y}_0=\operatorname{col}(\widetilde{u}_0,v_0,w_0)
\in D(\mathcal{A})$, we deduce after some considerations
(see also Luca \cite{L4}) that the sequence of the corresponding
strong solutions $\{y^j=\operatorname{col}(u^j,v^j,w^j)\}_{j\ge 1}$
converges as $j\to\infty$ to $y=\operatorname{col}(u,v,w)$,
which is a strong solution of our problem.

By  system \eqref{eS}, we deduce
$u_{xx},\,v_{xx}\in L^{\infty}(0,T;L^2(\mathbb{R}_+;\mathbb{R}^n))$
and, using the inequality
$$
\|z'\|_{L^2(\mathbb{R}_+)}\le C(\|z''\|_{L^2(\mathbb{R}_+)}
+\|z\|_{L^2(\mathbb{R}_+)}),\quad \text{for }
z\in H^2(\mathbb{R}_+),
$$
(see Adams \cite{A}), we get that
$u_x,\,v_x\in L^{\infty}(0,T;L^2(\mathbb{R}_+;\mathbb{R}^n))$.
So we obtain $u,\,v\in L^{\infty}(0,T;$ $H^2(\mathbb{R}_+;
\mathbb{R}^n))$.
\end{proof}

\begin{theorem}\label{theorem:6}
Assume that {\rm (A1)ab,  (A2)ac} hold. If
$f,g\in L^1(0,T;L^2(\mathbb{R}_+;\mathbb{R}^n))$
where $T$ is fixed positive value,
$b_k\in L^2(0,T)$, $k=\overline{1,2n+m}$,
$u_0,v_0\in L^2(\mathbb{R}_+;\mathbb{R}^n)$,
$w_0\in\overline{D(G_{12})\cap D(G_{22})}$, then  problem
\eqref{eS}, \eqref{BC}, \eqref{IC} has a unique weak
solution $\operatorname{col}(u,v,w)$ $\in C([0,T];Y)$.
\end{theorem}

\begin{proof}
By the assumptions of the theorem, it follows that
$y_0=\operatorname{col}(u_0,v_0,w_0)\in\overline{D(\mathcal{A})}$.
We consider $\{y_0^j\}_{j\ge 1}\subset Y$ such that
$\widetilde{y}_0^j\in D(\mathcal{A})$ and $y_0^j\to y_0$,
as $j\to\infty$, in $Y$. Also let the sequences
$\{f^j\}_{j\ge 1}$, $\{g^j\}_{j\ge 1}\subset
W^{1,1}(0,T;L^2(\mathbb{R}_+;\mathbb{R}^n))$
be such that $f^j\to f$, $g^j\to g$, as $j\to\infty$,
in $L^1(0,T;L^2(\mathbb{R}_+;\mathbb{R}^n))$ and the sequences
$\{b_k^j\}_{j\ge 1}\subset W^{1,2}(0,T)$ be such that $b_k^j\to b_k$,
 as $j\to\infty$, in $L^2(0,T)$, $k=\overline{1,2n+m}$.
Then the corresponding strong solutions
$y^j=\operatorname{col}(u^j,v^j,w^j)\in W^{1,\infty}(0,T;Y)$
of  problem \eqref{eS}, \eqref{BC}, \eqref{IC}, given
by Theorem \ref{theorem:5}, satisfy the inequality
\begin{align*}
\|y^j(t)-y^l(t)\|_Y
&\le \|y_0^j-y_0^l\|_Y+\sqrt{\frac{2}{\zeta_0}}
\Big(\int_0^T\|B^j(s)-B^l(s)\|^2_{\mathbb{R}^{2n+m}}ds\Big)^{1/2}\\
&\quad +\int_0^t\|F_0^j(s,\cdot)-F_0^l(s,\cdot)\|_X^2ds,\quad
0\le t\le T,\quad \forall j,l\in\mathbb{N},
\end{align*}
where $F_0^j=\operatorname{col}(f^j,g^j)$, $j\ge 1$, which leads
us to the conclusion.
\end{proof}


\begin{theorem}\label{theorem:7}
Assume that {\rm (A1)abc, (A2)ac} hold.
If $f,g\in L^1_{\rm loc}(\mathbb{R}_+;L^2(\mathbb{R}_+;
\mathbb{R}^n))$, $b_k\in L^2(\mathbb{R}_+)$,
$k=\overline{1,2n+m}$ such that
$\lim_{t\to\infty}f(t)=f^0$, $\lim_{t\to\infty}g(t)=g^0$, strongly in
$L^2(\mathbb{R}_+;\mathbb{R}^n)$ and
$\delta=\operatorname{col}(p,q,r)$ is the unique solution of
 \eqref{eq:1}. Then
$\lim_{t\to\infty}y(t)=\delta$, strongly in $Y$, where
$y(t)=\operatorname{col}(u(t),v(t),w(t))$, $t\ge 0$ is an
arbitrary weak solution of  \eqref{ePtil}.
\end{theorem}

\begin{proof}
 By Lemma \ref{lemma:5}, the operator $\mathcal{A}+\mathcal{B}$
is strongly monotone and equation \eqref{eq:1} has a unique solution
$\delta=\operatorname{col}(p,q,r)\in D(\mathcal{A})$.
We define for any $l\in\mathbb{N}$ the function
$$
B^l(t)=\begin{cases} B(t),&\text{for }0\le t\le l\\
0,&\text{for } t>l.
\end{cases}
$$
Let $y_0=\operatorname{col}(u_0,v_0,w_0)\in\overline{D(\mathcal{A})}$;
we denote by $y(t)$, $y^l(t)$, $t\ge 0$ the weak solutions of
problem \eqref{eS}, \eqref{BC}, \eqref{IC} corresponding to
data $\{B,f,g,y_0\}$, respectively
$\{B^l,f,g,y_0\}$, given by Theorem \ref{theorem:6}.
Then we have
\begin{equation}\label{eq:9}
\|y^l(t)-y(t)\|_Y\le \text{const.}
\Big(\int_l^{\infty}\|B(s)\|^2_{\mathbb{R}^{2n+m}}ds\Big)^{1/2},\quad
t>l.
\end{equation}

Because for $t>l$, $y^l$ is the weak solution corresponding to
$B(t)\equiv 0$, by Lemma \ref{lemma:6}, we deduce that
$y^l(t)\to\delta$, as $t\to\infty$ in $Y$, ($l\in\mathbb{N}$).
Hence this last conclusion with \eqref{eq:9} and the inequality
$$
\|y(t)-\delta\|_Y\le \|y(t)-y^l(t)\|_Y+\|y^l(t)-\delta\|_Y
$$
give us that $y(t)\to\delta$, as $t\to\infty$, in $Y$.
\end{proof}

\subsection*{Acknowledgments}
 The author wants to express her gratitude to the anonymous
referee for his/her valuable comments and suggestions.


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