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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2014 (2014), No. 127, pp. 1--12.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2014 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2014/127\hfil Elastic wave equations with critical damping]
{Energy decay for elastic wave equations with critical damping}

\author[J. L. Horbach, N. Nakabayashi \hfil EJDE-2014/127\hfilneg]
{Jaqueline Luiza Horbach, Naoki Nakabayashi}  % in alphabetical order

\address{Jaqueline Luiza Horbach \newline
Department of Mathematics,
Federal University of Santa Catarina,
88040-270 Florian\'opolis, Santa Catarina, Brazil}
\email{jaqueluizah@gmail.com}

\address{Naoki Nakabayashi \newline
Department of Mathematics,
Graduate School of Education, Hiroshima University,
Higashi-Hiroshima 739-8524, Japan}
\email{n.naoki2655@gmail.com} 

\thanks{Submitted January 17, 2014. Published May 16, 2014.}
\subjclass[2000]{35L52, 35B45, 35A25, 35B33}
\keywords{Elastic wave equation; critical damping; multiplier method; \hfill\break\indent 
 total energy; compactly supported initial data;  optimal decay}

\begin{abstract}
 We show that the total energy decays at the rate $E_u(t) = O(t^{-2})$,
 as $t \to +\infty$, for solutions to the Cauchy problem of a
 linear system of elastic wave with a variable damping term.
 It should be mentioned that the the critical decay satisfies
 $V(x) \ge C_0(1+|x|)^{-1}$ for  $C_0>2b$, where $b$ represents
 the speed of propagation of the P-wave.
\end{abstract}

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

\section{Introduction}

We consider the Cauchy problem for the linear system of elastic wave
equations with a critical potential type of damping $V(x)$ in 
$\mathbb{R}^2$:
\begin{gather}
\begin{gathered}
u_{tt}(t,x)-a^2\Delta u(t,x)-(b^2-a^2)\nabla(\operatorname{div}u(t,x))
+V(x)u_{t}(t,x)=0,\\
 (t,x)\in (0,\infty)\times\mathbb{R}^2,
\end{gathered} \label{eqn} \\
u(0,x)= u_0(x), \quad u_{t}(0,x) = u_{1}(x), \quad x\in \mathbb{R}^2, \label{initial}
\end{gather}
where the vector displacement $u=u(t,x)=(u_1(x,t), u_2(t,x))$ and
the coefficients  $a$ and $b$ are related to the Lam\'e coefficients
and the satisfy the condition of ellipticity $0 < a^2 \leq b^2$.
The initial data $u_0$ and $u_1$ are compactly supported from the
energy space; that is,
\begin{gather*}
u_0 \in (H^1(\mathbb{R}^2))^2, \quad u_1 \in (L^2(\mathbb{R}^2))^2,\\
\operatorname{supp} u_i \subset B(R_0):=\{x \in \mathbb{R}^2:|x|<R_0\}, 
\quad (i=0,1).
\end{gather*}
The system of elastic waves satisfies the property of  finite speed
of propagation given by coefficient $b$, which is
the speed of propagation of the longitudinal {\it P}-wave. The
coefficient $a$ is the speed of propagation of the transverse  
{\it S}-wave (cf. \cite{ruy}).

The damping coefficient $V(x)$ belongs to 
$C(\mathbb{R}^2) \cap L^\infty(\mathbb{R}^2)$ 
and satisfies
\begin{itemize}
\item[(A1)] 
${V(x) \ge \frac{C_0}{1+|x|}}$,
for all $x\in \mathbb{R}^2$ and for some $C_0>0$.
\end{itemize}
Under these conditions it is  standard to prove via semigroups
theory (cf. Horbach \cite[Theorem 2.1]{H}) that  problem
\eqref{eqn}-\eqref{initial} has a unique solution
\[
u\in C([0,+\infty);(H^1(\mathbb{R}^2))^2) \cap C^1([0,+\infty);
(L^2(\mathbb{R}^2))^2)
\]
  satisfying
\[
 E_u(t)+\int_0^t \int_{\mathbb{R}^2} V(x)|u_t(s,x)|^2 \,dx\,ds =E_u(0),
\]
where the total energy is 
\begin{equation} \label{e1.3}
E_u(t):=\frac{1}{2}\int_{\mathbb{R}^2}\{|u_{t}(t,x)|^2
+a^2|\nabla u(t,x)|^2+(b^2-a^2)(\operatorname{div}u(t,x))^2\}dx.
\end{equation}

Our main results in this article read as follows.

\begin{theorem} \label{thm1.1}
Let the damping coefficient  $V(x)$ satisfy   {\rm (A1)}
with $C_0> 2b$. Then the solution $u(t,x)$ to problem \eqref{eqn}-\eqref{initial}
satisfies
\begin{equation} \label{e1.4}
E_u(t) = O(t^{-2})
\end{equation}
as $t \to +\infty$.
\end{theorem}

\begin{proposition} \label{prop1.1}
Let $V(x)$ satisfy {\rm  (A1)} and $C_0$ satisfy
$0<C_0\leq2b$. Then for the solution $u(t,x)$ to 
\eqref{eqn}-\eqref{initial} one has the following two possibilities:
\begin{itemize}
\item[(I)] When $0 < C_0 \le b$ it holds that
\begin{equation} \label{e1.5}
E_u(t) = O(t^{-1+\delta}),\quad t \to +\infty
\end{equation}
for  any $\delta > 0$ satisfying $1-\frac{C_0}{b}<\delta <1$;

\item[(II)] when $b < C_0 \le 2b$ it holds that
\begin{equation} \label{e1.6}
E_u(t) = O(t^{-\frac{C_0}{b}+\delta}),\quad \text{as }t \to +\infty
\end{equation}
for any $\delta > 0$.
\end{itemize}
\end{proposition}


\begin{remark} \rm 
It follows from  Proposition \ref{prop1.1} part (I) 
that $E_u(t) = O(t^{-\frac{C_0}{b}+\varepsilon})$ for any small
 $\varepsilon > 0$ which is the same
decay rate as that  Proposition \ref{prop1.1} part (II), if one re-set
$\delta := 1-\frac{C_0}{b} + \varepsilon$ with any $\varepsilon
\in (0,C_0/b] \subset (0,1]$.
This implies that  when $0 < C_0 \leq 2b$ the obtained
decay rate is $E_u(t) = O(t^{-\frac{C_0}{b}+\delta})$ with any
small $\delta > 0$.
\end{remark}

\begin{remark} \rm 
When one compares these results with the one for the scalar wave equations
 due to \cite{ITY} the obtained decay rates are (almost) optimal.
\end{remark}

To begin we mention the motivation and some related 
results of this research.
There are a lot of results concerning the energy decay
estimates for the  scalar-valued wave equation
\begin{gather} \label{weqn}
w_{tt}(t,x)-\Delta w(t,x)+V(x)w_{t}(t,x)=0,\quad 
(t,x)\in (0,\infty)\times\mathbb{R}^n, \\
\label{winitial}
w(0,x)= w_0(x), \quad w_{t}(0,x) = w_{1}(x), \quad 
x\in \mathbb{R}^n, %\label{winitial}
\end{gather}
where the function $V(x)$ typically satisfies
\begin{equation} \label{e1.9}
V(x) = \frac{C_0}{(1+|x|)^{\alpha}}, \quad C_0 > 0,\quad \alpha \in [0,+\infty).
\end{equation}
In connection with problem \eqref{weqn}-\eqref{winitial}
 it is easy to proove the decreasing property of the total energy 
\[
E_{w}(t):=\frac{1}{2}\int_{\mathbb{R}^n}\{|w_{t}(t,x)|^2+|\nabla w(t,x)|^2\}dx.
\]

So a natural question arises whether $E_{w}(t)$ decays or not as $t \to +\infty$. 
 About this question Mochizuki \cite{Mochi}
first gave an answer that when $\alpha > 1$ (super-critical damping), 
the solution $w(t,x)$ to  \eqref{weqn}-\eqref{winitial}
is asymptotically free, and the corresponding energy satisfies 
${\lim_{t \to +\infty}}E_{w}(t) > 0$ (non-decay).
In this sense, the case $\alpha > 1$ shows a  hyperbolic aspect of the 
equation \eqref{weqn}.  On the other hand, Todorova-Yordanov \cite{TY} 
considered the case when $\alpha \in [0,1)$ (sub-critical damping), 
and they derived almost optimal decay estimates of the energy:
\[
E_{w}(t) = O(t^{\delta-\frac{n-\alpha}{2-\alpha}-1}),\quad \text{as }t \to +\infty
\]
for any $\delta > 0$. In connection with this, one can observe that the energy 
$E_{w}(t)$ has faster decay rates as $\alpha \to 1$.
This sub-critical damping case has a close relation to the so called diffusion 
phenomenon of the equation \eqref{weqn}.
Recently, Ikehata-Todorova-Yordanov \cite{ITY} announced a work on the 
critical damping case $\alpha = 1$, and roughly speaking, they derived 
the following results:
\[
E_{w}(t) = O(t^{\delta-\min\{C_0,n\}}),\quad\text{as } t \to +\infty,
\]
for any $\delta > 0$ (indeed, we can choose $\delta = 0$ in part). 
There exists a threshold concerning the decay rate from the
viewpoint of the dimension $n$ and the damping coefficient $C_0$. 

On the other hand,  for the elastic waves  it seems that there are
not so many results except for the references 
\cite{CDI,CI,CI-2,H,K}. That happens,  especially for
dissipative terms with variable coefficients. The local energy decay
property of the equation \eqref{eqn} without damping 
(i.e., $V(x) \equiv 0$) is studied by Kapitonov \cite{K}, and the result 
is an elastic
wave version of that derived by Morawetz \cite{M} to the scalar wave
equations \eqref{weqn} with $V(x) \equiv 0$. For the exterior mixed problem
of the elastic wave equation \eqref{eqn} with localized damping
coefficient $V(x)$ near spatial infinity Char\~ao and
Ikehata \cite{CI} derived the faster decay estimates of the total
energy and $L^2$-norm of solutions basing on a previous research
due to Ikehata \cite{I} about the scalar wave equations. In
Char\~ao and Ikehata \cite{CI-2} the equation \eqref{eqn}
with monotone nonlinearity and critical damping ($\alpha = 1$) has
been treated based on a method due to \cite{II}, however, the
obtained decay rate of the energy seems not to be sharp. Sharp
higher order energy decay estimates have been  recently studied by
Char\~ao-da Luz and Ikehata \cite{CDI} to the elastic wave equations 
with ``structural damping'', but the method in \cite{CDI} cannot be 
applied to the $x$-dependent variable coefficient case like \eqref{eqn}.

The purpose of this paper is to find sharp decay estimates of the energy 
to problem \eqref{eqn}-\eqref{initial} in the $2$-dimensional case,
and the strategy for the proof comes from the previous papers due 
to Ikehata-Inoue \cite{II} and the two 
dimensional Ikehata-Todorova-Yordanov \cite{ITY} method.
Several basic computations of the energy method have already been 
prepared by Horbach \cite{H}, so basing on this computations due 
to \cite{H} we will develop a method introduced in \cite{ITY} 
to the scalar wave equations.  Unlike the scalar wave equation, 
the elastic wave equation is vector valued and the equation itself 
has a quite complex form, so the treatment of the elastic wave equation 
is not so easy as compared with the scalar wave. 
The advantage is that the proof is very elementary in spite of the 
complexity of the equation.

The proof of Proposition \ref{prop1.1} part (I) above is already shown in Horbach \cite{H};
so that we restrict ourselves to prove Theorem \ref{thm1.1} and 
Proposition \ref{prop1.1} part (II) in the next section.

\subsection*{Open problem}
 Can one have the estimate  $E_u(t) = O(t^{-\min\{\frac{C_0}{b},n\}})$ 
(as $t \to +\infty$)  when $n \geq 3$. This will be an estimate, for 
a higher dimensional elastic wave version, with the same form as for
the scalar wave equations in \cite{ITY}.

\subsection{Notation}

We will use the following symbols:
\begin{gather*}
\|u\|^2 :=\int_{\mathbb{R}^2} |u(x)|^2dx 
= \int_{\mathbb{R}^2} \sum_{i=1}^2|u_i(x)|^2dx,
\\
\|\nabla u\|^2 :=\sum_{i=1}^2\|\nabla u_i\|^2
=\sum_{i=1}^2\int_{\mathbb{R}^2}\vert\nabla u_{i}(x)\vert^2dx 
=\sum_{i,j=1}^2\Big|\Big|\frac{\partial u_i}{\partial x_j}\Big|\Big|^2,
\\
(u,v) = \int_{\mathbb{R}^2} u(x)\cdot v(x)dx =\int_{\mathbb{R}^2} 
\sum_{i=1}^2 u_i(x)v_i(x) dx,
\\
u : \nabla v =u_1\nabla v_1+u_2 \nabla v_2 
=\Big(\sum_{i=1}^2u_i\frac{\partial v_i}{\partial x_1} , 
\sum_{i=1}^2u_i\frac{\partial v_i}{\partial x_2}\Big),
\\
\operatorname{div}(u : \nabla u) = u \cdot \Delta u+|\nabla u|^2,
\\
\operatorname{div}(u_t : \nabla u) = u_t \cdot \Delta u
 +\frac{1}{2}\frac{d}{dt}|\nabla u|^2,
\\
\operatorname{div}(u\ \operatorname{div}u) 
 = (\operatorname{div}u)^2+u \cdot \nabla (\operatorname{div}u),
\\
\operatorname{div}(u_t\ \operatorname{div}u) 
= \frac{1}{2}\frac{d}{dt}(\operatorname{div}u)^2
+u_t \cdot \nabla (\operatorname{div}u),
\end{gather*}
where $p\cdot q := p_{1}q_{1} + p_{2}q_{2}$ for 
$p =(p_{1},p_{2}) \in \mathbb{R}^2$ and  $q = (q_{1},q_{2}) \in \mathbb{R}^2$.

\section{Proof of the main results}

 First, we multiply the equation \eqref{eqn} by $f(t)u_{t}+g(t)u$, and integrate
 over $\mathbb{R}^2$ in order to get the following Lemma, where $f(t)$ and 
$g(t)$ are smooth functions specified later.

\begin{lemma} \label{lem3.1}
Let $u \in C([0,+\infty);(H^1(\mathbb{R}^2))^2) 
\cap C^1([0,+\infty);(L^2(\mathbb{R}^2))^2)$ be the solution to 
\eqref{eqn}-\eqref{initial}. Then
\begin{equation} \label{e3.1}
\frac{d}{dt}E(t)+F(t) = 0, \quad t \ge 0,
\end{equation}
where
\begin{equation} \label{e3.2}
\begin{aligned}
E(t)&:= \int_{\mathbb{R}^2}\frac{f(t)}{2}\{|u_{t}|^2
+a^2|\nabla u|^2+(b^2-a^2)(\operatorname{div}u)^2\}dx \\
&\quad + g(t)(u,u_{t})+\int_{\mathbb{R}^2}\frac{g(t)}{2}V(x)|u|^2dx
 -\int_{\mathbb{R}^2}\frac{g_{t}(t)}{2}|u|^2dx,
\end{aligned}
\end{equation}
and
\begin{equation} \label{e3.3}
\begin{aligned}
F(t)&:=\frac{1}{2}\int_{\mathbb{R}^2}\{2f(t)V(x)-2g(t)
 -f_{t}(t)\}|u_{t}|^2dx
 +\frac{1}{2}\int_{\mathbb{R}^2}\{g_{tt}(t)-g_{t}(t)V(x)\}|u|^2dx \\
&\quad +\frac{a^2}{2}\int_{\mathbb{R}^2}(2g(t)-f_{t}(t))|\nabla u|^2 dx
 +\frac{b^2-a^2}{2}\int_{\mathbb{R}^2}(2g(t)-f_{t}(t))(\operatorname{div}u)^2dx.
\end{aligned}
\end{equation}
\end{lemma}

\begin{proof}
 For the moment, one can assume that the corresponding solution $u(t,x)$ 
is sufficiently smooth and vanishes near infinity to proceed the computations below.
The general case follows from density arguments. 

We first multiply \eqref{eqn} by $f(t)u_{t}$ to get the equality
\[
f(t)(u_t\cdot u_{tt})-a^2f(t)(u_t \cdot \Delta u)-(b^2-a^2)
f(t)(u_t\cdot \nabla(\operatorname{div}u))+f(t)V(x)|u_t|^2=0, 
\]
so that
\begin{equation} \label{e3.4}
\begin{aligned}
&\frac{f(t)}{2}\frac{d}{dt}|u_t|^2-a^2f(t)
 \operatorname{div}(u_t:\nabla u)+\frac{a^2f(t)}{2}
\frac{d}{dt}|\nabla u|^2
 -(b^2-a^2)f(t) \operatorname{div}(u_t\ \operatorname{div}u)\\
&\quad +\frac{(b^2-a^2)f(t)}{2}\frac{d}{dt}(\operatorname{div}u)^2+f(t)V(x)|u_t|^2=0.
\end{aligned}
\end{equation}

Next, we multiply  \eqref{eqn} by $g(t)u$ to obtain
\[
g(t)(u\cdot u_{tt})-a^2g(t)(u \cdot \Delta u)
-(b^2-a^2)g(t)(u\cdot \nabla(\operatorname{div}u))+g(t)V(x)(u\cdot u_t)=0, 
\]
so that
\begin{equation} \label{e3.5}
\begin{aligned}
&g(t)\frac{d}{dt}(u\cdot u_t)-g(t)|u_t|^2-a^2g(t)
\operatorname{div}(u:\nabla u)+a^2g(t)|\nabla u|^2\\
&-(b^2-a^2)g(t)\operatorname{div}(u\ \operatorname{div}u) 
+(b^2-a^2)g(t)(\operatorname{div}u)^2\\
&+\frac{g(t)}{2}V(x)
\frac{d}{dt}|u|^2=0.
\end{aligned}
\end{equation}
Thus, by adding \eqref{e3.4} and \eqref{e3.5}, it follows that
\begin{align*}
& \Bigl\{\frac{f(t)}{2}\frac{d}{dt}|u_t|^2-a^2f(t)
 \operatorname{div}(u_t:\nabla u)+\frac{a^2f(t)}{2}\frac{d}{dt}|\nabla u|^2 
\\
&-(b^2-a^2)f(t)\ \operatorname{div}(u_t\ \operatorname{div}u)
 +\frac{(b^2-a^2)f(t)}{2}\frac{d}{dt}(\operatorname{div}u)^2+f(t)V(x)|u_t|^2\Bigl\} 
\\
& + \Bigl\{g(t)\frac{d}{dt}(u\cdot u_t)-g(t)|u_t|^2-a^2g(t)
 \operatorname{div}(u:\nabla u)+a^2g(t)|\nabla u|^2 
\\
&-(b^2-a^2)g(t)\ \operatorname{div}(u\ \operatorname{div}u)
+(b^2-a^2)g(t)\ (\operatorname{div}u)^2
+\frac{g(t)}{2}V(x)\frac{d}{dt}|u|^2\Bigl\}=0.
\end{align*}
This implies that
\begin{equation} \label{e3.6}
\begin{aligned}
&\frac{f(t)}{2}\frac{d}{dt}\Big(|u_t|^2+a^2|\nabla u|^2+(b^2-a^2)
(\operatorname{div}u)^2\Big)+g(t)\frac{d}{dt}(u\cdot u_t)
+\frac{g(t)}{2}V(x)\frac{d}{dt}|u|^2\\
&-g(t)|u_t|^2 - a^2g(t) \operatorname{div}(u:\nabla u)+a^2g(t)|\nabla u|^2
-(b^2-a^2)g(t) \operatorname{div}(u \operatorname{div}u)\\
&+(b^2-a^2)g(t)(\operatorname{div}u)^2
 - a^2f(t) \operatorname{div}(u_t:\nabla u)-(b^2-a^2)f(t)
 \operatorname{div}(u_t \operatorname{div}u)\\
&+f(t)V(x)|u_t|^2=0.
\end{aligned}
\end{equation}

If we define the density of energy 
\begin{equation} \label{e3.7}
\begin{aligned}
e(t,x)&:=\frac{f(t)}{2}\Big(|u_t|^2+a^2|\nabla u|^2
+(b^2-a^2)(\operatorname{div}u)^2\Big)\\
&\quad +g(t)(u\cdot u_t)+\frac{g(t)}{2}V(x)|u|^2-\frac{g_t(t)}{2}|u|^2,
\end{aligned}
\end{equation}
then we have
\begin{align*}
\frac{d}{dt}e(t,x)
&=\frac{f_t(t)}{2}\Big(|u_t|^2+a^2|\nabla u|^2+(b^2-a^2)
 (\operatorname{div}u)^2\Big)\\
&\quad +\frac{f(t)}{2}\frac{d}{dt}\Big(|u_t|^2+a^2|\nabla
u|^2+(b^2-a^2)(\operatorname{div}u)^2\Big)\\
&\quad + g(t)\frac{d}{dt}(u\cdot u_t)+\frac{g_t(t)}{2}V(x)|u|^2
+\frac{g(t)}{2}V(x)\frac{d}{dt}|u|^2-\frac{g_{tt}(t)}{2}|u|^2.
\end{align*}
The above identity can be written as
\begin{equation} \label{e3.8}
\begin{aligned}
&\frac{f(t)}{2}\frac{d}{dt}\Big(|u_t|^2+a^2|\nabla u|^2
 +(b^2-a^2)(\operatorname{div}u)^2\Big)+g(t)\frac{d}{dt}(u\cdot u_t)
 +\frac{g(t)}{2}V(x)\frac{d}{dt}|u|^2\\
&=\frac{d}{dt}e(t,x)-\frac{f_t(t)}{2}\Big(|u_t|^2+a^2|\nabla u|^2
 +(b^2-a^2)(\operatorname{div}u)^2\Big)\\
&\quad -\frac{g_t(t)}{2}V(x)|u|^2  +\frac{g_{tt}(t)}{2}|u|^2.
\end{aligned}
\end{equation}
Thus, by \eqref{e3.6} and \eqref{e3.8} we have the identity
\begin{equation} \label{e3.9}
\begin{aligned}
&\frac{d}{dt}e(t,x)+\Big(f(t)V(x)-g(t)
 -\frac{f_t(t)}{2}\Big)|u_t|^2+\Big(\frac{g_{tt}(t)}{2}
 -\frac{g_t(t)}{2}V(x)\Big)|u|^2\\
&+a^2\Big(g(t) -\frac{f_t(t)}{2}\Big)|\nabla u|^2
+ (b^2-a^2)\Big(g(t)-\frac{f_t(t)}{2}\Big)(\operatorname{div}u)^2-a^2g(t)
\operatorname{div}(u:\nabla u)\\
&-(b^2-a^2)g(t)  \operatorname{div}(u\operatorname{div}u) 
 -  a^2f(t)\ \operatorname{div}(u_t:\nabla u)\\
&-(b^2-a^2)f(t)  \operatorname{div}(u_t \operatorname{div}u)=0.
\end{aligned}
\end{equation}
 We integrate \eqref{e3.9} over $\mathbb{R}^2$ to obtain the identity
\begin{align*}
&\frac{d}{dt}\int_{\mathbb{R}^2}e(t,x)dx
 +\int_{\mathbb{R}^2}\Big(f(t)V(x)-g(t)-\frac{f_t(t)}{2}\Big)|u_t|^2dx\\
&+\int_{\mathbb{R}^2}\Big(\frac{g_{tt}(t)}{2}-\frac{g_t(t)}{2}V(x)\Big)|u|^2dx
 +a^2\int_{\mathbb{R}^2}\Big(g(t)-\frac{f_t(t)}{2}\Big)|\nabla u|^2dx\\
& +(b^2-a^2)\int_{\mathbb{R}^2}\Big(g(t)-\frac{f_t(t)}{2}\Big)
 (\operatorname{div}u)^2dx
-a^2\int_{\mathbb{R}^2}g(t) \operatorname{div}(u:\nabla u)dx\\
&-(b^2-a^2)\int_{\mathbb{R}^2}g(t) \operatorname{div}(u\operatorname{div}u)dx
 -a^2\int_{\mathbb{R}^2}f(t)\ \operatorname{div}(u_t:\nabla u)dx\\
&-(b^2-a^2)\int_{\mathbb{R}^2}f(t) \operatorname{div}(u_t \operatorname{div}u)dx=0.
\end{align*}
By applying the Gauss divergence theorem,  one notices that
\begin{gather*}
\int_{\mathbb{R}^2}\operatorname{div}(u:\nabla u)dx=0, \quad 
\int_{\mathbb{R}^2}\operatorname{div}(u \operatorname{div}u)dx=0,\\
\int_{\mathbb{R}^2}\operatorname{div}(u_t:\nabla u)dx=0, \quad 
\int_{\mathbb{R}^2}\operatorname{div}(u_t\ \operatorname{div}u)dx=0.
\end{gather*}
Therefore, one has arrived at the desired equality.
\[
\frac{d}{dt}E(t)+F(t) = 0.
\]
\end{proof}

Note that when one estimates the functions $E(t)$ and $F(t)$ it is sufficient 
to consider the spatial integration over the light cone
\begin{eqnarray*}
\Omega(t)=\{x \in \mathbb{R}^2:|x|\le R_0+bt\}
\end{eqnarray*}
since the finite speed of propagation property can be applied again 
to the solutions of the corresponding problem \eqref{eqn}-\eqref{initial}.

Now let us choose the functions $f(t)$ and $g(t)$ in
 Lemmas \ref{lem3.2} and \ref{lem3.3} as follows: 
When $C_0 > 2b$ we set
\begin{equation} \label{e3.10}
f(t)=(1+t)^2, \quad g(t)=(1+t),
\end{equation}
when $b<C_0 \le 2b$, for an arbitrarily fixed $\delta>0$ we choose
\begin{equation} \label{e3.11}
f(t)=(1+t)^{\frac{C_0}{b}-\delta}, \quad 
g(t)=\frac{C_0-b\delta}{2b}(1+t)^{\frac{C_0}{b}-1-\delta}.
\end{equation}
Then one has the following lemmas.

\begin{lemma} \label{lem3.2}
The smooth functions $f(t)$ and $g(t)$ defined by \eqref{e3.10} and \eqref{e3.11}
 satisfy the following properties: 
there exists a large $t_0 > 0$ such that for all $t \ge t_0 \ge 0$,
\begin{itemize}
\item[(i)]  $2f(t)V(x)-f_{t}(t)-2g(t) \ge 0$,  $x \in \Omega(t)$,

\item[(ii)] $2g(t)-f_{t}(t) = 0$.
\end{itemize}
\end{lemma}


\begin{proof}
 We will check only (i), because (ii) is quite easy.
First, we consider the case \eqref{e3.10} to check (i) when $C_0 > 2b$.  
In fact, for $x \in \Omega(t)$,
\begin{align*}
 2f(t)V(x)-f_{t}(t)-2g(t)
&= 2(1+t)^2V(x)-2(1+t)-2(1+t) \\
&= 2(1+t)\{(1+t)V(x)-2\}\\
&\ge 2(1+t)\Big\{(1+t)\frac{C_0}{1+|x|}-2\Big\} \\
&\ge 2(1+t)\Big\{(1+t)\frac{C_0}{1+bt+R_0}-2\Big\}.
\end{align*}
Here, we find that
\[
\lim_{t\to \infty}\Bigl((1+t)\frac{C_0}{1+bt+R_0}-2\Bigl)=\frac{C_0}{b}-2,
\]
so that there exists $t_0 > 0$ such that for all $t \in [t_0,+\infty)$,
\[
(1+t)\frac{C_0}{1+bt+R_0}-2 \ge \frac{1}{2}\Bigl(\frac{C_0}{b}-2\Bigl).
\]
Thus, we have the inequality
\[
 2(1+t)\Big\{(1+t)\frac{C_0}{1+bt+R_0}-2\Big\} 
\ge (1+t)\Big(\frac{C_0}{b}-2\Big), \quad 0\le t_0\le t.
\]

From  assumption of (I) of Theorem \ref{thm1.1}, since
$\frac{C_0}{b}-2 > 0$, one can check (i)  when $C_0 > 2b$.

On the other hand,  when $b < C_0 \le 2b$ it follows from
the definition of \eqref{e3.11} that
\begin{align*}
& 2f(t)V(x)-f_{t}(t)-2g(t)\\
&=  2(1+t)^{\frac{C_0}{b}-\delta}V(x)
 -\Big(\frac{C_0}{b}-\delta\Big)(1+t)^{\frac{C_0}{b}-\delta-1}
 -\Big(\frac{C_0}{b}-\delta\Big)(1+t)^{\frac{C_0}{b}-\delta-1}\\
&=2(1+t)^{\frac{C_0}{b}-\delta}V(x)-2\Big(\frac{C_0}{b}-\delta\Big)
 (1+t)^{\frac{C_0}{b}-\delta-1} \\
\\
&\ge 2(1+t)^{\frac{C_0}{b}-\delta-1}\Big\{(1+t)\frac{C_0}{1+|x|}
 -\Big(\frac{C_0}{b}-\delta\Big)\Big\}\\
& \ge 2(1+t)^{\frac{C_0}{b}-\delta-1}\Big\{(1+t)\frac{C_0}{1+bt +
R_0}-\Big(\frac{C_0}{b}-\delta\Big)\Big\},
\end{align*}
for all $x \in \Omega(t)$.

Here, we find that
\[
 \lim_{t\to \infty}\Bigl((1+t)\frac{C_0}{1+bt+R_0}
-\Big(\frac{C_0}{b}-\delta\Big)\Bigl)=\frac{C_0}{b}
-\Big(\frac{C_0}{b}-\delta\Big)=\delta > 0.
\]
So, there exists $t_0 > 0$ such that for all $t \in [t_0,+\infty)$
\[
(1+t)\frac{C_0}{1+bt+R_0}-\Big(\frac{C_0}{b}-\delta\Big) \ge \frac{1}{2}\delta.
\]
Thus, we have the  inequality
\[
 2(1+t)^{\frac{C_0}{b}-1-\delta}\Big\{(1+t)\frac{C_0}{1+bt+R_0}
 -\Big(\frac{C_0}{b}-\delta\Big)\Big\}
 \ge (1+t)^{\frac{C_0}{b}-1-\delta}\delta, \quad 0\le t_0\le t,
\]
which implies desired estimate.
\end{proof}

Based on Lemmas \ref{lem3.1} and \ref{lem3.2} we have the  inequality
\begin{equation} \label{e3.12}
\begin{aligned}
\frac{d}{dt}\{f(t)E_u(t)+g(t)(u,u_{t})\} 
&\le \frac{d}{dt}\Big\{\frac{1}{2}\int_{\mathbb{R}^2}(g_{t}(t)-g(t)V(x))|u|^2dx\Big\}\\
&\quad +\frac{1}{2}\int_{\mathbb{R}^2}(g_{t}(t)V(x)-g_{tt}(t))|u|^2dx
\end{aligned}
\end{equation}
for $t \ge t_0 \ge 0$.

We want to use the following lemma.
However, when $C_0 > 2b$ and $f(t)$, $g(t)$ are given by \eqref{e3.10}
the proof of this lemma is easily done.
We will prove the lemma  only for $b < C_0 \leq 2b$ and $f(t)$, 
$g(t)$  given by\eqref{e3.11}.

\begin{lemma} \label{lem3.3}
Assume the functions $f(t)$ and $g(t)$ defined by \eqref{e3.10} and 
\eqref{e3.11} satisfy the following three properties, for $t \ge t_0 \ge 0$:
\begin{itemize}
\item[(iii)]  $-g_{tt}(t) \le \frac{C_{1}}{1+bt}$,

\item[(iv)] $V(x)g_{t}(t) \le C_2V(x)$, for all $x \in \mathbb{R}^2$,

\item[(v)]  $g_{t}(t)-V(x)g(t) \le C_3$, for all $x \in \mathbb{R}^2$,
\end{itemize}
where $C_i$ $(i=1,2,3)$ are positive constants.
\end{lemma}

\begin{proof}
We proof only (iii) under the condition $b<C_0 \le 2b$ with $f(t)$, $g(t)$
given by \eqref{e3.11}. The other cases are easy to proof.
Note that 
\begin{align*}
-g_{tt}(t)&=-\Bigl(\frac{C_0-\delta b}{2b}\Bigl)
 \Bigl(\frac{C_0}{b}-\delta-1\Bigl)\Bigl(\frac{C_0}{b}-\delta-2\Bigl)
 (1+t)^{\frac{C_0}{b}-\delta-3}\\
& \le C(1+t)^{\frac{C_0}{b}-\delta-3}\\
&\le C(1+t)^{\frac{2b}{b}-\delta-3} \\
&= C(1+t)^{-1-\delta} \\
&\le C(1+t)^{-1}.
\end{align*}
Here, we find that
\[
\frac{1}{\max\{1,b\}(1+t)} \le \frac{1}{1+bt},
\]
so that
\[
\frac{1}{1+t} \le \frac{\max\{1,b\}}{1+bt} \le \frac{C}{1+bt}.
\]
Thus, we have the following inequality with some $C_1>0$:
\[
C(1+t)^{-1} \le \frac{C_1}{1+bt},
\]
which implies the desired inequality.
\end{proof}

By integrating both sides of \eqref{e3.12} over $[t_0,t]$, we find that
\begin{align*} 
& f(t)E_u(t) + g(t)(u(t,\cdot) , u_{t}(t,\cdot)) - \{f(t_0)E_u(t_0) 
 + g(t_0)(u(t_0,\cdot) , u_{t}(t_0,\cdot))\}\\
&\le \frac{1}{2}\int_{\mathbb{R}^2}(g_{t}(t)-V(x)g(t))|u(t,x)|^2dx 
 - \frac{1}{2}\int_{\mathbb{R}^2}(g_{t}(t_0)-V(x)g(t_0))|u(t_0,x)|^2dx \\
&\quad + \frac{1}{2}\int_{t_0}^{t}\int_{\mathbb{R}^2} V(x)g_{t}(s)|u(s,x)|^2\,dx\,ds
 + \frac{1}{2}\int_{t_0}^{t}\int_{\mathbb{R}^2}(-g_{tt}(s))|u(s,x)|^2\,dx\,ds.
\end{align*}
It follows from Lemma \ref{lem3.3} with large $t_0$ that
\begin{equation} \label{e3.13}
\begin{aligned}
&f(t)E_u(t)+g(t)(u(t,\cdot),u_{t}(t,\cdot)) \\
&\le C_{4}+\frac{C_{3}}{2}\int_{\mathbb{R}^2}|u(t,x)|^2dx
 +\frac{C_2}{2}\int_{t_0}^{t}\int_{\mathbb{R}^2}V(x)|u(s,x)|^2\,dx\,ds\\
&\quad +\frac{C_{1}}{2C_0}\int_{t_0}^{t}
 \int_{\mathbb{R}^2}\frac{C_0}{1+bs}|u(s,x)|^2\,dx\,ds,
\end{aligned}
\end{equation}
where
\[
C_4=f(t_0)E_u(t_0)+g(t_0)(u(t_0,\cdot),u_t(t_0,\cdot))
-\frac{1}{2}\int_{\mathbb{R}^2}(g_t(t_0)-V(x)g(t_0))|u(t_0,x)|^2dx.
\]
We shall rely on the following powerful lemma, 
motivated by results from Ikehata \cite{I} and  Char\~ao-Ikehata \cite{CI}.

\begin{lemma} \label{lem3.4}
Let $u \in C([0,+\infty);(H^1(\mathbb{R}^2))^2) 
\cap C^1([0,+\infty);(L^2(\mathbb{R}^2))^2)$ be the solution to
 \eqref{eqn}-\eqref{initial}. Then
\[
\|u(t,\cdot)\|^2 + \int_0^{t}\int_{\mathbb{R}^2}V(x)|u(s,x)|^2\,dx\,ds
\le \|u_0\|^2 + C\|d(\cdot)(V(\cdot)|u_0|+|u_{1}|)\|^2
\]
for all $t\ge 0$, where $C>0$ is a constant and
$d(x):=\{1+\log{(1+|x|)}\}(1+|x|)$.
\end{lemma}

\begin{proof} 
First, we define an auxiliary function
\[
\chi(t,x)=\int_0^{t}u(s,x)ds
\]
that satisfies
\begin{gather} \label{e3.14}
\chi_{tt}-a^2\Delta\chi-(b^2-a^2)\nabla(\operatorname{div}\chi)+V(x)\chi_{t}
 = V(x)u_0+u_{1}, \\ \label{e3.15} 
\chi(0,x)=0, \quad \chi_{t}(0,x)=u_0(x) \quad x \in \mathbb{R}^2.
\end{gather}
Multiplying \eqref{e3.14} by $\chi_{t}$ and integrating over 
$[0,t]\times\mathbb{R}^2$ we obtain
\begin{equation} \label{e3.16}
\begin{aligned}
&\frac{1}{2}\|\chi_{t}\|^2+\frac{a^2}{2}\|\nabla\chi\|^2
+\frac{(b^2-a^2)}{2}\int_{\mathbb{R}^2}(\operatorname{div}\chi)^2dx
+\int_0^{t}\int_{\mathbb{R}^2}V(x)|\chi_{t}|^2\,dx\,ds
\\
&= \frac{1}{2}\|u_0\|^2+\int_{\mathbb{R}^2}(V(x)u_0+u_{1})\cdot
\chi(t,x)dx.
\end{aligned}
\end{equation}
The next step is to use the two dimensional Hardy-Sobolev inequality \cite{MY},
\begin{equation} \label{e3.17}
\int_{\mathbb{R}^2}\frac{|v(x)|^2}{d(x)^2}dx 
\le C\int_{\mathbb{R}^2}|\nabla v(x)|^2dx, \quad v \in (H^1(\mathbb{R}^2))^2,
\end{equation}
where
\[
d(x):=\{1+\log{(1+|x|)}\}(1+|x|).
\]
The last term of \eqref{e3.16} can be estimated by using \eqref{e3.17}
 and the Schwarz inequality as follows.
\begin{align*}
& \int_{\mathbb{R}^2}(V(x)u_0+u_{1}) \chi(t,x)dx \\
& \le \int_{\mathbb{R}^2}(V(x)|u_0|+|u_{1}|)|\chi(t,x)|dx \\
& = \int_{\mathbb{R}^2}d(x)(V(x)|u_0|+|u_{1}|)\frac{|\chi(t,x)|}{d(x)}dx \\
&\le \Big\{\int_{\mathbb{R}^2}d(x)^2(V(x)|u_0|+|u_{1}|)^2 dx\Big\}^{1/2}
  \Big\{\int_{\mathbb{R}^2}\frac{|\chi(t,x)|^2}{d(x)^2}dx\Big\}^{1/2}\\
&\le \frac{1}{2\varepsilon}\|d(\cdot)(V(\cdot)|u_0|+|u_{1}|)\|^2 
 + \frac{\varepsilon}{2}\Big|\Big|\frac{\chi(t,\cdot)}{d(\cdot)}\Big|\Big|^2 \\
&\le \frac{1}{2\varepsilon}\|d(\cdot)(V(\cdot)|u_0|+|u_{1}|)\|^2 
 + \frac{\varepsilon C}{2}\|\nabla \chi\|^2,
\end{align*}
where $\varepsilon>0$ is an arbitrary real number and $C$ is the constant 
in the Hardy-Sobolev inequality.

Combining the above estimate with \eqref{e3.16}, we conclude that
\begin{equation} \label{e3.18}
\begin{aligned}
&\frac{1}{2}\|\chi_{t}\|^2+\frac{a^2}{2}(1-\frac{\varepsilon C}{a^2})\|\nabla\chi\|^2
+\frac{(b^2-a^2)}{2}\int_{\mathbb{R}^2}(\operatorname{div}\chi)^2 dx 
+\int_0^{t}\int_{\mathbb{R}^2}V(x)|\chi_{t}|^2\,dx\,ds
\\
&\le \frac{1}{2}\|u_0\|^2 + C\|d(\cdot)(V(\cdot)|u_0|+|u_{1}|)\|^2.
\end{aligned}
\end{equation}

Now, fixing  $\varepsilon>0$ such that $1-\frac{\varepsilon C}{a^2} > 0$, we obtain
\begin{equation} \label{e3.19}
\|\chi_{t}\|^2 + \int_0^{t}\int_{\mathbb{R}^2}V(x)|\chi_{t}|^2\,dx\,ds 
\le  \|u_0\|^2 + C\|d(\cdot)(V(\cdot)|u_0|+|u_{1}|)\|^2,
\end{equation}
which implies the desired statement of Lemma \ref{lem3.4},
with $\chi_t=u$:
\begin{equation} \label{e3.20}
\|u\|^2 + \int_0^{t}\int_{\mathbb{R}^2}V(x)|u|^2\,dx\,ds \ \le \
\|u_0\|^2 + C\|d(\cdot)(V(\cdot)|u_0|+|u_{1}|)\|^2.
\end{equation}
\end{proof}

As consequence of Lemma \ref{lem3.4}, since
\[
V(x) \ge \frac{C_0}{1+|x|} \ge \frac{C_0}{1+R_0+bt} 
\ge \frac{1}{(1+R_0)}\frac{C_0}{1+bt},\quad x \in \Omega(t)
\]
we  have
\begin{equation} \label{e3.21}
\frac{1}{(1+R_0)}\int_0^{t}\int_{\mathbb{R}^2}\frac{C_0}{1+bs}|u(s,x)|^2\,dx\,ds 
 \le  \|u_0\|^2 + C\|d(\cdot)(V(\cdot)|u_0|+|u_{1}|)\|^2,
\end{equation}
where $C>0$ is a constant.

After these preparations let us prove only Theorem \ref{thm1.1}.
The proof of Proposition \ref{prop1.1} part (II) is quite similar, 
using \eqref{e3.11} in stead of \eqref{e3.10}.

\begin{proof}[Proof of Theorem \ref{thm1.1}]
It follows from Lemma \ref{lem3.4}, \eqref{e3.13} and \eqref{e3.21} that
\begin{equation} \label{e3.22}
f(t)E_u(t)+g(t)(u(t,\cdot),u_{t}(t,\cdot))  \le  C_{R_0},
\end{equation}
where the constant $C_{R_0}>0$ depends on the $L^2$-norm of the initial 
data and $R_0$. By using the Schwarz inequality, the definition of
the total energy, and Lemma \ref{lem3.4} again we obtain
\[
f(t)E_u(t) \le g(t)\|u(t,\cdot)\|\ \|u_{t}(t,\cdot)\|+C_{R_0} 
\le Cg(t)\sqrt{E_u(t)}+C_{R_0}, \quad t\ge t_0,
\]
with some constant $C>0$ depending on $R_0$ and the initial data. 
Therefore, if we set $X(t)=\sqrt{E_u(t)}$ for $t \in[t_0,+\infty)$, then we get
\begin{equation} \label{e3.23}
f(t)X^2(t)-Cg(t)X(t)-C_{R_0} \le 0.
\end{equation}
By solving the quadratic inequality \eqref{e3.23} for $X(t)$ (cf. \cite{U}) 
we have
\[
\sqrt{E_u(t)} \le \frac{Cg(t)+\sqrt{C^2g^2(t)+4C_{R_0}f(t)}}{2f(t)}.
\]
This inequality leads to
\[
E_u(t) \le C\Big(\frac{g(t)}{f(t)}\Big)^2+C\Big(\frac{1}{f(t)}\Big), 
\quad t \ge t_0,
\]
which implies the desired estimate of the Theorem \ref{thm1.1}. 
Remember that $f(t) = (1+t)^2$ and $g(t) = (1+t)$ in the present case.
\end{proof}

\subsection*{Acknowledgments}
The second author (N. Nakabayashi) would like to thank Professors
 R. C. Char\~ao (UFSC) and C. R. da Luz (UFSC) for their kind
and warm hospitality during his stay at UFSC, Brazil  on January-February, 
2013. Substantial part of this work was done during the second author's 
stay at UFSC.


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\end{document}