text:
\section{Introduction}
\footnote[0]{2010 Mathematics Subject Classification. 35L20; 35B40; 47B25}

Let $N\ge 2$ and
let $\Omega$ be an exterior domain in $\mathbb{R}^N$
with a smooth boundary
$\partial \Omega$.
We assume that
$0 \notin \bar{\Omega}$
without loss of generality.
We study the initial-boundary value problem for
the damped wave equation
\begin{align}%
\label{dw}
	\left\{\begin{array}{ll}
	u_{tt} - \Delta u + a(x) u_t = 0,&x\in \Omega, t>0,\\
	u(x,t) = 0,&x\in \partial \Omega, t>0,\\
	u(x,0) = u_0(x), u_t(x,0)=u_1(x),&x\in \Omega.
	\end{array}\right.
\end{align}%
Here
$u = u(x,t)$ is a real-valued unknown function.
We assume that
$a(x) \in C^2(\bar{\Omega})$,
$a(x) > 0$ on $\bar{\Omega}$
and, there exist constants
$\alpha > 0$ and $a_0 > 0$
such that
\begin{align}%
\label{a}
	\lim_{|x|\to \infty} |x|^{-\alpha}a(x) = a_0,
\end{align}%
that is,
$a(x)$ is unbounded and diverges at the spatial infinity.
The initial data
$(u_0, u_1)$
is compactly supported, let us say
${\rm supp\,} (u_0,u_1) \subset B(0,R_0)$
with some $R_0 > 0$,
and
satisfies the compatibility condition of order $1$, namely,
\begin{align}%
\label{ini}
	(u_0, u_1) \in (H^2(\Omega) \cap H^1_0(\Omega) ) \times H^1_0(\Omega).
\end{align}%
Noting the compactness of the support of the initial data
and the finite propagation property,
we can show that
there exists a unique strong solution
\begin{align}%
\label{sol}
	u \in \bigcap_{i=0}^2 C^i([0,\infty);H^{2-i}(\Omega))
\end{align}%
in a standard way (see Ikawa \cite[Theorem 2]{Ika68}).

The term
$a(x)u_t$
describes the damping effect,
which plays a role in reducing the energy of the wave.
Our aim is to clarify
how the strength of the damping effects
the behavior of the solution.
Particularly, in this paper we study
the case where the strength of the damping increases at
the spatial infinity,
and as a typical example,
we consider the damping satisfying \eqref{a}.
We investigate
the asymptotic profile of the solution
and how the exponent
$\alpha$
is related to the decay rate of the energy of the solution.

For this purpose,
we also consider the corresponding parabolic problem
\begin{align}%
\label{h}
	\left\{\begin{array}{ll}
	v_t - a(x)^{-1}\Delta v = 0,&x\in \Omega, t>0,\\
	v(x,t) = 0,&x\in \partial \Omega, t>0,\\
	v(x,0) = v_0(x),&x\in \Omega.
	\end{array}\right.
\end{align}%
This equation is formally obtained
by dropping the term $u_{tt}$ from the equation \eqref{dw}
and dividing it by $a(x)$.

To observe the relation between the equations \eqref{dw} and \eqref{h},
we formally consider the case
$a(x) = |x|^{\alpha}$
and
$\Omega = \mathbb{R}^N$,
namely, we drop the boundary condition and consider
an initial value problem.
For the equation \eqref{dw}, we change the variables as
\begin{align*}%
	u(x,t) = \phi ( \lambda^{1/(2+\alpha)} x, \lambda^{1/2}t ),\quad
	y = \lambda^{1/(2+\alpha)} x,\ s = \lambda^{1/2}t
\end{align*
with a parameter $\lambda > 0$.
Then, the function $\phi = \phi(y,s)$ satisfies
\begin{align*}%
	\lambda^{2(1+\alpha)/(2+\alpha)} \phi_{ss} - \Delta_{y} \phi + |y|^{\alpha} \phi_s =0.
\end{align*}%
Thus, letting $\lambda \to 0$, we have the heat equation
\begin{align*}%
	|y|^{\alpha} \phi_s - \Delta_y \phi =0.
\end{align*}%
We note that
$\lambda \to 0$ with fixing $s$ is corresponding to $t \to \infty$.

The above observation suggests that
the solution of \eqref{dw} has the so-called {\em diffusion phenomena},
that is,
the solution of the damped wave equation \eqref{dw} is approximated by
a solution of the heat equation \eqref{h}
as time tends to infinity.
Moreover, in the constant damping case $\alpha =0$,
there is a derivation of the damped wave equation
from the heat balance law and the time-delayed Fourier's law
(Cattaneo-Vernotte law).
This derivation also indicates the diffusion phenomena.
For the detail, see for example, \cite{Ca58, Li97, Ve58}.

Indeed, for the damped wave equation with constant damping
in the whole space
\begin{align}%
\label{cdw}
	u_{tt} - \Delta u + u_t = 0,\quad x \in \mathbb{R}^N, t>0,
\end{align}%
many mathematicians studied the asymptotic behavior of solutions
and verified the diffusion phenomena.
We refer the reader to
\cite{Ma76, HsLi92, Ni96, Ni97, YaMi00, Kar00, Ni03MathZ, MaNi03, HoOg04, Na04, SaWaMZ}.
For an exterior domain $\Omega \subset \mathbb{R}^N$,
namely, in the case $a(x) \equiv 1$ in our problem \eqref{dw},
the diffusion phenomena was proved by
\cite{Ik02, IkNi03, ChHa03, RaToYo11}.

Asymptotic behavior of solutions to
the wave equation
with variable coefficient damping
\begin{align*}%
	\left\{ \begin{array}{ll}
		u_{tt} - \Delta u + b(x,t) u_t = 0,&x\in \Omega, t>0,\\
		u(x,t) =0, &x\in \partial\Omega,\\
		u(x,0) = u_0(x), \ u_t(x,0) = u_1(x),& x\in \Omega
	\end{array}\right.
\end{align*}%
has been also studied for a long time.
Concerning the behavior of the total energy
\begin{align*}%
	E(t;u) &= \frac12 \int_{\Omega} (u_t^2 + |\nabla u|^2 ) dx,
\end{align*}%
Mochizuki \cite{Mo76} considered the case $\Omega = \mathbb{R}^N$
with $N \neq 2$
and proved that if
\begin{align*}%
	0\le b(x,t) \le C (1+|x|)^{-1-\delta}
\end{align*}%
with some $\delta > 0$
and $|b_t(x,t)| \le C$,
then the M$\mathrm{\phi}$ller wave operator exists and
it is not identically zero.
Namely, there exists initial data
$(u_0, u_1)$ with finite energy and
for the associated solution $u$, the total energy $E(t;u)$ does not decay to zero
as $t \to \infty$,
and there also exists a solution $w$ of the free wave equation
\begin{align*}%
	w_{tt} - \Delta w = 0
\end{align*}%
such that $\lim_{t\to \infty} E(t; u-w) = 0$ holds
(see \cite{RaTa76} for the case $b = b(x) \in C_0^{\infty}(\mathbb{R}^N)$
and \cite{MoNa96, Mat02, Nak06, KaNaSo04, Nis09} for
further improvements).

On the other hand,
Matsumura \cite{Ma77} and Uesaka \cite{Ue79} studied the case
$b(x,t) \ge C(1+t)^{-1}$
and proved that the total energy decays to zero.
Mochizuki and Nakazawa \cite{MoNa96}
and Ueda \cite{Ueda16} gave a logarithmic improvement
of the assumption on the damping.
Also, Nakao \cite{Nakao01MathZ}, Ikehata \cite{Ik03JDE}
and Mochizuki and Nakao \cite{MoNak07}
proved the total energy decay for
the damping localized near infinity.

As for sharp decay estimates,
for the problem \eqref{dw} with $\Omega = \mathbb{R}^N$,
Todorova and Yordanov \cite{ToYo09} proved that
if the damping $a= a(x)$ is radially symmetric and
satisfies \eqref{a} with $\alpha \in (-1,0]$,
then the solution is estimated as
\begin{align*}%
	E(t;u) &= O(t^{-\frac{N+\alpha}{2+\alpha} -1 +\delta}),\\
	\| \sqrt{a} u(t) \|_{L^2(\Omega)}
	&= O(t^{-\frac{N+\alpha}{2(2+\alpha)} + \delta})
\end{align*}%
as $t\to \infty$.
Here $\delta$ is an arbitrary small number
(see also \cite{RaToYo09, RaToYo10} for higher order energy estimates
and see \cite{IkToYo13} for the case $\alpha = -1$).
For the asymptotic profile of solutions,
the second author \cite{Wa14} proved
that if
$\Omega = \mathbb{R}^N$ and $a(x) = (1+|x|^2)^{\alpha/2}$
with $\alpha \in (-1,0]$,
then the asymptotic profile of the solution is given by
that of the corresponding parabolic problem.
After that, in our previous results
\cite{SoWa16, SoWa17}, we extended the result of \cite{Wa14}
to exterior domains and more general (but bounded) damping
satisfying the condition \eqref{a} with $\alpha \in (-1,0]$.
Radu, Todorova and Yordanov \cite{RaToYo16} and Nishiyama \cite{Nis16}
proved the diffusion phenomena in an abstract setting.

However, for the damping term increasing at the spatial infinity,
there is no result about the diffusion phenomena,
while Khader \cite{Kh11, Kh13} studied
energy estimates and global existence of small solutions
for some nonlinear problems.

In this paper,
we establish sharp semigroup estimates for
the heat equation \eqref{h}
and prove the almost sharp weighted energy estimates
for the damped wave equation \eqref{dw}.
As their application, we show that
the solution of \eqref{dw} actually has the diffusion phenomena
in a suitable weighted $L^2$ space.

Our main result reads as follows:
\begin{theorem}\label{thm1}
Let
$u$
be the solution to \eqref{dw} and
let
$v$
be the solution to \eqref{h} with
$v_0(x) = u_0(x) + a(x)^{-1} u_1(x)$.
Then, for any $\delta>0$,
there exists $C = C(N, \alpha, a_0, R_0, \delta)>0$ such that we have for $t\ge 1$
\begin{align}%
\label{dp}
	\left\| \sqrt{a(\cdot)} (u(\cdot,t)-v(\cdot,t)) \right\|_{L^2}
	\le C (1+t)^{-\frac{N+\alpha}{2(2+\alpha)} -\frac{1+\alpha}{2+\alpha} + \delta}
		\left\| (u_0, u_1) \right\|_{H^2 \times H^1}.
\end{align}%
\end{theorem}

As a byproduct of Theorem \ref{thm1},
we obtain the almost optimal $L^2$-estimate for the solution.
\begin{corollary}\label{cor2}
The solution $u$ of \eqref{dw} satisfies
\begin{align*}%
	\left\| \sqrt{a(\cdot)} u(\cdot,t) \right\|_{L^2}
	\le C (1+t)^{-\frac{N+\alpha}{2(2+\alpha)}+\delta}
		\left\| (u_0, u_1) \right\|_{H^2 \times H^1},
\end{align*}%
where
$\delta > 0$ is an arbitrary small number
and
$C = C(N, \alpha, a_0, R_0, \delta)>0$.
When $N=2$, we may take $\delta = 0$.
\end{corollary}

Our strategy for the proof of Theorem \ref{thm1} is the following.
First, we treat the term $u_{tt}$ as a perturbation and write the equation \eqref{dw} as
\begin{align*}%
	u_t - a(x)^{-1} \Delta u = - a(x)^{-1}u_{tt}.
\end{align*}%
This is natural because we expect the diffusion phenomena,
and for the solution $v$ to the parabolic problem \eqref{h},
the term $v_{tt}$ decays faster than $v_t$ and $a(x)^{-1} \Delta v$.
Then, by Duhamel principle, the above equation
can be formally transformed into
the integral equation
\begin{align*}%
	u(t) = e^{t a(x)^{-1}\Delta} u_0 - \int_0^t e^{(t-s) a(x)^{-1}\Delta} [ a(x)^{-1} u_{ss}(s) ]\,ds.
\end{align*}%
We further apply the integration by parts to obtain
\begin{align*}%
	u(t) &= e^{t a(x)^{-1}\Delta} ( u_0 + a(x)^{-1}u_1) \\
		&\quad - \int_{t/2}^t e^{(t-s) a(x)^{-1}\Delta} [ a(x)^{-1} u_{tt}(s) ]\,ds \\
		&\quad - e^{\frac{t}{2} a(x)^{-1}\Delta} [ a(x)^{-1} u_{t}(t/2) ] \\
		&\quad - \int_0^{t/2}
				a(x)^{-1} \Delta e^{(t-s) a(x)^{-1}\Delta} [ a(x)^{-1} u_{t}(s) ]\,ds
\end{align*}%
By putting $v(t) = e^{t a(x)^{-1}\Delta} ( u_0 + a(x)^{-1}u_1)$,
it suffices to estimate the each term in the right-hand side.

To this end, in Section 2, we first investigate the heat semigroup
$e^{t a(x)^{-1}\Delta}$.
We let it make sense in an appropriate weighted space
via a suitable changing variable.
Moreover, by the Beurling-Deny criterion and the Gagliardo-Nirenberg inequality,
we derive the following optimal estimate for the semigroup
$e^{t a(x)^{-1}\Delta}$,
which is crucial for our argument:
\begin{align*}%
	\| e^{t a(x)^{-1}\Delta}v_0 \|_{L^2(\Omega, d\mu)} \le C
		t^{-(N/2)(1/q - 1/2)}
			\left( \int_{\Omega} |v_0(x)|^q |x|^{(N-2)\alpha(2-q)/4} \, d\mu \right)^{1/q},
\end{align*}%
where
$d\mu = a(x)dx$
and
$q \in [1,2] \ (N=2), \ q\in ( p_{\alpha}', 2]\ (N\ge 3)$
with
$p_{\alpha} = 2N(2+\alpha)/(\alpha(N-2))$.
In contrast with the decaying damping cases studied by \cite{SoWa16, SoWa17}
in which $q$ can be taken freely in $[1,2]$,
we cannot have $L^2$-$L^1$ estimate for $N\ge 3$.

After that, to estimate the terms including $u$ itself,
we apply energy estimates for
the damped wave equation \eqref{dw}
with a Ikehata-Todorova-Yordanov type weight function
$\Phi = \exp ( \beta A(x)/(1+t) )$,
where
$\beta$ is a positive constant and
$A(x)$ is a solution of the elliptic equation
\begin{align}%
\label{poi}
	\Delta A(x) = a(x).
\end{align}%
Roughly speaking, the weight function
$\Phi$ comes from the dual problem of the heat equation
\begin{align*}%
	a(x) v_t + \Delta v = 0.
\end{align*}%
We remark that Lions and Masmoudi \cite{LiMa98, LiMa01}
firstly introduced this type of weight functions
to study the uniqueness for the Navier-Stokes equations.

In Section 3, we discuss the construction of the auxiliary function $A(x)$.
When $a(x)$ is decaying at spatial infinity and radially symmetric,
Todorova and Yordanov \cite{ToYo09} solved the equation \eqref{poi}
by reducing the problem to an ordinary differential equation,
and applied it to obtain energy estimates for damped wave equations.
However, in our problem, we assume $a(x)$ is unbounded.
In this case we cannot directly construct a solution by the Newton potential.
Moreover, $a(x)$ is not radially symmetric as in \cite{ToYo09}
and we cannot reduce the problem to an ordinary differential equation.
Indeed, in \cite[Remark 3.1]{SoWa16},
we show a example of non-radial $a(x)$
satisfies $a(x) \to 1 \ (|x| \to \infty)$
but the corresponding $A(x)$ has bad behavior at $|x| \to \infty$.
To overcome these difficulties,
we follow our previous results \cite{SoWa17} and
weaken the problem \eqref{poi} to the inequality
\begin{align*}%
	(1-\varepsilon) a(x) \le \Delta A_{\varepsilon}(x) \le (1+\varepsilon) a(x)
\end{align*}%
with arbitrary small $\varepsilon >0$,
namely, we shall find a function $A_{\varepsilon}(x)$ satisfying the above
inequality and having good behavior as $|x| \to \infty$.

In Section 4, using this auxiliary function $A_{\varepsilon}(x)$,
we apply weighted energy methods developed by
\cite{ToYo01, Ik05IJPAM, IkTa05, RaToYo09, Ni10, Wa14, SoWa16, SoWa17}
and prove almost sharp higher order energy estimates of solutions.

Finally, in Section 5,
combining the heat semigroup estimates and the weighted energy estimates
for solutions to the damped wave equation \eqref{dw},
we have the desired estimate and finish the proof of Theorem \ref{thm1}.

We end up this section with introducing the notations and the terminologies
used throughout this paper.
We shall denote by
$C$ various constants,
which may change from line to line.
In particular,
we sometimes write
$C = C(\ast, \ldots, \ast)$
to emphasize the dependence on the parameters appearing in parentheses.

For $x_0 \in \mathbb{R}^N$ and $R>0$, we denote by
$B(x_0,R)$
the open ball centered at $x_0$ with the radius $R$.
Also, $\bar{B}(x_0, R)$ stands for the closure of $B(x_0,R)$.

We denote by
$L^p(\Omega) \, (1\le p \le \infty)$, $H^k(\Omega)\, (k\in \mathbb{Z}_{\ge 0})$
the usual Lebesgue space and Sobolev space with the norms
\begin{align*}%
	\| f \|_{L^p} &=
	\begin{cases}
		\displaystyle
		\left( \int_{\Omega} |f(x)|^p\,dx \right)^{1/p} &(1\le p <\infty),\\
		\displaystyle
		{\rm ess\,sup\,}_{x\in \Omega} |f(x)| &(p=\infty),
	\end{cases}\\
	\| f \|_{H^k} &=
		\left( \sum_{|\gamma|\le k}\| \partial_x^{\gamma} f \|_{L^2}^2 \right)^{1/2},
\end{align*}%
respectively.
For an interval $I$ and a Banach space $X$,
we define
$C^{r}(I;X)$
as the space of $r$-times continuously differentiable mapping from
$I$ to $X$ with respect to the topology in $X$.

\section{$L^p$-$L^q$ estimates for the parabolic problem}
In this section,
we consider the asymptotic behavior of solutions to
the parabolic problem \eqref{h}.
We note that the assumption \eqref{a} on $a(x)$ implies
that there exist constants
$0<a_1\le a_2$
such that
\begin{align}%
\label{a_est}
	a_1|x|^{\alpha} \le a(x) \le a_2|x|^{\alpha}
\end{align}%
holds for $x \in \bar{\Omega}$.
We remark that the results of this section
requires only \eqref{a_est} and
we do not use the property \eqref{a}.
For $1\le p < \infty$, we define the weighted Lebesgue spaces
\begin{align*}%
	L^p_{d\mu}(\Omega)
	:=
	\left\{ f \in L^p_{{\rm loc}}(\Omega) \,;\,
		\| f \|_{L^p_{d\mu}} = \left( \int_{\Omega} |f(x)|^p a(x) \,dx \right)^{1/p} < \infty
	\right\}.
\end{align*}%
We deal with the operator
\begin{align*}%
	L = - a(x)^{-1} \Delta
\end{align*}%
in $L^2_{d\mu}(\Omega)$.
The operator $L$ is regarded as
an associated operator of the symmetric sesquilinear form
\begin{align*}%
	\mathfrak{a}(u,v) := \int_{\Omega} \nabla u \cdot \nabla\bar{v} \,dx
\end{align*}%
with the domain
\begin{align*}%
	\mathcal{D} := D(\mathfrak{a})
	= \{ u\in C^{\infty}_c(\bar{\Omega}) ;\,
		u|_{\partial\Omega} = 0 \}.
\end{align*}%
The basic property of $L$ is given in the following.
\begin{lemma}\label{lem_A}
The sesquilinear form
$\mathfrak{a}$
is densely defined, continuous, accretive, symmetric and closable in
$L^2_{d\mu}(\Omega)$.
Therefore, there exists a realization
$L_{\ast}$ of $L$ such that $L_{\ast}$
is nonnegative and self-adjoint.
Moreover, $\mathcal{D}$ is a core for $L_{\ast}$.
\end{lemma}
While this lemma was already discussed in \cite{SoWa16},
we give a sketch of the proof for the reader's convenience.
\begin{proof}[Proof of Lemma \ref{lem_A}]
It is obvious that $\mathfrak{a}$
is densely defined, continuous, accretive, symmetric
in $L^2_{d\mu}(\Omega)$.
Moreover, by \cite[Proposition 1.3]{Ou},
we easily see that $\mathfrak{a}$ is closable in $L^2_{d\mu}(\Omega)$.
Let
$\mathfrak{a}_{\ast}$
be the closure of
$\mathfrak{a}$.
Then, \cite[Lemma 2.1]{SoWa16} shows that
the bilinear form $\mathfrak{a}_{\ast}$ is characterized as
\begin{align}%
\label{da}
	D(\mathfrak{a}_{\ast})
	&= \left\{ u \in L^2_{d\mu}(\Omega) \cap \dot{H}^1(\Omega) ;
	\int_{\Omega} \frac{\partial u}{\partial x_j} \varphi \,dx
	= - \int_{\Omega} u \frac{\partial \varphi}{\partial x_j}\,dx \right.\\
\nonumber
	&\left. \hspace{110pt}\ \mbox{for}\ \varphi \in C_c^{\infty}(\mathbb{R}^N),
	\ j=1,\ldots,N
	\right\},\\
\nonumber
	\mathfrak{a}_{\ast}(u,v)
	&= \int_{\Omega} \nabla u \cdot \nabla v\,dx.
\end{align}%
We define the Friedrichs extension
$L_{\ast}$
of $L$ by
\begin{align*}%
	D(L_{\ast}) &:= \left\{
		u \in D(\mathfrak{a}_{\ast}) ;
		\ \mbox{there exists}\ f\in L^2_{d\mu}(\Omega) \ \mbox{such that}\right.
		\\
	&\left. \hspace{70pt}\
	\mathfrak{a}_{\ast} (u,v) = (f,v)_{L^2_{d\mu}}
	\ \mbox{for any}\ v\in D(\mathfrak{a}_{\ast}) \right\},\\
	L_{\ast} u &= f.
\end{align*}%
Then, by \cite[Theorem X.23]{ReSi},
we see that
$L_{\ast}$
is a nonnegative self-adjoint operator on
$L^2_{d\mu}(\Omega)$.
Moreover, we have for $u, v \in \mathcal{D}$,
\begin{align*}%
	(-Lu, v)_{L^2_{d\mu}}
	= \int_{\Omega} (-L u) v d\mu
	= \int_{\Omega} (\Delta u ) v dx
	= \mathfrak{a}_{\ast}(u,v).
\end{align*}%
By a density argument, we can take $v \in D(\mathfrak{a}_{\ast})$
in the above equalities.
This implies $u \in D(L_{\ast})$ and $Lu = L_{\ast}u$,
and hence,
$L_{\ast}$ is an extension of $L$.
\end{proof}
\subsection{Transformation into a usual Lebesgue spaces}
We introduce a diffeomorphism
$\Psi \in C^{\infty}(\Omega ; \Omega_{\Psi})$
as
\begin{align*}%
	\Psi(x) := |x|^{\alpha/2}x,\quad
	\Omega_{\Psi} :=
		\left\{ y \in \mathbb{R}^N ; |y|^{-\alpha/(2+\alpha)}y \in \Omega \right\}.
\end{align*}%
Since we assume that
$\partial \Omega$
is smooth, so is $\partial \Omega_{\Psi}$.

The following lemma is a fundamental fact on changes of variables.
\begin{lemma}\label{lem_dpsi}
\begin{itemize}
\item[(i)]
One has
\begin{align*}%
	\frac{\partial \Psi}{\partial x}(x)
	&= |x|^{\alpha/2} \left( I + \frac{\alpha}{2} Q(x) \right),
	\quad Q(x) = \left( \frac{x_jx_k}{|x|^2} \right)_{j,k=1,\ldots,N},\\
	\det \left( \frac{\partial \Psi}{\partial x}(x) \right)
	&= \frac{2+\alpha}{2} |x|^{N \alpha/2}.
\end{align*}%
Here we denote by $I$ the identity matrix of the order $N$.
\item[(ii)]
Define $m(x) := |x|^{-\alpha}a(x)$
for $x \in \Omega$,
$\tilde{m}(y) := m( \Psi^{-1}(y) )$
for $y \in \Omega_{\Psi}$
and $d \nu = \tilde{m}(y)dy$.
Then, the norm of $L^2(\Omega_{\Psi}, d\nu)$ is equivalent to
$L^2$-norm with usual Lebesgue measure on $\Omega_{\Psi}$.
Moreover,
\begin{align*}%
	\int_{\Omega_{\Psi}} |v(y)|^2 \,d\nu
	= \frac{2+\alpha}{2}
		\int_{\Omega} |v \left( \Psi(x) \right) |^2 |x|^{(N-2)\alpha/2}\, d\mu
\end{align*}%
holds for $v \in C_c^{\infty} ( \Omega_{\Psi} )$.
\end{itemize}
\end{lemma}
\begin{proof}
(i) is straightforward.
For (ii), we note that the assumption \eqref{a_est} implies
\begin{align*}%
	a_1 \le m(x) \le a_2
\end{align*}%
for any $x\in \bar{\Omega}$.
From this, we obtain
$a_1 \le \tilde{m}(y) \le a_2$
for any $y \in \overline{\Omega_{\Psi}}$,
which leads to the equivalence between
$L^2(\Omega_{\Psi}, d\nu)$ and $L^2(\Omega_{\Psi})$.
Moreover, we have
\begin{align*}%
	\int_{\Omega_{\Psi}} |v(y)|^2\, d\nu
	&= \int_{\Omega_{\Psi}} |v(y)|^2 m \left( \Psi^{-1}(y) \right)\,dy\\
	&= \int_{\Omega} |v \left( \Psi(x) \right)|^2 m(x)
		\left| \det \left( \frac{\partial \Psi}{\partial x} \right) \right|\,dx \\
	&= \frac{2+\alpha}{2} \int_{\Omega} | v\left( \Psi(x) \right) |^2
			|x|^{(N-2)\alpha/2}\, d\mu,
\end{align*}%
which completes the proof.
\end{proof}

In view of the above lemma, we introduce an isometry from
$L^2(\Omega_{\Psi}, d\nu)$ to $L^2(\Omega, d\mu)$ as
\begin{align}%
\label{j}
	Jv(x) := \sqrt{ \frac{2+\alpha}{2} } |x|^{(N-2)\alpha/4}
		v \left( \Psi(x) \right)
\end{align}%
for $v \in L^2(\Omega_{\Psi}, d\nu)$.
It is easy to see that the inverse of $J$ is given by
\begin{align}%
\label{ji}
	J^{-1} u (y) =
		\sqrt{\frac{2}{2+\alpha}} |y|^{-(N-2)\alpha/(2(2+\alpha))}
		u \left( |y|^{-\alpha/(2+\alpha)} y \right)
\end{align}%
for $u \in L^2(\Omega, d\mu)$.
Then, the operator $L$ is transformed into the following uniformly elliptic operator
on $L^2(\Omega_{\Psi}, d\nu)$.

\begin{lemma}\label{lem_jiaj}
The operator
$B=J^{-1}LJ$ with the domain
$\mathcal{D}_{\Psi} := \{ u \circ \Psi^{-1} \,;\, u \in \mathcal{D} \}$
is given by
\begin{align}%
\label{B}
	B v(y) = \tilde{m}(y)^{-1}
		\left( - {\rm div\,}( b(y) \nabla v(y))
		- \frac{(N-2)^2 \alpha (4+\alpha)}{16|y|^2} v(y) \right)
\end{align}%
for $y \in \Omega_{\Psi}$ and $v \in \mathcal{D}_{\Psi}$,
where
\begin{align*}%
	b(y) = I + \alpha \left( 1+ \frac{\alpha}{4} \right) Q(\Psi^{-1}(y)).
\end{align*}%
Moreover, $B$ is a uniformly elliptic operator in $\Omega_{\Psi}$
with bounded coefficients.
\end{lemma}
\begin{proof}
Let $\varphi \in \mathcal{D}_{\Psi}$ be arbitrary fixed
and let $v \in \mathcal{D}_{\Psi}$.
Then, we obtain
\begin{align*}%
	\int_{\Omega_{\Psi}}
		\left( Bv(y) \right) \varphi(y)\,d\nu
	&= \frac{2+\alpha}{2} \int_{\Omega}
		\left( J^{-1} L J v(\Psi(x)) \right) \varphi(\Psi(x)) |x|^{(N-2)\alpha/2}\,d\mu \\
	&= \int_{\Omega} \left( AJv(x) \right) J \varphi(x)\,d\mu \\
	&= \int_{\Omega} \nabla Jv(x) \cdot \nabla J\varphi(x)\,dx.
\end{align*}%
Noting that
\begin{align*}%
	\sqrt{\frac{2}{2+\alpha}} \nabla J \varphi (x)
	&= \frac{(N-2)\alpha}{4} |x|^{(N-2)\alpha/4-2} x \varphi(\Psi(x)) \\
	&\quad + |x|^{N\alpha/4} \left( I + \frac{\alpha}{2}Q(x) \right)\nabla \varphi ( \Psi(x) )
\end{align*}%
and putting
$\beta = (N-2)\alpha/4$, we have
\begin{align*}%
	&\int_{\Omega} \nabla Jv(x) \cdot \nabla J \varphi(x)\,dx\\
	&= \int_{\Omega} \left( \beta \frac{y}{|y|^2}v(y)
		+ \left( I + \frac{\alpha}{2}Q(\Psi^{-1}(y)) \right)\nabla v(y) \right) \\
	&\qquad\quad \cdot \left( \beta \frac{y}{|y|^2} \varphi (y)
		+ \left( I + \frac{\alpha}{2}Q(\Psi^{-1}(y)) \right)\nabla \varphi (y) \right)\,dy \\
	&= \int_{\Omega}
		\left( I + \frac{\alpha}{2}Q(\Psi^{-1}(y)) \right)^2 \nabla v(y) \cdot \nabla \varphi(y)\,dy
		+ \beta \left( 1 + \frac{\alpha}{2} \right)
		\int_{\Omega} \frac{y}{|y|^2} \cdot \nabla ( v(y) \varphi(y) )\,dy\\
	&\quad + \beta^2 \int_{\Omega} \frac{1}{|y|^2} v(y) \varphi(y)\,dy \\
	&= - \int_{\Omega} {\rm div\,} (b(y)\nabla v(y) ) \varphi(y)\,dy
		+ \left( \beta^2 - (N-2)\beta \left( 1+\frac{\alpha}{2} \right) \right)
		\int_{\Omega} \frac{1}{|y|^2} v(y) \varphi(y)\,dy.
\end{align*}%
Finally, the uniform ellipticity of $B$ follows from
$Q \ge 0$ as a symmetric matrix.
The boundedness of the coefficients immediately shown
by the assumption $0 \notin \bar{\Omega}$ and the definition of $Q$.
\end{proof}

From the above lemma,
by the same procedure as \cite[pp.100--101]{Ou},
we can associate an operator
$B_2$ on $L^2(\Omega_{\Psi}, d\nu)$
with the sesquilinear form
\begin{align*}%
	\mathfrak{b}(u,v)
	&= \int_{\Omega_{\Psi}} \left(
		\left( b(y) \nabla u (y) \right) \cdot \nabla v(y)
		- \frac{ (N-2)^2 \alpha (4+\alpha)}{16|y|^2} u(y) v(y) \right)\, dy,\\
	D(\mathfrak{b})
	&= \{ u \circ \Psi^{-1} ; u \in D(\mathfrak{a}_{\ast}) \},
\end{align*}%
and $-B_2$ is the generator of an analytic semigroup
$T_2(t) = e^{-t B_2}$ on $L^2(\Omega, d\nu)$.
Therefore, it follows from \cite[Theorem 1.52]{Ou} that
$-B_2$ generates an analytic semigroup on $L^2(\Omega_{\Psi}, d\nu)$.

Furthermore,
applying \cite[Theorem 4.28]{Ou},
we extend it to an analytic semigroup
generated by $-B$
in $L^p(\Omega_{\Psi}, d\nu)$.
Summarizing the above, we have the following:
\begin{lemma}\label{lem_sg}
For every
$1<p<\infty$,
the operator
$-B_p$
defined by \eqref{B}
in
$L^p(\Omega_{\Psi}, d\nu)$
endowed with the domain
$D(B_p) = W^{2,p}(\Omega_{\Psi}, d\nu) \cap W_0^{1,p}(\Omega_{\Psi}, d\nu)$
generates an analytic semigroup
$( T_p(t) )_{t \ge 0}$
and
$\mathcal{D}_{\Psi}$
is a core for
$B_p$.
Moreover,
$T_p(t)$
is consistent for all
$1<p<\infty$.
\end{lemma}

By virtue of Lemma \ref{lem_sg},
we introduce $(T(t))_{t\ge 0}$ as a semigroup generated by $-B$ with
Dirichlet boundary condition.

Next, we consider
$L^2$-$L^q$ estimates for the semigroup $(T(t))_{t\ge 0}$.
As we said in the introduction,
this is crucial to prove the diffusion phenomena
for the damped wave equation \eqref{dw}.

\begin{proposition}\label{prop_lplq}
Let
$N\ge 2$
and
$\alpha > 0$.
Then we have the following
$L^{p}$-$L^2$ estimates:
\begin{itemize}
\item[(i)]
If $N=2$, then
the the semigroup
$(T(t))_{t\ge 0}$ is sub-Markovian and
satisfies the estimate
\begin{align*}%
	\| T(t) f \|_{L^{\infty}(\Omega_{\Psi}, d\nu)}
	\le C t^{-1/2} \| f \|_{L^2(\Omega_{\Psi}, d\nu)};
\end{align*}%
\item[(ii)]
If $N\ge 3$, then for every $2 \le p < p_{\alpha} := 2N(2+\alpha)/(\alpha(N-2))$,
then the semigroup
$(T(t))_{t\ge 0}$
satisfies the estimate
\begin{align*}%
	\| T(t) f \|_{L^p(\Omega_{\Psi}, d\nu)}
	\le C t^{-(N/2)(1/2 - 1/p)} \| f \|_{L^2(\Omega_{\Psi}, d\nu)}.
\end{align*}%
\end{itemize}
\end{proposition}
\begin{remark}\label{rem_lplq}
Let $N\ge 3, \Omega = \mathbb{R}^N\setminus B(0,1)$
and $\varepsilon > 0$.
Then, the function
\begin{align*}%
	\psi_0(y) = |y|^{-(N-2)\alpha/4} \left( 1 - |y|^{-N+2} \right)
\end{align*}%
belongs to $L^{p_{\alpha}+\varepsilon}(\Omega_{\Psi}, d\nu)$
and satisfies $B\psi_0 = 0$ in $\Omega_{\Psi}$.
This means that
$\psi_0$
is a stationary solution of the equation
\begin{align*}%
	\left\{ \begin{array}{ll}
		\psi_t + B_{p_{\alpha}+\varepsilon} \psi = 0,
			&y \in \Omega_{\Psi},\ t > 0,\\
		\psi (y,t) = 0,
			&y \in \partial \Omega_{\Psi},\ t>0,\\
		\psi (y,0) = \psi_0 (y),
			&y \in \Omega_{\Psi}.
		\end{array}\right.
\end{align*}%
By virtue of
$\psi_0$,
any decay estimate for the semigroup $(T(t))_{t\ge 0}$
in
$L^{p_{\alpha}+\varepsilon}(\Omega_{\Psi}, d\nu)$
cannot be expected.
We would expect that for $N \ge 3$
the $L^2$-$L^{p_{\alpha}}$ estimate is given by
\begin{align*}%
	\| T(t) f \|_{L^{p_{\alpha}}(\Omega_{\Psi}, d\nu)}
	\le C t^{-(N/2)(1/2-1/p_{\alpha})}(1+\log (1+t))^{1/p_{\alpha}}
		\| f \|_{L^2(\Omega_{\Psi}, d\nu)}.
\end{align*}%
\end{remark}

We postpone the proof of Proposition \ref{prop_lplq}
in the following subsections, and here
we give a practical version of it.
For the original variable $x$,
Proposition \ref{prop_lplq} and the identity
\begin{align*}%
	\left( \frac{2+\alpha}{2} \right)^{1-p/2}
	\int_{\Omega} | Jv(x) |^p |x|^{-(N-2)\alpha(p-2)/4} a(x)\, dx
	= \int_{\Omega_{\Psi}} |v(y)|^p\, d\nu
\end{align*}%
imply the following estimate.
\begin{proposition}\label{prop_lplq2}
Let $N \ge 2$ and $\alpha>0$ and let $v$ be a solution to \eqref{h}.
Then, we have the following.
\begin{itemize}
\item[(i)]
If $N=2$, then we have
\begin{align*}%
	\| v(t) \|_{L^2(\Omega, d\mu)} \le C t^{-1/2} \| v_0 \|_{L^1(\Omega, d\mu)};
\end{align*}%
\item[(ii)]
If $N\ge 3$, then for every $(p_{\alpha})^{\prime} < q \le 2$, we have
\begin{align*}%
	\| v(t) \|_{L^2(\Omega, d\mu)} \le C
		t^{-(N/2)(1/q - 1/2)}
			\left( \int_{\Omega} |v_0(x)|^q |x|^{(N-2)\alpha(2-q)/4} \, d\mu \right)^{1/q}.
\end{align*}%
\end{itemize}
\end{proposition}

\subsection{Proof of Proposition \ref{prop_lplq} for $N=2$}
In this subsection we use the following
Gagliardo--Nirenberg inequality with $N=2$.
\begin{lemma}[Gagliardo--Nirenberg inequality]\label{lem_gn}
Let $N=2$.
For every $2 \le q < \infty$, there exists a constant $C_q$ such that
\begin{align*}%
	\| w \|_{L^q(\Omega)} \le C_q
		\| \nabla w \|_{L^2(\Omega)}^{1-2/q} \| w \|_{L^2(\Omega)}^{2/q}
\end{align*}%
holds for any $w \in W^{1,2}_0(\Omega)$.
\end{lemma}
For the proof, see for example
\cite{Frbook}.
From this, we obtain the following estimate.
\begin{lemma}\label{lem_gn2}
For every $2 \le q < \infty$, we have
\begin{align}
\label{gn2}
	\| \varphi \|_{L^q(\Omega_{\Psi}, d\nu)}
		\le C  (B_2 \varphi, \varphi )_{d\nu}^{1/2-1/q}
			\| \varphi \|_{L^2(\Omega_{\Psi}, d\nu)}^{1/q}
\end{align}
for any $\varphi \in L^2(\Omega_{\Psi}, d\nu)$.
\end{lemma}
\begin{proof}
Since $\mathcal{D}_{\Psi}$ is a core for $B_2$,
it suffices to show \eqref{gn2} for $\varphi \in \mathcal{D}_{\Psi}$.
Let
$\varphi \in \mathcal{D}_{\Psi}$.
Noting that
\begin{align*}%
	b(y) \ge \min \left\{ 1, 1+\alpha + \frac{\alpha^2}{4} \right\} I = I
\end{align*}%
in the sense of symmetric matrix,
we have
\begin{align*}%
	( B_2 \varphi, \varphi )_{d\nu}
	= \int_{\Omega_{\Psi}}  ( b(y) \nabla \varphi) \cdot \nabla \varphi \,dy
	\ge \| \nabla \varphi \|_{L^2(\Omega_{\Psi})}^2.
\end{align*}%
Combining the above estimate with
$a_1 \le \tilde{m} \le a_2$
and Lemma \ref{lem_gn}, we obtain
\begin{align*}%
	\| \varphi \|_{L^q(\Omega_{\Psi}, d\nu)}
	&\le a_2^{1/q} \left( \int_{\Omega_{\Psi}} | \varphi (y) |^q\,dy \right)^{1/q}\\
	&\le a_2^{1/q} C_q
		\left( \int_{\Omega_{\Psi}} | \nabla \varphi(y) |^2\,dy \right)^{1/2-1/q}
		\left( \int_{\Omega_{\Psi}} | \varphi (y) |^2\, dy \right)^{1/q} \\
	&\le a_2^{1/q}a_1^{-1/2} C_q
		(B_2 \varphi, \varphi)_{d\nu}^{1/2-1/q}
		\| \varphi \|_{L^2(\Omega_{\Psi}, d\nu)},
\end{align*}%
which implies \eqref{gn2}.
\end{proof}

\begin{lemma}\label{lem_sm}
The semigroup $(T(t))_{t\ge 0}$ is sub-Markobian,
that is, $T(t)$ is positively preserving:
\begin{align*}%
	f \in L^2( \Omega_{\Psi}, d\nu),\ f\ge 0
	\ \Rightarrow\ T(t) f \ge 0
\end{align*}%
and $L^{\infty}$-contractive:
\begin{align*}%
	\| T(t) f \|_{L^{\infty}(\Omega_{\Psi}, d\nu)}
	\le \| f \|_{L^{\infty}(\Omega_{\Psi}, d\nu)}
	\quad \mbox{for}\quad f \in L^2(\Omega_{\Psi}, d\nu) \cap L^{\infty}(\Omega_{\Psi}, d\nu).
\end{align*}%
\end{lemma}
\begin{proof}
Let $v \in D(\mathfrak{b})$.
Using the characterization \eqref{da} of $D(\mathfrak{a}_{\ast})$,
we easily see that
$|v| \in D(\mathfrak{b})$.
Moreover, we have
\begin{align*}%
	\mathfrak{b}(|v|, |v|)
	= \int_{\Omega_{\Psi}}
		\left( b(y) \frac{v}{|v|} \nabla v (y) \right) \cdot \frac{v}{|v|} \nabla v(y) \, dy
	= \mathfrak{b}(v,v).
\end{align*}%
Thus, applying \cite[Theorem 2.7]{Ou}, we have
the positively preserving property of $T(t)$.
Next, we show the $L^{\infty}$-contractive property of $T(t)$.
By using the characterization \eqref{da} of $D(\mathfrak{a}_{\ast})$ again,
we also see that
$v \in D(\mathfrak{b})$
implies
$P v = (1 \wedge |v|) {\rm sign\,}v \in D(\mathfrak{b})$.
Furthermore, we have
\begin{align*}%
	\mathfrak{b}(Pv, v - Pv)
	= \int_{\Omega_{\Psi}}
	\left( b(y) \chi_{|v| < 1} \nabla v (y) \right)
		\cdot \left( (1 - \chi_{|v|<1}) \nabla v(y) \right) \, dy
	= 0.
\end{align*}%
Therefore, applying \cite[Theorem 2.13]{Ou}, we prove
the $L^{\infty}$-contractive property of $T(t)$.
\end{proof}

\begin{proof}[Proof of Proposition \ref{prop_lplq} for $N =2$ ]
By Lemma \ref{lem_sm}, we see that
the semigroup $T(t)$ is $L^{\infty}$-contractive.
Moreover, by Lemma \ref{lem_gn2},
$T(t)$ satisfies the second condition of \cite[Theorem 6.2]{Ou} with $d=2$.
Thus, applying \cite[Theorem 6.2]{Ou}, we conclude
\begin{align*}%
	\| T(t) f \|_{L^{\infty}(\Omega_{\Psi}, d\nu)}
	\le C t^{-1/2} \| f \|_{L^2(\Omega_{\Psi}, d\nu)},
\end{align*}%
which gives the assertion of Proposition \ref{prop_lplq} for $N=2$.
\end{proof}

\subsection{Proof of Proposition \ref{prop_lplq} for $N \ge 3$}
In this case we need to prepare some result for analytic semigroup
generated by $-B_p$ via $L^p$-theory.
First we prove that
$- B_p$
generates an analytic contraction semigroup when
\begin{align}%
\label{range.p}
	2-\frac{4}{4+\alpha} < p < 2+\frac{4}{\alpha}.
\end{align}%
\begin{lemma}\label{lem_sect}
Let
$N\ge 3$.
Assume that \eqref{range.p} is satisfied.
Then, there exists a constant
$\ell_{N,p,\alpha} \ge 0$ depending only on $N, p$ and $\alpha$
such that $B_p$ is m-sectorial of type
$S(\ell_{N,p,\alpha})$
in
$L^p(\Omega_{\Psi}, d\nu)$,
that is,
\begin{align}%
\label{sect}
	\left| {\rm Im\,} \int_{\Omega_{\Psi}} (B_p \varphi) \bar{\varphi} |\varphi|^{p-2} \,d\nu \right|
	\le \ell_{N,p,\alpha} \left(
		{\rm Re\,} \int_{\Omega_{\Psi}} (B_p \varphi) \bar{\varphi} |\varphi|^{p-2}\,d\nu
			\right)
\end{align}%
holds for
$\varphi \in D(B_p)$.
Moreover, there exists a constant
$M_{N,p,\alpha} \ge 1$
depending only on $N, p$ and $\alpha$ such that
\begin{align}%
\label{1/t}
	\| B_p T(t) f \|_{L^p(\Omega_{\Psi}, d\nu)}
	\le \frac{M_{N,p,\alpha}}{t} \| f \|_{L^p(\Omega_{\Psi}, d\nu)}
\end{align}%
holds for
$f \in L^p(\Omega_{\Psi}, d\nu)$ and $t >0$.
\end{lemma}
\begin{proof}
Since
$B_{p^{\prime}}$
is the adjoint operator of
$B_p$
and
$\mathcal{D}_{\Psi}$
is a core for $B_p$ by Lemma \ref{lem_sg},
it suffices to prove \eqref{sect} for
$p \ge 2$
and
$\varphi \in \mathcal{D}_{\Psi}$.
Indeed, if \eqref{sect} is true for $p\ge 2$ and $\varphi \in \mathcal{D}_{\Psi}$,
we deduce for $p \le 2$ and $\varphi \in \mathcal{D}_{\Psi}$ that
\begin{align*}%
	\left| {\rm Im\,} \int_{\Omega_{\Psi}} (B_p \varphi) \bar{\varphi} |\varphi|^{p-2} \,d\nu \right|
	&=
	\left| {\rm Im\,} \int_{\Omega_{\Psi}} \varphi ( B_{p'} (\bar{\varphi} |\varphi|^{p-2}) )
		\,d\nu \right| \\
	&=
	\left| {\rm Im\,} \int_{\Omega_{\Psi}} \bar{f}|f|^{p-2} ( B_{p'} (f) )
		\,d\nu \right| \\
	&\le \ell_{N,p,\alpha} \left(
		{\rm Re\,} \int_{\Omega_{\Psi}} \bar{f} |f|^{p-2} (B_{p'} f) \,d\nu
			\right) \\
	&= \ell_{N,p,\alpha} \left(
		{\rm Re\,} \int_{\Omega_{\Psi}} \varphi (B_{p'} (\bar{\varphi} |\varphi|^{p-2})) \,d\nu
			\right) \\
	&=\ell_{N,p,\alpha} \left(
		{\rm Re\,} \int_{\Omega_{\Psi}} (B_p \varphi) \bar{\varphi} |\varphi|^{p-2}\,d\nu
			\right),
\end{align*}%
where $f = \bar{\varphi} |\varphi|^{p-2}$.

Therefore, in what follows we assume
$p \ge 2$
and
$\varphi \in \mathcal{D}_{\Psi}$.
Setting $u = J \varphi \in \mathcal{D}$, we have
\begin{align*}%
	\int_{\Omega_{\Psi}} ( B_p \varphi) \bar{\varphi} |\varphi|^{p-2}\,d\nu
	&= \int_{\Omega_{\Psi}} (J^{-1}Lu(y)) \overline{J^{-1}u(y)} |J^{-1}u(y)|^{p-2} \, d \nu\\
	&= \left( \frac{2}{2+\alpha} \right)^{(p-2)/2}
		\int_{\Omega} ( - \Delta u(x)) \overline{u(x)} |u(x)|^{p-2} |x|^{-\beta}\, dx,
\end{align*}%
where
$\beta = (N-2)\alpha(p-2)/4$.
Moreover, integration by parts yields
\begin{align}%
\label{Delta}
	\int_{\Omega} (-\Delta u ) \bar{u}|u|^{p-2} |x|^{-\beta}\, dx
	&= \int_{\Omega} ( \nabla u \cdot \nabla (\bar{u}|u|^{p-2} )) |x|^{-\beta}\, dx\\
\nonumber
	&\quad + \int_{\Omega} ( \nabla u \cdot \nabla ( |x|^{-\beta} ))
			\bar{u} |u|^{p-2}\,dx.
\end{align}%
By integration by parts again and taking the real-part, we see that
\begin{align*}%
	&{\rm Re\,}\int_{\Omega} ( \nabla u \cdot \nabla (\bar{u}|u|^{p-2} )) |x|^{-\beta}\, dx \\
	&\quad = {\rm Re\,}
	\int_{\Omega} \left( |\nabla u|^2 |u|^{p-2}
		+ (p-2) \nabla u \cdot ( \bar{u} |u|^{p-4} {\rm Re\,}(\bar{u}\nabla u) ) \right)
			|x|^{-\beta} \, dx \\
	&\quad = {\rm Re\,}
	\int_{\Omega} \left( | \bar{u}\nabla u |^2
		+ (p-2) (\bar{u} \nabla u) {\rm Re\,} (\bar{u} \nabla u) \right) |u|^{p-4} |x|^{-\beta}\,dx \\
	&\quad= (p-1) \int_{\Omega} | {\rm Re\,}(\bar{u}\nabla u) |^2 |u|^{p-4} |x|^{-\beta} \, dx
		+ \int_{\Omega} | {\rm Im\,}(\bar{u}\nabla u) |^2 |u|^{p-4} |x|^{-\beta}\,dx,\\
	&{\rm Re\,}\int_{\Omega} ( \nabla u \cdot \nabla ( |x|^{-\beta} ))
			\bar{u} |u|^{p-2}\,dx \\
	&\quad =
		\frac{1}{p} \int_{\Omega}
			\nabla (|u|^p) \cdot \nabla (|x|^{-\beta})\,dx \\
	&\quad =
		- \frac{1}{p} \int_{\Omega} |u|^p \Delta ( |x|^{-\beta} )\,dx \\
	&\quad =
		\frac{\beta(N-2-\beta)}{p} \int_{\Omega} |u|^p |x|^{-\beta-2}\,dx
\end{align*}%
and hence, we eventually have
\begin{align}%
\label{Re}
	{\rm Re\,} \int_{\Omega} (B_p \varphi) \bar{\varphi} |\varphi|^{p-2}\,d\nu
	&= (p-1)\left( \frac{2}{2+\alpha} \right)^{(p-2)/2}
	 \int_{\Omega} | {\rm Re\,}(\bar{u}\nabla u) |^2 |u|^{p-4} |x|^{-\beta} \, dx\\
\nonumber
	&\quad + \left( \frac{2}{2+\alpha} \right)^{(p-2)/2}
	\int_{\Omega} | {\rm Im\,}(\bar{u}\nabla u) |^2 |u|^{p-4} |x|^{-\beta}\,dx\\
\nonumber
	&\quad + \frac{\beta(N-2-\beta)}{p} \left( \frac{2}{2+\alpha} \right)^{(p-2)/2}
	\int_{\Omega} |u|^p|x|^{-\beta -2}\, dx.
\end{align}%
Therefore, $B_p$ is accretive if
$\beta \le N-2$, that is, $p \le 2 + 4/\alpha$.
On the other hand, taking the imaginary part of \eqref{Delta}, we have
\begin{align*}%
	\left| {\rm Im\,} \int_{\Omega} (B_p \varphi) \bar{\varphi} |\varphi|^{p-2}\,d\nu \right|
	&\le |p-2|\left( \frac{2}{2+\alpha} \right)^{(p-2)/2}
		 \int_{\Omega} |x|^{-\beta} | {\rm Re\,} ( \bar{u} \nabla u ) |
			| {\rm Im\,} (\bar{u} \nabla u ) | | u |^{p-4} \, dx\\
	&\quad + \beta \left( \frac{2}{2+\alpha} \right)^{(p-2)/2}
		\int_{\Omega} |x|^{-\beta -1}
			| {\rm Im\,}(\bar{u}\nabla u)| |u|^{p-2}\,dx.
\end{align*}%
Therefore, choosing
$2 \le p < 2+ 4/\alpha$ and
\begin{align*}%
	\ell_{N,p,\alpha}
	&= \sqrt{ \frac{(p-2)^2}{4(p-1)} + \frac{\beta p}{4(N-2-\beta)}} \\
	&= \sqrt{ \frac{(p-2)^2}{4(p-1)} + \frac{p(p-2)}{4}
		\left( 2 + \frac{4}{\alpha} -p \right)^{-1}},
\end{align*}%
we obtain
\begin{align*}%
	\left| {\rm Im\,} \int_{\Omega} (B_p \varphi) \bar{\varphi} |\varphi|^{p-2}\,d\nu \right|
	\le \ell_{N,p,\alpha} \left(
		{\rm Re\,} \int_{\Omega} (B_p \varphi) \bar{\varphi} |\varphi|^{p-2}\,d\nu \right).
\end{align*}%
This gives the $m$-sectoriality of $B_p$ and
hence $-B_p$ generates an analytic contraction semigroup on
$L^p(\Omega_{\Psi}, d\nu)$.
The estimate \eqref{1/t} is an immediate consequence of
a property of an analytic semigroup.
\end{proof}

To deduce
$L^p$--$L^q$ estimates for
$T(t)$,
we use the following Gagliardo--Nirenberg inequality
(see for example \cite{Frbook}).
\begin{lemma}\label{lem_gn3}
Let $N \ge 3$.
Then, there exists a constant $C_N > 0$ depending only on $N$ such that
for every $w \in W_0^{1,2}(\Omega_{\Psi})$, we have
\begin{align*}%
	\| w \|_{L^{2N/(N-2)}(\Omega_{\Psi}, d\nu)}
		\le C_N \| \nabla w \|_{L^2(\Omega_{\Psi}, d\nu)}.
\end{align*}%
\end{lemma}
From this, we obtain the following estimate.
\begin{lemma}\label{lem_embed}
Let $N \ge 3$ and let $2 \le p < 2+ 4/\alpha$.
Then, there exists a constant
$C_{N,p}>0$ depending only on $N$ and $p$ such that
for every $\varphi \in \mathcal{D}_{\Psi}$, we have
\begin{align}%
\label{bp_embed}
	\| \varphi \|_{L^{pN/(N-2)}(\Omega_{\Psi}, d\nu)}
		\le C_{N,p} \left(
			{\rm Re\,} \int_{\Omega_{\Psi}} (B_p \varphi) \bar{\varphi} |\varphi|^{p-2}\,d\nu
				\right)^{1/p}.
\end{align}%
\end{lemma}
\begin{proof}
Here we use the differential expression of $B$.
\begin{align*}%
	{\rm Re\,} \int_{\Omega_{\Psi}} (B_p \varphi) \bar{\varphi} |\varphi|^{p-2}\,d\nu
	&= {\rm Re\,} \int_{\Omega_{\Psi}}
			b(y) \nabla \varphi \cdot \nabla (\bar{\varphi} |\varphi|^{p-2} )\,dy \\
	&\quad - \frac{(N-2)^2\alpha (4+\alpha)}{16}
			\int_{\Omega_{\Psi}} \frac{ |\varphi(y)|^p }{|y|^2} \,dy\\
	&\ge (p-1) \left( \frac{2+\alpha}{2} \right)^2
		 \int_{\Omega_{\Psi}} |{\rm Re\,} (\bar{\varphi} \varphi_r)|^2 |\varphi|^{p-4}\,dy\\
	&\quad - \left( \frac{N-2}{2} \right)^2
			\left( \left( \frac{2+\alpha}{2} \right)^2 -1 \right)
			\int_{\Omega_{\Psi}} \frac{ |\varphi(y)|^p}{|y|^2}\,dy,
\end{align*}%
where
$\varphi_r$
stands for the derivative of $\varphi$ with respect to the radial direction.
The above inequality can be proved by recalling
$b(y) = I + \alpha( 1+ \frac{\alpha}{4} ) Q(\Psi^{-1}(y))$,
$Q(\Psi^{-1}(y) ) = ( \frac{y_i y_j}{|y|^2} )_{i,j=1,\ldots,N}$,
and noting that
\begin{align*}%
	&{\rm Re\,} \left[ b(y) \nabla \varphi \cdot \nabla ( \bar{\varphi} |\varphi|^{p-2} ) \right]\\
	&= |\varphi|^{p-4} \left[ (p-1)  | {\rm Re\,}(\bar{\varphi} \nabla \varphi) |^2
		+  | {\rm Im\,}(\bar{\varphi} \nabla \varphi) |^2 \right] \\
	&\quad + \alpha \left( 1+ \frac{\alpha}{4} \right) |\varphi|^{p-4} \left[
		(p-1)  \left| {\rm Re\,} \left(\frac{y}{|y|}\cdot (\bar{\varphi}\nabla \varphi) \right) \right|^2
		+  \left| {\rm Im\,} \left(\frac{y}{|y|}\cdot (\bar{\varphi}\nabla \varphi) \right) \right|^2
		\right] \\
	&\ge \left( 1+ \alpha \left( 1+ \frac{\alpha}{4} \right) \right)(p-1) |\varphi|^{p-4}
		\left| {\rm Re\,} \left(\frac{y}{|y|}\cdot (\bar{\varphi}\nabla \varphi) \right) \right|^2 \\
	&= (p-1) \left( \frac{2+\alpha}{2} \right)^2
		|{\rm Re\,} (\bar{\varphi} \varphi_r)|^2 |\varphi|^{p-4}
\end{align*}%
Applying the Hardy's inequality
\begin{align*}%
	 \int_{\Omega_{\Psi}} |{\rm Re\,}(\bar{\varphi} \varphi_r) |^2 |\varphi|^{p-4}\, dy
	\ge \left( \frac{N-2}{2} \right)^2 \frac{4}{p^2}
		\int_{\Omega_{\Psi}} \frac{ |\varphi(y)|^p}{|y|^2}\, dy
\end{align*}%
and
\begin{align*}%
	(p-1) \left( \frac{2+\alpha}{2} \right)^2 \left( \frac{N-2}{2} \right)^2 \frac{4}{p^2}
		- \left( \frac{N-2}{2} \right)^2
			\left( \left( \frac{2+\alpha}{2} \right) - 1 \right) > 0,
\end{align*}%
we see that there exists $\delta \in (0,1)$ such that
\begin{align*}%
	{\rm Re\,} \int_{\Omega_{\Psi}} ( B_p \varphi) \bar{\varphi} |\varphi|^{p-2}\,d\nu
		&\ge \delta^2 \int_{\Omega_{\Psi}}
			|\bar{\varphi} \varphi_r |^2 |\varphi|^{p-4}\,dy\\
		&= \frac{4\delta^2}{p^2}
			\int_{\Omega_{\Psi}} | \nabla |\varphi|^{p/2} |^2\, dx.
\end{align*}%
By Lemma \ref{lem_gn3} with
$w = |\varphi|^{p/2}$, we obtain
\begin{align*}%
	\| \varphi \|_{L^{pN/(N-2)}(\Omega_{\Psi}, d\nu)}
	&= \left\| | \varphi|^{p/2} \right\|_{L^{2N/(N-2)}(\Omega_{\Psi}, d\nu)}^{2/p} \\
	& \le C_N \left\| \nabla | \varphi |^{p/2} \right\|_{L^2(\Omega_{\Psi}, d\nu)}^{2/p}\\
	&\le C_N \left( \frac{2\delta}{p} \right)^{-2/p}
		\left( {\rm Re\,} \int_{\Omega_{\Psi}} ( B_p \varphi)
				\bar{\varphi} |\varphi|^{p-2}\,d\nu \right)^{1/p}.
\end{align*}%
Since $a_1 \le m$, the proof is finished.
\end{proof}
\begin{proof}[Proof of Proposition \ref{prop_lplq} for $N\ge 3$]
Let
$2 \le r < 1+4/\alpha$ and $f \in L^r(\Omega_{\Psi})$.
Applying \eqref{1/t} and Lemma \ref{lem_embed}, we have
\begin{align*}%
	\| T(t) f \|_{L^{rN/(N-2)}(\Omega_{\Psi}, d\nu)}
	&\le C_{N,r,\alpha} \left(
		{\rm Re\,}\int_{\Omega_{\Psi}}
			(B_r T(t) f ) \overline{T(t)f} |T(t)f|^{r-2} \,d\nu \right)^{1/r} \\
	&\le C_{N,r,\alpha} \| B_p T(t) f \|_{L^r(\Omega_{\Psi}, d\nu)}^{1/r}
		\| T(t) f \|_{L^r(\Omega_{\Psi}, d\nu)}^{1-1/r} \\
	&\le C_{N,r,\alpha} \left( M_{N,r,\alpha} \right)^{1/r}
		t^{-1/r} \| f \|_{L^r(\Omega_{\Psi}, d\nu)}.
\end{align*}%
Fix
$2<p<(N/(N-2))(2+4/\alpha)$.
Then the case
$2<p \le 2N/(N-2)$ is already verified by the above result with
$r=2$ and $q=p$.

In the case $2N/(N-2) < p < (N/(N-2))(2+4/\alpha)$, we set
\begin{align*}%
	q_0 = 2,\quad
	q_m = \frac{N-2}{N}p,\quad
	q_j = \left( \frac{q_m}{q_0} \right)^{j/m} q_0\ (j=1,\ldots,m-1),
\end{align*}%
where $m$ is an integer satisfying
\begin{align*}%
	\left( 1 + \frac{2}{\alpha} \right)^{1/m} < \frac{N}{N-2}.
\end{align*}%
Since
\begin{align*}%
	2=q_0 < q_1 < \cdots < q_m < 2+\frac{4}{\alpha}
\end{align*}%
and
\begin{align*}%
	\frac{q_j}{q_{j-1}} = \left( \frac{q_m}{q_0} \right)^{1/m}
	= \left( \frac{N-2}{2N}p \right)^{1/m}
	< \left( 1+\frac{2}{\alpha} \right)^{1/m}
	< \frac{N}{N-2},
\end{align*}%
it follows from the previous discussion that
$L^{q_{j-1}}$-$L^{q_j}$ estimates are true for
$j=1,\ldots, m$ and
\begin{align*}%
	\| T(t/2) f \|_{L^{q_m}(\Omega_{\Psi}, d\nu)}
	&\le \left( \prod_{j=1}^m \| T(t/2m) \|_{L^{q_{j-1}}\to L^{q_j}} \right)
		\| f \|_{L^2(\Omega_{\Psi}, d\nu)}\\
	&\le C_{2, q_m} t ^{-(N/2)(1/2 - 1/q_m)} \| f \|_{L^2(\Omega_{\Psi}, d\nu)}.
\end{align*}%
Since we have
$\| T(t/2) \|_{L^{q_m} \to L^p} \le C_{q_m,p}t^{-1/p}$
in the previous step with
$r = (N-2)p/N$,
we obtain the desired assertion.
\end{proof}

\section{Related elliptic problem: construction of a weight function}
In this section, we construct a weight function,
which will be used for the weighted energy estimate of
the damped wave equation \eqref{dw}.
As we mentioned in the introduction,
in \cite[Remark 3.1]{SoWa16},
it is shown that the solution $A(x)$ of the Poisson equation
$\Delta A(x) = a(x)$
does not have the desired property in general
(in particular, the condition \eqref{a_est1} fails).
Therefore, following \cite{SoWa17},
we weaken the problem to the inequality
\begin{align}%
\label{ellip}
	(1-\varepsilon) a(x) \le \Delta A(x) \le (1+\varepsilon) a(x)
\end{align}%
for $x\in \Omega$, where $\varepsilon \in (0,1)$ is a parameter.
We construct a function $A_{\varepsilon}$
satisfying \eqref{ellip} and
\begin{align}%
\label{a_est1}
	&A_{1\varepsilon} \langle x \rangle^{2+\alpha}
		\le A_{\varepsilon}(x)
		\le A_{2\varepsilon} \langle x \rangle^{2+\alpha},\\
\label{a_est2}
	&\frac{|\nabla A_{\varepsilon}(x)|^2}{a(x)A(x)}
		\le \frac{2+\alpha}{N+\alpha} + \varepsilon
\end{align}%
for some $A_{1\varepsilon}, A_{2\varepsilon} > 0$.
\begin{lemma}\label{lem_ep}
For every $\varepsilon \in (0,1)$,
there exists
$A_{\varepsilon} \in C^2(\mathbb{R}^N)$
such that
\eqref{ellip}--\eqref{a_est2}.
\end{lemma}
\begin{proof}
While the proof is the same as that of \cite[Lemma 2.1]{SoWa17},
we give a proof for the reader's convenience.

First, noting $a(x) \in C^2(\bar{\Omega})$ and $a(x) > 0$,
we extend $a(x)$ as a positive function on the whole space $\mathbb{R}^N$
and denote it by $a(x)$ again.
We define
\begin{align*}%
	a_{\varepsilon}(x) := b_1(x) + \eta_{\varepsilon} b_2(x),
\end{align*}%
where
\begin{align*}%
	b_1(x) &= \Delta \left( \frac{a_0}{(N+\alpha)(2+\alpha)}\langle x \rangle^{2+\alpha} \right) \\
	&= a_0 \langle x \rangle^{\alpha}
		- \frac{a_0 \alpha}{N+\alpha} \langle x \rangle^{\alpha-2},
\end{align*}%
$b_2(x) = a(x) - b_1(x)$,
and
$\eta_{\varepsilon} \in C_c^{\infty}(\mathbb{R}^N, [0,1])$
is a cut-off function
such that $\eta_{\varepsilon} \equiv 1$ on $B(0,R_{\varepsilon})$
with some $R_{\varepsilon}>0$.
By the assumption \eqref{a},
for any $\varepsilon >0$,
we obtain
\begin{align}%
\label{a_ep}
	(1-\varepsilon) a(x) \le a_{\varepsilon} (x) \le (1+\varepsilon) a(x),
\end{align}%
provided that $R_{\varepsilon}$ is sufficiently large.

We take $A_{\varepsilon}$ as an appropriate solution of the Poisson equation
\begin{align}%
\label{ellip2}
	\Delta A_{\varepsilon}(x) = a_{\varepsilon}(x),
\end{align}%
that is, we define
\begin{align*}%
	A_{\varepsilon}(x)
	&:=
	\frac{a_0}{(N+\alpha)(2+\alpha)}\langle x \rangle^{2+\alpha}
	- \int_{\mathbb{R}^N} \mathcal{N}(x-y) \eta_{\varepsilon}(y) b_2(y)\,dy
	+ \lambda_{\varepsilon},
\end{align*}%
where
$\lambda_{\varepsilon}$ is a large constant determined later,
and
$\mathcal{N}$ is the Newton potential given by
\begin{align*}%
	\mathcal{N}(x) :=
		\left\{ \begin{array}{ll}
		\displaystyle \frac{1}{2\pi} \log \frac{1}{|x|}
		&\mbox{if}\ N=2,\\[8pt]
		\displaystyle \frac{\Gamma(N/2+1)}{N(N-2)\pi^{N/2}}|x|^{2-N}
		&\mbox{if}\ N\ge 3.
		\end{array}\right.
\end{align*}%
Then, clearly we have \eqref{ellip2} and
the property \eqref{ellip} is verified.
Since the leading term of $A_{\varepsilon}(x)$ for large $|x|$ is
$\frac{a_0}{(N+\alpha)(2+\alpha)}\langle x \rangle^{2+\alpha}$,
we have \eqref{a_est1},
provided that $\lambda_{\varepsilon} >0$ is sufficiently large.
Moreover, we easily compute
\begin{align*}%
	&\lim_{|x|\to \infty}
	\left( \frac{|\nabla A_{\varepsilon}(x)|^2}{a(x) A_{\varepsilon}(x)} \right) \\
	&\quad =
	\lim_{|x|\to\infty}
	\left( \frac{\langle x \rangle^{\alpha}}{a(x)}\cdot
		\frac{1}{\langle x \rangle^{-\alpha-2}A_{\varepsilon}(x)}
		\left| \frac{a_0}{N+\alpha} \frac{x}{\langle x \rangle}
			+ o(1) \right|^2 \right) \\
	&\quad = \frac{2+\alpha}{N+\alpha}.
\end{align*}%
Thus, retaking $\lambda_{\varepsilon}>0$ sufficiently large,
we have \eqref{a_est2}.
\end{proof}

\section{Weighted energy estimates for damped wave equations}

In this section, we give weighted energy estimates
for solutions to the damped wave equation \eqref{dw}.
As we described in the introduction,
we use the auxiliary function
$A_{\varepsilon}$ constructed in Section 3.

We first recall the finite propagation speed property for
the equation \eqref{dw}.
For the proof, see for example
\cite{Ikawa}.
\begin{lemma}[Finite speed of the propagation]\label{lem_fps}
Let $u$ be a solution of the equation \eqref{dw} with initial data
$(u_0, u_1)$ satisfying
${\rm supp\,}(u_0, u_1) \subset \bar{\Omega} \cap B(0,R_0)$.
Then, we have
\begin{align*}%
	{\rm supp\,}u(\cdot,t) \subset
	\bar{\Omega} \cap B(0,R_0 +t)
\end{align*}%
for $t \ge 0$.
\end{lemma}
\begin{remark}
From Lemma \ref{lem_fps}, we see that
$\langle x\rangle \le R_0 +1+ t$
for $x \in {\rm supp\,}u(\cdot,t)$.
We shall frequently use this inequality.
\end{remark}

We start with the following weighted energy identities.
In this step we do not need any special property
for the weight function.
\begin{lemma}[{\cite[Lemma 3.7]{SoWa16}}]
\label{e0-0}
Let $\Phi\in C^2(\overline{\Omega}\times [0,\infty))$
satisfy $\Phi>0$ and $\partial_t\Phi<0$
and let $u$ be a solution of \eqref{dw}. Then, we have
\begin{align}
\label{en_eq1}
	&\frac{d}{dt}\left[ \int_{\Omega} \Big(|\nabla u|^2+|u_t|^2\Big) \Phi\,dx \right]\\
\nonumber
	&\quad = \int_{\Omega} (\partial_t\Phi)^{-1} \big| \partial_t\Phi\nabla u -u_t\nabla\Phi \big|^2 \,dx\\
\nonumber
	&\qquad + \int_{\Omega}
		\Big( -2a(x)\Phi+\partial_t\Phi - (\partial_t\Phi)^{-1}|\nabla\Phi|^2 \Big)|u_t|^2 \,dx.
\end{align}
\end{lemma

\begin{lemma}[{\cite[Lemma 3.9]{SoWa16}}]
\label{e1-0}
Let $\Phi\in C^2(\overline{\Omega}\times [0,\infty))$
satisfy $\Phi>0$ and $\partial_t\Phi<0$
and let $u$ be a solution to \eqref{dw}.
Then, we have
\begin{align}
\label{en_eq2}
	\frac{d}{dt} \left[ \int_{\Omega} \Big(2uu_t+a(x)|u|^2\Big) \Phi\,dx  \right]
	&= 2\int_{\Omega} uu_{t} (\partial_t\Phi)\,dx
	+ 2\int_{\Omega} |u_t|^2 \Phi\,dx
	- 2\int_{\Omega} |\nabla u|^2 \Phi\,dx \\
\nonumber
	&\quad + \int_{\Omega} \big(a(x)\partial_t\Phi+\Delta\Phi\big)|u|^2\,dx.
\end{align}
\end{lemma

Using the function $A_{\varepsilon}(x)$ constructed
in the previous section,
we introduce our weight function.
\begin{definition}
Let $h:=\frac{2+\alpha}{N+\alpha}$
and $\varepsilon\in (0,1)$.
We define
\begin{equation}
\Phi_{\varepsilon}(x,t)
=
\exp\left(
\frac{1}{h+2\varepsilon}\,\frac{A_\varepsilon(x)}{1+t}
\right),
\end{equation}
where $A_{\varepsilon}$ is given in Lemma \ref{lem_ep}.
For $t \ge 0$, we also define the following energy
\begin{gather}
\label{en_xta}
E_{\partial x}(t;u)
:=
\int_{\Omega}|\nabla u|^2\Phi_\varepsilon\,dx,
\quad
E_{\partial t}(t;u)
:=
\int_{\Omega}|u_t|^2\Phi_\varepsilon\,dx,
\\
\label{en_*A}
E_{a}(t;u)
:=
\int_{\Omega}a(x)|u|^2\Phi_\varepsilon\,dx,
\quad
E_{*}(t;u)
:=
2\int_{\Omega}uu_t\Phi_\varepsilon\,dx,
\end{gather}
and
$E_1(t;u):=E_{\partial x}(t;u)+E_{\partial t}(t;u)$
and $E_2(t;u):=E_{*}(t;u)+E_{a}(t;u)$.
\end{definition}

The main result of this section is the following.
We give weighted energy estimates for solutions
of \eqref{dw}.

\begin{proposition}\label{main}
Assume that $(u_0,u_1)$ satisfies
${\rm supp}\,(u_0,u_1)\subset \overline{B}(0,R_0)$
and the compatibility condition of
order $k_0\geq 1$.
Let $u$ be a solution of the problem \eqref{dw}.
For every $\delta>0$ and $0 \leq k\leq k_0-1$,
there exist $\varepsilon>0$ and $M_{\delta,k,R_0}>0$ such that
for every $t\geq 0$,
\begin{align*}
&(1+t)^{\frac{N+\alpha}{2+\alpha}+2k+1-\delta}
\Big(E_{\partial x}(t;\partial_t^{k}u)+E_{\partial t}(t;\partial_t^{k}u)\Big)
+
(1+t)^{\frac{N+\alpha}{2+\alpha}+2k-\delta}E_a(t;\partial_t^{k}u) \\
&\quad \leq M_{\delta,k,R_0}\|(u_0,u_1)\|_{H^{k+1}\times H^k(\Omega)}^2.
\end{align*}
\end{proposition}

To prove this,
we need to prepare the following lemmas.
First, we calculate derivatives of the weight function $\Phi_{\varepsilon}$.
\begin{lemma}\label{lem_phi}
We have
\begin{align*}%
	\partial_t \Phi_{\varepsilon}(x,t) &= -\frac{1}{h+2\varepsilon} \frac{A_{\varepsilon}(x)}{(1+t)^2} \Phi_{\varepsilon}(x,t),\\
	\nabla \Phi_{\varepsilon}(x,t) &=
		\frac{1}{h+2\varepsilon} \frac{\nabla A_{\varepsilon}(x)}{1+t} \Phi_{\varepsilon}(x,t),\\
	\Delta \Phi_{\varepsilon}(x,t) &=
		\frac{1}{h+2\varepsilon} \frac{\Delta A_{\varepsilon}(x)}{1+t} \Phi_{\varepsilon}(x,t)
			+ \left| \frac{1}{h+2\varepsilon} \frac{\nabla A_{\varepsilon}(x)}{1+t} \right|^2
			\Phi_{\varepsilon}(x,t).
\end{align*}%
In particular, we obtain
\begin{align*}%
	-\Delta \Phi_{\varepsilon}(x,t) + \frac{|\nabla \Phi_{\varepsilon}(x,t)|^2}{\Phi_{\varepsilon}(x,t)}
		&= - \frac{1}{h+2\varepsilon} \frac{\Delta A_{\varepsilon}(x)}{1+t} \Phi_{\varepsilon}(x,t),\\
	 \frac{|\nabla \Phi_{\varepsilon}(x,t)|^2}{\partial_t \Phi_{\varepsilon}(x,t)}
	 	&= - \frac{1}{h+2\varepsilon} \frac{|\nabla A_{\varepsilon}(x)|^2}{A_{\varepsilon}(x)} \Phi_{\varepsilon}(x,t).
\end{align*}%
\end{lemma}
The proof is straightforward and we omit it.
The second tool is a Hardy type inequality.
\begin{lemma
\label{lem_ha}
For $t\geq 0$, we have
\begin{align}
\label{hardy}
\frac{1-\varepsilon}{h+2\varepsilon}\,\frac{1}{1+t}
E_{a}(t;u)
	\leq  E_{\partial x}(t;u).
\end{align}
\end{lemma}
\begin{proof}
As in the proof of \cite[Lemma 3.6]{SoWa16},
by integration by parts we have
\begin{align*}
	\int_{\Omega}
	\Delta(\log\Phi_\varepsilon)|u|^2\Phi_{\varepsilon}\,dx
	&=
	\int_{\Omega}\left(\Delta\Phi_\varepsilon-\frac{|\nabla\Phi_\varepsilon|^2}{\Phi_\varepsilon}\right)|u|^2\,dx \\
	&= \int_{\Omega}|\nabla u|^2\,\Phi_\varepsilon\,dx
		- \int_{\Omega} \Phi_{\varepsilon}^{-1} | \nabla ( \Phi_{\varepsilon} u ) |^2\, dx\\
	&\leq
	\int_{\Omega}|\nabla u|^2\,\Phi_\varepsilon\,dx.
\end{align*}
Using Lemma \ref{lem_phi} and \eqref{ellip}, we see that
\[
\Delta(\log\Phi_\varepsilon(x))=\frac{1}{h+2\varepsilon}\,\frac{\Delta A_{\varepsilon}(x)}{1+t}
\geq
\frac{1-\varepsilon }{h+2\varepsilon}\,\frac{a(x)}{1+t},
\]
which leads to \eqref{hardy}.
\end{proof}

Next, in order to clarify the effect of the finite propagation speed property,
we put
\[
a_1:=\inf_{x\in \Omega}\Big(\lr{x}^{-\alpha} a(x)\Big).
\]
Then we have the following.
\begin{lemma}\label{lem_appfps}
For $t\geq 0$, we have
\begin{align}
\label{E12-F}
E_{\partial t}(t;u)
&\leq \frac{1}{a_1} E_{a}(t;\partial_t u),
\\
\label{A/a}
\int_{\Omega} \frac{A_\varepsilon(x)}{a(x)}|u_{t}|^2 \Phi_\varepsilon \,dx
&\leq \frac{A_{2\varepsilon}}{a_1}(R_0+1+t)^{2}
E_{\partial t}(t;u),
\\
\label{E21-F}
|E_{*}(t;u)|
&\leq
\frac{2}{\sqrt{a_1}}
\sqrt{E_a(t;u)E_{\partial t}(t;u)}.
\end{align}
\end{lemma}

\begin{proof}
By $a(x)/ a_1 \ge \lr{x}^{\alpha} \ge 1$,
we have
\begin{align*}
\int_{\Omega}|u_t|^2\Phi_\varepsilon\,dx
\leq \int_{\Omega}\frac{a(x)}{a_1}|u_t|^2\Phi_\varepsilon\,dx
\leq \frac{1}{a_1} E_a(t;\partial_t u),
\end{align*}
which shows \eqref{E12-F}.
By \eqref{a_est1},
$1/a(x) \leq (a_1 \lr{x}^{\alpha} )^{-1}$
and Lemma \ref{lem_fps}, we have
\begin{align*}%
	\frac{A_{\varepsilon}(x)}{a(x)}
	\le \frac{A_{2\varepsilon} \lr{x}^{2+\alpha}}{a_1 \lr{x}^{\alpha}}
	\le \frac{A_{2\varepsilon} }{a_1} (R+t+1)^2,
\end{align*}%
which implies \eqref{A/a}.
Finally, using the Cauchy-Schwarz inequality and
the inequality $a(x)/ a_1 \ge \lr{x}^{\alpha} \ge 1$,
we see that
\begin{align*}
\left|\int_{\Omega}uu_t\Phi_\varepsilon\,dx\right|^2
&\leq
\frac{1}{a_1}\left(\int_{\Omega}a(x)|u|^2\Phi_\varepsilon\,dx\right)
E_{\partial t}(t;u)
\\
&\leq
\frac{1}{a_1}
E_a(t;u)E_{\partial t}(t;u),
\end{align*}
which yields \eqref{E21-F}.
The proof of \eqref{E21-F} is similar and we omit it.
\end{proof}

Applying lemmas proved above,
we have the following differential inequality
for $E_1(t;u)$ and $E_2(t;u)$.
\begin{lemma}
{\bf (i)}\ For every $t\geq 0$, we have
\begin{equation}\label{e0-1}
    \frac{d}{dt}E_1(t;u)
	\leq
    -E_{a}(t;\partial_t u).

\end{equation}
{\bf (ii)}\ For every $\varepsilon\in (0,1/3)$ and $t\geq 0$, we have
\begin{align}
    \frac{d}{dt}E_{2}(t;u)
	&\leq
	-\frac{1-3\varepsilon}{1-\varepsilon}E_{\partial x}(t;u)
\label{e1-1}
	\\
\nonumber
	&\quad +
	\left(\frac{2}{a_1}+\frac{A_{2\varepsilon}(R_0+1)^2}{\varepsilon a_1^2}\right)E_{a}(t;\partial_t u).
\end{align}
\end{lemma}

\begin{proof}
To prove \eqref{e0-1}, we use Lemma \ref{e0-0}
and estimate the last term of \eqref{en_eq1}.
By virtue of Lemma \ref{lem_phi}, \eqref{a_est2} and that $A_{\varepsilon} \ge 0$,
we have
\begin{align*}
&-2a(x)\Phi_\varepsilon+\partial_t\Phi_\varepsilon - (\partial_t\Phi_\varepsilon)^{-1}
|\nabla \Phi_\varepsilon|^2\\
&\quad=
\left(-2a(x)-\frac{A_{\varepsilon}(x)}{(h+2\varepsilon)(1+t)^2}
+\frac{1}{h+2\varepsilon}\frac{|\nabla A_\varepsilon(x)|^2}{A_\varepsilon(x)}\right)\Phi_\varepsilon
\\
&\quad\leq
\left(
-2a(x)
+\frac{h+\varepsilon}{h+2\varepsilon}a(x)\right)\Phi_\varepsilon
\\
&\quad\leq
-
a(x)\Phi_\varepsilon,
\end{align*}
which implies \eqref{e0-1}.
Next, we use Lemma \ref{e1-0} to verify \eqref{e1-1}
and estimate the last term of \eqref{en_eq2}.
It follows from Lemma \ref{lem_phi} and \eqref{ellip} that
\begin{align*}
a(x)\partial_t\Phi_\varepsilon+\Delta\Phi_\varepsilon
&=
\frac{1}{h+2\varepsilon}
\left(-\frac{a(x)A_{\varepsilon}(x)}{(1+t)^2}
+\frac{|\nabla A_\varepsilon(x)|^2}{(h+2\varepsilon)(1+t)^2}
+\frac{\Delta A_{\varepsilon}(x)}{1+t}
\right)\Phi_\varepsilon
\\
&
\leq
\left(
-\frac{\varepsilon}{(h+2\varepsilon)^2}\,\frac{a(x)A_{\varepsilon}(x)}{(1+t)^2}
+\frac{1+\varepsilon}{h+2\varepsilon}\,\frac{a(x)}{1+t}
\right)
\Phi_\varepsilon.
\end{align*}
and hence, with Lemma \ref{lem_ha},
we have
\begin{align*}
&\int_{\Omega} \big(a(x)\partial_t\Phi_\varepsilon+\Delta\Phi_\varepsilon\big)|u|^2\,dx\\
&\quad \leq
\frac{1+\varepsilon}{1-\varepsilon}\int_{\Omega} |\nabla u|^2\Phi_\varepsilon\,dx
-\frac{\varepsilon}{(h+2\varepsilon)^2}\,
\frac{1}{(1+t)^2}
\int_{\Omega} a(x)A_\varepsilon(x)|u|^2\Phi_\varepsilon\,dx.
\end{align*}
By using \eqref{A/a} and \eqref{E12-F},
the first term of the right-hand side of \eqref{en_eq2}
is also estimated as
\begin{align*}
&2\int_{\Omega} uu_{t} (\partial_t\Phi_\varepsilon)\,dx \\
&\quad =
-\frac{2}{h+2\varepsilon}\frac{1}{(1+t)^2}\int_{\Omega} uu_{t} A_\varepsilon(x)\Phi_\varepsilon\,dx
\\
&\quad \leq
\frac{2}{h+2\varepsilon}\frac{1}{(1+t)^2}
\left(\int_{\Omega} a(x)A_\varepsilon(x)|u|^2\Phi_\varepsilon\,dx\right)^\frac{1}{2}
\left(\int_{\Omega} \frac{A_\varepsilon(x)}{a(x)}|u_{t}|^2 \Phi_\varepsilon \,dx\right)^\frac{1}{2}
\\
&\quad \leq
\frac{\varepsilon}{(h+2\varepsilon)^2}\,\frac{1}{(1+t)^2}
\int_{\Omega} a(x)A_\varepsilon(x)|u|^2\Phi_\varepsilon\,dx
+
\frac{A_{2\varepsilon}(R_0+1)^2}{\varepsilon a_1^2}
E_a(t; \partial_tu).
\end{align*}
Combining them with \eqref{en_eq2},
we reach \eqref{e1-1}.
\end{proof}

Furthermore,
multiplying
$E_1(t;u)$ and $E_2(t;u)$
by the time weight functions
$(t_1+t)^m$ and $(t_2 + t)^{\lambda}$,
respectively,
we have the following time-weighted differential inequalities.

\begin{lemma}\label{est}
The following assertions hold:

\smallskip

\noindent{\bf (i)} %
Set
$t_*(\alpha,m):= 2m/a_1$.
Then for every $t,m\geq 0$ and $t_1\geq t_*(\alpha,m)$,
\begin{align}
\label{e1-m}
\frac{d}{dt}
\Big((t_1+t)^{m}E_1(t;u)\Big)
\leq
m
(t_1+t)^{m-1}E_{\partial x}(t;u)
- \frac{1}{2}(t_1+t)^{m}E_a(t;\partial_t u).
\end{align}
\noindent{\bf (ii)} %
For every $t,\lambda\geq 0$ and $t_2\geq 1$,
\begin{align}
\label{e2-l}
&\frac{d}{dt}\Big((t_2+t)^{\lambda}E_2(t;u)\Big) \\
\nonumber
&\quad\leq
\lambda(1+\varepsilon)(t_2+t)^{\lambda-1}
E_a(t;u)
-\frac{1-3\varepsilon}{1-\varepsilon}
(t_2+t)^{\lambda}E_{\partial x}(t;u)
\\
\nonumber
&\qquad
+
\left(
\frac{2}{a_1}+\frac{A_{2\varepsilon}(R_0+1)^2}{\varepsilon a_1^2}+
\frac{\lambda}{2\varepsilon a_1^2 t_2}
\right)(t_2+t)^{\lambda}E_a(t;\partial_t u).
\end{align}
\noindent{\bf (iii)} %
In particular, setting
\begin{align*}
\nu
&:=
\frac{4}{a_1}+\frac{2A_{2\varepsilon}(R_0+1)^2}{\varepsilon a_1^2}+
\frac{1}{\varepsilon a_1},
\\
t_{**}(\varepsilon,\alpha,\lambda)
&:=
\max\left\{
	\frac{(1-\varepsilon) \lambda \nu}{\varepsilon},
	\lambda, 1, t_{\ast}(\alpha,m)
	\right\},
\end{align*}
one has that for $t,\lambda\geq 0$ and
$t_3\geq t_{**}(\varepsilon,\alpha,\lambda)$,
\begin{align}
\nonumber
&\frac{d}{dt}
\Big(
\nu (t_3+t)^{\lambda}E_1(t;u)
+(t_3+t)^{\lambda}E_2(t;u)
\Big)
\\
\label{e3-l}
&\quad \leq
-\frac{1-4\varepsilon}{1-\varepsilon}
(t_3+t)^{\lambda}E_{\partial x}(t;u)
+
\lambda(1+\varepsilon)(t_3+t)^{\lambda-1}
E_a(t;u).
\end{align}
\end{lemma}

\begin{proof}
{\bf (i)}
Let $m\geq 0$ and $t_1\geq t_{*}(\alpha,m)$.
Using \eqref{e0-1} and \eqref{E12-F}, we have
\begin{align*}
&(t_1+t)^{-m}
\frac{d}{dt}
\Big((t_1+t)^{m}E_1(t;u)\Big) \\
&\quad =
\frac{m}{t_1+t}
E_{\partial x}(t;u)
+
\frac{m}{t_1+t}
E_{\partial t}(t;u)
+\frac{d}{dt}E_1(t;u)
\\
&\quad \leq
\frac{m}{t_1+t}
E_{\partial x}(t;u)
+
\frac{m}{t_1+t}
E_{\partial t}(t;u)-E_a(t;\partial_t u)
\\
&\quad \leq
\frac{m}{t_1+t}
E_{\partial x}(t;u)
+
\left(\frac{m}{a_1(t_1+t)} - 1\right) E_a(t;\partial_t u).
\end{align*}
Since
$t_{\ast} = 2m/a_1$,
we see that
\begin{align*}%
	\frac{m}{a_1(t_1+t)} - 1 \le - \frac12
\end{align*}%
holds for $t_1 \ge t_{\ast}$ and $t\ge 0$.
This implies \eqref{e1-m}.

\noindent{\bf (ii)}
By using \eqref{e2-l} and \eqref{e1-1},
$t\geq 0$ and $t_2 \geq 1$, we have
\begin{align*}%
&(t_2+t)^{-\lambda}
\frac{d}{dt}\Big((t_2+t)^{\lambda}E_2(t;u)\Big)
\\
&\quad \leq
\frac{\lambda}{t_2+t}E_*(t;u)
+
\frac{\lambda}{t_2+t}E_a(t;u)+\frac{d}{dt}E_2(t;u)
\\
&\quad \leq
\frac{\lambda}{t_2+t}E_*(t;u)
+
\frac{\lambda}{t_2+t}E_a(t;u)
-\frac{1-3\varepsilon}{1-\varepsilon}
E_{\partial x}(t;u) \\
&\qquad +\left(\frac{2}{a_1}+\frac{A_{2\varepsilon}(R_0+1)^2}{\varepsilon a_1^2}\right)
E_a(t;\partial_t u).
\end{align*}%
Noting that \eqref{E21-F} and \eqref{E12-F} lead to
\begin{align*}%
\frac{\lambda}{t_2+t}E_{*}(t;u)
&\leq
\frac{2\lambda}{\sqrt{a_1}(t_2+t)}
\sqrt{ E_a(t;u) E_{\partial t}(t;u) }
\\
&\leq
\frac{2\lambda}{a_1(t_2+t)}
\sqrt{ E_a(t;u) E_{a}(t; \partial_t u) }
\\
&\leq
\frac{\lambda\varepsilon}{t_2+t}
E_a(t;u)
+
\frac{\lambda}{\varepsilon a_1^2 (t_2+t)}
E_a(t;\partial_t u),
\end{align*}%
we deduce \eqref{e2-l}.

\smallskip

\noindent{\bf (iii)}
Combining \eqref{e1-m} with $m=\lambda$
and \eqref{e2-l}, we have
for $t_3\geq t_{**}(\varepsilon,\alpha,\lambda)$ and $t\geq 0$,
\begin{align*}%
&\frac{d}{dt}
\Big(
\nu(t_3+t)^{\lambda}E_1(t;u)
+(t_3+t)^{\lambda}E_2(t;u)
\Big)
\\
&\leq
\left(
\frac{\nu \lambda}{ t_3+t }
-\frac{1-3\varepsilon}{1-\varepsilon}
\right)(t_3+t)^{\lambda}E_{\partial x}(t;u)
+\lambda(1+\varepsilon)(t_3+t)^{\lambda-1}
E_a(t;u)
\\
&\quad
+\left(
\frac{2}{a_1}+\frac{A_{2\varepsilon}(R_0+1)^2}{\varepsilon a_1^2}+
\frac{\lambda}{2\varepsilon a_1^2 t_3}
- \frac{\nu}{2}\right)
(t_3+t)^{\lambda}E_a(t;\partial_t u)
\\
&\leq
-
\frac{1-4\varepsilon}{1-\varepsilon}
(t_3+t)^{\lambda}E_{\partial x}(t;u)
+\lambda(1+\varepsilon)(t_3+t)^{\lambda-1}
E_a(t;u),
\end{align*}%
which gives the assertion.
\end{proof}

\begin{proof}[Proof of Proposition \ref{main}]
By \eqref{E21-F}, we easily have
\begin{align}
\label{ea}
\nu E_1(t;u)+E_2(t;u)&\geq \frac{3}{4}E_a(t;u).
\end{align}
By using the above estimate,
we prove the assertion via mathematical induction.

\noindent
{\bf Step 1 ($k=0$).}\
From \eqref{e3-l}
and Lemma \ref{lem_ha}, we deduce that
\begin{align*}
&\frac{d}{dt}
\Big(
\nu (t_3+t)^{\lambda}E_1(t;u)
+(t_3+t)^{\lambda}E_2(t;u)
\Big) \\
&\quad \leq
\left(
-\frac{1-4\varepsilon}{1-\varepsilon}
+
\frac{\lambda(1+\varepsilon)(h+2\varepsilon)}{1-\varepsilon}
\right)
(t_3+t)^{\lambda}E_{\partial x}(t;u).
\end{align*}
Taking
$\lambda_0=\frac{(1-\varepsilon)(1-4\varepsilon)}{(1+\varepsilon)(h+2\varepsilon)}$,
integrating over $(0,t)$ with respect to $t$
and using \eqref{ea},
we have
\begin{align*}
&\frac{3}{4}(t_3+t)^{\lambda_0}E_a(t;u)
+
\frac{\varepsilon(1-4\varepsilon)}{1-\varepsilon}
\int_{0}^t(t_3+s)^{\lambda_0}E_{\partial x}(s;u)\,ds
\\
&
\leq
\nu t_3^{\lambda_0}E_1(0;u)
+t_3^{\lambda_0}E_2(0;u).
\end{align*}
This gives the desired estimate for $E_a(t;u)$.
Also, as a byproduct, we obtain
\begin{align*}%
	\frac{\varepsilon(1-4\varepsilon)}{1-\varepsilon}
	\int_{0}^t(t_3+s)^{\lambda_0}E_{\partial x}(s;u)\,ds
	&\le \nu t_3^{\lambda_0}E_1(0;u)
+t_3^{\lambda_0}E_2(0;u).
\end{align*}%
Using this inequality and \eqref{e1-m} with $m=\lambda_0+1$,
and integrating over $(0,t)$, we obtain
\begin{align*}
&(t_3+t)^{\lambda_0+1}E_1(t;u)
+\frac{1}{2}
\int_{0}^t(t_3+s)^{\lambda_0+1}E_a(s;\partial_t u)\,ds
\\
&\quad \leq
t_3^{\lambda_0+1}E_1(0;u)
+
(\lambda_0+1)
\int_0^t
(t_3+s)^{\lambda_0}E_{\partial x}(s;u)\,ds
\\
&\quad \leq
t_3^{\lambda_0+1}E_1(0;u)+\frac{(\lambda_0+1)(1-\varepsilon)}
{\varepsilon(1-4\varepsilon)}
\Big(
\nu t_3^{\lambda_0}E_1(0;u)
+t_3^{\lambda_0}E_2(0;u)
\Big).
\end{align*}
Finally, putting $\delta = 1/h - \lambda_0$,
we have the desired assertion with $k=0$.
Moreover, we also have
\begin{align*}%
	&\frac{1}{2}
	\int_{0}^t(t_3+s)^{\lambda_0+1}E_a(s;\partial_t u)\,ds \\
	&\le t_3^{\lambda_0+1}E_1(0;u)+\frac{(\lambda_0+1)(1-\varepsilon)}
	{\varepsilon(1-4\varepsilon)}
	\Big(
	\nu t_3^{\lambda_0}E_1(0;u)
	+t_3^{\lambda_0}E_2(0;u)
	\Big),
\end{align*}%
which will be used in the next step.
\smallskip

\noindent{\bf Step 2 ($1<k\leq k_0-1$).}\
Suppose that
for every $t\geq 0$,
\[
(1+t)^{\lambda_0+2k-1}
E_{1}(t;\partial_t^{k-1}u)
+
(1+t)^{\lambda_0+2k-2}E_a(t;\partial_t^{k-1}u)
\leq M_{\varepsilon,k-1}\|(u_0,u_1)\|_{H^{k}\times H^{k-1}(\Omega)}^2
\]
and additionally,
\[
\int_{0}^t(1+s)^{\lambda_0+2k-1}E_a(s;\partial_t^{k}u)\,ds
\leq M_{\varepsilon, k-1}'\|(u_0,u_1)\|_{H^{k}\times H^{k-1}(\Omega)}^2
\]
are valid.
Since the initial value $(u_0,u_1)$
satisfies the compatibility condition
of order $k$,
$\partial_t^{k} u$ is also a solution of
\eqref{dw} with the initial data
$(\partial_t^{k}u, \partial_t^{k+1}u)(x,0) = (u_{k-1},u_{k})(x)$.
Applying \eqref{e3-l} with $\lambda=\lambda_0+2k$,
putting $t_{3k}=t_{**}(\varepsilon,R_0,\alpha,\lambda_0+2k)$
(see Lemma \ref{est}\ {\bf (iii)}),
integrating over $(0,t)$,
and noting \eqref{ea},
we have
\begin{align*}
&
\frac{3}{4}
(t_{3k}+t)^{\lambda_0+2k}E_a(t;\partial_t^{k} u)
+\frac{1-4\varepsilon}{1-\varepsilon}
\int_0^t(t_{3k}+s)^{\lambda_0+2k}E_{\partial x}(s;\partial_t^{k} u)\,ds
\\
&\quad \leq
\nu t_{3k}^{\lambda_0+2k}E_1(0;\partial_t^{k} u)
+t_{3k}^{\lambda_0+2k-1}E_2(0;\partial_t^{k} u) \\
&\qquad +
(\lambda_0+2k)(1+\varepsilon)
M_{\varepsilon, k-1}'\|(u_0,u_1)\|_{H^{k}\times H^{k-1}(\Omega)}^2.
\end{align*}
In particular, we obtain the boundedness of
the second term of the left-hand side.
From this and \eqref{e1-m} with $m=\lambda_0+2k+1$
we conclude
\begin{align*}
&
(t_{3k}+t)^{\lambda_0+2k+1}E_1(t;\partial_t^{k}u)
+
\frac{1}{2}
\int_0^t(t_{3k}+s)^{\lambda_0+2k+1}E_a(s;\partial_t^{k+1}u)
\,ds
\\
&\quad \leq M''_{\varepsilon,k}
\Big(
E_1(0;\partial_t^{k} u)+E_2(0;\partial_t^{k} u)+\|(u_0,u_1)\|_{H^{k}\times H^{k-1}(\Omega)}^2
\Big)
\end{align*}
with some constant $M''_{\varepsilon,k}>0$.
Therefore, by induction, we obtain the desired inequalities for all $k\leq k_0-1$.
\end{proof}

\section{Proof of the diffusion phenomena}
In this section, we give a proof of Theorem \ref{thm1}.
We use the following lemma stated in
\cite[Section 4]{SoWa16}.
\begin{lemma}\label{Duhamel}
Assume that
$
	(u_0,u_1) \in (H^2\cap H^1_0(\Omega))
		\times H^1_0(\Omega)
$
and
suppose that ${\rm supp}\,(u_0,u_1)\subset \{ x\in \Omega ; |x| \le R_0 \}$.
Then for every $t\geq 0$,
\begin{align}
\nonumber
	u(x,t) - e^{-tL_{\ast}}[ u_0 + a(\cdot)^{-1}u_1]
	 &= - \int_{t/2}^t e^{-(t-s)L_{\ast}}[ a(\cdot)^{-1}u_{tt}(\cdot, s) ] ds\\
\nonumber
	&\quad -e^{-\frac{t}{2}L_{\ast}} [ a(\cdot)^{-1}u_{t}(\cdot, t/2) ] \\
\label{eq_du2}
	&\quad +\int_0^{t/2} L_{\ast}  e^{-(t-s)L_{\ast}}[ a(\cdot)^{-1}u_{t}(\cdot, s) ] ds,
\end{align}
where
$L_{*}$
is the Friedrichs extension of
$L=-a(x)^{-1}\Delta$ in $L^2_{d\mu}$.
\end{lemma}

\begin{proof}[Proof of Theorem \ref{thm1}]
First we show the assertion for $(u_0,u_1)$
satisfying the compatibility condition of order $2$.
Taking $L^2_{d\mu}$-norm of both side, we have
\[
\Big\|u(x,\cdot) - e^{-tL_{\ast}}[ u_0 + a(\cdot)^{-1}u_1]\Big\|_{L^2_{d\mu}}
\leq
\mathcal{J}_1(t)
+\mathcal{J}_2(t)
+\mathcal{J}_3(t),
\]
where
\begin{align*}
\mathcal{J}_1(t)&:=
\int_{t/2}^t
\big\|e^{-(t-s)L_{\ast}}[ a(\cdot)^{-1}u_{tt}(\cdot, s)]\big\|_{L^2_{d\mu}} ds,
\\
\mathcal{J}_2(t)&:=\big\|e^{-\frac{t}{2}L_{\ast}} [ a(\cdot)^{-1}u_{t}(\cdot, t/2)]\big\|_{L^2_{d\mu}},
\\
\mathcal{J}_3(t)&:=\int_0^{t/2} \big\|L_{\ast}  e^{-(t-s)L_{\ast}}[ a(\cdot)^{-1}u_{t}(\cdot, s)]\big\|_{L^2_{d\mu}} ds.
\end{align*}
Noting that
$a(x) \ge a_1$ and $\Phi(x,t) \ge 1$,
and applying Proposition \ref{main} with $k=1$ and $k=0$
we see that
\begin{align*}
\big\|a(\cdot)^{-1}\partial_t^{k+1}u(\cdot, s)\big\|_{L^2_{d\mu}}^2
&=
  \int_{\Omega}a(x)^{-1}|\partial_t^{k+1}u(\cdot, s)|^2\,dx
\\
&\leq
	\frac{1}{a_1} \int_{\Omega}|\partial_t^{k+1}u(\cdot, s)|^2\Phi_\varepsilon\,dx
\\
&\leq
  \frac{1}{a_1} E_{\partial t}(t,\partial_t^{k}u)
\\
&\leq
	\frac{M_{\delta,k}}{a_1} (1+t)^{-\frac{N+\alpha}{2+\alpha}-2k-1+\delta}
  \|(u_0,u_1)\|_{H^{k+1}\times H^k}^2.
\end{align*}
Therefore, we have
\begin{align*}
\mathcal{J}_1(t)
&\leq
\int_{t/2}^t
\big\|a(\cdot)^{-1}u_{tt}(\cdot, s)\big\|_{L^2_{d\mu}} ds
\\
&\leq
  \sqrt{\frac{M_{\delta,1}}{a_1}}
  \|(u_0,u_1)\|_{H^{2}\times H^1}
  \int_{t/2}^t (1+s)^{-\frac{N+\alpha}{2(2+\alpha)}-\frac32+\frac{\delta}{2}} ds
\\
&\leq
  C_{1,\delta}
  (1+t)^{-\frac{N+\alpha}{2(2+\alpha)}-\frac12+\frac{\delta}{2}}
  \|(u_0,u_1)\|_{H^{2}\times H^1},
\end{align*}
where $C_{1,\delta}$ is a positive constant depending on $\delta$.
In the same way, we deduce
\begin{align*}
\mathcal{J}_2(t)
&\leq \big\|a(\cdot)^{-1}u_{t}(\cdot, t/2)\big\|_{L^2_{d\mu}}
\leq
  \sqrt{\frac{M_{\delta,0}}{a_1}}
  	(1+t)^{-\frac{N+\alpha}{2(2+\alpha)}-\frac12 + \frac{\delta}{2}}
\|(u_0,u_1)\|_{H^{1}\times L^2}.
\end{align*}
Next, we estimate the term $\mathcal{J}_3(t)$.
First, we treat the case $N = 2$.
In this case, by Proposition \ref{prop_lplq2},
we have
\begin{align*}
	\mathcal{J}_3(t)
	&\leq
	C\int_0^{t/2}
	(t-s)^{-\frac32}
	\big
	\|a(\cdot)^{-1}u_{t}(\cdot, s)\big\|_{L^1_{d\mu}} ds \\
	&\leq
	C t^{-\frac32}
	\int_0^{t/2}
	\sqrt{\|\Phi_\varepsilon^{-1}(\cdot,s)\|_{L^1(\Omega)}E_{\partial t}(s;u)}\,
	ds.
\end{align*}
We compute
\begin{align*}
  \|\Phi_{\varepsilon}^{-1}(\cdot,t)\|_{L^1(\Omega)}
  &\leq
  \int_{\mathbb{R}^2}
  	\exp\left(-\frac{A_{1\varepsilon}}{h+2\varepsilon}\,\frac{|x|^{2+\alpha}}{1+t} \right)\,dx
  \\
  &=
  (1+t)^{\frac{2}{2+\alpha}}
  \int_{\mathbb{R}^2}\exp\left(-\frac{A_{1\varepsilon}}{h+2\varepsilon}|y|^{2+\alpha}\right)\,dy.
\end{align*}
From this and Proposition \ref{main}, we conclude
\begin{align*}%
	\mathcal{J}_3(t)
	&\leq
	C t^{-\frac32}
	\int_0^{t/2}
		(1+s)^{\frac{1}{2+\alpha}-\frac{2+\alpha}{2(2+\alpha)}-\frac12+\frac{\delta}{2}} ds
		\cdot \| (u_0,u_1) \|_{H^1 \times L^2}\\
	&\leq C t^{-\frac{2+\alpha}{2(2+\alpha)} - \frac{1+\alpha}{2+\alpha} + \frac{\delta}{2}}
		\| (u_0,u_1) \|_{H^1 \times L^2}.
\end{align*}%

When $N\ge 3$, Proposition \ref{prop_lplq2} leads to
\begin{align*}
	\mathcal{J}_3(t)
	&\leq
	C\int_0^{t/2}
	(t-s)^{-\frac{N}{2}\left(\frac{1}{q}-\frac{1}{2}\right)-1}
		\left(
		\int_{\Omega} | a(x)^{-1}u_t(x,s)|^q |x|^{(N-2)\alpha (2-q)/4} d\mu(x)
		\right)^{1/q}
	ds \\
	&\leq
	C
	t^{-\frac{N}{2}\left(\frac{1}{q}-\frac{1}{2}\right)-1}
	\int_0^{t/2}
		\left(
		\int_{\Omega} |u_t(x,s)|^q a(x)^{-(q-1)} |x|^{(N-2)\alpha (2-q)/4} dx
		\right)^{1/q}
	ds,
\end{align*}
where
$q \in ((p_{\alpha})^{\prime}, 2]$
and sufficiently close to $(p_{\alpha})^{\prime}$.
Using $a(x) \ge a_1 \langle x \rangle^{\alpha}$
and the Cauchy--Schwarz inequality,
we further calculate
\begin{align*}%
	J_3(t)
	&\le C t^{-\frac{N}{2}\left(\frac{1}{q}-\frac{1}{2}\right)-1}
		\int_{0}^{t/2}
		\left\| u_t(s) |x|^{%
			\alpha \left[ \frac{N}{2} \left( \frac{1}{q} - \frac{1}{2} \right)-\frac12 \right]}
		\right\|_{L^q} ds\\
	&\le C t^{-\frac{N}{2}\left(\frac{1}{q}-\frac{1}{2}\right)-1}
		\int_{0}^{t/2}
		\left\| u_t(s) \Phi_{\varepsilon}^{1/2} \right\|_{L^2}
		\left\| \Phi_{\varepsilon}^{-1/2}  |x|^{%
			\alpha \left[ \frac{N}{2} \left( \frac{1}{q} - \frac{1}{2} \right)-\frac12 \right]}
			\right\|_{L^{2q/(2-q)}} ds.
\end{align*}%
We also notice that
if $q$ is close to $(p_{\alpha})^{\prime}$,
then
$\frac{N}{2} \left( \frac{1}{q} - \frac{1}{2} \right)-\frac12 > 0$
holds.
Therefore,
by changing variables, we estimate
\begin{align*}%
	\left\| \Phi_{\varepsilon}^{-1/2}  |x|^{%
			\alpha \left[ \frac{N}{2} \left( \frac{1}{q} - \frac{1}{2} \right)-\frac12 \right]}
			\right\|_{L^{2q/(2-q)}}
	&\le C (1+t)^{%
		\frac{\alpha}{2+\alpha}%
		\left[ \frac{N}{2} \left( \frac{1}{q} - \frac{1}{2} \right)-\frac12 \right]%
		+ \frac{N}{2+\alpha} \left( \frac{1}{q} - \frac{1}{2} \right)} \\
	&\le C (1+t)^{%
		\frac{N}{2}\left(\frac{1}{q}-\frac{1}{2}\right)-\frac{\alpha}{2(2+\alpha)}}.
\end{align*}%
From this estimate and Proposition \ref{main}, we obtain
\begin{align*}%
	J_3(t)&\le
	C(1+t)^{-\frac{N}{2}\left(\frac{1}{q}-\frac{1}{2}\right)-1}
		\int_{0}^{t/2} (1+s)^{%
			\frac{N}{2}\left(\frac{1}{q}-\frac{1}{2}\right)-\frac{\alpha}{2(2+\alpha)}}
		(1+s)^{-\frac{N+\alpha}{2(2+\alpha)}-\frac{1}{2}+\frac{\delta}{2}}
		ds\cdot
		\| (u_0,u_1) \|_{H^1 \times L^2}\\
	&\le C
	(1+t)^{-\frac{N}{2}\left(\frac{1}{q}-\frac{1}{2}\right)-1%
		+\max\left\{ %
			0, \frac{N}{2}\left(\frac{1}{q}-\frac{1}{2}\right)%
				-\frac{N+2\alpha}{2(2+\alpha)}-\frac{1}{2}+\frac{\delta}{2}%
		\right\} + \delta^{\prime}}
		\| (u_0,u_1) \|_{H^1 \times L^2}
\end{align*}%
with arbitrary small $\delta^{\prime} > 0$.
Taking $q$ sufficiently close to $(p_{\alpha})^{\prime}$,
we conclude
\begin{align*}%
	J_3(t)&\le
	C (1+t)^{-\frac{N+\alpha}{2(2+\alpha)}%
		- \frac{1+\alpha}{2+\alpha} + \delta^{\prime \prime}}
		\| (u_0,u_1) \|_{H^1 \times L^2},
\end{align*}%
where
$\delta^{\prime\prime}>0$
depends on
$\delta$, $\delta^{\prime}$
and it can be taken arbitrary small.

Finally, combining the estimates for
$\mathcal{J}_1(t), \mathcal{J}_2(t), \mathcal{J}_3(t)$
and noting
$1/2 < (1+\alpha)/(2+\alpha)$,
we have the desired estimate.

Next we show the assertion for $(u_0,u_1)$
satisfying $(u_0,u_1)\in (H^2\times H^1_0(\Omega))\times H^1_0(\Omega)$
(the compatibility condition of order $1$)
via an approximation argument.
Fix $\phi\in C_c^\infty(\mathbb{R}^N,[0,1])$ such that $\phi\equiv1$ on $\overline{B}(0,R_0)$ and $\phi\equiv0$ on $\mathbb{R}^N\setminus B(0,R_0+1)$ and define
for $n\in\mathbb{N}$,
\[
\left(\begin{array}{c}
u_{0n}
\\
u_{1n}
\end{array}\right)
=
\left(\begin{array}{c}
\phi \tilde{u}_{0n}
\\
\phi \tilde{u}_{1n}
\end{array}\right),
\quad
\left(\begin{array}{c}
\tilde{u}_{0n}
\\
\tilde{u}_{1n}
\end{array}\right)
=
\left(1+\frac{1}{n}\mathcal{A}\right)^{-1}
\left(\begin{array}{c}
u_0
\\
u_1
\end{array}\right),
\]
where $\mathcal{A}$ is an quasi-$m$-accretive operator
in $\mathcal{H}=H^1_0(\Omega)\times L^2(\Omega)$
associated with \eqref{dw}, that is,
\[
\mathcal{A}=
\left(\begin{array}{cc}
0 & -1
\\
-\Delta & a(x)
\end{array}\right)
\]
endowed with domain
$D(\mathcal{A})=(H^2\cap H^1_0(\Omega))\times H_0^1(\Omega)$.
Then $(u_{0n},u_{1n})$ satisfies ${\rm supp} (u_{0n},u_{1n})\subset \overline{B}(0,R_0+1)$
and the compatibility condition of order $2$.
Let $v_n$ be a solution of \eqref{dw} with
$(u_{0n},u_{1n})$.
Observe that
\begin{align*}
\|(u_{0n},u_{1n})\|_{H^2\times H^1}^2
&\leq C^2\|\phi\|_{W^{2,\infty}}^2
\|(\tilde{u}_{0},\tilde{u}_{1})\|_{H^2\times H_1}^2
\\
&\leq C'^2\|\phi\|_{W^{2,\infty}}^2
(\|(\tilde{u}_{0},\tilde{u}_{1})\|_{\mathcal{H}}^2
+\|\mathcal{A}(\tilde{u}_{0},\tilde{u}_{1})\|_{\mathcal{H}}^2)
\\
&\leq C'^2\|\phi\|_{W^{2,\infty}}^2
(\|(u_{0},u_1)\|_{\mathcal{H}}^2
+\|\mathcal{A}(u_{0},u_{1})\|_{\mathcal{H}}^2)
\\
&\leq C''^2\|\phi\|_{W^{2,\infty}}^2
\|(u_{0},u_{1})\|_{H^2\times H^1}^2
\end{align*}
with suitable constants $C$, $C'$, $C''>0$,
and
\begin{gather*}
\left(\begin{array}{c}
u_{0n}
\\
u_{1n}
\end{array}\right)
\to
\left(\begin{array}{c}
\phi u_0
\\
\phi u_1
\end{array}\right)
=
\left(\begin{array}{c}
u_0
\\
u_1
\end{array}\right)
\quad
\text{in}\ \mathcal{H}
\end{gather*}
as $n\to \infty$ and also $u_{0n}+a^{-1}u_{1n}\to u_0+a^{-1}u_1$
in $L^2_{d\mu}$ as $n\to \infty$.
Using the result of the previous step, we deduce
\[
\Big\|v_n(\cdot,t) - e^{tL_{\ast}}[ u_{0n} + a(\cdot)^{-1}u_{1n}]\Big\|_{L^2_{d\mu}}
\leq
\tilde{C}(1+t)^{-\frac{N+\alpha}{2(2+\alpha)}-\frac{1+\alpha}{2+\alpha}+\delta}
\|(u_0,u_1)\|_{H^{2}\times H^1}
\]
with some constant $\tilde{C}>0$.
Letting $n\to\infty$, by continuity of the $C_0$-semigroup $e^{-t\mathcal{A}}$
in $\mathcal{H}$
we also obtain diffusion phenomena
for initial data in $(H^2\cap H^1_0(\Omega))\cap H^1_0(\Omega)$.
\end{proof}

\section*{Acknowledgments}

This work is partially supported
by Grant-in-Aid for Young Scientists Research (B),
No.\ 16K17619
and Grant-in-Aid for Young Scientists Research (B),
No.\ 16K17625.

redpajama_set_name: RedPajamaArXiv


FuturMaster specialises in the consumer goods, food and beverage, cosmetics and pharmaceutical industries as well as having a strong presence in other business sectors.
FuturMaster is a leading vendor of supply chain software solutions enabling our customers to sustainably improve their supply chain performance. For over 20 years, FuturMaster has developed solutions to enable our customers to achieve a global and integrated approach to their supply chain. From pure demand and supply planning to the entire global supply chain, we deliver scalable, attainable and cost effective solutions.
Pros: This platform allows for forecasting using a variety of different statistical techniques. It also separates out forecasts based on baseline, promo, launch, and free of charge. This separation of types of forecasts is a great benefit for a demand planner, as it helps to easily identify the sum of the forecasts as opposed to just seeing a single consolidated number. In addition, this platform allows demand planners to manage their forecasts via exception rather than having to actively manage each SKU within their catalogue. Managing via exception is made possible through some robust Alert Reporting that is build into the software.
Cons: While the functionality and reporting capabilities within the tool are great, one major negative is the time it takes to become well versed on the platform. This is not a highly intuitive software, as it requires dozens of hours of training before gaining a basic level of comfort. In addition, the platform forces you to accurately maintain most relevant master data, as elements such as material status directly impact the forecasting abilities within the platform.
Overall: Overall, this platform has helped to notably improve overall forecast accuracy for the demand planning team, while reducing the corresponding forecast biases. Once the initial learning curve and the master data were thoroughly cleansed, the demand planning tool proved to be much more effective than the previous solution.


The Limestone Coast is a perfect summer road trip destination. A couple of hours drive from Adelaide and you'll find yourself surrounded by a treasure trove of natural, sandy wonders with some of the bluest waters you've seen.
For the past three years, Dion has been living and breathing South Australia's best beaches, sharing not only his own photos but the best beach snaps taken by local photographers and fellow instagrammers. To date, the #SouthAustralianBeaches hashtag has been used on nearly 70,000 posts, so we're confident these five best beaches on the Limestone Coast will satisfy your summer cravings.
Tell me more: Surf Beach, Beachport is the perfect beach to hone your surfing skills. With consistent swells hitting the shoreline and with the Beachport Jetty, neighbouring headlands and islands as a backdrop, it not only makes a surfer's paradise but a photographer's too!
Thank you to Dion from @southaustralianbeaches for sharing his five best beaches on the Limestone Coast. Grab the beers, deck chairs and floaties and we'll meet you at the beach!
If you need more inspiration or temptation, check out our 5 best beaches on the Yorke Peninsula, on the Eyre Peninsula, on Kangaroo Island, on the Fleurieu Peninsula and in 5 of the Best Beaches in Adelaide.


Over recent decades, many efforts to realise the potential of digital technology in education have made two key errors. Collectively, they have put the technology above teaching and excitement above evidence. This means they have spent more time, effort and money looking to find the digital silver bullet that will transform learning than they have into evolving teaching practice to make the most of technology....."


GIRLS ON TOUR is a series of 5 interlinked e-novellas about four friends who find fun, romance and adventure in glamorous locations. The adventures commenced in July with stop offs in September, December and February 2015. In April 2015 the whole series will be published in one volume in paperback and ebook.
Poppy does Paris is the first in the series and is available now.
Wow, I loved this little novella. It is fun, flirty and fabulous … great start to a mini series.
The second in the series Lily does LA is available tomorrow, 25th September.
I loved this second instalment of Nicola Doherty Girls on Tour series.
I love Lily, she is feisty, fun and sensitive at the same time. Lily is an out of work actress, she's been struggling to find her big break and has filled her time by doing random jobs to pay the bills. When Lily is offered the chance to go to LA to help her cousin organise her wedding out there she jumps at the chance. What follows is a fun adventure in LA for Lily were she discovers a new found confidence, new friends, new allies, new love.
Nicola Doherty's writing is fresh, funny, honest and thoroughly enjoyable. I've really enjoyed books 1 and 2 in the Girls on Tour series, I bet Nicola had great fun writing these novellas. The stories are stand alone but the girls are inter-linked. Nicola has done a fab map on her website which explains the connections between the girls.
If you fancy a fun, frivolous, romantic read the Girls on Tour series is just perfect. Grab yourself a ticket and join the adventures of Poppy, Lily, Maggie and Rachel.
So excited for the lovely author Lisa J Hobman, her Christmas novella, Christmas Presence is now available to pre-order on amazon. The novella is due to be released on 6th November 2014.
Blurb: Mallory got her happily ever after, but haunting loss in the past has her running scared that she will lose her beloved yet again. Can anything…or anyone make her realise she needs to let go of the past and live for today?
I'm looking forward to this novella, I loved Mallory's story from Bridge over the Atlantic and the companion book Bridge of Hope. When I got to the end of both of these books I felt sad that I'd finished reading about Mallory, she had become like a friend to me so this novella is much needed fuel for my craving to know what follows for Mallory and Greg. I feel that the best thing an author can do is make you fall in love with the book characters and Lisa did just this in the above two novels.
The bestselling love story behind the massive Hollywood film starring James Marsden and Michelle Monaghan.
They were teenage sweethearts from opposite sides of the tracks – with a passion that would change their lives for ever. But life would force them apart.
This is the new epic love story from the multi-million-copy bestselling author of The Notebook, The Lucky One and The Last Song. Nicholas Sparks is one of the world's most beloved authors.
With over eighty million copies of his books sold, Nicholas Sparks is one of the world's most beloved storytellers. His novels include eleven No. 1 New York Times bestsellers. All Nicholas Sparks' books have been international bestsellers and have been translated into more than forty languages. Eight of his novels have been adapted into major films: Safe Haven, The Lucky One, The Last Song, Dear John, Nights in Rodanthe, Message in a Bottle, A Walk to Remember and The Notebook. Nicholas Sparks lives in North Carolina with his wife and family. You can visit him and sign up for the Nicholas Sparks eNewsletter at http://www.uk.nicholassparks.com, and follow him on Twitter: @NicholasSparks.
I found this story a little slow to start but was glad I stuck with it. It had the usual Nicholas Sparks complicated love story. This one spanned over 20 years. Childhood sweethearts Dawson and Amanda have not seen each other for over 20 years as they both went their separate ways after high school. Amanda has seemingly moved on and is married with children. Dawson's life has had to endure events which made his life take a different course altogether however, these events and the love he had for Amanda back in high school has made it difficult for Dawson to move on.
A dear close friend from their high school days has sadly passed away and Dawson and Amanda are brought back together. How will they cope with long time forgotten feelings they once had for each other?
What follows is a beautiful tale of friendship and love. Drama also gets thrown at them and it becomes quite gripping at times which left me sobbing and screaming 'no'!
Bloodthirsty druids and battle-hardened Iron Age warriors collide in the first volume of this action-packed historical fantasy trilogy which will appeal to fans of Joe Abercrombie and HBO's Game of Thrones.
I have to admit Age of Iron isn't a book I would normally read so I asked a family member to read the book and review it for me and they were very keen to get 'stuck in'.
For a start I'd never heard of Angus Watson but reading the synopsis I was intrigued enough to start reading the book. What a great novel, the characters were easy to picture and understand due to a well written tale. I picked the story up quickly and became engrossed in the characters, the scenario, the setting and I struggled to put it down.


The shortness of life prevents us from entertaining far-off hopes. — Horace, Odes (1.4.15) 23 BC.
And finally, love is magic, as is hatred, too, imprinting as they do upon the brain the image of a being whom we allow to haunt us. — Marguerite Yourcenar, L'Œuvre au noir/The Abyss. (Assimil Gmbh; Presumed to be 1st as edition is unstated edition June 25, 1976) Originally published 1968.
How you die out in me: down to the last worn-out knot of breath you're there, with a splinter of life. ― Paul Celan, Poems of Paul Celan. (Anvil Press Poetry November 9, 1995) Originally published 1972.


You can add cars, but how do you delete them? It turns out that most of the code you need for simple deletion was created when you added the CarTableViewController class. To enable deletion, all you really need to do is uncomment and modify one of the provided methods.
Note that there is another related method, tableView:canEditRowAtIndexPath:, that allows you to optionally enable or disable editing for an individual cell. Although you do not use it in the CarValet app, you should know how to use it. Uncomment the method, and for now, have it return NO for any odd row.


text:
Q: How do I detect hover to elements inside of each() in jQuery? Thanks for taking a look at my question.
I'm trying to be able to hover over portfolio items but I need to loop through them using each() because I need some way of identifying each item.
I'm trying to hover over .recent-work-item to show .recent-work-item__overlay the .show-none class does display:none;
Neither the hover nor the on.("mouseenter", function(){}) is working.
Here is the HMTL:
<section class="recent-work-item" data-portfolio-id="rwi-<?php echo $i;?>">

<?php the_title(); ?>

VIEW CASE

  <img src="<?php echo get_template_directory_uri();?>/assets/img/work1.jpg" class="portrait">

Here is the jQuery:
$.each($('.recent-work-item'), function(){

var thisid = $(this).attr("data-portfolio-id");
console.log(thisid);

$("[data-portfolio-id="+"'"+thisid+"']").on('mouseenter', function(){
   $(thisid).find('.recent-work-item__overlay').removeClass('show-none');
 });

$("[data-portfolio-id="+"'"+thisid+"']").on('mouseleave',function(){
  $(thisid).find('.recent-work-item__overlay').addClass('show-none');
});

});

This is not working, I can't get the hover to work and all I want to do is add or remove a class, can I not do this in each().
I've researched thoroughly in StackOverflow but can't find an answer. I would REALLY appreciate any help I can get on this.

A: I have test your code in my codepen, and the problem you should use $(this) than use $(thisid)
$.each($('.recent-work-item'), function(){

  var thisid = $(this).attr("data-portfolio-id");

  $("[data-portfolio-id="+"'"+thisid+"']").on('mouseenter', function(){
    $(this).find('.recent-work-item__overlay').removeClass('show-none');
  });

  $("[data-portfolio-id="+"'"+thisid+"']").on('mouseleave',function(){
    $(this).find('.recent-work-item__overlay').addClass('show-none');
  });

});

Here look at my codepen

A: Here I have added an example that shows how you could use CSS to show/hide elements. It might not give you exact answer to your problem, but it will help you change your stylesheets as per your requirement.
Essentially, as per the discussion in comments, I don't think you need javascript to design the page the way you need it.

.container {
  padding: 10px;
  border: 1px solid gray;
}

.container > .hideOnHover {
  display: block;
}

.container > .showOnHover {
  display: none;
}

.container:hover > .hideOnHover {
  display: none;
}

.container:hover > .showOnHover {
  display: block;
}

    This text will be hidden on hover.

    This text will be shown on hover.

redpajama_set_name: RedPajamaStackExchange


text:
The National Cricket Centre is a cricket ground in Couva, Trinidad and Tobago, located near the Ato Boldon Stadium and the National Cycling Centre.

History
Construction of the National Cricket Centre began in 1995, with the purchase of  of land in Couva by the Trinidad and Tobago Cricket Board from Caroni (1975) Ltd., a former sugar producing company. Construction began in 1997 and took five years, with the centre opening in June 2002. Facilities at the centre included the development centre, administrative hub and indoor cricket nets. Trinidad and Tobago first played at the ground in a first-class match against West Indies B in the 2002–03 Carib Beer Cup, with ground being intermittently used for first-class fixtures by Trinidad until 2010. In March 2003, the centre played host to a Women's One Day International between West Indies women and Sri Lanka women. The centre was one of three venues selected in Trinidad as practice venues for the 2007 Cricket World Cup. In November 2007, plans were unveiled to install floodlights and to erect a multipurpose pavilion with a seating capacity of 3,000. Major cricket returned to the centre in 2015, following a five-year hiatus. In January 2015, the centre hosted its first List A one-day match between the Leeward Islands and the West Indies under-19 team in the Regional Super50, with the centre hosting a further neutral match in that seasons competition. A further two neutral one-day matches were played there in the Regional Super50. First-class cricket too returned to the centre, with Trinidad and Tobago playing a further five first-class matches there in 2015 and 2016. An access road at the centre was named in honour of Sonny Ramadhin, following his death in February 2022.

Records

First-class
Highest team total: 406 all out (twice)
By Trinidad and Tobago v West Indies B, 2002–03
By Trinidad and Tobago v Leeward Islands, 2015–16
Lowest team total: 125 all out by Trinidad and Tobago v Windward Islands, 2014–15
Highest individual innings: 223 by Aneil Kanhai for West Indies B v Trinidad and Tobago, 2002–03
Best bowling in an innings: 6-59 by Kenroy Peters for Windward Islands v Trinidad and Tobago, 2014–15
Best bowling in a match: 9-109 by Amit Jaggernauth for Trinidad and Tobago v Leeward Islands, 2006–07

List A
Highest team total: 257 for 9 (50 overs) by Jamaica v ICC Americas, 2015–16
Lowest team total: 137 all out (47 overs) by West Indies under-19's v Jamaica, 2014–15
Highest individual innings: 113 not out by Rahkeem Cornwall for Leeward Islands v West Indies under-19's, 2014–15
Best bowling in an innings: 3 for 26 by Jonathan Carter for Barbados v Jamaica, 2015–16

See also
List of cricket grounds in the West Indies

Notes

References

External links
National Cricket Centre at ESPNcricinfo

1997 establishments in Trinidad and Tobago
Cricket grounds in Trinidad and Tobago

redpajama_set_name: RedPajamaWikipedia


A layered frozen dessert of ice cream sandwiches, whipped topping, peanuts and peanut butter chocolate sauce.
Line the bottom of a 9x13 inch dish with 8 1/2 ice cream sandwiches. Spread half the whipped topping over the sandwiches. Spoon half the chocolate mixture over that. Top with half the peanuts. Repeat layers. Freeze until firm, 1 hour. Cut into squares to serve.


text:
Employees With Benefits! Whitney Wolfe Herd And Bumble Leading The Pack Of Companies Hooking Up Their Workers With Extra Time Off.
Looks like it's a date.
When analyzing whether employers are doing the right thing for their workers, who wouldn't swipe right when they see a company that provides an extra paid week off for its employees' peace of mind?
Bumble CEO Whitney Wolfe Herd has done that for 700 employees, setting them free for vacation as Americans deal with burnout.
No strings, either.
A representative for the dating app told CNBC that the move is on top of employees' normal vacation allowance.
In a statement Bumble said that like most people, "our global team has had a very challenging time during the pandemic. …
"As vaccination rates have increased and restriction have begun to ease, we wanted to give our teams around the world and opportunity to shut off and focus on themselves for a week."
Bumble's head of editorial content, Clare O'Connor, tweeted that Wolfe Herd had "correctly intuited our collective burnout….
"In the U.S. especially, where vacation days are notoriously scarce, it feels like a big deal."
That Monday tweet, though widely reported, is no longer available.
Many companies are experimenting with similar options, including four-day work weeks, to combat increased stress levels.
Social media management platform Hootsuite is giving workers a week next month, too, though it will come over most of the month to ensure its customers don't have to deal with service interruptions.
In April, LinkedIn delivered its nearly 16,000 workers a week off, citing employee surveys that showed "clear burnout."
"I think the reality of the weight of the pandemic really took its toll," Teuila Hanson, the chief people officer for LinkedIn, told CNN this spring. "What we think is most valuable right now is time for all of us to collectively walk away."
Previous Tales Of The Crypt! Bitcoin Takes A Big Dip And Other Cryptocurrency Values Doing The Same Thing.
Next PBD Calls Out Mr. Olympia's Weak Marketing Strategy

redpajama_set_name: RedPajamaCommonCrawl


text:
Contest Inspires Winner to Try Even Harder
Lillie-Beth Sanger
A Sequoyah Elementary School fourth-grader believes one can't accomplish much without getting an education first.
And Gerald S. Fleming II's essay on what having an education can do for a person won him a $100 U.S. Savings Bond and a camera in the Kodak/Alan Page Challenge essay contest.
Gerald's essay, selected from among 1,030 entries, was the Oklahoma City school district's winner in the nationwide contest.
"Without an education, things won't go right for many people," Gerald said, adding that many of his friends believe that philosophy. His advice to peers? "Stay in school. " Gerald's essay is in the form of a poem, a form of writing he enjoys, he said. And among his favorite subjects in school are reading, spelling and science.
"Without an education you can't accomplish many things in life.
I hope to live somewhere nice and I'd like to have lots of money," the 10-year-old said.
Gerald also said he would like to be a scientist because he likes doing experiments and research.
"I'm interested in facts about the Earth, and I'm interested in plants and land forms," Gerald said.
It is important to set goals as well, he said, adding that sometimes his goals are as simple as trying to get as much homework done as he can in one sitting. "Goals are a part of everybody's life every day. " Gerald was surprised at winning the national contest for his district because there were other winners in a local contest, said his mother, Barbara Fleming, but winning the contest has inspired him to try harder "to be a winner. " "I think it has motivated him to do more," said Barbara, who is a teacher at Sequoyah. "I've always wanted to instill in him that the opportunities are there but you have to ... have the desire to say, 'I want to do this. ' " Gerald will represent his school in the district's upcoming spelling bee.
Gerald, who describes himself as shy, also is a member of Cub Scouts, plays basketball and attends Greater Shiloh Baptist Church.
Here is Gerald's winning essay:
With an Education With an education, the future is yours. Without an education, the future is a bore.
If you don't get an education, you'll be on the street. If you don't get an education, you'll be a dead beat.
You know the rules, so stay in school. Getting an education is real cool.
If you make good grades and pass the test, you can be a teacher or a rocket scientist.
You can be a doctor, a nurse, or even president. Getting an education is time well spent.
You can own your own business, a house or a car. With an education, you can go mighty far.
It may take awhile, and sometimes seem rough, but getting an education is a must.
So remember what I've said and follow the rules. Get an education or you lose. BIOG: NAME: