Read the Disclaimer before reading this post.
Basic knowledge about the HTTP is required.
If you have some programming knowledge, it helps too, because we'll be looking at some code in this post.
Redirect means allowing a website to forward the request for the resources to another URL/endpoint. Let's assume that you make a request to apple.com and apple.com can redirect you to another website(new-apple.com), so you'll end up at new-apple.com even though the original request was made for apple.com. This is called "redirection". There are different types of redirects in HTTP, check em out below.
The redirection can happen on the server-side or the client side.
Server-Side: Request to redirect is sent to the server, then the server notifies the browser to redirect to the url specified via the response.
Client-Side: Browser is notified to redirect to the url specified directly without the intervention of the server.
Open redirect is basically what the name says, Openly allow Redirects to any website.
Well this is bad right off the bat, think about it for a moment, what if apple.com, a TRUSTED website allows you to redirect to any other website. Then a malicious user can simply redirect apple.com to attacker.com, and people fall for it all the time believing that it's trusted, but infact, it's not. So allowing redirects to any website without a stop in the middle or without a proper notification for the user is Bad.
This link is to a sigup page, once you signup, you get redirected to https://example.com/login which is specified in the HTTP GET Parameter redirectUrl.
What happens if we change the example.com/login to attacker.com?
By visiting this url, if we get redirected to attacker.com after the signup, this means we have an open redirect vulnerablility. This is a classic open redirect and often used for phishing(Refer Exploitation section for more info).
Here, the php code blindly grabs the url from redirect_url parameter and redirects to that url using the Location HTTP header.
Here, a jsp page takes the url from the parameter u and blindly redirects it to the specified url.
We can assign the URL string to the location.href of window's object. This will cause a redirect. If there are no checks inplace, then it's a bug.
HTML Meta tags can refresh the site with the given url as it's content and also you can specify the refresh delay time.
Visit every endpoint of the target to find these "redirect" parameters.
View your proxy history, you might find something. Make sure to use filters.
Understand and analyze where the redirection is needed in the target application like redirecting to dashboard after login or something like that.
Test for basic modification of the url like target.com/?redirect_url=https://attacker.com.
Try with double forward slashes target.com//attacker.com.
Try target.com/@attacker.com. In this case the interpretation will be like, the target.com is the username and attacker.com will be the domain.
Try target.com/?image_url=attacker.com/.jpg if there's an image resource being loaded.
Try IP address instead of the domain name.
You can go further in terms of representing the IP in decimal, hex or octal.
You can also try target.com/?redirect_url=target.com.attacker.com to bypass weak regex implementations.
Chinese seperator 。 as the dot - https://attacker%E3%80%82com.
Test for String reverser unicode("\u202e") target.com@%E2%80%AE@attacker.com.
Back slashes http:/\/\attacker.com or https:/\attacker.com.
Different domain redirect_url=.jp resulting in redirection of target.com.jp which is not the same as target.com.
Try some unicode(including emojis) madness t𝐀rget.com or 𝐀ttacker.com('𝐀' is "\uD835\uDC00").
There are a lot more, please refer the Cool Finds & Cheat Sheets section of this article. If you want more tricks, DISCOVER EM!
This redirects the user to an evil login page instead if the original one and the user can be phished.
Then we can use this with the SSRF by adding the open Open Redirect url from above as the get-image parameter.
Now this will make the server make a request to one of it's own subdomain(test.example.com) and then it is redirected to attacker.com making the SSRF bug work.
Like this Open Redirects can be used in many ways, I can't discuss all of em, it just depends on the target.
Only use redirects if you really want em.
If you want to use them, make sure you properly check the whitelisted domains and allow the matched ones.
You can also use hmac if you wanted to, but it's quite easy to f*ck this up with it(length extension attacks), so dont over complicate things, use proper whitelisting.
These are the list of all unique Open Redirect Reports on HackerOne that I could find from the top 20 pages of Google results.
This list might not be unique, download em all, clean duplicates and sort em.


South Carolinians have soured on Nikki Haley, turning the relatively new governor from a national tea party favorite into a chief executive struggling to maintain support among members of her own party, the latest Winthrop University poll shows.
Only 34.6 percent of those surveyed – 1,073 registered S.C. Democrats, Republicans and independents – said they approved of Haley's job performance, according to the poll. Far more – 43 percent – said they disapprove of the way the Republican is handling her job as governor. The poll's margin of error was plus or minus 2.9 percent.
Haley's approval rating is lower than that of President Obama, a Democrat, according to the poll. Obama has a 44.8 percent approval rating in strongly Republican South Carolina, according to the Winthrop poll.


text:
\section{Introduction}
\label{sec:intro}
The paper is dedicated to the investigation of the time-decay of the solution of the following initial-value problem
\begin{equation}\label{eq:wave_1D}
\begin{cases}
\beta\left(x\right)\partial_{t}^{2}u\left(x,t\right)-\partial_x\left(\alpha\left(x\right)\partial_x u\left(x,t\right)\right)=0, & x\in\mathbb{R},\:t>0,\\
u\left(x,0\right)=u_0\left(x\right),\hspace{1em}\partial_{t}u\left(x,0\right)=u_1\left(x\right), & x\in\mathbb{R},
\end{cases}
\end{equation}
where the coefficients $\alpha$, $\beta$ are {positive,}
{Lipschitz}{, and constant outside a bounded domain,}
and the initial data $u_0$, $u_1$ are compactly supported.
Studying such time-decay is important for practical applications
and it shall be used in our follow-up paper \cite{paper-LAP}.

Most relevant works, such as the recent
ones \cite{Shapiro} and \cite{Boucl-Burq}, are focussed on proving the
decay in time of the solution derivatives in $L^2$-norm taken over
bounded sets. Such local energy decay is usually obtained in spatial
dimensions $d\geq 2$. Another recent paper \cite{Charao} contains
a time-decay result in the one-dimensional case under certain restrictions on $\alpha$, $\beta$ and localisation of the initial data $u_0$, $u_1$.

It is the purpose of the present work to improve such $L^2$-estimates
of the solution derivatives by showing that, in case of compactly
supported initial data, the decay of the solution in time is
exponential rather than algebraic. We also prove
that, in this case, an auxiliary condition on $\alpha$, $\beta$ used in \cite{Charao} can be avoided.
Moreover, we generalise the result to prove the decay not only of derivatives but also of the solution function itself. Namely, we conclude that the solution converges exponentially fast in time to a constant, which is explicitly computable in terms of $\alpha$, $\beta$ and $u_1$.

Let us denote $\mathbb{R}_+:=\left\{x\in\mathbb{R}: x\geq 0\right\}$
and start with a well-posedness result for
problem \eqref{eq:wave_1D}.

\begin{prop}\label{prop:WP}
Let $\alpha\in W^{1,\infty}\left(\mathbb{R}\right)$ and $\beta\in L^{\infty}\left(\mathbb{R}\right)$
with $\alpha\left(x\right)\geq\alpha_{\min}>0$ and $\beta\left(x\right)\geq\beta_{\min}>0$
for $x\in\mathbb{R}$. Furthermore,
let $u_{0}\in H^{1}\left(\mathbb{R}\right)$
and $u_{1}\in L^{2}\left(\mathbb{R}\right)$. Then, problem \eqref{eq:wave_1D} has
a unique solution $u\in C^{1}\left(\mathbb{R}_{+},L^{2}\left(\mathbb{R}\right)\right)\cap C\left(\mathbb{R}_{+},H^{1}\left(\mathbb{R}\right)\right)$, with the continuous dependence on the initial data, according to the following estimate
\begin{equation}\label{eq:WP_estim}
\left\Vert u\left(\cdot,t\right)\right\Vert _{H^{1}\left(\mathbb{R}\right)}+\left\Vert \partial_{t}u\left(\cdot,t\right)\right\Vert _{L^{2}\left(\mathbb{R}\right)}\leq C\left(\left\Vert u_{0}\right\Vert _{H^{1}\left(\mathbb{R}\right)}+\left\Vert u_{1}\right\Vert _{L^{2}\left(\mathbb{R}\right)}\right),\hspace{1em} 0\leq t\leq T,
\end{equation}
valid for any $T>0$, with some constant $C=C\left(T,\alpha,\beta\right)>0$.

\noindent Moreover, if $u_{0}\in H^{2}\left(\mathbb{R}\right)$ and $u_{1}\in H^{1}\left(\mathbb{R}\right)$,
then we have $u\in C^{2}\left(\mathbb{R}_{+},L^{2}\left(\mathbb{R}\right)\right)\cap C^{1}\left(\mathbb{R}_{+},H^{1}\left(\mathbb{R}\right)\right)\cap C\left(\mathbb{R}_{+},H^{2}\left(\mathbb{R}\right)\right)$.
\end{prop}

{For the long-time analysis}, we shall now make more restrictive assumptions.

\begin{assm}\label{assm:alph_bet}
Let $\alpha$, $\beta\in W^{1,\infty}\left(\mathbb{R}\right)$
be real-valued functions such that $\alpha(x)\geq\alpha_{\min}$, $\beta(x)\geq\beta_{\min}$ for $x\in\mathbb{R}$,  and $\alpha(x)\equiv\alpha_0$, $\beta(x)\equiv\beta_0$
for $x\in\mathbb{R}\backslash\Omega_{in}$, with some
open bounded interval
$\Omega_{in}\subset\mathbb{R}$ and constants $\alpha_{\min}$, $\beta_{\min}$, $\alpha_0$, $\beta_0>0$.
\end{assm}

\begin{assm}\label{assm:ICs}
Suppose that $u_{0}\in H^{1}\left(\mathbb{R}\right)$, $u_{1}\in
L^{2}\left(\mathbb{R}\right)$ are real-valued functions with the
supports $\mathrm{supp}\, u_{0}$, $\mathrm{supp}\, u_{1}\subset\Omega_0$ for some
open bounded interval
$\Omega_0\subset\mathbb{R}$.
\end{assm}

Our main results are given by the following two theorems.
\begin{thm}\label{thm:1D_decay}
Assume that $\alpha$, $\beta$, $\Omega_{in}$ are as in Assumption \ref{assm:alph_bet}, and $u_0$, $u_1$, $\Omega_0$  are as in Assumption \ref{assm:ICs}.
Then, for any
open bounded interval
$\Omega\subset\mathbb{R}$, the solution of \eqref{eq:wave_1D} obeys the decay estimate
	\begin{equation}
	\left\Vert u\left(\cdot,t\right)-u_{\infty}\right\Vert _{H^{1}\left(\Omega\right)}+\left\Vert \partial_{t}u\left(\cdot,t\right)\right\Vert _{L^{2}\left(\Omega\right)}\leq Ce^{-\Lambda t},\hspace{1em}t\geq 0,\label{eq:u_exp_conv}
	\end{equation}
	for some constant $C=C\left(u_{0},u_{1},\alpha,\beta,\left|\Omega_0\right|,\left|\Omega_{in}\right|\right){>0}$,
with $\left|\Omega_0\right|$, $\left|\Omega_{in}\right|$ denoting the Lebesgue measure of the sets $\Omega_0$, $\Omega_{in}$, respectively, and
	\begin{equation}
	u_{\infty}:=\frac{1}{2\left(\alpha_0\beta_0\right)^{1/2}}\int_{\Omega_0}u_{1}\left(x\right)\beta\left(x\right)dx,\hspace{1em}
	 \Lambda:=\frac{\gamma_{0}}{\sqrt2 \, t_0}e^{-2\gamma_{0}}\left(1-4\gamma_{0}e^{-4\gamma_{0}}\right)^{1/2},   \label{eq:u_infty_1D}
	\end{equation}
	\begin{equation}
\gamma_0:=\frac{t_0}{4}\left\Vert \frac{\alpha^\prime}{\left(\alpha\beta\right)^{1/2}}+\left(\frac{\alpha}{\beta^3}\right)^{1/2}\beta^\prime\right\Vert_{L^\infty\left(\Omega_{in}\right)},\hspace{1em} t_0:=\int_{\Omega_{in}}\left(\frac{\beta\left(x\right)}{\alpha\left(x\right)}\right)^{1/2}dx.\label{eq:gam0_t0}
	\end{equation}
\end{thm}

We remark that the steady state $u_{\infty}$ given in
\eqref{eq:u_infty_1D} generalises the one resulting from the d'Alembert
formula for constant $\alpha$ and $\beta$, and for $u_0$, $u_1$ with compact
support:
$$
u_\infty = \frac12 \sqrt{\frac{\beta_0}{\alpha_0}}
\int_{\mathbb R}
u_1(x)\, dx.
$$

Note that even though the decay \eqref{eq:u_exp_conv} is exponential, the decay rate $\Lambda$
might be be rather small in case of significantly varying coefficients $\alpha$, $\beta$ or large $\Omega_{in}$. In particular, the estimate for the decay rate given by the second equation of \eqref{eq:u_infty_1D} is exponentially small for large values of $\gamma_0$.
Therefore, an alternative estimate is provided by the following
theorem, which employs a different idea and yields a better decay
rate, but in return requires an
assumption on the variation of~$\alpha$ and~$\beta$.

\begin{thm}\label{thm:better_rate}
Assume that $\alpha$, $\beta$, $\Omega_{in}$ are as in Assumption
\ref{assm:alph_bet}, $u_0$, $u_1$, $\Omega_0$  are as in Assumption
\ref{assm:ICs}, and $t_0$ is
  defined in \eqref{eq:gam0_t0}.
Moreover, suppose
that
\begin{equation}\label{eq:b0_cond}
b_0:=\left\Vert \alpha\right\Vert _{L^{\infty}\left(\Omega_{in}\right)}^{3/2}\left\Vert \beta\right\Vert _{L^{\infty}\left(\Omega_{in}\right)}^{1/2}\left\Vert \left(\frac{1}{\left(\alpha\beta\right)^{1/2}}\right)^{\prime}\right\Vert _{L^{2}\left(\Omega_{in}\right)}^{2}<\frac{1}{t_0}.
\end{equation}
Then, for the solution of \eqref{eq:wave_1D}, estimate \eqref{eq:u_exp_conv} holds true,
and the decay rate is given by
\begin{equation}\label{Lambda}
\Lambda=\frac{1}{2t_0}\left|\log\left(b_{0}t_0\right)\right|.
\end{equation}
\end{thm}

\begin{rem}
Theorems~\ref{thm:1D_decay} and~\ref{thm:better_rate} are valid
  also for complex-valued $u_{0}$, $u_{1}$ in Assumption~\ref{assm:ICs}.
Indeed, due to the linearity of problem \eqref{eq:wave_1D} and
the real-valuedness of $\alpha$, $\beta$, both theorems could be
applied separately to the real and imaginary parts of the initial data.
\end{rem}

Theorems~\ref{thm:1D_decay} and~\ref{thm:better_rate} are to be compared with the following recent result from~\cite{Charao}.

\begin{prop}{\cite[Thm. 1.2 (for $d=1$)]{Charao}}
\label{prop:Charao}
Assume that $\alpha\left(x\right)\equiv
\alpha_0$ and that $\beta$ satisfies Assumption \ref{assm:alph_bet}. Moreover, suppose that $\beta$ is such that
\begin{equation}\label{eq:eta_cond}
\eta:=\left\Vert \beta\right\Vert _{L^{\infty}\left(\Omega_{in}\right)}^{1/2}\left\Vert \left(\frac{1}{\beta^{1/2}}\right)^\prime\right\Vert _{L^{\infty}\left(\Omega_{in}\right)}{\left|\Omega_{in}\right|}<1.
\end{equation}
Then, for any bounded $\Omega\subset\mathbb{R}$, the solution of \eqref{eq:wave_1D} satisfies
\begin{equation}\label{eq:Charao}
\left\Vert \partial_x u\left(\cdot,t\right)\right\Vert _{L^{2}\left(\Omega\right)}+\left\Vert \partial_{t}u\left(\cdot,t\right)\right\Vert _{L^{2}\left(\Omega\right)}\leq\frac{C}{t^{\left(1-\eta\right)/2}},\hspace{1em}t\geq t_1,
\end{equation}
for some constant $C>0$ and sufficiently large $t_1>0$.
\end{prop}

Note that, when the initial data are compactly supported, the
algebraic decay in \eqref{eq:Charao}  appears to be sub-optimal compared to both
Theorem~\ref{thm:1D_decay} and Theorem~\ref{thm:better_rate}, which
show the exponential decay of the solution derivatives. As {observed}
in \cite[Rem.~1.2]{Charao}, the smallness assumption
\eqref{eq:eta_cond} on $\eta$ measuring the relative perturbation of
the coefficients
is essential in the proof there,
but it is not needed in our proof of Theorem~\ref{thm:1D_decay}.

For the wave equation \eqref{eq:wave_1D} with \emph{constant}
coefficients and compactly supported initial data, the solution
converges on all bounded domains $\Omega$ to the constant $u_\infty$
in \emph{finite time}. However, for variable coefficients, convergence in finite time does not hold and exponential
decay is the generic scenario: as an example, consider
\eqref{eq:wave_1D} with $\alpha(x)\equiv \alpha_0$ in
$\mathbb{R}$ and $\alpha$ Lipschitz with $\beta(x)=\beta_1$ on
a closed subinterval of some open set $\Omega_{in}$, and $\beta(x)=\beta_0$ in
$\mathbb{R}\setminus\Omega_{in}$. A localised, travelling wave packet starting within $\Omega_{in}$ gets partly reflected and partly transmitted. Hence, its local norm $\left\Vert u\left(\cdot,t\right)-u_{\infty}\right\Vert _{H^{1}\left(\Omega_{in}\right)}$ decays by a constant factor at each of these reflections, giving rise to an exponential decay.
On the other hand, we note that, for non-compactly supported initial data, the decay of the derivatives of the solution in time is generally related to the decay of the initial data at infinity. This can be easily seen already for the constant-coefficient case: the solution furnished explicitly by the d'Alembert formula decays to some $u_\infty$ only algebraically fast if the assumed decay at infinity of $u_0$ or $u_1$ is algebraic.

The following Sections~\ref{sec:1D_WP},~\ref{sec:1D_decay_proof},
  and~\ref{sec:better_rate_proof} are devoted to the proof of
  Proposition \ref{prop:WP}, Theorem~\ref{thm:1D_decay}, and Theorem~\ref{thm:better_rate},
  respectively. Some conclusions are drawn in Section~\ref{sec:conc}, and
  technical lemmas used in the proof of Theorem~\ref{thm:1D_decay} are
  deferred to the Appendix.

\section{Proof of Proposition \ref{prop:WP}}\label{sec:1D_WP}
One standard approach {to prove}
the well-posedness of problem \eqref{eq:wave_1D} is
to construct the solution $u$, for coefficients $\alpha$ and $\beta$
sufficiently regular, by a Galerkin approximation (e.g.\ see \cite[Sect. 7.2]{Evans}, or \cite[Par. 11.2]{Ren-Rog}). Here, we shall, however,
use semigroup theory tools, since they require lower regularity assumptions
on $\alpha$ and $\beta$. {Although}
this approach is quite standard (e.g.
see \cite[Par. 2]{Reed},  \cite[Par. 3]{Strauss}, or
\cite[Sect. 12.3.2]{Ren-Rog}), we
still outline
it, adapting to our particular setting.

With the notation
\[
\phi\left(x,t\right):=\left(u\left(x,t\right),v\left(x,t\right)\right)^{T},\hspace{1em}\hspace{1em}v\left(x,t\right):=\partial_{t}u\left(x,t\right),
\]
\[
A=A\left(x\right):=i\left(\begin{array}{cc}
0 & 1\\
\frac{1}{\beta\left(x\right)}\partial_{x}\left(\alpha\left(x\right){\partial_{x}\,\cdot}\,\right)-\frac{1}{\beta\left(x\right)} & 0
\end{array}\right),\hspace{1em}\hspace{1em}C=C\left(x\right):=\left(\begin{array}{cc}
0 & 0\\
\frac{1}{\beta\left(x\right)} & 0
\end{array}\right),
\]
problem \eqref{eq:wave_1D} is equivalent to the first order linear system
\begin{equation}\label{eq:1storder}
\begin{cases}
  \partial_{t}\phi\left(x,t\right)={\left(-iA \left(x\right)+C \left(x\right)\right)}\phi\left(x,t\right),
  & x\in\mathbb{R},\hspace{1em}t>0,\\
\phi\left(x,0\right)=\phi_{0}\left(x\right):=\left(u_{0}\left(x\right),u_{1}\left(x\right)\right)^{T}, & x\in\mathbb{R}.
\end{cases}
\end{equation}
We split the proof into 2 steps.

\subsection*{Step 1. Functional setting.}
We {set
$L^{2}_{\beta}\left(\mathbb{R}\right):=L^{2}\left(\mathbb{R};\beta\left(x\right)dx\right)$}. Due to the assumption $\beta$, $1/\beta\in
L^{\infty}\left(\mathbb{R}\right)$, {the space} ${L^{2}_{\beta}\left(\mathbb{R}\right)}$
actually consists of the same functions as~$L^{2}\left(\mathbb{R}\right)$.
We define the operator $B^{2}:=-\frac{1}{\beta\left(x\right)}\partial_{x}\left(\alpha\left(x\right){\partial_{x}\,\cdot \,}\right)+\frac{1}{\beta\left(x\right)}$
in ${L^{2}_{\beta}\left(\mathbb{R}\right)}$.
Since $\alpha$, $\beta$ are assumed to be positive and bounded away from zero, the operator $B^2$ is strictly positive.
We claim that this operator is self-adjoint with its (maximal) domain
being $\Dom B^{2}=H^{2}\left(\mathbb{R}\right)$. Indeed, the
symmetry of the operator $B^{2}$ is evident, and {$w\in
  H^2\left(\mathbb{R}\right)$ implies $B^2w\in L^{2}_{\beta}\left(\mathbb{R}\right)$.}
By \cite[Lem 2.3]{Teschl}, it
remains to verify its surjectivity on ${L^{2}_{\beta}\left(\mathbb{R}\right)}$
or, equivalently, on
$L^{2}\left(\mathbb{R}\right)$.
The latter follows
from the fact that, for $\alpha\in W^{1,\infty}\left(\mathbb{R}\right)$,
the operator $\widetilde{B}^{2}:=-\partial_{x}\left(\alpha\left(x\right){\partial_{x}\,\cdot \,}\right)+1$
from $H^{2}\left(\mathbb{R}\right)$ into $L^{2}\left(\mathbb{R}\right)$
is strictly positive, and hence invertible.

Next, we define in ${L^{2}_{\beta}\left(\mathbb{R}\right)}$
the positive square
root $B:=\sqrt{B^{2}}$, whose
(maximal) domain is $\Dom B=H^{1}\left(\mathbb{R}\right)$. In fact,
for any $w\in H^{1}\left(\mathbb{R}\right)$, we have
\begin{align*}
\left\Vert Bw\right\Vert _{{L^{2}_{\beta}\left(\mathbb{R}\right)}}^{2} & =\left\langle Bw,Bw\right\rangle _{{L^{2}_{\beta}\left(\mathbb{R}\right)}}=\int_{\mathbb{R}}\left(B^{2}w\right)\left(x\right)w\left(x\right)\beta\left(x\right)dx\\
 & =\int_{\mathbb{R}}\left[\alpha\left(x\right)\left(w^{\prime}\left(x\right)\right)^{2}+w^{2}\left(x\right)\right]dx,
\end{align*}
which, due to positivity and boundedness of $\alpha$, shows the
equivalence {between $\left\Vert B\,\cdot\,\right\Vert _{{L^{2}_{\beta}\left(\mathbb{R}\right)}}$ and
$\left\Vert \,\cdot\,\right\Vert _{H^{1}\left(\mathbb{R}\right)}$}.

With this preparation, we can introduce the Hilbert space $X:=H^{1}\left(\mathbb{R}\right)\times L^{2}\left(\mathbb{R}\right)$
equipped with the
{inner product $\langle \phi,\widetilde \phi\rangle_X:=
  \langle Bu,B\widetilde
  u\rangle_{L^{2}_{\beta}\left(\mathbb{R}\right)}+
  \langle v,\widetilde
  v\rangle_{L^{2}_{\beta}\left(\mathbb{R}\right)}$}.
Note that we have $\phi_{0}\in X$.

\subsection*{Step 2. Evolution semigroup.}

Observe that the matrix operator $A$ is symmetric in
$X$, since for any $\phi$, $\widetilde{\phi}\in H^{2}\left(\mathbb{R}\right)\times H^{1}\left(\mathbb{R}\right)$, we have
\begin{align*}
\left\langle A\phi,\widetilde{\phi}\right\rangle _{X} & =i\left\langle Bv,B\widetilde{u}\right\rangle _{{L^{2}_{{\beta}}\left(\mathbb{R}\right)}}+i\left\langle -B^{2}u,\widetilde{v}\right\rangle _{{L^{2}_{{\beta}}\left(\mathbb{R}\right)}}\\
&=i\left\langle v,B^{2}\widetilde{u}\right\rangle_{{L^{2}_{{\beta}}\left(\mathbb{R}\right)}}-i\left\langle Bu,B\tilde{v}\right\rangle _{{L^{2}_{{\beta}}\left(\mathbb{R}\right)}}
 =\left\langle \phi,A\widetilde{\phi}\right\rangle _{X}.
\end{align*}
In fact, $A$ is self-adjoint in $X$ with (maximal) domain $\Dom A=H^{2}\left(\mathbb{R}\right)\times H^{1}\left(\mathbb{R}\right)$.
The latter statement follows from the maximality of $\Dom B^{2}=H^{2}\left(\mathbb{R}\right)$.

By Stone's theorem \cite[Ch.\ 1 Thm.\ 10.8]{Pazy}, the operator
$-iA$ generates a $C_{0}$-semigroup of unitary operators on $X$,
and the operator $C$ is a bounded perturbation. Therefore, using
results on the perturbation of semigroups \cite[Sect.\ 3.1 Thm.\ 1.1]{Pazy},
the operator $-iA+C${, which defines problem~\eqref{eq:1storder},} generates a $C_{0}$-semigroup of bounded
operators on $X$.
Consequently, we have the following two solution concepts {for
  problem~\eqref{eq:1storder}, and thus for problem~\eqref{eq:wave_1D}}, depending
on the regularity of $\phi_{0}=\left(u_{0},u_{1}\right)^{T}$ (see
\cite[p.\ 105]{Pazy} for this discussion):
\begin{itemize}
\item[ \emph{i)}] For $\phi_{0}\in X$,
problem~{\eqref{eq:1storder}} has a unique mild solution, i.e. $\phi\in C\left(\mathbb{R}_{+},X\right)$,
and hence
$u\in C\left(\mathbb{R}_{+},H^{1}\left(\mathbb{R}\right)\right)\cap C^{1}\left(\mathbb{R}_{+},L^{2}\left(\mathbb{R}\right)\right)$,
$v=\partial_{t}u\in C\left(\mathbb{R}_{+},L^{2}\left(\mathbb{R}\right)\right)$
\item[\emph{ii)}] For $\phi_{0}\in\text{Dom }A,$
  problem~{\eqref{eq:1storder}} has a unique classical
solution (in the semigroup sense), i.e. $\phi\in C^{1}\left(\mathbb{R}_{+},X\right)\cap C\left(\mathbb{R}_{+},\text{Dom }A\right)$,
and hence $u\in C\left(\mathbb{R}_{+},H^{2}\left(\mathbb{R}\right)\right)\cap C^{1}\left(\mathbb{R}_{+},H^{1}\left(\mathbb{R}\right)\right)$,
$v=\partial_{t}u\in C^{1}\left(\mathbb{R}_{+},L^{2}\left(\mathbb{R}\right)\right)$.
\end{itemize}
Finally, estimate \eqref{eq:WP_estim} is an automatic consequence of the semigroup approach due to \cite[Sect.\ 1.2 Thm.\ 2.2]{Pazy}.
\qed

\section{Proof of Theorem \ref{thm:1D_decay}}\label{sec:1D_decay_proof}

Without loss of generality, let us take
        $\Omega=\Omega_{in}=\Omega_0=\left(0,x_0\right)$ for some
        $x_0>0$. This is possible since the decay in
        the region of interest $\Omega$ can be deduced by enlarging it (if necessary) to {an interval}
         containing both $\Omega_0$ and $\Omega_{in}$.

{While the result of this theorem is stated for the mild solution
\begin{equation}\label{eq:u_reg}
  u\in C^{1}\left(\mathbb{R}_{+},L^{2}\left(\mathbb{R}\right)\right)\cap C\left(\mathbb{R}_{+},H^{1}\left(\mathbb{R}\right)\right),
\end{equation}
its proof requires
the regularity of classical solutions established in
Proposition~\ref{prop:WP}.
Hence we approximate the initial data $(u_0,\, u_1)^T\in
X{:=H^1\left(\mathbb{R}\right)\times L^2\left(\mathbb{R}\right)}$ by a sequence
{of functions} $(u_0^n,\, u_1^n)^T_{n\in\mathbb{N}}\subset H^2\left(\mathbb{R}\right)\times
H^1\left(\mathbb{R}\right)$ with support in $\Omega_0$
and converging in $X$ to $(u_0,\, u_1)^T$. Due to Proposition \ref{prop:WP}, each approximate initial condition $(u_0^n,\, u_1^n)$ gives rise to a classical solution
\begin{equation}\label{eq:un_reg}
  u^n\in C^{2}\left(\mathbb{R}_{+},L^{2}\left(\mathbb{R}\right)\right)\cap C^{1}\left(\mathbb{R}_{+},H^{1}\left(\mathbb{R}\right)\right)\cap C\left(\mathbb{R}_{+},H^{2}\left(\mathbb{R}\right)\right).
\end{equation}
By the linearity of problem \eqref{eq:wave_1D} and bound \eqref{eq:WP_estim}}, these classical solutions converge uniformly to {the
  mild solution} $u$ on any finite time interval $[0,T]$:
\begin{equation}\label{un-conv}
  u^n \:\stackrel{n\to \infty}{\longrightarrow}\: u \quad \mbox{in } C^1([0,T],L^2(\mathbb{R}))\cap C([0,T],H^1(\mathbb{R})).
\end{equation}

\noindent
We recall that the notions of classical and mild solutions are discussed in \cite[p.~105]{Pazy}.
\smallskip

The proof is performed in 3 steps.
In Step~1,
        {for any fixed $n\in\mathbb{N}$, we consider
          problem~\eqref{eq:wave_1D} for initial data $(u_0^n,\,
          u_1^n)\in H^2\left(\mathbb{R}\right)\times
H^1\left(\mathbb{R}\right)$ and classical solution $u^n$ as
in~\eqref{eq:un_reg}. W}e
        transform \eqref{eq:wave_1D} into an auxiliary
        problem, for which we construct a weighted, local energy
        functional that would {admit}
         an exponential decay under
        certain conditions on weight functions.
        In Step~2,
        we construct these weight functions such that the rate of decay of the energy functional can be estimated explicitly.
	In Step~3,
        we use the local energy decay {and the convergence in~\eqref{un-conv}}
        to prove that the {mild} solution {$u$} of~\eqref{eq:wave_1D}, for large
	times, converges
        to a constant uniformly on $\left[0,x_{0}\right]$. Moreover,
	we identify this constant as $u_{\infty}$ defined in \eqref{eq:u_infty_1D}.

\subsection*{Step 1. Auxiliary local energy functional.}
{{As anticipated,} here and in Step~2, we
  establish the exponential-in-time decay of any classical solution $u^n$, i.e.\ for fixed index $n\in\mathbb{N}$.}

Let us perform the
change of variable
	\begin{equation}
	x\mapsto y\left(x\right):=\dfrac{1}{t_0}\int_{0}^{x}\sqrt{\beta\left(\xi\right)/\alpha\left(\xi\right)}d\xi ,\hspace{1em}x\left(s\right)=y^{-1}\left(s\right),\label{eq:y_def}
	\end{equation}
	\begin{equation}
	t\mapsto \tau\left(t\right):=t/t_0,\hspace{1em}t\left(\tau\right)=t_0\tau,\hspace{1em} t_0:=\int_{0}^{x_0}\sqrt{\beta\left(\xi\right)/\alpha\left(\xi\right)}d\xi,\label{eq:tau_def}
	\end{equation}
and denote $v\left(y,\tau\right):={u^n}\left(x\left(y\right),t_0\tau\right)$, $v_{0}\left(y\right):={u^n_{0}}\left(x\left(y\right)\right)$,
$v_{1}\left(y\right):=t_0\, {u^n_{1}}\left(x\left(y\right)\right)$,
\begin{equation}\label{eq:rho_gam_def}
\rho\left(y\right):=\frac{1}{\sqrt{\alpha\left(x\left(y\right)\right)\beta\left(x\left(y\right)\right)}},\hspace{1em}
\gamma\left(y\right):=\frac{\rho^{\prime}\left(y\right)}{\rho\left(y\right)},\hspace{1em}\rho_0:=\rho\left(0\right)=\rho\left(1\right)=\frac{1}{\sqrt{\alpha_0\beta_0}}.
\end{equation}

Due to Assumption \ref{assm:alph_bet}, the map $x\mapsto y\left(x\right)$ is $W^{2,\infty}\left(\mathbb{R}\right)$, and hence $v$ inherits the regularity of {$u^n$} given by \eqref{eq:un_reg}:
\begin{equation}\label{eq:v_reg}
v\in C^{2}\left(\mathbb{R}_{+},L^{2}\left(\mathbb{R}\right)\right)\cap C^{1}\left(\mathbb{R}_{+},H^{1}\left(\mathbb{R}\right)\right)\cap C\left(\mathbb{R}_{+},H^{2}\left(\mathbb{R}\right)\right).
\end{equation}
We thus arrive at a problem equivalent to \eqref{eq:wave_1D}:
	\begin{equation}
	\begin{cases}
	\partial_{\tau}^{2}v\left(y,\tau\right)-\partial_{y}^{2}v\left(y,\tau\right)+\gamma\left(y\right)\partial_{y}v\left(y,\tau\right)=0, & y\in\mathbb{R},\hspace{1em}\tau>0,\\
	v\left(y,0\right)=v_{0}\left(y\right),\hspace{1em}\partial_{\tau}v\left(y,0\right)=v_{1}\left(y\right), & y\in\mathbb{R},
	\end{cases}\label{eq:v_pbm}
	\end{equation}
	where $\gamma\left(y\right)\equiv0$ for $y\in\mathbb{R}\backslash\left(0,1\right)$
	and $\text{supp }v_{0}$, $\text{supp
        }v_{1}\subseteq(0,1)$.

Motivated by the form of the global energy introduced in~\cite{Lewis} for the wave equation in~\eqref{eq:v_pbm},
we consider the local energy functional
	\begin{align}
	\mathcal{E}_{loc}\left(\tau\right):= & \frac{1}{2}\int_{0}^{1}\left[\left(\partial_{\tau}v\left(y,\tau\right)\right)^{2}+\left(\partial_{y}v\left(y,\tau\right)\right)^{2}\right]\frac{dy}{\rho\left(y\right)}\nonumber \\
	= & \frac{1}{4}\int_{0}^1\left[\left(\partial_{\tau}v\left(y,\tau\right)+\partial_{y}v\left(y,\tau\right)\right)^{2}+\left(\partial_{\tau}v\left(y,\tau\right)-\partial_{y}v\left(y,\tau\right)\right)^{2}\right]\frac{dy}{\rho\left(y\right)}>0.\label{eq:E_loc}
	\end{align}
This quantity is decaying since, by direct calculations employing
integration by parts, using \eqref{eq:rho_gam_def} and the wave
equation in \eqref{eq:v_pbm}, we have
\begin{align}\label{eq:E_aux_dt}
\mathcal{E}^\prime_{loc}\left(\tau\right)=&\int_{0}^{1}\left[
\partial_{\tau}v\left(y,\tau\right)\partial^2_{\tau}v\left(y,\tau\right)
+ \partial_{y}v\left(y,\tau\right){\partial_y\partial_{\tau}} v\left(y,\tau\right)
                                            \right]\frac{dy}{\rho\left(y\right)}
  \nonumber \\
  {=}& {\int_{0}^{1}\left[\partial_{\tau}v\left(y,\tau\right)
   \left(\partial^2_{\tau}v\left(y,\tau\right)
   -\partial_{y}^{2}v\left(y,\tau\right)+\frac{\rho^{\prime}\left(y\right)}{\rho\left(y\right)}\partial_{y}v\left(y,\tau\right)\right)
\right]\frac{dy}{\rho\left(y\right)}}\\
&{+\left[\partial_y v\left(1,\tau\right) \partial_\tau
        v\left(1,\tau\right)-\partial_y v\left(0,\tau\right)
        \partial_\tau v\left(0,\tau\right)\right]\frac{1}{\rho_0}} \nonumber\\
=&\left[\partial_y v\left(1,\tau\right) \partial_\tau v\left(1,\tau\right)-\partial_y v\left(0,\tau\right) \partial_\tau v\left(0,\tau\right)\right]\frac{1}{\rho_0}\le0. \nonumber
\end{align}
{Note that we have used here the regularity of a classical
  solution.} {Moreover}, in order to deduce the sign in the last line, we have used the
exact outflow
boundary conditions
(well-defined for any $t \geq 0$ due to \eqref{eq:v_reg}):
\begin{equation}\label{eq:BCs_y}
\left.\left(\partial_{y}-\partial_{\tau}\right)v\left(y,\tau\right)\right|_{y=0}=0=\left.\left(\partial_{y}+\partial_{\tau}\right)v\left(y,\tau\right)\right|_{y=1},
\end{equation}
which are due to the support properties of the functions $v_0$, $v_1$ and $\gamma$.

We note, in passing, that the global energy, defined in the same way as \eqref{eq:E_loc} but with the integration range replaced by $\mathbb{R}$, is conserved.

Even though the conventional local energy functional \eqref{eq:E_loc} can (and, in the proof of Theorem \ref{thm:better_rate}, will) be used to show the exponential decay of the solution derivatives under some quantitative restriction on the coefficients $\alpha$ and $\beta$, here, we proceed with an alternative strategy that does not require such a restriction.
To this effect, we consider now a modified local energy functional in the spirit of \cite[Sec.~3]{Arnold}:
	\begin{equation}
	E_{loc}\left(\tau\right):=\frac{1}{2}\int_{0}^{1}\left[\phi_{1}\left(y\right)\left(\partial_{\tau}v\left(y,\tau\right)+\partial_{y}v\left(y,\tau\right)\right)^{2}+\phi_{2}\left(y\right)\left(\partial_{\tau}v\left(y,\tau\right)-\partial_{y}v\left(y,\tau\right)\right)^{2}\right]\frac{dy}{\rho\left(y\right)},\label{eq:E_loc_mod}
	\end{equation}
	where $\phi_{1}\left(y\right)$, $\phi_{2}\left(y\right)$ are some
	strictly positive weight functions on $\left[0,1\right]$, which are yet
	to be chosen.

Similarly to \eqref{eq:E_aux_dt}, we differentiate under the integral sign to obtain (suppressing the arguments of all functions for the sake of brevity)
\begin{align}\label{eq:E_dt_estim_prelim}
E_{loc}^{\prime}\left(\tau\right)=&\int_{0}^{1}\left[\phi_{1}\left(\partial_{\tau}v+\partial_{y}v\right)\left(\partial_{\tau}^{2}v+{\partial_y\partial_{\tau}}v\right)+\phi_{2}\left(\partial_{\tau}v-\partial_{y}v\right)\left(\partial_{\tau}^{2}v-{\partial_y\partial_{\tau}}v\right)\right]\frac{dy}{\rho}\\
=&\int_{0}^{1}\left[\phi_{1}\left(\partial_{\tau}v+\partial_{y}v\right)\left(\partial_{y}^{2}v-\gamma\partial_{y}v+{\partial_y\partial_{\tau}}v\right)+\phi_{2}\left(\partial_{\tau}v-\partial_{y}v\right)\left(\partial_{y}^{2}v-\gamma\partial_{y}v-{\partial_y\partial_{\tau}}v\right)\right]\frac{dy}{\rho}\nonumber\\
=&{\frac12}\int_{0}^{1}\left[\phi_{1}\partial_{y}\frac{\left(\partial_{\tau}v+\partial_{y}v\right)^{2}}{\rho}-\phi_{2}\partial_{y}\frac{\left(\partial_{\tau}v-\partial_{y}v\right)^{2}}{\rho}+\left(\phi_{1}-\phi_{2}\right)\gamma\frac{\left(\partial_{\tau}v\right)^{2}-\left(\partial_{y}v\right)^{2}}{\rho}\right]dy,\nonumber
\end{align}
where we used the wave equation in~\eqref{eq:v_pbm} to eliminate
second-order time derivatives, and the definition of $\gamma$ in~\eqref{eq:rho_gam_def}.
Integrating \eqref{eq:E_dt_estim_prelim} by parts and employing boundary conditions \eqref{eq:BCs_y},
	we obtain
	\begin{align*}
	E_{loc}^{\prime}\left(\tau\right)= & \frac{1}{2}\int_{0}^{1}\left[-\phi_{1}^{\prime}\left(y\right)\left(\partial_{\tau}v\left(y,\tau\right)+\partial_{y}v\left(y,\tau\right)\right)^{2}+\phi_{2}^{\prime}\left(y\right)\left(\partial_{\tau}v\left(y,\tau\right)-\partial_{y}v\left(y,\tau\right)\right)^{2}\right.\\
	& \left.+\left(\phi_{1}\left(y\right)-\phi_{2}\left(y\right)\right)\gamma\left(y\right)\left(\left(\partial_{\tau}v\left(y,\tau\right)\right)^{2}-\left(\partial_{y}v\left(y,\tau\right)\right)^{2}\right)\right]\frac{dy}{\rho\left(y\right)}\\
	&-\left[\phi_{2}\left(1\right)\left(\partial_{y}v\left(1,\tau\right)\right)^{2}+\phi_{1}\left(0\right)\left(\partial_{y}v\left(0,\tau\right)\right)^{2}\right]\frac{2}{\rho_0}.
	\end{align*}
	We estimate
	\begin{align*}
	\left(\phi_{1}-\phi_{2}\right)\gamma\left(\left(\partial_{\tau}v\right)^{2}-\left(\partial_{y}v\right)^{2}\right) & =\left(\phi_{1}-\phi_{2}\right)\gamma\left(\partial_{\tau}v+\partial_{y}v\right)\left(\partial_{\tau}v-\partial_{y}v\right)\\
	& \leq\frac{1}{2}\left|\phi_{1}-\phi_{2}\right|\left|\gamma\right|\left(\left(\partial_{\tau}v+\partial_{y}v\right)^{2}+\left(\partial_{\tau}v-\partial_{y}v\right)^{2}\right),
	\end{align*}
	and therefore, since $\rho>0$, we arrive at
	\begin{align}
	E_{loc}^{\prime}\left(\tau\right)\leq & -\frac{1}{2}\int_{0}^{1}\left[\left(\phi_{1}^{\prime}\left(y\right)-\frac{\left|\gamma\left(y\right)\right|}{2}\left|\phi_{1}\left(y\right)-\phi_{2}\left(y\right)\right|\right)\left(\partial_{\tau}v\left(y,\tau\right)+\partial_{y}v\left(y,\tau\right)\right)^{2}\right.\nonumber \\
	& \left.+\left(-\phi_{2}^{\prime}\left(y\right)-\frac{\left|\gamma\left(y\right)\right|}{2}\left|\phi_{1}\left(y\right)-\phi_{2}\left(y\right)\right|\right)\left(\partial_{\tau}v\left(y,\tau\right)-\partial_{y}v\left(y,\tau\right)\right)^{2}\right]\frac{dy}{\rho\left(y\right)}.\label{eq:E_loc_dt}
	\end{align}

We aim to obtain
an estimate of the form $E_{loc}^{\prime}\left(\tau\right)\leq -\lambda_0 E_{loc}\left(\tau\right)$ with some $\lambda_0>0$,
and explore the possibility of choosing
the weight functions $\phi_1\left(y\right)$, $\phi_2\left(y\right)$
that would lead to
such an estimate. In \cite{Arnold} it is shown in
a more general setting that such weights do exist theoretically. Here,
we present an explicit
construction of $\phi_1$, $\phi_2$, which will permit us to obtain an explicit estimate of the decay rate of the solution.

\subsection*{Step 2. Construction of the weights $\phi_1$, $\phi_2$.}

Motivated by \eqref{eq:E_loc_dt}, we shall now construct functions $\phi_{1}\left(y\right)$, $\phi_{2}\left(y\right)>0$ in \eqref{eq:E_loc_mod}
	that satisfy, on $\left[0,1\right]$, the following differential inequalities
	\begin{equation}
	\begin{cases}
	\phi_{1}^{\prime}\left(y\right)-\frac{\left|\gamma\left(y\right)\right|}{2}\left|\phi_{1}\left(y\right)-\phi_{2}\left(y\right)\right|\geq\lambda\phi_{1}\left(y\right),\hspace{1em}y\in\left(0,1\right),\\
	-\phi_{2}^{\prime}\left(y\right)-\frac{\left|\gamma\left(y\right)\right|}{2}\left|\phi_{1}\left(y\right)-\phi_{2}\left(y\right)\right|\geq\lambda\phi_{2}\left(y\right),\hspace{1em}y\in\left(0,1\right),\\
	\phi_{1}\left(0\right)=\phi_{2}\left(0\right)=1,
	\end{cases}\label{eq:phi_syst}
	\end{equation}
	for some constant $\lambda>0$ to be chosen. To this end,
	we consider functions $\varphi_{1}\left(y\right)$, $\varphi_{2}\left(y\right)>0$ solving the linear ODE system
	\begin{equation}
	\begin{cases}
	\varphi_{1}^{\prime}\left(y\right)-\gamma_0\left(\varphi_{1}\left(y\right)-\varphi_{2}\left(y\right)\right)=\lambda\varphi_{1}\left(y\right),\hspace{1em}y\in\left(0,1\right),\\
	-\varphi_{2}^{\prime}\left(y\right)-\gamma_0\left(\varphi_{1}\left(y\right)-\varphi_{2}\left(y\right)\right)=\lambda\varphi_{2}\left(y\right),\hspace{1em}y\in\left(0,1\right),\\
	\varphi_{1}\left(0\right)=\varphi_{2}\left(0\right)=1,
	\end{cases}\label{eq:phi_syst_lin}
	\end{equation}
where $\gamma_{0}:=\frac{1}{2}\left\Vert \gamma\right\Vert _{L^{\infty}\left(0,1\right)}$.

It is easy to see that any solution of \eqref{eq:phi_syst_lin} automatically satisfies \eqref{eq:phi_syst}, provided that $\varphi_{1}\left(y\right)$, $\varphi_{2}\left(y\right)>0$ for all $y\in\left[0,1\right]$. Indeed, it suffices to verify that
\begin{equation}\label{eq:phis_diff}
0\leq\varphi_{1}\left(y\right)-\varphi_{2}\left(y\right)=\left|\varphi_{1}\left(y\right)-\varphi_{2}\left(y\right)\right|, \hspace{1em} y\in\left[0,1\right].
\end{equation}
Summing both ODEs of \eqref{eq:phi_syst_lin}, multiplying by
  $e^{-2\gamma_0y}$, and integrating on $(0,y)$,
we obtain
\[
\varphi_1\left(y\right)-\varphi_2\left(y\right)=\lambda\int_0^y e^{2\gamma_0\left(y-\xi\right)}\left[\varphi_1\left(\xi\right)+\varphi_2\left(\xi\right)\right]d\xi, \hspace{1em} y\in\left[0,1\right],
\]
and hence positivity of the integrand directly implies \eqref{eq:phis_diff}.

Note that it is possible to consider more general versions of \eqref{eq:phi_syst} (and hence \eqref{eq:phi_syst_lin}), with equations involving different parameters $\lambda_1$, $\lambda_2$ on the right-hand sides, and initial conditions given by a positive constant different from $1$. However, it can be shown that no advantage could be gained from such generalisations.

We shall now estimate the largest possible value $\lambda>0$ (or, more precisely, its supremum) such that $\varphi_1$, $\varphi_2>0$ in $\left[0,1\right]$.
	According to the first equation in \eqref{eq:phi_syst_lin} and \eqref{eq:phis_diff}, $\varphi_{1}\left(y\right)$
	is a monotonically increasing function of $y$ for $y\in\left(0,1\right)$ and thus $\varphi_1\left(0\right)=1$ implies $\varphi_1\left(y\right)>1$ for $y\in\left(0,1\right]$. Similarly, it follows from the second equation of \eqref{eq:phi_syst_lin} that  $\varphi_{2}\left(y\right)$ decreases monotonically in $\left[0,1\right]$ starting from the value $\varphi_{2}\left(0\right)=1$.
	Therefore, to guarantee the positivity of $\varphi_{2}\left(y\right)$ in $\left[0,1\right]$,
	it suffices to require that $\varphi_2\left(1\right)>0$.
We can write the solution of \eqref{eq:phi_syst_lin}
explicitly:
\begin{equation}\label{eq:phi1_sol}
\varphi_{1}\left(y\right)=\frac{e^{\gamma_{0}y}}{\sqrt{\lambda^{2}+\gamma_{0}^{2}}}\left[\sqrt{\lambda^{2}+\gamma_{0}^{2}}\cosh\left(\sqrt{\lambda^{2}+\gamma_{0}^{2}}y\right)+\left(\lambda-\gamma_{0}\right)\sinh\left(\sqrt{\lambda^{2}+\gamma_{0}^{2}}y\right)\right],
\end{equation}
\begin{equation}\label{eq:phi2_sol}
\varphi_{2}\left(y\right)=\frac{e^{\gamma_{0}y}}{\sqrt{\lambda^{2}+\gamma_{0}^{2}}}\left[\sqrt{\lambda^{2}+\gamma_{0}^{2}}\cosh\left(\sqrt{\lambda^{2}+\gamma_{0}^{2}}y\right)-\left(\lambda+\gamma_{0}\right)\sinh\left(\sqrt{\lambda^{2}+\gamma_{0}^{2}}y\right)\right],
\end{equation}

Then, the above requirement $\varphi_2\left(1\right)>0$ is equivalent to
\begin{equation}\label{eq:tanh_ineq1}
\tanh\left({\sqrt{\gamma_{0}^{2}+\lambda^{2}}}\right)<\frac{\sqrt{\gamma_0^2+\lambda^2}}{\gamma_0+\lambda}.
\end{equation}
To determine the interval of admissible values of $\lambda$
satisfying~\eqref{eq:tanh_ineq1}, we consider
the corresponding equality
\begin{equation}\label{eq:tanh_eq}
\tanh\left({\sqrt{\gamma_{0}^{2}+\lambda^{2}_{*}}}\right)=\frac{\sqrt{\gamma_0^2+\lambda^2_{*}}}{\gamma_0+\lambda_{*}},
\end{equation}
which implicitly defines a function $\lambda_{*}(\gamma_0)$ in $\mathbb{R}_+$. The proof of the unique solvability of \eqref{eq:tanh_eq} for
$\lambda_*$, as well as the fact that \eqref{eq:tanh_ineq1} then holds
for all $\lambda\in[0,\lambda_*)$, is actually deferred to Lemma \ref{lem:tanh_eq} in the Appendix.
The function $\lambda_{*}(\gamma_0)$ is illustrated in Figure \ref{fig:lmb_vs_gam}.
We see that $\lambda_{*}(\gamma_0)$ is monotonically decreasing with $\lambda_*\to0$ as $\gamma_0\to\infty$ (this can also be verified analytically by implicit differentiation of \eqref{eq:tanh_eq}).

While the parameter choice $\lambda=\lambda_*$ is not admissible
for the construction of the weight functions (since
$\varphi_2\left(1\right)=0$ if $\lambda=\lambda_*$), any value
$\lambda\in(0,\lambda_{*})$ is admissible, and therefore
  guarantees positivity of $\varphi_1$, $\varphi_2$.

\begin{figure}
\includegraphics[scale=0.8]{lmbs_vs_gamma}
\caption{Blue: function $\lambda_*(\gamma_0)$, implicitly defined by \eqref{eq:tanh_eq}. Red: function $\lambda_0(\gamma_0)$, defined in~\eqref{eq:lmb0_simpl1}.}\label{fig:lmb_vs_gam}
\end{figure}

Since the function $\lambda_{*}(\gamma_0)$ is not defined explicitly,
we give an example of an explicit lower bound that eventually yields the explicit (but suboptimal) decay rate stated in \eqref{eq:u_infty_1D}. The function
\begin{equation}
\lambda_0(\gamma_0):=\sqrt2 \gamma_{0}e^{-2\gamma_{0}}\sqrt{1-4\gamma_{0}e^{-4\gamma_{0}}}\label{eq:lmb0_simpl1}
\end{equation}
satisfies \eqref{eq:tanh_ineq1} and
$\lambda_0(\gamma_0)<\lambda_*(\gamma_0)$ for any $\gamma_0>0$,
and it is depicted in Figure \ref{fig:lmb_vs_gam}. The proof of
  the inequality $\lambda_0(\gamma_0)<\lambda_*(\gamma_0)$
is given in Lemma~\ref{lem:A2} in the Appendix.

Therefore, we can take $\lambda=\lambda_0:=\lambda_0(\gamma_0)$, which satisfies inequality \eqref{eq:tanh_ineq1}. Hence the weights $\phi_1\left(y\right):=\varphi_1\left(y\right)>0$, $\phi_2\left(y\right):=\varphi_2\left(y\right)>0$ satisfy \eqref{eq:phi_syst}.
Consequently, we obtain from \eqref{eq:E_loc_dt} and~\eqref{eq:E_loc_mod}
\begin{equation}\label{eq:E_loc_dec}
E_{loc}^{\prime}\left(\tau\right)\leq-\lambda_0 E_{loc}\left(\tau\right),\hspace{1em}\tau>0\hspace{1em}\Longrightarrow\hspace{1em}E_{loc}\left(\tau\right)\le E_{loc}\left(0\right)e^{-\lambda_0 \tau},\hspace{1em}\tau\geq 0.
\end{equation}

\subsection*{Step 3. Convergence of the solution and identification of the constant~$u_{\infty}$.}

	The local energy functional \eqref{eq:E_loc_mod} can be rewritten in the original variables $\left(x,t\right)$,
{and we {emphasise} here that it is a functional of the approximate classical solution $u^n$:}
	\begin{align*}
         {\widetilde{E}_{loc}[u^n]}\left(t\right):=E_{loc}\left(\frac{t}{t_0}\right)=&\frac{t_0}{2}\int_{0}^{x_{0}}\left[{\widetilde{\varphi}_1
\left(x\right)\left(\partial_{t}u^n\left(x,t\right)+\sqrt{\frac{\alpha\left(x\right)}{\beta\left(x\right)}}\partial_{x}u^n\left(x,t\right)\right)^{2}\right.\\
 &\left.+{\widetilde{\varphi}_2
 \left(x\right)\left(\partial_{t}u^n\left(x,t\right)-\sqrt{\frac{\alpha\left(x\right)}{\beta\left(x\right)}}\partial_{x}u^n\left(x,t\right)\right)^{2}\right]\beta\left(x\right)dx,
	\end{align*}
	where ${\widetilde{\varphi}_1
        \left(x\right):=\varphi_{1}\left(y\left(x\right)\right)$,
	${\widetilde{\varphi}_1
        \left(x\right):=\varphi_{2}\left(y\left(x\right)\right)$
	are given by \eqref{eq:phi1_sol}--\eqref{eq:phi2_sol}, with
        $\lambda=\lambda_0$ and the mapping $x\mapsto y\left(x\right)$
        given by \eqref{eq:y_def}. Then, from~\eqref{eq:E_loc_dec}, we obtain
$$
   {\widetilde{E}_{loc}[u^n]}\left(t\right)\le
    {\widetilde{E}_{loc}[u^n]}\left(0\right)e^{-\frac{\lambda_0}{t_0} t},\hspace{1em}t\geq 0,
$$
{where the decay rate $\lambda_0/t_0$ is independent of the index $n$. Due to the convergence~\eqref{un-conv}{,} we can pass to the limit $n\to\infty$ and obtain the same inequality for the mild solution $u$:
\begin{equation}\label{ineq:tE-decay}
  \widetilde{E}_{loc}[u]\left(t\right)\le
    \widetilde{E}_{loc}[u]\left(0\right)e^{-\frac{\lambda_0}{t_0} t},\hspace{1em}t\geq 0{,}
\end{equation}
and, due to the regularity of $u$ in~\eqref{eq:u_reg}, we have $\widetilde{E}_{loc}[u]\in C\left(\mathbb{R}_+\right)$.}

Setting
\[
C_{0}:={\min}\left\{ \underset{y\in\left(0,1\right)}{\inf}\varphi_{1}\left(y\right),\underset{y\in\left(0,1\right)}{\inf}\varphi_{2}\left(y\right)\right\}={\min}\left\{\varphi_1\left(0\right),\varphi_2\left(1\right)\right\}=\varphi_2\left(1\right)>0,
\]
we have
	\begin{equation*}
	{\widetilde{E}_{loc}[u]}\left(t\right)\geq C_{0}t_0\int_{0}^{x_{0}}\left[\beta\left(x\right)\left(\partial_{t}u\left(x,t\right)\right)^{2}+\alpha\left(x\right)\left(\partial_{x}u\left(x,t\right)\right)^{2}\right]dx,
	\end{equation*}
	and thus deduce that
	\begin{equation}
	\left\Vert \partial_{t}u\left(\cdot,t\right)\right\Vert _{L^{2}\left(0,x_{0}\right)}^{2}\leq\frac{\left\Vert 1/\beta\right\Vert _{L^{\infty}\left(0,x_{0}\right)}}{C_{0}t_0}
	 {\widetilde{E}_{loc}[u]}\left(t\right),\hspace{1em}\left\Vert \partial_{x}u\left(\cdot,t\right)\right\Vert _{L^{2}\left(0,x_{0}\right)}^{2}\leq\frac{\left\Vert 1/\alpha\right\Vert _{L^{\infty}\left(0,x_{0}\right)}}{C_{0}t_0}
	  {\widetilde{E}_{loc}[u]}\left(t\right).\label{eq:u_x_t_bnds}
	\end{equation}
This{, together with~\eqref{ineq:tE-decay},} yields the exponential
decay of $\left\Vert \partial_{t}u\right\Vert
_{L^{2}\left(0,x_{0}\right)}$ and $\left\Vert \partial_{x}u\right\Vert
_{L^{2}\left(0,x_{0}\right)}$ claimed in \eqref{eq:u_exp_conv}. In
  order to complete the proof, it remains to identify $u_\infty$,
{and prove that, for all $t\ge 0$, $u(\cdot,t)$ converges in
  $H^1(0,x_0)$
to $u_\infty$.}

Using the regularity of $u$ given by \eqref{eq:u_reg},
	we have, for $x\in\left(0,x_{0}\right)$, $t>0$,
	\[
	u\left(x,t\right)=u\left(0,t\right)+\int_{0}^{x}\partial_{x}u\left(s,t\right)ds=u\left(x_{0},t\right)-\int_{x}^{x_{0}}\partial_{x}u\left(s,t\right)ds,
      \]
      which implies
	\begin{equation}
          \left\Vert u\left(\cdot,t\right)-u\left(0,t\right)\right\Vert _{L^{2}\left(0,x_{0}\right)}^{2},\,\left\Vert u\left(\cdot,t\right)-u\left(x_{0},t\right)\right\Vert _{L^{2}\left(0,x_{0}\right)}^{2}\leq x_0^2\left\Vert \partial_{x}u\left(\cdot,t\right)\right\Vert _{L^{2}\left(0,x_{0}\right)}^{2}.\label{eq:u_bvp_bnds}
	\end{equation}

	Let us now integrate the wave equation in \eqref{eq:wave_1D} over
	the rectangle $\left(0,x_{0}\right)\times\left(0,t\right)$, $t>0$.
	Using the outflow boundary conditions \eqref{eq:u_reg}
	\[
	\left.\left(\partial_{x}-\sqrt{\frac{\beta_0}{\alpha_0}}\partial_{t}\right)u\left(x,t\right)\right|_{x=0}=0,\hspace{1em} \left.\left(\partial_{x}+\sqrt{\frac{\beta_0}{\alpha_0}}\partial_{t}\right)u\left(x,t\right)\right|_{x=x_{0}}=0,
	\]
	and the compact support of $u_{0}$, $u_{1}$, we obtain
	\begin{equation}
	\int_{0}^{x_{0}}\beta\left(s\right)\partial_{t}u\left(s,t\right)ds-\int_{0}^{x_{0}}\beta\left(s\right)u_{1}\left(s\right)ds=-\sqrt{\alpha_0\beta_0}\left(u\left(0,t\right)+u\left(x_{0},t\right)\right).\label{eq:u_IBVP_ident_prelim}
	\end{equation}
{Due to the weak regularity of the mild solution $u$,
{the same} identity actually first needs to be derived for the classical
  solutions $u^n$ from \eqref{eq:un_reg}, and then{~\eqref{eq:u_IBVP_ident_prelim} follows by} passing to the
  limit} {$n\to\infty$.}
{From~}\eqref{eq:u_IBVP_ident_prelim}, we {deduce}
	\begin{align}
	u\left(x,t\right)-\frac{1}{2\sqrt{\alpha_0\beta_0}}\int_{0}^{x_{0}}\beta\left(s\right)u_{1}\left(s\right)ds=&\frac{1}{2}\left[u\left(x,t\right)-u\left(0,t\right)\right]+\frac{1}{2}\left[u\left(x,t\right)-u\left(x_{0},t\right)\right]\nonumber\\
	&-\frac{1}{2\sqrt{\alpha_0\beta_0}}\int_{0}^{x_{0}}\beta\left(s\right)\partial_{t}u\left(s,t\right)ds\label{eq:u_IBVP_ident}
	\end{align}
{for all $x\in\left(0,x_{0}\right)$ and all $t\ge0$.}
For the first two terms on the right-hand side
  of~\eqref{eq:u_IBVP_ident}, we have
	\[
\left\Vert u\left(x,t\right)-u\left(0,t\right)\right\Vert_{H^{1}\left(0,x_0\right)}^2=\left\Vert u\left(x,t\right)-u\left(0,t\right)\right\Vert_{L^{2}\left(0,x_0\right)}^2+\left\Vert \partial_{x}u\left(x,t\right)\right\Vert_{L^{2}\left(0,x_0\right)}^2,
	\]
	\[
\left\Vert u\left(x,t\right)-u\left(x_0,t\right)\right\Vert_{H^{1}\left(0,x_0\right)}^2=\left\Vert u\left(x,t\right)-u\left(x_0,t\right)\right\Vert_{L^{2}\left(0,x_0\right)}^2+\left\Vert \partial_{x}u\left(x,t\right)\right\Vert_{L^{2}\left(0,x_0\right)}^2.
	\]
Moreover,
since the integral term on the right-hand side of
\eqref{eq:u_IBVP_ident} is independent of~$x$,
we also have
{
  \[
    \begin{split}
    \left\Vert
 \frac{1}{2\sqrt{\alpha_0\beta_0}}\int_{0}^{x_{0}}\beta\left(s\right)\partial_{t}u\left(s,t\right)ds
\right\Vert_{H^{1}\left(0,x_0\right)}
&=    \left\Vert
 \frac{1}{2\sqrt{\alpha_0\beta_0}}\int_{0}^{x_{0}}\beta\left(s\right)\partial_{t}u\left(s,t\right)ds
\right\Vert_{L^{2}\left(0,x_0\right)}\\
&\le \frac{1}{2}\sqrt{\frac{x_0}{\alpha_0\beta_0}}
\left\Vert\beta\right\Vert_{L^2\left(0,x_0\right)}\left\Vert\partial_t u\right\Vert_{L^2\left(0,x_0\right)}.
\end{split}
    \]}

\noindent
Therefore, taking the $H^{1}\left(0,x_{0}\right)$ norm of the identity
{in} \eqref{eq:u_IBVP_ident}, with $u_{\infty}$ defined as in~\eqref{eq:u_infty_1D},
we deduce from
\eqref{eq:u_bvp_bnds} that
	\begin{align*}
	\left\Vert u\left(\cdot,t\right)-u_{\infty}\right\Vert
          _{H^{1}\left(0,x_0\right)}
          \leq
          &\frac{1}{2}\left\Vert u\left(x,t\right)-u\left(0,t\right)\right\Vert_{H^{1}\left(0,x_0\right)}+\frac{1}{2}\left\Vert u\left(x,t\right)-u\left(x_0,t\right)\right\Vert_{H^{1}\left(0,x_0\right)}\\
          	&+{\frac{1}{2}\sqrt{\frac{x_0}{\alpha_0\beta_0}}\left\Vert\beta\right\Vert_{L^2\left(0,x_0\right)}\left\Vert\partial_t u\right\Vert_{L^2\left(0,x_0\right)}}\\
          \leq
          &\sqrt{1+x_0^2}\left\Vert\partial_x u\right\Vert_{L^2\left(0,x_0\right)}+\frac{1}{2}\sqrt{\frac{x_0}{\alpha_0\beta_0}}\left\Vert\beta\right\Vert_{L^2\left(0,x_0\right)}\left\Vert\partial_t u\right\Vert_{L^2\left(0,x_0\right)}.
	\end{align*}
	Finally, employing \eqref{eq:u_x_t_bnds}{,} we obtain
	\begin{align*}
	\left\Vert u\left(\cdot,t\right)-u_{\infty}\right\Vert _{H^{1}\left(0,x_{0}\right)}\leq&\frac{1}{2 \left(C_{0}t_0\right)^{1/2}}\left[2\left(\left(1+x_{0}^2\right)\left\Vert 1/\alpha\right\Vert _{L^{\infty}\left(0,x_{0}\right)}\right)^{1/2}\right.\\
	&+\left.\left(\frac{x_0}{\alpha_0\beta_0}\left\Vert 1/\beta\right\Vert _{L^{\infty}\left(0,x_{0}\right)}\right)^{1/2}\left\Vert \beta\right\Vert _{L^{2}\left(0,x_{0}\right)}\right
	 {\widetilde{E}_{loc}[u]^{1/2}}\left(t\right)
	\end{align*}
	which, together with~\eqref{ineq:tE-decay}, leads to the
          bound of $\left\Vert
            u\left(\cdot,t\right)-u_{\infty}\right\Vert
          _{H^{1}\left(0,x_{0}\right)}$ in
        \eqref{eq:u_exp_conv}.
{From~\eqref{eq:lmb0_simpl1} and~\eqref{ineq:tE-decay}, we find the claimed decay rate $\Lambda=\frac{\gamma_{0}}{\sqrt2 \, t_0}e^{-2\gamma_{0}}\left(1-4\gamma_{0}e^{-4\gamma_{0}}\right)^{1/2}$.}

Since $\left\Vert
          u\left(\cdot,t\right)-u_{\infty}\right\Vert
        _{H^{1}\left(0,x_{0}\right)}$ decays to zero, $u_\infty$ is
        uniquely defined as in~\eqref{eq:u_infty_1D}. This
          concludes the proof. \qed

        {
          In the proof of Theorem \ref{thm:1D_decay},
modified weight functions could also have been obtained by imposing the
initial condition for \eqref{eq:phi_syst} at an internal point $y_0$ in
$(0,1)$ instead of at $y_0=0$, and then solving the two  corresponding linear ODEs forward and backward.
Potentially, this may lead to
an increased maximal decay rate,
as compared to that we have deduced from~\eqref{eq:tanh_eq}.
}

We also remark that the decay rate given by $\lambda_0(\gamma_0)$ is
exponentially small for large~$\gamma_0$. This is not an artefact of
the approximation of $\lambda_*(\gamma_0)$ by $\lambda_0(\gamma_0)$
(see Figure~\ref{fig:lmb_vs_gam}), but is rather characteristic to the weight function approach in the modified local energy.

\section{Proof of Theorem \ref{thm:better_rate}}\label{sec:better_rate_proof}
As in the proof of Theorem~\ref{thm:1D_decay} in Section~\ref{sec:1D_decay_proof}, we assume, without loss of generality, that $\Omega=\Omega_{in}=\Omega_0=\left(0,x_0\right)$ for some $x_0>0$.

	To prove the claim, we shall modify Step 1 of the proof of Theorem \ref{thm:1D_decay}.
	Namely, instead of working with the modified local energy
        functional \eqref{eq:E_loc_mod}, we shall focus on its
        conventional counterpart \eqref{eq:E_loc} and estimate the quantities that enter it.
        {Since we shall not
          differentiate the energy functional in time, we
          work here directly with the mild solution $u$.}
        {We consider again equation~\eqref{eq:v_pbm} in the
          variables~$(y,\tau)$ given by~\eqref{eq:y_def}--\eqref{eq:tau_def}.
By treating the third term on the left-hand side as a (known)
          source term, it results in a wave equation with constant
          coefficients, to which we}
apply the d'Alembert  formula:
	\begin{equation}
	v\left(y,\tau\right)=\frac{1}{2}\left[v_{0}\left(y+\tau\right)+v_{0}\left(y-\tau\right)\right]+\frac{1}{2}\int_{y-\tau}^{y+\tau}v_{1}\left(z\right)dz-\frac{1}{2}\int_{0}^{\tau}\int_{y-s}^{y+s}\gamma\left(z\right)\partial_{y}v\left(z,\tau-s\right)dz ds.\label{eq:dAlemb}
	\end{equation}
	Therefore, we obtain
	\begin{align}
	\partial_{y}v\left(y,\tau\right)= & \frac{1}{2}\left[v_{0}^{\prime}\left(y+\tau\right)+v_{0}^{\prime}\left(y-\tau\right)\right]+\frac{1}{2}\left[v_{1}\left(y+\tau\right)-v_{1}\left(y-\tau\right)\right]\nonumber \\
	& +\frac{1}{2}\int_{0}^{\tau}\left[\gamma\left(y-s\right)\partial_{y}v\left(y-s,\tau-s\right)-\gamma\left(y+s\right)\partial_{y}v\left(y+s,\tau-s\right)\right]ds,\label{eq:dv_dy}
	\end{align}
	\begin{align}
	\partial_{\tau}v\left(y,\tau\right)= & \frac{1}{2}\left[v_{0}^{\prime}\left(y+\tau\right)-v_{0}^{\prime}\left(y-\tau\right)\right]+\frac{1}{2}\left[v_{1}\left(y+\tau\right)+v_{1}\left(y-\tau\right)\right]\nonumber \\
	& -\frac{1}{2}\int_{0}^{\tau}\left[\gamma\left(y-s\right)\partial_{y}v\left(y-s,\tau-s\right)+\gamma\left(y+s\right)\partial_{y}v\left(y+s,\tau-s\right)\right]ds.\label{eq:dv_dt}
	\end{align}
        Identity~\eqref{eq:dv_dt} becomes evident
by making a change of variable $s\mapsto \tau-s$ in the
        double-integral term in~\eqref{eq:dAlemb}
        before and after the differentiation with respect to $\tau$:
\begin{align*}
&\partial_\tau\int_{0}^{\tau}\int_{y-s}^{y+s}\gamma\left(z\right)\partial_{y}v\left(z,\tau-s\right)dz ds=\partial_\tau\int_{0}^{\tau}\int_{y-\tau+s}^{y+\tau-s}\gamma\left(z\right)\partial_{y}v\left(z,s\right)dz ds\\
&=\int_{0}^{\tau}\left[\gamma\left(y+\tau-s\right)\partial_{y}v\left(y+\tau-s,s\right)+\gamma\left(y-\tau+s\right)\partial_{y}v\left(y-\tau+s,s\right)\right]ds\\
&=\int_{0}^{\tau}\left[\gamma\left(y+s\right)\partial_{y}v\left(y+s,\tau-s\right)+\gamma\left(y-s\right)\partial_{y}v\left(y-s,\tau-s\right)\right]ds.
\end{align*}

	Note that{, for $\tau\ge1$,} the
        terms in \eqref{eq:dv_dy} and \eqref{eq:dv_dt}
{that do not contain the integral}
vanish
{for all $y$ in} $\left(0,1\right)$.
	To estimate $\mathcal{E}_{loc}$ for $\tau\ge1$, we
        insert~\eqref{eq:dv_dy} and~\eqref{eq:dv_dt} in {the
          definition of $\mathcal{E}_{loc}\left(\tau\right)$ in}~\eqref{eq:E_loc} and use that $\textrm{supp }\gamma\subset\left(0,1\right)$:
	\begin{align}
	\label{eq:E_int_estim}\mathcal{E}_{loc}\left(\tau\right)= & \frac{1}{4}\int_{0}^{1}\left[\left(\int_{y}^{1}\gamma\left(r\right)\partial_{y}v\left(r,\tau+y-r\right)dr\right)^{2}\right.\\
	&\left.+\left(\int_{0}^{y}\gamma\left(r\right)\partial_{y}v\left(r,\tau-y+r\right)dr\right)^{2} \right]\frac{dy}{\rho\left(y\right)}\nonumber\\
	\leq & \frac{1}{4}\int_{0}^{1}\left[\int_{y}^{1}\rho\left(\tilde{r}\right)\gamma^{2}\left(\tilde{r}\right)d\tilde{r}\int_{y}^{1}\left(\partial_{y}v\left(r,\tau+y-r\right)\right)^{2}\frac{dr}{\rho\left(r\right)}\right.\nonumber\\
&\left.+\int_{0}^{y}\rho\left(\tilde{r}\right)\gamma^{2}\left(\tilde{r}\right)d\tilde{r}\int_{0}^{y}\left(\partial_{y}v\left(r,\tau-y+r\right)\right)^{2}\frac{dr}{\rho\left(r\right)}                                                                          \right]\frac{dy}{\rho\left(y\right)}.\nonumber
        \end{align}
Denoting
	\begin{equation*}
	b_{0}:=\left\Vert \alpha\right\Vert
        _{L^{\infty}\left(0,x_{0}\right)}^{3/2}\left\Vert
          \beta\right\Vert
        _{L^{\infty}\left(0,x_{0}\right)}^{1/2}\left\Vert
          \left(1/\left(\alpha\beta\right)^{1/2}\right)^{\prime}\right\Vert
        _{L^{2}\left(0,x_{0}\right)}^{2},
	\end{equation*}
{as in}~\eqref{eq:b0_cond}, we have
\begin{equation*}
\max\left\{\sup_{y\in\left(0,1\right)}\frac{1}{\rho\left(y\right)}\int_{0}^{y}\rho\left(s\right)\gamma^{2}\left(s\right)ds,\sup_{y\in\left(0,1\right)}\frac{1}{\rho\left(y\right)}\int_{y}^{1}\rho\left(s\right)\gamma^{2}\left(s\right)ds\right\}\leq b_0t_0.
\end{equation*}
Therefore, we continue with~\eqref{eq:E_int_estim}, and estimate
\begin{align}\label{eq:E_loc_ineq_estim}
	\mathcal{E}_{loc}\left(\tau\right) &\leq\frac{b_{0}t_0}{4}\left[\int_{0}^{1}\int_{0}^{y}\left(\partial_{y}v\left(r,\tau-y+r\right)\right)^{2}\frac{1}{\rho\left(r\right)}drdy\right.\\
 &\phantom{=}\qquad\quad\left.+\int_{0}^{1}\int_{y}^{1}\left(\partial_{y}v\left(r,\tau+y-r\right)\right)^{2}\frac{1}{\rho\left(r\right)}drdy\right].\nonumber
\end{align}
{In the two terms on the right-hand side
  of~\eqref{eq:E_loc_ineq_estim}, the integration involves the values of $\partial_y v$ on two right-angled triangles with vertices $\left\{\left(0,\tau-1\right),\left(0,\tau\right),\left(1,\tau\right)\right\}$ and $\left\{\left(1,\tau-1\right),\left(0,\tau\right),\left(1,\tau\right)\right\}$. These triangles are parametrised by the sets of lines parallel to their hypotenuses, which are the two characteristic families of the wave equation in \eqref{eq:v_pbm}. We now change the integration variables to have parametrisation of these triangles by the sets of horizontal lines. Consequently, for the first integral, denoting $\sigma:=\tau-y+r$, we have that
  $\sigma\in(\tau-1,\tau)$ and $r\in(0,\sigma-\tau+1)$. Similarly, for
  the second integral, denoting $\sigma:=\tau+y-r$, we have that
  $\sigma\in(\tau-1,\tau)$ and $r\in(\tau-\sigma,1)$. Therefore, rewriting
  the integrals in~\eqref{eq:E_loc_ineq_estim} as integrals in the variables
  $\sigma$ and $r$, we obtain
  \[
    \begin{split}
     & \int_{0}^{1}\int_{0}^{y}\left(\partial_{y}v\left(r,\tau-y+r\right)\right)^{2}\frac{1}{\rho\left(r\right)}drdy
     +\int_{0}^{1}\int_{y}^{1}\left(\partial_{y}v\left(r,\tau+y-r\right)\right)^{2}\frac{1}{\rho\left(r\right)}drdy\\
     &\qquad=
     \int_{\tau-1}^{\tau}\int_{0}^{\sigma-\tau+1}\left(\partial_{y}v\left(r,\sigma\right)\right)^{2}\frac{1}{\rho\left(r\right)}drd\sigma+\int_{\tau-1}^{\tau}\int_{\tau-\sigma}^{1}\left(\partial_{y}v\left(r,\sigma\right)\right)^{2}\frac{1}{\rho\left(r\right)}drd\sigma.
      \end{split}
    \]
With this identity, taking into account the positivity of the
integrands, we obtain from~\eqref{eq:E_loc_ineq_estim} that
}
\begin{align}\label{eq:E_loc_ineq_estim1}
  \mathcal{E}_{loc}\left(\tau\right)
    \leq\frac{b_{0}t_0}{2}\int_{\tau-1}^{\tau}\int_{0}^{1}\left(\partial_{y}v\left(r,\sigma
\right)\right)^{2}\frac{1}{\rho\left(r\right)}drd\sigma
      \leq
  b_{0}t_0\int_{\tau-1}^{\tau}\mathcal{E}_{loc}\left(\sigma\right)d\sigma.
\end{align}
Since $\mathcal{E}_{loc}\left(\tau\right)$ is a positive non-increasing
function (recall \eqref{eq:E_aux_dt}), it follows
from~\eqref{eq:E_loc_ineq_estim1} that
	\[
	\mathcal{E}_{loc}\left(\tau\right)\leq b_{0}t_0\,\mathcal{E}_{loc}\left(\tau-1\right),\hspace{1em}\tau\ge1.
	\]
	Iterating this inequality $\left\lfloor \tau\right\rfloor $
	times (where $\left\lfloor X\right\rfloor $ denotes the integer part
	of $X$), {and using the assumption $b_{0}t_0<1$,} we obtain, for $\tau \geq 1$,
	\begin{eqnarray*}
	\mathcal{E}_{loc}\left(\tau\right)&\leq&
\mathcal{E}_{loc}\left(\lfloor\tau\rfloor\right)\leq
 (b_{0}t_0)^2\mathcal{E}_{loc}\left(\lfloor\tau\rfloor-2\right)\leq\ldots
\leq {(b_{0}t_0)^{\lfloor\tau\rfloor}\mathcal{E}_{loc}\left(0\right)}\nonumber\\
  &=&  \mathcal{E}_{loc}\left(0\right)e^{\lfloor\tau\rfloor\log (b_{0}t_0)} \leq
	 \mathcal{E}_{loc}\left(0\right)e^{(\tau-1)\log (b_{0}t_0)}
	=\mathcal{E}_{loc}\left(0\right)\frac{e^{\tau\log (b_{0}t_0)}}{b_0t_0},\label{eq:E_loc_dec_alt}
	\end{eqnarray*}
	which, due to
        {$\log (b_{0}t_0)<0$}, shows an exponential decay {in $\tau$}.

	When scaling back to the original variables $(x,t)$ we obtain
\[
  \tilde{\mathcal{E}}_{loc}\left(t\right) :=
  \mathcal{E}_{loc}\left(\frac{t}{t_0}\right)
  \le \tilde{\mathcal{E}}_{loc}\left(0\right)\frac{e^{\frac{t}{t_0}\log (b_{0}t_0)}}{b_0t_0}, \hspace{1em}t\geq t_0.
\]
	Step 3 of the proof of Theorem \ref{thm:1D_decay} can then be repeated (with a slight modification
	of the multiplicative constants), yielding the exponential convergence (with rate \eqref{Lambda})
	of the solution to the constant $u_{\infty}$.
\qed

\section{Conclusion}
\label{sec:conc}

For the homogeneous wave equation in $\mathbb R$ with
compactly supported initial data, in the case of constant coefficients,
the solution converges to the
steady state in finite time. However, exponential
convergence is the generic scenario in the case of
variable coefficients.

We proved two different results for
this convergence in local energy. The first one employs
a multiplier technique, constructing appropriate weight functions
to define a modified local energy that satisfies a first order
differential inequality. This allowed us to prove
the exponential decay.
Using the solution balance over space-time rectangles, we were
able to identify the constant steady state.
The second result improves the first one, under an additional
  assumption on the variation of the coefficients.
After rescaling the equation, its
integral form allowed us to estimate
the local energy by iterating in time, leading to an
exponential decay with improved rates.

Our results complement those of algebraic time-decay
presented in~\cite{Charao}, where
  the initial data were assumed to be
  sufficiently localised but not compactly supported. In that case, exponential decay is generally not expected.

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text:
New construction: Stadion Čair
23.11.2012 16:00 source: StadiumDB.com; author: michał
All over Serbia new stadiums are being built or renovated so it's about time first construction site from this country gets a presentation here, at StadiumDB.com. For starters we give you Niš, the country's third biggest city.
The centrally located stadium in Niš (Nish) was first opened in 1963. But since that time hasn't been through even one major renovation, so no wonder its condition began deteriorating to a stage at which the eastern stand had to be closed to the public in 1990s.
First concept for a revamped stadium in Niš was created back in 2004, but never came to fruition. One from 2010, very similar in terms of seating layout, was more lucky with demolition works on three stands (goal ends north and south and eastern stand) beginning in mid-2011, as planned.
View from summer of 2012. Photo: Novosti.rs
First phase left the main grandstand untouched, allowing for constant football and athletic use of the ground during construction process. The prefab structures on three sides of the field were quite easy to erect and cheap (with budget of RSD925m or some €10m), but were not finished in the initial timeframe – by the end of 2012. Both ends were opened for the 2012/13 season, but eastern stand was delayed to 2013.
After first phase capacity is 18,157 with curved ends having 9,037 and the double-tiered eastern stand 6,966. When (or if) the main stand gets revamped, it is to resemble the opposite one, raising capacity to some 23,000. Interestingly, after all works are done, the ground will still retain its slender floodlight masts dating back to late 1960s.
Related stadium designs
Stadion Čair
Related stadiums under construction

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However, research by the OECD and The MasterCard Foundation Youth Think Tank shows that there is a pervasive "skills mismatch" between what young people learn in secondary schools and universities and the skills they need to participate in an increasingly digitized workforce. Many Rwandans remain stuck in low-paying jobs due to limited skills.
Our TTW Workshop provides young people with training in foundational computer skills and relevant software programs like Microsoft Word, Excel, and PowerPoint. It also includes job application guidance, such as resume-writing, cover letters and interviews.
Further, the program emphasizes training in business soft skills, which are critically important but often not taught. Lessons in time management, public speaking, project prioritizing, and office communication etiquette, among others, help to provide young people with a holistic introduction to the kinds of skills they will need in the workplace.
Out of 50 spots available in the first round of the workshop, more than 400 young people applied—a statistic which highlights the incredible need for these kinds of programs in Rwanda, where many college graduates remain persistently unemployed.
We are happy to report, however, that by the end of our first Technology Training for the Workplace workshop, almost one-fourth of the class had secured a job! Many students cited the resume and interview skills lessons as crucial in helping them navigate and excel in the job application process.
Moving forward, we'll be staying in touch with our TTW graduates to find out how this program has impacted their experiences both in the job search and in the workplace. We're excited to see these aspiring entrepreneurs, accountants, NGO leaders, and social change-makers put their knowledge into action—paving the way for generations to come.


text:
Study of the Impact of Long-Term Space Travel on the Astronauts' Microbiome
The Microbiome experiment investigates the impact of space travel on both the human immune system and an individual's microbiome (the collection of microbes that live in and on the human body at any given time).
Martin Delahunty
Global Director, Nature Partner Journals, Nature Research
To monitor the status of the crewmembers' microbiome and immune system and their interaction with the unique environment of the International Space Station (ISS), we will take periodic samples from different parts of the body and the surrounding ISS environment. As part of this study, the likelihood and consequences of alterations in the microbiome due to extreme environments, and the related human health risk, will be assessed.
Click for further details
Working within Nature Research's Open Research Group, I am Global Director for Nature Partner Journals based in London. I have responsibility across five global office locations to develop Nature Partner Journals, a new series of online-only, open access journals, published in collaboration with world-renowned partners. Launched in April 2014, the portfolio now includes 14 titles including 'npj Biofilms and Microbiomes'. I have a Degree in Natural Sciences specializing in Microbiology from Trinity College Dublin and a Masters in Business Administration from The Open University. Within the broader medical publications community, I serve as Secretary and Trustee for the International Society for Medical Publication Professionals.


Owen is shorter than all of his friends. Don't blow it. I should be happy for you. The crowd began to chant. Did you give it to her? Will this change things? Tell Jean it's not his fault. That's a tough cookie. Even Ravi was surprised to hear that. Go only there.
They're going to find your stash of cigarettes, and then you'll be in trouble. We had to run the gauntlet of entrenched defenders. Sean has to work tomorrow. When people are killed, they die. Who has taken my handbag? And now he rests at home. My mom gets up early every morning. People will question all the good things they hear about you but believe all the bad without a second thought. The sons are loved by the parents. I told her nothing.
Cristi and Diana went potholing. He found that all his efforts amounted to nothing. Can you handle it? The bear would have liked to have eaten the apple. I thought you said you didn't want to talk about it. The two boys raced each other home. His explanation is unconvincing. We will call you when your order arrives. I've always paid the rent. Astronomy is the science of heavenly bodies.
He wore a coat too large for him. Now my daughter fears me. Drew shot himself in the foot. Jesus pretended to be angry even though he really wasn't. He is given to drink. I agree with you completely. Vidhyanath can come, too, if he wants to. I've done my part. A cucumber is related to a watermelon. I am rich.
We were left without gas when we were on our way. I'll tell him later. Republican leaders denied the charge. So who are we talking about? Our car is being repaired at the auto shop. The parents named the baby Akiyoshi. My wife and I prefer to spend time at home together. Hello, my name is Pekka. What is your name? We took it for granted that he would approve of the plan. I have something to tell Tal.
Jess can fix anything. I really like the technique called chiaroscuro. Pilot finished loading the truck. He has a lot of money. The whole crew was saved. How can you be sure it'll work? This shouldn't be allowed. We'd better wake Rolfe up. She declined to say more about it. My dad doesn't like soccer.
You'll never know. You'd better take another look. That bridge isn't strong enough to support so much weight. Where did he go yesterday? You're out of excuses. Would you like to go out to have a drink somewhere? I can't believe I forgot about that. It's a matter of either or. I did my best today. I didn't get an email from you today.
You shall have joy, or you shall have power, said God; you shall not have both. I have three years of experience. Today the ozone layer is thinning. I don't want to go away. What do you think Francis was running away from? Darci and I have fun together. These animals performed a service to their respective countries that no human could or would have performed. They gave their lives and/or their service in the name of technological advancement, paving the way for humanity's many forays into space. Suyog heard a shout. You should help your father. I have fewer students in my class this year than last year.
He is not to be trifled with. I can't tell for sure. What is right in one society can be wrong in another. Do you like your hot chocolate? I am not the captain of the new team. The capital of Morocco is Rabat. It tasted good. I'm having problems with him. This area is off limits to the public. We aren't friends.
Don't ask me. Ask them. Everybody still has a chance. I think you still love Srivatsan. He is quite tall. The magpie and the galah sat on the same tree branch. That made me happy. They are my favourite birds. Could you take off the price tag for me? My wristwatch is running a bit too fast. No other student in the class is as brilliant as he is. Mongo is a great admirer of yours. Your prepuce really bothers you. You think it's feminine.


In the very same year that Bowie's Ziggy landed to re-make rock in its image, Brian Ferry and his virtuoso band—including standouts Phil Manzanera on guitar and Brian Eno on synths, tape effects, and various "treatments"—prefigured a somehow even sexier, weirder, funkier, more disturbing future for pop, charting the territory for bands like Duran Duran, the Cars, Eurythmics, Pulp, and too many more to name. Roxy Music was so effortlessly original that once Bowie exhausted his space alien phase, he turned to Ferry and Eno for inspiration.
Like Bowie, Roxy Music favored saxophones, courtesy of Andy Mackay, who also played… the oboe? Manzanera's psychedelic flights were reminiscent of The Doors' Robby Krieger, with a Latin American flavor from his early days playing revolutionary Cuban folk songs. Paul Thompson's rhythmic pounding and smooth, country-ish grooves improbably married Moe Tucker and Kenny Buttrey.
Graham Simpson played the bass with "an exuberant rush," writes Kot. "They were specialists in their field," remarks Ferry," who himself drew from the rockers every British child of the 50s loved, but was also obsessed with Charlie Parker, Lester Young, Billie Holliday, Kurt Weill, the Beats, T.S. Eliot, Fred Astaire, and Cole Porter.
"Try to imagine," writes Gallagher, "how insane this TV footage looked" at the time. Imagine tuning in to Top of the Pops and catching them playing their debut single "Virginia Plain" (top), a song "named after a packet of cigarettes." (Read about how they recorded those motorcycle sounds.) Imagine seeing Mackay dressed like a Flash Gordon villain, playing oboe over Eno's sci-fi synth washes in the intro to "Ladytron" on the Old Grey Whistle Test, or seeing the band confidently stomp through "Re-make/Re-model," "Ladytron," and "Grey Lagoons," on the BBC's Full House, further up.
In that later 1972 live televised performance, Roxy Music was already delivering the sound of its future with "Grey Lagoons" from the following year's brilliant For Your Pleasure, the final album to feature Eno, who would go on to even stranger things in his solo work. Now imagine you happened to tune in to The Old Grey Whistle Test in '73 just in time to catch that album's "In Every Dream Home a Heartache," a warbly, sinister, Ballardian love song written for a blow-up doll.


text:
Outlook-Drshti Survey
Fits And Start-Ups: The A+ Schools
Although the start-up scene remains abuzz with activity, many B-schools are setting grads on the old-school employability path
By Arindam Mukherjee
Photograph by Mayur Bhatt
Arindam Mukherjee	September 22, 2016 00:00 IST Fits And Start-Ups: The A+ Schools
In the last two years, there has been considerable hype about start-ups and their potential to change the entrepreneurial landscape in the country. This has been accompanied by a frenzy among B-school graduates to run after start-ups, lured by bright prospects and the fat pay cheques they dished out. Accordingly, start-ups have been prominent first-day recruiters at all top B-schools, picking up the creme from campuses. But this year, things have changed rather drastically. With a sudden downtrend in this sunrise sector, start-ups have ceased to be the most favoured destination for B-school graduates. The reason is two-fold. First, the failure rate of start-ups has increased alarmingly, with a number of them biting the dust and news of lay-offs at others becoming the talk of town. Two, many start-ups have failed to keep their placement commitments and have either rescinded their offers or delayed joining dates. Recently the IITs together banned 31 companies from their placements for the same reasons. A majority of these were start-ups. Though start-up activity is still vibrant, there is both a slowdown in the pace at which they have picked up MBAs from campuses and a reluctance on part of students to join such ventures. The silver lining is that old-world brick-and-mortar companies in conventional businesses are recruiting from B-schools at a brisk pace, though not at the speed with which start-ups hired till last year. What is encouraging is a growing interest among students to look for these conventional companies, which assure stability in jobs as well as good growth prospects. In such a scenario, Outlook's ranking of India's 100 Best Business Schools, (in association with our research partner Drshti Strategic Research Services) assumes importance in helping students make an informed choice not just about colleges, but also about their careers. Like the last two years, there is little surprise in the top 15 in the Outlook ranking, with few being dislodged from the place they occupied last year. While there is a churn at the mid-level, the big difference is in the lower ranked colleges in the list, where over 30 new colleges have entered the charts by improving upon their parameters. While there was a noticeable improvement in academics of many colleges, there was also a genuine upliftment in the quality of research by faculty and students, which are reflected in the scores of many new entrants as well those that have risen in the ranks. What is not heartening is that the overall scores, which improved noticeably last year, have not shown much promise this year, with the overall total both at the topmost and the lowermost level dropping below last year's. But there has been encouraging results in the participation of colleges. Two years ago, looking at a good response in participation by B-schools to our survey, we had increased the list from 75 to 100. This year,...
Arindam Mukherjee	Business
More from Arindam Mukherjee
Why Jet Airways Nosedived And What's Next In The Turbulent Sky: A Dozen Issues
What iPhone, I Have The Blackberry!
Come Feb, New E-commerce Rules Could Shake Up Indian Market


text:
package com.backendless.persistence;

import com.backendless.commons.persistence.group.GroupingColumnValue;
import lombok.Getter;
import lombok.Setter;
import lombok.experimental.Accessors;

import java.util.ArrayList;
import java.util.List;

@Accessors( chain = true )
public class GroupDataQueryBuilder
{
  private final PagedGroupQueryBuilder<GroupDataQueryBuilder> pagedQueryBuilder;
  private final QueryOptionsBuilder<GroupDataQueryBuilder> queryOptionsBuilder;
  @Setter
  private boolean distinct = false;
  private List<GroupingColumnValue> groupPath;
  @Setter @Getter
  private int groupDepth;
  private final ArrayList<String> properties;
  private final ArrayList<String> excludeProperties;
  @Setter @Getter
  private String whereClause;
  private final List<String> groupBy;

  private GroupDataQueryBuilder()
  {
    properties = new ArrayList<>();
    excludeProperties = new ArrayList<>();
    pagedQueryBuilder = new PagedGroupQueryBuilder<>( this );
    queryOptionsBuilder = new QueryOptionsBuilder<>( this );
    groupBy = new ArrayList<>();
    groupPath = new ArrayList<>();
  }

  public static GroupDataQueryBuilder create()
  {
    return new GroupDataQueryBuilder();
  }

  public BackendlessGroupDataQuery build()
  {
    BackendlessGroupDataQuery dataQuery = pagedQueryBuilder.build();
    dataQuery.setDistinct( getDistinct() );
    dataQuery.setQueryOptions( queryOptionsBuilder.build() );
    dataQuery.setProperties( properties );
    dataQuery.setExcludeProperties( excludeProperties );
    dataQuery.setWhereClause( whereClause );
    dataQuery.setGroupBy( groupBy );
    dataQuery.setGroupPath( groupPath );
    dataQuery.setGroupDepth( groupDepth );

    return dataQuery;
  }

  /*--- Auto-generated code ---*/

  public GroupDataQueryBuilder setPageSize( int pageSize )
  {
    return pagedQueryBuilder.setPageSize( pageSize );
  }

  public GroupDataQueryBuilder setGroupPageSize( int groupPageSize )
  {
    return pagedQueryBuilder.setGroupPageSize( groupPageSize );
  }

  public GroupDataQueryBuilder setRecordsPageSize( int recordsPageSize )
  {
    return pagedQueryBuilder.setRecordsPageSize( recordsPageSize );
  }

  public GroupDataQueryBuilder setOffset( int offset )
  {
    return pagedQueryBuilder.setOffset( offset );
  }

  public GroupDataQueryBuilder prepareNextPage()
  {
    return pagedQueryBuilder.prepareNextPage();
  }

  public GroupDataQueryBuilder preparePreviousPage()
  {
    return pagedQueryBuilder.preparePreviousPage();
  }

  public boolean getDistinct()
  {
    return distinct;
  }

  public List<GroupingColumnValue> getGroupPath()
  {
    return new ArrayList<>( groupPath );
  }

  public GroupDataQueryBuilder setGroupPath( List<GroupingColumnValue> groupPath )
  {
    this.groupPath = new ArrayList<>( groupPath );
    return this;
  }

  public List<String> getProperties()
  {
    return (List<String>) this.properties.clone();
  }

  public GroupDataQueryBuilder setProperties( List<String> properties )
  {
    this.properties.clear();

    if (properties != null)
      for( String prop: properties )
        this.addProperty( prop );

    return this;
  }

  public GroupDataQueryBuilder setProperties( String... properties )
  {
    this.properties.clear();
    this.addProperties( properties );
    return this;
  }

  public GroupDataQueryBuilder addProperties( String... properties )
  {
    if( properties != null )
      for( String prop : properties )
        this.addProperty( prop );

    return this;
  }

  public GroupDataQueryBuilder addProperty( String property )
  {
    if( property != null && !property.equals( "" ) )
      properties.add( property );

    return this;
  }

  public GroupDataQueryBuilder addAllProperties()
  {
    this.addProperty( "*" );
    return this;
  }

  public ArrayList<String> getExcludedProperties()
  {
    return (ArrayList<String>) excludeProperties.clone();
  }

  public GroupDataQueryBuilder excludeProperties( ArrayList<String> excludeProperties )
  {
    this.excludeProperties.clear();

    if( excludeProperties != null )
      for( String exclProp: excludeProperties )
        this.excludeProperty( exclProp );

    return this;
  }

  public GroupDataQueryBuilder excludeProperties( String... excludeProperties )
  {
    this.excludeProperties.clear();

    if( excludeProperties != null )
      for( String exclProp: excludeProperties )
        this.excludeProperty( exclProp );

    return this;
  }

  public GroupDataQueryBuilder excludeProperty( String excludeProperty )
  {
    if( excludeProperty != null && !excludeProperty.isEmpty() )
      this.excludeProperties.add( excludeProperty );

    return this;
  }

  public List<String> getSortBy()
  {
    return queryOptionsBuilder.getSortBy();
  }

  public GroupDataQueryBuilder setSortBy( List<String> sortBy )
  {
    return queryOptionsBuilder.setSortBy( sortBy );
  }

  public GroupDataQueryBuilder setSortBy( String... sortBy )
  {
    return queryOptionsBuilder.setSortBy( sortBy );
  }

  public GroupDataQueryBuilder addSortBy( String sortBy )
  {
    return queryOptionsBuilder.addSortBy( sortBy );
  }

  public List<String> getRelated()
  {
    return queryOptionsBuilder.getRelated();
  }

  public GroupDataQueryBuilder setRelated( List<String> related )
  {
    return queryOptionsBuilder.setRelated( related );
  }

  public GroupDataQueryBuilder setRelated( String... related )
  {
    return queryOptionsBuilder.setRelated( related );
  }

  public GroupDataQueryBuilder setFileReferencePrefix( String fileReferencePrefix )
  {
    return queryOptionsBuilder.setFileReferencePrefix( fileReferencePrefix );
  }

  public GroupDataQueryBuilder addRelated( List<String> related )
  {
    return queryOptionsBuilder.addRelated( related );
  }

  public GroupDataQueryBuilder addRelated( String related )
  {
    return queryOptionsBuilder.addRelated( related );
  }

  public Integer getRelationsDepth()
  {
    return queryOptionsBuilder.getRelationsDepth();
  }

  public GroupDataQueryBuilder setRelationsDepth( Integer relationsDepth )
  {
    return queryOptionsBuilder.setRelationsDepth( relationsDepth );
  }

  public List<String> getGroupBy()
  {
    return new ArrayList<>( this.groupBy );
  }

  public GroupDataQueryBuilder setGroupBy( String... groupBy )
  {
    this.groupBy.clear();
    this.addGroupBy( groupBy );
    return this;
  }

  public GroupDataQueryBuilder addGroupBy( String... groupBy )
  {
    for( String grb : groupBy )
    {
      if( grb != null && !grb.equals( "" ) )
      {
        if( !grb.startsWith( "'" ) && !grb.endsWith( "'" ) )
          grb = "'" + grb + "'";
        this.groupBy.add( grb );
      }
    }

    return this;
  }

  public GroupDataQueryBuilder setRelationsPageSize( Integer relationsPageSize )
  {
    return queryOptionsBuilder.setRelationsPageSize( relationsPageSize );
  }

  public Integer getRelationPageSize()
  {
    return queryOptionsBuilder.getRelationsPageSize();
  }
}

redpajama_set_name: RedPajamaGithub


text:
Mind games: The battle of wits that unites snooker and Hearthstone
The agony and the ecstasy.
Article by	John Bedford, Editor in Chief, Metabomb
Updated on 6 May 2019
The final frames of this year's World Snooker Championship will take place later today, and a new Hearthstone meta is in full swing too. Republished from 2015, here's John Bedford's take on the curious mind games that unite the worlds of the Crucible and the tavern.
It's hard to imagine two more polarised moments in snooker's history. Back in 1997, Ronnie O'Sullivan - arguably the most talented player of all time, and alongside Jimmy White and Alex Higgins certainly one of the most popular - completes a maximum break of 147 in an unprecedented five minutes and 20 seconds.
A man so adept at the task before him that he appears almost bored by his own capabilities, O'Sullivan moves from shot to shot, sizing up each problem, divining the solution, and then executing it with almost reckless speed. He seems to almost dare himself to fail - and no one watching can quite bring themselves to believe he will.
Fast forward less than a decade, and O'Sullivan - now cutting an almost Syd Barrett-like figure with his closely shaven head and tired eyes - is playing against Peter Ebdon, a man who could charitably be called a master of the waiting game. Every angle before him is scrutinised, and every possibility is weighed, considered and checked against the rest. If O'Sullivan's a gambler, Ebdon is an accountant.
By the time the clock has ticked slowly beyond the five or so minutes it once took O'Sullivan to perform that record-breaking clearance, Ebdon has achieved a break of 12. Even worse, he finishes his turn by fumbling a red into the jaws of the pocket. As Ebdon settles into his chair - and in a rare break from snooker's gentlemanly code of conduct - Ronnie grins on behalf of the nation. As he leers at the back of Ebdon's head, I can almost hear it: "Sorry that happened..."
It's a pity both commentary tracks have been left running in this comparison video, but if a picture paints a thousand words...
I've recently come to realise that, throughout my life, I've been drawn towards a certain type of game - and nothing has come to highlight this quite like my passion for Hearthstone and for snooker. The differences between the two games are transparent enough, but it's what unifies them that makes them so captivating to watch and to play.
Both, for example, demand a high degree of forward planning based on incomplete information. Those who focus rigidly on the moment don't climb very high on Hearthstone's ladder, and they certainly don't win many frames of snooker. To win, you must instead grasp the problem at hand and play with it like a toy, turning objects of mathematical pleasure this way and that, exposing them to different lights, and determining the best possible course of action before committing to it. To win, you must be able to speculate and take risks.
Beyond that, I think there's a darker truth at work too. For someone to emerge victorious from these games of pause, reflection, consideration and execution, it's often not enough to simply win. The person who loses must also throw it away to one degree or another, whether through poor execution, inexperience or greed. When your turn comes, you have all the time and space in the world to be as terrible as you're capable of being.
Crucially, O'Sullivan's 147 - entertaining though it is to watch - is an anomaly in the game of snooker, much as a perfect starting hand against an impossibly weak one can occur between two otherwise equal and exceptional Hearthstone players. These occasions are inevitable given the rules of each game, but if they happened with any kind of regularity both would quickly lose their appeal.
Instead they derive their bite from the downtime that players endure after making a mistake or an inefficient play, the downtime during which the opponent is most likely capitalising on it. It's this lonely stretch of time you're given to reflect on your shortcomings that undoes you in the long run. Your opponent is mute while you wait for your next opportunity, and yet it's easy to sense their actions betraying a host of unspoken insults that gnaw and niggle at you.
The challenge of execution is very different in both games, of course, yet the smallest of mistakes under pressure can cause equally catastrophic outcomes.
Like Ebdon they sometimes stall, not because they're slowly thinking their play through, of course, but because they want to frustrate you. They toy endlessly with the delicious fruits of your mistakes, as though finding themselves spoilt for choice. Is that really the case though, or are your own devils gently whispering your inadequacies to you through the prism of your opponent?
Either way, you must be ready to capitalise on your opponent's next mistake, and do so without missing a beat. Most psychologically damning of all is that moment when their shoulders slump and the adrenaline surges through you with mean-spirited triumph, while they head off to endure the same unbearable fate you know only too well.
This guilty, unspoken sadism isn't the exclusive preserve of the players either - the audience in both games is complicit in every drop of it too. It's not just the technical appreciation of a game well played, and the illicit enjoyment of a psychological battle that draws spectators to both games. They also get to experience this agony and this ecstasy vicariously, and with none of the personal risk. For our collective sins, we at least veer towards cheering the underdog - we know how much more they suffer.
It's remarkable that on the surface neither game betrays the true depth of these emotions, and yet the concept of tilt is a crucial component of both. In this fractured state of mind, frustration takes hold and you begin playing with your heart, rather than your head. I've experienced it myself when it seems like nothing you do is good enough to climb higher on Hearthstone's ladder, and instead of taking a break you slide ever further down the rungs.
This state of tilt changes everything. The match, the world, and everything that takes place within it becomes a source of injustice, rather than the result of a personal oversight, or someone's transitory good fortune. Left unchallenged, these thoughts feed on your irrational beliefs until you slip further and further behind with each encounter. By contrast, when you're hot - like O'Sullivan picking apart the 147 - you're not even at the controls anymore.
When you slide, you must make peace with your past incompetence if you're to have any hope of rising above it. Only the task ahead of you can be allowed to exist, and you must silence the whispers of doubt that come at you, now armed with the hard evidence of your own inadequacy. So what if you're 9-2 down in a best of 19 match anyway? All you have to do now is not lose eight times and you're home free.
If you'd like to learn more about what a fascinating character Ronnie O'Sullivan is, I recommend reading the New Yorker's recent profile: The Unhappy King of Snooker - if only to picture a US audience pondering Ronnie's impossibly English 'My arsehole had gone.'
It all begs the question of why anyone would want to participate in these experiences that promise so much psychological punishment, and where supreme self-belief meets crushing self-doubt and both refuse to let go of the other. It would be crass to reduce the demons of some of snooker's most notably troubled players - White, Higgins, O' Sullivan - to a single cause, but it's at least interesting to wonder what could possibly draw such fragile personalities towards a sport that holds the potential for so much torment.
From my own perspective, these kind of games are intrinsically risky to participate in, and that's precisely what makes makes them so compelling to experience. You challenge the negative and positive beliefs about yourself in equal measures, and while it's unquestionably painful to misstep, to rise above your own failings is exquisite. You want to be better than the quiet voices of self-doubt would have you believe you are, and these games provide an arena within which to test them - with all the risk that entails.
Perhaps we simply play these games because we are the most interesting player any of us can ever hope to challenge - an opponent rich in belief, desire, pride, potential, competitiveness, and - ultimately - a capacity for failure despite all of these things. Hearthstone streamer Jeffrey "Trump" Shih once touched on this in a moment of insight so perfectly profound in its simplicity that he deserves the final word on the subject.
"You have an opponent," he says. "But really it's a game against yourself."
More about Hearthstone: Heroes of Warcraft
Hearthstone: Heroes of Warcraft Review
John Bedford
Editor in Chief, Metabomb
John is Metabomb's Editor in Chief, and is responsible for all of the Hearthstone news, features and guides content on the site.
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