Earlier this week, Apple changed its App Store policy to allow older versions to be downloaded. This policy has since been updated.
Apple's change of heart means older iDevices not capable of running the latest iOS can carry on being used. Check out my earlier article, 'Last Compatible' App Versions Now Offered for Older iDevices.
However, it would appear that not all developers were happy about this, claiming bug fixes fixed in newer versions or old APIs could cause problems and security issues.
Apple seem to have listened and has now given developers the option to offer old versions or not. A good move by Apple, meaning now if a problem occurs it will be down to the developer allowing that app version to be downloaded and not because of a 'free-for-all' policy by Apple.
Previous versions of your apps are now available for redownload by users who have already purchased them, allowing customers to use your apps with older devices which may no longer be supported by the current version of your app. If you do not wish to make these versions available, you can manage the availability of your apps' previous versions in the Rights and Pricing section of the Manage Your Apps module in iTunes Connect.
The only part I am a little unclear about is the 'redownload by users who have already purchased them'. This sounds like a major U-turn to me – or am I just reading it wrong?
This sounds like your Apple ID has to have already downloaded it previously, or does this only apply to paid for apps?
What about somebody who has never owned an iPhone and, because of the change in attitude towards older app versions, decides to pick up an iPhone 3G or 3GS. These wouldn't have downloaded any apps previously.
Would they still have access to free apps like Facebook, Twitter, YouTube, Shazam, eBay, Sky+, or XE Currency, for example – all of which work in iOS 6, will no doubt update to require iOS 7? If not, then there is no mileage in picking up an older iPhone or iPad, thus undoing the good news of the 'last compatible' app availability.
I hope this gets clarified.
'Purchased' in App Store parlance includes free apps that have been downloaded by the user. They were still 'purchased', even though they have no cost.
Accordingly, you will find free apps under the 'Purchased' section in the App Store app, both in iOS and OS X.


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Search results for "Dot To Dot Writing font".
Free Dot To Dot Writing fonts on the Web : No font previews found.
Free Dot Writing fonts overview. 21 fonts found. Preview, download and install the Dot Writing.ttf file.


Seven vivid, sensitive tales by author of The Red Badge of Courage. Fine title story plus "Miraculous Soldiers," "A Mystery of Heroism," "A Gray Sleeve," and three more.
Known as the Great Compromiser, Henry Clay earned his title by addressing sectional tensions over slavery and forestalling civil war in the United States. Today he is still regarded as one of the most important political figures in American history. As Speaker of the House of Representatives and secretary of state, Clay left an indelible mark on American politics at a time when the country's solidarity was threatened by inner turmoil, and scholars have thoroughly chronicled his political achievements. However, little attention has been paid to his extensive family legacy. In The Family Legacy of Henry Clay: In the Shadow of a Kentucky Patriarch, Lindsey Apple explores the personal history of this famed American and examines the impact of his legacy on future generations of Clays. Apple's study delves into the family's struggles with physical and emotional problems such as depression and alcoholism. The book also analyzes the role of financial stress as the family fought to reestablish its fortune in the years after the Civil War. Apple's extensively researched volume illuminates a little-discussed aspect of Clay's life and heritage, and highlights the achievements and contributions of one of Kentucky's most distinguished families.
The Rise of the Computer State is a comprehensive examination of the ways that computers and massive databases are enabling the nation's corporations and law enforcement agencies to steadily erode our privacy and manipulate and control the American people. This book was written in 1983 as a warning. Today it is a history. Most of its grim scenarios are now part of everyday life. The remedy proposed here, greater public oversight of industry and government, has not occurred, but a better one has not yet been found. While many individuals have willingly surrendered much of their privacy and all of us have lost some of it, the right to keep what remains is still worth protecting.
Explores the story of Federal contributions to dam planning, design, and construction.
"The log of the clay worker": v. 100, p. 188-193.


This is a really tasty summery drink. The coconut in the Malibu Rum mixes really well with the smoothie and I bet you can even claim that this is healthy because of the high fruit content. We tried this with cream and found it too sickly. It was quite nice with just a drop of lemonade added in but personally I thought it was best just as it was.
Place all the ingredients except chocolate in a blender.
Sprinkle grated chocolate over the top, if wanted.
Previous PostWhy I'm proud of my mummy tummy.
Next PostBlogger challenge day 1: My blogs name.
I would have never thought to add chocolate to this, but it sounds like an interesting twist! I'll have to give it a try. XOXO.


text:
Hidden treasure garden plays host to family sessions
Ailie's Garden plays host to NHS family sessions
By Rajmund Bakonyi @thecosmicraj Trainee Reporter
The project was one of many benefiting from funding
A HIDDEN treasure in Stirling has been playing host to sessions, supporting young families during the pandemic.
Play sessions have been going ahead at Ailie's Garden behind the Smith Art Gallery and Museum which serves as an open-air gallery, complete with musical instruments and games.
The sessions have been going ahead for children and families supported by NHS Forth Valley's Health Visiting Support Team in Stirling, which teamed up with the museum thanks to the help of a Covid-19 response grant.
The grant also helped promote virtual sessions after a decision was taken to postpone face-to-face appointments.
The aim was to provide an inclusive service, universally available and accessible to as many individuals and families as possible.
These sessions cover a wide range of subjects and include baby massage, benefits of playing with a child, sleep, managing behaviour and more.
Teresa McQuillian, nursery nurse with the support team, said: "Although we are still unable to reinstate face-to-face groups, we have been fortunate to be able to provide parents with the opportunity to attend our outdoor groups.
"We have worked with our colleagues from the Smith Art Gallery and Museum to offer groups for children under the age of one and those aged between one and three.
"Through the group for children aged under one, parents have been able to meet up in a Covid-safe environment and benefit from social interactions and discussions around health-related topics.
"The group for children aged one to three has focused on motor skills, balance, co-ordination and, of course, having fun.
"Despite the sometimes inclement weather parents and little ones have enjoyed meeting up and engaging in activities while benefiting from being in a lovely, peaceful outdoor environment.
"This partnership working with the local museum has been extremely positive."
Click here for more information on the community grants scheme, which aims to help organisations and community groups across the Forth Valley improve mental health and wellbeing.

redpajama_set_name: RedPajamaCommonCrawl


text:
\section{Introduction}

We consider small transversal vibrations of a uniform string, with a time
dependent length. The mechanical setting is sketched in Figure \ref{fig1}
where the left end of the string is fixed while the right moving end is also
allowed to move transversely and attached to a damping device (a dash-pot
with a damping factor $\eta \geq 0$).

\begin{figure}[tbph]
\centering\includegraphics[width=0.67\textwidth]{GhS3_fig1.jpg}
\caption{A string with one moving end subject to a dash-pot damping.}
\label{fig1}
\end{figure}

Denoting the displacement function by $u$, depending on the position $x$
along the string and the time $t$, the model can be stated as follows
\begin{equation}
\left\{
\begin{array}{ll}
u_{tt}-u_{xx}=0,\ \smallskip & \text{for }0<x<\ell (t)\text{ and }t>0, \\
\left( 1+\eta \ell ^{\prime }\left( t\right) \right) u_{x}\left( \ell \left(
t\right) ,t\right) +\left( \eta +\ell ^{\prime }\left( t\right) \right)
u_{t}\left( \ell \left( t\right) ,t\right) =0, \smallskip &
\text{for }t>0, \\
u\left( 0,t\right) =0\text{,}\smallskip & \text{for }t>0, \\
u(x,0)=u^{0}\left( x\right) ,\text{ }u_{t}\left( x,0\right) =u^{1}\left(
x\right) ,\text{ \ } & \text{for }0<x<L.%
\end{array}%
\right.  \tag{WP}  \label{waveq}
\end{equation}%
The
subscripts $t$ and $x$ in (\ref{waveq}) stand for the derivatives in time and space variables
respectively. The functions $u^{0}$ and $u^{1}$ represents the initial shape
and the initial transverse speed of the string, respectively. The initial
length of the string is denoted by $L=\ell (0)$.

We assume that $\ell \in C\left( \left[ 0,+\infty \right[ \right) $ and that%
\begin{equation}
\left\vert \ell ^{\prime }\left( t\right) \right\vert <1,\ \ \ \text{for }%
t\geq 0,  \label{tlike}
\end{equation}%
which means that the speed of variation of the length of the string $\ell
^{\prime }\left( t\right) $ is strictly less then the speed of propagation
of the wave (here simplified to $c=1$).

In this paper, we are mainly interested in the asymptotic behaviour in time
of the energy of the solution, defined as%
\begin{equation}
E_{\ell }\left( t\right) :=\frac{1}{2}\int_{0}^{\mathbf{\ell }\left(
t\right) }u_{t}^{2}\left( x,t\right) +u_{x}^{2}\left( x,t\right) dx,\ \ \
\text{for }t\geq 0.  \label{E}
\end{equation}

For the time-independent interval, i.e. when $\ell \left( t\right) =L$ for $%
t\geq 0$, it is well known that:

\begin{itemize}
\item If $\eta =0$, then it is the energy is constant, i.e. $E_{L}\left(
t\right) =E_{L}\left( 0\right) ,$ for $t\geq 0$.

\item If $\eta \geq 0$ with $\eta \neq 1\ $, then the energy decays
exponentially%
\begin{equation*}
E_{L}\left( t\right) \leq CE_{L}\left( 0\right) e^{-\frac{1}{L}\ln
\left\vert \gamma _{\eta }\right\vert t},\text{ for some constant }C>0,
\end{equation*}%
where $\gamma _{\eta }:=\frac{1+\eta }{1-\eta }$, see \cite%
{Vese1988,CoZu1995,Cher1994,QuRu1977}. In \cite{GhSe2022a}, the present
authors showed that%
\begin{equation}
\frac{1}{\gamma _{\eta }^{2}}E_{L}\left( 0\right) e^{-\frac{1}{L}\ln
\left\vert \gamma _{\eta }\right\vert t}\leq E_{L}\left( t\right) \leq
\gamma _{\eta }^{2}E_{L}\left( 0\right) e^{-\frac{1}{L}\ln \left\vert \gamma
_{\eta }\right\vert t},\ \ \ \text{for }t\geq 0.
\end{equation}%
\end{itemize}

The question of $E_{\ell }\left( t\right)$ behaviour in time is more
delicate if the interval depends on time. For instance, for the Dirichlet
boundary conditions (which corresponds to $\eta =+\infty $ in (\ref{waveq}%
)), the energy decays if the interval is expanding and increases if the
interval is shrinking, see \cite{BaCh1981}. See also \cite%
{Seng2018,Seng2018a,SuLL2015} when the variation of the length is uniform in
time.

Regarding the case with a velocity feedback at the moving endpoint $x=\ell
\left( t\right) $:

\begin{itemize}
\item Gugat \cite{Guga2008} considered the case $u_{x}\left( \ell \left(
t\right) ,t\right) +cu_{t}\left( \ell \left( t\right) ,t\right) =0$ where $c$
is as constant. See also \cite{LuFe2020} for the particular $\ell (t)=1+vt$
where $v$ is a constant, $0<v<1$.

\item Ammari et al. \cite{AmBE2018} considered the case $u_{x}\left( \ell
\left( t\right) ,t\right) +f\left( t\right) u_{t}\left( \ell \left( t\right)
,t\right) =0,$ where $f\in L^{\infty }\left( 0,+\infty \right) \ $and $\ell
\left( t\right) $ is periodic. See also \cite{HaHo2019}\ where $\ell \left(
t\right) $ is not necessarily periodic. Mokhtari \cite{Mokh2022} considered
the case with two moving endpoints.
\end{itemize}

\emph{For the special case }$\eta =1$, the boundary condition at $x=\ell
\left( t\right) $ reads
\begin{equation*}
u_{x}\left( \ell \left( t\right) ,t\right) +u_{t}\left( \ell \left( t\right)
,t\right) =0.
\end{equation*}%
This is a transparent condition, i.e. there is no reflections of waves from
the moving endpoint and consequently all the initial disturbances leave the
interval $\left( 0,\ell \left( t\right) \right) $ at most after a time
\begin{equation*}
T_{\ell }:=\beta ^{-1}\left( L\right) ,
\end{equation*}%
where $\beta \left( t\right) :=t-\ell \left( t\right) ,$ see Figure \ref%
{fig2}. Hence, wether the interval is expanding or shrinking, the linear
velocity feedback $-u_{t}\left( \ell ^{\prime }\left( t\right) ,t\right) $
steers the solution to the zero state in the finite time $T_{\ell }$. See
for instance \cite{Guga2008,HaHo2019}, and for the particular case $\ell
\left( t\right) =L$ see \cite{Vese1988,CoZu1995}. In the remaining of this
paper, we will assume that $\eta \geq 0$ and $\eta \neq 1.$
\begin{figure}[tbph]
\centering\includegraphics[width=0.3\textwidth]{GhS3_fig2.jpg}
\caption{An initial disturbance with a small support leaves the interval at
most after time $T_{\ell }$.}
\label{fig2}
\end{figure}

The approach used in the paper at hands is based on generalized Fourier
series. This is possible since the closed form for the solution of (\ref%
{waveq}) is given by the series%
\begin{equation}
u(x,t)=\sum_{n\in \mathbb{Z}}c_{n}\left( e^{\omega _{n}\varphi
(t+x)}-e^{\omega _{n}\varphi (t-x)}\right) ,\text{ \ \ \ for }0<x<\ell (t)%
\text{ and }t\geq 0,  \label{exact0}
\end{equation}%
where the coefficients $c_{n}$ can be explicitly computed in function of the
initial data $u^{0},u^{1}.\ $We denoted by $\omega _{n}$ a sequence of
complex number given by%
\begin{equation}
\omega _{n}:=-\frac{1}{2}\ln \left\vert \gamma _{\eta }\right\vert +\left\{
\begin{array}{ll}
\displaystyle\frac{2n+1}{2}i\pi , & \text{if }0\leq \eta <1,\smallskip \\
\displaystyle ni\pi , & \text{if }\eta >1,%
\end{array}%
\right.  \label{wn}
\end{equation}%
Observe that since $\left\vert \gamma _{\eta }\right\vert \geq 1$ for every $%
\eta \geq 0,$ the real part of $\omega _{n}$ is nonpositive.

By $\varphi ,$ we denoted a real function satisfying the functional equation
\begin{equation}
\varphi \left( t+\ell \left( t\right) \right) -\varphi \left( t-\ell \left(
t\right) \right) =2.  \label{Moor}
\end{equation}
This equation often called Moor's equation following his paper \cite{Moor1970}. Some pairs of solutions $(\ell, \phi)$ can be found in \cite{VePo1975}.

We will
assume that $\varphi $ is differentiable and increasing. More precisely,%
\begin{equation}
\varphi \in C^{1}\left( \left[ -L,+\infty \right[ \right) \text{ \ and \ }%
\varphi ^{\prime }>0,\text{ \ for }t\geq -L.  \label{mono}
\end{equation}%
The assumption $\varphi ^{\prime }>0$ is needed in the sequel since $\varphi
^{\prime }$ will serve as a weight for an $L^2$ space. Besides, a
deacreasing $\phi$ can not satisfy (\ref{Moor}) since $\ell \left(
t\right)>0 $. See \cite{HaHo2019}
for further discussions on the regularity of the solutions of (\ref{Moor}).

In this work, we demonstrate how the series formulas (\ref{exact0}) can be
used to achieve the following results:

\begin{itemize}
\item \emph{For the undamped case, i.e. $\eta =0$}, the energy of the
solution satisfies%
\begin{equation}
\frac{m\left( t\right) }{M\left( t_{0}\right) }E_{\ell }\left( t_{0}\right)
\leq E_{\ell }\left( t\right) \leq \frac{M\left( t\right) }{m\left(
t_{0}\right) }E_{\ell }\left( t_{0}\right) ,\text{ \ \ \ for }0\leq t_{0}<t,
\label{estim1}
\end{equation}%
where%
\begin{equation}
m\left( t\right) :=\min_{x\in \left[ 0,\ell \left( t\right) \right] }\left\{
\varphi ^{\prime }(t-x),\varphi ^{\prime }(t+x)\right\} \ \text{and }M\left(
t\right) :=\max_{x\in \left[ 0,\ell \left( t\right) \right] }\left\{ \varphi
^{\prime }(t-x),\varphi ^{\prime }(t+x)\right\} .  \label{m.M}
\end{equation}%
See Theorem \ref{th0damp} and its corollaries for sharper estimates.

\item \emph{For the damped case $\eta >0,$ with $\eta \neq 1$}, the energy $%
E_{\ell }\left( t\right) $ satisfies%
\begin{multline}
\left( \frac{e^{\ln \left\vert \gamma _{\eta }\right\vert \varphi
(t_{0}-\ell \left( t_{0}\right) )}}{M\left( t_{0}\right) }\right) m\left(
t\right) e^{-\ln \left\vert \gamma _{\eta }\right\vert \varphi (t+\ell
\left( t\right) )}E_{\ell }\left( t_{0}\right) \leq E_{\ell }\left( t\right)
\label{exp-decay} \\
\leq \left( \frac{e^{\ln \left\vert \gamma _{\eta }\right\vert \varphi
(t_{0}+\ell \left( t_{0}\right) )}}{m\left( t_{0}\right) }\right) M\left(
t\right) e^{-\ln \left\vert \gamma _{\eta }\right\vert \varphi (t-\ell
\left( t\right) )}E_{\ell }\left( t_{0}\right) ,\text{ \ for }0\leq t_{0}<t.
\end{multline}%
See Theorem \ref{th-stab1} and it corollaries for more estimates. The
estimate given in (\ref{exp-decay}), with explicit constants, is new to the
best to our knowledge.
\end{itemize}

After the present introduction, we derive the exact solution and the\
expression for the coefficients of the series formula (\ref{exact0}). In
Section 3, we establish upper and lower estimates for the energy $E_{\ell
}(t)$ of the undamped equation. In Section 4, we deal with the damped case.
Some examples will be included as a last section.
\section{Exact solution}

Let us introduce the following family of Hilbert spaces%
\begin{equation*}
V_{0}\left( 0,\ell (t)\right) :=\left\{ w\in H^{1}\left( 0,\ell (t)\right)
\text{, }w\left( 0\right) =0\right\} ,\ \ \ \text{\ for }t\geq 0
\end{equation*}%
and assume that the initial data satisfies%
\begin{equation}
u^{0}\in V_{0}\left( 0,L\right) ,\text{ \ }u^{1}\in L^{2}\left( 0,L\right) .
\label{ic}
\end{equation}

Let $T>0.$ Then, we have the following existence result for Problem (\ref%
{waveq}).

\begin{theorem}
\label{thexist1}Under the assumptions \emph{(\ref{tlike}), (\ref{mono}) }and
\emph{(\ref{ic}), }Problem \emph{(\ref{waveq})} has a unique solution
satisfying%
\begin{equation}
u\in C\left( [0,T];V_{0}\left( 0,\ell (t)\right) \right) \cap C^{1}\left(
[0,T];L^{2}\left( 0,\ell (t)\right) \right) ,  \label{regul}
\end{equation}%
\emph{\ } given by the series \emph{(\ref{exact0})} where the coefficients $%
c_{n}\in
\mathbb{C}
$ are computed as follows%
\begin{equation}
c_{n}=\frac{1}{4\omega _{n}}\int_{-L}^{L}\left( \tilde{u}_{x}^{0}+\tilde{u}%
^{1}\right) e^{-\omega _{n}\varphi (x)}dx\text{, \ \ \ for\ }n\in
\mathbb{Z}
,\   \label{cn}
\end{equation}%
where\emph{\ }$\tilde{u}_{x}^{0}$ is an even $($resp. $\tilde{u}^{1}$ is an
odd$)$ extension of the initial data $u^{0}$ $($resp. $u^{1})$ defined on
the interval $\left( -L,L\right) $. Moreover,%
\begin{equation}
\sum_{n\in \mathbb{%
\mathbb{Z}
}}\left\vert \omega _{n}c_{n}\right\vert ^{2}=\frac{1}{8}\int_{-L}^{L}\left(
\tilde{u}_{x}^{0}+\tilde{u}^{1}\right) ^{2}e^{\ln \left\vert \gamma _{\eta
}\right\vert \varphi (x)}\frac{dx}{\varphi ^{\prime }(x)}<+\infty .
\label{ncn}
\end{equation}
\end{theorem}

\begin{proof}
$\bullet $ \emph{The exact solution:} This part of solution is slightly
different from the approach in \cite{Vesn1971} where the author considered $1/\eta $ instead of $%
\eta $ in the boundary condition at $x=\ell(t)$. We include it here for the
sake clarity. The general solution of (\ref{waveq}) is given by D'Alembert's
formula
\begin{equation}
u(x,t)=f(t+x)+g\left( t-x\right) ,  \label{dlmbrt}
\end{equation}%
where $f$ and $g$ are arbitrary continuous functions. The boundary
conditions at the endpoint $x=0$, we have%
\begin{equation*}
f(t)=-g\left( t\right) .
\end{equation*}%
The condition at $x=\ell \left( t\right) \ $implies that%
\begin{equation*}
\left( 1+\eta \ell ^{\prime }\left( t\right) \right) \left[ f^{\prime
}(t+\ell \left( t\right) )-g^{\prime }\left( t-\ell \left( t\right) \right) %
\right] =-\left( \eta +\ell ^{\prime }\left( t\right) \right) \left[
f^{\prime }(t+\ell \left( t\right) )+g^{\prime }\left( t-\ell \left(
t\right) \right) \right] ,
\end{equation*}%
hence%
\begin{equation}
\left[ 1+\eta \ell ^{\prime }\left( t\right) +\eta +\ell ^{\prime }\left(
t\right) \right] f^{\prime }\left( \alpha \left( t\right) \right) =-\left[
1+\eta \ell ^{\prime }\left( t\right) -\eta -\ell ^{\prime }\left( t\right) %
\right] f^{\prime }\left( \beta \left( t\right) \right) ,  \label{+a1}
\end{equation}%
where $\beta \left( t\right) :=t-\ell \left( t\right) $ and $\alpha \left(
t\right) :=t+\ell \left( t\right) $. Then, noting that
\begin{equation}
\frac{1+\eta \ell ^{\prime }\left( t\right) -\eta -\ell ^{\prime }\left(
t\right) }{1+\eta \ell ^{\prime }\left( t\right) +\eta +\ell ^{\prime
}\left( t\right) }=\frac{1}{\gamma _{\eta }}\frac{\beta ^{\prime }\left(
t\right) }{\alpha ^{\prime }\left( t\right) },
\end{equation}%
we can rewrite (\ref{+a1}) as%
\begin{equation}
\alpha ^{\prime }\left( t\right) f^{\prime }\left( \alpha \left( t\right)
\right) =-\frac{1}{\gamma _{\eta }}\beta ^{\prime }\left( t\right) f^{\prime
}\left( \beta \left( t\right) \right) .
\end{equation}%
By integration, it follows that
\begin{equation}
f\left( \alpha \left( t\right) \right) =-\frac{1}{\gamma _{\eta }}f\left(
\beta \left( t\right) \right) +C.  \label{a+b}
\end{equation}

Let us assume for the moment that $C=0.$ Then, it is convenient to search
for $f$ in the form $f(\xi )=e^{\omega \varphi \left( \xi \right) }$, for a
constant $\omega $ and some function $\varphi $. Substituting $e^{\omega
\varphi \left( \xi \right) }$\ in (\ref{a+b}), we get
\begin{equation*}
e^{\omega \left[ \varphi \left( \alpha \left( t\right) \right) -\varphi
\left( \beta \left( t\right) \right) \right] }=-1/\gamma _{\eta }.
\end{equation*}%
Assuming that $\varphi $ satisfies (\ref{Moor}), we are led to the following
cases:

\begin{itemize}
\item[-] If $0\leq \eta <1$, then $\gamma _{\eta }\geq 1$ and we get
\begin{equation*}
e^{\omega \left[ \varphi \left( \alpha \left( t\right) \right) -\varphi
\left( \beta \left( t\right) \right) \right] }=e^{\left( 2n+1\right) i\pi
-\ln \gamma _{\eta }}.
\end{equation*}%
Solving this equation for $\omega $, we obtain a sequence of values $\omega
_{n},n\in
\mathbb{Z}
,$ where
\begin{equation*}
\omega _{n}=\frac{2n+1}{2}i\pi -\frac{1}{2}\ln \gamma _{\eta }.
\end{equation*}

\item[-] If $\eta >1$, we have $\gamma _{\eta }<-1$ and we obtain this time
\begin{equation*}
\omega _{n}=ni\pi -\frac{1}{2}\ln \left\vert \gamma _{\eta }\right\vert ,%
\text{ \ }n\in
\mathbb{Z}
.
\end{equation*}
\end{itemize}

Thus, if $\eta \geq 0$ and $\eta \neq 1$, we always have $\ln \left\vert
\gamma _{\eta }\right\vert \geq 1$ and $\omega _{n}$ given by (\ref{wn}).

Due to the superposition principal, it follows that $f$ can be written as%
\begin{equation*}
f\left( \xi \right) =\sum_{n\in \mathbb{Z}}c_{n}e^{\omega _{n}\varphi \left(
\xi \right) },\text{ \ \ \ }c_{n}\in
\mathbb{C}
,
\end{equation*}%
where $c_{n}$ are\ complex coefficients to be determined later. Since $%
f\left( \xi \right) =-g\left( \xi \right) ,$ then D'Alembert's formula for
the solution yields the series
\begin{equation}
u(x,t)=\sum_{n\in \mathbb{Z}}c_{n}\left( e^{\omega _{n}\varphi \left(
t+x\right) }-e^{\omega _{n}\varphi \left( t-x\right) }\right) ,\text{ \ for }%
0<x<\ell (t)\text{ and }t\geq 0.  \label{phi_cn}
\end{equation}

If $C\neq 0$ in (\ref{a+b}), then we can check that
\begin{equation*}
f\left( \xi \right) =\frac{C\gamma _{\eta }}{1+\gamma _{\eta }}+\sum_{n\in
\mathbb{Z}}c_{n}e^{\omega _{n}\varphi \left( \xi \right) },
\end{equation*}%
solves (\ref{a+b}). However, this will not affect the solution of (\ref%
{waveq}) since $f\left( \xi \right) =-g\left( \xi \right) $ and thus the
constant parts of $f$ and $g$ will be cancelled in the expression (\ref%
{phi_cn}).

$\bullet $ \emph{Computing the coefficients }$c_{n}$\emph{:} We extend $%
u(\cdot ,t)$ to an odd function $\tilde{u}(\cdot ,t)$ on the interval $%
\left( -\ell \left( t\right) ,\ell \left( t\right) \right) .$ This ensures
in particular that the boundary condition at $x=0$ is satisfied for $t\geq
0. $ It follows also that $\tilde{u}_{t}(\cdot ,t)$ and $\tilde{u}_{x}(\cdot
,t) $ are respectively an odd and an even function. Going back to (\ref%
{phi_cn}), we infer that
\begin{equation*}
\tilde{u}_{x}+\tilde{u}_{t}=2\varphi ^{\prime }(t+x)\sum_{n\in
\mathbb{Z}
}\omega _{n}c_{n}e^{\omega _{n}\varphi (t+x)}\smallskip ,
\end{equation*}%
for $x\in \left( -\ell \left( t\right) ,\ell \left( t\right) \right) $ and $%
t\geq 0$. Using the definition of $\omega _{n},$ we get%
\begin{equation}
\tilde{u}_{x}+\tilde{u}_{t}=\left\{
\begin{array}{ll}
\displaystyle2e^{\frac{1}{2}\left( i\pi -\ln \gamma _{\eta }\right) \varphi
(t+x)}\varphi ^{\prime }(t+x)\sum_{n\in
\mathbb{Z}
}\omega _{n}c_{n}e^{ni\pi \varphi (t+x)}\medskip , & \text{if }0\leq \eta <1,
\\
\displaystyle2e^{-\frac{1}{2}\ln \left\vert \gamma _{\eta }\right\vert
\varphi (t+x)}\varphi ^{\prime }(t+x)\sum_{n\in
\mathbb{Z}
}\omega _{n}c_{n}e^{ni\pi \varphi (t+x)}, & \text{if }1<\eta <+\infty ,%
\end{array}%
\right.  \label{29}
\end{equation}%
which implies that
\begin{equation}
\sum_{n\in
\mathbb{Z}
}\omega _{n}c_{n}e^{ni\pi \varphi (t+x)}=\left\{
\begin{array}{ll}
\frac{1}{2\varphi ^{\prime }(t+x)}\displaystyle e^{\frac{1}{2}\left( -i\pi
+\ln \gamma _{\eta }\right) \varphi (t+x)}\left( \tilde{u}_{x}+\tilde{u}%
_{t}\right) \medskip , & \text{if }0\leq \eta <1, \\
\frac{1}{2\varphi ^{\prime }(t+x)}\displaystyle e^{\frac{1}{2}\ln \left\vert
\gamma _{\eta }\right\vert \varphi (t+x)}\left( \tilde{u}_{x}+\tilde{u}%
_{t}\right) , & \text{if }1<\eta <+\infty .%
\end{array}%
\right.  \label{30}
\end{equation}

Taking into account that $\left\{ e^{ni\pi \varphi (t+x)}/\sqrt{2}\right\}
_{n\in \mathbb{%
\mathbb{Z}
}}$ is an orthonormal basis of the weighted space $L^{2}\left( -\ell \left(
t\right) ,\ell \left( t\right) ,\varphi ^{\prime }\left( t+x\right)
dx\right) ,$ we deduce that%
\begin{equation*}
\omega _{n}c_{n}=\left\{
\begin{array}{ll}
\displaystyle\frac{1}{4}\int_{-\ell \left( t\right) }^{\ell \left( t\right)
}e^{-\frac{1}{2}\left( i\pi -\ln \gamma _{\eta }\right) \varphi (t+x)}\left(
\tilde{u}_{x}+\tilde{u}_{t}\right) e^{-ni\pi \varphi (t+x)}dx\medskip , &
\text{if }0\leq \eta <1, \\
\displaystyle\frac{1}{4}\int_{-\ell \left( t\right) }^{\ell \left( t\right)
}e^{\frac{1}{2}\ln \left\vert \gamma _{\eta }\right\vert \varphi
(t+x)}\left( \tilde{u}_{x}+\tilde{u}_{t}\right) e^{-ni\pi \varphi (t+x)}dx,
& \text{if }1<\eta <+\infty ,%
\end{array}%
\right.
\end{equation*}%
for\ $n\in
\mathbb{Z}
$. Whether $0\leq \eta <1$ or $1<\eta <+\infty ,$ in both cases, we have%
\begin{equation}
c_{n}=\frac{1}{4\omega _{n}}\int_{-\ell \left( t\right) }^{\ell \left(
t\right) }\left( \tilde{u}_{x}+\tilde{u}_{t}\right) e^{-\omega _{n}\varphi
(t+x)}dx\text{, \ \ \ for\ }n\in
\mathbb{Z}
\text{.}  \label{cn=}
\end{equation}

Taking $t=0$, we obtain (\ref{cn}) as claimed.

$\bullet $ \emph{Regularity of the solution:} As a consequence of Parseval's
equality, we get%
\begin{equation*}
\sum_{n\in
\mathbb{Z}
}\left\vert \omega _{n}c_{n}\right\vert ^{2}=\left\{
\begin{array}{ll}
\displaystyle\frac{1}{8}\int_{-\ell \left( t\right) }^{\ell \left( t\right)
}\left\vert e^{-\frac{1}{2}\left( i\pi -\ln \gamma _{\eta }\right) \varphi
(t+x)}\right\vert ^{2}\left( \tilde{u}_{x}+\tilde{u}_{t}\right) ^{2}\frac{dx%
}{\varphi ^{\prime }(t+x)}\medskip , & \text{if }0\leq \eta <1 \\
\displaystyle\frac{1}{8}\int_{-\ell \left( t\right) }^{\ell \left( t\right)
}\left\vert e^{\frac{1}{2}\ln \left\vert \gamma _{\eta }\right\vert \varphi
(t+x)}\right\vert ^{2}\left( \tilde{u}_{x}+\tilde{u}_{t}\right) ^{2}\frac{dx%
}{\varphi ^{\prime }(t+x)}, & \text{if }1<\eta <+\infty .%
\end{array}%
\right.
\end{equation*}%
Whether $0\leq \eta <1$ or $1<\eta <+\infty ,$ the two cases of the
precedent identity can be written as%
\begin{equation}
\sum_{n\in
\mathbb{Z}
}\left\vert \omega _{n}c_{n}\right\vert ^{2}=\frac{1}{8}\int_{-\ell \left(
t\right) }^{\ell \left( t\right) }\left( \tilde{u}_{x}+\tilde{u}_{t}\right)
^{2}e^{\ln \left\vert \gamma _{\eta }\right\vert \varphi (t+x)}\frac{dx}{%
\varphi ^{\prime }(t+x)}.  \label{ncn+}
\end{equation}%
Due to (\ref{mono}) and (\ref{ic}), we have for $t=0,$%
\begin{equation*}
\frac{1}{\sqrt{\varphi ^{\prime }(x)}}e^{\ln \left\vert \gamma _{\eta
}\right\vert \varphi (t+x)}\left( \tilde{u}_{x}^{0}+\tilde{u}^{1}\right) \in
L^{2}\left( -L,L\right)
\end{equation*}%
and (\ref{ncn}) follows.

Due to the continuity and differentiability of $\varphi \left( t+x\right) $
and the exponential function, and since $\left\vert w_{n}\right\vert
=O\left( n\right) $ for large values of $n$, the regularity result (\ref%
{regul}) follows from the convergences of the series of the solution (\ref%
{phi_cn}) and its derivatives.
\end{proof}

\section{The undamped case}

In this section, we show some results for the undamped case, i.e. $\eta =0$
in Problem (\ref{waveq}). To know wether the energy is increasing or in
creasing, we compute $E_{\ell }^{\prime }\left( t\right) .$ Thus, using
Leibniz's rule for differentiation under the integral sign, we get%
\begin{equation*}
E_{\ell }^{\prime }\left( t\right) =\frac{1}{2}\ell ^{\prime }(t)\left[
u_{x}^{2}\left( \ell (t),t\right) +u_{t}^{2}\left( \ell (t),t\right) \right]
+\int_{0}^{\ell \left( t\right) }u_{t}u_{tt}+u_{x}u_{tx}dx.
\end{equation*}%
Since $u_{tt}=u_{xx}$ , then $u_{t}u_{tt}+u_{x}u_{tx}=\left(
u_{t}u_{x}\right) _{x}\ $and it follows%
\begin{equation}
E_{\ell }^{\prime }\left( t\right) =\frac{1}{2}\ell ^{\prime }(t)\left[
u_{x}^{2}\left( \ell (t),t\right) +u_{t}^{2}\left( \ell (t),t\right) \right]
+u_{t}u_{x}\left( \ell (t),t\right) .  \label{dE}
\end{equation}
Then, we have the following result.

\begin{lemma}
The energy of solution of Problem \emph{(\ref{waveq})}, with $\eta =0$,
satisfies%
\begin{equation*}
E_{\ell }^{\prime }\left( t\right) =-\frac{\ell ^{\prime }(t)}{2}\left(
1-\left\vert \ell ^{\prime }\left( t\right) \right\vert ^{2}\right)
u_{t}^{2}\left( \ell \left( t\right) ,t\right) ,\ \ \text{for }t>0.
\end{equation*}
\end{lemma}

\begin{proof}
The boundary condition at $x=\ell \left( t\right) ,$ with $\eta =0,$ reads
\begin{equation*}
u_{x}\left( \ell \left( t\right) ,t\right) =-u_{t}\left( \ell \left(
t\right) ,t\right) ,\ \ \text{for }t>0.
\end{equation*}%
Reporting this in (\ref{dE}), we get%
\begin{equation*}
E_{\ell }^{\prime }\left( t\right) =\frac{1}{2}\ell ^{\prime }(t)\left(
\left\vert \ell ^{\prime }\left( t\right) \right\vert ^{2}+1\right)
u_{t}^{2}\left( \ell \left( t\right) ,t\right) -\ell ^{\prime }\left(
t\right) u_{t}^{2}\left( \ell (t),t\right)
\end{equation*}%
and the lemma follows.
\end{proof}

\begin{remark}
The above lemma means that%
\begin{equation}
\Vert \text{ \ \ }E_{\ell }\left( t\right) \text{ is nondeacreasing if }%
-1<\ell ^{\prime }\left( t\right) \leq 0,\text{ and nonincreasing if }0\leq
\ell ^{\prime }\left( t\right) <1.  \label{E--ll}
\end{equation}%
The same result holds for the multidimensional wave equation, in a
time-dependent domain, with homogenous Dirichlet boundary conditions, see
\cite{BaCh1981}.
\end{remark}

The next theorem show that the asymptotic behaviour of $E_{\ell }\left(
t\right) $ is dictated by $\varphi ^{\prime }.$

\begin{theorem}
\label{th0damp}Under the assumptions \emph{(\ref{tlike}), (\ref{mono})} and
\emph{(\ref{ic}),} the solution of Problem \emph{(\ref{waveq})} satisfies%
\begin{multline}
\int_{0}^{\ell \left( t\right) }\left( \frac{1}{\varphi ^{\prime }(t+x)}+%
\frac{1}{\varphi ^{\prime }(t-x)}\right) \left( u_{x}^{2}+u_{t}^{2}\right) \\
+2\left( \frac{1}{\varphi ^{\prime }(t+x)}-\frac{1}{\varphi ^{\prime }(t-x)}%
\right) u_{x}u_{t}\ dx=4\mathcal{S}_{0},\text{ \ \ for }t\geq 0,
\label{E=4A}
\end{multline}%
\ \ where $\mathcal{S}_{0}:=\frac{\pi ^{2}}{2}\sum_{n\in
\mathbb{Z}
}\left\vert \left( 2n+1\right) c_{n}\right\vert ^{2}$. Moreover, it holds
that%
\begin{equation}
\mathcal{S}_{0}m\left( t\right) \leq E_{\ell }\left( t\right) \leq \mathcal{S%
}_{0}M\left( t\right) ,\text{ \ \ for}\ t\geq 0,  \label{E0}
\end{equation}%
where $m\left( t\right) $ and $M\left( t\right) $ are defined in \emph{(\ref%
{m.M})}.
\end{theorem}

\begin{proof}
If $\eta =0$ then $\gamma _{\eta }=1$ and $\omega _{n}=\left( 2n+1\right)
i\pi /2$. The identity (\ref{ncn+}) becomes%
\begin{equation}
\frac{1}{8}\int_{-\ell \left( t\right) }^{\ell \left( t\right) }\left(
\tilde{u}_{x}+\tilde{u}_{t}\right) ^{2}\frac{dx}{\varphi ^{\prime }(t+x)}=%
\frac{\pi ^{2}}{4}\sum_{n\in
\mathbb{Z}
}\left\vert \left( 2n+1\right) c_{n}\right\vert ^{2}=\frac{1}{2}\mathcal{S}%
_{0},\text{ \ \ for}\ t\geq 0.  \label{eq-per+}
\end{equation}%
Since $\tilde{u}_{x}$ is an even function of $x$ and that $\tilde{u}_{t}$ is
an odd one, then changing $x$ by $-x$ in the last formula, we also obtain%
\begin{equation}
\int_{-\ell \left( t\right) }^{\ell \left( t\right) }\left( \tilde{u}_{x}-%
\tilde{u}_{t}\right) ^{2}\frac{dx}{\varphi ^{\prime }(t-x)}=4\mathcal{S}_{0},%
\text{ \ \ for}\ t\geq 0.  \label{eq-per-}
\end{equation}

Taking the sum of (\ref{eq-per+}) and (\ref{eq-per-}),\ we obtain
\begin{equation*}
\int_{-\ell \left( t\right) }^{\ell \left( t\right) }\left( \tilde{u}_{x}+%
\tilde{u}_{t}\right) ^{2}\frac{dx}{\varphi ^{\prime }(t+x)}+\int_{-\ell
\left( t\right) }^{\ell \left( t\right) }\left( \tilde{u}_{x}-\tilde{u}%
_{t}\right) ^{2}\frac{dx}{\varphi ^{\prime }(t-x)}=8\mathcal{S}_{0}.
\end{equation*}%
Expanding $\left( u_{x}\pm u_{t}\right) ^{2}$ and collecting similar terms,
we get
\begin{multline}
\int_{-\ell \left( t\right) }^{\ell \left( t\right) }\left( \frac{1}{\varphi
^{\prime }(t+x)}+\frac{1}{\varphi ^{\prime }(t-x)}\right) \left( \tilde{u}%
_{x}^{2}+\tilde{u}_{t}^{2}\right)  \label{est} \\
+2\left( \frac{1}{\varphi ^{\prime }(t+x)}-\frac{1}{\varphi ^{\prime }(t-x)}%
\right) \tilde{u}_{x}\tilde{u}_{t}\ dx=8\mathcal{S}_{0},\text{ \ \ for}\
t\geq 0.
\end{multline}%
As the function under the integral sign is even, then (\ref{E=4A}) follows.

Next, we use the algebraic inequality $\pm 2u_{x}u_{t}\leq
u_{t}^{2}+u_{x}^{2}$ to obtain%
\begin{multline*}
\int_{0}^{\ell \left( t\right) }\left( \frac{1}{\varphi ^{\prime }(t+x)}+%
\frac{1}{\varphi ^{\prime }(t-x)}-\left\vert \frac{1}{\varphi ^{\prime }(t+x)%
}-\frac{1}{\varphi ^{\prime }(t-x)}\right\vert \right) \left(
u_{x}^{2}+u_{t}^{2}\right) \ dx\leq 4\mathcal{S}_{0} \\
\leq \int_{0}^{\ell \left( t\right) }\left( \frac{1}{\varphi ^{\prime }(t+x)}%
+\frac{1}{\varphi ^{\prime }(t-x)}+\left\vert \frac{1}{\varphi ^{\prime
}(t+x)}-\frac{1}{\varphi ^{\prime }(t-x)}\right\vert \right) \left(
u_{x}^{2}+u_{t}^{2}\right) \ dx,
\end{multline*}%
for$\ t\geq 0.$ Recalling that
\begin{equation}
\left( a+b\right) -\left\vert a-b\right\vert =2\min \left\{ a,b\right\} \ \
\ \text{and}\ \ \ \left( a+b\right) +\left\vert a-b\right\vert =2\max
\left\{ a,b\right\} ,  \label{max}
\end{equation}%
for $a,b\in
\mathbb{R}
,$ then%
\begin{multline*}
\int_{0}^{\ell \left( t\right) }\min \left\{ \frac{1}{\varphi ^{\prime }(t+x)%
},\frac{1}{\varphi ^{\prime }(t-x)}\right\} \left(
u_{x}^{2}+u_{t}^{2}\right) \ dx\leq 2\mathcal{S}_{0} \\
\leq \int_{0}^{\ell \left( t\right) }\max \left\{ \frac{1}{\varphi ^{\prime
}(t+x)},\frac{1}{\varphi ^{\prime }(t-x)}\right\} \left(
u_{x}^{2}+u_{t}^{2}\right) \ dx,
\end{multline*}%
for$\ t\geq 0.$ Recalling that $E_{\ell }\left( t\right) $ is defined by (%
\ref{E}), we deduce that%
\begin{equation*}
\min_{x\in \left[ 0,\ell \left( t\right) \right] }\left\{ \frac{1}{\varphi
^{\prime }(t+x)},\frac{1}{\varphi ^{\prime }(t-x)}\right\} E_{\ell }\left(
t\right) \leq \mathcal{S}_{0}\leq \max_{x\in \left[ 0,\ell \left( t\right) %
\right] }\left\{ \frac{1}{\varphi ^{\prime }(t+x)},\frac{1}{\varphi ^{\prime
}(t-x)}\right\} E_{\ell }\left( t\right) ,
\end{equation*}%
hence%
\begin{equation*}
\min_{x\in \left[ 0,\ell \left( t\right) \right] }\left\{ \varphi ^{\prime
}(t+x),\varphi ^{\prime }(t-x)\right\} \mathcal{S}_{0}\leq E_{\ell }\left(
t\right) \leq \max_{x\in \left[ 0,\ell \left( t\right) \right] }\left\{
\varphi ^{\prime }(t+x),\varphi ^{\prime }(t-x)\right\} \mathcal{S}_{0},
\end{equation*}%
which is (\ref{E0}).
\end{proof}
\begin{remark} An estimation analogue to (\ref{E0}) was obtained for the case of homogeneous boundary conditions at both ends, see \cite{HaHo2019}.
\end{remark}
If $\varphi ^{\prime }$ is monotone then the asymptotic behaviour is
dictated by $\ell \left( t\right) $ and we have the following refinements.

\begin{corollary}
\label{Cor1}Under the assumption of Theorem \ref{th0damp}, assume that \
\begin{equation}
\varphi ^{\prime }\text{ is monotone for }t\in \lbrack t_{0},t_{1}],\text{ }%
0\leq t_{0}<t_{1}.  \label{monot}
\end{equation}
\end{corollary}

\begin{itemize}
\item If $-1<\ell ^{\prime }\left( t\right) \leq 0$ on $[t_{0},t_{1}]$, then
$\varphi ^{\prime }$and $E_{\ell }\left( t\right) $ are nondeacreasing and%
\begin{equation}
\mathcal{S}_{0}\varphi ^{\prime }(t-\ell \left( t\right) )\leq E_{\ell
}\left( t\right) \leq \mathcal{S}_{0}\varphi ^{\prime }(t+\ell \left(
t\right) ),\text{ \ \ for}\ t\in \lbrack t_{0},t_{1}].  \label{ll-}
\end{equation}

\item If $0\leq \ell ^{\prime }\left( t\right) <1$ on $[t_{0},t_{1}]$, then $%
\varphi ^{\prime }$and $E_{\ell }\left( t\right) $ are nonincreasing and%
\begin{equation}
\mathcal{S}_{0}\varphi ^{\prime }(t+\ell \left( t\right) )\leq E_{\ell
}\left( t\right) \leq \mathcal{S}_{0}\varphi ^{\prime }\left( t-\ell \left(
t\right) \right) ,\text{ \ \ for}\ t\in \lbrack t_{0},t_{1}].  \label{ll+}
\end{equation}
\end{itemize}

\begin{proof}
We already have (\ref{E--ll}). The derivation of the identity (\ref{Moor})
yields
\begin{equation}
\frac{1+\ell ^{\prime }\left( t\right) }{1-\ell ^{\prime }\left( t\right) }%
\varphi ^{\prime }(t+\ell \left( t\right) )=\varphi ^{\prime }(t-\ell \left(
t\right) ).  \label{+t-t}
\end{equation}%
Taking into consideration the variation of the function $s\mapsto \left(
1+s\right) /\left( 1-s\right) $ on the interval $\left( -1,1\right) $, it
follows that%
\begin{equation*}
0<\frac{1+\ell ^{\prime }\left( t\right) }{1-\ell ^{\prime }\left( t\right) }%
\leq 1\text{ \ if }-1<\ell ^{\prime }\left( t\right) \leq 0\ \ \ \text{and \
\ }1\leq \frac{1+\ell ^{\prime }\left( t\right) }{1-\ell ^{\prime }\left(
t\right) } \ \text{ \ if }0\leq \ell ^{\prime }\left( t\right) <1,
\end{equation*}%
hence
\begin{gather}
\text{if \ }-1<\ell ^{\prime }\left( t\right) \leq 0,\text{ then \ }m\left(
t\right) =\varphi ^{\prime }(t-\ell \left( t\right) )\leq \varphi ^{\prime
}(t+\ell \left( t\right) )=M\left( t\right) ,  \label{ll--} \\
\text{if \ }0\leq \ell ^{\prime }\left( t\right) <1,\text{ then \ }M\left(
t\right) =\varphi ^{\prime }(t-\ell \left( t\right) )\geq \varphi ^{\prime
}(t+\ell \left( t\right) )=m\left( t\right) .  \label{ll++}
\end{gather}%
Of course, if $\varphi ^{\prime }$ is monotone and $-1<\ell ^{\prime }\left(
t\right) \leq 0$ then (\ref{ll--}) means that $\varphi ^{\prime }$ is
necessarily nondecreasing, and so is $\varphi ^{\prime }(t\pm \ell \left(
t\right) )\ $and $E_{\ell }\left( t\right) .$ The same argument can be made
when $0\leq \ell ^{\prime }\left( t\right) <1$. This shows the corollary.
\end{proof}

\begin{remark}
The assumption (\ref{monot}) is satisfied in all the examples of the last
section.
\end{remark}

Recall that $\mathcal{S}_{0}$ can be computed using the initial data by
setting $t=0$ in (\ref{E0}). If one needs to compare $E_{\ell }\left(
t\right) $ with $E_{\ell }\left( t_{0}\right) \ $for $0\leq t_{0}<t,$ then
we have the next result.

\begin{corollary}
Under the assumption of Theorem \ref{th0damp}, we have%
\begin{equation}
\frac{m\left( t\right) }{M\left( t_{0}\right) }E_{\ell }\left( t_{0}\right)
\leq E_{\ell }\left( t\right) \leq \frac{M\left( t\right) }{m\left(
t_{0}\right) }E_{\ell }\left( t_{0}\right) ,\text{ \ \ \ for }0\leq t_{0}<t%
\text{.}  \label{Et0}
\end{equation}%
Moreover, if $\varphi $ satisfies (\ref{monot}), then:
\end{corollary}

\begin{itemize}
\item If $-1<\ell ^{\prime }\left( t\right) \leq 0$ on $[t_{0},t_{1}]$, then
$\varphi ^{\prime }$and $E_{\ell }\left( t\right) $ are nondecreasing and
satisfy%
\begin{equation}
\frac{\varphi ^{\prime }\left( t-\ell \left( t\right) \right) }{\varphi
^{\prime }\left( t_{0}+\ell \left( t_{0}\right) \right) }E_{\ell }\left(
t_{0}\right) \leq E_{\ell }\left( t\right) \leq \frac{\varphi ^{\prime
}\left( t+\ell \left( t\right) \right) }{\varphi ^{\prime }\left( t_{0}-\ell
\left( t_{0}\right) \right) }E_{\ell }\left( t_{0}\right) ,\text{ \ \ for }%
t\in \lbrack t_{0},t_{1}].  \label{Et0-}
\end{equation}

\item If $0\leq \ell ^{\prime }\left( t\right) <1$ on $[t_{0},t_{1}]$, then $%
\varphi ^{\prime }$and $E_{\ell }\left( t\right) $ are nonincreasing and%
\begin{equation}
\frac{\varphi ^{\prime }\left( t+\ell \left( t\right) \right) }{\varphi
^{\prime }\left( t_{0}-\ell \left( t_{0}\right) \right) }E_{\ell }\left(
t_{0}\right) \leq E_{\ell }\left( t\right) \leq \frac{\varphi ^{\prime
}\left( t-\ell \left( t\right) \right) }{\varphi ^{\prime }\left( t_{0}+\ell
\left( t_{0}\right) \right) }E_{\ell }\left( t_{0}\right) ,\text{ \ \ for }%
t\in \lbrack t_{0},t_{1}].  \label{Et0+}
\end{equation}
\end{itemize}

\begin{proof}
Since (\ref{E0}) holds also for $t=t_{0}$, then the corollary follows from
the inequalities%
\begin{equation*}
\frac{E_{\ell }\left( t\right) }{M\left( t\right) }\leq \mathcal{S}_{0}\leq
\frac{E_{\ell }\left( t_{0}\right) }{m\left( t_{0}\right) }\text{ \ \ and \
\ }\frac{E_{\ell }\left( t_{0}\right) }{M\left( t_{0}\right) }\leq \mathcal{S%
}_{0}\leq \frac{E_{\ell }\left( t\right) }{m\left( t\right) }.
\end{equation*}
\end{proof}

\section{The damped case}

In this section, we investigate the case with boundary damping when%
\begin{equation*}
\eta >0\text{ \ and \ }\eta \neq 1
\end{equation*}%
in Problem (\ref{waveq}). Let us recall that we still have (\ref{dE}), i.e.
\begin{equation*}
E_{\ell }^{\prime }\left( t\right) =\frac{1}{2}\ell ^{\prime }(t)\left[
u_{x}^{2}\left( \ell (t),t\right) +u_{t}^{2}\left( \ell (t),t\right) \right]
+u_{t}u_{x}\left( \ell (t),t\right) ,\ \ \text{\ for}\ t\geq 0.
\end{equation*}%
but now the boundary condition at $x=\ell \left( t\right) $ is
\begin{equation*}
\left( 1+\eta \ell ^{\prime }\left( t\right) \right) u_{x}\left( \ell \left(
t\right) ,t\right) +\left( \eta +\ell ^{\prime }\left( t\right) \right)
u_{t}\left( \ell \left( t\right) ,t\right) =0,\ \ \text{\ for}\ t\geq 0
\end{equation*}

Let us discuss the sign of $E_{\ell }^{\prime }\left( t\right) $ for
different values of $\ell ^{\prime }\left( t\right) .$

\begin{itemize}
\item If the interval is shrinking $-1<\ell ^{\prime }\left( t\right) <0,$
for $t\in \left[ t_{1},t_{2}\right] \ $where $0\leq t_{1}<t_{2},$ then we
have the following cases:

- If $\ell ^{\prime }\left( t\right) =-1/\eta ,$ then the boundary condition
at $x=\ell \left( t\right) $ reads $u_{t}\left( \ell (t),t\right) =0$ and
thus
\begin{equation*}
E_{\ell }^{\prime }\left( t\right) =-\frac{1}{2\eta }u_{x}^{2}\left( \ell
(t),t\right) \leq 0,\text{ \ for }t\in \left[ t_{1},t_{2}\right] .
\end{equation*}

- If $\ell ^{\prime }\left( t\right) =-\eta ,$ then $u_{x}\left( \ell
(t),t\right) =0$ and thus
\begin{equation*}
E_{\ell }^{\prime }\left( t\right) =-\frac{1}{2}\eta u_{t}^{2}\left( \ell
(t),t\right) \leq 0,\text{ \ for }t\in \left[ t_{1},t_{2}\right] .
\end{equation*}

- Assume that $\ell ^{\prime }\left( t\right) \notin \left\{ -1/\eta
,0,-\eta \right\} ,$ then after some computation we can rewrite (\ref{dE}) as%
\begin{equation}
E_{\ell }^{\prime }\left( t\right) =-\frac{1}{2}\left( \ell ^{\prime }\left(
t\right) \eta ^{2}+2\eta +\ell ^{\prime }\left( t\right) \right) \frac{%
1-\left\vert \ell ^{\prime }\left( t\right) \right\vert ^{2}}{\left( 1+\eta
\ell ^{\prime }\left( t\right) \right) ^{2}}u_{t}^{2}\left( \ell \left(
t\right) ,t\right) ,\text{ \ for }t\in \left[ t_{1},t_{2}\right] .
\label{E'2}
\end{equation}%
The sign of $E_{\ell }^{\prime }\left( t\right) $ is opposite to the sign of
\begin{equation*}
P_{\ell }\left( \eta \right) =\ell ^{\prime }\left( t\right) \eta ^{2}+2\eta
+\ell ^{\prime }\left( t\right) .
\end{equation*}%
Due to (\ref{tlike}), the polynomial $P_{\ell }\left( \eta \right) $ has a
discriminant $\Delta =4(1-\left\vert \ell ^{\prime }\left( t\right)
\right\vert ^{2})>0$ and thus $P_{\ell }\left( \eta \right) $ has two real
roots
\begin{equation}
\eta _{1}:=\left( -1+\sqrt{1-\left\vert \ell ^{\prime }\left( t\right)
\right\vert ^{2}}\right) /\ell ^{\prime }\left( t\right) \text{ \ and \ }%
\eta _{2}:=\left( -1-\sqrt{1-\left\vert \ell ^{\prime }\left( t\right)
\right\vert ^{2}}\right) /\ell ^{\prime }\left( t\right) .  \label{eta1,2}
\end{equation}

\begin{figure}[tbph]
\centering\includegraphics[width=0.45\textwidth]{GhS3_fig4.jpg}
\caption{Variation of $\protect\eta_1 $ and $\protect\eta_2$ in function of $%
\ell ^{\prime }$ when $-1<\ell ^{\prime } <0$.}
\label{fig4}
\end{figure}

Both $\eta _{1}$ and $\eta _{2}$ have the opposite sign of $\ell ^{\prime
}\left( t\right) ,$ for $t\in \left[ t_{1},t_{2}\right] $ and in particular%
\begin{equation*}
0<\eta _{1}<1<\eta _{2}.
\end{equation*}%
We deduce from (\ref{E'2}) that:

\begin{itemize}
\item If $\eta _{1}<\eta <\eta _{2},$ for $t\in \left[ t_{1},t_{2}\right] ,$
then $P_{\ell }\left( \eta \right) >0$ and by consequence $E_{\ell }^{\prime
}\left( t\right) <0.$\label{shrink+decay}

\item If $\eta =\eta _{1}$ or $\eta =\eta _{2},$ for $t\in \left[ t_{1},t_{2}%
\right] ,$ then $P_{\ell }\left( \eta \right) =E_{\ell }^{\prime }\left(
t\right) =0,$ i.e. the energy is constant.

\item If $\eta \in \left]0, \eta _{1}\right[ \cup \left] \eta _{2},\infty%
\right[ ,$ for $t\in \left[ t_{1},t_{2}\right] ,$ then $P_{\ell }\left( \eta
\right) <0$ and we have $E_{\ell }^{\prime }\left( t\right) >0.$
\end{itemize}

\item If the interval is independent of time, i.e. $\ell ^{\prime }\left(
t\right) =0\ $for $t\in \left[ t_{1},t_{2}\right] ,$ then $u_{x}\left( \ell
\left( t\right) ,t\right) =-\eta u_{t}\left( \ell \left( t\right) ,t\right) $
and thus%
\begin{equation*}
E_{\ell }^{\prime }\left( t\right) =-\eta u_{t}^{2}\left( \ell \left(
t\right) ,t\right) \leq 0,\ \ \text{for }t\in \left[ t_{1},t_{2}\right] .
\end{equation*}

\item If the interval is expanding, i.e. $0<\ell ^{\prime }\left( t\right)
<1\ $for $t\in \left[ t_{1},t_{2}\right] ,$ then $P_{\ell }\left( \eta
\right) >0$ for $\eta >0.$ Taking into account (\ref{tlike}) and (\ref{E'2}%
), we deduce that $E_{\ell }^{\prime }\left( t\right) \leq 0,$ \ for $t\in %
\left[ t_{1},t_{2}\right] .$
\end{itemize}

\begin{remark}
\label{rmkEd}To summarize, under the assumption $(\ref{tlike}),$%
\begin{equation}
\left\Vert \text{\ \ \ \ }%
\begin{array}{l}
\text{If the interval is expanding, then }E_{\ell }\left( t\right) \text{ is
nonincreasing for any }\eta >0\text{.} \\
\text{If the interval is shrinking, then }E_{\ell }\left( t\right) \text{ is
nonincreasing if the damping} \\
\text{factor can be taken close enough to the optimal value }\eta =1.%
\end{array}%
\right.  \label{E3}
\end{equation}
\end{remark}

Let us now estimate $E_{\ell }\left( t\right) $ is using $\varphi ^{\prime }$
and $\exp \left( -\ln \left\vert \gamma _{\eta }\right\vert \varphi \right) $%
.

\begin{theorem}
\label{th-stab1}Under the assumptions \emph{(\ref{tlike}), (\ref{mono})} and
\emph{(\ref{ic}),} the solution of Problem \emph{(\ref{waveq})} satisfies%
\begin{multline}
\int_{0}^{\ell \left( t\right) }\left( \frac{e^{\ln \left\vert \gamma _{\eta
}\right\vert \varphi (t+x)}}{\varphi ^{\prime }(t+x)}+\frac{e^{\ln
\left\vert \gamma _{\eta }\right\vert \varphi (t-x)}}{\varphi ^{\prime }(t-x)%
}\right) \left( u_{x}^{2}+u_{t}^{2}\right) \\
+2\left( \frac{e^{\ln \left\vert \gamma _{\eta }\right\vert \varphi (t+x)}}{%
\varphi ^{\prime }(t+x)}-\frac{e^{\ln \left\vert \gamma _{\eta }\right\vert
\varphi (t-x)}}{\varphi ^{\prime }(t-x)}\right) u_{x}u_{t}dx=4\mathcal{S}%
_{\eta },  \label{E.D}
\end{multline}%
\ \ for$\ t\geq 0$, where $\mathcal{S}_{\eta }:=2\sum_{n\in
\mathbb{Z}
}\left\vert \omega _{n}c_{n}\right\vert ^{2}$. Moreover, it holds that%
\begin{equation}
\mathcal{S}_{\eta }\tilde{m}\left( t\right) \leq E_{\ell }\left( t\right)
\leq \mathcal{S}_{\eta }\tilde{M}\left( t\right) ,\text{ \ \ for}\ t\geq 0,
\label{stab}
\end{equation}%
where
\begin{eqnarray*}
\tilde{m}\left( t\right) &:&=\min_{x\in \left[ 0,\ell \left( t\right) \right]
}\left\{ \varphi ^{\prime }(t-x)e^{-\ln \left\vert \gamma _{\eta
}\right\vert \varphi (t-x)},\varphi ^{\prime }(t+x)e^{-\ln \left\vert \gamma
_{\eta }\right\vert \varphi (t+x)}\right\} , \\
\tilde{M}\left( t\right) &:&=\max_{x\in \left[ 0,\ell \left( t\right) \right]
}\left\{ \varphi ^{\prime }(t-x)e^{-\ln \left\vert \gamma _{\eta
}\right\vert \varphi (t-x)},\varphi ^{\prime }(t+x)e^{-\ln \left\vert \gamma
_{\eta }\right\vert \varphi (t+x)}\right\} .
\end{eqnarray*}
\end{theorem}

\begin{proof}
We argue as in the proof of Theorem \ref{th0damp}. Noting that $\tilde{u}%
_{x} $ is an even function of $x$ and that $\tilde{u}_{t}$ is an odd one,
then the identity (\ref{ncn+}) yields%
\begin{equation}
\int_{-\ell \left( t\right) }^{\ell \left( t\right) }e^{\ln \left\vert
\gamma _{\eta }\right\vert \varphi (t\pm x)}\left( \tilde{u}_{x}\pm \tilde{u}%
_{t}\right) ^{2}\frac{dx}{\varphi ^{\prime }(t\pm x)}=4\mathcal{S}_{\eta },%
\text{ \ \ for}\ t\geq 0.  \label{32}
\end{equation}%
Hence, summing, we get
\begin{equation*}
\int_{-\ell \left( t\right) }^{\ell \left( t\right) }e^{\ln \left\vert
\gamma _{\eta }\right\vert \varphi (t+x)}\left( \tilde{u}_{x}+\tilde{u}%
_{t}\right) ^{2}\frac{dx}{\varphi ^{\prime }(t+x)}+\int_{-\ell \left(
t\right) }^{\ell \left( t\right) }e^{\ln \left\vert \gamma _{\eta
}\right\vert \varphi (t-x)}\left( \tilde{u}_{x}-\tilde{u}_{t}\right) ^{2}%
\frac{dx}{\varphi ^{\prime }(t-x)}=8\mathcal{S}_{\eta }.
\end{equation*}%
Expanding squares, we get%
\begin{multline}
\int_{-\ell \left( t\right) }^{\ell \left( t\right) }\left( \frac{e^{\ln
\left\vert \gamma _{\eta }\right\vert \varphi (t+x)}}{\varphi ^{\prime }(t+x)%
}+\frac{e^{\ln \left\vert \gamma _{\eta }\right\vert \varphi (t-x)}}{\varphi
^{\prime }(t-x)}\right) \left( \tilde{u}_{x}^{2}+\tilde{u}_{t}^{2}\right)
\label{=E0} \\
+2\left( \frac{e^{\ln \left\vert \gamma _{\eta }\right\vert \varphi (t+x)}}{%
\varphi ^{\prime }(t+x)}-\frac{e^{\ln \left\vert \gamma _{\eta }\right\vert
\varphi (t-x)}}{\varphi ^{\prime }(t-x)}\right) \tilde{u}_{x}\tilde{u}_{t}%
\text{ }dx=8\mathcal{S}_{\eta },\text{ \ \ for}\ t\geq 0.
\end{multline}%
As the function under the integral sign is even, then (\ref{E.D}) follows.

For $0\leq x\leq \ell \left( t\right) $ and $t\geq 0,$ let us denote
\begin{equation*}
A\left( x,t\right) =\frac{e^{\ln \left\vert \gamma _{\eta }\right\vert
\varphi (t-x)}}{\varphi ^{\prime }(t-x)}\text{ and \ }B\left( x,t\right) =%
\frac{e^{\ln \left\vert \gamma _{\eta }\right\vert \varphi (t+x)}}{\varphi
^{\prime }(t+x)}.
\end{equation*}%
Then, we can rewrite (\ref{E.D}) as
\begin{equation*}
\int_{0}^{\ell \left( t\right) }\left( A\left( x,t\right) +B\left(
x,t\right) \right) \left( \tilde{u}_{t}^{2}+\tilde{u}_{x}^{2}\right)
dx+2\left( A\left( x,t\right) -B\left( x,t\right) \right) \tilde{u}_{t}%
\tilde{u}_{x}dx=4\mathcal{S}_{\eta }.
\end{equation*}%
Using the algebraic inequality%
\begin{equation*}
-\left\vert A-B\right\vert \left( u_{t}^{2}+u_{x}^{2}\right) \leq 2\left(
A-B\right) u_{t}u_{x}\leq \left\vert A-B\right\vert \left(
u_{t}^{2}+u_{x}^{2}\right) ,
\end{equation*}%
we get%
\begin{equation*}
\int_{0}^{\ell \left( t\right) }\left( \left( A+B\right) -\left\vert
A-B\right\vert \right) \left( \tilde{u}_{t}^{2}+\tilde{u}_{x}^{2}\right)
dx\leq 4\mathcal{S}_{\eta }\leq \int_{0}^{\ell \left( t\right) }\left(
\left( A+B\right) +\left\vert A-B\right\vert \right) \left( \tilde{u}%
_{t}^{2}+\tilde{u}_{x}^{2}\right) dx.
\end{equation*}%
Thanks to (\ref{max}), the precedent estimation yields%
\begin{multline*}
\int_{0}^{\ell \left( t\right) }\min \left\{ \frac{e^{\ln \left\vert \gamma
_{\eta }\right\vert \varphi (t-x)}}{\varphi ^{\prime }(t-x)},\frac{e^{\ln
\left\vert \gamma _{\eta }\right\vert \varphi (t+x)}}{\varphi ^{\prime }(t+x)%
}\right\} \left( \tilde{u}_{t}^{2}+\tilde{u}_{x}^{2}\right) dx\leq 2\mathcal{%
S}_{\eta } \\
\leq \int_{0}^{\ell \left( t\right) }\max \left\{ \frac{e^{\ln \left\vert
\gamma _{\eta }\right\vert \varphi (t-x)}}{\varphi ^{\prime }(t-x)},\frac{%
e^{\ln \left\vert \gamma _{\eta }\right\vert \varphi (t+x)}}{\varphi
^{\prime }(t+x)}\right\} \left( \tilde{u}_{t}^{2}+\tilde{u}_{x}^{2}\right) dx%
\text{,}
\end{multline*}%
\ for $t\geq 0.$ By consequence
\begin{multline*}
\min_{x\in \left[ 0,\ell \left( t\right) \right] }\left\{ \frac{e^{\ln
\left\vert \gamma _{\eta }\right\vert \varphi (t-x)}}{\varphi ^{\prime }(t-x)%
},\frac{e^{\ln \left\vert \gamma _{\eta }\right\vert \varphi (t+x)}}{\varphi
^{\prime }(t+x)}\right\} E_{\ell }\left( t\right) \leq \mathcal{S}_{\eta } \\
\leq \max_{x\in \left[ 0,\ell \left( t\right) \right] }\left\{ \frac{e^{\ln
\left\vert \gamma _{\eta }\right\vert \varphi (t-x)}}{\varphi ^{\prime }(t-x)%
},\frac{e^{\ln \left\vert \gamma _{\eta }\right\vert \varphi (t+x)}}{\varphi
^{\prime }(t+x)}\right\} E_{\ell }\left( t\right) .
\end{multline*}

This implies (\ref{stab}).
\end{proof}

Since $\ln \left\vert \gamma _{\eta }\right\vert \geq 0 $ for $\eta \geq 0,$
and $\varphi $ is nondecreasing, we have the following immediate corollary.

\begin{corollary}
Under the assumption of Theorem \ref{th-stab1}, it holds that%
\begin{equation}
\mathcal{S}_{\eta }m\left( t\right) e^{-\ln \left\vert \gamma _{\eta
}\right\vert \varphi (t+\ell \left( t\right) )}\leq E_{\ell }\left( t\right)
\leq \mathcal{S}_{\eta }M\left( t\right) e^{-\ln \left\vert \gamma _{\eta
}\right\vert \varphi (t-\ell \left( t\right) )},\text{ \ \ for}\ t\geq 0,
\label{stab0}
\end{equation}%
$m\left( t\right) $ and $M\left( t\right) $ are given by \emph{(\ref{m.M})}.
\end{corollary}

If $\varphi ^{\prime }$ is monotone, then (\ref{stab0}) can be replaced by
more explicit estimation.

\begin{corollary}
Under the assumption of Theorem \ref{th-stab1}, assume that $\varphi $
satisfies (\ref{monot}) on $[t_{0},t_{1}]$, then:
\end{corollary}

\begin{itemize}
\item If $-1<\ell ^{\prime }\left( t\right) \leq 0$ on $[t_{0},t_{1}]$, then
$\varphi ^{\prime }$ is nondecreasing and $E_{\ell }\left( t\right) $
satisfies
\begin{equation}
\mathcal{S}_{\eta }\varphi ^{\prime }(t-\ell \left( t\right) )e^{-\ln
\left\vert \gamma _{\eta }\right\vert \varphi (t+\ell \left( t\right) )}\leq
E_{\ell }\left( t\right) \leq \mathcal{S}_{\eta }\varphi ^{\prime }\left(
t+\ell \left( t\right) \right) e^{-\ln \left\vert \gamma _{\eta }\right\vert
\varphi (t-\ell \left( t\right) )},\text{ \ \ for }t\in \lbrack t_{0},t_{1}].
\label{stab1.}
\end{equation}

\item If $0<\ell ^{\prime }\left( t\right) <1$ on $[t_{0},t_{1}]$, then $%
\varphi ^{\prime }$and $E_{\ell }\left( t\right) $ are nonincreasing and%
\begin{equation}
\mathcal{S}_{\eta }\varphi ^{\prime }(t+\ell \left( t\right) )e^{-\ln
\left\vert \gamma _{\eta }\right\vert \varphi (t+\ell \left( t\right) )}\leq
E_{\ell }\left( t\right) \leq \mathcal{S}_{\eta }\varphi ^{\prime }\left(
t-\ell \left( t\right) \right) e^{-\ln \left\vert \gamma _{\eta }\right\vert
\varphi (t-\ell \left( t\right) )},\text{ \ \ for }t\in \lbrack t_{0},t_{1}].
\label{stab2.}
\end{equation}
\end{itemize}

\begin{proof}
It suffices to argue as in the proof of Corollary \ref{Cor1}.
\end{proof}

\begin{remark}
If the interval is shrinking, there is competition between the nondecreasing
$\varphi ^{\prime }$ and $e^{-\ln \left\vert \gamma _{\eta }\right\vert
\varphi }$ in estimation (\ref{stab1.}). The behaviour of $E_{\ell }\left(
t\right) $ depends on the value of the damping $\eta $ as stated in Remark %
\ref{rmkEd}.
\end{remark}

To compare $E_{\ell }\left( t\right) $ with the energy $E_{\ell }\left(
t_{0}\right) $ for $0\leq t_{0}<t,$ we have the following result.

\begin{corollary}
\label{CoroEt0}Under the assumption of Theorem \ref{th-stab1}, we have
\begin{multline}
\frac{m\left( t\right) e^{-\ln \left\vert \gamma _{\eta }\right\vert \varphi
(t+\ell \left( t\right) )}}{M\left( t_{0}\right) e^{-\ln \left\vert \gamma
_{\eta }\right\vert \varphi (t_{0}-\ell \left( t_{0}\right) )}}E_{\ell
}\left( t_{0}\right) \leq E_{\ell }\left( t\right)  \label{dEt0} \\
\leq \frac{M\left( t\right) e^{-\ln \left\vert \gamma _{\eta }\right\vert
\varphi (t-\ell \left( t\right) )}}{m\left( t_{0}\right) e^{-\ln \left\vert
\gamma _{\eta }\right\vert \varphi (t_{0}+\ell \left( t_{0}\right) )}}%
E_{\ell }\left( t_{0}\right) ,\text{ for }0\leq t_{0}<t.
\end{multline}%
Moreover, if $\varphi ^{\prime }$ satisfies (\ref{monot}) on $[t_{0},t_{1}]$%
, then:
\end{corollary}

\begin{itemize}
\item If $-1<\ell ^{\prime }\left( t\right) \leq 0$ on $[t_{0},t_{1}]$, then
$\varphi ^{\prime }$ is nondecreasing and $E_{\ell }\left( t\right) $
satisfies%
\begin{multline}
\frac{\varphi ^{\prime }\left( t-\ell \left( t\right) \right) e^{-\ln
\left\vert \gamma _{\eta }\right\vert \varphi (t+\ell \left( t\right) )}}{%
\varphi ^{\prime }\left( t_{0}+\ell \left( t_{0}\right) \right) e^{-\ln
\left\vert \gamma _{\eta }\right\vert \varphi (t_{0}-\ell \left(
t_{0}\right) )}}E_{\ell }\left( t_{0}\right) \leq E_{\ell }\left( t\right)
\label{stab3} \\
\leq \frac{\varphi ^{\prime }\left( t+\ell \left( t\right) \right) e^{-\ln
\left\vert \gamma _{\eta }\right\vert \varphi (t-\ell \left( t\right) )}}{%
\varphi ^{\prime }\left( t_{0}-\ell \left( t_{0}\right) \right) e^{-\ln
\left\vert \gamma _{\eta }\right\vert \varphi (t_{0}+\ell \left(
t_{0}\right) )}}E_{\ell }\left( t_{0}\right) ,\text{ \ \ for }t\in \lbrack
t_{0},t_{1}].
\end{multline}

\item If $0<\ell ^{\prime }\left( t\right) <1$ on $[t_{0},t_{1}]$, then $%
\varphi ^{\prime }$and $E_{\ell }\left( t\right) $ are nonincreasing and%
\begin{multline}
\frac{\varphi ^{\prime }\left( t+\ell \left( t\right) \right) e^{-\ln
\left\vert \gamma _{\eta }\right\vert \varphi (t+\ell \left( t\right) )}}{%
\varphi ^{\prime }\left( t_{0}-\ell \left( t_{0}\right) \right) e^{-\ln
\left\vert \gamma _{\eta }\right\vert \varphi (t_{0}-\ell \left(
t_{0}\right) )}}E_{\ell }\left( t_{0}\right) \leq E_{\ell }\left( t\right)
\label{stab4} \\
\leq \frac{\varphi ^{\prime }\left( t-\ell \left( t\right) \right) e^{-\ln
\left\vert \gamma _{\eta }\right\vert \varphi (t-\ell \left( t\right) )}}{%
\varphi ^{\prime }\left( t_{0}+\ell \left( t_{0}\right) \right) e^{-\ln
\left\vert \gamma _{\eta }\right\vert \varphi (t_{0}+\ell \left(
t_{0}\right) )}}E_{\ell }\left( t_{0}\right) ,\text{ \ \ for }t\in \lbrack
t_{0},t_{1}].
\end{multline}
\end{itemize}

\begin{proof}
Since (\ref{stab}) holds also for $t=t_{0}$, then the corollary follows by
combining the inequalities%
\begin{equation*}
\frac{e^{\ln \left\vert \gamma _{\eta }\right\vert \varphi (t-\ell \left(
t\right) )}}{M\left( t\right) }E_{\ell }\left( t\right) \leq \mathcal{S}%
_{\eta }\leq \frac{e^{\ln \left\vert \gamma _{\eta }\right\vert \varphi
(t_{0}+\ell \left( t_{0}\right) )}}{m\left( t_{0}\right) }E_{\ell }\left(
t_{0}\right)
\end{equation*}%
and%
\begin{equation*}
\frac{e^{\ln \left\vert \gamma _{\eta }\right\vert \varphi (t_{0}-\ell
\left( t_{0}\right) )}}{M\left( t_{0}\right) }E_{\ell }\left( t_{0}\right)
\leq \mathcal{S}_{\eta }\leq \frac{e^{\ln \left\vert \gamma _{\eta
}\right\vert \varphi (t+\ell \left( t\right) )}}{m\left( t\right) }E_{\ell
}\left( t\right) .
\end{equation*}
\end{proof}

\section*{Acknowledgements} The authors have been supported by the General
Direction of Scientific Research and Technological Development (Algerian
Ministry of Higher Education and Scientific Research) PRFU \#
C00L03UN280120220010.

\subsection*{ORCID} Abdelmouhcene Sengouga https://orcid.org/0000-0003-3183-7973.

redpajama_set_name: RedPajamaArXiv


text:
'Light 'em up': Minneapolis officers seen firing paint rounds at people on their porch
Joshua Bote
A video shared on social media appears to show Minneapolis law enforcement shooting marking rounds at people standing on their front porch this weekend.
During the video, which was shared by Minneapolis resident Tanya Kerssen on Saturday evening, a Humvee and an officer dressed in riot gear can be seen passing by her home to enforce curfew.
"Look at this," one unidentified woman says in the video clip. "They just keep coming."
A group of around 15 to 20 police officers and National Guardsmen then approach the front of the house, shouting "go inside!" five or six times and "get in your house now!" After around 10 seconds of commands, one officer yells, "light 'em up," and starts shooting green paint bullets at Kerssen and other unidentified people, according to the video.
George Floyd protests:Federal prisons locked down; social media goes dark for #BlackoutTuesday
She's a black female police chief. This is what she thinks about George Floyd's death. George W. Bush says George Floyd's death reveals America's 'tragic failures' Seth Rogen, Ashton Kutcher fire back at 'All Lives Matter' comments 10 top-rated face masks under $10 you can buy online right now Get the latest news straight to your phone: Download the USA TODAY app
The group on the porch runs inside. One person was hit with a bullet to their thigh, the video shows.
Paint rounds, or marking rounds, are used by police and military. They are for close-range training, counter-terrorism and urban warfare, among other purposes, according to one munitions vendor, Simunition,
It is unclear if the video was taken before or after curfew. A FAQ document posted by the Minnesota Department of Public Safety says that individuals can be on their porch during curfew but says people must listen to officers' orders.
"If a law enforcement officer or other public safety official asks you to go inside, or take any other action, you must follow the instruction," the document said.
The curfew for Minneapolis and Saint Paul is in place starting at 10 p.m. Tuesday until 4 a.m. Wednesday. Tuesday from 10 p.m. to 4 a.m.
Neither Kerssen nor the police department immediately responded to a request for comment from USA TODAY.
Follow Joshua Bote on Twitter: @joshua_bote


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text:
Actors + Performers
13 Ways to See Cheap Theater in New York City
By KC Wright | Last Updated: March 05, 2020
Photo Source: Shutterstock
Actors have a lot of necessary expenses, but seeing live theater doesn't have to be one of them! Here are 13 ways that New York actors can engage with the community and enjoy some of the city's best theatrical work—without breaking the bank.
Broadway Week
This annual two-week promotion features 2-for-1 tickets to dozens of shows, typically in September. So grab a friend and head to the theater.
If you've ever been to Times Square, you've likely come across TKTS' busy flagship booth. Located at the intersection of 47th and Broadway, tourists and New Yorkers alike line up as early as 10 a.m. for half-price tickets to matinees, and at 3 p.m. for discounted evening performances. But there's no need to fight through the crowds (and naked cowboys) to get these great deals; TKTS also offers discounted Broadway tickets at their South Street Seaport and Lincoln Center locations.
Off-Off and Beyond
In addition to running the TKTS booths, the Theatre Development Fund offers tickets to some of the city's most vibrant Off-Off Broadway performances for only $11. Members can sign up for free on the website and purchase these super-cheap tickets to dozens and dozens of plays, musicals, and events.
RUSH and Standing Room Only Tickets
Ever wonder about those lines outside "Hamilton" in the morning? Most Broadway shows offer a select number of discounted day-of tickets to savvy theatergoers. Dedicated patrons arrive hours before the box office opens to snatch up these great deals; though policies vary from theater to theater, most tickets are under $40 and available daily.
If waking up early isn't your thing, several popular shows offer a daily lottery two-and-a-half hours before curtain. Names are drawn at random and the lucky winners can purchase hugely discounted tickets to the evening's performance—without catching frostbite in the 8 a.m. rush line.
Broadway, Off-Broadway, Off-Off-Broadway: What to Know
HIPTIX at Roundabout
Theater lovers ages 18–40 can take advantage of this fantastic program from Roundabout Theatre Company; sign up online to purchase $30 advance tickets to any of Roundabout's Broadway, Off-Broadway, and Underground productions, and keep an eye out for Hiptix nights—special performances featuring post-show receptions (and refreshments) exclusively for members. Over 40? Check out ACCESS Roundabout for $10 tickets to the first four previews of any production.
30 Under 30 at Second Stage
This deal from Second Stage Theatre allows audience members aged 30 and under to purchase $30 tickets to most performances, and invites them to the 30 Under 30 After Party, where young audience members enjoy good conversation, free drinks, and snacks. Military discounts are also available online and at the box office, and students can buy inexpensive rush tickets one hour before curtain.
Pay What You Can and PS35 at Primary Stages
For select performances of Primary Stages' Off-Broadway productions, audience members can visit the box office at 4 p.m. and enter a lottery for Pay What You Can tickets. Under 35? Use the online code PS35 to purchase $35 tickets to any performance. Student rush tickets are also available!
Access25 at Atlantic Theater Company
Atlantic Theater Company offers this great deal to theatergoers of all ages; $25 tickets to every preview performance of the season. The Access25 tickets are first-come, first-serve and are available two weeks before the first performance of each show.
New Deal at Theatre for a New Audience
Patrons under 30 and full-time students of any age can visit this Brooklyn-based theater company—in their gorgeous space at the Polonsky Center—for only $20 a ticket to any performance. Just use the code NEWDEAL online or visit the box office with a valid ID.
CheapTix Sundays at New York Theatre Workshop
New York Theatre Workshop holds a select number of $25 tickets to the first two performances of every production, and sells them on a first-come-first-served basis. Can't make it to one of those two performances? NYTW also offers $25 CheapTix Rush for day-of tickets for young people, seniors, artists, and East Village and Lower East Side residents.
Hot Tix and Live for Five at Playwrights Horizons
Fans of new works have several opportunities to visit Playwrights Horizons on the cheap. With Hot Tix, patrons under 35 or full-time students can purchase $25 or $15 tickets. Similarly, audience members of all ages can submit their names for the Live for Five Lottery; winners get a $5 ticket to the first preview of new productions.
Cheap tickets are great, but free tickets are better! New York is one of the best places to find free outdoor theater, especially over the summer. Check out the events page on the New York City Department of Parks and Recreation website for ongoing listings of free shows all over the five boroughs.
Check out Backstage's theater audition listings!
Auditions & Open Casting Calls
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Auditions for Movies
See all Auditions
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'The Black Tent'
Production Company Crew, Discord Group
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text:
Moon May Harbor Ancient Pieces Of Venus' Surface
One of the Apollo 17 astronauts stands next to his lunar rover on the surface of the Moon. | … [+] Location: Taurus-Littrow Valley, Moon. (Photo by © Corbis/Corbis via Getty Images)
Two Yale University researchers have found a potential shortcut in sampling Venus' ancient surface. Instead of sending a probe on a costly and extraordinarily challenging Venus sample return mission, they propose simply finding a Venusian meteorite on our own Moon.
There's never been a bona fide detection of a Venusian meteorite on Earth. For one reason, that's because in the last several hundred million years at least, Venus' atmospheric pressures have been so intense that even a catastrophic impactor could not dislodge any Venusian rocks into space.
But before Venus underwent a runaway greenhouse and morphed into the climatic hellhole it is today, it may have had liquid water oceans as late as 700 million years ago. If so, its atmosphere would have been thin enough for surface rocks to have been dislodged by massive impactors and possibly have found their way to both the Earth and our Moon.
Due to weathering here on Earth, Venusian meteorites on Earth wouldn't survive long. But because our Moon has no atmosphere, the authors of a paper accepted by The Planetary Science Journal posit that the Moon may have be the ideal spot to preserve Venus meteorites.
Lead author Samuel Cabot and co-author Gregory Laughlin investigated the amount of material ejected from Venus when it suffered past impacts from asteroids and comets, and then traced the orbits of the rocks throughout the Solar System. They found that a small (but still significant) fraction of rocks ejected from Venus will be swept up by Earth's Moon.
Today, due to Venus' thick atmosphere, even a catastrophic impactor would not be capable of ejecting material with a force to overcome drag from Venus' thick atmospheric pressures.
1974: The cloud-covered planet Venus, the second planet from the Sun. (Photo by MPI/Getty Images)
But even if they found a Venus meteorite on the Moon or in Apollo rock samples, would researchers be able to offer a high degree of confidence that it originated on our sister planet? That is, in the same way that researchers have high confidence that the Allan Hills meteorite ALH84001 from Antarctica actually originated on Mars?
For an Allan Hills Martian meteorite, we can compare its composition to measurements made by Martian rovers, Cabot, a graduate student in Yale's Department of Astronomy, told me. A Venusian meteorite would lie in a completely new category from the rest, but require some extra work, he says. Several samples, plus some comparison to Venus' current atmosphere, would help single out Venus as the place of origin, Cabot explains.
However, the degree of confidence in a putative Venus meteorite would not be as high as for Martian meteorites, As Laughlin, a Yale planetary scientist, told me. That's because inclusions in the Martian meteorites contain gases that exactly match Mars' known atmosphere, he says.
As to where on the Moon to look?
The lunar highlands might be a good place to look since they host some of the oldest rocks, says Cabot.
Once identified, Laughlin says an analysis of grains in the rock would likely yield chemical evidence of existence of oceans. Similar strategies have been carried out on zircon minerals in ancient Earth rocks to date an early existence of oceans, he says. Radioactive dating would give an accurate age for a Venusian sample, which would place constraints on the presence or absence of oceans at a particular point in time, Laughlin notes.
How could we go about looking for Venus meteorites on our Moon now?
Upcoming Moon missions in NASA's Artemis program are a great opportunity to search for Venusian meteorites, says Cabot. In situ analysis might be possible, he says, with astronauts or robots picking out rocks to bring back to Earth for laboratory analysis.
Yet as the authors note, the largest and earliest Venusian rock fragments may lie in a zone beneath the regolith, known as the megaregolith. Finding such fragments will likely require excavating several meters below the lunar surface, they say.
Might there be microfossils in such Venusian surface fragments?
I don't think the odds are good for that, says Laughlin. Most Venusian rocks would have undergone some shock during the initial Venus impact, and then again later during the lunar impact, he says. We know that even with an extraordinary sample —- such as ALH84001 from Mars —- that identifying microfossils is fraught with controversy, says Laughlin. With Venus, it would almost certainly be far more controversial and far less clear-cut, he notes.
But would a Venus meteorite look like any other Moon rock?
The authors write that it would likely be made up of igneous rocks such as basalt, andesite and granite. But you would not be able to tell from looking at it, says Cabot, since a piece of Venus could be any type of rock. He maintains that a chemical test in a laboratory is probably going to be needed to pick out a fragment that is definitely from Venus.
As for how many pounds of lunar rocks researchers would need to sort through before hitting an actual Venus fragment?
Several hundred pounds of lunar rocks were returned from the Apollo missions, and there is a small chance a piece of Venus is within that collection, says Cabot. But based on our simulations and depending on when Venus lost its thin atmosphere, one would probably need to search 10 to 100 times that amount to find a fragment of Venus, he says.
Why not just go to Venus and do a sample return?
"Sample return from Venus or Mars is an extremely difficult and expensive undertaking," Stephen Kane, a planetary geophysicist at the University of California at Riverside, who did not take part in this research, told me. So, if samples from these bodies can be found in our locality then that is an enormous advantage, he says.
We could possibly acquire samples of Venus that predate the resurfacing of the planet, thought to have occurred 700 million to 1 billion years ago, at a time when Venus may have had temperate surface conditions, says Kane.
Tags: ancient, harbor, moon, pieces, Surface, Venus
Adopt best practices and emerging technologies
US judge rules Apple could bar Epic Games's 'Fortnite' from App Store


This page serves to provide a short collection of thoughts about various aspects of web design and best practices. It will be updated occasionally over time.
I originally added some of these to a Facebook business page, however Facebook is proving rather hopeless nowadays for reaching people as a business and somewhat counterproductive.
I've spent most of today studying CSS code for styling web pages, but not just any old CSS code. Instead I've been concentrating on really lean code, so that only the bare minimum is added to the page for ruthless efficiency and fast loading.
Very often you come across a web page that appears fine at first glance, but if you look closer you discover a plethora of errors, warnings and masses of unnecessary coding that is completely redundant. All of this clogs up your pages like a hamburger clogs up your arteries.
Good coding is fast to load for web browsers and easy for search engines to understand, so it also benefits your SEO. Good coding is a work of art.
After studying this site I just ran an online speed test to see how it performs and it's clear that lean code, properly optimised images and a fast web server all make a difference.
I have done a lot of photography over the years, stretching back long before digital started. Art directors love images with plenty of clean space because it gives them options to crop closer or place graphics to perfect effect.
Designing a website is not so different and I always start with a blank white page. I love empty space on a web page and contrary to what some might think this is not a lack of design, but rather allowing the design to breath.
It creates a much better viewing experience for the visitor when elements are clearly separated, rather than crammed into a small area as a jumbled mess.
Open spaces also allow room for clear, large fonts that are easily legible and creates a trust signal that improves the standing of your brand when everything is presented in a clean design.
Whenever I arrive at a busy web page that has no clear purpose I generally leave within 5 seconds. Whenever I arrive at a page where the site owner clearly doesn't care about the viewer experience I also leave.
In some ways it's not rocket science, however knowing when to stop is perhaps the greatest asset of any designer. Incidentally, it's also why you should hire a designer who understands these things. You start with a perfect clean page, so be careful before you mess it up.
It's not unusual for people to hire a web designer or try themselves, only to discover after a lot of work that it looks completely different online than it did on their computer, which can be frustrating and difficult to troubleshoot.
Since the 1990's web designers have hurled abuse at Internet Explorer in particular, which seemed to work in a parallel universe and require separate coding, though I'm glad to say problems are less common these days.
Most of the time a local host server like MAMP should indicate how a site will appear online. There are limits though to what this can achieve and it certainly won't tell you if there is something wrong with the setup of your chosen web server.
A lot of people spend a great deal of time and money building a site, but then simply opt for a cheap shared hosting plan believing they are all the same. You may be lucky, but equally it could open up a world of pain and frustration.
All of this is really just a way of explaining that your choice of web host is important and has an impact on your business, since slow speeds and downtime can affect both your direct sales and SEO ranking. Having a big burst of traffic only to find your server falls over because of limited resources is major false economy.
It makes sense to choose an option that is fast, reliable and compatible with your websites. I am happy to assist customers in choosing the right web host when I design them a website and deal with these points, so you simply order your design and watch your business grow. For most small businesses a basic shared hosting plan is adequate and I generally recommend Kualo.
Something I think about quite often is the choice of font, which can completely change the perception of your business when a visitor arrives at your website. There is very rarely a good reason to choose Comic Sans, but something like Gotham or Museo Sans creates a very professional impression.
At one time we were stuck with half a dozen web safe fonts like Verdana or Geneva and the internet was a pretty dull place. Nowadays the choice is vast, including many that are free, so there is no excuse for simply using any old font without a bit of thought.
There is a huge amount that can be learned about fonts, but the easiest option for many will simply be to opt for Google fonts. These are completely free and offer many choices that are close in appearance to some very expensive fonts.
There are also alternative services offering access to a huge range of stylish licensed fonts for just a few pounds a month, but without getting too technical my preferred method is to use web format fonts that are hosted on the server, then cached around the world on a CDN for extra speed.
Web fonts come in various different formats, so I use a rather neat system that automatically serves the fastest loading version that a browser can handle. This means that a modern version of Chrome should download Woff2, while Internet Explorer would download the slower Woff format. Both should look identical though.
This system ensures there is no faux rendering flash as the page opens or delays in connecting to external services. I hope the message from this is that there is a lot more to web design than simply placing a few pictures and a bit of text on a page before hitting the publish button.
Responsive websites represent a step change in the way that websites are designed, bringing them into line with the reality of modern web browsing. In essence, they reformat instantly to fit the size of the visitor's screen; providing a better browsing experience for the viewer.
If you went back to 2009 less than 1% of web traffic occurred on mobile devices but fast forward to 2016 and that figure has risen to a staggering 40% globally. It's even higher in the UK.
More importantly, it is set to grow every year for the simple reason that we now use our smartphones and tablets like iPads, rather than sitting chained to a table with a big desktop computer. Many people no longer need a traditional computer for online banking, social media or making purchases.
If you still have a website that isn't mobile friendly there is a good chance that visitors will leave immediately, but that is assuming they will even find it, since Google now penalises websites in their rankings if they are not mobile friendly.
There are of course designers who build one page for mobile and another for desktop, however this is an outdated approach in my opinion. First of all it requires updates on two separate versions, carrying greater risk of errors, such as spelling or worse. More importantly it only works reliably on two fixed sizes and there is too much device variation nowadays for this to work effectively.
There are various options for creating responsive websites and this site was created using Foundation, however there are other valid alternatives, such as Bootstrap that I also use.
I do not recommend WordPress though, because it has more security holes than a piece of Swiss cheese and they are a pain to maintain due to the necessity for constant plugin updates that often break things.
If the time has arrived when you need a responsive website with modern, SEO friendly coding, why not check out my website and contact me if you have any questions.


text:
"""
Reformat the given capa rule into a consistent style.
Use the -i flag to update the rule in-place.

Usage:

   $ python capafmt.py -i foo.yml

Copyright (C) 2020 Mandiant, Inc. All Rights Reserved.
Licensed under the Apache License, Version 2.0 (the "License");
 you may not use this file except in compliance with the License.
You may obtain a copy of the License at: [package root]/LICENSE.txt
Unless required by applicable law or agreed to in writing, software distributed under the License
 is distributed on an "AS IS" BASIS, WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
See the License for the specific language governing permissions and limitations under the License.
"""
import sys
import logging
import argparse

import capa.rules

logger = logging.getLogger("capafmt")

def main(argv=None):
    if argv is None:
        argv = sys.argv[1:]

    parser = argparse.ArgumentParser(description="Capa rule formatter.")
    parser.add_argument("path", type=str, help="Path to rule to format")
    parser.add_argument(
        "-i",
        "--in-place",
        action="store_true",
        dest="in_place",
        help="Format the rule in place, otherwise, write formatted rule to STDOUT",
    )
    parser.add_argument("-v", "--verbose", action="store_true", help="Enable debug logging")
    parser.add_argument("-q", "--quiet", action="store_true", help="Disable all output but errors")
    parser.add_argument(
        "-c",
        "--check",
        action="store_true",
        help="Don't output (reformatted) rule, only return status. 0 = no changes, 1 = would reformat",
    )
    args = parser.parse_args(args=argv)

    if args.verbose:
        level = logging.DEBUG
    elif args.quiet:
        level = logging.ERROR
    else:
        level = logging.INFO

    logging.basicConfig(level=level)
    logging.getLogger("capafmt").setLevel(level)

    rule = capa.rules.Rule.from_yaml_file(args.path, use_ruamel=True)
    reformatted_rule = rule.to_yaml()

    if args.check:
        if rule.definition == reformatted_rule:
            logger.info("rule is formatted correctly, nice! (%s)", rule.name)
            return 0
        else:
            logger.info("rule requires reformatting (%s)", rule.name)
            if "\r\n" in rule.definition:
                logger.info("please make sure that the file uses LF (\\n) line endings only")
            return 1

    if args.in_place:
        with open(args.path, "wb") as f:
            f.write(reformatted_rule.encode("utf-8"))
    else:
        print(reformatted_rule)

    return 0

if __name__ == "__main__":
    sys.exit(main())

redpajama_set_name: RedPajamaGithub


You can use the Test TCP utility (TTCP) to measure TCP throughput through an IP path. In order to use it, start the receiver on one side of the path, then start the transmitter on the other side. The transmitting side sends a specified number of TCP packets to the receiving side. At the end of the test, the two sides display the number of bytes transmitted and the time elapsed for the packets to pass from one end to the other. You can then use these figures to calculate the actual throughput on the link. For general information on TTCP, refer to Network Performance Testing with TTCP .
The TTCP utility can be effective in determining the actual bit rate of a particular WAN or modem connection. However, you can also use this feature to test the connection speed between any two devices with IP connectivity between them.
TTCP requires Cisco IOS® Software Version 11.2 or higher and Feature Sets IP Plus (is- images) or Service Provider (p- images).
Note: The ttcp command is a hidden, unsupported, privileged mode command. As such, its availability may vary from one Cisco IOS software release to another, such that it might not exist in some releases. Some platforms, for instance, require the Cisco IOS Enterprise feature set in order to perform this activity.
Ensure that there is IP connectivity between the two devices involved in the test.
Download and install the TTCP software for non-IOS clients, if necessary.
In the example shown below, we try to determine the connection speed of a modem connection between a Microsoft Windows PC and an AS5300 Access Server. Even though many of the topics and explanations that are included here are specific to modem connections, the TTCP utility can be used between any two devices.
Use the show modem operational-status command (for a modem link) to check the connection parameters. For other LAN or WAN scenarios, this step is not necessary.
This edited output shows that the client is connected in V.90 at a 45333 bps downlink rate and a 24000 BPS uplink rate. Data compression is disabled on the client modem. Since the TTCP test pattern is highly compressible, any data compression would skew our measure of true modem link throughput.
Start the ttcpw program on the PC (in a DOS window), running as a receiver. Refer to the Readme file provided with the windows TTCP software for the appropriate syntax.
Launch the TTCP sender (transmitter) on the AS5300. Leave most settings at the default, except for the number of buffers to transmit. The default number of buffers is 2048, with which the TTCP test would take a long time to complete. By reducing the number of buffers, we are able to finish the test in a reasonable timeframe.
Note: Try to get a snapshot of the modem (port) operational-status, as described above, just before you begin the TTCP test.
This causes the Cisco IOS TTCP to make a TCP connection to the TTCPW (on the Windows machine).
At this point, you may want to take another snapshot of the modem or port operational-status. This information can be useful during analysis to check whether, for example, the modem connection experienced any retrains or speedshifts.
Since it is most common to evaluate connect speeds in kbps (kilobits per second, or 1000 bits per second) rather that KBps (kilobytes per second, or 1024 bytes per second), we must use the information from TTCP to calculate the bit rate (in kbps). Use the number of bytes received and the transfer time to calculate the actual bit rate for the connection.
Calculate the bit rate by converting the number of bytes into bits and then divide this by the time for the transfer. In this example, the windows PC received 409600 bytes in 84.94 seconds. We can calculate the bit rate to be (409600 bytes * 8 bits per byte) divided by 84.94 seconds=38577 BPS or 38.577 kbps.
Note: The receiver-side results are slightly more accurate, since the transmitter might think it is finished after it performs the last write - that is, before the data has actually traversed the link.
This comes out as an uplink throughput of 141144 BPS - or almost a 6:1 compression ratio relative to the nominal uplink rate of 24 kbps. This is an interesting result considering hardware compression is disabled (which we determined from show modem operational-status). However, use the IOS command show compress to check whether any software compression is being used.
Cisco IOS treats data sourced by the router differently from data routed through the router. In our example above, although Microsoft Point-to-Point Compression (MPPC) compression was negotiated on the link under test, the data transmitted by the router did not use software compression, while the data transmitted by the PC did. This is why the uplink throughput was significantly greater than the downlink throughput. For performance testing of high bandwidth links, you should always test through the routers.
For IP paths with a large bandwidth * delay product, it is important to use a TCP window size sufficient to keep the pipe full. In the case of modem links, the default 4 KB window size is normally adequate. You can boost the IOS TCP window size with the command i p tcp window-size. Refer to the appropriate documentation for non-IOS systems.
Another easy way to test the throughput across a modem link is to use the open source tool Through-Putter . Install this tool on a web server behind the Access servers and have the Windows PC clients use a browser to call up the Java tool. It can then be used to quickly determine the data rate on a modem connection. This modem throughput applet is open source tool and is not supported by the Cisco Technical Assistance Center. Refer to the Readme file provided with the tool for further installation and operating instructions.