text:
\section{Introduction} \label{S1}

In the present paper we consider the reaction-diffusion equation
\begin{equation} \label{0.1}
T_t = \Delta T  + f(T)
\end{equation}
in the cylinder ${\mathbb{R}}\times\Omega$ where $\Omega$ is a domain in
${\mathbb{R}}^{n-1}$, with Neumann boundary conditions on
${\mathbb{R}}\times\partial\Omega$. The non-linear reaction term $f$ is
assumed to be Lipschitz continuous with $f(0)=f(1)=0$ and the
initial datum $T_0$ is between 0 and 1.

We will treat the case when $T_0$ is independent of the transversal
variable $y\in\Omega$, and so \eqref{0.1} becomes
\begin{equation} \label{0.2}
T_t = T_{xx}  + f(T)
\end{equation}
with $x\in{\mathbb{R}}$. This equation has been extensively studied in
mathematical, physical and other literature, starting with the
pioneering works of Fisher \cite{Fisher} and Kolmogorov, Petrovskii,
Piskunov \cite{KPP}. In these papers \eqref{0.2} was used to
describe the propagation of advantageous genes in a population. The
main object of study in these and many subsequent works was the
existence and stability of traveling fronts for \eqref{0.2} and
\eqref{0.1}. In the recent years most of the results have been
extended to include an advection term $u\cdot\nabla T$ in
\eqref{0.1}, and we refer to the reviews \cite{Berrev,Xin2} for an
extensive bibliography.

The above equations are used to model not only population genetics
phenomena. When $f(\theta)>0$ for $\theta\in(0,1)$, then $f$ is a {\it
combustion non-linearity} and \eqref{0.1}/\eqref{0.2} model an
exotermic chemical reaction in an infinite tube with a zero
heat-loss boundary, in particular, flame propagation in a premixed
combustible gas without advection (see Zel'dovich and
Frank-Kamenetskii \cite{ZFK}). In this setting $T$ is the normalized
temperature taking values in $[0,1]$. We note that \eqref{0.1} is
usually obtained from a system involving both the temperature and
the concentration of the reactants after the simplifying assumption
of equal thermal and material diffusivities.

A special case of positive $f$, used often in chemical and
biological literature, is the {\it KPP type} with $f''(\theta)\le c<0$
\cite{KPP}. In combustion models the non-linearity is often
considered to be of {\it Arrhenius type} with slow reaction rates at
low temperatures, modeled by $f(\theta)=e^{-A/\theta}(1-\theta)$. One
often approximates this situation by considering an {\it ignition
non-linearity} $f$ satisfying $f(\theta)=0$ for $\theta\in[0,\theta_0]$
and $f(\theta)>0$ for $\theta\in(\theta_0,1)$, with $\theta_0\in(0,1)$ the
{\it ignition temperature}.

The third prominent case is the {\it bistable non-linearity} with
$f(\theta)< 0$ for $\theta\in(0,\theta_0)$ and $f(\theta)>0$ for
$\theta\in(\theta_0,1)$, where one usually assumes $\int_0^1
f(\theta)\,d\theta>0$. This has been used to model signal propagation
along bistable transmission lines, in particular, nerve pulse
propagation \cite{NAY}. In biological context it is also called {\it
heterozygote inferior} (see Aronson and Weinberger \cite{AW2}).

In this paper we will consider all the above types. Our interest
here will not be in the question of traveling fronts, but in {\it
extinction} of reaction --- {\it quenching} of flames. We will
therefore assume the initial datum $T_0(x)$ for \eqref{0.2} to be
compactly supported, and will want to know when
\begin{equation} \label{0.3}
\|T(t,\cdot)\|_\infty \to 0 \text{ as } t\to\infty.
\end{equation}
For the sake of simplicity we will restrict ourselves to the case of
$T_0$ being the characteristic function of an interval,
\begin{equation} \label{0.4}
T_0(x)\equiv \chi_{[-L,L]}(x),
\end{equation}
and study how long-time behavior of $T$ depends on $L$. The methods
in this paper allow one to treat some other increasing 1-parameter
families of initial conditions, too.

Thus, we will study the competition of reaction and diffusion. The
former helps increasing the temperature, whereas the latter
(together with the compactness of the support of the initial datum)
works towards the extinction of the flame. This question was
originally addressed forty years ago by Kanel' \cite{Kanel} who
considered the case of ignition non-linearity and proved that if the
initial datum is large enough, then reaction wins, whereas if it is
small then diffusion manages to quench the flame. More precisely,
when $T$ solves \eqref{0.2}/\eqref{0.4} and $f$ is of ignition type,
Kanel' proved that there are two length scales $L_0$, $L_1$ such
that
\begin{align*}
& T(t,x)\to 0 \text{ as $t\to \infty$ uniformly in $x\in{\mathbb{R}}$ if
$L<L_0$,}
\\ & T(t,x)\to 1 \text{ as $t\to \infty$ uniformly on compacts if
$L>L_1$.}
\end{align*}
This has been extended to the case of bistable $f$ by
Aronson-Weinberger \cite{AW2}. Both results actually hold when
\eqref{0.4} is replaced by
\begin{equation} \label{0.5}
T_0(x)\equiv \alpha\chi_{[-L,L]}(x).
\end{equation}
for any $\alpha>\theta_0$, with $L_0$ and $L_1$ depending on $\alpha$
(in the ignition case this follows from \cite{Kanel}, in the
bistable case it was proved by Fife and McLeod \cite{FM}). A natural
question arises: does $L_0$ equal $L_1$? If this is true and if one
could determine the behavior of $T$ as $t\to\infty$ when $L=L_0$,
then one would be able to provide the complete ``phase portrait''
for the PDE \eqref{0.2} with respect to a 1-parameter family of
initial conditions.

Since these early works, particularly in the recent years, several
authors have studied quenching for \eqref{0.1}. The above results
have been extended to the case when \eqref{0.1} includes an
advection term $u\cdot\nabla T$, with $u$ a shear or periodic flow
(see \cite{Roq, Xin}), even for certain combustion non-linearities
\cite{ZlaArrh}. Quenching of large initial data by large amplitude
shear and cellular flows has been studied in
\cite{CKR,FKR,KZ,ZlaArrh}. However, the question whether $L_0=L_1$
remained open even in the simplest case of \eqref{0.2}. The
following two results provide the answer, including the treatment of
the critical case $L=L_0$.

The first of them holds for ignition and combustion non-linearities.

\begin{theorem} \label{T.1.1}
Let $\theta_0\in[0,1)$ and $f:[0,1]\to{\mathbb{R}}$ be Lipschitz with
$f(\theta)=0$ when $\theta\in[0,\theta_0]$, $f(\theta)>0$ when
$\theta\in(\theta_0,1)$, and $f(1)=0$. If $\theta_0>0$ then assume in
addition that $f$ is non-decreasing on $[\theta_0,\theta_0+\delta]$ for
some $\delta>0$. Let $T:[0,\infty)\times{\mathbb{R}}\to[0,1]$ solve
\begin{align}
& T_t = T_{xx}  + f(T) \label{1.1}
\\ & T_0(x) \equiv \chi_{[-L,L]}(x). \notag
\end{align}
Then there is $L_0\ge 0$ such that
\begin{SL}
\item[{\rm{(i)}}] if $L<L_0$, then $T\to 0$ uniformly on ${\mathbb{R}}$ as $t\to
\infty$;
\item[{\rm{(ii)}}] if $L=L_0$, then $T\to \theta_0$ uniformly on compacts as $t\to
\infty$;
\item[{\rm{(iii)}}] if $L>L_0$, then $T\to 1$ uniformly on compacts as $t\to
\infty$.
\end{SL}
\end{theorem}

{\it Remark.} The possibility of $L_0=0$ (so called {\it
hair-trigger effect}) cannot be excluded when $\theta_0=0$. More
precisely, using results from \cite{ZlaArrh} one can show that
$L_0=0$ when $f(\theta)\ge c\theta^p$ for some $p<3$ and all small
$\theta$, but $L_0>0$ when $f(\theta)\le c\theta^p$ for some $p>3$ and all
small $\theta$. We also note that if $\theta_0=0$, then the convergence
in (ii) is as in (i)
--- uniform on ${\mathbb{R}}$.
\smallskip

Our second result holds for bistable non-linearities. We define
$\theta_2\in(\theta_0,1)$ by $\int_0^{\theta_2} f(\theta)d\theta=0$ and let
$U$ be the unique function solving $0=U''+f(U)$ with $U(0)=\theta_2$
and $U'(0)=0$. Then $U$ is an even function and we will show in the
proof of the following theorem that it is positive on ${\mathbb{R}}$,
decreasing to 0 on $(0,\infty)$, and bell-shaped.

\begin{theorem} \label{T.1.2}
Let $\theta_0\in(0,1)$ and $f:[0,1]\to{\mathbb{R}}$ be Lipschitz with
$f(0)=f(\theta_0)=f(1)=0$, $f(\theta)< 0$ when $\theta\in(0,\theta_0)$, and
$f(\theta)>0$ when $\theta\in(\theta_0,1)$. Assume also that $\int_0^1
f(\theta)d\theta>0$. Let $T:[0,\infty)\times{\mathbb{R}}\to[0,1]$ solve the
problem \eqref{1.1}.
Then there is $L_0>0$ such that
\begin{SL}
\item[{\rm{(i)}}] if $L<L_0$, then $T\to 0$ uniformly on ${\mathbb{R}}$ as $t\to
\infty$;
\item[{\rm{(ii)}}] if $L=L_0$, then $T\to U$ uniformly on ${\mathbb{R}}$ as $t\to
\infty$;
\item[{\rm{(iii)}}] if $L>L_0$, then $T\to 1$ uniformly on compacts as $t\to
\infty$.
\end{SL}
\end{theorem}

{\it Remark.} Both results can be extended to some other increasing
families of initial conditions. In particular, to \eqref{0.5} with
$\alpha>\theta_0$.
\smallskip

The crux of the proofs of both theorems will be to show that there
is a single $L$ for which $T$ does not converge to 0 or 1 at $x=0$
as $t\to\infty$. In Theorem \ref{T.1.1} this will be achieved with
the help of Lemma~\ref{L.2.2} by comparing solutions of \eqref{1.1}
for two different initial conditions at differently rescaled times.
In Theorem \ref{T.1.2} it will follow from a detailed analysis of
the large time behavior of $T$ when the above limit is not 0 or 1,
and an application of the comparison principle.

We note here that Theorem \ref{T.1.1} is, in a sense, a limiting
case of Theorem \ref{T.1.2}. If one takes $f\to 0$ on $(0,\theta_0)$
keeping $f$ unchanged on $(\theta_0,1)$, one has $\theta_2\to\theta_0$ and
$U\to\theta_0$ on compacts. That is, the bell shaped solution $U$ from
Theorem \ref{T.1.2}(ii) converges to the constant solution $\theta_0$
from Theorem \ref{T.1.1}(ii).

The rest of the paper is devoted to the proofs of the two theorems.
Section \ref{S2} contains preliminary Lemmas \ref{L.2.1} and
\ref{L.2.2}. Sections \ref{S3} and \ref{S4} prove Theorems
\ref{T.1.1} and \ref{T.1.2}, respectively.

The author thanks Henri Berestycki, Peter Constantin, Fran\c cois
Hamel, Alexander Kiselev, Peter Pol\' a\v cik, Jean-Michel
Roquejoffre, and Lenya Ryzhik for encouragement and useful
discussions. He also acknowledges the hospitality of the Mathematics
Department of the University of Chicago where part of this work was
done, as well as partial support from the NSF through the grant
DMS-0314129.

\section{Preliminary Lemmas} \label{S2}

We start with

\begin{lemma} \label{L.2.1}
Let $f:[0,1]\to{\mathbb{R}}$ be Lipschitz with $f(0)=f(1)=0$. If
$T:[0,\infty)\times{\mathbb{R}}\to[0,1]$ solves \eqref{1.1},
then the following hold.
\begin{SL}
\item[{\rm{(i)}}] If $|x|\le|y|$ then $T(t,x)\ge T(t,y)$.
\item[{\rm{(ii)}}] There is $t_\ast>0$ (possibly $t_\ast=\infty$)
such that $T(t,0)$ as a function of $t$ is non-increasing on
$[0,t_\ast)$ and non-decreasing on $[t_\ast,\infty)$.
\item[{\rm{(iii)}}] If $f\ge 0$ then there is $\theta_*\in[0,1]$ such
that $f(\theta_*)=0$ and $T(t,x)\to\theta_*$ as $t\to\infty$, uniformly
on compacts.
\end{SL}
\end{lemma}

{\it Remarks.} 1. For sufficiently smooth $f$ this is essentially a
result of Kanel' \cite{Kanel}, although in (ii) he only proves
$T(t,0)$ to be eventually monotone.
\smallskip

2. In the case of \eqref{0.5} with $\alpha\in(\theta_0,1)$, part (ii)
has $0<t_\ast\le t_{\ast\ast}\le\infty$ such that $T(t,0)$ is
non-decreasing on $[0,t_\ast)$, non-increasing on
$[t_\ast,t_{\ast\ast})$ and non-decreasing on
$[t_{\ast\ast},\infty)$.
\smallskip

\begin{proof}
We first assume that $f$ is smooth and briefly recall main points of
the proofs of (i) and (ii) from \cite{Kanel}. Let $T^\varepsilon$ solve
\eqref{1.1} but with initial condition $T(0,x)\equiv \chi^\varepsilon (x)$
where $\chi^\varepsilon$ are smooth, symmetric, decreasing in $|x|$, and
converge to $\chi_{[-L,L]}$ in $L^1({\mathbb{R}})$ as $\varepsilon\to 0$. Then
$T^\varepsilon_x(0,x)\le 0$ for $x>0$, and by symmetry $T^\varepsilon_x(t,0)=0$.
Since
\[
(T^\varepsilon_x)_t = (T^\varepsilon_x)_{xx}  + f'(T^\varepsilon)T^\varepsilon_x,
\]
the maximum principle gives $T^\varepsilon_x(t,x)\le 0$ for $x>0$. Symmetry
then yields $T^\varepsilon_x(t,x)\ge 0$ for $x<0$. Since for any fixed
$t>0$ we have $T^\varepsilon(t,x)\to T(t,x)$ uniformly in $x$ as $\varepsilon\to
0$, this proves (i).

Now let $D^h(t,x)\equiv T(t+h,x)-T(t,x)$. By the mean value theorem,
\[
D^h_t = D^h_{xx} + f'(S)D^h
\]
for some $S(t,x)$. Let $\Delta_h$ be set of $(t,x)$ for which
$D_h(t,x)\le 0$. Then $\Delta_h\cap (\{0\}\times{\mathbb{R}}) =
\{0\}\times[-L,L]$. By the maximum principle and symmetry,
$\Delta_h$ is connected and its sections by lines parallel to the
$x$-axis are segments symmetric about the $t$-axis. Therefore there
is $0<t^h_*\le\infty$ such that $D^h(t,0)<0$ for $t\in[0,t^h_*)$ and
$D^h(t,0)\ge 0$ for $t\in[t^h_*,\infty)$. From $D^h(t,x) =
D^{h/2}(t+\tfrac h 2) + D^{h/2}(t)$ we obtain $t^{h/2}_*
\in[t^h_*,t^h_*+\tfrac h2]$, and (ii) follows with $t_*\equiv
\lim_{n\to\infty} t^{2^{-n}}_*$.

If $f$ is only Lipschitz, take smooth $f^\varepsilon$ such that $\|
f^\varepsilon-f\|_\infty \le \varepsilon$ and let $T^\varepsilon$ solve \eqref{1.1} with
$f^\varepsilon$ in place of $f$. One can then show that $V^\varepsilon\equiv
T^\varepsilon -T$ satisfies $|V^\varepsilon(t,x)|\le \tfrac \varepsilon c(e^{ct}-1)$ with
$c$ the Lipschitz constant for $f$ (we spell this argument out in
the proof of Theorem \ref{T.1.1} below). Therefore
$T(t,x)=\lim_{\varepsilon\to 0} T^\varepsilon(t,x)$ for all $t$ and $x$, and since
(i) and (ii) hold for each $T^\varepsilon$, they also hold for $T$.

Finally, assume that $f\ge 0$. By (ii), $\theta_*\equiv
\lim_{t\to\infty} T(t,0)$ is well defined. Let $\Phi$ solve
$\Phi_t=\Phi_{xx}$ on ${\mathbb{R}}^+$ with $\Phi(0,x)\equiv T(0,x)$ and
boundary condition $\Phi(t,0)\equiv T(t,0)$. Then
$\Phi(t,x)\to\theta_*$ as $t\to\infty$, uniformly on compacts. Since
by the comparison principle (e.g., \cite{Sm}) and (i), $\Phi(t,x)\le
T(t,x)\le T(t,0)$, the second claim in (iii) follows.

To prove the first claim, assume $f(\theta_*)>0$ and choose $\varepsilon>0$
such that for $\theta\le\theta_*+10\varepsilon$ we have $f(\theta)\ge
\theta-\theta_*+2\varepsilon$. Pick $t_0$ such that if $\Phi$ solves
$\Phi_t=\Phi_{xx}$ on ${\mathbb{R}}$ with initial condition
$\Phi(t_0,x)=T(t_0,x)$, then $\Phi(t,0)\ge \theta_*-\varepsilon$ and
$T(t,x)\le \theta_*+\varepsilon$ for $t\in[t_0,t_0+\ln 4]$ and $x\in{\mathbb{R}}$.
This is possible thanks to the second claim in (iii). Define
\[
S(t,x)\equiv \theta_*-2\varepsilon + (\Phi(t,x)-\theta_*+2\varepsilon)e^{t-t_0}.
\]
Then $S(t,x)\le \theta_*+10\varepsilon$ for $t\in[t_0,t_0+\ln 4]$ because
$\Phi(t,x)\le T(t,x)\le\theta_*+\varepsilon$ for these $t$. A simple
computation now shows that $S_t\le S_{xx} + f(S)$ for
$t\in[t_0,t_0+\ln 4]$. Hence, $S$ is a subsolution of \eqref{1.1}
with $S(t_0,x)=T(t_0,x)$, and so $S\le T$ for $t\in[t_0,t_0+\ln 4]$.
But $S(t_0+\ln 4,0)\ge \theta_*+2\varepsilon>T(t,0)$, which is a
contradiction. Therefore we have to have $f(\theta_*)=0$.
\end{proof}

Next, observe that we can use scaling to replace the variation in
the initial condition in \eqref{1.1} by variation in the reaction
strength. If $T$ solves \eqref{1.1} with $T(0,x)\equiv
\chi_{[-L,L]}(x)$, define $\tilde T(t,x) \equiv T(L^2 t,Lx)$, so that
we have
\[
\tilde T_t = \tilde T_{xx} + L^2f(\tilde T)
\]
and $\tilde T(0,x)=\chi_{[-1,1]}(x)$. Hence, Theorem \ref{T.1.1} will
be proved if we show that its conclusion holds for the $L$-dependent
family of problems
\begin{align}
& T_t  = T_{xx}  + Lf(T) \label{2.11}
\\ & T_0(x)  \equiv \chi_{[-1,1]}(x) \notag
\end{align}
instead of \eqref{1.1} (note that Lemma \ref{L.2.1} holds here,
too). This important observation motivates the following key lemma.

\begin{lemma} \label{L.2.2}
Let $\Omega$ be a connected open domain in ${\mathbb{R}}^n$ with a smooth
boundary (possibly $\Omega={\mathbb{R}}^n$)
and let $f,g:[0,\infty)\to{\mathbb{R}}$ be Lipschitz with $f(0)=g(0)=0$ and
$f\le g$. Let $T,S:[0,\infty)\times\Omega\to[0,\infty)$ be
continuous functions solving
\begin{align}
T_t &=\Delta T  + f(T) \label{2.1}
\\ S_t &=\Delta S  + g(S) \label{2.2}
\end{align}
in $\Omega$ with Dirichlet boundary conditions on $\partial\Omega$.
Assume $0\le T(0,x)\le S(0,x)$ for all $x\in\Omega$ and
$T(0,x_0)<S(0,x_0)$ for some $x_0$. Assume also that for any
$\theta>0$ the set $\Omega_{0,\theta}\equiv \{ x\in\Omega \,|\,
S(0,x)\ge\theta \}$ is compact. Finally, assume that there are
$\theta_1>0$ and $\varepsilon_1>0$ such that for any
$\theta\in[\theta_1,\|T\|_\infty)$ and $\varepsilon\in[0,\varepsilon_1]$ we have
\begin{equation} \label{2.3}
g \big( \theta + \varepsilon [\theta-\theta_1] \big) \ge (1+\varepsilon) f (\theta).
\end{equation}
Then
\begin{equation} \label{2.4}
\liminf_{t\to\infty} \inf_{T(t,x)>\theta_1} \frac {S(t,x)-\theta_1}
{T(t,x)-\theta_1} > 1
\end{equation}
with the convention that infimum over an empty set is $\infty$.
\end{lemma}

{\it Remark.} The result holds without change when we add a first
order term $u(x)\cdot\nabla$ with $u\in C^1$ to \eqref{2.1} and
\eqref{2.2}.

\begin{proof}
First notice that the assumptions imply that
\begin{equation*}
\Omega_{t,\theta}\equiv \{ x\in\Omega \,\big|\, S(t,x)\ge\theta \}
\end{equation*}
are compact. Indeed, by the maximum principle,
$\Omega_{t,\theta}\subseteq\tilde\Omega_{t,\delta\theta}$ where $\delta\equiv
e^{-tc}$ with $c$ the Lipschitz constant for $g$, and
$\tilde\Omega_{t,\theta}$ is defined as $\Omega_{t,\theta}$ but with
$\Phi$, the solution of
\[
\Phi_t = \Delta\Phi, \quad\quad \Phi(0,x)=S(0,x),
\]
in place of $S$. Compactness of $\tilde\Omega_{t,\theta}$ follows from
that of $\tilde\Omega_{0,\theta}$ and the Feynman-Kac formula.

The assumptions and the strong maximum principle also imply
$T(t,x)<S(t,x)$ for $t>0$ and $x\in\Omega$.
Let us define
\begin{align*}
\Omega(t) &\equiv \{ x\in\Omega \,\big|\, T(t,x)>\theta_1 \},
\\ \Omega'(t) &\equiv \{ x\in\Omega \,\big|\, T(t,x)=\theta_1 \},
\end{align*}
and let
\[
\omega(t)\equiv \min \bigg\{ 1+\varepsilon_1, \inf_{x\in\Omega(t)} \frac
{S(t,x)-\theta_1} {T(t,x)-\theta_1} \bigg\}.
\]
Since $\overline{\Omega(t)}$ is compact and $T<S$ continuous,
$\omega(t)>1$ for $t>0$. Hence the result will follow if we show
that $\omega$ is a non-increasing function. Since $\omega$ is
continuous (because $T$ and $S$ are), it is sufficient to show that
for any $t_0>0$ there is $\tau_0>0$ such that for all
$t\in[t_0,t_0+\tau_0]$ we have $\omega(t)\ge\omega(t_0)$.

Hence, fix $t_0>0$. Notice that uniform boundedness of the sets
$\Omega_{t,\theta_1}$ for $t\in[0,t_0+1]$ (by the maximum principle
and Feynman-Kac formula) and continuity of $T,S$ imply that $T,S$
are uniformly continuous on $[0,t_0+1]\times\Omega$, and the set
\[
\Sigma\equiv \{ (t,x) \,|\, t\in[t_0,t_0+1] \text{ and }
x\in\Omega(t)\cup\Omega'(t) \}
\]
is compact. Thanks to the uniform continuity of $T$ we only need to
consider the case $\Omega(t_0)\cup\Omega'(t_0)\neq\emptyset$, and
hence $\Sigma\neq\emptyset$. Since $S$ is continuous and $S>T$,
\[
\sigma\equiv \inf_{(t,x)\in\Sigma} \{S(t,x)-\theta_1 \}>0.
\]
We let $\delta\equiv \sigma/4(1+\varepsilon_1)$ and define
\begin{equation*}
\Delta \equiv \{ x \,\big|\, |T(t_0,x)-\theta_1|\le \delta \text{ and }
S(t_0,x)-\theta_1\ge \sigma-\delta \}
\end{equation*}
By the uniform continuity of $T,S$, there is $\tau_0\in(0,1)$ such
that for $t\in[t_0,t_0+\tau_0]$ and $x\in\Omega$ we have
\begin{equation} \label{2.5}
|T(t,x)-T(t_0,x)|\le\tfrac\delta 2 \quad \text{ and } \quad
|S(t,x)-S(t_0,x)|\le\tfrac\delta 2.
\end{equation}
So if $t\in[t_0,t_0+\tau_0]$, then
\begin{equation} \label{2.7}
\Omega(t)\subseteq \Omega(t_0)\cup\Delta
\end{equation}
(note that $S(t_0,x)-\theta_1\ge \sigma-\tfrac\delta 2$ for
$x\in\Omega(t)$ because then $(t,x)\in\Sigma$). Now if
$t\in[t_0,t_0+\tau_0]$ and $x\in\Delta$, then by \eqref{2.5},
\begin{equation} \label{2.6}
S(t,x)-\theta_1 > \frac \sigma 2 > (1+\varepsilon_1)|T(t,x)-\theta_1| \ge
\omega(t_0) (T(t,x)-\theta_1).
\end{equation}

Next let
\begin{equation} \label{2.8}
A\equiv  \{ x\in\Omega \,\big|\, T(t_0,x)>\theta_1+\delta \} =
\Omega(t_0)\smallsetminus\Delta
\end{equation}
and
\[
B\equiv \{ x\in\Omega \,\big|\, T(t_0,x)\ge\theta_1+\tfrac\delta 2 \}
\subseteq\Omega(t_0).
\]
Uniform continuity of $T$ shows that, $\dist(A,B^c)>0$, and so there
is an open set $\Gamma$ with a smooth boundary such that
$A\subseteq\Gamma\subseteq B$. Let $\tilde T\equiv T-\theta_1$, $\tilde
U\equiv \omega(t_0) \tilde T$, $\tilde S\equiv S-\theta_1$, $\tilde
f(\theta)\equiv f(\theta+\theta_1)$, and $\tilde g(\theta)\equiv
g(\theta+\theta_1)$. Then for $x\in\Gamma$ we have
\[
\tilde S(t_0,x)\ge \omega(t_0) \tilde T(t_0,x)=\tilde U(t_0,x)
\]
by the definition of $\omega(t_0)$, for $t\in[t_0,t_0+\tau_0]$ and
$x\in\partial\Gamma$ we have
\[
\tilde S(t,x) > \sigma-2\delta > \frac \sigma 2 \ge \omega(t_0) \tilde
T(t,x)=\tilde U(t,x)
\]
since $\partial\Gamma\subseteq B\smallsetminus A\subseteq\Delta$,
and for $t\in[t_0,t_0+\tau_0]$ and $x\in\Gamma$ we have
\begin{align*}
\tilde U_t &=\Delta\tilde U + \omega(t_0) \tilde f \big( \tfrac
1{\omega(t_0)} \tilde U \big),
\\ \tilde S_t &=\Delta\tilde S + \tilde g(\tilde S)
\end{align*}
by \eqref{2.1} and \eqref{2.2}. For these $(t,x)$ we have
$T(t,x)\ge\theta_1$ because of \eqref{2.5} and $\Gamma\subseteq B$,
and so by \eqref{2.3} and $\omega(t_0)-1 \in(0,\varepsilon_1]$,
\[
\omega(t_0) \tilde f \big( \tfrac 1{\omega(t_0)} \tilde U \big) =
\omega(t_0) f(T)\le g\big( \omega(t_0)[T-\theta_1]+\theta_1 \big)=\tilde
g(\tilde U).
\]
The comparison principle now shows that $\tilde S\ge\tilde U$ on
$[t_0,t_0+\tau_0]\times\Gamma$. Hence
\[
S(t,x)-\theta_1 \ge \omega(t_0) (T(t,x)-\theta_1)
\]
for $t\in[t_0,t_0+\tau_0]$ and $x\in A$, which together with
\eqref{2.6}, \eqref{2.8}, and \eqref{2.7} gives
$\omega(t)\ge\omega(t_0)$ for $t\in [t_0,t_0+\tau_0]$. The proof is
finished.
\end{proof}

\section{Proof of Theorem \ref{T.1.1}} \label{S3}

We can now complete the proof of Theorem \ref{T.1.1}. We will do
this for the formulation in \eqref{2.11}.

First assume $\theta_0>0$. We know from Lemma \ref{L.2.1}(iii) that
for every $L$ we have $T\to\theta^L_*$ uniformly on compacts, with
$\theta_*^L$ such that $f(\theta^L_*)=0$. Obviously
$\theta^L_*\notin(0,\theta_0)$ because in that case we would have
$T(t,x)\le\theta_0$ for all $t\ge t_0$ and consequently $T\to 0$
(since $\|T(t_0,\cdot)\|_1 <\infty$ and $T_t=T_{xx}$ for $t\ge
t_0$). So we are only left with $\theta^L_*\in\{0,\theta_0,1\}$.

Let $A$, $B$, and $C$ be the sets of $L\ge 0$ such that $\theta^L_*$
equals $0$, $\theta_0$, and $1$, respectively. Notice that since
$T(t,0)\ge T(t,x)$, the convergence of $T$ to $0$ for $L\in A$ is
actually uniform on ${\mathbb{R}}$. We have $A\cup B\cup C=[0,\infty)$ and
the comparison principle implies that the three sets are intervals
with $A$ lying to the left of $B$ and $B$ to the left of $C$.

Moreover, $A$ and $C$ are non-empty by Kanel' \cite{Kanel} and open.
The latter follows from the fact that if $T^L$ is the solution of
\eqref{2.11}, then for $L_1<L_2$ and $V\equiv T^{L_2}-T^{L_1}$ we
have $V\ge 0$ by comparison, and
\begin{align*}
V_t & = \Delta V +(L_2-L_1)f(T^{L_2}) + L_1[ f(T^{L_2})-f(T^{L_1}) ]
\\ & \le \Delta V + c(L_2-L_1) + cL_1 V
\end{align*}
with $c\ge\|f\|_\infty$ the Lipschitz constant for $f$. Since the
function $\tilde V(t,x)\equiv \tfrac{L_2-L_1}{L_1}(e^{cL_1 t}-1)$
satisfies
\[
\tilde V = \Delta \tilde V + c(L_2-L_1) + cL_1 \tilde V
\]
with $\tilde V(0)=0=V(0)$, the comparison principle gives $V\le\tilde
V$, that is,
\[
T^{L_2}(t,x)-T^{L_1}(t,x) \in [0, \tfrac{L_2-L_1}{L_1}(e^{cL_1
t}-1)].
\]
Therefore if $L_1\in A$, then $T^{L_1}(t_0,0)\le\tfrac 12 \theta_0$
for some $t_0>0$, and hence $T^{L_2}(t_0,0)<\theta_0$ (and so $L_2\in
A$) for $L_2<L_1+ \tfrac 12 \theta_0 L_1 (e^{cL_1 t_0}-1)^{-1}$. On
the other hand, Kanel's result \cite{Kanel} also holds for
\eqref{0.5}, and it says that for any $\alpha>\theta_0$ and $L>0$
there is $M=M(\alpha,L)<\infty$ such that if $T$ solves \eqref{2.11}
and $T(t_0,x)\ge \alpha\chi_{[-M,M]}(x)$, then $T\to 1$ uniformly on
compacts. Let $\theta_0<\alpha<\beta<1$ and if $L_1\in C$, let
$M=M(\alpha,\tfrac 12 L_1)$. For some $t_0$ we have
$T^{L_1}(t_0,x)\ge\beta\chi_{[-M,M]}(x)$ and so for any
$L_2>L_1-(\beta-\alpha)L_1(e^{cL_1 t_0}-1)^{-1}$ we have
$T^{L_2}(t_0,x)\ge\alpha\chi_{[-M,M]}(x)$. If in addition
$L_2>\tfrac 12 L_1$, we have $L_2\in C$. So $A,C$ are non-empty and
open, and hence $B$ is non-empty and closed.

The proof will be finished if we show that $B$ contains a single
element. Hence assume $L_1<L_2$ are both in $B$. Let $\theta_1\equiv
\tfrac 12 \theta_0 \in(0,\theta_0)$ and
\[
\varepsilon_1\equiv \min\{ L_2 L_1^{-1} -1,\, \delta(\delta+\theta_0)^{-1} \} >0
\]
(with $\delta$ from the statement of Theorem \ref{T.1.1}). Choose
$t_0>0$ such that $T^{L_1}(t,x)\le \theta_0+\tfrac\delta 2$ when $t\ge
t_0$. The comparison principle, $f\not\equiv 0$, and the strong
maximum principle yield $T^{L_1}< T^{L_2}$ for $t>0$ and both
$T^{L_1}$ and $T^{L_2}$ are obviously continuous for $t>0$.
Lipschitzness of $f$ and compact support of $T^{L_2}(0,\cdot)$ show
that for any $\theta>0$, the set of $x$ for which
$T^{L_2}(t_0,x)\ge\theta$ is compact. Finally, whenever
$\theta\in[\theta_1,\theta_0+\tfrac\delta 2]$ and $\varepsilon\in[0,\varepsilon_1]$, we
have $\theta+\varepsilon[\theta-\theta_1]\le \theta_0+\delta$. Thus by the
assumptions on $f$ (and the definition of $\varepsilon_1$),
\[
L_2 f(\theta+\varepsilon[\theta-\theta_1])\ge L_2 f(\theta) \ge (1+\varepsilon)L_1
f(\theta).
\]
Therefore Lemma \ref{L.2.2} applies to $T^{L_1}$ and $T^{L_2}$ (with
starting time $t_0$) and shows that for some $r>1$ and all large
enough $t$ we have
\[
T^{L_2}(t,x)-\theta_1 \ge r\big[ T^{L_1}(t,x)-\theta_1 \big]
\]
whenever $T^{L_1}(t,x)>\theta_1$. But since $\theta_1<\theta_0$, this
contradicts the assumption that both $T^{L_1}(t,0)$ and
$T^{L_2}(t,0)$ converge to $\theta_0$ as $t\to\infty$. Hence, $B=\{
L_0\}$ and we are done.

Now, consider $\theta_0=0$. We have $\theta^L_*\in\{0,1\}$, the sets
$A,C$ satisfy $A\cup C=[0,\infty)$, and by the comparison principle,
$A$ lies to the left of $C$. Moreover, $0\in A$ and $C$ is non-empty
and open by the same argument as above. Hence $A$ is closed and its
maximum is $L_0$ (possibly $L_0=0$). Lemma \ref{L.2.1}(iii) yields
(iii) of this theorem and $T(t,0)\ge T(t,x)$ gives (i) and (ii),
including the fact that the convergence in (ii) is uniform on
${\mathbb{R}}$. The proof is finished.

\section{Proof of Theorem \ref{T.1.2}} \label{S4}

The situation is somewhat more complicated here. Firstly, we do not
have Lemma \ref{L.2.1}(iii) at our disposal, and so the limit of $T$
as $t\to\infty$ need not always be a constant function. And
secondly, we cannot use Lemma \ref{L.2.2} and the scaling argument
preceding it in the way we did in the last section because it is not
anymore true that $L_2f\ge L_1f$ when $L_2>L_1$. We note that one
can still use the lemma without scaling, but then the argument
applies only to a restricted class of bistable $f$. Fortunately, it
turns out that the first of these difficulties actually cancels the
problems created by the second, as we shall see below.

Let us therefore go back to $T$ solving \eqref{1.1} rather than
\eqref{2.11}. We know from Lemma \ref{L.2.1}(ii) that
$\theta^L_*\equiv\lim_{t\to\infty} T(t,0)$ is well defined, and from
the comparison principle that it is non-decreasing in $L$.

First assume $\theta^L_*<\theta_2$, with $\theta_2$ defined in the
introduction. Choose $\varepsilon>0$ and a Lipschitz function $\tilde
f:[0,1]\to{\mathbb{R}}$ so that $\tilde f=0$ on $[0,\varepsilon]$, $\tilde f'(\varepsilon)<0$,
$\tilde f\ge f$ on $(\varepsilon,\tfrac 12(\theta^L_*+\theta_2)]$ and $\tilde f$
has a single zero there, $\tilde f>0$ on $(\tfrac
12(\theta^L_*+\theta_2),1)$, $\tilde f(1)=0>\tilde f'(1)$, and
\begin{equation} \label{3.1}
\int_0^1 \tilde f(\theta) d\theta<0.
\end{equation}
Let $t_0$ be such that for $t\ge t_0$ and all $x\in{\mathbb{R}}$ we have
$T(t,x)\le \tfrac 12(\theta^L_*+\theta_2)$. This is possible by Lemma
\ref{L.2.1}(i). Since $\tilde f\ge f$ on $[0,\tfrac
12(\theta^L_*+\theta_2)]$, starting from time $t_0$ one has $T_t\le
T_{xx}+\tilde f(T)$, that is, $T$ is a subsolution of the equation
\begin{equation} \label{3.2}
\Phi_t= \Phi_{xx}+\tilde f(\Phi).
\end{equation}

Let $\phi:{\mathbb{R}}\to[0,1]$ with $\phi(x)\to \varepsilon$ as $x\to\infty$ and
$\phi(x)\to 1$ as $x\to -\infty$ be the unique, up to translation,
traveling front profile (with speed $v$) for \eqref{3.2}
\cite{Kanel2}. That is, $\phi(x-vt)$ solves \eqref{3.2}. It follows
from \eqref{3.1} that in this case $v<0$.

From compactness of the support of $T(0,x)$ and Lipschitzness of
$f$, $T(t_0,x)\to 0$ as $|x|\to\infty$. This and
$\|T(t_0,\cdot)\|_\infty<1$ mean that there is $x_0$ such that
$T(t_0,x)\le\phi(x-x_0-vt_0)$, and since $\phi(x-x_0-vt)$ is a
solution and $T(t,x)$ a subsolution of \eqref{3.2},
\[
T(t,x)\le\phi(x-x_0-vt)
\]
for all $t\ge t_0$. But then $T(t,0)\le \phi(-x_0-vt) \to \varepsilon$ as
$t\to\infty$ because $v<0$. This holds for any small enough $\varepsilon>0$
and thus $\theta^L_*=0$.

Next assume $\theta^L_*>\theta_2$. Let $S$ be the solution of
\eqref{1.1} on ${\mathbb{R}}^+$ with $S(0,x)=0$ and $S(t,0)=s(t)$ a smooth
strictly increasing function with all derivatives bounded such that
$s(0)=0$, $s(t)\le T(t,0)$, and $\lim_{t\to\infty}s(t)=\theta^L_*$.
Then for any $h>0$ we have $S(h,x)> S(0,x)$ and so by comparison
$S(t+h,x)> S(t,x)$. Hence $\tilde S(x)\equiv \lim_{t\to\infty}
S(t,x)>0$ is well defined and $\tilde S(0)=\theta^L_*$. Since by
comparison again $S(t,x)\le T(t,x)\le T(t,0)$, we also have $\tilde
S(x)\le\theta^L_*$.

Standard parabolic regularity shows that $S(t,x)$ converges to $\tilde
S(x)$ along with its first two derivatives uniformly on compacts,
and so $\tilde S$ solves the stationary problem
\begin{equation} \label{3.3}
0=\tilde S''+f(\tilde S)
\end{equation}
on ${\mathbb{R}}^+$ (this can be found also in \cite{AW2}). But then for any
$y>0$
\[
\int_{\tilde S(y)}^{\theta^L_*} f(\theta) \,d\theta = \int_y^0 f(\tilde
S(x))\tilde S'(x) \,dx = \int_0^y \tilde S''(x)\tilde S'(x) \,dx = \tfrac
12 \big[ (\tilde S'(y))^2 - (\tilde S'(0))^2 \big].
\]
Assume there is $y>0$ such that $\tilde S(y)<\theta^L_*$, and then pick
one such that also $\tilde S'(y)<0$. Since $\int_{\omega}^{\theta^L_*}
f(\theta) d\theta$ is bounded below by a positive constant for all
$\omega\in[0,\tilde S(y)]$, so is $-\tilde S'(z)$ for all $z\ge y$
(because $\tilde S(z)\ge 0$). But that means $\tilde S$ is eventually
negative, a contradiction. Hence, we must have $\tilde
S\equiv\theta^L_*$, which is only possible if $\theta_*^L=1$. Moreover,
since $S$ converges to $\tilde S\equiv 1$ uniformly on compacts as
$t\to\infty$ (and $S\le T\le 1$), so does $T$.

The above shows that $\theta^L_*\in\{ 0,\theta_2,1\}$. As in the proof
of Theorem~\ref{T.1.1}, and using the equivalent of Kanel's result
for \eqref{0.5} and bistable $f$ \cite{FM}, one can show that the
intervals $A$, $C$ of $L$ for which $\theta_*^L=0,1$, respectively,
are non-empty and open. If $B$ is the interval of $L$ for which
$\theta_*^L=\theta_2$, then $B$ lies between $A$ and $C$ and again
$A\cup B\cup C=[0,\infty)$.

Next we need to prove that $B$ only contains one element.  We will
show below that if $L\in B$, then $T(t,x)\to U(x)$ uniformly on
${\mathbb{R}}$ as $t\to\infty$. Here $U$ solves \eqref{3.3} with
$U(0)=\theta_2$ and $U'(0)=0$. Assume now that $L_1<L_2$ are both in
$B$, with $T^{L_1}$ and $T^{L_2}$ the corresponding solutions of
\eqref{1.1}. We then have $T^{L_1}(t,0)\to\theta_2$, and since the
equation is translation invariant, we also have $\tilde
T(t,\varepsilon)\to\theta_2$ when $\tilde T$ solves \eqref{1.1} with initial
condition $\tilde T_0(x)\equiv \chi_{[-L_1+\varepsilon,L_1+\varepsilon]}(x)$. But if
$|\varepsilon|<L_2-L_1$, then $T^{L_2}_0(x)\ge \tilde T_0(x)$, and so by the
comparison principle,
\[
U(\varepsilon) = \lim_{t\to\infty} T^{L_2}(t,\varepsilon)\ge \theta_2.
\]
Since $U''(0)=-f(U(0))=-f(\theta_2)<0$, $U$ has a strict local maximum
at zero and therefore $U(\varepsilon)<U(0)=\theta_2$ for all small enough
$|\varepsilon|>0$. This is a contradiction and hence $B=\{ L_0\}$.

To complete the proof, we need to show that $T(t,x)\to U(x)$
uniformly as $t\to\infty$ when $L\in B$ (and hence
$\theta_*^L=\theta_2$).  Notice that the argument following \eqref{3.3}
applies to $U$ and we find for any $x>0$ such that $U(x)\ge 0$,
\begin{equation} \label{3.4}
\int^{\theta_2}_{U(x)} f(\theta)\,d\theta = \tfrac 12 (U'(x))^2.
\end{equation}
The definition of $\theta_2$ then shows that $U(x)\le\theta_2$, and
$U'(x)\neq 0$ when $U(x)\in(0,\theta_2)$. Since $U(x)$ cannot be
constant $\theta_2$ on any interval and $U'$ is continuous, we must
have $U'(x)<0$ for all $x>0$ such that $U(x)>0$. There is no $x$
with $U(x)=0$ because then \eqref{3.4} would give $U'(x)=0$,
contradicting uniqueness of solutions to initial value problems
associated to \eqref{3.3}. Hence $U(x)\in(0,\theta_2)$ and $U'(x)<0$
for $x>0$, with $U'(x)$ bounded away from zero when $U(x)$ is away
from zero (by \eqref{3.4} and the definition of $\theta_2$). This and
symmetry show that $U$ is indeed a symmetric bell-shaped solution
(with $U'$ decreasing on $[0,U^{-1}(\theta_0)]$ and increasing on
$[U^{-1}(\theta_0),\infty)$ by \eqref{3.4}) of the stationary problem
\eqref{3.3} such that $U(x)\to 0$ as $|x|\to\infty$.

If we now apply the argument involving $S$ and $\tilde S$ from the
case $\theta_*^L>\theta_2$, we find as above that $\tilde
S(x)\le\theta_2=\tilde S(0)$ and that $\tilde S>0$ is possible only if
$\tilde S'(0)= 0$. But then $\tilde S(0)=U(0)$ and $\tilde S'(0)=U'(0)$,
thus $\tilde S=U$. Moreover, uniform on compacts convergence of $S$ to
$U$ and $0\le S(t,x)\le U(x)\to 0$ as $|x|\to\infty$ yield uniform
on ${\mathbb{R}}$ convergence of $S$ to $U$. Since $T(t,x)\ge S(t,x)$, we
have $\liminf_{t\to\infty} T(t,x)\ge U(x)$ uniformly on ${\mathbb{R}}$.
Hence we are left with proving $\limsup_{t\to\infty} T(t,x)\le U(x)$
uniformly in $x>0$ (which suffices due to symmetry).

Let $0<x_0<\infty$ be such that if $S(0,x)\ge \theta_2
\chi_{[-x_0,x_0]}(x)$ and $S$ satisfies \eqref{1.1}, then $S\to 1$
uniformly on compacts. Such $x_0$ exists by \cite{FM} because
$\theta_2>\theta_0$. Then obviously for every $t\ge 0$ we have
$T(t,x_0)\le\theta_2$, because otherwise Lemma \ref{L.2.1}(i) would
imply $T\to 1$ uniformly on compacts. Since both $T(t,x)$ and
$V(x)=U(x-x_0)$ satisfy \eqref{1.1} on $(x_0,\infty)$,
$V(x_0)=\theta_2\ge T(t,x_0)$, and $V(x)>0=T(0,x)$ for $x>x_0$, the
comparison principle implies $T(t,x)\le V(x)$. Let us therefore
define
\[
x_1\equiv \min\{\tilde x\,|\, \limsup_{t\to\infty} T(t,x)\le U(x-\tilde
x) \text{ uniformly in $x>\tilde x$}\} \le x_0.
\]
The minimum is achieved because $U$ is uniformly continuous and by
Lemma \ref{L.2.1}(i), $\limsup_{t\to\infty} T(t,x)\le\theta_2=U(0)$.
We note that at this point one can derive $T\to U$ from $x_1<\infty$
and the results of \cite{BJP} when $f\in C^1$ and $f'(0)<0$. Our
non-linearity is more general and so \cite{BJP} is not applicable
here.

If $x_1=0$, then we are done, so assume $x_1>0$. First notice that
$\limsup_{t\to\infty} T(t,\tfrac 12 x_1)\le \theta_2-\delta_1$ for some
$\delta_1>0$. Indeed
--- in view of $\lim_{t\to\infty} T(t,0)=\theta_2$, Lemma
\ref{L.2.1}(i), and the comparison principle, it is sufficient to
show that there are $\delta_1,\delta_2>0$ such that if $S(0,x)\ge
(\theta_2-\delta_1)\chi_{[-x_1/2,x_1/2]}(x)$ and $S$ satisfies
\eqref{1.1}, then $S(t,0)\ge \theta_2+\delta_2$ for some $t>0$. This in
turn is true because it holds for $\delta_1=0$ and some $t,\delta_2>0$,
since then $S(0,0)=\theta_2$ and $S_t(0,0)=f(\theta_2)>0$, and because
$S(t,0)$ is continuous in $\delta_1$.

Now choose $x_2\in(\tfrac 12 x_1,x_1)$ such that
$U(x_1-x_2)\ge\theta_2-\delta_1$. The above and Lemma \ref{L.2.1}(i)
show that $\limsup_{t\to\infty} T(t,x)\le\theta_2-\delta_1$ uniformly in
$x\ge \tfrac 12 x_1$, and so
\begin{equation} \label{3.5}
\limsup_{t\to\infty} T(t,x)\le U(x-x_2)
\end{equation}
uniformly in $x\in[x_2,x_1]$. We will show that \eqref{3.5} also
holds uniformly in $x>x_1$, which will yield $x_1\le x_2$ by the
definition of $x_1$. This will be a contradiction and hence
necessarily $x_1=0$.

Let $s(t)$ be smooth, decreasing with all derivatives bounded, such
that $s(0)=\theta_2$ and $\lim_{t\to\infty} s(t)=U(x_1-x_2)$. Let
$S(t,x)$ satisfy \eqref{1.1} for $x>x_1$ with $S(0,x)=U(x-x_1)$ and
$S(t,x_1)=s(t)$. As above, one proves that this time $S$ is
non-increasing in $t$,
\begin{equation} \label{3.6}
S(t,x)\in[U(x-x_2),U(x-x_1)],
\end{equation}
and $\tilde S(x)\equiv \lim_{t\to\infty} S(t,x)$ satisfies \eqref{3.3}
and $\tilde S(x)\to 0$ as $x\to\infty$ (by \eqref{3.6}). Moreover,
$S\to \tilde S$ uniformly on compacts, which together with \eqref{3.6}
and $U(x)\to 0$ as $x\to\infty$ shows that $S\to\tilde S$ uniformly on
${\mathbb{R}}$. Since $\tilde S(x_1)=U(x_1-x_2)$ and $\tilde
S(\infty)=U(\infty)=0$, a formula similar to \eqref{3.4}, with the
integral from 0 to $U(x_1-x_2)$, gives $\tilde S'(x_1)=U'(x_1-x_2)$.
It follows that $\tilde S(x)=U(x-x_2)$.

Now pick any $\varepsilon>0$ and
choose $t_0$ such that
\begin{equation} \label{3.7}
S(t,x)-U(x-x_2)<\tfrac \varepsilon 2
\end{equation}
for $t\ge t_0$ and $x\ge x_1$. Then pick $t_1$ so that
\[
T(t,x)-U(x-x_1)<\varepsilon_0\equiv\tfrac \varepsilon 2 e^{-ct_0}
\]
for $t\ge t_1$ and $x\ge x_1$ (with $c$ the Lipschitz constant for
$f$). This is possible by the definition of $x_1$.

For any $t_2>t_1$ and $x>x_1$ we have
\[
T(t_2,x)-S(0,x)=T(t_2,x)-U(x-x_1)<\varepsilon_0,
\]
and for $t>t_2$ we have
\[
T(t,x_1)-S(t-t_2,x_1) = T(t,x_1)-s(t) \le T(t,x_1)-U(x_1-x_2)<\varepsilon_0
\]
by \eqref{3.5} if $t_2$ is large enough. Hence if we let
$R(t,x)\equiv S(t,x)+e^{ct}\varepsilon_0$, then $T(t_2,x)<R(0,x)$ for
$x>x_1$, $T(t,x_1)<R(t-t_2,x_1)$ for $t>t_2$, and
\[
R_t = S_t + ce^{ct}\varepsilon_0 = S_{xx} + f(S) + ce^{ct}\varepsilon_0 \ge R_{xx}
+ f(R).
\]
So $R$ is a supersolution of \eqref{1.1}, and by the comparison
principle $T(t,x)\le R(t-t_2,x)$ for $t>t_2$ and $x>x_1$. In
particular,
\[
T(t_2+t_0,x)\le R(t_0,x) = S(t_0,x)+\tfrac \varepsilon 2 < U(x-x_2)+\varepsilon
\]
for $x>x_1$ by \eqref{3.7}.
Since this holds for any large enough $t_2$, we have
$T(t,x)<U(x-x_2)+\varepsilon$ for all large $t$ and $x>x_1$. As $\varepsilon>0$
was arbitrary, this gives \eqref{3.5} uniformly in $x>x_1$. Hence
$x_1\le x_2<x_1$, a contradiction. Therefore we must have $x_1=0$
and the proof is finished.

redpajama_set_name: RedPajamaArXiv


Please find links below to postgame notes, stats and audio of Hurricanes players Nicolas Roy, Jordan Staal, Cam Ward and Justin Williams, as well as Hurricanes Head Coach Bill Peters and Tampa Bay Lightning Head Coach Jon Cooper, as they spoke to the media following the Hurricanes' 3-2 overtime win against the Lightning on Saturday at PNC Arena.
The Hurricanes will hold their end-of-season media availability on Monday, April 9, beginning at noon. All Hurricanes players will be available beginning at noon in the team's locker room at PNC Arena. At 1 p.m., Hurricanes Owner Tom Dundon will hold an end-of-season news conference in the media longue. Lunch will be available beginning at noon in the media workroom. Head Coach Bill Peters will hold his availability later in the week.


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This will decrease your monthly payment.
This will increase your fee by a small percent.
Additional Options: For Option D ONLY.
- Extent your plan start time an additional 1 to 2 months.
This allows you to work a month or two before you start paying your loan off.
* All the payment plans work with a down payment, the bigger, the better - for you.
Down Payments Usually begin at Only $1000.
* Some payment plans have a financing fee applied to them.
* Fees are NOT interest, so they cost you less!
* The plans allow you to begin your payments before, during, or after school.
* The fees vary depending on how much time you take to pay them off and when you start paying.
* Late payments are subject to a $25 per time fee.
* Fees, linens, oils, tables, books , posters, etc, are not included in tuition and payment plans.
* Students are qualified on an individual basis for the plans.
* Plans can be designed to fit your specific needs - examples are only examples. Let us know what your real needs are and we can design a plan for you!
* Please call & speak with Craig about which plan is best for you!
Many students are concerned about tuition, payments, loans, and their financial future.
Let me tell you a story about some past students. Both of these 2 students signed up for a payment plan that amounted to a down payment and a payment amount of about $7000. One was a man & one a woman. They both had set themselves up for a 2 year payment plan. That's about $280 per month payment. However, BOTH of them finished school & their situation changed for the better right away. This is common for our students because of what we do & teach here. The guy went straight to work and took as many jobs as he could (common in Colorado ski towns). He paid off the entire $7000 plus the interest in only 6 months & was DONE! The woman did the same, and when her parents saw that she was serious, they helped too! She also paid off the entire $7000 plus the interest in only 6 months & was DONE!
Let me give you some numbers to think about. If you go to a school that is worth going to AND they give federal loans, it could look like this: They will probably charge at least $10,000 in tuition if not more. (Sure, you could go to school for less, but in the long run, it will cost you way more.) If you set up a 10 year federal student loan, the payments will be low, but the overall cost of tuition will be $14000 or more. You will be locked into this payment for a long time. If you do what the 2 people above did, you will have a total tuition cost of less than $8500. If you stretched it out to 2 years after school, it would be under $9500. Now, we can keep our tuition lower because we do not have federal loans. Our classes will be smaller, more advanced, and more fun!
- Loans from friends or relatives are usually very flexible, have low or no interest rates, and are easy for some people to arrange. They, however, are not available to everyone. Check these out first!
- Banks or Credit Unions: provide loans at better rates than some places. Banks have personal loans available at rates based on your credit rating. They will provide you with a set payment plan. Having a cosigner with a good credit rating will increase your chances of getting a bank loan. Your best opportunities will be with a bank that you or a cosigner has a relationship with.
- Your local bank or Credit Union: Apply in person.
CALL US FOR MORE DETAILS!! - It is always best to discuss this with us to get the clearest picture of the best way for us to help you!!
- Loan Options: We have several loan options for students that may work for you if the things listed here do not, please call us for information on these.
- Credit cards may have a higher interest rate, but are more flexible with minimum payments. They also often offer low interest rates for up to a year when you transfer your balance to a new card. They are also the easiest way to for some people to get a loan.
- Get Creative: We also know of people who have raised money for their education in a "futures" sort of way. They would sell massages in the future to clients, friends, family, etc. at a discounted rate (less than the going rate for professional massage therapy in their location) if purchased before the student completed school.
- Talk to Us !: Financing opportunities and payment plans change often. Call and ask us for the current information that we have. When we work together, we can find a way that works for you!
Remember, you can earn from $40 to $130 or more per massage in many places.
We also suggest you keep in mind a few simple figures when you are considering how you will be paying back your loans & payment plans.
What other ways can you create for yourself to help more money flow more easily? Can you let go of some of your limiting reasons for not using money easily and create some new possibilities that work for you? Let us know what you dream up - we may be able to help you in ways you are not currently aware of!
Everyone who has fully committed to attending MountainHeart has made it work for them - one way or another!


text:
Will a magnet attract a nail through a piece of paper?
Can a magnet attract a nail through a piece of paper?
Do magnets work through paper?
Will a magnet attract a nail?
Which can be attracted by a magnet an iron nail or a piece of paper why?
Can a magnet attract a paper clip through materials?
What can magnets attract?
Are nails magnets?
Which type of force is the magnet using to move the nail?
What can magnets not go through?
What happens when you rub a nail with a magnet?
Several small objects that are attracted by a magnet, such as: paper clips, small nails, thumb tacks.
The magnetic force (or magnetism) can pass through thin sheets of non-magnetic objects such as paper, glass or wood. However, if the magnet is too weak and the material is too thick, the magnetic force may not be able to pass through.
The nail will stick to the bar magnet because it will become magnetized. The presence of the nearby north pole rearranges the magnetic domains inside the steel so that their south poles all point toward the north pole of the permanent magnet. As a result, the other end of the nail becomes a north pole.
As we observe, nails don't attract magnetic substances. Now we bring a pole if a bar magnet near the head of the nail. Then we bring the iron paper clips near the pointed end of the nail and observe that the iron paper clips now get attracted towards the nail, this happens because iron nail becomes a magnet.
THIS IS FUNNING: What will happen if I overstay my visa in USA?
Magnets attract paper clips because magnets have a magnetic field, which creates a force, that pulls on other ferromagnetic materials. A paper clip is usually made of steel wire. Steel is an alloy of iron and carbon. Iron is a ferromagnetic material that is attracted by magnets.
What Kind of Objects Are Attracted to Magnets?
Magnetic Elements: Metals. Iron, nickel and cobalt are strongly attracted to magnets. …
Magnetic Minerals. Certain minerals have attractions to magnetism, some weak, some very strong. …
Alloys: Metal Mixtures. …
Odd Everyday Things. …
Aurora Borealis.
But when you coil the wire around an iron nail, the magnetic domains inside the nail line up and make a strong, temporary magnet. If you disconnect the wire, the magnetic field disappears and the nail is no longer a magnet.
Since the iron nails and pins move towards the magnet, it means that the magnet exerts a force on them. The force exerted by a magnet is called magnetic force.
Metals That Don't Attract Magnets
In their natural states, metals such as aluminum, brass, copper, gold, lead and silver don't attract magnets because they are weak metals. However, you can add properties such as iron or steel to the weak metals to make them stronger.
Rubbing With a Magnet
Rubbing the nail with the permanent magnet causes a stronger, more lasting magnetism in the nail. … Stroking with one pole of a permanent magnet works because it aligns the atoms in the nail to "line up" in the same polar direction, giving the nail a north and a south magnetic pole.
THIS IS FUNNING: Your question: Do you have to work full time on a 482 visa?

redpajama_set_name: RedPajamaCommonCrawl


text:
Justin Pickard
Recent content by Justin Pickard
List of U.K. Prime Ministers 1945-2020
No Taff Vale Judgement 1902: Arthur Balfour (Conservative) [1] 1905: Henry Cambell-Bannerman (Liberal) [2] 1908: Herbert H. Asquith (Liberal) [3] 1912: Herbert H. Asquith (Liberal) [1] Balfour's government was dominated with foreign policy, and with only the 1902 Education Act as a domestic...
Simply a List Just a conventional list, with no precedent. 1945: Clement Attlee (Labour) [1] 1947: Clement Attlee (Labour) [2] 1951: Aneurin Bevan (Labour) [3] 1952: Aneurin Bevan (Labour) [4] 1955: Winston Churchill, Duke of London (Conservative) [5] 1958: Rab Butler (Conservative) [6] 1961...
AHC: Earliest Possible Digital Revolution
Give Liebniz another decade, and the patronage of George I of Great Britain? Lo! The mechanical calculators of the East India Company?
2007: Gordon Brown (Labour) [1] 2012: David Cameron (Conservative minority) [2] [1] Following his election as Labour leader following the resignation of Tony Blair, Gordon Brown calls a snap election. The sudden boost in support allows Labour to win a small majority, a mandate and another...
Skyscrapers in Totalitarian Country?
Oooh, also: stayed in the gently rotting, prostitute-riddled Hotel Ukraine (now the super-spiffy Radisson Royal) while travelling in Russia with college, back in 2004. The 'Seven Sisters', as they call them, are inhumanly vast.
And then there's this.
Princess Charlotte Lives 1818: Robert Banks Jenkinson, 2nd Earl of Liverpool (Tory) [1] 1822: Robert Banks Jenkinson, 2nd Earl of Liverpool (Tory) [2] 1824: Dudley Ryder, 1st Earl of Harrowby (Harrowby Tory leading Harrowby Tory-Whig Coalition) [3] 1826:Robert Stewart, 2nd Marquess of...
Princess Charlotte Lives 1818: Robert Banks Jenkinson, 2nd Earl of Liverpool (Tory) [1] 1822: Robert Banks Jenkinson, 2nd Earl of Liverpool (Tory) [2] 1824: Dudley Ryder, 1st Earl of Harrowby (Harrowby Tory leading Harrowby Tory-Whig Coalition) [3] 1826: Robert Stewart, Viscount Castlereagh...
Challenge: A British City-State
I'm not sure any of these are large enough, but Steelyard, King's Lynn, and Ipswich all had strong ties to the Hanseatic League. Maybe there's a pivot point somewhere in there?
Abandoned/Proposed Feasible Megaprojects
Flooding Egypt's Qatarra Depression for hydroelectricity, weather engineering, salt harvesting, and electrolysis.
Heathrow Airport (UK, west of London)
Gatwick expansion? Not in my (counterfactual) back yard! :mad:
AHC: Make Ireland's Population 50 Million by 2012
No potato famine, Dominion of Ireland, Irish Gandhi (allohistorical offspring of a famine victim) gains independence from the British Empire in the 1940s, Green Revolution boosted by Irish scientists' work on plant genetics in the 1950s, Irish Quiet Revolution in the late 1960s, Dublin...
Let's discuss the AH elements in "Fringe"
Some of the details are quite nice, and the existence of the observers indicates that there's something bigger at work here. It isn't a traditional "many-worlds" thing, and, even if it was intended as such, it'd already be better than Sliders. I do like the ID requirements for public...
Forum: Alternate History Books and Media
Alternate "Most Progressive" Nations
The multicultural aftermath of a Protestant Poland-Lithuania.
Great North Vote goes through 2005: Tony Blair (Labour) [1] 2007: Gordon Brown (Labour) [2] 2010: Lord Mandelson (Acting) [3] [1] In 2004, Prescott's office paved the way for the British federal state. The Northeast, then Yorkshire & Humber and finally the Northwest all agreed for a devolution...


I'm not a huge fan of research commissioned and released by vendors, since it is clearly self-serving. But that doesn't mean it's wrong (nor is it necessarily right). Just call me a bit skeptical.
Earlier this month, cloud file storage provider Egnyte (I don't know what the name means either) released research done by Forrester Consulting arguing that on-premise file servers should and will be killed off. Actually, the "killed off" part seems to be Egnyte's conclusion, which seems to be a reach.
Here's what the study found: Fifty-seven percent of IT pros surveyed think their file servers cost far less to run than they actually do. When IT runs the real numbers, they are a magnitude larger than what they initially imagined. A second piece of the puzzle is the fact that a bit over 40 percent of "information workers" use online storage for work purposes, services that IT did not specify or approve and don't manage.
Based on this, file servers are too expensive and workers are already used to online storage, which Egnyte claims is far cheaper. Putting two and two together, file servers should and will die. The replacement is a hybrid cloud where cloud file storage is synced with an on-premise NAS, which won't be so much a traditional file server as it is a "virtual appliance."
I have no doubt that plenty will move in this direction, and some may even use Egnyte's online storage services. (Did I mention that Egnyte sells the technology that will supposedly kill off the file server?) But certainly not everyone will, especially as server prices continue to plummet. Do you believe vendor-backed research? And in particular, how much sense does this report make? You tell me at dbarney@redmondmag.com.


text:
Fu Manchú y el beso de la muerte  es una película de crimen y aventuras estrenada en 1968 dirigida por Jesús Franco.

Coproducción hispano-alemana-británico-estadounidense se destaca por ser la primera de las películas del realizador junto al productor Harry Alan Towers. También fue la primera ocasión en que Christopher Lee encarnó el personaje creado por Sax Rohmer para una cinta dirigida por Jesús Franco.

Trama
Fu Manchú, un pérfido doctor que sueña con dominar el mundo y destruir la civilización occidental, se halla escondido junto a su malvada hija Lin Tang en una remota ciudad que ha descubierto en las selvas de Sudamérica. Tras descubrir un veneno mortal para los seres humanos, secuestra a diez mujeres para infectarlas con el mismo. Su plan consiste en que, para destruir a sus enemigos, esas mujeres besen a diferentes líderes mundiales para provocarles la muerte.

Una de las mujeres raptadas acude a Londres para besar al mayor enemigo de Fu Manchú: el agente de Scotland Yard Nayland Smith. Smith, cegado por el veneno, junto a su amigo, el Dr. Petrie, viajan a las selvas de Sudamérica para localizar a Fu Manchú y encontrar un antídoto. Se une a ellos el agente Carl Jansen y, tras conocer el plan de Fu Manchu para dominar el mundo, buscarán la manera de neutralizarlo.

Reparto

 Christopher Lee - Fu Manchú
 Richard Green - Nayland Smith
 Howard Marion-Crawford - Dr. Petrie
 Götz George - Carl Jansen
 Maria Rohm - Ursula Wagner
 Ricardo Palacios - Sancho López
 Loni von Friedl - Celeste
 Frances Khan - Carmen
 Tsai Chin - Lin Tang
 Isaura de Oliveira - Yuma
 Shirley Eaton - Viuda Negra

Recepción
La película obtiene críticas predominantemente negativas en los portales de información cinematográfica. En IMDb obtiene una puntuación de 4,6 sobre 10 con 925 valoraciones.

FilmAffinity, con una calificación de 4 sobre 10 calculada en 216 votos, también tiene críticas mayoritariamente negativas.

En Rotten Tomatoes Fu Manchú y el beso de la muerte obtiene una puntuación de 2,5 sobre 5 calculada sobre 213 valoraciones.

Referencias

Enlaces externos

 Ficha de la película en FilmAffinity

redpajama_set_name: RedPajamaWikipedia


text:
Q: slide in text on image hover css I'm trying to make a text appear on image hover, but will need it to slide in from top and cover only 50% of image height, but not sure how can I get this to work, here is the code

Traditional Pastry

and the CSS
.cags1{
    background-color: #3b283e;
    float:left;
}

.cags1:hover > img {
    opacity:0.5;
    padding: 10px;
    font-family: Tahoma,Arial,Helvetica;
    font-size: 14px;
    line-height: 30px;
    letter-spacing: 1px;
    text-align: center;
    color: #ffffff;
    text-shadow: 0 1px 3px #000000;
}

A: Here is a working example based on your code:

.img-container {
  position: relative;
  overflow: hidden;
  display: inline-block;
}
.img-container img {
  background-color:#3b283e;
  float:left
}
.cags1 {
    background-color: #3b283e;
    position: absolute;
    top: -20px;
    transition: 0.5s;
}

.img-container:hover > img {
    opacity:0.5;
    padding: 10px;
    font-family: Tahoma,Arial,Helvetica;
    font-size: 14px;
    line-height: 30px;
    letter-spacing: 1px;
    text-align: center;
    color: #ffffff;
    text-shadow: 0 1px 3px #000000;
}
.img-container:hover .cags1 {
  top: 20px;
}

Traditional Pastry

redpajama_set_name: RedPajamaStackExchange


P.C. 1986-499 1986-02-27Whereas the Minister of National Health and Welfare has consulted with the Ministers responsible for health care in the provinces respecting proposed Regulations prescribing the types of information that the Minister may require under paragraph 13(a) of the Canada Health Act in respect of extra-billing and user charges and the times at which and the manner in which such information shall be provided by the government of each province.
Therefore, Her Excellency the Governor General in Council, on the recommendation of the Minister of National Health and Welfare, pursuant to paragraph 22(1)(c) of the Canada Health ActFootnote *, is pleased hereby to make the annexed Regulations prescribing the types of information that the Minister of National Health and Welfare may require under paragraph 13(a) of the Canada Health Act in respect of extra-billing and user charges and the times at which and the manner in which such information shall be provided by the government of each province, effective April 1, 1986.
1 These Regulations may be cited as the Extra-billing and User Charges Information Regulations.
(b) a financial statement showing the aggregate amount actually charged through extra-billing, including an explanation regarding the method of determination of the aggregate amount.
(b) a financial statement showing the aggregate amount actually charged in respect of user charges to which section 19 of the Act applies, including an explanation regarding the method of determination of the aggregate amount.
(b) in respect of the financial statements referred to in paragraphs 3(b) and 4(b), before the sixteenth day of the twenty-first month following the end of the fiscal year to which they relate.
(2) The government of a province may, at its discretion, provide the Minister with adjustments to the estimates referred to in paragraphs 3(a) and 4(a) before February 16 of the fiscal year to which they relate.
(3) The information referred to in subsections (1) and (2) shall be transmitted to the Minister by the most practical means of communication.


text:
College football forecast: Harbaugh needs huge Big Ten win
By Matt Youmans (VSiN senior editor) Follow @mattyoumans247
After losses to Michigan State and Penn State, Jim Harbaugh still needs to coach Michigan to a big win in the Big Ten this season. The Wolverines are 7-point underdogs at Wisconsin on Saturday.
So far, the Jim Harbaugh era at Michigan is mostly hype. Too much talk and not enough action in the biggest games. Harbaugh has too often come up short against the Big Ten's top teams.
It's not enough to embarrass Rutgers and step outside of the conference to bully Florida and beat up an overmatched Jim McElwain when he was down and headed out.
The next two Saturdays, against Wisconsin and Ohio State, will define Harbaugh's third season. The Wolverines' first mission is to wreck the Badgers' realistic College Football Playoff plans. The game is in Madison, of course, because the Badgers (10-0) have faced zero quality opponents on the road this season.
When the Big Ten split into two seven-team divisions, Wisconsin won the lottery. The imbalance from East to West is absurd. Ohio State, Michigan, Michigan State and Penn State make for four heavyweights in the East. Meanwhile, the Badgers are barely challenged in the West with Iowa, Northwestern, Nebraska, Purdue, Minnesota and Illinois forming a group of mediocre contenders. It's not quite a bye to the conference championship game, but it's close.
While critics whine about a weak schedule, Wisconsin will have chances to prove its worth. Everything is breaking right for the Badgers, with the exception of injuries.
In a dominant showing against Iowa last week, Wisconsin was without safety D'Cota Dixon, the team's leading tackler, and top receiver Quintez Cephus, who is lost for the rest of the year. Now, starting center Tyler Biadasz is listed as questionable with a leg injury, and inside linebacker Chris Orr is out with a leg injury. The injury report is a long list, but football is a game of attrition and excuses are for losers.
The Badgers' biggest concern might be quarterback Alex Hornibrook, who sometimes is so bad it's a miracle the team can keep winning despite him. Hornibrook threw two interceptions that Hawkeyes cornerback Joshua Jackson returned for touchdowns. A week after a 55-24 blowout of the Buckeyes, Iowa's anemic offense totaled 66 yards in the 38-14 loss in Madison. In a crazy oddity, the Hawkeyes scored on two pick-sixes while their offense ran a total of two plays in Wisconsin territory.
The battle between the Badgers and Wolverines will be a black-and-blue defensive slugfest. Wisconsin ranks third in the nation in scoring defense (13.4 points per game), and Michigan ranks ninth (16.4).
The Badgers are banking on star freshman Jonathan Taylor, who rushed for 157 yards against Iowa and has 1,525 yards (7.0 per carry) and 12 touchdowns on the ground. Harbaugh has turned to inexperienced quarterback Brandon Peters after losing Wilton Speight to a back injury. Peters passed for four touchdowns with no interceptions in the past three games, but he's operating in a conservative offense.
In a game with a low total of 41, sharp money initially showed on Michigan, dropping the line from 9 early in the week to 7 today. Harbaugh might come up short again, but he should make the Badgers sweat, so this is a play on the Wolverines as 7-point 'dogs.
Saturday's plays: DUKE (plus-7) over Georgia Tech; Fresno State (-2) over WYOMING; IOWA (-7) over Purdue; Michigan (plus-7) over WISCONSIN; SOUTHERN CALIFORNIA (-15) over Ucla.
Last week: 4-5-1 against the spread
Season: 47-43-2


Saw imports of $260,579 and exports of $0 in Dec. This is a change of -22.47% and 0% respectively from the month Dec of 2016. This data is derived from Datamyne, using US import and export trade data released by the US Census Bureau in the Merchandise Trade dataset.
See company shipment details for HS Code 3204.15.3000 when you subscribe to Datamyne.


text:
Garden Media Guild
Exhibitors Hub
Follow @gardenpresevent
Founded by Richard Jackson Chair of Greenfingers and Guy Burtenshaw in 2006, the event was set up to bring together suppliers of garden products, with trade and consumer editors from both print and broadcast outlets. The only event of its kind for the industry, the Garden Press Event has grown in popularity, with over 110 exhibitors (including a host of international brands).
The show is now owned and run by GIMA & HTA who work together to create content and value for both visitors and exhibitors.
The Garden Press Event 2021 is an ideal opportunity to highlight your stories, plants, products and range early in the season and network directly with the leading writers from the National and Trade Press, Garden and Home Interest Magazines and TV and Radio Garden Broadcasters.
Since its beginnings the Garden Press Event has supported Greenfingers, a small national charity creating magical gardens in children's hospices around the UK. It continues to do so through the generosity of its exhibitors who donate the fee of exhibiting in the new product area to this worthy cause.
The Garden Press Event is held at the Business Design Centre. For more information about this venue please click Here.
This event is now firmly established as the launch of the Gardening Season, and a key date for all Gardening Journalists & Media. Get in touch with us to find out the benefits of attending this exclusive event.
Floorplan:
Garden Press Event c/o the HTA, Horticulture House, Chilton, Oxfordshire, OX11 0RN
jo.gildea@hta.org.uk
Wednesday 3rd March 2021 | Virtual Platform
email: events@hta.org.uk
© The Garden Press Event
Website Design by Nick Moore Software


Surface properties of high-density polyethylene and cross-linked polyethylene.
Surface properties of high density polyethylene (HDPE) and cross-linked polyethylene (XLPE) were studied in a number of ways. Firstly, the specimens of these materials were immersed in saline solutions of 0.005, 1, 10 and 100 mS/cm at 0 +/- 1, 23 +/- 3, 44 +/- 2, 73 +/- 2 and 98 +/- 2°C. Also, the specimens of these materials were subjected to the same temperatures in air. The contact angle of a 4∼5 mul sessile droplet of distilled water which is a measure of the surface hydrophobicity of the solid surface was measured as a function of time of immersion. The percentage increase in weight and increase in surface roughness of the specimens due to the adsorption of the saline solutions and partially due to deposit of salts, were measured. After the loss of hydrophobicity, the recovery of the contact angle in air at 50 +/- 14% humidity level and 23 +/- 3°C were also studied. The effects of (i) the volume of the droplet of water, (ii) the time to measure the contact angle after placing the droplet of water on the polymer surface, (iii) the applications of 12.5 kV/cm dc stress and (iv) RF discharges (∼250 kHz) at 23 +/- 3°C and 50 +/- 14% humidity level, for varying durations, on the contact angle were also studied. Scanning electron microscopy (SEM) and infrared spectroscopy (FTC) were carried out to detect the oxidation of the surface. The surface free energy per unit area of the virgin and aged specimens of HDPE was calculated. (Abstract shortened by UMI.)Dept. of Electrical and Computer Engineering. Paper copy at Leddy Library: Theses & Major Papers - Basement, West Bldg. / Call Number: Thesis1998 .K475. Source: Masters Abstracts International, Volume: 39-02, page: 0563. Adviser: Reuben Harkam. Thesis (M.A.Sc.)--University of Windsor (Canada), 1998.
Khan, Mohammad Ashraf., "Surface properties of high-density polyethylene and cross-linked polyethylene." (1998). Electronic Theses and Dissertations. 3678.


Dr Meissner is a clinical neurologist and basic neuroscientist with a specialization in movement disorders. He belongs to the local DBS team, is referring physician of the French National Reference Centre for MSA and is coordinator of the Bordeaux Competence Centre for Wilson's disease. Dr Meissner is member of Dr Bezard's research group at the Institute for Neurodegenerative Disorders.
He has authored or co-authored over 50 publications in the field of fundamental and clinical neurosciences. Most of his publications are dealing with the mechanisms of DBS in PD. He has also been interested in the pathophysiology of basal ganglia circuitry, with particular emphasis to the role of oscillatory activity in PD.
His current research interests in fundamental neurosciences include the development of innovative DBS techniques. During the last three years, he has set up an important clinical research activity in the area of PD and atypical parkinsonism.


text:
Q: viewing syslogs on kibana using logstash I have followed the steps as shown in the video. I am trying to run the below logstash3.conf file:
  input{
   file{
       path=>"/var/log/syslog"
       start_position=>"beginning"
       sincedb_path => "/dev/null"
  }
 }
output{
   elasticsearch{
   hosts=>["elasticsearch:9200"]
   index=>"pop"
   }
   stdout{}
}

I gave the following command to run logstash3.conf file
docker run -h logstash3 --name logstash3 --link elasticsearch:elasticsearch  --rm -v "$PWD":/config-dir logstash -f /config-dir/logstash3.conf

I am getting the below error(after which it does not print anything to the screen):
Sending Logstash's logs to /var/log/logstash which is now configured via log4j2.properties
10:00:19.082 [main] INFO  logstash.modules.scaffold - Initializing module {:module_name=>"netflow", :directory=>"/usr/share/logstash/modules/netflow/configuration"}
10:00:19.091 [main] INFO  logstash.modules.scaffold - Initializing module {:module_name=>"fb_apache", :directory=>"/usr/share/logstash/modules/fb_apache/configuration"}
10:00:19.163 [main] INFO  logstash.setting.writabledirectory - Creating directory {:setting=>"path.queue", :path=>"/var/lib/logstash/queue"}
10:00:19.164 [main] INFO  logstash.setting.writabledirectory - Creating directory {:setting=>"path.dead_letter_queue", :path=>"/var/lib/logstash/dead_letter_queue"}
10:00:19.275 [LogStash::Runner] INFO  logstash.agent - No persistent UUID file found. Generating new UUID {:uuid=>"81be107c-ad55-4efb-b7a9-873179a33b06", :path=>"/var/lib/logstash/uuid"}
10:00:20.737 [[main]-pipeline-manager] INFO  logstash.outputs.elasticsearch - Elasticsearch pool URLs updated {:changes=>{:removed=>[], :added=>[http://elasticsearch:9200/]}}
10:00:20.738 [[main]-pipeline-manager] INFO  logstash.outputs.elasticsearch - Running health check to see if an Elasticsearch connection is working {:healthcheck_url=>http://elasticsearch:9200/, :path=>"/"}
10:00:20.969 [[main]-pipeline-manager] WARN  logstash.outputs.elasticsearch - Restored connection to ES instance {:url=>"http://elasticsearch:9200/"}
10:00:21.342 [[main]-pipeline-manager] INFO  logstash.outputs.elasticsearch - Using mapping template from {:path=>nil}
10:00:21.345 [[main]-pipeline-manager] INFO  logstash.outputs.elasticsearch - Attempting to install template {:manage_template=>{"template"=>"logstash-*", "version"=>50001, "settings"=>{"index.refresh_interval"=>"5s"}, "mappings"=>{"_default_"=>{"_all"=>{"enabled"=>true, "norms"=>false}, "dynamic_templates"=>[{"message_field"=>{"path_match"=>"message", "match_mapping_type"=>"string", "mapping"=>{"type"=>"text", "norms"=>false}}}, {"string_fields"=>{"match"=>"*", "match_mapping_type"=>"string", "mapping"=>{"type"=>"text", "norms"=>false, "fields"=>{"keyword"=>{"type"=>"keyword", "ignore_above"=>256}}}}}], "properties"=>{"@timestamp"=>{"type"=>"date", "include_in_all"=>false}, "@version"=>{"type"=>"keyword", "include_in_all"=>false}, "geoip"=>{"dynamic"=>true, "properties"=>{"ip"=>{"type"=>"ip"}, "location"=>{"type"=>"geo_point"}, "latitude"=>{"type"=>"half_float"}, "longitude"=>{"type"=>"half_float"}}}}}}}}
10:00:21.364 [[main]-pipeline-manager] INFO  logstash.outputs.elasticsearch - New Elasticsearch output {:class=>"LogStash::Outputs::ElasticSearch", :hosts=>["//elasticsearch:9200"]}
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Kindly let me know how I can correct the error?


text:
Q: How can I disable an ng-click after a click inside a ng-if ng-switch statement? I have a flag button that when a user clicks on, flags the discussion, and afterwards the flag button is replaced with text 'successfully flagged'. Currently I am having trouble from disabling ng-click after clicking the flag button. ng-click still exists for the text 'successfully flagged' and i want to block clicks on this text to prevent errors on flagging the same discussion.
html:

            Successfully flagged

js:
$scope.isFlagging = false;
$scope.button_clicked = false;
    $scope.do_something = function() {
        $scope.button_clicked = true;
        this.isFlagging = true;
    }

By adding a lazy evaluation or by preventing propagation, I might be able to block the do_something() method from being called, but I am also looking to have the mouse cursor remain a pointer and not change to a 'click link' mouse icon, would this be possible? Looks like the mouse button image was a css issue i fixed
I've also tried just adding the ng-click to the ng-switch-when statement, such as the below, but after click, isFlagging is still false and I don't get the success text:

A: You can stop event bubbling by adding ng-click="$event.stopPropagation()" on "Successfully flagged" div. In this case click event will not reach parent container:

    Successfully flagged

A: Perhaps a simple conditional?
``$scope.do_something = function() {
  if ($scope.button_clicked === true) {
     return;
  }
  else {
  $scope.button_clicked = true;
    this.isFlagging = true;
  }
}``

A: Adding the button_clicked in the ng-click is a common pattern

        Successfully flagged


text:
The District Council of Light was a local government area in South Australia from 1977 to 1996, seated at Freeling.

History
The council was proclaimed on 1 March 1977 by the amalgamation of the District Council of Freeling and the District Council of Mudla Wirra. From 1 July 1977, it consisted of eight councillors, one representing each ward (Freeling, Gawler River, Greenock, Light, Para, Pinkerton, Roseworthy and Wasleys). As of 1977, its chambers were located in Freeling. On 13 March 1985, it lost areas around Gawler West and Willaston to the Town of Gawler.

In 1986, it covered an area of 662 square kilometres in an area described as "roughly bounded by the Light River to the north and the North Para and Gawler Rivers to the south", with a total population of 5,500. The main primary industries were cereal growing in the western, northern and central areas, market gardens and stud sheep in the south and vineyards and orchards in the east. The area included the Roseworthy Agricultural College and the Seppeltsfield winery.

It ceased to exist on 1 March 1996 when it was amalgamated with the District Council of Kapunda to form the District Council of Light and Kapunda (later renamed the Light Regional Council).

Chairmen
 Clarence Kenneth Tremlett (1977)
 Brian Eric Anders (1977–1983)
 Donald William Barkley (1983–1986)

References

Light, District Council of (1977–1996)
Light, District Council of (1977–1996)


text:
Home Science News
Humand lends reflexes to a stumbling robot
Let's face it, even the most cutting edge robots developed today are stumbling goofs. So, in an effort to make robots less blundering a team at MIT has borrowed human reflexes to one of their lab's creation: a robot called Hermes. When the bot is just about to stumble about an obstacle, a human operator who is strapped with all sorts of actuators and motors synced to bots' movements is alerted. In a split-second, Hermes is back on his feet thanks to the innate reflexes of good ol' humans.
Let's face it, even the most cutting edge robots developed today are stumbling goofs. So, in an effort to make robots less blundering a team at MIT has borrowed human reflexes to one of their lab's creation: a robot called Hermes. When the bot is just about to stumble about an obstacle, a human operator who is strapped with all sorts of actuators and motors synced to bots' movements is alerted. In a split-second, Hermes is back on his feet thanks to the innate reflexes of good ol' humans.
To be fair, Hermes isn't that impressive when you consider he isn't autonomous. Rather, Hermes is a sort of puppet – an avatar for the human operator. When the human says jump, the robot jumps. When the human picks up a hammer, the robot picks up a hammer. When the human uses said hammer to punch through a wall, the robot follows suit. Yes, MIT can be quite fun.
"For humanoid robots and legged robots in general, keeping balance is critical to being able to carry out any task," said PhD student Albert Wang. "We've decided to tackle this head-on by feeding the balance sensation back to the human via forces on his waist, and that way we can take advantage of the natural reflexes and learning capability of the human to be able to keep the robot balanced."
First, the team had to contrive a way to measure Hermes' balance and weight distribution. So, on each leg they fitted load sensors that measure how much pressure the robot is exerting on the ground. Based on this forces, the MIT engineers could calculate the momentary center of pressure or where it was shifting its weight. Past a critical level, the robot was in danger of falling. This is where the balance-feedback interface comes in. Basically, it's an exoskeleton packed with metal bars and motors that a person wears on his waist. When Hermes is about to fall, a signal is sent to the exoskeleton which pushes the person back and forth, mimicking the robot's weight distribution.
"The interface works by pushing harder on the operator as the robot's center of pressure approaches the edge of the support polygon," Wang explains. "If the robot is leaning too far forward, the interface will push the operator in the opposite direction, to convey that the robot is in danger of falling."
To test the interface, Wang performed a simple test: he hit Hermes with a hammer while PhD student Joao Ramos, acting as the operator, was unaware when the strike will fall. As Wang struck the robot, the platform exerted a similar jolt on Ramos, who reflexively shifted his weight to regain his balance, causing the robot to also catch itself, according to MIT News. Another fun test involved Hermes punching through a dry wall.
"These experiments show the versatility of the human operator. In one test, the robot unexpectedly got its arm stuck in the wall. But, because the human was in the loop, the operator could arrive at a creative solution which was translated directly to the robot," Wang says. "Our next goal is to try more complex coordinated movements such as swinging an axe or opening a spring-loaded door. These actions are difficult for many robots. If the robot stands stiff while pushing on a door, it tends to tip over. You have to lean your body weight into it and catch yourself as the because it's so natural to humans, you can have the human do it."
Hermes and his future cousins will prove handy in disaster relief situation where its just too dangerous to send humans and useless to send clumsy autonomous bots. Given time and enough punches in the gut, Hermes could grow to have its own reflexes.
"This interface likely won't even distract a person," Jonathan Hurst, associate professor of mechanical, industrial, and manufacturing engineering at Oregon State University, who was not involved in the research. "It's normal to keep your balance while focusing on a task. But perhaps more important than just a way to control a robot in the absence of knowing how to do it autonomously is being able to observe and collect data from the robot. Given hours of data recording the details of human strategies for balance and pose adjustment, I'd be willing to bet they will discover some relatively simple approaches for autonomous strategies."


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