The boat's named after me. No, it really is! The name's Bond... Neville Bond. It's an ex-hire boat and I changed the boat's name to mine. I bought it last year when my wife and I split our money after we got divorced.
My first thought was, of course, to cruise off to Llangollen! I'm from Cornwall originally but I've got five kids and my family has extended to Aberystwyth in Wales. On route to Llangollen I moored by that fabulous pub in Chirk, the Poachers, where my mother and sister came from Cornwall to join me for the next bit of the journey. Family from Aberystwyth came too and we all travelled across Chirk Aqueduct together. It was really windy, so it was quite exciting and we had an absolutely fantastic day!
Then they all left and I carried on my journey. Living alone on water did put me through a spell of feeling really down. It was November, the leaves were falling, and the canals were quiet. I was getting lonely, and in the end I thought "I'll sell the boat, I can't do this". So I rang Venetian Marina brokerage who said they'd sell it for me. But then, after I turned round to set off towards the brokerage, I was cruising along and thought, "No, I do like this, I really do like this!"
I booked into Overwater Marina at Audlem to moor for a while because it has wonderful facilities and company if I wanted it. Heading down Audlem locks, it was absolutely pouring with rain, then on the last lock the sun came out – I pulled into the marina and everything was lovely.
I'm back on the move again now summer's coming. I keep in touch with a big community of boaters on Facebook and we share tips about boat maintenance which is really helpful. It's hard to be lonely on the canals because there's such a friendly community and lots of passers-by. People ask where the sheep on the bow of my boat comes from – he's from the beer garden of a pub my ex-wife and I ran. We ran lots of different pubs so I guess I'm used to being sociable and that's why I quite like to chat to passers-by. I always get smiles from people who see the name on my boat, and they usually ask if it's true!


text:
In 1990, seven young male dancers - 6 gay,1 straight - joined Madonna on her most controversial tour. On stage and in the iconic film Truth or Dare, they showed the world how to express yourself. Now years later, they reveal the truth about life during and after the tour, when they were ambassadors to the world on behalf of the LGBT community during the height of the AIDS epidemic. STRIKE A POSE is a dramatic tale about overcoming shame and finding the courage to be who you are.
DIRECTED BY REIJER ZWAAN & ESTER GOULD
NETHERLANDS | 2016 | ENGLISH LANGUAGE
In 1990, seven young male dancers - 6 gay,1 straight - joined Madonna on her most controversial tour. On stage and in the iconic film Truth or Dare, they showed the world how to express yourself. Now years later, they reveal the truth about life during and after the tour, when they were ambassado...

redpajama_set_name: RedPajamaCommonCrawl


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text:
\section{Introduction} \setcounter{equation}{0}
\renewcommand{\theequation}{1.\arabic{equation}}

In this paper, we are concerned with the following reflected stochastic partial differential equations (SPDEs):
\begin{equation}\label{a01}
    \begin{cases}
\frac{\partial{X(x,t)}}{\partial{t}}=\frac{\partial^2{X(x,t)}}{\partial{x^2}}+f(x,t,X)+\frac{c_1}{(X-\Lambda^1)^{\vartheta}}- \frac{c_2}{(\Lambda^2-X)^{\vartheta}}
 \\
   \ \ \ \ \ \ \ \ \ \ \ +\chi\big(x,t,X\big)\dot{W}(x,t)+\Upsilon(x,t)-\Gamma(x,t),\ \ \ (x,t)\in{Q}:=[0,1]\times\mathbb{R}_+\\
    \Lambda^1(x,t)\leq{X}(x,t)\leq\Lambda^2(x,t).
                        \end{cases}
                        \end{equation}
Here ${W}$ is a space-time white noise and $\vartheta > 0$. Random measures $\Gamma$ and $\Upsilon$ are the additional forces
preventing $X$ from leaving $[\Lambda^1,\ \Lambda^2]$.

\vskip 0.3cm
Considering that SPDEs with reflection are natural extensions of the deterministic obstacle problems, there has been an upsurge of interest in
this topic.
The existence and uniqueness of the solutions have been well studied for the case of Lipchitz coefficient
with a single reflecting barrier
0, \emph{i.e.} $\Lambda^1=0,\Lambda^2=\infty, $ $c_1=c_2=0$, see Naulart and Pardoux \cite{np}, Donati-Martin and Pardoux \cite{dp1},  Xu and Zhang \cite{xz}, etc..
 Otobe \cite{os} and  Zhang and Yang \cite{zy} obtained the existence and uniqueness of the solutions of
SPDEs with double reflecting
barriers and Lipchitz coefficient, driven by an additive white noise and by multiplicative white noise respectively.

\vskip 0.3cm

All of the results mentioned above are devoted to the case
of Lipchitz coefficient. Mueller \cite{ml} proved that solutions of SPDEs $X_t=X_{xx}+g(X)\dot{W}(t,x)$, where  $g(X)$ is non Lipschiz, do not blow up in finite time, they did this by introducing a drift $X^{-\vartheta}$ into the former equation as a byproduct. It was proved that this drift forces solutions to stay positive with probability $1$.  Mueller and Pardoux \cite{mp} concentrated on the case when $0<\vartheta<3$ and showed that the solutions hit $0$ with positive probability.
Thus, $\vartheta=3$ is the critical exponent for $X$ to hit zero in finite time.
More precisely, if $\vartheta>3$, then $inf_x\ X(t,x)$ never reaches 0, and for $0<\vartheta<3$, $inf_x\ X(t,x)$
has a positive probability of reaching 0 in finite time. Existence and uniqueness for all time of a solution for $\vartheta=3$
was first verified by Zambotti \cite{zi}, and
it also showed the existence and uniqueness of the solution in the case of $0<\vartheta<3$ and a single reflecting barrier 0 in Zambotti \cite{zi}.
The hitting properties of the solution for $\vartheta=3$ was discussed in \cite{dmz}. All those discussions are under the assumption that SPDEs with the singular drift holds.

\vskip 0.3cm
Inspired by these, in this paper, we consider SPDEs
(\ref{a01}), which has double smooth reflecting walls $\Lambda^1$ and $\Lambda^2$ and two singular drifts $\frac{c_1}{(X-\Lambda^1)^{\vartheta}}$, $\frac{c_2}{(\Lambda^2-X)^{\vartheta}}$, for all $\vartheta>0$. As far as we know, there are few literatures that have studied the reflection problem with singularities.
To show the existence
and uniqueness of the solutions of Eq. \eqref{a01},
the strategy is
similar to that in Xu and Zhang \cite{xz} as well as \cite{zy}.
Due to the singular terms, we need to approximate the solution by introducing two monotone sequences.
As an extension from one wall to two walls case, we show that
for $\vartheta>3$, the solution of Eq. (\ref{a01}) stays in the interval
$\big(\Lambda^1,\ \Lambda^2\big)$ for all $t\geq 0$, a.s., that is $\Upsilon=\Gamma=0$;
for $0<\vartheta<3$, the solution of Eq. (\ref{a01}) hits $\Lambda^1$ or $\Lambda^2$ with positive probability in finite time.
Thus $\vartheta=3$ is the critical parameter for $X$ to hit reflecting walls. Various other
properties of the solutions have been studied, see
\cite{dmz}, \cite{dp2}, \cite{yz}, \cite{za} and \cite{zw}.
\vskip 0.3cm

The paper is organized as follows.
Definitions and assumptions are given in Section 2. Section 3 is to establish
the existence and uniqueness of the solution of Eq. \eqref{a01}. In Section 4 we
consider the pathwise properties of Eq. \eqref{a01},
Subsection 4.1 is devoted to the case $\vartheta>3$, and in
Subsection 4.2, we deal with the case when $0<\vartheta<3$.

\section{Preliminaries}
\setcounter{equation}{0}
\renewcommand{\theequation}{2.\arabic{equation}}

In this section, we state the precise assumptions on the coefficients and the concept of solution.

First, we formulate Eq. \eqref{a01} as follows:
\begin{equation}\label{a02}
\frac{\partial{X}}{\partial{t}}=\frac{\partial^2{X}}{\partial{x^2}}+f(x,t,X)+\frac{c_1}{(X-\Lambda^1)^{\vartheta}}- \frac{c_2}{(\Lambda^2-X)^{\vartheta}}+\chi(x,t,X)\dot{W}+\Upsilon-\Gamma;
\end{equation}
with the conditions
\begin{equation}\label{a02}
\left \{\begin{array}{ll}
        X(0,t)=0,\ X(1,t)=0,\ \ {\rm for}\ t\geq0;\\
        X(x,0)=X_0(x)\in{C}([0,1]);\\
        \Lambda^1(x,t)\leq{X}(x,t)\leq{\Lambda^2}(x,t),\ \ {\rm for}\ (x,t)\in{Q}:=[0,1]\times\mathbb{R}_+.
\end{array}
\right.
\end{equation}

Let $f, \chi:[0,1]\times\mathbb{R}_+\times{C}(Q)\rightarrow\mathbb{R}$ be given measurable maps. Set $\mathcal{A}_{M,T}=\{X\in C(Q):\ \underset{x\in[0,1],t\in[0,T]}{\sup}|X(x,t)|\leq{M}\}$.
Introduce the following conditions:\\
(H0) $\Lambda^i(x,t)\in C(Q)$ satisfy
$\Lambda^i(1,t)\leq{0}$,  $\Lambda^i(1,t)\geq{0}$, for $i=1,2$;\\
(H1)  $\Lambda^1(x,t)<\Lambda^2(x,t)$, $\forall(x,t)\in(0,1)\times \mathbb{R}_+$; \\
(H2)
$\frac{\partial{\Lambda^i}}{\partial{t}}+\frac{\partial^2{\Lambda^i}}{\partial{x^2}}\in
L^2([0,1]\times[0,T])$;\\
(H3)
$\frac{\partial{}}{\partial{t}}\Lambda^i(0,t)=\frac{\partial{}}{\partial{t}}\Lambda^i(1,t)=0$
for $t\geq0$;\\
(H4) $\frac{\partial{}}{\partial{t}}(\Lambda^2-\Lambda^1)\geq0$.\\
(F1) $f(\cdot,\cdot;0),\chi(\cdot,\cdot;0)\in L^2([0,1]\times[0,T])$, for $T>0$;\\
(F2) for every $X,\hat{X}\in\mathcal{A}_{M,T}$ there exists $C_{T,M}>0$ such that
$$|f(x,t;X)-f(x,t;\hat{X})|+|\chi(x,t;X)-\chi(x,t;\hat{X})|\leq{C_{T,M}}\underset{y\in[0,1],s\in[0,t]}{\sup}|{X}(y,s)-\hat{X}(y,s)|$$
for all $(x,t)\in[0,1]\times[0,T]$;\\
(F3) there exists $C_T>0$ such that
$$|f(x,t;X)|+|\chi(x,t;X)|\leq{C_T}(1+\underset{y\in[0,1],s\in[0,t]}{\sup}|X(y,s)|),\ \forall (x,t)\in[0,1]\times[0,T].$$

Now we recall the concept of solution of Eq. (\ref{a02}) in \cite{yz}.
\begin{definition} A triplet
$(X,\Upsilon,\Gamma)$ defined on a filtered probability space\\
$(\Omega,P,\mathcal {F};\{\mathcal {F}_t\})$ is a solution to the
SPDE \eqref{a02}, if\\
(i) $X\in{C(Q)}$ is a adapted random field
satisfying $\Lambda^1(x,t)\leq{X}(x,t)\leq{\Lambda^2}(x,t)$,
$X(0,t)=0$, and $X(1,t)=0$, a.s;\\
(ii) $\Upsilon(dx,dt)$ and $\Gamma(dx,dt)$ are positive and adapted random measures on $(0,1)\times\mathbb{R}_+$
satisfying
$$\Upsilon\big((\epsilon,1-\epsilon)\times[0,T]\big)<\infty,\ \Gamma\big((\epsilon,1-\epsilon)\times[0,T]\big)<\infty$$
for $\epsilon\in(0,\frac{1}{2})$ and $T>0$;\\
(iii) for every $t\geq0$ and $\psi\in{C_0^\infty}(0,1)$(the set of smooth
functions with compact supports)
\begin{eqnarray}\label{a03}
&&\big(X(t),\psi\big)-\big(X_0,\psi\big)-\int_0^t(X(s),\psi^{''})ds-\int_0^t\big(f(y,s,X),\psi\big)ds \nonumber\\
&=&\int_0^t\big(\frac{c_1}{(X-\Lambda^1)^{\vartheta}}
-\frac{c_2}{(\Lambda^2-X)^{\vartheta}},\psi\big)ds+\int_0^t\int_0^1(\chi(y,s,X),\psi)  W(dx,ds) \nonumber\\
&&+\int_0^t\int_0^1\psi\Upsilon(dx,ds)-\int_0^t\int_0^1\psi\Gamma(dx,ds),\ a.s,
\end{eqnarray}
where $(,)$ denotes the inner product in $L^2([0,1])$ and $X(t)$ denotes $X(\cdot,t)$;\\
(iv)$$\int_Q\big(X(x,t)-\Lambda^1(x,t)\big)\Upsilon(dx,dt)=\int_Q\big(\Lambda^2(x,t)-X(x,t)\big)\Gamma(dx,dt)=0.$$
\end{definition}

\section{The existence and uniqueness of solutions}
\setcounter{equation}{0}
\renewcommand{\theequation}{3.\arabic{equation}}
In this section, we prove the following theorem.

\begin{theorem}\label{t1}  Let $X_0\in{C}([0,1])$ satisfy
$\Lambda^1(x,0)\leq{X_0}(x)\leq{\Lambda^2}(x,0),\ X_0(0)=X_0(1)=0$. Assume conditions (H0)-(H4), (F1)-(F3) hold.
Then there exists a unique solution $(X, \Upsilon,\Gamma)$ to Eq. \eqref{a02}
 for all $\vartheta\geq0$.
\end{theorem}

As a first step we show that, for every $v(x,t)$ (sometimes denote by $v$) $\in{}C(Q)$, $v(x,0)=X_0(x)$ and
$v(0,t)=0,\ v(1,t)=0$, the deterministic PDE
\begin{equation}\label{a0001}
\left \{\begin{array}{ll}
       \frac{\partial{\Xi(x,t)}}{\partial{t}}=\frac{\partial^2{\Xi(x,t)}}{\partial{x^2}}+\frac{c_1}{(\Xi+v-\Lambda^1)^{\vartheta}}- \frac{c_2}{(\Lambda^2-\Xi-v)^{\vartheta}}+\Upsilon(x,t)-\Gamma(x,t);\\
        \Xi(0,t)=\Xi(1,t)=0,\ \ {\rm for}\ t\geq0;\\
        \Xi(x,0)=0,\ \ {\rm for}\ x\in[0,1];\\
        \Lambda^1(x,t)\leq{\Xi}(x,t)+v(x,t)\leq{\Lambda^2}(x,t),\ \ {\rm for}\ (x,t)\in{Q}
\end{array}
\right.
\end{equation}
has a unique continuous solution.

We now recall the precise definition of the solution for Eq. (\ref{a0001}) in \cite{yz}.
\begin{definition} A triplet $(\Xi,\Upsilon,\Gamma)$ is called a
solution to the PDE (\ref{a0001}) if\\
(i) $\Xi={\Xi(x,t);(x,t)\in{Q}}$ is a continuous function satisfying
$\Lambda^1(x,t)\leq{\Xi}(x,t)+v(x,t)\leq{\Lambda^2}(x,t),\ \Xi(x,0)=0,\
\Xi(0,t)=\Xi(1,t)=0$;\\
(ii) $\Upsilon(dx,dt)$ and $\Gamma(dx,dt)$ are measures on
$(0,1)\times\mathbb{R}_+$ satisfying
$$\Upsilon\big((\epsilon,1-\epsilon)\times[0,T]\big)<\infty,\  \Gamma\big((\epsilon,1-\epsilon)\times[0,T]\big)<\infty$$
for every $\epsilon\in(0,\frac{1}{2})$ and $T>0$;\\
(iii) for all $t\geq0$ and $\psi\in{C_0^\infty}(0,1)$ we have
\begin{eqnarray}\label{3.2}
     &&\big(\Xi(t),\psi\big)-\int_0^t(\Xi(s),\psi^{''})ds-\int_0^t\big(\frac{c_1}{(\Xi+v-\Lambda^1)^{\vartheta}}
-\frac{c_2}{(\Lambda^2-\Xi-v)^{\vartheta}},\psi\big)ds \nonumber\\
     &=&\int_0^t\int_0^1\psi\Upsilon(dx,ds)-\int_0^t\int_0^1\psi\Gamma(dx,ds),
\end{eqnarray}
where $\Xi(t)$ denotes $\Xi(\cdot,t)$.\\
(iv)
\begin{eqnarray*}
     &&\int_Q\big(\Xi(x,t)+v(x,t)-\Lambda^1(x,t)\big)\Upsilon(dx,dt) \\
          &=&\int_Q\big(\Lambda^2(x,t)-\Xi(x,t)-v(x,t)\big)\Gamma(dx,dt)\\
          &=&0.
\end{eqnarray*}
\end{definition}

\begin{prop}\label{p1}  Assume conditions (H0)-(H4) hold. Then
there exists a unique solution $(\Xi, \Upsilon,\Gamma)$ to Eq.
\eqref{a0001}.
Moreover,
denote $(\hat{\Xi},\hat\Upsilon,\hat\Gamma)$ by the solution to Eq. (\ref{a0001}) replacing $v$
with $\hat{v}$. Then
\begin{eqnarray}\label{a0002}
\|\Xi-\hat{\Xi}\|_\infty^T\leq\|v-\hat{v}\|_\infty^T,\ \forall T>0,
\end{eqnarray}
where
$\|J\|_\infty^T:=\underset{x\in[0,1],t\in[0,T]}{\sup}|J(x,t)|$.
\end{prop}

Before the proof of Proposition \ref{p1}, the following lemma for comparison is needed.
\begin{lemma}\label{l1}  Assume (H0)-(H4). Let $g^{\rho }:[0,1]\times \mathbb{R}\rightarrow \mathbb{R}$ be measurable,
$\rho >0$, such that $y\mapsto g^{\rho}(x,y)$ is monotone nonincreasing, Lipschitz-continuous uniformly in $x \in [0,1]$
and satisfies
$$|g^{\rho}(x,y)| \leq C(1+|y|),\ \ {\rm for}\  y\in \mathbb R, \ \rho>0,$$
for some $C\geq 0$. Consider the following PDE with reflection:
\begin{equation}\label{a05}
\left \{\begin{array}{ll}
        \frac{\partial{\Xi^\rho}}{\partial{t}}=\frac{\partial^2{\Xi^\rho}}{\partial{x^2}}+g^\rho(\Xi^{\rho})+\Upsilon^{\rho}-\Gamma^{\rho};\\
        \Xi^\rho(0,t)=\Xi^\rho(1,t)=0,\ \ {\rm for}\ t\geq0;\\
        \Xi^\rho(x,0)=0,\ \ {\rm for}\ x\in[0,1],
\end{array}
\right.
\end{equation}
where $\Upsilon^{\rho}$, $\Gamma^{\rho}$ satisfy $$\int_Q\big(\Xi^{\rho}(x,t)+v(x,t)-\Lambda^1(x,t)\big)\Upsilon^{\rho}(dx,dt)=0,$$
$$\int_Q\big(\Lambda^2(x,t)-\Xi^{\rho}(x,t)-v(x,t) \big)\Gamma^{\rho}(dx,dt)=0.$$
Then there exists a unique solution of Eq. \eqref{a05}, denoted by $\Xi^{\rho}$.
Moreover, if $\rho\mapsto g^{\rho}$ is monotone nonincreasing, then
$\rho\mapsto \Xi^{\rho}$ is monotone nonincreasing.
\end{lemma}

\begin{remark}\label{r1} To get the monotonicity of $\Xi^{\rho}$ w.r.t. $\rho$, the monotonicity of drift $g^{\rho}$ in
Eq. \eqref{a05} w.r.t. $\rho$ is required. If we assume that there exists the solution of Eq. \eqref{a05},
the Lipschitz-continuous and linear growth conditions of $g$ are not needed in Lemma \ref{l1}.
\end{remark}
\noindent\textbf{Proof.}\quad
The existence of the solution of Eq. \eqref{a05} can be obtained from the convergence of the penalized equation which is similarly to Theorem 2.1 in Otobe \cite{os}. The uniqueness of the solutions of \eqref{a05} can be proved in a similar way as Theorem 3.1 in Zhang and Yang \cite{zy}.
\vskip 0.3cm

Now we prove the comparison principle. Let $\rho_1\geq \rho_2>0$ and set $\phi=(\Xi^{\rho_1}- \Xi^{\rho_2})^+$, where $X^{\rho_1},\ X^{\rho_2}$ are the solutions of Eq. \eqref{a05} with drifts $g^{\rho_1}$,
$g^{\rho_2}$ respectively.
Consider the inner product between $\phi$ and $\Xi^{\rho_1}- \Xi^{\rho_2}$ in $L^2(0,1)$.
We have
\begin{eqnarray*}
     &&\| \phi(t) \|^2_{L^2}\\
     &=&2 \int_0^t <\phi,\ \frac{\partial^2}{\partial x^2}( \Xi^{\rho_1}-  \Xi^{\rho_2}  )>ds  +2 \int_0^t <\phi,\ g^{\rho_1}(\Xi^{\rho_1})-  g^{\rho_2}(\Xi^{\rho_2})>ds \nonumber\\
     && +  2 \int_0^t  <\phi,\ d\Upsilon^{\rho_1}- d\Upsilon^{\rho_2}  >  - 2 \int_0^t  <\phi,\ d\Gamma^{\rho_1}- d\Gamma^{\rho_2}  >  \nonumber\\
      &= & - 2 \int_0^t\| \frac{\partial \phi}{\partial x} \|^2_{L^2}ds + 2 \int_0^t <\phi,\ g^{\rho_1}(\Xi^{\rho_1})-  g^{\rho_1}(\Xi^{\rho_2})>ds\\
    && + 2 \int_0^t <\phi,\ g^{\rho_1}(\Xi^{\rho_2})-  g^{\rho_2}(\Xi^{\rho_2})>ds   \nonumber\\
     && +  2 \int_0^t  {1}_{\{\Xi^{\rho_1}\geq \Xi^{\rho_2}\}  } <\Xi^{\rho_1}+v-\Lambda^1- (\Xi^{\rho_2}+v-\Lambda^1)  ,\ d\Upsilon^{\rho_1}
    - d\Upsilon^{\rho_2}  > \\
     &&- 2 \int_0^t{1}_{\{\Xi^{\rho_1}\geq \Xi^{\rho_2} \} } <\Xi^{\rho_1}+v-\Lambda^2- (\Xi^{\rho_2}+v-\Lambda^2) ,\ d\Gamma^{\rho_1}- d\Gamma^{\rho_2}  > \\
      &=&  - 2 \int_0^t\| \frac{\partial \phi}{\partial x} \|^2_{L^2}ds + 2 \int_0^t <\phi,\ g^{\rho_1}(\Xi^{\rho_1})-  g^{\rho_1}(\Xi^{\rho_2})>ds\\
     &&+ 2 \int_0^t <\phi,\ g^{\rho_1}(\Xi^{\rho_2})-  g^{\rho_2}(\Xi^{\rho_2})>ds   \nonumber\\
     && +  2 \int_0^t {1}_{\{\Xi^{\rho_1}\geq \Xi^{\rho_2}\}  } (<\Xi^{\rho_1}+v-\Lambda^1  ,\ - d\Upsilon^{\rho_2}  > +<- (\Xi^{\rho_2}+v-\Lambda^1)  ,\ d\Upsilon^{\rho_1}  > )\\
      &&- 2 \int_0^t {1}_{\{  \Xi^{\rho_1}\geq \Xi^{\rho_2} \} }( <\Xi^{\rho_1}+v-\Lambda^2 ,\ - d\Gamma^{\rho_2}  >
      +<- (\Xi^{\rho_2}+v-\Lambda^2) ,\ d\Gamma^{\rho_1}  > )  \nonumber\\
     &\leq&  2 \int_0^t <\phi,\ g^{\rho_1}(\Xi^{\rho_2})-  g^{\rho_2}(\Xi^{\rho_2})>ds.
\end{eqnarray*}
\hfill$\Box$

\noindent\textbf{Proof of Proposition \ref{p1}}\quad  Existence: Consider the following reflected PDE with Lipschitz coefficients:
\begin{equation}\label{a04}
\left \{\begin{array}{ll}
       \frac{\partial{\Xi^{\epsilon_1,\epsilon_2}}}{\partial{t}}=\frac{\partial^2{\Xi^{\epsilon_1,\epsilon_2}}}{\partial{x^2}}
       +\frac{c_1}{[\epsilon_1+ (\Xi^{\epsilon_1,\epsilon_2}+v-\Lambda^1) ]^{\vartheta}}- \frac{c_2}{[\epsilon_2+ (\Lambda^2-\Xi^{\epsilon_1,\epsilon_2}-v) ]^{\vartheta}} +\Upsilon^{\epsilon_1,\epsilon_2}-\Gamma^{\epsilon_1,\epsilon_2};\\
       \Xi^{\epsilon_1,\epsilon_2}(0,t)=\Xi^{\epsilon_1,\epsilon_2}(1,t)=0;\\
       \Xi^{\epsilon_1,\epsilon_2}(x,0)=0,
\end{array}
\right.
\end{equation}
where $\epsilon_1,\epsilon_2>0$ and $\Upsilon^{\epsilon_1,\epsilon_2}$, $\Gamma^{\epsilon_1,\epsilon_2}$ satisfy $\int_Q\big(\Xi^{\epsilon_1,\epsilon_2}(x,t)+v(x,t)-\Lambda^1(x,t)\big) \\  \Upsilon^{\epsilon_1,\epsilon_2}(dx,dt)=0$, $\int_Q\big(\Lambda^2(x,t)-\Xi^{\epsilon_1,\epsilon_2}(x,t)-v(x,t)\big)\Gamma^{\epsilon_1,\epsilon_2}(dx,dt)=0$.
\vskip 0.3cm

The existence and uniqueness of the solutions of Eq. \eqref{a04} can be got from Lemma \ref{l1}.
Indeed, there is a unique solution ${\tilde{\Xi}}^{\epsilon_1,\epsilon_2}\in [\Lambda^1,\ \Lambda^2]$ of the following PDE
\begin{equation}\label{a041}
\left \{\begin{array}{ll}
       \frac{\partial{{\tilde{\Xi}}^{\epsilon_1,\epsilon_2}}}{\partial{t}}=\frac{\partial^2{{\tilde{\Xi}}^{\epsilon_1,\epsilon_2}}}{\partial{x^2}}
       +\frac{c_1}{[\epsilon_1+ ({\tilde{\Xi}}^{\epsilon_1,\epsilon_2}+v-\Lambda^1) \vee 0 ]^{\vartheta}}- \frac{c_2}{[\epsilon_2+ (\Lambda^2-{\tilde{\Xi}}^{\epsilon_1,\epsilon_2}-v)  \vee 0 ]^{\vartheta}} +\tilde\Upsilon^{\epsilon_1,\epsilon_2}-\tilde\Gamma^{\epsilon_1,\epsilon_2};\\
       {\tilde{\Xi}}^{\epsilon_1,\epsilon_2}(0,t)={\tilde{\Xi}}^{\epsilon_1,\epsilon_2}(1,t)=0;\\
       {\tilde{\Xi}}^{\epsilon_1,\epsilon_2}(x,0)=0,
\end{array}
\right.
\end{equation}
where $\tilde\Upsilon^{\epsilon_1,\epsilon_2}$, $\tilde\Gamma^{\epsilon_1,\epsilon_2}$ satisfy $\int_Q\big(\tilde \Xi^{\epsilon_1,\epsilon_2}(x,t)+v(x,t)-\Lambda^1(x,t)\big)  \tilde\Upsilon^{\epsilon_1,\epsilon_2}(dx,dt)=0$, $\int_Q\big(\Lambda^2(x,t)-\tilde \Xi^{\epsilon_1,\epsilon_2}(x,t)-v(x,t)\big)\tilde\Gamma^{\epsilon_1,\epsilon_2}(dx,dt)=0$.
Therefore, Eq.s \eqref{a04} and \eqref{a041} have the same solution.
Denote the solution of Eq. \eqref{a04} by $\Xi^{\epsilon_1,\epsilon_2}$.
\vskip 0.3cm

Fix $\epsilon_2$, by Lemma \ref{l1} we get that $\epsilon_1\mapsto \Xi^{\epsilon_1,\epsilon_2}$ is nonincreasing. Notice that
$\omega^{\epsilon_1,\epsilon_2}:=\epsilon_1 + \Xi^{\epsilon_1,\epsilon_2}$ is solution of
\begin{equation}\label{a07}
\left \{\begin{array}{ll}
       \frac{\partial{\omega^{\epsilon_1,\epsilon_2}}}{\partial{t}}=\frac{\partial^2{\omega^{\epsilon_1,\epsilon_2}}}{\partial{x^2}}
       +\frac{c_1}{ (\omega^{\epsilon_1,\epsilon_2}+v-\Lambda^1)^{\vartheta}}- \frac{c_2}{(\Lambda^2-\omega^{\epsilon_1,\epsilon_2}-v+  \epsilon_1+\epsilon_2)^{\vartheta}} +\Upsilon_\omega^{\epsilon_1,\epsilon_2}-\Gamma_\omega^{\epsilon_1,\epsilon_2};\\
       \omega^{\epsilon_1,\epsilon_2}(0,t)=\omega^{\epsilon_1,\epsilon_2}(1,t)=\epsilon_1;\\
       \omega^{\epsilon_1,\epsilon_2}(x,0)=\epsilon_1,
\end{array}
\right.
\end{equation}
where $\Upsilon_\omega^{\epsilon_1,\epsilon_2}$, $\Gamma_\omega^{\epsilon_1,\epsilon_2}$ satisfy $\int_Q\big(\omega^{\epsilon_1,\epsilon_2}(x,t)+v(x,t)-\Lambda^1(x,t)\big)\Upsilon_\omega^{\epsilon_1,\epsilon_2}(dx,dt)=0$, $\int_Q\big(\Lambda^2(x,t)-\omega^{\epsilon_1,\epsilon_2}(x,t)-v(x,t)\big)\Gamma_\omega^{\epsilon_1,\epsilon_2}(dx,dt)=0$.
\vskip 0.3cm

Let $g^{\epsilon_1}=\frac{c_1}{ (\omega^{\epsilon_1,\epsilon_2}+v-\Lambda^1)^{\vartheta}}- \frac{c_2}{(\Lambda^2-\omega^{\epsilon_1,\epsilon_2}-v+  \epsilon_1+\epsilon_2)^{\vartheta}}$. By Lemma \ref{l1}, we have for all $\epsilon_1''\geq \epsilon_1'>0$,
\begin{eqnarray}
     && \| (\omega^{\epsilon_1',\epsilon_2}-\omega^{\epsilon_1'',\epsilon_2}  )^+  \|^2_{L^2} \nonumber\\
     &\leq & \int_0^t\int_0^1[g^{\epsilon_1'}(  \omega^{\epsilon_1'',\epsilon_2}  )   -  g^{\epsilon_1''}(  \omega^{\epsilon_1'',\epsilon_2}  )](\omega^{\epsilon_1',\epsilon_2}-\omega^{\epsilon_1'',\epsilon_2}  )^+  dxds  \nonumber\\
     &= &\int_0^t\int_0^1 [  \frac{c_1}{ (\omega^{\epsilon_1'',\epsilon_2}+v-\Lambda^1)^{\vartheta}}- \frac{c_2}{(\Lambda^2-\omega^{\epsilon_1'',\epsilon_2}-v+  \epsilon_1'+\epsilon_2)^{\vartheta}} \nonumber\\
       &&- ( \frac{c_1}{ (\omega^{\epsilon_1'',\epsilon_2}+v-\Lambda^1)^{\vartheta}}- \frac{c_2}{(\Lambda^2-\omega^{\epsilon_1'',\epsilon_2}-v+  \epsilon_1''+\epsilon_2)^{\vartheta}}  )     ] \nonumber\\
       && (\omega^{\epsilon_1',\epsilon_2}-\omega^{\epsilon_1'',\epsilon_2}  )^+  dxds      \nonumber\\
        &\leq &0.
\end{eqnarray}

This means $\epsilon_1\mapsto\epsilon_1+ \Xi^{\epsilon_1,\epsilon_2}$ is nondecreasing. By the definition of \emph{uniformly continuous}, we obtain that $\Xi^{\epsilon_1,\epsilon_2}$ converges uniformly on $[0,1]\times[0, T]$ to a continuous function, denoted by $\Xi^{\epsilon_2}$, as $\epsilon_1 \downarrow 0$.
Moreover, $\epsilon_1\mapsto  \Upsilon^{\epsilon_1,\epsilon_2}-\Gamma^{\epsilon_1,\epsilon_2}$ is monotone nondecreasing and then $\Upsilon^{\epsilon_1,\epsilon_2}-\Gamma^{\epsilon_1,\epsilon_2}$ converges distributionally to some measure $\Upsilon^{\epsilon_2}-\Gamma^{\epsilon_2}$. From
the proof of existence of the solution in Theorem 3.1 in \cite{zy}, $\Upsilon^{\epsilon_1,\epsilon_2}$, $\Gamma^{\epsilon_1,\epsilon_2}$ converge distributionally to two measures $\Upsilon^{\epsilon_2}$, $\Gamma^{\epsilon_2}$ respectively, and denoted by $(\Xi^{\epsilon_2},\Upsilon^{\epsilon_2},\Gamma^{\epsilon_2} ) $ the solution of the following equation:
\begin{equation}\label{a08}
\left \{\begin{array}{ll}
       \frac{\partial{\Xi^{\epsilon_2}}}{\partial{t}}=\frac{\partial^2{\Xi^{\epsilon_2}}}{\partial{x^2}}
       +\frac{c_1}{ (\Xi^{\epsilon_2}+v-\Lambda^1)^{\vartheta}}- \frac{c_2}{[\epsilon_2+ (\Lambda^2-\Xi^{\epsilon_2}-v) ]^{\vartheta}}+\Upsilon^{\epsilon_2}-\Gamma^{\epsilon_2};\\
       \Xi^{\epsilon_2}(0,t)=\Xi^{\epsilon_2}(1,t)=0;\\
       \Xi^{\epsilon_2}(x,0)=0
\end{array}
\right.
\end{equation}
where $\Upsilon^{\epsilon_2}$, $\Gamma^{\epsilon_2}$ satisfy $$\int_Q\big(\Xi^{\epsilon_2}(x,t)+v(x,t)-\Lambda^1(x,t)\big)\Upsilon^{\epsilon_2}(dx,dt)=0,$$ $$\int_Q\big(\Lambda^2(x,t)-\Xi^{\epsilon_2}(x,t)-v(x,t)\big)\Gamma^{\epsilon_2}(dx,dt)=0.$$

In view of Remark
3.1
, we could again apply Lemma \ref{l1} to Eq. \eqref{a08}. It is then easy to get that
 $\epsilon_2\mapsto \Xi^{\epsilon_2}$ is nonincreasing. $\epsilon_2\mapsto\epsilon_2+ \Xi^{\epsilon_2}$ is also nondecreasing,
and thus $\Xi^{\epsilon_2}$ converges uniformly as $\epsilon_2 \downarrow 0$ on $[0,1]\times[0, T]$ to a continuous function denoted by $\Xi$.
Also, $\Upsilon^{\epsilon_2}$, $\Gamma^{\epsilon_2}$ converge distributionally to two measures $\Upsilon$, $\Gamma$, respectively, and $(\Xi,\Upsilon,\Gamma ) $ is the solution of Eq.
\eqref{a0001}.
\vskip 0.3cm

Uniqueness: suppose $(\Xi_1,\Upsilon_1,\Gamma_1)$ and $(\Xi_2,\Upsilon_2,\Gamma_2)$
are two solutions to Eq. \eqref{a0001}.
We can
choose an appropriate test function $\phi$ in Eq. (\ref{3.2}), following the same arguments as that in section 2.3 of Nualart and Pardoux \cite{np} and consequently get $\Xi_1=\Xi_2$. Using the same method of separating  the two measures in Theorem 3.1 of \cite{zy}, we deduce that
$\Upsilon_1=\Upsilon_2$, $\Gamma_1=\Gamma_2$ which completes the proof of uniqueness.
\vskip 0.3cm

We now prove the inequality \eqref{a0002}.
Define $k:=\|v-\hat{v}\|_\infty^T$ and
$\omega:=(\Xi^{\epsilon_1,\epsilon_2}-\hat{\Xi}^{\epsilon_1,\epsilon_2})-k$.
Then $\omega$ satisfies the following
PDE:
\begin{eqnarray*}
       \frac{\partial{\omega}}{\partial{t}}&=&\frac{\partial^2{\omega}}{\partial{x^2}}  +g^{\epsilon_1,\epsilon_2}(\Xi^{\epsilon_1,\epsilon_2})
       - g^{\epsilon_1,\epsilon_2}( \hat{\Xi}^{\epsilon_1,\epsilon_2}  )
       +\Upsilon^{\epsilon_1,\epsilon_2}-\hat\Upsilon^{\epsilon_1,\epsilon_2}-\big(\Gamma^{\epsilon_1,\epsilon_2}-\hat\Gamma^{\epsilon_1,\epsilon_2}\big)
\end{eqnarray*}
where
 \begin{eqnarray*}
      g^{\epsilon_1,\epsilon_2}(\Xi^{\epsilon_1,\epsilon_2}) & =&\frac{c_1}{[\epsilon_1+ (\Xi^{\epsilon_1,\epsilon_2}+v-\Lambda^1) ]^{\vartheta}}-  \frac{c_2}{[\epsilon_2+ (\Lambda^2-\Xi^{\epsilon_1,\epsilon_2}-v) ]^{\vartheta}}
\end{eqnarray*}
and
\begin{eqnarray*}
      g^{\epsilon_1,\epsilon_2}( \hat{\Xi}^{\epsilon_1,\epsilon_2}  ) & =&\frac{c_1}{[\epsilon_1+ (\hat{\Xi}^{\epsilon_1,\epsilon_2}+\hat{v}-\Lambda^1) ]^{\vartheta}}-  \frac{c_2}{[\epsilon_2+ (\Lambda^2-\hat{\Xi}^{\epsilon_1,\epsilon_2}-\hat{v}) ]^{\vartheta}}.
\end{eqnarray*}

Using Lemma \ref{l1},
we get $\omega_t^+=0$, i.e.
$\Xi^{\epsilon_1,\epsilon_2}-\hat{\Xi}^{\epsilon_1,\epsilon_2}\leq{k}$. By
symmetry,
$\hat{\Xi}^{\epsilon_1,\epsilon_2}-\Xi^{\epsilon_1,\epsilon_2}\leq{k}$. Then
we have for any $T>0$,
\begin{eqnarray*}
\|\Xi^{\epsilon_1,\epsilon_2}-\hat{\Xi}^{\epsilon_1,\epsilon_2}\|_\infty^T\leq\|v-\hat{v}\|_\infty^T.
\end{eqnarray*}
Let $\epsilon_1,\ \epsilon_2\downarrow0$, we conclude
that $\|{\Xi}-\hat \Xi\|_\infty^T\leq\|{v}-\hat v\|_\infty^T$. This is the same as Inequality \eqref{a0002}.
\hfill$\Box$
\vskip 0.3cm
Now we use Picard iteration to get the existence and uniqueness of the solution of Eq. \eqref{a01}.\\
\noindent\textbf{Proof of Theorem \ref{p1}}\quad Similar to Zhang and Yang \cite{zy}, we define
\begin{eqnarray*}
        v_1(x,t)&=&\int_0^1G_t(x,y)X_0(y)dy+\int_0^t\int_0^1G_{t-s}(x,y)f(y,s;X_0)dyds \nonumber\\
        &&+\int_0^t\int_0^1G_{t-s}(x,y)\chi(y,s;X_0)W(dy,ds),
\end{eqnarray*}
where $G_t(x,y)$ is Green's function of the heat equation. So $v_1(x,t)$ satisfies
\begin{eqnarray*}
\left \{\begin{array}{ll}
       \frac{\partial{v_1(x,t)}}{\partial{t}}=\frac{\partial^2{v_1(x,t)}}{\partial{x^2}}+f(x,t;X_0)+\chi(x,t;X_0)\dot{W}(x,t); \\
       v_1(0,t)=v_1(1,t)=0;  \\
      v_1(x,0)=X_0(x) .
\end{array}
\right.
\end{eqnarray*}
Let $(\Xi_1,\Upsilon_1,\Gamma_1)$ be the unique solution of Eq. (\ref{a0001}) with
$v=v_1$. Then
$(X_1,\Upsilon_1,\Gamma_1)$, where $X_1:=\Xi_1+v_1$, is the unique solution of the following
SPDE with two reflecting walls:
\begin{eqnarray*}
\left \{\begin{array}{ll}
       \frac{\partial{X_1(x,t)}}{\partial{t}}=\frac{\partial^2{X_1(x,t)}}{\partial{x^2}}+f(x,t;X_0)+\frac{c_1}{(X_1-\Lambda^1)^{\vartheta}}- \frac{c_2}{(\Lambda^2-X_1)^{\vartheta}}\\
       \ \ \ \ \ \ \ \ \ \ \ \ \ +\chi(x,t;X_0)\dot{W}(x,t)+\Upsilon_1-\Gamma_1; \\
      X_1(0,t)=X_1(1,t)=0;  \\
       X_1(x,0)=X_0(x);\\
       \Lambda^1(x,t)\leq{X}_1(x,t)\leq{\Lambda^2}(x,t).
\end{array}
\right.
\end{eqnarray*}
Iterating this procedure, suppose $X_{n-1}$ has been obtained. Let
\begin{eqnarray*}
      v_n(x,t)&=&\int_0^1G_t(x,y)X_0(y)dy+\int_0^t\int_0^1G_{t-s}(x,y)f(y,s;X_{n-1})dyds \nonumber\\
        &&+\int_0^t\int_0^1G_{t-s}(x,y)\chi(y,s;X_{n-1})W(dy,ds)
\end{eqnarray*}
and $(\Xi_n,\Upsilon_n,\Gamma_n)$ the unique solution of Eq. (\ref{a0001})
replacing $v$ by $v_n$. Then $(X_n,\Upsilon_n,\Gamma_n)$, where $X_n=\Xi_n+v_n$,
is the unique solution of the following SPDE:
\begin{eqnarray*}
\left \{\begin{array}{ll}
       \frac{\partial{X_n(x,t)}}{\partial{t}}=\frac{\partial^2{X_n(x,t)}}{\partial{x^2}}+f(x,t;X_{n-1})+\frac{c_1}{(X_n-\Lambda^1)^{\vartheta}}- \frac{c_2}{(\Lambda^2-X_n)^{\vartheta}}\\
       \ \ \ \ \ \ \ \ \ \ \ \ \
       +\chi(x,t;X_{n-1})\dot{W}(x,t)+\Upsilon_n-\Gamma_n;\\
        X_n(0,t)=X_n(1,t)=0;  \\
      X_n(x,0)=X_0(x);\\
       \Lambda^1(x,t)\leq{X}_n(x,t)\leq{\Lambda^2}(x,t).
\end{array}
\right.
\end{eqnarray*}
From Inequality \eqref{a0002},
\begin{eqnarray}
\|\Xi_n-\Xi_{n-1}\|_\infty^T\leq\|v_n-v_{n-1}\|_\infty^T
\end{eqnarray}
and therefore
\begin{eqnarray}
\|X_n-X_{n-1}\|_\infty^T\leq2\|v_n-v_{n-1}\|_\infty^T.
\end{eqnarray}
Under the assumption (F2,F3), there exists constant $C(k,T)$ depending on $k,T$, such that
\begin{eqnarray*}
\mathbb{E}\big(\|X_n-X_{n-1}\|_\infty^T\big)^k&\leq&2^k\mathbb{E}\big(\|v_n-v_{n-1}\|_\infty^T\big)^k   \nonumber\\
&\leq&{C^{k-1}}(k,T)\mathbb{E}\big(\|X_1-X_0\|_\infty^T\big)^k\frac{T^{n-1}}{(n-1)!}.
\end{eqnarray*}
Hence
there
exists $X(\cdot,\cdot)\in{C}(Q)$
such that
$\underset{n\rightarrow\infty}{lim}\mathbb{E}\big(\|X_n-X\|_\infty^T\big)^k=0$.
Going with the same argument as section 4 in \cite{zy}, we can obtain that
 $(X,\Upsilon,\Gamma)$ is a solution to the SPDE (\ref{a01}). Furthermore, the uniqueness is similar to that in \cite{zy} with the
help of Inequality \eqref{a0002}.
\hfill$\Box$

\section{Pathwise Properties}
\setcounter{equation}{0}
\renewcommand{\theequation}{4.\arabic{equation}}
Let S be the circle [0,1], with the endpoints identified, and let $\rho(x,y)$ be the distance from
$x$ to $y$ along the circle S. That is, let
$$
\rho(x,y)=\min_{k\in\mathbb{Z}}|x-y+k|.
$$
In this section, we
consider pathwise properties of the following SPDEs on $S\times\mathbb{R}_+$:
\begin{equation}\label{4.1}
\left \{\begin{array}{ll}
         \frac{\partial{X}}{\partial{t}}=\frac{\partial^2{X}}{\partial{x^2}}+f(x,t,X)+\frac{c_1}{(X-\Lambda^1)^{\vartheta}}- \frac{c_2}{(\Lambda^2-X)^{\vartheta}}+\chi(x,t,X)\dot{W}+\Upsilon-\Gamma;\\
       X(x,0)=X_0(x)\in{C}(S);\\
       \Lambda^1(x,t)\leq{X}(x,t)\leq{\Lambda^2}(x,t),\ \ {\rm for}\ (x,t)\in{S\times\mathbb{R}_+}.
\end{array}
\right.
\end{equation}\\
Here $\dot{W}(x,t)$ stands for the space-time white noise.

Set $\widehat{\mathcal{A}}_{M,T}=\{X\in C(S\times\mathbb{R}_+):\ \underset{x\in S,t\in[0,T]}{\sup}|X(x,t)|\leq{M}\}$. The reflecting walls $\Lambda^1,\ \Lambda^2$ and  coefficients: $f, \chi$ are assumed to be analogous with that in section 2, i.e.
\\
(H'1)$\Lambda^i(x,t)\in C(S\times \mathbb{R}_+)$ and $\Lambda^1(x,t)<\Lambda^2(x,t)$ for $(x,t)\in S\times\mathbb{R}_+$; \\
(H'2)
$\frac{\partial{\Lambda^i}}{\partial{t}}+\frac{\partial^2{\Lambda^i}}{\partial{x^2}}\in
L^2(S\times[0,T])$;\\
(H'3)
$\frac{\partial{}}{\partial{t}}\Lambda^i(0,t)=\frac{\partial{}}{\partial{t}}\Lambda^i(1,t)=0$
for $t\geq0$;\\
(H'4) $\frac{\partial{}}{\partial{t}}(\Lambda^2-\Lambda^1)\geq0$.\\
and\\
(F'1) $f(\cdot,\cdot;0),\chi(\cdot,\cdot;0)\in L^2(S\times[0,T])$;\\
(F'2) for every $X,\hat{X}\in \widehat{\mathcal{A}}_{M,T}$, there exists $C_{T,M}>0$ such that
$$|f(x,t;X)-f(x,t;\hat{X})|+|\chi(x,t;X)-\chi(x,t;\hat{X})|\leq{C_{T,M}}\underset{y\in S,s\in[0,t]}{\sup}|{X}(y,s)-\hat{X}(y,s)|,$$
for every $x\in S$ and
$t\in[0,T]$;\\
(F'3) there exists $C_T>0$ such that
$$|f(x,t;X)|+|\chi(x,t;X)|\leq{C_T}(1+\underset{y\in S,s\in[0,t]}{\sup}|X(y,s)|),\ \forall (x,t)\in S\times[0,T].$$

Here the precise definition of the solution to Eq. \eqref{4.1} is as follows.
\begin{definition} A triplet
$(X,\Upsilon,\Gamma)$  is a solution to the
Eq.
(\ref{4.1}) if\\
(i) $X=\{X(x,t);(x,t)\in{S\times\mathbb{R}_+}\}$ is a continuous, adapted random field
satisfying $\Lambda^1(x,t)\leq{X}(x,t)\leq{\Lambda^2}(x,t)$,
a.s;\\
(ii) $\Upsilon(dx,dt)$ and $\Gamma(dx,dt)$ are positive and adapted random measures on $S\times\mathbb{R}_+$
satisfying
$$\Upsilon\big(S\times[0,T]\big)<\infty,\ \Gamma\big(S\times[0,T]\big)<\infty$$
for $T>0$;\\
(iii) for all $t\geq0$ and $\psi\in{C^\infty}(S)$ we
have
\begin{eqnarray}
&&\big(X(t),\psi\big)-\big(X_0,\psi\big)-\int_0^t(X(s),\psi^{''})ds-\int_0^t\big(f(y,s,X),\psi\big)ds \nonumber\\
&=&\int_0^t\big(\frac{c_1}{(X-\Lambda^1)^{\vartheta}}
-\frac{c_2}{(\Lambda^2-X)^{\vartheta}},\psi\big)ds+\int_0^t\int_{S}(\chi(y,s,X),\psi)  W(dx,ds) \nonumber\\
&&+\int_0^t\int_{S}\psi\Upsilon(dx,ds)-\int_0^t\int_{S}\psi\Gamma(dx,ds),\ a.s,
\end{eqnarray}
where $(,)$ denotes the inner product in $L^2({S})$;\\
(iv)$$\int_{S\times\mathbb{R}_+}\big(X(x,t)-\Lambda^1(x)\big)\Upsilon(dx,dt)=\int_{S\times\mathbb{R}_+}\big(\Lambda^2(x)-X(x,t)\big)\Gamma(dx,dt)=0.$$
\end{definition}

The existence and uniqueness of the solution of Eq. (\ref{4.1}) is established in a similar way as Theorem \ref{t1}.
\begin{theorem}\label{t4.1}  Let $X_0\in{C}({S})$ satisfy
$\Lambda^1(x,0)\leq{X_0}(x)\leq{\Lambda^2}(x,0),$ for $x\in S$. Under the hypotheses (H'1)-(H'4), (F'1)-(F'3),
there exists a unique solution $(X, \Upsilon,\Gamma)$ to the Eq. \eqref{4.1}
 for all $\vartheta\geq0$.
\end{theorem}

Before studying the pathwise properties of SPDE \eqref{4.1},
we introduce some notation.
In view of assumption (H'2),
suppose that there exist $f_i\in L^2(Q_T)$, where $Q_T:=S\times[0,T]$, such that
$$
\frac{\partial \Lambda^i(x,t)}{\partial t}-\frac{\partial^2\Lambda^i(x,t)}{\partial x^2}=f_i(x,t).
$$
Let
$$
\bar{G}(t,x)=(4\pi t)^{-1/2}\exp(-\frac{x^2}{4t}).
$$
Recall that $\bar{G}(t,x)$ is the fundamental solution of the heat equation on $\mathbb{R}$.
Next, let $G(t,x,y)$ or $G_t(x,y)$ be the fundamental solution of the heat equation on $S$,
and recall that
$$
G(t,x,y)=\sum_{m\in\mathbb{Z}}\bar{G}(t,\rho(x,y)+m).
$$

We have the following
well-known result, the proof is similar to that in Walsh \cite{wa} and is hence omitted.
\begin{lemma}\label{semigroup}
Let $a\in(1,3)$. For any $x,y\in S$ and $t\in[0,T]$, there exists a constant $C_a$ such that
\begin{eqnarray}
\int_0^t\int_{S}|G(t-s,x,y)|^{a}dyds
\leq C_at^{\frac{3a-1}{2}}.
\end{eqnarray}
\end{lemma}

\subsection{The case $\vartheta>3$ }
We consider the special case $\vartheta>3$ in this subsection. We have the following theorem, which is a corollary of Lemma \ref{Lemma 1} below.
The proof of Lemma \ref{Lemma 1} is similar to that in \cite{ml}, but is nontrivial.

\begin{theorem}\label{t2} Let $X_0\in{C}(S)$ satisfy
$\Lambda^1(x,0)<{X_0}(x)<{\Lambda^2}(x,0), x\in S$ and suppose (H'1)-(H'4), (F'1)-(F'3) hold.
If $\vartheta>3$, then
the unique solution $(X, \Upsilon,\Gamma)$ to Eq. (\ref{4.1}) remains in $(\Lambda^1,\Lambda^2)$ for all time and $x\in S$,
hence $\Upsilon=\Gamma=0$.
\end{theorem}
Suppose that
$v(t,x)$ satisfies
\begin{equation}\label{estimate-v-epsilon}
\left \{\begin{array}{ll}
       v_t=v_{xx}+f(x,t,v)+g(x,t,v)+\chi(x,t,v)\dot{W}, \ t>0,\ x\in S;\\
        v(0,x)=X_0(x),
\end{array}
\right.
\end{equation}
where $f,\ \chi$ are given at the beginning of this section and
for every $\delta,\widetilde{\delta}>0$, denote
$$g(x,t,X(x,t))=\Big(|X(x,t)-\Lambda^1(x,t)|\vee(\delta/2)\Big)^{-\vartheta}-\Big(|\Lambda^2(x,t)-X(x,t)|\vee(\widetilde{\delta}/2)\Big)^{-\vartheta}.$$

\begin{lemma}\label{Lemma 1}
Given $T>0$, $\vartheta>3$ and $\epsilon>0$. There exists $\delta_0,\ \delta_1>0$ such that for any
$0<\delta\leq\delta_0$, $0<\widetilde{\delta}\leq\delta_1$, and $X_0(x)\in C(S)$
satisfying $\Lambda^2(x,0)-\widetilde{\delta}\geq X_0(x)\geq \Lambda^1(x,0)+\delta$ for all $x\in S$,
\begin{eqnarray}\label{estimate 1}
\mathbb{P}\Big(\inf_{0\leq t\leq T}\inf_{x\in S}\big( v(x,t)-\Lambda^1(x,t)\big)<\delta/2\Big)<\epsilon,
\end{eqnarray}
and
\begin{eqnarray}\label{estimate 2}
\mathbb{P}\Big(\inf_{0\leq t\leq T}\inf_{x\in S}\big( \Lambda^2(x,t)-v(x,t)\big)<\widetilde{\delta}/2\Big)<\epsilon.
\end{eqnarray}
Here $v(x,t)=\infty$ if $t$ is greater or equal to the blow-up time for $v(x,t)$.
\end{lemma}
\vskip 0.3cm

\noindent\textbf{Proof}\quad
At the beginning of the proof, we introduce some notations. Let $c:=\underset {Q_T} \inf(\Lambda^2(x,t)-\Lambda^1(x,t))$, $C:=\underset {Q_T}\sup(\Lambda^2(x,t)-\Lambda^1(x,t))$.
Denote
\begin{eqnarray*}
\Theta=\{X\in C(Q_T):\ \sup_{(y,s)\in Q_T}|X(y,s)-\Lambda^1(y,s)|\leq c/4\},
\end{eqnarray*}
$$\varphi:=\inf_{\Theta}\inf_{(x,t)\in Q_T}f(x,t,X(x,t))\ \text{and}\ \psi:=\sup_{\Theta}\sup_{(x,t)\in Q_T}f(x,t,X(x,t)).$$

By the assumptions in this section, $c,\ C, \varphi,\ \psi$ are finite constant. For any $l>0$, denote
$$
U_l=\Big\{X\in C(Q_T):\ l/2\leq X(y,s)-\Lambda^1(y,s) \leq 2l,\ \forall (y,s)\in Q_T\Big\}.
$$

Since $\vartheta>3$, it follows that
$
\frac{1}{\vartheta+1}<1/4.
$
Choose $\kappa$ such that $1/(\vartheta+1)<\kappa<1/4$.
Let $\delta_0,\delta_1\leq c/8,$ $0<\delta\leq\delta_0$ and $0<\widetilde{\delta}\leq\delta_1$.

Note that if $X\in U_\delta$, then
\begin{eqnarray*}
\Lambda^2(x,t)-X(x,t)&=&\Lambda^2(x,t)-\Lambda^1(x,t)+\Lambda^1(x,t)-X(x,t)\\
               &\geq&
                     c-2\delta
               \geq
                     c-2\delta_0
               \geq
                     3/4c,
\end{eqnarray*}
and
\begin{eqnarray*}
\Lambda^2(x,t)-X(x,t)&=&\Lambda^2(x,t)-\Lambda^1(x,t)+\Lambda^1(x,t)-X(x,t)\\
               &\leq&
                     \Lambda^2(x,t)-\Lambda^1(x,t)
               \leq
                    C.
\end{eqnarray*}
So we have
$
C\geq |\Lambda^2(x,t)-X(x,t)|\vee(\widetilde{\delta}/2)\geq 3/4 c.
$

Hence, for $a=\varphi-(3/4c)^{-\vartheta}$, $A=\psi-C^{-\vartheta}$,
\begin{eqnarray}\label{A1}
(2\delta)^{-\vartheta}+a\leq f(x,t,X(x,t))+g(x,t,X(x,t))\leq (\delta/2)^{-\vartheta}+A,\ \forall X\in U_\delta.
\end{eqnarray}

Let $w(t,x)$ satisfy
\begin{equation}\label{define w}
\left \{\begin{array}{ll}
w_t(x,t)=w_{xx}(x,t)+f(x,t,w(x,t))+g(x,t,w(x,t))\\
\ \ \ \ \  \ \  \ \ \ \ \ \ +\chi(x,t,w(x,t))\dot{W}(x,t);\\
w(x,k\beta)=\Lambda^1(x,k\beta)+\delta,\ k\beta\leq t< (k+1)\beta,\  x\in S,
\end{array}
\right.
\end{equation}
where
$\beta=2^{-\vartheta-2}\delta^{1+\vartheta}$,
$M\equiv T/\beta$ is an integer and $k=0,\cdots,M-1$.

To get Inequality (\ref{estimate 1}), it is enough to prove
\begin{eqnarray}\label{est-1}
&&\mathbb{P}\Big(
         \inf_{(x,t)\in Q_T}\big(v(x,t)-w(x,t)\big)\geq 0,\
         \inf_{(x,t)\in Q_T}\big(w(x,t)-\Lambda^1(x,t)\big)\geq \delta/2
         \Big)\nonumber \\
         &\geq& 1-\epsilon.
\end{eqnarray}

For $k\beta\leq t<(k+1)\beta$ and
$x\in S$, and assuming that $\Lambda^1(x,s)+\delta/2\leq w(s,x)\leq \Lambda^1(x,s)+2\delta$ for $k\beta\leq s\leq t$ and $x\in S$,
we have
\begin{eqnarray}\label{Eq. w}
w(x,t)=\int_{S}G(t-{k\beta},x,y)\Lambda^1(y,k\beta)dy+\delta+D_k(x,t)+N_{k,L}(x,t),
\end{eqnarray}
where
$L=\underset {X\in U_\delta}\sup \underset {(x,t)\in Q_T} \sup\chi(x,t,X(x,t)),$ and
\begin{eqnarray*}
D_k(x,t)&=&\int_{k\beta}^t\int_{S}G(t-s,x,y)g(y,s,w(y,s))dyds\\
&&+\int_{k\beta}^t\int_{S}G(t-s,x,y)[f(y,s,w(y,s))] dyds,
\end{eqnarray*}
\begin{eqnarray*}
N_{k,L}(x,t)=\int_{k\beta}^t\int_{S}G(t-s,x,y)[\chi(y,s,w(y,s))\wedge L]W(dy,ds).
\end{eqnarray*}
Denote by $\tau_w$ the blow-up time for $|w|$.
Let $w(x,t)=D_k(x,t)=N_{k,L}(x,t)=\infty$ if $t\geq\tau_w$.

Let
$$
\mathcal{N}_k
=\{\omega\in\Omega:
           \Big|\frac{N_{k,L}(x,t)}{(t-k\beta)^\kappa}\Big|\leq \delta^{1-\kappa(1+\vartheta)}2^{-3-2\vartheta+\kappa(\vartheta+2)},
           \forall (t,x)\in(k\beta,(k+1)\beta]\times S\}.
$$

Because of $1-\kappa(1+\vartheta)<0$, if $\delta_0$ is small enough, then from Lemma 2.3 in
\cite{ml}
$\mathbb{P}(\mathcal{N}^c_k)
\leq \frac{\epsilon}{M},\ \forall \delta\in(0,\delta_0],\ k=0,\cdots,M-1.$

Let
$\mathcal{N}=\bigcap_{k=0}^{M-1}\mathcal{N}_k.$ Since
\begin{eqnarray*}
      \mathbb{P}\Big(\inf_{0\leq t\leq T}\inf_{x\in S}\Big[v(x,t)-\Lambda^1(x,t)\Big]<\delta/2\Big)
&\leq&
      \mathbb{P}(\mathcal{N}^c)
\leq
      \sum_{k=0}^{M-1}\mathbb{P}(\mathcal{N}_k^c)
<
    \epsilon,
\end{eqnarray*}
it suffices to show that $w(x,t)\leq v(x,t)$ and $w(x,t)-\Lambda^1(x,t)\geq\delta/2$ for $0\leq t\leq T$
and $x\in S$. By comparison theorem (see Lemma 2.2 in \cite{ml}) and induction, $w(x,t)\leq v(x,t)$ for $0\leq t< k\beta$ and $x\in S$.

Therefore, our aim is to obtain that if the event $\mathcal{N}_k,\ k=0,\cdots,M$ occurs, then,
$
\Lambda^1(x,t)+\delta/2\leq w(x,t)\leq \Lambda^1(x,t)+2\delta,\ k\beta\leq t<(k+1)\beta,\ x\in S, k=0,\cdots,M-1
$ and $\Lambda^1(x,(k+1)\beta)+\delta\leq w((k+1)\beta-,x),\ x\in S$, where
$f(t-)=\underset{s\uparrow t} \lim f(s)$.

Let $t^*$ be the first time
$t\in [k\beta,(k+1)\beta)$ such that for some $x\in S$, $w(x,t^*)=\Lambda^1(x,t^*)+\delta/2$ or $\Lambda^1(x,t^*)+2\delta$. Define $t^*=(k+1)\beta$ if there is no such
time.

Thus, to get inequality (\ref{estimate 1}), we need only verify that on $\mathcal{N}_k$,\\
(i) $t^*=(k+1)\beta$; (ii) $\Lambda^1(x,(k+1)\beta)+\delta\leq w((k+1)\beta-,x),\ x\in S$.

For $x\in S$ and if $t^*<(k+1)\beta$, then by the definition of $\beta$,
\begin{eqnarray*}
      0\leq D_k(x,t^*)
&\leq&
      \beta
          \sup_{X\in U^1_\delta}
          \sup_{(x,t)\in Q_T}\Big(f(x,t,X(x,t))+g(x,t,X(x,t))\Big)\\
&\leq&
      \beta\Big[(\frac{\delta}{2})^{-\vartheta}+A\Big]\\
&=&
     2^{-\vartheta-2}\delta^{1+\vartheta}\Big[(\frac{\delta}{2})^{-\vartheta}+A\Big]=\frac{\delta}{4}+2^{-\vartheta-2}\delta^{1+\vartheta}A
\end{eqnarray*}
and $
    |N_{k,L}(x,t^*)|
\leq
     \delta2^{-3-2\vartheta}.$

Let $I_k(x,t)=\int_{S}G(t-{k\beta},x,y)\Lambda^1(y,k\beta)dyds$. Since $I_k$ and $\Lambda^1$ are the unique solutions of
the following PDEs respectively,
\begin{equation}
\left \{\begin{array}{ll}
\frac{\partial H(x,t)}{\partial t}=\frac{\partial^2H(x,t)}{\partial x^2},\ \ t\in[k\beta, (k+1)\beta),\ x\in S;\\
H(x,k\beta)=\Lambda^1(x,k\beta)
\end{array}
\right.
\end{equation}
and
\begin{equation}
\left \{\begin{array}{ll}
\frac{\partial\widetilde{H}(x,t)}{\partial t}=\frac{\partial^2\widetilde{H}(x,t)}{\partial x^2}+f_1(x,t),\ \ t\in[k\beta, (k+1)\beta),\ x\in S;\\
\widetilde{H}(x,k\beta)=\Lambda^1(x,k\beta),
\end{array}
\right.
\end{equation}
then, by Lemma \ref{semigroup},
\begin{eqnarray}\label{I_k}
  && |I_k(x,t^*)-\Lambda^1(x,t^*)|\nonumber\\
&=&
    |\int_{k\beta}^{t^*}\int_{S}G(t^*-s,x,y)f_1(y,s)dyds|\nonumber\\
&\leq&
    \Big[\int_{k\beta}^{t^*}\int_{S}|G(t^*-s,x,y)|^2dyds\Big]^{1/2}
    \Big[\int_{0}^{T}\int_{S}|f_1(y,s)|^2dyds\Big]^{1/2}\nonumber\\
&\leq&
     \widetilde{C}{(t^*-s)}^{1/4}
\leq \widetilde{C}{\beta}^{1/4}
=
     \widetilde{C}2^{\frac{-\vartheta-2}{4}}\delta^{\frac{1+\vartheta}{4}}.
\end{eqnarray}

Therefore, if $x\in S$ and if $t^*<(k+1)\beta$, then
\begin{eqnarray*}
    |w(x,t^*)-\Lambda^1(x,t^*)-\delta|
&\leq& |I_k(x,t^*)-\Lambda^1(x,k\beta)|+
    |D_k(x,t^*)|+|N_{k,L}(x,t^*)|\\
&<&
     \widetilde{C}2^{\frac{-\vartheta-2}{2}}\delta^{\frac{1+\vartheta}{2}}
    +
     \frac{\delta}{4}+2^{-\vartheta-2}\delta^{1+\vartheta}A
    +
      \delta2^{-3-2\vartheta}.
\end{eqnarray*}

Hence when $\delta_0$ is small enough,
$
|w(x,t^*)-\Lambda^1(x,t^*)-\delta|<\delta/2,\ \forall\delta\in(0,\delta_0],
$ and so $t^*=(k+1)\beta$.

To show (ii), in view of
\begin{eqnarray*}
  w(x,(k+1)\beta-)
&=&
   \int_{S}G((k+1)\beta-k\beta,x,y) \Lambda^1(y,k\beta)  dy  +\delta  \nonumber\\
   &&+D_k(x,(k+1)\beta)+N_{k,L}(x,(k+1)\beta),
\end{eqnarray*}
we have to check
\begin{eqnarray}\label{Eq. N111}
&&|I_k(x,(k+1)\beta)-\Lambda^1(x,(k+1)\beta)|+|N_{k,L}(x,(k+1)\beta)|\nonumber\\
&\leq& D_k(x,(k+1)\beta).
\end{eqnarray}

Therefore, inequality (\ref{Eq. N111}) is acquired by the following inequalities\\
$D_k(x,(k+1)\beta)
\geq
\delta2^{-2\vartheta-2}+a2^{-\vartheta-2}\delta^{1+\vartheta}$,
$|N_{k,L}(x,(k+1)\beta)|
\leq
\delta2^{-2\vartheta-3}$,\\ and
$|I_k(x,(k+1)\beta)-\Lambda^1(x,(k+1)\beta)|
       \leq
           \widetilde{C}2^{\frac{-\vartheta-2}{2}}\delta^{\frac{1+\vartheta}{2}}$.
This completes the proof of (\ref{estimate 1}). Similarly, we can get (\ref{estimate 2}).
\hfill$\Box$

\subsection{The case $0<\vartheta<3$}
In this subsection, we assume the diffusion coefficient $\chi$ is bounded away from zero.
 More precisely, we assume that
$$
\text{(A) There exists $c>0$ such that }c\leq|\chi(x,t,X)|,\ \forall(x,t)\in S\times\mathbb{R}_+,\ X\in\mathbb{R}.
$$
\vskip 0.3cm

Denote $X$ the solution of Eq. (\ref{4.1}), define
$$\tau_1:=  \inf \{ t>0;\ \underset{x\in S}\inf\big( X(x,t)-\Lambda^1(x,t)  \big)=0  \}, $$
$$\tau_2:=  \inf \{ t>0;\ \underset{x\in S}\inf\big( \Lambda^2(x,t)-X(x,t)  \big)=0  \},$$
and set $\tau:=\tau_1\wedge\tau_2$.
\vskip 0.3cm

The aim of this subsection is to show that the following proposition holds.
\begin{prop}\label{prop1}
Let the same assumptions as Theorem \ref{t4.1}, and (A) hold. Let $0<\vartheta<3$, and suppose that $X_0\in C(S)$ satisfies $\Lambda^1(x,0)<X_0(x)<\Lambda^2(x,0)$ for all
$x\in S$. If $X$ solves Eq. \eqref{4.1},
then $P(\tau<\infty)>0$, i.e, the solution of Eq. \eqref{4.1} hits $\Lambda^1(x,t)$ or $\Lambda^2(x,t)$ in finite time with positive probability.
\end{prop}

First, we introduce the following SPDE
\begin{equation}\label{b02}
\left \{\begin{array}{ll}
\frac{\partial{v (x,t)}}{\partial{t}}=\frac{\partial^2{v (x,t)}}{\partial{x^2}}+f(x,t,v )+\frac{c_1}{(v -\Lambda^1)^{\vartheta}} +\chi\big(x,t,v \big)\dot{W}(x,t);\\
v (x,0)=X_0(x).
\end{array}
\right.
\end{equation}
Then Eq. \eqref{b02} has a unique solution on the random time interval $[0,\tau_v]$, where
$$\tau_v:=\inf \{ t>0;\ \underset{x\in S}\inf\big( v (x,t)-\Lambda^1(x,t)  \big)\leq0  \}. $$

The next lemma is analogous to that in \cite{mp}.
\begin{lemma}\label{Lem1}
Let $0<\vartheta<3$ and suppose that $X_0\in C(S)$ and $\underset{x\in S}\inf \big(X_0(x)-\Lambda^1(x,0)\big)>0$. If $v$ solves Eq. \eqref{b02},
then $P(\tau_v<\infty)>0$.
\end{lemma}
\noindent\textbf{Proof.}\quad
Let $w(x,t)=v(x,t)-\Lambda^1(x,t)$ and $w(x,0)=X_0(x)-\Lambda^1(x,0)$.
Then $w(x,t)$ is the unique solution of the following equation on $[0,\tau_v]$,
\begin{equation}\label{b03}
\left \{\begin{array}{ll}
\frac{\partial{w (x,t)}}{\partial{t}}=\frac{\partial^2{w (x,t)}}{\partial{x^2}}-f_1(x,t)+f(x,t,w+\Lambda^1 )+\frac{c_1}{w^{\vartheta}}\\
\ \ \ \ \ \ \ \ \ \ \ \ \  +\chi\big(x,t,w+\Lambda^1 \big)\dot{W}(x,t),\ (x,t)\in S\times\mathbb{R}_+;\\
w(x,0)=X_0(x)-\Lambda^1(x,0).
\end{array}
\right.
\end{equation}

Now we use a Girsanov transformation to remove the drift term $-f_1(x,t)+f(x,t,w+\Lambda^1 )$ from Eq. (\ref{b03}).

Consider
\begin{equation}\label{b04}
\left \{\begin{array}{ll}
\frac{\partial{\Xi (x,t)}}{\partial{t}}=\frac{\partial^2{\Xi (x,t)}}{\partial{x^2}}+\frac{c_1}{\Xi^{\vartheta}} +\chi\big(x,t,\Xi+\Lambda^1 \big)\dot{W}(x,t),\ (x,t)\in S\times\mathbb{R}_+;  \\
\Xi(x,0)=X_0(x)-\Lambda^1(x,0).
\end{array}
\right.
\end{equation}
Eq. \eqref{b04} has a unique solution $\Xi$ on $[0,\tau_\Xi]$, where $$\tau_\Xi:=\inf \{ t>0;\ \underset{x\in S}\inf\big( \Xi (x,t)  \big)\leq0  \}.$$

Let $\mathbb{P}_w^{\tau_\Xi\wedge T},\ \mathbb{P}_\Xi^{\tau_\Xi\wedge T}$ denote the measures on the space of continuous function $g:S\times\mathbb{R}_+\rightarrow \mathbb{R}$ induced by $w^{\tau_\Xi\wedge T},\ \Xi^{\tau_\Xi\wedge T}$ respectively. Here
we denote $h^\tau$ the truncated function $h^\tau(x,t)=h(x,t)1_{\{t\leq\tau\}}$. Because of Condition (A), we have $\mathbb{P}_w^{\tau_\Xi\wedge T}$
is absolutely continuous with respect to $\mathbb{P}_\Xi^{\tau_\Xi\wedge T}$, and
$$
\frac{d\mathbb{P}_w^{\tau_\Xi\wedge T}}{d\mathbb{P}_\Xi^{\tau_\Xi\wedge T}}
  =
J
  :=
   \exp\big( \int_0^{T\wedge \tau_\Xi}\int_{S}  q(x,t) W(dx,dt) - 1/2\int_0^{T\wedge \tau_\Xi}\int_{S}  (q(x,t))^2  dxdt \big),
$$
where $q(x,t)=\chi^{-1}(x,t,\Xi+\Lambda^1) [-f_1(x,t)+f(x,t,\Xi+\Lambda^1)] $.

Note the fact that
$$
\int_0^{\tau_\Xi}\int_{S}(q(x,t))^2dxdt<\infty,\ a.s.,
$$
and the fact $\mathbb{P}(\tau_\Xi<\infty)>0$ by Theorem 1 of \cite{mp}, hence $\mathbb{P}(\tau_v<\infty)>0$.
\hfill$\Box$
\vskip 0.3cm
\noindent\textbf{Proof of Proposition \ref{prop1}}\quad
From Theorem 2.1 (comparison theorem) in \cite{dp1}, the solution $X$ of Eq. \eqref{4.1} is less than the solution $v$ of Eq. \eqref{b02} on the time
interval $[0,\tau_v\wedge\tau_1\wedge\tau_2]$, that is
$$
X(x,t)\leq v(x,t),\ \forall x\in S,\ t\in[0,\tau_v\wedge\tau_1\wedge\tau_2].
$$
Hence we have $\{\tau_v <\infty,\  \tau_2 =\infty \}\subset\{\tau_1 <\infty \}$.

By Lemma \ref{Lem1}, we have
\begin{eqnarray*}
0<P(\tau_v <\infty)
     &=&P(\{\tau_v <\infty,\  \tau_2 <\infty \}\cup \{\tau_v <\infty,\  \tau_2 =\infty \} )\\
     &\leq& P(\{\tau_2 <\infty \}\cup \{\tau_1 <\infty \} )\\
     &=& P(\tau_1\wedge\tau_2 <\infty  )\\
     &=& P(\tau <\infty  ).
\end{eqnarray*}
This proves Proposition \ref{prop1}.
\hfill$\Box$

\vskip 0.2cm {\small {\bf  Acknowledgements}\   The authors  thank
Prof. Zhao Dong for his encouragement and
valuable suggestions.

\vskip 0.2cm {\small {\bf  References}

redpajama_set_name: RedPajamaArXiv


text:
ASU Women's Basketball: Sun Devils struggle, Cal wins first game of the season
ASU struggles offensively yet again
By Mallory Schnell Feb 22, 2021, 7:56am MST
Share All sharing options for: ASU Women's Basketball: Sun Devils struggle, Cal wins first game of the season
Photo courtesy: Sun Devil Athletics
In a 67-55 loss, Arizona State (10-9, 5-9 Pac-12) handed the California Golden Bears (1-14, 1-11) its first win of the season.
Coming into the game, the Golden Bears had lost all 14 of their games played before Sunday, including a 56-53 loss to the Sun Devils back on Jan. 1.
After a slow first quarter where ASU scored eight points and Cal scored five, the Golden Bears outscored the Sun Devils 62-47 throughout the remaining three quarters.
ASU struggled offensively in yet another game, shooting 29 percent from the field, compared to Cal's 49 percent. The Sun Devils also struggled from the free throw line, only making 11 out of 20.
"We didn't make layups this weekend," coach Charli Turner Thorne said. "We didn't make free throws...We're a Division I program and we can't make easy shots and that's killing us."
Sophomore guard Leilani McIntosh recorded a season-high of 21 points in 40 minutes for Cal. McIntosh was 4-4 from three and 7-7 from the free throw line. Mia Mastrov and Evelien Lutje Schipholt each scored 12 points as well for the Golden Bears.
Taya Hanson had her second best scoring game of the season with 17 points and five rebounds in 34 minutes for ASU.
Other contributors for the Sun Devils were Jaddan Simmons with 10 points, five rebounds and four assists along with Jamie Loera and Maggie Besselink with six points each.
Being ninth in the Pac-12, Turner Thorne is not optimistic about the postseason.
"We were trying to get into postseason," said Turner Thorne. "That's not definitely an option anymore, so we've just got to try to play a little bit of basketball and not be so awful."
Arizona State will host its last regular season game next Sunday against the Arizona Wildcats at 12:00 p.m. MST.


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Your baby will just LOVE this, sleeping on the comfortable fur of Ghibli's classic giant character Totoro. And even if your little one is not familiar with the anime My Neighbor Totoro, they will instantly adore this cuddly sleeping bag and pillow. Never will your baby getting some sleep looked cuter!


text:
Q: How to get reference to aws api gateway in serverless framework provider:
  name: aws
  runtime: nodejs14.x

functions:
  hello:
    handler: handler.hello
    events:
      - httpApi:
          path: /
          method: get
          RestApiId:
            Ref: TestApi // How to get reference of AWS::Serverless::Api i.e. TestApi here

resources:
  Resources:
    authFunction:
      Type: AWS::Serverless::Function
      Properties:
        CodeUri: .
        Handler: handler.hello

    TestApi:
      DependsOn: AuthFunction
      Type: AWS::Serverless::Api
      Properties:
        StageName: dev
        Auth:
            DefaultAuthorizer: LambdaRequestAuthorizer
            Authorizers:
              LambdaRequestAuthorizer:
                FunctionPayloadType: REQUEST
                FunctionArn: !GetAtt AuthFunction.Arn

Getting error:
Configuration error at 'functions.hello': unrecognized property 'RestApiId'

A: Let's first clarify a few things.
The httpApi event is using HTTP API, not REST API from AWS Api Gateway.
You can set externally created HTTP API by specifying it in the following way:
provider:
  httpApi:
    id: <your http api reference>

If you'd like to use REST API, then you would need to use http event type and set it like this:
provider:
  apiGateway:
    restApiId: <your rest api reference>

redpajama_set_name: RedPajamaStackExchange


The state Supreme Court is considering changing Indiana's rules governing advertising by attorneys.
State Trial Court Services Director Tom Carusillo says many of the proposed changes are designed to clarify the rules' language or update the rules to fit modern communications technologies such as the Internet.
He says one proposed change would broaden a rule to allow more attorneys to advertise as specialists if they are certified in a particular field of law.
The Supreme Court's rules committee will be accepting comments on the proposed changes from judges, lawyers and the public until March 1.


text:
Posted by Kathleen Campion | Nov 7, 2017 | Reviews
JUNK LCT 10-4-17 247 JUNK, by Ayad Akhtar, directed by Doug Hughes, Lincoln Center Theater 10/14/17 Set Design: John Lee Beatty Costume Design: Catherine Zuber Lighting Design: Ben Stanton Sound Design: Mark Bennet Projections: 59 Productions Photo Credit: T Charles Erickson © T Charles Erickson Photography tcepix@comcast.net
Review by Kathleen Campion
T Charles Erickson Photography
It is 1989, the last gasp of the go-go eighties, when corporate raiders roam the landscape, attempting hostile takeovers; companies under siege employ "poison pill" defenses; and the occasional "white knights" prove less than constant. Oliver Stone had already unmasked (or glorified, depending on where you sit) the greed-is-good ethic that has always informed Wall Street.
At the Vivian Beaumont, Ayad Akhtar's latest, Junk, takes a deep dive into the morality of worshiping wealth. In the very first speech, reporter Judy Chu (Teresa Avia Lim) struts across the stage as our island guide:
This is a story of kings, or what passes for kings these days. Kings, then – bedecked in Brooks Brothers and Brioni, enthroned in sky-high castles, and embroiled in battles over, what else? Money. When did money become the thing – the only thing? Upgrade your place in line, or your prison cell, for a fee. Rent out your womb to carry someone else's child. Buy a stranger's life insurance policy – pay the premium until they die then collect the benefit. Oh, and cash. Whose idea was it to start charging us to get cash?
John Lee Beatty has rendered a Bloomberg-terminal screen as a theatrical set for Junk. It flashes, and moves, and reeks of self-important power. A huge lattice of one-dimensional cubicles stretches deck-to-grid and fill out as offices and trading floors, bedrooms and board rooms.
Two innovations drive the drama. Robert Merkin, here a thinly disguised Michael Milken, has figured out how to leverage less-than-investment-grade debt, called junk, and, in so doing, changes the takeover landscape. At the same time, law enforcement has a come-to-Jesus breakthrough. The U.S. District Attorney for the Southern District of New York, here a thinly disguised Rudolph Giuliani, realizes that instead of relying on the dusty SEC rules on insider trading, the RICO statutes, passed in 1970 to take down organized crime figures, could be employed to take down the Street's Gekkos and Milkens.
As a financial journalist, I covered this world. It was crazy dramatic, but until Junk, I'd not thought of it as theatrical, as in legit-stage drama (although Jerry Sterner's Other People's Money took a fair run at it in 1989).
This fictionalized account is largely on point — broad strokes, to be sure, but all-in-all accurate in spirit if not detail. That said, turning the "ripped from the financial pages" motif, (pages a fair share of your audience didn't read in the first place) into a cohesive drama? Yikes! (Does FINRA offer a best drama on financial markets award? The Boesky? The Bull? The Bernie?)
The show paints a realistic picture of how it was. The script is not shy on hate speech nor on misogyny. Much is made of the outsider status of the Jews vis-a-vis access to the white shoe firms. Add to that, little is made of a young woman's quid-pro-quo transaction of sex for access. Akhtar paints an accurate picture of "just the way things were." They were, and, regrettably, they still are.
Steven Pasquale plays Robert Merkin with a good deal of grit. He's the alpha male no one can effectively challenge. Yet, on some level he is beguiling; his conviction and confidence compel you to consider his "wisdom of the markets" position. He embodies the BSDs of the era.
There is a huge cast, 24 people in all, some with several roles. All are good, or bad, as they are written.
The direction is lively, the effects powerful, and, while set in the '80s, the moral desert of the money cult and the latter-day "kings" in towers hold a very current punch.
I'd say go.
PreviousCONQUEST OF THE UNIVERSE OR WHEN QUEENS COLLIDE
NextHello, Again — Film Review
BULL IN A CHINA SHOP
Just The Two Of Each Of Us


Today as I checked the arrival of the hatching of the bees in the nasty hive I saw everything going great in the morning, many bees at the entrance awaiting their first flight. I checked again later in the day and saw much fighting at the entrance. I had given them sugar water inside the hive to help them along and I guess now that the flowers are abating, thievery is about, so I closed down the entrance to help them defend the hive better and it seems to be working. If its still is bad tomorrow, I'll close it down some more or maybe even shut it off completely for a few days.


text:
Home » News » Blog » Reflections on the Murder of Sheikh Ahmed Yassin
Reflections on the Murder of Sheikh Ahmed Yassin
By: Jake Miller | mars 24, 2004
I think these are some of the saddest and by far the most frightening of days. Israel, together with this administration, has been trying since Sept. 11 to equate the Palestinian struggle for freedom with terrorism, trying to frighten the American people into believing that they are one and the same and that Israel is fighting the same « war on terror » as America. The assassination of Yassin is just one more step in this direction.
Sharon knew exactly what he was doing. Uri Avnery, a long-time leader of the Israeli peace movement, writes that the assassination is « the stupidest move », but it is not. It is, as usual, a very carefully planned strategy. Now the Arabs–and certainly the Palestinians–are saying that America is to blame this time and that Israel could not have done that without America's approval. Sharon's strong statement yesterday saying that Israel will continue to target and assassinate, and another statment made by one of his ministers following a meeting with Dick Cheney, is seen as further proof of that acceptance. America's refusal to condemn the attack while all other countries, including Britain, are comdemning it is yet another proof. Now it is rumored that Hamas is saying that since America is directly involved, they will turn their attention to America.
Thank God Rantissi, Yassin's successor, did not fall into the trap set for him, and he said emphatically that they would not. Funny that in spite of that, we see on the news today, March 23rd, that the US State Department issued an alert to its embassies and people around the world saying there is fear of targetting them by 'Hamas and Al Qaeda'. Is that wishful thinking?
Did anyone think that Sharon did not know this would happen? Attacking Americans would be a change of strategy for Hamas, who have never targeted anyone beyond Israel. If Amercians were, in fact, targeted by Palestinians there would be no sympathy for the Palestinian cause even for those who are currently sympathetic or on the border.
Needless to say, and from the reactions from the Arab world, the world is by far less safe now than it has been even over the past three years, even following the Iraq war. The talk now about the « Greater Middle East » and reform in the Middle East rings more hollow than ever.
What is « reform » when there are targeted assassinations that contradict all international laws – in addition to all other recent activities by the Israelis that are against international law? Once again, Israel commits a provocative act immediately on the eve of an Arab Summit and immediately when there is the slightest hint of progress towards peace. It is a constant Israeli strategy. This is not done lightly. And following the murder of Yassin and the collateral damage anyone around him, Israel attacked Gaza again, where three were killed yesterday morning… but we have not heard about that one in American media, have we?
It is a daily activity, and Yassin's murder was just Israel's way of sticking out its tongue at the Arabs and saying: « Here's one more outrage, show me what you can do now. » And they cannot do anything because they are impotent, and their governments are so hated by their own people, that they are rendered even more harmless.
For the record, Saudi Arabia just this week arrested a bunch of intellectuals as a – what else? – pre-emptive stirke, since they were PLANNING to write a petition to the King to ask for reform. They were released – pathetically! – only after Colin Powell flew there and expressed 'concern' and urged their release. They were released on condition that they sign a form saying they will never write a petition again asking for reform.
How's that? And that is only Saudi Arabia.
I have other instances throughout the Middle East too. Like the cross amputations in Sudan today, or Kuwait refusing yet again last week to give women the right to vote. All this brings more and more radicalized thoughts to otherwise ordinary and border-line individuals in the Arab street.
Congratulations to Sharon and the Bush Administration: they have succeeded in their strategy of promoting hate and provocation in the Arab World, and we are less safe because of it. Somehow, somewhere, there is a determination to bring the world, and certainly the Middle East, to the edge of annihilation. I just don't know why or in whose interest it could possibly be.
Waiting for the other shoe. . .
Why did Sharon commit this atrocity now? Besides the fact that there is an Arab Summit on the way, Sharon recognizes that the days of his freehand at madness may be numbered. He is rushing to build the wall' rushing to assassinate as many as he can, rushing to demolish homes, and rushing to confiscate as much land and destroy as much infrastructure as he can… It is a demolition and destructive frenzy, and here is the reason: three years ago he gellefully and in disbelief announced on a visit to the White House that this was the best administration that had ever been for Israel.
Indeed, with November coming up soon, he is now unsure of Bush's re-election and he has to maximize his madness before the elections in case Bush loses. Not that Kerry will be much better. From his statements, I am not optimistic, but at least I can only hope that the absolute freehand Sharon is given may be reduced, to say the least. Perhaps less children will die.
And yes, we are now waiting for the other shoe to fall, and we can all hold our breath because Israel will be attacked. No one doubts it. It is a question of when and where. And in addition to Israel, all those angered people around the world are rising to target anything remotely Israeli and also American. We are spinning now, looking over our shoulder every minute of everyday.
And of course, in the meantime, American media cannot help itself: it is already parading a frail 14 year old who was trying to blow himself up at a checkpoint yesterday. Why didn't they parade all the other Palestinian children killed over the previous month?
But they do not matter, you see. Notable mentions: Ala Al Nujili (13) and Mosbad Mwafi (14) [blown to smithereens on March 20th]; Raed Almoghary (12) killed when a shot entered his eye and exited from his left ear on March 6th; Bashir Abu Armana (13) killed while he was playing near his house in Hay Al Salam area on Feb.19th… and more. Here are statistics from a recent report by the UN Human Rights Commission: « Since the beginning of 2004, the IDF has killed more than 140 Palestinians and wounded more than 200, mostly children. Each month has seen an increase in the death toll. Thirty-two Palestinians were killed in January and 52 in February. In just the first two weeks of March, 44 have been killed, 30 of whom were killed in Gaza. »
http://www.wsws.org/articles/2004/mar2004/occu-m24.shtml
But you see, Israel is 'defending itself'. The Palestinians don't. But you see, the Israelis have a right to demolish 10 homes [only this morning in Khan Yunis] because they are close to settlements. Palestinians don't. But you see, they have a right to stand at the Human Rights Commission's meeting yesterday and say unabashedly that the Commission has descended to its lowest 'nadir' yet, because it convenes a session to 'bash Israel'. Palestinians cannot. But you see, they have a right to do anything they want to the Palestinian people, because they have an American President who flew by helicopter over refugee camps in 1998, looked down and later, after he became President, said dismissively: « Looked real bad down there. I don't see much we can do over there at this point. I think it's time to pull out of that situation. » And when warned about the 'dire consequences' to the Palestinians, he said: « Maybe that's the best way to get things back in balance. Sometimes a show of strength by one side can really clarify things. » [from Paul O'Neil's book The Price of Loyalty].
Are things clarified enough now, Mr. President?
That explains it, doesn't it?
And we are still waiting for the other shoe to fall.


It is now widely understood that all animals engage in complex interactions with bacteria (or microbes) throughout their various life stages. This ancient exchange can involve cooperation and has resulted in a wide range of evolved host-microbial interdependencies, including those observed in the gut. Ciona intestinalis, a filter-feeding basal chordate and classic developmental model that can be experimentally manipulated, is being employed to help define these relationships. Ciona larvae are first exposed internally to microbes upon the initiation of feeding in metamorphosed individuals; however, whether or not these microbes subsequently colonize the gut and whether or not Ciona forms relationships with specific bacteria in the gut remains unknown. In this report, we show that the Ciona gut not only is colonized by a complex community of bacteria, but also that samples from three geographically isolated populations reveal striking similarity in abundant operational taxonomic units (OTUs) consistent with the selection of a core community by the gut ecosystem.


text: import {CLIENT_ID} from '../constants/Config'

redpajama_set_name: RedPajamaGithub


Look no further than American Lock and Key Services for all of your home, business and auto lock and key needs. For over 40 years, American Lock and Key has been a local business that has been serving the greater Monterey Peninsula and Monterey County, CA area. We are conveniently located in Seaside, CA and have the knowledge, skills and equipment to provide top quality service at a reasonable price.
We are reliable, trustworthy, licensed, bonded and insured with multiple vehicles to come to your residence or business. Plus, we have a local convenient store location to service all of your lock and key needs.
Call us today at 831-899-3443 for an appointment.


Camp Alexander offers ten consecutive weekly camps for children ages 6-12. Additionally the Youth Leadership Program offers youth ages 12-14 the opportunity to enjoy all camp activities while also building upon their levels of responsibility and leadership. Families may sign up for a single week or the entire summer. There's enough fun for everyone!
Come see it all! Nature, swimming, arts & crafts, this is an action packed week full of all your favorite "classic" activities to fill your summer.
Join us as we endure a week full of sports! Do you have what it takes to be a Camp Olympian? From wacky to wonderful, all campers will fill their days with more fun than they knew could be had.
The name explains it all! That's right, a full week dedicated to water balloon fights, water guns and an epic skirmish between campers and counselors. Come join the battle.
What better way to exercise and have fun than Movin' and Groovin' at Camp Alexander! Obstacle courses, running through the trails, and maybe even a visit from a local gym! A week full of fun activities that will make exercise look like child's play.
Enjoy the outdoors? Then this is the week for you! Help your team build a tree fort and capture the Yeti! We'll spend the whole week in the woods and the beautiful wilderness that makes up Camp Alexander. Our annual summer Fish Fry will be held this week!
The summer is heating up, we're half way through, and there is lots of fun left to be had! Do you have a special talent? Come show off your singing, dancing, comedy, or any other skills at our annual Talent Show.
Join us in a fun-filled week of games that can't be done alone. Make new friends. Challenge yourself and enjoy the accomplishments of a team environment.
One week of water games won't cut it. Join us for round two as we build our own cardboard boats and steer our self-constructed watercraft to open water!
Let the games begin – Again! An entire week filled with team sports. Challenge yourself, challenge others and score one for team Camp Alexander.
Wrap up the summer with your Camp favorites.
Take advantage of the early bird specials by registering before May 15th!
$90 per week for campers if registered before May 15th (Early Bird Special).
$120 per week for campers if registered May 15th or later.


Which would be the best size to get?
I was just looking for this info. I have one coming next week and I'm not sure which to get. Or would the Ristretto work for this.
I finally figured this out. I'm posting this in case someone else has the Asus. It looks like it's the size 5 brain cell that's the best fit. It's not listed but the similar sized Samsung NC10 is.
Now I can place my order for the side effect.
Would this thing qualify as Checkpoint friendly?
Sorry for the late response. mrae is spot-on: the Size 5, with interior dimensions of 11.3 × 9.3 × 1.4, is the best fit we offer for the EEE PC 1000he. With the exception of the 11.3 of the Brain Cell vs. the 10.5 of the laptop, it's a great fit - and because the fit is so good in the other two dimensions, the EEE shouldn't move around inside of the Size 5.
The Brain Cell is not checkpoint friendly because it has an exterior pocket and metal snaps on the front. However, many customers have used it as a "checkpoint friendly" bag without issue - but again, it does not meet the TSA requirements.


Key Words: Fertilizer rates, Genotype, Transplanted aman rice, Yield, Gross return.
Int. J. Agr. Agri. Res. 9(1), 47-54, July 2016.
The field experiment was conducted at the BRAC Agricultural Research and Development Centre, Gazipur from June to October, 2012 to investigate the effect of different fertilizer rates on the growth and yield of transplanted aman rice genotypes. The experiment was carried out in split-plot design with three replications accommodating the genotypes (G1= CNI 9012, G2= GSR IRRI I 2, G3= BINA dhan7 and G4= Aloran) in sub plot and different fertilizer rates (F1= No fertilizer, F2= Half of recommendation rate, F3= Two third of recommendation rate and F4= Recommendation rate) in the main plot. Crop growth duration, plant height, tillers no. hill-1, panicle no. hill-1, panicle length, total grains panicle-1, fertility%, 1000 grain-wt. and grain yield were compared for different treatments. Genotypes showed significant results for all the traits except grain yield but fertilizer and their interaction exhibit insignificant results for most of the characters studied. The maximum mean for producing grain yield was observed in F2 (5.0 t ha-1) but the minimum in F1 (4.6 t ha-1). Therefore, application of no fertilizer had negative impact on grain yield. Apparently, the highest value for grain yield (5.2 t ha-1) was found in the combinations of G1 × F3, G3 × F2 and G4 × F2 as well as the lowest (4.2 t ha-1) from G1 × F1 combination. Maximum gross return (62500 Tk. ha-1) and gross margin (58255 Tk. ha-1) with F2 suggested that this fertilizer rate was best for profitable cultivation on tested transplanted aman rice genotypes.


text:
build:
	docker build -t rootbuilder .

release:
	docker tag rootbuilder quay.io/baseboxorg/rootbuilder
	docker push quay.io/baseboxorg/rootbuilder


text:
The 1962 Ford Falcon
One of the key factors in Ford's ability to produce cars in Australia was its establishment of a modern assembly plant.
Before the Ford Falcon's quality gremlins made themselves known, there were remarkably few gripes from motor critics. They disliked the ultra-light, twirly steering with its five turns lock-to-lock, the brakes had only marginal capabilities, handling was average at best, and the spare wheel sited on the trunk floor robbed the already shallow compartment of valuable space.
Some assembly glitches were noted, but they were not considered detrimental, an attitude fairly typical of the "forgiveness" practiced by the media in those days. "The Falcon is fundamentally a sound, beautiful-looking car, priced right for the big section of our market," Bryan Hanrahan wrote in the November 1960 issue of Modern Motor after testing a Deluxe sedan fitted with the two-speed automatic transmission.
Hanrahan managed to coax the car up to 90.2 mph and took 12 seconds for the 0-to-50-mph sprint. By comparison, the FB Holden Special sedan the magazine tested earlier the same year achieved a top speed of 82.3 mph and needed 13 seconds for the O-to-50 run.
In August 1962, Ford released an upgraded XL series under the marketing banner "The New Falcons: Trim, Taut, Terrific!" Coming barely one month after the release of the EJ Holden range, the Falcon face-lift was a minor makeover with a new convex grille, new taillights (they were still round, just sharper looking), and the much-lauded Thunderbird roofline. Roof pressing apart, all exterior panels were carryovers from the XK.
Under the skin, some strengthening had taken place as a result of local endurance testing, though. On the options list was a larger capacity 170-cid Pursuit engine with 101 bhp at 4400 rpm. The 3.5-inch bore remained the same as in the 144-cube engine, but the stroke was lengthened by .44 of an inch to 2.94. (In a four-way test of automatic transmission-equipped cars published in its November 1963 issue, Wheels was able to hit 50 mph with the enlarged Falcon engine in just 10.4 seconds.)
To broaden the range, increase its appeal, and compete with Holden's luxury Premier, Ford introduced "the bustling new Futura," a "limited edition...designed to bring you a whole new world of luxury, sizzle and fun -- and to set your pulses pounding." The Futura was easily distinguished from its siblings by the deluxe chromed full wheel trims, chromed edge top the fake hood air scoop, fender-top ornaments, a bold chromed side spear, Futura emblems on the C-pillars, and Futura spelled out in block letters across the rear.
Inside, there were individual front seats -- not quite bucket seats as we expect them today -- separated by a raised console with a brightly chromed lid, full carpeting, and a padded top for the dashboard. The interior was done in garish red.
Sales of the Ford Falcon pushed the company to the limit. Read about how the success of the 1962 Ford Falcon affected the future of the company on the next page.


Filta-Bac cream is a protective anti-bacterial cream and sun filter for animals. Use on damaged areas of the body to act as a protective second skin. Filta-Bac is suitable for most animals, it is particularly for use on horses, dogs and exotic animals. It can be used to treat cats however veterinary is recommended.
Filta-Bac cream is a protective anti-bacterial cream and sun filter for animals.
Use on damaged areas of the body to act as a protective second skin.
Filta-Bac is suitable for most animals, it is particularly for use on horses, dogs and exotic animals. It can be used to treat cats however veterinary advice is recommended as cats are sensitive to this product.
Covers wounds and acts like a second skin, keeping out dirt and keeping in moisture. Allows wounds to breathe, letting them heal naturally.
Correcting damaged or irritated skin.
Antibacterial properties prevent colonisation of the wound, thus less infections.
Filta-Bac has a SPF of 29-33.
Available in a 120g tube or 500g tub.


text:
package net.sf.jabref.logic.layout;

import java.io.File;
import java.io.IOException;
import java.io.StringReader;
import java.util.Collection;
import java.util.Collections;

import net.sf.jabref.logic.importer.ParserResult;
import net.sf.jabref.logic.importer.fileformat.BibtexParser;
import net.sf.jabref.logic.journals.JournalAbbreviationLoader;
import net.sf.jabref.logic.journals.JournalAbbreviationPreferences;
import net.sf.jabref.logic.layout.format.FileLinkPreferences;
import net.sf.jabref.logic.layout.format.NameFormatterPreferences;
import net.sf.jabref.model.entry.BibEntry;
import net.sf.jabref.preferences.JabRefPreferences;

import org.junit.Assert;
import org.junit.Before;
import org.junit.Test;

import static org.mockito.Mockito.mock;

public class LayoutTest {
    private LayoutFormatterPreferences prefs;

    /**
     * Initialize Preferences.
     */
    @Before
    public void setUp() {
        prefs = JabRefPreferences.getInstance().getLayoutFormatterPreferences(mock(JournalAbbreviationLoader.class));
    }

    /**
     * Return Test data.
     */
    public String t1BibtexString() {
        return "@article{canh05,\n" + "  author = {This\nis\na\ntext},\n"
                + "  title = {Effective work practices for floss development: A model and propositions},\n"
                + "  booktitle = {Hawaii International Conference On System Sciences (HICSS)},\n" + "  year = {2005},\n"
                + "  owner = {oezbek},\n" + "  timestamp = {2006.05.29},\n"
                + "  url = {http://james.howison.name/publications.html},\n" + "  abstract = {\\~{n} \\~n "
                + "\\'i \\i \\i}\n" + "}\n";
    }

    public static BibEntry bibtexString2BibtexEntry(String s) throws IOException {
        ParserResult result = new BibtexParser(JabRefPreferences.getInstance().getImportFormatPreferences()).parse(new StringReader(s));
        Collection<BibEntry> c = result.getDatabase().getEntries();
        Assert.assertEquals(1, c.size());
        return c.iterator().next();
    }

    public String layout(String layoutFile, String entry) throws IOException {
        BibEntry be = LayoutTest.bibtexString2BibtexEntry(entry);
        StringReader sr = new StringReader(layoutFile.replace("__NEWLINE__", "\n"));
        Layout layout = new LayoutHelper(sr, prefs)
                        .getLayoutFromText();

        return layout.doLayout(be, null);
    }

    @Test
    public void testLayoutBibtextype() throws IOException {
        Assert.assertEquals("Unknown", layout("\\bibtextype", "@unknown{bla, author={This\nis\na\ntext}}"));
        Assert.assertEquals("Article", layout("\\bibtextype", "@article{bla, author={This\nis\na\ntext}}"));
        Assert.assertEquals("Misc", layout("\\bibtextype", "@misc{bla, author={This\nis\na\ntext}}"));
    }

    @Test
    public void testHTMLChar() throws IOException {
        String layoutText = layout("\\begin{author}\\format[HTMLChars]{\\author}\\end{author} ",
                "@other{bla, author={This\nis\na\ntext}}");

        Assert.assertEquals("This is a text ", layoutText);

        layoutText = layout("\\begin{author}\\format[HTMLChars]{\\author}\\end{author}",
                "@other{bla, author={This\nis\na\ntext}}");

        Assert.assertEquals("This is a text", layoutText);
    }

    @Test
    public void testPluginLoading() throws IOException {
        String layoutText = layout("\\begin{author}\\format[NameFormatter]{\\author}\\end{author}",
                "@other{bla, author={Joe Doe and Jane, Moon}}");

        Assert.assertEquals("Joe Doe, Moon Jane", layoutText);
    }

    @Test
    public void testHTMLCharDoubleLineBreak() throws IOException {
        String layoutText = layout("\\begin{author}\\format[HTMLChars]{\\author}\\end{author} ",
                "@other{bla, author={This\nis\na\n\ntext}}");

        Assert.assertEquals("This is a text ", layoutText);
    }

    /**
     * [ 1495181 ] Dotless i and tilde not handled in preview
     *
     * @throws Exception
     */
    @Test
    public void testLayout() throws IOException {

        String layoutText = layout(
                "\\begin{abstract}

Abstract:  \\format[HTMLChars]{\\abstract}\\end{abstract}",
                t1BibtexString());

        Assert.assertEquals(
                "

Abstract:  ñ ñ í ı ı",
                layoutText);
    }

    @Test
    // Test for http://discourse.jabref.org/t/the-wrapfilelinks-formatter/172 (the example in the help files)
    public void testWrapFileLinksLayout() throws IOException {
        prefs = new LayoutFormatterPreferences(mock(NameFormatterPreferences.class),
                mock(JournalAbbreviationPreferences.class),
                new FileLinkPreferences(Collections.emptyList(), Collections.singletonList("src/test/resources/pdfs/")),
                mock(JournalAbbreviationLoader.class));
        String layoutText = layout("\\begin{file}\\format[WrapFileLinks(\\i. \\d (\\p))]{\\file}\\end{file}",
                "@other{bla, file={Test file:encrypted.pdf:PDF}}");
        Assert.assertEquals(
                "1. Test file (" + new File("src/test/resources/pdfs/encrypted.pdf").getCanonicalPath() + ")",
                layoutText);
    }
}


Day to day we live in a world where we are bombarded by technology. If it's not emails, its 'likes', tweets, snaps, shares, PM's, notifications, requests to link etc etc etc. It's all noise. Noise that we are all familiar with and by being used to it, we sometimes don't see it or even ignore it.
That's where promotional products come into their own. What I mean by this, is pens, mugs, key rings, rules, torches, coasters, mouse mats…. the list is endless. These items are loved by many, borrowed from others, picked up at an exhibition and given by companies. Why do we love this stuff? Cause it's free. And everyone loves something for free.
That free pen, as long as it's good quality, may well be your writing instrument of choice for years and the cost to you is zero. You didn't have to buy on Amazon or go to Smiths, it was given by someone who wanted to make YOU aware of THEIR brand. Every day you use and love that pen, you'll see them, subtly reminding you of who they are. You know what, when you are in the market for that service, you'll probably think of them! It's VERY powerful.
Why do people like Coca Cola advertise? we know who they are! They hope that when you are thirsty, there's will be the brand you think of and it's the same with that branded office mug, it's always there silently selling that company. It may well be used 6 times a day, that's 6 messages being burnt into your subconscious mind. Keep that mug a year… I'll leave you to work out the maths.
People often worry that the pen or items will 'go into the wrong hands', 'be given away', 'the kids will take it'. You know what, that's awesome. That's perfectly fine, as the product is still out there. The key is HOW it's been given, HOW it was followed up and what sort of quality it is.
The question of ROI is often asked of me. What will be my return on investment? I have no idea! But I have stopped counting the times I've spoken to clients who have kept my water bottles and a year later have ordered from me because the time is right!
In the next few years, the ability to cold call or cold email will be taken from us, and we need to get smart and be prepared! Lumpy mail sent to your prospect is the next big thing. An item, sent in a jiffy bag, too big for a letter box that has to be received I the hand. THAT'S THE KEY. A parcel that makes the recipient think 'what on earth is this'?'. You follow that up, and your hundreds of times more likely to be remembered than a SPAM email that's in the Junk folder.
You can't unsubscribe from direct mail, you can't put it in the junk folder, you can't block it. Sending promo items in the post is the new marketing goldmine and you need to get onboard. Promotional products are great; they make people smile, they show you are happy to invest in your prospects and existing clients. it's dead powerful and dead personal.
This entry was posted in Business Development, Lead generation, Marketing, Strategy on 20th July 2017 by Yafit Davis.
Conjure up an internal image of a PR agency, and you'd be forgiven for thinking sparkly office, blue sky thinking, expensive champagne lunches and power point presentations.
It's not something which many small businesses have the inclination or the budget for, especially when their marketing funds are limited and money can be better spent elsewhere.
But the problem is one of perception – the view that PR is the sole preserve of big businesses who can afford big agency fees. It's a self-perpetuating myth which drives the opportunity away from smaller companies. I'll never forget hearing Deborah Meaden on Dragon's Den saying that unless you have £30k to spend on PR it's a waste of time. QED! I have clients who have spent a tiny fraction of that amount and achieved new sales as a direct result of the work I have done for them.
Of course, the end result of an effective PR campaign is media coverage – media these days meaning online press as well as traditional print coverage. Work backwards, and how do you get the press to publish something about your business? The answer, by giving the journalist something which they and their readers are likely to find interesting. And what's that? It comes down to storytelling.
Yes, people like reading stories about big business, about who's doing what, about new developments and about scandals and controversy. But they also like stories about the little guy – inspiring success stories, human interest pieces about people doing amazing things, articles about those flexible and nimble companies at the cutting edge of innovation, about the start-ups which are going to become the next Renishaw, Facebook or Virgin.
Have you won any big contracts recently?
Won an award? Received any funding?
Achieved growth? Taken on an apprentice?
Merged with or acquired another business?
If the answer to any of those questions is yes, then you've got a story to tell! Either in your local business pages, or a trade title, or a small business publication. Apart from anything else, most journalists are swamped by the same old stories from the same old businesses put out by the same old agencies. They love to hear from new businesses! It makes their publications fresh and makes their readers – most of whom are involved in small businesses anyway – feel that they're relevant.
An extra tip – supply good photos! I'm not talking a lo-res job from your iPhone (although it may do for online use). If you're putting a proper press release together, spend a few quid and get a professional to take a good quality picture which helps tell the story. Or get a mate to do it for free. Either way, the combination of a good story and a great photo is something that journalists, who are a) pushed for time and b) inundated with the PR fluff which puts such a strain on the journalist/PR relationship, will be extremely grateful for.
The benefits? More than you can count. Even the much-derided local papers (tomorrow morning's fish and chip wrappers) not only have huge readerships in print, creating powerful word-of-mouth publicity, but these days have massive followings online too, greatly helping with your search engine presence. And more often than not they'll also have a business website which they're affiliated with, as well as sending out a digital daily newsletter. So you're getting much more than one hit, and a highly credible one too when compared with a paid-for advert.
What else? It's something you can boast about on your website and in your other marketing collateral, giving you a feather in your cap against your competitors. And it will be well-written copy which you can use for newsletters, blogs and so on, too.
PR's not a marketing panacea but it should definitely be considered as an important part of your marketing mix. And although online marketing and social media are changing the game there's still something about a dedicated news website which adds credibility to your cause and which also suggests a sense of permanence and gravitas which can be a problem with social media and its constant self-reinvention.
Ultimately, it all comes down to the story you have to tell. Don't put yourself down! Be proud of what you achieve in business and have the confidence to tell people about it. It's a competitive world out there and, if you want to get publicity, it's not a place for the shrinking violets.


text:
Schlossfestspiele
3rd International Music Festival
Hohenpriessnitz, Germany
Konzerte Masterclasses
Due to the measurements taken by the German government, The 3rd International Music Festival "Schlossfestspiele" will be postponed. The health of our students and employees is the most important and that is why we accept this measure to prevent the spread of the coronavirus.
We are currently working on offering live lessons online with our professors! Stay tuned for more substantial news very soon!
Alexander Vinnitsky
Price and application details
Alexander Vinnitsky has been granted the high titles of the Honorary Artist of The USSR (1989), People's Artist of Russia (2006), Laureate of International competitions, Professor of The Moscow Tchaikovsky Conservatoire, Professor of Sibelius Music Academy in Helsinki, Finland, Professor o the Turku Music Academy, "Suola di Musica di Fiesole" (Italy), a student and follower of the legendary violinist, David Oistrakh.
In 2007 he was awarded one of the most prestigious awards of Finnish musical association "Pro Musica Saatio" for contribution to the development of musical culture.
More than 40 musicians from different countries are studying in his tudio in Moscow Tchaikovsky Conservatory, among them are leaureates of international violin contests: Queen Elisabeth Musical Competition in Brussels, International Tchaikovsky Competition in Moscow, Russia, International Jean Sibelius Violin Competition in Helsinki, Finland, Paganini Competition in Genoa, italy, etc.
Alexander Vinntsky has served on the juries of many international competitions and was a Jury Chairman at all 3 David Oistrakh Violin competitions.
Alexander Vinnitsky performs at solo concerts, has performed in more than 50 countries all over the world and known for the wide-ranging repertoire, extanding from Baroque, Viennese Classics to the romantism and contemporary composers.
The 2017-2018 season in Moscow Conservatory Hall was highlighted by significant musical project by Alexander Vinnitsky: "All Violin Compositions by W.A. Mozart for violin with Orchestra" with his cadenzas which were published by prestigious Russian publishing house "Music".
Passive listener - 200 €
Day pass - 50 €
Price options 5 lessons: 749 € 4 lessons + 1 lesson with orchestra: 899 €
Included in price:
Individual lessons (open to the public)
Optional lesson with Taurida State Symphony Orchestra
An accompanist
Selected best participants will perform with Taurida State Symphony Orchestra
Accommodation is offered by Schloss Hohenpriessnitz for 29 € (hostel-type double room) and 49 € (hotel-type single room with private bathroom). There's a shared kitchen available for use.
© 2022 Schlossfestspiele Hohenpriessnitz


Well.. just recently upgraded the whole wordpress from 2.8 to 2.8.3 ..
Hmm.. the changelog doesn't tell much..
However.. there are some other thing in plugin that might also need upgrading..
such as wp-syntax , wp-useronline..
this time.. in cPanel provided by the DataKL.com .. just worked fine..
via the Admin Dashboard of wordpress interface..
can just click upgrade automatically..
and if everything goes fine.. it will do all the necessary thing accordingly..
except some part.. due to customization or hard-coding thing might have to re-do again..
one of the thing that I really want to do after upgrading the wordpress plugin is..
to minify all the java script file all over again.. it reduce the 'time to wait' drastically ..
and also this type of message ..
Always remember the birthday of the girl..
and also ALL the anniversary  date that you had together ..
" . $header . "
and somemore is .. to edit the css file to reflect something like this.
..that's all for this one..
And now your wp-syntax plugin would then able to show the header with just ..
then..extract the content.. put all the php script under a folder "jsmart"
changed the permission for jsmart/cache to be 777 ..
and all the php file as 755 ..
take a peek at the config.php in there..
mine just look something like this..
edited the jsmart/index.php to become as such..
edited the .htaccess to become as such..
p/s : i will be out-of-KL for a week.. starting tomorrow morning..
dunno if the internet access is available on-the-go..