\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2015 (2015), No. 311, pp. 1--13.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2015 Texas State University.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2015/311\hfil Quenching behavior of semilinear heat equations]
{Quenching behavior of semilinear heat equations with singular boundary conditions}

\author[B. Selcuk, N. Ozalp \hfil EJDE-2015/311\hfilneg]
{Burhan Selcuk, Nuri Ozalp}

\address{Burhan Selcuk \newline
Department of Computer Engineering,
Karabuk University, Bali klarkayasi  Mevkii 78050, Turkey}
\email{bselcuk@karabuk.edu.tr, burhanselcuk44@gmail.com}

\address{Nuri Ozalp \newline
Department of Mathematics, Ankara University, 
Besevler 06100, Turkey}
\email{nozalp@science.ankara.edu.tr}

\thanks{Submitted October 16, 2015. Published December 21, 2015.}
\subjclass[2010]{35K05, 35K15, 35B50}
\keywords{Heat equation; singular boundary condition; quenching; 
\hfill\break\indent maximum principle; monotone iteration}

\begin{abstract}
 In this article, we study the quenching behavior of solution to the semilinear
 heat equation
 $$
 v_t=v_{xx}+f(v),
 $$
 with  $f(v)=-v^{-r}$ or $(1-v)^{-r}$ and
 $$
 v_x(0,t)=v^{-p}(0,t), \quad v_x(a,t) =(1-v(a,t))^{-q}.
 $$
 For this, we utilize the quenching problem $u_t=u_{xx}$
 with $u_x(0,t)=u^{-p}(0,t)$, $u_x(a,t)=(1-u(a,t))^{-q}$.
 In the second problem, if $u_0$ is an upper solution (a lower solution)
 then we show that quenching occurs in a finite time, the
 only quenching point is $x=0$ ($x=a$)  and $u_t$ blows up at quenching
 time. Further, we obtain a local solution by using positive steady
 state. In the first problem, we first obtain a local solution by using
 monotone iterations. Finally, for $f(v)=-v^{-r}$ ($(1-v)^{-r}$), if $v_0$ is an
 upper solution (a lower solution) then we show that quenching occurs in a
 finite time, the only quenching point is $x=0$ ($x=a$) and $v_t$ blows
 up at quenching time.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{corollary}[theorem]{Corollary}
\allowdisplaybreaks



\section{Introduction}

In this article, we study the quenching behavior of solutions to the
semilinear heat equation with singular boundary conditions:
\begin{equation} \label{e1}
\begin{gathered}
v_t=v_{xx}+f(v),\quad 0<x<a,\; 0<t<T, \\
v_x(0,t)=v^{-p}(0,t),\quad  v_x(a,t) =(1-v(a,t))^{-q},\quad 0<t<T, \\
v(x,0)=v_0(x),\quad  0\leq x\leq a,
\end{gathered}
\end{equation}
where $p,q,r>0,T\leq \infty$, $f(u)=-v^{-r}$ or
 $f(u)=(1-v)^{-r}$. The initial function $v_0:[0,a]\to (0,1)$ satisfies the
compatibility conditions
\begin{equation*}
v_0'(0)=v_0^{-p}(0),\quad v_0'(a)=(1-v_0(a))^{-q}.
\end{equation*}
Our main purpose is to examine the quenching behavior of the solutions of
the problem \eqref{e1} having two singular heat sources. A solution $v(x,t)$ of
the problem \eqref{e1} is said to quench if there exists a finite time $T$ such
that
\begin{equation*}
\lim_{t\to T^{-}} \max \{v(x,t):0\leq x\leq a\}
\to 1\quad \text{or}\quad
\lim_{t\to T^{-}} \min \{v(x,t):0\leq x\leq a\} \to 0.
\end{equation*}
For the rest of this article, we denote the quenching time of  \eqref{e1} with $T$.

To study Problem \eqref{e1}, we utilize the following problem
\begin{equation} \label{e2}
\begin{gathered}
u_t=u_{xx},\quad 0<x<a,\; 0<t<T, \\
u_x(0,t)=u^{-p}(0,t),\quad  u_x(a,t)
=(1-u(a,t))^{-q},\quad  0<t<T, \\
u(x,0)=u_0(x),\quad  0\leq x\leq a,
\end{gathered}
\end{equation}
where $p,q$ are positive constants and $T\leq \infty $. The initial
function $u_0:[0,a]\to (0,1)$ satisfies the compatibility
conditions
\begin{equation*}
u_0'(0)=u_0^{-p}(0),\quad  u_0'(a)=(1-u_0(a))^{-q}.
\end{equation*}
Since 1975, quenching problems with various boundary conditions have been
studied extensively. Recently, the quenching problems which have been
studied with two nonlinear heat sources can be seen in 
\cite{c2,o1,o2,s1,z1}.
 For example, Chan and Yuen \cite{c2} considered the
problem
\begin{gather*}
u_t=u_{xx},\quad  \text{in }  \Omega , \\
u_x(0,t)=(1-u(0,t))^{-p},\quad  u_x(a,t) =(1-u(a,t))^{-q},\quad 0<t<T, \\
u(x,0)=u_0(x),\quad 0\leq u_0(x)<1,\quad \text{in }\bar{D},
\end{gather*}
where $a,p,q>0$, $T\leq \infty$, $D=(0,a)$, $\Omega =D\times (0,T)$. They showed
that $x=a$ is the unique quenching point in finite time if $u_0$ is a
lower solution, and $u_t$ blows up at quenching time. Further, they
obtained criteria for nonquenching and quenching by using the positive
steady states. Selcuk and Ozalp \cite{s1} considered the problem
\begin{gather*}
u_t=u_{xx}+(1-u)^{-p},\quad 0<x<1,\; 0<t<T, \\
u_x(0,t)=0,\quad u_x(1,t)=-u^{-q}(1,t),\quad  0<t<T, \\
u(x,0)=u_0(x),\quad 0<u_0(x)<1,\; 0\leq x\leq 1.
\end{gather*}
They showed that $x=0$ is the quenching point in finite time, 
$\lim_{t\to T^{-}}u(0,t)\to 1$, if $u(x,0)$ satisfies
 $u_{xx}(x,0)+(1-u(x,0))^{-p}\geq 0$ and 
$u_x(x,0)\leq 0$.  Further they showed that $u_t$ blows up at
quenching time. Furthermore, they obtained a quenching rate and a lower
bound for the quenching time.

Problems \eqref{e1} and  \eqref{e2} have two \ type of singularity
terms $(1-u)^{-q}$ and $u^{-p}$ on the boundaries. We discuss these two
situations in this article, 
$\lim_{t\to T^{-}}u(0,t)\to 0$ or $\lim_{t\to T^{-}}u(a,t)\to 1$. 
This article is organized as follows. 
In Section 2, we consider the problem \eqref{e2}. Firstly,
if $u_0$ is an upper solution (a lower solution) then we show that
quenching occurs in a finite time, the only quenching point is $x=0$ ($x=a$)
and $u_t$ blows up at quenching time. Further, we obtain a local
existence result by using positive steady state. 
In Section 3, we consider  problem \eqref{e1}. Firstly, we obtain 
local existence of  \eqref{e1} by using monotone iterations.
 Further, for $f(v)=-v^{-r}$ ($(1-v)^{-r}$), if $v_0$ is an upper solution 
(a lower solution) then we show that quenching
occurs in a finite time, the only quenching point is $x=0$ ($x=a$) and 
$v_t$ blows up at quenching time.

\section{Problem \eqref{e2}}

\subsection{Quenching on the boundary}

The proofs of the following lemma and theorem are analogous to those
by Chan and Yuen \cite[Section 2]{c2}.

\begin{definition} \label{def1} \rm
$\mu$ is called a lower solution of \eqref{e2} if 
$\mu \in C([0,a]\times [ 0,T))\cap C^{2,1}((0,a)\times (0,T))$
satisfies the following conditions:
\begin{gather*}
\mu _t\leq \mu _{xx},\quad  0<x<a,\; 0<t<T, \\
\mu _x(0,t)\geq \mu ^{-p}(0,t),\quad
\mu _x(a,t) \leq (1-\mu (a,t))^{-q},\quad <t<T, \\
\mu (x,0)\leq u_0(x),\quad 0\leq x\leq a.
\end{gather*}
It is an upper solution when the inequalities are reversed.
\end{definition}

\begin{theorem} \label{thm1}
Let $u(x,t,u_0)$ and$\ h(x,t,h_0)$ be
solutions of problem \eqref{e2} with data given by $u_0(x)$ and $h_0(x)$, 
respectively. If $u_0\leq h_0<1$, then $u(x,t,u_0)\leq h(x,t,h_0)$
on $[0,a]\times [ 0,T)$.
\end{theorem}

\begin{proof}
 For any $\tau <T$, let $w$ be a solution of the problem
\begin{gather*}
w_{xx}-w+w_t = 0\quad  \text{in }(0,a)\times (0,\tau ), \\
w(x,\tau ) = g(x)\quad  \text{on }[0,a], \\
w_x(0,t) = r(t)w(0,t),\quad w_x(a,t)=s(t)w(a,t),\quad 0<t<\tau ,
\end{gather*}
where $g\in C^{2}(\overline{D})$ has compact support in $D$,
$0\leq g\leq 1$, and $r$ and $s$ are smooth functions to be determined. 
By Lieberman \cite{l1}, $w$ exists. By Andersen \cite{a1}, there exists a constant 
$k$ (depending on the length of the interval $D$) such that $0\leq w\leq k$.
Now,
\begin{align*}
&\int_0^{a}[(u(x,\tau )-h(x,\tau ))w(x,\tau )-(u_0(x)-h_0(x))w(x,0)]dx
\\
&= \int_0^{\tau }\int_0^{a}\frac{\partial }{\partial \sigma }
[(u(x,\sigma )-h(x,\sigma ))w(x,\sigma )]\,dx\,d\sigma \\
&= \int_0^{\tau }\int_0^{a}[w(x,\sigma )\frac{\partial }{\partial \sigma
}(u(x,\sigma )-h(x,\sigma ))+(u(x,\sigma )-h(x,\sigma ))\frac{\partial }{
\partial \sigma }w(x,\sigma )]\,dx\,d\sigma \\
&= \int_0^{\tau }\int_0^{a}\Big[ w(x,\sigma )\frac{\partial ^{2}}{
\partial x^{2}}(u(x,\sigma )-h(x,\sigma ))+(u(x,\sigma )-h(x,\sigma ))\frac{
\partial }{\partial \sigma }w(x,\sigma )\Big] \,dx\,d\sigma \\
&= \int_0^{\tau }\{w(a,\sigma )[ (1-u(a,\sigma ))^{-q}-(1-h(a,\sigma
))^{-q}] -w(0,\sigma )[ u^{-p}(0,\sigma )-h^{-p}(0,\sigma )]
\\
&\quad -s(\sigma )[ u(a,\sigma )-h(a,\sigma )] w(a,\sigma )+r(\sigma )
 [ u(0,\sigma )-h(0,\sigma )] w(0,\sigma )\}d\sigma \\
&\quad +\int_0^{\tau }\int_0^{a}(u(x,\sigma )-h(x,\sigma ))(w_{\sigma
}(x,\sigma )+w_{xx}(x,\sigma ))\,dx\,d\sigma .
\end{align*}
Thus,
\begin{align*}
&\int_0^{a}[(u(x,\tau )-h(x,\tau ))g(x)-(u_0(x)-h_0(x))w(x,0)]dx \\
&= \int_0^{\tau }\{w(a,\sigma )\left[ (1-u(a,\sigma ))^{-q}-(1-h(a,\sigma
))^{-q}-s(\sigma )\left[ u(a,\sigma )-h(a,\sigma )\right] \right] \\
&\quad -w(0,\sigma )\left[ u^{-p}(0,\sigma )-h^{-p}(0,\sigma )-r(\sigma )\left[
u(0,\sigma )-h(0,\sigma )\right] \right] \}d\sigma \\
&\quad +\int_0^{\tau }\int_0^{a}(u(x,\sigma )-h(x,\sigma ))w(x,\sigma
)\,dx\,d\sigma .
\end{align*}
Let $r(\sigma )$ and $s(\sigma )$ be given by
\begin{gather*}
r(\sigma )(u(0,\sigma )-h(0,\sigma ))
= u^{-p}(0,\sigma)-h^{-p}(0,\sigma ), \\
s(\sigma )(u(a,\sigma )-h(a,\sigma ))
=(1-u(a,\sigma ))^{-q}-(1-h(a,\sigma ))^{-q}.
\end{gather*}
Since $u_0\leq h_0$ and $w(x,0)\geq 0$, we have
\begin{equation*}
\int_0^{a}(u(x,\tau )-h(x,\tau ))g(x)dx
\leq \int_0^{\tau}\int_0^{a}(u(x,\sigma )-h(x,\sigma ))w(x,\sigma )\,dx\,d\sigma .
\end{equation*}
Let
\begin{equation*}
(u(x,\sigma )-h(x,\sigma ))^{+}=\max \{ 0,u(x,\sigma )-h(x,\sigma )\} .
\end{equation*}
From, $0\leq w\leq k$, we obtain
\begin{equation*}
\int_0^{a}(u(x,\tau )-h(x,\tau ))g(x)dx
\leq 
k\int_0^{\tau}\int_0^{a}(u(x,\sigma )-h(x,\sigma ))^{+}\,dx\,d\sigma .
\end{equation*}
Since $g\in C^{2}(\overline{D})$ has compact support in $D$ and 
$0\leq g\leq 1$, we have
\begin{equation*}
\int_0^{a}(u(x,\sigma )-h(x,\sigma ))^{+}dx
\leq k\int_0^{\tau}\int_0^{a}(u(x,\sigma )-h(x,\sigma ))^{+}\,dx\,d\sigma .
\end{equation*}
By the Gronwall inequality,
\begin{equation*}
\int_0^{a}(u(x,\sigma )-h(x,\sigma ))^{+}dx\leq 0,
\end{equation*}
which gives $u(x,\tau )\leq h(x,\tau )$ for any $\tau >0$. Thus, the
theorem is proved. 
\end{proof}

\begin{lemma} \label{lem1} \begin{itemize}
\item[(i)] If $u_{xx}(x,0)\geq 0$ in $(0,a)$, then we obtain $u_t>0$ in 
$(0,a)\times (0,T)$.

\item[(ii)] 
If $u_{xx}(x,0)\leq 0$ in $(0,a)$, then we obtain $u_t<0$ in$ (0,a)\times (0,T)$.
\end{itemize}
\end{lemma}

\begin{proof} 
(i) Since $u_{xx}(x,0)\geq 0$ in $(0,a)$, $u_0'(0)=u_0^{-p}(0)$,
$u_0'(a)=(1-u_0(a))^{-q}$, it follows that $u_0(x)$ is a lower solution
of the problem \eqref{e1} from Definition \ref{def1}. 
The strong maximum principle implies that
\begin{equation*}
u(x,t)\geq u_0(x)\quad  \text{in }(0,a)\times (0,T).
\end{equation*}
Let $h$ be a positive number less than $T$, and
\begin{equation*}
z(x,t)=u(x,t+h)-u(x,t).
\end{equation*}
Then
\begin{gather*}
z_t = z_{xx}\quad  \text{in }(0,a)\times (0,T-h), \\
z(x,0) \geq 0\quad  \text{on }[0,a], \\
z_x(0,t) = -p\xi ^{-p-1}(t)z(0,t),z_x(a,t)
=q(1-\eta (t))^{-q-1}z(a,t),\quad 0<t<T-h,
\end{gather*}
where $\xi (t)$  between $u(0,t+h)$ and $u(0,t)$,
and $\eta (t)$ lies between $u(a,t+h)$ and $u(a,t)$. A proof similar to that of
Theorem \ref{thm1} shows that $z(x,t)\geq 0$. As $h\to 0$, we have 
$u_t\geq 0$ on $[0,a]\times (0,T)$.

Let $H=u_t$ in $[0,a]\times (0,T)$. Since
\begin{equation*}
H_t-H_{xx}=0\  \text{in }(0,a)\times (0,T),
\end{equation*}
it follows from the strong maximum principle that $H=u_t>0$ in 
$(0,a)\times (0,T)$.
\smallskip

(ii) If $u_{xx}(x,0)\leq 0$ in $(0,a)$, then from the above proof
we have $u_t\leq 0$ on $[0,a]\times (0,T)$ and $u_t<0$ in
$(0,a)\times (0,T)$. 
The proof is complete.
\end{proof}

 Now we show that, if $u_{xx}(x,0)\leq 0$ in $(0,a)$, namely, if 
$u_0$ is an upper solution, then we have quenching point at $x=0$.


\begin{theorem} \label{thm2} 
If $u_0$ is an upper solution, then there
exist a finite time $T$, such that the solution $u$ of the problem \eqref{e2}
quenches at time $T$.
\end{theorem}

\begin{proof}
Assume that $u_0$ is an upper solution. Then
\begin{equation*}
\omega =-(1-u(a,0))^{-q}+u^{-p}(0,0)>0.
\end{equation*}
Introduce a mass function; $m(t)=\int_0^{a}u(x,t)dx$, $0<t<T$. Then
\begin{equation*}
m'(t)=(1-u(a,t))^{-q}-u^{-p}(0,t)\leq -\omega ,
\end{equation*}
by Lemma \ref{lem1} (ii). Thus, $m(t)\leq m(0)-\omega t$; and so 
$m(T_0)=0$ for some $T_0$, $(0<T\leq T_0)$  which
means $u$ quenches in a finite time.
\end{proof}

\begin{theorem} \label{thm3} 
If $u_0$ is an upper solution, then $x=0$
is the only quenching point.
\end{theorem}

\begin{proof}
Since $u_x(a,t)=(1-u(a,t))^{-q}>1$ and 
$u_{xx}=u_t<0$ in $(0,a)\times (0,T)$, then $u_x$ is a decreasing
function and so, $u_x(x,t)>1$ in $(0,a)\times (0,T)$. 
Let $\eta \in (0,a)$. Integrating this with respect to $x$ from $0$ 
to $\eta $, we have
\begin{equation*}
u(\eta ,t)>u(0,t)+\eta >0.
\end{equation*}
So $u$ does not quench in $(0,a]$. The proof is complete.
\end{proof}  

\begin{theorem} \label{thm4}
If $u_0$ is an upper solution, then $u_t$
blows up at quenching time.
\end{theorem}

\begin{proof} 
Suppose that $u_t$ is bounded on $[0,a]\times[ 0,T)$. Then, there
exists a positive constant $M$ such that $ u_t>-M$. We have $u_{xx}>-M$. 
Integrating this twice with respect to $x$ from $0$ to $x$, and then 
from $0$ to $a$, we have
\begin{equation*}
\frac{-a}{u^{p}(0,t)}>-\frac{Ma^{2}}{2}-u(a,t)+u(0,t).
\end{equation*}
As $t\to T^{-}$, the left-hand side tends to negative\ infinity,
while the right-hand side is finite. This contradiction shows that $u_t$
blows up somewhere. 
\end{proof}

 Now, we show that, if $u_{xx}(x,0)\geq 0$ in $(0,a)$, namely 
$u_0$ is a lower solution then we have quenching point at $x=a$.


\begin{theorem} \label{thm5} 
If $u_0$ is a lower solution, then there exist a finite time $T$, 
such that the solution $u$ of the problem \eqref{e2}
quenches at time $T$.
\end{theorem}

\begin{proof} 
Assume that $u_0$ is a lower solution. Then, we obtain
\begin{equation*}
\omega =(1-u(a,0))^{-q}-u^{-p}(0,0)>0.
\end{equation*}
Introduce a mass function $m(t)=\int_0^{a}(1-u(x,t))dx$, $0<t<T$. Then
\begin{equation*}
m'(t)=-(1-u(a,t))^{-q}+u^{-p}(0,t)\leq -\omega ,
\end{equation*}
by Lemma \ref{lem1} (i). Thus, $m(t)\leq m(0)-\omega t$; and so 
$m(T_0)=0$ for some $T_0$, $(0<T\leq T_0)$ which means $u$
quenches in a finite time. 
\end{proof}

\begin{theorem} \label{thm6} 
If $u_0$ is a lower solution, then $x=a$
is the only quenching point.
\end{theorem}

\begin{proof}
Since $u_x(0,t)=u^{-p}(0,t)>1$ and 
$u_{xx}=u_t>0$ in $(0,a)\times (0,T)$. Then, $u_x$ is an increasing
function and so, $u_x(x,t)>1$ in $(0,a)\times (0,T)$. 
Let $\varepsilon \in (0,a)$. Integrating this with respect to $x$ from 
$a-\varepsilon $ to $a$, we have
\begin{equation*}
u(a-\varepsilon ,t)<u(a,t)-\varepsilon <1-\varepsilon .
\end{equation*}
So $u$ does not quench in $[0,a)$. 
\end{proof}


\begin{theorem} \label{thm7} 
If $u_0$ is a lower solution, then $u_t$
blows up at quenching time.
\end{theorem}

\begin{proof} 
Suppose that $u_t$ is bounded on $[0,1]\times [ 0,T)$. Then, there exists 
a positive constant $M$ such that $u_t<M $. We have $u_{xx}<M$. 
Integrating this twice with respect to $x$ from $x$ to $a$, and then 
from $0$ to $a$, we have
\begin{equation*}
\frac{a}{(1-u(a,t))^{q}}<\frac{Ma^{2}}{2}+u(a,t)-u(0,t).
\end{equation*}
As $t\to T^{-}$, the left-hand side tends to infinity, while the
right-hand side is finite. This contradiction shows that $u_t$ blows up
somewhere. 
\end{proof}

\begin{corollary} \label{coro1} 
We have the following results via Theorems \ref{thm2}--\ref{thm7}:

(i) If $u_0$ is an upper solution for the problem \eqref{e2}, then
the solution $u$ of the problem \eqref{e2} quenches in a finite time, $x=0$ is
the only quenching point, and $u_t$ blows up\ at quenching time.

(ii) If $u_0$ is a lower solution for the problem \eqref{e2}, then
the solution $u$ of the problem \eqref{e2} quenches in a finite time, $x=a$ is
the only quenching point, and $u_t$ blows up at quenching time.
\end{corollary}

\subsection{Steady state}
The proof of the following lemma and theorem is analogous to that
by Chan and Yuen \cite[Section 3]{c2}. Let us consider the positive steady
states of Problem \eqref{e2},
\begin{equation} \label{e3}
U_{xx}=0,\ U_x(0)=U^{-p}(0),\quad  U_x(a) =(1-U(a))^{-q}.  
\end{equation}
We have $U=I+nx$, where
\begin{equation*}
n=I^{-p},\quad n=(1-I-na)^{-q}.
\end{equation*}
From these, we have
\begin{equation} \label{e4}
U=I+I^{-p}x, 
\end{equation}
where
\begin{equation*}
I^{-p}=(1-I-I^{-p}a)^{-q},
\end{equation*}
which gives
\begin{equation*}
a(I)=I^{p}(1-(I+I^{p/q})).
\end{equation*}
If we let $p=q$, then we obtain
\begin{equation} \label{e5}
a(I)=I^{p}(1-2I)=I^{p}-2I^{p+1}. 
\end{equation}
Now, $a'(I)=0$ implies
\begin{equation} \label{e6}
I=\frac{p}{2(p+1)}.  
\end{equation}
We note that $a(I)>0$ for $0<l<1/2$. Since $a(0)=0$ and $a(1/2)=0$ and 
$a(I)>0$, it follows from \eqref{e6} that $\max_{0<l<1/2}a(I)$. We denote
this value by $A$. From \eqref{e5},
\begin{equation*}
A=\frac{p^{p}}{2^{p}(p+1)^{p+1}}.
\end{equation*}

\begin{lemma} \label{lem2} 
If $p=q$, then there is a solution $u$ if and
only if $0<a\leq A$. Furthermore, if $0<a<A$, then there exist two positive
solutions; if $a=A$, then there exists exactly one positive solution.
\end{lemma}

\begin{proof} 
Since $a(0)=0=a(1/2)$ and $a(I)>0$ for $0<l<1/2$,
the graph of $a(I)$ is concave downwards with maximum attained at $A$. Thus
for $p=q$, the problem $(3)$ has a solution if and only if $0<a\leq A$. To
each $a\in (0,A)$, there are exactly two values of $I$. If $a=A$, then $I$
is given by \eqref{e6}. 
\end{proof}

\begin{theorem} \label{thm8} 
If $p=q$ and $a\in (0,A)$, then $u$ exists
globally, provided $u_0\leq U(0)$.
\end{theorem}

\begin{proof} 
By Theorem \ref{thm1}, $u\leq U$. Hence $u$ exists
globally. 
\end{proof}

\section{Problem \eqref{e1}}

\subsection{Local solution}
It is well known that one of the most effective methods for obtaining
existence and uniqueness  of the solution of parabolic equations with
initial conditions is monotone iterative techniques
 (for details see \cite{c3,p1}). 
For applications of monotone iterative techniques in quenching
problem for a parabolic equation (see \cite{c1}).

Let $C^{m}(Q)$, $C^{\alpha }(Q)$ be the respective spaces of $m$-times
differentiable and H\"{o}lder continuous functions in $Q$ with exponent 
$\alpha \in (0,1)$, where $Q$ is any domain. Denote by 
$C^{2,1}([0,a]\times [ 0,T))$ the set of functions that are twice
continuously differentiable in $x$ and continuously differentiable in
 $t$ for $(x,t)\in [ 0,a]\times [ 0,T)$. It assumed that initial
function $u_0(x)$ is  in $C^{2+\alpha }$.


\begin{definition} \label{def2}
A function $\widetilde{u}$ is called an upper
solution of \eqref{e1}, if 
$\widetilde{u}\in C([0,a]\times [0,T))\cap C^{2,1}((0,a)\times (0,T))$ and 
$\widetilde{u}$ satisfies the following conditions:
\begin{gather*}
\widetilde{u}_t-\widetilde{u}_{xx}\geq f(\widetilde{u}),\quad 0<x<a,\; 0<t<T,
\\
\widetilde{u}_x(0,t)\leq \widetilde{u}^{-p}(0,t),\quad
 \widetilde{u}_x(a,t)\geq (1-\widetilde{u}(a,t))^{-q},\quad 0<t<T, \\
\widetilde{u}(x,0)\geq u_0(x),\quad  0\leq x\leq a\,.
\end{gather*}
A function $\widehat{u}$ is a lower solution of  \eqref{e1}, if 
$\widehat{u}\in C([0,a]\times [ 0,T))\cap C^{2,1}((0,a)\times (0,T))$,
satisfies the reversing inequalities.
\end{definition}

\begin{lemma} \label{lem3}
Let $\widetilde{u}$ and $\widehat{u}$ be a
positive upper solution and a nonnegative lower solution of \eqref{e1} in 
$[0,a]\times [ 0,T)$, respectively. Then, we obtain the
following results:
\begin{itemize}
\item[(a)] $\widetilde{u}\geq \widehat{u}$ in $[0,a]\times [ 0,T)$,
\item[(b)] if $u^{\ast }$ is a solution, then $\widetilde{u}\geq u^{\ast
}\geq \widehat{u}$ in $[0,a]\times [ 0,T)$.
\end{itemize}
\end{lemma}

\begin{proof} 
Let us prove it by utilizing \cite[Lemma 3.1]{f1}.
We select $f(v)=(1-v)^{-r}$ and we define 
$s(x,t)=\widetilde{u}(x,t)- \widehat{u}(x,t)$ in $[0,a]\times [ 0,T)$. 
Then $s(x,t)$ satisfies
\begin{gather*}
s_t\geq s_{xx}+r(1-\eta )^{-r-1}s,\quad 0<x<a,\; 0<t<T, \\
s_x(0,t)\leq -p\varphi ^{-p-1}s(0,t),\quad s_x(a,t)
\geq q(1-\xi (a,t))^{-q-1}s(a,t),\quad 0<t<T, \\
s(x,0)\geq 0,\quad 0\leq x\leq a,
\end{gather*}
where $\varphi (0,t)$ lies between 
$\widetilde{u}(0,t)$ and $\widehat{u}(0,t)$,
 $\eta (x,t)$ lies between $\widetilde{u}(x,t)$ and $\widehat{u}(x,t)$, 
and $\xi (a,t)$ lies between $\widetilde{u}(a,t)$ and $\widehat{u}
(a,t)$.


For any fixed $\tau \in (0,T)$, let
\begin{gather*}
L =\max_{0\leq x\leq a,\, 0\leq t\leq \tau }
(\frac{q}{2a} (1-\xi (x,t))^{-q-1}),\\
R=\max_{0\leq x\leq a,\, 0\leq t\leq \tau }
(\frac{p}{2a}\varphi ^{-p-1}(x,t)), \\
M =2L+2R+\max_{0\leq x\leq a} (2Lx-2R(a-x))^{2}
+\max_{0\leq x\leq a,\, 0\leq t\leq \tau } (r(1-\eta (x,t))^{-r-1}).
\end{gather*}
Set $w(x,t)=e^{-Mt-Lx^{2}-R(a-x)^{2}}s(x,t)$. Then $w$ satisfies
\begin{gather*}
w_t\geq w_{xx}+(4Lx-4R(a-x))w_x+cw,\quad 0<x<a,\; 0<t\leq \tau , \\
w_x(0,t)\leq kw(0,t),\quad w_x(a,t)\geq dw(a,t),\quad 0<t\leq \tau ,\\
w(x,0)\geq 0,\quad 0\leq x\leq a,
\end{gather*}
where $c=c(x,t)\leq 0$, $k=k(t)\geq 0$ and $d=d(t)\leq 0$. By the maximum
principle and Hopf's lemma for parabolic equations, we obtain that 
$w\geq 0$ in $[0,a]\times [ 0,\tau ]$.
 Thus, $\widetilde{u}\geq \widehat{u}$ in $[0,a]\times [ 0,T)$.
For $f(v)=-v^{-r}$, a similar process follows.
\smallskip

(b) It is clear from Definition \ref{def2} that every solution of the
problem \eqref{e1} is an upper solution as well as a lower solution of the
corresponding problem. If $u^{\ast }$ is a solution, then we obtain
\begin{gather*}
\widetilde{u}\geq u^{\ast },\quad
u^{\ast }\geq \widehat{u}, \quad 
\widetilde{u}\geq u^{\ast }\geq \widehat{u}
\end{gather*}
in $[0,a]\times [ 0,T)$ from Lemma \ref{lem3} (a). 
\end{proof}

For a given pair of ordered upper and lower solutions $\widetilde{u}$ and 
$\widehat{u}$ we set
\begin{equation*}
S=\{u\in C([0,a]\times [ 0,T)):\widehat{u}\leq u\leq \widetilde{u}\}.
\end{equation*}
Let
\begin{align*}
f(x,t,u(x,t)) = (1-u(x,t))^{-r}\quad  \text{or}\quad 
 f(x,t,u(x,t))=-u^{-r}(x,t), \\
g(x,t,u(x,t)) = u^{-p}(x,t),\ h(x,t,u(x,t))=(1-u(x,t))^{-q}
\end{align*}
Throughout this section we assume the following hypothesis on the
functions in Problem \eqref{e1}:
\begin{itemize}
\item[(H1)] 
\begin{itemize}
\item[(i)] The functions $f(x,t,\cdot)$ is in 
$C^{\alpha ,\alpha /2}([0,a]\times [ 0,T))$, $g(x,t,.)$ is in \\
 $C^{1+\alpha ,(1+\alpha )/2}(\{0\} \times (0,T))$ and
$h(x,t,.)$ is in $ C^{1+\alpha ,(1+\alpha )/2}(\{a\} 
\times (0,T))$, respectively.

\item[(ii)] Let $f(.,u)$, $g(.,u)$ and $h(.,u)$ are $C^{1}$-functions of
$u\in S$. Also,
\begin{equation} \label{e7}
\begin{gathered}
f_{u}(x,t,u)\geq 0\quad  \text{for }u\in S,(x,t)\in [ 0,a]\times [0,T), \\
g_{u}(x,t,u)\leq 0\quad  \text{for }u\in S,(x,t)\in \{0\} \times (0,T), \\
h_{u}(x,t,u)\geq 0\quad  \text{for }u\in S,(x,t)\in \{a\} \times (0,T).
\end{gathered}
\end{equation}
\end{itemize}
\end{itemize}
Condition \eqref{e7} implies that $f(.,u)$ and $h(.,u)$ are non-decreasing
in $u$, $g(.,u)$ is non-increasing in $u$, respectively, which is crucial
for the construction of monotone sequences.

Next, we construct monotone sequences of functions which give
the estimation of the solution $u$ of problem \eqref{e1}. Specifically, by
starting from any initial iteration $u^{0}$, we can construct a sequence
 $\{u^{(k)}\} $ from the linear iteration process
\begin{equation} \label{e8}
\begin{gathered}
u_t^{(k)}-u_{xx}^{(k)}=f(x,t,u^{(k-1)}),\quad 0<x<a,\; 0<t<T, \\
u_x^{(k)}(0,t)=g(0,t,u^{(k-1)}),\quad  u_x^{(k)}(a,t)=h(a,t,u^{(k-1)}),\quad
0<t<T, \\
u^{(k)}(x,0)=u_0(x),\quad 0\leq x\leq a.
\end{gathered}
\end{equation}
It is clear that the sequence governed by \eqref{e8} is well defined and can be
obtained by solving a linear initial boundary value problem. Starting from
initial iteration $u^{0}=\widetilde{u}$ and $u^{0}=\widehat{u}$, we define
two sequences of the functions 
$\{ \overline{u}^{(k)}\} $ and $\{ \underline{u}^{(k)}\} $ for 
$k=1,2,\dots $ respectively, and
refer to them as maximal and minimal sequences, respectively, where those
elements satisfy the above linear problem.

\begin{lemma} \label{lem4} 
The sequences $\{ \overline{u}^{(k)}\}$,
$\{ \underline{u}^{(k)} \} $ possess the
monotone property
\begin{equation*}
\widehat{u}\leq \underline{u}^{(k)}\leq \underline{u}^{(k+1)}\leq \overline{u
}^{(k+1)}\leq \overline{u}^{(k)}\leq \widetilde{u}
\end{equation*}
for $(x,t)\in [ 0,a]\times [ 0,T)$ and every $k=1,2,\dots$.
\end{lemma}

\begin{proof}
Let $\mu =\widetilde{u}-\overline{u}^{(1)}$. From \eqref{e8}
 and from Definition \ref{def2}, we obtain
\begin{gather*}
\mu _t-\mu _{xx}=\widetilde{u}_t-\widetilde{u}_{xx}-f(x,t,\widetilde{u}
)\geq 0,\quad 0<x<a,\; 0<t<T, \\
\mu _x(0,t)=\widetilde{u}_x(0,t)-g(0,t,\widetilde{u})\leq 0, \quad 0<t<T, \\
\mu _x(a,t)=\widetilde{u}_x(a,t)-h(a,t,\widetilde{u})\geq 0,
\quad 0<t<T, \\
\mu (x,0)=\widetilde{u}(x,0)-u_0(x)\geq 0,\quad 0\leq x\leq a.
\end{gather*}
From the Maximum principle and Hopf's Lemma for parabolic equations, we obtain 
$\mu \geq 0$ for $(x,t)\in [ 0,a]\times [ 0,T)$, i.e. 
$\overline{u}^{(1)}\leq \widetilde{u}$. Similarly, using the property of a lower
solution, we obtain $\underline{u}^{(1)}\geq \widehat{u}$.

Let $\mu ^{(1)}=\overline{u}^{(1)}-\underline{u}^{(1)}$. 
From \eqref{e7} and \eqref{e8}, we obtain
\begin{gather*}
\mu _t^{(1)}-\mu _{xx}^{(1)}=f(x,t,\widetilde{u})-f(x,t,\widehat{u})\geq 0,
\quad 0<x<a,\; 0<t<T, \\
\mu _x^{(1)}(0,t)=g(0,t,\widetilde{u})-g(0,t,\widehat{u})\leq
0,\quad 0<t<T,  \\
\mu _x^{(1)}(a,t)=h(a,t,\widetilde{u})-h(a,t,\widehat{u})\geq
0,\quad 0<t<T, \\
\mu ^{(1)}(x,0)=u_0(x)-u_0(x)=0, \quad 0\leq x\leq a.
\end{gather*}
From the Maximum principle and Hopf's Lemma for parabolic equations,
 we obtain $\mu^{(1)}\geq 0$ for $(x,t)\in [ 0,a]\times [ 0,T)$, i.e. 
$\underline{u}^{(1)}\leq \overline{u}^{(1)}$. Therefore,
\begin{equation*}
\widehat{u}\leq \underline{u}^{(1)}\leq \overline{u}^{(1)}\leq \widetilde{u}
\end{equation*}
for $(x,t)\in [ 0,a]\times [ 0,T)$.

Assume that
\begin{equation*}
\underline{u}^{(k-1)}\leq \underline{u}^{(k)}\leq \overline{u}^{(k)}\leq
\overline{u}^{(k-1)}
\end{equation*}
for $(x,t)\in [ 0,a]\times [ 0,T)$ and for some integer $k>1$.
Let $\mu ^{(k)}=\overline{u}^{(k)}-\overline{u}^{(k+1)}$. 
From \eqref{e7} and \eqref{e8}, we obtain
\begin{gather*}
\mu _t^{(k)}-\mu _{xx}^{(k)}=f(x,t,\overline{u}^{(k-1)})-f(x,t,\overline{u}
^{(k)})\geq 0,\quad 0<x<a,\; 0<t<T, \\
\mu _x^{(k)}(0,t)=g(0,t,\overline{u}^{(k-1)})
-g(0,t,\overline{u}^{(k)})\leq 0,\quad 0<t<T,  \\
\mu _x^{(k)}(a,t)=h(a,t,\overline{u}^{(k-1)})-h(a,t,\overline{
u}^{(k)})\geq 0,\quad 0<t<T, \\
\mu ^{(k)}(x,0)=0,\quad  0\leq x\leq a.
\end{gather*}
From the Maximum principle and Hopf's Lemma for parabolic equations,
 we obtain $\mu ^{(k)}\geq 0$ for $(x,t)\in [ 0,a]\times [ 0,T)$, i.e. 
$\overline{u}^{(k+1)}\leq \overline{u}^{(k)}$. A similar argument gives 
$\underline{u}^{(k+1)}\geq \underline{u}^{(k)}$ and 
$\overline{u}^{(k+1)}\geq \underline{u}^{(k+1)}$. Therefore, the result 
follows from the mathematical induction.
\end{proof}

\begin{lemma} \label{lem5}
For each positive integer $k$, $\overline{u}^{(k)}$ is an upper solution, 
$\underline{u}^{(k)}$ is a lower solution, 
$\underline{u}^{(k)}\leq $ $\overline{u}^{(k)}$ for $(x,t)\in [0,1]\times [ 0,T)$.
\end{lemma}

\begin{proof}
From \eqref{e7}, \eqref{e8}  and Lemma \ref{lem3}, $\overline{u}^{(k)}$
satisfies
\begin{gather*}
\begin{aligned}
\overline{u}_t^{(k)}-\overline{u}_{xx}^{(k)}
&=f(x,t,\overline{u} ^{(k-1)})\\
&=f(x,t,\overline{u}^{(k-1)})-f(x,t,\overline{u}^{(k)})+f(x,t,
\overline{u}^{(k)})\geq f(x,t,\overline{u}^{(k)}), 
\end{aligned} \\
\begin{aligned}
\overline{u}_x^{(k)}(0,t)
&=g(0,t,\overline{u}^{(k-1)})\\
&=g(0,t,\overline{u}
^{(k-1)})-g(0,t,\overline{u}^{(k)})+g(0,t,\overline{u}^{(k)})\leq g(0,t,
\overline{u}^{(k)}), 
\end{aligned}\\
\begin{aligned}
\overline{u}_x^{(k)}(a,t)
&=h(a,t,\overline{u}^{(k-1)})\\
&=h(a,t,\overline{u} ^{(k-1)})-h(a,t,\overline{u}^{(k)})
+h(a,t,\overline{u}^{(k)})\geq h(a,t,
\overline{u}^{(k)}), 
\end{aligned}\\
\overline{u}^{(k)}(x,0)=u_0(x),\quad 0\leq x\leq a,
\end{gather*}
and $\underline{u}^{(k)}$ satisfies
\begin{gather*}
\begin{aligned}
\underline{u}_t^{(k)}-\underline{u}_{xx}^{(k)}
&=f(x,t,\underline{u} ^{(k-1)})\\
&=f(x,t,\underline{u}^{(k-1)})-f(x,t,\underline{u}^{(k)})+f(x,t,
\underline{u}^{(k)})\leq f(x,t,\underline{u}^{(k)}),
\end{aligned} \\
\begin{aligned}
\underline{u}_x^{(k)}(0,t)
&=g(0,t,\underline{u}^{(k-1)})\\
&=g(0,t,\underline{u}
^{(k-1)})-g(0,t,\underline{u}^{(k)})+g(0,t,\underline{u}^{(k)})\geq g(0,t,
\underline{u}^{(k)}), 
\end{aligned}\\
\begin{aligned}
\underline{u}_x^{(k)}(a,t)
&=h(a,t,\underline{u}^{(k-1)})\\
&=h(a,t,\underline{u}
^{(k-1)})-h(a,t,\underline{u}^{(k)})+h(a,t,\underline{u}^{(k)})\leq h(a,t,
\underline{u}^{(k)}), 
\end{aligned}\\
\underline{u}^{(k)}(x,0)=u_0(x),0\leq x\leq a.
\end{gather*}
From Lemma \ref{lem4} and from the above inequalities, the functions 
$\overline{u}^{(k)}$ and $\underline{u}^{(k)}$ are ordered upper and 
lower solutions of  problem \eqref{e8}.
\end{proof}


We have the following existence theorem for problem \eqref{e1} via
Lemmas \ref{lem4} and \ref{lem5}.


\begin{theorem} \label{thm9}
Let $\widetilde{u},\widehat{u}$ be a pair of
ordered upper and lower solutions of the problem \eqref{e1}, and let Hypothesis 
(H1) hold. Then the sequences $\{ \overline{u}^{(k)}\}$,
$\{ \underline{u}^{(k)}\}$ given by Problem \eqref{e8} with 
$u^{0}=\widetilde{u}$ and $u^{0}=\widehat{u}$ converge monotonically to a maximal
solution $\overline{u}$ and minimal solution $\underline{u}$ of the
problem \eqref{e1}, respectively. Further,
\begin{equation} \label{e9}
\widehat{u}\leq \underline{u}^{(k)}
\leq \underline{u}^{(k+1)}\leq \underline{ u}
\leq \overline{u}\leq \overline{u}^{(k+1)}\leq \overline{u}^{(k)}
\leq \widetilde{u}  
\end{equation}
for $(x,t)\in [ 0,a]\times [ 0,T)$ and each positive integer $k$. 
Furthermore if $\underline{u}=\overline{u}\  (\equiv u^{\ast })$, 
then $u^{\ast }$ is the unique solution of the problem \eqref{e1} in $S$.
\end{theorem}

\begin{proof}
The pointwise limits
\begin{equation*}
\lim_{k\to \infty} \overline{u}^{(k)}(x,t)=\overline{u}(x,t),\quad
\lim_{k\to \infty} \underline{u}^{(k)}(x,t)=\underline{u}(x,t)
\end{equation*}
exist and satisfy the relation \eqref{e9}. Indeed, the sequence 
$\{\overline{u}^{(k)}\} $ is monotone non-increasing which is bounded
from below, while the sequence $\{ \underline{u}^{(k)}\} $ is
monotone nondecreasing and is bounded from above as in Lemma \ref{lem4}.

Let $\Theta =\underline{u}(x,t)-\overline{u}(x,t)$. 
From \eqref{e9}, we have $\underline{u}(x,t)\leq \overline{u}(x,t)$ 
for $(x,t)\in [ 0,a]\times[ 0,T)$. Also, $\Theta (x,t)$ satisfies
\begin{gather*}
\Theta _t-\Theta _{xx}=f(x,t,\underline{u})-f(x,t,\overline{u}),\quad
0<x<a,\; 0<t<T, \\
\Theta _x(0,t)=g(0,t,\underline{u})-g(0,t,\overline{u}),\quad 0<t<T, \\
\Theta _x(1,t)=h(a,t,\underline{u})-h(a,t,\overline{u}),\quad 0<t<T, \\
\Theta (x,0)=0,\quad 0\leq x\leq a.
\end{gather*}
By using the process of Lemma \ref{lem3} (a) and Lemma \ref{lem6}, we obtain 
$\Theta \geq 0$ for $(x,t)\in [ 0,a]\times [ 0,T)$, i.e. 
$\underline{u}(x,t)\geq \overline{u}(x,t)$, and so, we obtain 
$\underline{u}(x,t)=\overline{u}(x,t)$.

If $u^{\ast }$ is any other solution in $S$, then from Lemma \ref{lem5} we obtain 
\begin{gather*}
\overline{u}\geq u^{\ast }, \quad u^{\ast }\geq \underline{u},\\
\overline{u}\geq u^{\ast }\geq \underline{u}
\end{gather*}
in $[0,a]\times [ 0,T)$. This implies that
\begin{equation*}
\overline{u}=u^{\ast }=\underline{u}
\end{equation*}
and\ hence $u^{\ast }$ is the unique solution of the problem \eqref{e1}.
\end{proof}

\subsection{Quenching on the boundary}

In this subsection, we study quenching properties of the problem 
\eqref{e1} via Section 2.1.


\begin{lemma} \label{lem6} 
(i) $(f(v)=(1-v)^{-r})$ If $v_{xx}(x,0)+(1-v(x,0))^{-r}\geq 0$ in 
$(0,a)$ (i.e., if $v_0$ is a lower solution), then $v_t(x,t)\geq 0$ in
$ [0,a]\times [ 0,T)$.
Also, we obtain $v_t(x,t)>0$ in$ (0,a)\times [ 0,T)$ by strong
maximum principle.

(ii) $(f(v)=-v^{-r})$ If $v_{xx}(x,0)-v^{-r}(x,0)\leq 0$ in $(0,a)$ 
(i.e., if $v_0$ is an upper solution), then $v_t(x,t)\leq 0$ in
$ [0,a]\times [ 0,T)$. Also, we obtain $v_t(x,t)<0$ in$ (0,a)\times [ 0,T)$ 
by the  strong maximum principle.
\end{lemma}

\begin{proof} (i) 
Let us prove it by using \cite[Lemma 3.1]{f1}.
We let $f(v)=(1-v)^{-r}$ and we define $s(x,t)=v_t(x,t)$
in $[0,a]\times [ 0,T)$. Then $s(x,t)$ satisfies
\begin{gather*}
s_t=s_{xx}+r(1-v)^{-r-1}s,\quad 0<x<a,\; 0<t<T, \\
s_x(0,t)=-pv^{-p-1}s(0,t),\quad 
s_x(a,t)=q( 1-v(a,t))^{-q-1}s(a,t),\quad 0<t<T, \\
s(x,0)=v_{xx}(x,0)+(1-v(x,0))^{-r}\geq 0,\quad 0\leq x\leq a.
\end{gather*}
For a fixed $\tau \in (0,T)$, let
\begin{gather*}
L =\max_{0\leq x\leq a,\, 0\leq t\leq \tau }  
(\frac{q}{2a}(1-v(x,t))^{-q-1}),\\
R=\max_{0\leq x\leq a,\, 0\leq t\leq \tau } (\frac{p}{2a}v^{-p-1}(x,t)), \\
\begin{aligned}
M &=2L+2R+\underset{0\leq x\leq a}{\max }(2Lx-2R(a-x))^{2}
+\max_{0\leq x\leq a,\, 0\leq t\leq \tau } (r(1-\eta (x,t))^{-r-1}).
\end{aligned}
\end{gather*}
Set $w(x,t)=e^{-Mt-Lx^{2}-R(a-x)^{2}}s(x,t)$. Then $w$ satisfies
\begin{gather*} 
w_t=w_{xx}+(4Lx-4R(a-x))w_x+cw,\quad 0<x<a,\; 0<t\leq \tau , \\
w_x(0,t)=kw(0,t),\quad w_x(a,t)=dw(a,t),\; 0<t\leq \tau , \\
w(x,0)=0,\quad 0\leq x\leq a,
\end{gather*}
where $c=c(x,t)\leq 0$, $k=k(t)\geq 0$ and $d=d(t)\leq 0$. By the maximum
principle and Hopf's lemma for parabolic equations, we obtain that 
$w\geq 0$ in $[0,a]\times [ 0,\tau ]$.
Thus, $v_t\geq 0$ in $[0,a]\times [ 0,T)$. Also, we obtain 
$v_t(x,t)>0$ in $ (0,a)\times [ 0,T)$
by the strong maximum principle.
\smallskip

(ii) Now, if we let $f(v)=-v^{-r}$, and $v_{xx}(x,0)-v^{-r}(x,0)\leq 0$ 
in $(0,a)$, then using the same process above, we obtain 
$v_t(x,t)\leq 0$ in $ [0,a]\times [ 0,T)$. Also, we obtain 
$v_t(x,t)<0$ in $ (0,a)\times [ 0,T)$ by the strong maximum principle.
\end{proof} 

\begin{lemma} \label{lem7} 
If $v_x(x,0)\geq 0$, then $v_x\geq 0$ in $[ 0,a] \times (0,T)$.
\end{lemma}

\begin{proof} 
Let $H=v_x(x,t)$. Then 
\begin{gather*}
H_t=H_{xx}+f'(v)H,\quad 0<x<a,\; 0<t<T, \\
H(0,t)=v^{-p}(0,t)>0,\quad H(a,t)=(1-v(a,t))^{-q}>0,\; 0<t<T, \\
H(x,0)=v_x(x,0)\geq 0,\quad  0\leq x\leq a.
\end{gather*}
From the maximum principle, it follows that $H\geq 0$ and hence 
$v_x\geq 0$, in $[0,a]\times (0,T)$. 
\end{proof}

\begin{theorem} \label{thm10} (i)
 $(f(v)=-v^{-r})$ If $u_0(x)\geq v_0(x)$ and $u_0$ is an upper solution 
for  problem \eqref{e2}, then the solution $v$ of  problem \eqref{e1} 
quenches in a finite time and $x=0$ is the only quenching point.

(ii) $(f(v)=(1-v)^{-r})$ If $u_0(x)\leq v_0(x)$
and $u_0$ is a lower solution for the problem \eqref{e2}, then the solution $v$
of problem \eqref{e1} quenches in a finite time and $x=a$ is the only
quenching point.
\end{theorem}


\begin{proof}  (i) First, let $f(v)=-v^{-r}$. 
If $u_0(x)\geq v_0(x)$, then the solution $u$ of the problem \eqref{e2} 
is an upper solution of \eqref{e1} from Definition \ref{def2}.
 Further,  if $ u_0(x)$ is an upper solution for the problem \eqref{e2}, then 
$u$ quenches in a finite time, $\lim_{t\to T^{-}}u(0,t)\to 0$ from Corollary
\ref{coro2} (i). So, we obtain
\begin{equation*}
u_0\geq u\geq v
\end{equation*}
from Lemma \ref{lem3}. Thus, $v$ quenches in a finite time, 
$\lim_{t\to T^{-}}v(0,t)\to 0$.

(ii) Now, let $f(v)=(1-v)^{-r}$. If $u_0(x)\leq v_0(x)$,
then the solution $u$ of  \eqref{e2} is a lower solution of
\eqref{e1} from Definition \ref{def2}. Further, if $u_0(x)$ is a lower
solution of \eqref{e2}, then $u$ quenches in a finite time, 
$\lim_{t\to T^{-}}u(a,t)\to 1$ from Corollary \ref{coro2} (ii). So, we obtain
\begin{equation*}
u_0\leq u\leq v
\end{equation*}
from Lemma \ref{lem3}. As a result, $v$ quenches in a finite time, 
$\lim_{t\to T^{-}}v(a,t)\to 1$. 
\end{proof}


\begin{theorem} \label{thm11}
 (i) $(f(v)=-v^{-r})\ v_t$
blows up at the quenching time at the boundary $x=0$.

(ii) $(f(v)=(1-v)^{-r})v_t$ blows up at the
quenching time at the boundary $x=a$.
\end{theorem}

\begin{proof} (i) 
$(f(v)=-v^{-r})$ Suppose that $v_t$ is bounded on $[0,a]\times [ 0,T)$. 
Then, there exists a positive constant $M$ such that $v_t>-M$. That is
\begin{equation*}
v_{xx}-v^{-r}>-M.
\end{equation*}
Multiplying this inequality by $v_x$,  and integrating with respect to 
$x $ from $0$ to $x$, we have
\begin{equation*}
-\frac{1}{2}v^{-2p}(0,t)-\ln \big[ \frac{1}{v(0,t)}\big] >\frac{1}{2}
v_x^{2}-\ln \big[ \frac{1}{v(x,t)}\big] 
-M[ v(a,t)-v(x,t)]
\end{equation*}
for $r=1$, and
\begin{equation*}
-\frac{1}{2}v^{-2p}(0,t)+\frac{v^{-r+1}(0,t)}{-r+1}>-\frac{1}{2}v_x^{2}+
\frac{v^{-r+1}(x,t)}{-r+1}-M[ v(x,t)-v(0,t)]
\end{equation*}
for $r\neq 1$. We have, as $t\to T^{-}$, the left-hand side tends to
negative infinity, while the right-hand side is finite. This contradiction
shows that $v_t$ blows up at the quenching point $x=0$.

(ii) $(f(v)=(1-v)^{-r})$ Suppose\ that $v_t$ is
bounded on $[0,a]\times [ 0,T)$. Then, there exists a positive
constant $M$ such that $v_t<M$. That is,
\begin{equation*}
v_{xx}+(1-v)^{-r}<M.
\end{equation*}
Multiplying this inequality by $v_x$, and integrating with respect to 
$x $ from $x$ to $a$, we have
\begin{equation*}
\frac{1}{2}(1-v(a,t))^{-2q}+\ln [ \frac{1}{1-v(a,t)}] 
<\frac{1}{2} v_x^{2}+\ln [ \frac{1}{1-v(x,t)}] 
+M[ v(a,t)-v(x,t)]
\end{equation*}
for $r=1$, and
\begin{equation*}
\frac{1}{2}(1-v(a,t))^{-2q}+\frac{(1-v(a,t))^{-r+1}}{r-1}<\frac{1}{2}
v_x^{2}+\frac{(1-v(x,t))^{-r+1}}{r-1}
+M[ v(a,t)-v(x,t)]
\end{equation*}
for $r\neq 1$. As $t\to T^{-}$, the left-hand side tends to
infinity, while the right-hand side is finite. Hence, $v_t$ blows up at
the quenching point $x=a$. 
\end{proof}

\begin{corollary} \label{coro2} 
We have the following results via Theorems \ref{thm10} and \ref{thm11}:

(i) $(f(v)=-v^{-r})$ If $u_0(x)\geq v_0(x)$ and 
$u_0$ is an upper solution for the problem \eqref{e2}, then the solution $v$ of
the problem \eqref{e1} quenches in a finite time, $x=0$ is the only quenching
point, and $v_t$ blows up at the quenching time.

(ii) $(f(v)=(1-v)^{-r})$ If $u_0(x)\leq v_0(x)$
and $u_0$ is a lower solution for the problem \eqref{e2}, then the solution $v$
of the problem \eqref{e1} quenches in a finite time, $x=a$ is the only quenching
point, and $v_t$ blows up at the quenching time.
\end{corollary}

\begin{thebibliography}{99}

\bibitem{a1}  J. R. Anderson;
\emph{Local existence and uniqueness of
solutions of degenerate parabolic equations}, Comm. Partial Differential
Equations, 16 (1991) 105-143.

\bibitem{c1}  C. Y. Chan; N. Ozalp;
\emph{Beyond quenching for singular reaction-diffusion mixed-boundary value problems}, 
Advances in nonlinear dynamics (monograph), Stability and Control: 
Theory Methods and Appl., 5 (1997) 217-227.

\bibitem{c2}  C. Y. Chan, S. I. Yuen;
\emph{Parabolic problems with nonlinear absorptions and releases at the boundaries}, 
Appl. Math. Comput., 121 (2001) 203-209.

\bibitem{c3}  C. W. Chang, Y. H. Hsu, H. T. Liu;
\emph{Quenching behavior of parabolic systems with localized reaction term}, 
Mathematics and Statistics, 2 (1) (2014), 48-53.

\bibitem{f1}  S. C. Fu, J.-S. Guo, J. C. Tsai;
\emph{Blow up behavior for a semilinear heat equation with a nonlinear 
boundary condition}, Tohoku Math. J., 55 (2003) 565-581.

\bibitem{l1}  G. M. Lieberman;
\emph{Mixed boundary value problems for elliptic and parabolic differential 
equations of second order}, J. Math. Anal. Appl., 13 (1986), 422-440.

\bibitem{o1}  N. Ozalp, B. Selcuk;
\emph{Blow up and quenching for a problem with nonlinear boundary conditions}, 
Electron. J. Diff. Equ., Vol. 2015 (2015), No. 192, pp. 1-11.

\bibitem{o2}  N. Ozalp, B. Selcuk;
\emph{The quenching behavior of a nonlinear parabolic equation with a 
singular boundary condition}, Hacettepe Journal of Mathematics and Statistics, 
Vol 44 No.3 (2015) 615-621.

\bibitem{p1}  C. V. Pao;
\emph{Quasilinear parabolic and elliptic equations with nonlinear boundary conditions},
 Nonlinear Analysis, 66 (2007) 639-662.

\bibitem{s1}  B. Selcuk, N. Ozalp;
\emph{The quenching behavior of a semilinear heat equation with a singular 
boundary outflux}, Quart. Appl. Math., Vol. 72 No. 4 (2014), 747-752.

\bibitem{z1}  Y. Zhi, C. Mu;
\emph{The quenching behavior of a nonlinear parabolic equation with a nonlinear 
boundary outflux}, Appl. Math. Comput., 184 (2007) 624-630.

\end{thebibliography}

\end{document}