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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2015 (2015), No. 192, pp. 1--11.\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 - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2015/192\hfil Blow up and quenching]
{Blow up and quenching for a problem with nonlinear boundary conditions}

\author[N. Ozalp, B. Selcuk \hfil EJDE-2015/192\hfilneg]
{Nuri Ozalp, Burhan Selcuk}

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

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

\thanks{Submitted  May 27, 2015. Published July 20, 2015.}
\subjclass[2010]{35K20, 35K55, 35B50}
\keywords{Heat equation; nonlinear parabolic equation; blow up;
\hfill\break\indent nonlinear boundary condition;  quenching; maximum principle}

\begin{abstract}
 In this article, we study the blow up behavior of the heat equation
 $ u_t=u_{xx}$ with $u_x(0,t)=u^{p}(0,t)$, $u_x(a,t)=u^q(a,t)$.
 We also study the quenching behavior of the nonlinear parabolic equation
 $v_t=v_{xx}+2v_x^{2}/(1-v)$ with $v_x(0,t)=(1-v(0,t))^{-p+2}$,
 $ v_x(a,t)=(1-v(a,t))^{-q+2}$. In the blow up problem, if $u_0$
 is a lower solution then we get the blow up occurs in a finite time at the
 boundary $x=a$ and using positive steady state we give criteria for blow up
 and non-blow up. In the quenching problem, we show that the only quenching
 point is $x=a$ and $v_t$ blows up at the quenching time, under certain
 conditions and using positive steady state we give criteria for quenching
 and non-quenching. These analysis is based on the equivalence between the
 blow up and the quenching for these two equations.
\end{abstract}

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

\section{Introduction}

In this article, we study the  blow up and quenching problems
with nonlinear boundary conditions.

\subsection*{Blow up problem} We study the behavior of solutions to
the  heat equation, with nonlinear boundary conditions,
\begin{equation}  \label{e1.1}
\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) =u^q(a,t),\; 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 $u_0(x)$ is a non-negative smooth function satisfying the
compatibility conditions
\[
u_0'( 0) =u_0^{p}(0),\quad u_0'(a) =u_0^q(a).
\]
We are interested in the occurrence of finite-time blow-up, i.e, the
existence of a $T=T(u_0)<\infty$ such that
\[
\sup_{x\in [ 0,a]} u(x,t)\to \infty \quad  \text{as } t\to T.
\]

\subsection*{Quenching problem}
If we use the transform $u=\frac{1}{1-v}$ in the problem \eqref{e1.1},
then we obtain the  nonlinear parabolic equation, with nonlinear boundary conditions,
\begin{equation} \label{e1.2}
\begin{gathered}
v_t=v_{xx}+\frac{2v_x^{2}}{1-v},\quad 0<x<a,\; 0<t<T, \\
v_x( 0,t) =( 1-v(0,t)) ^{-p+2},\quad
 v_x(a,t) =( 1-v(a,t)) ^{-q+2},\quad  0<t<T, \\
v( x,0) =v_0( x) =1-\frac{1}{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 $v_0:[0,a]\to (0,1)$ satisfies the compatibility conditions
\[
v_0'( 0) =( 1-v_0(0)) ^{-p+2},\quad
v_0'( a) =( 1-v_0(a)) ^{-q+2}.
\]
A solution $v(x,t)$ of \eqref{e1.2} is said to quench if there exists
a finite time $T$ such that
\[
\lim_{t\to T^{-}}\max \{v(x,t):0\leq x\leq a\} = 1.
\]
For the rest of this article, we denote the quenching time of problem \eqref{e1.2} 
with $T$.
\smallskip

Blow up problems with various boundary conditions have been studied
extensively; see for example \cite{d1,f1,f2,g1,g2,l2,z1}.
Lin and Wang \cite{l2} considered the problem
\begin{gather*}
u_t=u_{xx}+u^{p},\quad  0<x<1,\; 0<t<T, \\
u_x( 0,t) =0,\quad u_x( 1,t) =u^q(1,t),\; 0<t<T, \\
u( x,0) =u_0( x) ,\quad 0\leq x\leq 1,
\end{gather*}
where $p,q>0,T\leq \infty $. They showed that the solutions have a finite
time blow-up and obtained the exact blow-up rates for the necessary and
sufficient conditions. They also proved that the blow-up will occur only at
the boundary $x=1$.  Fu et al. \cite{f3} studied the same problem.
Under certain conditions, they proved that the blow-up point occurs only at
the boundary $x=1$. Then, by applying the well-known method of Giga-Kohn,
they derived the time asymptotic of solutions near the blow-up time.
Finally, they proved that the blow-up was complete.

Since 1975, quenching problems with various boundary conditions have been
studied extensively \cite{c1,c2,d2,o1,s1,s2}.
 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 }\overline{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. Ozalp and Selcuk \cite{o1} 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) =(1-u(1,t))^{-q},\quad 0<t<T,\\
u( x,0) =u_0( x) ,\quad 0<u_0( x) <1,\quad 0\leq x\leq 1.
\end{gather*}
They showed that $x=1$ is the quenching point in finite time, if $u_0$
satisfies $u_{xx}(x,0)+( 1-u(x,0)) ^{-p}\geq 0$ and
$u_x(x,0)\geq 0$. 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.


In Section 2, we investigate the blow up behavior of the problem \eqref{e1.1}.
First, if $u_0$ is a lower solution, then we show that blow up occurs in
finite time at the boundary $x=a$. Also, using positive steady state we get
criteria for blow up and non-blow up. In Section 3, we investigate quenching
behavior of the problem \eqref{e1.2}. We show that the only quenching point is
$x=a$ and $v_t$ blows up at the quenching time, under certain conditions.
Also, using positive steady state we get criteria for quenching and
non-quenching.

\section{Blow up problem}

In this section, we assume that $u_{xx}(x,0)\geq 0$ in $(0,a)$. Now, we
give some auxiliary results for the problem \eqref{e1.1}.

\begin{definition} \label{def1} \rm
A function $\mu$ is called a lower solution of
\eqref{e1.1} if $\mu$ satisfies the following conditions:
\begin{gather*}
\mu _t-\mu _{xx}\leq 0,\quad 0<x<a,\; 0<t<T, \\
\mu _x( 0,t) \geq \mu ^{p}(0,t),\quad  \mu _x( a,t)
\leq \mu ^q(a,t),\quad 0<t<T, \\
\mu ( x,0) \leq u_0( x) ,\quad 0\leq x\leq a.
\end{gather*}
When reverse the inequalities, we have an upper solution.
\end{definition}

We have the following Theorem and Lemma for  problem \eqref{e1.1}
(see \cite{c2}).



\begin{theorem} \label{thm1}
Let $u(x,t,u_0)$ and $h(x,t,h_0)$ be
solutions of  \eqref{e1.1}\ with initial data given by $u_0(x)$ and
$h_0(x)$, respectively. If $u_0\leq h_0$, 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),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},  the solution $w$ 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 )\left[ u^q(a,\sigma )-h^q(a,\sigma )
\right] -w(0,\sigma )\left[ u^{p}(0,\sigma )-h^{p}(0,\sigma )\right] \\
&\quad -s(\sigma )\left[ u(a,\sigma )-h(a,\sigma )\right] w(a,\sigma )+r(\sigma )
\left[ u(0,\sigma )-h(0,\sigma )\right] 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[ u^q(a,\sigma )-h^q(a,\sigma
)-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 )) = u^q(a,\sigma
)-h^q(a,\sigma ).
\end{gather*}
Since $u_0\leq h_0$ and $w(x,0)\geq 0$, we have
\[
\int_0^{a}(u(x,\tau )-h(x,\tau ))g(x)dx\leq \int_0^{\tau
}\int_0^{a}(u(x,\sigma )-h(x,\sigma ))w(x,\sigma )\,dx\,d\sigma .
\]
Let
\[
(u(x,\sigma )-h(x,\sigma ))^{+}=\max  \{ 0,u(x,\sigma )-h(x,\sigma)\} .
\]
From, $0\leq w\leq k$, we obtain
\[
\int_0^{a}(u(x,\tau )-h(x,\tau ))g(x)dx\leq k\int_0^{\tau
}\int_0^{a}(u(x,\sigma )-h(x,\sigma ))^{+}\,dx\,d\sigma .
\]
Since $g\in C^{2}(\overline{D})$ has compact support in $D$ and
$0\leq g\leq 1$, we have
\[
\int_0^{a}(u(x,\sigma )-h(x,\sigma ))^{+}dx\leq k\int_0^{\tau
}\int_0^{a}(u(x,\sigma )-h(x,\sigma ))^{+}\,dx\,d\sigma .
\]
By the Gronwall inequality,
\[
\int_0^{a}(u(x,\sigma )-h(x,\sigma ))^{+}dx\leq 0,
\]
which gives $u(x,\tau )\leq h(x,\tau )$ for any $\tau >0$. Thus, the
theorem is proved.
\end{proof}


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

\item[(ii)]  $u_x>0$ in $(0,a)\times (0,T)$.
\end{itemize}
\end{lemma}

\begin{proof}
(i) Since $u_{xx}(x,0)\geq 0$ in $(0,a)$, $u_0'( 0) =u_0^{p}(0)$,
$u_0'(a) =u_0^q(a)$ it follows from Definition \ref{def1} that $u_0(x)$ is a
lower solution of the problem \eqref{e1.1}. The strong maximum principle implies
that
\[
u(x,t)\geq u_0(x)\quad  \text{in }(0,a)\times (0,T).
\]
Let $h$ be a positive number less than $T$, and
\[
z(x,t)=u(x,t+h)-u(x,t).
\]
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\eta ^{q-1}(t)z(a,t),\quad 0<t<T-h,
\end{gather*}
where $\xi (t)$ and $\eta (t)$ lie, respectively, between $u(0,t+h)$
and $u(0,t)$, and 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
\[
H_t-H_{xx}=0\quad  \text{in }(0,a)\times (0,T),
\]
it follows from the strong maximum principle that $H=u_t>0$ in
$(0,a)\times (0,T)$.

(ii) Since $u_x(0,t)=u^{p}(0,t)>0$ and $u_{xx}=u_t>0$ in
$(0,a)\times (0,T)$. Then, $u_x$ is an increasing function and so,
$u_x(x,t)>0$ in $(0,a)\times (0,T)$.
\end{proof}

\begin{theorem} \label{thm2}
 Let $u$ be a solution of the problem \eqref{e1.1},
$f(u)=u^q$ and $q>1$. We assume that
\begin{equation} \label{e2.1}
\int_0^{\infty } \frac{ds}{f(s)}<\infty .
\end{equation}
If $u_0$ is a lower solution, assumption \eqref{e2.1} is satisfied, $q\geq p$,
then
\begin{itemize}
\item[(a)] any positive solution of the problem \eqref{e1.1} must blow up in a
finite time $T$ such that there exists a positive constant $\delta$ with
\[
T\leq \frac{1}{\delta }\frac{M_0^{-q+1}}{q-1},
\]
where $M_0=\underset{x\in [ 0,a]}{\max }u_0(x)$,

\item[(b)] a blow up rate is obtained for $t$ sufficiently close to $T$ as
\[
\sup_{x\in [ 0,a]} u(x,t)\leq C(T-t)^{1/(-q+1)},
\]
where $C=(\delta (q-1))^{1/(-q+1)}$.
\end{itemize}
\end{theorem}

\begin{proof}
Let us prove it by using  \cite[Theorems 1 and 2]{z1}.
First, we define
\[
J(x,t)=u_t(x,t)-\delta u^q(x,t)\quad   \text{in }[0,a]\times [ \tau ,T),
\]
where $\tau \in (0,T)$ and $\delta $ is a positive constant to be
specified later. Then, $J(x,t)$ satisfies
\[
J_t-J_{xx}=\delta q(q-1)u^{q-2}u_x^{2}>0\quad  \text{in }
( 0,a) \times (\tau ,T),
\]
since $q\geq 1$. $J(x,\tau )\geq 0$ by Lemma \ref{lem1} (i),
if $\delta $ is small enough. Further,
\begin{gather*}
J_x(0,t) = pu^{p-1}(0,t)J(0,t)+(p-q)u^{p+q-1}(0,t)\leq
pu^{p-1}(0,t)J(0,t), \\
J_x(a,t) = qu^{q-1}(a,t)J(a,t),
\end{gather*}
since $q\geq p$\ and $t\in (\tau ,T)$. By the maximum principle and Hopf's
lemma for the parabolic equations, we obtain that $J(x,t)\geq 0$ for
$ (x,t)\in [ 0,a]\times [ \tau ,T)$. Thus, we get
\[
u_t(x,t)\geq \delta u^q(x,t),
\]
for $(x,t)\in [ 0,a]\times [ \tau ,T)$.

Integrating from $t$ to $T$ we obtain
\[
\int_t^{T}\frac{u_{s}(x,s)}{u^q(x,s)}ds\geq \delta (T-t).
\]
Let $u_0(x_0)=M_0=\max_{x\in [ 0,a]} u(x,0)$.
If $x_0$ is a blow up point and \\ $\sup_{x_0\in [ 0,a]} u(x_0,T)\to \infty $
as $T\to \infty $, then
\[
\int_{M_0}^{u(x_0,T)}\frac{ds}{f(s)}\geq \delta (T-t),
\]
where $f(s)=s^q$. But if assumption \eqref{e2.1} is satisfied, this leads
to a contradiction. Therefore, any positive solution of the problem
\eqref{e1.1} must blow up in finite time $T$. Further, we get an estimate for
finite blow up time as
\[
T\leq \frac{1}{\delta }\frac{M_0^{-q+1}}{q-1}.
\]
Furthermore, we get a blow up rate for $t$ sufficiently close to $T$ as
\[
\sup_{x\in [ 0,a]} u(x,t)\leq C(T-t)^{1/(-q+1)},
\]
where $C=(\delta (q-1))^{1/(-q+1)}$.
\end{proof}

\begin{theorem} \label{thm3}
 If $q>1$ and $u_0$ is a lower solution,
then $x=a$ is the only blow up point.
\end{theorem}

\begin{proof}
Define
\[
J(x,t)=u_x-\varepsilon ( x-b_{1}) u^q\  \text{in }
[b_{1},b_{2}]\times [ \tau ,T),
\]
where $b_{1}\in [ 0,a)$, $b_{2}\in (b_{1},a]$, $\tau \in (0,T)$
and $\varepsilon $ is a positive constant to be specified later.
Then, $J(x,t)$
satisfies
\[
J_t-J_{xx}=2\varepsilon qu^{q-1}u_x+\varepsilon q(q-1)(
x-b_{1}) u^{q-2}u_x^{2}>0
\]
in $( b_{1},b_{2}) \times [ 0,T)$. $J(x,\tau )\geq 0$ by
Lemma \ref{lem1} (ii), if $\varepsilon $ is small enough. Further
\begin{gather*}
J(b_{1},t) = u_x(b_{1},t)>0, \\
J(b_{2},t) = u_x(b_{2},t)-\varepsilon ( b_{2}-b_{1}) u^q>0,
\end{gather*}
for $t\in (\tau ,T)$. By the maximum principle, we obtain that $J(x,t)\geq 0$
for $(x,t)\in [ b_{1},b_{2}]\times [ 0,T)$. Namely,
$u_x\geq \varepsilon ( x-b_{1}) u^q$
for $(x,t)\in [b_{1},b_{2}]\times [ \tau ,T)$. Integrating this with respect to $x$
from $b_{1}$ to $b_{2}$, we have
\begin{gather*}
u^{-q+1}(b_{1},t) \geq u^{-q+1}(b_{2},t)+\frac{\varepsilon
(q-1)(b_{2}-b_{1})^{2}}{2}, \\
u(b_{1},t) \leq \big[ \frac{\varepsilon (q-1)(b_{2}-b_{1})^{2}}{2}\big]
^{\frac{1}{-q+1}}<\infty .
\end{gather*}
So $u$ does not blow up in $[0,a)$. The proof is complete.
\end{proof}

\begin{theorem} \label{thm4}
 If $u_0$ is a lower solution, $q>1$ and
$u_x(x,0)\geq xu^q(x,0)$ in $(0,a)$, then $x=a$ is the only blow up
point, $a\leq 1$.
\end{theorem}

\begin{proof}
Define $J(x,t)=u_x-xu^q$ in
$[0,a]\times [ 0,T)$. Then, $J(x,t)$ satisfies
\[
J_t-J_{xx}=2qu^{q-1}u_x+q(q-1)xu^{q-2}u_x^{2},
\]
since $u_x>0$, $J(x,t)$ cannot attain a negative interior minimum. On the
other hand, $J(x,0)\geq 0$ from $u_x(x,0)\geq xu^q(x,0)$ in
$(0,a)$ and
\begin{gather*}
J(0,t) = u^{p}(0,t)>0,  \\
J(a,t) = (1-a)u^q(a,t)\geq 0,
\end{gather*}
if $a\leq 1$, for $t\in (0,T)$. By the maximum principle, we obtain that
$ J(x,t)\geq 0$, i.e. $u_x\geq xu^q$ for
$(x,t)\in [ 0,a]\times [ 0,T)$. Integrating this with respect to $x$
from $x$ to $a$, we have
\begin{gather*}
u^{-q+1}(x,t)  \geq  u^{-q+1}(a,t)+(q-1)\frac{a^{2}-x^{2}}{2} \\
u(x,t) \leq  \big[ (q-1)\frac{a^{2}-x^{2}}{2}\big] ^{\frac{1}{-q+1}
}<\infty .
\end{gather*}
So $u$ does not blow up in $[0,a)$. The proof is complete.
\end{proof}

Now, we first obtain criteria for the blow up and non-blow up using positive
steady state. The proof of the following lemma and theorem is analogous to
that of Chan and Yuen \cite{c2}. Let us consider the positive steady states of
the problem \eqref{e1.1}:
\begin{equation} \label{e2.2}
U_{xx}=0,\quad U_x( 0) =U^{p}(0),\quad U_x( a) =U^q(a).
\end{equation}
We have $U=I+nx$, where
$n=I^{p}$, $n=(I+na)^q$.
From these, we have
\begin{equation} \label{e2.3}
U=I+I^{p}x,
\end{equation}
where
\[
I^{p}=(I+I^{p}a)^q,
\]
which gives
\begin{equation} \label{e2.4}
a(I)=I^{-p}( I^{p/q}-I) .
\end{equation}
We get
\[
\lim_{I\to 0} a(I)=\lim_{I\to 0} \frac{I^{p/q}-I}{I^{p}}=\infty .
\]
But, by using L'H\^{o}pital's rule two times, we obtain
\[
\lim_{I\to 0} a(I)= \lim_{I\to 0} \frac{( \frac{p}{q}) ( \frac{p}{q}-1) I^{p/q-2}}{
p(p-1)I^{p-2}}=0
\]
for $p\neq 1$ and $q\neq 1$. If $\alpha$ is a positive number, which is
very close to $0$, then we get $a(\alpha )=0$ and $a(1)=0$. Also, If we
select $p>q$, then we note that $a(I)>0$ for $\alpha <l<1$.
Now, $a'(I)=0$ implies
\begin{equation} \label{e2.5}
I=\big[ \frac{q(1-p)}{p(1-q)}\big] ^{\frac{q}{p-q}}.
\end{equation}
We denote this value by $A$. From \eqref{e2.4},
\[
A=\big[ \frac{q(1-p)}{p(1-q)}\big] ^{\frac{p(1-q)}{p-q}}
-\big[ \frac{q(1-p)}{p(1-q)}\big] ^{\frac{q(1-p)}{p-q}}.
\]

\begin{lemma} \label{lem2}
(i) If $q\geq p$, then the steady-state problem \eqref{e2.2}
does not have a positive solution.

(ii) If $p>q$, then \eqref{e2.2} has 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}
(i) For $a(I)>0$, we have
\[
a(I)=I^{-p+p/q}-I^{-p+1}
\]
which is impossible for $q\geq p$.

(ii) Since $a(\alpha )=0=a(1)$ and $a(I)>0$ for $\alpha<l<1$, the graph of
$a(I)$ is concave downwards with maximum attained at $A$. Thus for $p>q$,
the problem \eqref{e2.2} 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{e2.5}.
\end{proof}



\begin{theorem} \label{thm5}
(a) If $p>q$ and $a\in (0,A)$, then $u$ exists globally,
provided $u_0\leq U(0)$.

(b) Suppose that the assumptions of Theorem \ref{thm2} hold. Then,
$u$ blows up in a finite time and $x=a$ is the only blow up point.
Further, if $u_x(x,0)\geq xu^q(x,0)$ in $(0,a)$, then $a\leq 1$.
\end{theorem}

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

(b) By Lemma \ref{lem1} (i), $u_t>0$ in $(0,a)\times (0,T)$.
If $u$ does not blow up in a finite time, then $u$ converges to $U$ which by
Lemma \ref{lem2} (i), does not exist for $q\geq p$. This contradiction and Theorem \ref{thm2}
shows that $u$ blows up in a finite time for $q\geq p$. Further, from
Theorem \ref{thm3}, $x=a$ is the only blow up point.
 Furthermore, from Theorem \ref{thm4}, if
$u_x(x,0)\geq xu^q(x,0)$ in $(0,a)$, then $a\leq 1$. The proof is 
complete..
\end{proof}

\section{Quenching problem}

The equivalence between the blow-up problem and the quenching problem is
well known, for example see \cite{d2,s1}. Using transform $u=1/(1-v)$ in
problem \eqref{e1.1}, we obtain the quenching problem \eqref{e1.2}.
Then \eqref{e1.2} has three heat sources for $p,q>2$. We easily get
quenching properties this difficult problem via  \eqref{e1.1}.
 First, we give an auxiliary results for \eqref{e1.2}.



\begin{remark} \label{rmk1} \rm
(i) Let $u$ and $h$ be solutions of the problem \eqref{e1.1}  and $v$
and $k$ be solutions of the problem \eqref{e1.2}. We let
$u=\frac{1}{1-v}$, $h= \frac{1}{1-k}$. 
From Theorem \ref{thm1}, If $v_0\leq k_0<1$, then
$ v(x,t,v_0)\leq k(x,t,k_0)$ on $[0,a]\times [ 0,T]$.

(ii) Let $u$ be a solution of the problem \eqref{e1.1} and
$v$ be a solution of the problem \eqref{e1.2}. We define $u=\frac{1}{1-v}$.
If $u_0$ is a lower solution of the problem \eqref{e1.1}, then we known
the following results from Lemma \ref{lem1}:
\[
u_t>0, \quad u_x>0, \quad u_{xx}>0\quad  \text{in }(0,a)\times (0,T).
\]
Similarly, we obtain
\[
v_t>0,\quad v_x>0\quad \text{in }(0,a)\times (0,T).
\]
\end{remark}

\begin{theorem} \label{thm6}
 If $u_x(x,0)\geq u^{2}(x,0)$ in $[0,a]$ and
$p,q\geq 2$ in the problem \eqref{e1.1}, then $x=a$ is the only quenching
point of  problem \eqref{e1.2}.
\end{theorem}

\begin{proof}
Let $M(x,t)=u_x(x,t)-u^{2}(x,t)$ in
$[0,a]\times [ 0,T)$ and $M(x,t)$ satisfies
\begin{gather*}
M_t-M_{xx} = 2u_x^{2}(x,t)>0\quad   \text{in }( 0,a) \times [ 0,T), \\
M(0,t) = u^{p}(0,t)-u^{2}(0,t)\geq 0,\quad 0<t<T, \\
M(a,t) = u^q(a,t)-u^{2}(a,t)\geq 0,\quad 0<t<T, \\
M(x,0) \geq 0,\quad 0\leq x\leq a,
\end{gather*}
with $u_x(x,0)\geq u^{2}(x,0)$ in $[0,a]$. By the maximum principle, we
obtain $u_x(x,t)\geq u^{2}(x,t)$ in $[0,a]\times [ 0,T)$. Then, we
have
\[
v_x(x,t)=\frac{u_x(x,t)}{u^{2}(x,t)}\geq 1\quad  \text{in }[0,a]\times [ 0,T).
\]
Let $\varepsilon \in (0,a)$, integrating this with respect to $x$ from
$a-\varepsilon $ to $a$, we have
\[
v(a-\varepsilon ,t)\leq v(a,t)-\varepsilon \leq 1-\varepsilon .
\]
So $v$ does not quench in $[0,a)$. The proof is complete.
\end{proof}



\begin{theorem} \label{thm7}
If $\lim_{t\to T}v(a,t)=1$ for some
finite time $T$, then $v_t$ blows up.
\end{theorem}

\begin{proof}
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
\[
v_{xx}+\frac{2v_x^{2}}{1-v}<M.
\]
Integrating with respect to $x$ from $0$ to $a$, we have
\begin{gather*}
\int_0^{a}\frac{v_{xx}}{v_x}dx+\int_0^{a}\frac{2v_x}{1-v}dx
<\int_0^{a}\frac{M}{v_x}dx \\
\ln \frac{v_x(a,t)}{v_x(0,t)}-2\ln ( \frac{1-v(0,t)}{(1-v(a,t)}
) < \int_0^{a}\frac{M}{v_x}dx \\
\ln \frac{( 1-v(0,t)) ^{p}}{( 1-v(a,t)) ^q}
<\int_0^{a}\frac{M}{v_x}dx.
\end{gather*}
As $t\to T^{-}$, the left-hand side tends to infinity, while the
right-hand side is finite. This contradiction shows that $v_t$ blows up at
the quenching point $x=a$.
\end{proof}

As in Section 2, let us consider the positive steady states of problem
\eqref{e1.2}.
\begin{equation} \label{e3.1}
V_{xx}=-\frac{2V_x^{2}}{1-V},\quad
V_x( 0) =( 1-V(0)) ^{-p+2},\quad
V_x( a) =( 1-V(a)) ^{-q+2}.
\end{equation}
Dividing by $V_x$ and integrating with respect to $x$, we have
$V(x)=1-\frac{1}{cx+d}$, where
$c=d^{p}$.
From these, we have
\begin{equation} \label{e3.2}
V=1-\frac{1}{d^{p}x+d},
\end{equation}
where
$c=(ca+d)^q$,
which gives
\begin{equation} \label{e3.3}
a(d)=d^{-p}( d^{p/q}-d) .
\end{equation}
We obtain
\[
\lim_{d\to 0} a(d)=\lim_{d\to 0} \frac{d^{p/q}-d}{d^{p}}=\infty .
\]
But, by using L'H\^{o}pital's rule two times, we obtain
\[
\lim_{d\to 0} a(d)=\lim_{d\to 0}
\frac{( \frac{p}{q}) ( \frac{p}{q}-1) d^{p/q-2}}{p(p-1)d^{p-2}}=0
\]
for $p\neq 1$ and $q\neq 1$. If $\beta$ is a positive number, which is
very close to $0$, then we get $a(\beta )=0$ and $a(1)=0$. Also, If we
select $p>q$, then we note that $a(d)>0$ for $\beta <d<1$. Now,
$a'(d)=0$ implies
\begin{equation} \label{e3.4}
d=\big[ \frac{q(1-p)}{p(1-q)}\big] ^{\frac{q}{p-q}}.
\end{equation}
We denote this value by $A$. From \eqref{e3.3},
\[
A=\big[ \frac{q(1-p)}{p(1-q)}\big] ^{\frac{p(1-q)}{p-q}}
-\big[ \frac{q(1-p)}{p(1-q)}\big] ^{\frac{q(1-p)}{p-q}}.
\]

\begin{lemma} \label{lem3}
(i) If $q\geq p$, then the steady-state problem \eqref{e3.1}
does not have a positive solution.

(ii) If $p>q$, then it has a solution $v$ 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} (i) For $a(d)>0$, we have
\[
a(d)=d^{-p+p/q}-d^{-p+1}
\]
which is impossible for $q\geq p$.

(ii) Since $a(\beta )=0=a(1)$ and $a(d)>0$ for
$\beta <d<1$, the graph of $a(d)$ is concave downwards with maximum attained at
$A$. Thus for $p>q$, the problem \eqref{e3.1} has a solution if and only if
$0<a\leq A$. To each $a\in (0,A)$, there are exactly two values of $d$. If $a=A$,
then $d$ is given by \eqref{e3.4}.
\end{proof}


\begin{theorem} \label{thm8}
(a) If $p>q$ and $a\in (0,A)$, then $v$ exists globally,
provided $v_0\leq V(0)$.

(b) Suppose that the assumptions of Theorem \ref{thm6} holds. If
$q\geq p$, then $x=a$ is the only quenching point. Further, if
$\lim_{t\to T}v(a,t)=1$ for some finite time $T$, then $v_t$
blows up.
\end{theorem}

\begin{proof}  (a)
By Remark \ref{rmk1} (i), $v\leq V$. Hence $v$ exists globally.

(b) By Remark \ref{rmk1} (ii), $v_t>0$ on $(0,a)\times (0,T)$.
If $v$ does not quench in a finite time, then $v$ converges to $V$ which
does not exist for $q\geq p$ by Lemma \ref{lem3} (i). This contradiction shows that
$v$ quenches for $q\geq p$. Further, from Theorem \ref{thm6}, $x=a$ is the only
quenching point. Furthermore, from Theorem \ref{thm7}, if
$\lim_{t\to T}v(a,t)=1$ for some finite time $T$, then $v_t$ blows up. 
The proof is complete.
\end{proof}

\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{Singular reactions-diffusion mixed boundary value quenching problems},
Dynamical Systems and Applications, World Sci. Ser. Appl. Anal., 4,
World Sci. Publ., River Edge, NJ, (1995), 127-137.

\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{d1} K. Deng, M. Xu;
\emph{Remarks on Blow-Up Behavior for a
Nonlinear Diffusion Equation with Neumann Boundary Conditions}, Proceedings
of the American Mathematical Society, Vol. 127, No. 1. (Jan., 1999), pp.
167-172.

\bibitem{d2} K. Deng, C-L. Zhao;
\emph{Blow-up versus quenching},
Comm. Appl. Anal., 7 (2003), 87-100.

\bibitem{f1} R. Ferreira, A. D. Pablo, F. Quiros, J. D. Rossi;
\emph{The blow-up profile for a fast diffusion equation with a nonlinear boundary
condition}, Rocky Mountain Journal of Mathematics, Volume 33, Number 1,
Spring 2003.

\bibitem{f2} A. Friedman, B. Mcleod;
\emph{Blowup of positive solutions of semilinear heat equations},
Indiana Univ. Math. J. 34 (1985), 425-477.

\bibitem{f3} 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{g1} Y. Giga, R. V. Kohn;
\emph{Asymptotic self-similar blowup of semilinear heat equations},
Comm. Pure Appl. Math. 38 (1985), 297-319.

\bibitem{g2} J.-S. Guo;
\emph{Blow-up behavior for a quasilinear
parabolic equation with nonlinear boundary condition}, Discrete and
Continuous Dynamical Systems Vol. 18 No. 1 May (2007), 71--84.

\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{l2} Z. Lin, M. Wang;
\emph{The blow up properties of solutions to semilinear heat equation
with nonlinear boundary conditions},
Z. Angew. Math. Phys., 50 (1999) 361-374.

\bibitem{o1} N. Ozalp, B. Selcuk;
\emph{The quenching behavior of a nonlinear parabolic equation with a
singular boundary condition}, Hacettepe
Journal of Mathematics and Statistics, Doi: 10.15672/HJMS.2015449429.

\bibitem{s1} T. Salin;
\emph{On quenching with logarithmic singularity},
Nonlinear Anal., 52 (2003), 261-289.

\bibitem{s2} 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} H. Zang, Z. Liu, W. Zhan;
\emph{Growth estimates and blow-up in quasilinear parabolic problems},
Applicable Analysis, Vol. 86, 2 (2007), 261-268.

\end{thebibliography}

\end{document}