\documentclass[reqno]{amsart}
\usepackage{hyperref}
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\AtBeginDocument{{\noindent\small
Seventh Mississippi State - UAB Conference on Differential Equations and
Computational Simulations,
{\em Electronic Journal of Differential Equations},
Conf. 17 (2009),  pp. 39--49.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2009 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document} \setcounter{page}{39}
\title[\hfilneg EJDE-2009/Conf/17\hfil A boundary control problem]
{A boundary control problem with a nonlinear reaction term}

\author[J. R. Cannon, M. Salman \hfil EJDE/Conf/17 \hfilneg]
{John R. Cannon, Mohamed Salman}  %  in alphabetical order

\address{John R. Cannon \newline
University of Central Florida, Department of Mathematics, Orlando,
FL 32816, USA} 
\email{jcannon@pegasus.cc.ucf.edu}


\address{Mohamed Salman \newline
Tuskegee University, Department of Mathematics, Tuskegee, AL
36088, USA}
\email{msalmanz@gmail.com}


\thanks{Published April 15, 2009.}
\subjclass[2000]{35K57, 35K55}

\keywords{Reaction-diffusion, Parabolic, Nonlocal boundary conditions}

\begin{abstract}
 The authors study the problem $u_t=u_{xx}-au$,
 $0<x<1$, $t>0$; $u(x,0)=0$, and $-u_x(0,t)=u_x(1,t)=\phi(t)$,
 where $a=a(x,t,u)$, and $\phi(t)=1$ for $t_{2k} < t<t_{2k+1}$ and
 $\phi(t)=0$ for $t_{2k+1} <t<t_{2k+2}$, $k=0,1,2,\ldots$ with
 $t_0=0$ and the sequence $t_{k}$ is determined by the equations
 $\int_0^1 u(x,t_k)dx = M$, for $k=1,3,5,\dots$, and
 $\int_0^1 u(x,t_k)dx = m$, for $k=2,4,6,\dots$,  where $0<m<M$.
 Note that the switching points $t_k$, are unknown.
 A maximum principal argument has been used to prove that the solution
 is positive under certain conditions. Existence and uniqueness are
 demonstrated. Theoretical estimates of the $t_k$ and $t_{k+1}-t_k$
 are obtained and numerical verifications of the estimates are
 presented.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}

\section{Introduction}

In this paper, we consider the problem
\begin{equation}  \label{3.1}
\begin{gathered}
u_t=u_{xx}- a(x,t,u)u, \quad  0<x<1, \; t>0,   \\
-u_x(0,t)=u_x(1,t)=\phi(t), \quad t\ge 0,  \\
u(x,0)=0, \quad 0\le x\le 1,
\end{gathered}
\end{equation}
where $a(x,t,u)$ is a continuous function and
\begin{equation}
0\leq \alpha\leq a(x,t,u)\leq\beta  \label{H1}
\end{equation}
for $(x ,t)\in[0,1]\times [0,T]$ and $u\in \mathbb R$, and
\begin{equation}
\phi(t)= \begin{cases} 1, &t_{2n} \leq t\leq t_{2n+1},\\
0, &t_{2n+1} \leq t\leq t_{2n+2}, \end{cases}  \label{3.2}
\end{equation}
where $\{ t_n\}$ depends on
\begin{equation}
\mu (t)=\int_0^1 u(x,t)dx,  \label{3.3}
\end{equation}
where
\begin{gather*}
{2} \mu (t_{2n} )= m, \quad  n=1,2, \dots, \\
\mu (t_{2n+1} ) =M, \quad n=0,1, \dots,
\end{gather*}
with $0<m<M$.

\section{Existence of solutions}

We study the existence of a solution $u(x,t)$ to \eqref{3.1} for a
given stepwise boundary conditions $\phi (t)$. We assume that
\begin{equation}
\left| ua_{u}(x,t,u)\right| \leq C,  \label{H2}
\end{equation}
uniformly for all $x,t,u$, which ensures that the source term
$F(x,t,u)=-a(x,t,u)u$ is uniformly Lipschitz with respect to $u$;
i.e.,
\begin{equation*}
| F(x,t,u_{1})-F(x,t,u_{2})| \le C|u_{1}-u_{2}| ,
\end{equation*}
for all $x,t,u_{1},u_{2}$. The constants $C$'s in the above
inequalities or that come in the sequel aren't necessarily the
same.

Under certain smoothness conditions on $u(x,t)$, $a(x,t,u)$,
problem \eqref {3.1} is equivalent to
\begin{equation}
\begin{aligned}
u(x,t) &=2\int_{0}^{t}[\theta (x,t-\tau )+\theta (1-x,t-\tau
)]\phi (\tau )d\tau   \\
&\quad +\int_{0}^{t}\int_{0}^{1}[\theta (x-\xi ,t-\tau )
+\theta (x+\xi,t-\tau )]F(\xi ,\tau ,u(\xi ,\tau ))d\xi d\tau .
\end{aligned} \label{3.1i}
\end{equation}
Let us show that the integral equation (\ref{3.1i}) has a solution
by considering the operator
\begin{align*}
Hu &=2\int_{0}^{t}[\theta (x,t-\tau )+\theta (1-x,t-\tau )]\phi
(\tau )d\tau  \\
&\quad +\int_{0}^{t}\int_{0}^{1}[\theta (x-\xi ,t-\tau )+\theta (x+\xi
,t-\tau )]F(\xi ,\tau ,u(\xi ,\tau ))d\xi d\tau ,
\end{align*}
on the set of functions
\begin{equation*}
B_{\eta }=\{u(x,t)\in C([0,1]\times [0,\eta ]),\quad\|u\| _{\eta }< \infty \},
\end{equation*}
where
\begin{equation*}
\left\| u\right\| _{\eta }=
\sup_{0\le x\le 1,\, 0\le t\le \eta} \left| u(x,t)\right| .
\end{equation*}
The set $B_{\eta }$ is a Banach space. The mapping $H$ maps
$B_{\eta }$ into into itself \cite{can1}. Furthermore,
\begin{equation*}
\left| Hu_{1}-Hu_{2}\right| \le Ct\left\| u_{1}-u_{2}\right\|
_{t},
\end{equation*}
which implies
\begin{equation*}
\left\| Hu_{1}-Hu_{2}\right\| _{\eta }\le C\eta \left\|
u_{1}-u_{2}\right\| _{\eta }.
\end{equation*}
If we select $\eta <1/C$, then $H$ is a contraction map on
$B_{\eta }$, Thus $H$ has a unique fixed point $u\in B_{\eta }$,
which solves (\ref{3.1i}). Since $F$ is uniformly Lipschitz, the
solution $u$ can be extended on any time interval $[0,T]$ (see
\cite{can1}).

\section{Maximum Principle}

In this section, we use the maximum principle to prove that
problem \eqref{3.1} has a nonnegative solution. To achieve this,
we establish the following lemmas.

\begin{lemma}\label{lem3.1}
Let $D=\{(x,t):0<x<1;0< t\le T\}$ and $a(x,t,u)$
satisfy condition \eqref{H1}. The solution $u$ of
\begin{equation} \label{3.4}
\begin{gathered}
 u_{t}=u_{xx}-a(x,t,u)u, \quad \text{in } D,    \\
 u(x,0)\ge 0,\quad  u(x,0)\ge 0,\; 0\le x\le 1,  \\
 -u_{x}(0,t)=u_{x}(1,t)=1, \quad  0\le t\le T,
\end{gathered}
\end{equation}
is nonnegative on $\overline{D}$.
\end{lemma}

\begin{proof}
To prove that $u(x,t)\geq 0$ in $\overline D$, let us
assume the converse, i.e., $u(x,t)<0$ at some point in $\overline
D$. The continuity of $u(x,t)$ on $\overline D$ implies the
existence of a negative minimum in $\overline D$. If $\min_{\overline{D}}
u=u(0,\overline{t})$, for some $0<\overline t \leq T$, then the boundary
condition $u_x(0,t)=-1$ implies $u_x<0$ in neighborhood of
$(0,\overline t )$, so $u(x,\overline t )<u(0, \overline t )$ for
some small $x$, which contradicts the fact that $u(0,\overline{t} )$ is
the minimum.

A similar argument can be used to prove that the minimum can never
happen at $x=1$. So, $u$ has its negative minimum at
$(\overline x,\overline{t} )$ in the interior of $D$.
This implies $u_t(\overline{x} ,\overline{t} )\leq 0$ and $u_{xx}(\overline{x} ,\overline{t} )\geq 0$.
Therefore,
$u_t-u_{xx}+au$ is negative at $(\overline{x} ,\overline{t})$, which is a
contradiction.
Thus, we proved the lemma.
\end{proof}

Next, we consider the problem
\begin{equation} \label{3.6}
\begin{gathered}
u_t=u_{xx}-a(x,t,u)u,  \quad (x,t)=D,   \\
u_x(0,t)=u_x(1,t)=0, \quad 0\le t\leq T,   \\
u(x,0)\geq 0, \quad 0\le x\le 1,
\end{gathered}
\end{equation}
and $a(x,t,u)$ satisfies condition \eqref{H1}. We establish the
following lemma for a closely related problem.

\begin{lemma} \label{lem3.2}
For a positive constant $\gamma $, the solution $v(x,t;\gamma )$ of
\begin{gather*}
v_{t}=v_{xx}-\gamma v, \quad \text{in } D, \\
v_{x}(0,t)=v_{x}(0,t)=0, \quad  t\ge 0, \\
v(x,0)>0, \quad 0\leq x\leq 1
\end{gather*}
is positive for all $(x,t)\in \overline{D}$ where $\gamma $ is a
positive constant.
\end{lemma}

\begin{proof}
Let $w=e^{\gamma t}v$. Then
\begin{gather*}
w_t=w_{xx}, \quad \text{in } D,\\
w_x(0,t)=w_x(1,t)=0, \quad t\ge 0,\\
w(x,0)=v(x,0)>0, \quad 0\leq x\leq 1.
\end{gather*}
If $w\leq 0$, then $w$ has a minimum that's not positive either at
$x=0$ or $x=1$ for some $t=t_0\in (0,T]$, which implies by the
strong maximum principle \cite{prot}, $w_x(0,t_0)>0$ or
$w_x(1,t_0)<0$, which is a contradiction. Hence, $w>0$, and
therefore, $v>0$.
\end{proof}

\begin{lemma} \label{lem3.3}
The solution $z(x,t,\epsilon )$ of
\begin{gather*}
 z_{t}=z_{xx} \quad \text{in }D, \\
 z_{x}(0,t)=\epsilon,\quad  0\le t\le T, \\
 z_{x}(1,t)=-\epsilon, \quad  0\le t\le T, \\
 z(x,0)=0,\quad  0\le x\le 1,
\end{gather*}
satisfies the inequality
\begin{equation*}
-C\epsilon <z(x,t)\leq 0\quad \text{in }\overline{D},
\end{equation*}
where the positive constant $C$ is a linear function of $G$.
\end{lemma}

\begin{proof}
This is a straightforward application of the strong
minimum principle and a simple comparison with
$$
u(x,t)=-2\epsilon t+\epsilon x(1-x).
$$
\end{proof}

\begin{lemma}\label{lem3.4}
The solution $u$ of \eqref{3.6} satisfies the
inequality
\begin{equation*}
0<v(x,t;\beta )\leq u(x,t)\leq v(x,t;\alpha )\quad \text{in }\overline{D},
\end{equation*}
where $\alpha $ and $\beta $ are the lower and upper bound of $a(x,t,u)$,
respectively.
\end{lemma}

\begin{proof}
First consider $v(x,t;\beta )+z(x,t;\epsilon )$. For a fixed $T$,
we can chose $\epsilon$ sufficiently small so that $v+z>0$ in $\overline{D}$.
Consider $w=u-v-z$. Clearly, $w$ satisfies
\begin{gather*}
w_t=w_{xx}-aw-(a-\beta )v \quad \text{in } D,\\
w_x(0,t)=-\epsilon , \quad 0\le t\le T,\\
w_x(1,t)=\epsilon , \quad 0\le t\le T,\\
w(x,0)=0, \quad 0\le x \le 1.
\end{gather*}
Suppose $w<0$ somewhere in $\overline{D}$. Then the boundary conditions
force a negative minimum in $D$, where
$$
w_t-w_{xx}+aw+(a-\beta )v<0,
$$
which contradicts the equation
$$
w_t-w_{xx}+aw+(a-\beta )v=0 \quad\text{in}\;\;D.
$$
Thus, $w\geq 0$ which implies that
$$
u(x,t)\geq v(x,t,\beta )+z(x,t,\epsilon )
$$
for all $\epsilon >0$ sufficiently small.  Hence,
$$
u(x,t)\geq v(x,t;\beta ) .
$$
Likewise, by considering $w=v-z-u$, the inequality
$$
v(x,t;\alpha )\geq u(x,t),
$$
follows by a similar argument.
\end{proof}

\begin{theorem} \label{thm3.5}
The solution $u$ of \eqref{3.1} is nonnegative.
\end{theorem}

\begin{proof}
By applying, successively, Lemma \ref{lem3.1} and \ref{lem3.4} on each time
stage where we keep the flux $u_x$ either  zero or one, and the
conclusion follows.
\end{proof}

\section{Existence of the Time Switches}

If we formally differentiate \eqref{3.3}, we obtain
\begin{equation}
\mu '(t)=2\phi (t)- \int_0^1 a(x,t,u)udx .  \label{3.7}
\end{equation}
To prove the existence of $t_1$, let $\phi (t)=1$ for $t>0$. In
view of hypothesis \eqref{H1}, equation (\ref{3.7}) implies the estimate
\begin{equation*}
\mu '(t)\geq 2-\beta \int_0^1 u \, dx;
\end{equation*}
that is,
\begin{equation*}
\mu '(t)\geq 2-\beta\mu (t),\quad t\geq 0.
\end{equation*}
By applying Gronwal's inequality, we get
\begin{equation*}
\mu (t)\geq \frac{2}{\beta} [1-e^{-\beta t}].
\end{equation*}
Since $\mu (t)$ is continuous, then there exists a $t_1>0$ such that
\begin{equation*}
\mu (t_1)=M,
\end{equation*}
for any $0<M<\frac{2}{\beta}$.

Next, we prove the existence of $t_2$ by taking $\phi (t)=0$ for $t>t_1$.
This implies
\begin{equation*}
\mu '(t)=-\int_0^1 a(x,t,u)u(x,t)dx, \quad t>t_1.
\end{equation*}
Using the estimate on $a(x,t,u)$, we obtain
\begin{equation*}
\mu '(t)\leq -\alpha \mu (t),\quad t\geq t_1.
\end{equation*}
Gronwal's inequality implies
\[
\mu (t) \leq \mu (t_1)e^{-\alpha (t-t_1)}
= Me^{-\alpha (t-t_1)} , \quad t\geq t_1.
\]
Since $\mu (t)$ is continuous, then there exists a $t_2>t_1$ such
that
\begin{equation*}
\mu (t_2)=m,
\end{equation*}
where $0<m<M$.

For $t>t_2$, we take $\phi (t)=1$. This gives the estimate
\begin{equation*}
\mu '(t)\geq 2-\beta\mu (t),\quad t\geq t_2.
\end{equation*}
Using the condition $\mu (t_2)=m$ and Gronwal's inequality, we get
\begin{equation*}
\mu (t)\geq \frac{2}{\beta}-\big(\frac{2}{\beta} -m\big) e^{-\beta
(t-t_2)}, \quad t\geq t_2.
\end{equation*}
Note that the coefficient $\frac{2}{\beta}-m$ is positive, which
implies the existence of $t_3>t_2$ such that
\begin{equation*}
\mu(t_3)=M.
\end{equation*}
We inductively get for $t>t_{2n}$ and $\phi (t)=1$,
\begin{equation}
\mu (t)\geq \frac{2}{\beta} -\big(\frac{2}{\beta} -m\big)
e^{-\beta (t-t_{2n})} , \quad t\geq t_{2n},  \label{3.8}
\end{equation}
which implies the existence of $t_{2n+1}$ such that $\mu(t_{2n+1})=M$.

Also, for $t>t_{2n+1}$ and $\phi(t)=0$, we have,
\begin{equation}
\mu (t)\leq Me^{-\alpha (t-t_{2n+1})}, \quad t\geq t_{2n+1} ,  \label{3.9}
\end{equation}
which ensures the existence of $t_{2n+2}$ such that $\mu (t_{2n+2})=m$.

Estimate (\ref{3.8}) implies
\begin{equation*}
M=\mu (t_{2n+1})\geq \frac{2}{\beta}-\big(\frac{2}{\beta}
-m\big) e^{-\beta (t_{2n+1}-t_{2n})},
\end{equation*}
which gives rise to
\begin{equation}
t_{2n+1}-t_{2n}\leq\frac{1}{\beta} \ln \frac{2-m\beta}{2-M\beta} .
\label{3.10}
\end{equation}
Similarly, if we employ (\ref{3.9}), we can get
\begin{equation*}
t_{2n+2}-t_{2n+1}\leq \frac{1}{\alpha} \ln \frac{M}{m}.
\end{equation*}

\section{Numerical Example}

In this section, we consider a finite difference method to discretize the
problem
\begin{gather*}
u_t=c u_{xx}- \sin u, \quad 0<x<1, \; 0<t\le T,   \\
-u_x(0,t)=u_x(1,t)=\phi(t), \quad 0<t\le T,   \\
u(x,0)=0, \quad 0<x<1,
\end{gather*}
where the boundary control function is
\begin{equation}
\phi(t)= \begin{cases} 10, & t_{2n} \leq t\leq t_{2n+1},\\
0, & \text{elsewhere}, \end{cases}
\end{equation}
and $\{ t_n\}$ depends on
\begin{equation}
\mu (t)=\int_0^1 u(x,t)dx,
\end{equation}
where
\begin{gather*}
{2} \mu (t_{2n} )= 1, \quad  n=1,2, \ldots, \\
\mu (t_{2n+1} ) =2, \quad  n=0,1, \ldots.
\end{gather*}
The time limit and the diffusivity constant are taken as $T=40$
and $c =0.05$.

Let's consider the space and time discretization
\begin{itemize}
\item[(i)] $\Delta x=\frac{1}{J}$, $x_j=j\Delta x$, $j=0,1,\dots ,J$

\item[(ii)] $\Delta t=\frac{T}{N}$, $\tau_n =n\Delta t$, $n=0,\dots ,N$

\end{itemize}
where $J=50$ and $N=400$ . The integer $N$ is chosen large enough
so that the time step $\Delta t$ is much smaller than an estimated
differences between two consecutive values of the time switches.

We consider the backward implicit finite difference scheme
\begin{equation*}
\frac{U_j^{n+1}-U_j^{n}}{\Delta t} = c \frac{U_{j-1}^{n+1}
-2U_j^{n+1} +U_{j+1}^{n+1}}{(\Delta x)^2}- \sin U_j^{n},
\end{equation*}
which can be written as
\begin{equation}
-\nu U_{j-1}^{n+1} +(1+2\nu )U_j^{n+1} -\nu U_{j+1}^{n+1} =U_j^n
-\Delta t\sin U_j^n,
\end{equation}
where $\nu =c\Delta t/(\Delta x)^2$, $j=1,\dots ,J-1$ and
$n=0,1,\dots, N-1$. The initial condition is $U_j^0=0$ for
$j=1,\dots ,J-1$, and the boundary conditions are
\begin{equation}
-\frac{U_1^n-U_0^n}{\Delta x} =\frac{U_J^n-U_{J-1}^n}{\Delta x}
=\phi(\tau_n ),
\end{equation}
for $n=0,1,\dots ,N$. The total mass integral is calculated by the following
trapezoidal rule
\begin{equation}
\mu_n =\frac{h}{2} \sum_{j=0}^{N-1} \left( U_j^{n+1} +U_{j+1}^n \right).
\label{3.11}
\end{equation}
The numerical experiment is carried out in the following way. We start by
setting the flux at $\phi=10$ then we solve a tridiagonal system coming out
of the difference method. We evaluate the total mass $\mu_n$ and compare it
with the upper threshold $M=2$. We move to the next time step while keeping
the flux at $\phi =10$, if $\mu_n<M$, or switch it to $\phi=0$, if $\mu_n\ge
M $ . At the moment, say $\tau_{n_1}$, for some integer $n_1$, when the
total mass exceeds $M$ for the first time, we take $T_1=\tau_{n_1}$ as an
approximation for the first time switch. With $\phi=0$, we move on our
solution through the time, as long as $\mu_n$ does not fall below the
threshold $m=1$. By the moment, when $\mu_{n_2} \le m$, for some integer $%
n_2 $, we set $T_2=\tau_{n_2}$, and we switch the flux back to $\phi =10$ at
the next step. We keep switching the flux on ($\phi=10$) and off ($\phi=0$)
and calculating the time switches $T_k$ until the end of the run when $%
\tau_n = 40$.

Table (1) shows the times switches $T_n$. As we can see there, the
difference between any two consecutive time switches has a tendency to
alternate between 3.8, 3.9 and 1.2. For the same set of data, graphs (1)
through (5) show the concentration versus the space at consecutive time
steps. The graphs are obtained for different stages, where at each stage the
flux is kept constant at the end points. A profile of the concentrations at $%
x=0.5$ for various times is shown in graph (6) with the same specified data.
Graph (7) shows the total mass computed through (\ref{3.11}) versus the
time. Note the slow increase and the sharp fall in the graph due to the sink
term $\sin U^n_j$.

\begin{table}[tbp]
\begin{center}
\begin{tabular}{|c|c|c|c|c|c|}
\hline
$n$ & $T_n$ & $T_n - T_{n-1}$ & $n$ & $T_n$ & $T_n - T_{n-1}$ \\ \hline
1 & 5.2000 & 5.1000 & 8 & 21.6000 & 1.2000 \\
2 & 6.4000 & 1.2000 & 9 & 25.5000 & 3.9000 \\
3 & 10.2000 & 3.8000 & 10 & 26.7000 & 1.2000 \\
4 & 11.4000 & 1.2000 & 11 & 30.6000 & 3.9000 \\
5 & 15.3000 & 3.9000 & 12 & 31.8000 & 1.2000 \\
6 & 16.5000 & 1.2000 & 13 & 35.6000 & 3.8000 \\
7 & 20.4000 & 3.9000 & 14 & 36.8000 & 1.2000 \\ \hline
\end{tabular}
\end{center}
\caption{The Time switches $T_n$ and the differences $T_n-T_{n-1}$. Note the
differences between any two consecutive times tend to alternate between 1.2
and 3.8 or 3.9.}
\end{table}

%============================

\begin{figure}[htp]
\begin{center}
\includegraphics[width=0.6\textwidth]{fig1}
\end{center}
\caption{The first stage where the flux $\phi$ is held at 10 at
the end points. Each curve shows the concentration profile at
various discrete time steps $\tau_n=n \Delta t$. As the time goes
on, the level of concentrations gets higher}
\end{figure}

\begin{figure}[htp]
\begin{center}
\includegraphics[width=0.6\textwidth]{fig2}
\end{center}
\caption{The second stage where the flux $\phi$ is held at 0 at
the end points. As the time goes on, the level of concentrations
decreases. Notice the fluctuations when the concentration is
dropped suddenly to 0 at the beginning of the stage.}
\end{figure}

\begin{figure}[htp]
\begin{center}
\includegraphics[width=0.6\textwidth]{fig3}
\end{center}
\caption{The third stage where the flux $\phi$ is switched to 10
at the end points. Each curve shows the concentration profile at
various discrete time steps. Notice the fluctuations due to the
sudden change on the concentrations. After a little while, the
concentrations levels increase monotonically. }
\end{figure}

\begin{figure}[htp]
\begin{center}
\includegraphics[width=0.6\textwidth]{fig4}
\end{center}
\caption{The fourth stage where the flux $\phi$ is switched to 0
at the end points. Notice the similarity with the second stage.}
\end{figure}

\begin{figure}[htp]
\begin{center}
\includegraphics[width=0.6\textwidth]{fig5}
\end{center}
\caption{The fifth stage where the flux $\phi$ is switched to 10
at the end points. Notice the similarity with the third stage. }
\end{figure}

\begin{figure}[htp]
\begin{center}
\includegraphics[width=0.6\textwidth]{fig6}
\end{center}
\caption{The concentration profile $U$ at $x=0.5$ versus the time
shows periodic behavior due to the periodic change of the boundary
conditions }
\end{figure}

\begin{figure}[htp]
\begin{center}
\includegraphics[width=0.6\textwidth]{fig7}
\end{center}
\caption{The total mass computed via equation (\ref{3.11}) versus
the time. Note the slow increase and the sharp fall in the graph
due to the sink term $\sin U^n_j$.}
\end{figure}



%=========================================

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(1994), 227-243.

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\end{thebibliography}

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