\documentclass[reqno]{amsart} 

\AtBeginDocument{{\noindent\small 
{\em Electronic Journal of Differential Equations},
Vol. 2004(2004), No. 112, 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  (login: ftp)}
\thanks{\copyright 2004 Texas State University - San Marcos.} 
\vspace{9mm}}

\begin{document} 

\title[\hfilneg EJDE-2004/112\hfil Internal exact controllability]
{Internal exact controllability of the linear population dynamics 
with diffusion} 

\author[B. Ainseba, S. Ani\c{t}a\hfil EJDE-2004/112\hfilneg]
{Bedr'Eddine Ainseba, Sebastian Ani\c{t}a}  % in alphabetical order

\address{Bedr'Eddine Ainseba \hfill\break
Math\'{e}matiques Appliqu\'{e}es de Bordeaux, UMR CNRS 5466\\
Case 26, UFR Sciences et Mod\'elisation\\
Universit\'{e} Victor Segalen Bordeaux 2, 33076 Bordeaux Cedex, France}
\email{ainseba@sm.u-bordeaux2.fr}

\address{Sebastian Ani\c{t}a \hfill\break 
Faculty of Mathematics, University ``Al.I. Cuza'' and Institute of
Mathematics of the Romanian Academy, Ia\c{s}i 6600, Romania}
\email{sanita@uaic.ro}


\date{}
\thanks{Submitted January 4, 2004. Published September 29, 2004.}
\subjclass[2000]{93B05, 35K05, 46B70, 92D25}
\keywords{Exact controllability; age-structured population dynamics}


\begin{abstract}
 We consider the internal exact controllability of a linear age and space
 structured population model with nonlocal birth process. The control acts
 only in a spatial subdomain and only for small age classes. The methods we
 use combine the Carleman estimates for the backward adjoint system, some
 estimates in the theory of parabolic boundary value problems in $L^k$ and
 the Banach fixed point theorem.
\end{abstract}

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

\section{Introduction}

Let $\Omega $ be a bounded domain in $\mathbb{R}^n$ $(n\leq 3)$
with a smooth boundary $\partial \Omega $. Assume that a
biological population is free to move in the environment
$\Omega$. We denote by $y(a,t,x) $ the density of individuals of age
$a\geq 0$ at time $t\geq 0$ and location $x \in \overline{ \Omega}$
and assume that the flux of population takes the form $k\nabla y(a,t,x)$
with $k>0$, where $\nabla $ is the gradient vector with
respect to the spatial variable. Let $A$ be the life expectancy of
an individual and $T$ be a positive constant. Let $\beta (a)$ be
the natural fertility rate and $\mu (a)$ the natural mortality
rate corresponding to individuals of age $a$. The dynamics of the
population is described by the following model
\begin{equation}
\begin{gathered}
Dy+\mu (a)y-k\Delta y = f(a,x)+m(a,x)u(a,t,x), \quad (a,t,x)\in Q_T \\
 \frac{\partial y}{\partial \nu }(a,t,x) =0, \quad (a,t,x)\in \Sigma _T \\
 y(0,t,x)=\int_0^A\beta (a)y(a,t,x) da, \quad (t,x)\in (0,T)\times \Omega \\
 y(a,0,x)=y_0(a,x) , \quad (a,x)\in (0,A ) \times \Omega ,
\end{gathered} \label{e1}
\end{equation}
where $u$ is the control and $m$ is the characteristic function of
$(0,a^*)\times \omega $, $f$ is the density of an infusion of population and
$y_0$ is the initial population density. Here $a^*\in (0,A]$ and
 $\omega \subset \subset \Omega $ is a nonempty open subset, 
 $Q_T=(0,A)\times (0,T)\times \Omega $, 
$\Sigma_T=(0,A)\times (0,T)\times \partial \Omega $.

We  denote by
\[
Dy(a,t,x)={\lim_{\varepsilon \to 0} }\frac{y\left( a+\varepsilon
,t+\varepsilon, x \right) -y\left( a,t,x\right) }{\varepsilon }
\]
the directional derivative of $y$ with respect to the direction $\left(
1,1,0\right) $. If $y$ is smooth enough then
\[
Dy=\frac {\partial y}{\partial t}+\frac {\partial y}{\partial a}.
\]
The control acts only in the spatial set $\omega $ and for ages between $0$
and $a^*$.

Let $y_s$ be a nonnegative steady-state of \eqref{e1}, corresponding to 
$u \equiv 0$ and such that
\begin{equation}
y_s(a,x)\geq \rho _0>0\quad \mbox{\rm a.e. }(a,x)\in (0,a_1^*)\times \Omega ,
\label{e2}
\end{equation}
where $\rho _0>0$ is constant and $a_1^*\in (0,A)$ is a constant which will be
defined later.

The main goal of this paper is to prove the existence of a control $u$ such
that the solution $y$ of \eqref{e1} satisfies
\begin{equation}
\begin{gathered}
y(a,T,x)=y_s(a,x) \quad \mbox{\rm a.e. }(a,x)\in (0,A)\times \Omega,\\
y(a,t,x)\geq 0 \quad \mbox{\rm a.e. }(a,t,x)\in Q_T.
\end{gathered}\label{e3}
\end{equation}
Condition \eqref{e3} is natural because $y$ represents the density of a population.
We notice that if $y$ is the solution to \eqref{e1}, then $y-y_s$ is the solution
to
\begin{equation}
\begin{gathered}
Dz+\mu (a)z-k\Delta z=m(a,x)u(a,t,x), \quad (a,t,x)\in Q_T \\
 \frac{\partial z}{\partial \nu }(a,t,x) =0,\quad (a,t,x)\in \Sigma _T \\
 z(0,t,x)=\int_0^A\beta (a)z(a,t,x) da, \quad (t,x)\in (0,T)\times \Omega \\
 z(a,0,x)=z_0(a,x) ,  (a,x)\in (0,A)\times \Omega ,
\end{gathered}  \label{e4}
\end{equation}
where $z_0=y_0-y_s$.

The above formulated problem is equivalent to the exact null controllability
problem with state constraints for \eqref{e4}. Indeed, if we denote now by $z$ the
solution to \eqref{e4}, then condition \eqref{e3} becomes
\[
z(a,t,x)\geq -y_s(a,x) \quad \mbox{\rm a.e. }(a,t,x)\in Q_T .
\]
We recall that the internal null controllability of the linear heat
equation, when the control acts on a subset of the domain, was established
by G. Lebeau and L. Robbiano \cite{l4} and was later extended to some semilinear
equation by A.V. Fursikov and O.Yu. Imanuvilov \cite{f2}, in the sublinear case
and by V. Barbu \cite{b1} and E. Fernandez--Cara \cite{f1}, in the superlinear case. The
internal null controllability of the age-dependent population dynamics in
the particular case when the control acts in a spatial subdomain $\omega $
but for all ages $a$ (this is the particular case corresponding to $a^*=A$)
was investigated by B. Ainseba and S. Ani\c{t}a \cite{a2}.

This paper is organized as follows. We first give the hypotheses and state
the main result. The existence of a steady--state of \eqref{e1} with $u \equiv 0$
is established in Section 3. The proof of the local exact null
controllability is given in Section 4. The proof is based on Carleman's
inequality for the backward adjoint system associated with \eqref{e4}. 

\section{Assumptions and the main result}

Assume that the following hypotheses hold:

\begin{itemize}
\item[(H1)]   $\beta \in L^{\infty }(0,A)$,
$\beta (a)\geq 0$ a.e. $a\in (0,A)$ \\
 There exists $a_{0},a_{1}\in (0,A)$, $a_{0}<a_{1}$, such
that $\beta (a)=0$ a.e. \\
$a\in (0,a_{0})\cup (a_{1},A)$ and $\beta (a)>0$ a.e.
in $(a_{0},a_{1})$

\item[(H2)]  $\mu \in C([0,A))$, $\mu (a)\geq 0$
 a.e. $a\in (0,A)$, $ \int_{0}^{A}\mu (a)da=+\infty $

\item[(H3)]  $y_{0}\in L^{\infty }((0,A)\times \Omega )$,
$y_{0}(a,x)\geq 0$ a.e. in $(0,A)\times \Omega $\\
$f\in L^{\infty }((0,A)\times \Omega )$, $f(a,x)\geq 0
$ a.e. in $(0,A)\times \Omega $.
\end{itemize}

For the biological significance of the hypotheses and the basic existence
results for the solution to \eqref{e1} we refer to \cite{a3,g1,g2,i1,l2,w1}.

Let $y_s$ be a nonnegative steady-state of \eqref{e1}, corresponding to $u \equiv 0$ and such
that
\[
y_s(a,x)\geq \rho _0>0\quad \mbox{\rm a.e. }(a,x)\in (0,a_1)\times \Omega ,
\]
where $\rho _0>0$ is a constant.

Denote by $z_0=y_0-y_s$. Then we have the following internal controllability
result

\begin{theorem} \label{thm2.1}
Let $T>A-a^{*}$ be arbitrary but fixed. If $\| y_{0}-y_{s}\|
_{L^{\infty }((0,A)\times \Omega )}$ is small enough, then there exists
$u\in L^{2}(Q_{T})$ such that the solution $y$ of \eqref{e1} satisfies
\begin{equation}
\begin{gathered}
y(a,T,x)=y_{s}(a,x)\quad \mbox{\rm a.e. }(a,x)\in (0,A)\times \Omega \\
y(a,t,x)\geq 0\quad \mbox{\rm a.e. }(a,t,x)\in Q_{T}.
\end{gathered}    \label{e51}
\end{equation}
If $T<A-a^{*}$ and if $\| y_{0}-y_s\|_{L^{\infty }((a^*,A-T)\times \Omega )}>0$,
then there is no $u\in L^{2}(Q_{T})$ such that the solution $y$ of \eqref{e1} to
satisfy \eqref{e51}.
\end{theorem}

This result can be equivalently formulated as follows

\begin{theorem} \label{thm2.2}
Let $T>A-a^{*}$ be arbitrary but fixed. If $\| z_{0}
\|_{L^{\infty }((0,A)\times \Omega )}$ is small enough, then there
exists $u\in L^{2}(Q_{T})$ such that the solution $z$ of \eqref{e4} satisfies
\begin{equation}
\begin{gathered}
z(a,T,x)=0\quad \mbox{\rm a.e. }(a,x)\in (0,A)\times \Omega \\
z(a,t,x)\geq -y_{s}(a,x)\quad \mbox{\rm a.e. }(a,t,x)\in Q_{T}.
\end{gathered}\label{e5}
\end{equation}
If $T<A-a^{*}$ and if $\| z_0\|_{L^{\infty }((a^*,A-T)\times \Omega )}>0$,
then there is no $u\in L^{2}(Q_{T})$ such that the solution $z$ of \eqref{e4} to
satisfy \eqref{e5}.
\end{theorem}

\section{Existence of steady states for \eqref{e1}}

In this section we shall remind some results (see \cite{a2}) concerning the
existence of $y_s$, a nonnegative steady-state of \eqref{e1}, corresponding to $u \equiv 0$,
which satisfies \eqref{e2}. $y_s$ should be a solution to
\begin{equation}
\begin{gathered}
 \frac{\partial y_s}{\partial a}+\mu (a)y_s -k\Delta y_s=f(a,x),
\quad (a,x)\in (0,A)\times \Omega \\
\frac{\partial y_s}{\partial \nu }(a,x)=0, \quad (a,x)\in (0,A)
\times \partial \Omega \\
 y_s(0,x)=\int_0^{A }\beta (a) y_s(a,x) da, \quad  x\in \Omega\,.
\end{gathered} \label{e6}
\end{equation}
Denote by
$$ R= \int_0^{A }\beta (a)\exp\big({-\int_0^a\mu(s)ds}da\big)
$$
the reproductive number and consider $f_0$ a nonnegative constant.

\begin{theorem} \label{thm3.1}
\begin{itemize}
\item If $R<1$ and $f(a,x)\geq f_0>0$ a.e. $(a,x)\in (0,A)\times \Omega $,
then there exists a unique nonnegative solution to \eqref{e6},
which in addition satisfies \eqref{e2}.

\item  If $R=1$ and $f\equiv 0$, then there exist infinitely many
nonnegative solutions to \eqref{e6}, which satisfy \eqref{e2}.

\item  If $R>1$, then there is no nonnegative solution to \eqref{e6}, satisfying
\eqref{e2}.
\end{itemize}
\end{theorem}

\begin{proof}
 If $R<1$, then there exists a unique and nonnegative solution to \eqref{e6}
(this follows by Banach's fixed point theorem). Since $f(a,x)\geq f_0>0$
a.e. $(a,x)\in (0,A)\times \Omega $, then by the comparison result in
\cite{g1}(see also \cite{a3}) we get that
\[
y_s(a,x)\geq y_i(a,t,x) \quad \mbox{\rm a.e. }(a,t,x)\in Q= (0,A )\times
(0,+\infty )\times \Omega ,
\]
where $y_i$ is the solution to
\begin{gather*}
Dy_i+\mu y_i-k\Delta y_i=f_0, \quad (a,t,x) \in Q \\
 \frac{\partial y_i}{\partial \nu }=0, \quad (a,t,x)\in \Sigma \\
 y_i(0,t,x) =\int_0^{A }\beta (a)y_i(a,t,x) da, \quad (t,x)\in
(0,+\infty )\times \Omega \\
 y_i(a,0,x)=0, \quad (a,x)\in (0,A )\times \Omega
\end{gather*}
Note that $\Sigma =(0,A)\times (0,+\infty )\times \partial \Omega $); $y_i$
does not explicitly depend on $x$. So, we shall write $y_i(a,t)$ instead of $
y_i(a,t,x)$. It means that
\[
y_s(a,x)\geq y_i(a,t)\quad \forall t\in [0,+\infty ), \quad \mbox{\rm a.e.}
(a,x)\in (0,A)\times \Omega,
\]
and that $y_i$ is the solution of
\begin{gather*}
Dy_i+\mu y_i=f_0, \quad (a,t)\in (0,A)\times (0,+\infty ) \\
 y_i(0,t)=\int_0^{A}\beta (a)y_i(a,t)da, \quad  t\in
(0,+\infty ) \\
 y_i(a,0)=0, \quad a\in (0,A) .
\end{gather*}
For $t>A$ we have
$y_i(0,t)>0$ and $y_i(0,t)$ is continuous with respect to $ t$
(see \cite{a3}). As a consequence we obtain that there exists $\rho _0>0$ such
that, for $t$ large enough, and for any $a \in (0,a_1^*) $,
\[
y_i(a,t) >\rho _0,
\]
and in conclusion we get that $y_s$ satisfies \eqref{e2}. 

If $R=1$ and $f\equiv 0$, then all the solutions of \eqref{e6} which are satisfying
\eqref{e2} are given by
\[
y(a,x)= ce^{-\int_0^a\mu (s)ds},\quad (a,x)\in (0,A ) \times
\Omega,
\]
where $c\in \mathbb{R}^*_+$ is an arbitrary constant. The conclusion is now
obvious. 

If $R>1$ and if it would exist a nonnegative solution $y_s$ to \eqref{e6}
satisfying \eqref{e2}, then $y(a,t,x)=y_s(a,x)$, $(a,t,x)\in \overline{Q}$ is the
solution to
\begin{gather*}
Dy+\mu y-k\Delta y=f(a,x), \quad (a,t,x)\in Q \\
 \frac{\partial y}{\partial \nu }=0, \quad (a,t,x)\in \Sigma \\
 y(0,t,x)=\int_0^{A}\beta (a)y(a,t,x) ,\quad  (t,x)\in
(0,+\infty )\times \Omega \\
 y(a,0,x)=y_s(a,x) , \quad (a,x)\in (0,A)\times \Omega
\end{gather*}
and for $t\to +\infty $ we have (see \cite{a3,l3})
\[
\lim_{t\to +\infty }\| y(t) \|_{L^2((0,A)\times \Omega )}=+\infty .
\]
On the other hand
\[
\| y(t)\| _{L^2((0,A)\times \Omega )}= \| y_s
\|_{L^2((0,A)\times \Omega )},
\]
and so $\| y_s\| _{L^2((0,A)\times \Omega )}=+\infty $, which is
absurd.\end{proof}

\section{Proof of the main result}

We shall prove Theorem \ref{thm2.2} (which is equivalent to Theorem \ref{thm2.1}). We intend
to use the general Carleman inequality for linear parabolic equations given
in \cite{f2}. Namely, let $\widetilde{\omega }\subset \subset \omega $ be a
nonempty bounded set, $T_0\in (0,+\infty )$ and $\psi \in C^2(\overline{
\Omega })$ be such that
\[
\psi (x)>0, \ \forall x\in \Omega , \quad \psi (x)=0, \ \forall x\in
\partial \Omega , \quad \left| \nabla \psi (x)\right| >0, \ \forall x\in
\overline{\Omega }\setminus \widetilde{\omega }
\]
and set
\[
\alpha (t,x) = \frac{e^{\lambda \psi (x)}-e^{2\lambda \|
\psi \| _{C(\overline{\Omega }) }}}{t(T_0-t) },
\]
where $\lambda $ is an appropriate positive constant.
Denote by $D_{T_0}=(0,T_0)\times \Omega $.

\begin{lemma} \label{lm4.1}
There exist positive constants $C_{1},s_{1}$ such that
\begin{equation} \label{e7}
\begin{aligned}
&\frac{1}{s}\int_{D_{T_{0}}}t\left( T_{0}-t\right) e^{2s\alpha }
\left( \left| w_{t}\right| ^{2}+\left| \Delta w\right| ^{2}\right) dx\,dt\\
&+s\int_{D_{T_{0}}}\frac{e^{2s\alpha }}{t\left( T_{0}-t\right) }\left|
\nabla w\right| ^{2}dx\,dt
+s^{3}\int_{D_{T_{0}}}\frac{e^{2s\alpha }}{t^{3}\left( T_{0}-t\right) ^{3}}
\left| w\right| ^{2}dx\,dt\\
&\leq C_{1}\Big[ \int_{D_{T_{0}}}e^{2s\alpha }\left| w_{t}+\Delta w\right|
^{2}dx\,dt+s^{3}\int_{(0,T_{0})\times \omega }\frac{e^{2s\alpha }}{
t^{3}\left( T_{0}-t\right) ^{3}}\left| w\right| ^{2}dx\,dt\Big] ,
\end{aligned}
\end{equation}
for all $w\in C^{2}(\overline{D}_{T_{0}})$,
$\frac{\partial w}{\partial \nu }(t,x)=0$,
$\forall (t,x)\in (0,T_{0})\times\partial \Omega $ and $s\geq s_{1}$.
\end{lemma}

The proof of this result can be found in \cite{f2}. \smallskip

If $a^{*}=A$, the result has already been proved in \cite{a2}. We shall treat now
the case $a^*\in (0,A)$. Consider $a_1^*:=a^*$.
Let us choose $T_{0}\in (0,\min\{a_{0},a^{*},A-a^{*},T-A+a^{*},A-a_{1}\})$.
Define
\[
K=L^{\infty }\left( (0,A-a^{*}+T_{0})\times \Omega \right) .
\]
In what follows we shall denote by the same symbol $C$, several constants
independent of $z_0$ and all other variables.
For $b\in K$ arbitrary but fixed and for any $\varepsilon >0$, consider the
following optimal control problem:\\
Minimize
\begin{equation}
\Big\{ \int_G\int_\Omega \varphi (a,t,x)|u(a,t,x)|^2dx\,dt\,da+
\frac1\varepsilon \int_{\Gamma _0}\int_\Omega \left| z(a,t,x) \right|^2dx\,dl
\Big\} ,
\label{Pe} % (P_\varepsilon)
\end{equation}
subject to \eqref{e8} ($u\in L^2(G\times \Omega )$ and $z$ is the solution 
of \eqref{e8} corresponding to $u$). Here
\begin{gather*}
G=(0,a^*)\times (0,T_0)\cup (0,T_0)\times (0,A-a^*+T_0), \\
\Gamma _0=\{ T_0\} \times (T_0, A-a^*+T_0)\cup (T_0,a^*)\times \{ T_0\},\\
\varphi (a,t,x)=\begin{cases}
 e^{-2s\alpha (t,x) }t^3(T_0-t) ^3, &\mbox{if } t<a, \;(a,t)\in G \\
 e^{-2s\alpha (a,x)}a^3(T_0-a) ^3, &\mbox{if }a<t, \; (a,t)\in G 
\end{cases}
\end{gather*}
(See figure 1).
\begin{equation}
\begin{gathered}
Dz+\mu z-k\Delta z  =m(a,x)u(a,t,x) , \quad (a,t,x)\in G\times \Omega \\
 \frac{\partial z}{\partial \nu }=0, \quad (a,t,x)\in G\times
\partial \Omega \\
 z(0,t,x)=b(t,x) , \quad (t,x)\in (0,A-a^*+T_0)\times \Omega \\
 z(a,0,x)=z_0(a,x), \quad (a,x)\in (0,a^*)\times \Omega .
\end{gathered} \label{e8}
\end{equation}

\begin{figure}[ht]
\unitlength=0.8mm  
\begin{picture}(123.00,85.00)(10,10)
\put(37.00,14.00){\vector(0,1){71}}
\put(37.00,14.00){\vector(1,0){86}}
\put(30.00,45.00){\makebox(0,0)[cc]{$A-a^*$}}
\put(26.00,60.00){\makebox(0,0)[cc]{$A-a^*+T_0$}}
\put(33.00,10.00){\makebox(0,0)[cc]{$O$}}
\put(53.00,10.00){\makebox(0,0)[cc]{$T_0$}}
\put(58.00,10.00){\makebox(0,0)[cc]{$a_0$}}
\put(85.00,10.00){\makebox(0,0)[cc]{$a^*$}}
\put(95.00,10.00){\makebox(0,0)[cc]{$a_1$}}
\put(115.00,10.00){\makebox(0,0)[cc]{$A$}}
\put(70.00,10.00){\makebox(0,0)[cc]{$a^*-T_0$}}
\put(70,14.00){\line(1,1){45}} 
\put(115.00,14.00){\line(0,1){45}}
\put(37.00,44.00){\line(1,0){78}}
\put(85.00,14.00){\line(1,1){30}}
\put(37.00,59.00){\line(1,0){78}}
\put(123.00,10.00){\makebox(0,0)[cc]{$a$}}
\put(33.00,30.00){\makebox(0,0)[cc]{$T_0$}}
\put(33.00,85.00){\makebox(0,0)[cc]{$t$}}
\put(37.00,44.00){\line(1,1){15}}
\put(33.00,77.00){\makebox(0,0)[cc]{$T$}} 
\thicklines
\put(37.00,59.00){\line(1,0){15}}
\put(52.00,59.00){\line(0,-1){15}}
\put(52.00,44.00){\line(0,-1){15}}
\put(52.00,29.00){\line(1,0){33}}
\put(85.00,29.00){\line(0,-1){15}}
\end{picture}
\caption{}
\end{figure}

Denote by $\Psi _{\varepsilon }(u)$ the value of the cost function in $u $.
Since the cost function $\Psi _{\varepsilon }:L^2(G\times \Omega )
\to \mathbb{R}^+$ is convex, continuous and
\[
\lim _{\| u\| _{L^2{(G\times \Omega )}}\to +\infty } \Psi
_{\varepsilon }(u) =+\infty ,
\]
then it follows that there exists at least one minimum point for $
\Psi_\varepsilon$ and consequently an optimal pair $(u_{\varepsilon
},z_{\varepsilon})$ for $\mathbf{(P_{\varepsilon })}$. By standard arguments
we have
\begin{equation}
u_\varepsilon (a,t,x)=\tilde{m}(x)q_\varepsilon (a,t,x) {\varphi}^{-1} (a,t,x) \
\mbox{\rm a.e. }(a,t,x)\in G\times \Omega , \label{e9}
\end{equation}
where $\tilde{m}$ is the characteristic function of $\omega $ and
$q_\varepsilon $ is the solution of
\begin{equation}
\begin{gathered}
Dq-\mu q+k\Delta q=0, \quad (a,t,x)\in G\times \Omega \\
 \frac{\partial q}{\partial \nu }=0, \quad (a,t,x)\in G\times
\partial \Omega \\
 q(a,t,x)=0, \quad (a,t,x)\in (\Gamma \setminus \Gamma _0)\times
\Omega \\
 q(a,t,x)=-\frac 1\varepsilon z_\varepsilon (a,t,x) , \quad (a,t,x)
\in \Gamma _0\times \Omega .
\end{gathered}  \label{e10}
\end{equation}
Here $\Gamma =(0,T_0)\times \{A-a^*+T_0\} \cup \{a^* \} \times (0,T_0) \cup
\Gamma _0 $. 

Multiplying the first equation in \eqref{e10} by $z_\varepsilon $ and integrating
on $G\times \Omega $ we obtain after some calculation (and using \eqref{e8} and
\eqref{e9}) that
\begin{align*}
&\int_G\int_\omega \varphi (a,t,x)| u_\varepsilon(a,t,x)| ^2dx\,da\,dt +\frac
1\varepsilon \int_{\Gamma _0}\int_\Omega |z_\varepsilon(a,t,x)| ^2dx\,dl\\
&=-\int_0^{A-a^*+T_0}\int_\Omega b(t,x)q_\varepsilon (0,t,x)dx\,dt-
\int_0^{a^*}\int_\Omega z_0(a,x) q_\varepsilon (a,0,x)dx\,da.
\end{align*}
Let $S$ be an arbitrary characteristic line of equation
\[
S=\left\{ (\gamma +t,\theta +t) ; \ t\in (0,T_0) ,\,(\gamma ,\theta ) \in
(0,a^*-T_0)\times \{ 0\} \cup \{ 0\} \times (0,A-a^*)\right\} .
\]
Define
\begin{gather*}
\widetilde{u} (t,x)=u(\gamma +t,\theta +t,x), \quad
(t,x)\in (0,T_0)\times \Omega \\
\widetilde{z}_\varepsilon (t,x)=z_{\varepsilon }(\gamma +t,\theta +t,x), \quad
(t,x)\in (0,T_0)\times \Omega \\
\widetilde{q}_\varepsilon (t,x)=q_\varepsilon (\gamma +t,\theta +t,x) , \quad
(t,x)\in (0,T_0)\times \Omega \\
\widetilde{\mu }(t)=\mu (\gamma +t), \quad t\in (0,T_0).
\end{gather*}
Note that $(\widetilde{u}_\varepsilon,\widetilde{z}_\varepsilon )$ satisfies
\begin{equation}
\begin{gathered}
 (\widetilde{z}_\varepsilon ) _t +\widetilde{\mu } \widetilde{z}
_\varepsilon -k\Delta \widetilde{z}_\varepsilon =\tilde{m}(x)\widetilde{u}
_\varepsilon (t,x) , \quad (t,x)\in (0,T_0)\times \Omega
\\
 \frac{\partial \widetilde{z}_\varepsilon }{\partial \nu }=0, \quad
(t,x)\in (0,T_0)\times \partial \Omega
\\
 \widetilde{z}_\varepsilon (0,x)= \begin{cases}
b(\theta ,x)&\gamma =0,\; x\in \Omega \\
z_0(\gamma ,x) & \theta =0,\; x\in \Omega
\end{cases}
\end{gathered} \label{e11}
\end{equation}
By \eqref{e9} we get that
\begin{equation}
\widetilde{u}_\varepsilon (t,x)=\tilde{m}(x)\widetilde{q}_\varepsilon (t,x)\cdot
\frac {e^{2s\alpha (t,x) }}{t^3(T_0-t) ^3} \label{e12}
\end{equation}
a.e. $(t,x)\in (0,T_0)\times \Omega $,
\begin{equation}
\begin{gathered}
\left( \widetilde{q}_\varepsilon \right) _t+k\Delta \widetilde{q}
_\varepsilon =\widetilde{\mu }\widetilde{q}_\varepsilon , \quad (t,x)\in (0,T_0)
\times \Omega
\\
 \frac{\partial \widetilde{q}_\varepsilon }{\partial \nu }=0, \quad
(t,x)\in (0,T_0)\times \partial \Omega
\\
 \widetilde{q}_\varepsilon (T_0,x) =-\frac 1\varepsilon
\widetilde{z}_\varepsilon (T_0,x) \quad x\in \Omega .
\end{gathered} \label{e13}
\end{equation}
Multiplying the first equation in \eqref{e13} by $\widetilde{z}_\varepsilon$ and
integrating on $D_{T_0}$, we obtain that
\begin{equation}
\begin{aligned}
&\int_0^{T_0}\int_\omega e^{-2s\alpha (t,x)}t^3(T_0-t) ^3 | \widetilde{u}
_\varepsilon (t,x)| ^2dx\,dt+\frac 1{\varepsilon } \int_\Omega | \widetilde{z}
_\varepsilon (T_0,x)| ^2dx\\
&=-\int_\Omega \widetilde{z}_\varepsilon (0,x)\widetilde{q}_\varepsilon (0,x)
dx.
\end{aligned} \label{e14}
\end{equation}
By Carleman's inequality \eqref{e7} we infer that
\begin{align*}
&\int_0^{T_0}\int_\Omega e^{2s\alpha }[ \frac{t(T_0-t)}s \left( \left| \left(
\widetilde{q}_\varepsilon \right) _t\right| ^2+\left| \Delta \widetilde{q}
_\varepsilon \right| ^2\right) +\frac s{t(T_0-t) }\left| \nabla \widetilde{q}
_\varepsilon \right| ^2\\
&+\frac{s^3}{t^3(T_0-t)^3}\left| \widetilde{q}_\varepsilon \right| ^2]dx\,dt\\
&\leq C_1 \Big[  \int_0^{T_0}\int_\Omega
e^{2s\alpha }\| \widetilde{\mu }\| _{C\left( \left[ 0,T_0\right]
\right) }^2\cdot \left| \widetilde{q}_\varepsilon \right| ^2dx\,dt + s^3
 \int_{(0,T_0) \times \omega } \frac{e^{2s\alpha }}{t^3(T_0-t)^3
} \left| \widetilde{q}_\varepsilon \right| ^2dx\,dt \Big]
\end{align*}
and consequently
\begin{equation}
\begin{aligned}
&\int_0^{T_0}\int_\Omega e^{2s\alpha }[ \frac{t(T_0-t) } s\left( | \left(
\widetilde{q}_\varepsilon \right) _t | ^2+| \Delta \widetilde{q}_\varepsilon
| ^2\right) +\frac s{t(T_0-t)}| \nabla \widetilde{q}_\varepsilon | ^2\\
&+\frac{s^3}{t^3(T_0-t) ^3}| \widetilde{q}_\varepsilon | ^2] dx\,dt \\
&\leq C\int_0^{T_0}\int_\omega e^{2s\alpha }\frac{s^3}{t^3(T_0-t)^3} \left|
\widetilde{q}_\varepsilon \right| ^2dx\,dt,
\end{aligned}\label{e15}
\end{equation}
for $ s\geq \max(s_1,\,C\| \mu \| _{C\left( \left[
0,a^*\right]\right) }^{\frac 23}).$
Multiplying the first equation in \eqref{e13} by $\widetilde{q}_\varepsilon $ we
obtain that
\[
\frac 12\frac d{dt}\int_\Omega \left| \widetilde{q}_\varepsilon (t,x)
\right|^2dx-k\int_\Omega \left| \nabla \widetilde{q}_\varepsilon (t,x)
\right|^2dx-\int_\Omega \widetilde{\mu }(t)\left| \widetilde{q}_\varepsilon
(t,x)\right| ^2dx=0
\]
and
\[
\frac d{dt}\int_\Omega \left| \widetilde{q}_\varepsilon (t,x)\right| ^2dx
\geq 0 \quad \mbox{\rm a.e. }t\in (0,T_0).
\]
Integrating the last inequality we get that
\[
\int_\Omega \left| \widetilde{q}_\varepsilon (0,x)\right| ^2dx\leq
C\int_0^{T_0} \int_\Omega \left| \widetilde{q}_\varepsilon (t,x) \right| ^2
\frac{e^{2s\alpha \left( x,t\right) }}{t^3\left( T_0-t\right) ^3}dx.
\]
and by Carleman's inequality we have that
\begin{equation}
\int_\Omega \left| \widetilde{q}_\varepsilon (0,x)\right| ^2dx\leq
C\int_0^{T_0}\int_\omega \left| \widetilde{q}_\varepsilon (t,x)
\right|^2\cdot \frac{e^{2s\alpha \left( x,t\right) }}{t^3(T_0-t)^3}dx\,dt.
\label{e16}
\end{equation}
By Young's inequality, \eqref{e14}, \eqref{e16} and \eqref{e12} we obtain that
\begin{align*}
&\int_{(0,T_0)\times \omega }e^{-2s\alpha }t^3(T_0-t) ^3\left| \widetilde{u}
_\varepsilon (t,x)\right| ^2dx\,dt+\frac 1{\varepsilon } \int_\Omega \left|
\widetilde{z}_\varepsilon (T_0,x) \right| ^2dx\\
&\leq C\| \widetilde{z}_\varepsilon (0)\| _{L^2(\Omega )}^2,
\end{align*}
for $ s\geq \max (s_1,\,C\| \mu \|_{C\left( \left[
0,a^*\right]\right) }^{\frac 23})$.
Using now \eqref{e15} we get
\begin{align*}
&\int_0^{T_0}\int_\Omega e^{2s\alpha }[ \frac{t(T_0-t) }s\left( | \left(
\widetilde{q}_\varepsilon \right) _t | ^2+| \Delta \widetilde{q}_\varepsilon
| ^2\right) \\
&+\frac s{t(T_0-t)}| \nabla \widetilde{q} _\varepsilon | ^2
+\frac{s^3}{t^3(T_0-t) ^3}|\widetilde{q}_\varepsilon | ^2] dx\,dt \leq
C\| \widetilde{z}_\varepsilon (0) \| _{L^2\left( \Omega \right)
}^2,
\end{align*}
for any $\varepsilon >0$ and consequently
\[
\| \widetilde{v}_\varepsilon \| _{W_2^{1,2}\left( (0,T_0)\times
\Omega \right) }^2\leq C\| \widetilde{z}_\varepsilon (0) \|
_{L^2(\Omega )}^2,
\]
where $\widetilde{v}_\varepsilon (t,x)=\frac {e^{2s\alpha
(t,x)}}{t^{3}(T_0-t)^{3}} \widetilde{q}_\varepsilon$, $(t,x)\in
(0,T_0)\times \Omega $. As
\[
W_2^{1,2}\left( (0,T_0)\times \Omega \right) \subset L^l((0,T_0)\times
\Omega )
\]
(where $l=+\infty $ for $N=1,2$ and $ l=10$ for $N=3$), we may
infer that
\begin{equation}
\| \widetilde{u}_\varepsilon \| _{L^{10}\left( (0,T_0)\times
\Omega \right) }^2=\| m\widetilde{v}_\varepsilon \|_{L^{10}\left(
(0,T_0)\times \Omega \right) }^2\leq C\| \widetilde{z} _\varepsilon (0)
\| _{L^2(\Omega )}^2 ,\label{e17}
\end{equation}
for any $\varepsilon >0$ and $ s\geq \max(s_1,\,C\| \mu
\| _{C\left( \left[ 0,a^*\right] \right) }^{\frac 23}).$

The last estimate and the existence theory of parabolic boundary value
problems in $L^r$ (see \cite{l1}) imply that on a subsequence (also denoted by
$(\tilde{u}_{\varepsilon })$) we have that
\begin{gather*}
\widetilde{u}_\varepsilon \to \widetilde{u}\quad
\mbox{weakly in }L^{10}\left( (0,T_0)\times \Omega \right)
\\
\widetilde{z}_\varepsilon \to \widetilde{z}^{\widetilde{u}}\quad
\mbox{weakly in }W_{10}^{1,2}\left( (0,T_0)\times \Omega \right) ,
\end{gather*}
where $\left(\widetilde{u},\widetilde{z}^{\widetilde{u}}\right) $ satisfies
\eqref{e11} and
\[
\widetilde{z}^{\widetilde{u}}(T_0,x)=0\quad \mbox{\rm a.e. }x\in \Omega .
\]
By \eqref{e11} we get that
\[
\| \widetilde{z}^{\widetilde{u}}\| _{L^\infty \big((0,T_0)\times \Omega \big)}^2
\leq C\left( \| \widetilde{z}^{\tilde{u}} (0) \| _{L^\infty (\Omega )}^2 
+\| m\widetilde{u} \| _{L^3\left((0,T_0)\times \Omega \right)}^2\right)
\]
(we recall that 
$W_3^{1,2}\left( (0,T_0)\times \Omega \right) \subset
L^\infty \left( (0,T_0)\times \Omega \right)$ for $N\in \left\{
1,2,3\right\} $; see \cite{a1,l1}). So by \eqref{e17} we have
\[
\| \widetilde{z}^{\widetilde{u}}\| _{L^\infty \left(
(0,T_0)\times \Omega \right) }^2 \leq C\| \widetilde{z}^{\tilde{u}}(0)
\| _{L^\infty (\Omega )}^2 .
\]

We extend $u$ given by $\widetilde{u}$ (on each characteristic
line) by $0$. In this manner we get that $u\in L^2(Q_T) $.

Let $z^u$ be the solution to
\begin{gather*}
Dz+\mu z-k\Delta z  =m(a,x)u(a,t,x) , \quad(a,t,x)\in (0,A)\times
(0,A-a^*+T_0)\times \Omega \\
 \frac{\partial z}{\partial \nu }=0, \quad (a,t,x)\in (0,A)\times
(0,A-a^*+T_0)\times \partial \Omega \\
z(0,t,x)=b(t,x) , \quad (t,x)\in (0,A-a^*+T_0)\times \Omega \\
z(a,0,x)=z_0(a,x), \quad (a,x)\in (0,A )\times \Omega .
\end{gather*}
Since $z^u=0$ on $\Gamma_0\times \Omega $ and $u=0$ outside $G\times \Omega $
we conclude that $z^u(a,t,x)=0$ a.e. in $\{ (a,t,x); t\in (T_0,A-a^*+T_0)$,
$T_0<a<t+a^*-T_0, \ x\in \Omega \} $,
$z^u(a,A-a^*+T_0,x)=0$ a.e. $(a,x)\in (T_0,A) \times \Omega $
and that
\begin{equation}
\| z^u\| _{L^\infty (Q_{A-a^*+T _0}) }\leq C(\| z_0\|
_{L^\infty ((0,A )\times \Omega )} +\| b\| _{L^\infty
((0,A-a^*+T_0)\times \Omega )}). \label{e18}
\end{equation}

We are now ready to prove the exact null controllability result. For any
$b\in K$, we denote by $\Phi (b)\subset L^2((0,A-a^*+T_0)\times \Omega) $ the
set of all
$ \int_0^{A }\beta (a)z^u(a,t,x) da$, such that $u\in
L^2(Q_{A-a^*+T_0})$, $u=0$ outside $G\times \Omega $, where $z^u $ satisfies
\eqref{e18} and
$$z^u(a,t,x)=0$$
a.e. in $\{ (a,t,x); t\in (T_0,A-a^*+T_0),
 \ T_0<a<t+a^*-T_0, \ x\in \Omega \} , $
\[
z^u(a,A-a^*+T_0,x) =0, \quad \mbox{\rm a.e. }(a,x)\in (T_0,A)\times \Omega .
\]
There exists an element in $\Phi(b)$ which does not depend on $b$:
\\
If $t>T_{0}$, then $ \int_{0}^{A}\beta
(a)z^{u}(a,t,x)da=\int_t^A\beta (a)z^u(a,t,x)da$ and does not depend
on $b$.
\\
If $t\in (0,T_{0})$, then $\int_{0}^{A}\beta
(a)z^{u}(a,t,x)da=\int_{T_{0}}^{A-T_{0}}\beta (a)z^{u}(a,t,x)da$,
and this depends only on $z_{0}$ and not on $b$.
\smallskip

We also have that $z^u(a,A-a^*+T_0,x)=0$ a.e. $(a,x)\in (T_0,A )\times \Omega $
and
\begin{equation}
\big|\int_{T_0}^{A -T_0}\beta (a)z^u(a,t,x) da \big|\leq C\| \beta \|
_{L^\infty (0,A ) }\cdot \| z_0 \|_{ L^\infty \left( (0,A )
\times \Omega \right) }  \label{e19}
\end{equation}
a.e. in $(0,A-a^*+T_0)\times \Omega $.
It also follows that
$$
\int_0^A\beta (a)z^u(a,t,x)da=\int_0^{T_0}\beta (a)z^u(a,t,x)da
+\int_{A-T_0}^A\beta (a)z^u(a,t,x)da=0
$$
a.e. $(t,x)\in (A-a^*,A-a^*+T_0)$ (because $\beta (a)=0$ on
$(0,T_0)\cup (A-T_0,A)$).
So, for any $u$ as above we can take
\[
b(t,x)=\begin{cases}
0 & \mbox{a.e. } (t,x)\in (A-a^*,A-a^*+T_0)\times \Omega \\[3pt]
 \int_0^{A }\beta (a)z^u(a,t,x) da & \mbox{\rm a.e. }
(t,x)\in (0,A-a^*)\times \Omega
\end{cases}
\]
a fixed point of the multivalued function $\Phi$. In addition, by \eqref{e18} and
\eqref{e19} we have
\[
\| z^u\| _{L^\infty (Q_{A-a^*+T_0}) }\leq C \|z_0\|
_{L^\infty \left((0,A )\times \Omega \right) }.
\]

So, if $\| z_0\| _{L^\infty \left( (0,A ) \times \Omega \right) }$
is small enough, there exists $u\in L^2(Q_{A-a^*+T_0})$,
$u=0$ on $(a^*,A)\times (A-a^*,A-a^*+T_0)\times \Omega $, such that $z$, the
solution of \eqref{e4} (with $T:=A-a^*+T_0$) satisfies
\begin{gather*}
z(a,A-a^*+T_0,x)=0\quad \mbox{\rm a.e. }(a,x)\in (0,A)\times \Omega,\\
\| z\| _{L^\infty (Q_{A-a^*+T_0})}\leq C\|z_0 \|
_{L^\infty \left( (0,A)\times \Omega \right) } \leq \rho _0 \ .
\end{gather*}
In conclusion $z(a,t,x)\geq -\rho _0$ a.e. $(a,t,x)\in Q_{A-a^*+T_0}$.
This implies (via Theorem \ref{thm3.1}) that
$$z(a,t,x)\geq -y_s(a,x) \quad\mbox{a.e. } (a,t,x)\in
(0,a^*)\times (0,T)\times \Omega  .
$$
On the other hand $mu=0$ on
$(a^*,A)\times (0,T)\times \Omega $. The comparison principle for parabolic
equations allows us to conclude that
$$
z(a,t,x)\geq -y_s(a,x) \quad\mbox{a.e. } (a,t,x)\in
(a^*, A)\times (0,T)\times \Omega  . $$

For the second assertion of Theorem \ref{thm2.2} we assume by contradiction that
$T<A-a^*$ (this also implies that $a^*<A$), $\|z_0\|_{L^{\infty }((a^*,A-T)\times
\Omega )}>0$ and there exists $u\in L^2(Q_T)$ such that $z^u$ the solution of \eqref{e4}
satisfies \eqref{e5} (see figure 2).

\begin{figure}
\unitlength=0.6mm
\begin{picture}(123.00,75.00)(10,10)
%\put(37.00,14.00){\vector(0,1){71}}
\put(37.00,14.00){\vector(0,1){61}}
\put(37.00,14.00){\vector(1,0){86}}
\put(33.00,10.00){\makebox(0,0)[cc]{$O$}}
\put(30.00,45.00){\makebox(0,0)[cc]{$A-a^*$}}
\put(85.00,10.00){\makebox(0,0)[cc]{$a^*$}}
\put(115.00,10.00){\makebox(0,0)[cc]{$A$}}
 \put(115.00,14.00){\line(0,1){30}}
\put(37.00,44.00){\line(1,0){78}}
\put(37.00,29.00){\line(1,0){78}}
\put(85.00,14.00){\line(1,1){30}}
\put(123.00,10.00){\makebox(0,0)[cc]{$a$}}
\put(33.00,30.00){\makebox(0,0)[cc]{$T$}}
%\put(33.00,85.00){\makebox(0,0)[cc]{$t$}}
\put(33.00,75.00){\makebox(0,0)[cc]{$t$}}
\end{picture}
\caption{}
\end{figure}

Since $mu=0$ on $(a^{*},A)\times (0,T)\times \Omega $ we may
conclude that $z^{u}$ does not explicitly depend on $u$ on 
$\mathcal{S}\times \Omega $, where 
$\mathcal{S}= \{(a,t);a\in (a^{*},A),t\in (0,T),t<a-a^*\}$. 
However we have that $z^{u}$ satisfies
\begin{gather*}
Dz^u+\mu (a)z^u-k\Delta z^u=0, \quad (a,t,x)\in \mathcal{S}\times \Omega \\
 \frac{\partial z^u}{\partial \nu }(a,t,x)=0, \quad (a,t,x)\in
\mathcal{S}\times \partial \Omega \\
 z^u(a,0,x)=z_{0}(a,x), \quad (a,x)\in (a^*,A)\times \Omega ,
\end{gather*}
and since $\|z_0\|_{L^{\infty }((a^*,A-T)\times \Omega )}>0$, we conclude that
$\|z^u(\cdot ,T,\cdot )\|_{L^{\infty }((0,A)\times \Omega )}>0$ 
(this follows via the backward uniqueness theorem); which is in
contradiction to \eqref{e5}.
So, we get the conclusion.

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\end{document}