\documentclass[reqno]{amsart}
\usepackage{hyperref}
\usepackage{amssymb}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2014 (2014), No. 240, pp. 1--20.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2014 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2014/240\hfil Null controllability of a model]
{Null controllability of a model in population dynamics}

\author[Y. Echarroudi, L. Maniar \hfil EJDE-2014/240\hfilneg]
{Younes Echarroudi, Lahcen Maniar}  % in alphabetical order

\address{Younes Echarroudi \newline
D\'epartement de Math\'ematiques,
Facult\'e des Sciences Semlalia, Laboratoire LMDP,
UMMISCO (IRD-UPMC), B. P. 2390 Marrakech 40000, Maroc}
\email{yecharroudi@gmail.com}

\address{Lahcen Maniar\newline
D\'epartement de Math\'ematiques,
Facult\'e des Sciences Semlalia, Laboratoire LMDP,
UMMISCO (IRD-UPMC), B. P. 2390 Marrakech 40000, Maroc}
\email{maniar@uca.ma}

\thanks{Submitted May 7, 2014. Published November 17, 2014.}
\subjclass[2000]{35K65, 92D25, 93B05, 93B07}
\keywords{Population dynamics;
 Carleman estimate; observability inequality;
\hfill\break\indent null controllability}

\begin{abstract}
 In this article, we study the null controllability of a linear
 model with degenerate diffusion in population dynamics.
 We develop first a Carleman type  inequality for the adjoint system
 of an intermediate model, and then an  observability inequality.
 By a fixed point technique, we establish the  existence of a control
 acting on a subset of  the space domain that leads  the population
 of a certain age  to extinction in a finite time.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{proposition}[theorem]{Proposition}
\allowdisplaybreaks

\section{Introduction}

We consider the  linear population dynamics model
\begin{equation} \label{1}
\begin{gathered}
\frac{\partial y}{\partial t}  +\frac{\partial y}{\partial a}
-(k(x)y_{x})_{x}+\mu(t, a, x)y = \vartheta\chi_{\omega}  \quad \text{in } Q,\\
  y(t, a, 1)=y(t, a, 0)=0  \quad \text{on }(0,T)\times(0,A),\\
 y(0, a, x)=y_0(a, x) \quad \text{in }Q_A,\\
 y(t, 0, x)=\int_0^{A} \beta (t, a, x)y (t, a, x)da  \quad  \text{in } Q_T,
\end{gathered}
\end{equation}
where $Q=(0,T)\times(0,A)\times(0,1)$, $Q_A=(0,A)\times(0,1)$,
$Q_T=(0,T)\times(0,1)$ and we will denote  $q=(0,T)\times(0,A)\times \omega$.
The system \eqref{1} models the dispersion of a gene in a given population.
 In this case, $x$ represents the gene type and $y(t,a,x)$ is the distribution
of individuals of age $a$ at time $t$ and of gene type $x$ of the population.
The parameters $\beta(t,a,x)$,  $\mu(t,a,x)$ are respectively the natural
fertility and mortality rates of individuals of age $a$ at time $t$ and of
 gene type $x$, $A$ is the maximal age of life of population, and $k$ is
the gene dispersion coefficient. The subset
$\omega=(x_1, x_2) \Subset (0,1)$ is the region where a control $\vartheta$ is acting.
 This control corresponds to an external supply or to removal of individuals on
the subdomain  $\omega$. Finally, $\int_0^{A} \beta (t,a, x)y (t,a, x)da$
is the distribution of the newborns of population that are of gene type $x$
at time $t$. The variable $x$ can  also represent  a space variable,
 as in some diffusion population models studied in the literature.

 The question of null controllability is widely investigated in many papers,
among them we find \cite{Ain4,Ain3,Ain2,Ain1,Traore}
and the references therein. In \cite{Ain2,Ain1}, the authors proved
the existence of a control that leads the population to its steady state $y_{s}$.
This is equivalent to show the  null controllability for the system satisfied by
$y-y_{s}$. To reach this goal, the authors took the adjoint system as a collection
of parabolic equations along characteristic lines, and  used Carleman and
observability inequalities for the heat  equation proved in  \cite{Fursikov}.
In \cite{Ain4,Ain3,Traore}, following the same strategy of  \cite{Fursikov},
the authors showed   a direct Carleman estimate  for the backward adjoint system
of the population model  \eqref{1} and deduced its null  controllability by
showing adequate observability inequalities. Note that in \cite{Traore},
 Traore considered a nonlinear distribution of  newborns under the form
$F(\int_0^{A} \beta (t,a, x)y (t,a, x)da)$. In this contribution and contrary
to the previous works,  we consider that the dispersion coefficient $k$ in
our problem depend on $x$ and degenerate at the left boundary; i.e., $k(0)=0$,
e.g. $k(x)=x^{\alpha}$.
In this case, we say that the model \eqref{1} is a degenerate population dynamics
system. Genetically speaking, this assumption is naturel because it means
that if each population is not of a gene type, then this gene can not be
transmitted to its offspring.

 In this context of degeneracy, we will study the null controllability of
the degenerate model  \eqref{1}  at each fixed time $T>0$.  More exactly,
 we show that for all $y_0\in L^{2}(Q_A)$ and any $\delta\in (0,A)$,
there exists a control $\vartheta \in L^{2}(q)$ such that the associated solution
of \eqref{1}  verifies
\begin{equation} \label{2}
y(T,a, x)=0, \quad  \text{a.e. in } (\delta, A)\times(0,1).
\end{equation}
Such a control does not depend only on the initial distribution $y_0$,
but also on the parameter $\delta$. As in \cite{Ain3} and \cite{Traore},
we prove this result by developing a new Carleman estimate. This will be
obtained by  following the method of the work done in \cite{Bouss} for
degenerate heat equation.

The remainder of this article is organized as follows:
in Section \ref{sec-3}, we  give the functional framework in
which system \eqref{1} is wellposed and provide the proof of the Carleman
inequality for an intermediate trivial adjoint system. With the help
of this inequality, we establish the observability inequality and show
the null controllability of the intermediate system.
Using a generalization of the Leray-Schauder fixed point theorem,
we will deduce in Section \ref{last-section-3} the main result of null
controllability of \eqref{1}. The last section is an appendix which is
devoted to the proof of a Caccioppoli's inequality which plays a crucial
rule in the proof of the Carleman estimate.

\section{Well-posedness result}\label{sec3-1}

In this article, we assume that the dispersion coefficient  $k$ satisfies
the  hypotheses
\begin{equation}\label{5}
\begin{gathered}
k \in  C([0,1])\cap C^{1}((0,1]),\quad
 k>0 \text{ in }(0,1] \text{ and } k(0)=0,\\
\exists \gamma \in [0,1):  xk'(x)\leq\gamma k(x), \quad  x \in [0,1].
\end{gathered}
\end{equation}
The above hypothesis on $k$ means in the case of $k(x)=x^\alpha$
that $0\leq \alpha<1$.  Similarly,  all results of this chapter can be
obtained  also in the case of $1\leq \alpha<2$ taking,
instead of Dirichlet condition,   the Newmann condition $(k(x)u_x)(0)=0$
on $x=0$.

On the other hand, we assume that the  rates $\mu$ and $\beta$ satisfy
\begin{equation}\label{3}
\begin{gathered}
\mu \in L^{\infty}(Q), \quad \mu\geq0  \text{ a.e. in } Q, \\
\beta \in C^{2}([0,T]\times[0,A]\times[0,1]), \quad
\beta\geq0  \text{ a.e. in } Q.
\end{gathered}
\end{equation}

To prove the well-posedness of  \eqref{1}, we introduce the following
weighted Sobolev spaces
\begin{equation}\label{578}
\begin{gathered}
\begin{aligned}
H^{1}_{k}(0, 1):=\big\{&u \in L^{2}(0,1):u \text{ is abs.  cont. in } [0,1],\\
&\sqrt{k}u_{x} \in L^{2}(0,1), u(1)=u(0)=0\big\},
\end{aligned}\\
H^2_{k} (0, 1):=\big\{ u \in H^1_{k}(0, 1) :  k(x)u_x
\in H^1(0, 1)\big\},
\end{gathered}
\end{equation}
endowed respectively with the norms
\begin{equation}\label{579}
\begin{gathered}
\|u\|^{2}_{H^{1}_{k}(0, 1)}
:=\|u\|^{2}_{L^{2}(0,1)}+ \|\sqrt{k}u_{x}\|^{2}_{L^{2}(0,1)}, \quad u \in H^{1}_{k}(0, 1),\\
\|u\|^2_{H^2_{k}} := \|u\|^2 _{H^1_{k}(0, 1)}
+ \|(k(x)u_x)_x\|^2_{ L^2(0,1)}, \quad u \in H^{2}_{k}(0, 1).
\end{gathered}
\end{equation}
We recall from \cite{cmp, Can1} that the  operator
$Cu := (k(x)u_x)_x$, $u \in D(C) = H^2_{k}(0, 1)$,
is closed self-adjoint and  negative  with dense domain in $L^2(0, 1)$.

Using properties of the operator $C$, one can show as in \cite{Langlais,rh-sh,web}
the existence of a unique solution of the model \eqref{1} and that this solution
 is generated by a $C_0$-semigroup on the space  $L^{2}((0, A)\times(0, 1))$.
Moreover,  this solution has additional time, age and gene regularity.
More precisely, the following well-posedness result holds.

\begin{theorem}\label{410}
Under the assumptions \eqref{5} and \eqref{3} and for all $\vartheta\ \in L^{2}(Q)$
and $y_0 \in L^{2}(Q_A)$, the system \eqref{1} admits a unique solution $y$.
This solution belongs to
$E:=C([0, T], L^{2}((0, A)\times(0, 1)))\cap C([0, A], L^{2}((0, T)
\times(0, 1))) \cap L^{2}((0, T)\times(0, A), H^{1}_{k}(0, 1))$.
Moreover, the solution of \eqref{1} satisfies the  inequality
% \label{411}
\begin{align*}
&\sup_{t\in [0,T]}\|y(t)\|^{2}_{L^{2}(Q_A)}
 + \sup_{a\in [0,A]}\|y(a)\|^{2}_{L^{2}(Q_T)}
 +\int_0^1\int_0^{A}\int_0^{T}
(\sqrt{k(x)}y_x)^2\,dt\,da\,dx\\
&\leq C\Big(\int_{q}\vartheta^{2}+
\int_{Q_A}y_0^{2}\,da\,dx\Big).
\end{align*}
\end{theorem}

The properties of operator $C$ allow us also to define the root of the operator
$B=-C$  denoted by  $B^{1/2}$. On the other hand, by the definitions
\eqref{578} and \eqref{579} and following the same arguments used in
the proofs of \cite[Propositions 3.5.1, 3.6.1]{Tuc} one can show that
$D(B^{1/2})=H^{1}_{k}(0, 1)$. Moreover,  the following result is
 needed in the sequel. For the proof, see \cite[Corollary 3.4.6]{Tuc}.

\begin{proposition}\label{580}
The operator $B$ defined above has a unique extension
\begin{equation} \label{581}
B\in \mathcal{L}(H_{k}^{1}(0,1), H_{k}^{-1}(0,1)),
\end{equation}
where $H_{k}^{-1}(0,1)$ denotes the dual space of $H_{k}^{1}(0,1)$
with respect to the pivot space $L^{2}(0,1)$.
\end{proposition}

\section{Null controllability of an intermediate system}\label{sec-3}

In this section, we investigate the null controllability of the system
\begin{equation} \label{8}
\begin{gathered}
\frac{\partial y}{\partial t}  +\frac{\partial y}{\partial a}
-(k(x)y_{x})_{x}+\mu(t, a, x)y = \vartheta\chi_{\omega}  \quad \text{in } Q,\\
  y(t, a, 1)=y(t, a, 0)=0  \quad \text{in }(0,T)\times(0,A),\\
  y(0, a, x)=y_0(a, x) \quad  \text{in }Q_A,\\
  y(t, 0, x)=b(t,x)  \quad \text{in } Q_T,
\end{gathered}
\end{equation}
with $b \in L^{2}(Q_T)$. To reach this target, we show first a
Carleman estimate for the adjoint system of \eqref{8}.

\subsection{Carleman inequalities results}\label{sec-3-1}

Consider the  adjoint system  of \eqref{8},
\begin{equation} \label{12}
\begin{gathered}
\frac{\partial w}{\partial t} + \frac{\partial w}{\partial a}
+(k(x)w_{x})_{x}-\mu(t, a, x)w = 0,\\
  w(t, a, 1)=w(t, a, 0)=0, \\
  w(T, a, x)=w_T(a, x),\\
  w(t, A, x)=0.
\end{gathered}
\end{equation}

We assume that  $\mu$ satisfies \eqref{3}, $w_T \in L^{2}(Q_A)$
and that the coefficient of diffusion $k$ satisfies \eqref{5}.
Let us introduce the  weight functions
\begin{equation}\label{13}
\begin{gathered}
\Theta (t, a):= \frac{1}{(t(T-t))^{4}a^{4}},\quad
\psi(x):=c_1(\int_0^{x}\frac{r}{k(r)} dr-c_2), \\
 \varphi(t, a, x):=\Theta (t, a)\psi(x).
\end{gathered}
\end{equation}
For the moment, we suppose that $c_2> \frac {1}{k(1)(2-\gamma)}$ and
$c_1>0$. One can observe that $\psi(x)<0$, $x \in (0,1)$, or
$\Theta (a, t)\to +\infty$ as $t\to 0^{+}, T^{-}$ and $a\to 0^{+}$.
The first result of this paragraph is the following proposition.

\begin{proposition}\label{15}
Consider the two following systems with $h \in L^{2}(Q)$,
 \begin{gather} \label{14}
\begin{gathered}
\frac{\partial w}{\partial t} + \frac{\partial w}{\partial a} +(k(x)w_{x})_{x} = h,\\
  w(a, t, 1)=w(a, t, 0)=0, \\
  w(a, T, x)=w_T(a, x),\\
  w(A, t, x)=0,
\end{gathered} \\
\label{17} \begin{gathered}
  \frac{\partial w}{\partial t} + \frac{\partial w}{\partial a}
+(k(x)w_{x})_{x}-\mu(t, a, x)w = h,\\
  w(t, a, 1)=w(t, a, 0)= 0, \\
  w(T, a, x)= w_T(a, x),\\
  w(t, A, x)= 0.
\end{gathered}
\end{gather}
Then, there exist $C>0$ and $s_0>0$, such that every solutions of \eqref{14}
or  \eqref{17} satisfy, for $s\geq s_0$, the  inequality
\begin{equation} \label{16}
\begin{aligned}
&s^{3}\int_{Q}  \Theta^{3}\frac{x^{2}}{k(x)}w^{2}e^{2s\varphi} \,dt\,da\,dx
+ s\int_{Q}  \Theta k(x)w_{x}^{2}e^{2s\varphi} \,dt\,da\,dx  \\
 &\leq C\Big(\int_{Q}|h|^{2}e^{2s\varphi} \,dt\,da\,dx
+sk(1) \int_0^{A}\int_0^{T}\Theta w_{x}^{2}(a, t, 1)e^{2s\varphi(a, t, 1)}
\,dt\,da\Big).
\end{aligned}
\end{equation}
\end{proposition}

\begin{proof}
We establish the inequality \eqref{16} for every solution of system \eqref{14},
and  then  deduce the result for the model \eqref{17}.
Let  $w$ be  the solution of \eqref{14}.  The function
$\nu(t, a, x):=e^{s\varphi(t, a, x)}w(t, a, x)$
satisfies the  system
\begin{equation} \label{18}
\begin{gathered}
L^{+}_{s}\nu +  L^{-}_{s}\nu = e^{s\varphi}h, \\
 \nu(t, a, 1)=\nu(t, a, 0)=
 \nu(T, a, x)=\nu(0, a, x)=  \nu(t, A, x)=\nu(t, 0, x)=0,
\end{gathered}
\end{equation}
where
\begin{gather*}
L^{+}_{s}\nu:=(k(x)\nu_{x})_{x}-s(\varphi_{a}+\varphi_{t})\nu
+s^{2}\varphi^{2}_{x}k(x)\nu,\\
L^{-}_{s}\nu:=\nu_{t}+\nu_{a}-2sk(x)\varphi_{x}\nu_{x}-s(k(x)\varphi_{x})_{x}\nu.
\end{gather*}
Passing to the norm in \eqref{18}, one has
$$
\|L^{+}_{s}\nu\|^{2}_{L^{2}(Q)} +  \|L^{-}_{s}\nu\|^{2}_{L^{2}(Q)}
+ 2\langle L^{+}_{s}\nu,L^{-}_{s}\nu\rangle _{L^{2}(Q)}
= \|e^{s\varphi(a, t, x)}h\|^{2}_{L^{2}(Q)}.
$$
Then, the proof of step one is based on the calculus of the inner product
$\langle L^{+}_{s}\nu,L^{-}_{s}\nu\rangle $ whose a first expression is given
in the following lemma.

\begin{lemma}\label{19}
The identity
$\langle L^{+}_{s}\nu,L^{-}_{s}\nu\rangle =S_1+S_2$ holds with
\begin{align*} %\label{20}
S_1 &=s\int_{Q}(k(x)\nu_{x})^{2}\varphi_{xx}\,dt\,da\,dx
 -s^{3}\int_{Q}(k(x)\varphi_{x})_{x}k(x)\varphi^{2}_{x}\nu^{2}\,dt\,da\,dx\\
&\quad +s^{2}\int_{Q}(\varphi_{a}+\varphi_{t})(k(x)\varphi_{x})_{x}\nu^{2}
 \,dt\,da\,dx \\
&\quad +s\int_{Q}k(x)\nu_{x}((k(x)\varphi_{x})_{xx}\nu+(k(x)\varphi_{x})_{x}\nu_{x})
 \,dt\,da\,dx\\
&\quad +s^{3}\int_{Q}(k^{2}\varphi^{3}_{x})_{x}\nu^{2}\,dt\,da\,dx
 -s^{2}\int_{Q}(k(x)(\varphi_{a}+\varphi_{t})\varphi_{x})_{x}\nu^{2}\,dt\,da\,dx\\
&\quad +\frac{s}{2}\int_{Q}(\varphi_{at}+\varphi_{tt})\nu^{2}\,dt\,da\,dx
 -\frac{s^{2}}{2}\int_{Q}(\varphi^{2}_{x})_{t}k(x)\nu^{2}\,dt\,da\,dx\\
&\quad + \frac{s}{2}\int_{Q}(\varphi_{at}+\varphi_{aa})\nu^{2}\,dt\,da\,dx
 -\frac{s^{2}}{2}\int_{Q}(\varphi^{2}_{x})_{a}k(x)\nu^{2}\,dt\,da\,dx,
\end{align*}
and
\begin{align*}%\label{21}
S_2
&=\int_0^{A}\int_0^{T}[k(x)\nu_{x}\nu_{a}]_0^{1}\,dt\,da
 +\int_0^{A}\int_0^{T}[k(x)\nu_{x}\nu_{t}]_0^{1}\,dt\,da \\
&\quad + s^{2}\int_0^{A}\int_0^{T}[k(x)\varphi_{x}(\varphi_{a}
 +\varphi_{t})\nu^{2}]_0^{1}\,dt\,da-s^{3}\int_0^{A}\int_0^{T}[k^{2}(x)
 \varphi^{3}_{x}\nu^{2}]_0^{1}\,dt\,da \\
&\quad -s\int_0^{A}\int_0^{T}[k(x)\nu\nu_{x}(k(x)\varphi_{x})_{x}]_0^{1}\,dt\,da
 -s\int_0^{A}\int_0^{T}[(k(x)\nu_{x})^{2}\varphi_{x}]_0^{1}\,dt\,da.
\end{align*}
\end{lemma}

\begin{proof}
We have
\[
I_{11}=\int_{Q}(k(x)\nu_{x})_{x}\nu_{t}\,dt\,da\,dx
=\int_0^{A}\int_0^{T}[k(x)\nu_{x}\nu_{t}]_0^{1}\,dt\,da
 -\int_0^{1}\int_0^{A}[\frac{k(x)}{2}\nu_{x}^{2}]_0^{T}\,da\,dx.
\]
By the definition of $\nu$, one has
\begin{gather*}
I_{11}=\int_0^{A}\int_0^{T}[k(x)\nu_{x}\nu_{t}]_0^{1}\,dt\,da,\\
\begin{aligned}
I_{12}&=\int_{Q}(k(x)\nu_{x})_{x}\nu_{a}\,dt\,da\,dx\\
&=\int_0^{A}\int_0^{T}[k(x)\nu_{x}\nu_{a}]_0^{1}\,dt\,da
-\int_{Q}k(x)\nu_{x}\nu_{xa}\,dt\,da\,dx\\
&= \int_0^{A}\int_0^{T}[k(x)\nu_{x}\nu_{a}]_0^{1}\,dt\,da,
\end{aligned}\\
\begin{aligned}
 I_{13}&=\int_{Q} -2sk(x)\varphi_{x}\nu_{x}(k(x)\nu_{x})_{x} \,dt\,da\,dx\\
&= \int_{Q} -s\varphi_{x}((k(x)\nu_{x})^{2})_{x} \,dt\,da\,dx\\
&= -s\int_0^{A}\int_0^{T}[(k(x)\nu_{x})^{2}\varphi_{x}]_0^{1}\,dt\,da
+s\int_{Q}(k(x)\nu_{x})^{2}\varphi_{xx}\,dt\,da\,dx.
\end{aligned} \\
\begin{aligned}
I_{14}&=\int_{Q}(-s(k(x)\varphi_{x})_{x}\nu)(k(x)\nu_{x})_{x}\,dt\,da\,dx\\
&=-s\int_0^{A}\int_0^{T}[k(x)\nu_{x}\nu(k(x)\varphi_{x})_{x}]_0^{1}\,dt\,da\\
&\quad +s\int_{Q}k(x)\nu_{x}(\nu(k(x)\varphi_{x})_{xx}+\nu_{x}(k(x)\varphi_{x})_{x})
\,dt\,da\,dx.
\end{aligned}\\
\begin{aligned}
I_{21}&=-s\int_{Q}(\varphi_{a}+\varphi_{t})\nu \nu_{t}\,dt\,da\,dx
= \frac{-s}{2}\int_{Q}(\varphi_{a}+\varphi_{t})(\nu^{2})_{t}\,dt\,da\,dx \\
&=\frac{s}{2}\int_{Q}(\varphi_{ta}+\varphi_{tt})\nu^{2}\,dt\,da\,dx,
\end{aligned}\\
\begin{aligned}
 I_{22}&=-s\int_{Q}(\varphi_{a}+\varphi_{t})\nu \nu_{a}\,dt\,da\,dx
=\frac{-s}{2}\int_{Q}(\varphi_{a}+\varphi_{t})(\nu^{2})_{a} \,dt\,da\,dx\\
&=\frac{s}{2}\int_{Q}(\varphi_{aa}+\varphi_{ta})\nu^{2}\,dt\,da\,dx,
\end{aligned} \\
\begin{aligned}
 I_{23}&=\int_{Q} (2sk(x)\varphi_{x}\nu_{x})(s(\varphi_{a}+\varphi_{t})\nu)
\,dt\,da\,dx\\
&=-\int_{Q}s^{2}\nu^{2}(k(x)(\varphi_{a}+\varphi_{t})\varphi_{x})_{x}\,dt\,da\,dx\\
&\quad + s^{2}\int_0^{A}\int_0^{T}[k(x)(\varphi_{a}+\varphi_{t})\varphi_{x}
\nu^{2}]_0^{1}\,dt\,da.
\end{aligned} \\
\begin{aligned}
I_{24}&=\int_{Q}(s(\varphi_{a}+\varphi_{t})\nu)(s(k(x)\varphi_{x})_{x}\nu)\,dt\,da\,dx\\
&=\int_{Q}s^{2}(\varphi_{a}+\varphi_{t})(k(x)\varphi_{x})_{x}\nu^{2}\,dt\,da\,dx.
\end{aligned}\\
\begin{aligned}
I_{31}&=\int _{Q}s^{2}\varphi_{x}^{2}k(x)\nu\nu_{t} \,dt\,da\,dx\\
&=\int_0^{1}\int_0^{A}[\frac{s^{2}}{2}\varphi_{x}^{2}k(x)\nu^{2}]_0^{T}\,da\,dx
-\frac{s^{2}}{2}\int _{Q}(\varphi_{x}^{2}k(x))_{t}\nu^{2}\,dt\,da\,dx\\
& = \frac{-s^{2}}{2}\int _{Q}(\varphi_{x}^{2}k(x))_{t}\nu^{2}\,dt\,da\,dx,
\end{aligned}\\
I_{32}=s^{2}\int_{Q}\varphi_{x}^{2}k(x)\nu\nu_{a}\,dt\,da\,dx
=\frac{-s^{2}}{2}\int_{Q}(\varphi_{x}^{2}k(x))_{a}\nu^{2}\,dt\,da\,dx.
\\
\begin{aligned}
I_{33}&=\int_{Q}(-2sk(x)\varphi_{x}\nu_{x})(s^{2}\varphi_{x}^{2}k(x)\nu)\,dt\,da\,dx\\
&=\int_{Q}-s^{3}k^{2}(x)\varphi_{x}^{3}(\nu^{2})_{x} \,dt\,da\,dx\\
&= -s^{3}\int_0^{A}\int_0^{T}[k^{2}(x)\varphi_{x}^{3}\nu^{2}]_0^{1}\,dt\,da
+s^{3}\int_{Q}(k^{2}(x)\varphi_{x}^{3})_{x}\nu^{2}\,dt\,da\,dx.
\end{aligned}\\
\begin{aligned}
I_{34}&=\int_{Q}-(s(k(x)\varphi_{x})_{x}\nu)(s^{2}\varphi_{x}^{2}k(x)\nu)\,dt\,da\,dx\\
&=-s^{3}\int_{Q}(k(x)\varphi_{x})_{x}k(x) \varphi_{x}^{2}\nu^{2}\,dt\,da\,dx.
\end{aligned}
\end{gather*}
By adding all these identities, the result follows.
\end{proof}

Back to the proof of Proposition \ref{15}.
Now, using the definitions of $\varphi$ and $\psi$ given in \eqref{13},
the Dirichlet boundary conditions satisfied by $\nu$ and the assumption $k(0)=0$,
the expressions of $S_1$ and $S_2$ stated in the previous lemma can be
simplified  follows,
% \label{34}
\begin{align*}
S_1&=\frac{s}{2}\int_{Q}(\Theta_{aa}+\Theta_{tt})\psi\nu^{2}dx\,dt\,da
 +s\int_{Q}\Theta_{ta}\psi\nu^{2}\,dt\,da\,dx \\
&\quad +sc_1\int_{Q}\Theta (2k(x)-xk'(x))\nu_{x}^{2}\,dt\,da\,dx
-2s^{2}\int_{Q}\Theta c_1^{2}\frac{x^{2}}{k(x)}(\Theta_{a}+\Theta_{t})\nu^{2}
 \,dt\,da\,dx\\
&\quad+ s^{3}\int_{Q}\Theta^{3}c_1^{3}(\frac{x}{k(x)})^{2}(2k(x)-xk'(x))\nu^{2}
\,dt\,da\,dx,
\end{align*}
and
\[
S_2=-sc_1k(1)\int_0^{A}\int_0^{T}\Theta\nu_{x}^{2}(a,t,1)\,dt\,da
+2s^{3}\int_0^{A}\int_0^{T}\Theta^{3}c_1^{3}[\frac{x^{3}}{k(x)}\nu^{2}]_{x=0}
\,dt\,da.
\]
From  the third condition in assumptions \eqref{5}, we deduce that the function
$x\mapsto \frac{x^{3}}{k(x)}$ is nondecreasing in $(0,1]$, and
then, $0<\frac{x^{3}}{k(x)}\leq \frac{1}{k(1)}$,
$0\leq\frac{x^{3}}{k(x)}\nu^{2}\leq \frac{1}{k(1)}\nu^{2}$, $x \in (0,1]$.
Hence,  $ \lim_{x\to 0^{+}}\frac{x^{3}}{k(x)}\nu^{2}=0$.
Accordingly,
\begin{align*}
&\langle L^{+}_{s}\nu,L^{-}_{s}\nu\rangle \\
&= \frac{s}{2}\int_{Q}(\Theta_{aa}+\Theta_{tt})\psi\nu^{2}\,dt\,da\,dx+s
\int_{Q}\Theta_{ta}\psi\nu^{2}\,dt\,da\,dx\\
&\quad +sc_1\int_{Q}\Theta (2k(x)-xk'(x))\nu_{x}^{2}\,dt\,da\,dx
 -2s^{2}\int_{Q}\Theta c_1^{2}\frac{x^{2}}{k(x)}
 (\Theta_{a}+\Theta_{t})\nu^{2}\,dt\,da\,dx\\
&\quad + s^{3}\int_{Q}\Theta^{3}c_1^{3}(\frac{x}{k(x)})^{2}(2k(x)
 -xk'(x))\nu^{2}\,dt\,da\,dx \\
&\quad -sc_1k(1)\int_0^{A}\int_0^{T}\Theta\nu_{x}^{2}(a,t,1)\,dt\,da.
\end{align*}
Thanks to the third assumption in \eqref{5}, we have
\begin{equation} \label{37}
\begin{aligned}
S_1&\geq\frac{s}{2}\int_{Q}(\Theta_{aa}+\Theta_{tt})\psi\nu^{2}\,dt\,da\,dx
+ s\int_{Q}\Theta_{ta}\psi\nu^{2}\,dt\,da\,dx\\
&\quad +sc_1\int_{Q}\Theta k(x)\nu_{x}^{2}\,dt\,da\,dx
 -2s^{2}\int_{Q}\Theta c_1^{2}\frac{x^{2}}{k(x)}(\Theta_{a}+\Theta_{t})
 \nu^{2}\,dt\,da\,dx \\
&\quad+ s^{3}\int_{Q}\Theta^{3}c_1^{3}\frac{x^{2}}{k(x)}\nu^{2}\,dt\,da\,dx.
\end{aligned}
\end{equation}
Now, using the
$|\Theta(\Theta_{a}+\Theta_{t})|\leq c\Theta^{3},$ we infer for $s$ quite large that
\begin{equation} \label{133}
\begin{aligned}
&|-2s^{2}\int_{Q}\Theta c_1^{2}\frac{x^{2}}{k(x)}(\Theta_{a}
 +\Theta_{t})\nu^{2}\,dt\,da\,dx|\\
&  \leq 2s^{2}c_1^{2}c\int_{Q}\frac{x^{2}}{k(x)}\Theta^{3}\nu^{2}\,dt\,da\,dx
\leq \frac{c_1^{3}}{4}s^{3}\int_{Q}\frac{x^{2}}{k(x)}\Theta^{3}\nu^{2}\,dt\,da\,dx.
\end{aligned}
\end{equation}
On the other hand, we have
\begin{equation} \label{132}
 |\psi(x)|=|c_1l(x)-c_1c_2|\leq c_1|\int_0^{x}\frac{r}{k(r)}dr|+c_1c_2
\leq \frac{c_1}{(2-\gamma)k(1)}+c_1c_2,
\end{equation}
and this yields
\begin{equation} \label{38}
\begin{aligned}
&|\frac{s}{2}\int_{Q}(\Theta_{aa}+\Theta_{tt})\psi\nu^{2}\,dt\,da\,dx
 +s\int_{Q}\Theta_{ta}\psi\nu^{2}\,dt\,da\,dx| \\
&\leq s\Big(\frac{c_1}{(2-\gamma)k(1)}+c_1c_2\Big)
\int_{Q}\Big(\frac{\Theta_{aa}+\Theta_{tt}}{2}+|\Theta_{ta}|\Big)\nu^{2}\,dt\,da\,dx\\
&\leq M s\Big(\frac{c_1}{(2-\gamma)k(1)}+c_1c_2\Big)
\int_{Q}\Theta^{3/2}\nu^{2}\,dt\,da\,dx.
\end{aligned}
\end{equation}
By  H\"{o}lder, Young and Hardy-Poincar\'{e} inequalities (see \cite{Bouss})
and the fact that
\begin{equation} \label{110}
\exists M_1>0 \quad  \text{ such that } \Theta^{2}\leq M_1\Theta^{3},
\end{equation}
we conclude that
\begin{align*}
\int_0^{1}\Theta^{3/2}\nu^{2}dx
&=\int_0^{1}(\Theta^{1/2}\nu\frac{\sqrt{k}}{x})(\Theta\nu\frac{x}{\sqrt{k}})dx
\\
&\leq C\Big(\int_0^{1}\Theta k(x)\nu_{x}^{2}dx\Big)^{1/2}
\Big(\int_0^{1}\Theta^{2}\nu^{2}\frac{x^{2}}{k(x)}dx\Big)^{1/2}\\
& \leq C\epsilon\int_0^{1}\Theta k(x)\nu_{x}^{2}dx+
\frac{C_1} {4\epsilon}\int_0^{1}\Theta^{3}
\frac{x^{2}}{k(x)}\nu^{2}dx.
\end{align*}
By this  and \eqref{38}, we infer that
\begin{equation} \label{42}
\begin{aligned}
&\big|\frac{s}{2}\int_{Q}(\Theta_{aa}+\Theta_{tt})\psi\nu^{2}\,dt\,da\,dx
+s\int_{Q}\Theta_{ta}\psi\nu^{2}\,dt\,da\,dx\big|\\
&\leq sc_1C\epsilon\int_{Q}\Theta k(x)\nu_{x}^{2}\,dt\,da\,dx
+\frac{sc_1}{4}\frac{C_2} {\epsilon}\int_{Q}\Theta^{3}\frac{x^{2}}{k(x)}\nu^{2}
\,dt\,da\,dx.
\end{aligned}
\end{equation}
Taking $\epsilon$ small enough and $s$ quite large, we conclude that
\begin{equation} \label{43}
\begin{aligned}
&\big|\frac{s}{2}\int_{Q}(\Theta_{aa}+\Theta_{tt})\psi\nu^{2}dx\,dt\,da
+s\int_{Q}\Theta_{ta}\psi\nu^{2}\,dt\,da\,dx\big|\\
&\leq \frac{sc_1}{4}\int_{Q}\Theta k(x)\nu_{x}^{2}\,dt\,da\,dx
+\frac{c_1^{3}s^{3}}{4}\int_{Q}\Theta^{3}\frac{x^{2}}{k(x)}\nu^{2}\,dt\,da\,dx.
\end{aligned}
\end{equation}
This involves, combining with the inequalities \eqref{37} and \eqref{133} that
$$
S_1\geq K_1s^{3}\int_{Q}\Theta^{3}\frac{x^{2}}{k(x)}\nu^{2}\,dt\,da\,dx
+K_2s\int_{Q}\Theta k(x)\nu_{x}^{2}\,dt\,da\,dx.
$$
Therefore,
\begin{align*}
2\langle L^{+}_{s}\nu,L^{-}_{s}\nu\rangle
&\geq m\Big(s^{3}\int_{Q}\Theta^{3}\frac{x^{2}}{k(x)}\nu^{2}\,dt\,da\,dx
+s\int_{Q}\Theta k(x)\nu_{x}^{2}\,dt\,da\,dx\Big)\\
&\quad -2sc_1k(1)\int_0^{A}\int_0^{T}\Theta\nu_{x}^{2}(a,t,1)\,dt\,da.
\end{align*}
Hence, we obtain the following Carleman estimate for \eqref{18}
\begin{align*}
&s^{3}\int_{Q}\Theta^{3}\frac{x^{2}}{k(x)}\nu^{2}\,dt\,da\,dx
 +s\int_{Q}\Theta k(x)\nu_{x}^{2}\,dt\,da\,dx\\
&\leq C \Big(\int_{Q}h^{2}e^{2s\varphi}\,dt\,da\,dx
+sk(1)\int_0^{A}\int_0^{T}\Theta\nu_{x}^{2}(a,t,1)\,dt\,da\Big).
\end{align*}
To return  to system \eqref{14}, we use the function change
$\nu(t, a, x):=e^{s\varphi(t, a, x)}w(t, a, x)$. This implies that
$$
\nu_{x}=s\varphi_{x}e^{s\varphi}w+e^{s\varphi}w_{x},
e^{2s\varphi}w_{x}^{2}\leq2(\nu_{x}^{2}+s^{2}\varphi_{x}^{2}\nu^{2}).
$$
Then,  inequality \eqref{16} follows immediately for every solution of
system \eqref{14}. To show this  inequality  for the  solutions of \eqref{17},
we apply the last inequality for  the function $\overline{h}=h+\mu w$.
 Hence,  there are  two positive constants $C$ and $s_0$ such that, for
all $s\geq s_0$, the following inequality holds
\begin{equation} \label{44}
\begin{aligned}
&s^{3}\int_{Q}  \Theta^{3}\frac{x^{2}}{k(x)}w^{2}e^{2s\varphi} \,dt\,da\,dx
+ s\int_{Q}  \Theta k(x)w_{x}^{2}e^{2s\varphi} \,dt\,da\,dx \\
&\leq C\Big(\int_{Q}\mid\overline{h} \mid^{2}e^{2s\varphi} \,dt\,da\,dx+sk(1)
\int_0^{A}\int_0^{T}\Theta w_{x}^{2}(t, a, 1)e^{2s\varphi(t, a, 1)} \,dt\,da\big).
\end{aligned}
\end{equation}
On the other hand, we have
\[ % \label{130}
\int_{Q}\mid\overline{h} \mid^{2}e^{2s\varphi} \,dt\,da\,dx
\leq 2\Big(\int_{Q}|h|^{2}e^{2s\varphi}\,dt\,da\,dx
+\|\mu\|_{\infty}^{2}\int_{Q}|w|^{2}e^{2s\varphi}\,dt\,da\,dx\Big).
\]
Now, applying Hardy-Poincar\'{e} inequality to the function $\nu:=e^{s\varphi}w$,
we obtain
\begin{align*}
\int_{Q}|w|^{2}e^{2s\varphi}\,dt\,da\,dx
&\leq\frac{1}{k(1)}\int_{Q}\frac{k(x)}{x^{2}}|w|^{2}e^{2s\varphi}\,dt\,da\,dx
\\
&\leq\frac{C}{k(1)}\int_{Q}k(x)\nu_{x}^{2}\,dt\,da\,dx\\
&\leq\frac{C}{k(1)}\Big(\int_{Q}s^{2}c_1^{2}\Theta^{2}\frac{x^{2}}{k(x)}\nu^{2}
+\int_{Q}k(x)e^{2s\varphi}w_{x}^{2}\,dt\,da\,dx\Big).
\end{align*}
Thus,
\begin{align*}
\int_{Q}\mid\overline{h} \mid^{2}e^{2s\varphi} \,dt\,da\,dx
&\leq 2\Big[\int_{Q}|h|^{2}e^{2s\varphi}\,dt\,da\,dx
 + \|\mu\|_{\infty}^{2}\frac{C}{k(1)}\Big(\int_{Q}s^{2}c_1^{2}\Theta^{2}
\frac{x^{2}}{k(x)}\nu^{2} \\
&\quad +\int_{Q}k(x)e^{2s\varphi}w_{x}^{2}\,dt\,da\,dx\Big)\Big].
\end{align*}
This implies, using again \eqref{110} and taking $s$ quite large, that
\begin{align*}
&s^{3}\int_{Q}  \Theta^{3}\frac{x^{2}}{k(x)}w^{2}e^{2s\varphi} \,dt\,da\,dx
 + s\int_{Q}  \Theta k(x)w_{x}^{2}e^{2s\varphi} \,dt\,da\,dx \\
&\leq D(\int_{Q}| h|^{2}e^{2s\varphi} \,dt\,da\,dx
 +sk(1) \int_0^{A}\int_0^{T}\Theta w_{x}^{2}(t, a, 1)e^{2s\varphi(t, a, 1)}\,dt\,da\\
&\quad +\int_{Q}s^{2}c_1^{2}\Theta^{2} \frac{x^{2}}{k(x)}\nu^{2}\,dt\,da\,dx
+\int_{Q}k(x)e^{2s\varphi}w_{x}^{2}\,dt\,da\,dx)
\\
&\leq   C\Big(\int_{Q} |h|^{2}e^{2s\varphi} \,dt\,da\,dx+sk(1)
\int_0^{A}\int_0^{T}\Theta w_{x}^{2}(t, a, 1)e^{2s\varphi(t, a, 1)}\,dt\,da\Big).
\end{align*}
This completes the proof.
\end{proof}

Now, we can provide the main  result of this section,
namely an  $\omega$-Carleman estimate of the model \eqref{12}.

\begin{theorem}\label{47}
Assume that $k$ satisfies hypotheses \eqref{5} and let $A>0$ and $T>0$ be given.
 Then there exist two positive constants $C$ and $s_0$, such that every solution
 $w$ of \eqref{12} satisfies, for all $s\geq s_0$, the  inequality
\begin{equation} \label{48}
\int_{Q}(s\Theta k w_{x}^{2}+s^{3}\Theta^{3}\frac{x^{2}}{k} w^{2})e^{2s\varphi}
\,dt\,da\,dx\leq C \int_{\omega}\int_0^{A}\int_0^{T}w^{2}\,dt\,da\,dx.
\end{equation}
\end{theorem}

\begin{proof}
Let us introduce the smooth cut-off function $\xi:\mathbb{R}\to \mathbb{R}$
defined as follows
\begin{equation}\label{49}
\begin{gathered}
0\leq\xi(x)\leq1, \quad \forall x \in \mathbb{R},\\
\xi(x)=0, \quad   x \in [\frac{x_1+2x_2}{3},1],\\
\xi(x)=1, \quad   x \in [0,\frac{2x_1+x_2}{3}].
\end{gathered}
\end{equation}
We define the function $v:=\xi w$,  where $w$ is the solution of the system
\eqref{12}. Using the Carleman estimate obtained for the model \eqref{17}
 and Caccioppoli's inequality stated in Lemma \ref{Caccio}, one can prove the
existence of $C>0$ such the following estimate holds
\begin{equation} \label{50}
\int_0^{1}\int_0^{A}\int_0^{T}(s\Theta k v_{x}^{2}
+s^{3}\Theta^{3}\frac{x^{2}}{k}v^{2})e^{2s\varphi} \,dt\,da\,dx
\leq C \int_{\omega}\int_0^{A}\int_0^{T} w^{2}\,dt\,da\,dx.
\end{equation}
In $(x_1,1)$, let us consider the function $z:=\eta w$ with $\eta=1-\xi$.
Since $z$ is supported by $[0,T]\times[0,A]\times[x_1,1]$ and in this interval
the equation \eqref{17} is uniformly parabolic, then we can replace the
function $k$ by a positive function belonging to $C^{1}([0,1])$ and which
coincides with $k$ on $(x_1,1)$ denoted also by $k$ and this implies that
 \eqref{17} is nondegenerate. Moreover, we can prove in a similar manner
as in \cite{Ain3} the following result.

\begin{proposition}\label{569}
Let $z$ be the solution of
\begin{equation} \label{564}
\begin{gathered}
\frac{\partial z}{\partial t} + \frac{\partial z}{\partial a}
+(k(x)z_{x})_{x}-c(t, a, x)z = h \quad \text{in } Q_{b},\\
  z(t, a, 1)=z(t, a, 0)=0 \quad\text{on } (0,T)\times(0,A),
\end{gathered}
\end{equation}
with $h \in L^{2}(Q)$ and $k \in C^{1}([0,1])$ is a
positive function. Then, there exist two positive constants $c$ and $s_0$,
such that for any $s\geq s_0$, $z$ satisfies the estimate
\begin{equation} \label{570}
\begin{aligned}
&\int_{Q}(s^{3}\phi^{3}z^{2}+s\phi z_{x}^{2})e^{2s\Phi}\,dt\,da\,dx \\
&\leq c \Big(\int_{Q}h^{2}e^{2s\Phi}\,dt\,da\,dx
 +\int_{\omega}\int_0^{A}\int_0^{T}s^{3}\phi^{3}z^{2}e^{2s\Phi}\,dt\,da\,dx\Big),
\end{aligned}
\end{equation}
where $Q:=(0,T)\times(0,A)\times(0,1)$, the functions $\phi$ and $\Phi$ are
defined as follows
\begin{equation}\label{571}
\begin{gathered}
\phi(t,a,x)=\Theta(t,a)e^{\kappa\sigma(x)}, \quad
\Theta(t, a)= \frac{1}{t^{4}(T-t)^{4}a^{4}},\\
\Phi(a,t,x)=\Theta(t,a)\Psi(x), \quad
\Psi(x)=e^{\kappa\sigma(x)}-e^{2\kappa\|\sigma\|_{\infty}},
\end{gathered}
\end{equation}
$(t,a,x)\in Q$, $\kappa>0$, $\sigma$ is a function satisfying
 \begin{equation}\label{568}
\begin{gathered}
\sigma \in C^{2}([0,1]), \quad
\sigma(x)>0 \quad \text{in } (0,1),\quad  \sigma(0)=\sigma(1)=0, \\
\sigma_{x}(x)\neq0 \quad \text{in } [0,1]\backslash \omega_0,
\end{gathered}
\end{equation}
where $ \omega_0\Subset\omega $ is an open subset.
\end{proposition}

The existence of the function $\sigma$ is proved in \cite{Fursikov}.
Hence,  applying Proposition \ref{569}  to the function $z$ and
$h=(k\eta_{x}w)_{x}+k\eta_{x}w_{x}$, using the definitions of $\eta$,
$\sigma$, $\phi$ and $\Phi$ and thanks again to the Caccioppoli's
inequality we obtain the estimate
\begin{equation} \label{56}
\int_{Q}(s^{3}\phi^{3}z^{2}+s\phi z_{x}^{2})e^{2s\Phi}\,dt\,da\,dx
\leq C \int_{\omega}\int_0^{A}\int_0^{T}w^{2}dt \,da\,dx.
\end{equation}
Taking into account that $w=v+z$ and using the inequality \eqref{50}, we obtain
\begin{equation} \label{97}
\begin{aligned}
 &\int_{Q}(s\Theta k w_{x}^{2}+s^{3}\Theta^{3}\frac{x^{2}}{k}
 w^{2})e^{2s\varphi}\,dt\,da\,dx \\
&\leq 2\int_{Q}(s^{3}\Theta^{3}\frac{x^{2}}{k(x)}z^{2}
 +s\Theta k(x)z_{x}^{2})e^{2s\varphi}\,dt\,da\,dx \\
&\quad +2\int_{Q}(s^{3}\Theta^{3}\frac{x^{2}}{k(x)}v^{2}
 +s\Theta k(x)v_{x}^{2})e^{2s\varphi}\,dt\,da\,dx\\
&\leq C\int_{\omega}\int_0^{A}\int_0^{T} w^{2}\,dt\,da\,dx
+2\int_{Q}(s^{3}\Theta^{3}\frac{x^{2}}{k(x)}z^{2}
+s\Theta k(x)z_{x}^{2})e^{2s\varphi}\,dt\,da\,dx.
\end{aligned}
\end{equation}
On the other hand, by the definition of $\varphi$, taking
\[
c_1\geq \frac{k(1)(2-\gamma)(e^{2\kappa\|\sigma\|_{\infty}}-1)}{c_2k(1)(2-\gamma)-1},
\]
 one can prove the existence of $\varsigma>0$, such that,  for
all $(t,a,x)\in [0,T]\times[0,A]\times[x_1,1]$, we have
\[
\Theta k(x)e^{2s\varphi}\leq\varsigma\phi e^{2s\Phi},
\Theta^{3}\frac{x^{2}}{k(x)}e^{2s\varphi}\leq\varsigma\phi^{3}e^{2s\Phi}.
\]
Using this and the relation \eqref{56} it follows that
\begin{align*}
&\int_{Q}(s^{3}\Theta^{3}\frac{x^{2}}{k(x)}z^{2}+s\Theta k(x)z_{x}^{2})
e^{2s\varphi}\,dt\,da\,dx\\
&=\int_{x_1}^{1}\int_0^{A}\int_0^{T} (s^{3}\Theta^{3}\frac{x^{2}}{k(x)}z^{2}
 +s\Theta k(x)z_{x}^{2})e^{2s\varphi}\,dt\,da\,dx\\
& \leq\varsigma \int_{x_1}^{1}\int_0^{A}\int_0^{T}(s^{3}\phi^{3}z^{2}
+s\phi z_{x}^{2})e^{2s\Phi}\,dt\,da\,dx \\
& \leq \varsigma C \int_{\omega}\int_0^{A}\int_0^{T}w^{2}\,dt\,da\,dx.
\end{align*}
Finally, using the last inequality and \eqref{97} we obtain
the Carleman estimate \eqref{48}.
\end{proof}

\subsection{An observability inequality result}\label{sec3-2}

This paragraph is devoted to the observability inequality of the system \eqref{12}.
This inequality is obtained by using our  Carleman estimate \eqref{48}  and
Hardy-Poincar\'{e} inequality, see \cite{Bouss}.

\begin{proposition}\label{57}
Assume that $k$ satisfies the hypotheses \eqref{5}. Let $A>0$,  $T>0$ and
$0<\delta\leq\min(T,A)$. Take $w_T$ such  that
\begin{equation} \label{58}
w_T(a,x)=0 \quad \text{a.e. in } (0,\delta)\times(0,1).
\end{equation}
Then, there is $C_{\delta}>0$ such that every solution $w$ of \eqref{12}
satisfies the observability inequality
\begin{equation}
\begin{aligned}
&\int_0^{1}\int_0^{T} w^{2}(t,0,x)\,dt\,dx
 + \int_0^{1}\int_0^{A} w^{2}(0,a,x)\,da\,dx \\
&\leq C_{\delta} \int_{\omega}\int_0^{A}\int_0^{T}w^{2}(t,a,x)\,dt\,da\,dx.
\end{aligned}\label{59}
\end{equation}
\end{proposition}


For the proof, we need to show a crucial technical result.
For this, consider the following wholes, see  \cite{Traore},
\begin{equation}\label{61}
\begin{gathered}
N_1=\{(t,a)\in (0,T)\times(0,A); t\geq a+T-\delta\},\\
N_2=\{(t,a)\in (0,T)\times(0,A); t\leq a+\delta-A\},\\
D_1=\{(t,a)\in (0,T)\times(0,A);t\leq-\frac{T-\frac{\delta}{2}}{A-\frac{\delta}{2}}a
 +T-\frac{\delta}{2}\},\\
D_2=\{(t,a)\in (0,T)\times(0,A);a\geq-\frac{A-\frac{\delta}{2}}{T-\frac{\delta}{2}}t
 +A-\frac{\delta(\delta-2A)}{2(2T-\delta)}\},\\
D_{3}=(0,T)\times(0,A)-(D_1\cup D_2),
D_{4}=\{(t,a)\in D_{3}; (a,t)\notin N_1\cup N_2\}.
\end{gathered}
\end{equation}
See Figure \ref{fig1}.


\begin{figure}[ht] 
\begin{center}
\setlength{\unitlength}{1mm} 
\begin{picture}(70,68)(-9,-4)
\put(0,55){\line(1,0){55}}
\put(0,0){\line(1,0){60}}
\put(0,0){\line(0,1){60}}
\put(55,0){\line(0,1){55}}
\put(0,35){\line(1,1){20}}
\put(35,0){\line(1,1){20}}
\put(10,55){\line(1,-1){45}}
\put(0,45){\line(1,-1){45}}
\put(-.8,60){$\uparrow$}
\put(59.5,-.65){$\to$}
\put(-3,60){$t$}
\put(-4,54.5){$T$}
\put(-9.5,34.5){$T-\delta$}
\put(-3,17){$\delta$}
\put(4,48){$N_1$}
\put(38,40){$D_2$}
\put(24.5,27){$D_4$}
\put(13,15){$D_1$}
\put(47,4){$N_2$}
\put(29,-4){$A-\delta$}
\put(53,-4){$A$}
\put(61,-4){$a$}
\end{picture}
\end{center} 
\caption{Decomposition of the region $(0,T)\times(0,A)$} \label{fig1}
\end{figure}

\begin{lemma}\label{60}
Suppose that \eqref{58} holds. Then all solutions of \eqref{12} satisfy
$$
w(t,a,x)=0,  \quad\text{a.e. in } (N_1\cup N_2)\times (0,1).
$$
\end{lemma}

\begin{proof}
Let $(t_0,a_0)\in N_1$. Then,  we have $t_0=a_0+T-\delta+d$ with
$0\leq d<\delta$. Therefore $a_0<\delta-d$. Let
$S_{d}=\{(t_0+r,a_0+r), r\in(0,\delta-d-a_0)\}$ be a characteristic line in $N_1$.
Setting $\overline{w}(r,x)=w(t_0+r,a_0+r,x)$ and
$\widetilde{\mu}(r,x)=\mu(t_0+r,a_0+r,x)$, where $w$ is the solution of \eqref{12}.
Then, $\overline{w}$ solves
\begin{equation} \label{1007}
\begin{gathered}
\frac{\partial \overline{w}}{\partial r}
+(k(x)\overline{w}_{x})_{x}-\widetilde{\mu}(r, x)\overline{w}
= 0,\quad  \text{in } (0,\delta-d-a_0)\times(0,1), \\
  \overline{w}(r, 1)=\overline{w}(r, 0)=0, \quad \text{on } (0,\delta-d-a_0),\\
  \overline{w}(\delta-d-a_0,x)=w(T, \delta-d, x)= w_T(\delta-d, x),
\quad  \text{in } (0,1).
\end{gathered}
\end{equation}
Hence, $\overline{w}$ is given by
\begin{equation} \label{1008}
\overline{w}(r,\cdot)=L(\delta-d-a_0-r)\overline{w}(\delta-d-a_0,\cdot),
\end{equation}
where $(L(l))_{l\geq0}$ is the semigroup generated by the operator
 $C\overline{w}=(k\overline{w}_{x})_{x}-\widetilde{\mu}\overline{w}$.
Therefore,  \eqref{58}  and  \eqref{1008} lead to $\overline{w}=0$.
Thus, for  a. e. $d \in (0,\delta)$, $w=0$ on $S_{d}$. Subsequently,
$w=0$ in $N_1\times(0,1)$. Arguing in the same way for $N_2$ and the fact
that $w(t,A,x)=0$ in $(0,T)\times(0,1)$, we can show that $w=0$ in
$N_2\times(0,1)$ and this achieves the proof.
\end{proof}

\begin{proof}[Proof of Proposition \ref{57}]
Consider a smooth cut-off function
 $\rho_1 \in C_0^{\infty}(\mathbb{R}^{2}, [0,1])$ stated  as follows
$\rho_1(t,a)=1, (t,a) \in D_1,
 \rho_1(t,a)=0,   (t,a) \in D_2,
 \rho_1>0, (t,a)\in D_{3}$.
The function   $\widetilde{w}=\rho_1 w$ satisfies the  system
\begin{equation} \label{62}
\begin{gathered}
  \frac{\partial \widetilde{w}}{\partial t}
+ \frac{\partial \widetilde{w}}{\partial a}+(k(x)\widetilde{w}_{x})_{x}
-\mu(t, a, x)\widetilde{w}
=(\frac{\partial \rho_1}{\partial t} + \frac{\partial \rho_1}{\partial a})w
\quad \text{in } Q,\\
  \widetilde{w}(t, a, 1)=\widetilde{w}(t, a, 0)=0
\quad \text{ on }(0,T)\times(0,A),\\
 \widetilde{w}(T, a, x)= 0 \quad \text{in }Q_A,\\
 \widetilde{w}(t,A , x)=0 \quad   \text{in } Q_T.
\end{gathered}
\end{equation}
Multiplying \eqref{62} by $\widetilde{w}$, integrating over $Q$, using
the definition of $\rho_1$ and Lemma \ref{60},   we obtain
\begin{align*}
&\int_0^{1}\int_0^{A-\delta}w^{2}(0,a,x)\,da\,dx+\int_0^{1}\int_0^
{T-\delta}w^{2}(t,0,x)\,dt\,dx \\
&\leq - 2\int_{Q}(\frac{\partial \rho_1}{\partial t}
 + \frac{\partial \rho_1}{\partial a})\rho_1 w^{2}\,dt\,da\,dx\\
&\leq M_{\delta} \int_0^{1}\int_{D_{4}} w^{2}\,dt\,da\,dx.
\end{align*}
Thanks to Hardy-Poincar\'{e} inequality we conclude that
\begin{equation} \label{65}
\begin{aligned}
&\int_0^{1}\int_0^{A-\delta}w^{2}(0,a,x)\,da\,dx
 +\int_0^{1}\int_0^{T-\delta}w^{2}(t,0,x)\,dt\,dx\\
&\leq d_{\delta} \int_0^{1}\int_{D_{4}} k(x)w_{x}^{2}\,dt\,da\,dx,
\end{aligned}
\end{equation}
with $d_{\delta}=\frac{CM_{\delta}}{k(1)}$.
 Observe that $\Theta$ is bounded in $D_{4}$ to infer that
\begin{equation} \label{66}
\begin{aligned}
&\int_0^{1}\int_0^{A}w^{2}(0,a,x)\,da\,dx
 +\int_0^{1}\int_0^{T}w^{2}(t,0,x)\,dt\,dx\\
&\leq C_{{\delta}}\int_0^{1} \int_{D_{4}} \Theta k(x)w_{x}^{2}e^{2s\varphi}
\,dt\,da\,dx.
\end{aligned}
\end{equation}
Taking $s$  large and thanks to the Carleman inequality stated in
Theorem \ref{47}, we obtain the observability inequality of system \eqref{12}.
\end{proof}

\subsection{Null controllability of the intermediate system}\label{sec3-3}

This paragraph is devoted to study the null controllability of system \eqref{8}.
 For this, let $ \epsilon>0$ and  consider the following cost function
 $$
J_{\epsilon}(\vartheta)=\frac{1}{2\epsilon}
\int_0^{1}\int_{\delta}^{A}y^{2}(T, a, x)\,da\,dx
+\frac{1}{2}\int_{q}\vartheta^{2}(t,a,x)\,dt\,da\,dx.
$$
We can prove  that $J_{\epsilon}$ is continuous, convex and coercive.
Then, it admits at least one minimizer $\vartheta_{\epsilon}$ and,
arguing as in \cite{Ain6} or \cite[Chapter 5]{Ani1},   we have
\begin{equation} \label{671}
\vartheta_{\epsilon}=-w_{\epsilon}(t,a,x)\chi_{\omega}(x) \quad  \text{in } Q,
\end{equation}
with $w_{\epsilon}$ is a solution of the following system
\begin{equation} \label{681}
\begin{gathered}
\frac{\partial w_{\epsilon}}{\partial t} + \frac{\partial w_{\epsilon}}{\partial a}
+(k(x)(w_{\epsilon})_{x})_{x}-\mu(t, a, x)w_{\epsilon} = 0 \quad \text{in } Q,\\
w_{\epsilon}(t, a, 1)=w_{\epsilon}(t, a, 0)=0  \quad \text{on }(0,T)\times(0,A),\\
w_{\epsilon}(T, a, x)=\frac{1}{\epsilon}y_{\epsilon}(T, a, x)\chi_{(\delta, A)}(a)
\quad  \text{in }Q_A,\\
w_{\epsilon}(t, A, x)=0\quad  \text{in } Q_T,
\end{gathered}
\end{equation}
and $y_{\epsilon}$ is the solution of system \eqref{8} associated to the
control $\vartheta_{\epsilon}$.

Multiplying \eqref{681} by $y_{\epsilon}$, integrating over $Q$,
using \eqref{671} and Young inequality we obtain that
\begin{align*}
&\frac{1}{\epsilon}\int_0^{1}\int_{\delta}^{A}y_{\epsilon}^{2}(T, a, x)\,da\,dx
+\int_{q}\vartheta^{2}_{\epsilon}(t,a,x)dx\,dt\,da\\
&=\int_{Q_T}b(t,x)w_{\epsilon}(t, 0, x)\,dt\,dx
 +\int_{Q_A}y_0(a,x)w_{\epsilon}(0, a, x)\,da\,dx \\
& \leq \frac{1}{4C_{{\delta}}}\Big(\int_{Q_T}w_{\epsilon}
^{2}(t, 0, x)\,dt\,dx+\int_{Q_A}w_{\epsilon}^{2}(0, a, x)\,da\,dx\Big) \\
&\quad +C_{{\delta}}\Big(\int_{Q_T}b^{2}(t, x)\,dt\,dx
+\int_{Q_A}y_0^{2}(a,x)\,da\,dx\Big),
\end{align*}
with $C_{{\delta}}$ is the constant given in Proposition \ref{57}.
Hence, by the observability inequality \eqref{59}, we conclude that
\begin{align*}
 &\frac{1}{\epsilon}\int_0^{1}\int_{\delta}^{A}y_{\epsilon}^{2}(T, a, x)\,da\,dx
+ \int_{q}\vartheta^{2}_{\epsilon}(t,a,x)\,dt\,da\,dx \\
&\leq \frac{1}{4}\int_{q}w^{2}_{\epsilon}(t,a,x)\,dt\,da\,dx
+C_{{\delta}}\Big(\int_{Q_T}b^{2}(t, x)\,dt\,dx+\int_{Q_A}y_0^{2}(a,x)\,da\,dx\Big).
\end{align*}
Hence, \eqref{671} yields
\begin{equation} \label{72}
\begin{aligned}
&\frac{1}{\epsilon}\int_0^{1}\int_{\delta}^{A}y_{\epsilon}^{2}(T, a, x)\,da\,dx
+\frac{3}{4}\int_{q}\vartheta^{2}_{\epsilon}(t,a,x)\,dt\,da\,dx\\
&\leq C_{{\delta}}\Big(\int_{Q_T}b^{2}(t, x)\,dt\,dx+\int_{Q_A}y_0^{2}
(a,x)\,da\,dx\Big),
\end{aligned}
\end{equation}
and this yields
\begin{equation}
\begin{gathered}
\int_0^{1}\int_{\delta}^{A}y_{\epsilon}^{2}(T, a, x)\,dx\,da
\leq C_{{\delta}}\epsilon\Big(\int_{Q_T}b^{2}(t, x)\,dx\,dt
+\int_{Q_A}y_0^{2}(a,x)\,dx\,da\Big),\\
\int_{q}\vartheta^{2}_{\epsilon}(t,a,x)\,dt\,da\,dx
\leq \frac{4C_{{\delta}}}{3}\Big(\int_{Q_T}b^{2}(t, x)\,dt\,dx
+\int_{Q_A}y_0^{2}(a,x)\,da\,dx\Big).
\end{gathered}\label{73}
\end{equation}
 Then, we can extract two subsequences of $y_{\epsilon}$ and $\vartheta_{\epsilon}$
denoted also by $\vartheta_{\epsilon}$ and $y_{\epsilon}$ that converge weakly
towards $\vartheta$ and $y$ in $L^{2}(q)$ and
$L^{2}((0, T)\times(0, A); H^{1}_{k}(0, 1))$ respectively. Furthermore, $y$
is the unique solution of \eqref{8} that satisfies \eqref{2}.
In summary, we showed the following proposition.

\begin{proposition}\label{112}
For any $\delta>0$ assumed to be small enough, for all
$y_0 \in L^{2}(Q_A)$, there exists a control $\vartheta \in L^{2}(q)$ such
that the associated solution of system \eqref{8} verifies \eqref{2}.
\end{proposition}

\section{Main null controllability result}\label{last-section-3}

Now, after establishing the null controllability of system \eqref{8}
we are ready to provide the one of  the model \eqref{1}.
More precisely, we  have the following theorem.

\begin{theorem}\label{theorem-null controllability}
For any $\delta>0$ assumed to be small enough, for all $y_0 \in L^{2}(Q_A)$,
there exists a control $\vartheta \in L^{2}(q)$ such that the associated solution
of system \eqref{1} verifies \eqref{2}.
\end{theorem}

To prove this result, let $\lambda$ be a positive constant. A more precise
restriction will be given later. Put $\widetilde{y}=e^{-\lambda t}y$.
Then $\widetilde{y}$ solves
\begin{equation} \label{1027}
\begin{gathered}
  \frac{\partial \widetilde{y}}{\partial t}
+ \frac{\partial \widetilde{y}}{\partial a}
-(k(x)\widetilde{y}_{x})_{x}+\mu_1(t, a, x)\widetilde{y}
 = \widetilde{\vartheta}\chi_{\omega}  \quad \text{in } Q,\\
  \widetilde{y}(t, a, 1)=\widetilde{y}(t, a, 0)=0 \quad \text{on }(0,T)\times(0,A),\\
 \widetilde{y}(0, a, x)=y_0(a, x) \quad  \text{in }Q_A,\\
 \widetilde{y}(t, 0, x)=\int_0^{A} \beta (t, a, x)\widetilde{y} (t, a, x)da \quad
  \text{in } Q_T,
\end{gathered}
\end{equation}
with $\widetilde{\vartheta}=e^{-\lambda t}\vartheta$ and $\mu_1=\mu+\lambda$.
 Now, consider the system
\begin{equation} \label{1023}
\begin{gathered}
  \frac{\partial \widetilde{y}}{\partial t}
+ \frac{\partial \widetilde{y}}{\partial a}
-(k(x)\widetilde{y}_{x})_{x}+\mu_1(t, a, x)\widetilde{y}
= \widetilde{\vartheta}\chi_{\omega} \quad \text{in } Q,\\
  \widetilde{y}(t, a, 1)=\widetilde{y}(t, a, 0)=0  \quad \text{on }(0,T)\times(0,A),\\
 \widetilde{y}(0, a, x)=y_0(a, x) \quad  \text{in }Q_A,\\
 \widetilde{y}(t, 0, x)=b(t,x)  \quad  \text{in } Q_T,
\end{gathered}
\end{equation}
with $b\in L^{2}(Q_T)$. Thus, showing Theorem \ref{theorem-null controllability}
is equivalent to show the null controllability of system \eqref{1027}.
For this, we consider the following multi-valued mapping
$$
\Lambda_{\delta}:L^{2}(Q_T)\to \mathcal{P}(L^{2}(Q_T))
$$
defined, for every small $\delta>0$   and  $R \in L^{2}(Q_T)$, by
\[
\Lambda_{\delta}(R)
=\big\{\int_0^{A} \beta \widetilde{y} da:  \widetilde{y}
\text{ satisfies \eqref{2}  and \eqref{1023}  for $b=R$,   and
$\widetilde{\vartheta}$  satisfies \eqref{73}}\big\}.
\]
To prove that model \eqref{1027} is null controllable, it is sufficient
to prove that the multivalued mapping admits a
fixed point and this by using a generalization of the Leray-Schauder
fixed point theorem stated in \cite{Avramescu}.
To use this generalization, we introduce the  set
\begin{equation} \label{99}
N_{\delta}=\{R\in L^{2}(Q_T): \exists \rho\in(0,1), R \in \rho\Lambda_{\delta}(R)\}.
\end{equation}
The existence of a fixed point of the multi-valued mapping $\Lambda_{\delta}$
is an immediate consequence of the following proposition.

\begin{proposition}\label{100}
\begin{itemize}
\item[(i)] for all $R \in L^{2}(Q_T)$, $\Lambda_{\delta}(R)$ is a closed and
 convex set.

\item[(ii)] $\Lambda_{\delta}$ is upper semi-continuous on $L^{2}(Q_T)$.

\item[((iii)] $\Lambda_{\delta}: L^{2}(Q_T)\to P(L^{2}(Q_T))$
is a compact multivalued mapping.

\item[(iv)] $ N_{\delta}$ is bounded in $L^{2}(Q_T)$.
\end{itemize}
\end{proposition}

\begin{proof}
The proofs of (i) and (ii)  are similar to the ones of (ii) and (iv)
 in \cite{Traore}, with $R$ (respectively $R_{n}$) instead of
 $e^{-\lambda_0t}F(e^{\lambda_0t}R)$
(respectively $e^{-\lambda_0t}F(e^{\lambda_0t}R_{n})$)
and the convergence space of the subsequence of $\widetilde{y}_{n}$ is
$L^{2}((0,A)\times(0,T), H^{1}_{k}(0,1))$ instead of the space
$L^{2}((0,A)\times(0,T), H^{1}_0(0,1))$.

Now, we address the proof of (iii). Let $R \in L^{2}(Q_T)$ such that
$\|R\|_{L^{2}(Q_T)}\leq K$, $K>0$. We have to prove that any sequence of
elements of  $\Lambda_{\delta}(R)$ admits a convergent subsequence.
Let $(\rho_{n})_{n} \subseteq \Lambda_{\delta}(R)$. From the definition
of $\Lambda_{\delta}$, for all $n$ there exists
$(\widetilde{\vartheta}_{n}, \widetilde{y}_{n})\in L^{2}(q)\times L^{2}(Q)$
such that $\rho_{n}=\int_0^{A} \beta \widetilde{y}_{n} da$,
$\widetilde{\vartheta}_{n}$ verifies \eqref{73} and $\widetilde{y}_{n}$,
the associated solution of \eqref{1023}  verifies \eqref{2}.
Then, by \eqref{73} we have
\begin{equation} \label{77}
\begin{aligned}
\int_{q}\widetilde{\vartheta}^{2}_{n}(t,a,x)\,dt\,da\,dx
&\leq \frac{4 C_{\delta}}{3}\Big(\int_{Q_T}R^{2}(t, x)\,dt\,dx
+\int_{Q_A}y_0^{2}(a,x)\,da\,dx\Big)\\
&\leq \frac{4 C_{\delta}}{3}\Big(K^{2}+\int_{Q_A}y_0^{2}(a,x)\,da\,dx\Big).
\end{aligned}
\end{equation}
Hence, $\widetilde{\vartheta}_{n}$ is bounded in $L^{2}(q)$.
Thus, there exists a subsequence of $\widetilde{\vartheta}_{n}$ denoted by
$\widetilde{\vartheta}_{n_{k}}$ that converges weakly towards
 $\widetilde{\vartheta}$ in $L^{2}(q)$.
 On the other hand, multiplying \eqref{1023} by $\widetilde{y}_{n}$,
integrating over $Q$, using Young inequality, we infer that
\begin{equation} \label{78}
\begin{aligned}
&\int_{Q}k(x)(\widetilde{y}_{n})_{x}^{2}\,dt\,da\,dx
 + \lambda\int_{Q} \widetilde{y}_{n}^{2}\,dt\,da\,dx \\
&\leq \frac{1}{2\lambda}\int_{q}\widetilde{\vartheta}^{2}_{n}(t,a,x)\,dt\,da\,dx
+\frac{\lambda}{2}\int_{Q}\widetilde{y}_n^{2}\,dt\,da\,dx\\
&\quad+\frac{1}{2}\Big(\int_{Q_A}y_0^{2}(a,x)\,da\,dx+\int_{Q_T}R^{2}(t,x)\,da\,dx\Big).
\end{aligned}
\end{equation}
This implies
\begin{equation} \label{1024}
\begin{aligned}
&\int_{Q}k(x)(\widetilde{y}_{n})_{x}^{2}\,dt\,da\,dx
+ \frac{\lambda}{2}\int_{Q}\widetilde{y}^{2}\,dt\,da\,dx\\
&\leq \frac{1}{2\lambda}\int_{q}\widetilde{\vartheta}^{2}_{n}(t,a,x)\,dt\,da\,dx
+\frac{1}{2}\Big(\int_{Q_A}y_0^{2}(a,x)\,da\,dx+\int_{Q_T}R^{2}(t,x)\,da\,dx\Big).
\end{aligned}
\end{equation}
Taking $\lambda\geq2$ and using \eqref{77}, \eqref{1024} becomes
\begin{equation} \label{1025}
\int_{Q}k(x)(\widetilde{y}_{n})_{x}^{2}\,dt\,da\,dx
+\int_{Q}\widetilde{y}^{2}\,dt\,da\,dx
\leq \Big(\frac{1}{2}+\frac{C_{\delta}}{3}\Big)
\Big(K^2+\int_{Q_A}y_0^{2}(a,x)\,da\,dx\Big).
\end{equation}
 Therefore, $\widetilde{y}_{n}$ is bounded in
$L^{2}((0,T)\times(0,A), H^{1}_{k}(0,1))$.
Hence, we can extract a subsequence of $\widetilde{y}_{n}$ denoted by
$\widetilde{y}_{n_{k_1}}$ that converges weakly toward $\widetilde{y}$ in
$L^{2}((0,T)\times(0,A), H^{1}_{k}(0,1))$.
Now, we consider $\rho_{n_{k_1}}=\int_0^{A} \beta \widetilde{y}_{n_{k_1}} da$
the subsequence of $\rho_{n}$ associated to
$\widetilde{y}_{n_{k_1}}$. Using \eqref{3}, we conclude that $\rho_{n_{k_1}}$
satisfies the  system
\begin{equation} \label{79}
\begin{gathered}
\frac{\partial \rho_{n_{k_1}}}{\partial t}
-(k(x)(\rho_{n_{k_1}})_{x})_{x}+\int_0^{A} \beta \mu_1 \widetilde{y}_{n_{k_1}} da
=z_{n_{k_1}} \quad \text{in } Q_T,\\
\rho_{n_{k_1}}(t, 1)=\rho_{n_{k_1}}(t, 0)=0  \quad \text{on }(0,T),\\
 \rho_{n_{k_1}}(0, x)=\int_0^{A}\beta (0, a, x)y_0(a, x)da \quad  \text{in } (0,1),
\end{gathered}
\end{equation}
with,
\begin{align*}
z_{n_{k_1}}
& =\int_0^{A}\beta \widetilde{\vartheta}_{n_{k_1}}\chi_{\omega}da
+\int_0^{A}(\beta_{t}+\beta_{a})\widetilde{y}_{n_{k_1}}da\\
&-\Big(\int_0^{A}k(x)\beta_{x}(\widetilde{y}_{n_{k_1}})_{x}da
+\int_0^{A}(k(x)\beta_{x}\widetilde{y}_{n_{k_1}})_{x}da\Big).
\end{align*}
Taking into account the assumptions on $k$, using Hardy-Poincar\'{e}
and Minkowski's inequalities and exploiting the inequalities \eqref{77}
 and \eqref{1025} for $\widetilde{\vartheta}_{n_{k_1}}$ and
$\widetilde{y}_{n_{k_1}}$ respectively, we deduce that
\begin{equation} \label{1026}
\|z_{n_{k_1}}\|_{L^{2}(Q_T)}^{2}
\leq D_{\delta}(K^{2}+\int_{Q_A}y_0^{2}(a,x)\,da\,dx).
\end{equation}
Now, multiplying the first equation of system \eqref{79} by $\rho_{n_{k_1}}$,
integrating over $Q_T$ and using Young inequality, we obtain
\begin{equation} \label{105}
\begin{aligned}
&\int_{Q_T}k(x)(\rho_{n_{k_1}})_{x}^{2}\,dt\,dx
+ \frac{\lambda}{2}\int_{Q_T} \rho_{n_{k_1}}^{2}\,dt\,dx\\
&\leq\frac{1}{2\lambda}\int_{Q_T}z_{n_{k_1}}^{2}\,dt\,dx
+\frac{C_{\beta}}{2}\int_{Q_A}y_0^{2}(a,x)\,da\,dx.
\end{aligned}
\end{equation}
Taking again $\lambda\geq2$ in \eqref{105}, we conclude by \eqref{1026}
that $\rho_{n_{k_1}}$ is bounded in $L^{2}((0,T); H^{1}_{k}(0,1))$.
Now, thanks to Proposition \ref{580} we infer that
$\frac{\partial \rho_{n_{k_1}}}{\partial t}$ is bounded in
$L^{2}((0,T); H^{-1}_{k}(0,1))$.
Since $H^{1}_{k}(0,1)$ is compactly embedded in $L^{2}(0,1)$(see \cite{Bouss}),
we conclude by Aubin-Lions lemma the existence of a subsequence of
$\rho_{n_{k_1}}$ denoted by $\rho_{n_{j}}$ that converges strongly towards
$\rho$ in $L^{2}(Q_T)$.
This implies that $\rho_{n_{j}}$ converges weakly towards $\rho$ in $L^{2}(Q_T)$.
Thus,
\begin{equation} \label{106}
 \int_{Q_T}g\rho_{n_{j}}\,dt\,dx\to\int_{Q_T}g\rho \,dt\,dx, \quad
\forall g\in L^{2}(Q_T).
\end{equation}
On the other hand,  $\widetilde{y}_{n_{k_1}}$ converges weakly to
 $\widetilde{y}$ in $L^{2}((0,T)\times(0,A), H^{1}_{k}(0,1))$.
Then, $\widetilde{y}_{n_{k_1}}$ converges weakly toward $\widetilde{y}$ in
$L^{2}(Q)$ (because $L^{2}((0,T)\times(0,A), H^{1}_{k}(0,1)) \subseteq L^{2}(Q)$).
Subsequently, $\widetilde{y}_{n_{j}}$ the subsequence of
 $\widetilde{y}_{n_{k_1}}$ associated to $\rho_{n_{j}}$ converges weakly towards
$\widetilde{y}$ as well. The fact that $g\beta \in L^{2}(Q)$,  for all
$g\in L^{2}(Q_T)$, implies that
\begin{equation} \label{107}
 \int_{Q_T}g\rho_{n_{j}}\,dx\,dt\to
\int_{Q_T}g\Big(\int_0^{A}\beta \widetilde{y} da\Big) \,dx\,dt.
\end{equation}
Accordingly, by \eqref{106} and \eqref{107},  we infer that
 \begin{equation} \label{108}
\int_{Q_T}g\Big(\int_0^{A}\beta \widetilde{y} da-\rho\Big)\,dx\,dt=0,
\forall g\in L^{2}(Q_T) .
\end{equation}
Therefore,
\begin{equation} \label{109}
\rho(t,x)=\int_0^{A} \beta \widetilde{y} \,da \quad  \text{a.e. }  (t,x)\in Q_T.
\end{equation}
Furthermore, we can check by a standard argument that
 $\widetilde{y}$ satisfies \eqref{2} and solves \eqref{1023} with $R$
instead of $b$ and $\widetilde{\vartheta} \in L^{2}(q)$ verifies \eqref{73}
and this completes the proof of (iii).
The rest follows as in \cite{Traore}.
\end{proof}

\section{Appendix}

As is mentioned in the introduction, this section is devoted to a Caccioppoli's
type inequality which played  a crucial role to establish
the Carleman estimate \eqref{48}. This inequality is stated in the following lemma.

\begin{lemma}\label{51}
There exists a positive constant $C$ such that
\begin{equation}\label{Caccio}
\int_{\omega'}\int_0^{A}\int_0^{T}w_{x}^{2}e^{2s\varphi} \,dt\,da\,dx
\leq C\Big(\int_{q}s^{2}\Theta^{2}w^{2}e^{2s\varphi} \,dt\,da\,dx+\int_{q}h^{2}
e^{2s\varphi}\,dt\,da\,dx\Big),
\end{equation}
 where $\omega'=(\frac{2x_1+x_2}{3},\frac{x_1+2x_2}{3})$ and $w$ is the solution
of \eqref{14}.
\end{lemma}

\begin{proof}
Define the  smooth cut-off function $\zeta:\mathbb{R}\to \mathbb{R}$ by
\begin{equation}\label{81}
\begin{gathered}
0\leq\zeta(x)\leq1, \quad\text{if } x \in \mathbb{R},\\
\zeta(x)=0, \quad\text{if }   x<x_1 \text{ and } x>x_2,\\
\zeta(x)=1, \quad\text{if }  x \in \omega',\\
\zeta(x)>0, \quad\text{if }  x>x_1 \text{ and } x<x_2.
\end{gathered}
\end{equation}
For the solution  $w$ of \eqref{14}, we have
\begin{align*} %\label{131}
  0&=\int_0^{T}\frac{d}{dt}\Big[\int_0^{1}\int_0^{A}
\zeta^{2}e^{2s\varphi}w^{2}\,da\,dx\Big]dt\\
&=2s \int_0^{1}\int_0^{A}
\int_0^{T}\zeta^{2}\varphi_{t}w^{2}e^{2s\varphi}\,dt\,da\,dx
+2\int_0^{1}\int_0^{A} \int_0^{T}\zeta^{2}ww_{t}
 e^{2s\varphi}\,dt\,da\,dx\\
&=2s\int_0^{1}\int_0^{A}
\int_0^{T}\zeta^{2}\varphi_{t}w^{2}e^{2s\varphi}\,dt\,da\,dx \\
&\quad + 2\int_0^{1}\int_0^{A}\int_0^{T}\zeta^{2}w(-(kw_{x})_{x}-w_{a}
 +h+\mu w)e^{2s\varphi}\,dt\,da\,dx.
\end{align*}
Integrating by parts, we obtain
\begin{align*}
&2\int_{Q}k\zeta^{2}e^{2s\varphi}w_{x}^{2}\,dt\,da\,dx\\
&=-2s\int_{Q}\zeta^{2}w^{2}\psi(\Theta_{a}+
\Theta_{t})e^{2s\varphi}\,dt\,da\,dx-2\int_{Q}\zeta^{2}whe^{2s\varphi}\,dt\,da\,dx\\
&\quad -2\int_{Q}\zeta^{2}\mu w^{2}e^{2s\varphi}\,dt\,da\,dx
+\int_{Q}(k(\zeta^{2}e^{2s\varphi})_{x})_{x}w^{2}\,dt\,da\,dx.
\end{align*}
On the other hand, by the definitions of $\zeta$, $\psi$ and $\Theta$,
using Young inequality and taking $s$ quite large
there is a constant  $c$ such that
\begin{gather*}
2\int_{Q}k\zeta^{2}e^{2s\varphi}w_{x}^{2}\,dt\,da\,dx
\geq 2\min_{x\in\omega'}k(x)\int_{\omega'}\int_0^{A}
\int_0^{T}w_{x}^{2}e^{2s\varphi}\,dt\,da\,dx,
\\
\int_{Q}(k(\zeta^{2}e^{2s\varphi})_{x})_{x}w^{2}\,dt\,da\,dx
\leq c\int_{\omega'}\int_0^{A}\int_0^{T}s^{2}
\Theta^{2}w^{2}e^{2s\varphi}\,dt\,da\,dx,
\\
-2s\int_{Q}\zeta^{2}w^{2}\psi(\Theta_{a}+\Theta_{t})e^{2s\varphi}\,dt\,da\,dx
\leq c\int_{\omega'}\int_0^{A}\int_0^{T}s^{2}\Theta^{2}w^{2}
e^{2s\varphi}\,dt\,da\,dx,
\\
\begin{aligned}
&-2\int_{Q}\zeta^{2}whe^{2s\varphi}\,dt\,da\,dx\\
&\leq c\Big(\int_{\omega'}\int_0^{A}\int_0^{T}s^{2}\Theta^{2}w^{2}e^{2s\varphi}
\,dt\,da\,dx+\int_{\omega'}\int_0^{A}\int_0^{T}h^{2}e^{2s\varphi}\,dt\,da\,dx\Big),
\end{aligned}\\
-2\int_{Q}\zeta^{2}\mu w^{2}e^{2s\varphi}\,dt\,da\,dx
\leq c\int_{\omega'}\int_0^{A}\int_0^{T}s^{2}\Theta^{2}w^{2}e^{2s\varphi}\,dt\,da\,dx.
\end{gather*}
This all together imply that there is $ C>0$ such that
\[
\int_{\omega'}\int_0^{A}\int_0^{T}w_{x}^{2} e^{2s\varphi}\,dt\,da\,dx
\leq C\Big(\int_{q}s^{2}\Theta^{2}w^{2}e^{2s\varphi} \,dt\,da\,dx+\int_{q}h^{2}
e^{2s\varphi}\,dt\,da\,dx\Big).
\]
Thus, the proof is complete.
\end{proof}

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\end{document}
