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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2016 (2016), No. 219, pp. 1--25.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2016 Texas State University.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2016/219\hfil Controllability of cascade systems]
{Controllability of nonlinear degenerate parabolic cascade systems}

\author[M. Birba, O. Traore \hfil EJDE-2016/219\hfilneg]
{Mamadou Birba, Oumar Traore}

\address{Mamadou Birba \newline
Laboratoire LAMI, Universit\'a  Ouaga I,
 Professeur Joseph Ki Zerbo,
 01 BP 7021 Ouaga 01, Burkina Faso}
\email{birba12mamadou@gmail.com}

\address{Oumar Traore   \newline
D\'epartement de Math\'ematiques de la Decision, Universit\'e Ouaga2,
 12 BP 417 Ouaga 12, Laboratoire LAMI, Universit'a Ouaga I,
Professeur Joseph Ky Zerbo,
01 BP 7021 Ouaga 01, Burkina Faso}
\email{traore.oumar@univ-ouaga.bf}

\thanks{Submitted March 21, 2016. Published August 16, 2016.}
\subjclass[2010]{35K65, 93B05, 93B07, 93C20}
\keywords{Null controllability; nonlinear coupled systeml Carleman inequality;
\hfill\break\indent  observability inequality; degenerate parabolic system;
 Kakutani fixed point}

\begin{abstract}
 This article studies of null controllability property of  nonlinear
 coupled one dimensional degenerate parabolic equations.
 These equations form a cascade system, that is, the solution of the
 first equation acts as a control in the second equation and the
 control function  acts  only directly on the first equation.
 We prove positive null controllability results when the control and
 a coupling set have nonempty intersection.
\end{abstract}

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


\section{Introduction and statement of main results}

The control of coupled parabolic system is a challenging issue, which has
 attracted the interest of the control community in the last decade.
These parabolic systems arise, for example, in the study of chemical
reactions (see e.g. \cite{25,26}), and in a wide variety of mathematical
biology and physical situations (see e.g. \cite{28,27,29}).
For some examples involving degeneracy, let us recall some applications
arising in aeronautics (the Crocco equation, see, e.g., \cite{3}),
in physics (boundary layer models, see e.g. \cite{14}),
in genetics (Wright-Fisher and Fleming-Viot models, see e.g. \cite{15, 16})
 and in mathematical finance (Black-Merton-Scholes models, see e.g.
\cite{17, 18, 19}). In \cite{20, 21, 22}, the authors developed a functional
analytic approach to the construction of Feller semigroups generated by
degenerate elliptic operators with Wentzell boundary conditions.
In \cite{31}, the authors consider degenerate operators with several boundary
conditions.

In \cite{7}, the authors studied the null controllability properties for two
systems of coupled one dimensional degenerate parabolic equations.
The first system consists of two forward equations, while the second
one consists of one forward equation and one backward equation with
$k(x)=x^{\alpha}$, $0<\alpha<2$.


In this article, we study the null controllability of a cascade system
of nonlinear coupled one dimensional degenerate parabolic equations,
at each fixed time $T>0$. More exactly, we show that for all
 $y_0,z_0 \in L^2(\Omega)$ and $T>0$, there exists a control
$h\in L^2(\omega\times(0,T))$ such that the associated solution of
\eqref{syst1} satisfies
\[
y(\cdot,T)\in L^2(\Omega) \quad\text{and}\quad
z(x,T)=0 \quad \text{a.e. in } \Omega
\]

In this context of degeneracy and nonlinearity of the system, we will study
the null controllability of the linear degenerate coupled system by using
the method developed in \cite{3} and after, we will use Kakutani's fixed
point theorem to deduce the null controllability of the nonlinear system.

We consider the nonlinear coupled system
\begin{equation}\label{syst1}
\begin{gathered}
 y_t-(k(x)y_x)_x+f(y)=h1_{\omega}\quad\text{in }Q=\Omega\times(0,T),\\
 z_t-(k(x)z_x)_x+g(z)=y1_{\mathcal{O}}\quad\text{in }Q,\\
 y=z=0\quad\text{on }\Sigma=\{ 0,1\}\times(0,T),\\
 y(x,0)=y_0(x),\quad z(x,0)=z_0(x)\quad\text{in }\Omega=(0,1),
\end{gathered}
\end{equation}
where $f,g:\mathbb{R}\to\mathbb{R}$ are two locally
Lipschitz-continuous functions, the initial data $y_0,z_0$ are given in
 $L^2(\Omega)$, $h\in L^2(Q)$ is a control function to be determined,
 $k$ is a diffusion coefficient. Here, $\omega\subset\Omega$ is an arbitrarily
small open control set, $\mathcal{O}\subset\Omega$ is an arbitrarily small
open coupling set and  $1_{\omega}$ denotes the characteristic function of $\omega$.

In this article, we assume that the coefficient $k$ satisfies the following hypotheses:
\begin{equation} \label{k}
\parbox{10cm}{$k\in\mathcal{C}([0,1])\cap\mathcal{C}^1((0,1])$,
$k>0$  in $(0,1]$  and $k(0)=0$,
there exists $\lambda\in [0,1)$  such that $xk'(x)\leq\lambda k(x)$  for all
$x\in [0,1]$.}
\end{equation}

The remainder of this article is organized as follows:
In Section $2$, we establish the uniqueness of the  solution to  \eqref{syst2}.
Section $3$ is devoted to the proof of our general Carleman's inequality
of degenerate cascade system. This inequality is crucial for the proof of
the observability inequality that is used to prove the null controllability
of the linear system. In the next two Sections, we prove the null controllability
results for the linear system, and the last Section is devoted to the proof
of  Theorem \ref{t3'} and Theorem \ref{t3}, which are the main results
of this paper.

\begin{theorem}\label{t3'}
We assume that \eqref{k} holds, $\omega\cap\mathcal{O}\neq\emptyset$ and
$f,g:\mathbb{R}\to\mathbb{R}$ are two locally Lipschitz-continuous functions
such that $f(0)=g(0)=0$ and that:
\[
\lim_{|s|\to +\infty}\frac{f(s)}{|s|\log^{3/2}(1+|s|)}=0,\quad
\lim_{|\sigma|\to +\infty}\frac{g(\sigma)}{|\sigma|\log^{3/2}(1+|\sigma|)}=0.
\]
Then, for any  $(y_0,z_0)\in (H^2_k(\Omega))^2$ and $T>0$,  system \eqref{syst1}
admits (at least) a solution
$y,z\in X=\mathcal{C}([0,T]; L^2(\Omega))\cap L^2(0,T;H^1_k(\Omega))$
and there exists a control $h\in L^2(Q)$ such that the solution $(y,z)$
satisfies $z(x,T)=0$ a.e in $\Omega$.
\end{theorem}


\begin{theorem}\label{t3}
We assume that \eqref{k} holds and  $\omega\cap\mathcal{O}\neq\emptyset$.
For any subset $\omega'$ such that $\omega'\Subset\omega\cap\mathcal{O}$,
there exists two positive constants
$C=C(\Omega,\omega,k,\| a\|_{\infty},\| b\|_{\infty})$ and
 $s_0$ such that for every $s\geq s_0$, one has
\begin{align}
&\int_{Q}\Big(s\theta k(x)|v_x|^2+s^3\theta^3\frac{x^2}{k(x)}|v|^2
+s^4\theta^4 k(x)|w_x|^2+s^6\theta^6\frac{x^2}{k(x)}|w|^2\Big)e^{2s\Phi}
\,{\rm d}x\,{\rm d}t \nonumber \\
&\leq  Cs^7\int_0^T\int_{\omega'}\theta^7e^{2s\Phi}|w|^2. \label{p20}
\end{align}
\end{theorem}

Theorem \ref{t3}, improves the Carleman estimate established in \cite{8}
for parabolic  degenerate equations. Our proofs use the  Carleman estimate
established in \cite{8}.

\section{Well-posedness of the problem}

Now, we introduce the  Sobolev spaces
\begin{gather*}
\begin{aligned}
H^1_k(\Omega) = \big\{& u\in L^2(\Omega); u\text{ absolute continuous in }[0,1],\\
&\sqrt{k}u_x\in L^2(\Omega),\; u(0)=u(1)=0\big\},
\end{aligned}\\
H^2_k(\Omega) = \{ u\in H^1_k(\Omega);\;k(x)u_x\in H^1(\Omega)\},
\end{gather*}
respectively with the  norms
\begin{gather*}
\| u\|^2_{H^1_k(\Omega)}
= \| u\|^2_{L^2(\Omega)}+\| \sqrt{k} u_x\|^2_{L^2(\Omega)},\quad
\forall u\in H^1_k(\Omega);\\
\| u\|^2_{H^2_k(\Omega)} = \| u\|^2_{H^1_k(\Omega)}
+\| (ku_x)_x\|^2_{L^2(\Omega)},\quad\forall u\in H^2_k(\Omega).
\end{gather*}
Consider the unbounded operator $A : D(A)=H^2_k(\Omega)\to L^2(\Omega)$
defined by $Au=(k(x)u_x)_x$, $u\in  D(A)$.
Recall  that the operator $Au=(ku_x)_x$,
$u\in D(A)=H^2_k(\Omega)\subset H^1_k(\Omega)$, is a closed, symmetric,
self-adjoint and negative operator in addition, $D(A)$ is dense in
$L^2(\Omega)$ (see \cite{1,2,3}). Moreover, it is infinitesimal generator
of a strongly continuous semi-group. This permits to deduce the following result.

\begin{theorem}\label{uni}
Under  hypothesis  \eqref{k}, for all $a\in L^{\infty}(Q)$ and $f\in L^2(Q)$,
the  system
\begin{equation}\label{syst2}
\begin{gathered}
 y_t-(k(x)y_x)_x+ay=f\quad\text{in }Q,\\
 y=0\text{ on }\Sigma,\quad  y(x,0)=y_0(x) \text{ in }\Omega
\end{gathered}
\end{equation}
admits a unique solution
$y\in X=\mathcal{C}([0,T]; L^2(\Omega))\cap L^2(0,T;H^1_k(\Omega))$ and we have
\begin{equation}\label{l1'}
\sup_{[0,T]}\| y(t)\|^2_{ L^2(\Omega)}
+\int_0^T\| y(t)\|^2_{ H^1_k(\Omega)}\,{\rm d}t
\leq C\Big(\| f\|^2_{ L^2(Q)}+\| y_0\|^2_{ L^2(\Omega)}\Big).
\end{equation}
If moreover $y_0\in D(A)$ then, $y\in X_1= H^1(0,T;L^2(\Omega))
\cap L^2(0,T;H^2_k(\Omega))\cap\mathcal{C}([0,T]; H^1_k(\Omega))$ and
\begin{equation}\label{l2'}
\sup_{[0,T]}\| y(t)\|^2_{ H^1_k(\Omega)}
+\int_0^T(\| y_t\|^2_{ L^2(\Omega)}
+\|(k y_x)_x\|^2_{ L^2(\Omega)})\,{\rm d}t
\leq C(\| f\|^2_{ L^2(Q)}+\| y_0\|^2_{ H^1_k(\Omega)}).
\end{equation}
\end{theorem}

\begin{proof}
 The existence and uniqueness of the solution of system \eqref{syst1}
comes from the theory of continuous semi-group  see \cite{2} and \cite{11}.
Moreover, the solution is
\begin{equation}
y(t) =  T(t)y_0+\int_0^{t}T(t-s)(h(s)-f(s))\,{\rm d}s,
\end{equation}
where $(T(t))_{t\geq 0}$ is the semigroup generated by the operator $A$.

Multiplying the first equation of \eqref{syst2} by $y$ and  integrating
by parts then integrate on $\Omega$, we obtain
\begin{equation}
\frac{d}{2dt}\| y(t)\|^2_{L^2(\Omega)}+\|\sqrt{k} y_x(t)
\|^2_{L^2(\Omega)}=\int_{\Omega}y(f(t)-a(t)y(t))\,{\rm d}x.
\end{equation}
Thus,
\begin{equation}\label{u}
\begin{aligned}
&\frac{d}{2dt}\| y(t)\|^2_{L^2(\Omega)}
 +\|\sqrt{k} y_x(t)\|^2_{L^2(\Omega)} \\
&\leq(\frac{1}{2}+\| a(t)\|_{L^{\infty}(\Omega)})\| y(t)\|^2_{L^2(\Omega)}
+\frac{1}{2}\| f(t)\|^2_{L^2(\Omega)}.
\end{aligned}
\end{equation}
Now, integrating this inequality on $(0,t)$ we find
\begin{equation}
\begin{aligned}
&\frac{1}{2}\|y(t)\|^2_{L^2(\Omega)}+\int_0^{t}\|\sqrt{k} y_x(s)\|^2_{L^2(\Omega)}
\,{\rm d}s \\
&\leq(\frac{T}{2}+\|a\|_{L^{\infty}(Q)})\int_0^{t}
 \| y(s)\|^2_{L^2(\Omega)}\,{\rm d}s+\frac{1}{2}
\| f\|^2_{L^2(Q)}+\frac{1}{2}\| y_0\|^2_{L^2(\Omega)}.
\end{aligned}
\end{equation}
Using Gronwall's inequality, we obtain
\begin{equation}\label{l1}
\| y(t)\|^2_{L^2(\Omega)}\leq e^{( T(T+2\| a\|_{L^{\infty}(Q)}))}
(\| f(t)\|^2_{L^2(\Omega)}+\| y_0\|^2_{L^2(\Omega)}).
\end{equation}
From  \eqref{l1} and by integrating \eqref{u} on $(0,T)$, we obtain
\begin{equation}\label{l2}
\begin{aligned}
\|\sqrt{k} y_x\|^2_{L^2(Q)}
&\leq \frac{1}{2}
[1+(T+2\| a\|_{L^{\infty}(Q)})e^{( T(T+2\| a\|_{L^{\infty}(Q)}))}] \\
&\quad\times (\| f\|^2_{L^2(Q)}+\| y_0\|^2_{L^2(\Omega)}).
\end{aligned}
\end{equation}
Thus, \eqref{l1} and \eqref{l2} give \eqref{l1'}.

Now, multiplying the first equation of  system \eqref{syst2}, by $-(ky_x)_x$
and integrating by parts on $\Omega$, we have
\begin{align*}
&\frac{d}{2dt}\|\sqrt{k}y_x(t)\|^2_{L^2(\Omega)}
 +\| (ky_x)_x(t)\|^2_{L^2(\Omega)} \\
& =  \int_0^{1}((ky_x)_x(t))(a(t)y(t)-f(t))\\
&\leq \frac{1}{2}\| (ky_x)_x(t)\|^2_{L^2(\Omega)}+\| f(t)\|^2_{L^2(\Omega)}
 + \| a(t)\|^2_{L^{\infty}(\Omega)}\| y(t)\|^2_{L^2(\Omega)}.
\end{align*}
Integrating the last inequality on $(0,T)$ and using \eqref{l1}, one gets
\begin{equation}\label{l3}
\begin{aligned}
&\|\sqrt{k}y_x(T)\|^2_{L^2(\Omega)}
 + \frac{1}{2} \int_0^T\| (ky_x)_x(t)\|^2_{L^2(\Omega)} \\
&\leq \| f\|^2_{L^2(Q)}+\| \sqrt{k}y_x(0))\|^2_{L^2(\Omega)} \\
&\quad + \| a\|^2_{\infty}e^{( T(T+2\| a\|_{L^{\infty}(Q)}))}(\| f\|^2_{L^2(Q)}
 +\| y_0\|^2_{L^2(\Omega)})\\
&\leq (1+\| a\|^2_{L^{\infty}(Q)}e^{( T(T+2\| a\|_{L^{\infty}(Q)}))})
(\| f\|^2_{L^2(Q)}+\| y_0\|^2_{H^1_k(\Omega)}),
\end{aligned}
\end{equation}
with $\| y_0\|^2_{H^1_k(\Omega)}=\| y_0\|^2_{L^2(\Omega)}
+\| \sqrt{k}y_x(0))\|^2_{L^2(\Omega)}$.
From  \eqref{syst2}, we have
\begin{equation}\label{yt}
y^2_t(t) = ((ky_x(t))_x-a(t)y(t)+f(t))^2\leq 3[((ky_x(t))_x)^2
+(a(t)y(t))^2+f^2(t)].
\end{equation}
Hence, by integrating on $Q$ and using the inequalities \eqref{l1} and \eqref{l3},
we obtain
\begin{equation}\label{l4}
\int_0^T\| y_t(t)\|^2_{L^2(\Omega)}\,{\rm d}t
\leq  C (1+\| a\|^2_{\infty}e^{( T(T+2\| a\|_{L^{\infty}(Q)}))})
(\| f\|^2_{L^2(Q)}+\| y_0\|^2_{H^1_k(\Omega)}).
\end{equation}
Now, \eqref{l1}, \eqref{l3} and \eqref{l4} give \eqref{l2'}, for all
$y_0\in H^2_k(\Omega)$.
\end{proof}

\section{Carleman inequality for degenerate systems}

The main result of this section is the following.

For $\omega=(a,b)$ let $\alpha=\frac{2a+b}{3}$, $\beta=\frac{a+2b}{3}$ and let
$\rho\in\mathcal{C}^2(\mathbb{R})$ be such that $0\leq\rho\leq 1$ and
\[
\rho(x)=\begin{cases}
 1 & \text{if }x\in (0,\alpha)\\
 0 & \text{if }x\in (\beta,1)
\end{cases}
\]
Let us define
\[
\theta(t)=\frac{1}{(t(T-t))^4}\quad\forall t\in (0,T), \quad
 \psi(x)=c_1(\int_0^{x}\frac{r}{k(r)}\,{\rm d}r-c_2)
\]
 with $c_1>0$ and $ c_2>\frac{1}{k(1)(2-\lambda)}$. Let
$\Psi(x)=e^{r\sigma(x)} - e^{2r\|\sigma\|_{\infty}}$
be  where $r>0$ and $\sigma$ a function which satisfies:
$\sigma\in\mathcal{C}^2([0,1])$, $\sigma>0$ in $\Omega$,
$\sigma=0$ on $\partial\Omega$ and
$\sigma_x(x)\neq 0$ in $[0,1]\setminus\omega_0$ where $\omega_0$ is an open
set of $\Omega$ such that $\omega_0\Subset \omega$.

Let us define
$\Phi(x,t)=\theta(t)[\rho(x)\psi(x)+(1-\rho(x))\Psi(x)]$.
The existence of the function $\sigma$ is proved in \cite{4}.
Consider the  adjoint system where $G\in L^2(Q)$:
\begin{equation}\label{syst3}
\begin{gathered}
 w_t+(k(x)w_x)_x-aw=G\quad\text{in }Q,\\
 w=0\quad\text{on }\Sigma, \\
 w(x,T)=w_T(x)\quad \text{in }\Omega .
\end{gathered}
\end{equation}
We have the following result.

\begin{theorem}\label{t4}
Under  hypothesis \eqref{k}, for all $T>0$ and $l\in\mathbb{N}$,
there exists two constants $C=C(\Omega,\omega,\| a\|_{\infty})>0$ and
$s_0>0$ such that for all $s\geq s_0$ and all solutions $w$ of
\eqref{syst3}, we have
\begin{equation}\label{30'}
\begin{aligned}
&\int_{Q}\Big(s^{l+1}\theta^{l+1} k(x)|w_x|^2+s^{l+3}
\theta^{l+3}\frac{x^2}{k(x)}|w|^2\Big)e^{2s\Phi}\,{\rm d}x\,{\rm d}t\\
&\leq C\Big(s^l\int_{Q}\theta^l e^{2s\Phi}|G|^2\,{\rm d}x
\,{\rm d}t+s^{l+3}\int_0^T\int_{\omega}\theta^{l+3}
e^{2s\Phi}|w|^2\,{\rm d}x\,{\rm d}t\Big).
\end{aligned}
\end{equation}
\end{theorem}

The proof of Theorem \ref{t4} will be given at the end of this section as
a consequence of the following result.
Consider the adjoint system
\begin{equation}\label{syst4}
\begin{gathered}
 w_t+(k(x)w_x)_x=G\quad\text{in }Q,\\
 w=0\quad\text{on }\Sigma,\\
 w(x,T)=w_T(x)\quad\text{in }\Omega .
\end{gathered}
\end{equation}
We have

\begin{theorem}\label{4}
Under  hypothesis \eqref{k}, for all $T>0$ and $l\in\mathbb{N}$,
there exists two constants $C=C(\Omega,\omega)>0$ and $s_0>0$ such that for all
 $s\geq s_0$ and all solutions  $w$ of \eqref{syst4}, we have
\begin{equation}\label{4'}
\begin{aligned}
&\int_{Q}\Big(s^{l+1}\theta^{l+1} k(x)|w_x|^2+s^{l+3}\theta^{l+3}
\frac{x^2}{k(x)}|w|^2\Big)e^{2s\Phi}\,{\rm d}x\,{\rm d}t\\
&\leq C(s^l\int_{Q}\theta^l e^{2s\Phi}|G|^2\,{\rm d}x\,{\rm d}t+s^{l+3}
\int_0^T\int_{\omega}\theta^{l+3} e^{2s\Phi}|w|^2\,{\rm d}x\,{\rm d}t).
\end{aligned}
\end{equation}
\end{theorem}

For the proof of Theorem \ref{4}
we follow the ideas of \cite{1} and \cite{8}. We prove first a
Carleman inequality for the degenerate part and combine it with a
classical Carleman inequality for the non degenerate part.

\begin{proposition}\label{pro3}
Under the hypothesis of  theorem \ref{t4} and for all $l\in\mathbb{N}$,
there exists two constants $C=C(\Omega,l)$ and $s_0$ such that for all $s\geq s_0$
and all solutions $w$ of  \eqref{syst4}, we have the  inequality
\begin{equation}\label{p21}
\begin{aligned}
&\int_{Q}\Big(s^{l-1}\theta^{l-1}(|(k(x)w_x)_x|^2+|w_t|^2)
+s^{l+1}\theta^{l+1} k(x)|w_x|^2\\
&+s^{l+3}\theta^{l+3}
\frac{x^2}{k(x)}|w|^2\Big)e^{2s\varphi}\,{\rm d}x\,{\rm d}t\\
&\leq C\Big(s^l\int_{Q}\theta^le^{2s\varphi}|G|^2\,{\rm d}x\,
{\rm d}t+k(1)s^{l+1} \int_0^T\theta^{l+1} e^{2s\varphi(1,t)}|w_x(1,t)|^2
\,{\rm d}t\Big),
\end{aligned}
\end{equation}
where $\varphi(x,t)=\theta(t)\psi(x)$.
\end{proposition}

\begin{proof}
Let $ y=(s\theta)^{1/2}e^{s\varphi}w$.
 We obtain from the first equation of \eqref{syst4}:
\begin{equation} \label{p22}
 P^+_sy+P^-_sy=(s\theta)^{1/2}e^{s\varphi}f,
\end{equation}
with
\begin{gather*}
 P^+_sy=(ky_x)_x-s\varphi_ty+s^2\varphi^2_xky\\
 P^-_sy=y_t-s(k\varphi_x)_xy- 2s\varphi_xky_x-\frac{l}{2}\theta_t\theta^{-1}y
\end{gather*}
The inner product in $L^2(Q)$ gives
\[
\| P^+_sy\|^2+\| P^-_sy\|^2+2\langle P^+_sy,P^-_sy\rangle
=\|(s\theta)^{1/2}e^{s\varphi}f\|^2.
\]
The following result is useful for the proof of proposition \ref{pro3}.

\begin{lemma}\label{lem3}
There exists two constants $m>0$ and $m'>0$ such that
\begin{equation}\label{p23}
\begin{aligned}
\|(s\theta)^{1/2}e^{s\varphi}f\|^2
&\geq  \| P^+_sy\|^2+\| P^-_sy\|^2+ ms\int_{Q}\theta k(x)|y_x|^2 \\
&\quad  + ms^3\int_{Q}\theta^3\frac{x^2}{k(x)}|y|^2\,{\rm d}x\,{\rm d}t
- m'k(1)s\int_0^T\theta|y_x(1,t)|^2\,{\rm d}t
\end{aligned}
\end{equation}
\end{lemma}

\begin{proof}
We have
\begin{align*}
&\langle P^+_sy,P^-_sy\rangle \\
& =  \frac{s}{2}\int_{Q}\psi(x)\theta_{tt}|y|^2\,{\rm d}x\,{\rm d}t
 +sc_1\int_{Q}\theta(2k(x)-xk'(x))|y_x|^2\,{\rm d}x\,{\rm d}t\\
&\quad + \frac{l}{2}\int_{Q}k(x)\theta_t\theta^{-1}|y_x|^2\,{\rm d}x\,{\rm d}t
-c^2_1s(1+s+\frac{l}{2})\int_{Q}\theta_t\theta\frac{x^2}{k(x)}|y|^2
\,{\rm d}x\,{\rm d}t\\
&\quad + \frac{sl}{2}\int_{Q}\theta^2_t\theta^{-1}\psi(x)|y|^2
\,{\rm d}x\,{\rm d}t
+ c^3_1s^3\int_{Q}\theta^3(\frac{x}{k(x)})^2(2k(x)-xk'(x))|y|^2
\,{\rm d}x\,{\rm d}t\\
&\quad - c_1sk(1)\int_0^T\theta|y_x(1,t)|^2\,{\rm d}t.
\end{align*}
Using the conditions on $k$, one obtains
\begin{align*}
&\langle P^+_sy,P^-_sy\rangle \\
& \geq \frac{s}{2}\int_{Q}\psi(x)\theta_{tt}|y|^2\,{\rm d}x\,{\rm d}t
+sc_1\int_{Q}\theta k(x)|y_x|^2\,{\rm d}x\,{\rm d}t
+\frac{l}{2}\int_{Q}k(x)\theta_t\theta^{-1}|y_x|^2\,{\rm d}x\,{\rm d}t\\
&\quad - c^2_1s(1+s+\frac{l}{2})\int_{Q}\theta_t\theta\frac{x^2}{k(x)}|y|^2
\,{\rm d}x\,{\rm d}t
+\frac{sl}{2}\int_{Q}\theta^2_t\theta^{-1}\psi(x)|y|^2\,{\rm d}x\,{\rm d}t\\
&\quad + c^3_1s^3\int_{Q}\theta^3\frac{x^2}{k(x)}|y|^2\,{\rm d}x\,{\rm d}t
- c_1sk(1)\int_0^T\theta|y_x(1,t)|^2\,{\rm d}t.
\end{align*}
Notice that there exists a constant $C>0$ such that
$\theta_t\theta^{-1}\leq C\theta$,
$\theta^2_t\theta^{-1}\leq T\theta^{3/2}$,
$\theta_t\theta\leq C\theta^3$ and
$\theta_{tt}\leq C\theta^{3/2}$.
Then, we obtain the inequality
\begin{equation}
\big|\frac{s}{2}(1+l)\int_{Q}\psi(x)(\theta_{tt}+\theta^2_t\theta^{-1})y^2\big|
\,{\rm d}x\,{\rm d}t\leq sC(l,T,c_1)\int_{Q}\theta^{3/2}|y|^2
\,{\rm d}x\,{\rm d}t.
\end{equation}
This gives by H\"oder's and Hardy's type inequalities (see \cite{1})
\begin{align*}
&\big|\frac{s}{2}(1+l)\int_{Q}\psi(x)(\theta_{tt}
+\theta^2_t\theta^{-1})y^2\big|\,{\rm d}x\,{\rm d}t\\
&\leq  s\delta_0C(l,T,c_1)\int_{Q}\theta k(x)|y_x|^2\,{\rm d}x\,{\rm d}t
+ \frac{C(l,T,c_1)}{\delta_0}s^3\int_{Q}\theta^3\frac{x^2}{k(x)}|y|^2
\,{\rm d}x\,{\rm d}t.
\end{align*}
We have
\[
\big|-c^2_1s^2\big(1+\frac{1}{s}+\frac{l}{2s}\big)
\int_{Q}\theta_t\theta\frac{x^2}{k(x)} y^2\big|\,{\rm d}x\,{\rm d}t
\leq C(l,T,c_1)s^3\int_{Q}\theta^3\frac{x^2}{k(x)}|y|^2\,{\rm d}x\,{\rm d}t.
\]
Next,
\begin{equation}
\frac{l}{2}\int_{Q}\theta_t\theta^{-1}k(x)|y_x|^2\,{\rm d}x\,{\rm d}t
\leq sC(l,T,c_1)\int_{Q}\theta k(x)|y_x|^2\,{\rm d}x\,{\rm d}t.
\end{equation}
Using the inequalities above, we infer that there exists a constant $M>0$
such that
\begin{align*}
&2\langle P^+_sy,P^-_sy\rangle \\
&\geq   M\int_{Q}(s\theta k(x)|y_x|^2+s^3\theta^3\frac{x^2}{k(x)}|y|^2)
\,{\rm d}x\,{\rm d}t- 2c_1sk(1)\int_0^T\theta|y_x(1,t)|^2\,{\rm d}t.
\end{align*}
As $\|(s\theta)^{1/2}e^{s\varphi}f\|^2\geq2\langle P^+_sy,P^-_sy\rangle$,
one deduces inequality \ref{p23}.
\end{proof}

\begin{proof}[Proof of Proposition \ref{pro3} (continued)]
From \eqref{p22}, we obtain
\begin{equation}\label{p24}
\int_{Q}\frac{|(k(x)y_x)_x|^2}{s\theta}\,{\rm d}x\,{\rm d}t
\leq C_0\int_{Q}(\frac{|P^+_sy|^2}{s\theta}+s\theta k(x)|y_x|^2
+s^3\theta^3\frac{x^2}{k(x)}|y|^2)\,{\rm d}x\,{\rm d}t,
\end{equation}
with $C_0>0$, and
\begin{equation}\label{p25}
\int_{Q}\frac{|y_t|^2}{s\theta}\,{\rm d}x\,{\rm d}t
\leq C_1\int_{Q}(\frac{|P^-_sy|^2}{s\theta}
+s\theta k(x)|y_x|^2+s^3\theta^3\frac{x^2}{k(x)}|y|^2)\,{\rm d}x\,{\rm d}t,
\end{equation}
with $C_1>0$.
Then, the lemma \ref{lem3} the inequalities \eqref{p24} and \eqref{p25} give
\begin{equation}\label{p26}
\begin{aligned}
&C\|(s\theta)^{l/2}e^{s\varphi}f\|^2 \\
&\geq  \int_{Q}\frac{|(k(x)y_x)_x|^2}{s\theta}\,{\rm d}x\,{\rm d}t
+ \int_{Q}\frac{|y_t|^2}{s\theta}\,{\rm d}x\,{\rm d}t
+m'sk(1)\int_0^T\theta|y_x(1,t)|^2\,{\rm d}t \\
&\quad -  m\int_{Q}(s\theta k(x)|y_x|^2
 + s^3\theta^3\frac{x^2}{k(x)}|y|^2)\,{\rm d}x\,{\rm d}t,
\end{aligned}
\end{equation}
where, $C=\max(C_0,C_1)$, $m>0$ and $m'=2c_1C$.

Recalling now that $w=(s\theta)^{-l/2}e^{-s\varphi}y$, one obtains
\[
(s\theta)^{l-1}e^{2s\varphi}|w_t|^2
\leq 2\frac{|y_t|^2}{s\theta}+\frac{2}{s\theta}
\Big(\frac{l}{2}\theta_t\theta^{-1}+s\varphi_t\Big)^2|y|^2
\]
 and $(s\theta)^{l+1}e^{2s\varphi}|w_x|^2\leq 2s\theta|y_x|^2
+2s^3\theta\varphi^2_x|y|^2$.
Then, from the H\"oder's and Hardy's inequalities and the previous inequalities,
we obtain
\begin{gather}\label{p27}
\begin{aligned}
&s^{l-1}\int_{Q}\theta^{l-1}e^{2s\varphi}|w_t|^2\,{\rm d}x\,{\rm d}t\\
&\leq  2\int_{Q}\frac{|y_t|^2}{s\theta}\,{\rm d}x\,{\rm d}t
+c_0\int_{Q}\Big(s\theta k(x)|y_x|^2 + s^3\theta^3\frac{x^2}{k(x)}|y|^2\Big)
\,{\rm d}x\,{\rm d}t,
\end{aligned} \\
\label{p28}
\begin{aligned}
&s^{l+1}\int_{Q}\theta^{l+1}k(x)e^{2s\varphi}|w_x|^2\,{\rm d}x\,{\rm d}t\\
&\leq  c_1\int_{Q}\Big(s\theta k(x)|y_x|^2+ s^3\theta^3\frac{x^2}{k(x)}|y|^2\Big)
\,{\rm d}x\,{\rm d}t, 
\end{aligned}\\
\label{p29}
\begin{aligned}
&s^{l-1}\int_{Q}\theta^{l-1}e^{2s\varphi}|(k(x)w_x)_x|^2\,{\rm d}x\,{\rm d}t\\
&\leq  2\int_{Q}\frac{|(k(x)y_x)_x|^2}{s\theta}\,{\rm d}x\,{\rm d}t
+ c_2 s^3\int_{Q}\theta^3\frac{x^2}{k(x)}|y|^2 \,{\rm d}x\,{\rm d}t \\
&\quad + c_2s\int_{Q}\theta k(x)|y_x|^2 \,{\rm d}x\,{\rm d}t.
\end{aligned}
\end{gather}
We have $(s\theta)^{l+1}e^{2s\varphi}|w_x(1,t)|^2=s\theta|y_x(1,t)|^2$ since
$y(1,t)=0$.
Combining the inequalities  \eqref{p26} to  \eqref{p29}, we infer that the
estimate \eqref{p21} of the proposition \ref{pro3} holds.
\end{proof}


Back to the proof of Theorem \ref{4}.
We need the following result that we omit the proof (see \cite{4} for instance).

\begin{proposition}\label{pro4}
We assume that $k\in\mathcal{C}^1([0,1])$ is a strictly positive function
and $k(0)\neq 0$. Let $l\in\mathbb{N}$. Then, there exists two positive
constants $C>0$ and $s_0>0$ such that for all $s\geq s_0$ and every  solution
$z$ of  \eqref{syst4}, we have
\begin{align*}
&\int_{Q}\big((s\theta)^{l-1}|(k(x)z_x)_x|^2+(s\theta)^{l-1}|z_t|^2
 +(s\theta)^{l+1} |z_x|^2+(s\theta)^{l+3}|z|^2 \big)e^{2s\eta}
 \,{\rm d}x\,{\rm d}t\\
&\leq  C\Big(s^l\int_{Q}\theta^l e^{2s\eta}|g|^2\,{\rm d}x\,{\rm d}t
+s^{l+3}\int_0^T\int_{\omega}\theta^{l+3} e^{2s\eta}|z|^2
\,{\rm d}x\,{\rm d}t\Big),
\end{align*}
where $\eta(x,t)=\theta(t)\Psi(x)$.
\end{proposition}

Combining  Propositions \ref{pro3} and \ref{pro4}, it suffices to verify
that Theorem \ref{t4} is true for  system \eqref{syst4}.
Thus, by using these propositions, we find the following inequalities:
\begin{equation}\label{p30}
\begin{aligned}
&\int_0^T\int_0^{\alpha}\Big(s^{l-1}\theta^{l-1}(|(k(x)w_x)_x|^2+|w_t|^2)
+s^{l+1}\theta^{l+1} k(x)|w_x|^2 \\
&+s^{l+3}\theta^{l+3} \frac{x^2}{k(x)}|w|^2\Big)e^{2s\Phi}\,{\rm d}x\,{\rm d}t\\
&\leq C\Big(s^l\int_{Q}\theta^l e^{2s\varphi}|f|^2+s^{l+3}
\int_0^T\int_{\alpha}^{\beta}\theta^{l+3}e^{2s\varphi}(|w_x|^2+|w|^2)\Big)
\,{\rm d}x\,{\rm d}t,
\end{aligned}
\end{equation}
and
\begin{equation}\label{p31}
\begin{aligned}
& \int_0^T\int_{\beta}^{1}\Big(s^{l-1}\theta^{l-1}(|(k(x)w_x)_x|^2
+|w_t|^2)+s^{l+1}\theta^{l+1} k(x)|w_x|^2 \\
&+s^{l+3}\theta^{l+3}
\frac{x^2}{k(x)}|w|^2\Big)e^{2s\Phi}\,{\rm d}x\,{\rm d}t\\
&\leq
C'\Big(s^l\int_{Q}\theta^l e^{2s\eta}|f|^2+s^{l+3}
\int_0^T\int_{\alpha}^{\beta}\theta^{l+3}e^{2s\eta}
(|w_x|^2+|w|^2)\Big)\,{\rm d}x\,{\rm d}t.
\end{aligned}
\end{equation}
We recall that for every $x\in(\alpha,\beta)$, $\varphi$, $\eta$
and $\Phi$ are equivalent. As $(\alpha,\beta)\Subset\omega$,
Caccioppoli's type inequality (see \cite{4,6}) leads us to the
 inequality
\begin{equation}\label{p32}
s^{l+1}\int_{Q}\theta^{l+1}( |w_x|^2+|w|^2)\,{\rm d}x\,{\rm d}t
\leq C''(s^l\int_{Q}\theta^l |f|^2+s^{l+3}
\int_{\omega}\theta^{l+3}|w|^2)\,{\rm d}x\,{\rm d}t.
\end{equation}
Thus, combining  \eqref{p29} and \eqref{p30}, and adding
to both sides of the inequality the term
\begin{align*}
&\int_0^T\int_{\alpha}^{\beta}\Big(s^{l-1}\theta^{l-1}(|(k(x)w_x)_x|^2+|w_t|^2)
+s^{l+1}\theta^{l+1} k(x)|w_x|^2\\
&+s^{l+3}\theta^{l+3}\frac{x^2}{k(x)}|w|^2\Big)
e^{2s\Phi}\,{\rm d}x\,{\rm d}t,
\end{align*}
and using \eqref{p31} leads from
any solution $w$ of \eqref{syst4} to
\begin{equation}\label{p33}
\begin{aligned}
&\int_{Q}(s^{l-1}\theta^{l-1}(|(k(x)w_x)_x|^2+|w_t|^2)\,{\rm d}x\,{\rm d}t
+s^{l+1}\theta^{l+1} k(x)|w_x|^2 \\
&+s^{l+3}\theta^{l+3}\frac{x^2}{k(x)}|w|^2)
 e^{2s\Phi}\,{\rm d}x\,{\rm d}t\\
&\leq  C\Big(s^l\int_{Q}\theta^le^{2s\Phi}|f|^2+s^{l+3}
\int_0^T\int_{\omega}\theta^{l+3}e^{2s\Phi}|w|^2\Big)\,{\rm d}x\,{\rm d}t.
\end{aligned}
\end{equation}
This completes the proof of Theorem \ref{4}.
\end{proof}

\begin{proof}[Proof of Theorem \ref{t4}]
Now, applying  inequality \eqref{p33} to  \eqref{syst3} with $\bar{f}=f+aw$
as right hand term, one obtains from the proof of Theorem \ref{4},
the inequality \eqref{30'} of Theorem \ref{t4}.
\end{proof}

\section{Null controllability of a linear system}

The aim of this section is to prove the null controllability for the
 linear system
\begin{equation}\label{syst6'}
\begin{gathered}
 y_t-(k(x)y_x)_x+ay=h1_{\mathcal{\omega}}\quad \text{in }Q,\\
 z_t-(k(x)z_x)_x+bz=y1_{\mathcal{O}} \quad \text{in }Q,\\
 y=z=0\quad \text{on }\Sigma,\\
 y(x,0)=y_0(x)\quad z(x,0)=z_0(x)\quad \text{in }\Omega,
\end{gathered}
\end{equation}
where $a,b\in L^{\infty}(Q)$, $y_0,z_0\in L^2(\Omega)$ and $h\in L^2(Q)$
is the control of the system.
The adjoint of system \eqref{syst6'} is
\begin{equation}\label{syst6}
\begin{gathered}
 -w_t-(k(x)w_x)_x+aw=v1_{\mathcal{O}}\quad \text{in }Q,\\
 -v_t-(k(x)v_x)_x+bv=0\quad \text{in }Q,\\
 w=v=0\quad \text{on }\Sigma,\\
 w(x,T)=0\;\;v(x,T)=v_T(x)\quad \text{in }\Omega .
\end{gathered}
\end{equation}
The main results of this section reads as follows.

\begin{theorem}\label{nul}
Under  hypothesis \eqref{k}, $a,b\in L^{\infty}(Q)$ and for $y_0,z_0\in L^2(\Omega)$,
there exists $h\in L^2(Q)$ such that the corresponding solution to
 system \eqref{syst6'} satisfies $z(.,T)=0$ a.e. in $\Omega$.
\end{theorem}

First, we  give some results useful for the proof of Theorem \ref{nul}.
For any $\varepsilon>0$, we set
\begin{equation}
J_{\varepsilon}(h)=\frac{1}{2}\int_{q}\theta(t)^{-7}e^{-2s\Phi}h^2(x,t)
\,{\rm d}x\,{\rm d}t+\frac{1}{2\varepsilon}\int_{\Omega}z^2(x,T)\,{\rm d}x,
\end{equation}
where $q=\omega\times(0,T)$.
We have the following results.

\begin{lemma}\label{lem4}
For any $\varepsilon>0$, there exists a control
$ h_{\varepsilon}\in L^2(q_{\varepsilon})$ such that
$J_{\varepsilon}(h_{\varepsilon})\leq J_{\varepsilon}(h)$, for all
$h\in L^2(q)$. One has $h_{\varepsilon}(x,t)=\theta(t)^7e^{2s\Phi}w(x,t)1_{\omega}$.
Moreover, we have the  inequality
\begin{equation}\label{eqlem4}
\int_{q}\theta(t)^{-7}e^{-2s\Phi}h^2_{\varepsilon}(x,t)
\,{\rm d}x\,{\rm d}t
\leq\int_{\Omega}|y_0(x)w(x,0)|\,{\rm d}x
+\int_{\Omega}|z_0(x)v(x,0)|\,{\rm d}x.
\end{equation}
\end{lemma}

\begin{proof}
 The continuity and the strict convexity of the functional $J_{\varepsilon}$
are obvious. Moreover,
\[
 J_{\varepsilon}(h)\geq\frac{1}{2}\int_{q}\theta(t)^{-7}e^{-2s\Phi}h^2(x,t)
\,{\rm d}x\,{\rm d}t.
\]
Let, $h\in L^2((0,T)\times\Omega)$ and consider $(h_n)$ the sequence 
 $h_n=h1_{]1/n,T-1/n[}$.
 Then, $h_n(\cdot,t)\to h(\cdot,t)$, a.e. on $(0,1)\times(0,T)$ and we have
$\| h_n\|^2_{L^2((0,1)\times(0,T))}\leq\| h\|^2_{L^2((0,1)\times(0,T))}$,
for all $n\in\mathbb{N}$.
From the Lebesgue dominated convergence theorem,
\[
\int_0^T\int_{\omega}h^2_n(x,t)\,{\rm d}x\,{\rm d}t\to
\int_0^T\int_{\omega}h^2(x,t)\,{\rm d}x\,{\rm d}t.
\]
Then, there exists $n_0\in\mathbb{N}$ such that $n\geq n_0$ implies
\[
\int_0^T\int_{\omega}h^2_n(x,t)\,{\rm d}x\,{\rm d}t
\geq \frac{1}{2}\| h\|^2_{L^2(\omega\times(0,T))},
\]
 i.e.
\[
\int_{1/n}^{T-1/n}\int_{\omega}h^2_n(x,t)\,{\rm d}x\,{\rm d}t
\geq \frac{1}{2}\| h\|^2_{L^2(\omega\times(0,T))},\quad \forall n\geq n_0.
\]
 As, $\theta^{-1}(t)e^{-2s\Phi}\geq \eta>0$ on $\omega\times]1/n,T-1/n[$,
we have
\[
\frac{1}{2}\| h\|^2_{L^2(\omega\times(0,T))} \leq
\frac{1}{\eta}\int_0^T\int_{\omega}\theta^{-1}(t)e^{-2s\Phi} h^2(x,t)\,{\rm d}x
\,{\rm d}t,\quad \text{for }n\geq n_0.
\]
Therefore,
\[
J_{\varepsilon}(h)\geq\int_0^T\int_{\omega}\theta^{-1}(t)e^{-2s\Phi}
h^2(x,t)\,{\rm d}x\,{\rm d}t\geq\frac{\eta}{2}\| h\|^2_{L^2(q)}.
\]
The functional $J_{\varepsilon}$ is therefore coercive.
 As a consequence, there exists $ h_{\varepsilon}\in L^2(q)$
such that $J_{\varepsilon}(h_{\varepsilon})\leq J_{\varepsilon}(h)$ for all
$h\in L^2(q)$.

Now, using Euler's condition on $J_{\varepsilon}$, we obtain
\begin{equation}\label{euler}
\int_{q}\theta(t)^{-7}e^{-2s\Phi}h_{\varepsilon}(x,t)h(x,t)
\,{\rm d}x\,{\rm d}t
=-\frac{1}{\varepsilon}\int_{\Omega}z_{\varepsilon}(x,T)z(x,T)\,{\rm d}x.
\end{equation}
On the other hand, we consider the following two systems
\begin{equation}\label{sys}
\begin{gathered}
 \overline{y_t}-(k(x)\overline{y}_x)_x+a\overline{y}
=h1_{\mathcal{\omega}} \quad\text{in }Q=\Omega\times(0,T),\\
 \overline{z}_t-(k(x)\overline{z}_x)_x+b\overline{z}
=\overline{y}1_{\mathcal{O}} \quad\text{ in }Q,\\
 \overline{y}=\overline{z}=0\quad \text{ on }\Sigma=\{ 0,1\}\times(0,T),\\
 \overline{y}(x,0)=0\quad \overline{z}(x,0)=0\quad\text{in }\Omega=(0,1),
\end{gathered}
\end{equation}
and
\begin{equation}\label{sys'}
\begin{gathered}
 -w_t-(k(x)w_x)_x+aw=v1_{\mathcal{O}}\quad \text{in }Q=\Omega\times(0,T),\\
 -v_t-(k(x)v_x)_x+bv=0\quad \text{in }Q,\\
 w=v=0\quad \text{on }\Sigma=\{ 0,1\}\times(0,T),\\
 w(x,T)=0\quad  v(x,T)=-\frac{1}{\varepsilon}z(x,T)\quad\text{in }\Omega=(0,1).
\end{gathered}
\end{equation}
Multiplying the two equations of  system \eqref{sys} by $w$ and $v$ respectively,
integrating by parts on $Q$ and using \eqref{euler}, we infer that
\begin{equation}\label{h}
h_{\varepsilon}(x,t)=\theta(t)^7e^{2s\Phi}w_{\varepsilon}(x,t)1_{\omega}.
\end{equation}
Multiplying the two equations of  \eqref{sys'} by $y_{\varepsilon}$
and $z_{\varepsilon}$ associate to the control $h_{\varepsilon}$ respectively and integrating on $Q$ and using \eqref{h}, we infer that
\begin{equation}\label{h1}
\begin{aligned}
&\int_{q}\theta(t)^{-7}e^{-2s\Phi}h^2_{\varepsilon}(x,t)
\,{\rm d}x\,{\rm d}t+\frac{1}{\varepsilon}
\int_{\Omega}z^2(x,T)\,{\rm d}x \\
&= \int_{q}\theta(t)^7e^{2s\Phi}w^2_{\varepsilon}(x,t)\,{\rm d}x\,{\rm d}t
 +\frac{1}{\varepsilon}\int_{\Omega}z^2(x,T)\,{\rm d}x,\\
&= -\int_{\Omega}y_0(x)w(x,0)\,{\rm d}x-\int_{\Omega}z_0(x)v(x,0)\,{\rm d}x
\end{aligned}
\end{equation}
As a consequence, we obtain the estimate of the lemma \ref{lem4}.
\end{proof}

\begin{proposition}[Observability inequality]\label{p}
For any $T>0$ and $s_0\geq s$, there exists a constant
$C=C(T,\| a\|_{\infty}, \| b\|_{\infty})$ such that
\begin{equation}\label{observation}
\| w(\cdot,0)\|^2_{L^2(\Omega)}+\| v(\cdot,0)\|^2_{L^2(\Omega)}
\leq C\int_{q}|w(x,t)|^2\,{\rm d}x\,{\rm d}t.
\end{equation}
Moreover,
\begin{equation}\label{p1}
\| h_{\varepsilon}\|_{L^2(\Omega)}
\leq C(\| y_0\|_{L^2(\Omega)}+\| z_0\|_{L^2(\Omega)}).
\end{equation}
\end{proposition}

\begin{proof}
Multiplying the second equation of  system \eqref{sys'} by $v_t$ and
integrating on $(0,1)$, we obtain
\begin{equation}\label{I}
\begin{aligned}
&\int_{\Omega}v^2_t(x,t)\,{\rm d}x
 - \frac{d}{2dt}\int_{\Omega}k(x)v^2_x(x,t)\,{\rm d}x\\
&=\int_{\Omega}bv_t(x,t)v(x,t)\,{\rm d}x\\
&\leq  \frac{\| b\|^2_{\infty}}{2}\int_{\Omega}v^2(x,t)\,{\rm d}x
+\frac{1}{2}\int_{\Omega}v^2_t(x,t)\,{\rm d}x\quad\forall t\in [0,T].
\end{aligned}
\end{equation}
The function $x\mapsto \frac{k(x)}{x^2}$ is non increasing on
$(0,1]$, we obtain by using Hardy's inequality
\begin{equation}\label{I1}
\int_{\Omega}v^2(x,t)\,{\rm d}x
\leq\frac{1}{k(1)}\int_{\Omega}k(x)(\frac{v^2(x,t)}{x})^2\,{\rm d}x
\leq\frac{C}{k(1)}\int_{\Omega}k(x)v^2_x(x,t)\,{\rm d}x,
\end{equation}
with $C>0$.
Then, combining  inequalities \eqref{I} and \eqref{I1}, we obtain
\begin{equation}
\frac{d}{dt}\int_{\Omega}k(x)v^2_x(x,t)\,{\rm d}x
+\frac{C\| b\|^2_{\infty}}{k(1)}\int_{\Omega}k(x)v^2_x(x,t)\,{\rm d}x\geq 0 .
\end{equation}
This implies
\begin{equation}\label{I2}
\frac{d}{dt}\Big(e^{\frac{C\| b\|^2_{\infty}t}{k(1)}}
\int_{\Omega}k(x)v^2_x(x,t)\,{\rm d}x\Big)
\geq 0\quad\forall t\in [0,T].
\end{equation}
Thus, the function
\[
 t\mapsto e^{\frac{C\| b\|^2_{\infty}t}{k(1)}}
\int_{\Omega}k(x)v^2_x(x,t)\,{\rm d}x
\]
 is increasing on $[0,T]$.
We have
\begin{equation}
\frac{T}{4}\int_{\Omega}k(x)v^2_x(x,t)\,{\rm d}x 
\leq  C\int_{T/2}^{3T/4}\int_{\Omega}k(x)v^2_x(x,\tau)\,{\rm d}x\,{\rm d}\tau,
\quad\forall t\in [0,T/2].
\end{equation}
Then
\[
\int_{\Omega}k(x)v^2_x(x,t)\,{\rm d}x
\leq  C(T,\| b\|_{\infty})\int_{T/2}^{3T/4}\int_{\Omega}k(x)v^2_x(x,\tau)
\,{\rm d}x\,{\rm d}\tau.
\]
Taking $s\geq C(\| b\|_{\infty})T^8$ and $1\leq s\theta$, the latter inequality
and  \eqref{p20}  give
\begin{equation}\label{I3}
\int_{\Omega}v^2(x,t)\,{\rm d}x
\leq  Cs^7\int_0^T\int_{\omega'}\theta(\tau)^7e^{2s\Phi}|w(x,\tau)|^2
\,{\rm d}x\,{\rm d}\tau,\quad\forall t\in [0,T/2].
\end{equation}
Now, multiplying the first equation of  \eqref{sys'} by $w_t$ and integrating
on $(0,1)$, we obtain
\begin{equation}\label{I4}
\begin{aligned}
&\int_{\Omega}w^2_t(x,t)\,{\rm d}x
 - \frac{d}{2dt}\int_{\Omega}k(x)w^2_x(x,t)\,{\rm d}x\\
& =\int_{\Omega}w_t(x,t)( aw(x,t)-v(x,t))\,{\rm d}x\\
&\leq  \| a\|^2_{\infty}\int_{\Omega}w^2(x,t)\,{\rm d}x
+\frac{1}{2}\int_{\Omega}w^2_t(x,t)\,{\rm d}x
+ \int_{\Omega}v^2(x,t)\,{\rm d}x\quad\forall t\in [0,T].
\end{aligned}
\end{equation}
Using again that the function $x\mapsto \frac{k(x)}{x^2}$  decreases  on $(0,1]$,
the inequality \eqref{I3} and Hardy's inequality, we obtain that for all
$t\in [0,T/2]$
\begin{equation}\label{I5}
\begin{aligned}
&\int_{\Omega}w^2_t(x,t)\,{\rm d}x
 - \frac{d}{dt}\int_{\Omega}k(x)w^2_x(x,t)\,{\rm d}x \\
&\leq  Cs^7\int_0^T\int_{\omega'}\theta(\tau)^7e^{2s\Phi}|w(x,\tau)|^2
 \,{\rm d}x\,{\rm d}\tau
 + \frac{C\| a\|^2_{\infty}}{k(1)}\int_{\Omega}k(x)w^2_x(x,t)\,{\rm d}x.
\end{aligned}
\end{equation}
Hence,
\[
-\frac{d}{dt}\Big( e^{\frac{C\| a\|^2_{\infty}t}{k(1)}}
\int_{\Omega}k(x)w^2_x(x,t)\,{\rm d}x \Big)
\leq Ce^{\frac{C\| a\|^2_{\infty}t}{k(1)}}s^7
\int_0^T\int_{\omega'}\theta(\tau)^7e^{2s\Phi}|w(x,\tau)|^2
\,{\rm d}x\,{\rm d}\tau,
\]
for all $t\in [0,T/2]$.
Thus, for every $0\leq s\leq t\leq T/2$, we obtain
\begin{equation}\label{I6}
\begin{aligned}
&\int_{\Omega}k(x)w^2_x(x,s)\,{\rm d}x  \\
&\leq C\int_{\Omega}k(x)w^2_x(x,t)\,{\rm d}x 
 + Cs^7\int_0^T\int_{\omega'}\theta(\tau)^7e^{2s\Phi}|w(x,\tau)|^2
\,{\rm d}x\,{\rm d}\tau,
\end{aligned}
\end{equation}
for all $t\in [0,T/2]$.
Integrating on $[T/4,T/2]$ with respect to the variable $t$, we
obtain for all $s\leq T/4$
\begin{equation}
\begin{aligned}
&\frac{T}{4}\int_{\Omega}k(x)w^2_x(x,s)\,{\rm d}x \\
&\leq C\int_{T/4}^{T/2}\int_{\Omega}k(x)w^2_x(x,t)\,{\rm d}x\,{\rm d}t
+ Cs^7\int_0^T\int_{\omega'}\theta(\tau)^7e^{2s\Phi}|w(x,\tau)|^2
\,{\rm d}x\,{\rm d}\tau,
\end{aligned}
\end{equation}
for all $t\in [0,T/2]$.
Taking $1\leq (s\theta)^4$ for all $ s\geq CT^8$,
 we find by using the inequality \eqref{p20} that
\begin{equation}\label{I7}
\int_{\Omega}w^2(x,s)\,{\rm d}x
\leq  Cs^7\int_0^T\int_{\omega'}\theta(t)^7e^{2s\Phi}|w(x,t)|^2\,{\rm d}x\,{\rm d}t.
\end{equation}
Combining this result with the inequality \eqref{I3} for $s=t=0$ we can
conclude that
\begin{equation}\label{I8}
\int_{\Omega}(w^2(x,0)+ v^2(x,0))\,{\rm d}x
\leq  Cs^7\int_0^T\int_{\omega'}\theta(t)^7e^{2s\Phi}|w(x,t)|^2\,{\rm d}x\,{\rm d}t.
\end{equation}
\end{proof}

\begin{lemma}\label{lem5}
Let $g\in\mathcal{C}(\overline{\Omega})$ be such that
$-g(x)\geq m>0$ for all $x\in\overline{\Omega}$.
Let $\Phi(x,t)=\theta(t)g(x)$, for $(x,t)\in Q$ and
$ m_0=\min_{\overline{\Omega}}(-g)$. Then
\begin{equation}\label{I8'}
s^7\theta(t)^7e^{2s\Phi}\leq (\frac{7}{em_0})^7\quad
\text{for any }s>\frac{7T^8}{2^9m_0}\text{ and }(x,t)\in Q.
\end{equation}
\end{lemma}

\begin{proof}
We have $g(x)=\rho(x)\psi(x)+(1-\rho(x))\Psi(x)$, and for all
$x\in(0,1)$, $-g(x)>0$ and $g$ is a continuous function on $[0,1]$.
Thus, for every $x\in(0,1)$, there exists a constant $m>0$, such that
$-g(x)\geq m>0$. We have $\theta(t)\geq (\frac{4}{T^2})^4$.
Then $-g(x)\theta(t)\geq m_0(\frac{4}{T^2})^4$, for all $(x,t)\in Q$.
For $s>\frac{7T^8}{2^9m_0}$, we have $2s\Phi(x,t)\leq -7$ and we obtain
\begin{equation}\label{s}
e^{2s\Phi(x,t)}\leq e^{-7},\quad \text{for all }(x,t)\in Q.
\end{equation}
Now, if we suppose that there exists a constant $A>0$ such that
 $(s\theta(t))^7\leq A^7$ for all $t\in (0,T)$ then, we obtain
$s\leq A(\frac{T^2}{4})^4$ and as, $s>\frac{7T^8}{2^9m_0}$,
we can choose $A=7/m_0$. With this choice of $A$,  we obtain
\begin{equation}\label{s1}
(s\theta(t))^7\leq (\frac{7}{m_0})^7,\quad \text{for any } s>\frac{7T^8}{2^9m_0}.
\end{equation}
Using the inequalities \eqref{s} and \eqref{s1}, we obtain
\begin{equation}
(s\theta(t))^7e^{2s\Phi(x,t)}\leq (\frac{7}{em_0})^7, \quad
\text{for any } s>\frac{7T^8}{2^9m_0}\text{ and }(x,t)\in Q.
\end{equation}
This completes the proof of the lemma \ref{lem5}.
\end{proof}

\begin{proof}[Proof of Proposition \ref{p} (continued)]
From \eqref{I8} and lemma \ref{lem5}, we obtain
\begin{equation}\label{s2}
\int_{\Omega}(w^2(x,0)+ v^2(x,0))\,{\rm d}x \leq  C\big(\frac{7}{em_0}\big)^7
\int_0^T\int_{\omega'}|w(x,t)|^2\,{\rm d}x\,{\rm d}t.
\end{equation}
Now,  from inequalities \eqref{eqlem4}, \eqref{h}, \eqref{s2} and Cauchy-Schwarz's
inequality one obtains
\begin{align*}
&\int_{q}s^{-7}\theta(t)^{-7}e^{-2s\Phi}h^2_{\varepsilon}(x,t)
 \,{\rm d}x\,{\rm d}t \\
&\leq \| y_0\|_{L^2(\Omega)}\| w(\cdot,0)\|_{L^2(\Omega)}
 +\| z_0\|_{L^2(\Omega)}\| v(\cdot,0)\|_{L^2(\Omega)}\\
& \leq \frac{C}{2}(\| y_0\|_{L^2(\Omega)}+\| z_0\|_{L^2(\Omega)})^2
 +  \frac{1}{2}s^7\int_0^T\int_{\omega'}\theta(t)^7e^{2s\Phi}|w(x,t)|^2
 \,{\rm d}x\,{\rm d}t.
\end{align*}
This gives
\begin{equation}\label{I9}
\int_{q}s^{-7}\theta(t)^{-7}e^{-2s\Phi}|h_{\varepsilon}(x,t)|^2\,{\rm d}x\,{\rm d}t
 \leq
2C(\| y_0\|_{L^2(\Omega)}+\| z_0\|_{L^2(\Omega)})^2.
\end{equation}
We recall that
$w_{\varepsilon}(x,t)=s^{-7}\theta(t)^{-7}e^{-2s\Phi}h_{\varepsilon}(x,t)$.
Thus, using \eqref{p20}, \eqref{yt} and \eqref{I9}, we obtain
\begin{equation}
\| h_{\varepsilon}\|^2_{L^2(q)}
\leq 2C\big(\frac{7}{em_0}\big)^7(\| y_0\|_{L^2(\Omega)}+\| z_0\|_{L^2(\Omega)})^2.
\end{equation}
This completes the proof of the proposition \ref{p}.
\end{proof}

\begin{proof}[Proof of Theorem \ref{nul}]
Notice that the solution $(y,z)$ of  \eqref{syst6'}, can be decomposed
as follows
\[
 y=\overline{y}+y^o,\quad
 z=\overline{z}+y^o,
\]
where $(\overline{y},\overline{z})$, and $(y^o,z^o)$ are  the solutions of
the systems
\begin{equation}\label{sy}
\begin{gathered}
 \overline{y}_t-(k(x)\overline{y}_x)_x+a\overline{y}=h1_{\mathcal{\omega}}
\quad\text{in }Q=\Omega\times(0,T),\\
 \overline{z}_t-(k(x)\overline{z}_x)_x+b\overline{z}
=\overline{y}1_{\mathcal{O}} \quad\text{in }Q,\\
 \overline{y}=\overline{z}=0 \quad\text{on }\Sigma=\{ 0,1\}\times(0,T),\\
 \overline{y}(x,0)=0\quad \overline{z}(x,0)=0,\quad
\text{in }\Omega=(0,1)
\end{gathered}
\end{equation}
and
\begin{equation}\label{sy1}
\begin{gathered}
 y^o_t-(k(x)y^o_x)_x+ay^o=0\quad \text{in }Q=\Omega\times(0,T),\\
 z^o_t-(k(x)z^o_x)_x+bz^o=y^o1_{\mathcal{O}}\quad\text{in }Q,\\
 y^o=z^o=0\quad \text{on }\Sigma=\{ 0,1\}\times(0,T),\\
 y^o(x,0)=y_0(x)\quad z^o(x,0)=z_0(x) \quad\text{in }\Omega=(0,1).
\end{gathered}
\end{equation}
Let us define
$L: L^2(Q)\to L^2(\Omega)\times L^2(\Omega)$ by
$L(h)=(\overline{y}(x,T),\overline{z}(x,T))$
where $(\overline{y},\overline{z})$ is the solution corresponding
to \eqref{sy}. Also let
$M: L^2(\Omega)\times L^2(\Omega) \to L^2(\Omega)\times L^2(\Omega)$ be defined by
\[
M(y_0,z_0)=(y^o(x,T),z^o(x,T)),
\]
where $(y^o(x,T),z^o(x,T))$ is the corresponding solution to \eqref{sy1}.
Thus, Theorem \ref{nul} is equivalent to the inclusion
\begin{equation}\label{r}
R(M )\subset R(L).
\end{equation}
Both $M$ and $L$ are $L^2(\Omega)\times L^2(\Omega)$-valued, bounded
linear operators. Consequently \eqref{r} holds if and only if,
for every $(w_T,v_T)\in L^2(\Omega)\times L^2(\Omega)$ there exists a constant
$C>0$ such that
\begin{equation}\label{ri}
\| M^*(w_T,v_T)\|_{(L^2(\Omega))^2}\leq C\| L^*(w_T,v_T)\|_{L^2(Q)}.
\end{equation}

Now, by multiplying the two equations of  \eqref{sy} by $w$
and $v$ and integrating respectively, where $(w,v)$ solves the adjoint
system \eqref{syst6} and using the fact that these systems are duals, we obtain
\begin{equation}\label{r1}
L^*(w_T,v_T)= w1_{\omega}.
\end{equation}
On the other hand, multiply the two equations of  system \eqref{sy1} by $w$ and $v$
 where $(w,v)$ solves the adjoint system \eqref{syst6} and integrate
respectively on $Q$, and by using the fact that these systems are duals, we obtain
\begin{equation}\label{r2}
M^*(w_T,v_T)= (w(x,0),(v(x,0)).
\end{equation}
Hence, Theorem \ref{nul} is proved, since \eqref{r} is just
\eqref{observation}.
\end{proof}


\section{Proof of main results}

\begin{proof}[Proof of Theorem \ref{t3}]
Let $\mathcal{O}'\Subset\omega\cap\mathcal{O}$.
 Applying Theorem \ref{t4} and taking $l=3$ for the first equation of
\eqref{syst6}  and $l=0$ for the second equation of the same system;
 we find the following inequalities: 
\begin{equation}\label{d}
\begin{aligned}
&\int_{Q}(s^4\theta^4 k(x)|w_x|^2+s^6\theta^6\frac{x^2}{k(x)}|w|^2)e^{2s\Phi}
\,{\rm d}x\,{\rm d}t\\
&\leq  C(s^3\int_{Q}\theta^3 e^{2s\Phi}|v1_{\mathcal{O}}|^2\,{\rm d}x
\,{\rm d}t+s^6\int_0^T\int_{\mathcal{O}'}\theta^6 e^{2s\Phi}|w|^2
\,{\rm d}x\,{\rm d}t),
\end{aligned}
\end{equation}
for all $s\geq s_1$; and
\begin{equation}\label{d0}
\begin{aligned}
&\int_{Q}(s\theta k(x)|v_x|^2+s^3\theta^3\frac{x^2}{k(x)}|v|^2)
e^{2s\Phi}\,{\rm d}x\,{\rm d}t\\
&\leq  Cs^3\int_0^T\int_{\mathcal{O}'}\theta^3 e^{2s\Phi}|v|^2
\,{\rm d}x\,{\rm d}t,\quad\text{ for all }s\geq s_2.
\end{aligned}
\end{equation}
Adding these two inequalities, one obtains that for every $s\geq s_1+s_2$,
\begin{equation}\label{d1'}
\begin{aligned}
&\int_{Q}\Big(s\theta k(x)|v_x|^2+s^3\theta^3\frac{x^2}{k(x)}|v|^2
+s^4\theta^4 k(x)|w_x|^2+s^6\theta^6\frac{x^2}{k(x)}|w|^2\Big)
e^{2s\Phi}\,{\rm d}x\,{\rm d}t\\
&\leq Cs^3\int_0^T\int_{\mathcal{O}}\theta^3 e^{2s\Phi}|v|^2\,{\rm d}x
\,{\rm d}t+Cs^3\int_0^T\int_{\mathcal{O}'}\theta^3 e^{2s\Phi}|v|^2
\,{\rm d}x\,{\rm d}t\\
&\quad + Cs^6\int_0^T\int_{\mathcal{O}'}\theta^6 e^{2s\Phi}|w|^2
\,{\rm d}x\,{\rm d}t.
\end{aligned}
\end{equation}
We will absorb the term
$\int_0^T\int_{\mathcal{O}'}\theta^3 e^{2s\Phi}|v|^2\,{\rm d}x\,{\rm d}t $
 by the term $\int_0^T\int_{\mathcal{O}}\theta^3 e^{2s\Phi}|v|^2\,{\rm d}x\,{\rm d}t$
by using the properties of the integral.
We know that $\mathcal{O}'\Subset\mathcal{O}$ and therefore,
there exists a constant $C_1>0$ such that
\[
\int_0^T\int_{\mathcal{O}'}\theta^3 e^{2s\Phi}|v|^2\,{\rm d}x\,{\rm d}t
\leq C_1\int_0^T\int_{\mathcal{O}}\theta^3 e^{2s\Phi}|v|^2\,{\rm d}x\,{\rm d}t.
\]
Thus, the inequality \eqref{d1'} gives
\begin{equation}\label{d1}
\begin{aligned}
&\int_{Q}\Big(s\theta k(x)|v_x|^2+s^3\theta^3\frac{x^2}{k(x)}|v|^2
 +s^4\theta^4 k(x)|w_x|^2+s^6\theta^6\frac{x^2}{k(x)}|w|^2)e^{2s\Phi}
 \,{\rm d}x\,{\rm d}t\\
&\leq Cs^3\int_0^T\int_{\mathcal{O}}\theta^3 e^{2s\Phi}|v|^2\,{\rm d}x\,{\rm d}t+
Cs^6\int_0^T\int_{\mathcal{O}'}\theta^6 e^{2s\Phi}|w|^2\,{\rm d}x\,{\rm d}t.
\end{aligned}
\end{equation}
Now, let $\mathcal{O}'\Subset \omega'\Subset\omega\cap\mathcal{O}$,
we define and the function $\zeta\in\mathcal{C}^{\infty}(\Omega)$ as follows
\begin{equation}\label{d2}
\begin{gathered}
 0\leq\zeta(x)\leq 1\quad \forall x\in\Omega\\
 \zeta(x)=1\quad\text{if } x\in\mathcal{O}'\\
 \zeta(x)=0\quad\text{if } x\in\Omega\setminus\omega'
\end{gathered}
\end{equation}
In addition it is assumed that
\begin{equation}\label{d3}
\frac{\nabla\zeta}{\zeta^{1/2}}\in L^{\infty}(\Omega)\quad \text{and}\quad
\frac{\Delta\zeta}{\zeta^{1/2}}\in L^{\infty}(\Omega).
\end{equation}
Indeed, just take $\zeta=\zeta^4_0$ where $\zeta_0\in\mathcal{C}^{\infty}(\Omega)$,
to satisfy the two previous conditions on $\zeta$.

We set $\chi=s^3\theta^3 e^{2s\Phi}$ and multiply the first equation of
 \eqref{syst6} by $\zeta\chi v$ and integrate the result on $Q$.
We obtain
\begin{gather*}
\begin{aligned}
\int_{Q}v^21_{\mathcal{O}}\zeta\chi
&= 2\int_{Q}k(x)v_xw_x\zeta\chi-\int_{Q}vw(k(x)(\zeta\chi)_x)_x \\
&\quad +\int_{Q}(a+b)vw\zeta\chi+\int_{Q}vw\zeta\chi_t,
\end{aligned}\\
\int_{Q}v^21_{\mathcal{O}}\zeta\chi = \sum_{i=1}^{4}I_i.
\end{gather*}
Note that there exists a positive constant $C_2>0$ such that
\begin{equation}\label{d4}
|\chi_t|\leq C_2s^4\theta^5e^{2s\Phi}\quad \text{and}\quad
|\chi_x|\leq C_2s^4\theta^4\frac{x}{k(x)} e^{2s\Phi}
\end{equation}
and, using  H\"oder's and Young's inequalities, one obtains
\begin{align*}
I_1&= 2s^3\int_{Q}k(x)\theta v_xw_xe^{2s\Phi}\zeta \\
&\leq 2\delta_0  s\int_{Q}k(x)\theta |v_x|^2e^{2s\Phi}\zeta
 +\frac{2}{\delta_0}s^5\int_{Q}k(x)\theta^5 |w_x|^2e^{2s\Phi}\zeta,\quad
\delta_0>0.
\end{align*}
From  inequality \eqref{d0}, one obtains 
\begin{equation}
2\delta_0s\int_{Q}k(x)\theta |v_x|^2e^{2s\Phi}\zeta
\leq C\delta_0 s^3\int_0^T\int_{\mathcal{O}}\theta^3 |v|^2e^{2s\Phi}\zeta.
\end{equation}
It results that
\begin{gather}\label{d5}
I_1\leq C\delta_0 s^3\int_0^T\int_{\mathcal{O}}\theta^3 |v|^2e^{2s\Phi}\zeta
+\frac{2}{\delta_0}s^5\int_{Q}k(x)\theta^5 |w_x|^2e^{2s\Phi}\zeta, \\
I_3= \int_{Q}(a+b)vw\zeta\chi
\leq \delta_1s^3\int_{Q}\theta^3 |v|^2e^{2s\Phi}\zeta
 +\frac{\| a+b\|^2_{\infty}}{\delta_1}s^3\int_{Q}\theta^3 |w|^2e^{2s\Phi}\zeta,
\end{gather}
with $\delta_1>0$.
As the function $x\mapsto \frac{x^2}{k(x)}$ is increasing on $(0,1]$, we have
\[
\frac{\| a+b\|^2_{\infty}}{\delta_1}s^3\int_{Q}\theta^3 |w|^2e^{2s\Phi}\zeta
\leq \frac{\| a+b\|^2_{\infty}}{k(1)\delta_1}s^3
\int_0^T\theta^3[\int_0^{1}k(x)\zeta(\frac{we^{s\Phi}}{x})^2\,{\rm d}x]\,{\rm d}t.
\]
Using  Hardy's type inequality  to the function $we^{s\Phi}$
with Fubini-Tonelli's theorem, we obtain
\begin{align*}
&\frac{\| a+b\|^2_{\infty}}{\delta_1}s^3\int_{Q}\theta^3 |w|^2e^{2s\Phi}\zeta\\
&\leq \frac{C_4\| a+b\|^2_{\infty}}{k(1)\delta_1}s^3
 \int_{Q}\theta^3k(x)\zeta(s^2\Phi^2_x|w|^2+|w_x|^2)e^{2s\Phi})\,{\rm d}x\,{\rm d}t.
\end{align*}
Using that $\Phi_x=\frac{c_1\theta x}{k(x)}$, we have
\begin{align*}
&\frac{\| a+b\|^2_{\infty}}{\delta_1}s^3\int_{Q}\theta^3 |w|^2e^{2s\Phi}\zeta\\
&\leq \frac{C_5\| a+b\|^2_{\infty}}{k(1)\delta_1}
 \int_{Q}( s^5\theta^5 \frac{x^2}{k(x)}|w|^2+s^3\theta^3k(x)|w_x|^2)
 \zeta e^{2s\Phi}\,{\rm d}x\,{\rm d}t.
\end{align*}
From  inequality \eqref{d1}, we find
\begin{align*}
&\frac{\| a+b\|^2_{\infty}}{\delta_1}s^3\int_{Q}\theta^3 |w|^2e^{2s\Phi}\zeta \\
&\leq \frac{C_5C\| a+b\|^2_{\infty}}{k(1)\delta_1}
s^3\int_0^T\int_{\mathcal{O}}\theta^3 e^{2s\Phi}|v|^2\,{\rm d}x\,{\rm d}t \\
&\quad + \frac{C_5C\| a+b\|^2_{\infty}}{k(1)\delta_1}s^6
\int_0^T\int_{\mathcal{O}'}\theta^6 e^{2s\Phi}|w|^2\,{\rm d}x\,{\rm d}t.
\end{align*}
It results that
\begin{gather}\label{d5b}
\begin{aligned}
I_3 &\leq \delta_1s^3\int_{Q}\theta^3 |v|^2e^{2s\Phi}\zeta
 + \frac{C_5C\| a+b\|^2_{\infty}}{k(1)\delta_1}
s^3\int_0^T\int_{\mathcal{O}}\theta^3 e^{2s\Phi}|v|^2\zeta\,{\rm d}x\,{\rm d}t\\
&\quad + \frac{C_5C\| a+b\|^2_{\infty}}{k(1)\delta_1}s^6
\int_0^T\int_{\mathcal{O}'}\theta^6 e^{2s\Phi}|w|^2\zeta\,{\rm d}x\,{\rm d}t,
\end{aligned} \\
\begin{aligned}
I_4&=\int_{Q}vw\zeta\chi_t\,{\rm d}x\,{\rm d}t\leq \delta_2s^3
\int_{Q}\theta^3 |v|^2e^{2s\Phi}\zeta\,{\rm d}x\,{\rm d}t \\
&\quad +\frac{C^2_1}{\delta_2}s^5\int_{Q}k(x)\theta^7 |w|^2e^{2s\Phi}\zeta
\,{\rm d}x\,{\rm d}t,\quad \delta_2>0
\end{aligned} \\
\begin{aligned}
I_2&= -\int_{Q}vw(k(x)(\zeta\chi)_x)_x\,{\rm d}x\,{\rm d}t \\
& \leq\int_{Q}|k'(x)| |vw||(\chi\zeta)_x|\,{\rm d}x\,{\rm d}t
+\int_{Q}k(x)|vw||(\chi\zeta)_{xx}|\,{\rm d}x\,{\rm d}t.
\end{aligned}
\end{gather}
Notice that
\begin{equation}\label{wa}
\begin{gathered}
 k\in\mathcal{C}([0,1])\cap\mathcal{C}^1((0,1]),\\
 (\zeta\chi)_x=\zeta_x\chi+\zeta\chi_x\leq C_2s^4\theta^4e^{2s\Phi}\zeta^{1/2},\\
 (\chi\zeta)_{xx}=\zeta_{xx}\chi+2\zeta_x\chi_x+\zeta\chi_{xx}
\leq C_2(s\theta)^5e^{2s\Phi}\zeta^{1/2}.
\end{gathered}
\end{equation}
Using  inequality \eqref{wa}, one gets
\begin{align*}
I^1_2 &= \int_{Q}|k'(x)| |vw||(\chi\zeta)_x|\,{\rm d}x\,{\rm d}t\\
&\leq C_3s^4\int_{Q}\theta^4e^{2s\Phi}|wv|\zeta^{1/2}\,{\rm d}x\,{\rm d}t.\\
&\leq \delta_3s^3\int_{Q}\theta^3 e^{2s\Phi}|v|^2\zeta^{1/2}\,{\rm d}x\,{\rm d}t
+\frac{C^2_3}{\delta_3}s^5\int_{Q}\theta^5 |w|^2e^{2s\Phi}\zeta^{1/2}
\,{\rm d}x\,{\rm d}t, \quad \delta_3>0
\end{align*}
and 
\begin{align*}
I^2_2 &= \int_{Q}|k(x)| |vw||(\chi\zeta)_{xx}|\,{\rm d}x\,{\rm d}t\\
&\leq  C_3s^5\int_{Q}\theta^5 e^{2s\Phi}|wv|\zeta^{1/2}\,{\rm d}x\,{\rm d}t.\\
&\leq  \delta_4s^3\int_{Q}\theta^3 e^{2s\Phi}|v|^2\zeta^{1/2}\,{\rm d}x\,{\rm d}t
+\frac{C^2_3}{\delta_4}s^7\int_{Q}\theta^7 |w|^2e^{2s\Phi}\zeta^{1/2}\,{\rm d}x
\,{\rm d}t,\quad \delta_4>0.
\end{align*}
Therefore,
\begin{equation}
\begin{aligned}
I_2 &\leq (\delta_3+\delta_4 ) s^3\int_{Q}\theta^3 e^{2s\Phi}|v|^2\zeta^{1/2}
 \,{\rm d}x\,{\rm d}t \\
&\quad +\frac{C^2_3}{\delta_3}s^5  \int_{Q}\theta^5 |w|^2e^{2s\Phi}
 \zeta^{1/2}\,{\rm d}x\,{\rm d}t
+\frac{C^2_3}{\delta_4}s^7\int_{Q}\theta^7 |w|^2e^{2s\Phi}\zeta^{1/2}
 \,{\rm d}x\,{\rm d}t.
\end{aligned}
\end{equation}
By summing the $I_i$, for $1\leq i\leq 4$, we obtain 
\begin{align}
\sum_{i=1}^{4}I_i
&\leq (C\delta_0+\delta_1+\delta_2+\delta_3+\delta_4
+\frac{C_5C\| a+b\|^2_{\infty}}{k(1)\delta_1}) s^3
\int_{Q}\theta^3 |v|^2e^{2s\Phi}\zeta\,{\rm d}x\,{\rm d}t \nonumber \\
&\quad +\frac{2}{\delta_0}s^5\int_{Q}k(x)\theta^5 |w_x|^2e^{2s\Phi}\zeta
 \,{\rm d}x\,{\rm d}t
+\frac{C^2_1}{\delta_2}s^5\int_{Q}k(x)\theta^7 |w|^2e^{2s\Phi}
\zeta\,{\rm d}x\,{\rm d}t \nonumber \\
&\quad + \frac{C^2_3}{\delta_3}s^5\int_{Q}\theta^5 |w|^2e^{2s\Phi}\zeta^{1/2}
 \,{\rm d}x\,{\rm d}t
+ \frac{C^2_3}{\delta_4}s^7\int_{Q}\theta^7 |w|^2e^{2s\Phi}\zeta^{1/2}
 \,{\rm d}x\,{\rm d}t \nonumber \\
&\quad + \frac{C_5C\| a+b\|^2_{\infty}}{k(1)\delta_1}s^6
 \int_0^T\int_{\mathcal{O}'}\theta^6 e^{2s\Phi}|w|^2
 \,{\rm d}x\,{\rm d}t\zeta\,{\rm d}x\,{\rm d}t. \label{M1}
\end{align}
Now, we look to increase the term 
\[
\frac{2}{\delta_0}s^5\int_{Q}k(x)\theta^5 |w_x|^2e^{2s\Phi}\zeta\,{\rm d}x\,{\rm d}t.
\]
 To do this, we are going to multiply the first equation of  system \eqref{syst6}
 by $s^5\theta^5e^{2s\Phi}w\zeta$, and integrate on $Q$.
To simplify we set $\overline{\chi}=s^5\theta^5e^{2s\Phi}$,
\begin{align*}
&\int_{Q}v1_{\mathcal{O}}w\zeta\overline{\chi}\,{\rm d}x\,{\rm d}t\\
&= -\int_{Q}(w\overline{\chi}\zeta)w_t\,{\rm d}x\,{\rm d}t
 -\int_{Q}(w\overline{\chi}\zeta)(k(x)w_x)_x\,{\rm d}x\,{\rm d}t
 +\int_{Q}a\overline{\chi}w^2\zeta\,{\rm d}x\,{\rm d}t\\
&=\frac{1}{2}\int_{Q}w^2\zeta\overline{\chi}_t\,{\rm d}x\,{\rm d}t
 +\int_{Q}k(x)\overline{\chi}\zeta w^2_x\,{\rm d}x\,{\rm d}t
 -\frac{1}{2}\int_{Q}w^2(k(\zeta\overline{\chi})_x)_x
 +\int_{Q}a\overline{\chi}\zeta w^2\,{\rm d}x\,{\rm d}t.
\end{align*}
Therefore,
\begin{align*}
&\int_{Q}k(x)\overline{\chi}\zeta |w_x|^2\,{\rm d}x\,{\rm d}t \\
&\leq \int_{Q}v1_{\mathcal{O}}w\zeta\overline{\chi}\,{\rm d}x\,{\rm d}t
+\frac{1}{2}\int_{Q}|w|^2|\zeta\overline{\chi}_t|\,{\rm d}x\,{\rm d}t
+\frac{1}{2}\int_{Q}|w|^2\big|(k(\zeta\overline{\chi})_x)_x\big|\,{\rm d}x\,{\rm d}t\\
&\quad + \int_{Q}|a\overline{\chi}\zeta|| w|^2\,{\rm d}x\,{\rm d}t\\
&= \sum_{i=1}^{4}J_i.
\end{align*}
Using H\"oder-Young's inequality, one obtains 
\begin{equation}\label{j1}
\begin{aligned}
J_1 &= s^5\int_{Q}\theta^5e^{2s\Phi} vw\zeta\,{\rm d}x\,{\rm d}t \\
&\leq \gamma s^3\int_{Q}\theta^3e^{2s\Phi} |v|^2\zeta\,{\rm d}x
 \,{\rm d}t+\frac{1}{\gamma}s^7\int_{Q}\theta^7e^{2s\Phi} |w|^2\zeta
\,{\rm d}x\,{\rm d}t.
\end{aligned}
\end{equation}
Observe that $\overline{\chi}_t=(s^5\theta^4\theta_te^{2s\Phi}(5+2s\theta h(x))$
because by definition the function $\Phi$ is of the form $\Phi(x,t)=\theta(t)h(x)$
where $h(x)=\rho(x)\psi(x)+(1-\rho(x))\Psi(x)$.
Thus, $\overline{\chi}_t\leq Cs^6\theta^7e^{2s\Phi}$ for every $s\geq CT^8$.
Therefore, for all $s\geq 1$ we have
\begin{gather}\label{j2}
J_2 \leq  Cs^6\int_{Q}\theta^7e^{2s\Phi}|w|^2\zeta\,{\rm d}x\,{\rm d}t
\leq Cs^7\int_{Q}\theta^7e^{2s\Phi}|w|^2\zeta\,{\rm d}x\,{\rm d}t, \\
\label{j3}
J_4 = s^5\int_{Q}|a|\theta^5e^{2s\Phi}\zeta |w|^2\,{\rm d}x\,{\rm d}t
\leq \| a\|_{\infty}s^5\int_{Q}\theta^5e^{2s\Phi}\zeta |w|^2
\,{\rm d}x\,{\rm d}t.
\end{gather}
Observe also that 
$(\zeta\overline{\chi})_x\leq C(s\theta)^6\frac{x}{k(x)}e^{2s\Phi}\zeta^{1/2}$
 and from the linearity of the derivative function, we have
\begin{equation}\label{j4}
\begin{aligned}
J_3& =\int_{Q}w^2\left|( k(x)(\zeta\overline{\chi})_x)_x\right|\,{\rm d}x\,{\rm d}t\\
&\leq C\int_{Q}|w|^2k(x)|(\zeta\overline{\chi})_x|\,{\rm d}x\,{\rm d}t\\
&\leq  Cs^6\int_{Q}\theta^6|w|^2e^{2s\Phi}\zeta^{1/2}\,{\rm d}x\,{\rm d}t.
\end{aligned}
\end{equation}
Let $n,m$ be natural numbers such that for any $n\geq m$, one has
\begin{equation} \label{T}
s^m\theta^m\leq Cs^n\theta^n,\quad \forall s\geq C T^8.
\end{equation}
In fact 
\[
s^m\theta^m=s^m\theta^n(\theta^{-1})^{n-m}
\leq s^m\theta^n(\frac{T^8}{16})^{n-m}
\leq s^m\theta^n(\frac{C}{16}s)^{n-m}=Cs^n\theta^n.
\]
 Since  $s\geq C T^8$,
 using inequalities \eqref{j1} to \eqref{j4} and \eqref{T}, we find 
\begin{equation}\label{M2}
\begin{aligned}
&\int_{Q}k(x)\overline{\chi}\zeta |w_x|^2\,{\rm d}x\,{\rm d}t \\
&\leq
\gamma s^3\int_{Q}\theta^3e^{2s\Phi} |v|^2\zeta\,{\rm d}x\,{\rm d}t
+ (\frac{1}{\gamma}+C\| a\|_{\infty}+C) s^7\int_{Q}\theta^7e^{2s\Phi}
 |w|^2\zeta\,{\rm d}x\,{\rm d}t
\end{aligned}
\end{equation} 
Finally, combining inequalities \eqref{M1}, \eqref{T}, and \eqref{M2}, one gets
\begin{equation}\label{M3}
\begin{aligned}
 &s^3\int_{Q}\theta^3 |v|^2e^{2s\Phi}\zeta\,{\rm d}x\,{\rm d}t \\
 &\leq  \Big(C\delta_0+\delta_1+\delta_2+\delta_3+\delta_4
 +\frac{2\gamma}{\delta_0}+\frac{C\| a+b\|^2_{\infty}}{k(1)\delta_1}
 \Big) s^3\int_{Q}\theta^3 |v|^2e^{2s\Phi}\zeta\,{\rm d}x\,{\rm d}t \\
 &\quad +\Big(\frac{C\| a+b\|^2_{\infty}}{k(1)\delta_1}+\frac{C}{\delta_2}
 +\frac{1}{\gamma}+C\| a\|_{\infty}+C\Big) s^7\int_{Q}\theta^7 |w|^2
 e^{2s\Phi}\zeta\,{\rm d}x\,{\rm d}t\\
&\quad + \Big(\frac{C}{\delta_4}+\frac{C}{\delta_3}+\frac{2C}{\delta_0}\Big)
s^7\int_{Q}\theta^7 |w|^2e^{2s\Phi}\zeta^{1/2}\,{\rm d}x\,{\rm d}t.
\end{aligned}
\end{equation} 
We now set $\delta_2=\delta_3=\delta_4=\frac{C'\delta_0}{3}$,
$\gamma=\frac{C'\delta^2_0}{2}$ and $\delta_1=\frac{1}{\delta_0}$,
 with 
\[
C'=\max(\frac{1}{4}, C,\frac{C\| a+b\|^2_{\infty}}{k(1)},
C\| a\|_{\infty})\geq\frac{1}{4}.
\]
 Using  that
$\operatorname{supp}\zeta\subset\overline{\mathcal{O}'}$ we have
\begin{align*}
&(-4C'\delta^2_0+\delta_0-1
) s^3\int_0^T\int_{\mathcal{O}}\theta^3 |v|^2e^{2s\Phi} \\
&\leq \frac{C'\delta^3_0+2C'\delta^2_0+(6+2C')\delta_0+2}{\delta_0}
s^7\int_0^T\int_{\mathcal{O}'}\theta^7 |w|^2e^{2s\Phi}.
\end{align*}
The term $-4C'\delta^2_0+\delta_0-1>0$ is equivalent to
$ 16C'\geq 1$ and since $C'\geq\frac{1}{4}>\frac{1}{16}$,
for  $\delta_0$ fixed, we obtain by taking
$C=\frac{C'\delta^3_0+2C'\delta^2_0+(6+2C')
\delta_0+2}{-\delta(C'\delta^2_0+\delta_0-1)}$
\begin{equation}\label{M3'}
s^3\int_0^T\int_{\mathcal{O}}\theta^3 |v|^2e^{2s\Phi}
\leq   Cs^7\int_0^T\int_{\mathcal{O}'}\theta^7 |w|^2e^{2s\Phi}.
\end{equation}
Now, by multiplying the inequality \eqref{M3'} by $C_0$ and using
inequalities \eqref{d1} and \eqref{T}, we obtain the inequality
\eqref{p20} of Theorem \ref{t3}.
\end{proof}

\begin{proof}[Proof of Theorem \ref{t3'}] 
Consider the  functions 
\[
F(s)=\begin{cases}
 \frac{f(s)}{s}&\text{if }s\neq 0\\
 f'(0)&\text{if }s= 0
\end{cases}\quad\text{and}\quad
G(\sigma)=\begin{cases}
\frac{g(\sigma)}{\sigma} &\text{if }\sigma\neq 0\\
g'(0) &\text{if }\sigma= 0
\end{cases}
\]
We need the following result.

\begin{lemma}\label{lem6}
Under the hypothesis of Theorem \ref{t3'} and any $\varepsilon>0$, there exists 
a positive constant $C_{\varepsilon}$ such that
\begin{equation}
\begin{gathered}
|F(s)|^{2/3}\leq C_{\varepsilon}+\varepsilon log(1+|s|)\\
|G(\sigma)|^{2/3}\leq C_{\varepsilon}+\varepsilon log(1+|\sigma|).
\end{gathered}
\end{equation}
\end{lemma}

\begin{proof}
Indeed, it will be sufficient to prove that for each $\varepsilon>0$, one has
$|F(s)|\leq C_{\eta}+\varepsilon \log^{3/2}(1+|s|)$ for all $s\in\mathbb{R}$.
Let $\varepsilon>0$. From 
$\lim_{|s|\to +\infty}\frac{f(s)}{|s|\log^{3/2}(1+|s|)}=0$, there exists 
$s(\varepsilon)\geq 1$ such that 
\begin{equation}\label{dd}
|\frac{f(s)}{s}|\leq\varepsilon \log^{3/2}(1+|s|)\quad\text{for all } 
|s|>s(\varepsilon).
\end{equation}
On the other hand, using the fact that $f$ is a locally Lipschitz-continuous 
function, there exists a constant $C_{\eta}$ only depending on $\varepsilon$ 
and $f$, such that $|f(s)-f(s')|\leq C_{\eta}|s-s'|$, for all 
$(s,s')\in[-s(\varepsilon),s(\varepsilon)]^2$. In particular, 
for $s\neq 0$ and $s'=0$, we find
\begin{equation}\label{dd1}
|f(s)|\leq C_{\varepsilon}|s|\Longrightarrow |\frac{f(s)}{s}|\leq C_{\varepsilon}.
\end{equation}
From \eqref{dd} and \eqref{dd1}, we deduce that for all $s\in\mathbb{R}$, 
$|\frac{f(s)}{s}|\leq C_{\varepsilon}+\varepsilon \log^{3/2}(1+|s|)$.
Therefore, we obtain $|F(s)|^{2/3}\leq C_{\varepsilon}+ \varepsilon log(1+|s|)$, 
for all $s\in\mathbb{R}$ with $C_{\varepsilon}=( Ks(\varepsilon))^{2/3}$ 
depending only  on $\varepsilon$ and $f$, and $K>0$ is a constant. 

The same arguments, allow us to prove that 
$|G(s)|^{2/3}\leq C_{\varepsilon}+ \varepsilon log(1+|s|)$, for all 
$s\in\mathbb{R}$.
This completes the proof.
\end{proof}

Let $R>0$ be a constant whose value will be determined below. 
We will use the truncation continuous function $T_R:\mathbb{R}\to\mathbb{R}$, 
 given by
\[
T_R(s)=\begin{cases}
s& \text{if }|s|\leq R,\\
 R\operatorname{sgn}(s) &\text{ otherwise }.
\end{cases}
\]
For each $(\tilde{y},\tilde{z})\in X_1$, we will consider the 
 linear cascade system
\begin{equation}\label{l6}
\begin{gathered}
y_t-(ky_x)_x+F(T_R(\tilde{y}))y=h1_{\omega}\quad \text{in }Q,\\
z_t-(kz_x)_x+G(T_R(\tilde{z}))z=y1_{\mathcal{O}}\quad \text{in }Q,\\
y=z=0 \quad\text{on }\Sigma ,\\
y(x,0)=y_0(x),\quad z(x,0)=z_0(x)\quad\text{in }\Omega .
\end{gathered}
\end{equation}
Obviously, system \eqref{l6} is of the same form as system \eqref{syst6'}, 
with $a_{\tilde{y}}=F(T_R(\tilde{y}))\in L^{\infty}(Q)$ and
 $b_{\tilde{z}}=G(T_R(\tilde{z}))\in L^{\infty}(Q)$ since $F$ and $G$ 
are continuous functions.
Consequently, we can apply Theorem \ref{nul} to \eqref{l6}. 
In fact, we  apply this result in an adequate (eventually smaller)
 time interval $(0,T_R)$, where
\begin{equation}
T_R=\min(T,\| a_{\tilde{y}}\|^{-1/2},\| b_{\tilde{z}}\|^{-1/2}) .
\end{equation}
According to Theorem \ref{nul} and applying the density of $H^2_k(\Omega)$ 
in $L^2(\Omega)$, if $y_0,\,z_0\in H^2_k(\Omega)$ then, there exists 
a control $\tilde{h}\in L^2(\Omega\times(0,T_R))$ such that the solution 
of \eqref{l6} in $\Omega\times(0,T_R)$ satisfies $z(x,T_R)=0$ in 
$\Omega$ and  we have
\begin{gather}\label{l7}
\| \tilde{h}\|_{L^2(\omega\times(0,T_R)}
\leq C_0(\Omega,\omega,T_R,a_{\tilde{y}},b_{\tilde{z}})
(\| y_0\|_{H^1_k(\Omega)}+\| z_0\|_{H^1_k(\Omega)}), \\
\label{l8}
\| y\|_{X_R}+\| z\|_{X_R}\leq C_1(\Omega,\omega,T_R,a_{\tilde{y}},
b_{\tilde{z}})(\| y_0\|_{H^1_k(\Omega)}+\| z_0\|_{H^1_k(\Omega)}).
\end{gather}

We now extend the functions $\tilde{h}$, $y$ and $z$ by zero to the whole $Q$. 
Denote such extensions again $\tilde{h}$, $y$ and $z$ respectively. Then, 
$(y,z)$ lies in $X_1$ and solves the linearised system \eqref{l6} in $Q$
 with control term $\tilde{h}\in L^2(Q)$, and satisfies $z(x,T)=0$ in $\Omega$. 
Moreover, we have the estimates
\begin{gather}\label{l9}
\| \tilde{h}\|_{L^2(\omega\times(0,T)}
\leq C_0(\Omega,\omega,T_R,a_{\tilde{y}},b_{\tilde{z}})
(\| y_0\|_{H^1_k(\Omega)}+\| z_0\|_{H^1_k(\Omega)}), \\
\label{l10}
\| y\|_X+\| z\|_X\leq C_1(\Omega,\omega,T_R,a_{\tilde{y}},b_{\tilde{z}})
(\| y_0\|_{H^1_k(\Omega)}+\| z_0\|_{H^1_k(\Omega)}).
\end{gather}
For a fixed control $h\in L^2(Q)$, we now denote by $(y_h,z_h)$ 
the solution of  \eqref{l6} associated to $h$ and the potentials
 $a_{\tilde{y}},b_{\tilde{z}}$. For any 
$(\tilde{y},\tilde{z})\in X^2_1$, one defines the family of controls
\[
\mathcal{A}_R(\tilde{y},\tilde{z})
=\{ h\in L^2(Q) : (y_h,z_h)\in X^2_1, z_h(x,T)=0\text{ in 
 $\Omega$  and $h$ satisfies \eqref{l9}}\}.
\]
Thus, we can introduce the multi-valued mapping
\[
\Lambda_R\;(\tilde{y},\tilde{z})\in X^2_1
\mapsto\Lambda_R(\tilde{y},\tilde{z})\subset X^2_1,
\]
where
\[
\Lambda_R(\tilde{y},\tilde{z})
=\{ (y_h,z_h)\in X^2_1,\text{ solution of \eqref{l6}, satisfying \eqref{l10} and 
}h\in\mathcal{A}_R(\tilde{y},\tilde{z})\}.
\]
We will prove that this mapping admits at least one fixed point $(y,z)$. 
We will also prove that, for some $R$, every fixed point of $\Lambda_R$ 
satisfie\begin{equation}
\| y\|_{X_1}+\| z\|_{X_1}\leq R.
\end{equation}

Notice that Kakutani's fixed point theorem can be applied to $\Lambda_R$ thus, 
ensuring the existence of (at least) one fixed point of $\Lambda_R$ in $X^2_1$.

First, from  the inequalities \eqref{l9} and \eqref{l10}, we deduce that 
$\Lambda_R(\tilde{y},\tilde{z})$ is for every $(\tilde{y},\tilde{z})$
 a nonempty set. Moreover, $\Lambda_R(\tilde{y},\tilde{z})$ is a closed 
and convex subset for any $(\tilde{y},\tilde{z})$. In fact, if $(y,z)$ 
and $(y',z')$ are two elements of $\Lambda_R(\tilde{y},\tilde{z})$, 
solutions of \eqref{l6} associated to controls $h$ and $h'$ 
respectively. Let $\alpha\in [0,1]$. We are going to show that $(Y,Z)$ 
with $Y=\alpha y+(1-\alpha)y'$ and $Z=\alpha z+(1-\alpha )z'$ and 
$(Y,Z)$ is a solution of  \eqref{l6}. Consider the following two systems
\begin{equation}\label{l11}
\begin{gathered}
 y_t-(ky_x)_x+F(T_R(\tilde{y}))y=h1_{\omega}\quad \text{in }(0,T_0)\times\Omega,\\
 z_t-(kz_x)_x+G(T_R(\tilde{z}))z=y1_{\mathcal{O}}\quad \text{in }(0,T_0)\times\Omega,\\
 y=z=0 \quad\text{on }\Sigma ,\\
 y(x,0)=y_0(x),\quad z(x,0)=z_0(x)\quad\text{in }\Omega ,
\end{gathered}
\end{equation}
and
\begin{equation}\label{l12}
\begin{gathered}
 y_t'-(ky_x')_x+F(T_R(\tilde{y}))y'=h'1_{\omega}\quad \text{in }(0,T_1)\times\Omega,\\
 z_t'-(kz_x')_x+G(T_R(\tilde{z}))z'=y'1_{\mathcal{O}}\quad
 \text{in }(0,T_1)\times\Omega,\\
 y'=z'=0\quad \text{on }\Sigma ,\\
 y'(x,0)=y'_0(x),\quad z'(x,0)=z'_0(x)\quad\text{ in }\Omega .
\end{gathered}
\end{equation}
Since  systems \eqref{l11} and \eqref{l12} admit solutions at any times 
$T_0$ and $T_1$ respectively, we can deduce that $(Y,Z)$ is a solution of 
\begin{equation}\label{l13}
\begin{gathered}
 Y_t-(kY_x)_x+F(T_R(\tilde{y}))Y=H1_{\omega}\quad \text{in }Q,\\
 Z_t-(kZ_x)_x+G(T_R(\tilde{z}))Z=Y1_{\mathcal{O}}\quad \text{in }Q,\\
 Y=Z=0\quad\text{on }\Sigma ,\\
 Y(x,0)=Y_0(x),\quad Z(x,0)=Z_0(x)\quad\text{in }\Omega ,
\end{gathered}
\end{equation}
where $H=\alpha h+(1-\alpha)h'$, $Y_0(x)=y_0(x)+y_0'(x)$ and 
$Z_0(x)=z_0(x)+z_0'(x)$.
Consequently, with $T=\max(T_0,T_1)$, we have $Z(x,T)=0$ in $\Omega$ 
and the inequalities \eqref{l9} and \eqref{l10} are satisfied. 
We can deduce that $(Y,Z)\in\Lambda_R(\tilde{y},\tilde{z})$ for any 
$(\tilde{y},\tilde{z})\in X^2_1$. Then, $\Lambda_R(\tilde{y},\tilde{z})$ 
is the convex subset.
On the other hand, from \eqref{l10}, $\Lambda_R(\tilde{y},\tilde{z})$
 is bounded in $X^2_1$. Hence, $\Lambda_R$ maps the whole space $X^2_1$ 
in a bounded  subset of $X^2_1$. Now, let $K\subset X^2_1$ be a bounded set. 
Let us show that for any $(\tilde{y},\tilde{z})\in K$, 
$\Lambda_R(\tilde{y},\tilde{z})$ is a compact set of $X^2_1$. 
Thus, let $\{(y_n,z_n)\}$ be a sequence in $\Lambda_R(\tilde{y},\tilde{z})$. 
From  inequality \eqref{l10}, $(y_n,z_n)$ is bounded in $X^2_1$  there exists 
a sequence $\{ h_n\}$ in $\mathcal{A}_R(\tilde{y},\tilde{z})$ and from 
\eqref{l9}, $h_n$ is bounded in $L^2(Q)$. Thus, there exists a subsequence 
denoted again $\{(y_n,z_n)\}$ and $\{ h_n\}$ such that
\begin{equation}\label{l14}
\begin{gathered}
 h_n\to h\quad \text{weakly in }L^2(Q),\\
(y_n,z_n)\to (y,z)\quad \text{strongly in }
(\mathcal{C}([0,T];H^1_k))^2\text{ and weakly in }X^2_1.
\end{gathered}
\end{equation}
Since $(y_n,z_n)$ is a solution of the system
\begin{equation}\label{l15}
\begin{gathered}
 (y_n)_t-(k(y_n)_x)_x+F(T_R(\tilde{y}))y_n=(h_n)1_{\omega}\quad \text{ in }Q,\\
 (z_n)_t-(k(z_n)_x)_x+G(T_R(\tilde{z}))z_n=(y_n)1_{\mathcal{O}}\quad \text{ in }Q,\\
 y_n=z_n=0\quad \text{on }\Sigma ,\\
 y_n(x,0)=y_0(x),\quad z_n(x,0)=z_0(x)\quad\text{in }\Omega ,
\end{gathered}
\end{equation}
we conclude by passing to the limit that
 $(y,z)\in \Lambda_R(\tilde{y},\tilde{z})$ and is associated to the 
control $h\in\mathcal{A}_R(\tilde{y},\tilde{z})$ for any 
$(\tilde{y},\tilde{z})\in K$, as claimed. Thus, we can conclude that 
$\Lambda_R(K)=\cup\{\Lambda_R(\tilde{y},\tilde{z}):(\tilde{y},\tilde{z})\in K\}$ 
is relatively compact in $X^2_1$.

Let us now prove that the mapping 
$(\tilde{y},\tilde{z})\mapsto\Lambda_R(\tilde{y},\tilde{z})$ 
is upper hemicontinuous, i.e. that the real-valued function 
$(\tilde{y},\tilde{z})\in X^2_1\mapsto
\sup_{(y,z)\in\Lambda_R(\tilde{y},\tilde{z})}\langle\mu,(y,z)\rangle$ 
is upper semicontinuous for each bounded linear form $\mu\in (X^2_1)'$. 
In other words, let us see that
\begin{equation}\label{l16}
B_{\alpha,\mu}=\{ (\tilde{y},\tilde{z})\in X^2_1:\sup_{(y,z)\in\Lambda_R(\tilde{y},\tilde{z})}\langle\mu,(y,z)\rangle\geq\alpha\}
\end{equation}
is a closed subset of $X^2_1$ for every $\alpha\in\mathbb{R}$ and every 
$\mu\in (X^2_1)'$. Thus, let $((\tilde{y_n},\tilde{z_n}))_n $ be a sequence 
in $B_{\alpha,\mu}$ such that $(\tilde{y_n},\tilde{z_n})\to(\tilde{y},\tilde{z})$ 
in $X^2_1$. Our aim is to prove that $(\tilde{y},\tilde{z})\in B_{\alpha,\mu}$. 
In view of the continuity of on $F$, $G$ and $T_R$, we have
\begin{equation}\label{l17}
\begin{gathered}
 F(T_R(\tilde{y_n}))\to F(T_R(\tilde{y}))\quad \text{strongly in }L^{\infty}(Q),\\
G(T_R(\tilde{z_n}))\to G(T_R(\tilde{z}))\quad \text{strongly in }L^{\infty}(Q).
\end{gathered}
\end{equation}
Since all sets $\Lambda_R(\tilde{y_n},\tilde{z_n})$ are compact and 
satisfy \eqref{l10}, we deduce that
\begin{equation}\label{l18}
\alpha\leq\sup_{(y,z)\in\Lambda_R(\tilde{y_n},\tilde{z_n})}\langle\mu,(y,z)\rangle
= \langle\mu,(y_n,z_n)\rangle
\end{equation}
for some $(y_n,z_n)\in\Lambda_R(\tilde{y}_n,\tilde{z}_n)$. 
In fact, the mapping $(\tilde{y}_n,\tilde{z}_n)\mapsto\langle\mu,(y,z)\rangle$
 is continuous on the compact set $\Lambda_R(\tilde{y}_n,\tilde{z}_n)$, 
this upper boundary is achieved; i.e. there exists 
$(y_n,z_n)\in\Lambda_R(\tilde{y}_n,\tilde{z}_n)$, such that 
$\sup_{(y,z)\in\Lambda_R(\tilde{y}_n,\tilde{z}_n)}\langle\mu,(y,z)\rangle
= \langle\mu,(y_n,z_n)\rangle$, as claimed.
From the definitions of $\Lambda_R(\tilde{y}_n,\tilde{z}_n)$ and 
$\mathcal{A}_R(\tilde{y}_n,\tilde{z}_n)$, there must exist a sequence
 $(h_n)_n\subset L^2(\omega\times(0,T))$ solution of the system
\begin{equation}\label{l19}
\begin{gathered}
 (y_n)_t-(k(y_n)_x)_x+F(T_R(\tilde{y_n}))y_n=(h_n)1_{\omega}\quad \text{in }Q,\\
 (z_n)_t-(k(z_n)_x)_x+G(T_R(\tilde{z_n}))z_n=(y_n)1_{\mathcal{O}}\quad \text{in }Q,\\
 y_n=z_n=0\quad\text{on }\Sigma ,\\
 y_n(x,0)=y_0(x),\quad z_n(x,0)=z_0(x)\quad\text{in }\Omega .
\end{gathered},
\end{equation}
such that $h_n$ and $(y_n,z_n)$ satisfy the inequalities \eqref{l9} and \eqref{l10}
 respectively. Hence, $(y_n,z_n)$ and $h_n$ are uniformly bounded in $X^2_1$ and
 $L^2(Q)$ respectively. Therefore, we must write the following at least for 
a subsequence that we are going to denote $(y_n,z_n)$ and $h_n$ again 
respectively such that $(y_n,z_n)\to(\widehat{y},\widehat{z})$ strongly in 
$X^2_1$ and $h_n\to\widehat{h}$ weakly in $L^2(Q)$.
Since the subsequence $((y_n,z_n))_n$ is a solution of   \eqref{l19}, 
we check by passing to the limit that 
$(\widehat{y},\widehat{z})\in\Lambda_R(\tilde{y},\tilde{z})$ and 
$\widehat{h}\in\mathcal{A}_R(\tilde{y}_n,\tilde{z}_n)$. 
Consequently, we can take the limit in  \eqref{l18} and deduce that
$\alpha\leq\sup_{(y,z)\in\Lambda_R(\tilde{y},\tilde{z})}\langle\mu,(y,z)\rangle
= \langle\mu,(\widehat{y},\widehat{z})\rangle$, this is to say 
$(\widehat{y},\widehat{z})\in B_{\alpha,\mu}$. 
This proves that $(\tilde{y},\tilde{z})\mapsto\Lambda_R(\tilde{y},\tilde{z})$ 
is upper hemicontinuous on $X^2_1$.

As a consequence, for any fixed $R>0$, Kakutani's fixed point theorem can 
be applied ensuring the existence of a fixed point of $\Lambda_R$, i.e. 
that there exists $(y,z)\in X^2_1$ such that $(y,z)\in\Lambda_R(y,z)$.

Now, let us show  the existence $R>0$ such that 
$\| y\|_{L^{\infty}}+\| z\|_{L^{\infty}}\leq R$.
\begin{equation}\label{l20}
\begin{aligned}
&\| y\|_{X_1}+\| z\|_{X_1} \\
&\leq  C(1+\| a_{\tilde{y}}\|_{\infty}+\| b_{\tilde{z}}\|_{\infty})
e^{1/2CT(1+\| a_{\tilde{y}}\|^2_{\infty}+\| b_{\tilde{z}}\|^2_{\infty})}
(\| y_0\|_{H^1_k}+\| z_0\|_{H^1_k}),\\
&\leq  e^{1/2CT(1+a_{\tilde{y}}\|_{\infty}
+\| b_{\tilde{z}}\|_{\infty}+\| a_{\tilde{y}}\|^2_{\infty}
+\| b_{\tilde{z}}\|^2_{\infty})}(\| y_0\|_{H^1_k}+\| z_0\|_{H^1_k}).
\end{aligned}
\end{equation}
Since for any small $\varepsilon>0$, 
\begin{gather*}
 \| a_{\tilde{y}}\|_{\infty}=F(T_R(\tilde{y}))
\leq C_{\varepsilon}+\varepsilon \log(1+R),\\
\| a_{\tilde{y}}\|^2_{\infty}=F(T_R(\tilde{y}))^2
 \leq C_{\varepsilon}+\varepsilon \log(1+R),\\
 \| b_{\tilde{z}}\|_{\infty}= G(T_R(\tilde{y}))\leq C_{\varepsilon}+\varepsilon 
\log(1+R),\\
\| b_{\tilde{z}}\|^2_{\infty}=G(T_R(\tilde{y}))^2
\leq C_{\varepsilon}+\varepsilon \log(1+R).
\end{gather*}
Then \eqref{l20}, becomes
\begin{equation}\label{l21}
\begin{aligned}
\| y\|_{X_1}+\| z\|_{X_1}
&\leq  e^{C(1+C_{\varepsilon}+\varepsilon 
 \log(1+R))}(\| y_0\|_{H^1_k}+\| z_0\|_{H^1_k})\\
&\leq  e^{C(1+C_{\varepsilon})}(1+R)^{C_{\varepsilon}}(\| y_0\|_{H^1_k}
+\| z_0\|_{H^1_k})
\end{aligned}
\end{equation}
with $C>0$. The fact that $X_1\hookrightarrow L^{\infty}(Q)$ 
(a consequence of \cite[Theorem 5.4]{9}), with continuous embedding then,
 taking $\varepsilon=(2c)^{-1}$ like in \cite{8}, we infer that
\begin{equation}\label{l21b}
\| y\|_{\infty}+\| z\|_{\infty}\leq
C(1+R)^{1/2}\big(\| y_0\|_{H^1_k}+\| z_0\|_{H^1_k}\big),
\end{equation}
and we have $\| y\|_{\infty}+\| z\|_{\infty}\leq R$, for $R>0$ 
large enough. Thus, the proof of Theorem \ref{t3'} is complete.
\end{proof}

\subsection*{Acknowledgments}
This work has been supported in part by the Simons Foundation through the NLAGA
project that supports networks of resourceful young mathematicians
working in Sub-Saharan  Africa.

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\end{document}