\documentclass[reqno]{amsart}
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\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2014 (2014), No. 167, pp. 1--26.\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-2014/167\hfil Lipschitz stability  for linear  parabolic systems]
{Lipschitz stability for linear parabolic systems with interior degeneracy}

\author[I. Boutaayamou, G. Fragnelli, L. Maniar \hfil EJDE-2014/167\hfilneg]
{Idriss Boutaayamou, Genni Fragnelli, Lahcen Maniar}  % in alphabetical order

\address{Idriss Boutaayamou \newline
D\'epartement de Math\'ematiques, Facult\'e des Sciences Semlalia\\
LMDP, UMMISCO (IRD-UPMC),
Universit\'e Cadi Ayyad, Marrakech 40000, B.P. 2390, Morocco}
\email{dsboutaayamou@gmail.com}

\address{Genni Fragnelli \newline
Dipartimento di Matematica, Universit\`a degli Studi di Bari Aldo Moro,
Via E. Orabona 4, 70125 Bari - Italy}
\email{genni.fragnelli@uniba.it}

\address{Lahcen Maniar \newline
D\'epartement de Math\'ematiques, Facult\'e des Sciences Semlalia\\
LMDP, UMMISCO (IRD-UPMC),
Universit\'e Cadi Ayyad, Marrakech 40000, B.P. 2390, Morocco}
\email{maniar@uca.ma}

\thanks{Submitted June 28, 2014. Published July 30, 2014.}
\subjclass[2000]{35k65}
\keywords{Parabolic system; interior degeneracy; Carleman estimates;
\hfill\break\indent  Lipschitz stability}

\begin{abstract}
 In this article, we study an inverse problem for linear degenerate parabolic
 systems with one force. We establish Lipschitz stability for the source term
 from measurements of one component of the solution at a positive time and on
 a subset of the space domain, which contains degeneracy points.
 The key ingredient is the derivation of a Carleman-type estimate.
\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}

 The null controllability and inverse problems of parabolic equations and parabolic
coupled systems have attracted much interest in these last years, see
\cite{Ammar2,Ammar4, Ben,BCrGT,BukhgeimKlibanov,CrGR,CrGRY,FG,
FGPY,Gu,Yam,LR,RqCr1,RqCr2,Rus}. The main result in these papers is the development
 of suitable Carleman estimates, which are crucial tools to obtain observability
inequalities and Lipschitz stability for term sources, initial data, potentials
and diffusion coefficients. The above systems are considered to be non degenerate.
In other words, the diffusion coefficients are uniformly coercive. On the contrary,
the case of degenerate coefficients at the boundary is also considered in several
papers by developing adequate Carleman estimates. The null controllability and
inverse problems of degenerate parabolic equations  are studied in
\cite{Tort,Can1,Can2,Can3,21,Tort1}, and for the coupled degenerate parabolic
systems  in \cite{bahm,hajjaj,Bou,CanTer,Liu}.  In these papers, the degeneracy
considered is at the boundary of the spatial domain. \par
After the pioneering works \cite{Fra1,Fra2}, there has been substantial progress
in understanding the null controllability of parabolic equations with interior
degeneracy (see, e.g., \cite{fm}). In this scope, the goal of this paper is to
study an  inverse source problem of a $2 \times 2$ parabolic systems with interior
degeneracy and  different diffusion coefficients
\begin{equation} \label{system}
 \begin{gathered}
 u_t-(a_{1}u_x)_x+b_{11}u+b_{12}v=f, \quad (t,x)\in Q,\\
 v_t-(a_{2}v_x)_x+b_{22}v=0, \quad (t,x)\in Q,\\
 u(t,0)=u(t,1)=v(t,0)=v(t,1)=0, \quad t \in (0,T),\\
 u(0,x)=u_0(x), \quad v(0,x)=v_0(x), \quad x\in (0,1),
\end{gathered}
\end{equation}
where  $u_0, v_0 \in L^2 (0,1) $, $T> 0$
fixed, $Q:=(0,T)\times (0,1)$,  $b_{ij}\in L^\infty(0,1)$,
$i,j=1,2$, and every $a_i$, $i=1,2$, degenerates at an interior point $x_i$
of the spatial domain $(0,1)$
(for the precise assumptions we refer to Section 2).
For $t_0\in (0,T)$ given, let
$Q_{t_0}^T=(t_0,T)\times (0,1)$
 and $ T':=\frac{T+t_0}{2}$.
For a given $C_0>0$,  we denote by $S(C_0)$ the space
\begin{align*}
S(C_0):=\{f\in H^1(0,T;L^2(0,1)) : |f_t(t,x)|\leq C_0|f(T',x)|, \text{ a.e. }
(t,x)\in Q\}.
\end{align*}

More precisely, we want to establish Lipschitz stability for the source term $f$
from measurements of the component $u$ at  time $T'$ and on a subset
$\omega\subset (0,1)$, which contains the degeneracy points.

The main ingredient to obtain Lipschitz stability is  Carleman estimates for
degenerate equations.  For null controllability of a parabolic equation with
interior degeneracy,   Carleman estimates  were obtained in \cite{fm} and
in \cite{Fra2}. For  inverse problems,  these estimates are not sufficient,
and one needs   also  some additional estimates on the term $u$ with a special
weight and  the derivative term $u_t$. We prove first these for parabolic
equations with interior degeneracy similar to the ones obtained in
\cite{Tort, Tort1} in the case of a boundary degeneracy. This will lead to
obtain our Carleman estimates for system \eqref{system}. At the end  having
these Carleman estimates in hand, we follow the method developed in
\cite{Ben,BukhgeimKlibanov,Yam} to obtain the Lipschitz stability for the
source term $f$. The main task here is to estimate the source $f$ by the
measurements, on the domain $\omega$, of  the first component $u$ of the
solutions of system \eqref{system}.

To prove our Carleman estimates, we use the following Hardy-Poincar\'e
inequality proved in
\cite[Proposition 2.1]{Fra2}
\begin{equation}\label{hardy}
\int_0^1\frac{p(x)}{(x-x_0)^2}w^2(x)\,dx\leq C_{HP}\int_0^1p(x)|w_x(x)|^2\,dx
\end{equation}
for all functions $w$ such that
$$
w(0)=w(1)=0\quad\text{and}\quad \int_0^1p(x)|w_x(x)|^2\,dx<\infty.
$$
Here $p$ is any continuous function in $[0,1]$, with $p>0$ on
$[0,1]\setminus\{x_0\}$, $p(x_0)=0$, for some $x_0$ in $(0,1)$,
and such that there exists $\vartheta\in(1,2)$ so that the function
$ x\mapsto  p(x)/|x-x_0|^\vartheta$ is
non-increasing on the left of $x_0$ and nondecreasing on the right of $x_0$.

This article is organized as  follows:
in Section 2, we discuss the well-posedness of the system \eqref{system}.
Then, in Section 3,  we establish different Carleman estimates for parabolic
equations and parabolic systems \eqref{system}. Finally, in Section 4,
we apply the Carleman estimates to prove the Lipschitz stability result.

\section{Assumptions and well-posedness}

To study the well-posedness of system \eqref{system}, we consider two
situations, namely the weakly degenerate (WD) and the strongly degenerate
(SD) cases. The associated weighted spaces and  assumptions on diffusion
coefficients are the following:

\noindent\textbf{Case (WD):} for $i=1,2$, let
 $$
 H_{a_i}^1(0, 1):=\big\{ u \text{ abs. cont. in }[0, 1],\;
\sqrt{a_i} u_x \in L^2(0, 1),\; u(0)=u(1) = 0\big\},
$$
where the functions $a_i$ satisfy
\begin{equation} \label{assump}
\parbox{10cm}{there exists $x_i\in (0,1)$, $i=1,2$ such that
$a_i(x_i)=0$, $a_i>0$  in  $[0,1]\setminus \{x_i\}$,
$a_i\in C^1([0,1]\setminus\{x_i\})$;  \\
and there exists $K_i\in (0,1)$ such that $(x-x_i)a_i'\leq K_ia_i$,  a.e. in
$[0,1]$.}
\end{equation}

\noindent\textbf{Case (SD):}  for $i=1,2$, let
\begin{align*}
 H_{a_i}^1(0, 1)
:=\Big\{&u\in L^2(0,1): u \text{ is locally abs. cont. in }
[0, 1]\setminus\{x_i\},\\
&\sqrt{a_i} u_x \in L^2(0, 1),\; u(0)=u(1) = 0\Big\},
\end{align*}
where the functions $a_i$ satisfy
\begin{equation} \label{assump2}
\parbox{10cm}{there exists $x_i\in (0,1)$, $i=1,2$ such that
$a_i(x_i)=0$, $a_i>0$  in $[0,1]\setminus \{x_i\}$,
$a_i\in C^1([0,1]\setminus\{x_i\})\cap W^{1,\infty}(0,1)$;
there exists $K_i\in [1,2)$ such that $(x-x_i)a_i'\leq K_ia_i$ a.e. in
$[0,1]$, and
if $K_i > 4/3$, there exists $\gamma\in(0,K_i]$  such that
$a_i/|x-x_i|^\gamma$  is non-increasing on the left of $x_i$
 and nondecreasing on the right of $x_i$.}
\end{equation}
In both cases, for $i=1,2$, we consider the space
$$
H^2_{a_i} (0, 1):=\big\{ u \in H^1_{a_i}(0, 1) \,: \, a_iu_x
\in H^1(0, 1)\big\}
$$
with the norms
$$
\|u\|^2_{ H^1_{a_i}} := \|u\|^2_{L^2(0,1)} +
\|\sqrt{a_i}u_x\|^2_{ L^2(0,1)}, \quad
\|u\|^2_{H^2_{a_i}} := \|u\|^2 _{H^1_{a_i}} + \|(a_iu_x)_x\|^2_{ L^2(0,1)}.
$$
 We recall from \cite{fm} that, for $i=1,2$, the  operator
$(A_i,D(A_i))$ defined by $A_iu := (a_iu_x)_x$,
$u \in D(A_i) = H^2_{a_i}(0, 1)$
is closed negative self-adjoint  with dense domain in $L^2(0, 1)$.
 In the Hilbert space  $\mathbb{H}:= L^2(0,1) \times L^2(0, 1) $,
 the  system \eqref{system} can be
transformed into the  Cauchy problem
\begin{gather*}
X'(t) =\mathcal{A}X(t) -BX(t)+F(t), \quad  t\in (0,T),\\
X(0)=\begin{pmatrix}
u_0\\
v_0
\end{pmatrix},
\end{gather*}
where $X(t)=\begin{pmatrix}
u(t)\\
v(t)
\end{pmatrix}$,
 $\mathcal{A}=\begin{pmatrix}
A_1&0\\
0&A_2
\end{pmatrix}$,
$D(\mathcal{A})=D(A_1)\times D(A_2)$, $F(t)=\begin{pmatrix}
f(t)\\
0 \end{pmatrix}$ and $ B=\begin{pmatrix}
b_{11}&b_{12}\\
0&b_{22}
\end{pmatrix}$.

Since the operator $\mathcal{A}$
is diagonal and $B$ is a bounded perturbation, the following well-posedness
and regularity results hold.

\begin{proposition}\label{estimsemigroup}
(i)  The operator $\mathcal{A}$ generates a contraction strongly
continuous semigroup.

(ii) For all $(u_0,v_0)\in D(\mathcal{A})$ and  $f\in H^1(0,T;L^2(0,1))$, the
problem \eqref{system} has a unique solution
 $(u,v)\in C\big([0,T],D(\mathcal{A})\big)\cap C^1(0,T;\mathbb{H})$.

(iii)  For all $f\,\in L^2({Q})$, $u_0,v_0\in L^2(0,1)$,  and $\varepsilon\in (0,T)$,
 there exists a unique mild solution
$(u,v)\in X_T:=H^1\big( [\varepsilon,T],\mathbb{H} \big)
\cap L^2\big(\varepsilon,T;D(\mathcal{A})\big)$ of \eqref{system} satisfying
\begin{align*}
\|(u,v)\|_{X_T}\leq  C_T\Big(\|(u_0,v_0)\|_{\mathbb{H}}^2
+\|(F,G)\|_{\mathbb{H}}^2\Big).
\end{align*}
Moreover, for $f\,\in H^1(0,T;L^2(0,1))$ and  $\varepsilon\in (0,T)$,
we have $(u,v)\in Y_T:=C\big([\varepsilon,T],D(\mathcal{A})\big)
\cap C^1(\varepsilon,T;\mathbb{H})$.
\end{proposition}

\section{Carleman estimate}

 The main topic of this section is to  establish a Carleman estimate for a
degenerate parabolic single equation with a boundary observation
on the right hand side. Then, we will deduce the one for the degenerate system
 \eqref{system} with distributed
observation of $u$ on the subdomain $\omega$.

 Part of these estimates were obtained in \cite{Fra2, fm} under two different
assumptions on the degenerate diffusion coefficient for  a null controllability
 purpose.
In the forthcoming theorems we will prove additional estimates on $u$ and $u_t$,
that are crucial to prove Lipschitz stability results.

Throughout this section, we set $\omega=(\lambda,\beta)$ and assume,
without loss of generality, $x_1<x_2$.  Set also
$\omega':=\tilde{\omega}\cup\overline{\omega}:=(\lambda,\beta_1)\cup(\lambda_2,\beta)$
and
$\omega'':=\tilde{\tilde{\omega}}\cup\overline{\overline{\omega}}
:=(\lambda,\frac{\lambda_1+2\beta_1}{3})\cup(\frac{\lambda_2+2\beta_2}{3},\beta)$,
where $\lambda_i$ and $\beta_i$, for $i=1,2$, satisfy
$0<\lambda<\lambda_1<\beta_1<x_1<x_2<\lambda_2<\beta_2<\beta<1$.

\subsection{Carleman estimate for an equation}
In this subsection  we shall derive the Carleman estimate for the solution
of the  problem
\begin{equation} \label{problem}
 \begin{gathered}
 u_t-(a(x)u_x)_x+cu=h, \quad (t,x)\in Q,\\
 u(t,0)=u(t,1)=0, \quad  t \in (0,T),\\
 u(0,x)=u_0(x), \quad x\in (0,1),
\end{gathered}
\end{equation}
where the diffusion coefficient $a$ satisfies \eqref{assump} or \eqref{assump2}
in  $x_0\in (0,\,1)$ with $K$ in place of $K_i$, $i=1,2$,
$h\in L^2(Q)$ and $c\in L^\infty(Q)$.
As usual this aim relies on the introduction of some suitable weight functions.
Towards this end, as in \cite{Fra2,fm}, we define the following time and space
weight functions
\begin{gather*}
 \theta(t):=\frac{1}{[(t-t_0)(T-t)]^4},\quad \eta(t):= T+t_0-2t,\\
 \psi(x)=c_1\Big(\int_{x_0}^x\frac{y-x_0}{a(y)}\,dy-c_2\Big),\quad
\varphi(t,x):=\theta(t)\psi(x),
\end{gather*}
where $t_0\in (0,T)$ is given, $c_1>0$ and
$c_2>\max \big\{\frac{(1-x_0)^2}{a(1)(2-K)},\frac{x_0^2}{a(0)(2-K)}\big\}$.
For this choice it is easy to prove that $-c_1c_2 \le \psi(x)<0$ for every
$x\in [0,1]$, and that $\eta$ is positive if $0<t<{\frac{T+t_0}{2}}$ and
negative if  ${\frac{T+t_0}{2}<t<T}$.

Now we are ready to state the Carleman estimate related to \eqref{problem}.

\begin{theorem} \label{thm3.1}
 Assume \eqref{assump} or \eqref{assump2} and let $T>0$.
 Then there exist two positive constants $C$ and $s_0$ such that
the solution $u$ of \eqref{problem} in $H^1\left( [\varepsilon,T],L^2(0,1) \right)
\cap L^2\left(\varepsilon,T;H^2_{a}(0,1)\right)$ satisfies, for all $s\geq s_0$,
\begin{equation} \label{carlinv}
\begin{aligned}
&\int_{Q_{t_0}^T} \Big(s\theta a(x)u_x^2+s^3 \theta^3
\frac{(x-x_0)^2}{a(x)}u^2+s\theta^{3/2}
|\eta \psi|u^2+\frac{1}{s\theta}u_t^2 \Big)e^{2s\varphi}\,dx\,dt  \\
& \leq C\Big( \int_{Q_{t_0}^T} h^2 e^{2s\varphi}\,dx\,dt
+ sc_1\int_{t_0}^T \Big[\theta a(x-x_0)u_x^2e^{2s\varphi}\Big]_{x=0}^{x=1}dt\Big).
\end{aligned}
\end{equation}
\end{theorem}

\begin{proof}
Let $u$ be the solution of \eqref{problem}. For $s>0$,  the
function $w=e^{s\varphi}u$ satisfies
\[
\underbrace{-(aw_x)_x-s\varphi_tw-s^2a\varphi_x^2w}_{L_s^+w}
+\underbrace{w_t+2sa\varphi_xw_x+s(a\varphi_x)_xw}_{L_s^-w}
=\underbrace{he^{s\varphi}-cw}_{h_s}.
\]
Moreover, $w(t_0,x)=w(T,x)=0$. This property allows us to apply the
Carleman estimates established in \cite{fm} to $w$
with $Q_{t_0}^T$ in place of $(0,T)\times (0,1)$
\begin{equation}\label{3.4}
\begin{aligned}
& \|L_s^+w\|^2+\|L_s^-w\|^2+\int_{Q_{t_0}^T}
\Big(s^3 \theta^3\frac{(x-x_0)^2}{a(x)}w^2+s\theta a(x)w_x^2 \Big)\,dx\,dt  \\
& \leq C\Big( \int_{Q_{t_0}^T} h^2 e^{2s\varphi}\,dx\,dt
+ sc_1\int_{t_0}^T \Big[\theta a(x-x_0)w_x^2\Big]_{x=0}^{x=1}dt\Big).
\end{aligned}
\end{equation}
The operators $L_s^+$ and $L_s^-$ are not exactly the ones of \cite{Fra2,fm}.
However, one can prove that the Carleman estimates do not change.
Using the previous estimate we will bound the integral
$ \int_{Q_{t_0}^T}\big(\frac{1}{s\theta}u_t^2
+ s\theta^{3/2}|\eta \psi|u^2\big)e^{2s\varphi}\,dx\,dt$.
In fact, we have
\begin{align*}
&\int_{Q_{t_0}^T} s\theta^{3/2}|\eta \psi|w^2\,dx\,dt\\
&\leq C\int_{Q_{t_0}^T} s\theta^{3/2}w^2\,dx\,dt\\
& = sC \big|\int_{Q_{t_0}^T} \Big( \theta\frac{a^{1/3}}{|x-x_0|^{2/3}}w^2
\Big)^{3/4}\Big( \theta^3\frac{|x-x_0|^2}{a}w^2\Big)^{1/4}\,dx\,dt\big|\\
& \leq sC \frac32 \int_{Q_{t_0}^T} \theta\frac{a^{1/3}}{|x-x_0|^{2/3}}w^2\,dx\,dt
+ s^3C \frac12 \int_{Q_{t_0}^T}  \theta^3\frac{|x-x_0|^2}{a}w^2 \,dx\,dt,
\end{align*}
since $|\eta|\leq T+t_0$ and $|\psi|\leq c_1c_2$.
Now, if $ K \le  4/3$, we consider the function $p(x)=|x-x_0|^{4/3}$.
Obviously, there exists
$ q\in (1,4/3)$ such that the function $ \frac{p(x)}{|x-x_0|^q}$ is non-increasing
on the left of $x_0$ and nondecreasing on the right of $x_0$.
Then, we can apply the Hardy-Poincar\'e
inequality \eqref{hardy},  obtaining
\begin{align*}
 \int_0^1 \frac{a^{1/3}}{|x-x_0|^{2/3}}w^2\,dx
& \leq  \max_{x\in[0,1]}a^{1/3}(x)\int_0^1 \frac{1}{|x-x_0|^{2/3}}w^2\,dx\\
 & = \max_{x\in[0,1]}a^{1/3}(x)\int_0^1 \frac{p(x)}{|x-x_0|^2}w^2\,dx\\
 & \leq \max_{x\in[0,1]}a^{1/3}(x)C_{HP}\int_0^1  p(x)w_x^2\,dx\\
& = \max_{x\in[0,1]}a^{1/3}(x)C_{HP}\int_0^1 \ a \frac{|x-x_0|^{4/3}}{a}w_x^2\,dx\\
& = \max_{x\in[0,1]}a^{1/3}(x)C_{HP}C_1\int_0^1  aw_x^2\,dx,
\end{align*}
where
\[
C_1=\max\Big(\frac{x_0^{4/3}}{a(0)},\frac{(1-x_0)^{4/3}}{a(1)}\Big).
\]
In the previous inequality, we have used the property that the map
$ x \mapsto |x-x_0|^\gamma/a(x)$ is non-increasing on the left of $x_0$
and nondecreasing
on the right of $x_0$ for all $\gamma>K$,  see \cite[Lemma 2.1]{Fra2}.
If $K >4/3$, we can consider the function
$p(x) = (a(x)|x-x_0|^4)^{1/3}$. Then
$ p(x)=  a(x)\Big(\frac{(x-x_0)^2}{a(x)}\Big)^{2/3}\le C_1 a(x)$, where
\[
C_1:=\max\Big\{\Big(\frac{x_0^2}{a(0)}\Big)^{2/3},
\Big(\frac{(1-x_0)^2}{a(1)}\Big)^{2/3}\Big\}, \quad
 \frac{a^{1/3}}{|x-x_0|^{2/3}}=
\frac{p(x)}{(x-x_0)^2}.
\]
Moreover, using hypothesis \eqref{assump2}, one
has that the function $\frac{p(x)}{|x-x_0|^q}$, with
$ q: =\frac{4+\gamma}{3}$ in $(1,2)$, is non-increasing
on the left of $x_0$ and nondecreasing on the right of $x_0$.
The Hardy-Poincar\'{e} inequality implies
\begin{align*} %\label{hpappl}
\int_0^1 \frac{a^{1/3}}{|x-x_0|^{2/3}}w^2dx
&= \int_0^1 \frac{p}{(x-x_0)^2} w^2 dx \le C_{HP}\int_0^1  p (w_x)^2 dx\\
&\le C_{HP}C_1 \int_0^1  a (w_x)^2 dx.
\end{align*}
Thus, in every case,
\begin{equation}\label{step0}
 \int_{Q_{t_0}^T} \theta\frac{a^{1/3}}{|x-x_0|^{2/3}}w^2\,dx\,dt
\leq C\int_{Q_{t_0}^T} \theta a w_x^2\,dx\,dt
\end{equation}
for a positive constant $C$. Then, for $s$ large enough, we have
\begin{gather}\label{tag}
\int_{Q_{t_0}^T} s\theta^{3/2}|\eta \psi|w^2\,dx\,dt
\leq C\int_{Q_{t_0}^T} \big(s\theta a w_x^2+s^3\theta^3
\frac{|x-x_0|^2}{a}w^2\big)\,dx\,dt,
\\
\label{etoil}
\int_{Q_{t_0}^T} s\theta^{3/2}|\eta \psi|w^2\,dx\,dt
\leq C\Big( \int_{Q_{t_0}^T} h^2 e^{2s\varphi}\,dx\,dt
+ sc_1\int_{t_0}^T \Big[\theta a(x-x_0)w_x^2\Big]_{x=0}^{x=1}dt\Big).
\end{gather}
On the other hand, we have
\[
 \frac{1}{\sqrt{s\theta}}L_s^-w = \frac{1}{\sqrt{s\theta}}w_t+2c_1\sqrt{s\theta}(x-x_0)w_x+c_1\sqrt{s\theta}w.
\]
Therefore,
\begin{equation} \label{dis}
\begin{aligned}
 \int_{Q_{t_0}^T} \frac{1}{s\theta}w_t^2\,dx\,dt
& \leq C\Big( \|L_s^-w\|^2+ \int_{Q_{t_0}^T} s\theta
 \frac{|x-x_0|^2}{a} aw_x^2\,dx\,dt+\int_{Q_{t_0}^T} s\theta w^2\,dx\,dt\Big)\\
& \leq C\Big( \|L_s^- w\|^2+\int_{Q_{t_0}^T} s\theta a w_x^2\,dx\,dt
 +\int_{Q_{t_0}^T} s\theta w^2\,dx\,dt\Big),
\end{aligned}
\end{equation}
since  $1/\sqrt{\theta}$ is bounded and
\[
\frac{|x-x_0|^2}{a(x)}\leq \text{max}\Big\{\frac{x_0^2}{a(0)},
\frac{(1-x_0)^2}{a(1)}\Big\}
\]
(see \cite[Lemma 2.1]{Fra2}).
Proceeding as in the proof of  \eqref{step0}, we can estimate
$\int_{Q_{t_0}^T} s\theta w^2\,dx\,dt$ thanks to the Hardy-Poincar\'e inequality
\eqref{hardy},
\begin{align*}
\int_{Q_{t_0}^T} s\theta w^2\,dx\,dt
& = s \Big|\int_{Q_{t_0}^T} \Big( \theta\frac{a^{1/3}}{|x-x_0|^{2/3}}w^2
\Big)^{3/4}\Big( \theta\frac{|x-x_0|^2}{a}w^2\Big)^{1/4}\,dx\,dt\Big|\\
& \leq s \frac32 \int_{Q_{t_0}^T} \theta\frac{a^{1/3}}{|x-x_0|^{2/3}}w^2\,dx\,dt
 + \frac{s}{2} \int_{Q_{t_0}^T}  \theta\frac{|x-x_0|^2}{a}w^2 \,dx\,dt\\
&\leq C \int_{Q_{t_0}^T} \Big(s\theta aw_x^2+s^3\theta^3\frac{(x-x_0)^2}{a}w^2
\Big)dx\,dt.
\end{align*}
Hence, taking $s$ large enough, one has
\begin{equation}\label{w_t}
 \int_{Q_{t_0}^T} \frac{1}{s\theta} w_t^2\,dx\,dt
\leq C\Big( \int_{Q_{t_0}^T} h^2 e^{2s\varphi}\,dx\,dt
+ sc_1\int_{t_0}^T \Big[\theta a(x-x_0)w_x^2\Big]_{x=0}^{x=1}dt\Big).
\end{equation}
Now from \eqref{etoil} and \eqref{w_t} we can get the estimate of $u_t$ as follows:
from the definition of $w$, we have $w_t=u_te^{s\varphi}+s\varphi_tw$. Hence
\begin{equation*}
 \int_{Q_{t_0}^T} \frac{1}{s\theta}u_t^2e^{2s\varphi}\,dx\,dt
\leq 2\Big(\int_{Q_{t_0}^T} \frac{1}{s\theta} w_t^2\,dx\,dt
+\int_{Q_{t_0}^T} \frac{s^2\varphi_t^2}{s\theta} w^2\,dx\,dt\Big).
\end{equation*}
The second term in the above right-hand side is estimated as follows:
\begin{align*}
\int_{Q_{t_0}^T} \frac{s^2\varphi_t^2}{s\theta} w^2\,dx\,dt
&=16\int_{Q_{t_0}^T} s\theta^{3/2}\eta^2\psi^2 w^2\,dx\,dt\\
&\leq 16(T+t_0)c_1c_2 \int_{Q_{t_0}^T} s\theta^{3/2}|\eta\psi| w^2\,dx\,dt.
\end{align*}
Hence using \eqref{etoil} and \eqref{w_t} we conclude that
\begin{equation}\label{3.12}
\int_{Q_{t_0}^T} \frac{1}{s\theta}u_t^2e^{2s\varphi}\,dx\,dt
\leq C\Big( \int_{Q_{t_0}^T} h^2 e^{2s\varphi}\,dx\,dt
+ sc_1\int_{t_0}^T \Big[\theta a(x-x_0)w_x^2\Big]_{x=0}^{x=1}dt\Big).
\end{equation}
Thus \eqref{carlinv} follows by \eqref{3.4}, \eqref{step0} and \eqref{3.12}.
\end{proof}

\subsection{Carleman estimate for systems}

By the above Carleman estimate \eqref{carlinv}, we are able to show the main
result of this section,
which is the $\omega$-Carleman estimate for the system \eqref{system}.
For $x\in[0,1]$, let us define
\[
\varphi_i(t,x):=\theta(t)\psi_i(x),\quad
\theta(t):=\frac{1}{[(t-t_0)(T-t)]^4},\quad
\psi_i(x)=c_i[\int_{x_i}^x\frac{y-x_i}{a_i(y)}\,dy-d_i],
\]
and, for $x\in [-1,1]$,
$$
\Phi_i(t,x):=\theta(t)\Psi_i(x),\quad \Psi_i(x)=e^{2\rho_i}-e^{r_i\zeta_i(x)},
$$
where
\[
\zeta_i(x)=\int_x^1\frac{dy}{\sqrt{\tilde{a}_i(y)}},\quad
\rho_i=r_i\zeta_i(-1), \quad
\tilde{a}_i(x):= \begin{cases}
a_i(x), & x\in [0,1],\\
a_i(-x), & x\in [-1,0].
\end{cases}
\]
Here the functions $a_i$, $i=1,2$, satisfy hypothesis \eqref{assump}
or \eqref{assump2} and the positive constants $c_i,\,d_i,\text{ and }r_i$
are chosen such that
\begin{gather}\label{choice}
d_2>\frac{16A}{16A-15}\max\Big\{\frac{x_2^2}{(2-K_2)a_2(0)},
\frac{(1-x_2)^2}{(2-K_2)a_2(1)},d_2^\star\Big\},\quad \frac{15}{16}<A<1
\\
\begin{gathered}
d_1>\max\Big\{\frac{x_1^2}{(2-K_1)a_1(0)},\frac{(1-x_1)^2}{(2-K_1)a_1(1)}\Big\},\\
\rho_2>2\ln(2),\quad e^{2\rho_1}-e^{r_1\zeta_1(0)}\geq e^{2\rho_2}-1,
\end{gathered}\\
\begin{aligned}
&\max\Big\{\frac{e^{2\rho_2}-1}{d_2-d_2^\star},
 \frac{(2-K_2)a_2(1)(e^{2\rho_2}-1)}{(2-K_2)a_2(1)d_2-(1-x_2)^2},
 \frac{(2-K_2)a_2(0)(e^{2\rho_2}-1)}{(2-K_2)a_2(0)d_2-x_2^2}\Big\}\\
&\leq c_2 < \frac{4A}{3d_2}(e^{2\rho_2}-e^{r_2\zeta_{2}(0)})
\end{aligned} \\
\label{choice2}
\begin{aligned}
c_1\geq \max\Big\{&\frac{e^{2\rho_1}-1}{d_1-d_1^\star},
\frac{(2-K_1)a_1(1)(e^{2\rho_1}-1)}{(2-K_1)a_1(1)d_1-(1-x_1)^2},\\
&\frac{(2-K_1)a_1(0)(e^{2\rho_1}-1)}{(2-K_1)a_1(0)d_1-x_1^2},
 \frac{c_2d_2}{d_1-d_1^\star}\Big\},
\end{aligned}
\end{gather}
where
\[
A={\frac{\min \Psi_2(-x)}{\max \Psi_2(x)}},\quad
d_i^\star:=\max\Big\{\int_{x_i}^1\frac{y-x_i}{a_i(y)}\,dy,
\int_{x_i}^0\frac{y-x_i}{a_i(y)}\,dy\Big\}.
\]

\begin{remark} \label{rmk3.2} \rm
The interval
\begin{align*}
\Big[&\max\Big\{\frac{e^{2\rho_2}-1}{d_2-d_2^\star},
 \frac{(2-K_2)a_2(1)(e^{2\rho_2}-1)}{(2-K_2)a_2(1)d_2-(1-x_2)^2},
\frac{(2-K_2)a_2(0)(e^{2\rho_2}-1)}{(2-K_2)a_2(0)d_2-x_2^2}\Big\},\\
 &\frac{4A(e^{2\rho_2}-e^{r_2\zeta(0)})}{3d_2}\Big]
\end{align*}
is not empty. In fact, from $\rho_2>2\ln2$, $A>15/16$ and
$d_2> 16A d_2^\star /(16A-15)$, we have
\begin{align*}
\frac{d_2^\star}{d_2}<1-\frac{15}{16A}
&\Leftrightarrow \frac{5}{4}\leq \frac{4A}{3}(1-\frac{d_2^\star}{d_2})\\
& \Leftrightarrow 1+e^{-\rho_2}< \frac{4A}{3}(1-\frac{d_2^\star}{d_2})\\
& \Leftrightarrow \frac{e^{2\rho_2}-1}{d_2-d_2^\star}<\frac{4A}{3d_2}
(e^{2\rho_2}-e^{\rho_2})<\frac{4A}{3d_2}(e^{2\rho_2}-e^{r_2\zeta_2(0)}).
\end{align*}
Similarly for
\[
d_2>\frac{16A}{16A-15}\max\Big\{\frac{x_2^2}{(2-K_2)a_2(0)},
\frac{(1-x_2)^2}{(2-K_2)a_2(1)}\Big\}
\]
 one has
\begin{align*}
&\max\Big\{\frac{(2-K_2)a_2(1)(e^{2\rho_1}-1)}{(2-K_2)a_2(1)d_2-(1-x_2)^2},
\frac{(2-K_2)a_2(0)(e^{2\rho_1}-1)}{(2-K_2)a_2(0)d_2-x_2^2}\Big\}\\
&<\frac{4A}{3d_2}(e^{2\rho_2}-e^{r_2\zeta_2(0)}).
\end{align*}
\end{remark}

From \eqref{choice}-\eqref{choice2}, we have the following results.

\begin{lemma}\label{fcweiineq}
(i) For $(t,x)\in [0,T]\times[0,1]$,
\begin{equation} \label{3.24}
\varphi_1\leq\varphi_2,\quad -\Phi_1\leq -\Phi_2,\quad \varphi_i\leq-\Phi_i.
\end{equation}
(ii) For $(t,x)\in [0,T]\times[0,1]$,
\begin{equation} \label{3.25}
-\Phi_2(t,x)\leq -\Phi_2(t,-x),\quad 4\Phi_2(t,-x)+3\varphi_2(t,x)>0.
\end{equation}
\end{lemma}

\begin{proof}
 (i)
 \begin{enumerate}
  \item $\varphi_1\leq\varphi_2:$  since $\theta\geq 0$ it is sufficient
to prove $\psi_1\leq\psi_2$.
 By the choice of $c_1$, we have $c_1\geq \frac{c_2d_2}{d_1-d_1^\star}$.
Then $\max\{\psi_1(0),\psi_1(1)\}\leq -c_2d_2$.
  Hence, $\psi_1(x)\leq\psi_2(x)$.

\item $-\Phi_1\leq-\Phi_2:$\\ since $\Psi_i$ is increasing, it is sufficient
to prove that $\min \Psi_1(x)\geq \max \Psi_2(x)$.
  Indeed $\Psi_1(0)=e^{2\rho_1}- e^{r_1\zeta_1(0)}\geq e^{2\rho_2}-1=\Psi_2(1)$.

 \item $\varphi_i\leq-\Phi_i:$
  since $c_i\geq \frac{e^{2\rho_i}-1}{d_i-d_i^\star}$, then
$\max\{\psi_i(0),\psi_i(1)\}\leq -\Psi_i(1)$ and the thesis follows immediately.
  \end{enumerate}

(ii)  \begin{enumerate}
  \item The first inequality follows from $-\Psi_2(x)\leq -\Psi_2(-x)$
for all $x\in [0,1]$.

  \item $4\Psi_2(-x)+3\psi_2(x)>0:$
  by definition of $A$, we have $A\Psi_2(x)\leq \Psi_2(-x)$ and, obviously,
$4\Psi_2(-x)+3\psi_2(x)\geq 4A\Psi_2(x)+3\psi_2(x)$. Thus, to obtain the thesis,
it is sufficient to prove that $4A\Psi_2(x)+3\psi_2(x)>0$. This follows easily
observing that,  by the assumption $3c_2d_2<4A\Psi_2(0)$,
$-3\psi_2(x_0)< 4A\Psi_2(0)$. Hence
 $-3\psi_2(x)\le-3\psi_2(x_0)< 4A\Psi_2(0) \le 4A\Psi_2(x)$ for all $x \in [0,1]$.
 \end{enumerate}
\end{proof}

We show first an intermediate Carleman estimate with  distributed observation of
$u$ and $v$.

\begin{theorem} \label{theofun}
Let $T>0$. There exist two positive constants $C$ and $s_0$ such that,
for every $(u_0,v_0)\in \mathbb{H}$ and all
$s\geq s_0$, the solution of \eqref{system} satisfies
\begin{align*}
&\int_{Q_{t_0}^T} \Big(s\theta a_1u_x^2+s^3 \theta^3\frac{(x-x_1)^2}{a_1}u^2
 +s\theta^{3/2}|\eta \psi_1|u^2+\frac{1}{s\theta}u_t^2\Big)e^{2s\varphi_1}\,dx\,dt\\
&+\int_{Q_{t_0}^T} \Big(s\theta a_2v_x^2+s^3 \theta^3\frac{(x-x_2)^2}{a_2}v^2
 +s\theta^{3/2}|\eta \psi_2|v^2+\frac{1}{s\theta}v_t^2\Big)e^{2s\varphi_2}\,dx\,dt\\
& \leq C\Big(\int_{Q_{t_0}^T}f^2e^{-2s\Phi_2(t,-x)}dx\,dt
+\int_{t_0}^T\int_{\omega'}s^2\theta^2(u^2+v^2)e^{-2s\Phi_2(t,-x)}dx\,dt\Big).
\end{align*}
\end{theorem}

For the proof we shall  use the following classical Carleman estimate
(see \cite{Fra2}).

\begin{proposition}
 Let $z$ be the solution of
\begin{gather*} %\label{sysclassic}
 z_t-(az_x)_x=h, \quad x\in (A,B),\; t\in (0,T),\\
 z(t,A)=z(t,B)=0, \quad  t \in (0,T),
\end{gather*}
where $a\in C^1([A,B])$ is a strictly positive function.
Then there exist two positive constants $r$ and $s_0$
such that for any $s\geq s_0$,
\begin{align}\label{carclassic}
& \int_{t_0}^T\int_A^Bs\theta e^{r\zeta}z_x^2e^{-2s\Phi}\,dx\,dt
+\int_{t_0}^T\int_A^Bs^3\theta^3 e^{3r\zeta}z^2e^{-2s\Phi}\,dx\,dt\\
& \leq c\Big(\int_{t_0}^T\int_A^Bh^2e^{-2s\Phi}\,dx\,dt
-c\int_{t_0}^T\big[ \sigma(t,\cdot)z_x^2(t,\cdot)e^{-2s\Phi(t,\cdot)}\big]_{x=0}^{x=1}
\,dt\Big)
\end{align}
for some positive constant c. Here the functions $\Phi$, $\sigma$ and
$\zeta$ are defined, for $r,s>0$  and $(t,x)\in [0,T]\times[A,B]$, by
\begin{gather*}
\phi(t,x):=\theta(t)\Psi(x),\quad
\Psi(x):=e^{2r\zeta(A)}-e^{r\zeta(x)}>0, \\
\zeta(x):=\int_x^B\frac{1}{\sqrt{a(y)}}\,dy,\quad
\sigma(t,x):=rs\theta(t)e^{r\zeta(x)}.
\end{gather*}
\end{proposition}

\begin{proof}[Proof of Theorem \ref{theofun}]
Consider a cut-off function $\xi:[0,1]\to R$ such that
\begin{gather*}
 0\leq \xi(x)\leq1,\quad  \text{for all } x\in[0,1],\\
 \xi(x)=1,\quad   x\in[\lambda_1,\beta_2],\\
 \xi(x)=0, \quad  x\in [0,1]\setminus\omega.
\end{gather*}
Define $w:=\xi u$ and  $z:=\xi v$,  where $(u,v)$ is the solution of \eqref{system}.
Hence, $w$ and $z$ satisfy the system
\begin{gather*}
 w_t-(a_1w_x)_x+b_{11}w=\xi f-b_{12}z-(a_1\xi_x u)_x-\xi_xa_1u_x=:g, \quad
 (t,x) \in Q,\\
z_t-(a_2z_x)_x+b_{22}z=-(a_2\xi_x v)_x-\xi_xa_2v_x, \quad (t,x) \in Q,\\
 w(t,0)=w(t,1)=z(t,0)=z(t,1)=0, \quad t \in (0,T).
\end{gather*}
Applying the estimate \eqref{carlinv} and using $w=w_x=0$ in a neighborhood
of $x=0$ and $x=1$, from the definition of $\xi$, we have
\begin{align*}
&\int_{Q_{t_0}^T} \Big(s\theta a_1w_x^2+s^3 \theta^3\frac{(x-x_1)^2}{a_1}w^2
+s\theta^{3/2}|\eta \psi_1|w^2+\frac{1}{s\theta}w_t^2
 \Big)e^{2s\varphi_1}\,dx\,dt \\
&\leq C \int_{{Q_{t_0}^T}} g^2 e^{2s\varphi_1}\,dx\,dt
\end{align*}
for all $s\geq s_0$. Then using the fact that $\xi_x$ and $\xi_{xx}$
are supported in $\omega''$, we can write
$$
g^2\leq C(\xi^2f^2+b_{12}^2z^2+(u^2+u_x^2)\chi_{\omega''}).
$$
Hence, applying Cacciopoli inequality \eqref{cacciopoli} and the previous estimates,
 we obtain
\begin{equation} \label{estimw}
\begin{aligned}
&\int_{{Q_{t_0}^T}} \Big(s\theta a_1w_x^2+s^3 \theta^3\frac{(x-x_1)^2}{a_1}w^2
+s\theta^{3/2}|\eta \psi_1|w^2+\frac{1}{s\theta}w_t^2
 \Big)e^{2s\varphi_1}\,dx\,dt\\
& \leq C\Big(\int_{Q_{t_0}^T}\xi^2f^2e^{2s\varphi_1}
\,dx\,dt+\int_{Q_{t_0}^T}b_{12}^2z^2e^{2s
\varphi_1}\,dx\,dt\\
&\quad +\int_{t_0}^T\int_{\omega'} ((1+s^2\theta^2)u^2+f^2)e^{2s\varphi_1}\,
dx\,dt\Big).
\end{aligned}
\end{equation}
Arguing as before and applying the estimate \eqref{carlinv} to the second
component $z$ of the system, we obtain
\begin{align*}
&\int_{Q_{t_0}^T} \Big(s\theta a_2z_x^2+s^3 \theta^3
\frac{(x-x_2)^2}{a_2}z^2+s\theta^{3/2}|\eta \psi_2|z^2
+\frac{1}{s\theta}z_t^2\Big)e^{2s\varphi_2}\,dx\,dt\\
&\leq C\int_{t_0}^T\int_{\omega''} (v^2+v_x^2)e^{2s\varphi_2}\,dx\,dt.
\end{align*}
Hence,  Cacciopoli inequality \eqref{cacciopoli} yields
\begin{equation} \label{estimz}
\begin{aligned}
&\int_{Q_{t_0}^T} \Big(s\theta a_2z_x^2+s^3 \theta^3
\frac{(x-x_2)^2}{a_2}z^2+s\theta^{3/2}|\eta \psi_2|z^2
+\frac{1}{s\theta}z_t^2\Big)e^{2s\varphi_2}\,dx\,dt \\
&\leq C\int_{t_0}^T\int_{\omega'} (1+s^2\theta^2)v^2e^{2s\varphi_2}\,dx\,dt.
\end{aligned}
\end{equation}
On the other hand, using the Poincar\'e inequality and the fact that
$\varphi_1<\varphi_2$, we have
\begin{align*}%\label{new}
 \int_0^1b_{12}^2z^2e^{2s\varphi_1}dx
& \leq\|b_{12}\|_\infty^2\int_0^1(ze^{s\varphi_2})^2dx\\
&=\|b_{12}\|_\infty^2 \int_0^1\Big(\frac{|x-x_2|^2}{a_2(x)}z^2e^{2s\varphi_2}
 \Big)^{1/4}\Big(\frac{a_2^{1/3}(x)}{|x-x_2|^{2/3}}
z^2e^{2s\varphi_2}\Big)^{3/4}dx\\
&\le  \frac{\|b_{12}\|_\infty^2 }{4}\int_0^1\frac{|x-x_2|^2}{a_2(x)}z^2
e^{2s\varphi_2} dx \\
&\quad + \frac{3\|b_{12}\|_\infty^2 }{4}
\int_0^1\frac{a_2^{1/3}(x)}{|x-x_2|^{2/3}}z^2e^{2s\varphi_2}dx.
\end{align*}
Applying the Hardy-Poincar\'{e} inequality to $w(t,x):= e^{s\varphi(t,x)} z(t,x)$
and proceeding as in the proof of \eqref{step0}, one has
\begin{align*}
\int_0^1\frac{a_2^{1/3}(x)}{|x-x_2|^{2/3}}z^2e^{2s\varphi_2}dx
&= \int_0^1\frac{a_2^{1/3}(x)}{|x-x_2|^{2/3}}w^2dx\le C\int_0^1a(w_x)^2dx \\
&\le C\int_0^1 s^2\theta^2\frac{|x-x_2|^2}{a_2(x)}z^2e^{2s\varphi}dx
+ C\int_0^1a_2z_x^2e^{2s\varphi_2}dx,
\end{align*}
since $\psi_{2,x}(x)=c_2\frac{x-x_2}{a_2(x)}$. Hence, for a universal
 positive constant $C$, it results
\[
\int_{Q_{t_0}^T}b_{12}^2z^2e^{2s\varphi_1}dx\,dt
\leq C\int_{Q_{t_0}^T}(a_2z_x^2+s^2\theta^2\frac{(x-x_2)^2}{a_2}z^2)
e^{2s\varphi_2}dx\,dt.
\]
Taking $s$ such that $C \le \frac{s\theta}{2}$, we have
\begin{align}\label{absorbz}
\int_{Q_{t_0}^T}b_{12}^2z^2e^{2s\varphi_1}dx\,dt\leq\frac12\int_{Q_{t_0}^T}
 \Big(s\theta a_2z_x^2+s^3 \theta^3\frac{(x-x_2)^2}{a_2}z^2\Big)
e^{2s\varphi_2}\,dx\,dt.
\end{align}
Combining \eqref{estimw}, \eqref{estimz} and \eqref{absorbz} we obtain for
 $s$ large enough
\begin{align*}
&\int_{Q_{t_0}^T} \Big(s\theta a_1w_x^2+s^3 \theta^3\frac{(x-x_1)^2}{a_1}w^2
+s\theta^{3/2}|\eta \psi_1|w^2+\frac{1}{s\theta}w_t^2\Big)e^{2s\varphi_1}\,dx\,dt\\
&+\int_{Q_{t_0}^T} \Big(s\theta a_2z_x^2+s^3 \theta^3\frac{(x-x_2)^2}{a_2}z^2
+s\theta^{3/2}|\eta \psi_2|z^2+\frac{1}{s\theta}z_t^2\Big)e^{2s\varphi_2}\,dx\,dt\\
& \leq C\Big(\int_{Q_{t_0}^T}\xi^2f^2e^{2s\varphi_1}
\,dx\,dt+\int_{t_0}^T\int_{\omega'} s^2\theta^2v^2e^{2s\varphi_2}\,dx\,dt\\
&\quad +\int_{t_0}^T\int_{\omega'} (s^2\theta^2u^2+f^2)e^{2s\varphi_1}\,dx\,dt\Big).
\end{align*}
By the previous inequality, the definition of $w$ and $z$, it follows that
\begin{align}
&\int_{t_0}^T\int_{\lambda_1}^{\beta_1} \Big(s\theta a_1u_x^2
 +s^3 \theta^3\frac{(x-x_1)^2}{a_1}u^2+s\theta^{3/2}|\eta \psi_1|u^2
 +\frac{1}{s\theta}u_t^2\Big)e^{2s\varphi_1}\,dx\,dt \nonumber\\
&+\int_{t_0}^T\int_{\lambda_1}^{\beta_1} \Big(s\theta a_2v_x^2+s^3
 \theta^3\frac{(x-x_2)^2}{a_2}v^2+s\theta^{3/2}|\eta \psi_2|v^2
 +\frac{1}{s\theta}v_t^2\Big)e^{2s\varphi_2}\,dx\,dt \nonumber\\
&=\int_{t_0}^T\int_{\lambda_1}^{\beta_1} \Big(s\theta a_1w_x^2
 +s^3 \theta^3\frac{(x-x_1)^2}{a_1}w^2+s\theta^{3/2}|\eta \psi_1|w^2
 +\frac{1}{s\theta}w_t^2\Big)e^{2s\varphi_1}\,dx\,dt \nonumber\\
&\quad +\int_{t_0}^T\int_{\lambda_1}^{\beta_1} \Big(s\theta a_2z_x^2
 +s^3 \theta^3\frac{(x-x_2)^2}{a_2}z^2+s\theta^{3/2}|\eta \psi_2|z^2
 +\frac{1}{s\theta}z_t^2\Big)e^{2s\varphi_2}\,dx\,dt\\
&\leq\int_{Q_{t_0}^T} \Big(s\theta a_1w_x^2+s^3 \theta^3\frac{(x-x_1)^2}{a_1}w^2
 +s\theta^{3/2}|\eta \psi_1|w^2+\frac{1}{s\theta}w_t^2\Big)e^{2s\varphi_1}\,dx\,dt
\nonumber\\
&\quad+\int_{Q_{t_0}^T} \Big(s\theta a_2z_x^2+s^3 \theta^3\frac{(x-x_2)^2}{a_2}z^2
 +s\theta^{3/2}|\eta \psi_2|z^2+\frac{1}{s\theta}z_t^2\Big)e^{2s\varphi_2}\,dx\,dt
\nonumber\\
& \leq C\Big(\int_{Q_{t_0}^T}\xi^2f^2e^{2s\varphi_1}
\,dx\,dt+\int_{t_0}^T\int_{\omega'} s^2\theta^2v^2e^{2s\varphi_2}\,dx\,dt \nonumber\\
&\quad +\int_{t_0}^T\int_{\omega'} (s^2\theta^2u^2+f^2)e^{2s\varphi_1}\,dx\,dt\Big).
 \label{I}
\end{align}
Now define $U=\chi u$ and $V=\chi v$,  where $(u,v)$ is the solution of \eqref{system}
and $\chi:\,[0,1]\to \mathbb{R}$ is a cut-off function defined as
\begin{gather*}
 0\leq \chi(x)\leq1,\quad  x\in[0,1],\\
 \chi(x)=1, \quad x\in[\beta_2,1],\\
 \chi(x)=0, \quad  x\in [0,\frac{\lambda_2+2\beta_2}{3}].
\end{gather*}
Then $U$ and $V$ satisfy
 \begin{gather*}
 U_t-(a_1U_x)_x+b_{11}U=\chi f-b_{12}V-(a_1\chi_x u)_x-\chi_xa_1u_x, \quad (t,x) \in Q\\
V_t-(a_2V_x)_x+b_{22}V=-(a_2\chi_x v)_x-\chi_xa_2v_x, \quad (t,x) \in Q\\
 U(t,0)=U(t,1)=V(t,0)=V(t,1)=0, \quad t \in (0,T).
\end{gather*}
Using \eqref{carclassic} and a technique similar to the one used in
\eqref{dis}-\eqref{w_t}, one has
\begin{align*}
& \int_{Q_{t_0}^T}(s\theta e^{r_1\zeta_1}U_x^2+s^3\theta^3e^{3r_1\zeta_1}U^2
+\frac{1}{s\theta}U_t^2)e^{-2s\Phi_1}\,dx\,dt\\
& \leq C\int_{Q_{t_0}^T}\chi^2f^2e^{-2s\Phi_1}dx\,dt
+\tilde{C}\int_{Q_{t_0}^T}V^2e^{-2s\Phi_1}dx\,dt\\
&\quad +C\int_{t_0}^T\int_{\overline{\overline{\omega}}}(u^2+u_x^2)e^{-2s\Phi_1}
\,dx\,dt.
\end{align*}
Analogously, one can prove that $V$ satisfies
\begin{align*}
& \int_{Q_{t_0}^T}(s\theta e^{r_2\zeta_2}V_x^2+s^3\theta^3e^{3r_2\zeta_2}V^2
+\frac{1}{s\theta}V_t^2)e^{-2s\Phi_2}\,dx\,dt\\
&\leq C\int_{t_0}^T\int_{\overline{\overline{\omega}}}(v^2+v_x^2)e^{-2s\Phi_2}dx\,dt.
\end{align*}
Thus combining the last two inequalities,
\begin{align*}
& \int_{Q_{t_0}^T}(s\theta e^{r_1\zeta_1}U_x^2+s^3\theta^3e^{3r_1\zeta_1}U^2
+\frac{1}{s\theta}U_t^2)e^{-2s\Phi_1}\,dx\,dt\\
&+\int_{Q_{t_0}^T}(s\theta e^{r_2\zeta_2}V_x^2+s^3\theta^3e^{3r_2\zeta_2}V^2
+\frac{1}{s\theta}V_t^2)e^{-2s\Phi_2}\,dx\,dt\\
& \leq C\int_{Q_{t_0}^T}\chi^2f^2e^{-2s\Phi_1}dx\,dt
 +\tilde{C}\int_{Q_{t_0}^T}V^2e^{-2s\Phi_1}dx\,dt\\
&\quad +C\int_{t_0}^T\int_{\overline{\overline{\omega}}}(u^2+u_x^2)e^{-2s\Phi_1}dx\,dt
+C\int_{t_0}^T\int_{\overline{\overline{\omega}}}(v^2+v_x^2)e^{-2s\Phi_2}dx\,dt.
\end{align*}
Taking $s$  such that $\tilde{C}\leq \frac12 s^3\theta^3e^{3r_2\zeta_2}$,
using $-\Phi_1\leq-\Phi_2$ and Cacciopoli inequality \eqref{cacciopoli},  we obtain
\begin{align*}
& \int_{Q_{t_0}^T}(s\theta e^{r_1\zeta_1}U_x^2+s^3\theta^3e^{3r_1\zeta_1}U^2
+\frac{1}{s\theta}U_t^2)e^{-2s\Phi_1}\,dx\,dt\\
&+\int_{Q_{t_0}^T}(s\theta e^{r_2\zeta_2}V_x^2
 +\frac{1}{2}s^3\theta^3e^{3r_2\zeta_2}V^2+\frac{1}{s\theta}V_t^2)e^{-2s\Phi_2}
 \,dx\,dt\\
& \leq C\int_{Q_{t_0}^T}\chi^2f^2e^{-2s\Phi_1}dx\,dt+
C\int_{t_0}^T\int_{\overline{\omega}}(s^2\theta^2u^2+f)e^{-2s\Phi_1}dx\,dt\\
&\quad +C\int_{t_0}^T\int_{\overline{\omega}}s^2\theta^2v^2e^{-2s\Phi_2}dx\,dt.
\end{align*}
Then,  by Lemma \ref{fcweiineq},  one can prove that there exists a positive
constant $C$ such that for every $(t,x)\in [0,T]\times \operatorname{supp}(\chi)$,
\begin{align}\label{3.44}
 a_i(x)e^{2s\varphi_i(t,x)}\leq Ce^{r_i\zeta_i}e^{-2s\Phi_i}, \quad
\frac{(x-x_i)^2}{a_i(x)}e^{2s\varphi_i(t,x)}
\leq \frac{C}{2}e^{3r_i\zeta_i}e^{-2s\Phi_i}.
\end{align}
Consequently,
\begin{align*}
&\int_{Q_{t_0}^T} \Big(s\theta a_1U_x^2+s^3 \theta^3\frac{(x-x_1)^2}{a_1}U^2
 +s\theta^{3/2}|\eta \psi_1|U^2+\frac{1}{s\theta}U_t^2\Big)e^{2s\varphi_1}\,dx\,dt\\
&+\int_{Q_{t_0}^T} \Big(s\theta a_2V_x^2+s^3 \theta^3\frac{(x-x_2)^2}{a_2}V^2
+s\theta^{3/2}|\eta \psi_2|V^2+\frac{1}{s\theta}V_t^2\Big)e^{2s\varphi_2}\,dx\,dt\\
& \leq C\Big(\int_{Q_{t_0}^T}\chi^2f^2e^{-2s\Phi_1}
dx\,dt+\int_{t_0}^T\int_{\overline{\omega}}(s^2\theta^2u^2+f)e^{-2s\Phi_1}dx\,dt\\
&\quad +\int_{t_0}^T\int_{\overline{\omega}}s^2
\theta^2v^2e^{-2s\Phi_2}dx\,dt\Big).
\end{align*}
By the definitions of $U$ and $V$, we obtain
 \begin{align}
&\int_{t_0}^T\int_{\beta_2}^1 \Big(s\theta a_1u_x^2
 +s^3 \theta^3\frac{(x-x_1)^2}{a_1}u^2+s\theta^{3/2}|\eta \psi_1|u^2
 +\frac{1}{s\theta}u_t^2\Big)e^{2s\varphi_1}\,dx\,dt \nonumber\\
&+\int_{t_0}^T\int_{\beta_2}^1 \Big(s\theta a_2v_x^2
 +s^3 \theta^3\frac{(x-x_2)^2}{a_2}v^2+s\theta^{3/2}|\eta \psi_2|v^2
 +\frac{1}{s\theta}v_t^2\Big)e^{2s\varphi_2}\,dx\,dt \nonumber\\
&=\int_{t_0}^T\int_{\beta_2}^1 \Big(s\theta a_1U_x^2
 +s^3 \theta^3\frac{(x-x_1)^2}{a_1}U^2+s\theta^{3/2}|\eta \psi_1|U^2
 +\frac{1}{s\theta}U_t^2\Big)e^{2s\varphi_1}\,dx\,dt \nonumber\\
&\quad+ \int_{t_0}^T\int_{\beta_2}^1 \Big(s\theta a_2V_x^2
 +s^3 \theta^3\frac{(x-x_2)^2}{a_2}V^2+s\theta^{3/2}|\eta \psi_2|V^2
 +\frac{1}{s\theta}V_t^2\Big)e^{2s\varphi_2}\,dx\,dt \nonumber\\
&\leq\int_{Q_{t_0}^T} \Big(s\theta a_1U_x^2+s^3 \theta^3\frac{(x-x_1)^2}{a_1}U^2
 +s\theta^{3/2}|\eta \psi_1|U^2+\frac{1}{s\theta}U_t^2\Big)e^{2s\varphi_1}\,dx\,dt
 \nonumber\\
&\quad +\int_{Q_{t_0}^T} \Big(s\theta a_2V_x^2+s^3 \theta^3\frac{(x-x_2)^2}{a_2}V^2
 +s\theta^{3/2}|\eta \psi_2|V^2+\frac{1}{s\theta}V_t^2\Big)e^{2s\varphi_2}\,dx\,dt
 \nonumber\\
& \leq C\Big(\int_{Q_{t_0}^T}\chi^2f^2e^{-2s\Phi_1}dx\,dt
+\int_{t_0}^T\int_{\overline{\omega}}(s^2\theta^2u^2+f)e^{-2s\Phi_1}dx\,dt \nonumber\\
&\quad +\int_{t_0}^T\int_{\overline{\omega}}s^2
\theta^2v^2e^{-2s\Phi_2}dx\,dt\Big). \label{II}
\end{align}
To complete the proof it is sufficient to prove a similar inequality on the
interval $[0,\lambda_1]$. To this aim define the functions
$$
W(t,x):=\begin{cases}
u(t,x), & x\in [0,1],\\
-u(t,-x), & x\in [-1,0],
\end{cases},\quad
Z(t,x):= \begin{cases}
v(t,x), & x\in [0,1],\\
-v(t,-x), & x\in [-1,0],
\end{cases}
$$
where $(u,v)$ solves \eqref{system}, and
$$
\tilde{f}(t,x):=\begin{cases}
f(t,x), & x\in [0,1],\\
-f(t,-x), & x\in [-1,0],
\end{cases},\quad
\tilde{b}_{ij}(x):=
\begin{cases}
b_{ij}(x), & x\in [0,1],\\
b_{ij}(-x), & x\in [-1,0].
\end{cases}
$$
Therefore,  $(W,Z)$ solves
\begin{equation} \label{sysWZ}
 \begin{gathered}
 W_t-(\tilde{a}_1W_x)_x+\tilde{b}_{11}W=\tilde{f}-\tilde{b}_{12}Z,
 \quad x\in (-1,1),\; t\in (0,T),\\
Z_t-(\tilde{a}_2Z_x)_x+\tilde{b}_{22}Z=0, \quad x\in (-1,1),\; t\in (0,T),\\
 W(t,-1)=W(t,1)=Z(t,-1)=Z(t,1)=0, \quad t \in (0,T).
\end{gathered}
\end{equation}
Consider a cut-off function $\rho:[-1,1]\to \mathbb{R}$ such that
 \begin{gather*}
 0\leq \rho(x)\leq1, \quad x\in[-1,1],\\
 \rho(x)=1,  \quad x\in[-\lambda_1,\lambda_1],\\
 \rho(x)=0,  \quad x\in [-1,-\frac{\lambda_1+2\beta_1}{3}]
\cup[\frac{\lambda_1+2\beta_1}{3},1].
\end{gather*}
The functions $p=\rho W$ and $q=\rho Z$, where $(W,Z)$ is the solution of
\eqref{sysWZ}, satisfy
\begin{gather*}%\label{syspq}
 p_t-(\tilde{a}_1p_x)_x+\tilde{b}_{11}p=\rho\tilde{f}
 -\tilde{b}_{12}q-(\tilde{a}_1\rho_x W)_x-\rho_x\tilde{a}_1W_x, \quad
 x\in (-1,1),\; t\in (0,T),\\
q_t-(\tilde{a}_2q_x)_x+\tilde{b}_{22}q=-(\tilde{a}_2\rho_x Z)_x-\rho_x\tilde{a}_2Z_x,
\quad x\in (-1,1),\; t\in (0,T),\\
 p(t,-1)=p(t,1)=q(t,-1)=q(t,1)=0, \quad t \in (0,T).
\end{gather*}
Applying \eqref{carclassic} for the first component $p$ with $A=-\beta_1$,
 $B=\beta_1$ and proceeding as in \eqref{dis}-\eqref{w_t}, we obtain
\begin{align*}
& \int_{t_0}^T\int_{-\beta_1}^{\beta_1}(s\theta e^{r_1\zeta_1}p_x^2
 +s^3\theta^3e^{3r_1\zeta_1}p^2+\frac{1}{s\theta}p_t^2)e^{-2s\Phi_1}\,dx\,dt\\
& \leq C\int_{t_0}^T\int_{-\beta_1}^{\beta_1}\rho^2\tilde{f}^2e^{-2s\Phi_1}dx\,dt
+\tilde{C}\int_{t_0}^T\int_{-\beta_1}^{\beta_1}q^2e^{-2s\Phi_1}dx\,dt\\
&\quad +C\int_{t_0}^T\int_{\lambda_1}^{\frac{\lambda_1+2\beta_1}{3}}(W^2+W_x^2)e^{-2s\Phi_1}dx\,dt \\
&\quad +C\int_{t_0}^T\int_{-\frac{\lambda_1+2\beta_1}{3}}^{-\lambda_1}(W^2+W_x^2)e^{-2s\Phi_1}dx\,dt.
\end{align*}
Using the definition of $W$,  $-\Phi_1\leq-\Phi_2$ and
$e^{-2s\Phi_2(t,x)}\leq e^{-2s\Phi_2(t,-x)}$, a.e. $x\in [0,\beta_1]$, it follows
that
\begin{align*}
& \int_{t_0}^T\int_{-\beta_1}^{\beta_1}(s\theta e^{r_1\zeta_1}p_x^2
 +s^3\theta^3e^{3r_1\zeta_1}p^2+\frac{1}{s\theta}p_t^2)e^{-2s\Phi_1}\,dx\,dt\\
& \leq C\int_{t_0}^T\int_{0}^{\beta_1}\rho^2f^2e^{-2s\Phi_2(t,-x)}dx\,dt
 +\tilde{C}\int_{t_0}^T\int_{-\beta_1}^{\beta_1}q^2e^{-2s\Phi_1}dx\,dt\\
&\quad +C\int_{t_0}^T\int_{\lambda_1}^{\frac{\lambda_1+2\beta_1}{3}}(u^2+u_x^2)
e^{-2s\Phi_1(t,-x)}dx\,dt.
\end{align*}
Similarly, for the second component $q$, we obtain
\begin{align*}
&\int_{t_0}^T\int_{-\beta_1}^{\beta_1}(s\theta e^{r_2\zeta_2}q_x^2
+s^3\theta^3e^{3r_2\zeta_2}q^2+\frac{1}{s\theta}q_t^2)e^{-2s\Phi_2}\,dx\,dt\\
&\leq C\int_{t_0}^T\int_{\lambda_1}^{\frac{\lambda_1+2\beta_1}{3}}(v^2+v_x^2)
e^{-2s\Phi_2(t,-x)}dx\,dt.
\end{align*}
Thus it follows from  the last two inequalities that
\begin{align*}
& \int_{t_0}^T\int_{-\beta_1}^{\beta_1}(s\theta e^{r_1\zeta_1}p_x^2
 +s^3\theta^3e^{3r_1\zeta_1}p^2+\frac{1}{s\theta}p_t^2)e^{-2s\Phi_1}\,dx\,dt\\
&+\int_{t_0}^T\int_{-\beta_1}^{\beta_1}(s\theta e^{r_2\zeta_2}q_x^2
 +s^3\theta^3e^{3r_2\zeta_2}q^2+\frac{1}{s\theta}q_t^2)e^{-2s\Phi_2}\,dx\,dt \\
& \leq C\int_{t_0}^T\int_{0}^{\beta_1}\rho^2f^2e^{-2s\Phi_2(t,-x)}dx\,dt+\tilde{C}\int_{t_0}^T\int_{-\beta_1}^{\beta_1}q^2e^{-2s\Phi_1}dx\,dt \\
&\quad +C\int_{t_0}^T\int_{\lambda_1}^{\frac{\lambda_1+2\beta_1}{3}}(v^2+v_x^2)
 e^{-2s\Phi_2(t,-x)}dx\,dt\\
&\quad +C\int_{t_0}^T \int_{\lambda_1}^{\frac{\lambda_1+2\beta_1}{3}}(u^2+u_x^2)
 e^{-2s\Phi_1(t,-x)}dx\,dt.
\end{align*}
Taking $s$ such that $\tilde{C}\leq \frac12 s^3\theta^3e^{3r_2\zeta_2}$, using
 $-\Phi_1\leq-\Phi_2$ and Cacciopoli inequality \eqref{cacciopoli},  we obtain
\begin{equation} \label{estimpq}
\begin{aligned}
& \int_{t_0}^T\int_{-\beta_1}^{\beta_1}(s\theta e^{r_1\zeta_1}p_x^2
+s^3\theta^3e^{3r_1\zeta_1}p^2+\frac{1}{s\theta}p_t^2)e^{-2s\Phi_1}\,dx\,dt\\
&+\int_{t_0}^T\int_{-\beta_1}^{\beta_1}(s\theta e^{r_2\zeta_2}q_x^2
 +\frac{1}{2}s^3\theta^3e^{3r_2\zeta_2}q^2
 +\frac{1}{s\theta}q_t^2)e^{-2s\Phi_2}\,dx\,dt \\
& \leq C\int_{t_0}^T\int_{0}^{\beta_1}\rho^2f^2e^{-2s\Phi_2(t,-x)}dx\,dt
 +C\int_{t_0}^T\int_{\tilde{\omega}}(s^2\theta^2u^2+f)e^{-2s\Phi_1(t,-x)}dx\,dt\\
&\quad +C\int_{t_0}^T\int_{\tilde{\omega}}s^2\theta^2v^2e^{-2s\Phi_2(t,-x)}dx\,dt.
\end{aligned}
\end{equation}
Clearly, one can obtain the same estimates \eqref{3.44} in 
$[0,T]\times [0,\beta_1]$.
Hence, by \eqref{3.24}, \eqref{estimpq}, by the definitions of
$W$, $Z$, $p$ and $q$, and proceeding as in
\eqref{tag}, we obtain
\begin{align}
&\int_{t_0}^T\int_{0}^{\lambda_1} \Big(s\theta a_1u_x^2
 +s^3 \theta^3\frac{(x-x_1)^2}{a_1}u^2+s\theta^{3/2}|\eta \psi_1|u^2
 +\frac{1}{s\theta}u_t^2\Big)e^{2s\varphi_1}\,dx\,dt \nonumber\\
&+\int_{t_0}^T\int_{0}^{\lambda_1} \Big(s\theta a_2v_x^2
 +s^3 \theta^3\frac{(x-x_2)^2}{a_2}v^2+s\theta^{3/2}|\eta \psi_2|v^2
 +\frac{1}{s\theta}v_t^2\Big)e^{2s\varphi_2}\,dx\,dt \nonumber\\
&=\int_{t_0}^T\int_{0}^{\lambda_1} \Big(s\theta a_1W_x^2
 +s^3 \theta^3\frac{(x-x_1)^2}{a_1}W^2+s\theta^{3/2}|\eta \psi_1|W^2
 +\frac{1}{s\theta}W_t^2\Big)e^{2s\varphi_1}\,dx\,dt  \nonumber\\
&\quad +\int_{t_0}^T\int_{0}^{\lambda_1} \Big(s\theta a_2Z_x^2
 +s^3 \theta^3\frac{(x-x_1)^2}{a_2}Z^2+s\theta^{3/2}|\eta \psi_2|Z^2
 +\frac{1}{s\theta}Z_t^2\Big)e^{2s\varphi_2}\,dx\,dt  \nonumber\\
&=\int_{t_0}^T\int_{0}^{\lambda_1} \Big(s\theta a_1p_x^2
 +s^3 \theta^3\frac{(x-x_1)^2}{a_1}p^2+s\theta^{3/2}|\eta \psi_1|p^2
 +\frac{1}{s\theta}p_t^2\Big)e^{2s\varphi_1}\,dx\,dt  \nonumber\\
&\quad +\int_{t_0}^T\int_{0}^{\lambda_1} \Big(s\theta a_2q_x^2
 +s^3 \theta^3\frac{(x-x_2)^2}{a_2}q^2+s\theta^{3/2}|\eta \psi_2|q^2
 +\frac{1}{s\theta}q_t^2\Big)e^{2s\varphi_2}\,dx\,dt \nonumber\\
&\leq C\int_{t_0}^T\int_{-\beta_1}^{\beta_1}(s\theta e^{r_1\zeta_1}p_x^2
 +s^3\theta^3e^{3r_1\zeta_1}p^2+\frac{1}{s\theta}p_t^2)e^{-2s\Phi_1}\,dx\,dt
 \nonumber\\
&\quad +C\int_{t_0}^T\int_{-\beta_1}^{\beta_1}(s\theta e^{r_2\zeta_2}q_x^2
 +s^3\theta^3e^{3r_2\zeta_2}q^2+\frac{1}{s\theta}q_t^2)e^{-2s\Phi_2}\,dx\,dt
 \nonumber\\
& \leq C\int_{t_0}^T\int_{0}^{\beta_1}\rho^2f^2e^{-2s\Phi_2(t,-x)}dx\,dt
 +C\int_{t_0}^T\int_{\tilde{\omega}}(s^2\theta^2u^2+f)e^{-2s\Phi_1(t,-x)}dx\,dt
 \nonumber\\
&\quad +C\int_{t_0}^T\int_{\tilde{\omega}}s^2\theta^2v^2e^{-2s\Phi_2(t,-x)}dx\,dt.
\label{III}
\end{align}
Finally adding up \eqref{I}, \eqref{II} and \eqref{III}, the proof is complete.
\end{proof}

To estimate the term source with only the first  component of solutions
of \eqref{system}, we need to show a Carleman estimate with one force.
\begin{theorem} \label{caroneforce}
Let $T>0$. Moreover, assume that
\begin{align*}%\label{hypb12}
b_{12}\geqslant \mu>0 \quad \text{on } \omega'.
\end{align*}
There exist two positive constants $C$ and $s_0$ such that, for
every  $(u_0,v_0)\in \mathbb{H}$ and for all $s\geq s_0$, the
solution $(u,v)$ of \eqref{system} satisfies
\begin{equation} \label{Carineq2oneforce}
\begin{aligned}
I(u,v)&:=\int_{Q_{t_0}^T} \Big(s\theta a_1u_x^2+s^3 \theta^3
 \frac{(x-x_1)^2}{a_1}u^2+s\theta^{3/2}|\eta \psi_1|u^2
 +\frac{1}{s\theta}u_t^2\Big)e^{2s\varphi_1}\,dx\,dt\\
&+\int_{Q_{t_0}^T} \Big(s\theta a_2v_x^2+s^3 \theta^3\frac{(x-x_2)^2}{a_2}v^2
 +s\theta^{3/2}|\eta \psi_2|v^2+\frac{1}{s\theta}v_t^2\Big)e^{2s\varphi_2}\,dx\,dt\\
& \leq C\int_{Q_{t_0}^T}f^2e^{-2s\Phi_2(t,-x)}dx\,dt
 +C\int_{t_0}^T\int_{\omega}u^2dx\,dt:=I(f,u).
\end{aligned}
\end{equation}
\end{theorem}

The above theorem is a consequence of Theorem \ref{theofun}
and of the following lemma.

\begin{lemma}
 Let $\omega_2\Subset \omega_1$ such that $x_1,\,x_2 \notin\omega_1$.
Moreover, assume that
\[
b_{12}\geqslant \mu>0 \quad \text{on } \omega_1.
\]
There exists $C>0$ such that
\[
 \int_{t_0}^T\int_{\omega_2}s^2\theta^2v^2e^{-2s\Phi_2(t,-x)}dx\,dt
\leq\varepsilon J(v)+C \int_{Q_{t_0}^T}f^2e^{2s\varphi_2}\,dx\,dt
+C\int_{t_0}^T\int_{\omega} u^2\,dx\,dt,
\]
where $\varepsilon>0$ is small enough, $s$ is large enough and
$$
J(v)=\int_{Q_{t_0}^T} \Big(s\theta a_2v_x^2
+s^3 \theta^3\frac{(x-x_2)^2}{a_2}v^2\Big)e^{2s\varphi_2}\,dx\,dt.
$$
\end{lemma}

\begin{proof}
The choice of the weight functions given in Lemma \ref{fcweiineq}
 will play a crucial role.
We will adapt the technique used in \cite{hajjaj}.
Let $\chi\in C^\infty(0,1)$, such that $\operatorname{supp}(\chi)\subset\omega_1$
and $\chi\equiv1$ on $\omega_2$.
Multiplying the first equation of \eqref{system} by
$s^2\theta^2\chi e^{-2s\theta(t)\Psi_2(-x)}v$ and integrating over $Q_{t_0}^T$,
we obtain
\begin{equation} \label{step40}
\begin{aligned}
&\int_{Q_{t_0}^T}s^2\theta^2b_{12}\chi e^{-2s\theta(t)\Psi_2(-x)}v^2\,dx\,dt \\
&= \int_{Q_{t_0}^T}s^2\theta^2\chi e^{-2s\theta(t)\Psi_2(-x)}fv\,dx\,dt
-\int_{Q_{t_0}^T}s^2\theta^2\chi e^{-2s\theta(t)\Psi_2(-x)}u_tv\,dx\,dt\\
&\quad +\int_{Q_{t_0}^T}s^2\theta^2\chi e^{-2s\theta(t)\Psi_2(-x)}(a_1u_x)_xv\,dx\,dt\\
&\quad -\int_{Q_{t_0}^T}s^2\theta^2b_{11}\chi e^{-2s\theta(t)\Psi_2(-x)}uv\,dx\,dt.
\end{aligned}
\end{equation}
Integrating by parts and using the second equation in \eqref{system}, we obtain
\begin{equation} \label{step41}
\begin{aligned}
&\int_{Q_{t_0}^T}s^2\theta^2\chi e^{-2s\theta(t)\Psi_2(-x)}vu_t\,dx\,dt\\
&= \int_{Q_{t_0}^T}s^2\theta^2a_2\chi e^{-2s\theta(t)\Psi_2(-x)}u_xv_x\,dx\,dt
+\int_{Q_{t_0}^T}s^2\theta^2a_2(\chi e^{-2s\theta(t)\Psi_2(-x)})_xuv_x\,dx\,dt\\
&\quad + \int_{Q_{t_0}^T}(s^2\theta^2b_{22}+2s^3\theta^2\dot{\theta}\Psi_2(-x)
-2s^2\theta\dot{\theta})\chi e^{-2s\theta(t)\Psi_2(-x)}vu\,dx\,dt,
\end{aligned}
\end{equation}
and
\begin{equation} \label{step42}
\begin{aligned}
&\int_{Q_{t_0}^T}s^2\theta^2\chi e^{-2s\theta(t)\Psi_2(-x)}(a_1u_x)_xv\,dx\,dt \\
& = -\int_{Q_{t_0}^T}s^2\theta^2a_1\chi e^{-2s\theta(t)\Psi_2(-x)}v_xu_x\,dx\,dt\\
&\quad +\int_{Q_{t_0}^T}s^2\theta^2a_1(\chi e^{-2s\theta(t)\Psi_2(-x)})_xv_xu\,dx\,dt\\
&\quad + \int_{Q_{t_0}^T}s^2\theta^2(a_1(\chi e^{-2s\theta(t)\Psi_2(-x)})_x)_x
 vu\,dx\,dt.
\end{aligned}
\end{equation}
Combining \eqref{step40}-\eqref{step42}, we obtain
\[
\int_{Q_{t_0}^T}s^2\theta^2b_{12}\chi e^{-2s\theta(t)\Psi_2(-x)}v^2\,dx\,dt
 =I_1+I_2+I_3+I_4,
\]
where
\begin{gather*}
I_1= \int_{Q_{t_0}^T}s^2\theta^2\chi e^{-2s\theta(t)\Psi_2(-x)}vf\,dx\,dt,\\
I_2=-\int_{Q_{t_0}^T}s^2\theta^2(a_1+a_2)\chi e^{-2s\theta(t)\Psi_2(-x)}u_xv_x\,dx\,dt,\\
\begin{aligned}
I_3&=\int_{Q_{t_0}^T}s^2\theta^2(a_1-a_2)(\chi e^{-2s\theta(t)\Psi_2(-x)})_x
 uv_x\,dx\,dt \\
 &=\int_{Q_{t_0}^T}(s^2\theta^2\chi'+2s^3\theta^3\Psi_{2,x}(-x)\chi)(a_1-a_2)
 e^{-2s\theta(t)\Psi_2(-x)}uv_x\,dx\,dt,
\end{aligned}\\
\begin{aligned}
I_4&=\int_{Q_{t_0}^T}(2s^2\theta\dot{\theta}-s^2\theta^2(b_{11}+b_{22})
 -2s^3\theta^2\dot{\theta}\Psi_2(-x))\chi e^{-2s\theta(t)\Psi_2(-x)}uv\,dx\,dt\\
&\quad+\int_{Q_{t_0}^T}s^2\theta^2(a_1(\chi e^{-2s\theta(t)\Psi_2(-x)})_x)_x
 uv\,dx\,dt.
\end{aligned}
\end{gather*}
For  $\varepsilon>0$, we have
\begin{gather*}
\begin{aligned}
|I_1| &= \int_{Q_{t_0}^T}(fe^{s\varphi_2})(s^2\theta^2\chi
  e^{-s(2\theta(t)\Psi_2(-x)+\varphi_2)}v)\,dx\,dt\\
& \leq \frac12\int_{Q_{t_0}^T}f^2e^{2s\varphi_2}\,dx\,dt
+C\int_{t_0}^T\int_{\omega_1}s^4\theta^4e^{-2s(2\theta(t)\Psi_2(-x)
 +\varphi_2)}v^2\,dx\,dt,
\end{aligned}\\
\begin{aligned}
|I_2| & =\int_{Q_{t_0}^T}(\sqrt{s\theta a_2} e^{s\varphi_2}v_x)((s\theta)^{3/2}(a_2)^{-\frac12}(a_1+a_2)\chi e^{-s(2\theta(t)\Psi_2(-x)+\varphi_2)}u_x)\,dx\,dt\\
& \leq \varepsilon\int_{Q_{t_0}^T}s\theta a_2 e^{2s\varphi_2}v_x^2\,dx\,dt\\
&\quad + \frac{1}{\varepsilon}\underbrace{\int_{Q_{t_0}^T}s^3\theta^3
\frac{(a_1^2+a_2^2)}{a_2}\chi^2 e^{-2s(2\theta(t)\Psi_2(-x)+\varphi_2)}u_x^2
 \,dx\,dt}_{L}.
\end{aligned}
\end{gather*}
The integral $L$ should be estimated by an integral in $u^2$. For this, we multiply
the first equation in \eqref{system} by
$s^3\theta^3\frac{(a_1^2+a_2^2)}{a_1a_2}\chi^2e^{-2s(2\theta(t)\Psi_2(-x)+\varphi_2)}u$
and we integrate by parts, obtaining
$$
L=L_1+L_2+L_3+L_4+L_5,
$$
where
\begin{gather*}
L_1= \int_{Q_{t_0}^T}s^3\theta^3\frac{(a_1^2+a_2^2)}{a_1a_2}\chi^2
  e^{-2s(2\theta(t)\Psi_2(-x)+\varphi_2)}uf\,dx\,dt,\\
\begin{aligned}
L_2&=\frac12\int_{Q_{t_0}^T}s^3(3\theta^2-2s\theta^3(2\Psi_2(-x)
 +\psi_2))\dot{\theta}\frac{(a_1^2+a_2^2)}{a_1a_2}\chi^2\\
&\quad\times   e^{-2s(2\theta(t)\Psi_2(-x)+\varphi_2)}u^2\,dx\,dt,
\end{aligned}\\
L_3=\frac12\int_{Q_{t_0}^T}s^3\theta^3(a_1(\frac{(a_1^2+a_2^2)}{a_1a_2}\chi^2
 e^{-2s(2\theta(t)\Psi_2(-x)+\varphi_2)})_x)_xu^2\,dx\,dt,\\
L_4=-\int_{Q_{t_0}^T}s^3\theta^3\frac{(a_1^2+a_2^2)}{a_1a_2}\chi^2b_{11}
 e^{-2s(2\theta(t)\Psi_2(-x)+\varphi_2)}u^2\,dx\,dt,\\
L_5=-\int_{Q_{t_0}^T}s^3\theta^3\frac{(a_1^2+a_2^2)}{a_1a_2}\chi^2b_{12}
 e^{-2s(2\theta(t)\Psi_2(-x)+\varphi_2)}uv\,dx\,dt.
\end{gather*}
Since $|\dot{\theta}|\leq C\theta^2$ and $\operatorname{supp}(\chi)\subset\omega_1$,  we obtain
\begin{gather*}
|L_1|\leq\frac12\int_{Q_{t_0}^T}f^2e^{2s\varphi_2}\,dx\,dt
 +C\int_{t_0}^T\int_{\omega_1}s^6\theta^6 e^{-2s(4\theta(t)\Psi_2(-x)
 +3\varphi_2)}u^2\,dx\,dt,\\
|L_2|\leq C\int_{t_0}^T\int_{\omega_1}s^4\theta^5 e^{-2s(2\theta(t)\Psi_2(-x)
 +\varphi_2)}u^2\,dx\,dt,\\
|L_3|\leq C\int_{t_0}^T\int_{\omega_1}s^5\theta^5 e^{-2s(2\theta(t)\Psi_2(-x)
 +\varphi_2)}u^2\,dx\,dt,\\
|L_4|\leq C\int_{t_0}^T\int_{\omega_1}s^3\theta^3 e^{-2s(2\theta(t)\Psi_2(-x)
 +\varphi_2)}u^2\,dx\,dt,\\
\begin{aligned}
|L_5|&\leq \varepsilon\int_{Q_{t_0}^T}s^3\theta^3\frac{(x-x_2)^2}{a_2}
 e^{2s\varphi_2}v^2\,dx\,dt\\
&\quad +\frac{C}{\varepsilon}\int_{t_0}^T\int_{\omega_1}s^3\theta^3
 e^{-2s(4\theta(t)\Psi_2(-x)+3\varphi_2)}u^2\,dx\,dt.
\end{aligned}
\end{gather*}
Hence,
\begin{align*}
|L|&\leq C \int_{Q_{t_0}^T}f^2e^{2s\varphi_2}\,dx\,dt
 +C_{\varepsilon}\int_{t_0}^T\int_{\omega_1}s^6\theta^3 e^{-2s(4\theta(t)\Psi_2(-x)
 +3\varphi_2)}u^2\,dx\,dt\\
&\quad +\varepsilon\int_{Q_{t_0}^T}s^3\theta^3\frac{(x-x_2)^2}{a_2}
 e^{2s\varphi_2}v^2\,dx\,dt.
\end{align*}
Furthermore,
\[
|I_2|\leq C \int_{Q_{t_0}^T}f^2e^{2s\varphi_2}\,dx\,dt
+C_{\varepsilon}\int_{t_0}^T\int_{\omega_1}s^6\theta^6 e^{-2s(4\theta(t)\Psi_2(-x)
+3\varphi_2)}u^2\,dx\,dt+\varepsilon J(v).
\]
Using the fact that $\chi'$ and $\chi$ are supported in $\omega'$ and $x_2 \not \in \omega'$, proceeding as before, one has
\begin{align*}
|I_3|&\leq C\int_{Q_{t_0}^T}s^3\theta^3(\chi'+\chi)e^{-2s\theta(t)
 \Psi_2(-x)}uv_x\,dx\,dt\\
&=C\int_{Q_{t_0}^T}(\sqrt{s\theta a_2}e^{s\varphi_2}v_x)((s\theta)^{5/2}
 (a_2)^{-\frac12}(\chi'+\chi)e^{-s(2\theta(t)\Psi_2(-x)+\varphi_2)}u)\,dx\,dt\\
&\leq\varepsilon\int_{Q_{t_0}^T}s\theta a_2v_x^2e^{2s\varphi_2}\,dx\,dt
 +\frac{C}{\varepsilon}\int_{t_0}^T\int_{\omega_1} (s^5\theta^5
 e^{-2s(2\theta(t)\Psi_2(-x)+\varphi_2)}u^2\,dx\,dt,
\end{align*}
and
\begin{align*}
I_4
&\leq C\int_{Q_{t_0}^T}s^4\theta^4(\chi''+\chi'+\chi)
 e^{-2s\theta(t)\Psi_2(-x)}uv\,dx\,dt\\
&=C\int_{Q_{t_0}^T}(s^{3/2}\theta^{3/2} \frac{x-x_2}{\sqrt{a_2}}
 e^{s\varphi_2}v)((s\theta)^{5/2}\frac{\sqrt{a_2}}{x-x_2}(\chi''+\chi'+\chi)\\
&\quad\times  e^{-s(2\theta(t)\Psi_2(-x)+\varphi_2)}u)\,dx\,dt\\
&\leq\varepsilon\int_{Q_{t_0}^T}s^3\theta^3\frac{(x-x_2)^2}{a_2}
 v^2e^{2s\varphi_2}\,dx\,dt \\
&\quad +\frac{C}{\varepsilon}\int_{t_0}^T\int_{\omega_1}
 (s^5\theta^5e^{-2s(2\theta(t)\Psi_2(-x)+\varphi_2)}u^2\,dx\,dt,
\end{align*}
So, thanks to Lemma \ref{fcweiineq}, we have
\begin{gather*}
e^{-2s(2\theta(t)\Psi_2(-x)+\varphi_2)}\leq e^{-2s(4\theta(t)\Psi_2(-x)+3\varphi_2)},\\
 \sup_{(t,x)\in Q} s^r\theta^r(t)e^{-2s(4\theta(t)\Psi_2(-x)+3\varphi_2)}
 <\infty,\quad r\in \mathbb{R}.
\end{gather*}
Then, for $\varepsilon$ small enough and $s$ large enough, we have
\begin{align*}
&|\int_{Q_{t_0}^T}s^2\theta^2b_{12}\chi e^{-2s\theta(t)\Psi_2(-x)}v^2\,dx\,dt|\\
&\leq C \int_{Q_{t_0}^T}f^2e^{2s\varphi_2}\,dx\,dt
+C\int_{t_0}^T\int_{\omega} u^2\,dx\,dt+\varepsilon J(v).
\end{align*}
Finally, the definition of $\chi$, \eqref{assump} and the previous inequality give
\begin{align*}
&\mu \int_{t_0}^T\int_{\omega_2}s^2\theta^2 e^{-2s\theta(t)\Psi_2(-x)}v^2\,dx\,dt \\
& \leq |\int_{t_0}^T\int_{\omega_2}s^2\theta^2b_{12} e^{-2s\theta(t)\Psi_2(-x)}
 v^2\,dx\,dt|\\
& \leq \int_{Q_{t_0}^T} |s^2\theta^2b_{12}\chi e^{-2s\theta(t)\Psi_2(-x)}v^2|
\,dx\,dt\\
&\leq C \int_{Q_{t_0}^T}f^2e^{2s\varphi_2}\,dx\,dt+C\int_{t_0}^T\int_{\omega}
  u^2\,dx\,dt+\varepsilon J(v).
\end{align*}
This completes the proof.
\end{proof}

\section{The Lipschitz stability result}

The object of this section is to recover a source term $f$ from the knowledge
 of $(a_1u_x)_x(T',\cdot)$
and some additional observations of the component $u$.
The main result of this paper reads as follows.

\begin{theorem} \label{invprob}
 Let $C_0>0$.  Then, there exists $C=C(T,t_0,x_1,x_2,C_0)>0$ such that,
for all $f \in S(C_0)$ and $u_0\in L^2(0,1)$,
\begin{equation} \label{stability}
\begin{aligned}
\|f\|_{L^2(Q_{t_0}^T)}^2
&\leq    C \Big( \|u\|_{L^2(\omega_{t_0}^T)}^2+\|u_t\|_{L^2(\omega_{t_0}^T)}^2+
\|u(T',\cdot)\|_{L^2(0,1)}^2\\
&\quad +\|(a_1u_x)_x(T',\cdot)\|_{L^2(0,1)}^2\Big),
\end{aligned}
\end{equation}
where $\omega_{t_0}^T:= (t_0,T) \times \omega$.
\end{theorem}

\begin{proof}
 The functions $y=u_t$ and $z=v_t$,  where $(u,v)$ is the
solution of \eqref{system}, are solutions of the  system
\begin{gather*}%\label{sysyz}
 y_t-(a_1y_x)_x+b_{11}y+b_{12}z=f_t , \quad (t,x)\in Q,\\
z_t-(a_2z_x)_x+b_{22}z=0, \quad (t,x)\in Q,\\
 y(t,0)=y(t,1)=z(t,0)=z(t,1)=0, \quad t \in (0,T).
\end{gather*}
When we apply  Carleman estimate \eqref{Carineq2oneforce} to $(y,z)$, we obtain
\begin{equation} \label{Carineq2oneforcet}
\begin{aligned}
I(y,z)&:=\int_{Q_{t_0}^T} \Big(s\theta a_1y_x^2+s^3 \theta^3
 \frac{(x-x_1)^2}{a_1}y^2+s\theta^{3/2}|\eta \psi_1|y^2
 +\frac{1}{s\theta}y_t^2\Big)e^{2s\varphi_1}\,dx\,dt\\
&\quad +\int_{Q_{t_0}^T} \Big(s\theta a_2z_x^2+s^3 \theta^3\frac{(x-x_2)^2}{a_2}z^2
 +s\theta^{3/2}|\eta \psi_2|z^2+\frac{1}{s\theta}z_t^2\Big)e^{2s\varphi_2}\,dx\,dt\\
& \leq C\int_{Q_{t_0}^T}f_t^2e^{-2s\Phi_2(t,-x)}dx\,dt
+C\int_{t_0}^T\int_{\omega}u_t^2dx\,dt\\
&:=I(f_t,y).
\end{aligned}
\end{equation}
The terms appearing in \eqref{stability} are well defined, indeed,
by Proposition \ref{estimsemigroup}, we have $y\in L^2\left(t_0,T;D(A_1)\right)
\cap H^1\left(t_0,T;L^2(0,1)\right)$.

As in \cite{BukhgeimKlibanov}, we  divide the proof  into three steps.
\smallskip

\noindent\textbf{Step 1.} We show first that there exists a constant $C>0$ such that
\begin{equation}\label{step1}
\begin{aligned}
&I(f,u)+I(f_t,y)\\
&\leq C\left(\frac{1}{\sqrt{s}}\int_0^1f^2(T',x)e^{-2s\Phi_2(T',-x)}dx
+\|u\|_{L^2(\omega_{t_0}^T)}^2+\|u_t\|_{L^2(\omega_{t_0}^T)}^2\right),
\end{aligned}
\end{equation}
where, we recall, $T'= (T+t_0)/2$.
To obtain \eqref{step1}, it remains to prove that
$$
\int_{Q_{t_0}^T} (f^2+f_t^2) e^{-2s\Phi_2 (t,-x)}\,dx\,dt
\leq \frac{C}{\sqrt{s}}\int_0^1f^2(T',x)e^{-2s\Phi_2(T',-x)}dx.
$$
 Since $\Phi_{1,t}(T')=0$ and $\Phi_{1,tt}(t)\geq \mu_0>0$,
then Taylor's formula provides
$$
-\Phi_2(t,-x)\leq-\Phi_2(T',-x)-\frac{\mu_0 }{2}(t-T')^2,
$$
and then
$$
\int_{t_0}^Te^{-2s\Phi_2(t,-x)}dt\leq \frac{1}{\sqrt{\mu_0s}}
e^{-2s\Phi_2(T',-x)}\int_{-\infty}^\infty e^{-l^2} dl
\leq \frac{C}{\sqrt{s}}e^{-2s\Phi_2(T',-x)}.
$$
 So,
$$
\int_{Q_{t_0}^T}f^2(T',x)e^{-2s\Phi_2(t,-x)}dx\,dt
\leq \frac{C}{\sqrt{s}}\int_0^1f^2(T',x)e^{-2s\Phi_2(T',-x)}dx.
$$
For  $f\in S(C_0)$, one has
\begin{equation} \label{estimF}
|f(t,x)| \leq |f(T',x)|+\int_{T'}^t|f_t(s,x)|ds \leq C|f(T',x)|.
\end{equation}
Thus
\[
\int_{Q_{t_0}^T}(f^2+f_t^2)(t,x)e^{-2s\Phi_2(t,-x)}dx\,dt
\leq \frac{C}{\sqrt{s}}\int_0^1f^2(T',x)e^{-2s\Phi_2(T',-x)}dx.
\]
The purpose of the first step is acomplished.
\smallskip

\noindent\textbf{Step 2.}  Now, let us show that there exists a constant $C>0$
such that
\begin{equation} \label{step2}
\int_0^1(y(T',x)+b_{12}v(T',x))^2e^{2s\varphi_1(T',x)}dx\leq C(I(y,z)+I(u,v)).
\end{equation}
Since,  for a.e. $x\in(0,1)$,
$$
\lim_{t\to t_0} (y(t,x)+b_{12}v(t,x))^2e^{2s\varphi_1(t,x)}=0.
$$
Hence
\begin{equation} \label{step21}
\begin{aligned}
&\int_0^1(y(T',x)+b_{12}v(T',x))^2e^{2s\varphi_1(T',x)}dx\\
& = \int_0^1\int_{t_0}^{T'}\frac{\partial}{\partial t}
\big((y+b_{12}v)^2e^{2s\varphi_1(t,x)}\big) \,dt\,dx   \\
&= \int_{t_0}^{T'}\int_0^1(2(y+b_{12}v)(y_t+b_{12}z)
+2s\varphi_{1,t}(y+b_{12}v)^2)e^{2s\varphi_1(t,x)}dx\,dt.
\end{aligned}
\end{equation}
Using  Young inequality,  for $s$ large enough, we obtain
\begin{equation} \label{step22}
\begin{aligned}
&|\int_{t_0}^{T'}\int_0^1 2(y+b_{12}v)(y_t+b_{12}z)e^{2s\varphi_1}dx\,dt|\\
&\leq C\int_{Q_{t_0}^T}\big(s\theta y^2+s\theta z^2+s\theta v^2
+\frac{1}{s\theta}y_t^2\big)e^{2s\varphi_1}dx\,dt
\end{aligned}
\end{equation}
and, by the Hardy inequality,
\begin{align*}
&\int_{Q_{t_0}^T}s\theta y^2e^{2s\varphi_1}dx\,dt \\
&= s \Big|\int_{Q_{t_0}^T} \Big( \theta\frac{a_1^{1/3}}{|x-x_1|^{2/3}}y^2
e^{2s\varphi_1}\Big)^{3/4}\Big( \theta\frac{|x-x_1|^2}{a_1}y^2
e^{2s\varphi_1}\Big)^{1/4}\,dx\,dt\Big|\\
& \leq s \frac32 \int_{Q_{t_0}^T} \theta\frac{a_1^{1/3}}{|x-x_1|^{2/3}}y^2
e^{2s\varphi_1}\,dx\,dt + \frac{s}{2} \int_{Q_{t_0}^T}
\theta\frac{|x-x_1|^2}{a_1}y^2e^{2s\varphi_1} \,dx\,dt.
\end{align*}
Moreover, by the Hardy-Poincar\'e inequality applied to $ye^{s\varphi_1}$, one has
\begin{align*}
 \int_{Q_{t_0}^T}s\theta y^2e^{2s\varphi_1}dx\,dt
&\leq C \int_{Q_{t_0}^T} \Big(s\theta a_1[y_x+2s\varphi_{1,x}y]^2
 +s^3\theta^3\frac{(x-x_1)^2}{a_1}y^2\,\Big)e^{2s\varphi_1}dx\,dt\\
&\leq C \int_{Q_{t_0}^T} \Big(s\theta a_1y_x^2
+s^3\theta^3\frac{(x-x_1)^2}{a_1}y^2\,\Big)e^{2s\varphi_1}dx\,dt.
\end{align*}
Similarly, by $\varphi_1<\varphi_2$, we have
\begin{equation} \label{step24}
\begin{aligned}
&\int_{Q_{t_0}^T}s\theta(z^2+v^2)e^{2s\varphi_1}dx\,dt\\
&\leq C \int_{Q_{t_0}^T} \Big(s\theta a_2(z_x^2+v_x^2)+s^3\theta^3
\frac{(x-x_2)^2}{a_2}(z^2+v^2)\,\Big)e^{2s\varphi_2}dx\,dt.
\end{aligned}
\end{equation}
On the other hand, since $|\varphi_{1,t}|\leq C|\eta\psi_1|\theta^{3/2}$,
we have
\begin{equation} \label{step25}
\begin{aligned}
&\int_{t_0}^{T'}\int_0^1 s\varphi_{1,t}(y+b_{12}v)^2e^{2s\varphi}\\
&\leq C \int_{Q_{t_0}^T} s\theta^{3/2}|\eta \psi_1|y^2e^{2s\varphi_1}
 +s\theta^{3/2}|\eta \psi_2|v^2e^{2s\varphi_2}dx\,dt.
\end{aligned}
\end{equation}
Thus,  \eqref{step21}-\eqref{step25} yield  the  estimate \eqref{step2}.
\smallskip

\noindent\textbf{Step 3.}
Combining \eqref{Carineq2oneforce}, \eqref{Carineq2oneforcet}, \eqref{step1}
and \eqref{step2}, we deduce
\begin{equation} \label{step31}
\begin{aligned}
&\int_0^1(y(T',x)+b_{12}v(T',x))^2e^{2s\varphi_1(T',x)}dx \\
& \leq C \Big(\frac{1}{\sqrt{s}}\int_0^1f^2(T',x)e^{-2s\Phi_2(T',-x)}dx
+\|u\|_{L^2(\omega_{t_0}^T)}^2+\|u_t\|_{L^2(\omega_{t_0}^T)}^2\Big).
\end{aligned}
\end{equation}
Since $y+b_{12}v$ satisfies
\[
y(T',x)+b_{12}v(T',x)=(a_1u_x)_x(T',x)-b_{11}u(T',x)+f(T',x),
\]
it follows that
\begin{equation} \label{step32}
\begin{aligned}
\int_0^1f^2(T',x)e^{2s\varphi_1(T',x)}dx  
&\leq   C\Big(\int_0^1(y(T',x)+b_{12}v(T',x))^2e^{2s\varphi_1(T',x)}dx\\
&\quad + \|(a_1u_x)_x(T')\|_{L^2(0,1)}^2+ \|u(T')\|_{L^2(0,1)}^2\Big).
\end{aligned}
\end{equation}
Hence, by \eqref{step31} and \eqref{step32} we obtain, for $s$ large enough,
\begin{align*}
\int_0^1f^2(T',x)dx
&\leq  C\Big( \|u\|_{L^2(\omega_{t_0}^T)}^2+\|u_t\|_{L^2(\omega_{t_0}^T)}^2+
\|u(T')\|_{L^2(0,1)}^2 \\
&\quad +\|(au_x)_x(T')\|_{L^2(0,1)}^2\Big),
\end{align*}
which, together with \eqref{estimF}, give the thesis.
\end{proof}

\section{Appendix}

In this section, we show a Cacciopoli's inequality for inhomogenous degenerate
parabolic equations. This inequality is different from  the one shown in
\cite{Fra2} for homogenous case.

\begin{proposition}
 Let $\omega''\subset\omega'\Subset\omega\subset(0,1)$ and
$x_0\notin \overline{\omega'}$. Let $s\geq s_0>0$, then
there exists a positive constant
$C=C(s_0,T,\inf_{\omega''}a(x),\|c\|_{L^\infty(Q)})$ such that every solution
$u$ of \eqref{problem} satisfies
\begin{equation} \label{cacciopoli}
 \int_{t_0}^T\int_{\omega''}u_x^2e^{2s\varphi}\,dx\,dt
\leq C\int_{t_0}^T\int_{\omega'} (s^2\theta^2u^2+h^2)e^{2s\varphi}\,dx\,dt.
\end{equation}
\end{proposition}

\begin{proof}
Define a smooth cut-off function $\xi\in C^\infty([0,1])$ such that
$\xi\equiv 1$  in $\omega''$ and
$\operatorname{supp}(\xi)\subset \omega'$.
Since $u$ solves \eqref{problem}, we have
\begin{align*}
 0&=\int_{t_0}^T\frac{d}{dt}\Big(\int_0^1\xi^2e^{2s\varphi}u^2\,dx\Big)\,dt=\int_{t_0}^T\int_0^1\Big(2s\xi^2\varphi_te^{2s\varphi}u^2+2\xi^2e^{2s\varphi}uu_t\Big)\,dx\,dt\\
& = \int_{t_0}^T\int_0^1[2s\xi^2\varphi_te^{2s\varphi}u^2+2\xi^2e^{2s\varphi}u((au_x)_x+h-cu)]\,dx\,dt\\
& = \int_{t_0}^T\int_0^1[2\xi^2(s\varphi_t-c)e^{2s\varphi}u^2+2\xi^2e^{2s\varphi}uh-2(\xi^2e^{2s\varphi})_xauu_x-2\xi^2e^{2s\varphi}au_x^2]\,dx\,dt\\
& = -2\int_{t_0}^T\int_{\omega'} \xi^2e^{2s\varphi}au_x^2\,dx\,dt\\
&\quad +2\int_{t_0}^T\int_{\omega'} [\xi^2(s\varphi_t-c)e^{2s\varphi}u^2+\xi^2e^{2s\varphi}uh-(\xi^2e^{2s\varphi})_xauu_x]\,dx\,dt.
\end{align*}
Then, integrating by parts and using the Young inequality, we obtain
\begin{align*}
&2\int_{t_0}^T\int_{\omega'} \xi^2e^{2s\varphi}au_x^2\,dx\,dt \\
& =2\int_{t_0}^T\int_{\omega'} [\xi^2(s\varphi_t-c)e^{2s\varphi}u^2
 +\xi^2e^{2s\varphi}uh-(\xi^2e^{2s\varphi})_xauu_x]\,dx\,dt\\
& \leq\int_{t_0}^T\int_{\omega'} \Big[\Big(2\xi^2(s\varphi_t-c)e^{2s\varphi}
 +\xi^2e^{2s\varphi}+\Big(\sqrt{a}\frac{(\xi^2e^{2s\varphi})_x}{\xi e^{s\varphi}}
 \Big)^2\Big)u^2+\xi^2e^{2s\varphi}h^2\Big]\,dx\,dt\\
&\quad +\int_{t_0}^T\int_{\omega'} \Big(\sqrt{a}\xi e^{s\varphi}\Big)^2u_x^2\,dx\,dt.
\end{align*}
Hence, for $s$ large enough,
\begin{align*}
&\int_{t_0}^T\int_{\omega'} \xi^2e^{2s\varphi}au_x^2\,dx\,dt\\
&\leq \int_{t_0}^T\int_{\omega'} \Big[\Big(2\xi^2(s\varphi_t-c)e^{2s\varphi}+\xi^2e^{2s\varphi}+\Big(\sqrt{a}\frac{(\xi^2e^{2s\varphi})_x}{\xi e^{s\varphi}}\Big)^2\Big)u^2+\xi^2e^{2s\varphi}h^2\Big]\,dx\,dt\\
& \leq C\int_{t_0}^T\int_{\omega'} (s^2\theta^2u^2+h^2)e^{2s\varphi}\,dx\,dt.
\end{align*}
Since $x_0\notin \overline{\omega'}$,
\begin{align*}
 \inf_{\omega''}a(x) \int_{t_0}^T\int_{\omega''} e^{2s\varphi}u_x^2\,dx\,dt
&\leq \int_{t_0}^T\int_{\omega'} \xi^2e^{2s\varphi}au_x^2\,dx\,dt\\
&\leq C\int_{t_0}^T\int_{\omega'} (s^2\theta^2u^2+h^2)e^{2s\varphi}\,dx\,dt,
\end{align*}
and the proof is complete.
\end{proof}

\subsection*{Acknowledgments}
Genni Fragnelli was partially supported by the GNAMPA, project
Equazioni di evoluzione degeneri e singolari: controllo e applicazioni.


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\end{document}