\documentclass[reqno]{amsart}
%\usepackage{hyperref}
\usepackage{mathrsfs}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2018 (2018), No. 195, pp. 1--20.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2018 Texas State University.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2018/195\hfil Carleman estimate and null controllability]
{Carleman estimate and null controllability of a cascade degenerate parabolic
system with general convection terms}

\author[J. Xu, C. Wang, Y. Nie \hfil EJDE-2018/195\hfilneg]
{Jianing Xu, Chunpeng Wang, Yuanyuan Nie}

\address{Jianing Xu \newline
School of Mathematics,
Jilin University,
Changchun 130012, China}
\email{924751144@qq.com}

\address{Chunpeng Wang (corresponding author) \newline
School of Mathematics,
Jilin University,
Changchun 130012, China}
\email{wangcp@jlu.edu.cn}

\address{Yuanyuan Nie \newline
School of Mathematics,
Jilin University,
Changchun 130012, China}
\email{nieyy@jlu.edu.cn}

\thanks{Submitted April 2, 2018. Published December 6, 2018.}
\subjclass[2010]{93B05, 93C20, 35K67}
\keywords{Carleman estimate; null controllability; convection;
\hfill\break\indent  degenerate parabolic system}

\begin{abstract}
 This article shows Carleman estimate and null controllability of
 a cascade control system governed by the semilinear degenerate parabolic
 equations with the general convection terms.
 The semilinear parabolic equations are weakly degenerate on the boundary
 and the convection terms cannot be controlled by the diffusion terms.
 We establish the Carleman estimate and the observability inequality for
 the linear conjugate system, and prove that the control system is null
 controllable.
\end{abstract}

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


\section{Introduction}

In this article, we study the Carleman estimate and the null controllability
of the cascade semilinear degenerate parabolic system with convection terms
\begin{gather}
u_t-(x^\alpha u_x)_{x}+(P_1(x,t,u))_{x}+F_1(x,t,u)=h(x,t)\chi_\omega, \quad
 (x,t)\in Q_T, \label{a1} \\
v_t-(x^\alpha v_x)_{x}+(P_2(x,t,v))_{x}+F_2(x,t,u,v)=0, \quad
 (x,t)\in Q_T,\label{a2}\\
u(0,t)=v(0,t)=0,\quad u(1,t)=v(1,t)=0, \quad t\in (0,T), \label{a3}\\
u(x,0)=u_0(x),\quad v(x,0)=v_0(x), \quad x\in(0,1), \label{a4}
\end{gather}
where $0<\alpha<1/2$, $Q_T=(0,1)\times(0,T)$,
$h$ is the control function, $\omega\subset(0,1)$ is an interval,
$\chi_\omega$ is the characteristic function of $\omega$, $u_0,v_0\in L^2(0,1)$,
and $P_1,P_2,F_1,F_2$ are measurable functions.
It is noted that \eqref{a1} and \eqref{a2} are degenerate at the boundary $x=0$.
The cascade system \eqref{a1} and \eqref{a2} arises in some models from
mathematical biology and physics,
such as the Keller-Segel model \cite{by} and the Lotka-Volterra model \cite{sc}.

Controllability theory has been widely investigated for nondegenerate parabolic
equations and systems over the last forty years
and the known results are almost complete (see \cite{am1,am2,ba,fe,fu,go2,go3,go1,gu} and the references therein).
Recently, controllability theory for degenerate ones has been studied
and there have been many results
(see \cite{al,mu,ca1,Re1,R4,R8,R2,ca2,du2,du3,du1,fl,li,R3,w1,w2,PD,w3,w4}).
But it is far from being solved.
The null controllability of the following system, governed by
 a degenerate diffusion equation, is already investigated
\begin{gather}
u_t-(x^{\alpha}u_x)_x+c(x,t)u=h(x,t)\chi_\omega,
\quad (x,t)\in Q_T, \label{w.1}\\
\begin{gathered}
u(0,t)=u(1,t)=0 \quad\text{if } 0<\alpha<1,\; t\in(0,T),\\
(x^\alpha u_x)(0,t)=u(1,t)=0 \quad\text{if }\alpha\ge1,\; t\in(0,T),
\end{gathered} \label{w.2} \\
u(x,0)=u_0(x),\quad x\in(0,1), \label{w.3}
\end{gather}
where $\alpha>0$, $c\in L^\infty(Q_T)$.
It was shown that the system \eqref{w.1}--\eqref{w.3}
is null controllable if $0<\alpha<2$ \cite{al,R8,R2,R3},
while not if $\alpha\ge2$ \cite{R4}.
Although system \eqref{w.1}--\eqref{w.3}
is not null controllable for $\alpha\ge2$,
it was proved in \cite{du3,w1,w2,PD} and \cite{ca1,Re1,R4} that it
is approximately controllable in $L^2(0,1)$
and regional null controllable for each $\alpha>0$, respectively.
Flores and Teresa \cite{fl} studied the degenerate
convection-diffusion equation
\begin{equation}
u_t-(x^{\alpha}u_x)_x+x^{\alpha/2}b(x,t)u_x+c(x,t)u=h(x,t)\chi_\omega,
\quad (x,t)\in Q_T
\label{w.4}
\end{equation}
with $b\in L^\infty(Q_T)$
and proved that the system \eqref{w.4}, \eqref{w.2} and \eqref{w.3} is null
controllable for $0<\alpha<2$.
Clearly, the convection term can be controlled by the diffusion term in \eqref{w.4}.
 Wang and Du \cite{w3} considered the degenerate convection-diffusion equation
\begin{equation}
u_t-(x^{\alpha}u_x)_x+(b(x,t)u)_x+c(x,t)u=h(x,t)\chi_\omega,
\quad (x,t)\in Q_T, \label{w.5}
\end{equation}
and proved that system \eqref{w.5}, \eqref{w.2} and \eqref{w.3} is null
controllable if $0<\alpha<1/2$.
Here $0<\alpha<1/2$ is optimal when one establishes the Carleman estimate
in such a way as in \cite{w3}.
Since \eqref{w.5} is degenerate, the convection term can cause essential differences.
For example, problem \eqref{w.5}, \eqref{w.2} and \eqref{w.3}
is well-posed in the weakly degenerate case ($0<\alpha<1$),
while may be ill-posed in the strongly degenerate case ($\alpha\ge1$)
(see \cite{YW1,YW}).
Moreover, the system
\[
u_t-(x^{\alpha}u_x)_x+b(x,t)u_x+c(x,t)u=h(x,t)\chi_\omega,
\quad (x,t)\in Q_T
\]
with \eqref{w.2} and \eqref{w.3} was shown to be null controllable for
 $0<\alpha<1$ if $b,b_x,b_{xx},b_t\in L^\infty(Q_T)$ in \cite{w4}.
As for systems, the authors in \cite{mu} studied the null controllability of
system \eqref{a1}--\eqref{a4} without convection term, i.e.\
 the case that $P_1=P_2=0$.
It was shown in \cite{mu} that the system is null controllable if $0<\alpha<2$
(a different boundary condition at $x=0$ is prescribed if $1\le\alpha<2$).
 Du and Xu \cite{du1} considered the linear case of
 system \eqref{a1}--\eqref{a4} when the convection terms can be controlled
by the diffusion terms.
Furthermore, the Carleman estimate in \cite{du1} depends on the derivatives
of the coefficients of the convection terms
and cannot be used to treat the semilinear case.


In this article, we consider the more general system \eqref{a1}--\eqref{a4}
where the convection terms are independent of the diffusion terms.
More precisely, we assume that $P_1, P_2, F_1, F_2$ are measurable
functions satisfying
\begin{gather}
\begin{gathered}
P_1(x,t,0)=P_2(x,t,0)=F_1(x,t,0)=F_2(x,t,0,0)=0, \\
F_2(x,t,\cdot,\cdot)\in C^{1}(\mathbb{R}^2),\quad(x,t)\in Q_T,
\end{gathered}\label{a12} \\
\begin{gathered}
|P_1(x,t,y_1)-P_1(x,t,y_2)|\leq K|y_1-y_2|,\\
|P_2(x,t,z_1)-P_2(x,t,z_2)|\leq K|z_1-z_2|, \\
|F_1(x,t,y_1)-F_1(x,t,y_2)|\leq K|y_1-y_2|,\\
(x,t)\in Q_T,\, y_1,y_2,z_1,z_2\in\mathbb{R},
\end{gathered}\label{a14} \\
\big|\frac{\partial F_2(x,t,y,z)}{\partial y}\big|\leq K,\quad
\big|\frac{\partial F_2(x,t,y,z)}{\partial z}\big|\leq K,
\quad(x,t)\in Q_T,\, y,z\in\mathbb{R},
\label{a15}
\\
\frac{\partial F_2}{\partial y}\Big|_{(x_0,x_1)\times(0,T)\times\mathbb{R}^2}
\geq c_0\quad\text{or}\quad
\frac{\partial F_2}{\partial y}\Big|_{(x_0,x_1)\times(0,T)\times\mathbb{R}^2}
\leq -c_0,
\label{a16}
\end{gather}
where $K>0$, $c_0>0$ and $0<x_0<x_1<1$ such that $(x_0,x_1)\subset\omega$.
Since \eqref{a1} and \eqref{a2} are degenerate at the boundary $x=0$,
the classical solution may not exist and weak solution should be considered.
The key to prove the null controllability of system \eqref{a1}--\eqref{a4}
is the Carleman estimate for its linear conjugate problem.
The degeneracy of the equations and the existence of the general convection terms
cause essential difficulties for the Carleman estimate.
In order to establish the needed Carleman estimate for the degenerate parabolic system,
we use the Carleman estimate for a single degenerate parabolic equation in \cite{w3},
where the auxiliary functions for the Carleman estimate were chosen by the method of undetermined coefficients
and it was turned out that only the case $0<\alpha<1/2$ can be treated in such a way.
Using the Carleman estimate in \cite{w3}
and the classical Carleman estimate, together with some other energy estimates,
we establish the Carleman estimate for the linear conjugate problem of
problem \eqref{a1}--\eqref{a4}.
Some technical difficulties caused by the degeneracy of the equations and
the existence of the general convection terms have to be overcome.
After the Carleman estimate, we establish the observability inequality
and further prove the null controllability of system \eqref{a1}--\eqref{a4}.

This article is organized as follows.
In $\S$ 2, we introduce the well-posedness of problem \eqref{a1}--\eqref{a4}
and some a priori estimates. The Carleman estimate and the observability
inequality are established in $\S$ 3.
Finally, the null controllability of system \eqref{a1}--\eqref{a4} is shown in $\S$ 4.


\section{Well-posedness and some a priori estimates}

Consider the linear nondegenerate parabolic problem
\begin{gather}
\begin{gathered}
u^{\eta}_t-((x+\eta)^\alpha u^{\eta}_x)_{x}+({c_1}(x,t)u^{\eta})_{x}+{c_3}
(x,t)u^{\eta}+{c_4}(x,t)v^{\eta}=f_1(x,t), \\ 
(x,t)\in Q_T,
\end{gathered}\label{b1}\\
\begin{gathered}
v^{\eta}_t-((x+\eta)^{\alpha}v^{\eta}_x)_{x}+({c_2}(x,t)v^{\eta})_{x}
+{c_5}(x,t)u^{\eta}+{c_6}(x,t)v^{\eta}=f_2(x,t), \\
(x,t)\in Q_T,
\end{gathered}\label{b2}\\
u^{\eta}(0,t)=v^{\eta}(0,t)=0,\quad
u^{\eta}(1,t)=v^{\eta}(1,t)=0, \quad t\in (0,T),\label{b3}
\\
u^{\eta}(x,0)=u_0(x),\quad v^{\eta}(x,0)=v_0(x), \quad x\in(0,1),
\label{b4}
\end{gather}
where $0<\alpha<1$,
$0<\eta<1$, ${c_i}\in L^{\infty}(Q_T)~(1\le i\le 6)$,
$f_1,f_2\in L^{2}(Q_T)$, and $u_0,v_0\in L^{2}(0,1)$.
The problem \eqref{b1}--\eqref{b4} admits a unique solution $(u^{\eta},v^{\eta})$
with $u^{\eta},v^{\eta}\in L^{\infty}(0,T;L^{2}(0,1))\cap L^{2}(0,T;H^{1}(0,1))$.
Moreover, $(u^{\eta},v^{\eta})$ satisfies the following a priori estimates.

\begin{lemma}\label{lem1}
Assume that $0<\alpha<1$,
$0<\eta<1$, ${c_i}\in L^{\infty}(Q_T)$ with
$\|c_{i}\|_{L^{\infty}(Q_T)}\leq K$ $(1\leq i\leq6)$,
$f_1,f_2\in L^{2}(Q_T)$, and $u_0,v_0\in L^{2}(0,1)$.
Then the solution $(u^{\eta},v^{\eta})$ to problem \eqref{b1}--\eqref{b4} satisfies
\begin{gather}
\begin{aligned}
&\|u^{\eta}\|_{L^{\infty}(0,T;L^{2}(0,1))}
 +\|(x+\eta)^{\alpha/2}u^{\eta}_{x}\|_{L^{2}(Q_T)}
 +\|v^{\eta}\|_{L^{\infty}(0,T;L^{2}(0,1))}\\
&+\|(x+\eta)^{\alpha/2}v^{\eta}_{x}\|_{L^{2}(Q_T)}\\
&\leq N(\|f_1\|_{L^{2}(Q_T)}+\|f_2\|_{L^{2}(Q_T)}
+\|u_0\|_{L^{2}(0,1)}+\|v_0\|_{L^{2}(0,1)}),
\end{aligned}\label{b5} \\
\begin{aligned}
&\big|\int^{1}_0(u^{\eta}(x,t_2)-u^{\eta}(x,t_1))\varsigma(x) \,\mathrm{d}x\big|
 +\big|\int^{1}_0(v^{\eta}(x,t_2)-v^{\eta}(x,t_1))\varsigma(x) \,\mathrm{d}x\big|\\
&\leq N(t_2-t_1)^{1/2}(\|f_1\|_{L^{2}(Q_T)}+\|f_2\|_{L^{2}(Q_T)}
 +\|u_0\|_{L^{2}(0,1)} \\
&\quad +\|v_0\|_{L^{2}(0,1)})\|\varsigma\|_{H^{1}(0,1)},\quad
 0\leq t_1< t_2\leq T,\quad \varsigma\in H^{1}(0,1),
\end{aligned} \label{b6} \\
\begin{aligned}
&\int_0^{T-\delta} \int_0^1(u^{\eta}(x,\tau+\delta)-u^{\eta}(x,\tau))^{2}
 \,\mathrm{d}x\,\mathrm{d}\tau \\
&+\int_0^{T-\delta} \int_0^1(v^{\eta}(x,\tau+\delta)-v^{\eta}(x,\tau))^{2}
 \,\mathrm{d}x\,\mathrm{d}\tau \\
&\leq N\delta^{1/2}(\|f_1\|^{2}_{L^{2}(Q_T)}+\|f_2\|^{2}_{L^{2}(Q_T)}
 +\|u_0\|^{2}_{L^{2}(0,1)}+\|v_0\|^{2}_{L^{2}(0,1)}),\\
&\quad 0<\delta<T,
\end{aligned}\label{b7}
\end{gather}
where $N>0$ depends only on $K$, $T$, and $\alpha$.
\end{lemma}

\begin{proof}
Without loss of generality, it is assumed that $(u^\eta,v^\eta)$ is a smooth solution.
Otherwise, one can mollify $c_{i}~(1\leq i\leq6)$, $f_1$, $f_2$, $u_0$, $v_0$,
and prove the lemma by a standard limit process.
For convenience, $u^\eta$ and $v^\eta$ are abbreviated as $u$ and $v$,
respectively, in this proof.

For $0<s<T$, multiplying \eqref{b1} and \eqref{b2} by $u$ and $v$, respectively,
then integrating over $(0,1)\times(0,s)$ by parts and summing up, we obtain
\begin{align*}
&\frac{1}{2}\int^{s}_0 \int^{1}_0((u^2)_t+(v^2)_t)\,\mathrm{d}x\,\mathrm{d}t
+\int^{s}_0 \int^{1}_0(x+\eta)^\alpha (u_x^2+v_x^2)\,\mathrm{d}x\,\mathrm{d}t
\\
&=\int^{s}_0 \int^{1}_0(c_1uu_x+c_2vv_x)\,\mathrm{d}x\,\mathrm{d}t
-\int^{s}_0 \int^{1}_0(c_3u^2+(c_4+c_5)uv+c_6v^2)\,\mathrm{d}x\,\mathrm{d}t\\
&\quad +\int^{s}_0 \int^{1}_0(f_1u+f_2v)\,\mathrm{d}x\,\mathrm{d}t
\\
&\leq K\int^{s}_0 \int^{1}_0(|uu_x|+|vv_x|)\,\mathrm{d}x\,\mathrm{d}t
+{2}K\int^{s}_0 \int^{1}_0(u^2+v^2)\,\mathrm{d}x\,\mathrm{d}t \\
&\quad +\int^{s}_0 \int^{1}_0(|f_1u|+|f_2v|)\,\mathrm{d}x\,\mathrm{d}t
\\
&\leq \frac{1}{2}\int^{s}_0 \int^{1}_0(x+\eta)^\alpha (u_x^2+v_x^2)
 \,\mathrm{d}x\,\mathrm{d}t
+\frac12{K^2}\int^{s}_0 \int^{1}_0(x+\eta)^{-\alpha}(u^2+v^2)
 \,\mathrm{d}x\,\mathrm{d}t
\\
&\quad+\Big({2}K+\frac{1}{2}\Big)\int^{s}_0 \int^{1}_0(u^2+v^2)\,\mathrm{d}x\,\mathrm{d}t
+\frac{1}{2}\int^{s}_0 \int^{1}_0(f_1^2+f_2^2)\,\mathrm{d}x\,\mathrm{d}t.
\end{align*}
Therefore,
\begin{equation}
\begin{aligned}
&\int_0^1(u^2(x,s)+v^2(x,s))\,\mathrm{d}x
 +\int^{s}_0 \int^{1}_0(x+\eta)^\alpha (u_x^2+v_x^2)\,\mathrm{d}x\,\mathrm{d}t\\
&\leq \int_0^1(u_0^2(x)+v_0^2(x))\,\mathrm{d}x
 +K^2\int^{s}_0 \int^{\kappa}_0(x+\eta)^{-\alpha}(u^2+v^2)\,\mathrm{d}x\,\mathrm{d}t
\\
&\quad+(\kappa^{-\alpha}K^2+4K+1)\int^{s}_0 \int^{1}_0(u^2+v^2)
 \,\mathrm{d}x\,\mathrm{d}t
 +\int^{s}_0 \int^{1}_0(f_1^2+f_2^2)\,\mathrm{d}x\,\mathrm{d}t,
\end{aligned}\label{b8}
\end{equation}
where $0<\kappa<1$ will be determined. It follows from \eqref{b3} and
$0<\alpha<1$ that
\begin{equation}
\begin{aligned}
&\int^{s}_0 \int^{\kappa}_0(x+\eta)^{-\alpha} u^2(x,t)\,\mathrm{d}x\,\mathrm{d}t\\
&=\int^{s}_0 \int^{\kappa}_0(x+\eta)^{-\alpha}
 \Big(\int_0^xu_x(\tilde{x},t)\,\mathrm{d}\tilde{x}\Big)^2\,\mathrm{d}x\,\mathrm{d}t
\\
&\leq \int^{s}_0 \int^{\kappa}_0(x+\eta)^{-\alpha}
 \Big(\int_0^x(\tilde{x}+\eta)^{-\alpha}\,\mathrm{d}\tilde{x}\Big)
\Big(\int_0^x(\tilde{x}+\eta)^\alpha u_x^2(\tilde{x},t)\,\mathrm{d}\tilde{x}
 \Big)\,\mathrm{d}x\,\mathrm{d}t
\\
&\leq \frac{1}{1-\alpha}\int^{\kappa}_0(x+\eta)^{1-2\alpha}\,\mathrm{d}x
 \int^{s}_0 \int^{\kappa}_0(x+\eta)^{\alpha} u_x^2(x,t)\,\mathrm{d}x\,\mathrm{d}t
\\
&\leq \frac{(\kappa+\eta)^{2-2\alpha}-\eta^{2-2\alpha}}{2(1-\alpha)^2}
 \int^{s}_0 \int^{1}_0(x+\eta)^\alpha u_x^2(x,t)\,\mathrm{d}x\,\mathrm{d}t.
\end{aligned} \label{b9}
\end{equation}
Similarly, it holds that
\begin{equation}
\begin{aligned}
&\int^{s}_0 \int^{\kappa}_0(x+\eta)^{-\alpha} v^2(x,t)\,\mathrm{d}x\,\mathrm{d}t\\
&\leq\frac{(\kappa+\eta)^{2-2\alpha}-\eta^{2-2\alpha}}{2(1-\alpha)^2}
\int^{s}_0 \int^{1}_0(x+\eta)^\alpha v_x^2(x,t)\,\mathrm{d}x\,\mathrm{d}t.
\end{aligned} \label{b9-1}
\end{equation}
Take $0<\kappa<1$ so small that
\begin{equation}
K^2 \max _{0\leq\gamma\leq1}((\kappa+\gamma)^{2-2\alpha}
-\gamma^{2-2\alpha})\leq(1-\alpha)^2.
\label{b10}
\end{equation}
Substituting \eqref{b9}--\eqref{b10} into \eqref{b8} yields
\begin{equation}
\begin{aligned}
&\int_0^1(u^2(x,s)+v^2(x,s))\,\mathrm{d}x
 +\frac{1}{2}\int^{s}_0 \int^{1}_0(x+\eta)^\alpha (u_x^2+v_x^2)
 \,\mathrm{d}x\,\mathrm{d}t\\
&\leq \int_0^1(u_0^2(x)+v_0^2(x))\,\mathrm{d}x
+(\kappa^{-\alpha}K^2+4K+1)\int^{s}_0 \int^{1}_0(u^2+v^2)
\,\mathrm{d}x\,\mathrm{d}t \\
&\quad +\int^{s}_0 \int^{1}_0(f_1^2+f_2^2)\,\mathrm{d}x\,\mathrm{d}t.
\end{aligned} \label{b11}
\end{equation}
Then, \eqref{b5} follows from \eqref{b11} and the Gronwall inequality.
For $0\leq t_1<t_2\leq T$ and $\varsigma\in H^1(0,1)$, multiplying \eqref{b1}
and \eqref{b2} by $\varsigma$,
and then integrating over $(0,1)\times(t_1,t_2)$ by parts, we obtain
\begin{gather*}
\begin{aligned}
&\int_0^1(u(x,t_2)-u(x,t_1))\varsigma(x)\,\mathrm{d}x \\
&=\int^{t_2}_{t_1} \int^{1}_0(-(x+\eta)^\alpha u_x\varsigma_x
 +c_1u\varsigma_x-c_3u\varsigma-c_4v\varsigma+f_1\varsigma)\,\mathrm{d}x\,\mathrm{d}t,
\end{aligned}\\
\begin{aligned}
&\int_0^1(v(x,t_2)-v(x,t_1))\varsigma(x)\,\mathrm{d}x \\
&=\int^{t_2}_{t_1} \int^{1}_0(-(x+\eta)^\alpha v_x\varsigma_x+c_2v\varsigma_x
 -c_5u\varsigma-c_6v\varsigma+f_2\varsigma)\,\mathrm{d}x\,\mathrm{d}t,
\end{aligned}
\end{gather*}
which, together with the H\"older inequality and \eqref{b5}, lead to \eqref{b6}.
It remains to prove \eqref{b7}. For $0<\delta<T$ and $0<\tau<T-\delta$,
multiplying \eqref{b1} and \eqref{b2} by $u(x,\tau+\delta)-u(x,\tau)$
and $v(x,\tau+\delta)-v(x,\tau)$, respectively, then integrating over
$(\tau,\tau+\delta)$ with respect to $t$
and summing up, we obtain
\begin{align*}
&(u(x,\tau+\delta)-u(x,\tau))^2+(v(x,\tau+\delta)-v(x,\tau))^2
\\
&=\int_\tau^{\tau+\delta}((x+\eta)^\alpha u_x)_x(u(x,\tau+\delta)-u(x,\tau))
\,\mathrm{d}t \\
&\quad+\int_\tau^{\tau+\delta}((x+\eta)^\alpha v_x)_x(v(x,\tau+\delta)-v(x,\tau))
 \,\mathrm{d}t
\\
&\quad-\int_\tau^{\tau+\delta}(c_1u)_x(u(x,\tau+\delta)-u(x,\tau))\,\mathrm{d}t
-\int_\tau^{\tau+\delta}(c_2v)_x(v(x,\tau+\delta)-v(x,\tau))\,\mathrm{d}t
\\
&\quad+\int_\tau^{\tau+\delta}(f_1-c_3u-c_4v)(u(x,\tau+\delta)-u(x,\tau))\,\mathrm{d}t
\\
&\quad+\int_\tau^{\tau+\delta}(f_2-c_5u-c_6v)(v(x,\tau+\delta)-v(x,\tau))\,\mathrm{d}t.
\end{align*}
Integrating this equality by parts, over $(0,1)\times(0,T-\delta)$,  yields
\begin{align}
&\int^{T-\delta}_0 \int^{1}_0(u(x,\tau+\delta)-u(x,\tau))^2
\,\mathrm{d}x\,\mathrm{d}\tau
+\int^{T-\delta}_0 \int^{1}_0(v(x,\tau+\delta)-v(x,\tau))^2\,\mathrm{d}x\,\mathrm{d}\tau
\nonumber \\
&=-\int^{T-\delta}_0 \int^{1}_0 \int_\tau^{\tau+\delta}(x+\eta)^\alpha u_x(x,t)
(u_x(x,\tau+\delta)-u_x(x,\tau))\,\mathrm{d}t\,\mathrm{d}x\,\mathrm{d}\tau
\nonumber\\
&\quad-\int^{T-\delta}_0 \int^{1}_0 \int_\tau^{\tau+\delta}
(x+\eta)^\alpha v_x(x,t)(v_x(x,\tau+\delta)-v_x(x,\tau))
\,\mathrm{d}t\,\mathrm{d}x\,\mathrm{d}\tau
\nonumber\\
&\quad+\int^{T-\delta}_0 \int^{1}_0 \int_\tau^{\tau+\delta}c_1(x,t)u(x,t)
(u_x(x,\tau+\delta)-u_x(x,\tau))\,\mathrm{d}t\,\mathrm{d}x\,\mathrm{d}\tau
\nonumber\\
&\quad+\int^{T-\delta}_0 \int^{1}_0 \int_\tau^{\tau+\delta}c_2(x,t)v(x,t)
(v_x(x,\tau+\delta)-v_x(x,\tau))\,\mathrm{d}t\,\mathrm{d}x\,\mathrm{d}\tau
\nonumber\\
&\quad+\int^{T-\delta}_0 \int^{1}_0 \int_\tau^{\tau+\delta}(f_1(x,t)-c_3(x,t)u(x,t)
-c_4(x,t)v(x,t)) \nonumber \\
&\quad\times  (u(x,\tau+\delta)-u(x,\tau))\,\mathrm{d}t\,\mathrm{d}x
\,\mathrm{d}\tau
\nonumber\\
&\quad+\int^{T-\delta}_0 \int^{1}_0 \int_\tau^{\tau+\delta}(f_2(x,t)
-c_5(x,t)u(x,t)-c_6(x,t)v(x,t)) \nonumber \\
&\quad\times (v(x,\tau+\delta)-v(x,\tau))\,\mathrm{d}t
\,\mathrm{d}x\,\mathrm{d}\tau
\nonumber\\
&\leq \Big(T\int_\tau^{\tau+\delta} \int^{1}_0(x+\eta)^\alpha u_x^2
\,\mathrm{d}x\,\mathrm{d}t\Big)^{1/2} \nonumber\\
&\quad\times \Big(\delta\int^{T-\delta}_0 \int^{1}_0(x+\eta)^\alpha (u_x(x,\tau
+\delta)-u_x(x,\tau))^2\,\mathrm{d}x\,\mathrm{d}\tau\Big)^{1/2}
\nonumber\\
&\quad+\Big(T\int_\tau^{\tau+\delta} \int^{1}_0(x+\eta)^\alpha v_x^2\,\mathrm{d}
x\,\mathrm{d}t\Big)^{1/2}\nonumber\\
&\quad\times \Big(\delta\int^{T-\delta}_0
\int^{1}_0(x+\eta)^\alpha (v_x(x,\tau+\delta)-v_x(x,\tau))^2
\,\mathrm{d}x\,\mathrm{d}\tau\Big)^{1/2}
\nonumber\\
&\quad+K\Big(T\int_\tau^{\tau+\delta} \int^{1}_0(x+\eta)^{-\alpha}
u^2\,\mathrm{d}x\,\mathrm{d}t\Big)^{1/2}\nonumber\\
&\quad\times\Big(\delta\int^{T-\delta}_0
\int^{1}_0(x+\eta)^\alpha (u_x(x,\tau+\delta)-u_x(x,\tau))^2
\,\mathrm{d}x\,\mathrm{d}\tau\Big)^{1/2}
\nonumber\\
&\quad+K\Big(T\int_\tau^{\tau+\delta} \int^{1}_0(x+\eta)^{-\alpha} v^2\,\mathrm{d}x
\,\mathrm{d}t\Big)^{1/2}\nonumber\\
&\quad\times\Big(\delta\int^{T-\delta}_0
\int^{1}_0(x+\eta)^\alpha (v_x(x,\tau+\delta)-v_x(x,\tau))^2\,\mathrm{d}x
\,\mathrm{d}\tau\Big)^{1/2}
\nonumber\\
&\quad+\Big(T\int_\tau^{\tau+\delta} \int^{1}_0(|f_1|+K|u|+K|v|)^{2}
\,\mathrm{d}x\,\mathrm{d}t\Big)^{1/2}\nonumber\\
&\quad\times\Big(\delta\int^{T-\delta}_0
\int^{1}_0 (u(x,\tau+\delta)-u(x,\tau))^2\,\mathrm{d}x\,\mathrm{d}\tau\Big)^{1/2}
\nonumber\\
&\quad+\Big(T\int_\tau^{\tau+\delta} \int^{1}_0(|f_2|+K|u|+K|v|)^{2}
 \,\mathrm{d}x\,\mathrm{d}t\Big)^{1/2}\nonumber\\
&\quad\times \Big(\delta\int^{T-\delta}_0 \int^{1}_0 (v(x,\tau+\delta)-v(x,\tau))^2
\,\mathrm{d}x\,\mathrm{d}\tau\Big)^{1/2}
\nonumber\\
&\leq 2(T\delta)^{1/2}\int^{T}_0 \int^{1}_0(x+\eta)^\alpha u_x^2
\,\mathrm{d}x\,\mathrm{d}t+2(T\delta)^{1/2}\int^{T}_0
\int^{1}_0(x+\eta)^\alpha v_x^2\,\mathrm{d}x\,\mathrm{d}t
\nonumber\\
&\quad+2K(T\delta)^{1/2}\Big(\int^{T}_0 \int^{1}_0(x+\eta)^{-\alpha}
u^2\,\mathrm{d}x\,\mathrm{d}t\Big)^{1/2}\Big(\int^{T}_0 \int^{1}_0
(x+\eta)^\alpha u_x^2\,\mathrm{d}x\,\mathrm{d}t\Big)^{1/2}
\nonumber\\
&\quad+2K(T\delta)^{1/2}\Big(\int^{T}_0 \int^{1}_0(x+\eta)^{-\alpha} v^2
\,\mathrm{d}x\,\mathrm{d}t\Big)^{1/2}\Big(\int^{T}_0
\int^{1}_0(x+\eta)^\alpha v_x^2\,\mathrm{d}x\,\mathrm{d}t\Big)^{1/2}
\nonumber\\
&\quad+2(3T\delta)^{1/2}\Big(\int^{T}_0 \int^{1}_0(f_1^2+K^2u^2+K^2v^2)
\,\mathrm{d}x\,\mathrm{d}t\Big)^{1/2}
\Big(\int^{T}_0 \int^{1}_0u^2\,\mathrm{d}x\,\mathrm{d}t\Big)^{1/2}
\nonumber\\
&\quad+2(3T\delta)^{1/2}\Big(\int^{T}_0 \int^{1}_0(f_2^2+K^2u^2+K^2v^2)
\,\mathrm{d}x\,\mathrm{d}t\Big)^{1/2}
\Big(\int^{T}_0 \int^{1}_0v^2\,\mathrm{d}x\,\mathrm{d}t\Big)^{1/2}.
\label{b12}
\end{align}
For $0<\kappa<1$ satisfying \eqref{b10}, it follows from \eqref{b9}
 and \eqref{b9-1} that
\begin{gather}
\begin{aligned}
&\int^{T}_0 \int^{1}_0(x+\eta)^{-\alpha}u^2\,\mathrm{d}x
 \,\mathrm{d}t \\
&\leq\frac{1}{2K^2}\int^{T}_0 \int^{1}_0(x+\eta)^{\alpha}u_x^2\,\mathrm{d}x
 \,\mathrm{d}t
+\kappa^{-\alpha}\int^{T}_0 \int^{1}_0u^2\,\mathrm{d}x\,\mathrm{d}t,
\end{aligned} \label{b13-0}\\
\begin{aligned}
&\int^{T}_0 \int^{1}_0(x+\eta)^{-\alpha}v^2\,\mathrm{d}x\,\mathrm{d}t\\
&\leq\frac{1}{2K^2}\int^{T}_0 \int^{1}_0(x+\eta)^{\alpha}v_x^2
 \,\mathrm{d}x\,\mathrm{d}t +\kappa^{-\alpha}\int^{T}_0 \int^{1}_0v^2\,\mathrm{d}x
 \,\mathrm{d}t.
\end{aligned} \label{b13}
\end{gather}
Then, \eqref{b7} follows from \eqref{b12}--\eqref{b13} and \eqref{b5}.
\end{proof}

Using the a priori estimates in Lemma \ref{lem1}, one can prove the
well-posedness of  problem \eqref{a1}--\eqref{a4}.

\begin{proposition} \label{prop12}
Assume that $0<\alpha<1$, and
$P_1,P_2,F_1,F_2$ satisfy \eqref{a12}--\eqref{a16}.
For $h\in L^{2}(Q_T)$ and $u_0, v_0\in L^{2}(0,1)$,
problem \eqref{a1}--\eqref{a4}
admits a unique solution $(u,v)\in {\mathscr H}_\alpha(Q_T)
\times {\mathscr H}_\alpha(Q_T)$.
Furthermore, $u, v\in L^\infty(0,T;L^2(0,1))\cap C_w([0,T];L^2(0,1))$.
Here, ${\mathscr H}_\alpha(Q_T)=\big\{w\in L^2(Q_T):
x^{\alpha/2}w_x\in L^2(Q_T)\big\}$ and a function $\xi\in C_w([0,T];L^2(0,1))$ 
means that
$\int_0^1\xi(x,t)\gamma(x)\,\mathrm{d}x\in C([0,T])$ for each $\gamma\in L^2(0,1)$.
\end{proposition}

The proof of Proposition \ref{prop12} is standard and is omitted here.
We refer to \cite[Theorem 2.1]{w3} for the single equation case.


\section{Uniform Carleman estimate and observability inequality}

In this section, we prove the uniform Carleman estimate and the observability 
inequality for the linear nondegenerate conjugate system
\begin{gather}
\begin{gathered}
U^\eta_t+((x+\eta)^\alpha U^\eta_x)_{x}+{c_1}(x,t)U^\eta_{x}
-{c_3}(x,t)U^\eta-{c_4}(x,t)V^\eta=0, \\
(x,t)\in Q_T, \end{gathered} \label{a.1}
\\
V^\eta_t+((x+\eta)^\alpha V^\eta_x)_{x}+{c_2}(x,t)V^\eta_{x}-{c_5}(x,t)V^\eta=0,
\quad (x,t)\in Q_T,\label{a.2}
\\
U^\eta(0,t)=V^\eta(0,t)=0,\quad U^\eta(1,t)=V^\eta(1,t)=0, \quad t\in (0,T),
\label{a.3}
\\
U^\eta(x,T)=U_T(x),\quad V^\eta(x,T)=V_T(x), \quad x\in(0,1), \label{a.4}
\end{gather}
where $0<\alpha<1/2$, $0<\eta<1$, ${c_i}\in L^{\infty}(Q_T)\,(1\le i\le 5)$, 
and $U_T, V_T\in L^2(0,1)$.

Let $\psi\in C^\infty([0,1])$ satisfy
$$
\psi\begin{cases}
=1, & x\in[0,{(3x_0+2x_1)}/{5}],
\\
\in [0,1], & x\in\tilde{\omega},
\\
=0,& x\in[{(2x_0+3x_1)}/{5},1],
\end{cases}
$$
where $\tilde{\omega}=({(3x_0+2x_1)}/{5},{(2x_0+3x_1)}/{5})$ and
 $\hat{\omega}=({(4x_0+x_1)}/{5},{(x_0+4x_1)}/{5})$.
Set
\begin{gather*}
\varphi(x,t)=\theta(t)g(x),\quad
\Psi(x,t)=\theta(t)\big(\mathrm{e}^{2\zeta(0)}-\mathrm{e}^{\zeta(x)}\big),
\quad(x,t)\in Q_T,
\\
\Phi(x,t)=\psi(x)\varphi(x,t)-(1-\psi(x))\Psi(x,t),\quad(x,t)\in Q_T,
\end{gather*}
where
\begin{gather*}
\theta(t)=\frac{1}{(t(T-t))^{4}},\quad t\in(0,T),
\\
g(x)=8((x+\eta)^{(4-2\alpha)/3}-8),
\quad
\zeta(x)=(1+\eta)^{1-\alpha/2}-(x+\eta)^{1-\alpha/2},\quad x\in(0,1).
\end{gather*}

The Carleman estimate for the solution to  problem \eqref{a.1}--\eqref{a.4} 
is as follows.

\begin{theorem} [Uniform Carleman estimate] \label{thm1}
Assume that $0<\alpha<1/2$, $0<\eta<1$, $c_{i}\in{L^\infty(Q_T)}$ with
$\|c_{i}\|_{L^\infty(Q_T)}\leq K\,(1\leq i\leq5)$, and 
$c_4\big|_{(x_0,x_1)\times(0,T)}\geq c_0$ or
$c_4\big|_{(x_0,x_1)\times(0,T)}\leq-c_0$.
There exist two constants $s_0>0$ and $M_0>0$ depending only on $x_0$, $x_1$,
$K$, $c_0$, $T$, and $\alpha$,
such that for each $U_T, V_T\in L^{2}(0,1)$,
the solution $(U^\eta,V^\eta)$ to problem \eqref{a.1}--\eqref{a.4} satisfies
\begin{align*}
&\int^{T}_0 \int^{1}_0(s\theta (U^\eta_{x})^{2}+s^{3} \theta^{3} (U^\eta)^{2}
+s\theta (V^\eta_{x})^{2}+s^{3} \theta^{3} (V^\eta)^{2})\mathrm{e}^{2s\Phi}
\,\mathrm{d}x\,\mathrm{d}t \\
&\leq M_0\int^{T}_0 \int_{\omega}(U^\eta)^{2}\,\mathrm{d}x\,\mathrm{d}t,\quad 
s\geq s_0.
\end{align*}
\end{theorem}

\begin{proof}
For convenience, $U^\eta$ and $V^\eta$ are abbreviated by $U$ and $V$, respectively,
 in the proof. Without loss of generality,
it is assumed that $U, V\in C^2(\overline Q_T)$.
Set
\begin{equation}
w(x,t)=\psi(x)U(x,t),\quad W(x,t)=\psi(x)V(x,t),\quad (x,t)\in Q_T. \label{c5}
\end{equation}
Then, $(w,W)$ solves
\begin{align}
w_t+((x+\eta)^{\alpha}w_{x})_{x}=\rho_1,\quad
W_t+((x+\eta)^{\alpha}W_{x})_{x}=\rho_2,\quad (x,t)\in Q_T,
\label{c9}
\end{align}
where
\begin{gather}
\rho_1=\varrho_1-{c_1}w_{x}+{c_3}w,\quad \rho_2=\varrho_2-{c_2}W_{x}+{c_5}W,
\quad (x,t)\in Q_T, \label{c8}
\\
\varrho_1=((x+\eta)^{\alpha}\psi'U)_{x}+\psi'(x+\eta)^{\alpha}U_{x}
+{c_1}\psi'U+{c_4}\psi V,\quad (x,t)\in Q_T, \label{f7}
\\
\varrho_2=((x+\eta)^{\alpha}\psi'V)_{x}+\psi'(x+\eta)^{\alpha}V_{x}+{c_2}\psi'V,\quad
(x,t)\in Q_T. \label{c7}
\end{gather}
Set
\begin{equation}
Y(x,t)=\mathrm{e}^{s\varphi(x,t)}w(x,t),\quad Z(x,t)
=\mathrm{e}^{s\varphi(x,t)}W(x,t),\quad (x,t)\in Q_T. \label{c10}
\end{equation}
From \eqref{c9} it follows that
\begin{gather*}
\mathrm{e}^{s\varphi}((\mathrm{e}^{-s\varphi}Y)_t
 +((x+\eta)^{\alpha}(\mathrm{e}^{-s\varphi}Y)_{x})_{x})
=\mathrm{e}^{s\varphi}\rho_1,\quad (x,t)\in Q_T,
\\
\mathrm{e}^{s\varphi}((\mathrm{e}^{-s\varphi}Z)_t+((x+\eta)^{\alpha}
(\mathrm{e}^{-s\varphi}Z)_{x})_{x})=\mathrm{e}^{s\varphi}\rho_2,\quad (x,t)\in Q_T.
\end{gather*}
From \cite[Proposition 3.1]{w3}, there exist three positive constants
 $M_1$, $M_2$, and $s_1$ depending only on $T$ and $\alpha$,
such that for each $s\geq s_1$,
\begin{align*}
&M_1s\int^{T}_0 \int^{1}_0(x+\eta)^{(4\alpha-2)/3}\theta (Y_x^2+Z_x^2)
 \,\mathrm{d}x\,\mathrm{d}t
+M_2s^3\int^{T}_0 \int^{1}_0\theta^3(Y^2+Z^2)\,\mathrm{d}x\,\mathrm{d}t \\
&\leq\int^{T}_0 \int^{1}_0(\rho_1^2+\rho_2^2)\mathrm{e}^{2s\varphi}
 \,\mathrm{d}x\,\mathrm{d}t.
\end{align*}
This formula, together with \eqref{c8} and \eqref{c10}, leads to that for each
 $s\geq s_1$,
\begin{align*}
&M_1s\int^{T}_0 \int^{1}_0(x+\eta)^{(4\alpha-2)/3}
 \theta (Y_x^2+Z_x^2)\,\mathrm{d}x\,\mathrm{d}t
+M_2s^3\int^{T}_0 \int^{1}_0\theta^3(Y^2+Z^2)\,\mathrm{d}x\,\mathrm{d}t
\\
&\leq 2\int^{T}_0 \int^{1}_0(\varrho_1^2+\varrho_2^2)\mathrm{e}^{2s\varphi}
 \,\mathrm{d}x\,\mathrm{d}t \\
&\quad +M_3\int^{T}_0 \int^{1}_0(Y_x^2+Z_x^2+Y^2+Z^2
 +s^2(x+\eta)^{2(1-2\alpha)/3}\theta^2(Y^2+Z^2))\,\mathrm{d}x\,\mathrm{d}t,
\end{align*}
where $M_3>0$ depends only on $K$, $T$, and $\alpha$.
Therefore, there exist $s_2\ge s_1$ and $M_4>0$ depend only on $K$, $T$,
and $\alpha$, such that for each $s\geq s_2$,
\begin{equation} \label{new1}
\begin{aligned}
&s\int^{T}_0 \int^{1}_0(x+\eta)^{(4\alpha-2)/3}\theta((\mathrm{e}^{-s\varphi}Y)_x^2
+(\mathrm{e}^{-s\varphi}Z)_x^2)\mathrm{e}^{2s\varphi}\,\mathrm{d}x\,\mathrm{d}t\\
&+s^3\int^{T}_0 \int^{1}_0\theta^3 (Y^2+Z^2)\,\mathrm{d}x\,\mathrm{d}t
\\
&\leq M_4\int^{T}_0 \int^{1}_0(\varrho_1^2
 +\varrho_2^2)\mathrm{e}^{2s\varphi}\,\mathrm{d}x\,\mathrm{d}t.
\end{aligned}
\end{equation}
From \eqref{new1}, $0<\alpha<1/2$, \eqref{f7}--\eqref{c10} and \eqref{c5}, we obtain
\begin{align*}
&\int^{T}_0 \int^{1}_0(s\theta w_{x}^{2}+s^{3} \theta^{3} w^{2}
 +s\theta W_{x}^{2}+s^{3} \theta^{3} W^{2})\mathrm{e}^{2s\varphi}
 \,\mathrm{d}x\,\mathrm{d}t
\\
&\leq M_{5}\Big(\int^{T}_0 \int_{\tilde{\omega}}(U^{2}+U_x^{2}+V^{2}+V_x^{2})\mathrm{e}^{2s\varphi}\,\mathrm{d}x\,\mathrm{d}t
+\int^{T}_0 \int^{1}_0W^2\mathrm{e}^{2s\varphi}\,\mathrm{d}x\,\mathrm{d}t\Big),
\quad s\ge s_2,
\end{align*}
where $M_{5}>0$ depends only on $K$, $T$, $x_0$, $x_1$, and $\alpha$.
Therefore, there exist $s_3\ge s_2$ and $M_6>0$ depend only on $K$, $T$, $x_0$,
$x_1$, and $\alpha$, such that for each $s\geq s_3$,
\begin{align*}
&\int^{T}_0 \int^{1}_0(s\theta w_{x}^{2}+s^{3} \theta^{3} w^{2}
+s\theta W_{x}^{2}+s^{3} \theta^{3} W^{2})\mathrm{e}^{2s\varphi}
\,\mathrm{d}x\,\mathrm{d}t \\
&\leq M_{6}\int^{T}_0 \int_{\tilde{\omega}}(U^{2}+U_x^{2}
+V^{2}+V_x^{2})\mathrm{e}^{2s\varphi}\,\mathrm{d}x\,\mathrm{d}t.
\end{align*}
This formula, together with \eqref{c5} and the definition of $\psi$, leads
 to that for each $s\geq s_3$,
\begin{equation}
\begin{aligned}
&\int^{T}_0 \int^{{(3x_0+2x_1)}/{5}}_0(s\theta U_{x}^{2}+s^{3} \theta^{3} U^{2}
+s\theta V_{x}^{2}+s^{3} \theta^{3} V^{2})\mathrm{e}^{2s\Phi}
\,\mathrm{d}x\,\mathrm{d}t
\\
&\leq M_{6}\int^{T}_0 \int_{\tilde{\omega}}(U^{2}+U_x^{2}+V^{2}
+V_x^{2})\mathrm{e}^{2s\varphi}\,\mathrm{d}x\,\mathrm{d}t.
\end{aligned}\label{c13}
\end{equation}
Set
$$
q(x,t)=(1-\psi(x))U(x,t),\quad Q(x,t)=(1-\psi(x))V(x,t),
\quad(x,t)\in Q_T.
$$
From the classical Carleman estimate \cite[Proposition 4.2]{al},
there exist $s_4>0$ and $M_7>0$ depend only on $K$, $T$, $x_0$, $x_1$, and
$\alpha$, such that for each $s\geq s_4$,
\begin{align*}
&\int^{T}_0 \int^{1}_0(s\theta \mathrm{e}^{\zeta}q_{x}^{2}+s^{3}
 \theta^{3}\mathrm{e}^{3\zeta} q^{2}
+s\theta \mathrm{e}^{\zeta}Q_{x}^{2}+s^{3} \theta^{3}
 \mathrm{e}^{3\zeta}Q^{2})\mathrm{e}^{-2s\Psi} \,\mathrm{d}x\,\mathrm{d}t
\\
&\leq M_{7}\int^{T}_0 \int_{\tilde{\omega}}(U^{2}+U_x^{2}
 +V^{2}+V_x^{2})\mathrm{e}^{-2s\Psi}\,\mathrm{d}x\,\mathrm{d}t.
\end{align*}
This formula, and the definition of $\psi$, leads to
\begin{equation}
\begin{aligned}
&\int^{T}_0 \int^{1}_{{(2x_0+3x_1)}/{5}}(s\theta \mathrm{e}^{\zeta}U_{x}^{2}
+s^{3} \theta^{3} \mathrm{e}^{3\zeta}U^{2}+s\theta \mathrm{e}^{\zeta}V_{x}^{2}
+s^{3} \theta^{3}\mathrm{e}^{3\zeta} V^{2})\mathrm{e}^{2s\Phi}
\,\mathrm{d}x\,\mathrm{d}t
\\
&\leq M_{7}\int^{T}_0 \int_{\tilde{\omega}}(U^{2}
+U_x^{2}+V^{2}+V_x^{2})\mathrm{e}^{-2s\Psi}\,\mathrm{d}x\,\mathrm{d}t,
\quad s\geq s_4.
\end{aligned} \label{c14}
\end{equation}
It follows from \eqref{c13}, \eqref{c14} and the definition of
$\Phi,\varphi,\Psi,\psi$ that
\begin{equation}
\begin{aligned}
&\int^{T}_0 \int^{1}_0(s\theta U_{x}^{2}+s^{3} \theta^{3} U^{2}
+s\theta V_{x}^{2}+s^{3} \theta^{3} V^{2})\mathrm{e}^{2s\Phi}
 \,\mathrm{d}x\,\mathrm{d}t
\\
&\leq M_{8}\int^{T}_0 \int_{\tilde{\omega}}(U^{2}+U_x^{2}
+V^{2}+V_x^{2})(\mathrm{e}^{2s\varphi}+\mathrm{e}^{-2s\Psi}
+\mathrm{e}^{2s\Phi})\,\mathrm{d}x\,\mathrm{d}t
\\
&\leq 3M_{8}\int^{T}_0 \int_{\tilde{\omega}}(U^{2}+U_x^{2}+V^{2}+V_x^{2})
 \mathrm{e}^{2s\Phi}\,\mathrm{d}x\,\mathrm{d}t,\quad s\geq\max\{s_3, s_4\},
\end{aligned}\label{c15}
\end{equation}
where $M_{8}>0$ depends only on $K$, $T$, $x_0$, $x_1$, and $\alpha$.
Let ${\xi_1}\in C^\infty([0,1])$ such that
$\operatorname{supp}{\xi_1}\subset\hat\omega$, $0\le {\xi_1}\le1$ in $(0,1)$
and ${\xi_1}\equiv1$ in $\tilde{\omega}$.
For $s>0$, \eqref{a.1}--\eqref{a.3} show
\begin{align*}
0&=\int_0^T\frac{d}{dt}\int_0^1\xi_1^2 (U^2+V^{2})
\mathrm{e}^{2s\Phi}\,\mathrm{d}x\,\mathrm{d}t
\\
&=2s\int_0^T \int_0^1\xi_1^2\Phi_t (U^2+V^{2})\mathrm{e}^{2s\Phi}
 \,\mathrm{d}x\,\mathrm{d}t \\
&\quad +2\int_0^T \int_0^1\xi_1^2 ((x+\eta)^{\alpha} U_x^2+(x+\eta)^{\alpha}
 V_{x}^{2})\mathrm{e}^{2s\Phi}\,\mathrm{d}x\,\mathrm{d}t
\\
&\quad+4\int_0^T \int_0^1{\xi_1}{\xi'_1} (U(x+\eta)^{\alpha} U_x+V(x+\eta)^{\alpha} V_x)\mathrm{e}^{2s\Phi}\,\mathrm{d}x\,\mathrm{d}t
\\
&\quad-2\int_0^T \int_0^1\xi_1^2 ({c_1}UU_x+{c_2}VV_x)\mathrm{e}^{2s\Phi}
 \,\mathrm{d}x\,\mathrm{d}t
\\
&\quad
+4s\int_0^T \int_0^1\xi_1^2\Phi_x (U(x+\eta)^{\alpha}
 U_x+V(x+\eta)^{\alpha} V_x)\mathrm{e}^{2s\Phi}\,\mathrm{d}x\,\mathrm{d}t
\\
&\quad+2\int_0^T \int_0^1\xi_1^2 ({c_3}U^2+{c_5}V^2+{c_4}UV)
\mathrm{e}^{2s\Phi}\,\mathrm{d}x\,\mathrm{d}t,
\end{align*}
which leads to
\begin{align*}
&\int_0^T \int_0^1\xi_1^2 ((x+\eta)^{\alpha} U_x^2+(x+\eta)^{\alpha}
V_{x}^{2})\mathrm{e}^{2s\Phi}\,\mathrm{d}x\,\mathrm{d}t
\\
&\leq \frac12\int_0^T \int_0^1\xi_1^2 ((x+\eta)^{\alpha} U_x^2
 +(x+\eta)^{\alpha} V_{x}^{2})\mathrm{e}^{2s\Phi}\,\mathrm{d}x\,\mathrm{d}t\\
&\quad +M_9(1+s^2)\int_0^T \int_{\hat{\omega}}\theta^2 (U^2+V^{2})
 \mathrm{e}^{2s\Phi}\,\mathrm{d}x\,\mathrm{d}t,
\end{align*}
with $M_{9}>0$ depending only on $K$, $T$, $x_0$, $x_1$, and $\alpha$.
Hence, for $s>0$,
\begin{equation}
\int^{T}_0 \int_{\tilde{\omega}}(U_x^2+V^2_x)\mathrm{e}^{2s\Phi}
\,\mathrm{d}x\,\mathrm{d}t
\leq M_{10}(1+s^2)\int_0^T \int_{\hat{\omega}}\theta^2 (U^2+V^{2})
\mathrm{e}^{2s\Phi}\,\mathrm{d}x\,\mathrm{d}t,
\label{new2}
\end{equation}
where $M_{10}>0$ depends only on $K$, $T$, $x_0$, $x_1$, and $\alpha$.
From \eqref{c15} and \eqref{new2}, there exist $s_5\ge\max\{s_3, s_4\}$
and $M_{11}>0$ depend only on $K$, $T$, $x_0$, $x_1$, and $\alpha$,
such that for each $s\geq s_5$,
\begin{equation}
\begin{aligned}
&\int^{T}_0 \int^{1}_0(s\theta U_{x}^{2}+s^{3} \theta^{3} U^{2}
 +s\theta V_{x}^{2}+s^{3} \theta^{3} V^{2})\mathrm{e}^{2s\Phi}
 \,\mathrm{d}x\,\mathrm{d}t \\
&\leq M_{11}\int^{T}_0 \int_{\hat{\omega}}(U^{2}+V^{2})
 \mathrm{e}^{2s\Phi}\,\mathrm{d}x\,\mathrm{d}t.
\end{aligned} \label{c16}
\end{equation}
Let $\xi_2\in C^\infty([0,1])$ such that
$\operatorname{supp}\xi_2\subset(x_0,x_1)$, $0\le \xi_2\le1$ in
$(0,1)$ and $\xi_2\equiv1$ in $\hat{\omega}$.
Multiplying \eqref{a.1} by $\xi_2 V\mathrm{e}^{2s\Phi}$, then integrating
by parts and using \eqref{a.2}, one gets
\begin{equation}
\begin{aligned}
&\int^{T}_0 \int^{1}_0c_4\xi_2 V^2\mathrm{e}^{2s\Phi}\,\mathrm{d}x\,\mathrm{d}t
\\
&=-2\int^{T}_0 \int^{1}_0(x+\eta)^\alpha\xi_2 U_xV_x\mathrm{e}^{2s\Phi}
 \,\mathrm{d}x\,\mathrm{d}t
+\int^{T}_0 \int^{1}_0c_1\xi_2 U_xV\mathrm{e}^{2s\Phi}\,\mathrm{d}x
 \,\mathrm{d}t \\
&\quad +\int^{T}_0 \int^{1}_0c_2\xi_2 UV_x\mathrm{e}^{2s\Phi}
\,\mathrm{d}x\,\mathrm{d}t
-\int^{T}_0 \int^{1}_0\Big((c_3+c_5+2s\Phi_t)\xi_2 \\
&\quad -((x+\eta)^\alpha (\xi_2\mathrm{e}^{2s\Phi})_x)_x\mathrm{e}^{-2s\Phi}\Big)
UV\mathrm{e}^{2s\Phi}\,\mathrm{d}x\,\mathrm{d}t.
\end{aligned}\label{c17}
\end{equation}
The H\"older inequality gives
\begin{gather}
\begin{aligned}
&\big|\int^{T}_0 \int^{1}_0(x+\eta)^\alpha\xi_2 U_x V_x\mathrm{e}^{2s\Phi}
 \,\mathrm{d}x\,\mathrm{d}t\big| \\
&\leq\frac12\int^{T}_0 \int^{1}_0\theta U_x^2\mathrm{e}^{2s\Phi}\,\mathrm{d}x
 \,\mathrm{d}t
 +\frac{1}{2}\int^{T}_0 \int^{1}_0\theta^{-1} V_x^2\mathrm{e}^{2s\Phi}
 \,\mathrm{d}x\,\mathrm{d}t,
\end{aligned} \label{c18}\\
\begin{aligned}
&\big|\int^{T}_0 \int^{1}_0c_1\xi_2 U_x V\mathrm{e}^{2s\Phi}
 \,\mathrm{d}x\,\mathrm{d}t\big| \\
&\leq\frac12\int^{T}_0 \int^{1}_0\theta U_x^2\mathrm{e}^{2s\Phi}
 \,\mathrm{d}x\,\mathrm{d}t
+\frac12{K^2}\int^{T}_0 \int^{1}_0\theta^{-1} V^2\mathrm{e}^{2s\Phi}
\,\mathrm{d}x\,\mathrm{d}t,
\end{aligned} \label{c19}\\
\begin{aligned}
&\big|\int^{T}_0 \int^{1}_0c_2\xi_2 UV_x\mathrm{e}^{2s\Phi}\,\mathrm{d}x
 \,\mathrm{d}t\Big| \\
&\leq\frac12\int^{T}_0 \int^{1}_0\theta V_x^2\mathrm{e}^{2s\Phi}
 \,\mathrm{d}x\,\mathrm{d}t 
 +\frac12{K^2}\int^{T}_0 \int_{\omega}\theta^{-1}U^2
 \mathrm{e}^{2s\Phi}\,\mathrm{d}x\,\mathrm{d}t,
\end{aligned}\label{c20}
\end{gather}
and
\begin{equation}
\begin{aligned}
&\big|\int^{T}_0 \int^{1}_0((c_3+c_5+2s\Phi_t)\xi_2
 -((x+\eta)^\alpha(\xi_2\mathrm{e}^{2s\Phi})_x)_x\mathrm{e}^{-2s\Phi})
 UV\mathrm{e}^{2s\Phi}\,\mathrm{d}x\,\mathrm{d}t\big|
\\
&\leq\frac12(1+s^2)\int^{T}_0 \int^{1}_0\theta^3 V^2\mathrm{e}^{2s\Phi}
 \,\mathrm{d}x\,\mathrm{d}t \\
&\quad +M_{12}(1+s^2)\int^{T}_0 \int_{\omega}\theta^{-3}U^2\mathrm{e}^{2s\Phi}\,\mathrm{d}x\,\mathrm{d}t,
\end{aligned} \label{c21}
\end{equation}
where $M_{12}>0$ depends only on $K$, $T$, $x_0$, $x_1$, and $\alpha$.
Since $c_4\big|_{(x_0,x_1)\times(0,T)}\geq c_0$ or
$c_4\big|_{(x_0,x_1)\times(0,T)}\leq-c_0$,
it follows from \eqref{c17}--\eqref{c21} and the definition of $\xi_2$ that
\begin{equation}
\begin{aligned}
& c_0\int^{T}_0 \int_{\hat\omega}V^2\mathrm{e}^{2s\Phi}\,\mathrm{d}x\,\mathrm{d}t\\
&\leq \frac32\int^{T}_0 \int^{1}_0\theta U_x^2\mathrm{e}^{2s\Phi}
 \,\mathrm{d}x\,\mathrm{d}t
+\frac12\int^{T}_0 \int^{1}_0\theta V_x^2\mathrm{e}^{2s\Phi}
 \,\mathrm{d}x\,\mathrm{d}t \\
&\quad +2\int^{T}_0 \int^{1}_0\theta^{-1} V_x^2\mathrm{e}^{2s\Phi}
 \,\mathrm{d}x\,\mathrm{d}t 
 +\frac12{K^2}\int^{T}_0 \int^{1}_0\theta^{-1} (U^2+V^2)\mathrm{e}^{2s\Phi}
 \,\mathrm{d}x\,\mathrm{d}t \\
&\quad +\frac12(1+s^2)\int^{T}_0 \int^{1}_0\theta^3 V^2\mathrm{e}^{2s\Phi}
 \,\mathrm{d}x\,\mathrm{d}t \\
&\quad +M_{12}(1+s^2)\int^{T}_0 \int_{\omega}\theta^{-3}U^2\mathrm{e}^{2s\Phi}
 \,\mathrm{d}x\,\mathrm{d}t.
\end{aligned} \label{c22}
\end{equation}
Then the theorem follows from \eqref{c16} and \eqref{c22}.
\end{proof}

Below we prove the observability inequality for the solution to 
 problem \eqref{a.1}--\eqref{a.4}.

\begin{theorem}[Uniform observability inequality] \label{thm2}
Assume that $0<\alpha<1/2$, $0<\eta<1$, $c_{i}\in{L^\infty(Q_T)}$ with
$\|c_{i}\|_{L^\infty(Q_T)}\leq K\,(1\leq i\leq5)$, and 
$c_4\big|_{(x_0,x_1)\times(0,T)}\geq c_0$ or 
$c_4\big|_{(x_0,x_1)\times(0,T)}\leq-c_0$.
There exists $M>0$ depending only on $x_0$, $x_1$, $K$, $c_0$, $T$, 
and $\alpha$, but independent of $\eta$,
such that for each $U_T, V_T\in L^{2}(0,1)$, the solution $(U^\eta,V^\eta)$ 
to problem \eqref{a.1}--\eqref{a.4} satisfies
\[
\int^{1}_0((U^\eta)^{2}(x,0)+(V^\eta)^{2}(x,0))dx 
\leq M\int^{T}_0\int_{\omega}(U^\eta)^{2}\,\mathrm{d}x\,\mathrm{d}t.
\]
\end{theorem}

\begin{proof}
As in Theorem \ref{thm1}, $U^\eta$ and $V^\eta$ are abbreviated by $U$ and $V$, 
respectively, and it is assumed that $U, V\in C^2(\overline Q_T)$.
Multiplying \eqref{a.1} and \eqref{a.2} by $U$ and $V$, respectively, and 
then integrating over $(0,1)$ with respect to $x$, we obtain
\begin{gather*}
\begin{aligned}
&\frac{1}{2}\frac{d}{dt}\int^{1}_0U^{2}\,\mathrm{d}x
-\int^{1}_0(x+\eta)^{\alpha }U^{2}_{x}\mathrm {d}x
+\int^{1}_0{c_1}U U_{x}\,\mathrm{d}x-\int^{1}_0{c_3}U^{2}
\mathrm {d}x-\int^{1}_0{c_4}UV\mathrm {d}x \\
&=0,
\end{aligned}\\
\frac{1}{2}\frac{d}{dt}\int^{1}_0V^{2}\,\mathrm{d}x
-\int^{1}_0(x+\eta)^{\alpha}V^{2}_{x}\mathrm {d}x
+\int^{1}_0{c_2}V V_{x}\,\mathrm{d}x-\int^{1}_0{c_5}V^2\,\mathrm{d}x=0,
\end{gather*}
for $t\in (0,T)$.
Using the H\"older inequality and the Hardy inequality yields
\[
-\frac{d}{dt}\int^{1}_0(U^{2}+V^{2})\,\mathrm{d}x
\leq\tilde M\int^{1}_0(U^{2}+V^{2})\,\mathrm{d}x,\quad t\in (0,T),
\]
where $\tilde M>0$ depends only on $K$ and $\alpha$.
Hence
\begin{equation} \label{c23}
\int^{1}_0(U^{2}(x,0)+V^{2}(x,0))\,\mathrm{d}x
\leq\mathrm{e}^{\tilde{M} t}\int^{1}_0(U^{2}(x,t)+V^{2}(x,t))
\,\mathrm{d}x,\quad t\in (0,T).
\end{equation}
Integrating \eqref{c23} over $[T/4,3T/4]$ leads to
\begin{align}
\label{c24}
\frac{T}{2}\int^{1}_0(U^{2}(x,0)+V^{2}(x,0))\,\mathrm{d}x\leq
\mathrm{e}^{3\tilde M T/4}\int^{3T/4}_{T/4}
\int^{1}_0(U^{2}+V^{2})\,\mathrm{d}x \,\mathrm{d}t.
\end{align}
The theorem follows from \eqref{c24}, the Hardy inequality and Theorem \ref{thm1}.
\end{proof}

By a standard limit process, one can get the Carleman estimate and the
 observability inequality for the degenerate parabolic system
\begin{gather}
U_t+(x^{\alpha}U_x)_{x}+{c_1}(x,t)U_{x}-{c_3}(x,t)U-{c_4}(x,t)V=0, 
\quad (x,t)\in Q_T, \label{c1}
\\
V_t+(x^{\alpha}V_x)_{x}+{c_2}(x,t)V_{x}-{c_5}(x,t)V=0, \quad (x,t)\in Q_T,
\label{c2}
\\
U(0,t)=V(0,t)=0,\quad U(1,t)=V(1,t)=0, \quad t\in (0,T), \label{c3}
\\
U(x,T)=U_T(x),\quad V(x,T)=V_T(x), \quad x\in(0,1), \label{c4}
\end{gather}
where $0<\alpha<1/2$, ${c_i}\in L^{\infty}(Q_T)\,(1\le i\le 5)$, and 
$U_T, V_T\in L^2(0,1)$.

\begin{theorem} \label{thm3}
Assume that $0<\alpha<1/2$, $0<\eta<1$, $c_{i}\in{L^\infty(Q_T)}$ with
$\|c_{i}\|_{L^\infty(Q_T)}\leq K\,(1\leq i\leq5)$, and 
$c_4\big|_{(x_0,x_1)\times(0,T)}\geq c_0$ or 
$c_4\big|_{(x_0,x_1)\times(0,T)}\leq-c_0$.
There exist three positive constants $s_0$, $M_0$ and $M$ depending only on 
$x_0$, $x_1$, $K$, $c_0$, $T$, and $\alpha$,
such that for each $U_T,V_T\in L^{2}(0,1)$,
the solution $(U,V)$ to problem \eqref{c1}--\eqref{c4} satisfies
\begin{gather*}
\begin{aligned}
&\int^{T}_0 \int^{1}_0(s\theta U_{x}^{2}+s^{3} \theta^{3} U^{2}
+s\theta V_{x}^{2}+s^{3} \theta^{3} V^{2})\mathrm{e}^{2s\tilde\Phi} 
\,\mathrm{d}x\,\mathrm{d}t \\
&\leq M_0\int^{T}_0 \int_{\omega}U^{2}\,\mathrm{d}x\,\mathrm{d}t,\quad s\geq s_0,
\end{aligned} \\
\int^{1}_0(U^{2}(x,0)+V^{2}(x,0))dx 
\leq M\int^{T}_0\int_{\omega}U^{2}\,\mathrm{d}x\,\mathrm{d}t,
\end{gather*}
where
\[
\tilde\Phi(x,t)=8\theta(t)\psi(x)(x^{(4-2\alpha)/3}-8)
-\theta(t)(1-\psi(x))\big(\mathrm{e}^{2}
-\mathrm{e}^{1-x^{1-\alpha/2}}\big),
\]
for $(x,t)\in Q_T$.
\end{theorem}

\section{Null controllability}

In this section, we study the null controllability of system \eqref{a1}--\eqref{a4}.
First consider the nondegenerate parabolic system
\begin{gather}
u^{\eta}_t-((x+\eta)^{\alpha}u^\eta_x)_{x}+(P_1(x,t,u^\eta))_{x}
+F_1(x,t,u^\eta)=h^\eta\chi_\omega, \quad (x,t)\in Q_T, \label{d1}
\\
v^\eta_t-((x+\eta)^{\alpha}v^\eta_x)_{x}+(P_2(x,t,v^\eta))_{x}
+F_2(x,t,u^\eta,v^\eta)=0, \quad (x,t)\in Q_T, \label{d2}
\\
u^\eta(0,t)=v^\eta(0,t)=0,\quad u^\eta(1,t)=v^\eta(1,t)=0, \quad t\in (0,T),
\label{d3}
\\
u^\eta(x,0)=u_0(x),\quad v^\eta(x,0)=v_0(x), \quad x\in(0,1), \label{d4}
\end{gather}
where $0<\alpha<1/2$, $0<\eta<1$, and $u_0, v_0\in L^2(0,1)$.

\begin{lemma} \label{lem51}
Assume that $0<\alpha<1/2$, $0<\eta<1$,
and $P_1,P_2,F_1,F_2$ satisfy \eqref{a12}--\eqref{a16}. For each 
$u_0, v_0\in L^2(0,1)$,
there exists $h^\eta\in L^2(Q_T)$, such that the solution $(u^\eta,v^\eta)$ 
to problem \eqref{d1}--\eqref{d4} satisfies
\begin{align}
u^\eta(x,T)=v^\eta(x,T)=0,\quad x\in (0,1). \label{111}
\end{align}
Furthermore, there exists $M>0$ depending only on $x_0$, $x_1$, $K$, $c_0$, $T$, 
and $\alpha$, such that
\begin{equation}
\|h^\eta\|_{L^2(Q_T)}\leq M(\|u_0\|_{L^2(0,1)}+\|v_0\|_{L^2(0,1)}).\label{222}
\end{equation}
\end{lemma}

\begin{proof}
For $(x,t,y,z)\in Q_T\times\mathbb{R}^2$,
set
\begin{gather*}
c_1(x,t,y)=\begin{cases}
\frac{P_1(x,t,y)-P_1(x,t,0)}{y}, &y\neq0,\\
0,& y=0,
\end{cases}\\
c_2(x,t,z)=\begin{cases}
\frac{P_2(x,t,z)-P_2(x,t,0)}{z}, &z\neq0,\\
0,& z=0,
\end{cases}\\
c_3(x,t,y)=\begin{cases}
\frac{F_1(x,t,y)-F_1(x,t,0)}{y}, &y\neq0,\\
0,& y=0,
\end{cases} \\
c_4(x,t,y,z)=\int^1_0\frac{\partial F_2}{\partial y}(x,t,\lambda y,\lambda z)
\,\mathrm{d}\lambda,
\quad
c_5(x,t,y,z)=\int^1_0\frac{\partial F_2}{\partial z}(x,t,\lambda y,\lambda z)
\,\mathrm{d}\lambda.
\end{gather*}
Then, \eqref{a14}--\eqref{a16} show that
$c_1, c_2, c_3\in L^\infty(Q_T\times\mathbb{R})$ and 
$c_4, c_5\in L^\infty(Q_T\times\mathbb{R}^2)$ satisfy
\begin{gather*}
\|c_{i}\|_{L^\infty(Q_T\times\mathbb{R})}\leq K\quad(i=1,2,3)\\
\|c_{j}\|_{L^\infty(Q_T\times\mathbb{R}^2)}\leq K\quad (j=4,5),\\
c_4\big|_{(x_0,x_1)\times(0,T)\times\mathbb{R}^2}\geq c_0
\quad\text{or}\quad 
c_4\big|_{(x_0,x_1)\times(0,T)\times\mathbb{R}^2}\leq-c_0.
\end{gather*}
Let $y$, $z\in L^2(Q_T)$. For $\varepsilon>0$, consider the problem
\begin{equation}
\min \Big\{\int^{T}_0 \int^{1}_0h^2\,\mathrm{d}x\,\mathrm{d}t
+\frac{1}{\varepsilon}\int^{1}_0u^2(x,T)\,\mathrm{d}x
+\frac{1}{\varepsilon}\int^{1}_0v^2(x,T)\,\mathrm{d}x:
h\in L^2(Q_T)\Big\},
\label{d5}
\end{equation}
where $(u,v)$ is the solution to the problem
\begin{gather}
\begin{aligned}
&u_t-((x+\eta)^{\alpha}u_x)_{x}+({c_1}(x,t,y(x,t))u)_{x}+{c_3}(x,t,y(x,t))u\\
&=h\chi_\omega, \quad (x,t)\in Q_T,
\end{aligned} \label{d6} \\
\begin{aligned}
&v_t-((x+\eta)^{\alpha}v_x)_{x}+({c_2}(x,t,z(x,t))v)_{x}
+{c_4}(x,t,y(x,t),z(x,t))u \\
&+{c_5}(x,t,y(x,t),z(x,t))v=0, \quad (x,t)\in Q_T,
\end{aligned} \label{d7}
\\
u(0,t)=v(0,t)=0,\quad u(1,t)=v(1,t)=0, \quad t\in (0,T), \label{d8}
\\
u(x,0)=u_0(x),\quad v(x,0)=v_0(x), \quad x\in(0,1).
\label{d9}
\end{gather}
As shown in \cite{ba}, one can prove that problem \eqref{d5}--\eqref{d9}
admits a unique solution
$h^{\eta,\varepsilon,y,z}=U^{\eta,\varepsilon,y,z}\chi_\omega$,
where $(U^{\eta,\varepsilon,y,z},V^{\eta,\varepsilon,y,z})$
is the solution to the problem
\begin{gather}
\begin{aligned}
&U^{\eta,\varepsilon,y,z}_t+((x+\eta)^{\alpha}U^{\eta,\varepsilon,y,z}_x)_{x}
+{c_1}(x,t,y(x,t))U^{\eta,\varepsilon,y,z}_{x}\\
&-{c_3}(x,t,y(x,t))U^{\eta,\varepsilon,y,z}
 -{c_4}(x,t,y(x,t),z(x,t))V^{\eta,\varepsilon,y,z}=0, \quad (x,t)\in Q_T,
\end{aligned}\label{d10} \\
\begin{aligned}
&V^{\eta,\varepsilon,y,z}_t+((x+\eta)^{\alpha}V^{\eta,\varepsilon,y,z}_x)_{x}
+{c_2}(x,t,z(x,t))V^{\eta,\varepsilon,y,z}_{x}\\
&-{c_5}(x,t,y(x,t),z(x,t))V^{\eta,\varepsilon,y,z}=0, \quad (x,t)\in Q_T,
\end{aligned}\label{d11}
\\
U^{\eta,\varepsilon,y,z}(0,t)=V^{\eta,\varepsilon,y,z}(0,t)=0,\quad
U^{\eta,\varepsilon,y,z}(1,t)=V^{\eta,\varepsilon,y,z}(1,t)=0, \quad t\in (0,T),
\label{d12}
\\
U^{\eta,\varepsilon,y,z}(x,T)=-\frac{1}{\varepsilon}u^{\eta,\varepsilon,y,z}(x,T),
\quad V^{\eta,\varepsilon,y,z}(x,T)
=-\frac{1}{\varepsilon}v^{\eta,\varepsilon,y,z}(x,T), \quad x\in(0,1),
\label{d13}
\end{gather}
with $(u^{\eta,\varepsilon,y,z},v^{\eta,\varepsilon,y,z})$ solving
 problem \eqref{d6}--\eqref{d9} for $h=h^{\eta,\varepsilon,y,z}$.
Multiplying \eqref{d6} with $h=h^{\eta,\varepsilon,y,z}$, \eqref{d7},
\eqref{d10} and \eqref{d11}
by $U^{\eta,\varepsilon,y,z}$, $V^{\eta,\varepsilon,y,z}$,
$u^{\eta,\varepsilon,y,z}$ and $v^{\eta,\varepsilon,y,z}$, respectively,
and then integrating over $Q_T$ by parts, we obtain
\begin{equation}
\begin{aligned}
&\int^{T}_0 \int^{1}_0h^{\eta,\varepsilon,y,z}\chi_\omega U^{\eta,\varepsilon,y,z}
\,\mathrm{d}x\,\mathrm{d}t
+\frac{1}{\varepsilon}\int_0^1(u^{\eta,\varepsilon,y,z}(x,T))^2\,\mathrm{d}x \\
&+\frac{1}{\varepsilon}\int_0^1(v^{\eta,\varepsilon,y,z}(x,T))^2\,\mathrm{d}x\\
&=-\int_0^1U^{\eta,\varepsilon,y,z}(x,0)u_0(x)\,\mathrm{d}x
 -\int_0^1V^{\eta,\varepsilon,y,z}(x,0)v_0(x)\,\mathrm{d}x.
\end{aligned} \label{new3}
\end{equation}
It follows from \eqref{new3},
$h^{\eta,\varepsilon,y,z}=U^{\eta,\varepsilon,y,z}\chi_\omega$,
the H\"older inequality and Theorem \ref{thm2} that
\begin{align*}
&\int^{T}_0 \int^{1}_0(h^{\eta,\varepsilon,y,z})^2\,\mathrm{d}x\,\mathrm{d}t
+\frac{1}{\varepsilon}\int_0^1(u^{\eta,\varepsilon,y,z}(x,T))^2\,\mathrm{d}x
+\frac{1}{\varepsilon}\int_0^1(v^{\eta,\varepsilon,y,z}(x,T))^2\,\mathrm{d}x
\\
&\leq \int_0^1|U^{\eta,\varepsilon,y,z}(x,0)u_0(x)|\,\mathrm{d}x
 +\int_0^1|V^{\eta,\varepsilon,y,z}(x,0)v_0(x)|\,\mathrm{d}x
\\
&\leq \frac{M}{2}\int_0^1(u_0^2(x)+v^2_0(x))\,\mathrm{d}x
 +\frac{1}{2M}\int_0^1((U^{\eta,\varepsilon,y,z}(x,0))^2
 +(V^{\eta,\varepsilon,y,z}(x,0))^2)\,\mathrm{d}x
\\
&\leq \frac{M}{2}\int_0^1(u_0^2(x)+v^2_0(x))\,\mathrm{d}x
 +\frac{1}{2}\int^{T}_0 \int^{1}_0(h^{\eta,\varepsilon,y,z})^2
 \,\mathrm{d}x\,\mathrm{d}t,
\end{align*}
where $M>0$ depending only on $x_0$, $x_1$, $K$, $c_0$, $T$ and $\alpha$,
is given in Theorem \ref{thm2}.
Hence
\begin{equation}
\begin{aligned}
&\int^{T}_0 \int^{1}_0(h^{\eta,\varepsilon,y,z})^2\,\mathrm{d}x\,\mathrm{d}t
+\frac{2}{\varepsilon}\int_0^1(u^{\eta,\varepsilon,y,z}(x,T))^2
\,\mathrm{d}x+\frac{2}{\varepsilon}\int_0^1(v^{\eta,\varepsilon,y,z}(x,T))^2
\,\mathrm{d}x
\\
&\leq M\int^{1}_0(u_0^2(x)+v_0^2(x))\,\mathrm{d}x,
\end{aligned}\label{d14}
\end{equation}
which, together with Lemma \ref{lem1}, leads to
\begin{gather}
\begin{aligned}
&\|u^{\eta,\varepsilon,y,z}\|_{L^{\infty}(0,T;L^{2}(0,1))}
+\|(x+\eta)^{\alpha/2}u^{\eta,\varepsilon,y,z}_{x}\|_{L^{2}(Q_T)} \\
&+\|v^{\eta,\varepsilon,y,z}\|_{L^{\infty}(0,T;L^{2}(0,1))}
 +\|(x+\eta)^{\alpha/2}v^{\eta,\varepsilon,y,z}_{x}\|_{L^{2}(Q_T)}
\\
&\leq N(\|h^{\eta,\varepsilon,y,z}\|_{L^{2}(Q_T)}+\|u_0\|_{L^{2}(0,1)}
+\|v_0\|_{L^{2}(0,1)})
\\
&\leq N(M^{1/2}+1)(\|u_0\|_{L^{2}(0,1)}+\|v_0\|_{L^{2}(0,1)}),
\end{aligned} \label{d15} \\
\begin{aligned}
&\int^{T-\delta}_0 \int^{1}_0(u^{\eta,\varepsilon,y,z}(x,\tau+\delta)
-u^{\eta,\varepsilon,y,z}(x,\tau))^2 \,\mathrm{d}x\,\mathrm{d}\tau \\
&+\int^{T-\delta}_0 \int^{1}_0(v^{\eta,\varepsilon,y,z}(x,\tau+\delta)
 -v^{\eta,\varepsilon,y,z}(x,\tau))^2 \,\mathrm{d}x\,\mathrm{d}\tau\\
&\leq N\delta^{1/2}(\|h^{\eta,\varepsilon,y,z}\|^2_{L^{2}(Q_T)}
 +\|u_0\|^2_{L^{2}(0,1)}+\|v_0\|^2_{L^{2}(0,1)}) \\
&\leq N\delta^{1/2}(M+1)(\|u_0\|^2_{L^{2}(0,1)}+\|v_0\|^2_{L^{2}(0,1)}),
 \quad 0<\delta<T,
\end{aligned} \label{d16}
\end{gather}
where $N>0$ depending only on $K$, $T$ and $\alpha$, is given in Lemma \ref{lem1}.
Define
$$
\Lambda_\varepsilon:(y,z)\mapsto(u^{\eta,\varepsilon,y,z},v^{\eta,\varepsilon,y,z}),
\quad y,z\in B_R=\{w\in L^2(Q_T):\|w\|_{L^2(Q_T)}\leq R\}
$$
with
$$
R=NT^{1/2}(M^{1/2}+1)(\|u_0\|_{L^{2}(0,1)}+\|v_0\|_{L^{2}(0,1)}).
$$
It follows from \eqref{d15} and \eqref{d16} that $\Lambda_\varepsilon$
 is a mapping from $B_R$ to $B_R$,
and $\Lambda_\varepsilon$ is compact and continuous.
Therefore, the Schauder fixed point theorem yields that $\Lambda_\varepsilon$
admits a fixed point $(u^{\eta,\varepsilon},v^{\eta,\varepsilon})\in B_R\times B_R$,
which solves problem \eqref{d1}--\eqref{d4} with
$h=h^{\eta,\varepsilon}=h^{\eta,\varepsilon,u^{\eta,\varepsilon},
v^{\eta,\varepsilon}}$.
Moreover, \eqref{d14}--\eqref{d16} yield
\begin{gather*}
\begin{aligned}
&\int^{T}_0 \int^{1}_0(h^{\eta,\varepsilon})^2\,\mathrm{d}x\,\mathrm{d}t
+\frac{2}{\varepsilon}\int_0^1(u^{\eta,\varepsilon}(x,T))^2\,\mathrm{d}x
+\frac{2}{\varepsilon}\int_0^1(v^{\eta,\varepsilon}(x,T))^2\,\mathrm{d}x\\
&\leq M\int^{1}_0(u_0^2(x)+v_0^2(x))\,\mathrm{d}x,
\end{aligned} \\
\begin{aligned}
&\|u^{\eta,\varepsilon}\|_{L^{\infty}(0,T;L^{2}(0,1))}
 +\|(x+\eta)^{\alpha/2}u^{\eta,\varepsilon}_{x}\|_{L^{2}(Q_T)}
 +\|v^{\eta,\varepsilon}\|_{L^{\infty}(0,T;L^{2}(0,1))} \\
&+\|(x+\eta)^{\alpha/2}v^{\eta,\varepsilon}_{x}\|_{L^{2}(Q_T)}\\
&\leq N(M^{1/2}+1)(\|u_0\|_{L^{2}(0,1)}+\|v_0\|_{L^{2}(0,1)}),
\end{aligned} \\
\begin{aligned}
&\int^{T-\delta}_0 \int^{1}_0(u^{\eta,\varepsilon}(x,\tau+\delta)
 -u^{\eta,\varepsilon}(x,\tau))^2 \,\mathrm{d}x\,\mathrm{d}\tau \\
&+\int^{T-\delta}_0 \int^{1}_0(v^{\eta,\varepsilon}(x,\tau+\delta)
 -v^{\eta,\varepsilon}(x,\tau))^2 \,\mathrm{d}x\,\mathrm{d}\tau \\
&\leq N\delta^{1/2}(M+1)(\|u_0\|^2_{L^{2}(0,1)}+\|v_0\|^2_{L^{2}(0,1)}).
\end{aligned}
\end{gather*}
Then there is $\varepsilon_n\in(0,1)$ with
 $\lim_{n\to\infty}\varepsilon_n=0$,
$E_m\subset Q_T$ with $\lim_{m\to\infty}\operatorname{meas}E_m=0$,
$h^\eta\in L^2(Q_T)$
and $u^\eta,v^\eta\in L^\infty(0,T;L^2(0,1))\cap L^2(0,T;H^1(0,1))$, such that
\begin{gather}
\begin{gathered}
 h^{\eta,\varepsilon_n}\rightharpoonup h^\eta,\quad
 u^{\eta,\varepsilon_n}\rightharpoonup u^\eta,\quad
 v^{\eta,\varepsilon_n}\rightharpoonup v^\eta, \quad
 u^{\eta,\varepsilon_n}_x\rightharpoonup u_x^\eta,\quad
 v^{\eta,\varepsilon_n}_x\rightharpoonup v_x^\eta\\
\text{in $L^2(Q_T)$ as } n\to\infty,
\end{gathered} \label{d17}\\
\begin{gathered}
u^{\eta,\varepsilon_n}\to u^\eta,\quad  v^{\eta,\varepsilon_n}\to v^\eta \\
\text{uniformly in $Q_T\backslash E_m$  as $n\to\infty$  for each positive integer }m,
\end{gathered}\label{d18} \\
u^\eta(x,T)=v^\eta(x,T)=0,\quad x\in(0,1), \label{d19} \\
\int^{T}_0 \int^{1}_0(h^{\eta})^2\,\mathrm{d}x\,\mathrm{d}t
\leq M\int^{1}_0(u_0^2(x)+v_0^2(x))\,\mathrm{d}x.\label{d19-0}
\end{gather}
It follows from \eqref{a14}, \eqref{a15} and \eqref{d18} that
\begin{equation}
\begin{gathered}
P_1(x,t,u^{\eta,\varepsilon_n})\rightharpoonup P_1(x,t,u^{\eta}),\quad
P_2(x,t,v^{\eta,\varepsilon_n})\rightharpoonup P_2(x,t,v^{\eta}), \\
F_1(x,t,u^{\eta,\varepsilon_n})\rightharpoonup F_1(x,t,u^{\eta}),\\
F_2(x,t,u^{\eta,\varepsilon_n},v^{\eta,\varepsilon_n})\rightharpoonup
F_2(x,t,u^{\eta},v^{\eta}) \quad \text{in $L^2(Q_T)$  as } n\to\infty.
\end{gathered} \label{d20}
\end{equation}
From \eqref{d17} and \eqref{d20}, one can show that $(u^\eta,v^\eta)$
is the solution to problem \eqref{d1}--\eqref{d4}.
Finally, \eqref{111} and \eqref{222} follow from \eqref{d19} and \eqref{d19-0}.
\end{proof}

Now we are ready to prove the null controllability of the degenerate 
parabolic system \eqref{a1}--\eqref{a4}.

\begin{theorem} \label{thm4}
Assume that $0<\alpha<1/2$, and $P_1,P_2,F_1,F_2$ satisfy \eqref{a12}--\eqref{a16}.
The system \eqref{a1}--\eqref{a4} is null controllable.
More precisely, for each $u_0$, $v_0\in L^2(0,1)$, there exists 
$h\in L^2(Q_T)$, such that the solution $(u,v)$ to problem 
\eqref{a1}--\eqref{a4} satisfies
\begin{equation}
u(x,T)=v(x,T)=0, \quad x\in(0,1). \label{333}
\end{equation}
Furthermore, there exists $M>0$ depending only on $x_0$, $x_1$, $K$, $c_0$, $T$,
 and $\alpha$, such that
\begin{equation}
\|h\|_{L^2(Q_T)}\leq M(\|u_0\|_{L^2(0,1)}+\|v_0\|_{L^2(0,1)}). \label{d21}
\end{equation}
\end{theorem}

\begin{proof}
For each $0<\eta<1$, Lemma \ref{lem51} shows that there exists 
$h^\eta\in L^2(Q_T)$ with
\begin{equation}
\|h^\eta\|_{L^2(Q_T)}\leq M(\|u_0\|_{L^2(0,1)}+\|v_0\|_{L^2(0,1)}), \label{d22}
\end{equation}
such that the solution $(u^\eta,v^\eta)$ to problem \eqref{d1}--\eqref{d4} satisfies
\begin{align}
u^\eta(x,T)=v^\eta(x,T)=0,\quad x\in (0,1),
\label{d23}
\end{align}
where $M>0$ depends only on $x_0$, $x_1$, $K$, $c_0$, $T$, and $\alpha$.
Rewrite \eqref{d1} and \eqref{d2} into
\begin{gather*}
u^{\eta}_t-((x+\eta)^{\alpha}u^\eta_x)_{x}
+(c_1(x,t)u^\eta)_{x}+c_{3}(x,t)u^\eta=h^\eta\chi_\omega, \quad (x,t)\in Q_T,
\\
v^\eta_t-((x+\eta)^{\alpha}v^\eta_x)_{x}+(c_2(x,t)v^\eta)_{x}
+c_{4}(x,t)u^\eta+c_{5}(x,t)v^\eta=0, \quad (x,t)\in Q_T,
\end{gather*}
where for $(x,t)\in Q_T$,
\begin{gather*}
c_1(x,t)=\begin{cases}
\frac{P_1(x,t,u^\eta(x,t))-P_1(x,t,0)}{u^\eta(x,t)}, &u^\eta(x,t)\neq0,\\
0,& u^\eta(x,t)=0,
\end{cases} \\
c_2(x,t)=\begin{cases}
\frac{P_2(x,t,v^\eta(x,t))-P_2(x,t,0)}{v^\eta(x,t)}, &v^\eta(x,t)\neq0, \\
0,& v^\eta(x,t)=0,
\end{cases}\\
c_3(x,t)=\begin{cases}
\frac{F_1(x,t,u^\eta(x,t))-F_1(x,t,0)}{u^\eta(x,t)}, &u^\eta(x,t)\neq0, \\
0,& u^\eta(x,t)=0,
\end{cases} \\
c_4(x,t)=\int^1_0\frac{\partial F_2}{\partial y}(x,t,\lambda u^\eta(x,t),
 \lambda v^\eta(x,t))\,\mathrm{d}\lambda, \\
c_5(x,t)=\int^1_0\frac{\partial F_2}{\partial z}(x,t,\lambda u^\eta(x,t),
 \lambda v^\eta(x,t))\,\mathrm{d}\lambda.
\end{gather*}
Then, $c_{i}\in{L^\infty(Q_T)}$ with
$\|c_{i}\|_{L^\infty(Q_T)}\leq K$ $(1\leq i\leq5)$.
Lemma \ref{lem1} yields
\begin{gather}
\begin{aligned}
&\|u^{\eta}\|_{L^{\infty}(0,T;L^{2}(0,1))}
 +\|(x+\eta)^{\alpha/2}u^{\eta}_{x}\|_{L^{2}(Q_T)}
 +\|v^{\eta}\|_{L^{\infty}(0,T;L^{2}(0,1))} \\
& +\|(x+\eta)^{\alpha/2}v^{\eta}_{x}\|_{L^{2}(Q_T)} \\
&\leq N(\|h^\eta\|_{L^{2}(Q_T)}+\|u_0\|_{L^{2}(0,1)}+\|v_0\|_{L^{2}(0,1)}),
\end{aligned} \label{d24} \\
\begin{aligned}
&\int_0^{T-\delta} \int_0^1(u^{\eta}(x,\tau+\delta)-u^{\eta}(x,\tau))^{2}
 \,\mathrm{d}x\,\mathrm{d}\tau \\
&+\int_0^{T-\delta} \int_0^1(v^{\eta}(x,\tau+\delta)
 -v^{\eta}(x,\tau))^{2}\,\mathrm{d}x\,\mathrm{d}\tau\\
&\leq N\delta^{1/2}(\|h^\eta\|^{2}_{L^{2}(Q_T)}
 +\|u_0\|^{2}_{L^{2}(0,1)}+\|v_0\|^{2}_{L^{2}(0,1)}),\quad 0<\delta<T,
\end{aligned}\label{d25}
\end{gather}
where $N>0$ depending only on $K$, $T$ and $\alpha$, is given in
Lemma \ref{lem1}.
By \eqref{d22}--\eqref{d25}, there exist $\eta_n\in(0,1)$ with
$\lim_{n\to\infty}\eta_n=0$,
$E_m\subset Q_T$ with \\
$\lim_{m\to\infty}\operatorname{meas}E_m=0$, $h\in L^2(Q_T)$
and $u, v\in L^\infty(0,T;L^2(0,1))\cap {\mathscr H}_\alpha$, such that
\begin{gather}
\begin{gathered}
h^{\eta_n}\rightharpoonup h,\quad u^{\eta_n}\rightharpoonup u,\quad
 v^{\eta_n}\rightharpoonup v,\quad
(x+\eta)^{\alpha/2}u^{\eta_n}_x\rightharpoonup x^{\alpha/2}u_x,\\\
 (x+\eta)^{\alpha/2}v^{\eta_n}_x\rightharpoonup x^{\alpha/2}v_x \quad
\text{in $L^2(Q_T)$ as } n\to\infty,
\end{gathered} \label{d26} \\
\begin{gathered}
u^{\eta_n}\to u,\quad v^{\eta_n}\to v \\
\text{uniformly in $Q_T\backslash E_m$ as $n\to\infty$ for each positive integer } m,
\end{gathered}\label{d27}
\\
u(x,T)=v(x,T)=0,\quad x\in(0,1), \label{d28}\\
\int^{T}_0 \int^{1}_0h^2\,\mathrm{d}x\,\mathrm{d}t
\leq M\int^{1}_0(u_0^2(x)+v_0^2(x))\,\mathrm{d}x.
\label{d28-0}
\end{gather}
It follows from \eqref{a14}, \eqref{a15} and \eqref{d27} that
\begin{equation}
\begin{gathered}
P_1(x,t,u^{\eta_n})\rightharpoonup P_1(x,t,u),\quad
P_2(x,t,v^{\eta_n})\rightharpoonup P_2(x,t,v),\\
F_1(x,t,u^{\eta_n})\rightharpoonup F_1(x,t,u),\quad
F_2(x,t,u^{\eta_n},v^{\eta_n})\rightharpoonup F_2(x,t,u,v)
\end{gathered}\label{d29}
\end{equation}
in $L^2(Q_T)$  as $ n\to\infty$.
From \eqref{d26} and \eqref{d29}, one can show that $(u,v)$ solves
problem \eqref{a1}--\eqref{a4}.
Finally, \eqref{333} and \eqref{d21} follow from \eqref{d28} and \eqref{d28-0}.
\end{proof}


\subsection*{Acknowledgments}
This research was supported by the National Natural Science Foundation of China
(No. 11571137 and 11601182),
the Natural Science Foundation for Young Scientists of Jilin Province
(No. 20180520213JH), and by the Scientific and Technological project 
of Jilin Provinces Education
Department in Thirteenth Five-Year (No. JJKH20180114KJ).

The authors would like to express their sincerely thanks
to the referees and to the editor
for their helpful comments on the original version of the paper.


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\end{document}