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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2009(2009), No. 73, pp. 1--21.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2009 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2009/73\hfil Controllability of parabolic equations]
{Controllability of 1-d coupled degenerate parabolic equations}

\author[P. Cannarsa, L. de Teresa\hfil EJDE-2009/73\hfilneg]
{Piermarco Cannarsa, Luz de Teresa}  % in alphabetical order

\address{Piermarco Cannarsa \newline
Dipartimento di Matematica \\
Universit\`a di Roma ``Tor Vergata" \\
00133, Roma, Italy}
\email{cannarsa@mat.uniroma2.it}

\address{Luz de Teresa \newline
Instituto de Matem\'aticas  \\
Universidad Nacional Aut\'onoma de M\'exico\\
Circuito Exterior, C.U. \\
C. P. 04510 D.F., Mexico}
\email{deteresa@matem.unam.mx}

\thanks{Submitted October 21, 2008. Published June 3, 2009.}
\subjclass[2000]{35K65, 93C20}
\keywords{Degenerate parabolic systems; controllability}

\begin{abstract}
 This article is devoted to the study of null controllability
 properties for two systems of  coupled one dimensional
 degenerate parabolic equations. The first system consists
 of two forward equations, while the second one consists
 of one  forward equation and one backward equation.
 Both systems are in cascade, that is, the solution of the first
 equation acts as a control for the second equation and the
 control function only acts directly in the first equation.
 We prove positive null controllability results when the
 control and coupling sets have nonempty intersection and $0$
 does not belong to the coupling set.
\end{abstract}

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


 \section{Statement of the problem}

In this paper we are concerned with the controllability properties
of systems of coupled degenerate parabolic equations. We are going
to consider two different kind of systems: the first one consists
of two forward equations and the second one, consists of one
forward equation and one backward equation. More precisely, given
two non empty open sets $\omega\subset (0,1)$ and $\mathcal{O} \subset (0,1)$
and a number $\alpha \in [0,2)$, we consider the system of equations
\begin{equation}\label{eq:1}
\begin{gathered}
y_t-(x^\alpha y_x)_x +c(t,x)y=\xi +h \mathbb{I}_\omega  \quad\text{in }
 Q=(0,T)\times(0,1)\,,     \\
y(t,1)=0\quad t\in (0,T)\,,  \\
y(t,0)=0 \quad \text{if }  0\leq \alpha<1, \; t\in (0,T)\,,\\
(x^\alpha y_x)(t,0)=0 \quad \text{if  } 1\leq \alpha<2,
\; t\in (0,T)\,, \\
y(0,\cdot)=y^0 \quad\text{in } (0,1) \,,
\end{gathered}
\end{equation}
and
\begin{equation} \label{eq:2}
\begin{gathered}
u_t-(x^\alpha u_x)_x+d(t,x)u=y
\mathbb{I}_{\mathcal O}  \quad\text{in } Q\,, \\
u(t,1)=0 \quad t\in (0,T)\,, \\
u(t,0)=0 \quad \text{if }  0\leq \alpha<1,\;t\in (0,T)\,, \\
(x^\alpha u_x)(t,0)=0 \quad \text{if }  1\leq \alpha<2,\;
t\in (0,T)\,,  \\
u(0,\cdot)=u^0  \quad\text{in } (0,1) \,,
\end{gathered}
\end{equation}
 or the system
\begin{equation} \label{eq:3}
 \begin{gathered}
y_t-(x^\alpha y_x)_x +c(t,x)y=\xi
    +h\mathbb{I}_\omega  \quad\text{in } Q \,, \\
y(t,1)=0\quad t \in     (0,T)\,, \\
y(t,0)=0 \quad \text{if  }  0\leq   \alpha<1,\; t\in (0,T) \,,  \\
(x^\alpha y_x)(t,0)=0 \quad \text{if  }  1\leq \alpha<2,
\;  t\in (0,T) \,,    \\
 y(0,\cdot)=y^0 \quad\text{in } (0,1) \,,
\end{gathered}
 \end{equation}
and
 \begin{equation} \label{eq:4}
 \begin{gathered}
-q_t-(x^\alpha q_x)_x+d(t,x)q=y
    \mathbb{I}_{\mathcal O}  \quad\text{in } Q\,, \\
q(t,1)=0\quad t\in (0,T)\,, \\
q(t,0)=0 \quad \text{if   } 0\leq \alpha<1,\; t\in (0,T) \,,\\
(x^\alpha q_x)(t,0)=0 \quad \text{if  }  1\leq \alpha<2,
\;  t\in(0,T) \,, \\
 q(T,\cdot )=0 \quad \text{in } (0,1)\,,
\end{gathered}
\end{equation}
where $y^0\in L^2(0,1)$, $\xi\in L^2(Q)$,
$c(t,x),d(t,x)\in L^\infty(Q)$ are
given, $h$ denotes a control function to be determined,
and $\mathbb{I}_A$ denotes the
characteristic function of the set $A$.

Models of type \eqref{eq:1}-\eqref{eq:2} are the linear  version of more
complex models that appear in mathematical biology and in a wide
variety of physical situations (see e.g. \cite{kal,NSS,CPZ}). The
controllability properties of nondegenerate parabolic cascade
systems have been studied in different contexts in the last
fifteen years or so (see \cite{BF,deT,BGP1,BGP2,GP,G,KdeT}).
However, as far as we know, the degenerate case has not been
analyzed in the literature.

On the other hand, coupled systems like \eqref{eq:3}-\eqref{eq:4} arise in a
natural way when treating ``insensitizing problems" (see
\cite{Lio} for the original formulation). To be more specific,
consider the system of equations
\begin{equation}\label{eq:6}
\begin{gathered}
 \bar y_t-(x^\alpha \bar y_{x})_x+c(t,x)\bar
    y=\xi+h\mathbb{I}_\omega     \quad\text{in } Q\,,  \\
\bar y(t,1)=0\quad t\in (0,T)\,,    \\
\bar y(t,0)=0 \quad \text{if  }  0\leq\alpha<1,\; t\in(0,T)\,, \\
(x^\alpha \bar y_x)(t,0)=0 \quad \text{if   }  1\leq  \alpha<2,
\;  t\in(0,T)\,,  \\
\bar y(0,\cdot )=y_0+\tau\bar y_0 \quad \text{in } (0,1) \,.
 \end{gathered}
\end{equation}
In this system, $\xi\in L^2(Q)$ and $y_0\in L^2(\Omega)$ are given,
$h\in L^2(\omega\times (0,T))$ is a control to be determined and
$\bar y_0\in L^2(\Omega)$ is unknown but $\tau$ is small and
$\|\bar y_0\|_2=1$. Let $\mathcal{O}\subset \Omega$ be a nonempty set, and consider
the functional
$$
\Phi(h,\tau)=\frac12\int_0^T\!\!\!\int_\mathcal{O} |\bar y|^2 \,dx\,dt.
$$
We will say that $h$ insensitizes $\Phi$ if
\begin{equation}\label{eq:ins}
\frac{\partial \Phi}{\partial\tau}\big|_{\tau=0}=0 .
\end{equation}
%
It is not difficult to see (e.g.\cite{BF}) that condition
\eqref{eq:ins} is equivalent to obtain a  control $h$ such that
system \eqref{eq:3}-\eqref{eq:4} satisfies
$q(0,\cdot)=0$.

In this paper we extend the Carleman estimates obtained in one
dimensional domains   by the first author and collaborators
\cite{CMV,ACF} to the case of cascade systems as specified before,
and recover controllability results similar to those obtained in
\cite{deT} and \cite{GdeT}.

We introduce the weight $e_M (t) = \exp (M t^{-4 })$, and define
the Hilbert space
 $$
L^2(e_M(t)) =\big\{ f: \int_0^T\!\!\int_\Omega f^2(t,x) e_M(t)
\,dx\,dt<\infty\big\} .
$$
The main results in this paper are as follows.

\begin{theorem}\label{thm1}
Assume that $0\not\in\overline {\mathcal O}$ and that  $\omega\cap
{\mathcal O}\not =\emptyset.$ There exists a positive constant
$M=M(\omega,T)$ such that,  if  $\xi \in L^2(e_M(T-t)) $  and
$y^0, u^0\in L^2(\Omega)$, then there exists $h\in L^2(Q)$ such that
the corresponding solution to \eqref{eq:1}-\eqref{eq:2} satisfies
$y(T,\cdot)=u(T,\cdot)=0$.
\end{theorem}

\begin{theorem}\label{thm2}
Assume that $0\not\in\overline {\mathcal O}$ and that $\omega\cap
{\mathcal O}\not =\emptyset.$ There exists a positive constant
$M=M(\omega,T)$ such that,  if  $\xi \in L^2(e_M(t)) $ and
$y^0=0$,  then there exists $h\in L^2(Q)$ such that the
corresponding solution to \eqref{eq:3}-\eqref{eq:4} satisfies
$q(0,\cdot)=0$.
\end{theorem}

\begin{remark}\rm
Observe that in Theorem \ref{thm2}, we require $y_0$ to be equal to zero.
In \cite{deT}, for the non degenerate case, it is proved that
there exists initial data $y^0\in L^2(\Omega)$ such that the solution
$q$ to \eqref{eq:4}   does not vanish at $t=0$ for any $h\in
L^2(\omega\times(0,T))$. In other words, system
\eqref{eq:3}-\eqref{eq:4} is not null controllable for general
initial data in $L^2$. This situation is due to the fact that
equation \eqref{eq:3} is forward in time and equation \eqref{eq:4}
is backward. A more complete analysis of this phenomenon (in the
non degenerate case) can be found in \cite{deT} and in
\cite{deTz-ins}.
\end{remark}

It is by now well understood that the null controllability of
systems is  equivalent to the validity of an observability
inequality for the adjoint system. To be more specific, instead of
proving Theorems \ref{thm1} and \ref{thm2}  directly, we will prove equivalent
results.  That is, we consider  the adjoint system to
\eqref{eq:1}-\eqref{eq:2},
\begin{equation} \label{eq:28M}
 \begin{gathered}
z_t+(x^\alpha  z_x)_x-c(t,x)z=v \mathbb{I}_{\mathcal O}  \quad\text{in }
Q\,, \\
z(t,1)=0\quad t\in (0,T)\,, \\
z(t,0)=0 \quad \text{if   }  0\leq \alpha<1,\; t\in (0,T) \,,\\
(x^\alpha z_x)(t,0)=0 \quad \text{if   }  1\leq \alpha<2,
\;  t\in (0,T) \,, \\
z(T,\cdot)=z^0 \quad \text{in } (0,1)
 \end{gathered}
\end{equation}
and
\begin{equation} \label{eq:27M}
 \begin{gathered}
v_t+(x^\alpha v_x)_x -d(t,x)v=0 \quad\text{in } Q \,, \\
v(t,1)=0\quad t\in (0,T)\,, \\
v(t,0)=0 \quad \text{if   }  0\leq \alpha<1,\;t\in (0,T) \,,\\
(x^\alpha v_x)(t,0)=0 \quad \text{if }  1\leq \alpha<2,
\; t\in (0,T), \\
v(T,\cdot)=v^0 \quad\text{in } (0,1) \,,
\end{gathered}
 \end{equation}
and the adjoint system to \eqref{eq:3}-\eqref{eq:4}:
\begin{equation} \label{eq:28}
 \begin{gathered}
z_t+(x^\alpha z_x)_x-c(t,x)z=p \mathbb{I}_{\mathcal O}  \quad\text{in } Q\,,\\
z(t,1)=0\quad  t\in (0,T)\,, \\
z(t,0)=0 \quad \text{if }  0\leq \alpha<1,\;t\in (0,T) \,,\\
(x^\alpha z_x)(t,0)=0 \quad \text{if }  1\leq \alpha<2,
\; t\in (0,T) ,\\
z(T,\cdot)=0 \quad \text{in }\ (0,1) .
\end{gathered}
\end{equation}
and
\begin{equation} \label{eq:27}
 \begin{gathered}
 p_t-(x^\alpha p_x)_x +d(t,x)p=0 \quad\text{in } Q \,, \\
p(t,1)=0\quad t\in (0,T),\; t\in (0,T) \,,\\
p(t,0)=0 \quad \text{if }  0\leq \alpha<1, \;  t\in (0,T),\\
(x^\alpha p_x)(t,0)=0 \quad \text{if }  1\leq \alpha<2,
\;  t\in (0,T), \\
p(0,\cdot)=p^0 \quad\text{in } (0,1) \,.
\end{gathered}
 \end{equation}

Then we have the following observability inequalities.

\begin{proposition}\label{prop:main3}
Suppose $\mathcal O\cap \omega\neq \emptyset$ and suppose that
$0\not \in \overline{\mathcal O}$. Then, there exist constants $M>0$
large enough and $C>0 $ such that for
every solution to \eqref{eq:28M}-\eqref{eq:27M} the following holds
\begin{equation}\label{eq:36}
\int_\Omega( v^2(0)+z^2(0))dx  + \iint_Q
e^{-M/(T-t)^4}z^2\,dx\,dt \leq C\int_0^T\!\!\!\int_\omega
z^2\,dx\,dt\,.
\end{equation}
Moreover,  there exist positive constants $M$  and $C$ such that for every
solution to \eqref{eq:28}-\eqref{eq:27} the following holds
\begin{equation}\label{eq:37}
\iint_Q e^{-M/t^4}z^2\,dx\,dt
\leq  C\int_0^T\!\!\!\int_\omega  z^2\,dx\,dt .
\end{equation}
\end{proposition}

The rest of the paper is structured in the following way. In the next section
we prove a Carleman inequality for a single parabolic degenerate heat equation.
This inequality will be used in Section 3 to prove Carleman
inequalities for the cascade  systems \eqref{eq:28M}-\eqref{eq:27M} and \eqref{eq:28}-\eqref{eq:27}. In
the last section we prove  \eqref{eq:36} and \eqref{eq:37}, and sketch a proof
of Theorem \ref{thm1}, the proof of Theorem \ref{thm2} being similar.

\section{Degenerate parabolic equations}

In this section we are concerned with the solutions of a degenerate
parabolic equation of the form
\begin{equation} \label{eq:7}
 \begin{gathered}
v_t+(x^\alpha v_x)_x +c(t,x)v=F \quad\text{in } Q \,, \\
v(t,1)=0\quad t\in (0,T)\,,     \\
v(t,0)=0 \quad \text{if   }  0\leq  \alpha<1, t\in (0,T)\,,\\
 (x^\alpha v_x)(t,0)=0 \quad \text{if   }  1\leq \alpha<2,
\;  t\in (0,T)\,,     \\
 v(0,\cdot)=v^0 \quad\text{in } (0,1) \,.
\end{gathered}
 \end{equation}
In the first part of this chapter we prove existence and uniqueness
and, in the second part, we prove the Carleman inequality
for \eqref{eq:7} that we will use in Chapter 3.

\subsection{Well-posedness}

First, we briefly describe the weighted spaces where the  above
problem is well-posed. Let us set $a(x)=x^\alpha$.  For $0 \leq \alpha < 1$,
define the Hilbert space
\begin{align*} %\label{H1a1}
 H^1_{a} (0,1)&:=\big\{ u \in L^2(0,1) :u
\text{ is absolutely continuous in } [0,1],\\
&\qquad \sqrt{a} u_x \in  L^2(0,1) \text{ and } u(0)=u(1)=0 \big\},
\end{align*}
and  the unbounded operator $A:D(A)\subset L^2(0,1)\to L^2(0,1)$ by
\begin{gather*}%\label{D(A)-1}
\forall u \in D(A), \quad  Au:= (au_x)_x,   \\
D(A) :=\{ u \in H^1_{a}(0,1)  : au_x \in  H^1(0,1) \}.
\end{gather*}
Notice that, if $u \in D(A)$ (or even $u \in H^1_{a}(0,1)$),
then $u$ satisfies the Dirichlet boundary conditions
$u(0)=u(1)=0$.

For  $1 \leq \alpha < 2$, let us change the definition of $H^1_{a} (0,1)$ to
\begin{align*} %\label{H^1_a-2}
 H^1_{a} (0,1)&:=\{ u \in L^2(0,1) :u
\text{ is locally absolutely continuous in } (0,1],\\
&\qquad \sqrt{a} u_x \in  L^2(0,1) \text{ and } u(1)=0 \}\,.
\end{align*}
Then, the operator $A:D(A)\subset L^2(0,1)\to L^2(0,1)$ will be
defined by
\begin{gather*}%\label{D(A)-2}
\forall u \in D(A), \quad    Au:= (au_x)_x,  \\
\begin{aligned}
D(A) &:=  \big\{ u \in L^2(0,1) :u \text{ is locally absolutely continuous
 in } (0,1],\\
&\qquad au \in H^1_0(0,1), \;  au_x \in  H^1(0,1) \text{ and }
(au_x) (0) =0 \big\}.
\end{aligned}
\end{gather*}
In fact, it can be proved (see, e.g., \cite{cpaa}) that
\begin{equation*}
D(A) =   \{ u \in H^1_{a}(0,1)  :au_x \in  H^1(0,1) \}\,.
\end{equation*}
Notice that when $u \in D(A)$, then $u$ satisfies the Neumann boundary
condition $(au_x) (0) =0$ and the  Dirichlet boundary condition $u(1)=0$.

In both cases $0 \leq \alpha < 1$ and  $1 \leq \alpha < 2$, the following
results hold, (see, e.g., \cite{campiti} and \cite{CMV}).

\begin{proposition} \label{prop-A}
The operator  $A: D(A) \subset L^2(0,1)\to L^2(0,1)$ is  closed
self-adjoint negative, with dense domain.
\end{proposition}

Hence, $A$ is the infinitesimal generator of a strongly continuous
semigroup $e^{tA}$ on $L^2(0,1)$. Consequently, we have the
following well-posedness result.

\begin{theorem} \label{thm-wp}
Let $F$ be given in  $L^2(Q_T)$.
For all $v_0 \in  L^2(0,1)$, problem
\eqref{eq:7}  has a unique solution
\begin{equation}
\label{reg1} v \in \mathcal U:=\mathcal C ^0 ([0,T]; L^2(0,1))  \cap L^2(0,T;
H^1_a(0,1)) .
\end{equation}
Moreover, if $v_0 \in D(A)$, then
\begin{equation}
\label{reg2}
v \in  \mathcal C ^0 ([0,T]; H^1_a(0,1)) \cap    L^2 (0,T; D(A))
\cap   H^1(0,T; L^2(0,1)).
\end{equation}
\end{theorem}


\begin{remark} \rm
Most of the results of this paper hold (and will be stated) for
solutions in the above class \eqref{reg1}.
However, in the proofs, we will assume--often without further
notice--that solutions belong to the stronger class \eqref{reg2}.
This can yields no loss of generality, since the general
result can always be recovered by a standard density argument.
\end{remark}

\subsection{Carleman inequalities}

For $\omega =(a,b)$  let us call $ \kappa=\frac{2a+b}{3}$,
$\lambda=\frac{a+2b}{3}$,
and let $\xi\in C^2(\mathbb{R})$ be such that $0\leq \xi\leq 1$  and
$$
\xi (x) =\begin{cases}
1 &\text{if } x\in (0,\kappa )\\
0 &\text{if } x\in(\lambda,1).
\end{cases}
$$
 Let us define
\begin{gather*}
\theta(t)=\frac{1}{(t(T-t))^4} \quad\forall t\in(0,T),\\
\psi(x)= \begin{cases}
(x^{2-\alpha} -c_1), & 0\leq \alpha<2, \; \alpha\not = 1,\; \forall x\in [0,1]\\
(e^x-c_1), &\alpha=1, \;\forall x\in [0,1]
\end{cases}
\end{gather*}
where $c_1$ is  such that $\psi(x)<0$ for every $x\in [0,1]$.
Now, let us set
\begin{gather*}
\zeta (x)=\frac{1-x^{\alpha/2}}{1-\alpha/2}, \\
\Psi(x)=e^{2r\zeta(0)}-e^{r\zeta(x)}\\
\Phi(t,x)=\theta(t)[\xi(x)\psi(x)-(1-\xi(x))\Psi(x)].
\end{gather*}
The main result of this section is as follows.

\begin{theorem}\label{teor:main1I}
Let $0\leq \alpha<2$ and $T>0$ be given. Then there exists two
positive constants $C,s_0$ such that for all $s\geq s_0$ and for
every solution $v \in \mathcal U$ to \eqref{eq:7},
\begin{equation}\label{eq:8I}
\begin{aligned}
&\iint_Q( s \theta x^\alpha v_x^2+s^3\theta^3x^{2-\alpha
}v^2)e^{2s\Phi}\,dx\,dt   \\
&\leq C\Big( \iint_Q
e^{2s\Phi}F^2\,dx\,dt+\int_0^T\!\!\!\int_\omega e^{2s\Phi}v^2\,dx\,dt \Big)
\end{aligned}
\end{equation}
\end{theorem}

\begin{remark}\rm
This inequality was basically proved in \cite{CMV,ACF,SICON}.
The reason why we provide the proof is that, here, we need the
locally distributed term in the right-hand side of \eqref{eq:8I}
to appear with the same exponential weight as in the left-hand
side of the inequality. In \cite{CMV,ACF,SICON} such a term was
replaced by a boundary term involving the normal derivative of
the solution.
\end{remark}

The proof of Theorem \ref{teor:main1I} will be given at the end of
this section as a consequence of the following result.
Let us consider any solution  $v$ to the system
\begin{equation} \label{eq:10}
 \begin{gathered}
v_t+(x^\alpha v_x)_x  =F \quad\text{in } Q\,, \\
v(t,1)=0\quad t\in (0,T)\,, \\
 v(t,0)=0 \quad \text{if  } 0\leq \alpha<1,\; t\in (0,T)\,,\\
(x^\alpha v_x)(t,0)=0 \quad \text{if }  1\leq \alpha<2,
\; t\in (0,T)\,, \\
v(0,\cdot)=v^0 \quad\text{in } (0,1) \,.
\end{gathered}
\end{equation}

\begin{theorem}\label{teor:main2I}
Let $0\leq \alpha<2$ and $T>0$ be given. Then
there exists two positive constants $C,s_0$ such that for all
$s\geq s_0$ and
for every solution $v \in \mathcal U$ to \eqref{eq:10},
\begin{equation}\label{eq:11I}
\begin{aligned}
&\iint_Q( s \theta x^\alpha v_x^2+s^3\theta^3x^{2-\alpha
}v^2)e^{2s\Phi}\,dx\,dt   \\
&\leq C\Big( \iint_Q
e^{2s\Phi}F^2\,dx\,dt+\int_0^T\!\!\!\int_\omega e^{2s\Phi}v ^2
\Big)\end{aligned}
\end{equation}
\end{theorem}

The proof of Theorem \ref{teor:main2I} follows the ideas of \cite{ACF}.
That is, we prove first a Carleman inequality for the degenerate part
and combine it with a classical Carleman inequality for the non
degenerate part. We will see that the appropriate combination of both
inequalities drives to \eqref{eq:11I}.

Let $\varphi(t,x)=\psi(x)\theta(t)$. Then we will prove the following result.

\begin{theorem}\label{teor:main1}
Let $0\leq \alpha<2$ and $T>0$ be given. Then there exists  two positive
constants $C,s_0$ such that for all $s\geq s_0$ and for every
solution $v \in \mathcal U$ to \eqref{eq:10},
\begin{equation}\label{eq:8}
\begin{aligned}
&\iint_Q\Big(\frac{|(x^\alpha v_x)_x|^2}{s\theta} +\frac{|
    v_t|^2}{s\theta}+s^3\theta^3x^\alpha v_x^2+s^3\theta^3x^{2-\alpha
    }v^2\Big)e^{2s\varphi}\,dx\,dt   \\
&\leq C\Big( \iint_Q e^{2s\varphi}F^2\,dx\,dt+\int_0^T{s\theta
  e^{2s\varphi}v_x^2}|_{x=1}\Big).
\end{aligned}
\end{equation}
\end{theorem}

For the proof of Theorem \ref{teor:main1} we follow the ideas in
\cite{CMV,ACF}, that is we use an appropriate change of variables
and the following Hardy type inequality.


\begin{lemma}
\begin{enumerate}
\item Let $0\leq \alpha^*<1$. Then, for all locally absolutely continuous
function $u\in (0,1)$ satisfying
$$
u(x)\to 0 \text{ as }x\to 0^+ \quad  \text{and}\quad
\int_0^1x^{\alpha^*}u_x^2dx<\infty ,
$$
the following inequality holds
\begin{equation}\label{eq:9}
\int_0^1x^{\alpha^*-2}u^2dx\leq\frac{4}{(1-\alpha^*)^2}\int_0^1x^{\alpha^*}u_x^2dx.
\end{equation}

\item Let $1<\alpha^*<2, $ then the above inequality \eqref{eq:9}
still holds for all locally absolutely continuous function $u$
in $(0,1) $ satisfying
$$
u(x)\to 0 \text{ as }{x\to 1^-}\quad  \text{and}\quad
\int_0^1x^{\alpha^*}u_x^2dx<\infty .
$$
\end{enumerate}
\end{lemma}

\begin{remark}\rm
Observe that \eqref{eq:9} is false for $\alpha^*=1$.
\end{remark}

\begin{proof}[Sketch of the proof of Theorem \ref{teor:main1}]
%

Let us define $w(t,x)=e^{s\varphi(t,x)}v(t,x)$ where $v$ satisfies
\eqref{eq:10}. Then $w$ solves
\begin{equation} \label{eq:12}
 \begin{gathered}
(e^{-s\varphi} w)_t+(x^\alpha (e^{-s\varphi}w)_x)_x  =F \quad\text{in } Q\,, \\
w(t,1)=0\quad t\in (0,T)\,, \\
w(t,0)=0 \quad \text{if  } 0\leq \alpha<1,\; t\in (0,T)\,,\\
(x^\alpha w_x)(t,0)=0 \quad \text{if }  1\leq \alpha<2,
\; t\in (0,T)\,,\\
w(0,\cdot)=w(T,\cdot)=0 \quad\text{in } (0,1) \,,
\end{gathered}
 \end{equation}
We can rewrite the above system as
 $$
P_sw=P_s^+w +P_s^-w=Fe^{ s\varphi}
$$
where
\begin{gather*}
P_s^+w=-s\varphi_t w+s^2x^\alpha\varphi^2_xw+(x^\alpha w_x)_x,\\
P_s^- = w_t-s(x^\alpha \varphi_x)_xw-2sx^\alpha\varphi_xw_x .
\end{gather*}
We observe that, for $\alpha\neq 1$,
\begin{equation}\label{eq:13}
(x^\alpha  w_x)_x=P_s^+w+s\theta_t (x^{2-\alpha}-c_1)  w
 -s^2c_2x^{2-\alpha} \theta^2 w
\end{equation}
with $c_2$ a generic constant, whereas,   for $\alpha =1$,
\begin{equation}\label{eq:13.1}
(x^\alpha w_x)_x=P_s^+w+s\theta_t(e^x-c_1)w -s^2xe^{2x}\theta^2 w.
\end{equation}
Observe that
 $$
\|Fe^{ s\varphi}\|^2\geq \|P^+_sw\|^2+ \|P^-_sw\|^2
+ 2\langle P_s^+w,P_s^-w\rangle .
$$
 Following  \cite{CMV}, we conclude that, for every $0\leq \alpha <2$,
\begin{equation}\label{eq:des}
\begin{aligned}
\|Fe^{ s\varphi}\|^2
&\geq  \|P^+_sw\|^2+
\|P^-_sw\|^2 + 2\langle P_s^+w,P_s^-w\rangle\\
&\geq \|P^+_sw\|^2+ \|P^-_sw\|^2   +
Cs^3 \iint_Q\theta^3x^{2-\alpha}w^2+Cs \iint \theta x^\alpha w_x^2\\
&\quad -C' \int_0^T\{ s\theta w_x^2\}\Big|_{x=1}.
\end{aligned}
\end{equation}

Now, we consider the case $\alpha\not = 1$.
 From \eqref{eq:13} and the fact that
$|\theta_t|\leq C\theta^{5/4}\leq C\theta^2$ we obtain
\begin{equation}\label{eq:14}
\begin{aligned}
 &\iint_Q \frac{|(x^\alpha w_x)_x|^2}{\theta s }\,dx\,dt\\
&\leq C\Big( \iint_Q  \frac{|P_s^+|^2}{\theta s}
 +s\frac{\theta_t^2}{\theta}w^2
 +s\frac{\theta_t^2}{\theta}x^{2(2-\alpha)}w^2+s^3\theta^3
    x^{2(2-\alpha)}w^2 \,dx\,dt\Big).
\end{aligned}
\end{equation}
Observe that
\begin{align*}
\iint_Qs\frac{\theta_t^2}{\theta}w^2
&\leq C  \iint_Q s\theta^{3/2}w^2\,dx\,dt\\
&=C \iint s^{1/2}\theta^{1/2}wx^{\frac{\alpha -2}{2}}\theta wx^{-(\frac{\alpha-2}{2})}s^{1/2} \,dx\,dt \\
&\leq C\Big[ \iint_Qs\theta w^2x^{\alpha-2} + \iint s\theta^2
w^2x^{2-\alpha} \,dx\,dt\Big]
\end{align*}
and, since $x\leq 1$ and $\theta^{3/2}\leq C(T)\theta^2$,
\begin{align*}
\iint_Qs\frac{\theta_t^2}{\theta}x^{2(2-\alpha)}w^2
&\leq C  \iint_Q s\theta^{3/2}x^{2(2-\alpha)}w^2\,dx\,dt\\
&=C \iint s \theta^{ 2}w^2x^{2-\alpha  } \,dx\,dt .
\end{align*}
In conclusion,
\[
\iint_Q \frac{|(x^\alpha w_x)_x|^2}{\theta s }\,dx\,dt
\leq C\Big( \iint_Q \frac{|P_s^+|^2}{\theta s}
 +s^3 \theta^{3}w^2x^ {2-\alpha  } \,dx\,dt
 + \iint_Qs\theta w^2x^{\alpha-2}\,dx\,dt\Big).
\]
Applying Hardy's inequality, we obtain
\begin{equation}\label{eq:15}
\begin{aligned}
&\iint_Q \frac{|(x^\alpha w_x)_x|^2}{\theta s }\,dx\,dt\\
&\leq C\Big( \iint_Q \frac{|P_s^+|^2}{\theta  s}\,dx\,dt
 + \iint_Q s^3 \theta^{^3}w^2x^{2-\alpha  } \,dx\,dt
 + \iint_Qs\theta w_x^2x^{\alpha}\,dx\,dt\Big).
\end{aligned}
\end{equation}
Proceeding as before, it is not difficult to prove that
\begin{equation}\label{eq:16}
\begin{aligned}
&\iint_Q \frac{|w_t|^2}{\theta s }\,dx\,dt\\
&\leq  C\Big( \iint_Q \frac{|P_s^-|^2}{\theta s}\,dx\,dt
+ \iint_Q s^3 \theta^{ 3}w^2x^{2-\alpha  } \,dx\,dt  + \iint_Qs\theta
w_x^2x^{\alpha}\,dx\,dt\Big).
\end{aligned}
\end{equation}
Combining \eqref{eq:des}, \eqref{eq:15} and \eqref{eq:16}
we conclude that, for $s$ large enough,
\begin{equation}\label{eq:17}
\begin{aligned}
C\|Fe^{s\varphi}\|^2
&\geq \iint_Q \frac{|w_t|^2}{\theta s
}\,dx\,dt+ \iint_Q \frac{|(x^\alpha w_x)_x|^2}{\theta s }\,dx\,dt  \\
&\quad +  s^3 \iint_Q\theta^3x^{2-\alpha}w^2+ s \iint \theta x^\alpha w_x^2
-C'\int_0^T\{ s\theta w_x^2\}\Big|_{x=1}.
\end{aligned}
\end{equation}

For $\alpha\not =1$ recall that $\varphi =\theta (t)\psi(x)$ with
$$
\psi_x=c_1(2-\alpha) x^{1-\alpha}\quad\text{and}\quad
\psi_{xx}=c_1(2-\alpha)(1-\alpha) x^{-\alpha} .
$$
Then
$x^{2\alpha}\psi_x^4=Cx^{2(2-\alpha)}$ and $x^\alpha\psi_x^2=Cx^{2-\alpha}$.
Moreover,
$v(t,x)=e^{-s\varphi}w(t,x)$, $v_t=-s\theta_t\psi e^{-s\varphi}w+e^{-s\varphi}w_t$ and
$v_x(t,x)=-s\theta\psi_xe^{-s\varphi}w+e^{-s\varphi}w$.
Therefore,
\begin{align*}
&\iint_Q \Big(s^3\theta^3x^{2-\alpha}v^2+s\theta x^\alpha
v_x^2+\frac{v^2_t}{\theta s}+\frac{(x^\alpha v_x)_x^2}{\theta s}\Big)\,dx\,dt
\\
&\leq \iint_Q
\Big(s^3\theta^3x^{2-\alpha}e^{-2s\varphi}w^2+s\theta
x^\alpha(2s^2\theta^2\psi_x^2e^{-2s\varphi}w^2)+2e^{-2s\varphi}w_x^2\Big)\,dx\,dt
\\
&\quad+ \iint_Q\Big(2\frac{e^{-2s\varphi}w_t^2}{\theta s}+2\frac{s^2\theta^2_t\psi^2e^{-2s\varphi}w^2}{\theta s}+\frac{2}{\theta s}(x^\alpha
w_x)_x^2e^{-2s\varphi} \Big) \,dx\,dt
\\
&\quad+ \iint_Q\Big(2\frac{s^2\theta^2}{\theta s}\alpha
x^{2(\alpha-1)}\psi_x^2e^{-2s\varphi}w^2+2\frac{s^2\theta^2}{s\theta}x^{2\alpha}
\psi_{xx}^2e^{-2s\varphi}w^2\Big)\,dx\,dt
\\
&\quad+ \iint_Q
\Big(2\frac{s^4\theta^4}{s\theta}x^{2\alpha}\psi_x^2e^{-2s\varphi}w^2
+4\frac{s^2\theta^2}{s\theta}x^{2\alpha}\psi_x^2e^{-2s\varphi}w_x^2\Big)\,dx\,dt.
\end{align*}
Using several times the Hardy type estimate and the bounds on $\varphi$
and on its derivatives, it is not difficult to conclude that
\begin{equation}\label{eq:20}
\begin{aligned}
&\iint_Q e^{2s\varphi}\Big(s^3\theta^3x^{2-\alpha}v^2+s\theta
x^\alpha v_x^2+\frac{v^2_t}{\theta s}
+\frac{(x^\alpha v_x)_x^2}{\theta s}\Big)\\
&\leq C \iint_Q \Big(s^3\theta^3x^{2-\alpha} w^2+  {s\theta}x^{\alpha}
w_x^2+\frac{ w_t^2}{\theta s} +\frac{(x^\alpha w_x)_x^2}{\theta s} \Big).
\end{aligned}
\end{equation}
Observe that $v|_{x=1}=0$ and then $v_x|_{x=1}=e^{ s\varphi}w_x|_{x=1}$.
The latter combined with \eqref{eq:17} and \eqref{eq:20} leads
to \eqref{eq:8}.

We now consider the case $\alpha=1$. From \eqref{eq:13.1} we have
\begin{equation}\label{eq:14bis}
\begin{aligned}
 &\iint_Q \frac{|(x^\alpha w_x)_x|^2}{\theta s }\,dx\,dt \\
 &\leq C\Big( \iint_Q
    \frac{|P_s^+|^2}{\theta
    s}+s\frac{\theta_t^2}{\theta}w^2 +s\frac{\theta_t^2}{\theta}xe^{2x}w^2+s^3\theta^3
    x^{2 }e^{4x}w^2 \,dx\,dt\Big).
\end{aligned}
\end{equation}
Observe that
\begin{align*}
\big| \int_0^1 s\frac{\theta_t^2}{\theta}w^2dx\big|
& \leq C\int_0^1 s\theta^{3/2} \left(x^{-1/4}w^{3/2})(x^{1/4}w^{1/2}\right)
 dx\\
& \leq C\int_0^1 s \left(\theta x^{-1/3}w^2\right)^{3/4}
  \left(\theta^{3 }x w^{ 2}\right)^{1/4}dx\\
& \leq C\left(\int_0^1 s  \theta x^{-1/3}w^2dx \right)^{3/4}
  \left(\int_0^1\theta^{3 }x w^{2}dx\right)^{1/4}.
\end{align*}
We now use Hardy's inequality with $\alpha= 5/3$ to obtain
\begin{equation}\label{eq:haruno}
\big|\int_0^1 s\frac{\theta_t^2}{\theta}w^2dx \big|
\leq C\Big(\int_0^1 s  \theta x^{5/3}w_x^2dx\Big)^{3/4}
\Big(\int_0^1\theta^{3 }x w^{ 2}dx\Big)^{1/4}.
\end{equation}
Since $5/3>1$, using Young's inequality we get, by integrating in time,
$$
\big| \iint_Qs\frac{\theta_t^2}{\theta}xe^{2x}w^2dx dt\big|
\leq C\Big( \iint_Q s\theta x w_x^2\,dx\,dt
+ \iint_Qs^3\theta^{3 }x w^{ 2}\,dx\,dt\Big).
$$
Proceeding as before it is not difficult to see that
$$
\iint_Q \frac{|(x w_x)_x|^2}{\theta s }\,dx\,dt
\leq C\Big( \iint_Q  \frac{|P_s^+|^2}{\theta s}
 +  \iint_Q s\theta x w_x^2\,dx\,dt + \iint_Qs^3\theta^{3 }x w^{
2}\,dx\,dt\Big).
$$
In a similar way the following inequality can be proved
 $$
\iint_Q \frac{|w_t|^2}{\theta s }\,dx\,dt
\leq C\Big( \iint_Q  \frac{|P_s^-|^2}{\theta s}
 +  \iint_Q s\theta xw_x^2\,dx\,dt + \iint_Qs^3\theta^{3 }x w^{
2}\,dx\,dt\Big).
$$
The last part of the proof is similar to the
case $\alpha \neq 1$, the only difference being the use of Hardy's
inequality (false if $\alpha=1$) with the same exponent as in
\eqref{eq:haruno}. \end{proof}

We will also need the following Carleman estimates, valid
in the nondegenerate case.

\begin{proposition}[Classical Carleman Estimates]
Let $z$ be solution of
\begin{equation} \label{eq:21}
 \begin{gathered}
z_t+(a(x) z_x)_x -c(t,x)z =h  \quad\text{in } Q\,, \\
z(t,1)=0,\quad z(t,0)=0\quad  t\in (0,T)\,,
\end{gathered}
 \end{equation}
where $a\in C^1([0,1])$ is a strictly positive function.
Let us define $\varrho(t,x)=\theta(t)\Psi(x)$. Then there exist
 two positive constants $r$ and $s_0$ such that for any $s>s_0$, the
solution of \eqref{eq:21} satisfies
\begin{equation}\label{eq:22}
\begin{aligned}
&\iint_Q(\frac{|(a(x) z_x)_x|^2}{s\theta} +\frac{| z_t|^2}{s\theta}+s e^{
r\zeta(x)}\theta  z_x^2+s^3\theta^3e^{ 3r\zeta(x)}
z^2)e^{-2s\varrho}\,dx\,dt   \\
&\leq C\Big( \iint_Q
e^{-2s\varrho}h^2\,dx\,dt+\int_0^T\!\!\!\int_\omega
e^{-2s\varrho}z^2 \,dx\,dt\Big)
\end{aligned}
\end{equation}
for some positive constant $C$.
\end{proposition}

The proof of the above result is by now classical and can be found,
e.g., in \cite{FI}.
We are now almost ready to prove Theorem \ref{teor:main2I}.
First, we recall Caccioppoli's inequality. For completeness,
we give a sketch of its proof in the appendix at the end of the
paper. A complete proof can be found in
\cite{ACF}.

\begin{lemma}[Caccioppoli's inequality] \label{le:caccioppoli}
Suppose $\omega'\subset\subset \omega$, then
there exists a constant $C>0$ such that, for every solution
of \eqref{eq:10}, the following inequality holds
$$
\int_0^T\!\!\!\int_{\omega'}  v_x^2e^{2s\Phi}\,dx\,dt\leq
C\Big(\int_0^T\!\!\!\int_{\omega}v^2e^{2s\Phi}\,dx\,dt+
\iint_Q F^2\,dx\,dt\Big) .
$$
\end{lemma}

\begin{proof}[Proof of Theorem \ref{teor:main2I}]
 Observe that $v=\xi v+(1-\xi)v$. Define $w=\xi v$, clearly $w $
is solution of equation \eqref{eq:10}
with second member $G=\xi F+(x^{\alpha}\xi_xv)_x+\xi_xx^\alpha v_x$. We can then
apply inequality \eqref{eq:11I} to $w$. Observe that, by construction,
$w_x|_{x=1}=0$.
Then
\begin{align*}
&\iint_Q\Big(\frac{|(x^\alpha w_x)_x|^2}{s\theta}
    +\frac{| w_t|^2}{s\theta}+s^3\theta^3x^\alpha w_x^2
+s^3\theta^3x^{2-   \alpha}w^2\Big)e^{2s\varphi }\,dx\,dt   \\
&\leq C\Big( \iint_Q e^{2s\varphi }F^2\,dx\,dt+
\int_0^T\!\!\!\int_{\omega'}e^{2s\varphi}(v_x^2+v^2)\,dx\,dt \Big).
\end{align*}
Since, for $x\in(0,\kappa)$, $\varphi(x)=\Phi(x)$ and $w=v$, we have
\begin{equation}\label{eq:24}
\begin{aligned}
&\int_0^T\!\!\!\int_0^\kappa \Big(\frac{|(x^\alpha
v_x)_x|^2}{s\theta} +\frac{| v_t|^2}{s\theta}+s^3\theta^3x^\alpha
v_x^2+s^3\theta^3x^{2-\alpha }v^2\Big)e^{2s\Phi }\,dx\,dt\\
&\leq C\Big( \iint_Q e^{2s\varphi}F^2\,dx\,dt+ \int_0^T\!\!\!\int_{\kappa
}^\lambda e^{2s\varphi}(v_x^2+v^2)\,dx\,dt
 \Big)\,.\end{aligned}
\end{equation}
Define $z=(1-\xi)v$, then $z$ is solution to \eqref{eq:21}
(in fact in an smaller set $Q_\delta=(\delta, 1)\times (0,T)$)
with $h= (1-\xi)F-(x^{\alpha}\xi_xv)_x-\xi_x x^\alpha v_x$ and inequality
\begin{equation}\label{eq:25}
\begin{aligned}
&\iint_{Q_\delta}\Big(\frac{|(a(x) z_x)_x|^2}{s\theta} +\frac{|
z_t|^2}{s\theta}+s e^{ r\zeta(x)}\theta  z_x^2+s^3\theta^3e^{ 3r\zeta(x)}
z^2\Big)e^{-2s\varrho}\,dx\,dt   \\
&\leq C  \iint_Q e^{-2s\varrho}F^2\,dx\,dt+C\int_0^T\!\!\!
 \int_\kappa^\lambda e^{-2s\varrho}(v^2+v_x^2)\,dx\,dt\\
&\quad +C \int_0^T\!\!\!\int_\omega e^{-2s\varrho}z^2 \,dx\,dt .
\end{aligned}
\end{equation}
Again, since $-\varrho(t,x)=\Psi(t,x)$ and $z=v$  for
$x\in (\lambda, 1)$, we obtain
\begin{equation}\label{eq:26}
\begin{aligned}
&\int_0^T\!\!\!\int_\lambda^1\left(\frac{|(x^\alpha v_x)_x|^2}{s\theta}
+\frac{|v_t|^2}{s\theta}+s e^{ r\zeta(x)}\theta  v_x^2
+s^3\theta^3e^{3 r\zeta(x)} v^2\right)e^{2s\Phi}\,dx\,dt   \\
&\leq C\Big( \iint_Q
e^{-2s\varrho}F^2\,dx\,dt+\int_0^T\!\!\!\int_\kappa^\lambda
e^{-2s\varrho}(v^2+v_x^2)\,dx\,dt \Big)\,.
\end{aligned}
\end{equation}
Observe that, for $x\in (\kappa,1)$, $x^\alpha\leq Ce^{ r\zeta(x)}$ and
$x^{2-\alpha}\leq Ce^{ 3r\zeta(x)}$. So, combining inequalities \eqref{eq:26}
and \eqref{eq:25}, and adding to both sides of the inequality
the term
\[
 \int_0^T\!\!\!\int_\kappa^\lambda
e^{2s\Phi}\left(s^3\theta^3x^{2-\alpha}v^2+s\theta x^\alpha
v_x^2\right)\,dx\,dt
\]
we obtain
\begin{align*}
& \iint_Q\Big(\frac{|(x^\alpha v_x)_x|^2}{s\theta}
+\frac{| v_t|^2}{s\theta}+sx^\alpha\theta  v_x^2+s^3\theta^3x^{2-\alpha}
v^2\Big)e^{2s\Phi}\,dx\,dt   \\
&\leq C\Big( \iint_Q
(e^{-2s\varrho}+e^{2s\varphi})F^2\,dx\,dt+\int_0^T\!\!\!\int_\kappa^\lambda
(e^{-2s\varrho}+e^{2s\varphi}+e^{2s\Phi})(v^2+v_x^2)\,dx\,dt
\Big)
\end{align*}
Observe that $-\varrho$ , $\varphi$ and $\Phi$ are
equivalent  for $x\in (\kappa,\lambda)$, which means that, for some $C>0$,
\begin{align*}
&\iint_Q\Big(\frac{|(x^\alpha v_x)_x|^2}{s\theta}
+\frac{| v_t|^2}{s\theta}+sx^\alpha\theta  v_x^2+s^3\theta^3x^{2-\alpha}
v^2\Big)e^{2s\Phi}\,dx\,dt   \\
&\leq C\Big( \iint_Q  e^{2s\Phi}
F^2\,dx\,dt+\int_0^T\!\!\!\int_\kappa^\lambda
 e^{2s\Phi} (v^2+v_x^2)\,dx\,dt\Big).
\end{align*}
We conclude the proof of Theorem \ref{teor:main2I} combining this last
inequality with Cacciopoli's inequality.
\end{proof}

\begin{proof}[Proof of Theorem  \ref{teor:main1I}]
Apply Theorem  \ref{teor:main1} to \eqref{eq:10} for
$\overline{F}=F-c(t,x)v$. Then, clearly $v$ the solution to \eqref{eq:7}
satisfies
\begin{equation} \label{eq:ref1}
\begin{aligned}
&\iint_Q( s \theta x^\alpha v_x^2+s^3\theta^3x^{2-\alpha
}v^2)e^{2s\Phi}\,dx\,dt   \\
&\leq C\Big( \iint_Q
e^{2s\Phi}(F^2+c^2(t,x)v^2)\,dx\,dt+\int_0^T\!\!\!\int_\omega
e^{2s\Phi}v ^2 \Big).
&\end{aligned}
\end{equation}
Observe that $x^{\alpha-2}$ is a decreasing function in $(0,1)$
and $\lim_{x\to 0^+}x^{\alpha-2}=\infty$. That means that
 $$
c^2(t,x)\leq \|c\|_{\infty}^2x^{\alpha-2}\quad \forall (t,x)\in Q,
$$
 so
\begin{equation}\label{eq:ref2}
\iint_Q e^{2s\Phi}c^2(t,x)v^2\,dx\,dt\leq
C\|c\|_{\infty}^2\iint_Q e^{2s\Phi}
x^{\alpha-2}v^2\,dx\,dt\,.
\end{equation}
For $\alpha\neq 1$ we apply Hardy inequality to $w=e^{s\Phi}v$. Then,
$$
\iint_Q e^{2s\Phi} x^{\alpha-2}v^2\,dx\,dt
\leq C\Big(\iint_Qx^\alpha s^2 \Phi_x^2v^2 e^{2s\Phi} +x^\alpha
v^2_x e^{2s\Phi}\,dx\,dt\Big).
$$
Observe that for $x\in (0, \kappa)$,
$\Phi_x=(2-\alpha)x^{1-\alpha}\theta(t)$ and for $1\ge x\ge \kappa$
there exists $C$ such that $\Phi_x\le C(2-\alpha)x^{1-\alpha}\theta(t)$.
Then, the last inequality with \eqref{eq:ref2} implies that there
exists $C>0$ such that
% \label{eq:ref3}
\begin{align*}
& \iint_Q( s \theta x^\alpha v_x^2+s^3\theta^3x^{2-\alpha
}v^2)e^{2s\Phi}\,dx\,dt   \\
&\leq C\Big( \iint_Q e^{2s\Phi} F^2
\,dx\,dt+\iint_Q(x^{2-\alpha}  s^2\theta^2 v^2   +x^\alpha
v^2_x) e^{2s\Phi}\,dx\,dt +\int_0^T\!\!\!\int_\omega e^{2s\Phi}v
^2 \Big).
\end{align*}
 Observe that in the right hand side we have smaller exponents of $s$
so for $s$ large enough we obtain \eqref{eq:8I}.

 The proof for $\alpha=1$ is similar but, instead of \eqref{eq:ref2},
observe that
 \begin{equation}\label{eq:ref4}
\iint_Q e^{2s\Phi}c^2(t,x)v^2\,dx\,dt\leq
C\|c\|_{\infty}^2\iint_Q e^{2s\Phi} x^{-1/3}v^2\,dx\,dt
\end{equation}
to obtain
\begin{align*}
&\iint_Q( s \theta x  v_x^2+s^3\theta^3x v^2)e^{2s\Phi}\,dx\,dt   \\
&\leq C\Big( \iint_Q e^{2s\Phi} F^2 \,dx\,dt+\iint_Q(x^{ 5/3}    s^2\theta^2 v^2   +x^{5/3} v^2_x) e^{2s\Phi}\,dx\,dt +\int_0^T\!\!\!\int_\omega
    e^{2s\Phi}v ^2 \Big).
\end{align*}
 The conclusion is then straightforward.
\end{proof}

\section{Carleman inequality for cascade systems}

In this section we will prove a Carleman inequality that is valid for
both: the adjoint system to \eqref{eq:1}-\eqref{eq:2}, i.e.,
\begin{equation} \label{eq:28M'}
 \begin{gathered}
z_t+(x^\alpha z_x)_x-c(t,x)z=v \mathbb{I}_{\mathcal O} \quad\text{in } Q\,, \\
z(t,1)=0\quad t\in (0,T)\,, \\
z(t,0)=0 \quad \text{ if  }  0\leq \alpha<1,\; t\in (0,T)\,,\\
(x^\alpha z_x)(t,0)=0 \quad \text{if }  1\leq \alpha<2,
\;  t\in (0,T)\,, \\
 z(T,\cdot)=z^0 \quad\text{in } (0,1) \,,
\end{gathered}
and
\end{equation}
\begin{equation} \label{eq:27M'}
 \begin{gathered}
v_t+(x^\alpha v_x)_x -d(t,x)v=0 \quad\text{in } Q \,, \\
v(t,1)=0 \quad t\in (0,T)\,, \\
v(t,0)=0 \quad \text{if }  0\leq \alpha<1,\; t\in (0,T)\,,\\
(x^\alpha v_x)(t,0)=0 \quad \text{if }  1\leq \alpha<2,
\;  t\in (0,T)\,, \\
v(T,\cdot)=v^0 \quad\text{in } (0,1) \,,
\end{gathered}
\end{equation}
and the adjoint system to \eqref{eq:3}-\eqref{eq:4}, i.e.,
\begin{equation} \label{eq:28'}
\begin{gathered}
z_t+(x^\alpha z_x)_x-c(t,x)z=p \mathbb{I}_{\mathcal O}
  \quad\text{in }\ Q\,, \\
z(t,1)=0\quad  t\in (0,T)\,, \\
z(t,0)=0 \quad \text{if  }  0\leq \alpha<1,\; t\in (0,T)\,,\\
(x^\alpha z_x)(t,0)=0 \quad \text{if  }  1\leq \alpha<2,
\;  t\in (0,T) \,,\\
z(T,\cdot)=z^0 \quad \text{in }\ (0,1) \,.
\end{gathered}
\end{equation}
and
\begin{equation} \label{eq:27'}
\begin{gathered}
p_t-(x^\alpha p_x)_x +d(t,x)p=0 \quad\text{in } Q \,, \\
p(t,1)=0\quad t\in (0,T)\,, \\
p(t,0)=0 \quad \text{if }  0\leq \alpha<1,\; t\in (0,T)\,,\\
(x^\alpha p_x)(t,0)=0 \quad \text{if }  1\leq \alpha<2,
\; t\in (0,T) \,,\\
p(0,\cdot)=p^0 \quad\text{in } (0,1) \,.
\end{gathered}
\end{equation}

\begin{remark} \rm
Observe that in \eqref{eq:28'} we have allowed for $z(T)$ any
value $z^0 $ in $L^2(0,1)$. This can be so since the Carleman
inequality is valid for general data. However, in the next
section, where the observability inequality is proved, it is
necessary to consider $z(T)=0$.
\end{remark}

 We have the following result.

\begin{theorem}\label{teor:main3p}
Assume $\mathcal O\cap \omega\neq \emptyset$ and suppose that
$0\not \in \overline{\mathcal
O}$. Then there exist two positive constants $C, s_0$ such that,
for all $s\ge s_0$ and every solution to
\eqref{eq:28M'}-\eqref{eq:27M'}, the following holds
\begin{equation}\label{eq:29M}
\begin{aligned}
&\iint_Q\left( s \theta x^\alpha v_x^2+s^3\theta^3x^{2-\alpha }v^2
+ s \theta x^\alpha z_x^2+s^3\theta^3x^{2-\alpha}z^2\right)e^{2s\Phi}\,dx\,dt
 \\
&\leq  C\int_0^T\!\!\!\int_\omega e^{2s\Phi}z^2\,dx\,dt.
\end{aligned}
\end{equation}
Moreover, there exist two positive constants $C, s_0$ such that, for all $s\ge s_0$ and every solution to
\eqref{eq:28'}-\eqref{eq:27'}, the following holds
\begin{equation}\label{eq:29}
\begin{aligned}
&\iint_Q\left( s \theta x^\alpha p_x^2+s^3\theta^3x^{2-\alpha }p^2+s \theta x^\alpha z_x^2+s^3\theta^3x^{2-\alpha
}z^2\right) e^{2s\Phi}\,dx\,dt  \\
&\leq  C\int_0^T\!\!\!\int_\omega e^{2s\Phi}z^2\,dx\,dt.
\end{aligned}
\end{equation}
\end{theorem}

\begin{proof}
 We will prove only \eqref{eq:29}. Indeed, the proof
of \eqref{eq:29M} is similar because  the  boundary conditions at
$t=0,T$ are made irrelevant by
the fact that the weight $\theta^j e^{2s\Phi}$, with $j=1,3$,
 vanishes as
$t\to 0$ and $t\to T$. Let us define $p(t)=v(T-t)$, with $v$
solution to \eqref{eq:27M'}, and observe that $p$ solves
\eqref{eq:27'} (with an appropriate choice of $\tilde d$).

The proof is to be completed in several steps.

\noindent{\bf Step 1.}
Take $\mathcal O'\subset\subset \omega\cap \mathcal O$. Observe that
$w(t,x):=p(T-t,x)$ solves \eqref{eq:7} and  apply
Theorem \ref{teor:main1I} to $p$, which is a solution of \eqref{eq:27'}.
Then, for   $s> s_1 $, we get
\begin{equation}\label{eq:30}
 \iint_Q\left( s \theta x^\alpha
p_x^2+s^3\theta^3x^{2-\alpha }p^2\right)e^{2s\Phi}\,dx\,dt
\leq C \int_0^T\!\!\!\int_{\mathcal O'}
e^{2s\Phi}p^2\,dx\,dt  .
\end{equation}
Theorem \ref{teor:main1I} can also be applied to $z$ yielding
\begin{align*}
&\iint_Q \left(s \theta x^\alpha
p_x^2+s^3\theta^3x^{2-\alpha }p^2\right)e^{2s\Phi}\,dx\,dt +
 \iint_Q\left( s \theta x^\alpha
z_x^2+s^3\theta^3x^{2-\alpha }z^2\right)e^{2s\Phi}\,dx\,dt  \\
&\leq C \Big[\int_0^T\!\!\!\int_{\mathcal O}
e^{2s\Phi}p^2\,dx\,dt +\int_0^T\!\!\!\int_{\mathcal O'}
e^{2s\Phi}(p^2 +    z^2)\,dx\,dt \Big].
\end{align*}
Now, observe that, since $0\not\in \overline{\mathcal O }$,
\begin{equation}
\label{eq:cero}\int_0^T\!\!\!\int_{\mathcal O}
e^{2s\Phi}p^2\,dx\,dt\leq C \iint_Q
s^3\theta^3x^{2-\alpha}p^2e^{2s\Phi}\,dx\,dt \leq
C\int_0^T\!\!\!\int_{\mathcal O'}  e^{2s\Phi}p^2\,dx\,dt \ .
\end{equation}
All together, we
obtain
\begin{equation}\label{eq:31}
\begin{aligned}
& \iint_Q( s \theta x^\alpha
p_x^2+s^3\theta^3x^{2-\alpha }p^2)e^{2s\Phi}\,dx\,dt +
\iint_Q( s \theta x^\alpha z_x^2+s^3\theta^3x^{2-\alpha
}z^2)e^{2s\Phi}\,dx\,dt  \\
&\leq C \Big[\int_0^T\!\!\!\int_{\mathcal O'}  e^{2s\Phi}(p^2 +
   z^2)\,dx\,dt \Big]\,.
\end{aligned}
\end{equation}

\noindent{\bf Step 2.}
 Take $\mathcal O'\subset
\subset\omega'\subset \subset \omega \cap {\mathcal O}$. Let $\xi_1\in
C^\infty_0(\Omega)$ be such that
 \begin{equation}\label{eq:32}
 1\geq \xi_1\geq 0, \quad \xi_1(x) =1 \text{ if } x\in \mathcal O',\quad
\xi_1(x)=0\text{ if } x\in \Omega\backslash\omega'.
\end{equation}
 Furthermore, we shall require  $\xi_1$ to satisfy
 \begin{equation}\label{eq:33}
 \frac{\Delta \xi_1}{\xi_1^{1/2}}\in L^\infty (\Omega),\quad
\frac{\nabla \xi_1}{\xi_1^{1/2}}\in L^\infty (\Omega).
\end{equation}
Observe that condition \eqref{eq:33} is easy to obtain: it suffices
to  take $\xi \in C^\infty_0(\Omega)$ satisfying \eqref{eq:32},
and define $\xi_1 =\xi^4$. Then  $\xi_1$
  will satisfy both \eqref{eq:32} and \eqref{eq:33}.

Let us multiply \eqref{eq:28} by $  \xi_1 p e^{2s\Phi}$.
To simplify notation, set $u =   e^{2s \Phi}$.
  Then
  \begin{equation}\label{eq:34}
\begin{aligned}
&  \iint_Q z_t\xi_1 up\,\,dx\,dt
  \iint_Q(x^\alpha z_x)_x \xi_1 up\, \,dx\,dt- \iint_Q c(t,x) z \xi_1 up
\,\,dx\,dt\\
&= \int_0^T\!\!\!\int_{\mathcal O}\xi_1p^2 u\,
\,dx\,dt \,.
\end{aligned}
\end{equation}
  We observe that $u(T) =u(0)=0$. Integrating by parts in \eqref{eq:34},
we obtain
\begin{equation}\label{eq:35}
\begin{aligned}
&\iint_Q zu\xi_1\left[ p_t - (x^\alpha p_x)_x
+d(t,x)p\right]\,\,dx\,dt  -  \iint_Q  (c+d)z \xi_1
 up\,\,dx\,dt \\
& + \iint_Q  z\left[p (x^\alpha u\xi_1)_x +2p_x x^\alpha
(u\xi_1)_x\right] \,\,dx\,dt +  \iint_Q  zp\xi_1u_t\,\,dx\,dt \\
&= \int_0^T\!\!\!\int_{\mathcal O}\xi_1p^2 u\, \,dx\,dt .
\end{aligned}
\end{equation}
Let us rewrite \eqref{eq:35} as $I_1 +I_2+I_3+I_4 =\int_0^T\int_{\mathcal{O}} \xi_1 p^2 u.$ We
observe that $I_1 =0$ since $p$ satisfies \eqref{eq:27'}. By H\"older's and Young's
inequalities, we get
$$
I_2 \leq \frac{\delta_1}{2} \iint_Q \xi_1p^2 u \,dx\,dt
 +\frac1{\delta_1}(\|c\|^2_\infty +\|
d\|^2_\infty )\int_0^T\int_\Omega \xi_1 z^2 u \,dx\,dt
$$
with $\delta_1$ to be chosen later.

 Let us estimate $I_3$. First, we have
\begin{align*}
I^1_3 &:=   \iint_Q  z p (x^\alpha u\xi_1)_x  \,dx\,dt  \\
& = \iint_Q  z \left[p \alpha x^{\alpha-1} u\xi_1+p x^\alpha
u\xi_{1,x}+px^\alpha u_x\xi_1 \right] \,dx\,dt\\
&\leq \frac{\delta_2}{2} \iint_Q \xi_1p^2 u \,dx\,dt\\
&\quad + \frac{1}{2\delta_2} \iint_Q
z^2\Big(x^{2(\alpha-1)}u\xi_1+x^{2\alpha}
\frac{|\xi_{1,x}|^2}{\xi_1}u+x^{2\alpha} \frac{|u_{ x}|^2}{u}\xi_1
\Big) \,dx\,dt\,.
\end{align*}
Observe that $\frac{|u_{ x}|^2}{u}=4s^2u\Phi_x^2$. Then
$$
\iint_Q z^2\Big(x^{2(\alpha-1)}u\xi_1+x^{2\alpha}
\frac{|\xi_{1,x}|^2}{\xi_1}u+x^{2\alpha} \frac{|u_{ x}|^2}{u}\xi_1
\Big) \,dx\,dt
\leq C\int_0^T\!\!\!\int_{\omega'}uz^2\,dx\,dt\,.
$$
So, for $I^1_3$ we conclude that
$$
|I^1_3| \leq \frac{\delta_2}{2} \iint_Q \xi_1p^2 u \,dx\,dt
+C\int_0^T\!\!\!\int_{\omega'}uz^2\,dx\,dt\,.
$$
 We now proceed to estimate the other term in $I_3$:
\begin{align*}
I^2_3 &:=  2 \iint_Q  z p_x x^\alpha( u\xi_{1,x}+u_x\xi_{1} ) \,dx\,dt  \\
& \leq \frac{\delta_3}{2} \iint_Q  s\theta x^\alpha p_x^2u\,dx\,dt+
\frac{1}{2\delta_3} \iint_Q
z^2x^\alpha\Big(\frac{ u_x^2 \xi_1^2}{u\theta}+ \frac{u
\xi_{1,x}^2}{\theta} \Big) \,dx\,dt\,.
\end{align*}
Observe that the term in $p_x^2$ can be estimated using Carleman's
inequality for $p$, while  the coefficient of $z^2$ in  the other
integral is  bounded above. Thus,
$$
 I^2_3   \leq \frac{\delta_3}{2}\int_0^T\!\!\!\int_{\mathcal{O}'}  p^2u\,\,dx\,dt+
C\int_0^T\!\!\!\int_{\omega'} z^2 e^{2s\Phi}\,\,dx\,dt\,.
$$
Finally, we get for $I_4$,
$$
I_4 =  \iint_Q  zp\xi_1u_t\,dx\,dt
\leq   \frac{\delta_4}{2} \iint_Q  \xi_1 p^2u\,dx\,dt
+ \frac{1}{2\delta_4} \iint_Q  z^2 \xi_1 \frac{|u_t|^2}{u}\,dx\,dt\,.
$$
Observe that $\frac{|u_t|^2}{u}=4s^2\Phi_t^2e^{2s\Phi}$ to conclude that
$$
I_4\leq\frac{\delta_4}{2} \iint_Q  \xi_1 p^2u+
C\int_0^T\!\!\!\int_{\omega'} z^2 e^{2s\Phi}\,dx\,dt \,.
$$
Putting the above estimates together and choosing convenient
$\delta_i$'s, we obtain,
since the support of $\xi_1$ is contained in $\mathcal{O}$,
$$
 \int_0^T\!\!\!\int_{\mathcal O'}  e^{2s\Phi}p^2\,dx\,dt
 \leq C\int_0^T\!\!\!\int_{\omega'} z^2
 e^{2s\Phi}\,dx\,dt.
$$
The last inequality together with \eqref{eq:31} completes the proof.
\end{proof}

\section{Proof of the main results}

\begin{proof}[Proof of Proposition \ref{prop:main3}]
 Multiplying equation \eqref{eq:27M} by $v_t$ and integrating
on $(0,1)$, we obtain
\begin{equation} \label{eq:38}
\begin{aligned}
&\int_0^1 v_t^2(t,x)dx-\frac12\frac{d}{dt}\int_0^1x^\alpha v_x^2(t,x)dx
\\
&\leq \frac{\|d\|_\infty^2}{2}\int_0^1v^2(t,x)dx +\frac12\int_0^1v_t^2(t,x)dx\quad\forall t\in[0,T]\,.
\end{aligned}
\end{equation}
By Hardy's inequality,
\begin{equation}\label{eq:39}
\int_0^1v^2(t,x)dx\leq  \int_0^1 x^{\alpha-2}v^2(t,x)dx \leq C\int_0^1x^\alpha v_x^2(t,x)dx \,.
\end{equation}
Then, combining \eqref{eq:38} and \eqref{eq:39}, we get
$$
0\leq \frac{d}{dt}\Big( e^{Ct}\int_0^1x^\alpha v_x^2(t,x)dx\Big)
\quad\forall t\in[0,t]\,.
$$
The above estimate implies that, for all $0\leq t\le T/2$,
$$
\frac T4\;\int_0^1x^\alpha v_x^2(t,x)dx
\leq C\int_{T/2}^{3T/4}\!\!\!\int_0^1 x^\alpha
v_x^2(\tau,x)dxd\tau\,.
$$
The latter inequality, combined with Hardy's inequality
and \eqref{eq:29M}, yields
 \begin{equation}\label{eq:40}
\begin{aligned}
 \int_0^1 v^2(t,x)dx
&\leq C(T)\int_{T/2}^{3T/4}\!\!\!\int_0^1 x^\alpha v_x^2(\tau,x)dxd\tau\\
&\leq C  \iint_Qs\theta x^\alpha v_x^2(\tau,x) e^{2s\Phi}dxd\tau \\
&\leq C\int_0^T\!\!\!\!\int_\omega z^2(\tau,x)dxd\tau
\end{aligned}
\end{equation}
for all $0\leq t\le T/2$.
Now, multiplying \eqref{eq:28M} by $z_t$ we get
\begin{equation}\label{eq:ceroprim}
\begin{aligned}
\int_0^1z_t^2(t,x)dx-\frac{d}{dt}\int_0^1x^\alpha z_x^2(t,x)dx\\\leq
2\|c\|_\infty^2\int_0^1z^2(t,x)dx + 2\int_0^1v^2(t,x)dx\quad\forall t\in[0,T]\,.
\end{aligned}
\end{equation}
Combining the latter with
\eqref{eq:40} and Hardy's inequality,  we obtain
\begin{align*}
&\int_0^1z_t^2(t,x)dx-\frac{d}{dt}\int_0^1x^\alpha z_x^2(t,x)dx\\
&\leq C\int_0^1x^\alpha
z_x^2(t,x)dx+C\int_0^T\!\!\!\int_\omega z^2(t,x)\,dx\,dt
\quad\forall t\in[0,T/2]\,.
\end{align*}
Hence,
$$
-\frac{d}{dt}\Big( e^{Ct}\int_0^1x^\alpha z_x^2(t,x)dx\Big)
\le Ce^{Ct}\int_0^T\!\!\!\int_\omega z^2(t,x)\,dx\,dt \quad
\forall t\in[0,T/2]\,.
$$
Thus, for every $0\le s\le t\le T/2$,
$$
\int_0^1x^\alpha z_x^2(s,x)dx\leq C\int_0^1x^\alpha z_x^2(t,x)dx
+C\int_0^T\!\!\!\int_\omega z^2(t,x)\,dx\,dt\,.
$$
 So, integrating
in $t$ over $[T/4,T/2]$ we get, for every $s\le T/4$,
\begin{align*}
\frac{T}{4}\int_0^1x^\alpha z_x^2(s,x)dx
&\leq C\int_{T/4}^{T/2}\!\!\!\int_0^1x^\alpha
z_x^2(t,x)\,dx\,dt +C \int_0^T\!\!\!\int_\omega z^2(t,x)\,dx\,dt\\
& \leq C  \iint_Q s\theta x^\alpha z_x^2(t,x) e^{2s\Phi}\,dx\,dt
 +C\int_0^T\!\!\int_\omega z^2(t,x)\,dx\,dt\\
&\leq C\int_0^T\!\!\int_\omega z^2(t,x)\,dx\,dt\,.
\end{align*}
By Hardy's inequality we conclude that, for every $s\le T/4$,
\begin{equation}\label{eq:45}
\begin{aligned}
\int_0^1 z^2(s,x)dx
&\leq\int_0^1x^{\alpha-2}z^2(s,x)dx\\
&\leq C\int_0^1x^\alpha z_x^2(s,x)dx\\
&\leq C\int_0^T\!\!\!\int_\omega z^2(t,x)\,dx\,dt \,.
\end{aligned}
\end{equation}
 Combining this result with \eqref{eq:40}, for $s=0=t$, we
obtain
\begin{equation}\label{eq:46}
\int_0^1(v^2(x,0)+z^2(x,0))dx \leq
C\int_0^T\!\!\!\int_\omega z^2(t,x)\,dx\,dt\,.
\end{equation}
On the other hand, \eqref{eq:45} and Carleman's inequality also yield
$$
\int_0^{T/4}\!\!\!\int_0^1 x^{\alpha}z_x^2(t,x)\,dx\,dt
+\iint_Q \theta x^{\alpha }z_x^2(t,x)e^{2s\Phi}\,dx\,dt \leq
 C \int_0^T\!\!\!\!\int_\omega z^2(t,x)\,dx\,dt \,.
$$
Therefore, by Hardy's inequality and  the definition of $\Phi$,
we conclude that there exists $M>0$ such that
$$
\iint_Q e^{-M/(T-t)^4} z^2(t,x)\,dx\,dt\leq C \int_0^T\!\!\!\!\int_\omega
z^2(t,x)\,dx\,dt \,.
$$
The above estimate, together with
\eqref{eq:46}, implies \eqref{eq:36}.

We now briefly describe how to prove \eqref{eq:37}. Proceeding as
in the proof of \eqref{eq:36} it is not difficult to see that
for all $3T/4\le s\leq T$ we have that
$$
\frac T4\;\int_0^1x^\alpha p_x^2(s,x)dx
\leq C\int_{T/2}^{3T/4}\!\!\!\int_0^1 x^\alpha p_x^2(\tau,x)dxd\tau\,.
$$
Then, for all $s\in [ 3T/4, T]$,
$$
\int_0^1z_t^2(s,x)dx-\frac{d}{dt}\int_0^1x^\alpha z_x^2(s,x)dx\leq
C\int_0^1x^\alpha z_x^2(s,x)dx+C\int_0^T\!\!\!\int_\omega
z^2(t,x)\,dx\,dt\,.
$$
Following the steps of the above  proof, since $z(T,\cdot)=0$ we
easily get that
$$
\int_{\frac{3T}{4}}^T\!\int_0^1 x^{\alpha-2}z^2(t,x)\,dx\,dt\leq
C \int_0^T\!\!\!\int_\omega z^2(t,x)\,dx\,dt \,.
$$
Combining this result with the Carleman
inequality for cascade systems we obtain, for $M$ large enough,
$$
\iint_Q e^{-M/ t ^4} z^2(t,x)\,dx\,dt
\leq C_T \int_0^T\!\!\!\!\int_\omega z^2(t,x)\,dx\,dt\,.
$$
The proof is thus complete.\end{proof}

\begin{proof}[Proof of Theorem \ref{thm1}]
The fact that Proposition \ref{prop:main3} implies Theorem \ref{thm1}
can be proved in several ways.
The most direct argument is the following.

Let $H=L^2(\Omega)\times L^2(\Omega)\times L^2(e_M(T-t))$, and let $M$ and
$L$ be the following linear mappings:
\begin{gather*}
L: L^2(Q)\to L^2(0,1)\times L^2(0,1)\\
h\mapsto (y(T),u(T))
\end{gather*}
where $(y(\cdot),u(\cdot))$ is the solution corresponding
to \eqref{eq:1}-\eqref{eq:2}
with $(y^0,u^0, \xi)=(0,0,0)$, and
\begin{gather*}
M: H\to L^2(0,1)\times L^2(0,1)\\
(y^0,u^0,\xi)\mapsto(y(T),u(T))
\end{gather*}
where $(y(\cdot),u(\cdot))$ now solves \eqref{eq:1}-\eqref{eq:2}
with $h=0$. Then Theorem \ref{thm1} is equivalent to the inclusion
\begin{equation}\label{eq:rangos}
R(M)\subset R(L) .
\end{equation}
Both $M$ and $L$ are $L^2(0,1)\times L^2(0,1)$-valued, bounded linear
operators. Consequently \eqref{eq:rangos} holds if and only, for every
$(z^0,v^0)\in L^2(0,1)\times L^2(0,1)$,
 \begin{equation}\label{eq:des2}
\| M^*(z^0,v^0)\|_H\leq C\|L^*(z^0,v^0)\|_{L^2(Q)}
\end{equation}
for some constant $C>0$. Now, a simple computation shows that
$$
M^*(z^0,v^0)= (z(x,0),v(x,0), z(t,x)),\quad L^*(z^0,v^0)=z1_\omega
$$
where $z$ and $v$ solve the adjoint system \eqref{eq:27M}-\eqref{eq:28M}.
Hence \eqref{eq:des2} is just \eqref{eq:36} and Theorem \ref{thm1}
is proved.\end{proof}

\begin{remark}\rm\hfill

$\bullet$ The results of this paper can be generalized to systems with more
general (degenerate) coefficients than $a(x)=x^\alpha$ (see for example
\cite{ACF} and \cite{SICON}).

$\bullet$ The null controllability problem when $\mathcal{O}\cap \omega=\emptyset$
is open even in the non-degenerate case. Approximate controllability
results for the linear case ($c(t,x)=d(t,x)=0$) can be found in \cite{KdeT}.

$\bullet$ Another interesting problem is to dispense with the condition
$0\not \in\overline {\mathcal{O}}$. However, it is not difficult to see that
the controllability results of this paper are valid for any  open
$\mathcal{O}$ such that $\mathcal{O}\cap\omega\not =\emptyset$ when  the coupling term
$y\mathbb{I}_\mathcal{O}$ in \eqref{eq:2} and \eqref{eq:4} is replaced by
$x^{\beta/2}y\mathbb{I}_\mathcal{O}$  with $\beta>2-\alpha$. Observe that
the fact that $0\not  \in\overline{\mathcal{O}}$ is used only in
\eqref{eq:cero}. Under the conditions given for $\beta$, such an
estimate reduces to
$$
\int_0^T\!\!\!\int_{\mathcal O}  e^{2s\Phi}x^{\beta}p^2\,dx\,dt
\leq C \iint_Q  s^3\theta^3x^{2-\alpha}p^2e^{2s\Phi}\,dx\,dt.
$$
The rest of the proof of the Carleman inequality remains the same.
The energy estimates are easily checked just noting that  the
term
$$
\int_0^1x^\beta v^2\,dx\,dt\,,
$$
that now replaces $\int_0^1 v^2\,dx\,dt$ in \eqref{eq:ceroprim},
can be easily bounded as follows
 $$
\int_0^1x^\beta v^2\,dx\,dt\le C\int_0^1 x^{\alpha-2}v^2dx
 \leq C\int_0^1x^\alpha v_x^2dx .
$$
\end{remark}

\section{Appendix}

In this appendix we give a sketch of the proof of
Lemma~\ref{le:caccioppoli} (Caccioppoli's inequality). Let
us set $\omega=(a,b)$ and $\omega'=(a',b')$ with $a<a'<b'<b$. We can suppose,
without loss  of generality, that $a\neq 0$. Let $\eta:\mathbb{R}\to\mathbb{R}$ be a smooth function satisfying $\eta_x^2/\eta\in L^\infty(\mathbb{R})$ such that $0\leq\eta\leq 1$, $\eta\equiv 1$ on $(a',b')$, and
$\eta\equiv 0$ on $[0,a)\cup(b,1]$. Then, in view of \eqref{eq:10},
\begin{align*}
 0&=\int_0^T\frac{d}{dt}\int_0^1\eta \,v^2e^{2s\Phi} \,dx\,dt\\
 &= 2\iint_Q\eta \,vv_te^{2s\Phi} \,dx\,dt+2s\iint_Q\Phi_t\eta
v^2e^{2s\Phi} \,dx\,dt \\
&=  2\iint_Q \left(\eta \,x^\alpha v_x^2
+ \eta_x x^\alpha v_x v+2s\Phi_x \eta \,x^\alpha v_xv
\right)e^{2s\Phi}\,dx\,dt\\
&\quad + 2 \iint_QF\,\eta\, ve^{2s\Phi}
\,dx\,dt+ 2s\iint_Q\Phi_t\eta\, v^2e^{2s\Phi} \,dx\,dt\,.
\end{align*}
Now, observe that, for every $\varepsilon>0$,
$$
\iint_Q \eta_xx^\alpha v_x v e^{2s\Phi}\,dx\,dt
\leq \frac\varepsilon2\,\iint_Q \eta \,x^\alpha v_x^2
e^{2s\Phi}\,dx\,dt +\frac 1{2\varepsilon}
\iint_Q\frac{\eta_x^2}{\eta}\,x^\alpha v^2 e^{2s\Phi}\,dx\,dt\,,
$$
and
$$
\iint_Q \Phi_x \eta \,x^\alpha v_xve^{2s\Phi}\,dx\,dt
\leq\frac\varepsilon2 \iint_Q \eta\, x^\alpha v_x^2
e^{2s\Phi}\,dx\,dt+\frac 1{2\varepsilon}\iint_Q \Phi_x^2
\eta\,x^\alpha v^2 e^{2s\Phi}\,dx\,dt \,.
$$
Proceeding in the same way with the other terms, and choosing
$\varepsilon$ small enough, we obtain that
$$
\iint_Q \eta \,x^\alpha v_x^2 e^{2s\Phi}\,dx\,dt
\leq C\Big(\iint_Q \lambda_\eta
v^2e^{2s\Phi}\,dx\,dt+\iint_Q\eta\, F^2\,dx\,dt\Big),
$$
where $\lambda_\eta$ is a bounded function with support in
$\omega=(a,b)$, defined in terms of $\eta$. Since $a\neq 0$ and
$a'\neq 0$, Caccioppoli's inequality follows.

\subsection*{Acknowledgments}
Piermarco Cannarsa wass partially supported by the Italian PRIN 2005
Program ``Metodi di viscosit\`a, metrici e di teoria
del controllo in equazioni alle derivate parziali nonlineari''.

Luz de Teresa was partially supported by project IN102799 of  DGAPA
 and CONACyT, Mexico.


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