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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2016 (2016), No. 333, pp. 1--13.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2016 Texas State University.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2016/333\hfil Global Carleman estimate]
{Global Carleman estimate for the plate equation and applications
to inverse problems}

\author[P. Gao \hfil EJDE-2016/333\hfilneg]
{Peng Gao}

\address{Peng Gao \newline
 School of Mathematics and Statistics, and
Center for Mathematics and Interdisciplinary Sciences,
Northeast Normal University,
 Changchun 130024,  China}
\email{gaopengjilindaxue@126.com}

\thanks{Submitted November 12, 2016. Published December 28, 2016.}
\subjclass[2010]{35R30, 35G15}
\keywords{Carleman estimate; plate equation; inverse problem}

\begin{abstract}
 In this article, we establish a Carleman estimate for
 the plate equation in order to solve an inverse problem
 retrieving the zeroth-order term for a plate equation
 from boundary measurements. We prove the local stability result 
 for this inverse problem. Our proof relies on Carleman estimate.
\end{abstract}

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

\section{Introduction}

In this article, we discuss the local Lipschitz stability in
determining a coefficient of the zeroth-order term for a plate equation
from boundary measurements.
Physically speaking, we are required to determine a coefficient
$p(x)$ from measurements of boundary displacement. The stability
result for the inverse problem is based on a  global Carleman
estimate for a plate equation.
\smallskip
 
\noindent\textbf{Inverse Problem:}
Determine $p(x)$ for $x \in I$ such that
\begin{equation} \label{10}
\begin{gathered}
u_{tt}+u_{xxxx}+pu=0 \quad \text{in }Q, \\
u(0,t)=0=u(1,t) \quad \text{in }(0,T),\\
u_{x}(0,t)=0=u_{x}(1,t)\quad \text{in }(0,T),\\
u(x,0)=a(x) \quad \text{in }I, \\
u_{t}(x,0)=b(x) \quad \text{in }I,
\end{gathered}
\end{equation}
from the observed data $u|_{\{1\}\times(0,T)}$, where
$I=(0,1)$, $T>0$ and $Q=I\times(0,T)$.

The inverse problem for the wave equation has drawn the attention of
many authors. In \cite{I1} it is discussed the global Lipschitz stability in
determining a coefficient of the zeroth-order term for a wave
equation from data of the solution in a sub-domain over a time
interval. In \cite{B2}, uniqueness and Lipschitz stability are
obtained for the inverse problem of retrieving a stationary
potential for the wave equation with Dirichlet data and
discontinuous principal coefficient from a single time dependent
Neumann boundary measurement.
Doubonva  cite{D1} solved  an inverse problem of
retrieving a stationary potential for the wave equation with
Dirichlet data from a single time-dependent Neumann boundary
measurement on a suitable part of the boundary. The
uniqueness and the stability are also proved for this problem when a Neumann
measurement is only located on a part of the boundary satisfying a
rotated exit condition.
Bellassoued \cite{B3} studied the global logarithmic
stability in determination of a coefficient of the zeroth-order term
in a wave equation from data of the solution in a subboundary over a
time interval.
 The similar inverse problems for parabolic equations can be found in the
 survey paper by Yamamoto \cite{Y1} and the references therein.
The inverse problem for dispersive equations is also interesting
and was studied in \cite{B4,C1,M1}
for Schr\"{o}dinger equations.
Baudouin \cite{B1} studied the same problem for the Korteweg-de Vries equation.
 To our best knowledge, very little work is concerned with the
inverse problem for plate equations, our work
is motivated by \cite{I1}. The key
ingredient we follow here to determine the coefficient relies on a
Carleman estimate for the plate operator.

In this article we use the following assumptions:
\begin{itemize}
\item[(H1)]  $x_0<0$, $\beta \in (0,1)$,
$T>\sup_{x\in I}|x-x_0|$.
\end{itemize}
 We define the functions
\begin{gather*}
\psi(x,t)=(x-x_0)^2-\beta t^2,\quad \varphi(x,t)=e^{\mu \psi(x,t)},\\
l(x,t)=\lambda \varphi(x,t), \quad  \theta=e^{l(x,t)}.
\end{gather*}
 Let 
\[
L_{M}^{\infty}(I)=\{p \in L^{\infty}(I)| 
\|p\|_{L^{\infty}(I)}\leq M\}.
\]
Let $P$ be the operator $Py:=y_{tt}+y_{xxxx}+py$  with domain
\begin{align*}
\mathcal {U}:=\Big\{&y\in C^{1}(-T,T;H^2(I))\cap C^2(-T,T;L^2(I))\cap
C(-T,T;H^4(I)):\\
&y(t,0)=y(t,1)=y_{x}(t,0)=y_{x}(t,1)=0,\;t\in(-T,T),\\
&  y(\pm T,x)=y_{t}(\pm T, x)=0,\; x\in I,\; Py \in
L^2(-T,T;L^2(I))\Big\},
\end{align*}
where $p\in L^{\infty}(I)$.
Throughout this paper, $C$ stands for a generic positive
constant whose value can change from line to line.
\begin{itemize}
\item[(H2)]  $a \in H^2_0(I)$ satisfies $|a(x)|\geq r_0>0$ for all
$x \in I$, $b\in L^2(I)$.
\end{itemize}

The main result in this paper is the following local stability for
the inverse problem.

\begin{theorem}\label{T2} 
Let assumptions {\rm (H1), (H2)} be satisfied.
Then there exists a constant $C^{\star}=C^{\star}(T,x_0,a,b,M)>0$
such that for all $p,q\in L_{M}^{\infty}(I)$, we have
\begin{equation}\label{*}
\|p-q\|_{L^2(I)}\leq C^{\star}\int_0^T[(u(p)-u(q))_{xxt}^2
+(u(p)-u(q))_{xxxt}^2](1,t)dt
\end{equation}
where $u(p),u(q)$ are the solutions of \eqref{10} depending on $p,q$.
\end{theorem}


\begin{remark} \label{rmk1.2} \rm
Stability estimates play a special role in the theory of inverse
problems of mathematical physics that are ill posed in the classical
sense. They determine the choice of regularization parameters and
the rate at which solutions of regularized problems converge to an
exact solution.
\end{remark}

\begin{remark} \label{rmk1.3} \rm
From Theorem \ref{T2}, we can obtain the mapping $p\to u(p)|_{\{1\}\times(0,T)}$ 
is one to one, and the mapping
$u(p)|_{\{1\}\times(0,T)}\to p$ is continuous.
\end{remark}

To obtain the stability of \eqref{*}, the following Carleman
estimate is essential.

\begin{theorem}\label{T1}
There exist three constants $\mu_0>1$, $\lambda_0>0$ and $C_1>0$
 such that for $\mu=\mu_0$ and for
every $\lambda\geq \lambda_0$ and $y\in \mathcal {U}$, we have
\begin{equation}\label{2}
\begin{aligned}
&\int_{-T}^T\int_{I}(\lambda \mu^2 \varphi
\theta^2y_{xxx}^2+\lambda^3 \mu^4 \varphi^3
\theta^2y_{xx}^2+\lambda^5 \mu^6 \varphi^5
\theta^2y_{x}^2+\lambda^7 \mu^8 \varphi^7
\theta^2y^2 \\
&+\lambda^3 \mu^4 \varphi^3\theta^2y_{t}^2
 +\lambda \mu^2 \varphi \theta^2y_{xt}^2)\,dx\,dt\\
&\leq C_1[\int_{-T}^T\int_{I}\theta^2|Py|^2\,dx\,dt
+\int_0^T(\lambda^3 \mu^3 \varphi^3 \theta^2y_{xx}^2+\lambda \mu \varphi
\theta^2y_{xxx}^2)(1,t)dt].
\end{aligned}
\end{equation}
\end{theorem}

Carleman estimate is an $L^2$-weighted estimate with large
parameter for a solution to a PDE and it is one of the major tools
used in the study of unique continuation, observability, and
controllability problems for various kinds of PDEs. Its history may
date back to Carleman \cite{C3} for a two-dimensional elliptic
equation. Then, many authors have considered this estimate such as
Egorov \cite{E1}, H\"{o}rmander \cite{H1,H2}, Isakov
\cite{I4,I5,I6}, Tataru \cite{T2}, Taylor \cite{T3} and Tr\`{e}ves
\cite{T4}.  There have already been rich amounts of work for the
Carleman estimates of second order parabolic equations, see
\cite{F1,I2,Y1}. 
Global Carleman estimate
for fourth order parabolic equation we established in\cite{G1,G2,G3,G4,G5,Z1}.
A similar estimate for the
hyperbolic equation can be found in \cite{F2,I3}. 
In \cite{G0,R1} there is a Carleman estimate for the KdV equation without the
interior observation. Then a global Carleman estimate for the KdV
equation was established in \cite{C2}. In \cite{G6} there are
global Carleman estimates for forward stochastic fourth order parabolic 
equation and backward stochastic fourth order parabolic equation.

The remainder of this article is organized as follows. In Section 2, we
establish the Carleman estimate \eqref{2}, Section 3 we prove 
 Theorem \ref{T2}.

\section{Proof of Theorem \ref{T1}}

As in \cite{M2}, it is sufficient to prove \eqref{2} for
$\widetilde{P}y=y_{tt}+y_{xxxx}$ with $y\in\mathcal {U}$. In fact,
assume that we have proved \eqref{2} for $\widetilde{P}y$, we have
$$
\int_{Q}\theta^2|\widetilde{P}y|^2\,dx\,dt
\leq \int_{Q}\theta^2|Py|^2\,dx\,dt
+\|p\|^2_{L^{\infty}(I)}\int_{Q}\theta^2y^2\,dx\,dt.
$$
By choosing $\lambda_0=\lambda_0(\mu,T)>0$ large, when
 $\lambda>\lambda_0$, it is possible to absorb
$\|p\|^2_{L^{\infty}(I)}\int_{Q}\theta^2y^2\,dx\,dt$ with the
left-hand side of \eqref{2}, concluding that \eqref{2} also holds
for $Py$.

Set $u=\theta y, \widetilde{P}y=f$. Direct computations show that
\begin{align*}\theta
(y_{tt}+y_{xxxx})=u_{tt}+A_0u_{t}+A_1u+A_2u_{x}+A_3u_{xx}
+A_4u_{xxx}+u_{xxxx}
\end{align*}
where
\begin{gather*}
A_0=-2l_{t},\quad
A_1=l_{t}^2-l_{tt}+l_{x}^4+4l_{x}l_{xxx}-l_{xxxx}-6l_{x}^2l_{xx}+3l_{xx}^2,\\
A_2=12l_{x}l_{xx}-4l_{x}^3-4l_{xxx},\quad 
A_3=6l_{x}^2-6l_{xx}, \quad A_4=-4l_{x}.
\end{gather*}
Set
\begin{gather*}
I_1=u_{xxxx}+B_1u+B_3u_{xx}+u_{tt},\\
I_2=B_2u_{x}+B_4u_{xxx}+au+B_0u_{t},\\
Ru= \theta f-I_1-I_2=S_0u+S_1u_{x}+S_2u_{xx},
\end{gather*}
where
\begin{gather*}
a=-12l_{x}^2l_{xx},\quad B_0=-2l_{t}, \quad B_1=l_{x}^4, \\
B_2=-4l_{x}^3, \quad B_3=6l_{x}^2, \quad B_4=-4l_{x},\\
S_0=l_{t}^2-l_{tt}+4l_{x}l_{xxx}-l_{xxxx}+6l_{x}^2l_{xx}+3l_{xx}^2,\\
S_1=12l_{x}l_{xx}-4l_{xxx},\quad S_2=-6l_{xx}.
\end{gather*}
\smallskip

\noindent\textbf{Step 1.}
We shall prove the  equality
 \begin{equation}\label{5}
\begin{aligned}
I_1\cdot I_2
 &=u^2\{\cdots\}+u_{x}^2\{\cdots\}
+u_{xx}^2\{\cdots\}+u_{xxx}^2\{\cdots\}+u_{t}^2\{\cdots\}\\
&\quad +u_{xt}^2\{\cdots\}+u_{xt}u_{xxx}\{\cdots\}+u_{t}u_{x}\{\cdots\}
 +u_{t}u_{xxx}\{\cdots\}\\
&\quad +\{\cdots\}_{x}+\{\cdots\}_{xx}+\{\cdots\}_{xxx}+\{\cdots\}_{xxxx}
+\{\cdots\}_{t}+\{\cdots\}_{tt},
\end{aligned}
\end{equation}
where
\begin{gather*}
\begin{aligned}
\{\cdots\}_{x}
&=\{\frac{3}{2}B_{2xx}u_{x}^2-\frac{3}{2}B_2u_{xx}^2
 +\frac{1}{2}B_4u_{xxx}^2+4a_{x}u_{x}^2-2a_{xxx}u^2\\
&\quad +B_0u_{t}u_{xxx}+\frac{1}{2}B_1B_2u^2+\frac{3}{2}(B_1B_4)_{xx}u^2
 -\frac{3}{2}B_1B_4u_{x}^2+\frac{1}{2}B_2B_3u_{x}^2\\
&\quad +\frac{1}{2}B_3B_4u_{xx}^2-(aB_3)_{x}u^2+B_0B_3u_{t}u_{x}
 -\frac{1}{2}B_2u_{t}^2-\frac{3}{2}B_{4xx}u^2_{t}
 +\frac{3}{2}B_4u^2_{xt}\}_{x},
\end{aligned}\\
\{\cdots\}_{xx}=\{-\frac{3}{2}B_{2x}u_{x}^2+3a_{xx}u^2-2au_{x}^2
-\frac{3}{2}(B_1B_4)_{x}u^2+\frac{1}{2}aB_3u^2+\frac{3}{2}B_{4x}u^2_{t}\}_{xx},
\\
\{\cdots\}_{xxx}=\{\frac{1}{2}B_2u_{x}^2-2a_{x}u^2
+\frac{1}{2}B_1B_4u^2-\frac{1}{2}B_4u^2_{t}\}_{xxx},
\\
\{\cdots\}_{xxxx}=\{\frac{1}{2}au^2\}_{xxxx},
\\
\{\cdots\}_{t}=\{\frac{1}{2}B_0B_1u^2-\frac{1}{2}B_0B_3u^2_{x}
+B_2u_{t}u_{x}+\frac{1}{2}B_0u^2_{t}-a_{t}u^2+B_4u_{t}u_{xxx}\}_{t},
\\
\{\cdots\}_{tt}=\{\frac{1}{2}au^2\}_{tt},
\\
\begin{aligned}
u^2\{\cdots\}&=u^2\{\frac{1}{2}a_{xxxx}-\frac{1}{2}(B_1B_2)_{x}
-\frac{1}{2}(B_1B_4)_{xxx}+aB_1-\frac{1}{2}(B_0B_1)_{t}\\
&\quad+\frac{1}{2}(aB_3)_{xx}+\frac{1}{2}a_{tt}\},
\end{aligned} \\
u_{x}^2\{\cdots\}=u_{x}^2\{-\frac{1}{2}B_{2xxx}-2a_{xx}
+\frac{3}{2}(B_1B_4)_{x}-\frac{1}{2}(B_2B_3)_{x}-aB_3+\frac{1}{2}(B_0B_3)_{t}\},
\\
u_{xx}^2\{\cdots\}=u_{xx}^2\{\frac{3}{2}B_{2x}+a-\frac{1}{2}(B_3B_4)_{x}\},
\\
u_{xxx}^2\{\cdots\}=u_{xxx}^2\{-\frac{1}{2}B_{4x}\},
\\
u_{t}^2\{\cdots\}=u_{t}^2\{\frac{1}{2}B_{2x}-\frac{1}{2}B_{0t}-a
+\frac{1}{2}B_{4xxx}\},
\\
u_{xt}^2\{\cdots\}=u_{xt}^2\{-\frac{3}{2}B_{4x}\},
\\
u_{xt}u_{xxx}\{\cdots\}=u_{xt}u_{xxx}\{-B_0\},
\\
u_{t}u_{x}\{\cdots\}=u_{t}u_{x}\{-(B_0B_3)_{x}-B_{2t}\},
\\
u_{t}u_{xxx}\{\cdots\}=u_{t}u_{xxx}\{-B_{4t}-B_{0x}\}.
\end{gather*}
Indeed, \eqref{5} can be obtained from the following equations
\begin{gather*}
\begin{aligned}
u_{xxxx}B_2u_{x}
&=\frac{1}{2}(B_2u_{x}^2)_{xxx}-\frac{3}{2}(B_{2x}u_{x}^2)_{xx}
+\frac{3}{2}(B_{2xx}u_{x}^2-B_2u_{xx}^2)_{x} \\
&\quad +\frac{3}{2}B_{2x}u_{xx}^2-\frac{1}{2}B_{2xxx}u_{x}^2,
\end{aligned}\\
u_{xxxx}B_4u_{xxx}=\frac{1}{2}[(B_4u_{xxx}^2)_{x}-B_{4x}u_{xxx}^2],\\
\begin{aligned}\\
u_{xxxx}au
&=\frac{1}{2}(au^2)_{xxxx}-2(a_{x}u^2)_{xxx}+(3a_{xx}u^2-2au^2_{x})_{xx}\\
&\quad +(4a_{x}u^2_{x}-2a_{xxx}u^2)_{x}+au_{xx}^2
-2a_{xx}u_{x}^2+\frac{1}{2}a_{xxxx}u^2,
\end{aligned} \\
u_{xxxx}bu_{xx}=\frac{1}{2}(bu_{xx}^2)_{xx}-(b_{x}u_{xx}^2)_{x}
 -bu_{xxx}^2+\frac{1}{2}b_{xx}u_{xx}^2,
\\
u_{xxxx}B_0u_{t}=(B_0u_{t}u_{xxx})_{x}-B_{0x}u_{t}u_{xxx}-B_0u_{xt}u_{xxx},
\\
B_1uB_2u_{x} =\frac{1}{2}[(B_1B_2u^2)_{x}-(B_1B_2)_{x}u^2], \\
\begin{aligned}
B_1uB_4u_{xxx}&=\frac{1}{2}(B_1B_4u^2)_{xxx}-\frac{3}{2}[(B_1B_4)_{x}u^2]_{xx}\\
&\quad +\frac{3}{2}[(B_1B_4)_{xx}u^2-B_1B_4u^2_{x}]_{x}
  +\frac{3}{2}(B_1B_4)_{x}u_{x}^2-\frac{1}{2}(B_1B_4)_{xxx}u^2,
\end{aligned} \\
B_1ubu_{xx}=\frac{1}{2}(bB_1u^2)_{xx}-[(bB_1)_{x}u^2]_{x}
 -bB_1u_{x}^2+\frac{1}{2}(bB_1)_{xx}u^2,
\\
B_1uB_0u_{t}=\frac{1}{2}[(B_0B_1u^2)_{t}-(B_0B_1)_{t}u^2],
\\
B_3u_{xx}B_2u_{x}=\frac{1}{2}[(B_2B_3u_{x}^2)_{x}-(B_2B_3)_{x}u_{x}^2],
\\
B_3u_{xx}B_4u_{xxx}=\frac{1}{2}[(B_3B_4u_{xx}^2)_{x}-(B_3B_4)_{x}u_{xx}^2],
\\
B_3u_{xx}au=\frac{1}{2}(aB_3u^2)_{xx}-[(aB_3)_{x}u^2]_{x}-
aB_3u_{x}^2+\frac{1}{2}(aB_3)_{xx}u^2,
\\
B_3u_{xx}B_0u_{t}=-(B_0B_3)_{x}u_{t}u_{x}+\frac{1}{2}(B_0B_3)_{t}u^2_{x}
-\frac{1}{2}(B_0B_3u^2_{x})_{t}+(B_0B_3u_{x}u_{t})_{x},
\\
u_{tt}B_2u_{x}=(B_2u_{t}u_{x})_{t}-B_{2t}u_{t}u_{x}
 -\frac{1}{2}[(B_2u_{t}^2)_{x}-B_{2x}u_{t}^2],
\\
u_{tt}B_0u_{t}=\frac{1}{2}[(B_0u_{t}^2)_{t}-B_{0t}u_{t}^2],
\\
u_{tt}au=\frac{1}{2}(au^2)_{tt}-(a_{t}u^2)_{t}-au_{t}^2+\frac{1}{2}a_{tt}u^2,
\\
\begin{aligned}
u_{tt}B_4u_{xxx}&=(B_4u_{t}u_{xxx})_{t}-B_{4t}u_{t}u_{xxx}
 -\frac{1}{2}(B_4u_{t}^2)_{xxx}+\frac{3}{2}(B_{4x}u_{t}^2)_{xx}\\
&\quad -\frac{3}{2}[B_{4xx}u_{t}^2-B_4u_{xt}^2]_{x}
 -\frac{3}{2}B_{4x}u_{xt}^2+\frac{1}{2}B_{4xxx}u_{t}^2,
\end{aligned} \\
\begin{aligned}
u_{tt}bu_{xx}
&=(bu_{t}u_{xx})_{t}+b_{xt}u_{t}u_{x} -\frac{1}{2}b_{tt}u_{x}^2
 +\frac{1}{2}(b_{t}u_{x}^2)_{t}\\
&\quad -(b_{t}u_{t}u_{x})_{x}-\frac{1}{2}(bu_{t}^2)_{xx}+(b_{x}u_{t}^2)_{x}
+bu^2_{xt}-\frac{1}{2}b_{xx}u_{t}^2.
\end{aligned}
\end{gather*}
\smallskip

\noindent\textbf{Step 2.} We shall prove the  estimate
\begin{align*}
&\int_{-T}^T\int_{I}(\lambda \mu^2 \varphi
\theta^2y_{xxx}^2+\lambda^3 \mu^4 \varphi^3
\theta^2y_{xx}^2+\lambda^5 \mu^6 \varphi^5
\theta^2y_{x}^2+\lambda^7 \mu^8 \varphi^7 \theta^2y^2\\
&+\lambda^3 \mu^4 \varphi^3 \theta^2y_{t}^2
 +\lambda \mu^2 \varphi \theta^2y_{xt}^2)\,dx\,dt\\
&\leq C_1\Big[\int_{-T}^T\int_{I}\theta^2|Py|^2\,dx\,dt+\int_{-T}^T(\lambda^3 \mu^3
\varphi^3 \theta^2y_{xx}^2+\lambda \mu \varphi
\theta^2y_{xxx}^2)(1,t)dt\Big].
\end{align*}
\par
Indeed,
\begin{gather*}
u_{xxx}^2\{\cdots\}=u_{xxx}^2\{2l_{xx}\}, \quad
u_{xx}^2\{\cdots\}=u_{xx}^2\{6l^2_{x}l_{xx}\}, \\
u_{x}^2\{\cdots\}=u_{x}^2\{102l^4_{x}l_{xx}+r_1\}, \quad
u^2\{\cdots\}=u^2\{2l^6_{x}l_{xx}+r_2\}, \\
u_{t}^2\{\cdots\}=u_{t}^2\{6l^2_{x}l_{xx}+l_{tt}-2l_{xxxx}\}, \quad
u_{xt}^2\{\cdots\}=u_{xt}^2\{6l_{xx}\}, \\
u_{xt}u_{xxx}\{\cdots\}=u_{xt}u_{xxx}\{2l_{t}\}, \quad
u_{t}u_{x}\{\cdots\}=u_{t}u_{x}\{24l_{xt}l_{x}^2+24l_{t}l_{x}l_{xx}\}, \\
u_{t}u_{xxx}\{\cdots\}=u_{t}u_{xxx}\{6l_{xt}\},
\end{gather*}
where
\begin{gather*}
r_1=60l^3_{xx}+180l_{x}l_{xx}l_{xxx}+30l_{x}^2l_{xxxx}-6l_{tt}l_{x}^2
 -12l_{t}l_{x}l_{xt},\\
 r_2=-6(l_{x}^2l_{xx})_{xxxx}+2(l^5_{x})_{xxx}+(l_{t}l_{x}^4)_{t}
 -36(l_{x}^4l_{xx})_{xx}-6(l_{x}^2l_{xx})_{tt}.
\end{gather*}
Now, we estimate the term
$\int_{-T}^T\int_{I}(\{\cdots\}_{x}+\{\cdots\}_{xx}+\{\cdots\}_{xxx}
+\{\cdots\}_{xxxx}+\{\cdots\}_{t}+\{\cdots\}_{tt})\,dx\,dt$
in \eqref{5}.
Indeed, noting that
\begin{gather*}
y(0,t)=y(1,t)=y_{x}(0,t)=y_{x}(1,t)=0\quad \forall t\in(-T,T), \\
y(x,\pm T)=y_{t}(x,\pm T)=0\quad \forall x\in I,
\end{gather*}
this implies
\begin{gather*}
u(0,t)=u(1,t)=u_{x}(0,t)=u_{x}(1,t)=0 \quad \forall t\in(-T,T)\\
u(x,\pm T)=u_{t}(x,\pm T)=0\quad \forall x\in I.
\end{gather*}
Thus
\begin{equation}\label{d}
\begin{gathered}
\int_{-T}^T\int_{I}(\{\cdots\}_{t}+\{\cdots\}_{tt})\,dx\,dt=0,\\
\begin{aligned}
&\int_{-T}^T\int_{I}(\{\cdots\}_{x}+\{\cdots\}_{xx}+\{\cdots\}_{xxx}
 +\{\cdots\}_{xxxx})\,dx\,dt\\
&=\int_{-T}^T\{u_{xx}^2(-10l_{x}^3)+u_{xxx}^2(-2l_{x})\}(\cdot,t)|_0^{1}dt
=: V(1)-V(0).
\end{aligned}
\end{gathered}
\end{equation}
If we choose $\lambda\geq \lambda_0=\lambda_0(\mu,T)$ with $\lambda_0$ 
large enough such that $\lambda\mu^{-1}\varphi\geq1$, then it holds
\begin{equation}\label{a}
\begin{gathered}
|r_1|\leq C(\lambda^3\mu^6\varphi^3+\lambda^3\mu^4\varphi^3
 +\lambda^3\mu^5\varphi^3)
\leq C\lambda^5\varphi^5(\mu^2+\mu^4+\mu^3)
\leq C\lambda^5\mu^5\varphi^5,
\\
|r_2|\leq C(\lambda^3\mu^6\varphi^3+\lambda^3\mu^8\varphi^3
 +\lambda^5\mu^8\varphi^5)
\leq C\lambda^7\varphi^7(\mu^2+\mu^4+\mu^6)
\leq C\lambda^7\mu^7\varphi^7,
\\
\begin{aligned}
|S_0|^2&\leq C(\lambda^2\mu^2\varphi^2+\lambda^2\mu^4\varphi^2
 +\lambda^4\mu^8\varphi^4
+\lambda^6\mu^8\varphi^6+\lambda^4\mu^4\varphi^4+\lambda^2\mu^8\varphi^2)\\
&\leq C\lambda^7\varphi^7(\mu^{-3}+\mu^{-1}+\mu^5+\mu^7+\mu+\mu^3)\\
&\leq C\lambda^7\mu^7\varphi^7,
\end{aligned}\\
|S_1|^2\leq C(\lambda^4\mu^6\varphi^4+\lambda^2\mu^6\varphi^2)
 \leq C\lambda^5\mu^5\varphi^5,
\\
|S_2|^2\leq C\lambda^2\mu^4\varphi^2\leq C\lambda^3\mu^3\varphi^3
\end{gathered}
\end{equation}
and
\begin{equation}\label{b}
\begin{gathered}
u_{xt}^2\{\cdots\}+u_{xxx}^2\{\cdots\}+u_{xt}u_{xxx}\{\cdots\}
\geq C(\lambda \mu^2\varphi u_{xxx}^2+\lambda \mu^2\varphi u_{xt}^2),
\\
u_{x}^2\{\cdots\}+u_{t}^2\{\cdots\}+u_{t}u_{x}\{\cdots\}
\geq C(\lambda^5 \mu^6\varphi^5u_{x}^2+\lambda^3
\mu^4\varphi^3u_{t}^2),
\\
u_{xxx}^2\{\cdots\}+u_{t}^2\{\cdots\}+u_{t}u_{xxx}\{\cdots\}
\geq C(\lambda \mu^2\varphi u_{xxx}^2+\lambda^3
\mu^4\varphi^3u_{t}^2).
\end{gathered}
\end{equation}
Since
$I_1+I_2=\theta f-S_0u-S_1u_{x}-S_2u_{xx}$,
it is clear that
\begin{equation}\label{c}
\begin{aligned}
2\int_{-T}^T\int_{I}I_1I_2~\,dx\,dt
&\leq \|I_1+I_2\|^2_{L^2(-T,T;L^2(I))}\\
&=\|\theta f+S_0u+S_1u_{x}+S_2u_{xx}\|^2_{L^2(Q)}\\
&\leq C \int_{-T}^T\int_{I}(\theta^2f^2+S_0^2u^2
+S_1^2u^2_{x}+S_2^2u^2_{xx})\,dx\,dt.
\end{aligned}
\end{equation}
From \eqref{d}-\eqref{c}, we can obtain
\begin{align*}
&\int_{-T}^T\int_{I}[\lambda \mu^2 \varphi
u_{xxx}^2+\lambda^3 \mu^4 \varphi^3 u_{xx}^2+\lambda^5
\mu^6 \varphi^5 u_{x}^2+\lambda^7 \mu^8 \varphi^7
u^2+\lambda^3 \mu^4 \varphi^3 u_{t}^2 \\
&+\lambda \mu^2 \varphi u_{xt}^2]\,dx\,dt+V(1)-V(0) \\
&\leq C \int_{-T}^T\int_{I}(\theta^2f^2+\lambda^7\mu^7\varphi^7u^2
+\lambda^5\mu^5\varphi^5u^2_{x}+\lambda^3\mu^3\varphi^3u^2_{xx})\,dx\,dt.
\end{align*}
Noting that $-V(0)\geq 0$ and that
\[
 |V(1)|\leq C\int_0^T(\lambda^3 \mu^3
\varphi^3 u_{xx}^2+\lambda \mu \varphi u_{xxx}^2)(1,t)dt,
\]
and choosing $\lambda_0$  large enough, when $\lambda\geq\lambda_0$, 
we conclude that
\begin{align*}
&\int_{-T}^T\int_{I}(\lambda \mu^2 \varphi
u_{xxx}^2+\lambda^3 \mu^4 \varphi^3 u_{xx}^2+\lambda^5
\mu^6 \varphi^5 u_{x}^2+\lambda^7 \mu^8 \varphi^7u^2 \\
&+\lambda^3 \mu^4 \varphi^3 u_{t}^2+\lambda \mu^2\varphi u_{xt}^2)\,dx\,dt
\\
&\leq C[\int_{-T}^T\int_{I}\theta^2f^2\,dx\,dt+\int_0^T(\lambda^3 \mu^3
\varphi^3 u_{xx}^2+\lambda \mu \varphi u_{xxx}^2)(1,t)dt].
\end{align*}
Returning $u$ to $\theta y$, we can obtain \eqref{2}.

\section{Proof of Theorem \ref{T2}}

In this section, we use the Bukhgeim-Klibanov method to study
the inverse problem. We are now in a position to prove the stability
result Theorem \ref{T2}, the proof follows the ideas used in
\cite{I1,W1}.
The idea is to reduce the nonlinear inverse problem to some perturbed 
inverse problem which will be solved with the help of a global Carleman estimate.
Firstly, we consider the system
\begin{equation}\label{1}
\begin{gathered}
y_{tt}+y_{xxxx}+py=G \quad \text{in }Q,\\
y(0,t)=0=y(1,t) \quad \text{in }(0,T), \\
y_{x}(0,t)=0=y_{x}(1,t)\quad \text{in }(0,T),\\
y(x,0)=0 \quad \text{in }I,\\
y_{t}(x,0)=0\quad \text{in }I\,.
\end{gathered}
\end{equation}

\begin{proposition}\label{pro2}
Let assumptions {\rm (H1), (H2)} be satisfied. 
Assume that there exists a function $g_0\in L^2(0,T)$ such that
\begin{equation}\label{8}
\begin{gathered}
G\in H^{1}(0,T;L^2(I)),\quad \|G_{t}\|_{L^2(Q)}\leq M,\\
g_0(t)|G(x,0)|\geq |G_{t}(x,t)|, \quad (x,t)\in Q.
\end{gathered}
\end{equation}
Then there exists a constant $C_1=C_1(M,T,r_0)$ such that
\begin{equation}\label{12}
\|G\|_{H^{1}(0,T;L^2(I))}\leq C_1 \int_0^T(y_{xxt}^2+y_{xxxt}^2)(1,t)dt.
\end{equation}
\end{proposition}

\begin{proof}
Setting $y_1=y_{t}$, we have
\begin{gather*}
y_{1tt}+y_{1xxxx}+py_1=G_{t} \quad \text{in }Q,\\
y_1(0,t)=0=y_1(1,t) \quad \text{in }(0,T), \\
y_{1x}(0,t)=0=y_{1x}(1,t) \quad \text{in }(0,T), \\
y_1(x,0)=0 \quad \text{in }I, \\
y_{1t}(x,0)=G(x,0) \quad \text{in }I\,.
\end{gather*}
Since $y(x,0)=y_{t}(x,0)=0$, from \cite{L1} we have
\[
y\in C([0,T],H^4(I))\cap C^{1}(0,T;H^2(I))\cap C^2(0,T;L^2(I))
\] 
and there exists a constant $C=C(T,M)>0$ such that
\[
\|y\|_{H^2(I\times (0,T))}\leq C\|G\|_{H^{1}(0,T;L^2(I))}.
\]
We extend the function $y$ from $I\times (0,T)$ by the formula 
$y(x,t)=y(x,-t), (x,t)\in I\times (-T,0)$ and
denote the extension by the same symbol $y$. 
We know $y\in C([-T,T],H^4(I))\cap C^{1}([-T,T];H^2(I))\cap C^2([-T,T];L^2(I))$ 
and there exists a constant $C=C(T,M)>0$ such that
\[
\|y\|_{H^2(I\times (-T,T))}\leq C\|G\|_{H^{1}(0,T;L^2(I))}.
\]
We extend the function $G_{t}$ on $I\times (-T,T)$ as the even function in $t$
and denote the extension by the same symbol $G_{t}$. 
Then $G_{t}\in L^2(-T,T;L^2(I))$.

By assumption (H1), there exists a $\beta \in (0,1)$ such that 
$\beta>\sup_{x\in I}|x-x_0|/T^2$.
Therefore, by definition of $\psi,\varphi$, we have
\[
\varphi(x,0)\geq 1,\quad \varphi(x,-T)=\varphi(x,T)<1, \quad x\in [0,1].
\]
Therefore for given $\varepsilon>0$, we can choose a sufficiently small 
$\delta=\delta(\varepsilon)>0$, such that
\begin{gather*}
\varphi(x,t)\geq 1-\varepsilon,\quad (x,t)\in [0,1]\times[-\delta,\delta],\\
\varphi(x,t)\leq 1-2\varepsilon, \quad 
(x,t)\in [0,1]\times([-T,-T+2\delta]\cup [T-2\delta,T]).
\end{gather*}
To apply Theorem \ref{T1}, we introduce a cut-off function $\chi$ satisfying
$0\leq \chi \leq 1$, $\chi \in C^{\infty}(\mathbb{R})$, and
\[
\chi(t)=\begin{cases}
0, &[-T,-T+\delta]\cup [T-\delta,T]\\
1, & [-T+2\delta,T-2\delta].
\end{cases}
\]
We set
\begin{align*}
u=e^{\lambda \varphi}\chi y_{t},w=\chi y_{t},D(y)=\int_0^T(y_{xxt}^2+y_{xxxt}^2)(1,t)dt.
\end{align*}
Direct computations show that
\begin{align*}
Pu&=u(\lambda\varphi_{tt}-\lambda^2\varphi_{t}^2
 -\lambda^4\varphi_{x}^4+6\lambda^3\varphi_{x}^2\varphi_{xx}
-3\lambda^2\varphi_{xx}^2+\lambda\varphi_{xxxx} \\
&\quad -4\lambda^2\varphi_{x}\varphi_{xxx})+2\lambda\varphi_{t}u_{t}
 +u_{x}(4\lambda^3\varphi^3_{x}-12\lambda^2\varphi_{x}\varphi_{xx}
 +4\lambda\varphi_{xxx}) \\
&\quad +u_{xx}(-6\lambda^2\varphi_{x}^2+6\lambda\varphi_{xx})
  +u_{xxx}\cdot 4\lambda\varphi_{x}
  +e^{\lambda \varphi}\chi_{tt}y_{t}
  +2e^{\lambda \varphi}\chi_{t} y_{tt}+e^{\lambda \varphi}\chi G_{t}.
\end{align*}
Multiplying $Pu$ by $u_{t}$ and integrating it over $I\times (-T,0)$, 
it follows that
\begin{equation}\label{3}
\begin{aligned}
\int_{-T}^{0}\int_{I} Pu\cdot u_{t}\,dx\,dt
&=\frac{1}{2}\int_{I}u_{t}(x,0)^2dx+\frac{1}{2}\int_{I}u_{xx}^2(0,x)dx\\
&\geq\frac{1}{2}\int_{I}u_{t}(x,0)^2dx
 =\frac{1}{2}\int_{I}G(x,0)^2e^{2\lambda\varphi(x,0)}dx.
\end{aligned}
\end{equation}
On the other hand, by the Cauchy inequality, we have
\begin{align*}
&\int_{-T}^{0}\int_{I} Pu\cdot u_{t}\,dx\,dt \\
&\leq C[\int_{-T}^{0}\int_{I}(G_{t}^2\chi^2\theta^2
 +\theta^2\chi^2_{tt}y_{t}^2+\theta^2\chi^2_{t}y_{tt}^2)\,dx\,dt\\
&\quad +\int_{-T}^{0}\int_{I}(\lambda^7\mu^8\varphi^7u^2
 +\lambda^5\mu^6\varphi^5u_{x}^2+\lambda^3\mu^4\varphi^3u_{xx}^2
 +\lambda\mu^2\varphi u_{xxx}^2\\
&\quad +\lambda^3\mu^4\varphi^3u_{t}^2 +\lambda\mu^2\varphi u_{xt}^2)\,dx\,dt]
=: J_1+J_2.
\end{align*}
It is easy to see that
\begin{align*}
J_1&\leq C\Big[\int_{-T}^T\int_{I}G_{t}^2\chi^2\theta^2\,dx\,dt
 +\Big(\int_{-T+\delta}^{-T+2\delta}
 +\int_{T-2\delta}^{T-\delta}\Big)(y_{t}^2+y_{tt}^2)\theta^2\,dx\,dt\Big]\\
&\leq C\Big[\int_{-T}^T\int_{I}G_{t}^2\chi^2\theta^2\,dx\,dt
 +e^{2\lambda(1-2\varepsilon)}\|y\|^2_{H^2(I\times(-T,T))}\Big]\\
&\leq C\Big[\int_{-T}^T\int_{I}G_{t}^2\chi^2\theta^2\,dx\,dt
 +e^{2\lambda(1-2\varepsilon)}\|G\|^2_{H^{1}(0,T;L^2(I))}\Big].
\end{align*}
Noting $Pw=\chi G_{t}+2y_{tt}\chi_{t}+\chi_{tt}y_{t}$, 
from Theorem \ref{T1} it follows that
\begin{align*}
& \int_{-T}^T\int_{I}(\lambda \mu^2 \varphi
\theta^2w_{xxx}^2+\lambda^3 \mu^4 \varphi^3
\theta^2w_{xx}^2+\lambda^5 \mu^6 \varphi^5
\theta^2w_{x}^2+\lambda^7 \mu^8 \varphi^7 \theta^2w^2\\
&+\lambda^3 \mu^4 \varphi^3 \theta^2w_{t}^2
 +\lambda \mu^2 \varphi \theta^2w_{xt}^2)\,dx\,dt\\
&\leq C_1\Big[\int_{-T}^T\int_{I}\theta^2|Pw|^2\,dx\,dt
 +\int_0^T(\lambda^3 \mu^3 \varphi^3 \theta^2w_{xx}^2
 +\lambda \mu \varphi \theta^2w_{xxx}^2)(1,t)dt\Big]\\
&\leq C[\int_{-T}^T\int_{I}\theta^2|\chi G_{t}
 +2y_{tt}\chi_{t}+\chi_{tt}y_{t}|^2\,dx\,dt \\
&\quad +\int_0^T(\lambda^3 \mu^3 \varphi^3 \theta^2y_{xxt}^2
 +\lambda \mu \varphi \theta^2y_{xxxt}^2)(1,t)dt]\\
&\leq C\Big[ \int_{-T}^T\int_{I}G_{t}^2\theta^2\,dx\,dt
 +e^{2\lambda(1-2\varepsilon)}\|G\|^2_{H^{1}(0,T;L^2(I))}+D(y)\Big];
\end{align*}
thus
\begin{align*}
J_2 & \leq \int_{-T}^{0}\int_{I}(\lambda^7\mu^8\varphi^7u^2
 +\lambda^5\mu^6\varphi^5u_{x}^2+\lambda^3\mu^4\varphi^3u_{xx}^2
 +\lambda\mu^2\varphi u_{xxx}^2 \\
&\quad +\lambda^3\mu^4\varphi^3u_{t}^2+\lambda\mu^2\varphi u_{xt}^2)\,dx\,dt\\
&\leq\int_{-T}^T\int_{I}(\lambda^7\mu^8\varphi^7u^2+\lambda^5\mu^6\varphi^5u_{x}^2
 +\lambda^3\mu^4\varphi^3u_{xx}^2 +\lambda\mu^2\varphi u_{xxx}^2\\
&\quad +\lambda^3\mu^4\varphi^3u_{t}^2+\lambda\mu^2\varphi u_{xt}^2)\,dx\,dt\\
&\leq C\int_{-T}^T\int_{I}(\lambda \mu^2 \varphi
\theta^2w_{xxx}^2+\lambda^3 \mu^4 \varphi^3
\theta^2w_{xx}^2+\lambda^5 \mu^6 \varphi^5
\theta^2w_{x}^2+\lambda^7 \mu^8 \varphi^7 \theta^2w^2 \\
&\quad +\lambda^3 \mu^4 \varphi^3 \theta^2w_{t}^2
 +\lambda \mu^2 \varphi \theta^2w_{xt}^2)\,dx\,dt \\
&\leq C\Big[ \int_{-T}^T\int_{I}G_{t}^2\theta^2\,dx\,dt
 +e^{2\lambda(1-2\varepsilon)}\|G\|^2_{H^{1}(0,T;L^2(I))}+D(y)\Big].
\end{align*}
From the above estimates, we know that
\begin{equation}\label{4}
\begin{aligned}
&\int_{-T}^{0}\int_{I} Pu\cdot u_{t}\,dx\,dt\\
&\leq C\Big[\int_{-T}^T\int_{I}G_{t}^2\theta^2\,dx\,dt
+e^{2\lambda(1-2\varepsilon)}\|G\|^2_{H^{1}(0,T;L^2(I))}+D(y)\Big].
\end{aligned}
\end{equation}
Consequently, \eqref{3} and \eqref{4} imply
\begin{equation}\label{6}
\begin{aligned}
&\frac{1}{2}\int_{I}G(x,0)^2e^{2\lambda\varphi(x,0)}dx\\
&\leq C\Big[ \int_{-T}^T\int_{I}G_{t}^2\theta^2\,dx\,dt
 +e^{2\lambda(1-2\varepsilon)}\|G\|^2_{H^{1}(0,T;L^2(I))}+D(y)\Big].
\end{aligned}
\end{equation}
Next we estimate the first term of the right-hand side of \eqref{6}.
\begin{align*}
&\int_{-T}^T\int_{I}G_{t}^2\theta^2\,dx\,dt\\
&=\int_{I}\Big(\int_{-T}^T|G_{t}(x,t)|^2e^{2\lambda e^{\mu \psi(x,t)}}dt\Big)dx\\
&\leq \int_{I}\Big(\int_{-T}^T|g_0(t)G(x,0)|^2e^{2\lambda 
e^{\mu [(x-x_0)^2-\beta t^2]}}dt\Big)dx\\
&=\int_{I}\Big(\int_{-T}^T|g_0(t)|^2e^{2\lambda e^{\mu [(x-x_0)^2-\beta t^2]}} 
e^{-2\lambda \varphi(x,0)}dt|G(x,0)|^2 e^{2\lambda \varphi(x,0)}\Big)dx \\
&=\int_{I}\Big(\int_{-T}^T|g_0(t)|^2e^{2\lambda e^{\mu [(x-x_0)^2-\beta t^2]}} 
e^{-2\lambda e^{ \mu (x-x_0)^2}}dt|G(x,0)|^2 e^{2\lambda \varphi(x,0)}\Big)dx\\
&=\int_{I}\Big(\int_{-T}^T|g_0(t)|^2e^{2\lambda e^{\mu (x-x_0)^2}
 (e^{-\beta t^2}-1)}dt|G(x,0)|^2 e^{2\lambda \varphi(x,0)}\Big)dx\\
&\leq\int_{I}\Big(\int_{-T}^T|g_0(t)|^2e^{2\lambda (e^{-\beta t^2}-1)}dt
|G(x,0)|^2 e^{2\lambda \varphi(x,0)}\Big)dx\\
&=: \int_{I}(\int_{-T}^Th_{\lambda}(t)dt|G(x,0)|^2 e^{2\lambda \varphi(x,0)})dx.
\end{align*}
We have that $h_{\lambda}(t)$ is in $L^{1}(-T,T)$, and 
$\lim_{\lambda\to\infty} h_{\lambda}(t)=0$ for $t\neq 0$ and 
$|h_{\lambda}(t)|\leq |g_0|^2\in L^{1}(-T,T)$.
Hence the Lebesgue theorem implies
\[
\int_{-T}^T|g_0(t)|^2e^{2\lambda (e^{-\beta t^2}-1)}dt=o(1)
\]
as $\lambda\to \infty$, so that
\begin{equation}\label{7}
\int_{-T}^T\int_{I}G_{t}^2\theta^2\,dx\,dt
=o(1)\int_{I}G(x,0)^2e^{2\lambda\varphi(x,0)}dx.
\end{equation}
By the assumption in \eqref{8}, we have
\begin{equation}\label{9}
\|G\|^2_{H^{1}(0,T;L^2(I))}\leq C\int_{I}G(x,0)^2dx.
\end{equation}
Therefore \eqref{8}, \eqref{6} and \eqref{9} yield
$\|G\|^2_{H^{1}(0,T;L^2(I))}\leq CD(y)$.
\end{proof}

Now we can prove Theorem \ref{T2}.
In Proposition \ref{pro2}, we set $y=u(p)-u(q)$ and $G(x,t)=(p-q)u(q)$.
Recalling that $H^{1}(0,T;L^2(I))\subset C([0,T];L^2(I))$, 
it is easy to see that assumption \eqref{8} holds,
and that \eqref{12} implies
\begin{align*}
\|p-q\|_{L^2(I)}
&\leq C\|(p-q)(x)u(q)(0,x)\|_{L^2(I)}\\
&\leq C\|(p-q)u(q)\|_{C([0,T];L^2(I))}\\
&\leq C\|(p-q)u(q)\|_{H^{1}(0,T;L^2(I))}\\
&=C\|G\|_{H^{1}(0,T;L^2(I))}\\
&\leq C\int_0^T(y_{xxt}^2+y_{xxxt}^2)(1,t)dt\\
&\leq C\int_0^T[(u(p)-u(q))_{xxt}^2+(u(p)-u(q))_{xxxt}^2](1,t)dt,
\end{align*}
this completes the proof.


\subsection*{Acknowledgements} 
This research is supported by NSFC Grant (11601073).
I am grateful to the anonymous referee and editor for 
their very helpful comments and suggestions.
I sincerely  thank Professor Yong Li for many useful suggestions and help.


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