\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small {\em Electronic Journal of Differential Equations}, Vol. 2007(2007), No. 06, pp. 1--18.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu (login: ftp)} \thanks{\copyright 2007 Texas State University - San Marcos.} \vspace{8mm}} \begin{document} \title[\hfilneg EJDE-2007/06\hfil Stabilization of solutions] {Stabilization of solutions to higher-order nonlinear Schr\"{o}dinger equation with localized damping} \author[E. Bisognin, V. Bisognin, O. P. Vera V.\hfil EJDE-2007/06\hfilneg] {Eleni Bisognin, Vanilde Bisognin, Octavio Paulo Vera Villagr\'{a}n} % in alphabetical order \address{Eleni Bisognin \newline Centro Universit\'{a}rio Franciscano-UNIFRA\\ Rua dos Andrade 1614, CEP: 97010-032\\ Santa Maria, R. S., Brasil} \email{eleni@unifra.br} \address{Vanilde Bisognin \newline Centro Universit\'{a}rio Franciscano-UNIFRA\\ Rua dos Andrade 1614, CEP: 97010-032\\ Santa Maria, R. S., Brasil} \email{vanilde@unifra.br} \address{Octavio Paulo Vera Villagr\'{a}n \newline Departamento de Matem\'{a}tica \\ Universidad del B\'{\i}o-B\'{\i}o \\ Collao 1202, Casilla 5-C \\ Concepci\'{o}n, Chile} \email{overa@ubiobio.cl \; octavipaulov@yahoo.com} \thanks{Submitted August 2, 2006. Published January 2, 2007.} \subjclass[2000]{35K60, 93C20} \keywords{Higher order nonlinear Schr\"{o}dinger equation; stabilization; \hfill\break\indent localized damping} \begin{abstract} We study the stabilization of solutions to higher-order nonlinear Schr\"{o}dinger equations in a bounded interval under the effect of a localized damping mechanism. We use multiplier techniques to obtain exponential decay in time of the solutions of the linear and nonlinear equations. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \section{Introduction} In this work we consider the initial-value problem of the higher-order nonlinear Schr\"{o}dinger equation with localized damping \begin{gather}\label{e101} i u_{t} + \alpha u_{xx} + i \beta u_{xxx} + |u|^{2} u + i a(x) u=0\\ \label{e102} u(x, 0) = u_{0}(x) \end{gather} where $00$, and with boundary conditions \begin{gather}\label{e103} u(0, t) = u(L, t)=0 \quad \mbox{for all } t>0\\ \label{e104} u_{x}(L, t)=0 \quad \mbox{for all } t>0 \end{gather} with $\alpha, \beta\in\mathbb{R}$, $\beta\neq 0$, $u=u(x,t)$ a complex valued function, $a=a(x)$ a nonnegative everywhere function such that $a(x)\in C^{\infty}((0, L))$ and $a(x)\geq a_{0}>0$. Equation \eqref{e101} is a particular case of the equation \begin{equation} \label{R} \begin{gathered} i u_{t} + \omega u_{xx} + i \beta u_{xxx} + \gamma |u|^{2} u + i \delta |u|^{2} u_{x} + i \epsilon u^{2} \overline{u}_{x}=0 \quad x, t\in \mathbb{R}\\ u(x, 0) = u_{0}(x) \end{gathered} \end{equation} where $\omega , \beta , \gamma , \delta $ are real numbers and $\beta \neq 0$. This equation was first proposed by Hasegawa and Kodama \cite{ha2} as a model for the propagation of a signal in a optic fiber (see also \cite{ko1}). The equation \eqref{R} can be reduced to other well known equations. For instance, setting $\omega =1$, $\beta = \delta =\epsilon =0$ in \eqref{R} we have the semilinear Schr\"{o}dinger equation, \begin{equation} i u_{t} + u_{xx} + \gamma |u|^{2} u =0. \label{R1} \end{equation} If we let $\beta = \gamma =0$ and $\omega =1$ in \eqref{R} we obtain the nonlinear Schr\"{o}dinger equation \begin{equation} i u_{t} + u_{xx} + i \delta |u|^{2} u_{x} + i \epsilon u^{2} \overline{u}_{x}=0. \label{R2} \end{equation} Letting $\alpha = \gamma = \epsilon =0$ in \eqref{R} arises is the complex modified Korteweg-de Vries (KdV) equation \begin{equation} i u_{t} + i \beta u_{xxx} + i \delta |u|^{2} u_{x} =0.\label{R3} \end{equation} The initial value problem for the equations \eqref{R1}, \eqref{R2} and \eqref{R3} has been extensively studied in the last few years, see for instance \cite{bo1,co1,ka1,sa1} and references therein. In 1992, Laurey \cite{la1} considered the equation \eqref{R} and proved local well-posedness of the initial value problem associated for data in $H^{s}(\mathbb{R})$, $s>3/4$, and global well-posedness in $H^{s}(\mathbb{R})$, $s\geq 1$. In 1997, Staffilani \cite{s1} established local well-posedness for data in $H^{s}(\mathbb{R})$, $s\geq 1/4$ in \eqref{R} improving Laurey's result. A similar result was given in \cite{ca1,ca2} with $w(t)$, $\beta(t)$ real functions. Recently, Sep\a'{u}lveda and Vera \cite{se1} showed that $C^{\infty}$ solutions $u(x, t)$ are obtained for all $t>0$ if the initial data $u_{0}(x)$ decays faster than polynomially on $\mathbb{R}^{+}=\{x\in\mathbb{R}:x>0\}$ and has a certain initial Sobolev regularity. In \cite{bi1} Bisognin and Vera considered the equation \eqref{R} with $\delta=\epsilon=0$ and proved the unique continuation property. This paper concerns the exponential stabilization of the solution of \eqref{e101} when the damping $a=a(x)$ is effective only on a subset of the interval $(0, L)$. This problem was extensively studied in the context of wave equations, see Dafermos \cite{da1}, Haraux \cite{ha1}, Slemrod \cite{sl1}, Zuazua \cite{zu1} and Nakao \cite{na1}. The same problem has been also studied for the KdV equation. Here we can mention the works of Komornik, Russell and Zhang \cite{ko3}. Using a different damping mechanism they obtained the exponential decay with periodic boundary conditions. In \cite{pe1}, Menzala, Vasconcellos and Zuazua studied the nonlinear KdV equation inspired in the work of Rosier \cite{ro1}. They studied the stabilization of solutions for the KdV equation in a bounded interval under the effect of a localized damping mechanism. Using compactness arguments, the smoothing effect of the KdV equation on the line and the unique continuation results, the authors deduced the exponential decay in time of the solutions of the linear equation and a local uniform stabilization result of the solutions of the nonlinear equation when the localized damping is active simultaneously only in a neighborhood of both extremes $x=0$, $x=L$. The same result was obtained by the KdV coupled system by Menzala, Bisognin and Bisognin \cite{pe2}. The main result of this paper says that the total energy $E(t)$ associated to \eqref{e101} decays exponentially as $t\to +\infty$, for bounded sets of initial data. In order to prove the result we use multipliers together with compactness arguments and smoothing properties proved by M. Sep\a'{u}veda and Vera \cite{se1} and the Unique Continuation Principle valid for this problem, see Bisognin and Vera \cite{bi1}. This paper is organized as follows: In section two, we study the existence of global solution to the linear and nonlinear problem. In section three, we study the stabilization result of the problem. First we prove the exponential decay in the linear problem and of at end we prove the stabilization of the solution of the nonlinear problem. The notation that we use in this article is standard and can be found in Temam \cite{te1}. \section{Stabilization of solutions of the linear problem} In this section we are interested in proving the global existence and uniqueness of the solution and the exponential decay of the solution of the linear problem associated to \eqref{e101}-\eqref{e104}. We consider the problem \begin{gather} \label{e201} i u_{t} + \alpha u_{xx} + i \beta u_{xxx} + i a(x) u=0\\ \label{e202} u(x, 0)=u_{0}(x),\quad \mbox{for all } x\in I\\ \label{e203} u(0, t)=u(L, t)=0,\quad \mbox{for all } t>0\\ \label{e204} u_{x}(L, t)=0,\quad \mbox{for all } t>0 \end{gather} where $I=(0, L)$, $a\in C^{\infty}(I)$, $a(x)\geq a_{0}>0$ is assumed to be nonnegative everywhere in an open non empty proper subset $\omega$ of $I$ and we will prove the global existence of the problem \eqref{e201}-\eqref{e204}. We consider $a\equiv 0$ and the operator $A=-\beta \partial_{x}^{3} + i \alpha \partial_{x}^{2}$ with domain \[ \mathcal{D}(A)=\{v\in H^{3}(I):v(0)=v(L)=0,\;v_{x}(L)=0\}\subseteq L^{2}(I) \] \begin{lemma} \label{lem2.1} Let $a\equiv 0$ and $\beta>0$. Then, the operator $A$ is the infinitesimal generator of a strongly continuous semigroup $\{S(t)\}_{t\geq 0}$ of contractions in $L^{2}(I)$. \end{lemma} \begin{proof} It is easy to prove that $A$ is closed. Let us to prove that $A$ is dissipative. Integration by parts give us \[ (Av, v)_{L^{2}(I)} = \int_{0}^{L}(-\beta v_{xxx} + i \alpha v_{xx}) \overline{v}\,dx = - \frac{\beta}{2} |v_{x}^{2}(0, t)|^{2} - i \alpha\int_{0}^{L}|v_{x}|^{2}dx. \] Hence, \[ \mathop{\rm Re}(Av, v)_{L^{2}(I)} = - \frac{\beta}{2} |v_{x}^{2}(0, t)|^{2} \leq0, \] where $A$ is dissipative. On the other hand, the adjoint of the operator $A$ is given by \[ H^{*}v=\beta v_{xxx} - i \alpha v_{xx} \] with domain \[ \mathcal{D}(H^{*})=\{v\in H^{3}(I):v(0, t)=v(L, t)=0,\;v_{x}(0, t)=0\}\subseteq L^{2}(I). \] A similar calculation shows that \[ (H^{*}v, v)_{L^{2}(I)} = \int_{0}^{L}(\beta v_{xxx} - i \alpha v_{xx}) \overline{v}\,dx = - \frac{\beta}{2} |v_{x}^{2}(0, t)|^{2} + i \alpha\int_{0}^{L}|v_{x}|^{2}\,dx. \] Hence \[ \mathop{\rm Re}(Av, v)_{L^{2}(I)}= - \frac{\beta}{2} |v_{x}^{2}(0, t)|^{2} \leq 0. \] The conclusion of Lemma \ref{lem2.1} follows from the Stone Theorem \cite{pa1} of semigroup theory. \end{proof} The above discussion proves the following result. \begin{theorem} \label{thm2.1} Let $u_{0}\in \mathcal{D}(A)$, $a\equiv 0$ and $\beta>0$. Then, there exists a unique function $u$ such that $u\in C(0, +\infty: H^{3}(I))\cap C^{1}(0, +\infty: L^{2}(I))$ which solves \eqref{e201}-\eqref{e204}. \end{theorem} The well-posedness of system \eqref{e201}-\eqref{e204}, when $a\not\equiv 0$ can be handled in a similar way by considering the term $a(x) u$ as a linear perturbation of the case $a\equiv 0$. Now, we will prove the exponential decay of the total energy $E(t)$ associated to \eqref{e201}-\eqref{e204} under suitable assumptions on the open subset $\omega$ of $I$. We denote by $\{S(t)\}_{t\geq 0}$ the semigroup of contractions associated with $A$, and by $\mathcal{H}$ the Banach space $C([0, T]: L^{2}(I))\cap L^{2}(0, T: H^{1}(I))$ with the norm \begin{equation} \label{e205}\|v\|_{\mathcal{H}} = \sup_{[0, T]}\|v( \cdot , t)\|_{L^{2}(I)} + \Big(\int_{0}^{T}\|v( \cdot , t)\|_{H^{1}(I)}^{2}\Big)^{1/2} \end{equation} \begin{theorem} \label{thm2.2} Consider the solution of the problem \eqref{e201}-\eqref{e204}. Then there exist $c>0$ and $\mu>0$ such that \begin{equation} \label{e206}\|u( \cdot , t)\|_{L^{2}(I)}^{2}\leq c \|u_{0}\|_{L^{2}(I)}^{2} e^{-\mu t} \end{equation} for all $t\geq 0$ and $u_{0}\in L^{2}(I)$. \end{theorem} For the proof of the above theorem we need the following result. \begin{lemma} \label{lem2.2} Let $|\alpha|<3 \beta$. Then \begin{enumerate} \item The map $u_{0}\in L^{2}(I)\to S(t)u_{0}\in \mathcal{H}$ is continuous. \item For $u_{0}\in L^{2}(I)$, $\partial_{x}u(0, \cdot )$ makes sense in $L^{2}(I)$ and \begin{gather} \label{e207} \|u_{x}(0, \cdot )\|_{L^{2}(0, T)}\leq \frac{1}{\sqrt{\beta}}\;\|u_{0}\|_{L^{2}(I)} \\ \label{e208}\int_{0}^{T}|u_{0}|^{2}\,dx \leq \frac{1}{T}\int_{0}^{T}\int_{0}^{L}|u|^{2}\,dx\,dt + \beta\int_{0}^{T}|u_{x}(0, t)|^{2}\,dt + 2\int_{0}^{T}\int_{0}^{L}a(x) |u|^{2}\,dx. \end{gather} \end{enumerate} \end{lemma} \begin{proof} (1) For $u_{0}\in L^{2}(I)$, let $u=S(t)u_{0}$ be the mild solution of \eqref{e201}-\eqref{e204}. By Theorem \ref{thm2.1}, $u\in C(0, T: L^{2}(I))$ and \begin{equation} \label{e209}\|u\|_{C(0, T: L^{2}(I))} \leq \|u_{0}\|_{L^{2}(I)}. \end{equation} To see that $u\in L^{2}(0, T: H^{1}(I))$ we first assume that $u_{0}\in \mathcal{D}(A)$. Let $\xi=\xi(x, t)\in C^{\infty}([0, L]\times [0, T])$. Multiplying the equation \eqref{e201} by $\xi \overline{u}$ we have \begin{equation} \label{e210}i \xi \overline{u} u_{t} + \alpha \xi \overline{u} u_{xx} + i \beta \xi \overline{u} u_{xxx} + i \xi a(x) |u|^{2}=0. \end{equation} Applying the conjugate, we have \begin{equation} \label{e211}- i \xi u \overline{u}_{t} + \alpha \xi u \overline{u}_{xx} - i \beta \xi u \overline{u}_{xxx} - i \xi a(x) |u|^{2}=0. \end{equation} Subtracting \eqref{e210} and \eqref{e211} and integrating over $x\in (0, L)$, we have \begin{equation} \label{e212} \begin{aligned} & i \frac{d}{dt}\int_{0}^{L}\xi |u|^{2}\,dx - i\int_{0}^{L}\xi_{t} |u|^{2}\,dx + i \beta\int_{0}^{L}\xi \overline{u} u_{xxx}\,dx + i \beta\int_{0}^{L}\xi u \overline{u}_{xxx}\,dx \\ & + \alpha\int_{0}^{L}\xi \overline{u} u_{xx}\,dx - \alpha\int_{0}^{L}\xi u \overline{u}_{xx}\,dx + 2 i\int_{0}^{L}\xi a(x) |u|^{2}\,dx = 0. \end{aligned} \end{equation} Each term in the above equation is treated separately, integrating by parts and using the boundary conditions we obtain \begin{gather*} \begin{aligned} \int_{0}^{L}\xi \overline{u} u_{xxx}\,dx & = \int_{0}^{L}\xi_{xx} \overline{u} u_{x}\,dx + 2\int_{0}^{L}\xi_{x} |u_{x}|^{2}\,dx + \int_{0}^{L}\xi u_{x} \overline{u}_{xx}\,dx\\ &\quad +\xi(0, t) |u_{x}(0, t)|^{2} \end{aligned}\\ \int_{0}^{L}\xi u \overline{u}_{xxx}\,dx = \int_{0}^{L}\xi_{xx} u \overline{u}_{x}\,dx + \int_{0}^{L}\xi_{x} |u_{x}|^{2}\,dx - \int_{0}^{L}\xi u_{x} \overline{u}_{xx}\,dx \\ \int_{0}^{L}\xi \overline{u} u_{xx}\,dx = -\int_{0}^{L}\xi_{x} \overline{u} u_{x}\,dx - \int_{0}^{L}\xi |u_{x}|^{2}\,dx \\ \int_{0}^{L}\xi u \overline{u}_{xx}\,dx = -\int_{0}^{L}\xi_{x} u \overline{u}_{x}\,dx - \int_{0}^{L}\xi |u_{x}|^{2}\,dx. \end{gather*} Replacing these expression in \eqref{e212} and performing straightforward calculations, \begin{align*} & i \frac{d}{dt}\int_{0}^{L}\xi |u|^{2}\,dx - i\int_{0}^{L}\xi_{t} |u|^{2}\,dx + i \beta\int_{0}^{L}\xi_{xx} (|u|^{2})_{x}\,dx + 3 i \beta\int_{0}^{L}\xi_{x} |u_{x}|^{2}\,dx \\ & + i \beta \xi(0, t) |u_{x}(0, t)|^{2} - 2 i \alpha \mathop{\rm Im} \int_{0}^{L}\xi_{x} \overline{u} u_{x}\,dx + 2 i\int_{0}^{L}\xi a(x) |u|^{2}\,dx=0 \end{align*} and \begin{equation} \begin{aligned} & \frac{d}{dt}\int_{0}^{L}\xi |u|^{2}\,dx - \int_{0}^{L}\xi_{t} |u|^{2}\,dx + \beta\int_{0}^{L}\xi_{xx} (|u|^{2})_{x}\,dx + 3 \beta\int_{0}^{L}\xi_{x} |u_{x}|^{2}\,dx \\ & + \beta \xi(0, t) |u_{x}(0, t)|^{2} - 2 \alpha \mathop{\rm Im} \int_{0}^{L}\xi_{x} \overline{u} u_{x}\,dx + 2\int_{0}^{L}\xi a(x) |u|^{2}\,dx=0. \end{aligned} \label{e213} \end{equation} Let $\xi(x, t)\equiv x$, then in \eqref{e213}, we obtain \[ \frac{d}{dt}\int_{0}^{L}x |u|^{2}\,dx + 3 \beta\int_{0}^{L}|u_{x}|^{2}\,dx - 2 \alpha \mathop{\rm Im} \int_{0}^{L}\overline{u} u_{x}\,dx + 2\int_{0}^{L}x a(x) |u|^{2}\,dx=0. \] Hence, \begin{equation} \label{e2014} \begin{aligned} & \frac{d}{dt}\int_{0}^{L}x |u|^{2}\,dx + 3 \beta\int_{0}^{L}|u_{x}|^{2}\,dx + 2\int_{0}^{L}x a(x) |u|^{2}\,dx \\ & =2 \alpha \mathop{\rm Im} \int_{0}^{L}\overline{u} u_{x}\,dx\leq |\alpha|\int_{0}^{L}|u|^{2}\,dx + |\alpha| \int_{0}^{L}|u_{x}|^{2}\,dx \end{aligned} \end{equation} then \[ \frac{d}{dt}\int_{0}^{L}x |u|^{2}\,dx + (3 \beta - |\alpha|)\int_{0}^{L}|u_{x}|^{2}\,dx + 2\int_{0}^{L}x a(x) |u|^{2}\,dx \leq |\alpha|\int_{0}^{L}|u|^{2}\,dx \] and \begin{equation}\label{e214} \frac{d}{dt}\int_{0}^{L}x |u|^{2}\,dx + (3 \beta - |\alpha|)\int_{0}^{L}|u_{x}|^{2}\,dx + 2\int_{0}^{L}x a(x) |u|^{2}\,dx \leq |\alpha| \|u\|_{L^{2}(0, L)}^{2}. \end{equation} Integrating \eqref{e214} over $t\in [0, T]$ we have \begin{align*} & \int_{0}^{L}x |u|^{2}\,dx + (3 \beta - |\alpha|)\int_{0}^{T}\int_{0}^{L}|u_{x}|^{2}\,dx\,dt + 2\int_{0}^{T}\int_{0}^{L}x a(x) |u|^{2}\,dx\,dt \\ & \leq \int_{0}^{L}x |u_{0}|^{2}\,dx + |\alpha|\int_{0}^{T}\|u\|_{L^{2}(0, L)}^{2}\,dt\\ & \leq L \|u_{0}\|_{L^{2}(0, L)}^{2} + T |\alpha| \|u_{0}\|_{L^{2}(0, L)}^{2}\\ & = [L + T |\alpha| ] \|u_{0}\|_{L^{2}(0, L)}^{2} \end{align*} Hence, \begin{align*} & \int_{0}^{L}x |u|^{2}\,dx + (3 \beta - |\alpha|) \|u_{x}\|_{L^{2}(0, T: L^{2}(0, L))}^{2} + 2\int_{0}^{T}\int_{0}^{L}x a(x) |u|^{2}\,dx\,dt \\ & \leq [L + T |\alpha| ] \|u_{0}\|_{L^{2}(0, L)}^{2}. \end{align*} Using that $a(x)\geq a_{0}>0$, we obtain \begin{equation}\label{e215} \int_{0}^{L}x |u|^{2}\,dx + (3 \beta - |\alpha|) \|u_{x}\|_{L^{2}(0, T: L^{2}(0, L))}^{2} \leq [L + T |\alpha| ] \|u_{0}\|_{L^{2}(0, L)}^{2}. \end{equation} In particular, using that $|\alpha|<3 \beta$, \[ (3 \beta - |\alpha|) \|u_{x}\|_{L^{2}(0, T: L^{2}(0, L))}^{2} \leq [L + T |\alpha| ] \|u_{0}\|_{L^{2}(0, L)}^{2}. \] and \begin{equation} \label{e216}\|u_{x}\|_{L^{2}(0, T: L^{2}(0, L))}^{2} \leq \frac{1}{(3 \beta - |\alpha|)}\left[L + T |\alpha| \right] \|u_{0}\|_{L^{2}(0, L)}^{2}. \end{equation} By the density of $\mathcal{D}(A)$ in $L^{2}(I)$ the result extends to arbitrary $u_{0}\in L^{2}(I)$. We remark that: (a) The estimate \eqref{e215} gives a smoothing effect. (b) In \eqref{e2014} using Young's estimate and assuming that $\beta>0$ we have \[ 2 \alpha \mathop{\rm Im}\int_{0}^{L}\overline{u} u_{x}\,dx\leq \frac{|\alpha|^{2}}{2 \beta}\int_{0}^{L}|u|^{2}\,dx + 2 \beta\int_{0}^{L}|u_{x}|^{2}\,dx. \] Then, in we obtain \eqref{e214}, \[ \frac{d}{dt}\int_{0}^{L}x |u|^{2}\,dx + \beta\int_{0}^{L}|u_{x}|^{2}\,dx + 2\int_{0}^{L}x a(x) |u|^{2}\,dx \leq \frac{|\alpha|^{2}}{2 \beta} \|u\|_{L^{2}(0, L)}^{2} \] and the assumption that $|\alpha|<3 \beta$ can be removed. (2) We also assume $u_{0}\in \mathcal{D}(A)$ and taking $\xi(x, t)=1$ in \eqref{e213}, we have \begin{equation} \label{e217}\frac{d}{dt}\int_{0}^{L}|u|^{2}\,dx + \beta |u_{x}(0, t)|^{2} + 2\int_{0}^{L}a(x) |u|^{2}\,dx = 0. \end{equation} Hence, integrating \eqref{e217} over $t\in [0, T]$ we have \[ \int_{0}^{L}|u|^{2}\,dx + \beta\int_{0}^{T}|u_{x}(0, t)|^{2}\,dt + 2\int_{0}^{T}\int_{0}^{L}a(x) |u|^{2}\,dx\,dt = \int_{0}^{L}|u_{0}|^{2}\,dx \] Using $a(x)\geq a_{0}>0$, we obtain \begin{gather*} \int_{0}^{L}|u|^{2}\,dx + \beta\int_{0}^{T}|u_{x}(0, t)|^{2}\,dt \leq \int_{0}^{L}|u_{0}|^{2}\,dx \\ \beta\int_{0}^{T}|u_{x}(0, t)|^{2}\,dt \leq \int_{0}^{L}|u_{0}|^{2}\,dx - \int_{0}^{L}|u|^{2}\,dx\leq \int_{0}^{L}|u_{0}|^{2}\,dx; \end{gather*} therefore, \eqref{e207} is proved. \end{proof} On the other hand, taking $\xi(x, t)=T - t$ in \eqref{e213} we have \begin{equation} \label{e218} \begin{aligned} & \frac{d}{dt}\int_{0}^{L}(T - t) |u|^{2}\,dx + \int_{0}^{L}|u|^{2}\,dx \\ & + \beta (T - t) |u_{x}(0, t)|^{2} + 2\int_{0}^{L}(T - t) a(x) |u|^{2}\,dx=0. \end{aligned} \end{equation} Integrating \eqref{e218} over $t\in[0, T]$ we have \begin{align*} & -T\int_{0}^{L}|u_{0}|^{2}\,dx + \int_{0}^{T}\int_{0}^{L}|u|^{2}\,dx\,dt \\ & + \beta \int_{0}^{T}(T - t) |u_{x}(0, t)|^{2}\,dt + 2\int_{0}^{T}\int_{0}^{L}(T - t) a(x) |u|^{2}\,dx\,dt=0. \end{align*} Then \begin{align*} &T\int_{0}^{L}|u_{0}|^{2}\,dx \\ & = \int_{0}^{T}\int_{0}^{L}|u|^{2}\,dx\,dt + \beta \int_{0}^{T}(T - t) |u_{x}(0, t)|^{2}\,dt + 2\int_{0}^{T}\int_{0}^{L}(T - t) a(x) |u|^{2}\,dx\,dt\\ & \leq \int_{0}^{T}\int_{0}^{L}|u|^{2}\,dx\,dt + \beta \int_{0}^{T}T |u_{x}(0, t)|^{2}\,dt + 2\int_{0}^{T}\int_{0}^{L}T a(x) |u|^{2}\,dx\,dt \end{align*} and \begin{equation} \label{e219} \begin{aligned} \int_{0}^{L}|u_{0}|^{2}\,dx & \leq \frac{1}{T}\int_{0}^{T}\int_{0}^{L}|u|^{2}\,dx\,dt + \beta \int_{0}^{T}|u_{x}(0, t)|^{2}\,dt \\ &\quad + 2\int_{0}^{T}\int_{0}^{L}a(x) |u|^{2}\,dx\,dt. \end{aligned} \end{equation} Equation \eqref{e219} holds trivially for any $u_{0}\in L^{2}(0, L)$. %\end{proof} \begin{proof}][Proof of Theorem \ref{thm2.2}] To show the result,from \eqref{e219}, it suffices to prove \begin{equation} \label{e220}\frac{1}{T}\int_{0}^{T}\int_{0}^{L}|u|^{2}\,dx\,dt \leq \beta\int_{0}^{T}|u_{x}(0, t)|^{2}\,dt + 2\int_{0}^{T}\int_{0}^{L}a(x) |u|^{2}\,dx\,dt. \end{equation} Let us argue by contradiction. Suppose that \eqref{e220} is not valid. Then, there will exist a sequence of solutions $\{u_{n}\}$ of \eqref{e201}-\eqref{e204} such that \[ \lim_{n\to\infty}\frac{\|u_{n}\|_{L^{2}(0, T: L^{2}(I))}^{2}} {\beta\int_{0}^{T}|u_{n}'(0, t)|^{2}\,dt + 2\int_{0}^{T}\int_{0}^{L}a(x) |u_{n}|^{2}\,dx\,dt}=+\infty. \] Let \[ \lambda_{n}=\|u_{n}\|_{L^{2}(0, T: L^{2}(I))}\quad\mbox{and}\quad v_{n}(x, t) = \frac{u_{n}(x, t)}{\lambda_{n}}. \] We have that $v_{n}$ solves the \eqref{e201}-\eqref{e204} problem with initial data $v_{n}(x, 0) = \frac{u_{n}(x, 0)}{\lambda_{n}}$. Furthermore, \begin{gather} \label{e221}\|v_{n}\|_{L^{2}(0, T: L^{2}(I))}=1, \\ \label{e222}\beta\int_{0}^{T}|v_{n}'(0, t)|^{2}\,dt + 2\int_{0}^{T}\int_{0}^{L}a(x) |v_{n}|^{2}\,dx\,dt\to 0,\quad\mbox{as}\quad n\to\infty . \end{gather} In view of \eqref{e219}, it follows that $v_{n}(x, 0)$ is bounded in $L^{2}(I)$. Thus \[ \|v_{n}( \cdot , t)\|_{L^{2}(I)}\leq c,\quad\mbox{for all } 0\leq t\leq T. \] According to \eqref{e207} \begin{equation} \label{e223}\|v_{n}\|_{L^{2}(0, T: H^{1}(I))}\leq c(T) \|v_{n}( \cdot , t)\|_{L^{2}(I)}\leq \text{constant, for all } n\in\mathbb{N}. \end{equation} Estimates \eqref{e223} and \eqref{e221} tell us that \[ i (v_{n})_{t} = -\alpha (v_{n})_{xx} - i \beta (v_{n})_{xxx} - i a(x) v_{n} \] is bounded in $L^{2}(0, T: H^{-2}(I))$. Since the embedding $H^{1}(I)\hookrightarrow L^{2}(I)$ is compact it follows that $(v_{n})$ is relatively compact in $L^{2}(0, T: L^{2}(I))$. By extracting a subsequence we may deduce that \begin{gather*} v_{n}\rightharpoonup v\quad\mbox{weakly in }L^{2}(0, T: H^{-2}(I)), \\ v_{n}\to v\quad\mbox{strongly in }L^{2}(0, T: L^{2}(I)). \end{gather*} Since \begin{equation} \label{e224}\|v_{n}\|_{L^{2}(0, T: L^{2}(I))}=1 \end{equation} it follows that \begin{equation} \label{e225}\|v\|_{L^{2}(0, T: L^{2}(I))}=1. \end{equation} By the weak lower semicontinuity, we have \begin{align*} 0 &= \lim_{n\to\infty}\inf\big\{\beta\int_{0}^{T}|v_{n}'(0, t)|^{2}\,dt +2\int_{0}^{T}\int_{0}^{L}a(x) |v_{n}|^{2}\,dx\,dt\big\}\\ & \geq \beta\int_{0}^{T}|v'(0, t)|^{2}\,dt + 2\int_{0}^{T}\int_{0}^{L}a(x) |v|^{2}\,dx\,dt \end{align*} which guarantees that $a(x) v\equiv 0$, and in particular, $v\equiv 0$ in $\omega\times (0, T)$. On the other hand, the limit $v$ satisfies \[ i v_{t} + \alpha v_{xx} + i \beta v_{xxx} + i a(x) v=0. \] Using Holmgren's Uniqueness Theorem (see \cite{re1}) we deduce that $v\equiv 0$ in $I\times (0, T)$. This contradicts \eqref{e221}. Consequently, \eqref{e220} has to be true. On the other hand, we have \[ \frac{d}{dt} \|u( \cdot , t)\|_{L^{2}(I)}^{2} + 2\int_{0}^{L}a(x) |u|^{2}\,dx \leq -\beta |u_{x}(0, t)|^{2}\leq 0 \] and \[ \|u( \cdot , T)\|_{L^{2}(I)}^{2} = \|u_{0}\|_{L^{2}(I)}^{2} - \beta\int_{0}^{T}|u_{x}(0, t)|^{2}\,dt - 2\int_{0}^{T}\int_{0}^{L}a(x) |u|^{2}\,dx\,dt \] which together with \eqref{e219} give us the inequality \begin{align*} &(1 + c) \|u( \cdot , T)\|_{L^{2}(I)}^{2} \\ & \leq (1 +c) \big[\|u_{0}\|_{L^{2}(I)}^{2} - \beta\int_{0}^{T}|u_{x}(0, t)|^{2}\,dt - 2\int_{0}^{T}\int_{0}^{L}a(x) |u|^{2}\,dx\,dt\big]\\ & \leq c \|u_{0}\|_{L^{2}(I)}^{2} - \big[\beta\int_{0}^{T}|u_{x}(0, t)|^{2}\,dt + 2\int_{0}^{T}\int_{0}^{L}a(x) |u|^{2}\,dx\,dt\big]\\ & \leq c \|u_{0}\|_{L^{2}(I)}^{2}. \end{align*} Consequently, \[ \|u( \cdot , T)\|_{L^{2}(I)}^{2}\leq \mu \|u_{0}\|_{L^{2}(I)}^{2}\quad\mbox{with } \mu=\frac{c}{1. + c}<1 \] Therefore, by a semigroup property, the conclusion of the Theorem follows. \end{proof} \section{Stabilization of the solution of the non-linear problem} In this section we prove the existence of a global solution (and uniqueness) and the exponential decay of the solution of the nonlinear problem \eqref{e101}-\eqref{e104}. The proof of the result needs of the Unique Continuation Principle since we are dealing with a nonlinear equation. \begin{theorem}[local existence and uniqueness] \label{thm3.1} Let $|\alpha|<3 \beta$ and $u_{0}\in L^{2}(I)$. Then, there exist $T_{0}>0$ and a unique function $u\in L^{\infty}(0, T_{0}: L^{2}(I))\cap L^{2}(0, T_{0}: H_{0}^{1}(I))$ that satisfies \eqref{e101}-\eqref{e104}. \end{theorem} \begin{proof} Let $T>0$ and consider the set of functions $X(T)=\{u:u\in L^{2}(0, T: H_{0}^{1}(I))\}$ with the norm \[ \|u\|_{X(T)} = \Big(\int_{0}^{T}\|u(s)\|_{H^{1}(I)}^{2}\,ds\Big)^{1/2}. \] We define the map $P:X(T)\to L^{\infty}(0, +\infty : L^{2}(I))$ given by \begin{equation} \label{e301}P(u)(t) = S(t)u_{0} + \int_{0}^{t}S(t - \tau) g(u)(\tau)\,d\tau \end{equation} where $g(u)=|u|^{2} u$. In order to prove local existence (and uniqueness) it is sufficient to prove that $P$ maps $X(T)$ into itself continuously and it is a contraction for $T>0$ sufficiently small. According to the results of section two it follows that the semigroup of contractions $\{S(t)\}_{t\geq 0}$ corresponding to the linear system satisfies the following properties: \begin{gather} \label{e302} \|S(t)u\|_{L^{2}}\leq \|u_{0}\|_{L^{2}}\\ \label{e303} \|S(t)u_{0}\|_{L^{2}(0, T: H_{0}^{1}(I))}\leq c(T) \|u_{0}\|_{L^{2}} \end{gather} for all $T>0$, where $c(T) = \frac{|\alpha| T + L}{3 \beta - |\alpha|}$. It follows that $S(t)u_{0}\in X(T)$. On the other hand, the function \[ J(t) = \int_{0}^{t}S(t - s) g(u) \] is a solution to the problem \[ i J_{t} + \alpha J_{xx} + i \beta J_{xxx} + i a(x) J = F, \] where $F=-g(u)$. We can follow the same idea due to L. Rosier \cite{ro1} (Proposition 4.1) to prove that $J\in X(T)$ and $F\to J$ which maps $L^{2}(0, T: L^{2}(I))$ to $X(T)$ is continuous. Furthermore, the map that associates to each $u$ in $L^{2}(0, T: H_{0}^{1}(I))$ the element $g(u)$ in $L^{1}(0, T: L^{2}(I))$ is also continuous. Consequently, $P$ maps $X(T)$ into $X(T)$ continuously. Now, let us prove that $P$ is a contraction in a suitable ball of $X(T)$ provided that $T>0$ is chosen sufficiently small. Let $u$ and $v$ be elements of $X(T)$, then \[ P(u) - P(v) = -\int_{0}^{t}S(t - \tau) [g(u) - g(v)]\,d\tau . \] Direct calculation, \eqref{e301}, \eqref{e303} and Holder's inequality yield \begin{align*} &\|P(u)(t) - P(v)(t)\|_{H^{1}(I)}^{2}\\ & = \big\|\int_{0}^{t}S(t - \tau) [g(u) - g(v)]\,d\tau\big\|_{H^{1}(I)}^{2} \\ & \leq \big[\int_{0}^{t}\|S(t - \tau) [g(u) - g(v)]\|_{H^{1}(I)}d\tau\big]^{2} \\ & \leq \Big[\Big(\int_{0}^{t}d\tau\Big)^{1/2}\Big(\int_{0}^{t}\|S(t - \tau) [g(u) - g(v)]\|_{H^{1}(I)}^{2}d\tau\Big)^{1/2}\Big]^{2} \\ & \leq \Big(\int_{0}^{t}d\tau\Big)\Big(\int_{0}^{t}\|S(t - \tau) [g(u) - g(v)]\|_{H^{1}(I)}^{2}d\tau\Big) \\ & \leq M^{2} T\int_{0}^{t}\|g(u) - g(v)\|_{L^{2}(I)}^{2}d\tau \\ & \leq T \big(\frac{|\alpha| T + L}{3 \beta - |\alpha|}\big)^{2}\int_{0}^{T}\|g(u) - g(v)\|_{L^{2}(I)}^{2}\,dt. \end{align*} Hence, \begin{equation} \label{e304} \begin{aligned} &\|P(u)(t) - P(v)(t)\|_{L^{\infty}(0, T:H^{1}(I))}^{2}\\ & \leq T \left(\frac{|\alpha| T + L}{3 \beta - |\alpha|}\right)^{2}\int_{0}^{T}\|g(u) - g(v)\|_{L^{2}(I)}^{2}\,dt. \end{aligned} \end{equation} On the other hand, using that $| |u| - |v| |\leq |u - v|$, we have \begin{align*} &\|g(u) - g(v)\|_{L^{2}(I)} \\ & = \| |u|^{2} u - |v|^{2} v\|_{L^{2}(I)} \\ & = \| |u|^{2} (u - v) + (|u|^{2} - |v|^{2}) v \|_{L^{2}(I)} \\ & = \| |u|^{2} (u - v) + (|u| + |v|) (|u| - |v|) v \|_{L^{2}(I)} \\ & \leq \| |u|^{2} (u - v)\|_{L^{2}(I)} + \|(|u| + |v|) (|u| - |v|) v \|_{L^{2}(I)} \\ & \leq \|u\|_{L^{\infty}(I)}^{2}\|u - v\|_{L^{2}(I)} + \left(\|u\|_{L^{\infty}(I)} + \|v\|_{L^{\infty}(I)}\right) \|v\|_{L^{\infty}(I)}\|u - v \|_{L^{2}(I)} \\ & \leq \|u\|_{L^{\infty}(I)}^{2}\|u - v\|_{L^{2}(I)} + \left(\|u\|_{L^{\infty}(I)}\|v\|_{L^{\infty}(I)} + \|v\|_{L^{\infty}(I)}^{2}\right)\|u - v \|_{L^{2}(I)} \\ & \leq \frac{3}{2}\;\left(\|u\|_{L^{\infty}(I)}^{2} + \|v\|_{L^{\infty}(I)}^{2}\right)\|u - v \|_{L^{2}(I)}. \end{align*} Thus \begin{equation} \label{e305}\|g(u) - g(v)\|_{L^{2}(I)}^{2} \leq \frac{3}{2}\;\left(\|u\|_{L^{\infty}(I)}^{2} + \|v\|_{L^{\infty}(I)}^{2}\right)^{2}\|u - v \|_{L^{2}(I)}^{2}. \end{equation} Therefore, from \eqref{e304}, we have \begin{align*} &\|P(u)(t) - P(v)(t)\|_{L^{\infty}(0, T:H^{1}(I))}^{2}\\ & \leq c T \big(\frac{|\alpha| T + L}{3 \beta - |\alpha|}\big)^{2}\int_{0}^{t}\big(\|u\|_{H^{1}(I)}^{2} + \|v\|_{H^{1}(I)}^{2}\big)^{2}\|u - v \|_{L^{2}(I)}^{2}dt. \end{align*} Then \[ \|P(u)(t) - P(v)(t)\|_{X(T)}^{2} \leq c T \big(\frac{|\alpha| T + L}{3 \beta - |\alpha|}\big)^{2}\big(\|u\|_{X(T)}^{2} + \|v\|_{X(T)}^{2}\big)^{2} \|u - v\|_{X(T)}^{2} \] and \[ %begin{align*} \|P(u)(t) - P(v)(t)\|_{X(T)} \leq c T^{1/2} \big(\frac{|\alpha| T + L}{3 \beta - |\alpha|}\big)\big(\|u\|_{X(T)}^{2} + \|v\|_{X(T)}^{2}\big) \|u - v\|_{X(T)}. \] %end{align*} This shows that $P$ is a contraction in the ball $\mathbb{B}_{R}=\{u\in X(T):\|u\|_{X(T)}\leq R\}$ with \begin{equation} \label{e306}2 c T^{1/2} \big(\frac{|\alpha| T + L}{3 \beta - |\alpha|}\big) R^{2}<1. \end{equation} Therefore, the proof will be complete if we show that for a suitable choice of $R$ and $T$ satisfying \eqref{e306}, the map $P$ maps $\mathbb{B}_{R}$ into itself. Putting all the previous estimates together, we have \[ \|P(u)\|_{X(T)}^{2} = \int_{0}^{T}\|P(u)\|_{H^{1}(I)}^{2}\,dt. \] From \eqref{e305}, \begin{equation} \label{e307}\|P(u)\|_{X(T)}^{2} \leq c T \Big(\frac{|\alpha| T + L}{3 \beta - |\alpha|}\Big)^{2} \|u_{0}\|_{X(T)}^{6}\leq c T \Big(\frac{|\alpha| T + L}{3 \beta - |\alpha|}\Big)^{2} R^{6} \end{equation} for all $u\in \mathbb{B}_{R}$. Choosing $R=\|u_{0}\|_{L^{2}(I)}$ from \eqref{e307} we deduce that \[ \|P(u)\|_{X(T)}^{2}\leq \big[c T^{1/2} \big(\frac{|\alpha| T + L}{3 \beta - |\alpha|}\big) \|u_{0}\|_{L^{2}(I)}^{2}\big] \|u_{0}\|_{L^{2}(I)}. \] Let us choose $T>0$ sufficiently small and such that \begin{equation}\label{e308} c T^{1/2} \big(\frac{|\alpha| T + L}{3 \beta - |\alpha|}\big) \|u_{0}\|_{L^{2}(I)}^{2}<1. \end{equation} Hence, $P$ map $\mathbb{B}_{R}$ into itself. \end{proof} \begin{theorem}[Global existence and uniqueness] \label{thm3.2} Let $|\alpha|<3 \beta$ and $u_{0}\in L^{2}(I)$. Then, there exists a unique function $u\in L^{\infty}(0, T: L^{2}(I))\cap L^{2}(0, T: H_{0}^{1}(I))$ that satisfies the problem \eqref{e101}-\eqref{e104}. \end{theorem} \begin{proof} It follows from Theorem \ref{thm3.1} that we can extend the solution $u$ to the maximal interval of existence $0\leq t0$ such that $00$. Let $u\in L^{\infty}(0, T: L^{2}(I)))\cap L^{2}(0, T: H^{1}(I)))$ be the global solution of the problem \begin{gather} \label{e317} i u_{t} + i \beta u_{xxx} + \alpha u_{xx} + \delta |u|^{2} u + i a(x) u = 0\quad \mbox{in } I\times (0, T)\\ \label{e318} u(0, t)=u(L, t),\quad t\in (0, T) \\ \label{e319} u_{x}(L, t)=0,\quad t\in (0, T)\\ \label{e320} u(x, t)\equiv 0\quad\mbox{in } \omega\times (0, T) \end{gather} with $\epsilon\geq 0$ and $T>0$, then necessarily $u\equiv 0$ in $I\times (0, T)$. \end{theorem} For the moment, let us assume that $\omega$ satisfies the (UCP). Then we have the following result. \begin{theorem} \label{thm3.4} Let $|\alpha|<3 \beta$, $a=a(x)$ a non-negative function, $a\in C^{\infty}(I)$ such that $a(x)\geq a_{0}>0$ is assumed to be nonnegative everywhere in an open non empty proper subset $ \omega$. Let $u$ be the global solution of the problem \eqref{e101}-\eqref{e104}. Then, for any $L>0$ and $R>0$ there exist positive constants $c>0$ and $\mu>0$ such that \[ E(t)\leq c \|u_{0}\|_{L^{2}(I)}^{2} e^{-\mu t} \] for any $t\geq 0$ and any solution of \eqref{e101}-\eqref{e104} with $u_{0}\in L^{2}(I)$ such that $\|u_{0}\|_{L^{2}(I)}\leq R$. \end{theorem} \begin{proof} We proceed as in he proof of Theorem \ref{thm2.2}. From \eqref{e311} we have \begin{equation} \label{e321}\int_{0}^{L}|u|^{2}\,dx + \beta |u_{x}(0, t)|^{2} + 2\int_{0}^{T}\int_{0}^{L}a(x) |u|^{2}\,dx\,dt = \|u_{0}\|_{L^{2}(I)}^{2}. \end{equation} Next, we multiply the equation in \eqref{e101}-\eqref{e104} by $(T -t) \overline{u}$, \begin{equation} i (T - t) \overline{u} u_{t} + i \beta (T - t) \overline{u} u_{xxx} + \alpha (T - t) \overline{u} u_{xx} + (T - t) |u|^{4} + i (T - t) a(x) |u|^{2}=0. \label{e322} \end{equation} Applying conjugate we have \begin{equation} -i (T - t) u \overline{u}_{t} - i \beta (T - t) u \overline{u}_{xxx} + \alpha (T - t) u \overline{u}_{xx} + (T - t) |u|^{4} - i (T - t) a(x) |u|^{2}=0. \label{e323} \end{equation} Subtracting \eqref{e322} with \eqref{e323} and performing straightforward calculations we obtain \begin{equation} \frac{d}{dt}\int_{0}^{L}(T - t) |u|^{2}\,dx + \int_{0}^{L}|u|^{2}\,dx + \beta (T - t) |u_{x}(0, t)|^{2} + 2\int_{0}^{L}(T - t) a(x) |u|^{2}\,dx =0. \label{e324} \end{equation} Integrating over $t\in [0, T]$, we have \begin{align*} &-T\int_{0}^{L}|u_{0}|^{2}\,dx + \int_{0}^{T}\int_{0}^{L}|u|^{2}\,dx\,dt + \beta\int_{0}^{T}(T - t) |u_{x}(0, t)|^{2}\,dt\\ & + 2\int_{0}^{T}\int_{0}^{L}(T - t) a(x) |u|^{2}\,dx\,dt =0. \end{align*} and \begin{align*} \int_{0}^{L}|u_{0}|^{2}\,dx & = \frac{1}{T}\int_{0}^{T}\int_{0}^{L}|u|^{2}\,dx\,dt + \frac{\beta}{T}\int_{0}^{T}(T - t) |u_{x}(0, t)|^{2}\,dt \\ &\quad + \frac{2}{T}\int_{0}^{T}\int_{0}^{L}(T - t) a(x) |u|^{2}\,dx\,dt. \end{align*} Consequently, \begin{equation} \int_{0}^{L}|u_{0}|^{2}\,dx \leq \frac{1}{T}\int_{0}^{T}\int_{0}^{L}|u|^{2}\,dx\,dt + \beta\int_{0}^{T}|u_{x}(0, t)|^{2}\,dt + 2\int_{0}^{T}\int_{0}^{L}a(x) |u|^{2}\,dx\,dt. \label{e325} \end{equation} To show the result it is sufficient to prove \begin{equation} \label{e326}\int_{0}^{T}\int_{0}^{L}|u|^{2}\,dx\,dt\leq c \big\{\beta\int_{0}^{T}|u_{x}(0, t)|^{2}\,dt + 2\int_{0}^{T}\int_{0}^{L}a(x) |u|^{2}\,dx\,dt\big\} \end{equation} for some positive constant $c$ independent of the solution $u$. \\ Let us argue by contradiction. Suppose that \eqref{e326} is not true. Then, there will exist a sequence of solutions $u^{n}$ of \eqref{e101}-\eqref{e104} such that \[ \lim_{n\to\infty}\frac{\|u^{n}\|_{L^{2}(0, T: L^{2}(I)))}^{2}} {\beta\int_{0}^{T}|u_{x}^{n}(0, t)|^{2}\,dt + 2\int_{0}^{T}\int_{0}^{L}a(x) |u^{n}|^{2}\,dx\,dt}=+ \infty. \] Let $\lambda^{n}=\|u^{n}\|_{L^{2}(0, T: L^{2}(I))}$ and $v^{n}(x, t)=\frac{u^{n}}{\lambda_{n}}$. For each $n\in \mathbb{N}$ the function $v^{n}$ satisfies \begin{gather}\label{e327} i (v^{n})_{t} + \alpha (v^{n})_{xx} + i \beta (v^{n})_{xxx} + (\lambda^{n})^{2} |v^{n}|^{2} v^{n} + i a(x) v^{n}=0\quad\mbox{in } I\times (0, T)\\ \label{e328} v_{x}^{n}(L, t)=0,\quad\mbox{for all } t>0\\ \label{e329} v^{n}(0, t)=v^{n}(L, t)=0,\quad\mbox{for all } t>0\\ \label{e330} v^{n}(x, 0)=\frac{u^{n}(x, 0)}{\lambda^{n}}, \quad\mbox{for all } x\in I. \end{gather} We have \begin{gather} \label{e331}\|v^{n}\|_{L^{2}(0, T: L^{2}(I))}=1,\\ \label{e332}\beta\int_{0}^{T}|u_{x}^{n}(0, t)|^{2}\,dt + 2\int_{0}^{T}\int_{0}^{L}a(x) |u^{n}|^{2}\,dx\,dt\to 0\quad\mbox{as } n\to + \infty. \end{gather} In view of \eqref{e325} it follows that $v^{n}(x, 0)$ is bounded in $L^{2}(I)$. Thus, \begin{equation} \label{e333}\|v^{n}( \cdot , t)\|_{L^{2}(I)}\leq c\quad\mbox{for all}\quad 0\leq t\leq T. \end{equation} According to \eqref{e216} \begin{equation} \label{e334}\|v^{n}\|_{L^{2}(0, T: H^{1}(I))}\leq c(T) \|v^{n}( \cdot , t)\|_{L^{2}(I)},\quad \forall n\in \mathbb{N}. \end{equation} On the other hand, $|v^{n}|^{2}v^{n}$ belongs to $L^{2}(0, T: L^{1}(I))$ and \begin{equation} \label{e335}\||v^{n}|^{2}v^{n}\|_{L^{2}(0, T: L^{1}(I))}\leq \|v^{n}\|_{L^{\infty}(0, T: L^{2}(I))}^{2} \|v^{n}\|_{L^{2}(0, T: H^{1}(I))} \end{equation} and by \eqref{e334} we obtain a constant $c>0$ such that \begin{equation} \label{e336}\||v^{n}|^{2}v^{n}\|_{L^{2}(0, T: L^{1}(I))}\leq c. \end{equation} Since $(\lambda^{n})$ is a bounded sequence, because $\|u^{n}( \cdot , 0)\|_{L^{2}(I)}\leq R$, it follows by \eqref{e317}-\eqref{e320}, \eqref{e334} and \eqref{e336} that \[ (v^{n})_{t} = -\alpha (v^{n})_{xx} - i \beta (v^{n})_{xxx} - (\lambda^{n})^{2} |v^{n}|^{2} v^{n} - i a(x) v^{n} \] is bounded in $L^{2}(0, T: H^{-2}(I))$. Since the embedding $H^{1}(I)\hookrightarrow L^{2}(I)$ is compact it follows that $(v^{n})$ is relatively compact in $L^{2}(0, T: L^{2}(I))$. By extracting a subsequence we can deduce that \begin{gather*} v^{n}\rightharpoonup v \quad\mbox{weakly in } L^{2}(0, T: H^{-2}(I)), \\ v^{n}\to v \quad\mbox{strongly in } L^{2}(0, T: L^{2}(I)). \end{gather*} Since $\|v_{n}\|_{L^{2}(0, T: L^{2}(I))}=1$, then \begin{equation} \label{e337}\|v\|_{L^{2}(0, T: L^{2}(I))}=1. \end{equation} By lower semicontinuity, we have \begin{align*} 0 & = \lim_{n\to\infty}\inf\big\{\beta\int_{0}^{T}|v_{x}^{n}(0, t)|^{2}\,dt + 2\int_{0}^{T}\int_{0}^{L}a(x) |v^{n}|^{2}\,dx\,dt\big\}\\ & \geq \beta\int_{0}^{T}|v_{x}(0, t)|^{2}\,dt + 2\int_{0}^{T}\int_{0}^{L}a(x) |v|^{2}\,dx\,dt \end{align*} which guarantees that $a v\equiv 0$, and in particular $v\equiv 0$ in $\omega\times (0, T)$. We now distinguish the following two situations: (1) There exists a subsequence of $(\lambda_{n})$ also denoted by $(\lambda_{n})$ such that \[ \lambda_{n}\to 0\quad\mbox{as } n\to\infty. \] In this case, the limit $v$ satisfies the linear problem \begin{gather*} i v_{t} + i \beta v_{xxx} + \alpha v_{xx} + \delta |v|^{2} v + i a(x) v = 0\\ v(0, t)=v(L, t)\quad\mbox{for all } t\in (0, T) \\ v_{x}(L, t)=0\quad\mbox{for all } t\in (0, T)\\ v(x, t)= 0\quad\mbox{in } \omega\times (0, T) \end{gather*} Then, by Holmgren's uniqueness Theorem (see \cite{re1}), $v\equiv 0$ in $I\times (0, T)$ and this contradicts \eqref{e337}. (2) There exists a subsequence of $(\lambda_{n})$ also denoted by $(\lambda_{n})$ and $\lambda>0$ such that $\lambda_{n}\to \lambda$. In this case, the limit function $v$ solves \eqref{e328}-\eqref{e330} and, by the (UCP) assumed to hold for the subset $\omega$, we have that $v\equiv 0$ in $I\times (0, T)$, and again, in this case, we have a contradiction. In the cases (1) and (2) we have a contradiction. Hence, \eqref{e326} holds and the proof is complete. \end{proof} \begin{proof}[Proof of Theorem \ref{thm3.3}] From Theorem \ref{thm3.2} we obtain that if $u_{0}\in L^{2}(I)$ then \[ u\in L^{\infty}(0, T: L^{2}(I))\cap L^{2}(0, T: H_{0}^{1}(I)) \] and $u_{t}\in L^{2}(0, T: H_{0}^{-2}(I))$. Consequently, we know that $u$ is weakly continuous from $[0, T]$ into $L^{2}(I)$. According to the structure of $\omega$, $u\equiv 0$ in $\{(0, \delta)\times (L - \delta, L)\}\times (0, T)$. Let us define the extended function \[ u(x, t) = \begin{cases} u(x, t) &\mbox{if } (x, t)\in (\delta, L - \delta)\times (0, T)\\ 0 & \mbox{if } (x, t)\in \{\mathbb{R}- (\delta, L -\delta)\}\times (0, T). \end{cases} \] Then, $u$ satisfies \begin{gather} \label{e338} i u_{t} + \alpha u_{xx} + i \beta u_{xxx} + \lambda |u|^{2} u + i a(x) u=0\quad\mbox{in }\mathbb{R}\times (0, T)\\ \label{e339} u(x, 0)=\phi(x)\quad\mbox{in }\mathbb{R}. \end{gather} and \[ \phi(x) = \begin{cases} u_{0}(x) & \mbox{if } x\in (\delta, L - \delta)\\ 0 & \mbox{if } x\in \{\mathbb{R}- (\delta, L - \delta)\}. \end{cases} \] If we consider $v(x, t)=u(x + t, t)$, then $v$ solves \begin{gather} \label{e340} i v_{t} + \alpha v_{xx} + i \beta v_{xxx} + \lambda |v|^{2} v + i a(x) v=0\quad\mbox{in }\mathbb{R}\times (0, T)\\ \label{e341} u(x, 0)=\phi(x)\quad\mbox{in }\mathbb{R}. \end{gather} Since $\phi$ has compact support and belongs to $L^{2}(\mathbb{R})$, we have \begin{equation} \label{e342}\int_{\mathbb{R}}\phi^{2}(x) e^{2 b x}\,dx<\infty,\quad \forall b>0. \end{equation} Thus, by regularizing properties (see \cite{se1}), $v\in C^{\infty}(I\times (0, T))$. Therefore, $v$ is smooth as well, and applying the unique continuation result we have that $v\equiv 0$ and $u\equiv 0$, $x\in I$, $t\in(0, T)$. \end{proof} \subsection*{Remark about the hypothesis $|\alpha|<3 \beta$} We consider the Gauge transformation \[ u(x, t) = e^{i d_{2} x + i d_{3} t} v(x - d_{1} t, t)\equiv e^{\theta} v(\eta, \xi) \] and $\theta = i d_{2} x + i d_{3} t$, $\eta=x - d_{1} t$, $\xi =t$. Then \begin{gather*} u_{t} = i d_{3} e^{\theta} v - d_{1} e^{\theta} v_{\eta} + e^{\theta} v_{\xi},\\ u_{x} = i d_{2} e^{\theta} v + e^{\theta} v_{\eta},\\ u_{xx} = - d_{2}^{2} e^{\theta} v + 2 i d_{2} e^{\theta} v_{\eta} + e^{\theta} v_{\eta \eta},\\ u_{xxx} = - i d_{2}^{3} e^{\theta} v - 3 d_{2}^{2} e^{\theta} v_{\eta} + 3 i d_{2} e^{\theta} v_{\eta\eta} + e^{\theta} v_{\eta\eta\eta}. \end{gather*} Replacing in \eqref{R}, we have \begin{align*} & - d_{3} e^{\theta} v - i d_{1} e^{\theta} v_{\eta} + i e^{\theta} v_{\xi} - \omega d_{2}^{2} e^{\theta} v + 2 i \omega d_{2} e^{\theta} v_{\eta} + \omega e^{\theta} v_{\eta\eta} \\ & \beta d_{3}^{3} e^{\theta} v - 3 i \beta d_{2}^{2} e^{\theta} v_{\eta} - 3 \beta d_{2} e^{\theta} v_{\eta\eta} + i \beta e^{\theta} v_{\eta\eta\eta} + \gamma |v|^{2} e^{\theta} v\\ &- \delta d_{2} |v|^{2} e^{\theta} v + i \delta |v|^{2} e^{\theta} v_{\eta} + \epsilon d_{2} e^{\theta} v^{2}\overline{v} + i \epsilon e^{\theta} v^{2} v_{\eta}=0 \end{align*} and \begin{align*} i v_{\xi} + (\omega - 3 \beta d_{2}) v_{\eta\eta} + i \beta v_{\eta\eta\eta} + (2 i \omega d_{2} - 3 i \beta d_{2}^{2} - i d_{1} + i \delta |v|^{2} + i \epsilon v^{2}) v_{\eta}\\ (\beta d_{2}^{3} - \omega d_{2}^{2} - d_{3} + \gamma |v|^{2} - \delta d_{2} |v|^{2}) v + \epsilon d_{2} v^{2}\overline{v}=0\,. \end{align*} Then \[ d_{1}=\frac{\omega^{2}}{3 \beta},\quad d_{2}=\frac{\omega}{3 \beta},\quad d_{3}=\frac{- 2\omega^{3}}{27 \beta^{2}}. \] This way in \eqref{R} we obtain \[ i v_{\xi} + i \beta v_{\eta\eta\eta} + i (\delta |v|^{2} + \epsilon v^{2}) v_{\eta} + \big(\gamma - \frac{\omega \delta}{3 \beta}\big)|v|^{2}v + \frac{\epsilon \delta}{3 \beta} v^{2}\overline{v}=0, \] but $v^{2} \overline{v}=v v \overline{v}=|v|^{2}v$, then using the Gauge transformation we have the equivalent problem to \eqref{R} \begin{equation} \label{Q} \begin{gathered} i v_{\xi} + i \beta v_{\eta\eta\eta} + i \delta |v|^{2} v_{\eta} + i \epsilon v^{2} v_{\eta} + \big(\gamma + \frac{\epsilon \delta}{3 \beta} - \frac{\omega \delta}{3 \beta}\big)|v|^{2}v =0 \quad \eta, \xi\in \mathbb{R}\\ v(\eta, 0) = e^{- i \frac{\omega}{3 \beta} \eta}u_{0}(\eta). \end{gathered} \end{equation} Here, rescaling the equation, we take $\beta =1$. \begin{equation} \label{Qtilde} \begin{gathered} i v_{t} + i v_{xxx} + i \delta |v|^{2} v_{x} + i \epsilon v^{2} v_{x} + \big(\gamma + \frac{\epsilon \delta}{3} - \frac{\omega \delta}{3}\big)|v|^{2}v =0 \quad x, t\in \mathbb{R}\\ v(x, 0) = e^{- i \frac{\omega}{3} x}u_{0}(x). \end{gathered} \end{equation} The above Gauge's transformation is a bicontinuous map from $L^{p}([0, T]: H^{s}(\mathbb{R})$ to itself, as far as $0