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

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2005(2005), No. 122, pp. 1--31.\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 2005 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2005/122\hfil Exact boundary controllability]
{Exact boundary controllability for higher order nonlinear
Schr\"{o}dinger equations with constant coefficients}
\author[J. C. Ceballos V., R. Pavez F., O. P. Vera V.\hfil EJDE-2005/122\hfilneg]
{Juan Carlos Ceballos V., Ricardo Pavez F.,\\
 Octavio Paulo Vera Villagr\'{a}n}

\address{Juan Carlos Ceballos V. \hfill\break
Departamento de Matem\'{a}tica,
Universidad del B\'{\i}o-B\'{\i}o, Collao 1202,
Casilla 5-C, Concepci\'{o}n, Chile}
\email{jceballo@ubiobio.cl \quad Fax 56-41-731018}

\address{Ricardo Pavez F.\hfill\break
Departamento de Matem\'{a}tica,
Universidad del B\'{\i}o-B\'{\i}o, Collao 1202, Casilla 5-C,
Concepci\'{o}n, Chile}
\email{rpavez@ubiobio.cl \quad  Fax 56-41-731018}

\address{Octavio Paulo Vera Villagr\'{a}n \hfill\break
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 \quad octavipaulov@yahoo.com\quad
 Fax 56-41-731018}


\date{}
\thanks{Submitted January 20, 2005. Published October 31, 2005.}
\thanks{J. C. Ceballos was supported by grant 0528081/R from
Proyectos de Investigacion Internos, \hfill\break\indent
Universidad del B\'{\i}o-B\'{\i}o.
Concepci\'{o}n. Chile}
\subjclass[2000]{35K60, 93C20}
\keywords{KdVK equation; boundary control;
 Hilbert uniqueness method; \hfill\break\indent
Ingham's inequality; smoothing properties}

\begin{abstract}
 The exact boundary controllability of the higher order
 nonlinear Schr\"{o}dinger equation with constant coefficients
 on a bounded domain with various boundary conditions is studied.
 We derive the exact boundary controllability for this equation for
 sufficiently small initial and final states.
\end{abstract}

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

\section{Introduction}

We consider the initial-value problem
\begin{equation} \label{eHSCHROD}
\begin{gathered}
i u_{t} + \alpha  u_{xx} + i \beta  u_{xxx} + |u|^{2} u =0,
\quad x, t\in \mathbb{R}\\
u(x, 0)  = u_{0}(x)
\end{gathered}
\end{equation}
 where $\alpha , \,\beta \in \mathbb{R},$ $\beta \neq 0 $
and $u$ is a complex valued function. The above equation is a
particular case of the equation
\begin{equation} \label{eQ}
\begin{gathered}
i u_{t} + \alpha  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 $\alpha ,\, \beta ,\,  \gamma , \,\delta $, with
$\beta \neq 0 $ and $u$ is a complex valued function. This
equation was first proposed by  Hasegawa and  Kodama \cite{h1} as
a model for the propagation of a signal in a fiber optic   (see
also \cite{k1}). The equation \eqref{eQ} can be reduced to other
well known equations. For instance, setting $\alpha =1$, $\beta =
\epsilon =\gamma =0$ in \eqref{eQ} we have the semi linear
Schr\"{o}dinger equation, i. e.,
\begin{equation}
u_{t} - i   u_{xx} - i \gamma  |u|^{2} u =0.\label{eQ1}
\end{equation}
If we let $\beta = \gamma =0$ and $\alpha =1$ in \eqref{eQ} we
obtain the derivative nonlinear Schr\"{o}dinger equation
\begin{equation}
u_{t} - i   u_{xx} - \delta  |u|^{2} u_{x} - \epsilon  u^{2}
\overline{u}_{x}=0.\label{eQ2}
\end{equation}
Letting $\alpha = \gamma = \epsilon =0$ in \eqref{eQ},
the equation that arises is the complex modified Korteweg-de Vries equation,
\begin{equation}
u_{t} + \beta  u_{xxx} + \delta  |u|^{2} u_{x} =0.\label{eQ3}
\end{equation}
The initial-value problem for the equations \eqref{eQ1},
\eqref{eQ2} and \eqref{eQ3} has been extensively studied, see for
instance \cite{b1,c3,k2,l2,l4,l5,m1,r1} and references therein. In
1992,  Laurey \cite{l1} considered the equation \eqref{eQ} and
proved local well-posedness of the initial-value problem
associated for data in $H^{s}(\mathbb{R})$ with $s>3/4$, and
global well-posedness in $H^{s}(\mathbb{R})$ where $s\geq 1$. In
1997, Staffilani \cite{s2} established local well-posedness for
data in $H^{s}(\mathbb{R})$ with $s\geq 1/4$, improving Laurey's
result. Similar results were given in \cite{c1,c2} for \eqref{eQ}
where $w(t)$, $\beta(t)$ are real functions.

For the case of the \eqref{eHSCHROD} if we consider the Gauge
transformation
\begin{equation*}
u(x, t)  =  e^{i \frac{\alpha}{3} x + i 2 \frac{\alpha^{3}}{27}}
v(x - \frac{\alpha^{2}}{3} t, t)
 \equiv  e^{\theta} v(\eta, \xi)
\end{equation*}
where $\theta=i \frac{\alpha}{3} x + i 2 \frac{\alpha^{3}}{27},$
$\eta=x - \frac{\alpha^{2}}{3} t$ and $\xi =t,$ then
\begin{gather*}
u_{t}  =  i 2 \frac{\alpha^{3}}{27} e^{\theta} v -
\frac{\alpha^{2}}{3} e^{\theta} v_{\eta} + e^{\theta} v_{\xi}\\
u_{xx}  =  - \frac{\alpha^{2}}{9} e^{\theta} v + i \frac{2}{3}
\alpha  e^{\theta} v_{\eta} + e^{\theta} v_{\eta
 \eta}\\
u_{xxx}  =  - i \frac{\alpha^{3}}{27} e^{\theta} v - \frac{1}{3}
\alpha^{2} e^{\theta} v_{\eta} + i \alpha
 e^{\theta} v_{\eta \eta} + e^{\theta} v_{\eta \eta \eta}.
\end{gather*}
Replacing in \eqref{eHSCHROD} and considering $\beta=1$(rescaling
the equation) we obtain
\begin{equation} \label{eKdVm}
\begin{gathered}
i v_{\xi} + i v_{\eta \eta \eta} + |v|^{2} v - \frac{4}{27}
\alpha^{3} v=0,
\quad x, t\in \mathbb{R}\\
v(x, 0)  = v_{0}(x)\equiv u_{0}(x) e^{- i \frac{\alpha}{3}}
\end{gathered}
\end{equation}
 Thus \eqref{eHSCHROD} is reduced to a complex
modified Korteweg-de Vries type equation.
In this paper, we consider
the boundary control of the Schr\"{o}dinger equation
\begin{equation}
\label{e101} i u_{t} + \alpha  u_{xx} + i \beta  u_{xxx} +
|u|^{2} u + i \delta u_{x}=0
\end{equation}
where $\alpha , \,\beta , \,\delta \in \mathbb{R}$, $\beta \neq 0
$ and $u$ is a complex valued function on the domain $(a, b),$
$t>0,$ and with the boundary condition
\begin{equation}
\label{e102}u(a, t)=h_{0},\quad u(b, t)=h_{1},\quad
u_{x}(a, t)=h_{2},\quad u_{x}(b, t)=h_{3}.
\end{equation}
In this
paper we want to study directly the exact boundary controllability
problem for the higher order Schr\"{o}dinger equation by adapting
the   method of  \cite{l5} which combines the
Hilbert Uniqueness Method (HUM) and multiplier techniques. This
method has been successfully applied to study controllability of
wave and plate equations, Schr\"{o}dinger and KdV equations (see for
instance \cite{b1,c3,f1,h2,k2,k3,l2,l4,l6,m1} and references
therein). The first result of this paper concerns boundary
controllability of the
higher order linear Schr\"{o}dinger equation.

\begin{theorem} \label{thm1.1}
Let $H_{p}^{2}=\{w\in H^{2}(0, 2 \pi): w(0)=w(2 \pi), \,w'(0)=w'(2
\pi)\}$ and $T>0.$ Then, for any $y_{0}, y_{T}\in (H_{p}^{2})'$
(the dual space of $H_{p}^{2}$), there exist $h_{k}\in L^{2}(0,
T)$ ($k=0, 1, 2$) such that the solution $y\in C([0, T]:
(H_{p}^{2})')$ of the boundary initial-value higher order
Schr\"{o}dinger equation
\begin{gather}
\label{e103} i y_{t} + i \beta  y_{xxx} + \alpha  y_{xx}=0,
\quad (x, t)\in (0, 2 \pi)\times (0, T);\\
\label{e104} \partial_{x}^{k}y(2 \pi , t) -
\partial_{x}^{k}y(0, t)=h_{k}(t),\quad k=0, 1, 2;\\
\label{e105} y( . , 0)=y_{0}
\end{gather}
 satisfies $y( . , T)=y_{T}$.
\end{theorem}

We see that explicit controls may be given. Unfortunately, the state
$y$ is only known to belong to $C([0, T]: (H_{p}^{2})')$ so it
seems quite difficult to deduce from Theorem \ref{thm1.1} controllability
results for higher order nonlinear Schr\"{o}dinger equation \eqref{e101}.

The second result relates exact boundary controllability for the
linear higher order Schr\"{o}dinger equation with boundary control
on $y_{x}$ at $x=L$. In this part  a condition on the coefficients
$\alpha $ and $\beta$ given by the second and the third order
derivatives that appear in $(HSCHROD)$ is needed. A condition on
the length $L$ of the domain appears.

\begin{theorem} \label{thm1.2}
Let $|\alpha |<3 \beta $, $\delta>0$ and
\[
\mathcal{N} =\Big\{2 \pi  \beta  \sqrt{\frac{k^{2} + k l + l^{2}}
{3 \beta  \delta + \alpha^{2}}} : k, l\in \mathbb{N}^{*}
\Big\}.
\]
Then for any $T>0$ and $L\in (0, +\infty )\setminus \mathcal{N}$,
and for any $y_{0}, y_{T}\in L^{2}(0, L)$, there exists $h\in
L^{2}(0, T)$ such that the mild solution $y\in C([0, T]: L^{2}(0,
L)) \cap L^{2}(0, T: H^{1}(0, L))$ of the system
\begin{gather}
\label{e106} i y_{t} + i \beta  y_{xxx} + \alpha  y_{xx} + i \delta  y_{x}=0\\
\label{e107} y(0, t)=y(L, t)=0\\
\label{e108} y_{x}(L, t)=h(t)\\
\label{e109} y(x, 0)=y_{0}(x)
\end{gather}
 satisfies $y( . , T)=y_{T}$.
\end{theorem}

To prove this we use the Hilbert uniqueness method and the multiplier method.
It turns out
that the study of \eqref{e106}-\eqref{e109} as a boundary
initial-value problem is more delicate than the study of
\eqref{e103}-\eqref{e105}, and -because of the extra term
$y_{x}$ in \eqref{e108}- the observability result holds true if
and only if $L\notin \mathcal{N}$. On the other hand, the solution $y$
belongs this time to a functional space in which we may give a sense
to the nonlinear term $|y|^{2} y$ in \eqref{eHSCHROD}. By means of the
Banach Contraction Fixed Point Theorem and Theorem \ref{thm1.2} we get the
main result of the paper, that is the exact boundary controllability
of the higher order nonlinear
Schr\"{o}dinger equation on a bounded domain.

\begin{theorem} \label{thm1.3}
 Let $|\alpha |<3 \beta $, $\delta>0$, $T>0$
and $L>0$. Then, there exists $r_{0}>0$ such that for any $y_{0}$,
$y_{T}\in L^{2}(0, L)$ with $\|y_{0}\|_{L^{2}(0, L)}<r_{0}$,
$\|y_{T}\|_{L^{2}(0, L)}<r_{0}$, there is function $y$ in
\begin{equation}
\label{e110} C([0, T]: L^{2}(0, L))\cap
L^{2}([0, T]: H^{1}(0, L)) \cap
W^{1, 1}([0, T]: H^{-2}(0, L))
\end{equation}
which is a solution of
\begin{gather}
\label{e111} i y_{t}= - ( i \beta  y_{xxx} + \alpha
 y_{xx} + |y|^{2} y + i \delta  y_{x} )
\quad \mbox{in } \mathcal{D}^{,}(0, T: H^{-2}(0, L))\\
\label{e112}  y(0, . )=0\quad \mbox{in } L^{2}(0, L)
\end{gather}
and such that $y( . , 0)=y_{0}$, $y( . , T)=y_{T}$. If
moreover $L\notin \mathcal{N}$, then in addition, it is possible to
assume that $y(L, . )=0$ in $L^{2}(0, T)$ and take
$y_{x}(L, . )$ in
$L^{2}(0, T)$ as a control function.
\end{theorem}

In a forthcoming paper we study the case $|\alpha |\geq 3 \beta $
for Theorems \ref{thm1.2} and  \ref{thm1.3} using the Gauge
transformation (KdVm described above) and following the same idea
shown here.

This paper is organized as follows: Section 2 outlines briefly the
notation and terminology to be used subsequently and some previous
result. Section 3 we derive from the Hilbert uniqueness method
a direct proof of the exact
controllability result for the higher order  linear Schr\"{o}dinger
equation. In section 4, we consider another boundary controllability
problem for the higher order linear Schr\"{o}dinger equation, in which
only the value of the first spatial derivative (at $x=L$) of the
state function is assumed to be controlled: this boundary
initial-value problem is first shown to admit solutions, later on,
an observability result is given and used to show using the
Hilbert uniqueness method the
exact boundary controllability for higher order linear Schr\"{o}dinger
equation with these boundary conditions. Finally, in section 5, we
prove the main result of this paper, that is, the exact local
boundary controllability of the higher order nonlinear Schr\"{o}dinger
equation on a bounded domain.

 \section{Preliminaries}

For an arbitrary Banach space $X$, the associated norm will be denoted by
$\| \cdot \|_{X}$. If $\Omega =(a, b)$ is a bounded open interval
and $k$ a non-negative integer, we denote by $C^{k}(\Omega) =
C^{k}(a, b)$ the functions that, along with their first $k$ ones,
are continuous on $[a, b]$ with the norm
\begin{equation}
\label{e201}\|f\|_{C^{k}(\Omega)} = \sup_{x\in \Omega ,\, 0\leq
j\leq k}|f^{(j)}(x)|.
\end{equation}
As usual, $\mathcal{D}(\Omega)$ is the subspace of
$C^{\infty}(\overline{\Omega})$ consisting of functions with compact
support in $\Omega$. Its dual space $\mathcal{D}'$ is the space of
Schwartz distributions on $\Omega$. For $1\leq p <\infty $,
$L^{p}(\Omega)$ denotes those functions $f$ which are $p$-power
absolutely integrable on $\Omega $ with the usual modification n
case $p=\infty $. If  $s\geq 0$ is an integer and $1\leq p \leq
\infty $, $W^{s, p}(\Omega)$ is the Sobolev space consisting of
those $L^{p}(\Omega)$-functions whose first $s$ generalized
derivatives lie in $L^{p}(\Omega)$, with the usual norm
\begin{equation}
\label{e202}\|f\|_{W^{s, p}(\Omega)}^{p} =
\sum_{k=0}^{s}\|f^{(k)}\|_{L^{p}(\Omega)}^{p}.
\end{equation}
If $p=2$ we write $H^{2}(\Omega)$ for $W^{s, 2}(\Omega)$. The
notation $H^{s}(\Omega)$ is frequent where $s$ is a positive
integer.
\begin{equation}
\label{e203}\| \cdot \|_{s}=\| \cdot \|_{H^{s}(a, b)}.
\end{equation}
For $s\geq 1$, $H_{0}^{s}((a, b))$ is the closed linear subspace of
$H^{s}((a, b))$ of functions $f$ such that $f(a)=f'(a)=\cdots =f^{s
- 1}(a)=0$. $H_{\rm loc}^{s}(\Omega)$ is the set of real-valued
functions $f$ defined on $\Omega$ such that, for each $\varphi \in
\mathcal{D}(\Omega)$, $\varphi  f\in H^{s}(\Omega)$. This space is
equipped with the weakest topology such that all of the mapping
$f\mapsto \varphi  f$, for $\varphi \in \mathcal{D}(\Omega)$, are
continuous from $H^{s}(\Omega)$ into  $H_{\rm loc}^{s}(\Omega)$. With
this topology, $H_{\rm loc}^{s}(\Omega)$ is a Fr\'echet space. If $X$ is a
Banach space, $T$ a positive real number and $1\leq p\leq +\infty$,
we will denote by $L^{p}(0, T; X)$ the Banach space of all
measurable functions $u:(0, T)\mapsto X$, such that
$t\mapsto \|u(t)\|_{X}$ is in $L^{p}(0, T), $ with the norm
\[
\|u\|_{L^{p}(0, T; X)}=\Big(\int_{0}^{T}\|u(t)\|_{X}^{p}\,dx\Big)^{1/p}
\quad \mbox{if } 1\leq p<+\infty ,
\]
and if $p=\infty $, then
\[
\|u\|_{L^{\infty}(0, T;\;X)}=\sup_{0<t<T}\;\|u\|_{X}.
\]
Similarly, if $k$ is a positive integer, then $C^{k}(0, T: X)$
denote the space of all continuous functions $u:[0, T]\mapsto X$,
such that their derivatives up to the $k$ order exist and are continuous.

For notation, we write $\partial = \partial /\partial x$,
$\partial_{t} = \partial/\partial t$ and
$u_{j}=\partial_{x}^{j}u = \partial^{j}u/\partial x^{j}$.
\smallskip

\noindent\textbf{Definition.} % 2.1.}
For $k=\{2, 3\}$, we define the space
\[
H_{p}^{k}=\Big\{u\in H^{k}(0, 2 \pi): \frac{d^{j}u}{dx^{j}}(0)
= \frac{d^{j}u}{dx^{j}}(2 \pi)\mbox{ for } 0\leq j\leq k -
1\Big\}
\]
We remark that $H^{k}(0, 2 \pi)$ denotes the classical Sobolev
space on the interval $(0, 2 \pi)$.
\smallskip

\noindent\textbf{Definition.} % 2.2.}
For $n\in \mathbb{Z}$, let the $n$-th Fourier
coefficient of $u\in L^{2}(0, 2 \pi)$,
\begin{equation}
\label{e204}\widehat{u}(n) =
\frac{1}{2 \pi}\int_{0}^{2 \pi}e^{- i n x} u(x)\,dx
\end{equation}

\begin{lemma} \label{lem2.1}
For $n\in \mathbb{Z}$, we have
\begin{equation}
\label{e205}\sum_{n\in \mathbb{Z}}|\widehat{u}(n)|^{2} =
\frac{1}{2 \pi}\int_{0}^{2 \pi}|u(x)|^{2}\,dx
\end{equation}
\end{lemma}
The proof of the above lemma is straightforward.
We remark that for $k=2$ (similarly for $k=3$) we have
\[
\widehat{u}(n)  =
\frac{1}{2\pi}\int_{0}^{2 \pi}e^{- i n x} u(x)\,dx
 =  - \frac{1}{n^{2}} \partial^{2}\widehat{u}(x)
\]
then $- n^{2} \widehat{u}(n) = \partial^{2}\widehat{u}(n)$.
Applying $| \cdot |$ and squaring we obtain
$[ n^{2} |\widehat{u}(n)|^{2} ]^{2} = |\partial
^{2}\widehat{u}(n)|^{2}$ where by applying ${\sum_{n\in
\mathbb{Z}}}$ and using \eqref{e202} it follows that
\[
\sum_{n\in \mathbb{Z}} [ n^{2} |\widehat{u}(n)|^{2} ]^{2} =
\sum_{n\in \mathbb{Z}}|\partial^{2}\widehat{u}(n)|^{2}=
\frac{1}{2 \pi}\int_{0}^{2\pi}|\partial^{2}u(x)|^{2}\,dx< \infty.
\]
Hence, we have that for all $u\in L^{2}(0, 2 \pi)$, $k\in
\{2, 3\}$
\begin{equation}
\label{e206}u\in H_{p}^{k} \quad \mbox{if and only if} \quad
\sum_{n\in
\mathbb{Z}} [ n^{k} |\widehat{u}(n)|^{2} ]^{2}<\infty,
\end{equation}
and the Sobolev norm
\[
\|u\|_{H^{k}(0, 2 \pi)}=\Big[\sum_{j=0}^{k}\int_{0}^{2 \pi}|
\partial^{j}u(x)|^{2}\,dx\Big]^{1/2}=
\Big[\sum_{j=0}^{k}\|\partial
^{j}u\|_{L^{2}(0, 2 \pi)}^{2}\Big]^{1/2}
\]
reduces to
\begin{equation}
\label{e207}\|u\|_{H^{k}(0, 2 \pi)}=\Big[\sum_{n\in
\mathbb{Z}}(1 + n^{2} + \ldots + n^{2 k})
|\widehat{u}(n)|^{2}\Big]^{1/2}\quad \mbox{for } u\in
H_{p}^{k}.
\end{equation}
In what follows, the Hilbert space $H_{p}^{k}$ is endowed with the
norm
$\|u\|_{H^{k}(0, 2\pi)}$.

\begin{lemma}[Ingham's Inequality \cite{i1}] \label{lemIn}
Assume  the strictly increasing sequence
$\{\lambda_{k}\}_{k\in \mathbb{Z}}$ of
real numbers satisfies the ``gap" condition
$\lambda_{k + 1} -\lambda_{k}\geq \gamma$, for all
$k\in \mathbb{Z}$, for some $\gamma>0$. Then, for all
 $T>2\pi/\gamma $ there are two positive
constants $C_{1}, C_{2}$ depending only on $\gamma $ and $T$ such
that
\begin{equation}
\label{e208}C_{1}(T, \gamma)\sum_{k=-\infty}^{\infty}|a_{k}|^{2}\leq
\int_{0}^{T}
\big|\sum_{k=-\infty}^{\infty}a_{k} e^{i t \lambda_{k}}\big|\,dx
\leq C_{2}(T, \gamma)\sum_{k=-\infty}^{\infty}|a_{k}|^{2}
\end{equation}
for every complex sequence $(a_{k})_{k\in \mathbb{Z}}\in l^{2}$,
where
\begin{equation}\label{e209}
\begin{gathered}
C_{1}(T, \gamma)=\frac{2 T}{\pi}  \big( 1 -
\frac{4 \pi^{2}}{T^{2} \gamma^{2}} \big)>0, \\
C_{2}(T, \gamma)=\frac{8 T}{\pi}  \big( 1 +
\frac{4 \pi^{2}}{T^{2} \gamma^{2}} \big)>0
\end{gathered}
\end{equation}
and $l^{2}$ is the Hilbert space of square summable sequences,
sequences $\{a_{k}\}$ such that $\sum_{k\in \mathbb{N}}|a_{k}|^{2}<\infty$.
\end{lemma}

Finally, we denote by $c$, a generic constant, not necessarily the
same at each occasion, which depends in an increasing way on the
indicated quantities.

 \section{Exact boundary controllability of
the higher order linear Schr\"{o}dinger equation by means of control on
data $[\partial^{k}y( . , t)]_{0}^{2 \pi}$ for
$k=0, 1, 2$}

 For simplicity, in  this section, we restrict
ourselves to the case where the space domain
$[0, L]$ is $[0, 2 \pi]$; although Theorem \ref{thm1.1} holds for arbitrary $L>0$.

\begin{lemma} \label{lem3.1}
Let $A$ denote the operator
$Au=(- \beta \partial^{3} + i \alpha  \partial^{2})u$ on the domain
$D(A)=H_{p}^{3}\subseteq L^{2}(0, 2 \pi )$. Then $A$ generates a
strongly continuous unitary group $(S(t))_{t\in \mathbb{R}}$
on $L^{2}(0, 2 \pi )$.
\end{lemma}

\begin{proof} Let $A:D(A)\subseteq L^{2}(0, 2 \pi ) \mapsto
L^{2}(0, 2 \pi )$ such that $u \mapsto Au=- \beta
 \partial^{3}u + i \alpha  \partial^{2}u$. We have
\begin{align*}
\langle Au, v\rangle
& =  \langle - \beta  \partial^{3}u + i \alpha
 \partial^{2}u, v\rangle\\
&=  - \beta  \langle \partial^{3}u, v\rangle
+ i \alpha  \langle \partial^{2}u, v\rangle \\
& =  \beta  \langle u, \partial^{3}v\rangle
+ i \alpha  \langle u, \partial^{2}v\rangle\\
&=  \langle u, \beta  \partial^{3}v\rangle
+ \langle u, -i \alpha  \partial^{2}v\rangle \\
& =  \langle u, - (-\beta  \partial^{3}v + i \alpha  \partial^{2}v)\rangle \\
& =  \langle u, - Av\rangle
\end{align*}
then $A^{*}=- A$. Hence, by the Stone theorem \cite{p1}, $A$
is the infinitesimal generator of a unitary group of class $C_{0}$
(all groups of class $C_{0}$ are strongly continuous) on $L^{2}(0, 2 \pi)$.
\end{proof}

\noindent\textbf{Definition.} % 3.1.}
Let $T>0$. For
$u_{T}={\sum_{n\in\mathbb{Z}}c_{n} e^{i n t }\in L^{2}(0, 2 \pi)}$,
the mild solution of the uncontrolled problem
\begin{equation} \label{eP}
 \begin{gathered}
\partial_{t}u + \beta  \partial^{3}u - i \alpha  \partial^{2}u = 0,\quad
 x\in (0, 2 \pi),\; t\in \mathbb{R} ;\\
\partial^{k}u(0, t) =  \partial^{k}u(2 \pi, t),\quad  k=0, 1, 2;\\
u( . , T)= u_{T}( . )
\end{gathered}
\end{equation}
is given by
\begin{equation} \label{e301}
u(x, t)= \sum_{n\in \mathbb{Z}}c_{n} e^{i (\beta
 n^{3} - \alpha  n^{2}) (t - T) + i n x}
\end{equation}

\begin{remark} \label{rmk3.1} \rm
 Let $u(x, t) = {\sum_{n\in
\mathbb{Z}}\widehat{u}(n, t) e^{i n x}}$, then
\[
u(x, t)= \sum_{n\in \mathbb{Z}}c_{n} e^{i [ (\beta  n^{3} -
\alpha  n^{2}) (t - T) + n x]}.
\]
In fact,
\begin{gather*}
\partial _{t}u(x, t)  =
\sum_{n\in \mathbb{Z}}\partial _{t}\widehat{u}(n, t) e^{i n x}\\
\partial^{2}u(x, t)  =
\sum_{n\in \mathbb{Z}}(i n)^{2} \widehat{u}(n, t) e^{i n x}=
-\sum_{n\in \mathbb{Z}}n^{2} \widehat{u}(n, t) e^{i n x}\\
\partial^{3}u(x, t)  =
\sum_{n\in \mathbb{Z}}(i n)^{3} \widehat{u}(n, t) e^{i n x}=
- i\sum_{n\in \mathbb{Z}}n^{3} \widehat{u}(n, t) e^{i n x},
\end{gather*}
hence, if $u$ is the solution of \eqref{eP}, we obtain
\[
\sum_{n\in \mathbb{Z}}\partial_{t}\widehat{u}(n, t) e^{i n x} -
i \beta \sum_{n\in
\mathbb{Z}}n^{3} \widehat{u}(n, t) e^{i n x} + i \alpha
\sum_{n\in \mathbb{Z}}n^{2} \widehat{u}(n, t) e^{i n x} = 0.
\]
Multiplying by $e^{- i m x}$($m\in \mathbb{Z}$) and integrating
over $x\in (0, 2 \pi)$ we obtain
\[
\sum_{n\in \mathbb{Z}}\partial_{t}\widehat{u}(n, t) - i (\beta
 n^{3} - \alpha
n^{2}) \widehat{u}(n, t)\int_{0}^{2 \pi}e^{i (n - m) x}dx =0.
\]
Using that
\[
\int_{0}^{2 \pi}e^{i (n - m) x}dx =  \begin{cases}
0, & \mbox{if } n\neq m \\
2 \pi , & \mbox{if } n=m
\end{cases}
\]
we have that ${\sum_{n\in\mathbb{Z}}\partial_{t}\widehat{u}(n, t)
- i (\beta  n^{3} - \alpha   n^{2}) \widehat{u}(n, t)=0}$, then
$\partial_{t}\widehat{u}(n, t) - i (\beta  n^{3} - \alpha
n^{2}) \widehat{u}(n, t)=0$ where
\[
\partial_{t}\left[e^{- i (\beta  n^{3} - \alpha   n^{2}) t}
\widehat{u}(n, t)\right ]=0.
\]
Integrating over $t\in [0, T]$ yields
\[
\widehat{u}(n, t) = \widehat{u}(n, 0)
e^{ i (\beta  n^{3} - \alpha   n^{2}) t}
\]
multiplying by $e^{i n x}$ and applying ${\sum_{n\in\mathbb{Z}}}$ we obtain
\begin{align*}
u(x, t)&=\sum_{n\in \mathbb{Z}}\widehat{u}(n, t)e^{i n x} \\
& = \sum_{n\in \mathbb{Z}}\widehat{u}(n, 0)
e^{ i [ (\beta  n^{3} - \alpha   n^{2}) t + n x ]}\\
& =  \sum_{n\in \mathbb{Z}}\widehat{u}(n, 0)
e^{i (\beta  n^{3} - \alpha   n^{2}) T}
e^{ i [ (\beta  n^{3} - \alpha   n^{2}) (t - T) + n x ]}\\
& =  \sum_{n\in \mathbb{Z}}c_{n} e^{ i (\beta  n^{3} - \alpha
  n^{2}) (t - T) + i n x}.
\end{align*}
where $c_{n}=\widehat{u}(n, 0) e^{i (\beta  n^{3} - \alpha
n^{2}) T}$ and $u(x, T)=u_{T}={\sum_{n\in
\mathbb{Z}}c_{n} e^{i n x}}$.
\end{remark}

For the rest of this article, $u$ will denote the
solution of \eqref{eP} associated with $u_{T}$.
We show the following result for the non-homogeneous problem.

\begin{theorem} \label{thm3.1}
Let $H_{p}^{2}=\{w\in H^{2}(0, 2 \pi): w(0)=w(2 \pi),  w'(0)=w'(2 \pi)\}$
and $T>0$. Then for any $y_{0}$, $y_{T}\in ( H_{p}^{2} )'$
(the dual space of $H_{p}^{2}$), there exist
$h_{k}\in L^{2}(0, T)$ ($k=0, 1, 2$) such that the solution
$y\in C([0, T]: ( H_{p}^{2} )')$ of the
boundary initial-value higher order Schr\"{o}dinger equation
\begin{equation} \label{eP1}
\begin{gathered}
\partial_{t}y + \beta  \partial^{3}y - i \alpha  \partial^{2}y = 0,\quad
(x, t)\in (0, 2 \pi)\times (0, T);\\
\partial^{k}y(2 \pi, t) - \partial^{k}y(0, t) = h_{k}(t),\quad k=0, 1, 2;\\
y( . , 0)= y_{0}
\end{gathered}
\end{equation}
 satisfies $y( . ,T)=y_{T}$.
\end{theorem}

\begin{remark} \label{rmk3.2} \rm
 Given $y_{0}\in (H_{p}^{2})'$, $h_{k}\in L^{2}(0, T)$
($k=0, 1, 2$), we want to find $y$ such that it
satisfies \eqref{eP1}. We first prove that \eqref{eP1} admits a unique
solution $y\in C([0, T]: (H_{p}^{2})')$ in a certain sense, and
this solution is the classical one whenever $y\in D(A)$,
and $h_{k}(k=0, 1, 2)$ are smooth enough and vanish at 0.
\end{remark}

\begin{lemma} \label{lem3.2}
\textbf{(1)}  Assume that $h_{k}\in
C_{0}^{2}([0, T])=\{h\in C^{2}([0, T]: \mathbb{C}): h(0)=0\}$
and $y_{0}\in H_{p}^{3}$. Then there exists a unique solution $y\in
C([0, T]: H^{3}(0, 2 \pi))\cap
C^{1}([0, T]: L^{2}(0, 2 \pi))$ of \eqref{eP1}. Moreover, for any
$u_{T}\in H_{p}^{3}$ and any $t\in [0, T]$ we have
\begin{equation}
\begin{aligned}
&\int_{0}^{2 \pi}u(x, t) \overline{y(x, t)}\,dx \\
& = \int_{0}^{2 \pi}u(x, 0) \overline{y_{0}(x)}\,dx - (\beta -
i \alpha)\int_{0}^{t}\partial^{2}u(0, s) \overline{h_{0}(s)}\,ds \\
&\quad +\beta \int_{0}^{t}\partial
u(0, s) \overline{h_{1}(s)}\,ds +
\int_{0}^{t}u(0, s) (\overline{\beta  h_{2}(s) + i \alpha
 h_{1}(s)})\,ds.
\end{aligned} \label{e302}
\end{equation}
\textbf{(2)}  For $u_{T}\in H_{p}^{2}$, $u\in
C([0, T]: H_{p}^{2})$ and $\partial^{2}u(0, . )$
makes sense in $L^{2}(0, T)$.

\noindent\textbf{(3)}  Assume now that $y_{0}\in (H_{p}^{2})'$
and $h_{k}\in L^{2}(0, T)$($k=0, 1, 2$). Then, there exists a unique
$y \in C([0, T]: (H_{p}^{2})')$ such that for all $u_{T}\in H_{p}^{2}$
and for all $t\in [0, T]$,
\begin{equation} \label{e303}
\begin{aligned}
&\langle u( . , t), y(t)\rangle _{H_{p}^{2}\times (H_{p}^{2})'} \\
& = \langle u( . , 0), y_{0}\rangle _{H_{p}^{2}\times (H_{p}^{2})'}
- (\beta - i \alpha)\int_{0}^{t}\partial^{2}u(0, s) \overline{h_{0}(s)}\,ds
 \\
&\quad + \beta \int_{0}^{t}\partial u(0, s) \overline{h_{1}(s)}\,ds +
\int_{0}^{t}u(0, s) (\overline{\beta  h_{2}(s) + i \alpha
 h_{1}(s)})\,ds
\end{aligned}
\end{equation}
\end{lemma}

\begin{proof} {\bf (1)} Let $\phi_{i}\in
C^{\infty}([0, 2 \pi])$($i=0, 1, 2$) be such that
\[
\phi_{i}^{(k)}(0) = 0\quad \mbox{and}\quad
{\phi_{i}^{(k)}(2 \pi)}= \begin{cases}
-1,& i=k \\
0, & i\neq k.
\end{cases}
\]
We consider the change of function $z(x, t) =
\sum_{i=0}^{2} [ h_{i}(t) \phi_{i}(x) + (S(t)y_{0})(x) +
y(x, t) ]$, then
\begin{align*}
z(2 \pi , t) - z(0, t)
& = \sum_{i=0}^{2}h_{i}(t) \phi_{i}(2 \pi) + (S(t)y_{0})(2 \pi) +
y(2 \pi, t) \\
&\quad - \sum_{i=0}^{2}h_{i}(t) \phi_{i}(0) + (S(t)y_{0})(0) + y(0, t)\\
& =  -h_{0}(t) + (S(t)y_{0})(2 \pi) - (S(t)y_{0})(0) +
y(2 \pi, t) -
y(0, t)\\
& =  -h_{0}(t) + (S(t)y_{0})(2 \pi) - (S(t)y_{0})(0) + h_{0}(t)\\
& =  (S(t)y_{0})(2 \pi) - (S(t)y_{0})(0)
\end{align*}
using that $y_{0}\in H_{p}^{3}$ we obtain
$z(2 \pi , t) =z(0, t)$. The other initial conditions are
calculated in a similar
way. Hence, this change of the function yields an equivalent problem
to \eqref{eP1}: Find $z$ such that
\begin{equation} \label{eP2}
\begin{gathered}
\partial _{t}z + \beta  \partial^{3}z - i \alpha  \partial^{2}z
=f(x, t)\\
=  \sum_{i=0}^{2}\left[h_{i}'(t)
\phi_{i}(x) + \beta  h_{i}(t)  \phi_{i}^{(3)}(x)
- i \alpha  h_{i}(t) \phi_{i}^{(2)}(x)\right ]  \\
\partial^{k}z(2 \pi, t) = \partial ^{k}z(0, t),\quad  k=0, 1, 2 \\
z( . , 0)= 0
\end{gathered}
\end{equation}
Since $f\in C^{1}([0, T]: L^{2}(0, 2 \pi))$, this
non-homogeneous problem admits a unique solution (see \cite{p1}),
$z\in C([0, T]: H_{p}^{3})\cap
C^{1}([0, T]: L^{2}(0, 2 \pi))$.
This proves the first assertion in (1).

Let $u_{T}\in H_{p}^{3}$, then $u\in C([0, T]: H_{p}^{3})\cap
C^{2}([0, T]: L^{2}(0, 2 \pi))$. Multiplying the equation \eqref{eP}
by $\overline{y}$ and integrating in $x\in [0, 2 \pi ]$ and
$t\in[0, T]$ we have
\[
\int_{0}^{t}\int_{0}^{2 \pi}\overline{y}[\partial_{s}u]\,dx\,ds +
\beta
\int_{0}^{t}\int_{0}^{2 \pi}\overline{y}[\partial^{3}u]\,dx\,ds -
i \alpha
\int_{0}^{t}\int_{0}^{2 \pi}\overline{y}[\partial^{2}u]\,dx\,ds=0.
\]
Each term is treated separately. Integrating by parts,
\begin{align*}
&\int_{0}^{t}\int_{0}^{2 \pi}\overline{y}[\partial_{s}u]\,dx\,ds \\
&=  \int_{0}^{2 \pi}\overline{y(x, t)} u(x, t)\,dx -
\int_{0}^{2 \pi}\overline{y(x, 0)} u(x, 0)\,dx -
\int_{0}^{t}\int_{0}^{2 \pi}[\partial_{s}\overline{y}] u\,dx\,ds\,,
\end{align*}
\begin{align*}
\beta \int_{0}^{t}\int_{0}^{2 \pi}\overline{y}[\partial^{5}u]\,dx\,ds
&=  \beta
\int_{0}^{t}\overline{h_{0}(s)}[\partial^{2}u(0, s)]\,ds
- \beta \int_{0}^{t}\overline{h_{1}(s)}[\partial u(0, s)]\,ds\\
&\quad +\beta \int_{0}^{t}\overline{h_{2}(s)} u(0, s)\,ds -
\int_{0}^{t}\int_{0}^{2 \pi}[\partial^{3}\overline{y}] u\,dx\,ds\,,
\end{align*}
\begin{align*}
&- i \alpha
\int_{0}^{t}\int_{0}^{2 \pi}\overline{y}[\partial^{2}u]\,dx\,ds \\
&= - i \alpha
\int_{0}^{t}\overline{h_{0}(s)}[\partial^{2}u(0, s)]\,ds
+ i \alpha \int_{0}^{t}\overline{h_{1}(s)} u(0, s)\,ds
 - i \alpha
\int_{0}^{t}\int_{0}^{2 \pi}[\partial^{2}\overline{y}] u\,dx\,ds\,.
\end{align*}
Therefore,
\begin{align*}
& \int_{0}^{2 \pi}\overline{y(x, t)}\;u(x, t)\,dx -
\int_{0}^{2 \pi}\overline{y(x, 0)} u(x, 0)\,dx -
\int_{0}^{t}\int_{0}^{2 \pi}[\partial_{s}\overline{y}] u\,dx\,ds\\
&+ \beta \int_{0}^{t}\overline{h_{0}(s)}[\partial^{2}u(0, s)]\,ds
- \beta \int_{0}^{t}\overline{h_{1}(s)}[\partial
u(0, s)]\,ds +  \beta \int_{0}^{t}\overline{h_{2}(s)}\;u(0, s)\,ds\\
&- \int_{0}^{t}\int_{0}^{2 \pi}[\partial^{3}\overline{y}] u\,dx\,ds
- i \alpha \int_{0}^{t}\overline{h_{0}(s)}[\partial^{2}u(0, s)]\,ds
 +i \alpha \int_{0}^{t}\overline{h_{1}(s)} u(0, s)\,ds\\
&-i \alpha \int_{0}^{t}\int_{0}^{2 \pi}[\partial^{2}\overline{y}] u\,dx\,ds
=0\,,
\end{align*}
where
\begin{align*}
&\int_{0}^{2 \pi}u(x, t) \overline{y(x, t)}\,dx\\
& =
\int_{0}^{2 \pi}u(x, 0) \overline{y_{0}(x)}\,dx - (\beta -
i \alpha)\int_{0}^{t}[\partial^{2}u(0, s)] \overline{h_{0}(s)}\,ds\\
&\quad +\beta \int_{0}^{t}[\partial
u(0, s)] \overline{h_{1}(s)}\,ds +
\int_{0}^{t}u(0, s) (\overline{\beta  h_{2}(s) + i \alpha
 h_{1}(s)})\,ds.
\end{align*}
Result (1) follows.

Now, we proof (2). By \eqref{e301}, for $t_{1}, t_{2}\in [0, T]$
\begin{gather*}
u(x, t_{1})=\sum_{n\in \mathbb{Z}}c_{n} e^{i (\beta  n^{3} -
\alpha  n^{2}) (t_{1} - T) + i n x }, \\
u(x, t_{2})=\sum_{n\in \mathbb{Z}}c_{n} e^{i (\beta  n^{3} -
\alpha  n^{2}) (t_{2} - T) + i n x};
\end{gather*}
hence
\begin{align*}
&u(x, t_{1}) - u(x, t_{2})  \\
&=  \sum_{n\in \mathbb{Z}}c_{n} e^{i (\beta  n^{3}
- \alpha  n^{2}) T}\big(e^{i (\beta  n^{3} - \alpha  n^{2}) t_{1}}
- e^{i (\beta
 n^{3} -  \alpha  n^{2}) t_{2}}\big) e^{i n x}.
\end{align*}
 From \eqref{e203}, if $u_{T}\in H_{p}^{2}$ then
${\sum_{n\in \mathbb{Z}}|n^{2} c_{n}|^{2}<\infty }$
and ${\sum_{n\in \mathbb{Z}}|n c_{n}|^{2}<\infty }$.
Using Lebesgue's Theorem \cite{r2},
\[
|u(x, t_{1}) - u(x, t_{2})|  =  \sum_{n\in \mathbb{Z}}\big|
(n^{2} + n) c_{n} \big(e^{i (\beta  n^{3} - \alpha
 n^{2}) t_{1}} - e^{i (\beta  n^{3} -  \alpha
 n^{2}) t_{2}}\big)\big|^{2}
\]
which approaches $0$ as $t_{1}\to t_{2}$.
We conclude that $u\in C([0, T]: H_{p}^{2})$. Hence
$u(0, . )$, $\partial u(0, . )$  exist in $C([0, T])\subseteq
L^{2}(0, T)$. The same argument shows that if $u_{T}\in H_{p}^{3}$,
$u\in C([0, T]: H_{p}^{3})$ and
\begin{equation}
\label{e304}\partial^{2}u(0, t) = \sum_{n\in \mathbb{Z}}
\big(- n^{2} c_{n} e^{- i (\beta  n^{3} -  \alpha
 n^{2}) T} \big)  e^{i (\beta  n^{3} -  \alpha  n^{2}) t}.
\end{equation}
The sum in \eqref{e304} makes sense in $L^{2}(0, T)$ wherever
${\sum_{n\in \mathbb{Z}} ( n^{2} |c_{n}| )^{2}<
\infty }$, that is, $u_{T}\in H_{p}^{2}$. From now on,
$\partial^{2}u(0, . )$ denotes for $u_{T}\in H_{p}^{2}$,
the sum in \eqref{e304}.
\end{proof}

\begin{remark} \label{rmk3.3} \rm
The linear map $u_{T}\mapsto \partial^{2}u(0, . )$ is continuous since
\begin{equation}
\label{e305}
\big\| \sum_{n\in \mathbb{Z}}
\big( n^{2} c_{n} e^{- i (\beta  n^{3} -  \alpha
 n^{2}) T}\big) e^{i (\beta  n^{3} -  \alpha
 n^{2}) t}\big\| \leq  \big([
 \frac{T}{2 \pi}]   + 1\big )
\sum_{n\in \mathbb{Z}}[ n^{2} |c_{n}| ]^{2}
\end{equation}
where $[x]$ denotes the integral
part of a real number $x$. Identifying $L^{2}(0, 2 \pi)$ with its
dual by means of the conjugate linear map
$y\mapsto \langle  . , y\rangle _{L^{2}(0, 2 \pi )}$,
we have the following dense and
compact embedding (see \cite{l7})
\begin{equation}
\label{e306}H_{p}^{2}\hookrightarrow L^{2}(0, 2 \pi)
\hookrightarrow (L^{2}(0, 2 \pi))'\hookrightarrow (H_{p}^{2})'.
\end{equation}
Moreover,
\begin{equation}
\label{e307}\langle u, y\rangle _{H_{p}^{2}\times (H_{p}^{2})'} =
\langle u, y\rangle _{L^{2}(0, 2 \pi)}=\int_{0}^{2 \pi}u \overline{y}\,dx
\end{equation}
for $u\in H_{p}^{2}$ and $y\in L^{2}(0, 2 \pi)$. Then
\begin{align*}
&\langle u( . , t), y(t)\rangle _{H_{p}^{2}\times (H_{p}^{2})'} \\
& = \langle u( . , 0), y_{0}\rangle _{H_{p}^{2}\times (H_{p}^{2})'}
- (\beta - i \alpha)\int_{0}^{t}[\partial^{2}u(0, s)] \overline{h_{0}(s)}\,ds
\\
&\quad +\beta \int_{0}^{t}\partial u(0, s) \overline{h_{1}(s)}\,ds
+ \int_{0}^{t}u(0, s) (\overline{\beta  h_{2}(s) + i \alpha
 h_{1}(s)})\,ds
\end{align*}
for $h_{k}\in C_{0}^{2}([0, T]) $ ($k=0, 1, 2$) and
$y_{0}, u_{T}\in H_{p}^{3}$. Since $H_{p}^{3}$ is dense in
$H_{p}^{2}$, using (2),
we see that \eqref{e303} also is true for $u_{T}\in H_{p}^{2}$.
\end{remark}

\noindent\textbf{Definition.}% 3.2.}
 For $y_{0}\in (H_{p}^{2})'$ and $h_{k}\in L^{2}(0, T)$ ($k=0, 1, 2$),
we define a weak solution of
\eqref{eP1} as a function $y\in C([0, T]: (H_{p}^{2})')$ such that
\eqref{e303} holds for all $u_{T}\in H_{p}^{2}$ and all $t\in [0, T]$.
\smallskip

\noindent\textbf{Claim.}
For $t$ fixed in $[0, T]$, \eqref{e303} defines
$y(t)\in (H_{p}^{2})'$ in a unique manner.

In fact, from the proof of (2) the map
$\Xi : H_{p}^{2}\to  \mathbb{C}$, $u_T\mapsto \Xi(u_T)$, given by
\begin{align*}
\Xi (u_{T})
& =- (\beta -i \alpha)\int_{0}^{t}\overline{h_{0}(s)}
[\partial^{2}u(0, s)]\,ds
+ \beta \int_{0}^{t}\overline{h_{1}(s)}[\partial u(0, s)]\,ds \\
&\quad +\int_{0}^{t}( \beta  \overline{h_{2}(s)} - i \alpha
 \overline{h_{1}(s)} ) u(0, s)\,ds
\end{align*}
is a continuous linear form. On the other hand, the map
$\Phi :H_{p}^{2}\mapsto H_{p}^{2}$ with $u_{T}\to \Phi
(u_{T}) =u( . , t)$ is an automorphism of the Hilbert space,
hence, for each $t\in [0, T], $ $y(t)$ is uniquely defined in
$(H_{p}^{2})'$. Moreover, for $t\in [0, T]$,
\begin{align*}
 \|y(t)\|_{(H_{p}^{2})'}
&=\sup_{\|u( . , t)\|_{H_{p}^{2}}\leq 1}|\langle u( . , t), y(t)\rangle| \\
&=\sup_{\|u( . , t)\|_{H_{p}^{2}}\leq 1}| \langle u( . , 0), y_{0}
\rangle _{H_{p}^{4}\times (H_{p}^{2})'} - (\beta -
 i \alpha)\int_{0}^{t}[\partial^{2}u(0, s)] \overline{h_{0}(s)}\,ds\\
&\quad + \beta \int_{0}^{t}[\partial u(0, s)] \overline{h_{1}(s)}\,ds
 +\int_{0}^{t}u(0, s) (\overline{\beta  h_{2}(s) + i \alpha
 h_{1}(s)})\,ds| \\
& \leq  \sup_{\|u( . , t)\|_{H_{p}^{2}}\leq 1}| \langle u( .
 , 0), y_{0}\rangle _{H_{p}^{2}\times (H_{p}^{2})'}| \\
&\quad + (|\beta| +|\alpha |)\;\sup_{\|u( . , t)\|_{H_{p}^{2}}\leq 1}
\int_{0}^{t}| \overline{h_{0}(s)} [\partial^{2}u(0, s)] |\,ds \\
&\quad +|\beta | \sup_{\|u( . , t)\|_{H_{p}^{2}}\leq 1}
\int_{0}^{t} | \overline{h_{1}(s)} [\partial
u(0, s)]  |\,ds \\
&\quad+ \sup_{\|u( . , t)\|_{H_{p}^{2}}\leq 1}
\int_{0}^{t}|( \beta  \overline{h_{2}(s)} - i \alpha
 \overline{h_{1}(s)} ) u(0, s) |\,ds  \\
& \leq  \sup_{\|u( . , t)\|_{H_{p}^{2}}\leq
1}\|u( . , 0)\|_{(H_{p}^{2})'}  \|y_{0}\|_{H_{p}^{2}} \\
&\quad + (|\beta| + |\alpha |)\; \sup_{\|u( . , t)\|_{H_{p}^{2}}\leq 1}
\|\overline{h_{0}(t)}\|_{L^{2}(0, T)} \|\partial^{2}u(0, t)\|_{L^{2}(0, T)}
 \\
&\quad +|\beta |\sup_{\|u( . , t)\|_{H_{p}^{2}}\leq 1}
\|\overline{h_{1}(t)}\|_{L^{2}(0, T)} \|\partial
u(0, t)\|_{L^{2}(0, T)}\\
&\quad +\sup_{\|u( . , t)\|_{H_{p}^{2}}\leq 1}
\|( \beta  \overline{h_{2}(s)} - i \alpha
 \overline{h_{1}(s)} ) \|_{L^{2}(0, T)} \|u(0, t)\|_{L^{2}(0, T)} \\
& \leq  c \big( \|y_{0}\|_{(H_{p}^{2})'} +
\|h_{0}\|_{L^{2}(0, T)} + \|h_{1}\|_{L^{2}(0, T)} +
\|h_{2}\|_{L^{2}(0, T)} \big)
\end{align*} % \label{e308}
where $c$ is a positive constant which does not depend on
$t$ or on $y_{0}$, $h_{0}$, $h_{1}$, $h_{2}$. Since
\begin{equation}
\label{e309} y\in C([0, T]: L^{2}(0, 2 \pi))\subseteq
C([0, T]: (H_{p}^{2})')
\end{equation}
for $y\in H_{p}^{3}$ and $(h_{0}, h_{1}, h_{2})\in
[C_{0}^{2}([0, T])]^{3}$, and since $H_{p}^{3}$ is dense in
$L^{2}(0, T)$ and $C_{0}^{2}([0, T])$ is dense in $L^{2}(0, L)$,
it follows from \eqref{e309} that
$y\in C([0, T]: (H_{p}^{2})')$.
%\end{proof}

\begin{lemma}[Observability result] \label{lem3.3}
Let $T>0$. There exist positive numbers $C_{1}^{T}$, $C_{2}^{T}$
 such that for every $u_{T}\in H_{p}^{2}$
\begin{equation} \label{e310}
\begin{aligned}
C_{1}^{T} \|u_{T}\|_{H_{p}^{2}(0, 2 \pi)}^{2}
& \leq \|u(0, . )\|_{L^{2}(0, T)}^{2} + \|\partial
u(0, . )\|_{L^{2}(0, T)}^{2}
+ \|\partial^{2}u(0, . )\|_{L^{2}(0, T)}^{2}  \\
& \leq C_{2}^{T} \|u_{T}\|_{H_{p}^{2}(0, 2 \pi)}^{2}
\end{aligned}
\end{equation}
\end{lemma}

\begin{proof} In $L^{2}(0, T)$ we have that
\begin{gather*}
u(0, t) = \sum_{n\in \mathbb{Z}}c_{n}
e^{i (\beta  n^{3} - \alpha  n^{2}) (t - T)}\\
\partial u(0, t) = \sum_{n\in \mathbb{Z}}i n c_{n} e^{i (\beta  n^{3}
- \alpha  n^{2}) (t - T)}\\
\partial^{2}u(0, t)
 =  \sum_{n\in \mathbb{Z}}- n^{2} c_{n} e^{i (\beta  n^{3}
- \alpha  n^{2}) (t - T)}.
\end{gather*}
Hence
\begin{equation} \label{e311}
\begin{aligned}
&\|u(0, t)\|_{L^{2}(0, T)}^{2} + \|\partial
u(0, t)\|_{L^{2}(0, T)}^{2} +
\|\partial^{2}u(0, t)\|_{L^{2}(0, T)}^{2}  \\
& \leq  \big([ \frac{T}{2 \pi} ] + 1 \big)
\sum_{n\in \mathbb{Z}}( 1 + n^{2} + n^{4} )|c_{n}|^{2} \\
& \leq C_{2}^{T}\;\|u_{T}\|_{H_{p}^{2}(0, 2 \pi)}^{2}\quad
\mbox{for } u_{T}\in H_{p}^{2}
\end{aligned}
\end{equation}
where $C_{2}^{T}=( [\frac{T}{2 \pi}]+ 1 )$. To prove the left inequality
we first take $T'\in (0, T)$ and $\gamma >2 \pi /T'$. Let
$N\in \mathbb{N}^{*}$ be such that
\[
n\in \mathbb{Z},\quad |n|\geq N\Rightarrow [\beta  (n + 1)^{5}
- \alpha  (n + 1)^{3}] - [\beta  n^{5} - \alpha  n^{3}]\geq
\gamma .
\]
By Ingham's inequality \cite{i1} there exists $c^{T'}>0$ such
that for all sequences $(a_{n})_{n\in \mathbb{Z}}$ in
$l^{2}(\mathbb{Z})$,
\begin{equation} \label{e312}
 \sum_{|n|\geq N}|a_{n}|^{2}\leq
c^{T'}\int_{0}^{T'}\Big|\sum_{|n|\geq N}a_{n}  e^{i (\beta
 n^{3} - \alpha  n^{2}) (t - T)}\Big|^{2}\,dt.
\end{equation}
Let $\mathcal{Z}_{n}=\mathop{\rm Span} (e^{i n x})$ for
$n\in \mathbb{Z}$ and
$\mathcal{Z}={\oplus_{n\in \mathbb{Z}}\mathcal{Z}_{n}\subseteq H_{p}^{2}}$.
 We define a semi-norm $p$ in $\mathcal{Z}$
by: $\forall  u\in \mathcal{Z}$,
\begin{equation} \label{e313}
\begin{aligned}
p(u) & =  ( |u(0)|^{2} + |\partial u(0)|^{2}
+ |\partial^{2}u(0)|^{2} )^{1/2} \\
& =  \Big(\big|\sum_{n\in \mathbb{Z}}\widehat{u}(n)\big|^{2}
+ \big|\sum_{n\in \mathbb{Z}}i n \widehat{u}(n)\big|^{2}
+ \big|\sum_{n\in \mathbb{Z}}-n^{2}\widehat{u}(n)\big|^{2} \Big)^{1/2}
\end{aligned}
\end{equation}
(For $u\in \mathcal{Z}$, $\widehat{u}(n)=0$ for $|n|$ large enough).

Let $u_{T}\in \mathcal{Z}\cap ( {\oplus_{|n|<N}\mathcal{
Z}_{n} )^{\bot}}$, that is, $c_{n}=0$ for $|n|<N$ or for $|n|$
large enough. Using \eqref{e301} and \eqref{e312} we have
\begin{equation} \label{e314}
\|u_{T}\|_{H_{p}^{2}((0, 2 \pi))}^{2}  =
\sum_{n\geq N}( 1 + n^{2} + n^{4} ) |c_{n}|^{2}
 \leq  c^{T'}\int_{0}^{T}[p(u( . , t))]^{2}\,dt.
\end{equation}
Since $T>T'$, it follows from \eqref{e311}, \eqref{e314} and a
result by Komornik (see \cite{k2}) that there exists a constant
$C_{1}^{T}>0$ such that for all $u_{T}$ in $\mathcal{Z}$,
\begin{equation} \label{e315}
\begin{aligned}
C_{1}^{T} \|u_{T}\|_{H_{p}^{2}(0, 2 \pi)}^{2}
& \leq  \int_{0}^{T}[ p(u( . , t)) ]^{2}\,dt  \\
& =  \|u(0, . )\|_{L^{2}(0, T)}^{2} + \|\partial
u(0, . )\|_{L^{2}(0, T)}^{2} +
\|\partial^{2}u(0, . )\|_{L^{2}(0, T)}^{2}
\end{aligned}
\end{equation}
and the result follows.
\end{proof}

We remark that by a density argument we obtain the left inequality in
\eqref{e310} in the general case ($u_{T}\in H_{p}^{2}$).

\begin{proof}[Proof of Theorem \ref{thm3.1}]
 Without loss of generality we may assume
that $y_{0}=0$. In fact, if $y_{0}$, $y_{T}\in (H_{p}^{2})'$, if
there exist $h_{k}\in L^{2}(0, T)$ ($k=0, 1, 2$) such that the
weak solution $\widetilde{y}$ of \eqref{eP1} and
$\widetilde{y}( . , 0)=0$ satisfies
$\widetilde{y}( . , T)=y_{T} - S(T)y_{0}$,
then $y_{T}=S(T)y_{0} + \widetilde{y}( . ,t)$ is the weak solution
of \eqref{eP1} with the same control functions and its such that
$y( . ,T)=y_{T}$. In what follows we assume that $y_{0}=0$. For
$u_{T}\in H_{p}^{2}$ we let
$\Lambda :H_{p}^{2}\mapsto (H_{p}^{2})'$,
\[
u_{T}\mapsto  \Lambda (u_{T})= y_{T}.
\]
where $y$ is the weak solution of \eqref{eP1} and
$h_{k}$($k=0, 1, 2)$ are chosen the following way:
\begin{gather*}
\overline{h_{0}(t)}=\frac{-1}{(\beta + i \alpha)}\partial^{2}u(0, t),\quad
\overline{h_{1}(t)}=\frac{1}{\beta}\partial u(0, t),\\
\overline{h_{2}(t)}=i \frac{1}{\beta} u(0, t) + i \frac{\alpha
}{\beta^{2}} \partial u(0, t)
\end{gather*}
As above $u$ stands for the solutions of \eqref{eP} associated with
$u_{T}$. Clearly $\Lambda :H_{p}^{2}\mapsto (H_{p}^{2})'$ is a
conjugate linear continuous map. Moreover
\begin{align*}
\langle u_{T}, \Lambda(u_{T})\rangle _{H_{p}^{2}\times (H_{p}^{2})'}
& = \int_{0}^{T}( |u(0, t)|^{2} + |\partial u(0, t)|^{2}
+ |\partial^{2}u(0, t)|^{2})\,dt\\
& \geq  C_{1}^{T} \|u_{T}\|_{H_{p}^{2}(0, 2 \pi)}^{2}.
\end{align*}
By Lemmas \ref{lem3.2} and \ref{lem3.3} it follows from Lax-Milgram's Theorem
(see \cite{y1}) that $\Lambda $ is invertible.
Then the theorem follows.
\end{proof}

\begin{remark} \label{rmk3.5}\rm
  If $T=2 \pi$,  Lemma \ref{lem3.3} is trivial. Indeed,
for any $u_{T}\in H_{p}^{2}$,
\[
\|u_{T}\|_{H_{p}^{4}(0, 2 \pi)}^{2} = \|u(0, .
 )\|_{L^{2}(0, 2 \pi)}^{2} + \|\partial
u(0, . )\|_{L^{2}(0, 2 \pi)}^{2} +
\|\partial^{2}u(0, . )\|_{L^{2}(0, 2 \pi)}^{2}.
\]
\end{remark}

 \section{Exact boundary controllability of
the higher order linear Schr\"{o}dinger equation by means of the control
$\partial y(L, t)$}

 We consider now, the scalar space $\mathbb{R}$.
In this section, $L$ stands for some positive number. We shall prove
the controllability in $L^{2}(0, L)$ of
\begin{equation} \label{eR1}
\begin{gathered}
\partial _{t}y + \beta  \partial^{3}y - i \alpha  \partial^{2}y
+ \delta  \partial y= 0 \\
y(0, t) = y(L, t) = 0 \\
\partial y(L, t)=h(t) \\
y( . , 0)= y_{0}
\end{gathered}
\end{equation}
where $h\in L^{2}(0, T)$ stands for the control
function. More precisely we shall prove that, for any $L>0$, $T>0$,
$y_{0}$, $y_{T}\in L^{2}(0, L)$ there exists $h\in L^{2}(0, T)$
such that a mild solution
\begin{equation}
y\in C([0, T]: L^{2}(0, L))\cap L^{2}(0, T: H^{1}(0, L))\cap
H^{1}(0, T: H^{-2}(0, L))
\end{equation}
of \eqref{eR1} which verifies the equation \eqref{eR1} in
$\mathcal{D}'(0, T: H^{-2}(0, L))$ and $y_{0}$ in $L^{2}(0, L)$ may be
found such that $y( . ,T)=y_{T}$.

We begin by showing the well-posedness of the initial-value
homogeneous problem with $|\alpha | <3 \beta $
\begin{equation} \label{eR2}
\begin{gathered}
\partial _{t}y + \beta  \partial^{3}y - i \alpha  \partial^{2}y
+ \delta  \partial y= 0 \\
y(0, t) = y(L, t) = 0 \\
\partial y(L, t)=0 \\
y( . , 0)= y_{0}\,.
\end{gathered}
\end{equation}
Let $A$ denote the operator
$Aw=- \beta  w''' +i \alpha  w'' - \delta  w'$ on the (dense)
domain $D(A)\subseteq L^{2}(0, L)$, defined by
\[
\label{e401}D(A)=\{w\in H^{3}(0, L): w(0)=w(L)=w'(L)=0\}
\]

\begin{lemma} \label{lem4.1}
Operator $A$ generates a strongly continuous semigroup of
contractions on $L^{2}(0, L)$.
\end{lemma}

\begin{proof} $A$ is closed. Let $w\in D(A)$. Then
\begin{align*}
&\mathop{\rm Re} \langle w, Aw\rangle _{L^{2}(0, L)} \\
& = \mathop{\rm Re} \int_{0}^{L}[ - \beta  w''' + i \alpha  w'' - \delta  w' ]
w(x)\,dx \\
& = \mathop{\rm  Re} \Big[- \beta \int_{0}^{L}w'''(x) w(x)\,dx + i \alpha
\int_{0}^{L}w'' w(x)\,dx - \delta
\int_{0}^{L}w'(x) w(x)\,dx\Big].
\end{align*}
Each term is treated separately. Integrating by parts,
\begin{gather*}
\int_{0}^{L}w'''(x) w(x)\,dx  =  \frac{1}{2} [w'(0)]^{2},\\
\int_{0}^{L}w''(x) w(x)\,dx  =  -\int_{0}^{L}[w'(x)]^{2}\,dx
\end{gather*}
Then
\[
\mathop{\rm Re} \langle w, Aw\rangle _{L^{2}(0, L)}
 = -  \frac{\beta}{2} [ w'(0) ]^{2}
\leq 0\quad \mbox{if}\quad \beta > \frac{1}{3} |\alpha |
\]
hence, $A$ is dissipative. It can be seen that
$A^{*}(w) =\beta  w''' - i \alpha  w'' + \delta  w'$ with
domain $D(A^{*})=\{w\in H^{5}(0, L): w(0)=w(L)=w'(0)=0\}$, so that
\[
\mathop{\rm Re} \langle w, A^{*}w\rangle _{L^{2}(0, L)}
=- \frac{\beta}{2} [ w'(L) ]^{2}\leq 0 ,
\quad \mbox{if } \beta > \frac{1}{3} |\alpha |
\]
and $A^{*}$ is dissipative. Hence, by the Lumer-Phillips Theorem,
$A$ is the infinitesimal generator of a $C_{0}$ semigroup of
contractions on $L^{2}(0, L)$. The result follows.
\end{proof}

We denote by $(S(t))_{t\geq 0}$ the semi-group of contractions
associated with $A$, and we let $\mathbb{H}$ denote the Banach space
$C([0, T]: L^{2}(0, L))\cap L^{2}([0, T]: H^{1}(0, L))$
endowed with the norm
\begin{equation} \label{e402}
\begin{aligned}
\|y\|_{\mathbb{H}} & =
\sup_{t\in[0, T]}\|y( . , t)\|_{L^{2}(0, L)}
+ \Big(\int_{0}^{T}\|y( . , t)\|_{H^{1}(0, L)}^{2}\,dt\Big)^{1/2}\\
& = \sup_{t\in[0, T]}\|y( . , t)\|_{L^{2}(0, L)} +
\|y( . , t)\|_{L^{2}(0, T: H^{1}(0, L))}.
\end{aligned}
\end{equation}
Using the multiplier method, we get useful estimates
for the mild solutions of \eqref{eR2}.

\begin{lemma} \label{lem4.2}
Let $|\alpha |<3 \beta $. Then
\begin{itemize}
\item[(1)] The map $y_{0}\in L^{2}(0, L)\mapsto
S( \cdot )y_{0}\in \mathbb{H}$ is continuous.
\item[(2)]  For $y_{0}\in L^{2}(0, L)$, $\partial y(0, . )$
makes sense in $L^{2}(0, L)$, and for all $y_{0}\in L^{2}(0, L)$,
\begin{gather}
\label{e403}\|\partial y( . , t)\|_{L^{2}(0, T)}
 \leq   \|y_{0}\|_{L^{2}(0, L)}\\
\label{e404}\|y_{0}\|_{L^{2}(0, L)}^{2}  \leq
\frac{1}{T} \|S( \cdot   )y_{0}\|_{L^{2}((0, T)\times
(0, L))}^{2} + \|\partial y(0, . )\|_{L^{2}(0, T)}^{2}
\end{gather}
\end{itemize}
\end{lemma}

\begin{proof} (1) For $y_{0}\in L^{2}(0, L)$ we write $y$ the
mild solution $S( \cdot  )y_{0} $ of $ (R_{2})$.
 By Lemma \ref{lem4.1}, $ y\in C([0, T]: L^{2}(0, L))$ and
\begin{equation}
\label{e405}\|y\|_{C([0, T]: L^{2}(0, L))}\leq
\|y_{0}\|_{L^{2}(0, L)}
\end{equation}
To see that $y\in L^{2}(0, T: H^{2}(0, L))$ we first assume that
$y\in D(A)$. Let $\xi = \xi(x, t)\in C^{\infty}([0, T]\times
[0, L])$. Then, multiplying the equation \eqref{eR2} by $i \xi  y$
we have
\begin{gather*}
 i \xi  \overline{y} \partial _{t}y + i \xi
 \overline{y} \partial ^{3}y
+ \alpha  \xi  \overline{y} \partial ^{2}y + i \delta  \xi  \overline{y} \partial y =0\\
- i \xi  y \partial_{t}\overline{y} - i \xi  y \partial
^{3}\overline{y} + \alpha  \xi  y \partial ^{2}\overline{y} -
i \delta  \xi  y \partial \overline{y} =0
\end{gather*}
(applying conjugates).
Subtracting, integrating over $x\in (0, L)$ and using straightforward
calculus, we obtain
\begin{gather*}
i \partial_{t}\int_{0}^{L}\xi  |y|^{2}\,dx -
i\int_{0}^{L}\partial_{t}\xi |y|^{2}\,dx + i \beta \int_{0}^{L}\xi
 \overline{y} \partial ^{3}y\,dx
+ i \beta \int_{0}^{L}\xi  y \partial ^{3}\overline{y}\,dx  \\
+ \alpha \int_{0}^{L}\xi  \overline{y} \partial ^{2}y\,dx -
\alpha \int_{0}^{L}\xi  y \partial ^{2}\overline{y}\,dx -
i \delta \int_{0}^{L}\partial \xi |y|^{2}\,dx =0.
\end{gather*}
Each term is treated separately. Integrating by parts
\begin{gather*}
\begin{aligned}
\int_{0}^{L}\xi  \overline{y} \partial ^{3}y\,dx
& = \int_{0}^{L}\partial ^{2}\xi \overline{y} \partial y\,dx +
2\int_{0}^{L}\partial \xi |\partial y|^{2}\,dx -
\xi(0, t) |\partial y(0, t)|^{2} \\
&\quad + \int_{0}^{L}\xi  \partial y \partial ^{2}\overline{y}\,dx
\end{aligned}\\
\int_{0}^{L}\xi  y \partial ^{3}\overline{y}\,dx
 = \int_{0}^{L}\partial ^{2}\xi y \partial \overline{y}\,dx +
\int_{0}^{L}\partial \xi |\partial y|^{2}\,dx - \int_{0}^{L}\xi
 \partial y \partial ^{2}\overline{y}\,dx,\\
\int_{0}^{L}\xi  \overline{y} \partial ^{2}y\,dx
 = -\int_{0}^{L}\partial \xi \overline{y} \partial y\,dx -
\int_{0}^{L}\xi  |\partial y|^{2}\,dx,\\
\int_{0}^{L}\xi  y \partial ^{2}\overline{y}\,dx
 = -\int_{0}^{L}\partial \xi y \partial \overline{y}\,dx -
\int_{0}^{L}\xi  |\partial y|^{2}\,dx\,.
\end{gather*}
Then
\begin{align*}
&i \partial_{t}\int_{0}^{L}\xi  |y|^{2}\,dx -
i\int_{0}^{L}\partial_{t}\xi |y|^{2}\,dx + i \beta
\int_{0}^{L}\partial ^{2}\xi \overline{y} \partial y\,dx +
2 i \beta
\int_{0}^{L}\partial \xi |\partial y|^{2}\,dx\\
& - i \beta  \xi(0, t) |\partial y(0, t)|^{2} + i \beta
\int_{0}^{L}\xi  \partial y \partial ^{2}\overline{y}\,dx +
i \beta \int_{0}^{L}\partial ^{2}\xi y \partial \overline{y}\,dx
+ i \beta
\int_{0}^{L}\partial \xi |\partial y|^{2}\,dx\\
&- i \beta \int_{0}^{L}\xi  \partial
y \partial^{2}\overline{y}\,dx - \alpha \int_{0}^{L}\partial
\xi \overline{y} \partial y\,dx - \alpha \int_{0}^{L}\xi
 |\partial y|^{2}\,dx +
\int_{0}^{L}\partial \xi y \partial \overline{y}\,dx \\
&+ \int_{0}^{L}\xi  |\partial y|^{2}\,dx - i \delta
\int_{0}^{L}\partial \xi  |y|^{2}\,dx =0\,.
\end{align*}
Hence,
\begin{align*}
& i \partial_{t}\int_{0}^{L}\xi  |y|^{2}\,dx -
i\int_{0}^{L}\partial_{t}\xi |y|^{2}\,dx + i \beta
\int_{0}^{L}\partial ^{2}\xi \partial (|y|^{2})\,dx + 3 i \beta
\int_{0}^{L}\partial \xi |\partial y|^{2}\,dx\\
& - i \beta  \xi(0, t) |\partial y(0, t)|^{2} -
2 i \alpha \mathop{\rm Im}\int_{0}^{L}\partial \xi \overline{y} \partial
y\,dx - i \delta \int_{0}^{L}\partial \xi  |y|^{2}\,dx =0\,.
\end{align*}
Thus
\begin{align*}
& \partial_{t}\int_{0}^{L}\xi  |y|^{2}\,dx -
\int_{0}^{L}\partial_{t}\xi |y|^{2}\,dx - \beta
\int_{0}^{L}\partial ^{3}\xi |y|^{2}\,dx + 3 \beta
\int_{0}^{L}\partial \xi |\partial y|^{2}\,dx\\
& - \beta  \xi(0, t) |\partial y(0, t)|^{2} - \delta
\int_{0}^{L}\partial \xi  |y|^{2}\,dx \\
&= 2 \alpha \mathop{\rm Im}
\int_{0}^{L}\partial \xi \overline{y} \partial y\,dx \\
& \leq |\alpha |\int_{0}^{L}\partial \xi |y|^{2}\,dx + |\alpha
|\int_{0}^{L}\partial \xi |\partial y|^{2}\,dx,
\end{align*}
where
\begin{equation} \label{e406}
\begin{aligned}
&\partial_{t}\int_{0}^{L}\xi  |y|^{2}\,dx +
\int_{0}^{L}[ 3 \beta - |\alpha | ] \partial \xi |\partial
y|^{2}\,dx - \int_{0}^{L}\partial_{t}\xi |y|^{2}\,dx - \beta
\int_{0}^{L}\partial ^{3}\xi |y|^{2}\,dx \\
& - \beta  \xi(0, t) |\partial y(0, t)|^{2} -
\delta \int_{0}^{L}\partial \xi  |y|^{2}\,dx - |\alpha
|\int_{0}^{L}\partial \xi |y|^{2}\,dx \leq 0.
\end{aligned}
\end{equation}
Choosing $\xi(x, t) = x$ leads to
\[
\partial_{t}\int_{0}^{L}x |y|^{2}\,dx +
\int_{0}^{L}[ 3 \beta - |\alpha | ] |\partial y|^{2}\,dx -
( \delta + |\alpha | )\int_{0}^{L}|y|^{2}\,dx \leq 0.
\]
Integrating over $t\in [0, T]$ we obtain
\begin{align*}
&\int_{0}^{L}x |y|^{2}\,dx + [ 3 \beta - |\alpha | ]
\int_{0}^{T}\int_{0}^{L}|\partial y|^{2}\,dx\,dt  \\
&\leq  ( \delta +|\alpha | )
\int_{0}^{T}\int_{0}^{L}|y|^{2}\,dx\,dt + \int_{0}^{L}x |y_{0}|^{2}\,dx\\
& \leq  ( \delta + |\alpha
| )\int_{0}^{T}\int_{0}^{L}|y|^{2}\,dx\,dt +
L\int_{0}^{L}|y_{0}|^{2}\,dx.
\end{align*}
Using that $|\alpha |<3 \beta, $ the second and the third terms in
the left hand on the above equation are positive, thus we obtain
\begin{align*}
&[ 3 \beta - |\alpha | ] \|\partial
y\|_{L^{2}(0, T: L^{2}(0, L))}^{2}\\
&\leq \big[ ( |\delta| +
|\alpha | ) \|y\|_{L^{2}(0, T: L^{2}(0, L))}^{2} +
L \|y_{0}\|_{L^{2}(0, L)}^{2} \big],
\end{align*}
where
\begin{equation}\label{e407}
\begin{aligned}
&\|\partial y\|_{L^{2}(0, T: L^{2}(0, L))}^{2}\\
&\leq \frac{1}{( 3 \beta - |\alpha | )}\big[( |\delta| + |\alpha
| ) \|y\|_{L^{2}(0, T: L^{2}(0, L))}^{2} +
L \|y_{0}\|_{L^{2}(0, L)}^{2} \big]\,.
\end{aligned}
\end{equation}
Then, using \eqref{e405},
\begin{equation} \label{e408}
\begin{aligned}
\|y\|_{L^{2}(0, T: H^{1}(0, L))} & \leq  \Big[ T +
\frac{1}{( 3 \beta - |\alpha | )} [( |\delta| +
|\alpha | ) T + L ]\Big]^{1/2} \|y_{0}\|_{L^{2}(0, L)} \\
& \leq  \big[ \frac{1}{( 3 \beta - |\alpha
| )}[ (|\delta| + 3 \beta ) T + L  ]\big]^{1/2}
\|y_{0}\|_{L^{2}(0, L)}
\end{aligned}
\end{equation}
By the density of $D(A)$ in $L^{2}(0, L)$ the result extends to arbitrary
$y_{0}\in L^{2}(0, L)$. \smallskip

\noindent(2) We also assume $y_{0}\in D(A)$ and taking $\xi(x, t)=1$ in
\eqref{e406}, we get
\begin{equation}
\label{e409}\beta  |\partial y(0, t)|^{2}  \leq
\int_{0}^{L}|y_{0}|^{2}\,dx - \int_{0}^{L}|y|^{2}\,dx\leq
\int_{0}^{L}|y_{0}|^{2}\,dx.
\end{equation}
On the other hand the choice $\xi(x, t)=T - t$ yields
\begin{equation}
\label{e410}\partial_{t}\int_{0}^{L}(T - t) |y|^{2}\,dx +
\int_{0}^{L}|y|^{2}\,dx - \beta  (T - t) |\partial y(0, t)|^{2}
\leq 0.
\end{equation}
Integrating over $t\in [0, T]$ we have
\[
- T\int_{0}^{L} |y_{0}|^{2}\,dx +
\int_{0}^{L}\int_{0}^{L}|y|^{2}\,dx\,dt - \beta \int_{0}^{L}(T -
t) |\partial y(0, t)|^{2}\,dt \leq 0.
\]
Hence
\begin{equation} \label{e411}
\begin{aligned}
\int_{0}^{L} |y_{0}|^{2}\,dx
& \leq  \frac{1}{T}\int_{0}^{L}\int_{0}^{L}|y|^{2}\,dx\,dt
- \frac{\beta}{T}\int_{0}^{L}(T - t) |\partial y(0, t)|^{2}\,dt  \\
& \leq
\frac{1}{T}\int_{0}^{L}\int_{0}^{L}|y|^{2}\,dx\,dt + \beta
\int_{0}^{L}|\partial y(0, t)|^{2}\,dt.
\end{aligned}
\end{equation}
By \eqref{e410} there exists a unique continuous (linear)
extension of the map $y_{0}\in D(A)\mapsto
\partial y(0, . )\in L^{2}(0, T)$ to the whole space
$L^{2}(0, L)$. In what follows we also will denote by $\partial
y(0, . )$ the value of this map at any $y_{0}\in L^{2}(0, L)$. It
is trivial to see that \eqref{e410} and \eqref{e411}
are true for any $y_{0}\in L^{2}(0, L)$.
\end{proof}

\begin{lemma}[Observability result] \label{lem4.3}
 Let $|\alpha |<3 \beta $, $\delta>0$ and
\begin{equation}
\mathcal{N}=\Big\{2 \pi \beta  \sqrt{\frac{k^{2} + k l + l^{2})}
{3 \beta \delta + \alpha^{2}}} : k, l\in
\mathbb{N}^{*}\Big\}.
\end{equation}
Then, for all $L\in (0, + \infty )\backslash \mathcal{N}$,
for all $T>0$, there exists $C=C(L, T)>0$ such that for all
$y_{0}\in L^{2}(0, L)$,
\begin{equation}
\label{e412}\|y_{0}\|_{L^{2}(0, L)}\leq C\;\|\partial
y(0, . )\|_{L^{2}(0, T)}.
\end{equation}
\end{lemma}

\begin{proof} (By contradiction) If the statement is false, there
exists a sequence $(y_{0}^{n})_{n\geq 0}\in L^{2}(0, L)$ such that
$\|y_{0}^{n}\|_{L^{2}(0, L)}=1$ for any $n$, but $\|\partial
y^{n}(0, . )\|_{L^{2}(0, T)}\to 0$ as $n\to \infty$, where
 $y^{n}=S( \cdot  )y_{0}^{n}$. Using \eqref{e409}
have that $\{y^{n}\}$ is bounded in
$L^{2}(0, T: H^{2}(0, L))\;( \hookrightarrow
L^{2}(0, T: H^{1}(0, L)\;)$. On the other hand,
\begin{equation}
\partial_{t}y^{n}=- (\beta  \partial ^{3}y^{n}
- i \alpha  \partial ^{2}y^{n} + \delta  \partial y^{n})\quad
\mbox{is bounded in } L^{2}(0, T: H^{-2}(0, L)).
\end{equation}
But $H^{1}(0, L)\stackrel{c}\hookrightarrow
L^{2}(0, L)\hookrightarrow H^{-2}(0, L)$, then from Lions-Aubin's
Theorem (see \cite{l7}), the set $\{y^{n}\}$ is relatively compact in
$L^{2}(0, T: L^{2}(0, L))$. Without loss of generality, we may
assume that the sequence $\{y^{n}\}$ is convergent in
$L^{2}(0, T: L^{2}(0, L))$. We infer from \eqref{e404} that
$\{y_{0}^{n}\}$ is a Cauchy sequence in $L^{2}(0, T)$. Let
$ y_{0}={\lim_{n\to 0}y_{0}^{n}} $ and
$ y=S( \cdot  )y_{0}$. By Lemma \ref{lem4.2}, $\partial y^{n}(0, .
 )\to \partial y(0, . ) $ in $ L^{2}(0, T)$. Thus,
$\|y_{0}\|_{L^{2}(0, L)}=1$ and $\partial y(0, . )=0$,
but such function does not exist because of the following lemma.
\end{proof}

\begin{lemma} \label{lem4.4}
For $T>0$ let $\mathcal{F}_{T}$ denote the space
of the initial states $y_{0}\in L^{2}(0, L)$ such that the mild
solution $y=S( \cdot  )y_{0}$ of \eqref{eR2} satisfies
$\partial y(0, . )=0$ in $L^{2}(0, T)$.
Then, for $L\in (0, \infty)\backslash \mathcal{N}$,
$\mathcal{F}_{T}=\{0\}$, for all $T>0$.
\end{lemma}

\begin{proof} It is obvious that if $T<T'$ then
$\mathcal{F}_{T'}\subseteq \mathcal{F}_{T}$.

\noindent\textbf{Claim.} For any $T>0$, $\mathcal{F}_{T}$ is a
finite-dimensional vector space.
In fact, if $\{y_{0}^{n}\}$ is a sequence in the unit ball
$\mathbb{B}_{\mathcal{F}_{T}}
=\{y\in \mathcal{F}_{T}: \|y\|_{L^{2}(0, L)}\leq 1\}$
the same argument as above
shows that there exist a convergent subsequence. Since the unit ball
is compact, by the Riesz Theorem (see \cite{r2})
$\mathcal{F}_{T}$ is finite dimensional and the claim follows.

Let $T'>0$ be given. To prove that $\mathcal{F}_{T'}=\{0\}$, it is
sufficient to find $0<T<T'$ such that $\mathcal{F}_{T}=\{0\}$. Since
the map $T\mapsto \dim(\mathcal{F}_{T})_{n\in \mathbb{N}}$ is
non-increasing, there exist $T, \epsilon>0$ such that
$T<T +\epsilon <T'$ and $\dim  \mathcal{F}_{T}=\dim  \mathcal{F}_{T +
\epsilon}$, where we obtain that $\mathcal{F}_{t}=\mathcal{F}_{T}$ for
$T\leq t\leq T + \epsilon$. Let $y_{0}\in \mathcal{F}_{T}$,
$y=S( \cdot  )y_{0}$ and $0<t<\epsilon$. Since $S(\tau
)(S(t)y_{0})=S(\tau + t)y_{0}$ for $0\leq \tau \leq T$ and $y_{0}\in
\mathcal{F}_{T + \epsilon}$, we see that
\begin{equation}
\label{e413}\frac{S(t)y_{0} - y_{0}}{t}\in \mathcal{F}_{T}
\end{equation}
Let $\mathcal{M}_{T} =\{\widetilde{y}=S(\tau)\widetilde{y}_{0}: 0\leq
\tau \leq T, \widetilde{y}_{0}\in \mathcal{F}_{T}\}\subseteq
C([0, T]: L^{2}(0, L))$. Since $y\in H^{1}(0, T + \epsilon
: H^{-2}(0, L))$,
\begin{equation}
\label{e414}\lim_{t\to 0^{+}}\frac{y(t + \cdot  ) -
y}{t}=y' \quad \mbox{in } L^{2}(0, T: H^{-2}(0, L)).
\end{equation}
On the other hand, by \eqref{e413},
$\frac{y(t + \cdot  ) - y}{t}\in \mathcal{M}_{T}$
for $0<t<\epsilon $ and  $\mathcal{M}_{T}$ is
closed in $L^{2}(0, T: H^{-2}(0, L))$, since
$\dim  \mathcal{M}_{T}<\infty $. It follows that
$y'\in C([0, T]: L^{2}(0, L))$
and $y\in C^{1}([0, T]: L^{2}(0, L))$. Hence, we may write
\[
y'(0) =\lim_{t\to 0^{+}}\frac{S(t)y_{0} - y_{0}}{t} \quad
\mbox{in } L^{2}(0, L).
\]
Then
\begin{equation}
\label{e415}y_{0}\in D(A),\quad A(y_{0})=y'(0)\in \mathcal{
F}_{T}\quad \mbox{and}\quad \partial y(0, . )\in C([0, T]).
\end{equation}
Hence,
\[
\big( \frac{dy_{0}}{dx}\big)_{x=0}=\partial y(0, 0)=0.
\]
If $\mathcal{F}_{T}\neq \{0\}$, the map
$y_{0}\in \mathbb{C}  \mathcal{F}_{T}\mapsto A(y_{0})\in \mathbb{C}
 \mathcal{F}_{T}$ (where $\mathbb{C}  \mathcal{F}_{T}$
 denote the complexification of $\mathcal{F}_{T}$) has at least
one eigenvalue, thus there exist $\lambda \in \mathbb{C}$,
$y_{0}\in H^{3}(0, L)\backslash \{0\}$ such that
\begin{equation}
\lambda  y_{0} = - \beta  y_{0}''' + i \alpha  y_{0}'' - i \delta  y_{0}'\\
y_{0}(0) = y_{0}(L)=y_{0}'(0) = y_{0}'(L)=0.
\end{equation}
We prove in the following Lemma that this does not hold if
$L\in \mathcal{N}$.
\end{proof}

\begin{lemma} \label{lem4.5}
Let $|\alpha |<3 \beta $, $L\in (0, + \infty )$ and
\begin{equation} \label{eK}
\begin{gathered}
\exists  \lambda \in \mathbb{C},\, \exists y_{0}\in H^{3}(0, L)\backslash
\{0\}\quad \mbox{such that}\\
\lambda  y_{0} + \beta  y_{0}''' - i \alpha  y_{0}'' + \delta  y_{0}' = 0, \\
y_{0}(0)=y_{0}(L)=y_{0}'(0)=0,
\end{gathered}
\end{equation}
Then \eqref{eK} is satisfied if and only if $L\in \mathcal{N}$.
\end{lemma}

\begin{proof}
Let $y_{0}\in (\mathbb{K})$, we denote by
$u\in H^{2}(\mathbb{R})$ its prolongation by $0$; i. e.,
\[
u(x) =\begin{cases}
y_{0},& \mbox{if } x\in (0, L)\\
0, & \mbox{if } x\in  (0, L)^{c}.
\end{cases}
 \]
Then
\begin{equation}
\label{e416}\lambda  u + \beta  u''' - i \alpha  u'' + \delta
 u' = \beta  y_{0}''(0) \delta _{0} - \eta  y_{0}''(L) \delta
_{L}\quad \mbox{in } \mathcal{D}'(\mathbb{R}),
\end{equation}
where $\delta_{x_{0}}$ denotes the Dirac measure at $x_{0}$. Is easy
to see that \eqref{eK} is equivalent to the existence of complex
numbers $\mu , \eta ,\lambda$ (with $(\mu, \eta)\neq (0, 0)$) and
of a function $u\in H^{2}(\mathbb{R})$ with compact support in
$[-L, L]$ such that
\begin{equation}
\label{e417}\lambda  u + \beta  u''' - i \alpha  u'' + \delta
 u' = \eta  \delta _{0} - \mu  \delta _{L}\quad \mbox{in }
\mathcal{D}'(\mathbb{R}).
\end{equation}
Taking Fourier transform we have
\[
( \lambda + \beta  (i \xi)^{3} - i \alpha  (i \xi)^{2} +
\delta  (i \xi) ) \widehat{u}(\xi)=\eta  - \mu
 e^{- i L \xi};
\]
hence setting $\lambda = - i p$, we obtain
\[
\widehat{u}(\xi)=i\; \big[ \frac{\eta  - \mu
 e^{- i L \xi}}{\beta  \xi^{3} - \alpha  \xi^{2} - \delta
 \xi + p} \big].
\]
Using Paley-Wiener's theorem (see \cite{r2}) and the usual
characterization of $H^{2}(\mathbb{R})$ functions by means of their
Fourier transform, we see that \eqref{eK} is equivalent to the
existence of $p\in \mathbb{C}$ and $(\eta , \mu)\in
\mathbb{C}^{2}\backslash \{(0, 0)\}$ such that the map
\[
f(\xi)=\Big[ \frac{\eta  - \mu  e^{- i L \xi}}{\beta
 \xi^{3} -  \alpha  \xi^{2} - \delta  \xi + p} \Big]
\]
satisfies
\begin{itemize}
\item[(1)] $f$ is an entire function in $\mathbb{C}$
\item[(2)] $\int_{\mathbb{R}}| f(\xi) |^{2} ( 1 + | \xi  |^{2} )^{2}\,d\xi
<\infty $
\item[(3)] For all $\xi \in \mathbb{C}$,
$| f(\xi) | \leq C( 1 + | \xi  | )^{N} e^{L | \mathop{\rm Im} \xi  |}$
for some positive constants $C, N$.
\end{itemize}
Since the roots of $\eta - \mu  e^{- i L \xi}$ are simple unless
$\eta = \mu = 0$, (1) holds provided that the roots of
$\beta \xi^{3} - \alpha  \xi^{2} - \delta  \xi + p$ are
simple, and the roots of
 $\eta - \mu e^{- i L \xi}$. We have that if (1) holds,
 then (2) and (3) are satisfied. It follows that \eqref{eK}
 is equivalent to the existence of complex number $p, \mu_{0}$
 and of positive integers $k$, $l$, $m$, and $n$ such that, if we set
\[
\mu_{1}=\mu_{0} + k \frac{2 \pi}{L},\quad
\mu_{2}=\mu_{1} + l \frac{2 \pi}{L} = \mu_{0} + (k + l)
\frac{2 \pi}{L}
\]
we have
\[
\xi^{3} - \frac{\alpha}{\beta} \xi^{2} - \frac{\delta }{\beta} \xi
+ \frac{1}{\beta} p = (\xi - \mu_{0}) (\xi - \mu_{1}) (\xi -
\mu_{2})
\]
that is
\begin{gather*}
\mu_{0} + \mu_{1} + \mu_{2} = \frac{\alpha}{\beta}, \\
\mu_{0} \mu_{1} + \mu_{0} \mu_{2} + \mu_{1} \mu_{2} =- \frac{\delta }{\beta}, \\
\mu_{0} \mu_{1} \mu_{2} = \frac{1}{\beta} p\,.
\end{gather*}
Straightforward calculus leads to
\begin{gather*}
L =2 \pi \beta
\sqrt{\frac{k^{2} + k l + l^{2}}{3 \beta  \delta + \alpha^{2}}},\\
\mu_{0}=\frac{1}{3}\left[\frac{\alpha}{\beta} - ( 2 k + l  )
\frac{2 \pi}{L}\right],\\
p=\beta  \mu_{0} \big(\mu_{0} + k \frac{2 \pi}{L}\big)
\big(\mu_{0} + (k + l) \frac{2 \pi}{L}\big).
\end{gather*}
Hence, \eqref{eK} is satisfied if and only if $L\in \mathcal{N}$.
This complete the proof of Lemmas \ref{lem4.3}, \ref{lem4.4}, and
\ref{lem4.5}.
\end{proof}

\begin{remark} \label{rmk4.2}
For $L\in \mathcal{N}$, if $p$ is given as above and
$y_{0}$ (with $\mathop{\rm Re}y_{0}\neq 0$) is as in
\eqref{eK} with $\lambda = - i p$, then
$y(x, t)=\mathop{\rm Re} ( e^{- i p t} y_{0}(x) )$ is a nontrivial
smooth solution of \eqref{eR2} such that
 $\partial y(0, . )\equiv 0$.
Thus, the result in Lemma \ref{lem4.3}. holds if and only if $L\notin \mathcal{N}$.
\end{remark}

The goal of the following lemma is to define in a certain weak sense
a solution of the non-homogeneous problem $(R_{1})$.

\begin{lemma} \label{lem4.6}
Let $|\alpha|<3 \beta $. There exists a unique linear continuous
map $\Pi :L^{2}(0, L)\times L^{2}(0, T)\mapsto \mathbb{H} $ such that,
for $y_{0}\in D(A)$ and $h\in C^{2}([0, T])$ with $h(0)=0$,
$\Pi(y_{0}, h)$ is the unique classical solution of \eqref{eR1}.
\end{lemma}

\begin{proof} We assume here that $y_{0}\in D(A)$ and
$h\in C_{0}^{2}([0, T])=\{h\in C^{2}([0, T]: \mathbb{R}): h(0)=0\}$.
Let $\phi \in C^{\infty}([0, L])$ be such that $\phi(0)=\phi(L)=0$
and $\phi'(L)=-1$. Then the change of function $z(x, t)=y(x, t) -
(S(t)y_{0})(x) + h(t) \phi(x)$ transforms \eqref{eR1} into
\begin{equation} \label{eR3}
 \begin{gathered}
\partial _{t}z + \beta  \partial^{3}z - i \alpha  \partial^{2}z +
\delta  \partial z= h'(t) \phi(x) + h(t) [\beta  \phi''' -
i \alpha  \phi'' + \delta  \phi' ]
=f(x, t) \\
z(0, t) = z(L, t) = 0\\
\partial z(L, t)=0\\
z( . , 0)= 0
\end{gathered}
\end{equation}
Using Lemma \ref{lem4.1}. and that $f\in C^{1}([0, T]: L^{2}(0, L))$, we obtain
 that there exists a unique solution (see \cite{p1}) for the
non-homogeneous problem $z\in C([0, T]: D(A))\cap C^{1}([0, T]: L^{2}(0, L))$
of \eqref{eR3}.
Hence, for smooth data $y_{0}\in D(A)$, $h\in C_{0}^{2}([0, 1])$,
\eqref{eR1} admits a unique classical solution
\[
\label{e418}y\in C([0, T]: H^{3}(0, L))\cap
C^{1}([0, T]: L^{2}(0, L)).
\]
On the other hand, we assume that $y_{0}\in D(A)$, $h\in
C_{0}^{2}([0, T])$. Let $\xi =\xi(x, t)\in
C^{\infty}([0, T]\times [0, L])$. From equation \eqref{eR1} we have
(multiplying by $i$)
\begin{equation}
i \partial_{t}y + i \beta  \partial ^{3}y + \alpha  \partial
^{2}y + i \delta  \partial y=0.
\end{equation}
Multiplying by $\xi  \overline{y}$ we obtain
\begin{gather*}
i \xi  \overline{y} \partial_{t}y + i \beta  \xi
 \overline{y} \partial ^{3}y + \alpha  \xi
 \overline{y} \partial
^{2}y + i \delta  \xi  \overline{y} \partial y  =  0,\\
- i \xi  y \partial_{t}\overline{y} - i \beta  \xi
 y \partial ^{3}\overline{y} + \alpha  \xi  y \partial
^{2}\overline{y} - i \delta  \xi  y \partial \overline{y}  =
0,
\end{gather*}
(applying conjugate).
Subtracting and integrating over $x\in [0, L]$ we obtain
\begin{equation} \label{e419}
\begin{aligned}
& i \partial_{t}\int_{0}^{L}\xi  |y|^{2}\,dx -
i \int_{0}^{L}\partial_{t}\xi |y|^{2}\,dx + i \beta
\int_{0}^{L}\xi  \overline{y} \partial ^{3}y\,dx
+ i \beta \int_{0}^{L}\xi  y \partial ^{3}\overline{y}\,dx  \\
& +\alpha \int_{0}^{L}\xi
 \overline{y} \partial ^{2}y\,dx - \alpha \int_{0}^{L}\xi
 y \partial ^{2}\overline{y}\,dx - i \delta \int_{0}^{L}\partial
\xi  |y|^{2}\,dx = 0.
\end{aligned}
\end{equation}
Each term is treated separately. Integrating by parts
\begin{gather*}
\begin{aligned}
\int_{0}^{L}\xi  \overline{y} \partial ^{3}y\,dx
& = \int_{0}^{L}\partial ^{2}\xi \overline{y} \partial y\,dx +
2\int_{0}^{L}\partial \xi |\partial y|^{2}\,dx \\
&\quad +\xi(L, t) |h(t)|^{2}
 - \xi(0, t) |\partial y(0, t)|^{2}
+ \int_{0}^{L}\xi  \partial y \partial ^{2}\overline{y}\,dx,
\end{aligned}
\\
\int_{0}^{L}\xi  y \partial ^{3}\overline{y}\,dx
 = \int_{0}^{L}\partial ^{2}\xi y \partial \overline{y}\,dx +
\int_{0}^{L}\partial \xi |\partial y|^{2}\,dx - \int_{0}^{L}\xi
 \partial y \partial ^{2}\overline{y}\,dx,\\
\int_{0}^{L}\xi  \overline{y} \partial ^{2}y\,dx
 = -\int_{0}^{L}\partial \xi \overline{y} \partial y\,dx -
\int_{0}^{L}\xi
 |\partial y|^{2}\,dx,\\
\int_{0}^{L}\xi  y \partial ^{2}\overline{y}\,dx
 = -\int_{0}^{L}\partial \xi y \partial \overline{y}\,dx -
\int_{0}^{L}\xi  |\partial y|^{2}\,dx\,.
\end{gather*}
Then
\begin{align*}
& i \partial_{t}\int_{0}^{L}\xi  |y|^{2}\,dx -
i\int_{0}^{L}\partial_{t}\xi |y|^{2}\,dx + i \beta
\int_{0}^{L}\partial ^{2}\xi \overline{y} \partial y\,dx +
2 i \beta
\int_{0}^{L}\partial \xi |\partial y|^{2}\,dx\\
&  +\beta  \xi(L, t) |h(t)|^{2} - i \beta
 \xi(0, t) |\partial y(0, t)|^{2} + i \beta \int_{0}^{L}\xi
 \partial y \partial ^{2}\overline{y}\,dx +  i \beta
\int_{0}^{L}\partial ^{2}\xi y \partial \overline{y}\,dx \\
&  +i \beta \int_{0}^{L}\partial \xi |\partial y|^{2}\,dx -
i \beta \int_{0}^{L}\xi  \partial y \partial ^{2}\overline{y}\,dx
- \alpha \int_{0}^{L}\partial \xi \overline{y} \partial y\,dx -
\alpha
\int_{0}^{L}\xi  |\partial y|^{2}\,dx \\
&  +\int_{0}^{L}\partial \xi y \partial \overline{y}\,dx +
\int_{0}^{L}\xi  |\partial y|^{2}\,dx - i \delta
\int_{0}^{L}\partial \xi  |y|^{2}\,dx =0\,.
\end{align*}
Hence,
\begin{align*}
& i \partial_{t}\int_{0}^{L}\xi  |y|^{2}\,dx -
i\int_{0}^{L}\partial_{t}\xi |y|^{2}\,dx + i \beta
\int_{0}^{L}\partial ^{2}\xi \partial (|y|^{2})\,dx + 3 i \beta
\int_{0}^{L}\partial \xi |\partial y|^{2}\,dx\\
& +\beta  \xi(L, t) |h(t)|^{2} - i \beta
 \xi(0, t) |\partial y(0, t)|^{2} - 2 i \alpha
 \mathop{\rm Im}\int_{0}^{L}\partial \xi \overline{y} \partial y\,dx -
i \delta \int_{0}^{L}\partial \xi  |y|^{2}\,dx\\
& =0\,.
\end{align*}
Thus
\begin{align*}
&  \partial_{t}\int_{0}^{L}\xi  |y|^{2}\,dx -
\int_{0}^{L}\partial_{t}\xi |y|^{2}\,dx -  \beta
\int_{0}^{L}\partial ^{3}\xi |y|^{2}\,dx + 3 \beta
\int_{0}^{L}\partial \xi |\partial y|^{2}\,dx\\
&  +\beta  \xi(L, t) |h(t)|^{2} - \beta
 \xi(0, t) |\partial y(0, t)|^{2} - \delta \int_{0}^{L}\partial
\xi
 |y|^{2}\,dx\\
&  = 2 \alpha  \mathop{\rm Im} \int_{0}^{L}\partial
\xi \overline{y} \partial y\,dx\leq |\alpha |\int_{0}^{L}\partial
\xi |y|^{2}\,dx + |\alpha |\int_{0}^{L}\partial \xi |\partial
y|^{2}\,dx\,,
\end{align*}
where
\begin{align*}
&\partial_{t}\int_{0}^{L}\xi  |y|^{2}\,dx +
\int_{0}^{L}[ 3 \beta - |\alpha | ] \partial \xi |\partial
y|^{2}\,dx - \int_{0}^{L}\partial_{t}\xi |y|^{2}\,dx - \beta
\int_{0}^{L}\partial ^{3}\xi |y|^{2}\,dx\\
&+\beta  \xi(L, t) |h(t)|^{2} - \beta
 \xi(0, t) |\partial y(0, t)|^{2} - \delta \int_{0}^{L}\partial
\xi  |y|^{2}\,dx - |\alpha |\int_{0}^{L}\partial \xi |y|^{2}\,dx
\leq 0.
\end{align*}
Integrating over $t\in [0, T]$ we have
\begin{equation} \label{e420}
\begin{aligned}
& \int_{0}^{L}\xi  |y|^{2}\,dx +
\int_{0}^{t}\int_{0}^{L}[ 3 \beta - |\alpha | ] \partial
\xi |\partial y|^{2}\,dx\,ds
 -\int_{0}^{t}\int_{0}^{L}\partial_{t}\xi |y|^{2}\,dx\,ds \\
&  - \beta \int_{0}^{t}\int_{0}^{L}\partial^{3}\xi |y|^{2}\,dx\,ds
 + \beta \int_{0}^{t}\xi(L, s) |h(s)|^{2}\,ds \\
&- \beta\int_{0}^{t}\xi(0, s) |\partial y(0, s)|^{2}\,ds  \\
&- \delta \int_{0}^{t}\int_{0}^{L}\partial \xi
 |y|^{2}\,dx\,ds - |\alpha |\int_{0}^{t}\int_{0}^{L}\partial
 \xi |y|^{2}\,dx\,ds \\
& \leq \int_{0}^{L}\xi(x, 0)  |y_{0}|^{2}\,dx.
\end{aligned}
\end{equation}
Choosing $\xi(x, t)=-1$ leads to
\[
 \int_{0}^{L}|y|^{2}\,dx - \beta \int_{0}^{t}|h(s)|^{2}\,ds +
\beta \int_{0}^{t}|\partial y(0, s)|^{2}\,ds \leq
\int_{0}^{L}|y_{0}|^{2}\,dx,
\]
where
\[
\|y\|_{L^{2}(0, L)}^{2} + \beta  \|\partial
y(0, . )\|_{L^{2}(0, T)}^{2}\leq \|y_{0}\|_{L^{2}(0, L)}^{2} +
\beta  \|h\|_{L^{2}(0, T)}^{2}.
\]
Setting $\|(y_{0}, h)\|=\big[ \|y_{0}\|_{L^{2}(0, L)}^{2} +
\beta  \|h\|_{L^{2}(0, T)}^{2} \big ]^{1/2}$, we get
\begin{equation}
\label{e421}\|y\|_{C([0, T]: L^{2}(0, L))} \leq \|(y_{0}, h)\|
\end{equation}
which yields
\begin{equation}
\label{e422}\|y\|_{L^{2}([0, T]\times (0, L))} \leq
\sqrt{T}\;\|(y_{0}, h)\|.
\end{equation}
Now, we take $\xi(x, t)=x$, and $t=T$ in \eqref{e420}
\begin{align*}
&\int_{0}^{L}x |y(x, T)|^{2}\,dx +
\int_{0}^{T}\int_{0}^{L}[ 3 \beta - |\alpha
| ] |\partial y|^{2}\,dx\,dt \\
&+ \beta  L\int_{0}^{T}|h(s)|^{2}\,ds
 - \delta \int_{0}^{T}\int_{0}^{L}|y|^{2}\,dx\,dt - |\alpha
|\int_{0}^{T}\int_{0}^{L}|y|^{2}\,dx\,dt \\
& \leq \int_{0}^{L}x  |y_{0}|^{2}\,dx\,,
\end{align*}
where
\begin{align*}
&\int_{0}^{L}x |y(x, T)|^{2}\,dx +
\int_{0}^{T}\int_{0}^{L}[ 3 \beta - |\alpha | ] |\partial
y|^{2}\,dx\,dt   \\
&   \leq  \int_{0}^{L}x  |y_{0}|^{2}\,dx  - \beta
 L\int_{0}^{T}|h(s)|^{2}\,ds + ( \delta
 + |\alpha | )\int_{0}^{T}\int_{0}^{L}|y|^{2}\,dx\,dt  \\
&  \leq ( \delta
 + |\alpha | )\int_{0}^{T}\int_{0}^{L}|y|^{2}\,dx\,dt +
 L\Big(\int_{0}^{L}|y_{0}|^{2}\,dx + \beta
\int_{0}^{T}|h(s)|^{2}\,ds\Big) \\
& =( \delta
 + |\alpha | ) \|y\|_{L^{2}([0, T]\times (0, L))}^{2}
 + L\big(\|y_{0}\|_{L^{2}(0, L)}^{2} +\beta  \|h\|_{L^{2}(0, T)}^{2}\big) \\
& =( \delta  + |\alpha | ) T \|(y_{0}, h)\|^{2}
  + L\big(\|(y_{0}, h)\|^{2}\big) \\
& =[( \delta + |\alpha | ) T +  L]\|(y_{0}, h)\|^{2}\,.
\end{align*}
Then
\[
[ 3 \beta - |\alpha | ]\int_{0}^{T}\int_{0}^{L}|\partial
y|^{2}\,dx\,dt\leq [( \delta
 + |\alpha | ) T +  L]\|(y_{0}, h)\|^{2},
\]
where
\begin{equation} \label{e423}
\|\partial y\|_{L^{2}(0, T: L^{2}(0, L))}^{2}
\big[\frac{( \delta  + |\alpha | ) T + L}{3 \beta
- |\alpha|}]\|(y_{0}, h)\|^{2}\,.
\end{equation}
Adding \eqref{e422} and \eqref{e423} and using the fact that
$|\alpha|<3 \beta$ we obtain
\begin{equation}
\label{e424}\|y\|_{L^{2}(0, T: H^{1}(0, L))}\leq
\big[\frac{( \delta  + 3 \beta ) T + L}{3 \beta - |\alpha
|}\big]^{1/2}\|(y_{0}, h)\|.
\end{equation}
Using \eqref{e421} and \eqref{e424}, and the density of $D(A)$
in $L^{2}(0, L)$ and of $C_{0}^{2}([0, T])$ in $L^{2}(0, T)$, we
see that the linear map $(y_{0}, h)\in D(A)\times
C_{0}^{2}([0, T])\mapsto y\in \mathbb{H}$ may be extended in a
unique manner to the whole space $L^{2}(0, T)\times L^{2}(0, L)$
to give a linear map
$\Pi :L^{2}(0, T)\times L^{2}(0, L)\mapsto \mathbb{H}$.
\end{proof}

\begin{remark} \label{rmk4.3}
(a) For $y_{0}\in L^{2}(0, L)$ and $h\in
L^{2}(0, T)$, the weak solution $\Pi(y_{0}, h)$ is solution of
\eqref{eR1} in $\mathcal{D}'(0, T: H^{-2}(0, L))$. Moreover,
$\Pi(y_{0}, h)( . , 0)=y_{0}$ and $\Pi(y_{0}, h)( . , T)$
are well-defined in $L^{2}(0, L)$, since
$\Pi(y_{0}, h)\in C([0, T]: L^{2}(0, L))$.

\noindent(b) $\Pi(y_{0}, 0)=S( \cdot  )y_{0}$, hence,
$\Pi(y_{0}, 0)=S( \cdot  )y_{0} + \Pi(0, h)$.
\end{remark}

To apply the Hilbert uniqueness method, we need some observability result concerning the
following backward well-posed homogeneous problem:
 For $|\alpha |<3 \beta$ and $\delta >0$
\begin{equation} \label{eR4}
\begin{gathered}
\partial _{t}u + \beta  \partial^{3}u - i \alpha  \partial^{2}u
+ \delta  \partial u=0, \\
u(0, t) = u(L, t) = 0, \\
\partial u(0, t)=0 ,\\
u(T, 0)= u_{T}(x)\,.
\end{gathered}
\end{equation}
The change of variables $\tau T - t$ and $\zeta = L - x$
transform \eqref{eR4} into \eqref{eR2} and vice-versa.
Using Lemmas \ref{lem4.1}, \ref{lem4.2} and \ref{lem4.3},
we readily get the following result.

\begin{lemma}]Observability result] \label{lem4.7}
Let $L, T>0$, $|\alpha |<3 \beta$ and $\delta>0$. For any
$u_{T}\in L^{2}(0, L)$
the mild solution of \eqref{eR4} belongs to $\mathbb{H}$, the function
$\partial u(L, . )$ makes sense in $L^{2}(0, T)$. If moreover,
$L\notin \mathcal{N}$, there exists a constant $C=C(L, T)>0$ such that
for any $u_{T}\in L^{2}(0, L)$ we have that
\begin{equation} \label{e425}
\|\partial u(L, . )\|_{L^{2}(0, T)} \leq
\|u_{T}\|_{L^{2}(0, T)}\leq C\;\|\partial
u(L, . )\|_{L^{2}(0, T)}.
\end{equation}
\end{lemma}

It remains to apply the Hilbert uniqueness method.

\begin{theorem} \label{thm4.1}
Let $|\alpha |<3 \beta $, $\delta>0$ and
\[
\mathcal{N}=\big\{2 \pi \beta   \sqrt{\frac{k^{2} + k l +
l^{2})}{3 \beta  \delta + \alpha^{2}}} : k, l\in
\mathbb{N}^{*}\big\}.
\]
Then, for any $T>0$ and $L\in (0, + \infty )\backslash \mathcal{N}$,
and for any $y_{0}$, $y_{T}\in L^{2}(0, L)$, there exists $h\in
L^{2}(0, T)$ such that the mild solution $y\in
C([0, T]: L^{2}(0, L))\cap L^{2}(0, T: H^{1}(0, L))$ of
\begin{gather}
\label{e426} \partial_{t}y + \beta  \partial^{3}y - i \alpha
 \partial^{2}y + \delta  \partial y = 0\\
\label{e427} y(0, t)=y(L, T)=0\\
\label{e428} \partial y(L, t)=h(t)\\
\label{e429} y(x, 0)=y_{0}(x)
\end{gather}
 satisfies $y( . , T)=y_{T}$.
\end{theorem}

\begin{proof} By Remark \ref{rmk4.3} (b) we may assume, without loss of
generality, that $y_{0}=0$. (see the proof of Theorem \ref{thm3.1}) Let
$(u_{T}, h)\in C_{c}^{\infty}(0, L)\times C_{c}^{\infty}(0, L)$,
let $u$ (resp. $y$) be the classical solution of
\eqref{e426}-\eqref{e429} (resp. $(R_{1}))$. Multiplying
\eqref{e426} by $u$ and integrating over $x\in [0, L]$ we have
\begin{equation}
\label{e430}\int_{0}^{L}u \partial_{t}y\,dx + \beta
\int_{0}^{L}u \partial^{3}y\,dx - i \alpha
\int_{0}^{L}u \partial^{2}y\,dx + \delta \int_{0}^{L}u \partial
y\,dx = 0
\end{equation}
Each term is treated separately. Integrating by parts,
\begin{gather*}
\int_{0}^{L}u \partial _{t}y\,dx
 = \partial_{t}\int_{0}^{L}u y\,dx - \int_{0}^{L}\partial
_{t}u y\,dx,\\
\int_{0}^{L}u \partial^{3}y\,dx  =  - \partial u(L, t) h(t) -
\int_{0}^{L}\partial^{3}u y\,dx,\\
\int_{0}^{L}u \partial^{2}y\,dx  =
\int_{0}^{L}\partial^{2}u y\,dx,\\
\int_{0}^{L}u \partial y\,dx  =  -\int_{0}^{L}\partial u y\,dx\,.
\end{gather*}
Then in \eqref{e430} we obtain
\begin{align*}
&\partial_{t}\int_{0}^{L}u y\,dx - \int_{0}^{L}\partial _{t}u y\,dx -
\beta  \partial u(L, t) h(t) - \beta
\int_{0}^{L}\partial^{3}u y\,dx \\
&- i \alpha \int_{0}^{L}\partial^{2}u y\,dx - \delta \int_{0}^{L}\partial
u y\,dx = 0
\end{align*}
where
\begin{equation} \label{e431}
\begin{aligned}
\partial_{t}\int_{0}^{L}u y\,dx - \beta  \partial u(L, t) h(t)
&= - 2 i \alpha\int_{0}^{L}\partial u \partial y\,dx\\
&\leq  |\alpha |\int_{0}^{L}|\partial u|^{2}\,dx + |\alpha
|\int_{0}^{L}|\partial y|^{2}\,dx\,.
\end{aligned}
\end{equation}
Integrating over $t\in [0, T]$ and using that $y_{0}=0$ we obtain
\begin{equation} \label{e432}
\begin{aligned}
&\int_{0}^{L}u_{T}(x) y(x, T)\,dx \\
& \leq  \beta \int_{0}^{T}\partial u(L, t) h(t)\,dt + |\alpha
|\int_{0}^{T}\int_{0}^{L}|\partial u|^{2}\,dx\,dt
+|\alpha |\int_{0}^{T}\int_{0}^{L}|\partial
y|^{2}\,dx\,dt
\end{aligned}
\end{equation}
By a density argument we see that \eqref{e432} holds for
$u_{T}\in L^{2}(0, L)$ and $h\in L^{2}(0, T)$. Let $\Lambda $ denote the
linear continuous map
$\Lambda :L^{2}(0, L)\mapsto L^{2}(0, L)$ with
$u_{T}\mapsto \Lambda (u_{T}) = y( . ,T)$
and $y$ standing for the solution of \eqref{eR1} associated with the
data $h( . )=\partial  u(L, . )\in L^{2}(0, T)$. It follows
 \eqref{e432} and by Lemma \ref{lem4.7} that
\[
\langle \Lambda (u_{T}), u_{T}\rangle _{L^{2}(0, L)}=\|\partial
 u(L, . )\|_{L^{2}(0, T)}^{2}\geq
C^{-2}\;\|u_{T}\|_{L^{2}(0, L)}^{2}.
\]
Therefore, by Lax-Milgram's theorem (see \cite{y1}), $\Lambda $ is
invertible. The proof is complete.
\end{proof}

\begin{remark} \label{rmk4.4}
When $y_{0}=0$, the Hilbert uniqueness  method yields $u$, a linear
continuous selection of the control, namely the map
$\Gamma_{0}:L^{2}(0, L)\mapsto L^{2}(0, T)$ with
$y_{T}\mapsto \Gamma_{0}(y_{T}) =\partial u(L, . )$ where $u$
denotes the solution of \eqref{eR4} associated with $u_{T}=\Lambda
^{-1}(y_{T})$.
\end{remark}

 \section{Exact boundary controllability for
a higher order nonlinear Schr\"{o}dinger equation with constant
coefficients on a bounded domain}

In this section we prove that the
following boundary-control system (for $|\alpha |<3 \beta $ and
$\delta>0$)
\begin{equation} \label{eQ11}
\begin{gathered}
\partial _{t}y + \beta  \partial^{3}y
- i \alpha  \partial^{2}y -  i |y|^{2} y + \delta  \partial y= 0 \\
y(0, t) = y(L, t) = 0\\
\partial y(L, t)=h(t),\quad h\in L^{2}(0, T)\\
y( . , 0)= y_{0}
\end{gathered}
\end{equation}
is exactly controllable in a neighborhood of the null
state. More precisely we show that for any $L>0$ and $T>0$ there
exists a radius $r_{0}>0$ such that for every
$y_{0}, y_{T}\in L^{2}(0, L)$ with $\|y_{0}\|_{L^{2}(0, L)}<r_{0}$,
$\|y_{T}\|_{L^{2}(0, L)}<r_{0}$ we may find
$y\in \mathbb{H}=C([0, T]: L^{2}(0, L))\cap L^{2}(0, T: H^{1}(0, L))$
such that
\begin{itemize}
\item[(1)] $\partial _{t}y =- ( \beta  \partial^{3}y - i \alpha
 \partial^{2}y -  i |y|^{2} y +
\partial y )$ in $\mathcal{D}'(0, T: H^{-2}(0, L))$.
\item[(2)] $y( . , 0)=y_{0}$, $y( . ,T)=y_{T}$.
\end{itemize}

\begin{remark} \label{rmk5.1} \rm
 For $ y\in \mathbb{H}$,
$ \partial y \in L^{2}(0, T: L^{2}(0, L))$,
 $ \partial^{3}y\in L^{2}(0, T: H^{-2}(0, L)) $,
and $ |y|^{2} y\in L^{1}(0, T: L^{2}(0, L)) $.
 Hence,
\[
\partial_{t}y=- (\beta
 \partial^{3}y - i \alpha  \partial^{2}y - i |y|^{2} y + \delta
 \partial y)\in L^{1}(0, T: H^{-2}(0, L));
\]
i. e., $ y\in W^{1, 1}(0, T: H^{-2}(0, L))$.
\end{remark}

To solve \eqref{eQ11}, we write $y=S(t)y_{0} + y_{1} + y_{2}$ where
$(S(t))_{t\geq 0}$ denotes the semi-group associated with the
operator $A$ of section 4, $y_{1}$ and $y_{2}$ are respectively
solutions of the two nonhomogeneous problems:
\begin{equation} \label{eQ22}
\begin{gathered}
\partial _{t}y_{1} + \beta  \partial^{3}y_{1}
- i \alpha  \partial^{2}y_{1} + \delta  \partial y_{1}= 0 \\
y_{1}(0, t) = y_{1}(L, t) = 0\\
\partial y_{1}(L, t)=h(t)\\
y_{1}( . , 0)= y_{0}
\end{gathered}
\end{equation}
and
\begin{equation} \label{eQ33}
\begin{gathered}
\partial _{t}y_{2} + \beta  \partial^{3}y_{2}
- i \alpha  \partial^{2}y_{2} + \delta  \partial y_{2}= f \\
y_{2}(0, t) = y_{2}(L, t) = 0 \\
\partial y_{2}(L, t)=0 \\
y_{2}( . , 0)= 0 \,.
\end{gathered}
\end{equation}
 In \eqref{eQ33} we have the set $f=- |y|^{2} y$.
Let $\Gamma_{1}:h\in L^{2}(0, T)\mapsto y_{1}\in \mathbb{H}$ be the
map which associates the weak solution of \eqref{eQ22} with $h$.
By Lemma \ref{lem4.6}, $\Gamma_{1}$ is a linear continuous map.

\begin{lemma} \label{lem5.1}
 For $ |\alpha |<3 \beta$, we have
\begin{itemize}
\item[(1)] If $y\in L^{2}(0, T: H^{1}(0, L))$, $|y|^{2} y\in
L^{1}(0, T: L^{2}(0, L))$ and the map
$y\mapsto |y|^{2} y$ is continuous.
\item[(2)] For $f\in L^{1}(0, T: L^{2}(0, L))$ the mild solution
$y_{2}$ of \eqref{eQ33} belong to $\mathbb{H}$. Moreover the linear map
$\Gamma_{2}:f\mapsto y_{2}$ is continuous.
\end{itemize}
\end{lemma}

We remark that for $f\in L^{1}(0, T: L^{2}(0, L))$ the mild
solution $y_{2}$ of \eqref{eQ33} is given by
\[
y_{2}( . , t)=\int_{0}^{t}S(t - s) f( . , s)\,ds
\]
\begin{proof}[Proof of Lemma \ref{lem5.1}]
 (1) Let $y, z\in L^{2}(0, T: H^{1}(0, L))$.
Let $\mathcal{H}_{1}$ be the norm of the Sobolev embedding
$H^{1}(0, L)\hookrightarrow L^{2}(0, L)$. We have
\[
|y|^{2} y - |z|^{2} z = (|y|^{2} - |z|^{2}) y + |z|^{2} (y - z)=
(|y| - |z|) (|y| + |z|) y + |z|^{2} (y - z);
\]
hence
\begin{align*}
\big||y|^{2} y - |z|^{2} z\big|
& =  \big| |y| - |z|\big| (|y| + |z|) |y| + |z|^{2} |y - z|\\
& \leq |y - z| (|y| + |z|) |y| + |z|^{2} |y - z|\,.
\end{align*}
Applying the triangular inequality and Holder's inequality,
\begin{equation} \label{e501}
\begin{aligned}
\| |y|^{2} y - |z|^{2} z\|_{L^{1}(0, T: L^{2}(0, L))}
&\leq \int_{0}^{T}\| |(y - z)( . , t)| (|y| + |z|) |y|\|_{L^{2}(0, L)}\,dt\\
&\quad +\int_{0}^{T}\|\;|z|^{2} |(y - z)( . , t)|\;\|_{L^{2}(0, L)}\,dt \\
& \leq \int_{0}^{T}\|y\|_{L^{\infty}(0, L)}^{2}
\|z\|_{L^{\infty}(0, L)} \| (y - z)( . , t) \|_{L^{2}(0, L)}\,dt \\
&\quad +\int_{0}^{T}\|z\|_{L^{\infty}(0, L)}^{2}
\| (y - z)( . , t) \|_{L^{2}(0, L)}\,dt \\
& \leq  \mathcal{H}_{1}\int_{0}^{T}\| (y - z)( . , t) \|_{L^{2}(0, L)}\,dt \\
& \leq  \mathcal{H}_{1} \| (y -z)( . , t) \|_{L^{2}(0, T: H^{1}(0, L))}.
\end{aligned}
\end{equation}
Choosing $z=0$ yields $|y|^{2} y\in L^{1}(0, T: L^{2}(0, L))$, and
\eqref{e501} with $z$ tending to $y$ gives the continuity of the map
$|y|^{2} y$. \smallskip

\noindent(2) Since
\[
\|1_{[0, t]}(s) S(t - s) f( . , s)\|_{L^{2}(0, L)}\leq
\|f( . , s)\|_{L^{2}(0, L)},
\]
using Lebesgue's Theorem, the mild solution
$ y_{2}( . , t)=\int_{0}^{t}S(t - s) f( . , t)\,ds $
belongs to $ C([0, T]: L^{2}(0, L))$.
Moreover, for every $t\in [0, T]$,
\begin{equation} \label{e502}
\|y_{2}( . , t)\|_{L^{2}(0, L)}\leq
\int_{0}^{t}\|f( . , s)\|_{L^{2}(0, L)}\,ds \leq
\|f\|_{L^{1}(0, T: L^{2}(0, L))}
\end{equation}
so the linear map $f\in L^{1}(0, T: L^{2}(0, L))
\mapsto y_{2}\in C([0, T]: L^{2}(0, L))$ is continuous.
To show that this map is well-defined and continuous from
$L^{1}(0, T: L^{2}(0, L))$ into $L^{2}(0, T: H^{1}(0, L))$, it
is clearly sufficient to prove that
there exists $c_{2}>0$ such that for for all
$f\in C^{1}([0, T]: L^{2}(0, L))$,
\[
\|\partial y_{2}\|_{L^{2}((0, T)\times (0, L))} \leq
c_{2} \|f\|_{L^{1}(0, T: L^{2}(0, L))}.
\]
In fact, multiplying \eqref{eQ33} by $i x \overline{y}_{2}$,
\begin{gather*}
i x \overline{y}_{2} \partial_{t}y_{2} + i \beta
 x \overline{y}_{2} \partial^{3}y_{2} + \alpha
 x \overline{y}_{2} \partial^{2}y_{2} + i \delta
 x \overline{y}_{2} \partial y_{2}
 =  i x \overline{y}_{2} f ,\\
- i x y_{2} \partial_{t}\overline{y}_{2} - i \beta
 x y_{2} \partial^{3}\overline{y}_{2} + \alpha
 x y_{2} \partial^{2}\overline{y}_{2}
 - i \delta  x y_{2} \partial \overline{y}_{2}
  =  - i x y_{2} f
\end{gather*}
(applying conjugate).
Subtracting and integrating over $x\in [0, L]$ we have
\begin{align*}
& i\partial_{t}\int_{0}^{L}x |y_{2}|^{2}\,dx + i \beta
\int_{0}^{L}x \overline{y}_{2} \partial^{3}y_{2}\,dx + i  \beta
\int_{0}^{L}x y_{2} \partial^{3}\overline{y}_{2}\,dx \\
&+ \alpha \int_{0}^{L}x \overline{y}_{2} \partial^{2}y_{2}\,dx
- \alpha \int_{0}^{L}x y_{2} \partial^{2}\overline{y}_{2}\,dx - i \delta
\int_{0}^{L}|y_{2}|^{2}\,dx \\
&= 2 i \mathop{\rm Re}\int_{0}^{L}x \overline{y}_{2} f\,dx\,.
\end{align*}
Each term is treated separately. Integrating by parts,
\begin{gather*}
\int_{0}^{L}x \overline{y}_{2} \partial^{3}y_{2}\,dx =
2\int_{0}^{L}|\partial y_{2}|^{2}\,dx + \int_{0}^{L}x \partial
y_{2} \partial \overline{y}_{2}\,dx\,,
\\
\int_{0}^{L}x y_{2} \partial^{3}\overline{y}_{2}\,dx =
\int_{0}^{L}|\partial y_{2}|^{2}\,dx - \int_{0}^{L}x \partial
y_{2} \partial \overline{y}_{2}\,dx\,,
\\
\int_{0}^{L}x \overline{y}_{2} \partial^{2}y_{2}\,dx =
-\int_{0}^{L}\overline{y}_{2} \partial y_{2}\,dx -
\int_{0}^{L}x |\partial y_{2}|^{2}\,dx \,,
\\
\int_{0}^{L}x y_{2} \partial^{2}\overline{y}_{2}\,dx =
-\int_{0}^{L}y_{2} \partial \overline{y}_{2}\,dx -
\int_{0}^{L}x |\partial y_{2}|^{2}\,dx\,.
\end{gather*}
Then
\begin{align*}
&i\partial_{t}\int_{0}^{L}x |y_{2}|^{2}\,dx + 3 i \beta
\int_{0}^{L}|\partial y_{2}|^{2}\,dx - 2 i \alpha
\int_{0}^{L}\overline{y}_{2} \partial y_{2}\,dx - i \delta
\int_{0}^{L}|y_{2}|^{2}\,dx \\
&= 2 i \mathop{\rm Re}\int_{0}^{L}x \overline{y}_{2} f\,dx\,,
\end{align*}
or
\begin{align*}
&\partial_{t}\int_{0}^{L}x |y_{2}|^{2}\,dx + 3 \beta
\int_{0}^{L}|\partial y_{2}|^{2}\,dx - 2 \alpha
\int_{0}^{L}\overline{y}_{2} [\partial y_{2}]\,dx - \delta
\int_{0}^{L}|y_{2}|^{2}\,dx \\
&= 2 i Re\int_{0}^{L}x \overline{y}_{2} f\,dx\,.
\end{align*}
Hence
\begin{align*}
&  \partial_{t}\int_{0}^{L}x |y_{2}|^{2}\,dx + 3 \beta
\int_{0}^{L}|\partial y_{2}|^{2}\,dx \\
&= 2 i Re\int_{0}^{L}x \overline{y}_{2} f\,dx +
\int_{0}^{L}|y_{2}|^{2}\,dx
+ 2 \alpha \int_{0}^{L}\overline{y}_{2} [\partial y_{2}]\,dx\\
& \leq  2\int_{0}^{L}x |y_{2}| |f|\,dx + \delta
\int_{0}^{L}|y_{2}|^{2}\,dx + |\alpha |\int_{0}^{L}|y_{2}|^{2}\,dx +
|\alpha |\int_{0}^{L}|\partial y_{2}|^{2}\,dx\,.
\end{align*}
Thus
\begin{align*}
&\partial_{t}\int_{0}^{L}x |y_{2}|^{2}\,dx + \int_{0}^{L}( 3 \beta -
|\alpha | )  |\partial y_{2}|^{2}\,dx \\
&\leq  2\int_{0}^{L}x |y_{2}| |f|\,dx
 + ( |\delta| + |\alpha | )\int_{0}^{L}|y_{2}|^{2}\,dx\,.
\end{align*}
Integrating over $t\in [0, T]$ we obtain
\begin{align*}
& \int_{0}^{L}x |y_{2}|^{2}\,dx +
\int_{0}^{T}\int_{0}^{L}( 3 \beta - |\alpha | )  |\partial
y_{2}|^{2}\,dx\,dt  \\
&\leq 2\int_{0}^{T}\int_{0}^{L}x |y_{2}| |f|\,dx\,dt
+( |\delta| + |\alpha | )\int_{0}^{T}\int_{0}^{L}|y_{2}|^{2}\,dx\,dt
+ \int_{0}^{L}x |y_{0 2}|^{2}\,dx\,.
\end{align*}
Using \eqref{e502} the result follows.
\end{proof}

\begin{theorem} \label{thm5.1}
Let $|\alpha |<3 \beta$, $\delta>0$, $T>0$
and $L>0$. Then, there exists $r_{0}>0$ such that for any $y_{0}$,
$y_{T}\in L^{2}(0, L)$ with $\|y_{0}\|_{L^{2}(0, L)}<r_{0}$,
$\|y_{T}\|_{L^{2}(0, L)}<r_{0}$, there exists
\begin{equation} \label{e503}
y\in C([0, T]: L^{2}(0, L))\cap
L^{2}(0, T: H^{1}(0, L))\cap W^{1, 1}(0, T: H^{-2}(0, L))
\end{equation}
solution of
\begin{gather}
\label{e504}i \partial_{t}y_{t}=- ( i \beta
 \partial^{3}y + \alpha  \partial^{2}y + |y|^{2} y + i \delta
 \partial y )\quad
\mbox{in } \mathcal{D}'(0, T: H^{-2}(0, L))\\
\label{e505} y(0, . )=0\quad \mbox{in } L^{2}(0, T)
\end{gather}
such that $y( . , 0)=y_{0}$, $y( . , T)=y_{T}$. Moreover, if
$L\notin \mathcal{N}$, then in addition it can be assumed that
$y(L, . )=0$ in $L^{2}(0, T)$ and take $\partial y(L, . )$ in $L^{2}(0, T)$
 as control function.
\end{theorem}

\begin{proof} We first assume that $L\notin \mathcal{N}$. We show that
for $T>0$ there exists $r_{0}>0$ small enough such that if
$\|y_{0}\|_{L^{2}(0, L)}<r_{0}$, $\|y_{T}\|_{L^{2}(0, L)}<r_{0}$,
the state $y_{T}$ may be reached from $y_{0}$ for  a
higher order  nonlinear Schr\"{o}dinger equation. Let $y_{0}$, $y_{T}$
be states in
$L^{2}(0, L)$ such that $\|y_{0}\|_{L^{2}(0, L)}<r$,
$\|y_{T}\|_{L^{2}(0, L)}<r$, $r>0$ to be chosen later. Let
$\Theta :L^{2}(0, T: H^{1}(0, L))\mapsto \mathbb{H}$, defined by
\[
\Theta (y)  =  S( \cdot  )y_{0} + (\Gamma_{1}\circ
\Gamma_{0}) ( y_{T} - S(T)y_{0} + \Gamma_{2}(|y|^{2} y)( .
 , T) ) + \Gamma_{2}(-|y|^{2} y)
\]
where $\Gamma_{0}$ is well-defined in Remark \ref{rmk4.4}, $\Gamma_{1}$ and
$\Gamma_{2}$ are defined in this section. $\Theta $ is well-defined
and continuous by Lemmas \ref{lem4.2}, \ref{lem4.6}, and
Remark \ref{rmk4.4}. We have that each
fixed point of $\Theta$ verifies \eqref{eQ11} in
$\mathcal{D}'(0, T: H^{-2}(0, L))$ and $u( . , T)=y_{T}$. To prove the
existence of a fixed-point for $\Theta$ we apply the Banach
contraction fixed-point theorem to the restriction of $\Theta$
to some closed ball $\overline{\mathbb{B}}(0, R)$ in
$L^{2}(0, T: H^{1}(0, L))$ ($R$ will be chosen later). We need that
\begin{gather}
\label{e506}
\Theta (\overline{\mathbb{B}}(0, R))\subseteq
\overline{\mathbb{B}}(0, R), \\
\label{e507}\exists \;C_{3}\in ]0, 1[\, \forall y, z\in
\overline{\mathbb{B}}(0, R):\quad \|\Theta(y) - \Theta(z)\|\leq
C_{3} \|y - z\|,
\end{gather}
where $\| \cdot  \|$  stands for the norm
$L^{2}(0, T: H^{1}(0, L))$. Let $\kappa_{1}$ (resp.
$\kappa_{2}, \kappa_{2}'$) denotes the norm of $\Gamma_{1}$ (resp.
$\Gamma_{2}, \Gamma_{2}$) as a map from $L^{2}(0, T)$(resp.
$L^{1}(0, T: L^{2}(0, L))$ into
$L^{2}(0, T: H^{1}(0, L))$ (resp. $L^{2}(0, T: H^{1}(0, L))$,
$C([0, T]: L^{2}(0, L))$), and $\kappa$ denote the norm of
$\Gamma_{0}$ as a map from $L^{2}(0, L)$ into $L^{2}(0, L)$. Set
$\kappa_{3}=\sqrt{\frac{( \delta  + 3 \beta ) T +
 L}{3 \beta - |\alpha |}}$.
 Let $y, z\in L^{2}(0, T: H^{1}(0, L))$. Assume that
$\|y\|\leq R$, $\|z\|\leq R$. Then by \eqref{e408} and \eqref{e501},
\begin{equation} \label{e508}
\begin{aligned}
\|\Theta (y)\| & \leq  \sqrt{\frac{( \delta
 + 3 \beta ) T +  L}{3 \beta - |\alpha |}} \|y_{0}\|_{L^{2}(0, L)} \\
&\quad +\kappa_{1} \kappa (\|y_{T}\|_{L^{2}(0, L)} +
\|y_{0}\|_{L^{2}(0, L)} + \kappa_{2}' C_{1} \|y\|^{2})
+ \kappa_{2} C_{1} \|y\|^{2} \\
& \leq  C_{1} (\kappa_{2} + \kappa
 \kappa_{1} \kappa_{2}') R^{2} + (2 \kappa \kappa_{1} +\kappa_{3}) r.
\end{aligned}
\end{equation}
Hence, we have the first condition on $R$ and $r: $
\begin{equation}\label{e509}
C_{1} (\kappa_{2} + \kappa
 \kappa_{1} \kappa_{2}') R^{2} + (2 \kappa \kappa_{1} +
\kappa_{3}) r\leq R.
\end{equation}
Now write
\begin{equation}
\label{e510}\Theta(y) - \Theta(z) = \Gamma_{2} (|z|^{2} z -
|y|^{2} y) + (\Gamma_{1}\circ \Gamma_{0}) (\Gamma_{2}(|y|^{2} y -
|z|^{2} z)( . , T)).
\end{equation}
Therefore, by \eqref{e501},
\begin{equation}
\label{e511}\|\Theta(y) - \Theta(z)\| = 2 C_{1} (\kappa_{2} +
\kappa \kappa_{1} \kappa_{2}') R \|y - z\|
\end{equation}
Condition \eqref{e506} will hold provided that
\begin{equation}
\label{e512}2 C_{1} (\kappa_{2} + \kappa
 \kappa_{1} \kappa_{2}') R<1.
\end{equation}
Let $R$ be some positive number verifying \eqref{e511}. Then
\eqref{e508} holds true if we take
$r=R/\big(2 (2 \kappa  \kappa_{1} + \kappa_{3})\big)$.
Setting
\[
r_{0}=\frac{1}{4 C_{1} (2 \kappa  \kappa_{1} + \kappa_{3})
(\kappa_{2} + \kappa  \kappa_{1} \kappa_{2}')}\,,
\]
we see that $ r\to r_{0} $ as $ R\to
1/(2 C_{1} (\kappa_{2} + \kappa
 \kappa_{1} \kappa_{2}'))$. It follows that if
$\|y_{0}\|_{L^{2}(0, L)}<r_{0}$ every $y_{T}$ with
$\|y_{T}\|_{L^{2}(0, L)}<r_{0}$ may be reached by a solution of the
higher order nonlinear Schr\"{o}dinger equation coming from $y_{0}$. The
proof of the theorem is completed when $L\notin \mathcal{N}$. If now
$L\in \mathcal{N}$, it is sufficient to consider some $\widetilde{L}>L$
such that $\widetilde{L}\notin \mathcal{N}$ and to apply the theorem to
the functions $\widetilde{y}_{0}, \widetilde{y}_{T}\in
L^{2}(0, \widetilde{L})$, where
$\widetilde{y}_{0}, \widetilde{y}_{T}$ denote the prolongations by
zero of the given states $\widetilde{y}_{0}, \widetilde{y}_{T}\in
L^{2}(0, L)$, and then to restrict the solution $\widetilde{y}$ to
the domain $(0, T)\times (0, L)$.
The proof follows.
\end{proof}

\subsection*{Acknowledgement}
 The authors want to thank Prof. Carlos Picarte (Universidad del B\'\i o-B\'\i o)
for his help in the typesetting the original manuscript.

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\end{document}