\documentclass[12pt]{article}
\title{ Exact Boundary Controllability for the Korteweg- \\
de Vries Equation on a Bounded Domain}
\author{Lionel Rosier}
\date{}
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\begin{document}
\maketitle
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%\begin{center}{\Large {Exact Boundary Controllability for the Korteweg-de
%Vries Equation on a Bounded Domain}}
%\end{center}
%\vskip 4. cm
%\maketitle
%\vspace*{-1cm}
\begin{center}
Laboratoire d'Analyse Num\'erique et EDP,\\
Universit\'e Paris 11,\\
B\^at. 425, F91405 Orsay Cedex, France.\\
{\footnotesize
e-mail: Lionel.Rosier@math.u-psud.fr}\\
\vspace{0.5cm}
%\setcounter{page}{159}
\begin{abstract}
The exact boundary controllability of the linear
Korteweg-de Vries equation on bounded domains with various
boundary conditions is studied. When boundary conditions bear on spatial
derivatives up to order 2, the exact controllability result by
Russell-Zhang is directly proved by means of Hilbert Uniqueness Method.
When only the first spatial derivative at the right endpoint is assumed to be
controlled, a quite different analysis shows that the exact controllability
also holds true. From this last result we derive the exact boundary
controllability of the nonlinear KdV equation on bounded domains, for
sufficiently small initial and final states.\\
\vspace{0.5cm}
{\bf\sc Key Words:} {\sl Korteweg-de Vries equation, boundary
controllability, Hilbert Uniqueness Method.}
\end{abstract}
\end{center}
\va
%\setcounter{chapter}{1}
\renewcommand{\theequation}{\thesection.\arabic{equation}}
%\setcounter{page}{160}
\section{Introduction and main results.}
\par
The Korteweg-de Vries (KdV) equation
\begin{equation}
y_t+yy_x+y_{xxx}=0,
\end{equation}
is a well known instance of a nonlinear dispersive partial differential
equation, which may serve as a model for (among other things)
propagation of small amplitude long water waves in a uniform channel.
\par
Recently stabilizability and controllability results have been obtained in
[8] for the forced Korteweg - de Vries equation
with periodic boundary conditions
\begin{eqnarray}
y_t+yy_x+y_{xxx} & = & f(t,x),\quad \quad 0<x<1,\ 0<t<T  \\
{\frac{\partial ^k y}{\partial x^k}}(t,0) & =
& {\frac{\partial ^k y}{\partial x^k}}(t,1),\quad \quad 0\le k\le 2,
\end{eqnarray}
the control action $f(t,x)$ being localized:
\begin{equation}
\hbox{supp}\ f(t,.)\subset [a,b]\subset [0,1].
\end{equation}
\newpage
%\setcounter{page}{158}
On the other hand, there are to date few results
concerning the boundary controllability of KdV equation. In [16] boundary
controllability of linear KdV equation
\begin{eqnarray}
y_t+y_{xxx} &=& 0, \quad 0<x<2\pi ,\quad 0<t<T  \\
{\frac{\partial ^k y}{\partial x^k}}(t,0) &=&
{\frac{\partial ^k y}{\partial x^k}}(t,2\pi ), \quad k\in\{ 0,2 \} ,\\
{\frac{\partial y}{\partial x}}(t,2\pi ) -
{\frac{\partial y}{\partial x}}(t,0)&=&h(t),\quad 0<t<T,
\end{eqnarray}
is obtained as a consequence of the {\em uniform stabilizability} of
this
time-reversible (linear) system in both t-directions, whose proof is
rather long and technical. Moreover only initial and terminal states
$y_0$ and $y_T$ for which
\begin{equation}
\int_0^{2\pi}\big( y_T(x)-y_0(x)\big )dx =0
\end{equation}
are concerned and the control $h(t)$ is not explicitly given. The first
goal
of this paper is to directly study the (exact) boundary controllability
problem for linear KdV equation by means of the Hilbert Uniqueness Method
(H.U.M). This
approach has been successfully applied for studying controllability
of wave and plate equations, and more recently of Schr\"{o}dinger equation
(see for instance [1], [5], [7], [11], [12], [13], [14]). 
The first result of this
paper concerning boundary controllability of linear KdV equation is as
follows: \va\va
{\bf Theorem 1.1:}
{\it 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 $y_0,y_T\in  (H^2_p)'$ (the dual space of $H^2_p$),
 there exist $h_0,\ h_1,\ h_2
\in L^2(0,T )$ such that the solution $y\in C([0,T],(H^2_p)')$
of the
boundary initial-value KdV equation: }
\begin{eqnarray}
y_t+y_{xxx} & = & 0,\quad \quad \quad (x,t)\in
(0,2\pi )\times (0,T) \\
y(t,2\pi )-y(t,0) & = & h_0(t),                         \\
y_x(t,2\pi )-y_x(t,0) & = & h_1(t),                     \\
y_{xx}(t,2\pi )-y_{xx}(t,0) & = & h_2(t),                \\
y(0,.) & = & y_0
\end{eqnarray}
satisfies $y(T,.)=y_T$.
\va\par
Notice 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 1.1 controllability results for nonlinear KdV equation
(1.1).
Lastly it should be more realist to set boundary conditions on
the value or the
first  (spatial)
derivative of the function $y$ rather than on the second derivative
$y_{xx}$, if we have in mind to test such controllability result
or the validity of KdV equation as a model with an experimental
device: For instance we may observe the propagation of water waves of small
amplitude in a bounded channel, the level -or the slope- of the surface
of the fluid
being controlled at the ends of the channel. Notice that, as it
was suggested in
[2], the extra term $y_x$ should be in this case
incorporated in the equation in order to
obtain an appropriate model for water waves in a uniform channel when
coordinates $x,\ y$ are taken with respect to a {\it fixed} frame.
This inclusion
is not without consequence in controllability results: Condition on the
lenght $L$ of the domain appears. The main result concerning (exact) boundary
controllability for linear KdV equation (with
boundary control on $y_x$ at $x=L$) is
the following:\va\va
{\bf Theorem 1.2:} {\it Let
${\cal{N}}=\{ 2\pi \sqrt{ \frac{k^2+kl+l^2}{3} }; \ k,l\in \N ^* \}$.
Then for any $T>0$ and $L\in (0,+\infty )\setminus {\cal N}$, 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{eqnarray}
y_t+y_x+y_{xxx} & = & 0  \\
y(t,0)=y(t,L) & = & 0 \\
y_x(t,L) & = & h(t)   \\
y(0,x) & = &  y_0
\end{eqnarray}
satisfies $y(T,.)=y_T$.}
\va\par
This theorem is proved by means of H.U.M and multiplier method.
It turns out that the study of (1.14)-(1.17) as a boundary
initial-value problem is more delicate than for (1.9)-(1.13), and that
- because of the extra term $y_x$ in (1.14) - the
(key) observability result holds true iff $L\not\in {\cal N}$. 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 $yy_x$ in (1.1).
\par
 As a consequence of previous controllability result
we get the main result of the paper, that is the exact boundary
controllability of nonlinear KdV equation on a bounded domain:\va\va
{\bf Theorem 1.3:}
{\it Let $T>0$ and $L>0$.
Then there exists $r_0>0$ such that for any
$y_0,y_T\in L^2(0,L)$ with $\Vert y_0\Vert _{L^2(0,L)} <r_0$,
$\Vert y_T\Vert _{L^2(0,L)} <r_0$, there
exists  $y\in C([0,T],L^2(0,L))\cap L^2(0,T,$
$H^1(0,L)) \cap W^{1,1}(0,T,H^{-2}(0,L))$  solution of
\begin{eqnarray}
y_t\, & = &\, -(y_x+yy_x+y_{xxx}) \hbox{ in } {\cal D}'(0,T,H^{-2}(0,L)) \\
y(.,0)\, & = &\, 0 \hbox{ in } L^2(0,T),
\end{eqnarray}
and such that $y(0,.)=y_0,\ y(T,.)=y_T$. If moreover 
$L\not\in {\cal N}$, then we may in addition assume that 
$y(.,L) = 0 \hbox{ in } L^2(0,T)$ and take
$y_x(.,L)\hbox{ in } L^2(0,T)$ as control function.
}
\va\par
Theorem 1.3 is proved by means of Banach contraction fixed point theorem
and Theorem 1.2.
\par
The paper is organized as follows: In Section 2 we derive from H.U.M. a
direct proof of the exact boundary controllability result by
Russell-Zhang for (linear) KdV equation (Th. 1.1). 
In Section 3 we consider
another boundary controllability problem 
for (linear) KdV equation, in which only the
value of 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
(Prop. 3.6). Next an observability result is given (Prop. 3.7)
and used to show
(applying H.U.M) the exact boundary controllability for linear KdV equation with
these boundary conditions (Th. 1.2). Finally in Section 4 we derive from
Th. 1.2 the main result of this paper, that is the exact
(local) boundary controllability of (nonlinear) KdV equation on a
bounded domain (Th. 1.3).
%\setcounter{chapter}{2}
\setcounter{equation}{0}
\section{Exact boundary controllability of linear KdV equation by means of
control on data $\left[ {\frac{\partial ^k y}{\partial x^k}}(t,.)
\right] _0^{2\pi},\ k=0,\ 1,\ 2$.}
\par
In this section, for reasons of simplicity, we restrict ourselves to the case
where the space domain $[0,L]$ is $[0,2\pi ]$, although Theorem 1.1 holds true
for arbitrary $L>0$. On the other hand all the functions of this section
are assumed
to take complex values. For $k\in\{ 2,3\}$ we set
\begin{equation}
H_p^k:=\left\{ u\in H^k(0,2\pi );\ {\frac{d ^j u}{d x^j}}(0)=
{\frac{d ^j u}{d x^j}}(2\pi ) \hbox{ for } 0\le j\le k-1\right\} .
\end{equation}
$(H^k(0,2\pi )$ denotes classical Sobolev space on the interval $(0,2\pi )$.)
For $n\in \Z$, let
\begin{equation}
\hat{u}(n)=\frac{1}{2\pi}\int_0^{2\pi}u(t) e^{-int}dt
\end{equation}
be the nth Fourier coefficient of $u\in L^2(0,2\pi )$. It is easy
to see that, for all $u\in L^2(0,2\pi )$, $k\in\{ 2,3 \}$:
\begin{equation}
u\in H_p^k \iff
\sum_{n\in \Z}
(n^k\vert \hat{u}(n)\vert )^2<\infty ,
\end{equation}
and that  the Sobolev norm
$ \Vert u\Vert _k:=
\big( \sum_{j=0}^k \int_0^{2\pi}
\vert u^{(j)}(x) \vert ^2\, dx \big) ^{ \frac{1}{2} }$ 
reduces to
\begin{equation}
\Vert u\Vert _k=\big( 2\pi \sum_{n\in \Z}(1+n^2+\cdots +n^{2k})
\vert \hat{u} (n)\vert ^2 \big) ^{\frac{1}{2}}
\end{equation}
for $u\in H_p^k$. In what follows the (Hilbert) space
$H_p^k$ is endowed with the norm
$\Vert .\Vert _k$.\par
Let $A$ denote the operator $Au=-u'''$ on the domain ${\cal D}(A)=
H_p^3\subset L^2(0,2\pi )$. As it is mentioned in [16], $A$ generates a
strongly continuous unitary group $\big( S(t) \big) _{t\in \R}$
on $L^2(0,2\pi )$. Let $T>0$. For
$u_T=\sum_{n\in\Z}c_ne^{inx}\in L^2(0,2\pi )$, the (mild) solution of
the uncontrolled problem:
\begin{eqnarray}
u_t+u_{xxx} & = & 0,\quad \quad x\in (0,2\pi ), t\in\R   \\
u(t,0) & = & u(t,2\pi ),                                  \\
u_x(t,0) & = & u_x(t,2\pi ),                                \\
u_{xx}(t,0) & = & u_{xx}(t,2\pi ),                        \\
u(T,.) & = & u_T
\end{eqnarray}
is given by
\begin{equation}
 u(t,x)=\sum_{n\in\Z}c_ne^{i\big( n^3(t-T)+nx\big)}.
\end{equation}
(In the following, $u$ will denote the solution of (2.5)-(2.9)
associated with $u_T$.) \va
We now consider the following non-homogeneous problem: Given
$y_0\in (H_p^2)'$,
$h_0,\ h_1,
\ h_2$ in $L^2(0,T )$, find $y$ such that:
\begin{eqnarray}
y_t+y_{xxx} & = & 0,\quad \quad x\in (0,2\pi ),\  t\in (0,T)             \\
y(t,2\pi )-y(t,0) & = & h_0(t),                                        \\
y_x(t,2\pi )-y_x(t,0) & = & h_1(t),                                    \\
y_{xx}(t,2\pi )-y_{xx}(t,0) & = & h_2(t),                               \\
y(0,.) & = & y_0.
\end{eqnarray}
We first prove that 1) equations (2.11)-(2.15) admit a unique solution
$y\in C([0,T],(H_p^2)')$ in a sense to be made precise, and 2)
this solution is the classical one whenever $y_0\in {\cal D}(A)$,
 and $h_0,\ h_1,\ h_2$ are smooth enough
and vanish at $0$. More precisely,
we prove the following:\va\va
{\bf Proposition 2.1:} {\em 1) Assume
$h_0,\ h_1,\ h_2 \in C_0^2([0,T])
:=\{ h\in C^2([0,T],\C );\ h(0)=0\}$ 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 (2.11)-(2.15). Moreover for any $u_T\in H_p^3$
and any $S\in [0,T]$ we have
\begin{equation}
\begin{array}{l}
\int_0^{2\pi }u(S,x)\overline{y(S,x)}dx\\
\quad =\int_0^{2\pi }u(0,x)\overline{y_0(x)}dx
-\int_0^S\bigg( u_{xx}(t,0)\overline{h_0(t)}-u_x(t,0)\overline{h_1(t)}
+u(t,0)\overline{h_2(t)}\bigg) dt.
\end{array}
\end{equation}
2) For $u_T\in H_p^2$, $u\in C([0,T],H_p^2)$ and $u_{xx}(.,0)$ makes
sense in $L^2(0,T)$.\va
3) Assume now only that $y_0\in (H_p^2)'$ and $h_0,\ h_1,\ h_2\in L^2(0,T)$.
 Then there exists a unique
$y\in C([0,T],(H_p^2)')$ such that
\begin{equation}
\begin{array}{c}
\forall u_T\in H_p^2,\ \forall S\in [0,T]\\
\\
<y(S),u(S,.)>_{(H_p^2)',H_p^2}
= <y_0,u(0,.)>_{(H_p^2)',H_p^2}\\
\\
\ -\int_0^S\big( u_{xx}(t,0)\overline{h_0(t)}-u_x(t,0)\overline{h_1(t)}
+u(t,0)\overline{h_2(t)}\big) dt.
\end{array}
\end{equation}
}
\va
{\it Proof.}
1) Let $\phi _0,\phi _1,\phi _2\in C^{\infty}([0,2\pi ])$ be such that
$\phi _i^{(k)}(0)=0$ and $\phi _i^{(k)}(2\pi )=-\delta _i^k$ for
$i,k=0,1,2$. The change of function
\begin{equation}
z(t,x):=h_0(t)\phi _0(x)+h_1(t)\phi _1(x)+h_2(t)\phi _2(x)+
\big( S(t)y_0 \big) (x)+ y(t,x)
\end{equation}
yields an equivalent problem: Find $z$ such that
\begin{eqnarray}
z_t+z_{xxx} & = & f(t,x): = \sum_{i=0}^2\big( h_i'(t)\phi _i(x)+
h_i(t)\phi _i'''(x)\big)                            \\
z(t,2\pi ) & = &  z(t,0),                          \\
z_x(t,2\pi ) & = & z_x(t,0),                      \\
z_{xx}(t,2\pi ) & = & z_{xx}(t,0),                \\
z(0,.) & = & 0 .
\end{eqnarray}

Since $f\in C^1([0,T],L^2(0,2\pi ))$, this non-homogeneous
problem admits (see [15]) a unique solution
 $z\in C([0,T],H_p^3)\cap C^1([0,T],L^2(0,2\pi ))$. 
This proves the first assertion in 1).
Let $u_T\in H_p^3$. Then
$u\in C([0,T],H_p^3)\cap C^1([0,T],L^2(0,2\pi ))$. Integrations
by parts in
\begin{equation}
\int_0^S\int_0^{2\pi}(u_t+u_{xxx})\overline{y}dxdt=0
\end{equation}
lead to
\begin{eqnarray}
&-&\int_0^S\int_0^{2\pi}u(\overline{y_t}+\overline{y_{xxx}})dxdt
 +\int_0^{2\pi }[\big( u\overline{y} \big) (.,x)]_0^{S}dx\nonumber \\
&+&\int_0^S[(u_{xx}\overline{y}-u_x\overline{y_x}+u\overline{y_{xx}})
(t,.)]_0^{2\pi}dt =0.
\end{eqnarray}
Taking into account (2.6)-(2.8) and (2.11)-(2.15), we get (2.16).\va
2) By (2.10), for $t_1,t_2\in [0,T]$
\begin{equation}
u(t_1,x)-u(t_2,x)=\sum_{n\in\Z}c_ne^{-in^3T}
{\big( e^{in^3t_1}-e^{in^3t_2}}\big) e^{inx}.
\end{equation}
If $u_T\in H_p^2$,
$\sum_{n\in\Z}\vert n^2c_n\vert ^2<\infty$
and it follows from Lebesgue's theorem that
\begin{equation*}
\sum_{n\in\Z} \vert n^2c_ne^{-in^3T}
\big( e^{in^3t_1}-e^{in^3t_2} \big) \vert ^2\to 0 
\quad\hbox{ as } t_1\to t_2,
\end{equation*}
hence
$u\in C([0,T],H_p^2)$. Hence $u(.,0)$ and $u_x(.,0)$ exist in
$C([0,T])\subset L^2(0,T)$. The same argument shows that if
 $u_T\in H_p^3$, $u\in C([0,T],H_p^3)$ and
\begin{equation}
u_{xx}(t,0)=\sum_{n\in\Z} \big( -n^2e^{-in^3T}c_n\big) e^{in^3t}.
\end{equation}
The sum in (2.27) makes sense in $L^2(0,T)$ whenever
$\sum_{n\in\Z}\big( n^2\vert c_n\vert\big) ^2<\infty$, that is
$u_T\in H_p^2$. From now on $u_{xx}(.,0)$ denotes, for $u_T\in H_p^2$,
the sum in (2.27). Notice that the linear map $u_T\mapsto u_{xx}(.,0)$ is
continuous since
\begin{equation}
\Vert \sum_{n\in\Z} \big( -n^2e^{-in^3T}c_n\big) e^{in^3t}
\Vert _{L^2(0,T)}^2
\le 2\pi \big( \left[ {\frac{T}{2\pi }} \right]+1 \big)
\sum_{n\in\Z} \big( n^2\vert c_n\vert\big) ^2,
\end{equation}
where $[x]$ denotes the integral part of a real number $x$.
\va
3) Identifying $L^2(0,2\pi )$ with its dual by means of the
(conjugate linear) map $y\mapsto (.,y)_{L^2(0,2\pi )}$,
 we have the following diagram:
\begin{equation}
H_p^2\subset L^2(0,2\pi )= L^2(0,2\pi )'\subset (H_p^2)'
\end{equation}
where each embedding is dense and compact. Moreover
\begin{equation}
<y,u>_{(H_p^2)',H_p^2}\ =(u,y)_{L^2(0,2\pi)}\
(=\int_0^{2\pi}u\overline{y}dx ) \ \
\end{equation}
for $u\in H_p^2$ and $y\in L^2(0,2\pi )$.  It follows that (2.17) holds
true for
$h_0,\ h_1, \ h_2\in C_0^2([0,T])$ and $y_0,u_T\in H_p^3$. Since $H_p^3$
 is dense in
$H_p^2$, using 2) we see that (2.17) also be true for $u_T\in H_p^2$.
For $y_0\in (H_p^2)'$ and
$h_0,\ h_1, \ h_2\in L^2(0,T)$, we define a "weak"  solution of
(2.11)-(2.15)
as a function $y\in C([0,T],(H_p^2)')$ such that (2.17) holds true
for all $u_T\in H_p^2$ and all $S\in [0,T]$. Let $S\in [0,T]$ be fixed.
We first show that (2.17) defines $y(S)$ in $(H_p^2)'$ in a unique manner.
It follows from the proof of 2) that the map
\begin{equation}
u_T\in H_p^2 \mapsto -\int_0^S {\bigg(}
u_{xx}(t,0)\overline{h_0(t)}-u_x(t,0)\overline{h_1(t)}
+u(t,0)\overline{h_2(t)} {\bigg)} dt \in \C
\end{equation}
is a continuous linear form. On the other hand (see (2.3),
(2.4) and (2.10)) the
map $u_T\in H_p^2\mapsto u(S,.)\in H_p^2$ is clearly an
automorphism of Hilbert space. Hence for each $S\in [0,T]$, $y(S)$ is
uniquely defined in $(H_p^2)' $. Moreover for $S\in [0,T]$,
\begin{eqnarray}
\Vert y(S)\Vert_{(H_p^2)'}
&=& \sup_{\Vert u(S,.)\Vert _{2}
\le 1}\vert <y(S),u(S,.)>\vert  \nonumber \\
&=& \sup_{\Vert u_T\Vert _{2}\le 1}
\left\vert <y_0,u(0,.)>_{(H_p^2)',H_p^2}\right.
\nonumber\\
& & \left. -\int_0^S\big( u_{xx}(t,0)\overline{h_0(t)}
-u_x(t,0)\overline{h_1(t)}
+u(t,0)\overline{h_2(t)}\big) dt \right\vert    \nonumber \\
&\le &  \sup_{\Vert u_T\Vert _{2}\le 1}
\left( \Vert y_0 \Vert _{(H_p^2)'} .\Vert u(0,.)\Vert _{H_p^2}
+ \Vert u_{xx}(.,0)\Vert_{L^2(0,T)}.\Vert h_0\Vert_{L^2(0,T)}
\right. \nonumber \\
&& \left. \quad\quad\quad +
\Vert u_{x}(.,0)\Vert_{L^2(0,T)}.\Vert h_1\Vert_{L^2(0,T)}
 +\Vert u(.,0)\Vert_{L^2(0,T)}.\Vert h_2\Vert_{L^2(0,T)} \right)
 \nonumber \\
&\le & C\left(
\Vert y_0 \Vert _{(H_p^2)'}+
\Vert h_0\Vert_{L^2(0,T)}+
\Vert h_1\Vert_{L^2(0,T)}+
\Vert h_2\Vert_{L^2(0,T)} \right) ,
\end{eqnarray}
where $C$ is a positive constant which does not depend on $S$ or on
$y_0,\ h_0,\ h_1,\ h_2$.
Since
\begin{equation}
y\in C([0,T],L^2(0,2\pi ))\subset C([0,T],(H_p^2)')
\end{equation}
for $y_0\in H_p^3$ and
$(h_0,h_1,h_2)\in C_0^{2}([0,T])^3$, and since
$H_p^3$ is dense in $L^2(0,L)$ and
$C_0^{2}([0,T])$ is dense in $L^2(0,T)$,
it follows  from (2.32)
that $y\in C([0,T],(H_p^2)')$ in the general case too. \qed
\par
For applying H.U.M. we need the following (key) observability
result.
\va\va
{\bf Proposition 2.2:} {\it Let $T>0$. Then there exist positive numbers
$C_1^T,C_2^T$
such that for every
$ u_T\in H_p^2 $
\begin{equation}
C_1^T \, \Vert u_T\Vert _2^2
\le \Vert u(.,0)\Vert _{L^2(0,T)} ^2 +
\Vert u_x(.,0)\Vert _{L^2(0,T)} ^2 +\Vert u_{xx}(.,0)\Vert _{L^2(0,T)} ^2
\le C_2^T\, \Vert u_T\Vert _2^2.
\end{equation} }
\va
{\it Proof.} We have in $L^2(0,T):$
\begin{eqnarray}
u(t,0) & = & \sum_{n\in\Z}c_ne^{in^3(t-T)},                          \\
u_x(t,0) & = & \sum_{n\in\Z}inc_ne^{in^3(t-T)},                      \\
u_{xx}(t,0) & = & \sum_{n\in\Z}-n^2c_ne^{in^3(t-T)}.
\end{eqnarray}
Hence
\begin{eqnarray}
\Vert u(.,0)\Vert _{L^2(0,T)} ^2 &+&  \Vert u_x(.,0)\Vert _{L^2(0,T)} ^2 
+\Vert u_{xx}(.,0)\Vert _{L^2(0,T)} ^2 \nonumber\\
&\le &
2\pi \left( \left[ \frac{T}{2\pi }\right] +1 \right) \sum_{n\in\Z}
(1+n^2+n^4)\, \vert c_n \vert ^2   \nonumber \\
&\le & C_2^T\, \Vert u_T\Vert _2^2.
\end{eqnarray}
if we set $C_2^T:= \left[ \frac{T}{2\pi }\right] +1$. To prove the left
inequality we first take $T'\in (0,T)$ and $\gamma  > \frac{2\pi}{T'}$. \va
Let $N\in \N ^*$ be such that $N^3-(-N)^3 \ge \gamma$ and 
\begin{equation}
n\in \Z,\ \vert n\vert \ge N \ \ \Rightarrow \ \
(n+1)^3 - n^3 \ge \gamma .
\end{equation}
By Ingham's inequality (see [6]) there exists $C^{T'}>0$ such that
\begin{equation}
\forall (a_n)_{n\in \Z}\in l^2(\Z ) \ \ \
\sum_{\vert n\vert \ge N}  \, \vert a_n\vert ^2
\le C^{T'}\, \int_0^{T'} \left\vert \sum_{\vert n\vert \ge N}
a_n\, e^{in^3(t-T)}\, \right\vert ^2 \, dt.
\end{equation}
Let ${\cal Z}_n :=\hbox{Span} (e^{inx})$ for $n\in \Z$ and
${\cal Z}:=\oplus_{n\in \Z} {\cal Z}_n \subset H_p^2$. We define a
semi-norm $p$ in ${\cal Z}$ by
\begin{eqnarray}
\forall u\in {\cal Z} \  \  \  \
p(u)&=&\left( \vert u(0)\vert ^2 + \vert u'(0)\vert ^2+
\vert u''(0)\vert ^2\right) ^{\frac{1}{2}}  \\
&=&\left( \vert \sum_{n\in \Z}{\hat u}(n) \vert ^2 +
\vert \sum_{n\in \Z}in{\hat u}(n) \vert ^2 +
\vert \sum_{n\in \Z}-n^2{\hat u}(n) \vert ^2 \right) ^{\frac{1}{2}}.
\end{eqnarray}
(For $u\in {\cal Z}$, $\hat u(n)=0$ for $\vert n\vert$ large enough.)
Let  $u_T\in {\cal Z}\cap (\oplus_{\vert n\vert <N} {\cal Z}_n )^{\perp}$,
that is $c_n=0$ for $\vert n\vert <N$ or for $\vert n\vert$ large enough.
Using (2.10), (2.40) we get
\begin{equation}
\Vert u_T\Vert _2^2=2\pi \sum_{\vert n\vert \ge N}
(1+n^2+n^4)\, \vert c_n \vert ^2 
\le 2\pi C^{T'}\, \int_0^{T'}p\big( u(t,.)\big) ^2\, dt.
\end{equation}
Since $T>T'$, it follows from (2.38), (2.43) and a result by Komornik
(see [7; Th. 5.2]) that there exists a constant $C_1^T>0$ such that
for any $u_T\in {\cal Z}$
\begin{equation}
C_1^T \, \Vert u_T\Vert _2^2 \le \int_0^T
p\big( u(t,.)\big) ^2\, dt =
\Vert u(.,0)\Vert _{L^2(0,T)} ^2 +  \Vert u_x(.,0)\Vert _{L^2(0,T)} ^2 +
\Vert u_{xx}(.,0)\Vert _{L^2(0,T)} ^2\cdot
\end{equation}
We get the left inequality (in (2.34)) in the general case ($u_T\in H_p^2$)
by a density argument. \qed
\par
We may now complete the proof of Theorem 1.1 by applying H.U.M..\va
{\it Proof of Theorem 1.1.}
We first remark that we may assume $y_0=0$. Indeed if $y_0,y_T$ are given
in $(H_p^2)'$, if there exist $h_0,\ h_1, \ h_2\in L^2(0,T)$ such that the
(weak) solution $\tilde y$ of (2.11)-(2.14) and ${\tilde y}(0,.)=0$
satisfies $\tilde y(T,.)=y_T-S(T)y_0$,
then $y(t,.):=S(t)y_0 + {\tilde y}(t,.)$
is the (weak) solution of (2.11)-(2.15) (with the same control functions)
and is such that
$y(T,.)=y_T$. In what follows we assume $y_0=0$.
For $u_T\in H_p^2$ we let $\Lambda (u_T)$
denote $y(T)\in (H_p^2)'$, where $y$ is the (weak) solution of
(2.11)-(2.15) and $h_0$, $h_1$ and $h_2$ are chosen in the following
way:
\begin{eqnarray}
h_0(t) & := & -u_{xx}(t,0) \ ,                    \\
h_1(t) & := & u_{x}(t,0) \ ,                      \\
h_2(t) & := & -u(t,0) \ .
\end{eqnarray}
(As above u stands for the solution of (2.5)-(2.9) associated with $u_T$.)
 Clearly $\Lambda :H_p^2\to (H_p^2)'$ is a (conjugate linear) continuous map.
Moreover
\begin{equation}
<\Lambda (u_T),u_T>_{(H_p^2)',H_p^2} \ =\int_0^T
\bigg( \vert u(t,0)\vert ^2 +\vert u_x(t,0)\vert ^2 +
 \vert u_{xx}(t,0)\vert ^2 \bigg) dt
\ge C_1^T \Vert u_T\Vert _{2}^2
\end{equation}
by Propositions 2.1 and 2.2. It follows from Lax-Milgram theorem
(see [19; p.92])
that  $\Lambda$ is invertible. The proof
of Theorem 1.1 is complete. \qed
\va
{\it Remark 2.1.} (i) If $T=2\pi$ the observability result is obvious. Indeed
for any $u_T\in H_p^2$
\begin{equation}
{\Vert u_T\Vert ^2_2} =
\Vert u(.,0)\Vert _{L^2(0,2\pi )} ^2 +
\Vert u_x(.,0)\Vert _{L^2(0,2\pi )} ^2 +
\Vert u_{xx}(.,0)\Vert _{L^2(0,2\pi )} ^2.
\end{equation}
(ii) Setting $h_1(t)=0$ instead of $u_x(t,0)$ we also
get (for some constant $\bar C_1^T >0$)
\begin{equation}
\forall u_T\in H_p^2 \ \ \ <\Lambda (u_T),u_T>_{(H_p^2)',H_p^2}\  \ge \
\bar C_1^T \Vert u_T\Vert _{2}^2 ,
\end{equation}
so $\Lambda $ is also onto.\va
(iii) If we assume $h_0(t)=h_2(t)=0$, the same arguments as above yield
the following result: Let ${\tilde H}_p^1:=\{ u\in H_p^1;\ \hat u (0)=0\}$.
Then for every $T>0$ and every $y_0,y_T\in ({\tilde H}_p^1)'$ there
exists $h_1\in L^2(0,T)$ such that the solution 
$y\in C([0,T],({\tilde H}_p^1)')$
of (1.9)-(1.13) (with $h_0=h_2=0$) satisfies $y(T,.)=y_T$. Indeed it may be
shown that for any $u_T\in {\tilde H}_p^1$,  $u_x(.,0)$ 
makes sense in $L^2(0,T)$ and that
$C_3^T \Vert u_T\Vert _1^2\le \Vert u_x(.,0)\Vert _{L^2(0,T)}^2
\le C_4^T \Vert u_T\Vert _1^2$
for some constants $C_3^T,C_4^T>0$.  When we
compare these results with those by Russell-Zhang, we remark that 1) the
spaces in which we get exact controllability are larger than $L^2(0,L)$
($L^2(0,L) \subset ({\tilde H}_p^1)'$ and 
$L^2(0,L)\subset (H_p^2)'$); 2) the assumption (1.8) is
not used here and 3) the control may be explicitly given :
For reaching $y_T\in (H_p^2)'$
we may choose $h_0,h_1,h_2$ given by (2.45)-(2.47), with
$u_T=\Lambda ^{-1}(y_T-S(T)y_0)$. However, even if $y_0,y_T\in L^2(0,2\pi )$,
$y$ is only known to belong to $C([0,T],(H_p^2)')$. So this method is
not appropriate for starting the study of
controllability of nonlinear KdV equation.
In the following section, another (more appropriate) exact controllability
result for linear KdV equation with {\it different} boundary conditions
is given.
\va
\section{Exact boundary controllability of linear KdV equation by means
of control on $y_{x}(t,L)$.}
%\setcounter{chapter}{3}
\setcounter{equation}{0}
\par
>From now on, the scalar space is $\R$.
In this section $L$ stands for some positive number. We shall prove the
controllability in $L^2(0, L )$ of
\begin{eqnarray}
y_t+y_x+y_{xxx} & = & 0     \\
y(t,0)=y(t,L) & = & 0       \\
y_x(t,L) & = & h(t)         \\
y(0,x) & = & y_0
\end{eqnarray}
where $h\in L^2(0,T)$ stands for the control function.
More precisely we shall prove that, for any $L>0$, $T>0$,
for any $y_0,y_T\in L^2(0,L)$ there exists $h\in L^2(0,T)$ such that a mild
solution $y\in C([0,T],L^2(0,L))\cap L^2(0,T,H^1(0,L))
\cap H^1(0,T,H^{-2}(0,L))$ of (3.1)-(3.4) (
which verifies (3.1) in ${\cal D}'(0,T,H^{-2})$ and (3.4) in
$L^2(0,L)$ ), may be found such that $y(T,.)=y_T$.
\par
We begin by showing  the well-posedness of the (initial-value) homogeneous
problem:
\begin{eqnarray}
y_t+y_x+y_{xxx} & = & 0                          \\
y(t,0) = y(t,L)& = & 0                         \\
y_x(t,L) & = & 0                              \\
y(0,x) & = & y_0(x).
\end{eqnarray}

Let $A$ denote the operator $Aw=-w'''-w'$ on the (dense) domain
${\cal D}(A)\subset L^2(0,L)$ defined by
\begin{equation}
{\cal D}(A)=\{ w\in H^3(0,L);\ w(0)=w(L)=w'(L)=0 \} .
\end{equation}
The following result holds: \va\va
{\bf Proposition 3.1:} {\it $A$ generates a strongly continuous semigroup
of contractions on $L^2(0,L)$. }
\va\va
{\it Proof.} It is easy to see that
$A$ is closed. Let $w\in {\cal D}(A)$.
Then
\begin{equation}
\begin{array}{l@{=}l}
(w,Aw)_{L^2(0,L)} & \int_0^L w(x)\big( -w'''(x)-w'(x)\big) dx  \\
\  & \int_0^Lw'(x)w''(x)dx - [ww'']_0^L -[\frac{w^2}{2}]_0^L \\
\  & [\frac{{w'}^2}{2}]_0^L    \\
\  &-{\frac{ w'(0) ^2}{2} }\le 0.
\end{array}
\end{equation}
Hence $A$ is dissipative. It may be seen that $A^*(w)=w'+w'''$ with domain
${\cal D}(A^* )=\{ w\in H^3(0,L);\ w(0)=w(L)=w'(0)=0 \}$, so that
$(w,A^* (w))=-{ \frac{w'(L)^2 }{2} } \le 0$ and $A^*$ is dissipative too.
Now the result follows from [15; cor. 4.4 chapter 1]. \qed
\par
>From now on we let $\left( S(t)_{t\ge 0} \right)$ denote
the semi-group of contractions associated with $A$, and we let
$B$ 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}
\Vert y \Vert _{B}:=\sup_{t\in [0,T]}\Vert y(t,.)\Vert _{L^2(0,L)} +
\left( \int_0^T\Vert y(t,.)\Vert_{H^1(0,L)}^2 dt\right) ^{\frac{1}{2}}.
\end{equation}
Using the multiplier method, we get useful estimates for
the mild solutions of (3.5)-(3.8).\va\va
{\bf Proposition 3.2:} {\it 1) The map $y_0\in L^2(0,L)\mapsto
S(.)y_0\in B$ is continuous.\va
2) For $y_0\in L^2(0,L)$, $y_x(.,0)$ makes sense in $L^2(0,L)$, and
\begin{eqnarray}
\forall y_0\in L^2(0,L)\ \ \ \
&&\Vert y_x(.,0)\Vert _{L^2(0,T)}\le \Vert y_0\Vert  _{L^2(0,L)}, \\
&&\Vert y_0\Vert ^2 _{L^2(0,L)} \le
\frac{1}{T} \Vert S(\cdot )y_0\Vert ^2_{L^2((0,T)\times (0,L))}
+\Vert y_x(\cdot ,0)\Vert ^2 _{L^2(0,T)}.
\end{eqnarray} }
\va
{\it Proof.} 1) For $y_0\in L^2(0,L)$ we write $y$ the mild solution
$S(\cdot )y_0$ of (3.5)-(3.8). By Proposition 3.1 $y\in C([0,T],L^2(0,L))$ and
\begin{equation}
\Vert y\Vert _{C([0,T],L^2(0,L))} \le \Vert y_0\Vert _{L^2(0,L)}.
\end{equation}
To see that $y\in L^2(0,T,H^1(0,L))$ we first assume $y\in {\cal D}(A)$.
Let $q\in C^{\infty}([0,T]\times [0,L]).$ Integrating
by parts in
\begin{equation}
\int_0^T\int_0^L qy(y_t+y_x+y_{xxx})dxdt=0,
\end{equation}
using (3.5)-(3.8) (and Fubini's theorem) we get:
\begin{eqnarray}
&-&\int_0^T\int_0^L(q_t+q_x+q_{xxx})\frac{y^2}{2}dxdt +
\int_0^L (q{\frac{y^2}{2}})(T,x)dx \nonumber \\
&-& \int_0^L q(0,x){\frac{y^2_0\, (x)}{2}}dx
+ \frac{3}{2}\int_0^T\int_0^L q_x{y_x^2}dxdt\nonumber \\
&+& \int_0^T (q{\frac{y_x^2}{2}})(t,0) dt=0.
\end{eqnarray}
Choosing $q(t,x)=x$ leads to
\begin{equation}
-\int_0^T\int_0^L y^2dxdt +
\int_0^L x y(T,x)^2dx - \int_0^L x y_0(x)^2dx +
3\int_0^T\int_0^L  y_x^2dxdt\, =\, 0.
\end{equation}
Hence
\begin{equation}
\int_0^T\int_0^L{y_x^2}\, dxdt  \le \frac{1}{3}
\left(  \int_0^T\int_0^L{y^2}\, dxdt +L\int_0^L y_0(x)^2\, dx \right)
\end{equation}
and then, using (3.14),
\begin{equation}
\Vert y\Vert _{L^2(0,T,H^1(0,L))} \le \left( \frac{4T+L}{3}\right)
^{\frac{1}{2}}\ \Vert y_0\Vert _{L^2(0,L)}.
\end{equation}
By the density of ${\cal D}(A)$ in $L^2(0,L)$ the result extends to
arbitrary $y_0\in L^2(0,L)$.\va
2) We also assume  $y_0\in  {\cal D}(A)$ and take $q=1$ in (3.16). We get
\begin{equation}
\int_0^T y_x^2(t,0)\, dt =\int_0^L y_0^2(x)\, dx -
\int_0^L y^2(T,x)\, dx \le \int_0^L y_0^2(x)\, dx.
\end{equation}
On the other hand the choice $q(t,x)=T-t$ yields
\begin{eqnarray}
\int_0^T\int_0^L y^2 \, dxdt
- \int_0^L Ty_0^2(x)\, dx  +
\int_0^T (T-t)y_x^2(t,0)\, dt=0.
\end{eqnarray}
Hence
\begin{equation}
\int_0^L y_0^2(x)\, dx  \le  \frac{1}{T} \int_0^T\int_0^L y^2 \, dxdt
+ \int_0^T y_x^2(t,0)\, dt.
\end{equation}
Thanks to (3.20) there exists a unique continuous (linear) extension of the
map $y_0\in {\cal D}(A)\mapsto y_x(.,0) \in L^2(0,T)$ to the whole
space $L^2(0,L)$. In the following we also will denote by $y_x(.,0)$
the value of this map at any $y_0\in L^2(0,L)$. Obviously (3.20) and
(3.22) hold
true for any $y_0\in L^2(0,L)$. \qed \va
We now prove the following observability result.
\va\va
{\bf Proposition 3.3:}
{\it Let
${\cal{N}}=\{ 2\pi \sqrt{ \frac{k^2+kl+l^2}{3} }; \ k,l\in \N ^* \}$.
Then
$\forall L\in (0,+\infty ) \setminus {\cal N}$, $\forall T>0$,
$\exists C=C(L,T)>0$ such that
\begin{equation}
\forall y_0\in L^2(0,L) \ \
\Vert y_0 \Vert _{L^2(0,L)}\le C \Vert y_x(.,0) \Vert _{L^2(0,T)}.
\end{equation} }
\va
{\it Proof.} We proceed as in [12: Appendix I] (See also
[1]). If the statement is false, there exists a sequence
$(y_0^n)_{n\ge 0}$ in $L^2(0,L)$ such that
$\Vert y_0^n\Vert _{L^2(0,L)} =1$ for any $n$ but
$\Vert y_x^n(.,0)\Vert _{L^2(0,T)} \to 0$ as $n\to \infty$,
where $y^n:=S(\cdot )y_0^n$. Clearly $y^n$ is bounded in $L^2(0,T,H^1(0,L))$
(see (3.19)). On the other hand $y_t^n=-(y_x^n+y_{xxx}^n)$ is bounded in
$L^2(0,T,H^{-2}(0,L))$. Since the first embedding in
\begin{equation}
H^1(0,L)\hookrightarrow L^2(0,L)\hookrightarrow H^{-2}(0,L)
\end{equation}
is compact, it follows
from a result by Simon (see [18; cor. 4]) that 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 (3.13)
that $(y_0^n)$ is a Cauchy sequence
in $L^2(0,L)$. Let $y_0=\lim_{n\to \infty}\, y_0^n$ and $y=S(\cdot )y_0$.
By Proposition 3.2 $y_x^n(.,0)\to y_x(.,0)$ in $L^2(0,T)$. Thus
$\Vert y_0\Vert _{L^2(0,L)} =1$ and $y_x(.,0)=0$, but such a function
does not exist because of the following: \va\va
{\bf Lemma 3.4:} {\it For $T>0$ let $N_T$ 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 (3.5)-(3.8) satisfies
$y_x(\cdot ,0)=0 \hbox{ in } L^2(0,T)$. Then for
$L\in (0,+\infty )\setminus {\cal N}$, $N_T=(0) \ \ \forall T>0$. }
\va\va
{\it Proof of Lemma 3.4.}  Obviously $T<T'\Rightarrow N_{T'}\subset N_T$.
On the other hand for any $T>0$ $N_T$ is a finite-dimensional
vector space : Indeed if $(y_0^n)$ is a sequence in the unit ball
$\{ y\in N_T;\ \Vert y\Vert _{L^2(0,L)}\le 1\}$ the same argument as above
shows that there exists a convergent subsequence. Since the unit ball is
compact, $N_T$ is finite dimensional by Riesz theorem. Let
$T'>0$ be given. To prove $N_{T'}=(0)$, it is sufficient to find
$0<T<T'$ such that $N_T=(0)$. Since the map
$T\mapsto \hbox{dim}(N_T) \in \N$
is nonincreasing, there exist $T,\epsilon >0$ such that $T<T+\epsilon <T'$
and $\hbox{dim}N_T=\hbox{dim}N_{T+\epsilon}$. (Hence $N_t=N_T$ for
$T\le  t\le T+\epsilon$.) Let $y_0\in N_T$, $y=S(\cdot )y_0$
and $0<t<\epsilon$. Since
$S(\tau )\big( S(t)y_0\big)=S(\tau +t)y_0$ for $0\le\tau \le T$ and
$y_0\in N_{T+\epsilon}$, we see that
\begin{equation}
{\frac{S(t)y_0-y_0}{t}}\in N_T.
\end{equation}
Let
\begin{equation}
M_T:=\{ {\tilde y}=S(\tau ) {\tilde y_0} ;\
0\le \tau \le T,\ {\tilde y_0} \in N_T\} \subset C([0,T],L^2(0,L)).
\end{equation}
Since
$y\in H^1(0,T+\epsilon ,H^{-2}(0,L))$,
\begin{equation}
\lim_{t\to 0^+} {\frac{y(t+\cdot  )-y}{t}}=y'
\ \hbox{ in } L^2(0,T,H^{-2}(0,L)).
\end{equation}
On the other hand (thanks to (3.25))
${\frac{y(t+\cdot  )-y}{t}}\in M_T$ for $0<t<\epsilon$
and $M_T$ is closed in $L^2(0,T,H^{-2}(0,L))$ (since $dim(M_T)<\infty$).
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
\begin{equation}
y'(0)=\lim_{t\to 0^+} {\frac{S(t)y_0-y_0}{t}} \ \hbox{ in } L^2(0,L).
\end{equation}
This in turn implies
\begin{equation}
y_0\in {\cal D}(A),\ \  A(y_0)=y'(0)\in N_T \ \hbox{ and } \
y_x(.,0)\in C([0,T]).
\end{equation}
Thus
\begin{equation}
\left({\frac{dy_0}{dx}}\right) _{\big\vert x=0} =y_x(0,0)=0.
\end{equation}
If $N_T\ne (0)$, the map
$y_0\in \C N_T \mapsto A(y_0)\in \C N_T$ (where $\C N_T$ denote the
complexification of $N_T$) has (at least) one eigenvalue, hence there
exist $\lambda \in \C$, $y_0\in H^3(0,L) \setminus \{ 0\}$
such that (prime denoting here spatial derivative)
\begin{eqnarray}
&&\lambda \, y_0=-{y_0}'-{y_0}''', \\
&&y_0(0)=y_0(L)=y_0'(0)=y_0'(L)=0.
\end{eqnarray}
We prove in the following lemma that this does not hold if
$L\not\in {\cal N}$.
\va\va
{\bf Lemma 3.5:} {\it Let $L\in (0,+\infty )$. Consider the following
assertion:\va
\begin{equation}
({\cal A}) \ \ \
\exists \lambda\in \C ,\ \exists y_0\in H^3(0,L) \setminus \{ 0\}\ \
\hbox{ s.t. }\
\left\{
\begin{array}{cl}
&{\lambda}\, y_0+{y_0}'+{y_0}'''=0, \\
&y_0(0)=y_0(L)=y_0'(0)=y_0'(L)=0.
\end{array}
\right.
\end{equation}
Then $({\cal A}) \iff  L\in {\cal N}$. }
\va\va
{\it Proof of Lemma 3.5.} Assume $y_0$ is as in the assertion $({\cal A})$,
let us denote by $u\in H^2(\R )$ its prolongation by $0$.
Then
\begin{equation}
\lambda u+u'+u'''={y_0}''(0)\delta _0
-{y_0}''(L)\delta _L\  \hbox{ in } \ {\cal D}'(\R ),
\end{equation}
where $\delta _{x_0}$ denotes the
Dirac measure at $x_0$. It is easy to see that $({\cal A})$ is
equivalent to the existence of complex numbers
$\alpha ,\beta ,\lambda$ (with
$(\alpha , \beta )\ne (0,0)$) and of a function
$u\in H^2(\R)$ with compact support in $[-L,L]$ such that
\begin{equation}
\lambda u+u'+u'''=\alpha\delta _0 -\beta\delta _L
\hbox{ in } {\cal D}'(\R ).
\end{equation}
Taking Fourier transform we get
\begin{equation}
\big( \lambda +(i\xi )+(i\xi )^3 \big) \hat u(\xi )=
\alpha -\beta e^{-iL\xi },
\end{equation}
hence (setting $\lambda =-ip$)
\begin{equation}
\hat u (\xi )= i{\frac{\alpha -\beta e^{-iL\xi } }{ {\xi}^3-{\xi}+p} }.
\end{equation}
Using Paley-Wiener theorem (see [19]) and the usual characterization of
$H^2(\R )$ functions by means of their Fourier transforms, we see that
$({\cal A})$ is equivalent to the existence of $p\in \C$ and of
$(\alpha ,\beta )\in \C ^2\setminus \{ (0,0)\}$  such that the map
\begin{equation}
f(\xi ):=
{\frac{\alpha -\beta e^{-iL\xi } }{ {\xi}^3-{\xi}+p} }
\end{equation}
satisfies :
\begin{eqnarray}
&(1)&f \hbox{ is an entire function in } \C ; \nonumber\\
&(2)&\int_{\R}\vert f(\xi )\vert ^2
\big( 1+\vert \xi \vert ^2\big) ^2\, d\xi <\infty ;\nonumber \\
&(3)& \forall \xi\in \C \ \ \
\vert f(\xi )\vert \le C(1+\vert \xi \vert )^N \,
e^{L\vert \hbox{Im}\, \xi \vert} \hbox{ for some positive constants } C,N.
\end{eqnarray}
Since the roots of $\alpha -\beta e^{-iL\xi }$ are simple (unless
$\alpha =\beta =0$), (1) holds provided that the roots of
${\xi}^3-{\xi}+p$ are simple and also roots of  $\alpha -\beta e^{-iL\xi }$.
Notice that if (1) holds true, then (2) and (3) are satisfied. It
follows that $({\cal A })$ is equivalent to the existence of complex
numbers $p,\mu _0$ and of positive integers $k,l$ such that, if we set
\begin{equation}
\mu _1:=\mu _0+k{\frac{2\pi}{L}} \hbox{ and }
\mu _2:=\mu _1+l{\frac{2\pi}{L}},
\end{equation}
we have
\begin{equation}
{\xi}^3-{\xi}+p\equiv (\xi -\mu _0)\, (\xi -\mu _1)\, (\xi -\mu _2),
\end{equation}
that is
\begin{eqnarray}
\mu _0+\mu _1 +\mu _2&=&0,  \\
\mu _0\mu _1 + \mu _0\mu _2  +\mu _1\mu _2 &=&-1, \\
\mu _0\mu _1\mu _2&=&-p.
\end{eqnarray}
Easy calculations lead to
\begin{eqnarray}
L&=&2\pi \sqrt{\frac{k^2+kl+l^2}{3}}, \\
\mu_0&=&-{\frac{1}{3}}(2k+l){\frac{2\pi}{L}}, \\
p&=&-\mu _0(\mu _0+k{\frac{2\pi}{L}} )(\mu _0+ (k+l){\frac{2\pi}{L}}).
\end{eqnarray}
Hence $({\cal A})\iff L\in {\cal N}$. This completes the proof of Lemmas
3.5, 3.4 and of Proposition 3.3. \qed \va
{\it Remark 3.1.} (i) For $L\in {\cal N}$, if $p$ is given by (3.47) and
$y_0$ (with $\hbox{Re}\, y_0\ne 0$) is as in $({\cal A})$
(with $\lambda =-ip$), then
$y(t,x):=\hbox{Re}\, \big( e^{-ipt}y_0(x) \big) $ is a
nontrivial
(smooth) solution
of (3.5)-(3.8) such that $y_x(.,0)\equiv 0$. Thus the result in
Proposition 3.3 holds true iff $L\not\in {\cal N}$. On the other hand
it may be shown (as in [7; remark 4.4, p.56]) that the
existence of unobservable solutions of (3.5)-(3.8)
leads to uncontrollability of equations (3.1)-(3.4). \va
(ii) In the particular situation where $k=l$, the system (3.45)-(3.47)
reduces to
$L=2k\pi ,\ \mu _0=-1,\ p=0$ (hence $\lambda =0$). This yields the
(unobservable) steady solutions of (3.5)-(3.8):
$y(t,x)=a(\cos x -1)$, $a\in \R$.\va
(iii) If the term $y_x$ is removed from (3.5), it may be shown that
Proposition 3.3 is true for every $L>0$.
\va\par
The goal of the following proposition is to define -in a certain weak sense-
solutions of the non-homogeneous problem (3.1)-(3.4).
\va\va
{\bf Proposition 3.6:} {\it There exists a unique linear continuous map
$\Psi :L^2(0,L)\times L^2(0,T)\to B$ such that, for
$y_0\in {\cal D}(A)$ and $h\in C^2([0,T])$
with $h(0)=0$, $\Psi (y_0,h)$
is the unique (classical) solution of (3.1)-(3.4). }
\va\va
{\it Proof.} Let $C_0^2([0,T]):=\{ h\in C^2([0,T],\R );\ h(0)=0\}$. \va
{\bf First step:} We assume here
$y_0\in {\cal D}(A)$ and $h\in C_0^2([0,T])$. Let $\phi \in
C^{\infty }([0,L])$ be such that $\phi (0)=\phi (L)=0,\ \phi '(L)=-1$. The
change of function $z(t,x)=y(t,x)-\big( S(t)y_0\big) (x)+h(t)\phi (x)$
transforms (3.1)-(3.4)
into
\begin{eqnarray}
z_t+z_x+z_{xxx}&=&f(t,x):=h'(t)\phi (x)+h(t)
\big( \phi '(x)+\phi '''(x)\big)              \\
z(t,0)=z(t,L) & = & 0                         \\
z_x(t,L) & = & 0                              \\
z(0,x) & = & 0.
\end{eqnarray}
Since $f\in C^1([0,T],L^2(0,L))$, there exists by Proposition 3.1
and a classical result concerning such non-homogeneous problems (see
[15]) a unique solution
$$z\in C([0,T],{\cal D}(A))\cap C^1([0,T],L^2(0,L))$$ of (3.48)-(3.51).
Hence for (smooth) data $y_0\in {\cal D}(A)$, $h\in C_0^2([0,1])$
(3.1)-(3.4)
 admits a unique (classical) solution
\begin{equation}
y\in C([0,T],H^3(0,L))\cap C^1([0,T],L^2(0,L)).
\end{equation}
 \va
{\bf Second step:} (A priori estimates by multiplier method.)
We also assume $y_0\in {\cal D}(A)$, $h\in C_0^2([0,T])$.
Let $q\in C^{\infty}([0,T]\times [0,L]).$ By (3.52) we may integrate
by parts in
\begin{equation}
\int_0^S\int_0^Lqy(y_t+y_x+y_{xxx})dxdt=0.
\end{equation}
($S$ denotes some arbitrary number in $[0,T]$.) Using (3.1)-(3.4)
(and Fubini's theorem) we get:
\begin{eqnarray}
&-&\int_0^S\int_0^L(q_t+q_x+q_{xxx})\frac{y^2}{2}dxdt +
\int_0^L {\frac{(qy^2)(S,x)}{2}}dx  \nonumber\\
&-& \int_0^L {\frac{(qy^2)(0,x)}{2}}dx  
- \frac{1}{2}\int_0^Sq(t,L)h(t)^2dt  \nonumber \\
&+&\frac{3}{2}\int_0^S\int_0^L q_x{y_x^2}dxdt +
\frac{1}{2}\int_0^S(qy_x^2)(t,0)dt=0.
\end{eqnarray}
Choosing $q=1$ leads to
\begin{equation}
\int_0^L y(S,x)^2dx +\int_0^S y_x(t,0)^2dt=\int_0^L y_0(x)^2dx +
\int_0^S h(t)^2dt
\le \Vert y_0\Vert _{L^2(0,L)}^2 +\Vert h\Vert _{L^2(0,T)}^2.
\end{equation}
Setting $\Vert(y_0,h)\Vert :=
\left( \Vert y_0\Vert _{L^2(0,L)}^2 +\Vert h\Vert _{L^2(0,T)}^2 \right)
^{\frac{1}{2}}$, we get
\begin{equation}
\Vert y\Vert_{C([0,T],L^2(0,L))}\le \Vert (y_0,h)\Vert
\end{equation}
which yields
\begin{equation}
\Vert y\Vert_{L^2((0,T)\times (0,L))}
\le \sqrt{T}\Vert (y_0,h)\Vert ,
\end{equation}
and
\begin{equation}
\Vert y_x(.,0)\Vert _{L^2(0,T)} \le \Vert (y_0,h)\Vert .
\end{equation}
Now take $q(t,x)=x$, and $S=T$. (3.54) gives
\begin{eqnarray}
&-&\int_0^T\int_0^L\frac{y^2}{2}dxdt +
\int_0^Lx\frac{y(T,x)^2}{2}dx - \int_0^L x{\frac{y_0(x)^2}{2}}dx
\nonumber \\
& +& \frac{3}{2}\int_0^T\int_0^L{y_x^2}dxdt
-\frac{1}{2}\int_0^T Lh(t)^2dt =0.
\end{eqnarray}
Hence
\begin{equation}
\int_0^T\int_0^L{y_x^2}dxdt  \le \frac{1}{3}
\left(  \int_0^T\int_0^L{y^2}dxdt  + \int_0^L x y_0(x)^2 dx
+L\int_0^T h(t)^2dt \right)
\end{equation}
and then, using (3.57),
\begin{equation}
\Vert y\Vert _{L^2(0,T,H^1(0,L))} \le \left( \frac{4T+L}{3}\right)
^{\frac{1}{2}}\ \Vert (y_0,h)\Vert .
\end{equation}
Using (3.56), (3.61), and the density of
${\cal D}(A)$ in $L^2(0,L)$ and of
$C_0^2([0,T])$ in $L^2(0,T)$,
we see that the linear map
$(y_0,h)\in {\cal D}(A)\times C_0^2([0,T])\mapsto y\in B$ may be
extended in a unique manner to the whole space
$L^2(0,L)\times L^2(0,T)$ to give a (linear) continuous map
$\Psi :L^2(0,L)\times L^2(0,T)\to B$.
\qed
\va
{\it Remark 3.2.} i) For $y_0\in L^2(0,L)$ and
$h\in L^2(0,T)$, the (weak) solution $\Psi (y_0,h)$
is solution of (3.1) in ${\cal D}'(0,T,H^{-2}(0,L))$. Moreover
$\Psi (y_0,h)(0,.)(=y_0)$ and $\Psi (y_0,h)(T,.)$ are well-defined
in $L^2(0,L)$
since $\Psi (y_0,h)\in C([0,T],L^2(0,L))$.\va
ii) Obviously $\Psi (y_0,0)=S(\cdot )y_0$, hence
$\Psi (y_0,h)=S(\cdot )y_0+\Psi(0,h)$.
\va\par
To apply H.U.M. we need some observability result concerning the following
(backward well-posed) homogeneous problem:
\begin{eqnarray}
u_t+u_x+u_{xxx} & = & 0                      \\
u(t,0)=u(t,L) & = & 0                        \\
u_x(t,0) & = & 0                             \\
u(T,x) & = & u_T(x).
\end{eqnarray}
(Notice that (3.64) (resp. (3.65)) differs from (3.7) (resp. (3.8)) .)
The change of variables $\tau=T-t$, $\xi =L-x$ transforms (3.62)-(3.65)
into
(3.5)-(3.8) and vice-versa. Using Propositions 3.1, 3.2 and 3.3
we readily get the following result. \va\va
{\bf Proposition 3.7: } {\it Let $L,T>0$. For any $u_T\in L^2(0,L)$ the
mild solution of (3.62)-(3.65) belongs to $B$, the function $u_x(.,L)$
makes sense in $L^2(0,T)$. If moreover $L\not\in {\cal N}$, there exists
a constant $C=C(L,T)>0$ such that for any $u_T\in L^2(0,L)$ we have
\begin{equation}
\Vert u_x(.,L) \Vert _{L^2(0,T)} \le
\Vert u_T \Vert _{L^2(0,L)} \le
C \, \Vert u_x(.,L) \Vert _{L^2(0,T)}.
\end{equation} }
\par
It remains to apply H.U.M..\va
{\it Proof of Theorem 1.2.}  Thanks to Remark 3.2 ii),
we may assume without loss of
generality that $y_0=0$. (See the proof of Th. 1.1 for the details.) Let
$(u_T,h)\in C_c^{\infty} (0,L)\times C_c^{\infty} (0,T)$,
let $u$ (resp. $y$) be
the (classical) solution of (3.62)-(3.65) (resp. (3.1)-(3.4)).
Integrating by parts in
\begin{equation}
\int_0^T\int_0^L u(y_t+y_x+y_{xxx})dxdt=0,
\end{equation}
we readily get
\begin{equation}
\int_0^L u_T(x)y(T,x)dx=\int_0^T u_x(t,L)h(t)dt.
\end{equation}
By an argument of density we see that (3.68) holds true for $u_T\in L^2(0,L)$
and $h\in L^2(0,T)$. Let $\Lambda$ denote the (linear) continuous map
\begin{equation}
u_T\in L^2(0,L)\mapsto y(T,.)\in L^2(0,L),
\end{equation}
$y$ standing for the solution of (3.1)-(3.4) associated with the data
$h:=u_x(.,L)\in L^2(0,T)$. It follows from (3.68) and Prop. 3.7
that
\begin{equation}
\big(\Lambda (u_T),u_T\big) _{L^2(0,L)}=
\Vert u_x(.,L)\Vert _{L^2(0,T)}^2 \ge
C^{-2} \Vert u_T \Vert _{L^2(0,L)}^2.
\end{equation}
Hence $\Lambda$ is invertible by Lax-Milgram theorem. The proof of
Theorem 1.2 is complete. \qed
\va
{\it Remark 3.3.} When $y_0=0$, H.U.M. yields a 
(linear) continuous selection of the
control, namely the map
\begin{equation}
\Gamma :\ y_T\in L^2(0,L)\mapsto u_x(.,L) \in L^2(0,T),
\end{equation}
where $u$ denotes the solution of (3.62)-(3.65) associated with
$u_T:=\Lambda ^{-1}(y_T)$.
%\setcounter{chapter}{4}
\setcounter{equation}{0}
\va
\section{Exact boundary controllability of nonlinear KdV equation on
a bounded domain.}
\par
In this section we prove that the following boundary-control system
\begin{eqnarray}
y_t+y_x+yy_x+y_{xxx} & = & 0   \\
y(t,0)=y(t,L) & = & 0           \\
y_x(t,L) & = & h(t)\quad \quad (h\in L^2(0,T))  \\
y(0,x) & = & y_0(x)
\end{eqnarray}
is exactly controllable in a neighbourhood of the nul 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 $\Vert y_0\Vert _{L^2(0,L)}<r_0$, $\Vert y_T\Vert _{L^2(0,L)} <r_0$
we may find
$y\in B=C([0,T],L^2(0,L))\cap L^2(0,T,H^1(0,L))$
such that 1) $y_t=-(y_x+yy_x+y_{xxx})$ in ${\cal D}'(0,T,H^{-2}(0,L))$ and 2)
$y(0,.)=y_0,\ y(T,.)=y_T$. (Notice that for $y\in B$,
$y_x\in L^2(0,T,L^2(0,L))$,  $y_{xxx}\in L^2(0,T,H^{-2}(0,L))$  and
$yy_x\in L^1(0,T,L^2(0,L))$ - see below Prop. 4.1 -, hence
$y_x+yy_x+y_{xxx}\in  L^1(0,T,H^{-2}(0,L))$.)
If moreover $L\not\in {\cal N}$, then (4.2) is true in $L^2(0,T)$ and
(4.3) holds in a weak sense (that we will not specify) for some
control function $h\in L^2(0,T)$.
\par
To solve (4.1)-(4.4), we write $y=S(t)y_0+y_1+y_2$
where $(S(t))_{t\ge 0}$ denotes the semi-group associated with the operator
$A$ of section 3,
$y_1$ and
$y_2$ are (respective) solutions of two nonhomogeneous problems:
\begin{eqnarray}
y_{1t}+y_{1x}+y_{1xxx}& = & 0                       \\
y_1(t,0)=y_1(t,L)& = & 0                            \\
y_{1x}(t,L) & = & h(t)                               \\
y_1(0,x) & = & 0
\end{eqnarray}
and
\begin{eqnarray}
y_{2t}+y_{2x}+y_{2xxx} & = & f                          \\
y_2(t,0)=y_2(t,L) & = & 0                                   \\
y_{2x}(t,L) & = & 0                                         \\
y_2(0,x) & = & 0.
\end{eqnarray}
(In (4.9) we have set $f=-yy_x$.)
Let $\psi_1 :h\in L^2(0,T)\mapsto y_1\in B$ be the map which associates
with
$h$ the weak solution of (4.5)-(4.8). By Proposition 3.6, $\psi _1$
is a (linear) continuous map. The following proposition deals with
equations (4.9)-(4.12):  \va\va
{\bf Proposition 4.1:} {\it 1) If $y\in L^2(0,T,H^1(0,L))$,
$yy_x\in L^1(0,T,L^2(0,L))$ and the map $y\mapsto yy_x$ is continuous.\va
2) For $f\in L^1(0,T,L^2(0,L))$ the mild solution $y_2$ of (4.9)-(4.12)
belongs to $B$. Moreover the linear map $\psi_2 :f\mapsto y_2$ is continuous.}
\va\va
{\it Remark 4.1.}
Recall that for
 $f\in L^1(0,T,L^2(0,L))$ the mild solution $y_2$ of (4.9)-(4.12)
is given by
\begin{equation}
y_2(t,.):=\int_0^t S(t-s)f(s,.)ds.
\end{equation}
\va\va
{\it Proof.} 1) Let $y,\ z\in L^2(0,T,H^1(0,L))$. Let $C_1$ denote
the norm of the Sobolev embedding
$H^1(0,L)\hookrightarrow L^{\infty} (0,L)$.
Applying triangular inequality and H\"{o}lder's inequality we get
\vfill\eject
\begin{eqnarray}
\Vert yy_x &-& zz_x\Vert _{L^1(0,T,L^2(0,L))} \nonumber \\
&\le& \int_0^T \Vert (y-z)y_x(t,.)\Vert _{L^2(0,L)}dt +
\int_0^T \Vert z(y_x-z_x)(t,.)\Vert _{L^2(0,L)}dt  \nonumber \\
&\le& \int_0^T \Vert (y-z)(t,.)\Vert _{L^{\infty} (0,L)}
\Vert y_x(t,.)\Vert _{L^2(0,L)}dt \nonumber \\
&\ &\quad +
\int_0^T \Vert z(t,.)\Vert _{ L^{\infty} (0,L) }
\Vert (y_x-z_x)(t,.)\Vert _{L^2(0,L)}dt    \nonumber \\
&\le& C_1\left( \int_0^T \Vert (y-z)(t,.)\Vert _{H^1 (0,L)}
\Vert y(t,.)\Vert _{H^1(0,L)}dt  \right. \nonumber \\ 
&\ &\quad\  +\left.
\int_0^T \Vert z(t,.)\Vert _{ H^1(0,L) }
\Vert (y-z)(t,.)\Vert _{H^1(0,L)}dt \right) \nonumber \\
&\le& C_1\left( \Vert y\Vert _{L^2(0,T,H^1(0,L))} \right. \nonumber\\
&\ &\quad\  +\left. 
\Vert z\Vert _{L^2(0,T,H^1(0,L))} \right)
\Vert y-z\Vert _{L^2(0,T,H^1(0,L))}.
\end{eqnarray}
(4.14) with $z=0$ yields $yy_x\in L^1(0,T,L^2(0,L))$, and (4.14)
with $z$ tending to $y$ gives the continuity of the map $y\mapsto yy_x$.\va\va
2) Since
\begin{equation}
\Vert 1_{[0,t]}(s)S(t-s)f(s,.)\Vert _{L^2(0,L)}
\le  \Vert f(s,.)\Vert _{L^2(0,L)}
  \in L^1(0,T),
\end{equation}
it follows from Lebesgue's theorem that the mild solution
$y_2(t,.)=\int_0^t S(t-s)f(s,.)ds$ belongs to $C([0,T],L^2(0,L))$.
Moreover for every $t\in [0,T]$
\begin{equation}
\Vert y_2(t,.)\Vert_{L^2(0,L)} \le
\int_0^t\Vert f(s,.)\Vert _{L^2(0,L)}ds\le
\Vert f\Vert _{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 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:
\begin{equation}
\exists C_2 >0,\ \forall f\in {C^1}([0,T],L^2(0,L))\quad \quad
\Vert y_{2x}\Vert _{L^2((0,T)\times (0,L))}
\le C_2\Vert f\Vert _{L^1(0,T,L^2(0,L))}.
\end{equation}
Integrating by parts in
\begin{equation}
\int_0^T\int_0^L x y_2(y_{2t}+y_{2x}+y_{2xxx})dxdt=
\int_0^T\int_0^L x y_2f\, dxdt
\end{equation}
we get
\begin{eqnarray}
\int_0^L x\frac{y_2(T,x)^2}{2}dx
&-& \int_0^T\int_0^L\frac{y_2(t,x)^2}{2}dxdt
+\frac{3}{2}\int_0^T\int_0^L y_{2x}(t,x)^2dxdt \nonumber \\
&=& \ \int_0^T\int_0^L x(fy_2)(t,x)dxdt.
\end{eqnarray}
Hence, using (4.16)
\begin{eqnarray}
\int_0^T\int_0^L y_{2x}(t,x)^2dxdt
&\le & \frac{1}{3}\int_0^T\int_0^L {y_2(t,x)}^2dxdt +
\frac{2}{3}L\int_0^T
\Vert f(t,.)\Vert _{L^2(0,L)}.\Vert y_2(t,.)\Vert _{L^2(0,L)}dt
\nonumber \\
&\le & \frac{1}{3} T\Vert f\Vert _{L^1(0,T,L^2(0,L))}^2
+\frac{2}{3}L\Vert f\Vert _{L^1(0,T,L^2(0,L))}.
\Vert y_2\Vert _{C([0,T],L^2(0,L))}  \nonumber \\
&\le & \frac{1}{3}(T+2L)\Vert f\Vert _{L^1(0,T,L^2(0,L))}^2 .
\end{eqnarray}
The proof of Proposition 4.1 is complete. \qed
\par
We may now prove the main result of this paper.\va
{\it Proof of Theorem 1.3.}
We first assume that $L\not\in {\cal N}$.
We show that for $T>0$ there exists $r_0>0$
(small enough) such that if
$\Vert y_0\Vert _{L^2(0,L)}$,
$\Vert y_T\Vert _{L^2(0,L)} <r_0$, the state $y_T$ may
be reached from $y_0$ for nonlinear KdV equation.
Let $y_0,\ y_T$ be states in $L^2(0,L)$ such that
$\Vert y_0\Vert _{L^2(0,L)}$,
$\Vert y_T\Vert _{L^2(0,L)}\le r$, $r>0$ to be chosen later.
Let $F$ denote the nonlinear map
\begin{equation}
\begin{array}{l}
y\in L^2(0,T,H^1(0,L))\\
\ \ \mapsto F(y):=S(.)y_0+
\psi _1 \circ \Gamma(y_T-S(T)y_0+\psi _2(yy_x)(T,.))+
\psi _2(-yy_x)  \in B.
\end{array}
\end{equation}
($\Gamma$ is defined in Remark 3.3, $\psi_1$ and $\psi _2$ are defined
at the beginning of this section.)
$F$ is well-defined and continuous
by Propositions 3.2, 3.6, 4.1 and Remark 3.3. Clearly each fixed point of
$F$ verifies (4.1) -in ${\cal D}'(0,T,H^{-2}(0,L))$ -, (4.4) and
$y(T,.)=y_T$. For proving existence of fixed-point for $F$ we
apply Banach contraction fixed-point theorem to the restriction
of $F$ to some closed ball $\overline{B}(0,R)$ in $L^2(0,T,H^1(0,L))$
($R$ will be chosen later.) We need:
\begin{equation}
F\big( \overline{B}(0,R)\big) \subset \overline{B}(0,R),
\end{equation}
and
\begin{equation}
\exists C_3\in ]0,1[,\ \forall y,z\in \overline{B}(0,R)
\ \ \Vert F(y)-F(z)\Vert \le C_3\Vert y-z\Vert,
\end{equation}
where $\Vert .\Vert$ from now on stands for the norm $L^2(0,T,H^1(0,L))$.
Let $K_1$ (resp. $K_2,\ K_2'$) denote the norm of $\psi _1$ (resp.
$\psi _2, \ \psi _2$) 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 let $K$ denote the norm of $\Gamma$ as a map from $L^2(0,L)$ into
$L^2(0,T)$. Set $K_3=\sqrt{\frac{4T+L}{3}}$.
Let $y,z\in L^2(0,T,H^1(0,L))$. Assume
\begin{equation}
\Vert y\Vert \le R,\ \Vert z\Vert \le R.
\end{equation}
Then by (3.19) and (4.14)
\begin{eqnarray}
\Vert F(y)\Vert &\le & \sqrt{\frac{4T+L}{3}} \Vert y_0\Vert _{L^2(0,L)} +
K_1K(\Vert y_T\Vert _{L^2(0,L)}+\Vert y_0\Vert _{L^2(0,L)}+
K_2'C_1\Vert y\Vert ^2) +
K_2C_1\Vert y\Vert ^2  \nonumber \\
&\le & C_1(K_2+KK_1K_2')R^2+(2KK_1+K_3)r.
\end{eqnarray}
(4.22) leads to the first condition on $R,\ r$:
\begin{equation}
C_1(K_2+KK_1K_2')R^2+(2KK_1+K_3)r \le R.
\end{equation}
Now write
\begin{equation}
F(y)-F(z)=\psi _2(zz_x-yy_x)+\psi _1\circ \Gamma
\big( \psi _2(yy_x-zz_x)(T,.)\big) .
\end{equation}
We get, by (4.14),
\begin{equation}
\Vert F(y)-F(z)\Vert \le 2C_1(K_2+KK_1K_2')R\Vert y-z\Vert .
\end{equation}
(4.23) will hold provided that
\begin{equation}
2C_1(K_2+KK_1K_2')R <1.
\end{equation}
Let $R$ be some positive number verifying (4.29). Then (4.26) holds true
if we take
\begin{equation}
r:= \frac{R}{2(2KK_1+K_3)}.
\end{equation}
Setting
\begin{equation}
r_0:=\big( 4C_1(2KK_1+K_3)(K_2+KK_1K_2')\big) ^{-1},
\end{equation}
we see that $r\to r_0$ as $R\to \big( 2C_1(K_2+KK_1K_2') \big) ^{-1}$. It
follows that if $\Vert y_0\Vert _{L^2(0,L)}<r_0$
every $y_T$ with $\Vert y_T\Vert _{L^2(0,L)}<r_0$
may be reached by a solution of nonlinear KdV equation coming
from $y_0$. The proof of Theorem 1.3 is achieved when $L\not\in {\cal N}$.
If now $L\in {\cal N}$, it is sufficient to consider some ${\bar L}>L$
such that
${\bar L}\not\in {\cal N}$ and to apply the theorem to the functions
${\bar y}_0,\ {\bar y}_T\in L^2(0,{\bar L} )$, where
${\bar y}_0,\ {\bar y}_T$ denote
the prolongations (by $0$)
of (given) states $y_0$, $y_T\in L^2(0,L)$, and then to restrict the
solution $\bar y$ to the domain $(0,T)\times (0,L)$.
The proof of Theorem 1.3 is complete. \qed
\va
\sou{Acknowledgments}\va
The author thanks the Dipartimento di Matematica of Politecnico Di Torino
for its hospitality. The author also like to thank J.M. CORON for bringing
his attention to the boundary controllability of KdV and E. ZUAZUA for
pointing to him techniques permitting to drop useless assumptions.
\va\va
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\end{document}
