\documentclass[12pt]{article} 
\title{ Exact boundary controllability for the linear\\
Korteweg-de Vries equation on the half-line} 
\author{Lionel Rosier}
\date{}
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\def\O{\Omega}
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\def\proof{\noindent {\em Proof:}\qquad}
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\newtheorem{thm}{Theorem}
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\begin{document} 
\maketitle
\vspace*{-1cm}
\begin{center}

Laboratoire d'Analyse Num\'erique et EDP,\\
Universit\'e Paris-Sud,\\
B\^atiment 425,\\ 
91405 Orsay Cedex, France.\\
\vspace{0.5cm}
Phone: (33) 1 69 15 66 57\\
Fax:   (33) 1 69 15 67 18\\
{\footnotesize
e-mail: Lionel.Rosier@math.u-psud.fr}\\
\vspace{2.cm}
{\large Submitted to {\em SIAM J. Control and Optimization}}\\
\end{center}
\vspace{1.cm}
\begin{abstract}
%\baselineskip1.cm
This paper is concerned with the controllability of the linear
Korteweg-de Vries equation
on the domain $\O =(0,+\infty )$, the control being applied at the
left endpoint $x=0$. It is shown that the {\em exact} boundary
controllability holds true in $L^2(0,+\infty )$ provided  that the
solutions are not required to be in $L^{\infty }(0,T,L^2(0,+\infty ))$.
The proof rests on some Carleman's estimate
and some approximation theorem.
A similar result is obtained for
the heat equation and for the Schr\"odinger equation.
\va\va
{\bf Key words.} Exact boundary controllability, 
unbounded domain, Carleman's estimate
\va
{\bf AMS subject classifications.} 35Q20, 93B05, 93C20
\end{abstract}
\renewcommand{\theequation}{\thesection.\arabic{equation}}
\vfill\eject
\setcounter{equation}{0}
\section{Introduction and main results}
\va
%\baselineskip1.cm
The Korteweg-de Vries (KdV) equation 
\begin{equation}
\label{eq:100}
u_t+u_x+uu_x+u_{xxx}=0,\quad t\ge 0,\  x\in \O \subset\R
\end{equation}
may serve as a model for (among other things) propagation of small
amplitude long water waves in a uniform channel. In this context,
$t$ is time, $x$
is the space variable and $u$ stands for the deviation of the liquid's
surface from the equilibrium position. 
The boundary (resp. internal) controllability of (\ref{eq:100}) has been
extensively studied
(see \cite{RZ1}, \cite{RZ3}, \cite{R1}, \cite{R2}, and also
\cite {Mi} for the BBM equation) 
when $\O$ is bounded, say $\O =(0,L)$. 
The (local) exact boundary  
controllability of (\ref{eq:100}) follows in \cite{R1} from the {\em exact} 
boundary controllability  of the associated linear KdV equation, namely
\begin{equation}
\label{eq:101}
u_t+u_x+u_{xxx}=0.
\end{equation} 
Up to date, there is no result as far as the boundary controllability
of (\ref{eq:100}) or (\ref{eq:101}) on some 
{\em unbounded} domain (say $\Omega =(0,+\infty )$)
is concerned. The aim of this paper is to
fill this gap in providing a study
of the exact boundary controllability of (\ref{eq:101}) on $(0,+\infty )$,
which may be seen as a first step in the knowledge of the control theory
for (\ref{eq:100}) on unbounded domains.
It should be observed that the {\em approximate} boundary controllability of
({\ref{eq:101}) in $L^2(0,+\infty )$
is quite easy to be proved, whereas the {\em exact} boundary 
controllability requires a more sophisticated analysis, due to a lack of
compactness. An enlightening example of the difference between exact and
approximate (internal) controllabilities for linear PDE in unbounded domains is 
provided by next result, whose (simple) proof is sketched in appendix.
\begin{prop}
Consider a (real) constant coefficients differential operator
$Au=\sum_{i=0}^n a_i\frac{d^i\, u}{d\, x^i}$, with domain
${\cal D}(A)=\{ u\in L^2(\R );\ Au\in L^2(\R )\}$.
Assume that $n\ge 2$ (with $a_n\ne 0$) and that $A$ generates
a continuous semigroup $\big( S(t)\big) _{t\ge 0}$ on $L^2(\R )$.
Let $T>0$ and  $L_1<L_2$ be some numbers. Set 
\begin{equation*} 
{\cal R}=\left\{ \int_0^T\, S(T-t)f(t,\cdot )dt;\ \ 
f\in L^2(\R ^2),\ \hbox{supp}\, f\subset [0,T]\times [L_1,L_2]\right\}
\end{equation*}
where $\hbox{supp}\, f$ denotes the support of $f$.
Then ${\cal R}$ is a {\em strict dense} subspace of $L^2(\R )$.
\end{prop}
In other words, when considering {\em mild} solutions (which belong to
$C([0,T],L^2(\R ))\,$) of the forced initial value problem
\begin{equation*}
\left\{
\begin{array}{rl}
\frac{du}{dt}-Au&=f\\
u(0)&=0,
\end{array}
\right.
\end{equation*}
where $f$ denotes any square integrable function supported in
$[0,T]\times [L_1,L_2]$, then the space ${\cal R}$ of all reachable states is
dense in (but different from) $L^2(\R )$. Notice that
for $Au=-u_{xxx}-u_x$, letting
$L_1=-1<L_2=0$ and taking the restrictions
to $(0,+\infty )$ of the mild solutions, we readily
infer the approximate {\em boundary} controllability of (\ref{eq:101})
in $L^2(0,+\infty )$.
It turns out that the {\em exact} boundary controllability of (\ref{eq:101})
in $L^2(0,+\infty )$
also fails to be true if we restrict ourselves to solutions with bounded
energy, that is which belong to $L^{\infty}(0,T,L^2(0,+\infty ))$. An 
{\em implicit} formulation (that is without specification of the boundary 
conditions) of this fact is given in next theorem,
to be proved in this paper.
\begin{thm}
Let $T>0$. Then there exists $u_0\in L^2(0,+\infty )$ such that if $u$
is any function in $L^{\infty}(0,T,L^2(0,+\infty ))$ satisfying
\begin{equation}
\label{eq:102}
\left\{
\begin{array}{rl}
u_t+u_x+u_{xxx}&=0\ 
\hbox{ in } {\cal D}'\big( (0,T)\times (0,+\infty )\big) \\
u\at{t=0}&=u_0,
\end{array}
\right.
\end{equation}
then $u\at{t=T}\ne 0$. 
\end{thm}
(Notice that $u\at{t=0}$ and $u\at{t=T}$ are
meaningful in $H^{-3}(0,+\infty )$
for any $u\in L^{\infty }(0,T,L^2(0,+\infty ))$ satisfying (\ref{eq:102}):
Indeed, such a function belongs to $W^{1,\infty}(0,T,H^{-3}(0,+\infty ))$.)
Theorem 1 tells us that even the (boundary) {\em null-controllability} fails
to be true for solutions with bounded energy. Nevertheless, when the
bounded energy condition ($u\in L^{\infty }(0,T,L^2(0,+\infty ))\, $) is
dropped, the exact boundary controllability of KdV holds true, as it is
shown in the following theorem, which is the main result of the paper.
\begin{thm}
Let $T,\epsilon ,b$ be positive numbers, with $\epsilon <\frac{T}{2}$. Let 
$L^2((0,+\infty ),e^{-2bx}dx)$ 
denotes the space of (class of) measurable functions 
$u:(0,+\infty )\to \R$ such that 
$\int_0^{+\infty}u^2(x)e^{-2bx}\, dx <\infty$. Let $u_0\in L^2(0,+\infty )$
and $u_T\in L^2((0,+\infty ),e^{-2bx}dx)$. Then  there exists a function 
\begin{equation*}
u\in L^2 _{loc}\big( [0,T]\times [0,+\infty )\big)
\cap C\big( [0,\epsilon ],L^2(0,+\infty )\big)
\cap  C\big( [T-\epsilon ,T],L^2((0,+\infty ),e^{-2bx}dx)\big)
\end{equation*}
which solves 
\begin{equation}
\label{eq:1000}
\left\{
\begin{array}{rl}
u_t+u_x+u_{xxx}&=0\
\hbox{ in } {\cal D}'\big( (0,T)\times (0,+\infty )\big) ,\\
u\at{t=0}&=u_0,\\
u\at{t=T}&=u_T.
\end{array}
\right.
\end{equation}
\end{thm}
\noindent
Let us do some comments. \va
1) The proof of Theorem 2 combines Fursikov-Imanuvilov's approach
(see \cite{FI})
for the boundary controllability of Burgers equation on bounded 
domains,  which is based on 
some global Carleman's estimate, and, for the extension to some 
unbounded domain,
Rosay's clever proof of Malgrange-Ehrenpreis' theorem (see
\cite{Rosay}),  which uses
some approximation theorem. Roughly speaking, the approximation
theorem allows to modify a sequence of solutions of $u_t+u_x+u_{xxx}=f$,
defined on an increasing sequence of domains, in such a way that it
converges (strongly) in $L^2_{loc}(\R ^2 )$. It should be
emphasized that our approach allows to
consider initial and final states in {\em different} spaces of functions,
exploiting thus some {\em asymmetric} property of the KdV equation, namely
the (forward) wellposedness of (\ref{eq:101})
in the asymmetric space $L^2(\R, e^{2bx}dx)$ for any
$b>0$ (see \cite{Ka}). Notice 
that we may require that $u\in C([T-\epsilon ,T],L^2(0,+\infty ))$ if
$u_T$ is also assumed to be in $L^2(0,+\infty )$.\va
2) As in \cite{C} and \cite{L}, the formulation of previous 
boundary controllability
result is {\em implicit}. Nevertheless, setting $h_0=u\at{x=0}$, 
$h_1={u_x}\at{x=0}$ and $h_2={u_{xx}}\at{x=0}$, it may be seen that 
$h_0,h_1,h_2\in H^{-1}(0,T)$ and, thanks to Holmgren's uniqueness
theorem, that $u$ is the {\em only} solution (in the same space as above)
of the initial-value boundary problem
\begin{equation*}
\left\{
\begin{array}{rl}
u_t+u_x+u_{xxx}&=0\ 
\hbox{ in } {\cal D}'\big( (0,T)\times (0,+\infty )\big) ,\\
u\at{x=0}&=h_0,\ {u_x}\at{x=0}=h_1,\ {u_{xx}}\at{x=0}=h_2,\\
u\at{t=0}&=u_0,\\
\end{array}
\right.
\end{equation*}
Moreover $u$ satisfies $u\at{t=T}=u_T$.
\va
3) The method described in 1) applies also to many other linear P.D.E.
for which the characteristic hyperplans take the form  $\{ t=Const.\}$: For
instance the heat equation $u_t-\Delta u=0$ and
the Schr\"odinger equation $iu_t+\Delta u=0$ are concerned. (See Section 5.)

The paper is outlined as follows. The proof of Theorem 1 is given in
Section 2. It rests on some duality argument
and on the behaviour of the traces ${u_x}\at{x=0}$,
${u_{xx}}\at{x=0}$ of exponential solutions for 
(\ref{eq:101}) with the boundary condition $u\at{x=0}=0$. 
Some global Carleman's estimate for the KdV equation (which is
subsequently used) is stated and proved in Section 3. The proof of Theorem 2
is given in Section 4, together with the proof of the approximation theorem
(Lemma 5). In the last section we sketch the proof of similar results
for the heat equation and the Schr\"odinger equation.

>From now on, for the sake of brevity, we shall write $P$ for the operator
$\frac{\partial}{\partial t}+ \frac{\partial}{\partial x} +
\frac{\partial ^3}{\partial x^3}$.
\setcounter{equation}{0}
\section{Proof of Theorem 1}

The proof of Theorem 1 rests on the following key result. 
\begin{lem}
There exists a family $(v^{\lambda})_{\lambda >0}$ of functions in 
$\cap_{n\ge 0} C^{\infty}([0,T],H^n(0,+\infty ))$ such that
for every $\lambda >0$ 
\begin{equation}
\label{eq:301}
P\, v^{\lambda } =0 \quad \hbox{in } (0,T)\times(0,+\infty ),
\end{equation}
\begin{equation}
\label{eq:302}
{v^{\lambda }}\at{x=0} =0 \quad \hbox{on } (0,T),
\end{equation}
\begin{equation}
\label{eq:303}
\Vert {v^{\lambda }}\at{t=0}\Vert _{L^2(0,+\infty )}=1,
\end{equation}
and
\begin{equation}
\label{eq:304}
\Vert {v^{\lambda }_x}\at{x=0}\Vert _{H^n(0,T)}+
\Vert {v^{\lambda }_{xx}}\at{x=0}\Vert _{H^n(0,T)}\to 0 \hbox{ as }
\lambda  \to 0\  (\hbox{ for every }n\ge 1).
\end{equation}
\end{lem}
\noindent{\em Proof.}
Let us consider the operator $Av:=-v_{xxx}-v_x$ with domain 
\begin{equation*}
D(A)=H^3(0,+\infty )\cap H^1_0(0,+\infty )\subset L^2(0,+\infty ).
\end{equation*}
Then
$A$ generates a strongly continuous semigroup $\big( S(t) \big) _{t\ge 0}$
on $L^2(0,+\infty )$, and we shall search for $v^{\lambda }$ in the form 
of an exponential solution:  
\begin{equation*}
v^{\lambda }(t,\cdot )=S(t)v^{\lambda }_0
=e^{-{\lambda }t}v^{\lambda }_0, 
\end{equation*}
where 
$v^{\lambda }_0\in D(A)$ solves $A v^{\lambda }_0
=-\lambda  v^{\lambda }_0$ and 
$\lambda  \in (0, +\infty )$. The roots of the equation $-z^3-z=- \lambda  $ 
may be written in the form 
$r, -\frac{r}{2}\pm i\mu$, where $0<r\sim \lambda $ as 
$\lambda \to 0^+$
and $\mu =\left( 1+\frac{3}{4}r^2 \right) ^{\frac{1}{2}}$.
Let
$w^{\lambda }_0(x):=\Im m\left( e^{(-\frac{r}{2}+i\mu )x} \right) =
e^{-\frac{r}{2} x}\sin (\mu x)$ (hence $w^{\lambda }_0\in D(A)$ and 
$A w^{\lambda }_0=-{\lambda }w^{\lambda }_0$). Easy calculations give 
\begin{equation*}
\Vert w^{\lambda }_0 \Vert _{L^2(0,+\infty )}=
\left( \frac{2\mu ^2}{r(r^2+4\mu ^2)} \right) ^{\frac{1}{2}} \cdot
\end{equation*}
Set  $c_{\lambda }:= \left( \frac{r(r^2+4\mu ^2)}{2\mu ^2} 
\right) ^{\frac{1}{2}}$,
$v^{\lambda }_0:=c_{\lambda }w^{\lambda }_0$ and
$v^{\lambda }(t,x):=e^{-{\lambda }t}v^{\lambda }_0(x)$.
Since
we know that (\ref{eq:301})-(\ref{eq:303}) be true, it remains to prove
(\ref{eq:304}). Obviously $v^{\lambda }_x(t,0)=c_{\lambda }\mu
e^{-{\lambda }t}$
and
$v^{\lambda }_{xx}(t,0)=-c_{\lambda }\mu\,  r\,
e^{-{\lambda }t}$
and since $c_{\lambda }\to 0$
as ${\lambda }\to 0^+$, (\ref{eq:304}) follows. \qed
\va\indent
It will result from next lemma that the traces $u\at{x=0}$,
${u_x}\at{x=0}$ and ${u_{xx}}\at{x=0}$ of a bounded energy solution 
$u=u(t,x)$
of (\ref{eq:101}) belong to the dual space to $H^1(0,T)$ (which does not be
confused with $H^{-1}(0,T)=H^1_0(0,T)'$). 
\begin{lem}
Let $T$ and $L$ be positive numbers and let
$u\in L^{\infty}(0,T,L^2(0,L))$ be such that $P\, u=0$  in
${\cal D}'\big( (0,T)\times (0,L)\big)$.
Then $u\in H^3(0,L,H^1(0,T)')$ and we have
for some constant $C=C(L,T)>0$
\begin{equation}
\label{eq:306}
\Vert u(\cdot ,0)\Vert _{H^1(0,T)'} + 
\Vert u_x(\cdot ,0)\Vert _{H^1(0,T)'} +          
\Vert u_{xx}(\cdot ,0)\Vert _{H^1(0,T)'} \le C
\Vert u\Vert _{L^{\infty }(0,T,L^2(0,L))}.          
\end{equation}
\end{lem}
\noindent{\em Proof.} 
Since $u_t=-(u_{xxx}+u_x)\in L^2(0,T,H^{-3}(0,L))$ we see that 
$u\in H^1(0,T,H^{-3}(0,L))$, hence for every $f\in H^1(0,T,H^3_0(0,L))$
\begin{equation}
\label{eq:3060}
\int_0^T <u_t,f>\, dt 
=-\int_0^T\!\! \int_0^L uf_t\, dxdt + [<u,f>]_{t=0}^T,
\end{equation}
where $<\cdot ,\cdot >$ denotes the duality pairing
$<\cdot ,\cdot >_{H^{-3}(0,L),H^3_0(0,L)}$.
Since $u\in C([0,T],H^{-3}(0,L))\cap L^{\infty}(0,T,L^2(0,L))$,
$u$, as a function of $t$, is weakly continuous in $L^2(0,L)$. Hence 
\begin{equation}
\begin{array}{rl}
\label{eq:307}
\left\vert \int_0^T <u_t,f>\, dt\right\vert 
&\le
\Vert u\Vert_{L^2((0,T)\times (0,L))}\cdot 
\Vert f_t\Vert_{L^2((0,T)\times (0,L))}\\
&\quad +\Vert u\Vert _{L^{\infty}(0,T,L^2(0,L))}\cdot \big(
\Vert f(0,\cdot )\Vert _{L^2(0,L)} + \Vert f(T,\cdot )\Vert _{L^2(0,L)} \big)
\\ 
&\le C_1\Vert u\Vert _{L^{\infty}(0,T,L^2(0,L))}\cdot
\Vert f\Vert _{L^2(0,L,H^1(0,T))}
\end{array}
\end{equation}
for some constant $C_1=C_1(T,L)>0$. Since $H^1(0,T,H^3_0(0,L))$ is dense
in $L^2(0,L,H^1(0,T))$, we infer from (\ref{eq:307}) that
$u_t\in L^2(0,L,H^1(0,T)')$.
Integrating three times w.r.t. $x$ in the
following equation 
\begin{equation*}
(u_{xx}+u)_x=-u_t\qquad \hbox{ in }{\cal D}'(0,L,H^1(0,T)'),
\end{equation*}
($u_t\in L^2(0,L,H^1(0,T)')$ being given by (\ref{eq:3060})),  
we deduce that $u\in H^3(0,L,H^1(0,T)')$ and (\ref{eq:306}) follows. \qed
\va\indent
We now proceed to the proof of Theorem 1. Arguing by contradiction, 
we assume that for every $u_0\in L^2(0,+\infty )$ there exists a 
function $u\in L^{\infty}(0,T,L^2(0,+\infty ))$ such that  $P\, u=0$ in
${\cal D}'((0,T)\times (0,+\infty ))$, 
$u\at{t=0}=u_0$ and $u\at{t=T}=0$. Let $E$ denote
the space 
\begin{equation*}
\{ u\in L^{\infty} (0,T,L^2(0,+\infty )),\  P\, u=0 
\hbox{ in }{\cal D}'((0,T)\times (0,+\infty ))
\hbox{ and }
u\at{t=T}=0 \} ,
\end{equation*}
endowed with the norm 
$\Vert u\Vert _E=\Vert u\Vert _{L^{\infty}(0,T,L^2(0,+\infty ))}$. It is a 
Banach space, since 
\begin{equation*}
\Vert u\at{t=T} \Vert _{H^{-3}(0,+\infty )} \le C 
\Vert u\Vert _{H^1(0,T,H^{-3}(0,+\infty ))} \le C'\Vert u\Vert _E
\end{equation*}
for all $u\in E$ and some constants $C,C'>0$.
Also, the linear map
$\Lambda :u\in E\mapsto u\at{t=0}\in H^{-3}(0,+\infty )$ is
continuous. Actually, thanks to \cite[Lemme 8.1]{LM}, $\Lambda$
takes values in
$L^2(0,+\infty )$ and we readily infer from the closed graph theorem that
$\Lambda$ is continuous as a map from $E$ into $L^2(0,+\infty )$. Let
$N=\hbox{ker}\, \Lambda$, let $\tilde E$ stand for the quotient space of $E$
by $N$ and let $\pi$ denote the natural projection of $E$ onto $\tilde E$. Then
$\tilde E$ is a Banach space for the norm
$\Vert\pi (u)\Vert _{\tilde E}:=\inf_{w\in \pi (u)}\Vert w\Vert _E$ and the
induced map  $\tilde \Lambda :\tilde E\to L^2(0,+\infty )$ (defined by
$\tilde \Lambda (\pi (u))=\Lambda (u)$ for any $u\in E$)
has a continuous inverse by the
open mapping theorem. For every $\lambda  >0$
we pick $u^{\lambda } \in E$ such that
$\pi (u^{\lambda })= \tilde \Lambda ^{-1}(v^{\lambda }(0,\cdot ))$
(with $v^\lambda$ as in Lemma 1) and
\begin{equation}
\label{eq:308}
\Vert u^{\lambda } \Vert _E\le 2\Vert \pi (u^{\lambda })\Vert _{\tilde E}
\le 2\Vert \tilde \Lambda ^{-1}\Vert .
\end{equation}
Let $L$ be a positive number. Integrations by part in
\begin{equation*}
\int_0^T\!\! \int_0^L  P(u^{\lambda}) v^{\lambda}\, dxdt=0
\end{equation*}
result in 
\begin{equation}
\label{eq:309}
-\int_0^L v^{\lambda }(0,x)^2\, dx + 
\left[ <u^{\lambda }_{xx} + u^{\lambda} ,v^{\lambda }>
-<u^{\lambda }_x,v^{\lambda }_x>+
<u^{\lambda },v^{\lambda }_{xx}> \right] _{x=0}^L =0,
\end{equation}
where $<\cdot ,\cdot >$ denotes here the duality pairing 
$<\cdot ,\cdot >_{H^1(0,T)',H^1(0,T)}$. Since 
\begin{equation*}
\begin{array}{rl}
\Vert u^{\lambda }(\cdot ,L)\Vert _{H^1(0,T)'} + 
\Vert u^{\lambda }_x(\cdot ,L)\Vert _{H^1(0,T)'} +           
\Vert u^{\lambda }_{xx}(\cdot ,L)\Vert _{H^1(0,T)'} 
&\le C \Vert u\Vert _{L^{\infty}(0,T,L^2(L,L+1))} \\ 
&\le C \Vert u\Vert _{L^{\infty}(0,T,L^2(0,+\infty ))}
\end{array}
\end{equation*}
(where $C=C(1,T)$ is as in Lemma 2) and since $v^{\lambda }(\cdot ,L)$,
$v^{\lambda }_{x}(\cdot ,L)$ and $v^{\lambda }_{xx}(\cdot ,L)\to 0$ in 
$H^1(0,T)$ as $L\to +\infty$, letting $L\to +\infty$ in (\ref{eq:309})
and using (\ref{eq:302})-(\ref{eq:303}) we get
\begin{equation*}
1=\int_0^{+\infty}v^{\lambda }(0,x)^2\, dx
=<u^{\lambda }_x,v^{\lambda }_x>\at{x=0}
-<u^{\lambda },v^{\lambda }_{xx}>\at{x=0}.
\end{equation*}
Hence, by (\ref{eq:306}) and (\ref{eq:308}) (also with $C=C(1,T)$)
\begin{equation}
\label{eq:310}
1\le 2C\Vert \tilde \Lambda ^{-1}\Vert \left( 
\Vert {v^{\lambda }_x}\at{x=0}\Vert _{H^1(0,T)} +
\Vert {v^{\lambda }_{xx}}\at{x=0}\Vert _{H^1(0,T)} \right) \cdot
\end{equation}
Letting ${\lambda }\to 0$ in (\ref{eq:310}) and using 
(\ref{eq:304}) we get a contradiction. The proof of Theorem
1 is complete.\qed
\setcounter{equation}{0}
\section{A Carleman's estimate}

Let $T$ and $L$ be positive numbers. Set
\begin{equation*}
{\cal Z}=\{ q\in C^3([0,T]\times [-L,L]);\ q(t,\pm L)=q_x(t,\pm L)
=q_{xx}(t,\pm L)=0\hbox{ for }0\le t\le T \} .
\end{equation*}
This section is devoted to the proof of the following global Carleman's 
estimate for the KdV equation. 
\begin{prop} There exists a smooth positive function $\psi$ on 
$[-L,L]$ (which depends on $L$) and there exist constants $s_0=s_0(L,T)$ 
and $C=C(L,T)$ such that for all $s\ge s_0$ and all $q\in {\cal Z}$
\begin{equation}
\label{eq:601}
\begin{array}{rl}
\displaystyle
\int_0^T\!\! \int_{-L}^L \{ \frac{s^5}{t^5(T-t)^5}\vert q\vert ^2 
&\displaystyle
+\frac{s^3}{t^3(T-t)^3}\vert q_x\vert ^2 + 
\frac{s}{t(T-t)}\vert q_{xx}\vert ^2 \} e^{-\frac{2s\psi (x)}{t(T-t)}}
\, dxdt \\
&\displaystyle
\le C \int_0^T\!\! \int_{-L}^L\vert q_t+q_x+q_{xxx}\vert ^2 
e^{-\frac{2s\psi (x)}{t(T-t)}} \, dxdt.
\end{array}
\end{equation}
\end{prop}
\noindent{\em Proof.} 
Let $\psi=\psi (x)$ be a positive function (to be specified later)
of class $C^3$ in $[-L,L]$ and let 
$\varphi (t,x):=\frac{\psi (x)}{t(T-t)}$. Let $q$ be given in ${\cal Z}$ and 
let $s>0$. Set $u:=e^{-s\varphi }q$ and $w:= e^{-s\varphi }P( e^{s\varphi }u)$. 
We readily get
\begin{equation}
\label{eq:1999}
w=Au+Bu_x+Cu_{xx}+u_{xxx}+u_t,
\end{equation}
with 
\begin{equation*}
\begin{array}{rl}
A&:=s(\varphi _t+\varphi _x+ \varphi _{xxx})+3s^2\varphi _x
\varphi _{xx}+(s\varphi _x)^3,\\
B&:=1+3s\varphi _{xx}+3(s\varphi _x)^2,\\
C&:=3s\varphi _x.
\end{array}
\end{equation*}
Set $M_1(u):=u_t+u_{xxx}+Bu_x$ and $M_2(u):=Au+Cu_{xx}$. We deduce the
following inequality 
\begin{equation}
\label{eq:602}
2\int\!\!\!\int M_1(u)\, M_2(u) \le\int\!\!\!\int \big( M_1(u)+M_2(u)\big )^2 
=\int\!\!\!\int w^2.
\end{equation}
(Here and in what follows, the integrals are extended to
$(0,T)\times (-L,L)$.) To compute the integral in the left hand side of
(\ref{eq:602}) we perform integrations by part w.r.t. $x$ or $t$.  We readily
get 
\begin{equation}
\label{eq:603}
\int\!\!\!\int M_1(u)Au =-\frac{1}{2}\int\!\!\!\int (A_t+A_{xxx}+(AB)_x)u^2 +
\frac{3}{2}\int\!\!\!\int A_x u_x^2,
\end{equation}
and
\begin{equation}
\label{eq:604}
\int\!\!\!\int (u_{xxx}+Bu_x)Cu_{xx} =
-\frac{1}{2}\int\!\!\!\int C_xu_{xx}^2 -\frac{1}{2}\int\!\!\!\int (BC)_x u_x^2.
\end{equation}
Finally, using (\ref{eq:1999}),
\begin{equation}
\label{eq:605}
\begin{array}{rl}
\int\!\!\!\int u_tCu_{xx}
&=-\int\!\!\!\int C_xu_tu_x -\int\!\!\!\int Cu_{tx}u_x \\
&=\int\!\!\!\int C_x\big( Au+Bu_x+Cu_{xx}+u_{xxx}-w\big) u_x 
+\frac{1}{2}\int\!\!\!\int C_tu_x^2 \\
&=-\frac{1}{2}\int\!\!\!\int (C_xA)_xu^2 +
\frac{1}{2}\int\!\!\!\int (2BC_x-(CC_x)_x+C_{xxx}+C_t)u_x^2 -
\int\!\!\!\int C_xu_{xx}^2\\
&\ -\int\!\!\!\int C_xwu_x\cdot
\end{array}
\end{equation}
Combining (\ref{eq:603}), (\ref{eq:604}) and (\ref{eq:605}) we get 
\begin{equation}
\label{eq:1900}
\begin{array}{l}
2\int\!\!\!\int M_1(u) M_2(u) =-\int\!\!\!\int 
\big( A_t+A_{xxx}+(AB)_x+(C_xA)_x\big) u^2 \\
+\int\!\!\!\int \big( 3A_x-(BC)_x+2BC_x-(CC_x)_x+C_{xxx}+C_t\big) u_x^2 
-3\int\!\!\!\int C_xu_{xx}^2 -2\int\!\!\!\int C_xwu_x\cdot
\end{array}
\end{equation}
If $\epsilon$ is any number in $(0,1)$, then by Cauchy-Schwarz inequality 
\begin{equation*}
2\int\!\!\!\int C_xwu_x \le \epsilon
\int\!\!\!\int C_x^2u_x^2 +{\epsilon}^{-1} \int\!\!\!\int w^2 \cdot 
\end{equation*}
Hence, setting 
\begin{equation*}
\begin{array}{rl}
D&:=-\big( A_t+A_{xxx}+(AB)_x+(C_xA)_x\big) , \\
E&:=3A_x+BC_x-B_xC-(CC_x)_x+C_{xxx}+C_t-\epsilon C_x^2, \\
F&:=-3C_x,
\end{array}
\end{equation*}
 and using (\ref{eq:602}), (\ref{eq:1900}) we get 
\begin{equation}
\label{eq:607}
\int\!\!\!\int Du^2 +\int\!\!\!\int Eu_x^2 + \int\!\!\!\int Fu_{xx}^2 \le 
(1+{\epsilon}^{-1}) \int\!\!\!\int w^2.
\end{equation}
The function $\psi$ will be chosen in such a way that $D$, $E$ and $F$ 
be positive. Clearly
\begin{equation*}
\begin{array}{rl}
\displaystyle
D&\displaystyle
=-(AB)_x +\frac{1}{t^4(T-t)^4}O(s^4)\qquad (\hbox{as } s\to +\infty )\\
&\displaystyle
=-\big( 3(s\varphi _x)^5\big) _x + \frac{O(s^4)}{t^4(T-t)^4}\\ 
&\displaystyle
=-15s^5\frac{\psi '(x)^4\psi ''(x)}{t^5(T-t)^5} + 
\frac{ O(s^4) }{t^4(T-t)^4}\cdot
\end{array}
\end{equation*}
It follows that for $s$ large enough, if
\begin{equation}
\label{eq:608}
\vert \psi '(x)\vert >0 \hbox{ and } \psi ''(x)<0 \hbox{ for }
x\in [-L,L],
\end{equation}
we have 
\begin{equation}
\label{eq:609}
D\ge C_1\frac{s^5}{t^5(T-t)^5}
\end{equation}
for some constant $C_1>0$. On the other hand, expanding $E$ in a series of
powers of $s$, it is easy to see that there is no term in $s^3$ (because 
of cancellations) and that 
\begin{equation*}
\begin{array}{rl}
\displaystyle
E&\displaystyle
= 9s^2\big( (1-\epsilon )\varphi_{xx}^2-\varphi _x\varphi _{xxx}\big) +
\frac{O(s)}{t^2(T-t)^2} \\
&\displaystyle
=9s^2\frac{(1-\epsilon )\psi ''(x)^2-\psi '(x)
\psi '''(x)}{t^2(T-t)^2} 
+\frac{O(s)}{t^2(T-t)^2} \cdot 
\end{array}
\end{equation*}
Hence for $s$ large enough, if
\begin{equation}
\label{eq:610} 
(1-\epsilon )\psi ''(x)^2-\psi '(x)\psi '''(x) >0\quad \forall x\in [-L,L]
\end{equation}
we get for some constant $C_2>0$ 
\begin{equation}
\label{eq:611}
E \ge C_2\frac{s^2}{t^2(T-t)^2} \cdot
\end{equation}
Finally, for some constant $C_3>0$ 
\begin{equation}
\label{eq:612}
F=-\frac{9\psi ''(x)s}{t(T-t)} \ge C_3\frac{s}{t(T-t)}
\end{equation}
provided that (\ref{eq:608}) holds true. Pick now some smooth positive 
function $\psi$ on $[-L,L]$ such that (\ref{eq:608}) and (\ref{eq:610}) are
fulfilled for some $\epsilon >0$. (For instance, 
picking any $\epsilon$ in $(0,1)$,   
$\psi (x)=-x^2+(2L+1)(x+2L)$ is convenient.) We
infer from (\ref{eq:607}), (\ref{eq:609}), (\ref{eq:611}) and 
(\ref{eq:612}) that for $s$ large enough,
\begin{equation}
\label{eq:613}
\int\!\!\!\int \{
\frac{s^5}{t^5(T-t)^5} u^2 +\frac{s^2}{t^2(T-t)^2}u_x^2 + 
\frac{s}{t(T-t)}u_{xx}^2 \} \le C_4\int\!\!\!\int w^2
\end{equation}
for some constant $C_4>0$. Actually (\ref{eq:613}) may be slightly improved
by observing that 
\begin{equation*}
\begin{array}{rl}
\displaystyle
\int\!\!\!\int \frac{s^3}{t^3(T-t)^3}u_x^2 
&\displaystyle
=-\int\!\!\!\int \frac{s^3}{t^3(T-t)^3}uu_{xx} \\
&\displaystyle
\le\frac{1}{2}\left( \int\!\!\!\int \frac{s^5}{t^5(T-t)^5}u^2 +
\int\!\!\!\int \frac{s}{t(T-t)}u_{xx}^2 \right) \\
&\displaystyle
\le \frac{C_4}{2} \int\!\!\!\int w^2
\end{array}
\end{equation*}
(thanks to (\ref{eq:613})), hence, for $s$ large enough, 
\begin{equation}
\label{eq:614}
\int\!\!\!\int \{ \frac{s^5}{t^5(T-t)^5}u^2 +\frac{s^3}{t^3(T-t)^3}u_x^2+
\frac{s}{t(T-t)}u_{xx}^2 \} \le \frac{3}{2}C_4\int\!\!\!\int w^2\cdot
\end{equation}
Replacing $u$ by $e^{-s\varphi}q$ in (\ref{eq:614}) we readily get 
(\ref{eq:601}) for some constant $C>0$ and $s$ large enough. The 
proof of proposition 2 is complete. \qed
\begin{cor}
Let $L>0$ and let $f=f(t,x)$ be any function in 
$L^2\big( \R _t\times (-L,L)_x \big)$ such that 
$\hbox{supp}\, f\subset [t_1,t_2]\times (-L,L)$, where 
$-\infty <t_1<t_2<\infty$. 
Then for every $\epsilon >0$ there exist a positive number 
$C=C(L,t_1,t_2,\epsilon )$ ($C$ does
not depend on $f$) and a function 
$v\in L^2 \big( \R \times (-L,L)\big)$ 
such that 
\begin{equation}
\label{eq:615}
v_t+v_x+v_{xxx}=f \hbox{ in } {\cal D}'(\R\times (-L,L)),
\end{equation}
\begin{equation}
\label{eq:616}
\hbox{supp}\, v\subset [t_1-\epsilon ,t_2+\epsilon ]\times (-L,L),
\end{equation}
\begin{equation}
\label{eq:6160}
\Vert v\Vert _{L^2\big( \R \times (-L,L)\big)}
\le C\Vert f\Vert _{L^2\big( \R \times (-L,L)\big)}\cdot
\end{equation}
\end{cor}
\noindent{\em Proof.} 
Applying a translation w.r.t. time if needed we may assume without 
loss of generality that $0=t_1-\epsilon <t_1<t_2<t_2+\epsilon = : T$. 
We readily infer
from (\ref{eq:601}) that for some constants $k,C_1>0$ and for every 
$q\in {\cal Z}$ 
\begin{equation}
\label{eq:617}
\int_0^T\!\! \int_{-L}^L\vert q\vert ^2 e^{-\frac{k}{t(T-t)}}\, dxdt
\le C_1\int_0^T\!\! \int_{-L}^L\vert P q\vert ^2\, dxdt.
\end{equation}
Thus the following bilinear form
\begin{equation*}
(p,q):=\int_0^T\!\!\int_{-L}^L P p\, P q\, dxdt
\end{equation*}
is a scalar product on ${\cal Z}$. Let $H$ denote the completion of 
${\cal Z}$ for $(\cdot ,\cdot )$. Obviously 
$\vert q\vert ^2e^{-\frac{k}{t(T-t)}}$ is integrable on $(0,T)\times (-L,L)$
if $q\in H$, and (\ref{eq:617}) holds true, as well. On the other hand the
linear form 
\begin{equation*}
l(q):=-\int_0^T\!\! \int_{-L}^L fq\, dxdt
\end{equation*}
is well-defined and continuous on $H$. Indeed, using (\ref{eq:617})
and the assumption on the support of $f$, we get
\begin{equation}
\int_0^T\!\! \int_{-L}^L \vert fq \vert \, dxdt = 
\int_{t_1}^{t_2}\!\! \int_{-L}^L \vert fq\vert \, dxdt\le C_2 
\Vert f\Vert _{L^2\big( (t_1,t_2)\times (-L,L)\big) }\cdot
(q,q) ^{\frac{1}{2}}
\end{equation}
for some constant $C_2>0$. It follows from Riesz representation theorem 
that there exists a unique $p\in H$ such that 
\begin{equation}
\label{eq:618}
\forall q\in H\quad (p,q)=l(q).
\end{equation}
We set $v:=P(p)\in L^2\big( (0,T)\times (-L,L)\big)$. Taking 
$q\in{\cal D}\big( (0,T)\times (-L,L)\big)$ 
as a test function in (\ref{eq:618}) we get
\begin{equation*}
<P^*(v),q>_{{\cal D}'(Q),{\cal D}(Q)}=<-f,q>_{{\cal D}'(Q),{\cal D}(Q)}
\end{equation*}
where $Q=(0,T)\times(-L,L)$ and
$P^*=-P$ is the (formal) adjoint to the operator $P$. Hence
$Pv=f$ in ${\cal D}'(Q)$. Notice that
$v\in H^1(0,T,H^{-3}(-L,L))$, since $v$ and $v_t=f-v_{xxx}-v_x$ belong to
$L^2(0,T,H^{-3}(-L,L))$, hence $v\at{t=0}$ and $v\at{t=T}$ are meaningful
in $H^{-3}(-L,L)$. Let now $q\in {\cal Z}\subset H^1(0,T,H^3_0(-L,L))$. 
It follows from (\ref{eq:618})  
that
\begin{equation*}
\begin{array}{rl}
-\int_0^T\!\!\int_{-L}^L fq\, dxdt 
&= \int_0^T\!\!\int_{-L}^L v(q_t+q_x+q_{xxx})\, dxdt \\
&= -\int_0^T <v_t+v_x+v_{xxx},q>\, dt +[<v,q>]_{t=0}^T \\
&= -\int_0^T\!\! \int_{-L}^L fq\, dxdt +[<v,q>]_{t=0}^T
\end{array}
\end{equation*}
where $<\cdot ,\cdot >$ denotes the duality pairing
$<\cdot ,\cdot > _{H^{-3}(-L,L),H^3_0(-L,L)}$. 
Since $q\at{t=0}$ and $q\at{t=T}$ may be arbitrarily 
chosen in ${\cal D}(-L,L)$, we infer that $v\at{t=0}=v\at{t=T}=0$ in 
$H^{-3}(-L,L)$. Extending $v$ by setting $v(t,x)=0$ for 
$(t,x)\not\in (0,T)\times (-L,L)$, 
we see that (\ref{eq:615}), (\ref{eq:616}) and (\ref{eq:6160}) hold true
(with $C=C_2$). \qed
\begin{rmq}
Using \cite[Thm 1.2]{R1} instead of Proposition 2, one may prove that
the result in Corollary 1 also holds true for $\epsilon =0$ and that
the weight $e^{-\frac{k}{t(T-t)}}$ may be dropped 
in the integral term of the l.h.s. of (\ref{eq:617}). Nevertheless, the proof 
given here is direct, shorter and it leads to a self-contained paper.
Moreover, this proof works also for the heat equation (see below the
proof of Theorem 4).
\end{rmq}
\setcounter{equation}{0}
\section{The proof of Theorem 2}

For the comfort of the reader, we first give an outline
of the proof of Theorem 2.
In the first step, we show that we are done if for any
$f\in L^2_{loc}(\R ^2)$ with support in some strip $[0,T]\times \R$
and any $\epsilon >0$,
there exists a function $u\in L^2_{loc}(\R ^2)$ supported
in $[-\epsilon ,T+\epsilon ]\times \R$,
which solves $P\, u=f$.
This problem has already been solved when the whole domain $\R ^2$ is
replaced by $\R _t\times (-n,n)_x$, $n\ge 1$  (Corollary 1, Section 3).
At this stage, we are given a sequence of solutions  of $P\, u=f$, which
are defined on an increasing sequence of domains and are supported in
$[-\epsilon , T+\epsilon ]\times \R$. To ensure the convergence of this
sequence in $L^2_{loc}(\R ^2 )$, we need some approximation theorem (Lemma
5), which differs from the one in \cite{Rosay} by a careful control on the
growth of the support in time. Two technical lemmas  (namely, Lemmas 3 and 4)
are needed to prove the approximation theorem.
The final step is a standard Mittag-Leffler's procedure.
\va

Let $u_0\in L^2(0,+\infty )$ and
$u_T\in L^2\big( (0,+\infty ),e^{-2bx}dx\big)$. 
It is well known (see \cite{Ka}) that the operator 
$Av=-v_{xxx}$ with domain $H^3(\R )$ (resp. 
$\{ v\in L^2(\R , e^{2bx}dx),\  Av\in L^2(\R , e^{2bx}dx)\}$) generates
a continuous semigroup on $L^2(\R )$ (resp. $L^2(\R , e^{2bx}dx)$). Thanks
to the standard change of functions
\begin{equation}
\label{eq:400}
v(t,x)=u(t,t+x)
\end{equation}
we easily get two functions $u_1(t,x)$, $u_2(t,x)$ such that
$u_1\in C([0,T],L^2(\R ))$,
$u_2\in C([0,T],$\va
$L^2(\R ,e^{2bx}dx))$
and
\begin{equation*}
Pu_1=Pu_2=0\quad \hbox{on } (0,T)\times\R ,
\end{equation*}
\begin{equation*}
u_1(0,x)=
\left\{
\begin{array}{ll}
u_0(x)&\hbox{ for a.e. } x>0, \\
0 &\hbox{ for a.e. }x<0,
\end{array}
\right.
\quad 
u_2(0,x)=
\left\{
\begin{array}{ll}
u_T(-x)&\hbox{ for a.e. } x<0, \\
0 &\hbox{ for a.e. } x>0.
\end{array}
\right. 
\end{equation*}
Set now ${\tilde u}_2(t,x)=u_2(T-t,-x)$. Obviously
$P\, {\tilde u}_2=0$ and
${\tilde{u_2}}_{\vert _{t=T}}={\tilde u}_T$
on $(0,+\infty )$.
Let  $\epsilon '$ be any number in $(\epsilon ,\frac{T}{2})$ and let
$\varphi\in C^{\infty}([0,T])$ be such that $\varphi (t)=1$ 
for $t\le \epsilon '$ and
$\varphi (t)=0$ for $t\ge T- \epsilon '$. The change of functions 
\begin{equation*}
u(t,x)=\varphi (t)u_1(t,x)+
\big( 1-\varphi (t)\big) {\tilde u}_2(t,x)+w(t,x)
\end{equation*}
transforms (\ref{eq:1000}) into
\begin{equation*}
\left\{
\begin{array}{rl}
P\, w&=\frac{d\varphi}{dt}\, ({\tilde u}_2-u_1)\quad \hbox{in } 
{\cal D}'\big( (0,T)\times (0,+\infty )\big) \\
w\at{t=0}&=w\at{t=T}=0 \quad\hbox{on } (0,+\infty ).
\end{array}
\right.
\end{equation*}
Setting 
$f(t,x)=\varphi '(t)\big( {\tilde u}_2(t,x)-u_1(t,x)\big)$, 
it is clear that we are done 
if the following result is proved. 
\begin{prop} 
\label{prop100}
Let $f=f(t,x)$ be any function in $L^2_{loc}(\R ^2)$ such that
\begin{equation*}
\hbox{supp} \, f\subset [t_1,t_2]\times \R 
\end{equation*}
where $0 <t_1<t_2< T$. Let $\epsilon \in (0,\min (t_1,T-t_2))$. 
Then there exists 
$u\in L^2_{loc}(\R ^2)$ such that  
\begin{equation}
\label{eq:10000}
P\, u=f \hbox{ in }{\cal D}'(\R ^2) \ \hbox{ and } \ 
\hbox{supp}\ u\subset [t_1-\epsilon ,t_2+\epsilon ]\times\R .
\end{equation}
\end{prop}
\begin{rmq}
The question whether Proposition 3 remains valid with 
$\epsilon =0$ is open. Notice that the answer is negative for the heat
equation. (See below Remark 3.)
\end{rmq}
As in \cite{Rosay} the proof of Proposition 3 rests on an approximation 
theorem (Lemma 5), which in turn is obtained as a consequence of two 
preliminary lemmas. In what follows $S_L$ will denote the unitary group
in $L^2(-L,L)$ generated by the operator $Au=-u_{xxx}-u_x$ with domain
\begin{equation*}
{\cal D}(A)=\{ u\in H^3(-L,L);\ u(-L)=u(L),\ u_x(-L)=u_x(L),\ 
u_{xx}(-L)=u_{xx}(L) \} \cdot
\end{equation*}
Set $e_n(x)=\frac{1}{\sqrt{2L}}e^{in\frac{\pi}{L}x}$ for $n\in\Z$. $e_n$ is
an eigenvector for $A$ associated with the eigenvalue 
$\omega _n=i\lambda _n$, with
\begin{equation}
\label{eq:401}
\lambda _n=\left( n\frac{\pi}{L}\right) ^3 -  n\frac{\pi}{L} \cdot
\end{equation}
If $u_0$ is any complex-valued function in $L^2(-L,L)$,
decomposed as $u_0=\sum_{n\in\Z}c_ne_n$, 
we have for every $t\in\R$
\begin{equation}
\label{eq:402}
S_L(t)\, u_0=\sum_{n\in\Z}e^{i\lambda _n t} c_n e_n.
\end{equation}
We are now ready to state the first lemma, which may be seen as a preliminary 
version to the approximation theorem. 
\begin{lem}                                                      
Let $l_1,l_2,L,t_1,t_2,T$ be numbers such that $0<l_1<l_2<L$ and
$0<t_1<t_2<T$. Let $u\in L^2\left( (0,T)\times (-l_2,l_2)\right)$ be
such that 
\begin{equation}
\label{eq:404}
P\, u=0 \ \hbox{ in } (0,T)\times (-l_2,l_2)\quad \hbox{ and }
\hbox{supp}\ u\subset [t_1,t_2]\times (-l_2,l_2).
\end{equation}
Let $\delta >0$ with $2\delta <\min (t_1,T-t_2)$ and $\eta >0$ be given. 
Then there exist $v_1,v_2\in L^2(-L,L)$ and 
$v\in L^2\left( (0,T)\times (-L,L)\right)$ such that 
\begin{equation}
\label{eq:406}
P\, v=0\ \hbox{ in } (0,T)\times (-L,L),
\end{equation}
\begin{equation}
\label{eq:408}
v(t,\cdot )=S_L(t-t_1+2\delta )v_1\ 
\hbox{ for }\  t_1-2\delta <t<t_1-\delta ,
\end{equation}
\begin{equation}
\label{eq:410}
v(t,\cdot )=S_L(t-t_2-\delta )v_2\ 
\hbox{ for }\  t_2+\delta <t<t_2+2\delta , 
\end{equation}
\begin{equation}
\label{eq:412}
\Vert v-u\Vert _{L^2\left( (t_1-2\delta ,t_2+2\delta )
\times (-l_1,l_1)\right) } <\eta .
\end{equation}
\end{lem}
Roughly speaking, equations (\ref{eq:408})-(\ref{eq:410}) mean that 
for $t\in (t_1-2\delta ,t_1-\delta )\cup (t_2+\delta , t_2+2\delta )$ $v$
satisfies (in addition to $P\, v=0$) the boundary conditions 
$v(-L)=v(L)$, $v_x(-L)=v_x(L)$ and $v_{xx}(-L)=v_{xx}(L)$. 
\va
{\em Proof of Lemma 3.} Set                    
$Q=(0,T)\times (-L,L)$, $Q_{\delta}=(t_1-2\delta, t_2+2\delta )
\times (-l_1,l_1)$.
Smoothing $u$ by convolution and multiplying the regularized function 
by a cut-off function (of $x$)  we easily get a function 
$u'\in {\cal D}(\R ^2)$ such that 
\begin{equation}
\label{eq:415}
\left\{
\begin{array}{l}
supp\ u'\subset [t_1-\delta ,t_2+\delta ]\times[-l_2,l_2],\\
\\
P\, u'=0\ \hbox{ in }(0,T)\times (-l_1,l_1) \hbox{ and}\\
\\
\Vert u'-u\Vert_{L^2\big( (0,T)\times (-l_1,l_1)\big) }
< \frac{\eta}{2}\cdot
\end{array}
\right.
\end{equation}
Let  
\begin{equation*}
{\cal E}=\{ v\in L^2 (Q) ; \
\exists v_1,v_2\in L^2(-L,L)\hbox{ s.t. }
(\ref{eq:406}),\  (\ref{eq:408}) \hbox{ and } (\ref{eq:410}) 
\hbox{ hold true} \} .
\end{equation*}
The lemma is proved if we may find $v\in{\cal E}$ such that 
$\Vert v- u'\Vert _{L^2(Q_{\delta})} <\frac{\eta}{2}$. We are done
if we prove $u'\in \overline{{\cal E}}={\cal E}^{\perp\, \perp}$, where
closure and orthogonal complement are taken in the space $L^2(Q_{\delta})$.
Fix a function $g\in {\cal E}^{\perp}\subset L^2(Q_{\delta })$. 
Before proving 
$(u',g)_{L^2(Q_{\delta})}=0$, we begin with the 
\va
\noindent{\sc Claim 1:} Let ${\cal T}=\{ \varphi\in C^{\infty}(\R ^2);\
\hbox{\it supp}\ \varphi\subset [t_1-\delta ,t_2+\delta ]\times \R \}$. 
Then there 
exists $C>0$ such that 
\begin{equation}
\label{eq:413}
\forall \varphi\in {\cal T} \quad \vert (\varphi ,g) _{L^2(Q_{\delta})} \vert
\le C\Vert P\, \varphi \Vert _{L^2(Q )}.
\end{equation}
\noindent{\em Proof of Claim 1.} Let $\varphi\in {\cal T}$, set
$\psi (t):=\int_0^t S_L(t-\tau )P \varphi (\tau )\, d\tau$ 
for $0\le t\le T$, that is
$\psi$ is the (strong) solution of the following boundary initial-value
problem:
\begin{equation*}
\begin{array}{rl}
P\, \psi &=P\, \varphi\hbox{ in } Q,\\
\psi (t,-L)&=\psi (t,L),\\
\psi _x (t,-L)&=\psi _x(t,L),\\
\psi _{xx}(t,-L)&=\psi _{xx}(t,L),\\
\psi (0,\cdot )&=0.
\end{array}
\end{equation*}
Clearly $v:=\psi -\varphi\in {\cal E}$ ((\ref{eq:408})-(\ref{eq:410}) hold
true with $v_1=0$, $v_2=\psi (t_2+\delta )$) hence
$(\psi -\varphi ,g)_{L^2(Q_{\delta })}=0$. On the other hand, it is clear that
\begin{equation*}
\forall t\in [0,T]\quad \Vert\psi (t)\Vert _{L^2(-L,L)} \le 
\Vert P\, \varphi \Vert _{L^1(0,t,L^2(-L,L))} \le \sqrt{T} 
\Vert P\, \varphi\Vert _{L^2(Q)}
\end{equation*}
hence 
\begin{equation*}
\vert (\varphi ,g)\vert _{L^2 (Q_{\delta }) } 
=\vert (\psi ,g)\vert _{L^2(Q_{\delta }) }
\le T\Vert g\Vert _{L^2(Q_{\delta })}\cdot \Vert P\, \varphi\Vert _{L^2(Q)}.
\end{equation*}
This completes the proof of Claim 1. We now proceed to the \va
\noindent{\sc Claim 2: } 
There exists a function $w\in L^2(Q)$ such that
\begin{equation}
\label{eq:414}
\forall \varphi\in{\cal T} \quad (\varphi ,g)_{L^2(Q_{\delta })}
=(P\, \varphi ,w)_{L^2(Q)} .
\end{equation}
\noindent{\em Proof of Claim 2. }
Let ${\cal Z}:=\{ (P\, \varphi )\at{Q};\ \varphi\in {\cal T} \}$. Notice first
that for any $\zeta\in {\cal Z}$, if $\zeta = (P\, \varphi _1)\at{Q}=
(P\, \varphi _2)\at{Q}$
for two functions $\varphi _1,\varphi _2\in {\cal T}$, then
$\varphi _1-\varphi _2\in {\cal E}$, hence
$(\varphi _1-\varphi _2,g)_{L^2(Q_{\delta })}=0$. It follows that the (linear)
map $\Lambda :\zeta\in {\cal Z}\mapsto (\varphi ,g)_{L^2(Q_{\delta })} \in\R$
(if $\zeta =(P\, \varphi )\at{Q}$, $\varphi\in {\cal T}$) is well-defined. Let
$H$ denote the closure of ${\cal Z}$ in $L^2(Q)$. We infer
from (\ref{eq:413}) that $\Lambda$ may be extended to $H$ in such a 
way that $\Lambda$ be a continuous linear form on $H$. It follows from
Riesz representation theorem that there exists $w\in H$ such that  
$\Lambda (\zeta )=(\zeta ,w)_{L^2(Q)}$ for all $\zeta\in H$. Then 
(\ref{eq:414}) holds true.
\va\par
We are now ready to prove 
$(u',g)_{L^2(Q_{\delta })} =0$. Extend $g$ and $w$ on $\R ^2$ to 
$\tilde g$, $\tilde w$ by setting 
\begin{equation*}
\begin{array}{rl}
\tilde g(t,x)&=0 \hbox{ for } (t,x)\in \R ^2\setminus Q_{\delta },\\
\tilde w(t,x)&=0 \hbox{ for } (t,x)\in \R ^2\setminus Q.
\end{array}
\end{equation*}
Set $\Omega =(t_1-\delta ,t_2+\delta )\times \R$ and let 
$\varphi\in {\cal D}(\Omega )\subset{\cal T}$. Obviously 
\begin{equation*}
(\varphi ,g)_{L^2(Q_{\delta })} = (\varphi ,\tilde g)_{L^2(\Omega )}
\ \hbox{ and }\ 
(P\, \varphi ,w)_{L^2(Q)} =(P\, \varphi ,\tilde w)_{L^2(\Omega )},
\end{equation*}
hence it follows from (\ref{eq:414}) that
\begin{equation*}
<P^*\, \tilde w,\varphi > _{{\cal D}'(\Omega ),{\cal D}(\Omega )} =
<\tilde g,\varphi > _{{\cal D}'(\Omega ),{\cal D}(\Omega )} \cdot
\end{equation*}
Thus
\begin{equation*}
P^* \, \tilde w = \tilde g \ \hbox{ in }\ {\cal D}'(\Omega )
\end{equation*}
and 
\begin{equation*}
P^*\, \tilde w=0 \ \hbox{ for }\ t_1-\delta <t<t_2+\delta 
\ \hbox{ and } \ \vert x\vert >l_1.
\end{equation*}
Since
\begin{equation*}
\tilde w(t,x)=0\ \hbox{ for } t_1-\delta <t<t_2+\delta \hbox{ and }
\vert x\vert >L
\end{equation*}
we infer from Holmgren's uniqueness theorem (see \cite[Thm 8.6.8]{H}) that
\begin{equation}
\label{eq:416}
\tilde w(t,x)=0 \hbox{ for } t_1-\delta <t<t_2+\delta 
\ \hbox{ and }\ \vert x\vert >l_1.
\end{equation}
Applying (\ref{eq:414}) to $u'\in {\cal T}$ and using 
(\ref{eq:415}), (\ref{eq:416}) we get
\begin{equation*}
\begin{array}{rl}
(u',g)_{L^2(Q_{\delta })} &=(P\, u',w)_{L^2(Q)}      \\
&=(P\, u',w)_{L^2\big( (t_1-\delta 
,t_2+\delta )\times (-l_1,l_1)\big) }  \\
&=0.
\end{array}
\end{equation*}
The proof of Lemma 3 is complete. \qed
\va\indent
Next result is an observability result. 
\begin{lem}
Let $l,L,T$ be positive numbers such that $l<L$. Then there exists a 
constant $C>0$ such that for every $u_0\in L^2(-L,L)$, if $u$ denotes
$S_L(\cdot )u_0$, we have 
\begin{equation}
\label{eq:4160}
\Vert u_0\Vert _{L^2(-L,L)} \le C 
\Vert u\Vert _{L^2\big( (0,T)\times (-l,l)\big) } .
\end{equation}
(Hence 
\begin{equation}
\label{eq:417}
\Vert u\Vert _{L^2\big( (0,T)\times (-L,L)\big) } \le \sqrt{T}C 
\Vert u\Vert _{L^2\big( (0,T)\times (-l,l)\big) } .)
\end{equation}
\end{lem}
\noindent{\em Proof.} 
Pick $T'\in (0,\frac{T}{2})$ and $\gamma >\frac{\pi}{T'}$. Let
$N\in \N$ be such that 
\begin{equation*}
\lambda _N-\lambda _{-N}\, =2\lambda _N \ge \gamma \ \hbox{ and }
(n\in\Z,\ \vert n\vert \ge N)\im \lambda_{n+1}-\lambda _n\ge\gamma .
\end{equation*}
By Ingham's inequality (see \cite{In}) there exists 
a constant $C^{T'}>0$ such that for
every sequence $(a_n)_{\vert n\vert \ge N}$ of complex numbers, with
$a_n=0$ for $\vert n\vert$ large enough, the following inequality holds
true
\begin{equation}
\label{eq:418}
\sum_{\vert n \vert \ge N} \vert a_n\vert ^2 
\le C^{T'} \int_{-T'}^{T'} \left\vert
\sum_{\vert n \vert \ge N} a_ne^{-i\lambda _nt}\right\vert ^2 dt .
\end{equation}
Let ${\cal Z }_n:=\hbox{Span} (e_n)$ for $n\in \Z$ and 
${\cal Z}=\bigoplus _{n\in\Z}{\cal Z}_n \subset L^2(-L,L)$. We define 
a semi-norm $p$ in ${\cal Z}$ by
\begin{equation*}
p(u):=\left( \int_{-l}^l \vert u(x)\vert ^2\, dx \right) ^{\frac{1}{2}}
\ \hbox{ for } u\in {\cal Z}.
\end{equation*}
$p$ is clearly a norm in each ${\cal Z}_n$. On the other hand, if 
$u_0\in {\cal Z}\cap 
\left( \bigoplus _{\vert n\vert <N} {\cal Z}_n \right) ^{\perp}$
(i.e.  $u_0$ may be written in the form
$u_0=\sum_{\vert n\vert \ge N}c_ne_n$ with $c_n=0$ for $\vert n\vert$
large enough), then applying (\ref{eq:418}) (with
$\displaystyle a_n=\frac{c_n}{\sqrt{2L}}
e^{ i(\lambda _nT'+n\frac{\pi}{L}x)}$)
and integrating w.r.t. $x$ on $(-l,l)$ we get
\begin{equation*}
2l\sum_{\vert n\vert\ge N} \frac{\vert c_n\vert ^2}{2L} \le
C^{T'} \int_{-l}^l\int_0^{2T'}
\left\vert \sum_{\vert n\vert \ge N} e^{i\lambda _n\tau} c_n e_n(x)
\right\vert ^2d\tau dx,
\end{equation*}
hence, by Fubini's theorem,  
\begin{equation*}
\Vert u_0\Vert ^2_{L^2(-L,L)} \le \frac{L}{l} C^{T'} 
\int_0^{2T'} p\big( S_L(\tau )u_0 \big) ^2 \, d\tau .
\end{equation*}
Finally, for any $u_0\in L^2(-L,L)$, we have
\begin{equation*}
\int_0^{2T'} p\big( S_L(\tau )u_0 \big) ^2\, d\tau \le
\Vert S_L(\cdot )u_0\Vert ^2_{L^2\big( (0,2T')\times (-L,L)\big) }
=2T' \Vert u_0\Vert ^2_{L^2(-L,L)}.
\end{equation*}
Since $T>2T'$, it follows from \cite[Thm 5.2]{Ko} that there exists a
constant $C>0$ such that (\ref{eq:4160}) holds true for all 
$z_0\in {\cal Z}$. We get (\ref{eq:4160}) for all $u_0\in L^2(-L,L)$
by a density argument. \qed
\va\indent
We now proceed to the proof of the following approximation theorem, 
which differs from the one in \cite{Rosay} by an additional property
on the
support. 
\begin{lem} Let $n\in\N \setminus \{ 0,1\}$ 
and let $t_1,t_2,T$ be numbers such that 
$0<t_1<t_2<T$. Let $u\in L^2\big( (0,T)\times (-n,n)\big)$ be such that
$P\, u=0$ in $(0,T)\times (-n,n)$ and 
$\hbox{supp}\ u\subset [t_1,t_2]\times (-n,n)$. Let 
$0<\epsilon <\min (t_1,T-t_2)$. Then there exists 
$v\in L^2\big( (0,T)\times (-n-1,n+1)\big)$ such that  
\begin{equation}
\label{eq:419}
P\, v=0\ \hbox{ in } (0,T)\times (-n-1,n+1),
\end{equation}
\begin{equation}
\label{eq:420}
\hbox{supp}\ v\subset [t_1-\epsilon ,t_2+\epsilon ]\times (-n-1,n+1),
\end{equation}
\begin{equation}
\label{eq:421}
\Vert v-u\Vert _{L^2\big( (0,T)\times (-n+1,n-1)\big) } <\epsilon .
\end{equation}
\end{lem}
\noindent{\em Proof.}   Let $\eta >0$ (to be chosen later). 
By Lemma 3, applied with $L=n+1$,  there 
exists $\tilde v\in L^2\big( (0,T)\times (-n-1,n+1)\big)$ such that
\begin{equation*}
P\, \tilde v=0 \ \hbox{ in }\ (0,T)\times (-n-1,n+1), 
\end{equation*}
\begin{equation}
\label{eq:450}
\tilde v(t,\cdot )=S_{n+1}(t-t_1+\frac{\epsilon}{2})
v_1\ \hbox{ for } \ 
t_1-\frac{\epsilon}{2}<t<t_1-\frac{\epsilon}{4},
\end{equation}
\begin{equation}
\label{eq:451}
\tilde v(t,\cdot )=S_{n+1}(t-t_2-\frac{\epsilon}{4})
v_2 \ \hbox{ for }\  
t_2+\frac{\epsilon}{4}<t<t_2+\frac{\epsilon}{2}   
\end{equation}
for some $v_1,v_2\in L^2(-n-1,n+1)$ and
\begin{equation*}
\Vert \tilde v -u\Vert _{\textstyle \, L^2
\big( (t_1-\frac{\epsilon}{2},t_2+\frac{\epsilon}{2} )
\times (-n+1,n-1)\big) } <\eta \cdot
\end{equation*}
In order that (\ref{eq:420}) be fulfilled, we multiply $\tilde v$ 
by a cut-off function. Let $\varphi\in {\cal D}(0,T)$ be such that 
$0\le\varphi \le 1$, $\varphi (t)=1$ for all 
$t\in [t_1-\frac{\epsilon}{4},t_2+\frac{\epsilon}{4}]$ and 
$\hbox{supp}\, (\varphi )
\subset [t_1-\frac{\epsilon}{2}, t_2+\frac{\epsilon}{2}]$. 
Set $\bar v(t,x)=\varphi (t)\tilde v(t,x)$. It follows that
\begin{equation*}
supp\ \bar v \subset 
[t_1-\frac{\epsilon}{2}, t_2+\frac{\epsilon}{2}]\times (-n-1,n+1)\cdot
\end{equation*}
Hence
\begin{equation*}
\begin{array}{rl}
\Vert \bar v-u\Vert _{\textstyle \,
L^2\big( (0,T)\times (-n+1,n-1)\big) }
&=\Vert \bar v-u\Vert 
_{\textstyle \, L^2\big( (t_1-\frac{\epsilon}{2},t_2
+\frac{\epsilon}{2})\times (-n+1,n-1)\big) }\\  
&\le \Vert (\varphi -1)\tilde v\Vert 
_{\textstyle \, L^2\big( (t_1-\frac{\epsilon}{2},t_2
+\frac{\epsilon}{2})\times (-n+1,n-1)\big) }\\
&\ +
\Vert \tilde v - u \Vert 
_{\textstyle \, L^2\big( (t_1-\frac{\epsilon}{2},t_2
+\frac{\epsilon}{2})\times (-n+1,n-1)\big) }.
\end{array}
\end{equation*}
Since $\hbox{\it supp}\ u\subset [t_1,t_2]\times (-n,n)$ and 
$\varphi (t)=1$ for $t_1-\frac{\epsilon}{4}\le t\le t_2+\frac{\epsilon}{4}$, 
we get
\begin{equation}
\label{eq:422}
\begin{array}{rl}
\Vert (\varphi -1)\tilde v
\!\!\!\!  &\Vert ^2
_{\textstyle \, L^2\big( (t_1-\frac{\epsilon}{2},
t_2+\frac{\epsilon}{2})\times (-n+1,n-1)\big) }  \\
&\\
&\le \Vert \tilde v\Vert ^2
_{\textstyle \, 
L^2\big( \{ (t_1-\frac{\epsilon}{2},t_1-\frac{\epsilon}{4})
\cup (t_2+\frac{\epsilon}{4},t_2
+\frac{\epsilon}{2})\} \times (-n+1,n-1)\big) } \\
&\le \Vert \tilde v -u\Vert ^2 
_{\textstyle \, 
L^2\big( (t_1-\frac{\epsilon}{2},t_2+\frac{\epsilon}{2})
\times (-n+1,n-1)\big) }\\
&<\eta ^2\cdot
\end{array}
\end{equation}
Hence 
\begin{equation}
\label{eq:423}
\Vert \bar v-u\Vert _{
\textstyle \, L^2\big( (0,T)\times (-n+1,n-1)\big)} \le 2\eta \cdot
\end{equation}
Finally $P\, \bar v=\frac{d\varphi}{dt}\tilde v$ in $(0,T)\times (-n-1,n+1)$,
hence
\begin{equation*}
\begin{array}{rl}
\Vert P\, \bar v \!\!\!\! &\Vert ^2 _{
\textstyle \, L^2\big( (0,T)\times (-n-1,n+1)\big)} \\
&\\
& \le
\Vert\frac{d\varphi}{dt}\Vert ^2_{\textstyle \, L^{\infty}(0,T)} \cdot
\Vert \tilde v\Vert ^2
_{\textstyle \, L^2\big( \{ (t_1-\frac{\epsilon}{2},t_1-\frac{\epsilon}{4})
\cup (t_2+\frac{\epsilon}{4},t_2
+\frac{\epsilon}{2})\} \times (-n-1,n+1)\big) }.
\end{array}
\end{equation*}
Since (\ref{eq:450}), (\ref{eq:451}) hold true, we infer from 
Lemma 4 that there exists a constant $C=C(n,\epsilon )>0$ such that
\begin{equation*}
\Vert \tilde v \Vert 
_{\textstyle \, L^2\big( (t_1-\frac{\epsilon}{2},t_1-\frac{\epsilon}{4})
\times (-n-1,n+1)\big) }
\le C
\Vert \tilde v \Vert 
_{\textstyle \, L^2\big( (t_1-\frac{\epsilon}{2},t_1-\frac{\epsilon}{4})
\times (-n+1,n-1)\big) }
\end{equation*}
and also
\begin{equation*}
\Vert \tilde v \Vert 
_{\textstyle \, L^2\big( (t_2+\frac{\epsilon}{4},
t_2+\frac{\epsilon}{2})\times (-n-1,n+1)\big) }
\le C
\Vert \tilde v \Vert 
_{\textstyle \, L^2\big( (t_2+\frac{\epsilon}{4},
t_2+\frac{\epsilon}{2})\times (-n+1,n-1)\big) }.
\end{equation*}
Hence, by (\ref{eq:422}),
\begin{equation}
\label{eq:424}
\Vert P\, \bar v\Vert _{\textstyle \, L^2\big( (0,T)\times (-n-1,n+1)\big) }
\le C\Vert \frac{d\varphi}{dt}\Vert _{\textstyle \, 
L^{\infty}(0,T)}\, \eta \cdot
\end{equation}
We finally modify $\bar v$ in order that (\ref{eq:419}) be satisfied.
By Corollary 1 there exists a constant $C'>0$ (which depends on
$n$, $t_1$, $t_2$ and $\epsilon$) and a function 
$w\in L^2\big( (0,T)\times (-n-1,n+1)\big) $ such that 
\begin{equation*}
P\, w=P\, \bar v\ \hbox{ in } (0,T)\times (-n-1,n+1),
\end{equation*}
\begin{equation*}
\hbox{\it supp}\ w\subset [t_1-\epsilon ,t_2+\epsilon ]\times (-n-1,n+1)
\end{equation*}
and also
\begin{equation}
\label{eq:425}
\Vert w\Vert _{L^2 \big( (0,T)\times (-n-1,n+1)\big) }
\le C'
\Vert P\, \bar v\Vert _{L^2 \big( (0,T)\times (-n-1,n+1)\big) } \cdot
\end{equation}
Set $v:=\bar v-w$. Then (\ref{eq:419}) and (\ref{eq:420}) are obvious and 
we infer from (\ref{eq:423}), (\ref{eq:424}) and (\ref{eq:425}) that
\begin{equation*}
\Vert v-u\Vert _{L^2 \big( (0,T)\times (-n+1,n-1)\big) }
\le \big( 2+CC'\Vert \frac{d\varphi}{dt}\Vert _{L^{\infty}(0,T)} 
\big) \eta \cdot
\end{equation*}
Hence (\ref{eq:421}) holds true provided that $\eta$ be small enough. \qed
\va\indent
We now turn to the proof of Proposition 3, which is carried out as in 
\cite{Rosay}.\va
\noindent{\em Proof of Proposition 3.} Let $(t_1^n)_{n\ge 2}$ and 
$(t_2^n)_{n\ge 2}$ be two sequences of numbers such that 
\begin{equation}
\label{eq:426}
\forall n\ge 2\qquad t_1-\epsilon <t_1^{n+1}<t_1^n<t_1
<t_2<t_2^n<t_2^{n+1}<t_2+\epsilon \cdot
\end{equation}
We construct (by induction on $n$) a sequence $(u_n)_{n\ge 2}$ of 
functions such that, for every $n\ge 2$, 
\begin{equation}
\label{eq:427}
u_n\in L^2\big( (0,T)\times (-n,n)\big) ,
\end{equation}
\begin{equation}
\label{eq:428}
\hbox{\it supp}\ u_n\subset [t_1^n,t_2^n]\times (-n,n),
\end{equation}
\begin{equation}
\label{eq:429}
P\, u_n=f \ \hbox{ in }\  (0,T)\times (-n,n)
\end{equation}
and, if $n> 2$, 
\begin{equation}
\label{eq:430}
\Vert u_n-u_{n-1}\Vert _{L^2\big( (0,T)\times (-n+2,n-2)\big)}  
<2^{-n} \cdot
\end{equation}
$u_2$ is given by Corollary 1. Let now $n\ge 2$ and assume that 
$u_2,\hdots ,u_n$ have been constructed in such a way that 
(\ref{eq:427})-(\ref{eq:430}) hold true. By Corollary 1 there exists 
$w\in L^2\big( (0,T)\times (-n-1,n+1)\big)$ such that  
\begin{equation*}
\hbox{\it supp}\ w\subset [t_1^2,t_2^2]\times (-n-1,n+1)
\ \hbox{ and } \ P\, w=f \ \hbox{ in }\ (0,T)\times (-n-1,n+1)\cdot
\end{equation*}
Since $P (u_n-w)=0$ in $(0,T)\times (-n,n)$ and 
\begin{equation*}
\hbox{\it supp}\, (u_n-w\at{(0,T)\times (-n,n)} )\subset 
[t_1^n,t_2^n]\times (-n,n),
\end{equation*}
with $t_1^{n+1}<t_1^n<t_2^n<t_2^{n+1}$, it follows from Lemma 5 that
there exists a function $v\in L^2\big( (0,T)\times (-n-1,n+1)\big)$ such that 
\begin{equation*}
\hbox{\it supp}\ v\subset [t_1^{n+1},t_2^{n+1}]\times (-n-1,n+1), \ 
\ P\, v=0\hbox{ in } (0,T)\times (-n-1,n+1)
\end{equation*}
and also
\begin{equation*}
\Vert v-(u_n-w)\Vert _{L^2\big( (0,T)\times (-n+1,n-1)\big)} <2^{-n-1} \cdot 
\end{equation*}
We set $u_{n+1}:=v+w$. Then (\ref{eq:427})-(\ref{eq:430}) are fulfilled. 
Extending the $u_n$'s by setting $u_n(t,x)=0$ for $(t,x)\not\in 
(0,T)\times (-n,n)$, 
we infer from (\ref{eq:430}) that the sequence 
$(u_n)_{n\ge 2}$ converges in $L^2_{loc}(\R ^2)$ towards
 a function $u$ such that 
\begin{equation*}
\hbox{\it supp}\ u\subset [t_1-\epsilon ,t_2+\epsilon ]\times \R ,
\end{equation*}
by (\ref{eq:426}) and (\ref{eq:428}). Finally 
$P\, u =f$ in $\R ^2$, because of (\ref{eq:429}). 
This completes the proof
of Proposition 3 and also the proof of Theorem 2. \qed
\renewcommand{\theequation}{\thesection.\arabic{equation}}
\setcounter{equation}{0}
\section{The heat equation and the Schr\"odinger equation}

In  this section, we are concerned with the control 
of the heat equation and of the Schr\"odinger equation in unbounded domains.
Let us first briefly discuss the controllability of the heat equation 
\begin{equation}
\label{eq:H1}
u_t-\Delta u=0,
\end{equation}
where $\Delta u=\sum_{i=1}^N \frac{\partial ^2 u}{\partial x_i^2}$, $N\ge 1$.
By Proposition 1, the (boundary or internal) {\em approximate}
controllability of (\ref{eq:H1}) in unbounded domains is obvious. Notice
that this result is still valid (but not so obvious) 
for a {\em semilinear} heat equation, see
\cite{TZ}. As far as the boundary null-controllability of (\ref{eq:H1})
is concerned, it has been proved in \cite{MZ} that no (non trivial)
function in ${\cal D}(\Omega )$ (the space of test functions)
can be driven to $0$, when $\Omega =\R _+^N
:=\{ (x',x_N):\ x'\in \R ^{N-1},\ x_N>0\}$ 
and solutions of (\ref{eq:H1}) are taken in some {\em transposition} sense. 
However, if {\em all} 
the solutions of (\ref{eq:H1}) are taken into consideration, then
the null-controllability is recovered, thanks to the following result
by Jones (see \cite{J}, \cite{L}).
\begin{thm}
\label{thm3}
Let $g\in C^0(\R ^N)$ and $T>0$. Then there exists
a function $u\in C^0([0,T]\times \R ^N)$ which solves (\ref{eq:H1}) in the
distributional sense for $t>0$ and satisfies
$u_{\vert _{t=0}}=g$,  $u_{\vert _{t=T}}=0$.
\end{thm}
\noindent Notice, as in \cite{L}, that the boundary control problem is solved
once and for all 
without reference to any {\em specific} domain or set of boundary conditions.
Theorem \ref{thm3} is derived in \cite{J} from the existence,
for any $\epsilon >0$, of a
fundamental solution of the heat equation which is supported in the strip
$[0,\epsilon ]\times \R ^N$. A result close to Theorem 3 
may be proved along the same lines as in Section 4.
\begin{thm}
\label{thm4}
Let $\big( S(t)\big) _{t\ge 0}$ denotes the continuous semigroup
on $L^2(\R ^N)$ generated by the operator $Au=\Delta u$ with
domain $H^2(\R ^N)$.
Let $T,\epsilon$ be positive numbers with $\epsilon <\frac{T}{2}$
and let $u_0,u_1\in L^2(\R ^N)$.
Then  there exists a function
\begin{equation*}
u\in L^2 _{loc}\big( [0,T]\times \R ^N)\big)
\,  \cap \,  C\big( [0,\epsilon ] \cup [T-\epsilon ,T],L^2(\R ^N )\big)
\end{equation*}
which solves
\begin{equation}
\label{eq:1500}
\left\{
\begin{array}{rl}
u_t-\Delta u&=0\qquad
\hbox{ in }\, {\cal D}'\big( (0,T)\times \R ^N\big) ,\\
u\at{t=0}&=u_0,\\
u\at{t=T}&=S(T)u_1.
\end{array}
\right.
\end{equation}
\end{thm}
\noindent{\em Sketch of the proof.}
Set, for any $L>0$, $\Omega _L:=(-L,L)^N$. 
Let $P_1$ 
denotes the operator
$\frac{\partial}{\partial t}-\Delta$. Let
$\epsilon '\in (\epsilon ,\frac{T}{2})$ and let
$\varphi\in C^{\infty}([0,T])$ be such that $\varphi (t)=1$
for $t\le \epsilon '$ and
$\varphi (t)=0$ for $t\ge T- \epsilon '$. The change of functions
\begin{equation*}
u(t,\cdot )=\varphi (t)S(t)u_0  +
\big( 1-\varphi (t)\big) S(t)u_1 + w(t,\cdot )
\end{equation*}
transforms (\ref{eq:1500}) into
\begin{equation*}
\left\{
\begin{array}{rl}
P_1\, w&=\frac{d\varphi}{dt}\, S(t)(u_1-u_0)\quad \hbox{ in }
{\cal D}'\big( (0,T)\times \R ^N\big) \\
w\at{t=0}&=w\at{t=T}=0.
\end{array}
\right.
\end{equation*}
Once again, it is clear that we are done if Proposition 3 holds
true with $P_1$ instead of $P$. The following estimate
(see \cite[Lem. 5.2]{F}, \cite[Thm 4.1]{FI}): 
\begin{equation*}
\exists k>0,\ \exists C>0 \hbox{ s.t.}\quad
\int_0^T\!\! \int_{\Omega _L} \vert q\vert ^2 
e^{-\frac{k}{t(T-t)}} dxdt 
\le C
\int_0^T\!\! \int_{\Omega _L} \vert q_t+\Delta q\vert ^2\, dxdt 
\end{equation*}
for any $q\in C^2([0,T]\times [-L,L]^N)$ such that
$q(t,x)=\partial _n q(t,x)=0$ for $(t,x)\in [0,T]\times \partial \Omega _L$, 
shows that Corollary 1 holds true
with $P_1$ instead of $P$. The other key ingredient in
the proof of Proposition 3, namely the internal observability
result (Lemma 4), may be found in the literature (see
\cite[Cor. 2]{LeRo}). If $\big( S_L(t)\big) _{t\ge 0}$ denotes now
the continuous {\it semigroup} on $L^2(\Omega _L)$ generated by the
operator $Au=\Delta u$ with domain
$H^2_{per}(\Omega _L)$, then the 
proofs of Lemmas 3 and 5 and of Proposition 3  are word for 
word the same as those given above for the KdV equation.\qed
\begin{rmq}
For the heat equation, the results in Corollary 1 and
in Proposition 3 are no longer true if we do $\epsilon =0$.
Indeed, if we assume that Corollary 1 is true for the 1-d heat equation 
with $\epsilon =0$ 
(for any $f$), then an argument similar 
to the one used in the proof of Theorem 1
shows that for some constant $C>0$ we have
\begin{equation}
\label{eq:3000}
\int_0^T\!\! \int_{-L}^L \vert q\vert ^2\, dxdt
\le C
\int_0^T\!\! \int_{-L}^L \vert q_t-q_{xx}\vert ^2\, dxdt
\end{equation}
for any $q\in C^2([0,T]\times [-L,L])$ such that
$q(t,\pm L)=q_x(t,\pm L)=0$. Let $E$ be the classical 
fundamental solution of the
1-d heat equation, namely 
\begin{equation*}
E(t,x)=
\left\{
\begin{array}{ll}
(4\pi t)^{-\frac{1}{2}}\exp (-\frac{x^2}{4t}) &\hbox{ if } t>0,\ x\in \R ,\\
0                                             &\hbox{ otherwise.}
\end{array}
\right.
\end{equation*}
Let $\rho\in {\cal D}(-L,L)$  be such that $\rho (x)= 1$ for
$\vert x\vert < \frac{L}{2}$.
Set $q(t,x)=\rho (x) \frac{\partial E}{\partial x}(t,x)$ for
$x\in (-L,L)$ and $t>0$. Direct computations show that
$\Vert q\Vert _{L^2((\epsilon, T+\epsilon )\times (-L,L))}\to +\infty$
as $\epsilon\to 0^+$,
whereas
$\Vert q_t-q_{xx}\Vert _{L^2((\epsilon, T+\epsilon )\times (-L,L))}=O(1)$, 
contradicting (\ref{eq:3000}).\end{rmq}
We now turn to the Schr\"odinger equation
\begin{equation}
\label{eq:S1}
iu_t+\Delta u=0.
\end{equation}
The following result is proved in the same way as Theorem 2.
\begin{thm}
Let $T,\epsilon$ be positive numbers with $\epsilon <\frac{T}{2}$
and let $u_0,u_T\in L^2(\R ^N)$.
Then  there exists a function
\begin{equation*}
u\in L^2 _{loc}\big( [0,T]\times \R ^N\big)
\, \cap \, C\big( [0,\epsilon ] \cup [T-\epsilon ,T],L^2(\R ^N )\big)
\end{equation*}
which solves
\begin{equation}
\label{eq:2000}
\left\{
\begin{array}{rl}
iu_t+\Delta u&=0\quad
\hbox{ in }\  {\cal D}'\big( (0,T)\times \R ^N\big) ,\\
u\at{t=0}&=u_0,\\
u\at{t=T}&=u_T.
\end{array}
\right.
\end{equation}
\end{thm}
\noindent{\em Sketch of the proof.}
For the sake of simplicity, we assume $N=1$.
We shall write $P_2$ for the operator
$\frac{\partial}{\partial t}-i\frac{\partial ^2}{\partial  x^2}$.
Let $\big( S(t)\big) _{t\in\R}$  denotes  the unitary group
on $L^2(\R )$ generated by the operator $Au=iu_{xx}$ with domain
$H^2(\R )$.
Let $\epsilon '$ and $\varphi$ be as in the proof of 
Theorem 4. The change of functions
\begin{equation*}
u(t,\cdot )=\varphi (t)S(t)u_0  +
\big( 1-\varphi (t)\big) S(t-T)u_T + w(t,\cdot)
\end{equation*}
transforms (\ref{eq:2000}) into
\begin{equation*}
\left\{
\begin{array}{rl}
P_2\, w&=\frac{d\varphi}{dt}\, S(t)\big( S(-T)u_T-u_0\big)
\quad \hbox{in }
{\cal D}'\big( (0,T)\times \R \big) \\
w\at{t=0}&=w\at{t=T}=0.
\end{array}
\right.
\end{equation*}
We are again led to prove Proposition 3, but  with $P_2$ instead
of $P$. Let $S_L$ denote here the unitary group on $L^2(-L,L)$  generated
by the operator $Au=iu_{xx}$ with domain
$\{ u\in H^2(-L,L);\ u(-L)=u(L),\ u_x(-L)=u_x(L) \}$. Let
$e_n(x)=\frac{1}{\sqrt{2L}}e^{i n\frac{\pi}{L}x}$ and
$\lambda _n=(n\frac{\pi}{L}) ^2$ for $n\in \Z$. If
$u_0\in L^2(-L,L)$ is decomposed as 
$u_0=\sum_{n\in\Z}c_n e_n$ then
$S(t)u_0=\sum_{n\in\Z}e^{-i\lambda _n t} c_n e_n$
for all $t\in\R$. 
\va

A proof of Corollary 1 using a controllability result
in the literature instead of a  Carleman's estimate is as follows.
Let $w(t):= \int_{t_1}^t S_L(t-\tau )f(\tau )\, d\tau \in L^2(-L,L)$
for $t_1\le t\le t_2$. By \cite[Thm 1.2]{M} (the result being in fact
true for any final state $y_T\in L^2(\Omega )$ instead of $0$) there
exists some (internal) control function
$h\in L^2((t_1,t_2)\times (L,L+1))$ such that the solution
$y\in C([t_1,t_2],L^2(-L,L+1))$ of 
\begin{equation*}
\left\{
\begin{array}{rll}
y_t-iy_{xx}&=h\chi _{(L,L+1)}&\hbox{ in }(t_1,t_2)\times (-L,L+1),\\
y&=0  &\hbox{ on }(t_1,t_2)\times\{ -L,L+1\} ,\\
y\at{t=t_1}&= 0,&
\end{array}
\right.
\end{equation*}
satisfies $y\at{t=t_2}=-w(t_2)$ on $(-L,L)$. Clearly, the function 
\begin{equation*}
v(t):=
\left\{
\begin{array}{ll}
w(t)+y(t)&\hbox{ if }t_1\le t\le t_2, \\
0 &\hbox{ otherwise,}
\end{array}
\right.
\end{equation*}
fulfills $v_t-iv_{xx}=f$ in ${\cal D}'( \R\times (-L,L))$, 
(\ref{eq:616}) (with $\epsilon =0$) and (\ref{eq:6160}).

As for the KdV equation the proof of Lemma 4 rests on Ingham's
inequality and on \cite[Thm 5.2]{Ko}. Here
${\cal Z}=\bigoplus _{n\in \N}{\cal Z}_n$, with
${\cal Z}_0=\hbox{Span}(e_0)$ and
${\cal Z}_n=\hbox{Span}(e_n,e_{-n})$ for $n\ge 1$. To properly
handle the l.h.s. in Ingham's estimate
\begin{equation}
\sum_{n\ge N} \vert a_n\vert ^2
\le C^{T'} \int_{-T'}^{T'} \left\vert
\sum_{n\ge N} a_n e^{-i\lambda _nt}\right\vert ^2 dt
\end{equation}
(which is applied with
$a_n=(c_ne_n+c_{-n}e_{-n})e^{-i\lambda _nT'}$) it is sufficient
to observe that the following estimate
\begin{equation*}
\Vert c_ne_n + c_{-n}e_{-n}\Vert ^2 _{L^2(-l,l)}
\ge  \frac{1}{2} \big( \Vert c_n e_n\Vert ^2_{L^2(-l,l)}
+ \Vert c_{-n} e_{-n}\Vert ^2_{L^2(-l,l)} \big)
\end{equation*}
holds true provided that $\vert n\vert$ be large enough, which implies
(for $N$ large enough)
\begin{equation*}
\int_{-l}^l \sum_{n\ge N}
\vert c_ne_n+c_{-n}e_{-n}\vert ^2\, dx \ge
\frac{l}{2L} \sum_{\vert n\vert \ge N}\vert c_n\vert ^2 =
\frac{l}{2L} \Vert u_0\Vert ^2_{L^2(-L,L)}.
\end{equation*}
The rest of the proof of Lemma 4 and of Proposition 3 is 
as above for the  KdV equation.\qed
\va
{\sc Appendix: Proof of proposition 1}
\va\va\indent
We argue by contradiction and assume that ${\cal R}=L^2(\R )$. 
Consider the
map 
\begin{equation*}
\Lambda :f\in L^2\big( (0,T)\times (L_1,L_2)\big)\to 
\int_0^T S(T-t)\tilde f(t,\cdot )dt \in L^2(\R )
\end{equation*}
(where
$\tilde f$ is the prolongation of $f$ by 0 on $\R ^2$). Let
$N=\hbox{ker}(\Lambda )$. Then the restriction of $\Lambda $ to the orthogonal
complement of $N$ in $L^2\big( (0,T)\times (L_1,L_2)\big)$
is a one-to-one continuous linear map which is onto
$L^2(\R)$, hence its inverse $(\Lambda \at{N^{\perp}})^{-1}$ is continuous. Let
$f\in {\cal D}\big( (0,T)\times (L_1,L_2)\big)$ and let $w_T\in D(A^*)$,
where $A^*$ denotes the adjoint of the operator $A$.
(Clearly $A^* w=\sum_{i=0}^n (-1)^ia_i\frac{d^iw}{dx^i}$.)
Recall that $A^*$ generates
the continuous semigroup $\big( S^* (t)\big) _{t\ge 0}$ on $L^2(\R )$.
Set $w(t):=S^* (T-t)w_T$ for $t\in [0,T]$.
Then $w$ solves
\begin{equation*}
\left\{
\begin{array}{rl}
\frac{dw}{dt}&=-A^* w\\
w\at{t=T}&=w_T.
\end{array}
\right.
\end{equation*}
Let $u(t)=\int_0^t S(t-s)\tilde f(s,\cdot )ds$. Integrating by part in
\begin{equation*}
\int_0^T\!\!  \int_{\R}\left( \frac{dw}{dt}+A^*w\right) u\, dxdt=0,
\end{equation*}
we get 
\begin{equation}
\label{eq:501}
\int_0^T\!\! \int_{L_1}^{L_2}f(t,x)w(t,x)\, dxdt = \int_{\R}w_T(x)u(T,x)\, dx.
\end{equation}
The same equation holds true (by density) for 
$f\in L^2\big( (0,T)\times (L_1,L_2)\big)$ and
$w_T\in L^2(\R )$.
Letting
$f=(\Lambda \at{N^{\perp}})^{-1}(w_T)$, where $w_T$ is any function in 
$L^2(\R )\setminus \{ 0\}$, we get
\begin{equation*}
\Vert w_T \Vert ^2_{L^2(\R )}\le 
\Vert f\Vert _{L^2 \big( (0,T)\times (L_1,L_2)\big)} \cdot
\Vert w\Vert _{L^2 \big( (0,T)\times (L_1,L_2)\big)} ,
\end{equation*}
hence
\begin{equation*}
\Vert w_T \Vert _{L^2(\R )}\le 
\Vert (\Lambda \at{N^{\perp}})^{-1} \Vert \cdot
\Vert w\Vert _{L^2 \big( (0,T)\times (L_1,L_2)\big)} \cdot
\end{equation*}
Replacing $w_T$ by $w_T(\cdot +n)$ (and also $w(t,x)$ by $w(t,x+n)$), 
we get
\begin{equation*}
\Vert w_T \Vert _{L^2(\R )}=\Vert w_T(\cdot +n)\Vert _{L^2(\R )}\le 
\Vert (\Lambda \at{N^{\perp}})^{-1} \Vert \cdot
\Vert w\Vert _{L^2 \big( (0,T)\times (L_1+n,L_2+n)\big)}.
\end{equation*}
Letting $n\to\infty$, we get (by Lebesgue's theorem)
$w_T=0$, a contradiction. It follows that
${\cal R}\ne L^2(\R )$. Let now $w_T\in {\cal R}^{\perp}$. We infer from
(\ref{eq:501}) that $\int_0^T\!\! \int_{L_1}^{L_2}f(t,x)w(t,x)\, dxdt=0$
for all $f\in L^2\big( (0,T)\times (L_1,L_2)\big)$, hence
$w\at{(0,T)\times (L_1,L_2)}=0$. Since $n\ge 2$ (with $a_n\ne 0$) 
it follows from Holmgren's uniqueness theorem 
that $w=0$ in $(0,T)\times \R$, hence $w_T=0$, and we 
infer that $\cal R$ is dense in $L^2(\R )$.
\va\va
{\bf Acknowledgement.} The author wishes to thank E. Zuazua who introduced
to him Jones' and his work.
\va
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\end{document}

