# Local controllability of the Korteweg-de Vries equation with the right
Dirichlet control

Hoai-Minh Nguyen  Laboratoire Jacques Louis Lions,
Sorbonne Université
Paris, France<EMAIL_ADDRESS>
###### Abstract.

The Korteweg-de Vries (KdV) equation with the right Dirichlet control was
initially investigated more than twenty years ago. It was shown that this
system is small time, locally, exactly controllable for all non-critical
lengths and its linearized system is not controllable for all critical
lengths. Even though the controllability of the KdV system has been studied
extensively in the last two decades, the local controllability of this system
for critical lengths remains an open question. In this paper, we give a
definitive answer to this question. First, we characterize all critical
lengths and the corresponding unreachable space for the linearized system. In
particular, we show that the unreachable space is always of dimension 1.
Second, we prove that the KdV system with the right Dirichlet control is not
locally null controllable in small time. Third, we give a criterion to
determine whether the system is locally exactly controllable in finite time or
not locally null controllable in any positive time for all critical lengths.
Consequently, we show that there exist critical lengths such that the system
is not locally null controllable in small time but is locally exactly
controllable in finite time. These facts are surprising and distinct in
comparison with related known results. First, it is known that the
corresponding KdV system with the right zero Dirichlet is locally exactly
controllable in small time using internal controls. Second, the unreachable
space of the linearized system of the corresponding KdV system with the right
Neumann control might be of arbitrary dimension. Third, the KdV system with
the right Neumann control is locally exactly controllable in small time if the
corresponding unreachable space is of dimension 1.

Key words: Controllability, KdV equations, critical lengths, unreachable
space, power series expansion, Hilbert uniqueness method

AMS subject classification: 35C20, 35Q53, 93B05, 93B07, 93C20.

###### Contents

  1. 1 Introduction and the statement of the main results
    1. 1.1 State of the art
    2. 1.2 Statement of the main results
    3. 1.3 Ideas of the proofs
    4. 1.4 Organisation of the paper
  2. 2 Linear KdV-Burgers equations
    1. 2.1 A useful lemma
    2. 2.2 Proof of Proposition 2.1
  3. 3 Linearized KdV equations
    1. 3.1 Proof of Proposition 3.1
      1. 3.1.1 Existence and estimates
      2. 3.1.2 Uniqueness
    2. 3.2 Proof of Proposition 3.2
    3. 3.3 Proof of Proposition 3.3
  4. 4 The set of the critical lengths
  5. 5 The unreachable space of the linearized system for a critical length
  6. 6 Properties of controls which steer 0 at time $0$ to $0$ at time $T$
  7. 7 Attainable directions in the unreachable space in small time
  8. 8 Small time local controllability properties of the KdV system - Proof of Theorem 1.2
  9. 9 Local, exact controllability of the KdV equation in a positive time - Proof of Theorem 1.3
    1. 9.1 Proof of Assertion $i)$ of Theorem 1.3
    2. 9.2 Proof of Assertion $ii)$ of Theorem 1.3
  10. A Hardy type inequalities
  11. B On the zeros of $\det_{Q}$
  12. C A lemma related to the moment method
  13. D Scilab program for computing $L_{n}$ and checking the local controllability property for $n=0,1,2,3$.

##  1\. Introduction and the statement of the main results

This paper is devoted to the local controllability of the Korteweg-de Vries
(KdV) equation using the right Dirichlet control. More precisely, we consider
the following control problem, for $T>0$,

(1.1) $\left\\{\begin{array}[]{cl}y_{t}+y_{x}+y_{xxx}+yy_{x}=0&\mbox{ in
}(0,T)\times(0,L),\\\\[6.0pt] y(\cdot,0)=y_{x}(\cdot,L)=0&\mbox{ in
}(0,T),\\\\[6.0pt] y(\cdot,L)=u&\mbox{ in }(0,T),\\\\[6.0pt]
y(0,\cdot)=y_{0}&\mbox{ in }(0,L).\end{array}\right.$

Here $y$ is the state, $y_{0}\in L^{2}(0,L)$ is an initial datum, and $u$ is a
control, belonging to an appropriate functional space. The KdV equation has
been introduced by Boussinesq [11] and Korteweg and de Vries [27] as a model
for the propagation of surface water waves along a channel. This equation also
furnishes a very useful nonlinear approximation model including a balance
between weak nonlinearity and weak dispersive effects, see e.g. [45, 33, 25].
The KdV equation has been investigated from various aspects of mathematics,
including the well-posedness, the existence, and stability of solitary waves,
the integrability, the long-time behavior, etc., see e.g. [45, 33, 25, 43,
29].

###  1.1. State of the art

The local controllability for the KdV equation has been studied extensively in
the literature, see, e.g., the surveys [40, 14] and the references therein. We
briefly review here some results concerning boundary controls. When the
controls are $y(\cdot,0)$, $y(\cdot,L)$, $y_{x}(\cdot,L)$, Russell and Zhang
[42] proved that the KdV equation is small time, locally exactly controllable.
The case of left boundary control ($y(\cdot,L)=y_{x}(\cdot,L)=0$) was
investigated by Rosier [39] (see also [22]). The small-time local, null
controllability holds in this case. The exact controllability does not hold
for initial and final data in the $L^{2}(0,L)$ due to the regularisation
effect but holds for a subclass of infinitely smooth initial and final data
[31].

A very close setting to the one considered here is the setting in which one
controls the right Neumann boundary, i.e., $y(\cdot,0)=y(\cdot,L)=0$ and
$y_{x}(\cdot,L)$ is a control. For initial and final data in $L^{2}(0,L)$, and
controls in $L^{2}(0,T)$, Rosier [38] proved that the KdV system with the
right Neumann control is small time, locally, exactly controllable provided
that the length $L$ is not critical, i.e., $L\notin{\mathcal{N}}_{N}$ 111The
letter $N$ stands for the Neumann boundary control., where
222$\mathbb{N}_{*}=\mathbb{N}\setminus\\{0\\}.$

(1.2)
${\mathcal{N}}_{N}:=\left\\{2\pi\sqrt{\frac{k^{2}+kl+l^{2}}{3}};\,k,l\in\mathbb{N}_{*}\right\\}.$

To this end, Rosier studied the controllability of the corresponding
linearized system and showed that the linearized system is exactly
controllable if $L\not\in{\mathcal{N}}_{N}$. He as well established that when
$L\in{\mathcal{N}}_{N}$, the linearized system is not controllable. More
precisely, Rosier showed that there exists a non trivial, finite dimensional
subspace ${\mathcal{M}}_{N}$ of $L^{2}(0,L)$ such that its orthogonal space is
reachable from $0$ for small time whereas ${\mathcal{M}}_{N}$ is not for any
time. To tackle the control problem for a critical length
$L\in{\mathcal{N}}_{N}$ with initial and final data in $L^{2}(0,L)$ and
controls in $L^{2}(0,T)$, Coron and Crépeau introduced the power series
expansion method [19]. The idea is to take into account the effect of the
nonlinear term $yy_{x}$ absent in the corresponding linearized system. Using
this method, Coron and Crépeau showed [19] (see also [18, section 8.2]) that
the KdV system is small time, locally, exactly controllable when
$\dim{\mathcal{M}}_{N}=1$. Cerpa [13] developed the analysis in [19] to prove
that the KdV system is finite time, locally, exactly controllable in the case
$\dim{\mathcal{M}}_{N}=2$. Later, Crépeau and Cerpa [15] succeeded to extend
the ideas in [13] to obtain the local, exact controllability in finite time
for all other critical lengths. Recently, with Coron and Koenig [20], we prove
that such a system is not small time, locally, null controllable for a class
of critical lengths. This fact is surprising when compared with known results
on internal controls for the KdV equation. It is known, see [12, 32, 37], that
the KdV system (1.1) with $u=0$ is small time, locally controllable using
internal controls whenever the control region contains an arbitrary, open
subset of $(0,L)$. The power expansion method is also a starting point of the
analysis in [20]. Part of the analysis is to characterize all controls that
bring 0 at 0 to 0 at time T for the corresponding linearized system. This idea
is then used in the study of the water tank problem [21]. It is interesting to
note that there are other types of boundary controls for the KdV equation for
which there is no critical length, see [38, 39, 23, 14]. There are also
results on the internal controllability for the KdV equation, see [42, 28, 12]
and the references therein. A minimal time of the null controllability is also
required for some linear partial differential equations. This is the case for
equations with a finite speed of propagation, such as the transport equation,
the wave equation, or the hyperbolic system, see, e.g., [18] and the
references therein. But this can also happen for equations with the infinite
speed of propagation, such as some parabolic systems [7], Grushin-type
equations [2, 6, 26, 3], Kolmogorov-type equations [4], and the references
therein.

We now turn back to the control problem (1.1). This control problem was first
investigated by Glass and Guerrero [23]. To this end, in the spirit of
Rosier’s work mentioned above, they introduced the corresponding set of
critical lengths 333The letter $D$ stands for the Dirichlet boundary control.

(1.3) ${\mathcal{N}}_{D}=\Big{\\{}L\in\mathbb{R}_{+};\,\exists
z_{1},z_{2}\in\mathbb{C}:\eqref{cond-z1z2}\mbox{ holds}\Big{\\}},$

where

(1.4)
$z_{1}e^{z_{1}}=z_{2}e^{z_{2}}=-(z_{1}+z_{2})e^{-(z_{1}+z_{2})}\quad\mbox{ and
}\quad L^{2}=-(z_{1}^{2}+z_{1}z_{2}+z_{2}^{2}).$

They proved that the set ${\mathcal{N}}_{D}$ is infinite and has no
accumulation point. Concerning (1.1), Glass and Guerrero proved that the
corresponding linearized KdV system is small time, exactly controllable with
initial and final data in $H^{-1}(0,L)$ using controls in $L^{2}(0,T)$ if
$L\not\in{\mathcal{N}}_{D}$. Developing this result, they also established
that the KdV system (1.1) is small-time locally controllable for initial and
final data in $L^{2}(0,L)$ and controls in $H^{1/6_{-}}$ 444Controls in
$H^{1/6_{-}}$ means controls in $H^{1/6-\varepsilon}$ for all $\varepsilon>0$.
for non-critical lengths, i.e., $L\not\in{\mathcal{N}}_{D}$.

Even the local controllability of the KdV system using boundary controls has
been investigated extensively in the last two decades, to our knowledge, there
is no result on the local controllability of system (1.1) for critical
lengths. In comparison with the set of critical lengths ${\mathcal{N}}_{N}$,
the set ${\mathcal{N}}_{D}$ is less explicit. Moreover, the unreachable space
for the linearized system related to (1.1) has not been determined due to the
lack of an appropriate observability inequality for the linearized system in a
suitable functional setting. As revealed later in this paper, the control
properties related to the unreachable space for the right Dirichlet control
are quite distinct from the one for the right Neumann control.

###  1.2. Statement of the main results

The main goal of this paper is to give a rather complete picture of the local
controllability of the control KdV system (1.1). We first characterize all the
critical lengths and describe the corresponding unreachable space
$\mathcal{M}_{D}$ for the corresponding linearized system. The first result in
this direction is.

######  Theorem 1.1.

Let $L\in{\mathcal{N}}_{D}$. There exists a unique pair
$(a,b)\in\mathbb{R}^{2}$ such that $a<0$, $b>0$,

(1.5) $(a+ib)e^{a+ib}=(a-ib)e^{a-ib}=-2ae^{-2a},$

and

(1.6) $L^{2}=b^{2}-3a^{2}.$

Set

(1.7) $q=-\frac{2a(a^{2}+b^{2})}{L^{3}}>0,$ (1.8) $\phi(x)=-\beta e^{\alpha
x}\cos(\beta x)+\beta e^{-2\alpha x}+3\alpha e^{\alpha x}\sin(\beta x)\mbox{
for }x\in[0,L],$

with

(1.9) $\mbox{$\alpha=-a/L\quad$ and $\quad\beta=-b/L$},$

and

(1.10) $\Phi(t,x)=e^{qt}\phi(x)\mbox{ for }(t,x)\in\mathbb{R}\times[0,L].$

Then $\Phi$ is a solution of the system

(1.11) $\left\\{\begin{array}[]{cl}\Phi_{t}+\Phi_{x}+\Phi_{xxx}=0&\mbox{ in
}\mathbb{R}\times(0,L),\\\\[6.0pt]
\Phi(\cdot,0)=\Phi_{x}(\cdot,0)=\Phi(\cdot,L)=\Phi_{xx}(\cdot,L)=0&\mbox{ in
}\mathbb{R},\end{array}\right.$

and the unreachable space $\mathcal{M}_{D}$ of the linearized system of (1.1)
is given by

(1.12) $\mathcal{M}_{D}=\mbox{span }\Big{\\{}\phi\Big{\\}}.$

Consequently,

(1.13) $\dim\mathcal{M}_{D}=1.$

######  Remark 1.1.

The main part of Theorem 1.1 is to show that the complex numbers $z_{1}$,
$z_{2}$ from the definition of ${\mathcal{N}}_{D}$ in (1.3) can be chosen as

$z_{1}=a+ib\quad\mbox{ and }\quad z_{2}=a-ib,$

for some $a<0$ and $b>0$. This is non-trivial, see Section 4.

######  Remark 1.2.

Implicit information given in Theorem 1.1 is that the function $\phi$, defined
in (1.8), satisfies the following boundary condition:

$\phi(0)=\phi_{x}(0)=\phi(L)=\phi_{xx}(L)=0.$

Using Theorem 1.1, we can precisely describe the set ${\mathcal{N}}_{D}$ in
the following result.

######  Proposition 1.1.

Let $L>0$. Then $L\in{\mathcal{N}}_{D}$ if and only if $L=L_{n}$ for some
$n\in\mathbb{N}$ where $L_{n}$ is given by

$L_{n}^{2}=4a_{n}^{2}(e^{-6a_{n}}-1),$

with $a_{n}<0$ being uniquely determined by

$b_{n}\cos b_{n}+a_{n}\sin b_{n}=0,$

where $b_{n}$ satisfies

$\pi+2n\pi<b_{n}<3\pi/2+2n\pi\quad\mbox{ and }\quad
b_{n}^{2}=4a_{n}^{2}(e^{-6a_{n}}-1/4).$

After presenting the results on the unreachable space of the linearized system
and the characterization of the critical lengths, we next discuss the local
controllability of (1.1). Set, for $T>0$,

(1.14) $X_{T}:=C\big{(}[0,T];L^{2}(0,L)\big{)}\cap
L^{2}\big{(}(0,T);H^{1}(0,L)\big{)}$

equipped with the corresponding norm:

$\|y\|_{X_{T}}=\|y\|_{C\big{(}[0,T];L^{2}(0,L)\big{)}}+\|y\|_{L^{2}\big{(}(0,T);H^{1}(0,L)\big{)}}.$

We are ready to state the local controllability property of (1.1) in small
time.

######  Theorem 1.2.

Let $L\in{\mathcal{N}}_{D}$. There exist $T=T_{0}>0$ and $\varepsilon_{0}>0$
such that for all solutions $y\in X_{T}$ of the system

(1.15) $\left\\{\begin{array}[]{cl}y_{t}+y_{x}+y_{xxx}+yy_{x}=0&\mbox{ in
}(0,T)\times(0,L),\\\\[6.0pt] y(\cdot,0)=y_{x}(\cdot,L)=0&\mbox{ in
}(0,T),\\\\[6.0pt] y(\cdot,L)=u&\mbox{ in }(0,T),\\\\[6.0pt]
y(0,\cdot)=\varepsilon\phi&\mbox{ in }(0,L),\end{array}\right.$

with $0<\varepsilon<\varepsilon_{0}$ and
$\|u\|_{H^{1/2}(0,T)}<\varepsilon_{0}$, we have

$y(T,\cdot)\neq 0.$

Recall that $\phi$ is defined in (1.8).

We next present a criterion on the local controllability property of (1.1) in
finite time. Set

(1.16)
$\Omega(z)=\int_{0}^{L}\left|\sum_{j=1}^{3}\big{(}\lambda_{j}e^{\lambda_{j}L}-\lambda_{j+1}e^{\lambda_{j+1}L}\big{)}e^{\lambda_{j+2}x}\right|^{2}\phi_{x}(x)\,dx,$

where $\lambda_{j}=\lambda_{j}(z)$ with $j=1,2,3$ are the three solutions of
the equation

$\lambda^{3}+\lambda+iz=0$

and the convention $\lambda_{j+3}=\lambda_{j}$ for $j\geq 1$ is used. It is
clear that $\Omega$ is continuous with respect to $z$. One can show that, for
$c\in\mathbb{R}$,

$\lim_{|z|\to\infty,z\in\mathbb{R}}\Omega(z+ic)=+\infty.$

The criterion on the local controllability property of (1.1) in finite time is
stated in the following result.

######  Theorem 1.3.

Let $L\in{\mathcal{N}}_{D}$ and let $q>0$ be defined by (1.7). Set

$\omega=\min_{z\in\mathbb{R}}\Omega(z+iq/2),$

where $\Omega$ is defined by (1.16). The following two facts hold.

$i)$ If $\omega\geq 0$, then system (1.1) is not locally null controllable in
any positive time with controls in $H^{1/2}$. More precisely, given $T>0$
arbitrary, there exists $\varepsilon_{T}>0$ such that for all solution of
$y\in X_{T}$ of (1.15) with $0<\varepsilon<\varepsilon_{T}$ and
$\|u\|_{H^{1/2}(0,T)}<\varepsilon_{T}$, we have

$y(T,\cdot)\neq 0.$

$ii)$ If $\omega<0$, then system (1.1) is locally exactly controllable in
finite time with controls in $H^{1/3}$ and initial and final data in
$L^{2}(0,L)$. More precisely, there exist $T_{0}>0$ and $\varepsilon_{0}>0$
such that for all $y_{0},y_{1}\in L^{2}(0,L)$ with
$\|y_{0}\|_{L^{2}(0,L)}\leq\varepsilon_{0}$ and
$\|y_{1}\|_{L^{2}(0,L)}\leq\varepsilon_{0}$, there exists $u\in
H^{1/3}(0,T_{0})$ with $\|u\|_{H^{1/3}(0,T_{0})}\leq
C(\|y_{0}\|_{L^{2}(0,L)}+\|y_{1}\|_{L^{2}(0,L)})^{1/2}$ such that

$y(T_{0},\cdot)=y_{1},$

where $y\in X_{T_{0}}$ is the unique solution of (1.1). Here $C$ denotes a
positive constant independent of $y_{0}$ and $y_{1}$.

######  Remark 1.3.

Using Scilab, the program is given in Appendix D, one can show that

  * •

if $n=0$, then $L=4.5183604$, $a=-0.5065520$, $b=4.6027563$, $q=0.2354919$,
and $\Omega(iq/2)=-0.0140641$.

  * •

if $n=1$, then $L=10.866906$, $a=-0.6903700$, $b=10.932497$, $q=0.1291104$,
and $\Omega(iq/2)=-0.0061256$.

  * •

if $n=2$, then $L=17.177525$, $a=-0.7947960$, $b=17.232599$, $q=0.0933315$,
and $\Omega(iq/2)=-0.0036196$.

  * •

if $n=3$, then $L=23.476776$, $a=-0.8687610$, $b=23.524949$, $q=0.0744156$,
and $\Omega(iq/2)=-0.0024525$.

Parameters of the program ($k$, $j$, $j0$) are given for $n=0$. The parameters
for $n=1,2,3$ are also given there.

As a consequence of Theorem 1.2 and Theorem 1.3 (see also Remark 1.3), there
exist critical lengths for which system (1.1) is not locally null controllable
in small time but locally exactly controllable in finite time even the
unreachable space of the corresponding linearized system is of dimension 1.
This result is surprising and distinct when compared with known related
results on the local controllability of the KdV equation. First, it is known
that the corresponding KdV system with the right zero Dirichlet is locally
exactly controllable in small time using internal controls. Second, the
unreachable space of the linearized system of the corresponding KdV system
with the right Neumann control might be of arbitrary dimension. Third, the KdV
system with the right Neumann control is locally exactly controllable in small
time if the corresponding unreachable space is of dimension 1.

Finally, in the case $L\not\in{\mathcal{N}}_{D}$, we can prove the following
result, which sharpens the results in [23] mentioned previously.

######  Theorem 1.4.

Let $L\not\in{\mathcal{N}}_{D}$ and $T>0$. There exist $C>0$ and
$\varepsilon_{0}>0$ depending only on $L$ and $T$ such that for all
$y_{0},\,y_{1}\in L^{2}(0,L)$ with $\|y_{0}\|_{L^{2}(0,L)}\leq\varepsilon_{0}$
and $\|y_{1}\|_{L^{2}(0,L)}\leq\varepsilon_{0}$, there exists $u\in
H^{1/3}(0,T)$ with $\|u\|_{H^{1/3}(0,T)}\leq
C(\|y_{0}\|_{L^{2}(0,L)}+\|y_{1}\|_{L^{2}(0,L)})$ such that

$y(T,\cdot)=y_{1},$

where $y\in X_{T}$ is the unique solution of (1.1).

###  1.3. Ideas of the proofs

The characterization of the critical lengths and the unreachable space is
based on the study of system (1.4). In comparison with the characterization of
the set ${\mathcal{N}}_{N}$, the study of ${\mathcal{N}}_{D}$ is more complex.
The main goal and the difficulty in this direction, see Section 4, are to
obtain/characterize all solutions of (1.4). The existence of a class of
solutions is easier as observed previously in [23]. As a consequence, we
derive that $\dim\mathcal{M}_{D}=1$ which is completely different from what is
known for the right Neumann control.

Our approaches to Theorem 1.2 and Theorem 1.3 are inspired by the power series
expansion method introduced by Coron and Crépeau [19]. The idea of this method
is to search for/understand a control $u$ of the form

$u=\varepsilon u_{1}+\varepsilon^{2}u_{2}+\cdots.$

The corresponding solution then formally has the form

$y=\varepsilon y_{1}+\varepsilon^{2}y_{2}+\cdots,$

and the non-linear term $yy_{x}$ can be written as

$yy_{x}=\varepsilon^{2}y_{1}y_{1,x}+\cdots.$

One then obtains the following systems for $y_{1}$ and $y_{2}$:

(1.17) $\left\\{\begin{array}[]{cl}y_{1,t}+y_{1,x}+y_{1,xxx}=0&\mbox{ in
}(0,T)\times(0,L),\\\\[6.0pt] y_{1}(\cdot,0)=y_{1,x}(\cdot,L)=0&\mbox{ in
}(0,T),\\\\[6.0pt] y_{1}(\cdot,L)=u_{1}&\mbox{ in }(0,T),\end{array}\right.$
(1.18)
$\left\\{\begin{array}[]{cl}y_{2,t}+y_{2,x}+y_{2,xxx}+y_{1}y_{1,x}=0&\mbox{ in
}(0,T)\times(0,L),\\\\[6.0pt] y_{2}(\cdot,0)=y_{2,x}(\cdot,L)=0&\mbox{ in
}(0,T),\\\\[6.0pt] y_{2}(\cdot,L)=u_{2}&\mbox{ in }(0,T).\end{array}\right.$

The idea is then to find the corresponding controls $u_{1}$ and $u_{2}$ such
that, if $y_{1}(0,\cdot)=y_{2}(0,\cdot)=0$, then $y_{1}(T,\cdot)=0$ and the
$L^{2}$-orthogonal projection of $y_{2}(T,\cdot)$ on ${\mathcal{M}}_{D}$ is a
given (non-zero) element in ${\mathcal{M}}_{D}$. To this end, in [19, 13, 15],
the authors used delicate contradiction arguments to capture the structure of
their studied KdV system.

We here use the ideas in the spirit of the joint work with Coron and Koenig
[20]. The starting point of the analysis is also the power series expansion
method. The strategy is to characterize all possible $u_{1}$ which steers 0 at
time 0 to $0$ at time $T$ (Proposition 6.1). This is done by taking the
Fourier transform with respect to time of the solution $y_{1}$ and applying
Paley-Wiener’s theorem. We then investigate the projection of $y_{2}$ into the
unreachable space $\mathcal{M}_{D}$ (in Section 6). This is where our analysis
deviates from [20] to take the advantage of the fact that $q$ is real and to
use the characterization via Paley-Wiener’s theorem to construct $u_{1}$ for
which the projection of $y_{2}$ into $\mathcal{M}_{D}$ has a desired direction
(see the proof of Assertion $ii)$ of Theorem 1.3). Concerning the projection
of $y_{2}$ into $\mathcal{M}_{D}$, we use the information of
$|\hat{u}(z+iq/2)|^{2}$ instead of $\hat{u}(z)\overline{\hat{u}(z+iq)}$ as
suggested in [20]. This new proposal has two advantages. First, the way to
obtain the projection of $y_{2}$ into the unreachable space $\mathcal{M}_{D}$
is less complex. More importantly, it allows us to obtain the criterion on the
local controllability of (1.1) in finite time given in Theorem 1.3 by taking
into account the fact that $|\hat{u}(z+iq/2)|^{2}\geq 0$. Another important
part in the proof of the local controllability in Theorem 1.2 and Theorem 1.3
is to obtain the observability for the linearized system with initial/final
data in $\mathcal{M}_{D}^{\perp}$. To this end, one needs to establish the
optimal results on the stability and the well-posedness of the linearized
system of (1.1) and related ones in which the boundary conditions are in
fractional Sobolev spaces, see Section 3. The proof of the local
controllability in finite time given in Assertion $ii)$ of Theorem 1.3 is
based on a new strategy to handle the lack of symmetry in our control system
in comparison with the one using the right Neumann control. Concerning the KdV
system using the right Neumann control, one can bring a state forward or
backward easily thanks to the symmetry of the boundary conditions. This is not
the case for (1.1) since the Dirichlet condition on the right is controlled
and the Dirichlet condition on the left is imposed to be 0. To overcome this,
we introduce a new approach that involves a KdV system with a new type
boundary condition, investigated in Section 3, in a backward way. Our analysis
uses the Banach fixed point arguments while the Brouwer fixed point theorem is
usually applied to related contexts. It is worth noting that the analysis in
this paper is based on the expansion up to the second order even if the
unreachable space is of dimension 1. This explains the cost of the control
given in Assertion $ii)$ of Theorem 1.3. This way is different from what has
been done for the KdV system with the right Neumann control where the
expansion up to the third order is required.

Here are other comments on the analysis.

$i)$ Various interpolation inequalities involving fractional Sobolev spaces
and their dual spaces are established and used in the proof of Theorem 1.2 and
Theorem 1.3.

$ii)$ The analysis for the well-posedness of the KdV equation with various
boundary conditions is in the spirit of the joint work with Coron and Koenig
[20]. This partly involves a connection between the linear KdV equation and
the linear KdV-Burgers equation as previously used by Bona et al. [9] and
inspired by the work of Bourgain [10], and Molinet and Ribaud [34]. In this
work, to deal with the boundary conditions containing the information of
$y_{xx}$ and derive optimal estimates on $y$, $y_{x}$, and $y_{xx}$ in
fractional Sobolev spaces and their dual spaces, new ideas and technique are
implemented (see Remark 3.6). It is worth noting that similar estimates for
$y_{xx}$ are not known for the KdV equation in the real line setting and it is
not clear whether such estimates hold for that setting. Our estimates
particularly allow us to get the observability inequality for the critical
lengths, which plays a role in determining the unreachable space for the
linearized KdV system (see Section 3). The unreachable space is not known
previously and this determination is also a contribution of the paper.

$iii)$ The arguments used in this paper to disprove the small time, local
controllability improve the ones in [20]. One of the key steps in the proof is
to establish new (optimal) estimates for solutions of the KdV equations with
various boundary conditions (see Section 3). The proof also relies on the
positivity of a scalar product which comes naturally in the study of the
controllability of the KdV system for small time. The positivity of a scalar
product for small time was also a crucial point of several lacks of small
time, local controllability results for systems with infinite propagation
speed. Nevertheless, the previous ways used to derive the positivity are
different, see, e.g., [17, 30, 5].

###  1.4. Organisation of the paper

The paper is organized as follows. Section 2 is devoted to the study of the
linear KdV-Burgers equation with the periodic boundary condition. Section 3 is
devoted to the study of the (linear and nonlinear) KdV equations with various
boundary conditions. Section 4 is devoted to the study of critical lengths and
some properties of the unreachable space $\mathcal{M}_{D}$. We there prove
(1.5), (1.6), and (1.11), and establish Proposition 1.1. In Section 5, we
prove (1.12) and establish the corresponding observability inequality (Lemma
5.2). The proof of Theorem 1.4 is given at the end of this section. In Section
6, we study properties of the controls which steer 0 at time 0 to 0 at time T.
Using results in Section 6, we study attainable directions for small time in
Section 7. The proofs of Theorem 1.2 and Theorem 1.3 are given in Section 8
and Section 9, respectively. Several technical results are stated and proved
in the appendix. The Scilab program is also given there.

##  2\. Linear KdV-Burgers equations

This section is devoted to the study of the linear KdV-Burgers equations with
the periodic boundary condition. Here is the main result of this section,
which plays an essential role in the study of the linearized KdV equation with
various boundary conditions investigated in Section 3.

######  Proposition 2.1.

Let $g\in L^{1}\big{(}\mathbb{R}_{+};L^{2}(0,L)\big{)}$ with
$\int_{0}^{L}g(t,x)\,dx=0$ for a.e. $t>0$, and let $y_{0}\in L^{2}(0,L)$ be
such that $\int_{0}^{L}y_{0}(x)\,dx=0$. There exists a unique solution $y\in
C\big{(}[0,+\infty);L^{2}(0,L)\big{)}\cap
L^{2}\big{(}[0,+\infty);H^{1}([0,L])\big{)}$ which is periodic in space of the
system

(2.1) $\left\\{\begin{array}[]{cl}y_{t}+4y_{x}+y_{xxx}-3y_{xx}=g&\mbox{ in
}\mathbb{R}_{+}\times(0,L),\\\\[6.0pt] y(0,\cdot)=y_{0}&\mbox{ in
}(0,L).\end{array}\right.$

Moreover, for $t\in\mathbb{R}_{+}$ and $x\in[0,L]$,

(2.2)
$\|y(t,\cdot)\|_{L^{2}(0,L)}+\|y(\cdot,x)\|_{H^{1/3}(\mathbb{R}_{+})}+\|y_{x}(\cdot,x)\|_{L^{2}(\mathbb{R}_{+})}+\|y_{xx}(\cdot,x)\|_{[H^{1/3}(\mathbb{R}_{+})]^{*}}\\\\[6.0pt]
\leq
C\Big{(}\|y_{0}\|_{L^{2}(0,L)}+\|g\|_{L^{1}\big{(}\mathbb{R}_{+};L^{2}(0,L)\big{)}}\Big{)},$

and

(2.3)
$\|y(\cdot,x)\|_{L^{2}(\mathbb{R}_{+})}+\|y_{x}(\cdot,x)\|_{[H^{1/3}(\mathbb{R}_{+})]^{*}}\leq
C\Big{(}\|y_{0}\|_{H^{-1}(0,L)}+\|g\|_{L^{1}\big{(}\mathbb{R}_{+};H^{-1}(0,L)\big{)}}\Big{)}$

for some positive constant $C$ depending only on $L$.

The rest of this section consisting of two subsections is organized as
follows. In the first subsection, we present and prove a useful lemma in the
spirit of Bourgain’s. The second subsection is devoted to the proof of
Proposition 2.1.

###  2.1. A useful lemma

The following result in the spirit of Bourgain’s is used in the proof of
Proposition 2.1.

######  Lemma 2.1.

Let $(a_{n})_{n\in\mathbb{Z}\setminus\\{0\\}}\subset\mathbb{R}$ be such that
$\sum_{n\in\mathbb{Z}\setminus\\{0\\}}a_{n}^{2}<+\infty$. For $-2<s\leq 3$,
there exists a positive constant $C=C_{s}$ such that

(2.4)
$\int_{\mathbb{R}}(1+|z|)^{2s/3}\left(\sum_{n\in\mathbb{Z}\setminus\\{0\\}}\frac{|a_{n}||n|^{3-s}}{(z+4n-n^{3})^{2}+n^{4}}\right)^{2}\,dz\leq
C\sum_{n\in\mathbb{Z}\setminus\\{0\\}}a_{n}^{2}.$

###### Proof.

For $s>-5/2$ and for $-1\leq z\leq 1$, one has

$\sum_{n\in\mathbb{Z}\setminus\\{0\\}}\frac{|a_{n}||n|^{3-s}}{(z+4n-n^{3})^{2}+n^{4}}\,\leq
C\left(\sum_{n\in\mathbb{Z}\setminus\\{0\\}}a_{n}^{2}\right)^{1/2}.$

It hence suffices to prove that, for $-2<s\leq 3$,

(2.5)
$\int_{z\in\mathbb{R};|z|>1}(1+|z|)^{2s/3}\left(\sum_{n\in\mathbb{Z}\setminus\\{0\\}}\frac{|a_{n}||n|^{3-s}}{(z+4n-n^{3})^{2}+n^{4}}\right)^{2}\,dz\leq
C\sum_{n\in\mathbb{Z}\setminus\\{0\\}}a_{n}^{2}.$

We have

(2.6)
$\int_{z>1}(1+|z|)^{2s/3}\left(\sum_{n\in\mathbb{Z}\setminus\\{0\\}}\frac{|a_{n}||n|^{3-s}}{(z+4n-n^{3})^{2}+n^{4}}\right)^{2}\,dz\\\\[6.0pt]
\leq
C\sum_{m\in\mathbb{N}_{*}}\int_{m^{3}}^{(m+1)^{3}}m^{2s}\left(\sum_{n\in\mathbb{Z}\setminus\\{0\\}}\frac{|a_{n}||n|^{3-s}}{(z+4n-n^{3})^{2}+n^{4}}\right)^{2}\,dz.$

For $m^{3}\leq z\leq(m+1)^{3}$, one gets

(2.7)
$\sum_{n\in\mathbb{Z}\setminus\\{0\\}}\frac{|a_{n}||n|^{3-s}}{(z+4n-n^{3})^{2}+n^{4}}\\\\[6.0pt]
=\sum_{k;\,m+k\in\mathbb{Z}\setminus\\{0\\}}\frac{|m+k|^{3-s}|a_{m+k}|}{(z+4(m+k)-(m+k)^{3})^{2}+(m+k)^{4}}\\\\[6.0pt]
\leq C\mathop{\sum_{k;|k|\leq
2m}}_{\,m+k\in\mathbb{Z}\setminus\\{0\\}}\frac{|m+k|^{3-s}|a_{m+k}|}{m^{4}(|k|+1)^{2}}+C\mathop{\sum_{k;\;|k|\geq
2m+1}}_{m+k\in\mathbb{Z}\setminus\\{0\\}}\frac{|k|^{3-s}|a_{m+k}|}{k^{4}(|k|+1)^{2}}.$

Combining (2.6) and (2.7) yields

(2.8)
$\int_{z>1}(1+|z|)^{2s/3}\left(\sum_{n\in\mathbb{Z}\setminus\\{0\\}}\frac{|a_{n}||n|^{3-s}}{(z+4n-n^{3})^{2}+n^{4}}\right)^{2}\,dz\\\\[6.0pt]
\leq C\sum_{m\in\mathbb{N}_{*}}\left(\mathop{\sum_{k;|k|\leq
2m}}_{m+k\in\mathbb{Z}\setminus\\{0\\}}\frac{|m+k|^{3-s}|a_{m+k}|}{m^{3-s}(|k|+1)^{2}}+C\mathop{\sum_{k;\;|k|\geq
2m+1}}_{m+k\in\mathbb{Z}\setminus\\{0\\}}\frac{|m|^{1+s}|a_{m+k}|}{|k|^{1+s}(|k|+1)^{2}}\right)^{2}\\\\[6.0pt]
\leq
C\sum_{m\in\mathbb{N}_{*}}\left(\mathop{\sum_{k}}_{m+k\in\mathbb{Z}\setminus\\{0\\}}\frac{|a_{m+k}|^{2}}{(|k|+1)^{1+\varepsilon}}\right)\left(\mathop{\sum_{k;|k|\leq
2m}}_{m+k\in\mathbb{Z}\setminus\\{0\\}}\frac{|m+k|^{6-2s}}{m^{6-2s}(|k|+1)^{3-\varepsilon}}+\mathop{\sum_{k;|k|\geq
2m+1}}_{m+k\in\mathbb{Z}\setminus\\{0\\}}\frac{m^{2+2s}}{|k|^{2+2s}(|k|+1)^{3-\varepsilon}}\right).$

Here $0<\varepsilon\leq 1$ is fixed such that $s>(\varepsilon-4)/2$ (the fact
$s>-2$ is used). We have, for $m\in\mathbb{N}_{*}$, $|k|\leq 2m$, and
$m+k\in\mathbb{Z}\setminus\\{0\\}$,

$\frac{|m+k|^{6-2s}}{m^{6-2s}}\leq C\mbox{ for }s\leq 3,$

and, for $m\in\mathbb{N}_{*}$,

$\mathop{\sum_{k;|k|\geq
2m+1}}_{m+k\in\mathbb{Z}\setminus\\{0\\}}\frac{m^{2+2s}}{|k|^{2+2s}(|k|+1)^{3-\varepsilon}}=m^{2+2s}\mathop{\sum_{k;|k|\geq
2m+1}}_{m+k\in\mathbb{Z}\setminus\\{0\\}}\frac{1}{k^{2+2s}(|k|+1)^{3-\varepsilon}}\leq\frac{C}{m^{2-\varepsilon}}\leq
C\mbox{ for }s>(\varepsilon-4)/2.$

Using the fact, for $m\in\mathbb{N}_{*}$,

$\sum_{k;\,m+k\in\mathbb{Z}\setminus\\{0\\}}\frac{1}{(|k|+1)^{3-\varepsilon}}<+\infty\mbox{
for }0<\varepsilon<1,$

we derive from (2.8) that

(2.9)
$\int_{z>1}(1+|z|)^{2s/3}\left(\sum_{n\in\mathbb{Z}\setminus\\{0\\}}\frac{|a_{n}||n|^{3-s}}{(z+4n-n^{3})^{2}+n^{4}}\right)^{2}\,dz\\\\[6.0pt]
\leq
C\sum_{m\in\mathbb{N}_{*}}\sum_{k;\,m+k\in\mathbb{Z}\setminus\\{0\\}}\frac{|a_{m+k}|^{2}}{(|k|+1)^{2}}\leq
C\sum_{k\in\mathbb{Z}}\frac{1}{(|k|+1)^{2}}\sum_{m;\,m+k\in\mathbb{Z}\setminus\\{0\\}}|a_{m+k}|^{2},$

which yields

(2.10)
$\int_{z>1}(1+|z|)^{2s/3}\left(\sum_{n\in\mathbb{Z}\setminus\\{0\\}}\frac{|a_{n}||n|^{3-s}}{(z+4n-n^{3})^{2}+n^{4}}\right)^{2}\,dz\leq
C\sum_{n\in\mathbb{Z}\setminus\\{0\\}}|a_{n}|^{2}.$

Similarly, we have

(2.11)
$\int_{z<-1}(1+|z|)^{2s/3}\left(\sum_{n\in\mathbb{Z}\setminus\\{0\\}}\frac{|a_{n}||n|^{3-s}}{(z+4n-n^{3})^{2}+n^{4}}\right)^{2}\,dz\leq
C\sum_{n\in\mathbb{Z}\setminus\\{0\\}}|a_{n}|^{2}.$

Estimate (2.5) now follows from (2.10) and (2.11). The proof is complete. ∎

###  2.2. Proof of Proposition 2.1

We only derive the estimates. The uniqueness follows from the estimates and
the existence follows from the proof of these estimates as well.

Multiplying the equation of $y$ by $y$ and integrating by parts, we have

$\frac{1}{2}\frac{d}{dt}\int_{0}^{L}|y(t,x)|^{2}\,dx+3\int_{0}^{L}|y_{x}(t,x)|^{2}\,dx=\int_{0}^{L}g(t,x)y(t,x)\,dx.$

This yields

$\frac{1}{2}\frac{d}{dt}\int_{0}^{L}|y(t,x)|^{2}\,dx\leq\|g(t,\cdot)\|_{L^{2}(0,L)}\|y(t,\cdot)\|_{L^{2}(0,L)}.$

Applying the Grönwall lemma, we obtain the desired estimate for
$\|y(t,\cdot)\|_{L^{2}(0,L)}$.

We next establish the estimates for
$\|y(\cdot,x)\|_{H^{1/3}(\mathbb{R}_{+})}$,
$\|y_{x}(\cdot,x)\|_{L^{2}(\mathbb{R}_{+})}$, and
$\|y_{xx}(\cdot,x)\|_{[H^{1/3}(\mathbb{R}_{+})]^{*}}$. For notational ease, we
assume that $L=2\pi$. It suffices to consider the case $g\equiv 0$ and the
case $y_{0}\equiv 0$ separately.

We first consider the case $g\equiv 0$. Write the solution under the form

$y(t,x)=\sum_{n\in\mathbb{Z}}a_{n}(t)e^{inx}\mbox{ in
}\mathbb{R}_{+}\times(0,L)=\mathbb{R}_{+}\times(0,2\pi).$

We then derive that

$y_{0}(x)=\sum_{n\in\mathbb{Z}}a_{n}(0)e^{inx}\mbox{ for }x\in[0,2\pi]$

and

$a_{n}^{\prime}(t)=\big{(}-3n^{2}-i(4n-n^{3})\big{)}a_{n}(t)\mbox{ for
}t\in\mathbb{R}_{+}.$

We thus have

$a_{n}(t)=e^{\big{(}-3n^{2}-i(4n-n^{3})\big{)}t}a_{n}(0)\mbox{ for
}t\in\mathbb{R}_{+},n\in\mathbb{Z}.$

Since $\int_{0}^{L}y_{0}\,dx=0$, it follows that $a_{0}=0$. This in turn
implies that $a_{0}(t)=0$ for $t\in\mathbb{R}_{+}$. We hence obtain

$y(t,x)=\sum_{n\in\mathbb{Z}\setminus\\{0\\}}e^{inx}e^{\big{(}-3n^{2}-i(4n-n^{3})\big{)}t}a_{n}(0)\mbox{
in }\mathbb{R}_{+}\times(0,2\pi).$

Extend $y(t,x)$ for $t<0$ by

(2.12)
$y(t,x)=\sum_{n\in\mathbb{Z}\setminus\\{0\\}}e^{inx}e^{\big{(}3n^{2}-i(4n-n^{3})\big{)}t}a_{n}(0)\mbox{
in }\mathbb{R}_{-}\times(0,2\pi),$

and still denote this extension by $y(t,x)$. We have, for $z\in\mathbb{R}$ and
$n\in\mathbb{Z}\setminus\\{0\\}$,

(2.13)
$\int_{0}^{\infty}e^{\big{(}-3n^{2}-i(4n-n^{3})\big{)}t}e^{-itz}\,dt+\int_{-\infty}^{0}e^{\big{(}3n^{2}-i(4n-n^{3})\big{)}t}e^{-itz}\,dt\\\\[6.0pt]
=\frac{1}{3n^{2}+i(4n-n^{3}+z)}+\frac{1}{3n^{2}-i(4n-n^{3}+z)}=\frac{6n^{2}}{9n^{4}+(4n-n^{3}+z)^{2}}.$

This implies

(2.14) $\|y(\cdot,x)\|_{H^{1/3}(\mathbb{R}_{+})}^{2}\leq
C\int_{z\in\mathbb{R}}(1+|z|)^{2/3}\left(\sum_{n\in\mathbb{Z}\setminus\\{0\\}}\frac{n^{2}|a_{n}(0)|}{n^{4}+(z+4n-n^{3})^{2}}\right)^{2}\,dz.$

Applying Lemma 2.1 with $s=1$, we obtain

(2.15) $\|y(\cdot,x)\|_{H^{1/3}(\mathbb{R}_{+})}^{2}\leq
C\|y_{0}\|_{L^{2}(0,2\pi)}^{2}.$

Similarly, we have

(2.16) $\|y_{x}(\cdot,x)\|_{L^{2}(\mathbb{R}_{+})}^{2}\leq
C\int_{z\in\mathbb{R}}\left(\sum_{n\in\mathbb{Z}\setminus\\{0\\}}\frac{|n|^{3}|a_{n}(0)|}{(z+4n-n^{3})^{2}+n^{4}}\right)^{2}$

and

(2.17) $\|y_{xx}(\cdot,x)\|_{[H^{1/3}(\mathbb{R}_{+})]^{*}}^{2}\leq
C\int_{z\in\mathbb{R}}(1+|z|)^{-2/3}\left(\sum_{n\in\mathbb{Z}\setminus\\{0\\}}\frac{n^{4}|a_{n}(0)|}{(z+4n-n^{3})^{2}+n^{4}}\right)^{2}.$

Applying Lemma 2.1 with $s=0$ and $s=-1$, we get

(2.18) $\|y_{x}(\cdot,x)\|_{L^{2}(\mathbb{R}_{+})}^{2}\leq
C\|y_{0}\|_{L^{2}(0,2\pi)}^{2}.$

and

(2.19) $\|y_{xx}(\cdot,x)\|_{[H^{1/3}(\mathbb{R}_{+})]^{*}}^{2}\leq
C\|y_{0}\|_{L^{2}(0,2\pi)}^{2}.$

The proof of (2.2) in the case $g\equiv 0$ is complete.

We next deal with $\eqref{lem-kdvB-cl4}$. From (2.12) and (2.13), we have

$\|y(\cdot,x)\|_{L^{2}(\mathbb{R}_{+})}^{2}\leq
C\int_{z\in\mathbb{R}}\left(\sum_{n\in\mathbb{Z}\setminus\\{0\\}}\frac{|n|^{3}(|a_{n}(0)|/|n|))}{n^{4}+(z+4n-n^{3})^{2}}\right)^{2}\,dz$

and

(2.20) $\|y_{x}(\cdot,x)\|_{[H^{1/3}(\mathbb{R}_{+})]^{*}}^{2}\leq
C\int_{z\in\mathbb{R}}(1+|z|)^{-2/3}\left(\sum_{n\in\mathbb{Z}\setminus\\{0\\}}\frac{|n|^{4}(|a_{n}(0)|/|n|)}{(z+4n-n^{3})^{2}+n^{4}}\right)^{2}.$

Applying Lemma 2.1 with $s=0$ and $s=-1$, we obtain

(2.21) $\|y(\cdot,x)\|_{L^{2}(\mathbb{R}_{+})}^{2}\leq
C\sum_{n\in\mathbb{Z}\setminus\\{0\\}}\frac{|a_{n}(0)|^{2}}{n^{2}}\leq
C\|y_{0}\|_{H^{-1}(0,2\pi)}^{2}$

and

(2.22) $\|y_{x}(\cdot,x)\|_{[H^{1/3}(\mathbb{R}_{+})]^{*}}^{2}\leq
C\sum_{n\in\mathbb{Z}\setminus\\{0\\}}\frac{|a_{n}(0)|^{2}}{n^{2}}\leq
C\|y_{0}\|_{H^{-1}(0,2\pi)}^{2}.$

This yields (2.3).

We next deal with the case $y_{0}\equiv 0$. The proof, in this case, can be
derived from the previous case as follows. For $t>0$, let $W(t)$ be the
mapping which maps $y_{0}$ to $y(t,\cdot)$ with $g\equiv 0$. We then have
555This identity can be derived from (2.28) and (2.29) in Remark 2.3.

(2.23) $y(t,x)=\int_{0}^{t}W(t-s)g(s,x)\,ds.$

This implies

(2.24)
$y_{x}(t,x)=\int_{0}^{t}\partial_{x}\Big{(}W(t-s)g(s,x)\Big{)}\,ds=\int_{0}^{\infty}\mathds{1}_{(0,t)}(s)\partial_{x}\Big{(}W(t-s)g(s,x)\Big{)}\,ds.$

Hence

$\displaystyle\|y_{x}(\cdot,x)\|_{L^{2}_{t}(\mathbb{R}_{+})}$
$\displaystyle\leq\int_{0}^{+\infty}\|\mathds{1}_{(0,t)}(s)\partial_{x}\big{(}W(t-s)g(s,x)\big{)}\|_{L^{2}_{t}(\mathbb{R}_{+})}\,ds$
$\displaystyle=\int_{0}^{+\infty}\left(\int_{0}^{\infty}|\mathds{1}_{(0,t)}(s)\partial_{x}\big{(}W(t-s)g(s,x)\big{)}|^{2}\,dt\right)^{1/2}\,ds$
$\displaystyle=\int_{0}^{+\infty}\left(\int_{s}^{\infty}|\partial_{x}\big{(}W(t-s)g(s,x)\big{)}|^{2}\,dt\right)^{1/2}\,ds.$

By applying the results in the previous case, we have

$\left(\int_{s}^{\infty}|\partial_{x}\big{(}W(t-s)g(s,x)\big{)}|^{2}\,dt\right)^{1/2}\leq
C\|g(s,x)\|_{L^{2}_{x}(0,L)}.$

We thus obtain

(2.25) $\|y_{x}(\cdot,x)\|_{L^{2}_{t}(\mathbb{R}_{+})}\leq
C\int_{0}^{+\infty}\|g(s,x)\|_{L^{2}_{x}}\,ds.$

By the same arguments, we have

(2.26)
$\|y_{xx}(\cdot,x)\|_{[H^{1/3}_{t}(\mathbb{R}_{+})]^{*}}\leq\int_{0}^{+\infty}\|\mathds{1}_{(0,t)}(s)\partial_{xx}\big{(}W(t-s)g(s,x)\big{)}\|_{[H^{1/3}_{t}(\mathbb{R}_{+})]^{*}}\,ds\\\\[6.0pt]
=\int_{0}^{+\infty}\|\partial_{xx}\big{(}W(t-s)g(s,x)\big{)}\|_{[H^{1/3}_{t}(s,+\infty)]^{*}}\,ds\leq
C\int_{0}^{+\infty}\|g(s,x)\|_{L^{2}_{x}(0,L)}\,ds.$

Similarly, we obtain

(2.27)
$\|y(\cdot,x)\|_{H^{1/3}_{t}(\mathbb{R}_{+})}\leq\int_{0}^{+\infty}\|\mathds{1}_{(0,t)}(s)W(t-s)g(s,x)\|_{H^{1/3}_{t}(\mathbb{R}_{+})}\,ds\\\\[6.0pt]
=\int_{0}^{+\infty}\|W(t-s)g(s,x)|\|_{H^{1/3}_{t}(s,+\infty)}\,ds\leq
C\int_{0}^{+\infty}\|g(s,x)\|_{L^{2}_{x}(0,L)}\,ds.$

Assertion (2.2) in the case $y_{0}\equiv 0$ now follows from (2.25), (2.26),
and (2.27).

Assertion (2.3) in the case $y_{0}\equiv 0$ follows similarly and the details
are omitted.

The proof is complete. ∎

######  Remark 2.1.

The proof gives as well that

$y\in C([0,L];H^{1/3}(0,+\infty)),\;\;y_{x}\in
C([0,L];L^{2}(0,+\infty)),\;\;\mbox{ and }\;\;y_{xx}\in
C([0,L];[H^{1/3}(0,+\infty)]^{*}).$

######  Remark 2.2.

Assume that $g=0$ in $\mathbb{R}_{+}\times(0,L)$ and $y_{0}\in
C^{\infty}([0,L])$ is such that $\int_{0}^{L}y_{0}(x)\,dx=0$, $y_{0}-c\in
C^{\infty}_{c}((0,L))$ for some constant $c$. Using (2.12), one can show that
$y\in C^{\infty}([0,+\infty)\times[0,L])$. Moreover, using the equation of
$y$, one can show that $\partial_{t}^{k}y(0,x)=0$ for all $k\geq 1$.

######  Remark 2.3.

Assume that $y_{0}=0$ in $(0,L)$ and $g\in
C^{\infty}_{c}\big{(}(0,+\infty)\times[0,L]\big{)}$ being such that
$\int_{0}^{L}g(t,x)\,dx=0$ for $t>0$. One can prove that the solution is then
smooth and is 0 around the time $0$. Indeed, for notational ease, assume that
$L=2\pi$. One then can show that

(2.28) $y(t,x)=\sum_{n\in\mathbb{Z}\setminus
0}e^{inx}\int_{0}^{t}e^{\big{(}-3n^{2}-i(4n-n^{3})\big{)}(t-s)}g_{n}(s)\,ds,$

where

(2.29) $g(t,x)=\sum_{n\in\mathbb{Z}\setminus\\{0\\}}g_{n}(t)e^{inx}.$

The properties of $y$ follow.

######  Remark 2.4.

There is no term $\|y_{xx}\|_{[H^{2/3}(\mathbb{R}_{+})]^{*}}$ in (2.3) since
we applied Lemma 2.1, which requires $s>-2$.

######  Remark 2.5.

In [20, Lemma 4.1], we proved that, when $y_{0}=0$,

(2.30) $\|y(\cdot,x)\|_{L^{2}(\mathbb{R}_{+})}\leq
C\|g\|_{L^{1}\big{(}\mathbb{R}_{+}\times(0,L)\big{)}}.$

Estimate (2.3) improves this.

##  3\. Linearized KdV equations

In this section, we establish several well-posedness results (Proposition 3.1,
Proposition 3.2, and Proposition 3.3) for the linearized KdV equation equipped
with various boundary conditions in the energy space $X_{T}$ defined in (1.14)
and present some of their consequences. Proposition 3.1 implies the local
well-posedness of (1.1) in $X_{T}$ (Proposition 3.4). Proposition 3.1 and
Proposition 3.2 are the starting points of our analysis in deriving the
unreachable space and establishing the corresponding observability inequality
in Section 5 (Proposition 5.1). Proposition 3.3 and its consequence
(Proposition 3.5) will be used in the proof of Theorem 1.3.

Here is the first main result of this section.

######  Proposition 3.1.

Let $L>0$, $T>0$, $(h_{1},h_{2},h_{3})\in H^{1/3}(0,T)\times
H^{1/3}(0,T)\times L^{2}(0,T)$, $f\in L^{1}\big{(}(0,T);L^{2}(0,L)\big{)}$,
and $y_{0}\in L^{2}(0,L)$. There exists a unique solution $y\in X_{T}$ of the
system

(3.1) $\left\\{\begin{array}[]{cl}y_{t}+y_{x}+y_{xxx}=f&\mbox{ in
}(0,T)\times(0,L),\\\\[6.0pt]
y(\cdot,0)=h_{1},\;y(\cdot,L)=h_{2},\;y_{x}(\cdot,L)=h_{3}&\mbox{ in
}(0,T),\\\\[6.0pt] y(0,\cdot)=y_{0}&\mbox{ in }(0,L).\end{array}\right.$

Moreover, for $x\in[0,L]$,

(3.2)
$\|y\|_{X_{T}}+\|y(\cdot,x)\|_{H^{1/3}(0,T)}+\|y_{x}(\cdot,x)\|_{L^{2}(0,T)}+\|y_{xx}(\cdot,x)\|_{[H^{1/3}(0,T)]^{*}}\\\\[6.0pt]
\leq
C_{T,L}\Big{(}\|y_{0}\|_{L^{2}(0,L)}+\|f\|_{L^{1}\big{(}(0,T);L^{2}(0,L)\big{)}}+\|(h_{1},h_{2})\|_{H^{1/3}(0,T)}+\|h_{3}\|_{L^{2}(0,T)}\Big{)},$

and

(3.3)
$\|y(\cdot,x)\|_{L^{2}(0,T)}+\|y_{x}(\cdot,x)\|_{[H^{1/3}(0,T)]^{*}}\\\\[6.0pt]
\leq
C_{T,L}\Big{(}\|y_{0}\|_{[H^{1}(0,L)]^{*}}+\|f\|_{L^{1}\big{(}(0,T);[H^{1}(0,L)]^{*}\big{)}}+\|(h_{1},h_{2})\|_{L^{2}(0,T)}+\|h_{3}\|_{[H^{1/3}(0,T)]^{*}}\Big{)},$

where $C_{T,L}$ denotes a positive constant independent of $x$, $y_{0}$, $f$,
and $h_{1},\,h_{2},\,h_{3}$.

Here and in what follows, $[H^{s}]^{*}$ denotes the dual space of $H^{s}$ for
$s>0$ and it is equipped with the standard corresponding norm.

######  Remark 3.1.

In Proposition 3.1, we implicitly admit that

$y\in C([0,L];H^{1/3}(0,T)),\quad y_{x}\in C([0,L];L^{2}(0,T)),\quad\mbox{ and
}\quad y_{xx}\in C([0,L];[H^{1/3}(0,T)]^{*}).$

These facts are derived from the proof. Results and analysis which are related
to Proposition 3.1 will be discussed in Remark 3.6.

The next result is on the well-posedness of the linearized KdV system for
which the Dirichlet condition and the second derivative in $x$ on the right
are described.

######  Proposition 3.2.

Let $L>0$, $T>0$, $(h_{1},h_{2},h_{3})\in H^{1/3}(0,T)\times
H^{1/3}(0,T)\times[H^{1/3}(0,T)]^{*}$,
$f\in L^{1}\big{(}(0,T);L^{2}(0,L)\big{)}$, and $y_{0}\in L^{2}(0,L)$. There
exists a unique solution $y\in X_{T}$ of the system

(3.4) $\left\\{\begin{array}[]{cl}y_{t}+y_{x}+y_{xxx}=f&\mbox{ in
}(0,T)\times(0,L),\\\\[6.0pt]
y(\cdot,0)=h_{1},\;y(\cdot,L)=h_{2},\;y_{xx}(\cdot,L)=h_{3}&\mbox{ in
}(0,T),\\\\[6.0pt] y(0,\cdot)=y_{0}&\mbox{ in }(0,L).\end{array}\right.$

Moreover, for $0\leq x\leq L$,

(3.5)
$\|y\|_{X_{T}}+\|y(\cdot,x)\|_{H^{1/3}(0,T)}+\|y_{x}(\cdot,x)\|_{L^{2}(0,T)}+\|y_{xx}(\cdot,x)\|_{[H^{1/3}(0,T)]^{*}}\\\\[6.0pt]
\leq
C_{T,L}\Big{(}\|y_{0}\|_{L^{2}(0,T)}+\|f\|_{L^{1}\big{(}(0,T);L^{2}(0,L)\big{)}}+\|(h_{1},h_{2})\|_{H^{1/3}(0,T)}+\|h_{3}\|_{[H^{1/3}(0,T)]^{*}}\Big{)},$

and

(3.6)
$\|y(\cdot,x)\|_{L^{2}(0,T)}+\|y_{x}(\cdot,x)\|_{[H^{1/3}(0,T)]^{*}}\\\\[6.0pt]
\leq
C_{T,L}\Big{(}\|y_{0}\|_{[H^{1}(0,L)]^{*}}+\|f\|_{L^{1}\big{(}(0,T);[H^{1}(0,L)]^{*}\big{)}}+\|(h_{1},h_{2})\|_{L^{2}(0,T)}+\|h_{3}\|_{[H^{2/3}(0,T)]^{*}}\Big{)},$

for some positive constant $C_{T,L}$ independent of $x$, $y_{0}$, $f$, and
$(h_{1},h_{2},h_{3})$.

Here is the third main result of this section.

######  Proposition 3.3.

Let $L>0$, $T>0$, $(h_{1},h_{2},h_{3})\in L^{2}(0,T)\times H^{1/3}(0,T)\times
L^{2}(0,T)$, $f\in L^{1}\big{(}(0,T);L^{2}(0,L)\big{)}$, and $y_{0}\in
L^{2}(0,L)$. There exists a unique solution $y\in X_{T}$ of the system

(3.7) $\left\\{\begin{array}[]{cl}y_{t}+y_{x}+y_{xxx}=f&\mbox{ in
}(0,T)\times(0,L),\\\\[6.0pt]
y_{x}(\cdot,0)=h_{1},\;y(\cdot,L)=h_{2},\;y_{x}(\cdot,L)=h_{3}&\mbox{ in
}(0,T),\\\\[6.0pt] y(0,\cdot)=y_{0}&\mbox{ in }(0,L).\end{array}\right.$

Moreover, for $0\leq x\leq L$,

(3.8)
$\|y\|_{X_{T}}+\|y(\cdot,x)\|_{H^{1/3}(0,T)}+\|y_{x}(\cdot,x)\|_{L^{2}(0,T)}+\|y_{xx}(\cdot,x)\|_{[H^{1/3}(0,T)]^{*}}\\\\[6.0pt]
\leq
C_{T,L}\Big{(}\|y_{0}\|_{L^{2}(0,T)}+\|f\|_{L^{1}\big{(}(0,T);L^{2}(0,L)\big{)}}+\|h_{2}\|_{H^{1/3}(0,T)}+\|(h_{1},h_{3})\|_{L^{2}(0,T)}\Big{)}$

and

(3.9)
$\|y(\cdot,x)\|_{L^{2}(0,T)}+\|y_{x}(\cdot,x)\|_{[H^{1/3}(0,T)]^{*}}\\\\[6.0pt]
\leq
C_{T,L}\Big{(}\|y_{0}\|_{[H^{1}(0,L)]^{*}}+\|f\|_{L^{1}\big{(}(0,T);[H^{1}(0,L)]^{*}\big{)}}+\|(h_{1},h_{2})\|_{L^{2}(0,T)}+\|h_{3}\|_{[H^{1/3}(0,T)]^{*}}\Big{)},$

for some positive constant $C_{T,L}$ independent of $x$, $y_{0}$, $f$, and
$(h_{1},h_{2},h_{3})$.

As a consequence of Proposition 3.1, we obtain the following well-posedness
result, which particularly yields the well-posedness for system (1.1).

######  Proposition 3.4.

Let $L>0$ and $T>0$. There exists $\varepsilon_{0}>0$ such that for
$(h_{1},h_{2},h_{3})\in H^{1/3}(0,T)\times H^{1/3}(0,T)\times L^{2}(0,T)$,
$f\in L^{1}\big{(}(0,T);L^{2}(0,L)\big{)}$, and $y_{0}\in L^{2}(0,L)$ with

$\|y_{0}\|_{L^{2}(0,L)}+\|f\|_{L^{1}\big{(}(0,T);L^{2}(0,L)\big{)}}+\|(h_{1},h_{2})\|_{H^{1/3}(0,T)}+\|h_{3}\|_{L^{2}(0,T)}\leq\varepsilon_{0},$

there exists a unique solution $y\in X_{T}$ of the system

(3.10) $\left\\{\begin{array}[]{cl}y_{t}+y_{x}+y_{xxx}+yy_{x}=f&\mbox{ in
}(0,T)\times(0,L),\\\\[6.0pt]
y(\cdot,0)=h_{1},\;y(\cdot,L)=h_{2},\;y_{x}(\cdot,L)=h_{3}&\mbox{ in
}(0,T),\\\\[6.0pt] y(0,\cdot)=y_{0}&\mbox{ in }(0,L).\end{array}\right.$

Moreover, we have

(3.11)
$\|y\|_{X_{T}}+\|y(\cdot,x)\|_{H^{1/3}(0,T)}+\|y_{x}(\cdot,x)\|_{L^{2}(0,T)}+\|y_{xx}(\cdot,x)\|_{[H^{1/3}(0,T)]^{*}}\\\\[6.0pt]
\leq
C_{T,L}\Big{(}\|y_{0}\|_{L^{2}(0,L)}+\|f\|_{L^{1}\big{(}(0,T);L^{2}(0,L)\big{)}}+\|(h_{1},h_{2})\|_{H^{1/3}(0,T)}+\|h_{3}\|_{L^{2}(0,T)}\Big{)},$

for some positive constant $C_{T,L}$ independent of $x$, $y_{0}$, $f$, and
$(h_{1},h_{2},h_{3})$.

As a consequence of Proposition 3.3, we obtain the following well-posedness
result which will be used in the proof of Assertion $ii)$ of Theorem 1.3.

######  Proposition 3.5.

Let $L>0$ and $T>0$. There exists $\varepsilon_{0}>0$ such that for
$(h_{1},h_{2},h_{3})\in L^{2}(0,T)\times H^{1/3}(0,T)\times L^{2}(0,T)$, $f\in
L^{1}\big{(}(0,T);L^{2}(0,L)\big{)}$, and $y_{0}\in L^{2}(0,L)$ with

$\|y_{0}\|_{L^{2}(0,L)}+\|f\|_{L^{1}\big{(}(0,T);L^{2}(0,L)\big{)}}+\|h_{2}\|_{H^{1/3}(0,T)}+\|(h_{1},h_{3})\|_{L^{2}(0,T)}\leq\varepsilon_{0},$

there exists a unique solution $y\in X_{T}$ of the system

(3.12) $\left\\{\begin{array}[]{cl}y_{t}+y_{x}+y_{xxx}+yy_{x}=f&\mbox{ in
}(0,T)\times(0,L),\\\\[6.0pt]
y_{x}(\cdot,0)=h_{1},\;y(\cdot,L)=h_{2},\;y_{x}(\cdot,L)=h_{3},&\mbox{ in
}(0,T),\\\\[6.0pt] y(0,\cdot)=y_{0}&\mbox{ in }(0,L).\end{array}\right.$

Moreover, for $0\leq x\leq L$,

(3.13)
$\|y\|_{X_{T}}+\|y(\cdot,x)\|_{H^{1/3}(0,T)}+\|y_{x}(\cdot,x)\|_{L^{2}(0,T)}+\|y_{xx}(\cdot,x)\|_{[H^{1/3}(0,T)]^{*}}\\\\[6.0pt]
\leq
C_{T,L}\Big{(}\|y_{0}\|_{L^{2}(0,L)}+\|f\|_{L^{1}\big{(}(0,T);L^{2}(0,L)\big{)}}+\|h_{2}\|_{H^{1/3}(0,T)}+\|(h_{1},h_{3})\|_{L^{2}(0,T)}\Big{)},$

for some positive constant $C_{T,L}$ independent of $x$, $y_{0}$, $f$, and
$u$.

######  Remark 3.2.

The meaning of the solutions considered in Proposition 3.1, Proposition 3.2,
and Proposition 3.3 are given in Definition 3.1, Definition 3.2, Definition
3.3, respectively. These are motivated by the integration by parts arguments.
The meaning of the solutions in Proposition 3.4 and Proposition 3.5 are
understood in the same manner where the nonlinear term $yy_{x}$ plays as a
part of the source $f$.

Proposition 3.4 and Proposition 3.5 follows from Proposition 3.1 and
Proposition 3.3 by standard arguments. We just mention here that

$\|yy_{x}\|_{L^{1}\big{(}(0,T);L^{2}(0,L)\big{)}}\leq\int_{0}^{T}\sup_{x\in[0,L]}|y(t,x)|\left(\int_{0}^{L}|y_{x}(t,x)|^{2}\right)^{\frac{1}{2}}\\\\[6.0pt]
\leq
C\int_{0}^{T}\left(\int_{0}^{L}|y(t,x)|^{2}+|y_{x}(t,x)|^{2}\right)^{\frac{1}{2}}\left(\int_{0}^{L}|y_{x}(t,x)|^{2}\right)^{\frac{1}{2}}\leq
C\|y\|_{X_{T}}^{2}.$

The details of the proof are left to the reader. A sharper estimate can be
derived from Lemma 8.1

The rest of this section is organized as follows. The proof of Proposition
3.1, Proposition 3.2, and Proposition 3.3 are given in Section 3.1, Section
3.2, and Section 3.3, respectively.

###  3.1. Proof of Proposition 3.1

We first give the meaning of the solutions considered in Proposition 3.1.

######  Definition 3.1.

Let $L>0$, $T>0$, $(h_{1},h_{2},h_{3})\in H^{1/3}(0,T)\times
H^{1/3}(0,T)\times L^{2}(0,T)$, $f\in L^{1}\big{(}(0,T);L^{2}(0,L)\big{)}$,
and $y_{0}\in L^{2}(0,L)$. A solution $y\in X_{T}$ of the system (3.1) is a
function $y\in X_{T}$ such that

(3.14)
$\int_{0}^{T}\int_{0}^{L}f(t,x)\varphi(t,x)\,dx\,dt+\int_{0}^{L}y_{0}(x)\varphi(0,x)\,dx+\int_{0}^{T}h_{3}(t)\varphi_{x}(t,L)\,dt\\\\[6.0pt]
-\int_{0}^{T}h_{2}(t)\varphi_{xx}(t,L)\,dt+\int_{0}^{T}h_{1}(t)\varphi_{xx}(t,0)\,dt=-\int_{0}^{T}\int_{0}^{L}y(\varphi_{t}+\varphi_{x}+\varphi_{xxx})\,dx\,dt$

for all $\varphi\in C^{3}([0,T]\times[0,L])$ with $\varphi(T,\cdot)=0$ and
$\varphi(\cdot,0)=\varphi(\cdot,L)=\varphi_{x}(\cdot,0)=0$.

######  Remark 3.3.

One can check that if $y$ is a smooth solution of (3.1) then $y$ is a solution
of (3.1) in the sense of Definition 3.1 by standard integration by parts
arguments. In fact, Definition 3.1 is motivated by such arguments.

The proof of Proposition 3.1 is divided into two parts given in the following
two subsections. The first part is on the existence and the estimates for a
constructed solution. The second part is on the uniqueness.

####  3.1.1. Existence and estimates

The proof of (3.2) for a solution constructed below is given in two steps.

Step 1: We first consider the case where $y_{0}=0$ and $f=0$. By the
linearity, it suffices to consider the three cases
$(h_{1},h_{2},h_{3})=(0,0,h_{3})$, $(h_{1},h_{2},h_{3})=(h_{1},0,0)$, and
$(h_{1},h_{2},h_{3})=(0,h_{2},0)$ separately. In what follows, we extend
$h_{1},h_{2},h_{3}$ by $0$ for $t>T$ and still denote $y$ the corresponding
solution and these extensions by $h_{1},h_{2},h_{3}$.

In what follows in this proof, for an appropriate function $v$ defined on
$\mathbb{R}\times(0,L)$, we denote by $\hat{v}$ its Fourier transform with
respect to $t$, i.e., for $z\in\mathbb{C}$,

$\hat{v}(z,x)=\frac{1}{\sqrt{2\pi}}\int_{0}^{+\infty}v(t,x)e^{-izt}\mathop{}\mkern-3.0mu{d}t.$

Extend $y$ and $h_{1},h_{2},h_{3}$ by $0$ for $t<0$ and still denote these
extensions by $y$, and $h_{1},h_{2},h_{3}$. Then

(3.15) $y_{t}+y_{x}+y_{xxx}=0\mbox{ in }\mathbb{R}\times(0,L).$

Taking the Fourier transform with respect to $t$, we obtain, for
$z\in\mathbb{R}$,

(3.16) $iz\hat{y}+\hat{y}_{x}+\hat{y}_{xxx}=0\mbox{ in }(0,L).$

For $z\in\mathbb{C}$, let $\lambda_{j}=\lambda_{j}(z)$ with $j=1,2,3$ be the
three solutions of the equation $\lambda^{3}+\lambda+iz=0$. Set

$Q(z):=\begin{pmatrix}1&1&1\\\
e^{\lambda_{1}L}&e^{\lambda_{2}L}&e^{\lambda_{3}L}\\\
\lambda_{1}e^{\lambda_{1}L}&\lambda_{2}e^{\lambda_{2}L}&\lambda_{3}e^{\lambda_{3}L}\end{pmatrix},$

and

$\Xi=\Xi(z):=\det\begin{pmatrix}1&1&1\\\
\lambda_{1}&\lambda_{2}&\lambda_{3}\\\
\lambda_{1}^{2}&\lambda_{2}^{2}&\lambda_{3}^{2}\end{pmatrix},$

with the convention $\lambda_{j+3}=\lambda_{j}$ for $j\geq 1$. It is useful to
note that

(3.17) $H(z):=\det Q(z)\Xi(z)\mbox{ in }\mathbb{C}.$

is an analytic function, see, e.g., [20, Lemma A1]. Moreover, $H$ only has a
finite number of zeros on the real line since $H(z)\neq 0$ for
$z\in\mathbb{R}$ with large $|z|$.

We first consider the case $(h_{1},h_{2},h_{3})=(0,0,h_{3})$. Taking into
account the equation of $\hat{y}$, we search for the solution of the form

(3.18) $\hat{y}(z,\cdot)=\sum_{j=1}^{3}a_{j}e^{\lambda_{j}x},$

where $a_{j}=a_{j}(z)$ for $j=1,2,3$. Taking the boundary condition, we then
have

$\begin{array}[]{c}\sum_{j=1}^{3}a_{j}=0,\\\\[6.0pt]
\sum_{j=1}^{3}a_{j}e^{\lambda_{j}L}=0,\\\\[6.0pt]
\sum_{j=1}^{3}a_{j}\lambda_{j}e^{\lambda_{j}L}=\hat{h}_{3},\\\\[6.0pt]
\end{array}$

This implies, with the convention $\lambda_{j+3}=\lambda_{j}$,

(3.19) $a_{j}=\frac{e^{\lambda_{j+2}L}-e^{\lambda_{j+1}L}}{\det
Q}\hat{h}_{3}\mbox{ for }j=1,2,3.$

To estimate the solution, we proceed as follows. By Lemma C.1 in the appendix,
there exists $g_{3}\in C^{\infty}(\mathbb{R})$ with
$\operatorname{supp}g_{3}\subset[T,3T]$ such that if $z$ is a real solution of
the equation $H(z)=0$ of order $m$ then $z$ is also a real solution of order
$m$ of $\hat{h}_{3}(z)-\hat{g}_{3}(z)$, and, for $k\geq 1$,

(3.20) $\|g_{3}\|_{H^{k}(\mathbb{R})}\leq
C_{k}\|h_{3}\|_{H^{-1/3}(\mathbb{R})}.$

We now establish (3.2). Let $y_{3}$ be the solution of (3.1) where
$(h_{1},h_{2},h_{3})$ are $(0,0,h_{3}-g_{3})$, and $f=0$, and $y_{0}=0$. We
have, by applying (3.19) to $y_{3}$,

(3.21) $\hat{y}_{3}(z,x)=\frac{\hat{h}_{3}(z)-\hat{g}_{3}(z)}{\det
Q(z)}\sum_{j=1}^{3}\big{(}e^{\lambda_{j+2}L}-e^{\lambda_{j+1}L}\big{)}e^{\lambda_{j}x}\mbox{
for a.e. }x\in(0,L).$

We derive that, by the choice of $g_{3}$, for $z\in\mathbb{R}$ and
$|z|\leq\gamma$,

(3.22)
$\left|\frac{\hat{h}_{3}(z)-\hat{g}_{3}(z)}{H(z)}\right|\left|\Xi(z)\sum_{j=1}^{3}\big{(}e^{\lambda_{j+2}L}-e^{\lambda_{j+1}L}\big{)}e^{\lambda_{j}x}\right|\leq
C_{T,\gamma}\|h_{3}-g_{3}\|_{H^{-1/3}(\mathbb{R})},$

and, by Lemma 3.1 below on the behaviors of $\lambda_{j}(z)$ for large $z$
(the case large negative $z$ can be obtained by considering the conjugate), we
obtain, for $z\in\mathbb{R}$, $|z|\geq\gamma$ with sufficiently large
$\gamma$,

(3.23) $\left|\frac{1}{\det
Q}\sum_{j=1}^{3}\big{(}e^{\lambda_{j+2}L}-e^{\lambda_{j+1}L}\big{)}e^{\lambda_{j}x}\right|\leq\frac{C}{(1+|z|)^{1/3}}.$

Combining (3.22) and (3.23) yields

$\|y_{3}(\cdot,x)\|_{H^{1/3}(\mathbb{R})}\leq
C_{T,L}\|h_{3}-g_{3}\|_{L^{2}(\mathbb{R})}$

and

$\|y_{3}(\cdot,x)\|_{L^{2}(\mathbb{R})}\leq
C_{T,L}\|h_{3}-g_{3}\|_{H^{-1/3}(\mathbb{R})}.$

Similarly, we have

$\|y_{3,x}(\cdot,x)\|_{L^{2}(\mathbb{R})}+\|y_{3,xx}(\cdot,x)\|_{H^{-1/3}(\mathbb{R})}\leq
C_{T,L}\|h_{3}-g_{3}\|_{L^{2}(\mathbb{R})}$

and

$\|y_{3,x}(\cdot,x)\|_{H^{-2/3}(\mathbb{R})}\leq
C_{T,L}\|h_{3}-g_{3}\|_{H^{-1/3}(\mathbb{R})}.$

The estimates for $\|y(\cdot,x)\|_{H^{1/3}(0,T)}$,
$\|y_{x}(\cdot,x)\|_{L^{2}(0,T)}$, and
$\|y_{xx}(\cdot,x)\|_{[H^{1/3}(0,T)]^{*}}$ and the estimate for
$\|y(\cdot,x)\|_{L^{2}(0,T)}$, $\|y_{x}(\cdot,x)\|_{[H^{1/3}(0,T)]^{*}}$
follow by noting that $y=y_{3}$ in $(0,T)\times(0,L)$ 666More precisely, one
can take $y=y_{3}$ in $(0,T)\times(0,L)$..

We also have, by integration by parts, for $0\leq\tau_{1}<\tau_{2}\leq T$,

(3.24)
$\frac{1}{2}\int_{0}^{L}|y(\tau_{2},x)|^{2}\,dx-\frac{1}{2}\int_{0}^{L}|y(\tau_{1},x)|^{2}\,dx+\frac{1}{2}\int_{\tau_{1}}^{\tau_{2}}\Big{(}|y(t,L)|^{2}-|y(t,0)|^{2}\Big{)}\,dt\\\\[6.0pt]
+\int_{\tau_{1}}^{\tau_{2}}\Big{(}y_{xx}(t,L)y(t,L)-y_{xx}(t,0)y(t,0)\Big{)}\,dt-\frac{1}{2}\int_{\tau_{1}}^{\tau_{2}}\Big{(}|y_{x}(t,L)|^{2}-|y_{x}(t,0)|^{2}\Big{)}\,dt=0.$

Using the estimates for $\|y(\cdot,x)\|_{H^{1/3}(0,T)}$,
$\|y_{x}(\cdot,x)\|_{L^{2}(0,T)}$, and
$\|y_{xx}(\cdot,x)\|_{[H^{1/3}(0,T)]^{*}}$, we obtain the one for
$\|y\|_{X_{T}}$.

The proof in the case $(h_{1},h_{2},h_{3})=(h_{1},0,0)$ or in the case
$(h_{1},h_{2},h_{3})=(0,h_{2},0)$ is similar after noting Lemma A.1 in the
appendix. We mention here that the solution corresponding to the triple
$(h_{1},0,0)$ is given by

$\hat{y}(z,x)=\frac{\hat{h}_{1}(z)}{\det
Q}\sum_{j=1}^{3}(\lambda_{j+2}e^{-\lambda_{j}L}-\lambda_{j+1}e^{-\lambda_{j}L})e^{\lambda_{j}x}\mbox{
for a.e. }x\in(0,L),$

and the solution corresponding to the triple $(0,h_{2},0)$ is given by

$\hat{y}(z,x)=\frac{\hat{h}_{2}(z)}{\det
Q}\sum_{j=1}^{3}(\lambda_{j+1}e^{\lambda_{j+1}L}-\lambda_{j+2}e^{\lambda_{j+2}L})e^{\lambda_{j}x}\mbox{
for a.e. }x\in(0,L).$

The details are left to the reader.

######  Remark 3.4.

If $h_{1},h_{2},h_{3}\in C^{\infty}_{c}(\mathbb{R})$, the constructed solution
is also smooth.

Step 2: We now deal with the general case. The starting point of the proof is
a connection between the linearized KdV equation and the linear KdV-Burgers
equation. Set $v(t,x)=e^{-2t+x}y(t,x)$, which is equivalent to
$y(t,x)=e^{2t-x}v(t,x)$. One can check that if $y$ satisfies the equation

$y_{t}+y_{x}+y_{xxx}=f\mbox{ in }\mathbb{R}_{+}\times(0,L),$

if and only if

$v_{t}+4v_{x}+v_{xxx}-3v_{xx}=fe^{-2t+x}\mbox{ in }\mathbb{R}_{+}\times(0,L).$

Set, in $\mathbb{R}_{+}\times(0,L)$,

(3.25)
$\psi(t,x)=\psi(t):=\frac{1}{L}\int_{0}^{L}f(t,\xi)e^{-2t+\xi}\mathop{}\mkern-3.0mu{d}\xi\quad\mbox{
and }\quad g(t,x):=f(t,x)e^{-2t+x}-\psi(t,x).$

Then

$\int_{0}^{L}g(t,x)\mathop{}\mkern-3.0mu{d}x=0\mbox{ for }t>0.$

Let $y_{1}\in C\big{(}[0,+\infty);L^{2}(0,L)\big{)}\cap
L^{2}_{\operatorname{loc}}\big{(}[0,+\infty);H^{1}(0,L)\big{)}$ be the unique
solution periodic in space of the system

(3.26) $y_{1,t}+4y_{1,x}+y_{1,xxx}-3y_{1,xx}=g\mbox{ in
}(0,+\infty)\times(0,L),$

and

(3.27) $y_{1}(0,\cdot)=y_{0}e^{x}\mbox{ in }(0,L).$

Set

$\alpha=\frac{1}{L}\int_{0}^{L}y_{1}(0,x)\,dx.$

By Proposition 2.1, we have, for $x\in[0,L]$,

(3.28)
$\|y_{1}(\cdot,x)-\alpha\|_{H^{1/3}(\mathbb{R}_{+})}+\|y_{1,x}(\cdot,x)\|_{L^{2}(\mathbb{R}_{+})}+\|y_{1,xx}(\cdot,x)\|_{[H^{1/3}(\mathbb{R}_{+})]^{*}}\\\\[6.0pt]
\leq
C\Big{(}\|f\|_{L^{1}(\mathbb{R}_{+};L^{2}(0,L))}+\|y_{0}\|_{L^{2}(0,L)}\Big{)}$

and

(3.29)
$\|y_{1}(\cdot,x)-\alpha\|_{L^{2}(\mathbb{R}_{+})}+\|y_{1,x}(\cdot,x)\|_{[H^{1/3}(\mathbb{R}_{+})]^{*}}\leq
C\Big{(}\|y_{0}\|_{[H^{1}(0,L)]^{*}}+\|f\|_{L^{1}(\mathbb{R}_{+};[H^{1}(0,L)]^{*})}\Big{)}.$

Let $y_{2}\in X_{T}$ be the unique solution of

(3.30) $\left\\{\begin{array}[]{cl}y_{2,t}+y_{2,x}+y_{2,xxx}=0&\mbox{ in
}(0,T)\times(0,L),\\\\[6.0pt]
y_{2}(t,0)=h_{1}(t)-e^{2t}\Big{(}y_{1}(t,0)+\int_{0}^{t}\psi(s)\,ds\Big{)}&\mbox{
in }(0,T),\\\\[6.0pt]
y_{2}(t,L)=h_{2}(t)-e^{2t-L}\Big{(}y_{1}(t,L)+\int_{0}^{t}\psi(s)\,ds\Big{)}&\mbox{
in }(0,T),\\\\[6.0pt]
y_{2,x}(t,L)=h_{3}(t)-\big{(}e^{2t-\cdot}(y_{1}(t,\cdot)+\int_{0}^{t}\psi(s)\,ds)\big{)}_{x}(t,L)&\mbox{
in }(0,T),\\\\[6.0pt] y_{2}(t=0,\cdot)=0&\mbox{ in }(0,L).\end{array}\right.$

Applying the results of Step 1 to $y_{2}$, and using (3.28) and (3.28), we
derive that, for $x\in[0,L]$,

$\|y_{2}\|_{X_{T}}+\|y_{2}(\cdot,x)\|_{H^{1/3}(0,T)}+\|y_{2,x}(\cdot,x)\|_{L^{2}(0,T)}+\|y_{2,xx}(\cdot,x)\|_{[H^{1/3}(0,T)]^{*}}\\\\[6.0pt]
\leq
C_{T}\Big{(}\|(h_{1},h_{2})\|_{H^{1/3}(\mathbb{R}_{+})}+\|h_{3}\|_{L^{2}(\mathbb{R}_{+})}+\|f\|_{L^{1}([0,T];L^{2}(0,L))}\Big{)}$

and

$\|y_{2}(\cdot,x)\|_{L^{2}(0,T)}+\|y_{2,x}(\cdot,x)\|_{[H^{1/3}(0,T)]^{*}}\\\\[6.0pt]
\leq
C_{T}\Big{(}\|(h_{1},h_{2})\|_{L^{2}(\mathbb{R}_{+})}+\|h_{3}\|_{[H^{1/3}(\mathbb{R}_{+})]^{*}}+\|f\|_{L^{1}([0T];[H^{1}(0,L)]^{*})}\Big{)}.$

The conclusion follows by noting that
$y=e^{2t-x}\Big{(}y_{1}+\int_{0}^{t}\psi(s)\,ds\Big{)}+y_{2}$ in
$(0,T)\times(0,L)$.

The proof of the existence of one solution and its estimates is complete if
one can show that using this process, one can construct a solution which
verifies Definition 3.1. To this end, one first notes that if $h_{1}$,
$h_{2}$, $h_{3}$, $y_{0}$, $g$ are smooth and
$\operatorname{supp}h_{1},\operatorname{supp}h_{2},\operatorname{supp}h_{3}\Subset(0,T)$,
$\operatorname{supp}y_{0}\Subset(0,L)$ and
$\operatorname{supp}g\Subset(0,T)\times(0,L)$ then one can construct a
solution with the desired estimates. Indeed, the solution $y_{1}$ given by
(3.26) is smooth and has the property that $y_{1}(t,0)$ and $y_{1}(t,L)$ are 0
for $t$ close to $0$ (see Remark 2.2 and Remark 2.3). One can then construct a
smooth solution $y_{2}$ of (3.30) in the interval $(0,2T)$ where the the
expression of $y_{2}(t,0)$, $y_{2}(t,L)$, and $y_{2,x}(t,L)$ are replaced by
the expressions given in (3.30) multiplied by a cutoff function $\chi$, which
is 1 in $(0,T)$ and is $0$ for $t>3T/2$. The solution $y_{2}$ is also smooth
(see Remark 3.4). Then
$e^{2t-x}\Big{(}y_{1}+\int_{0}^{t}\psi(s)\,ds\Big{)}+y_{2}$ is still a
solution in the time interval $(0,T)$ with the desired estimates. Using this
observation, by standard approximation arguments, one can construct a solution
in the sense of Definition 3.1 with the required estimates. ∎

In the proof of Proposition 3.1, we used the following elementary result.

######  Lemma 3.1.

For $p\in\mathbb{C}$ and $z$ in a sufficiently small conic neighborhood of
${\mathbb{R}}_{+}$, let $\lambda_{j}$ with $j=1,\,2,\,3$ be the three
solutions of the equation $\lambda^{3}+\lambda+iz=0$. Consider the convention
$\Re(\lambda_{1})<\Re(\lambda_{2})<\Re(\lambda_{3})$ and similarly for
$\bar{\lambda}_{j}$. We have, in the limit $|z|\to\infty$,

(3.31)
$\lambda_{j}=\mu_{j}z^{1/3}-\frac{1}{3\mu_{j}}z^{-1/3}+O(z^{-2/3})\quad\text{
with }\mu_{j}=e^{-i\pi/6-2ji\pi/3},$

Here $z^{1/3}$ denotes the cube root of $z$ with the real part positive.

Here and in what follows, for $s\in\mathbb{R}$, $O(z^{s})$ denotes a quantity
bounded by $Cz^{s}$ for large positive $z$. Similar convention is used for
$O(|z|^{s})$ for $z\in\mathbb{C}$.

$\mu_{1}$$\mu_{2}$$\mu_{3}$

Figure 1. The roots $\lambda_{j}$ of $\lambda^{3}+\lambda+iz=0$ satisfy, when
$z>0$ is large, $\lambda_{j}\sim\mu_{j}z^{1/3}$ where $\mu_{j}^{3}=-i$.

The same proof as in Step 1 of the proof of Proposition 3.1 gives the
following result.

######  Lemma 3.2.

Let $T>0$ and $(h_{1},h_{2},h_{3})\in H^{1/3}(0,T)\times H^{1/3}(0,T)\times
L^{2}(0,T)$. Then the unique solution $y\in X_{T}$ of the system (3.1) with
$y_{0}=0$ and $f=0$ satisfy, for $x\in[0,L]$,

(3.32) $\|y(\cdot,x)\|_{[H^{2/3}(0,T)]^{*}}\leq
C_{T,L}\Big{(}\|(h_{1},h_{2})\|_{[H^{2/3}(0,T)]^{*}}+\|h_{3}\|_{[H^{1}(0,T)]^{*}}\Big{)},$

where $C_{T,L}$ denotes a positive constant independent of $x$, $y_{0}$, and
$h_{1},\,h_{2},\,h_{3}$.

####  3.1.2. Uniqueness

In this section, we establish the uniqueness of the solutions given in
Proposition 3.1. The uniqueness is somehow known for a close definition of the
solution which requires $\varphi$ less regular. The definition considered here
is somehow more advantageous, see Remark 3.5. We shall highlight the analysis
in such a way that one can extend it to other settings considered in this
paper; thus we do not need to repeat the arguments as much as possible. Let
$y\in X_{T}$ be a solution with the zero data, i.e., $f=0$, $y_{0}=0$, and
$(h_{1},h_{2},h_{3})=(0,0,0)$. Fix $\psi\in
C^{\infty}_{c}\big{(}(0,T)\times(0,L)\big{)}$ (arbitrarily). Let
$\widetilde{y}\in X_{T}$ be a solution of the backward system

$\left\\{\begin{array}[]{cl}\widetilde{y}_{t}+\widetilde{y}_{x}+y_{xxx}=\psi&\mbox{
in }(0,T)\times(0,L),\\\\[6.0pt]
\widetilde{y}(\cdot,0)=0,\;\widetilde{y}_{x}(\cdot,0)=0,\;\widetilde{y}(\cdot,L)=0&\mbox{
in }(0,T),\\\\[6.0pt] \widetilde{y}(T,\cdot)=0&\mbox{ in
}(0,L).\end{array}\right.$

Using the construction given in Section 3.1.1, one can assume that
$\widetilde{y}$ is smooth (see the last part of Step 2 of Section 3.1.1).
Taking $\varphi=\widetilde{y}$ in the definition, we derive that

$\int_{0}^{T}\int_{0}^{L}\psi(t,x)y(t,x)\,dt\,dx=0.$

Since $\psi\in C^{\infty}_{c}\big{(}(0,T)\times(0,L)\big{)}$ is arbitrary, we
deduce that

$y=0\mbox{ in }(0,T)\times(0,L).$

The uniqueness is proved. ∎

######  Remark 3.5.

The proof of the uniqueness also gives the uniqueness of the solutions in
$L^{1}\big{(}(0,T)\times(0,L)\big{)}$, i.e., one requires $y\in
L^{1}\big{(}(0,T)\times(0,L)\big{)}$ instead of $y\in X_{T}$ in Definition
3.1.

######  Remark 3.6.

We end this section with some comments on Proposition 3.1 and its proof.

i) The well-posedness of (3.1) in $X_{T}$ is proved in [8, Theorem 2.10 and
Proposition 2.16] for $L=1$ when $(0,T)=\mathbb{R}_{+}$ and the estimate for
$\|y_{x}(\cdot,x)\|_{L^{2}(0,T)}$ in this case is a consequence of their
results. For $L=1$, their results imply that, when
$(h_{1},h_{2},h_{3})\equiv(0,0,0)$,

(3.33) $\|y_{x}(\cdot,x)\|_{L^{2}(\mathbb{R}_{+})}\leq
C\|y_{0}\|_{L^{2}(0,1)}\mbox{ for }x\in[0,1].$

Note that $L=1$ is smaller than the smallest critical length in
${\mathcal{N}}_{N}$ which is $2\pi$. The estimate as (3.33) cannot hold for
arbitrary $L$. Such an estimate is not valid for any critical length by
considering a non-zero initial datum in $\mathcal{M}_{N}$.

ii) Similar estimates for $\|y(\cdot,x)\|_{H^{1/3}(0,T)}$ and
$\|y_{x}(\cdot,x)\|_{L^{2}(0,T)}$ as in (3.2) in the real line space setting
can be found in [16, 24]. To our knowledge, variants of the estimate for
$\|y_{xx}(\cdot,x)\|_{[H^{1/3}(0,T)]^{*}}$ in (3.2) are not known even in the
real line space setting. Our proof of Proposition 3.1 is in the spirit of
[20], which involves the Fourier transform with respect to time of the
solution, as in [8], and a connection between the linearized KdV and the
linear KdV-Burger equations. However, in the study of the linear KdV-Burger
equation with periodic boundary conditions, the singularity of the kernel is
appropriately compensated (see the proof of Proposition 2.1), which is one of
the novelties of the analysis. The proof given here is self-contained and is
different from the ones in [16, 24] for the whole line setting, which are
based on the Riemann-Liouville fractional integrals and the theory of Airy
functions.

###  3.2. Proof of Proposition 3.2

We first give the definition of the solutions considered in Proposition 3.2.

######  Definition 3.2.

Let $L>0$, $T>0$, $(h_{1},h_{2},h_{3})\in H^{1/3}(0,T)\times
H^{1/3}(0,T)\times[H^{1/3}(0,T)]^{*}$, $f\in
L^{1}\big{(}(0,T);L^{2}(0,L)\big{)}$, and $y_{0}\in L^{2}(0,L)$. A solution
$y\in X_{T}$ of the system (3.4) is a function $y\in X_{T}$ such that

(3.34)
$\int_{0}^{T}\int_{0}^{L}f(t,x)\varphi(t,x)+\int_{0}^{L}y_{0}(x)\varphi(0,x)\,dx-\int_{0}^{T}h_{3}(t)\varphi(t,L)\,dt\\\\[6.0pt]
-\int_{0}^{T}h_{2}(t)\varphi_{xx}(t,L)\,dt+\int_{0}^{T}h_{1}(t)\varphi_{xx}(t,0)\,dt=-\int_{0}^{T}\int_{0}^{L}y(\varphi_{t}+\varphi_{x}+\varphi_{xxx})\,dx\,dt$

for all $\varphi\in C^{3}([0,T]\times[0,L])$ with $\varphi(T,\cdot)=0$,
$\varphi(\cdot,0)=\varphi_{x}(\cdot,0)=\varphi_{x}(\cdot,L)=0$.

The proof of Proposition 3.1 is similar to the one of Proposition 3.1. The
details are left to the reader. We just present here the formula of the
solution of the system when $y_{0}=0$, $h_{1}=0$, $h_{2}=0$, and $f=0$ and
discuss the uniqueness.

Extend $h_{3}$ by $0$ for $t\not\in[0,T]$ and still denote this extension by
$h_{3}$. Then $h_{3}\in H^{-1/3}(\mathbb{R})$. Denote $y$ be the corresponding
solution for $t\geq 0$ and extend $y$ by 0 for $t<0$. Still denote this
extension by $y$. One then has

(3.35) $\left\\{\begin{array}[]{cl}y_{t}+y_{x}+y_{xxx}=0&\mbox{ in
}\mathbb{R}\times(0,L),\\\\[6.0pt]
y(\cdot,0)=0,\;y(\cdot,L)=0,\;y_{xx}(\cdot,L)=h_{3}&\mbox{ in
}\mathbb{R}.\end{array}\right.$

Taking the Fourier transform with respect to $t$, from the system of $y$, we
have

(3.36)
$\left\\{\begin{array}[]{cl}iz\hat{y}(z,x)+\hat{y}_{x}(z,x)+\hat{y}_{xxx}(z,x)=0&\mbox{
in }\mathbb{R}\times(0,L),\\\\[6.0pt] \hat{y}(z,0)=\hat{y}(z,L)=0&\mbox{ in
}\mathbb{R},\\\\[6.0pt] \hat{y}_{xx}(z,L)=\hat{h}_{3}(z)&\mbox{ in
}\mathbb{R}.\end{array}\right.$

Taking into account the equation of $\hat{y}$, we search for the solution of
the form

(3.37) $\hat{y}(z,\cdot)=\sum_{j=1}^{3}a_{j}e^{\lambda_{j}x},$

where $\lambda_{j}=\lambda_{j}(z)$ with $j=1,2,3$ are the three solutions of
the equation $\lambda^{3}+\lambda+iz=0$, and $a_{j}=a_{j}(z)$ for $j=1,2,3$.
We then have

$\begin{array}[]{c}\sum_{j=1}^{3}a_{j}=0,\\\\[6.0pt]
\sum_{j=1}^{3}a_{j}e^{\lambda_{j}L}=0,\\\\[6.0pt]
\sum_{j=1}^{3}a_{j}\lambda_{j}^{2}e^{\lambda_{j}L}=\hat{h}_{3},\\\\[6.0pt]
\end{array}$

This implies, with the convention $\lambda_{j+3}=\lambda_{j}$,

(3.38)
$a_{j}=\frac{e^{\lambda_{j+2}L}-e^{\lambda_{j+1}L}}{\sum_{k=1}^{3}e^{-\lambda_{k}L}(\lambda_{k+2}^{2}-\lambda_{k+1}^{2})}\hat{h}_{3}\mbox{
for }j=1,2,3.$

We now deal with the uniqueness. Let $y\in X_{T}$ be a solution with the zero
data, i.e., $f=0$, $y_{0}=0$, and $(h_{1},h_{2},h_{3})=(0,0,0)$. Let $\psi\in
C^{\infty}_{c}\big{(}(0,T)\times(0,L)\big{)}$. Let $\widetilde{y}\in X_{T}$ be
a solution of the backward system

$\left\\{\begin{array}[]{cl}\widetilde{y}_{t}+\widetilde{y}_{x}+\widetilde{y}_{xxx}=\psi&\mbox{
in }(0,T)\times(0,L),\\\\[6.0pt]
\widetilde{y}(\cdot,0)=0,\;\widetilde{y}_{x}(\cdot,0)=0,\;\widetilde{y}_{x}(\cdot,L)=0&\mbox{
in }(0,T),\\\\[6.0pt] \widetilde{y}(T,\cdot)=0&\mbox{ in
}(0,L).\end{array}\right.$

One can show that $\widetilde{y}$ is smooth (see the construction of the
solutions given in Proposition 3.3 and the last part of Step 2 of Section
3.1.1; in fact, the arguments are simpler since the data on the boundary are 0
and the initial datum is 0). Taking $\varphi=\widetilde{y}$ in Definition 3.1,
we derive that

$\int_{0}^{T}\int_{0}^{L}\psi(t,x)y(t,x)\,dt\,dx=0.$

Since $\psi\in C^{\infty}_{c}\big{(}(0,T)\times(0,L)\big{)}$ is arbitrary, we
deduce that

$y=0\mbox{ in }(0,T)\times(0,L).$

The uniqueness is proved. ∎

######  Remark 3.7.

The proof of the uniqueness also gives the uniqueness of the solutions in
$L^{1}\big{(}(0,T)\times(0,L)\big{)}$, i.e., one requires $y\in
L^{1}\big{(}(0,T)\times(0,L)\big{)}$ instead of $y\in X_{T}$ in Definition
3.2.

###  3.3. Proof of Proposition 3.3

We first introduce the notion of the solutions considered in Proposition 3.3.

######  Definition 3.3.

Let $L>0$, $T>0$, $(h_{1},h_{2},h_{3})\in L^{2}(0,T)\times H^{1/3}(0,T)\times
L^{2}(0,T)$,
$f\in L^{1}\big{(}(0,T);L^{2}(0,L)\big{)}$, and $y_{0}\in L^{2}(0,L)$. A
solution $y\in X_{T}$ of the system (3.7) is a function $y\in X_{T}$ such that

(3.39)
$\int_{0}^{T}\int_{0}^{L}f(t,x)\varphi(t,x)+\int_{0}^{L}y_{0}(x)\varphi(0,x)\,dx+\int_{0}^{T}h_{3}(t)\varphi_{x}(t,L)\,dt\\\\[6.0pt]
-\int_{0}^{T}h_{1}(t)\varphi_{x}(t,0)\,dt-\int_{0}^{T}h_{2}(t)\varphi_{xx}(t,L)\,dt=-\int_{0}^{T}\int_{0}^{L}y(\varphi_{t}+\varphi_{x}+\varphi_{xxx})\,dx\,dt$

for all $\varphi\in C^{3}([0,T]\times[0,L])$ with $\varphi(T,\cdot)=0$,
$\varphi(\cdot,0)=\varphi(\cdot,L)=\varphi_{xx}(\cdot,0)=0$.

The proof of Proposition 3.3 is similar to the one of Proposition 3.1. We only
give here the formula of the solutions in the case $f=0$, $y_{0}=0$, and
$(h_{1},h_{2},h_{3})=(0,0,h_{3})$. Taking the Fourier transform with respect
to $t$, from the system of $y$, we have

(3.40)
$\left\\{\begin{array}[]{cl}iz\hat{y}(z,x)+\hat{y}_{x}(z,x)+\hat{y}_{xxx}(z,x)=0&\mbox{
in }\mathbb{R}\times(0,L),\\\\[6.0pt] \hat{y}_{x}(z,0)=\hat{y}(z,L)=0&\mbox{
in }\mathbb{R},\\\\[6.0pt] \hat{y}_{x}(z,L)=\hat{h}_{3}(z)&\mbox{ in
}\mathbb{R}.\end{array}\right.$

Taking into account the equation of $\hat{y}$, we search for the solution of
the form

(3.41) $\hat{y}(z,\cdot)=\sum_{j=1}^{3}a_{j}e^{\lambda_{j}x},$

where $\lambda_{j}=\lambda_{j}(z)$ with $j=1,2,3$ are the three solutions of
the equation $\lambda^{3}+\lambda+iz=0$, and $a_{j}=a_{j}(z)$ for $j=1,2,3$.
We then have

$\begin{array}[]{c}\sum_{j=1}^{3}\lambda_{j}a_{j}=0,\\\\[6.0pt]
\sum_{j=1}^{3}a_{j}e^{\lambda_{j}L}=0,\\\\[6.0pt]
\sum_{j=1}^{3}a_{j}\lambda_{j}e^{\lambda_{j}L}=\hat{h}_{3},\\\\[6.0pt]
\end{array}$

This implies, with the convention $\lambda_{j+3}=\lambda_{j}$,

(3.42)
$a_{j}=\frac{\lambda_{j+1}e^{\lambda_{j+2}L}-\lambda_{j+2}e^{\lambda_{j+1}L}}{\sum_{k=1}^{3}\lambda_{k}e^{\lambda_{k}L}(\lambda_{k+1}e^{\lambda_{k+2}L}-\lambda_{k+2}e^{\lambda_{k+1}L})}\hat{h}_{3}\mbox{
for }j=1,2,3.$

The rest of the proof is almost the same as the one of Proposition 3.1 and is
left to the reader. ∎

##  4\. The set of the critical lengths

The main goal of this section is to establish Theorem 1.1, except for (1.12)
and its consequence (1.13), and Proposition 1.1. The results in this section,
in particular in (4.1) will be complemented with the analysis in the next
section (Section 5) to derive (1.12) and (1.13). We begin with

######  Proposition 4.1.

Assume that $L\in{\mathcal{N}}_{D}$. There exists $a<0$ and $b>0$ such that

(4.1) $(a+ib)e^{(a+ib)}=(a-ib)e^{(a-ib)}=-2ae^{-2a}$

and

(4.2) $(a+ib)^{2}+(a-ib)^{2}+(a+ib)(a-ib)=-L^{2}.$

These conditions are equivalent to

(4.3) $b^{2}=L^{2}+3a^{2},\quad a^{2}+b^{2}=4a^{2}e^{-6a},\quad b\cos b+a\sin
b=0,$ (4.4) $a\cos b-b\sin b>0.$

Moreover,

(4.5) $a$ and $b$ are uniquely determined by $L$.

###### Proof.

The proof is divided into two steps.

Step 1: We prove (4.1) - (4.4).

Let $z_{1}=a+ib$ and $z_{2}=c+id$ where $a,b,c,d\in\mathbb{R}$. Set

$z_{3}=-(z_{1}+z_{2})=-(a+c)-i(b+d).$

Note that, if $z_{1}+z_{2}+z_{3}=0$, then

(4.6) $z_{1}^{2}+z_{2}^{2}+z_{1}z_{2}=-(z_{1}z_{2}+z_{2}z_{3}+z_{3}z_{1}).$

Condition (1.4) can be then written as

(4.7) $z_{1}+z_{2}+z_{3}=0,\quad
z_{1}e^{z_{1}}=z_{2}e^{z_{2}}=z_{3}e^{z_{3}},\quad\mbox{ and }\quad
z_{1}z_{2}+z_{2}z_{3}+z_{3}z_{1}=L^{2}.$

In the three numbers $a$, $c$, and $-(a+c)$, there are two numbers of the same
sign. Note also that if $(z_{1},z_{2},z_{3})$ is a solution of (4.7), then
$(\bar{z}_{1},\bar{z}_{2},\bar{z}_{3})$ is also a solution. Without loss of
generality, one might assume from now on that

(4.8) $ac\geq 0,\quad|c|\geq|a|,$

and

(4.9) $b\geq 0.$

From (1.4), we have

$(a+ib)^{2}+(c+id)^{2}+(a+ib)(c+id)=-L^{2}.$

This yields

$a^{2}-b^{2}+c^{2}-d^{2}+ac-bd+i(2ab+2cd+ad+bc)=-L^{2}.$

We thus obtain

(4.10) $a^{2}+ac+c^{2}+L^{2}=b^{2}+bd+d^{2}$

and

(4.11) $2ab+2cd+ad+bc=0.$

From (1.4), we have

(4.12)
$(a+ib)e^{a+ib}=(c+id)e^{c+id}=-\big{(}a+c+i(b+d)\big{)}e^{-(a+c)-i(b+d)}.$

Considering the modulus in (4.12), we get

(4.13)
$(a^{2}+b^{2})e^{2a}=(c^{2}+d^{2})e^{2c}=\big{(}(a+c)^{2}+(b+d)^{2}\big{)}e^{-2(a+c)}.$

Considering the imaginary parts in (4.12), we obtain

(4.14) $(b\cos b+a\sin b)e^{a}=(d\cos d+c\sin
d)e^{c}=\big{(}-(b+d)\cos(b+d)+(a+c)\sin(b+d)\big{)}e^{-(a+c)}.$

This implies that $c\neq 0$ since otherwise $a$ is also 0 (since $|a|\leq|c|$)
and one reaches, by (4.13),

$|b|=|d|=|b+d|,$

which yields $b=d=0$ and $L=0$. This is impossible since $L>0$.

Using (4.11) and the facts $ac\geq 0$ and $c\neq 0$, we have

(4.15) $d=-\frac{2a+c}{2c+a}b.$

This implies

(4.16) $b+d=\frac{c-a}{2c+a}b.$

Combining (4.9) and (4.15) yields that

(4.17) $b>0$

since $b^{2}+bd+d^{2}\geq L^{2}$.

The proof now is divided into four cases.

$\bullet$ Case 1: $a\geq 0$. Thus $c\geq a\geq 0$. Since
$(a+c)^{2}e^{-2(a+c)}\leq e^{-2}$ for $a+c\geq 0$, it follows that

$(b+d)^{2}+e^{-2}\geq(b+d)^{2}+(a+c)^{2}e^{-2(a+c)}\\\\[6.0pt]
\geq\big{(}(a+c)^{2}+(b+d)^{2}\big{)}e^{-2(a+c)}\mathop{=}^{\eqref{thmCha-L-
cond-modulus}}(a^{2}+b^{2})e^{2a}\geq b^{2}e^{2a}\geq b^{2}.$

In summary, one reaches

(4.18) $(b+d)^{2}+e^{-2}\geq b^{2}.$

One has, for $c\geq a\geq 0$,

(4.19) $(b+d)^{2}\mathop{=}^{\eqref{thmCha-L-
cond-b+d}}\left(\frac{c-a}{2c+a}\right)^{2}b^{2}\leq b^{2}/4.$

Combining (4.18) and (4.19) yields

(4.20) $e^{-2}\geq\frac{3}{4}b^{2};\mbox{ thus, by \eqref{thmCha-L-sign-b2},
}0<b\leq 0.425.$

Since $a\geq 0$, it follows from (4.20) that

(4.21) $b\cos b+a\sin b>0.$

We have

$a+c\leq(2a^{2}+2c^{2}+2ac)^{1/2}\mathop{\leq}^{\eqref{thmCha-L-
cond1}}\left(2(b^{2}+d^{2}+bd)\right)^{1/2}=\left(b^{2}+d^{2}+(b+d)^{2}\right)^{1/2}\\\\[6.0pt]
\mathop{\leq}^{\eqref{thmCha-L-cond-d-tb},\eqref{thmCha-L-
bd-p1}}\left(2b^{2}+b^{2}/4\right)^{1/2}=(9b^{2}/4)^{1/2}\mathop{\leq}^{\eqref{thmCha-
L-b-p1}}(3e^{-2})^{1/2}\leq 0.638.$

Since $0\leq b+d\leq b/2\leq 0.225$ by (4.16), (4.19), and (4.20), it follows
that

(4.22)
$-(b+d)\cos(b+d)+(a+c)\sin(b+d)\leq-(b+d)\cos(0.225)+(a+c)(b+d)\\\\[6.0pt]
\leq-0.97(b+d)+0.638(b+d)\leq 0.$

Combining (4.14), (4.21), and (4.22), we derive that the case $a\geq 0$ does
not happen.

$\bullet$ Case 2: $a<0$ and $a\neq c$. Thus $c<a<0$ since $|c|\geq|a|$.

$\ast$ Case 2.1: $a<0$ and $c\leq 2a$. We have

$d^{2}\mathop{=}^{\eqref{thmCha-L-cond-d-
tb}}\left(\frac{2a+c}{2c+a}\right)^{2}b^{2}\leq(4/5)^{2}b^{2}.$

It follows that

$b^{2}\leq(a^{2}+b^{2})\mathop{=}^{\eqref{thmCha-L-cond-
modulus}}e^{2(c-a)}(c^{2}+d^{2})\leq c^{2}e^{c}+(4/5)^{2}b^{2}.$

This yields, since $c\leq 0$,

$(9/25)b^{2}\leq c^{2}e^{c}\leq 2^{2}e^{-2}.$

We obtain

(4.23) $b^{2}\leq\frac{100}{9}e^{-2}\leq 1.51,\mbox{ which yields, by
\eqref{thmCha-L-sign-b2}, }0<b\leq 1.23.$

We have

(4.24) $a^{2}+b^{2}\mathop{=}^{\eqref{thmCha-L-cond-
modulus}}\big{(}(a+c)^{2}+(b+d)^{2}\big{)}e^{-4a-2c}\geq(a+c)^{2}e^{-2a-2c}$

since $a<0$. It follows from (4.23) that

$1.51\geq
b^{2}\geq(a+c)^{2}e^{-2a-2c}-a^{2}\geq(a+c)^{2}\Big{(}e^{-2a-2c}-\frac{1}{9}\Big{)}.$

Since $f_{1}(t)=t^{2}(e^{2t}-1/9)$ is an strictly increasing function for
$t>0$ and

$f_{1}(0.66)\geq 1.55,$

it follows that

(4.25) $0.66\geq|a+c|=|c|+|a|\geq 3|a|.$

Using (4.23) and (4.25), we obtain

(4.26) $b\cos b+a\sin b\geq b\cos 1.23+ab>0.33b+ab>0.$

and, since $0<b+d\leq b/2\leq 0.64$ by (4.16) and (4.23),

(4.27) $-(b+d)\cos(b+d)+(a+c)\sin(b+d)\\\\[6.0pt] \leq-(b+d)\cos
0.64+(a+c)(b+d)\cos 0.64\mathop{<}^{\eqref{thmCha-L-p3}}0.$

Here we used the fact $\sin x\geq x\cos\theta_{0}$ for $0\leq
x\leq\theta_{0}\leq 1$.

Combining (4.14), (4.26), and (4.27), we derive that Case 2.1 does not happen.

$\ast$ Case 2.2: $a<0$, $2a\leq c<a$ with $|a|\leq 1$. We have

$a^{2}+b^{2}\mathop{=}^{\eqref{thmCha-L-cond-
modulus}}(c^{2}+d^{2})e^{2(c-a)}\mathop{\leq}^{\eqref{thmCha-L-cond-d-
tb}}(c^{2}+b^{2})e^{2(c-a)}.$

This implies, with $s=c-a<0$,

$b^{2}(1-e^{2s})=b^{2}(1-e^{2(c-a)})\leq c^{2}e^{2(c-a)}-a^{2}\\\\[6.0pt]
=(a+s)^{2}e^{2s}-a^{2}=a^{2}e^{2s}+2ase^{2s}+s^{2}e^{2s}-a^{2}\leq(2as+s^{2})e^{2s}.$

We derive that

(4.28) $b^{2}\leq\frac{|2as+s^{2}|}{|1-e^{2s}|}e^{2s}.$

Set

$g(t)=2te^{-2t}-1+e^{-2t}\mbox{ for }t\geq 0.$

Then

$g^{\prime}(t)=(2-4t-2)e^{-2t}=-4te^{-2t}<0\mbox{ for }t>0,$

which yields $g(t)<g(0)=0$ for $t>0$. This implies

(4.29) $|s|e^{2s}\leq\frac{1}{2}(1-e^{2s}).$

Combining (4.28) and (4.29) yields

(4.30) $b^{2}\leq\frac{1}{2}(2|a|+|s|)=\frac{1}{2}(|a|+|c|)\leq 3/2,$

where in the last inequality we used the fact $|c|\leq 2|a|$ and $|a|\leq 1$.

We have

(4.31) $a^{2}+b^{2}\mathop{=}^{\eqref{thmCha-L-cond-
modulus}}\big{(}(a+c)^{2}+(b+d)^{2}\big{)}e^{-4a-2c}\geq(a+c)^{2}e^{-2a-2c}.$

It follows from (4.30) that

$\frac{1}{2}(|a|+|c|)\geq
b^{2}\geq(a+c)^{2}e^{-2a-2c}-a^{2}\geq(a+c)^{2}(e^{-2a-2c}-\frac{1}{4}),$
(4.32) $\frac{1}{2}\geq|a+c|(e^{-2a-2c}-\frac{1}{4}),$

Since $f_{2}(t)=t(e^{2t}-1/4)$ is an increasing function for $t\geq 0$ and

$f_{2}(0.32)\geq 0.52,$

it follows from (4.32) that

(4.33) $|a|+|c|\leq 0.32,\mbox{ which yields }|a|\leq 0.16.$

Using (4.30) and (4.32), we obtain

(4.34) $b\cos b+a\sin b\geq b\cos\sqrt{3/2}+ab>0.33b+ab>0,$

and, since $0<b+d\leq b/2\leq\sqrt{3/2}/2\leq 0.62$ by (4.16) and (4.30),

(4.35) $-(b+d)\cos(b+d)+(a+c)\sin(b+d)\\\\[6.0pt] \leq-(b+d)\cos
0.62+(a+c)(b+d)\cos 0.62\mathop{<}^{\eqref{thmCha-L-2.2-p4}}0.$

Here we used the fact $\sin x\geq x\cos\theta_{0}$ for $0\leq
x\leq\theta_{0}\leq 1$.

Combining (4.14), (4.34), and (4.35), we derive that Case 2.2 does not happen.

$\ast$ Case 2.3: $a<0$ and $2a\leq c<a$ with $|a|>1$. We have, by (4.13),

$(a^{2}+b^{2})e^{2a}=(c^{2}+d^{2})e^{2c}$

Since $|c|\geq|a|\geq 1$ and $d^{2}\leq b^{2}$ by (4.15), and the function
$f_{3}(t)=t^{2}e^{-2t}$ is strictly decreasing in $[1,+\infty)$, we derive
that

$(a^{2}+b^{2})e^{2a}>(c^{2}+d^{2})e^{2c}.$

Thus Case 2.3 does not happen.

$\bullet$ Case 3: $a<0$ and $a=c$. We derive from (4.16) that

$b+d=0.$

Assertion (4.1)-(4.4) then follow from (4.13), (4.14) and the fact that

$\Re(a+ib)e^{(a+ib)}=e^{a}(a\cos b-b\sin b)>0.$

$\bullet$ Case 4: $a=0$ and $c<0$. It follows that

$d=-b/2.$

This case does not happen by (4.13).

Step 2: We prove (4.5). Since

$L^{2}=b^{2}-3a^{2}=4a^{2}(e^{-6a}-1)$

and the function $h(t)=4t^{2}(e^{6t}-1)$ is strictly increasing in
$(0,+\infty)$, it follows that

$a\mbox{ is uniquely determined by }L.$

This in turn implies that

$b\mbox{ is uniquely determined by }L$

since $L^{2}=b^{2}-3a^{2}$ and $b>0$.

The proof is complete. ∎

We are ready to give

###### Proof of Proposition 1.1.

By (4.3) and (4.4), we have

$L^{2}=4a^{2}(e^{-6a}-1),$

where $a<0$ is determined by, with $b^{2}=4a^{2}(e^{-6a}-1/4)$, $b>0$,

$b\cos b+a\sin b=0\quad\mbox{ and }\quad a\cos b-b\sin b>0.$

Since $b>0$ and $a<0$, it follows that

$\cos b<0\quad\mbox{ and }\quad\sin b<0.$

Hence $b=b_{n}$ for some $\pi+2n\pi<b_{n}<3\pi/2+n2\pi$ with $n\in\mathbb{N}$.
Since $b_{n}\geq\pi$ and $4a_{n}^{2}(e^{-6a_{n}}-1/4)=b_{n}^{2}$, it follows
that

$|a_{n}|\geq 0.43.$

This yields

(4.36) $|\cot b_{n}|=|a_{n}/b_{n}|=\frac{1}{4|a_{n}|(e^{6|a_{n}|}-1/4)}\leq
0.045.$

We thus derive that

$|b_{n}-3\pi/2|\leq 0.045.$

Set

$t_{n}=b_{n}-\pi-n2\pi.$

Since

$|b_{n}\cos b_{n}|=|a_{n}\sin b_{n}|,$

it follows that

(4.37) $(t_{n}+\pi+n2\pi)\cos t_{n}=|a_{n}|\sin t_{n}.$

For $t_{n}\in(\pi/4,\pi/2)$, the LHS of (4.37) is a strictly decreasing
function of $t_{n}$ and the RHS of (4.37) is a strictly increasing function of
$t_{n}$ since $|a_{n}|$ is a increasing function of $t_{n}$. Hence there
exists a unique $t_{n}\in(\pi/4,\pi/2)$ such that the identity holds. One can
check that the corresponding $a_{n}$ fulfills all the requirements. ∎

######  Remark 4.1.

It is clear, as $n\to+\infty$, that $a_{n}/b_{n}\to 0$ so $b_{n}-3\pi/2-2\pi
n\to 0$ as $n\to+\infty$.

We end this section with the following result, which yields (1.8) and (1.11).
Recall that $\alpha$ and $\beta$ are defined by (1.9) and $q$ is defined by
(1.7).

######  Lemma 4.1.

Let $L\in{\mathcal{N}}_{D}$. Set

(4.38) $\eta_{1}=\alpha+i\beta,\quad\eta_{2}=\alpha-i\beta,\quad\mbox{ and
}\quad\eta_{3}=-2\alpha,$ (4.39)
$\varphi(x)=(\eta_{3}-\eta_{2})e^{\eta_{1}x}+(\eta_{1}-\eta_{3})e^{\eta_{2}x}+(\eta_{2}-\eta_{1})e^{\eta_{3}x}\mbox{
in }[0,L],$

and

$\Psi(t,x)=e^{qt}\psi(x)\mbox{ in }\mathbb{R}\times[0,L].$

Then $\varphi(x)=-2i\phi$ and $\Psi$ satisfies (1.11). Consequently, (1.8) and
(1.11) hold.

###### Proof.

Since

$\eta_{1}e^{-\eta_{1}L}=\eta_{2}e^{-\eta_{2}L}=\eta_{3}e^{-\eta_{3}L}$
$\eta_{1}+\eta_{2}+\eta_{3}=0,\quad\eta_{1}\eta_{2}+\eta_{2}\eta_{3}+\eta_{3}\eta_{1}=1,\quad\mbox{
and }\quad\eta_{1}\eta_{2}\eta_{3}=-q,$

it follows that

$\left\\{\begin{array}[]{cl}\varphi_{xxx}+\varphi_{x}+q\varphi=0&\mbox{ in
}(0,L),\\\\[6.0pt]
\varphi(0)=\varphi_{x}(0)=\varphi(L)=\varphi_{xx}(L)=0&\mbox{ in
}[0,L].\end{array}\right.$

We derive that $\Psi$ is a solution of (1.11). A computation gives

$\displaystyle\varphi(x)=$
$\displaystyle(-3\alpha+i\beta)e^{(\alpha+i\beta)x}+(3\alpha+i\beta)e^{(\alpha-i\beta)x}-2i\beta
e^{-2\alpha x}$ $\displaystyle=$ $\displaystyle-6i\alpha e^{\alpha
x}\sin(\beta x)+2i\beta e^{\alpha x}\cos(\beta x)-2i\beta e^{-2\alpha x}$
$\displaystyle=$ $\displaystyle 2i\Big{(}\beta e^{\alpha x}\cos(\beta x)-\beta
e^{-2\alpha x}-3\alpha e^{\alpha x}\sin(\beta x)\Big{)}=-2i\phi.$

The proof is complete. ∎

##  5\. The unreachable space of the linearized system for a critical length

In this section, we prove that the unreachable space is given by (1.12) and
study its controllability properties of the linearized KdV system:

(5.1) $\left\\{\begin{array}[]{cl}y_{t}+y_{x}+y_{xxx}=0&\mbox{ in
}(0,T)\times(0,L),\\\\[6.0pt] y(\cdot,0)=y_{x}(\cdot,L)=0&\mbox{ in
}(0,T),\\\\[6.0pt] y(\cdot,L)=u&\mbox{ in }(0,T),\end{array}\right.$

for a critical length. The main result of this section is Proposition 5.1
which is based on an observability inequality with initial data and final data
in $\mathcal{M}_{D}^{\perp}$, where $\mathcal{M}_{D}$ is defined by (1.12). In
comparison with the right Neumann boundary control system, this part in the
Dirichlet setting is more complex and technical. The proof of Proposition 5.1
involves the well-posedness and the estimates for solutions in $X_{T}$ given
in Proposition 3.1 and Proposition 3.2.

We begin with the following simple but useful result.

######  Lemma 5.1.

Let $L\in{\mathcal{N}}_{D}$ and $T>0$. For $y_{0}\in L^{2}(0,L)$ and $u\in
H^{1/3}(0,T)$, let $y\in X_{T}$ be the unique solution of the system

$\left\\{\begin{array}[]{cl}y_{t}+y_{x}+y_{xxx}=0&\mbox{ in
}(0,T)\times(0,L),\\\\[6.0pt] y(\cdot,0)=y_{x}(\cdot,L)=0&\mbox{ in
}(0,T),\\\\[6.0pt] y(\cdot,L)=u&\mbox{ in }(0,T),\\\\[6.0pt]
y(0,\cdot)=y_{0}&\mbox{ in }(0,L).\end{array}\right.$

Then

$\int_{0}^{L}y(t,x)\Phi(t,x)\,dx\mbox{ is constant on }[0,T].$

Recall that $\Phi$ is defined by (1.10).

###### Proof.

Using the fact, by integration by parts,

$\frac{d}{dt}\int_{0}^{L}y(t,x)\Phi(t,x)\,dx=0,$

the conclusion follows. ∎

We are ready to state the main result of this section on the unreachable space
for the linearized KdV system (5.1).

######  Proposition 5.1.

Let $L\in{\mathcal{N}}_{D}$ and $T>0$. We have

$i)$ for $\varphi\in\mathcal{M}_{D}\setminus\\{0\\}$, there does not exist
$u\in H^{1/3}(0,T)$ such that $y(T,\cdot)=\varphi$ where $y\in X_{T}$ is the
unique solution of (5.1) with $y(0,\cdot)=0$.

$ii)$ there exists a linear continuous operator
$\mathcal{L}:\mathcal{M}_{D}^{\perp}\to H^{1/3}(0,T)$ such that
$y(T,\cdot)=\varphi$ where $y\in X_{T}$ is the unique solution of (5.1) with
$y(0,\cdot)=0$ and $u=\mathcal{L}(\varphi)$.

$iii)$ There exists a linear continuous operator
$\hat{\mathcal{L}}:\mathcal{M}_{D}^{\perp}\to H^{1/3}(0,T)$ such that
$y(T,\cdot)=0$ where $y\in X_{T}$ is the unique solution of (5.1) with
$y(0,\cdot)=\varphi$ and $u=\hat{\mathcal{L}}(\varphi)$.

###### Proof.

Assertion $i)$ is just a consequence of Lemma 5.1.

We next deal with assertions $ii)$ and $iii)$. Let
$\psi\in{\mathcal{M}}_{D}^{\perp}$. Let $y^{*}\in X_{T}$ be the unique
solution of the backward linear KdV system

(5.2) $\left\\{\begin{array}[]{cl}y^{*}_{t}+y^{*}_{x}+y^{*}_{xxx}=0&\mbox{ in
}(0,T)\times(0,L),\\\\[6.0pt]
y^{*}(\cdot,0)=y^{*}_{x}(\cdot,0)=y^{*}(\cdot,L)=0&\mbox{ in
}(0,T),\\\\[6.0pt] y^{*}(T,\cdot)=\psi&\mbox{ in }(0,L).\end{array}\right.$

Applying the observability inequality in Lemma 5.2 below to
$y^{*}(T-\cdot,L-\cdot)$, we obtain

(5.3)
$\lambda^{-1}\|\psi\|_{L^{2}(0,L)}\leq\|y^{*}_{xx}(\cdot,L)\|_{[H^{1/3}(0,T)]^{*}}\leq\lambda\|\psi\|_{L^{2}(0,L)}$

for some constant $\lambda\geq 1$. Fix a continuous linear mapping

(5.4) $\mathcal{L}_{1}:[H^{1/3}(0,T)]^{*}\to H^{1/3}(0,T)$

such that

(5.5)
$\langle\mathcal{L}_{1}(h),h\rangle_{H^{1/3}(0,T);[H^{1/3}(0,T)]^{*}}\geq
C\|h\|_{[H^{1/3}(0,T)]^{*}}^{2}\quad\mbox{ for all }h\in[H^{1/3}(0,T)]^{*},$

for some positive constant $C$. This can be done using the Fourier series or
the Fourier transform appropriately.

We first prove assertion $ii)$. Equipped $\mathcal{M}_{D}^{\perp}$ with the
$L^{2}(0,L)$-scalar product. Define

${\mathcal{A}}:\mathcal{M}_{D}^{\perp}\to\mathcal{M}_{D}^{\perp}$

by

${\mathcal{A}}(\psi)=y(T,\cdot),$

where $y\in X_{T}$ is the unique solution of the following system

(5.6) $\left\\{\begin{array}[]{cl}y_{t}+y_{x}+y_{xxx}=0&\mbox{ in
}(0,T)\times(0,L),\\\\[6.0pt] y(\cdot,0)=y_{x}(\cdot,L)=0&\mbox{ in
}(0,T),\\\\[6.0pt] y(\cdot,L)=h&\mbox{ in }(0,T),\\\\[6.0pt]
y(0,\cdot)=0&\mbox{ in }(0,L),\end{array}\right.$

with $h=\mathcal{L}_{1}(y_{xx}^{*}(\cdot,L))$ where $y^{*}$ is determined by
(5.2) ($y(T,\cdot)\in\mathcal{M}_{D}^{\perp}$ since $y(0,\cdot)=0$ by Lemma
5.1). An integration by part yields

(5.7)
$\int_{0}^{L}y(T,x)z^{*}(T,x)\,dx=-\int_{0}^{T}y(t,L)z^{*}_{xx}(t,L)\,dt,$

for all solutions $z^{*}\in X_{T}$ of (5.2) with
$z^{*}(T,\cdot)\in\mathcal{M}_{D}^{\perp}$.

Using (5.3) and applying the Lax-Milgram theory, we derive that
${\mathcal{A}}$ is linear continuous and its inverse is also linear
continuous. The conclusion of $ii)$ follows by taking
$\mathcal{L}(\psi)=\mathcal{L}_{1}(y^{*}_{xx}(\cdot,L))$ and $y^{*}$ is the
solution of (5.2) with $\psi$ being replaced by ${\mathcal{A}}^{-1}(\psi)$.

We now deal with assertion $iii)$. Set

$H=\Big{\\{}y^{*}_{xx}(\cdot,L)\in[H^{1/3}(0,T)]^{*};\mbox{ where $y^{*}$ is
determined by \eqref{Sys-KdV-BW} with
$\psi\in\mathcal{M}_{D}^{\perp}$}\Big{\\}}.$

It follows from (5.3) that $H$ is a closed subspace of $[H^{1/3}(0,T)]^{*}$ so
is a Hilbert space. We next consider the following bilinear form on $H$:

$\hat{a}(u,v)=\langle\mathcal{L}_{1}(u),v\rangle_{H^{1/3}(0,T);[H^{1/3}(0,T)]^{*}}.$

Using (5.3), we derive from the Lax-Milgram theorem that there exists a
continuous linear application $\hat{\mathcal{A}}:\mathcal{M}_{D}^{\perp}\to H$
such that

(5.8)
$\int_{0}^{L}\varphi(x)y^{*}(0,x)\,dx=\hat{a}(\hat{\mathcal{A}}\varphi,y^{*}_{xx}(\cdot,L))$

for all solution $y^{*}\in X_{T}$ of (5.2) with
$\psi\in\mathcal{M}_{D}^{\perp}$.

Set

$\hat{\mathcal{L}}=\mathcal{L}_{1}\circ\hat{\mathcal{A}}.$

The conclusion of $iii)$ then follows after noting that if
$y(0,\cdot)\in\mathcal{M}_{D}^{\perp}$ then
$y(T,\cdot)\in\mathcal{M}_{D}^{\perp}$ by Lemma 5.1, and the fact

$\int_{0}^{L}y(T,x)y^{*}(T,x)\,dx-\int_{0}^{L}y(0,x)y^{*}(0,x)\,dx=-\int_{0}^{T}y(t,L)y^{*}_{xx}(t,L)\,dt.$

The proof is complete. ∎

Here is the observability inequality used in the proof of Proposition 5.1.

######  Lemma 5.2.

Let $L\in{\mathcal{N}}_{D}$, $T>0$, and let $y\in X_{T}$ be a solution of the
linearized KdV equation in $(0,T)\times(0,L)$ with
$y(0,L-\cdot)\in\mathcal{M}_{D}^{\perp}$ and with
$y(\cdot,0)=y(\cdot,L)=y_{x}(\cdot,L)=0$. Then there exists $\Lambda\geq 1$
depending only on $L$ and $T$ such that

(5.9)
$\Lambda^{-1}\|y(0,\cdot)\|_{L^{2}(0,L)}\leq\|y_{xx}(\cdot,0)\|_{[H^{1/3}(0,T)]^{*}}\leq\Lambda\|y(0,\cdot)\|_{L^{2}(0,L)}.$

###### Proof.

By Proposition 3.1, we have

(5.10) $\|y_{xx}(\cdot,0)\|_{[H^{1/3}(0,T)]^{*}}\leq
C\|y(0,\cdot)\|_{L^{2}(0,L)}.$

We next prove

(5.11) $\|y(0,\cdot)\|_{L^{2}(0,L)}\leq
C\|y_{xx}(\cdot,0)\|_{[H^{1/3}(0,T)]^{*}}$

by contradiction. Assume that there exists a sequence $(\varphi_{n})$ such
that $\varphi_{n}(L-\cdot)\in\mathcal{M}_{D}^{\perp}$ and

(5.12)
$\|y_{n,xx}(\cdot,0)\|_{[H^{1/3}(0,T)]^{*}}\leq\frac{1}{n}\|y_{n}(0,\cdot)\|_{L^{2}(0,L)}=\frac{1}{n},$

where $y_{n}\in X_{T}$ is the unique solution of the linearized KdV equation
in $(0,T)\times(0,L)$ with $y_{n}(0,\cdot)=\varphi_{n}$ and
$y_{n}(\cdot,0)=y_{n}(\cdot,L)=y_{n,x}(\cdot,L)=0$. Set

(5.13) ${\bf y}_{n}(t,x)=y_{n}(T-t,L-x).$

Then ${\bf y}_{n}\in X_{T}$ is a solution of the equation ${\bf y}_{n,t}+{\bf
y}_{n,x}+{\bf y}_{n,xxx}=0$ in $(0,T)\times(0,L)$. By the regularizing effect
of the linearized KdV equation, without loss of generality, one might assume
that ${\bf y}_{n}(0,\cdot)=y_{n}(T,L-\cdot)$ converges in $L^{2}(0,L)$.
Applying Proposition 3.2 to ${\bf y}_{n}$, one derives that
$y_{n}(0,L-\cdot)={\bf y}_{n}(T,\cdot)$ is a Cauchy sequence in $L^{2}(0,L)$.
In other words, $\varphi_{n}=y_{n}(0,\cdot)$ is a Cauchy sequence in
$L^{2}(0,L)$. Let $\varphi$ be the limit of $\varphi_{n}$ in $L^{2}(0,L)$ and
denote $y\in X_{T}$ be the corresponding solution of the linearized KdV
system. Then $\|\varphi\|_{L^{2}(0,L)}=1$ and
$\varphi(L-\cdot)\in\mathcal{M}_{D}^{\perp}$.

Set ${\bf y}(t,x)=y(T-t,L-x)$. Then ${\bf y}\in X_{T}$ is a solution of the
system

(5.14) ${\bf y}_{t}+{\bf y}_{x}+{\bf y}_{xxx}=0\mbox{ in }(0,T)\times(0,L),$
(5.15) ${\bf y}(\cdot,0)={\bf y}_{x}(\cdot,0)={\bf y}(\cdot,L)={\bf
y}_{xx}(\cdot,L)=0\mbox{ in }(0,T),$

and

${\bf y}(0,\cdot)\in\mathcal{M}_{D}^{\perp}\;(\mbox{by
\lx@cref{creftypecap~refnum}{lem-projection}})\quad\mbox{ and
}\quad\quad\|{\bf
y}(0,\cdot)\|_{L^{2}(0,L)}\mathop{\geq}^{\lx@cref{creftypecap~refnum}{pro-
kdv2}}C\|\varphi\|_{L^{2}(0,L)}=C.$

By Lemma 5.3 below, ${\bf y}(0,\cdot)=c\phi$ in $(0,L)$ for some
$c\in\mathbb{R}$ where $\phi$ is defined by (1.8). Since ${\bf
y}(0,\cdot)\in\mathcal{M}_{D}^{\perp}$, it follows that $c=0$ and hence ${\bf
y}(0,\cdot)=0$ in $(0,L)$ We obtain a contradiction. Thus (5.11) is proved.

The conclusion now follows from (5.10) and (5.11). ∎

In Lemma 5.2, we used the following result, which is also helpful in the proof
of Assertion ii) of Theorem 1.3.

######  Lemma 5.3.

Let $L\in{\mathcal{N}}_{D}$ and $T>0$. Assume that $y\in X_{T}$ is a solution
of the system

(5.16) $\left\\{\begin{array}[]{cl}y_{t}+y_{x}+y_{xxx}=0&\mbox{ in
}(0,T)\times(0,L),\\\\[6.0pt]
y(\cdot,0)=y_{x}(\cdot,0)=y(\cdot,L)=y_{xx}(\cdot,L)=0&\mbox{ in
}(0,T).\end{array}\right.$

There exists some $c\in\mathbb{R}$ such that

(5.17) $y=c\Phi\mbox{ in }(0,T)\times(0,L).$

###### Proof.

Set

(5.18) ${\mathcal{V}}=\Big{\\{}\varphi\in
L^{2}(0,T)\cap\mathcal{M}_{D}^{\perp};\exists\,y\in X_{T}\mbox{ satisfying
}\eqref{lem-Phi-st1}\mbox{ and }y(0,\cdot)=\varphi\Big{\\}}\subset
L^{2}(0,L).$

We claim that

(5.19) ${\mathcal{V}}=\\{0\\}.$

Admitting the claim, the conclusion now follows from the claim as follows. Set

$\varphi=\mbox{proj}_{\mathcal{M}_{D}}y(0,\cdot).$

Then

$\varphi=c\phi\mbox{ for some $c\in\mathbb{R}$}.$

Set

$\widetilde{y}=y-c\Phi\mbox{ in }(0,T)\times(0,L),$

where $\Phi$ is defined by (1.10). Then $\widetilde{y}\in X_{T}$ is a solution
of (5.16) and $\widetilde{y}(0,\cdot)\in\mathcal{M}_{D}^{\perp}$. By (5.19),
one has

$\widetilde{y}(0,\cdot)=0.$

It follows that

$y=c\Phi\mbox{ in }(0,T)\times(0,L),$

which is the conclusion.

It remains to prove (5.19). We will prove that ${\mathcal{V}}=\\{0\\}$ by
contradiction. Assume that ${\mathcal{V}}\neq\\{0\\}$. Using the regularity
theory for the linear KdV equation and Proposition 3.2, one can show that

${\mathcal{V}}\subset C^{\infty}([0,L]).$

For the same reason, one can show that any bounded sequence in ${\mathcal{V}}$
(equipped $L^{2}(0,L)$-norm) has a subsequence converging in ${\mathcal{V}}$.
Thus ${\mathcal{V}}$ is of finite dimension and is not $\\{0\\}$.

We can now involve the arguments via spectral theory in the spirit of [23] or
even simpler (see also [1, 38]) to show that there exists
$\varphi\in{\mathcal{V}}\setminus\\{0\\}$ and $\lambda\in\mathbb{C}$ such that

(5.20) $\lambda\varphi+\varphi_{x}+\varphi_{xxx}=0\mbox{ in }(0,L),$ (5.21)
$\varphi(\cdot,0)=\varphi(\cdot,0)=\varphi_{x}(\cdot,L)=\varphi_{xx}(\cdot,L)=0.$

Indeed, this can be done by considering

(5.22)
$\begin{array}[]{ccc}{\mathcal{A}}:{\mathcal{V}}&\to&{\mathcal{V}}\\\\[6.0pt]
\psi&\mapsto&-(\psi_{x}+\psi_{xxx})\end{array}$

and taking $\lambda\in\mathbb{C}$ and
$\varphi\in{\mathcal{V}}\setminus\\{0\\}$ such that
${\mathcal{A}}\varphi=\lambda\varphi$. The only point required to be checked
is the fact that $\psi_{x}+\psi_{xxx}\in{\mathcal{V}}$ for
$\psi\in{\mathcal{V}}$. To this end, one just notes that
$-(\psi_{x}+\psi_{xxx})=y_{t}(0,\cdot)$ where $y\in X_{T}$ is the
corresponding solution (thus $y(0,\cdot)=\psi$).

Let $\eta_{j}$ with $j=1,2,3$ be the three solutions of the equations

$\eta^{3}+\eta+\lambda=0.$

Then $\varphi$ has the form

$\varphi=\gamma_{1}e^{\eta_{1}x}+\gamma_{2}e^{\eta_{2}x}+\gamma_{3}e^{\eta_{3}x}\mbox{
in }[0,L].$

Using the boundary conditions of $\varphi$, we obtain

$\left\\{\begin{array}[]{c}\gamma_{1}+\gamma_{2}+\gamma_{3}=0,\\\\[6.0pt]
\gamma_{1}\eta_{1}+\gamma_{2}\eta_{2}+\gamma_{3}\eta_{3}=0,\\\\[6.0pt]
\gamma_{1}e^{\eta_{1}L}+\gamma_{2}e^{\eta_{2}L}+\gamma_{3}e^{\eta_{3}L}=0,\\\\[6.0pt]
\gamma_{1}\eta_{1}^{2}e^{\eta_{1}L}+\gamma_{2}\eta_{2}^{2}e^{\eta_{2}L}+\gamma_{3}\eta_{3}^{2}e^{\eta_{3}L}=0.\end{array}\right.$

As in [23, (30)], we then derive that

$\eta_{1}e^{-\eta_{1}L}=\eta_{2}e^{-\eta_{2}L}=\eta_{3}e^{-\eta_{3}L}.$

Without loss of generality, one can assume that

$-\eta_{1}L=z_{1}\quad\mbox{ and }\quad-\eta_{2}L=z_{2},$

(here $z_{1}$ and $z_{2}$ are the complex numbers in the definition of
${\mathcal{N}}_{D}$) which yields

$\varphi=c\phi$

for some constant $c\in\mathbb{C}$. In other words,

$\varphi\in\mathcal{M}_{D}.$

We have a contradiction since $\varphi\neq 0$ and
$\varphi\in\mathcal{M}_{D}^{\perp}$.

The proof is complete. ∎

By the same arguments used in the proof of Lemma 5.2, we also have the
following observability inequality in the case $L\not\in{\mathcal{N}}_{D}$,
which is the key point of the proof of Theorem 1.4.

######  Lemma 5.4.

Let $L\not\in{\mathcal{N}}_{D}$, $T>0$, and let $y\in X_{T}$ be a solution of
the linearized KdV equation with $y(0,\cdot)\in L^{2}(0,L)$ and
$y(\cdot,0)=y(\cdot,L)=y_{x}(\cdot,0)=0$. Then, for some $\Lambda\geq 1$,

(5.23)
$\Lambda^{-1}\|y(0,\cdot)\|_{L^{2}(0,L)}\leq\|y_{xx}(\cdot,0)\|_{[H^{1/3}(0,T)]^{*}}\leq\Lambda\|y(0,\cdot)\|_{L^{2}(0,L)}.$

Here is a variant of Proposition 5.1 in the case $L\not\in{\mathcal{N}}_{D}$.

######  Proposition 5.2.

Let $L\not\in{\mathcal{N}}_{D}$ and $T>0$. Then

a) There exists a linear continuous operator $\mathcal{L}:L^{2}(0,L)\to
H^{1/3}(0,T)$ such that $y(T,\cdot)=\varphi$ where $y\in X_{T}$ is the unique
solution of (5.1) with $y(0,\cdot)=0$ and $u=\mathcal{L}(\varphi)$.

b) There exists a linear continuous operator $\hat{\mathcal{L}}:L^{2}(0,L)\to
H^{1/3}(0,T)$ such that $y(T,\cdot)=0$ where $y\in X_{T}$ is the unique
solution of (5.1) with $y(0,\cdot)=\varphi$ and
$u=\hat{\mathcal{L}}(\varphi)$.

We are ready to give

###### Proof of Theorem 1.4.

Theorem 1.4 follows from Proposition 3.4 and Lemma 5.4 as usual. The details
are omitted. ∎

##  6\. Properties of controls which steer 0 at time $0$ to $0$ at time $T$

In this section, we study controls that steer 0 at time $0$ to $0$ at time $T$
for the linearized KdV system of (1.1). To this end, it is convenient to
introduce the following quantities.

######  Definition 6.1.

For $z\in\mathbb{C}$, let $(\lambda_{j})_{1\leq j\leq
3}=\big{(}\lambda_{j}(z)\big{)}_{1\leq j\leq 3}$ be the three solutions of

(6.1) $\lambda^{3}+\lambda+iz=0.$

Set

(6.2) $Q=Q(z):=\begin{pmatrix}1&1&1\\\
e^{\lambda_{1}L}&e^{\lambda_{2}L}&e^{\lambda_{3}L}\\\
\lambda_{1}e^{\lambda_{1}L}&\lambda_{2}e^{\lambda_{2}L}&\lambda_{3}e^{\lambda_{3}L}\end{pmatrix},$
(6.3)
$P_{D}=P_{D}(z):=\sum_{j=1}^{3}\lambda_{j}^{2}\big{(}\lambda_{j+1}e^{\lambda_{j+1}L}-\lambda_{j+2}e^{\lambda_{j+2}L}\big{)},$

and

(6.4) $\Xi=\Xi(z):=\det\begin{pmatrix}1&1&1\\\
\lambda_{1}&\lambda_{2}&\lambda_{3}\\\
\lambda_{1}^{2}&\lambda_{2}^{2}&\lambda_{3}^{2}\end{pmatrix},$

with the convention $\lambda_{j+3}=\lambda_{j}$ for $j\geq 1$.

######  Remark 6.1.

The matrix $Q$ and the quantities $P_{D}$ and $\Xi$ are antisymmetric with
respect to $\lambda_{j}$ ($j=1,2,3$), and their definitions depend on a choice
of the order of $(\lambda_{1},\lambda_{2},\lambda_{3})$. Nevertheless, we
later consider a product of either $P_{D}$, $\Xi$, or $\det Q$ with another
antisymmetric function of $(\lambda_{j})$, or deal with $|\det Q|$, and these
quantities therefore make sense. The definitions of $P$, $\Xi$, and $Q$ are
only understood in these contexts.

Given $u\in H^{1/3}(0,+\infty)$, let $y\in C([0,+\infty);L^{2}(0,L))\cap
L^{2}_{\operatorname{loc}}([0,+\infty);H^{1}(0,L))$ be the unique solution of
the system

(6.5) $\left\\{\begin{array}[]{cl}y_{t}+y_{x}+y_{xxx}=0&\mbox{ in
}(0,+\infty)\times(0,L),\\\\[6.0pt] y(\cdot,0)=y_{x}(\cdot,L)=0&\mbox{ in
}(0,+\infty),\\\\[6.0pt] y(\cdot,L)=u&\mbox{ in }(0,+\infty),\\\\[6.0pt]
y(0,\cdot)=0&\mbox{ in }(0,L).\end{array}\right.$

In what follows in this section, we extend $y$ and $u$ by $0$ for $t<0$ and
still denote these extensions by $y$ and $u$, respectively. For an appropriate
function $v$ defined on $\mathbb{R}\times(0,L)$, let $\hat{v}$ denote its
Fourier transform with respect to $t$, i.e.,

$\hat{v}(z,x)=\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}v(t,x)e^{-izt}\,dt.$

From the system of $y$, we have

(6.6)
$\left\\{\begin{array}[]{cl}iz\hat{y}(z,x)+\hat{y}_{x}(z,x)+\hat{y}_{xxx}(z,x)=0&\mbox{
in }\mathbb{R}\times(0,L),\\\\[6.0pt] \hat{y}(z,0)=\hat{y}_{x}(z,L)=0&\mbox{
in }\mathbb{R},\\\\[6.0pt] \hat{y}(z,L)=\hat{u}(z)&\mbox{ in
}\mathbb{R}.\end{array}\right.$

This system has a unique solution outside a discrete set of $z$ in
$\mathbb{C}$, see [20, Lemma 2.1].

Taking into account the equation of $\hat{y}$, we search for the solution of
the form

$\hat{y}(z,\cdot)=\sum_{j=1}^{3}a_{j}e^{\lambda_{j}x},$

where $\lambda_{j}=\lambda_{j}(z)$ with $j=1,2,3$ determined by (6.1).

Using the boundary conditions for $\hat{y}$, we require that

$\left\\{\begin{array}[]{cl}\sum_{j=1}^{3}a_{j}=0,\\\\[6.0pt]
\sum_{j=1}^{3}e^{\lambda_{j}L}a_{j}=\hat{u},\\\\[6.0pt]
\sum_{j=1}^{3}\lambda_{j}e^{\lambda_{j}L}a_{j}=0.\end{array}\right.$

This implies

$Q(a_{1},a_{2},a_{3})^{\mathsf{T}}=(0,\hat{u},0)^{\mathsf{T}},$

where $Q=Q(z)$ is given in Definition 6.1. We thus obtain

$a_{j}=\frac{\hat{u}}{\det
Q}\big{(}\lambda_{j+1}e^{\lambda_{j+1}L}-\lambda_{j+2}e^{\lambda_{j+2}L}\big{)}.$

This yields

(6.7) $\hat{y}(z,x)=\frac{\hat{u}}{\det
Q}\sum_{j=1}^{3}\big{(}\lambda_{j+1}e^{\lambda_{j+1}L}-\lambda_{j+2}e^{\lambda_{j+2}L}\big{)}e^{\lambda_{j}x}.$

From (6.7), we derive that

(6.8) $\partial_{xx}\hat{y}(z,0)=\frac{\hat{u}(z)P_{D}(z)}{\det Q(z)},$

where $P_{D}(z)$ is given in Definition 6.1.

Set

(6.9) $G(z)=P_{D}(z)/\Xi(z)\quad\mbox{ and }\quad H(z)=\det Q(z)/\Xi(z),$

where $\Xi(z)$ is given in Definition 6.1. It is convenient to consider
$\partial_{xx}\hat{y}(z,0)$ under the form

(6.10) $\partial_{xx}\hat{y}(z,0)=\frac{\hat{u}(z)G(z)}{H(z)},$

By [20, Lemmas A1 and B1], $\partial_{xx}\hat{y}(z,0)$ is a meromorphic
function, and

(6.11) $G(z)$ and $H(z)$ are entire functions.

We thus have just established the following result.

######  Lemma 6.1.

Let $u\in H^{1/3}(\mathbb{R}_{+})$ and let $y\in
C\big{(}[0,+\infty);L^{2}(0,L)\big{)}\cap
L^{2}_{\operatorname{loc}}\big{(}[0,+\infty);H^{1}(0,L)\big{)}$ be the unique
solution of

(6.12) $\left\\{\begin{array}[]{cl}y_{t}+y_{x}+y_{xxx}=0&\mbox{ in
}(0,+\infty)\times(0,L),\\\\[6.0pt] y(\cdot,0)=y_{x}(\cdot,L)=0&\mbox{ in
}(0,+\infty),\\\\[6.0pt] y(\cdot,L)=u&\mbox{ in
}(0,+\infty),\end{array}\right.$

with

(6.13) $y(0,\cdot)=0\mbox{ in }(0,L).$

Outside a discrete set $z\in\mathbb{R}$, we have

(6.14) $\hat{y}(z,x)=\frac{\hat{u}}{\det
Q}\sum_{j=1}^{3}\big{(}\lambda_{j+1}e^{\lambda_{j+1}L}-\lambda_{j+2}e^{\lambda_{j+2}L}\big{)}e^{\lambda_{j}x}\mbox{
for a.e. }x\in(0,L).$

######  Remark 6.2.

Assume that $\hat{u}(z,\cdot)$ is well-defined for $z\in\mathbb{C}$ (e.g. when
$u$ has a compact support). Then the conclusions of Lemma 6.1 hold outside of
a discrete set $z\in\mathbb{C}$.

We end this section with the following result, which is the starting point of
our approach, and follows from Lemma 6.1 and Paley-Wiener’s theorem, see e.g.,
[41].

######  Proposition 6.1.

Let $L>0$, $T>0$, and $u\in H^{1/3}(\mathbb{R}_{+})$. Assume that $u$ has a
compact support in $[0,T]$, and $u$ steers $0$ at the time $0$ to $0$ at the
time $T$, i.e., the unique solution $y$ of (6.12) and (6.13) satisfies
$y(T,\cdot)=0$ in $(0,L)$. Then $\hat{u}$ and $\hat{u}G/H$ satisfy the
assumption of Paley-Wiener’s theorem concerning the support in $[-T,T]$, i.e.,

(6.15) $\hat{u}\mbox{ and }\hat{u}G/H\mbox{ are entire functions},$

and

(6.16) $|\hat{u}(z)|+\left|\frac{\hat{u}G(z)}{H(z)}\right|\leq Ce^{T|z|},$

for some positive constant $C$.

######  Remark 6.3.

The computations in this section are in the spirit of the ones [20].
Nevertheless, in the conclusions of Proposition 6.1, we have/require that

$\hat{y}(z,L)\mbox{ and }\partial_{xx}\hat{y}(z,0)\mbox{ are entire functions
satisfying \eqref{pro-Gen-cl2}}.$

This is different with the one used in [20, Proposition 3.1] where one obtains
that

$\partial_{x}\hat{y}(z,L)\mbox{ and }\partial_{x}\hat{y}(z,0)\mbox{ are entire
functions satisfying a variant of \eqref{pro-Gen-cl2}}.$

These differences are important to take into account different boundary
conditions, see the proof of Assertion $ii)$ of Theorem 1.3 in Section 9.2.

##  7\. Attainable directions in the unreachable space in small time

In this section, we investigate whether or not directions in
$\mathcal{M}_{D}$, defined in (1.12), can be reached in small time. The
starting point comes from the power series expansion approach. Let $y_{1}$ and
$y_{2}$ be the solutions of

(7.1) $\left\\{\begin{array}[]{cl}y_{1,t}+y_{1,x}+y_{1,xxx}=0&\mbox{ in
}(0,T)\times(0,L),\\\\[6.0pt] y_{1}(\cdot,0)=y_{1,x}(\cdot,L)=0&\mbox{ in
}(0,T),\\\\[6.0pt] y_{1}(\cdot,L)=u_{1}&\mbox{ in }(0,T),\\\\[6.0pt]
y_{1}(0,\cdot)=0&\mbox{ in }(0,L),\end{array}\right.$ (7.2)
$\left\\{\begin{array}[]{cl}y_{2,t}+y_{2,x}+y_{2,xxx}+y_{1}y_{1,x}=0&\mbox{ in
}(0,T)\times(0,L),\\\\[6.0pt]
y_{2}(\cdot,0)=y_{2}(\cdot,L)=y_{2,x}(\cdot,L)=0&\mbox{ in }(0,T),\\\\[6.0pt]
y_{2}(0,\cdot)=0&\mbox{ in }(0,L),\end{array}\right.$

for some control $u_{1}$. The key point of this approach is to first
understand how one can choose $u_{1}$ so that

$y_{1}(T,\cdot)=0$

and then analyse what the behavior of
$\mbox{proj}_{\mathcal{M}_{D}}y_{2}(T,\cdot)$ is. To this end, we compute the
quantity

(7.3) $\int_{0}^{L}y_{2}(T,x)\Phi(t,x)\,dx,$

where $\Phi$ is defined in (1.10). Multiplying the equation of $y_{2}$ by
$\Phi$, integrating by parts in $(0,T)\times(0,L)$, we obtain, after using the
boundary conditions and the initial conditions,

(7.4)
$\int_{0}^{L}y_{2}(T,x)\Phi(t,x)\,dx=\frac{1}{2}\int_{0}^{T}\int_{0}^{L}y_{1}^{2}(t,x)\Phi_{x}(t,x)\,dx.$

The goal is then to understand the value of the RHS of (7.4).

We will study the value of the RHS of (7.4) in a more general setting.
Motivated by the definition of $y_{1}$, we consider the unique solution $y\in
X_{T}$ of the system, for $u\in H^{1/3}(0,T)$,

(7.5) $\left\\{\begin{array}[]{cl}y_{t}+y_{x}+y_{xxx}=0&\mbox{ in
}(0,T)\times(0,L),\\\\[6.0pt] y(\cdot,0)=y_{x}(\cdot,L)=0&\mbox{ in
}(0,T),\\\\[6.0pt] y(\cdot,L)=u&\mbox{ in }(0,T),\\\\[6.0pt]
y(0,\cdot)=0&\mbox{ in }(0,L).\end{array}\right.$

Guided by the definition of $\phi$, as suggested in [20, 35], for
$\eta_{1},\,\eta_{2},\,\eta_{3}\in\mathbb{C}\setminus\\{0\\}$, we set

(7.6) $\varphi(x)=\sum_{j=1}^{3}(\eta_{j+1}-\eta_{j})e^{\eta_{j+2}x}\mbox{ for
}x\in[0,L],$

with the convention $\eta_{j+3}=\eta_{j}$ for $j\geq 1$. The following
assumptions on $\eta_{j}$ are used repeatedly throughout this section:

(7.7)
$\eta_{1}+\eta_{2}+\eta_{3}=0,\quad\eta_{1}\eta_{2}+\eta_{1}\eta_{3}+\eta_{2}\eta_{3}=1,$

and

(7.8) $\eta_{1}e^{-\eta_{1}L}=\eta_{2}e^{-\eta_{2}L}=\eta_{3}e^{-\eta_{3}L}.$

Extend $y$ and $u$ by $0$ for $t>T$ and still denote these extensions by $y$
and $u$, respectively. Then, by Lemma A.1 in the appendix,

$\|u\|_{H^{1/3}(\mathbb{R}_{+})}\leq C\|u\|_{H^{1/3}(0,T)}.$

Assume that

$y(T,\cdot)=0.$

Then the extension $y\in C([0,+\infty);L^{2}(0,L))\cap
L^{2}((0,+\infty);H^{1}(0,L))$ is also a solution of the linearized KdV system
in $[0,+\infty)\times(0,L)$ using the control which is the extension of $u$
(by $0$ outside $(0,T)$), i.e.,

(7.9) $\left\\{\begin{array}[]{cl}y_{t}+y_{x}+y_{xxx}=0&\mbox{ in
}(0,+\infty)\times(0,L),\\\\[6.0pt] y(\cdot,0)=y_{x}(\cdot,L)=0&\mbox{ in
}(0,+\infty),\\\\[6.0pt] y(\cdot,L)=u&\mbox{ in }(0,+\infty),\\\\[6.0pt]
y(0,\cdot)=0&\mbox{ in }(0,L).\end{array}\right.$

In what follows in this section, we study this quantity, for $p\in\mathbb{R}$:

(7.10)
$\int_{0}^{T}\int_{0}^{L}y^{2}(t,x)\varphi_{x}(x)e^{pt}\,dt\,dx\quad\left(=\int_{0}^{+\infty}\int_{0}^{L}y^{2}(t,x)\varphi_{x}(x)e^{pt}\,dt\,dx\right).$

We have, by Lemma 6.1 (see also Remark 6.2), for $z\in\mathbb{C}$ outside a
discrete set,

(7.11)
$\hat{y}(z,x)=\hat{u}(z)\frac{\sum_{j=1}^{3}\big{(}\lambda_{j}e^{\lambda_{j}L}-\lambda_{j+1}e^{\lambda_{j+1}L}\big{)}e^{\lambda_{j+2}x}}{\sum_{j=1}^{3}(\lambda_{j+1}-\lambda_{j})e^{-\lambda_{j+2}L}},$

where $\lambda_{j}=\lambda_{j}(z)$ for $j=1,\,2,\,3$ are determined by (6.1).

We begin with

######  Lemma 7.1.

Let $p\in\mathbb{R}$,
$\eta_{1},\,\eta_{2},\,\eta_{3}\in\mathbb{C}\setminus\\{0\\}$, and let
$\varphi$ be defined by (7.6). Set, for $(z,x)\in\mathbb{R}\times(0,L)$,

$B_{D}(z,x)=\left|\frac{\sum_{j=1}^{3}\big{(}\lambda_{j}e^{\lambda_{j}L}-\lambda_{j+1}e^{\lambda_{j+1}L}\big{)}e^{\lambda_{j+2}x}}{\sum_{j=1}^{3}(\lambda_{j+1}-\lambda_{j})e^{-\lambda_{j+2}L}}\right|^{2}\varphi_{x}(x),$

where $\lambda_{j}=\lambda_{j}(z)$ with $j=1,2,3$ are the three solutions of
$\lambda^{3}+\lambda+iz=0$. Let $u\in H^{1/3}(0,+\infty)$ with compact support
in $[0,+\infty)$ and let $y\in C\big{(}[0,+\infty);L^{2}(0,L)\big{)}\cap
L^{2}_{\operatorname{loc}}\big{(}[0,+\infty);H^{1}(0,L)\big{)}$ be the unique
solution of (7.9). Assume that $y(t,\cdot)=0$ for large $t$. We have

(7.12)
$\int_{0}^{L}\int_{0}^{+\infty}|y(t,x)|^{2}\varphi_{x}(x)e^{pt}\,dt\,dx=\int_{\mathbb{R}}|\hat{u}(z+ip/2)|^{2}\int_{0}^{L}B_{D}(z+ip/2,x)\,dx\,dz.$

###### Proof.

The conclusion is a direct consequence of Parseval’s theorem and (7.11). ∎

We next investigate the behavior of

$\int_{0}^{L}B_{D}(zn+ip/2,x)\,dx$

for $z\in\mathbb{R}$ with large $|z|$. We have

######  Lemma 7.2.

Let $p\in\mathbb{R}$ and
$\eta_{1},\eta_{2},\eta_{3}\in\mathbb{C}\setminus\\{0\\}$. Assume (7.8). We
have

(7.13) $\int_{0}^{L}B_{D}(z+ip/2,x)\,dx=E_{D}|z|^{-1/3}+O(|z|^{-2/3})\mbox{
for $z\in\mathbb{R}$ with large $|z|$},$

where $E_{D}=E_{D}(\eta_{1},\eta_{2},\eta_{3})$ is defined by

(7.14)
$E_{D}=\frac{1}{\sqrt{3}A}\sum_{j=1}^{3}\eta_{j+2}^{2}(\eta_{j+1}-\eta_{j}),$

with 777$A$ does not depend on $j$ by (7.8).

$A=A(\eta_{1},\eta_{2},\eta_{3}):=\eta_{j}e^{-\eta_{j}L}.$

Here and in what follows, for $s\in\mathbb{R}$, $O(z^{s})$ denotes a quantity
bounded by $Cz^{s}$ for large positive $z$. Similar convention is used for
$O(|z|^{s})$ for $z\in\mathbb{C}$.

###### Proof.

We first consider the case where $z$ is positive and large. We use the
following convention $\Re(\lambda_{1})<\Re(\lambda_{2})<\Re(\lambda_{3})$.

We first look at the denominator of $B_{D}(z+ip/2,x)$. We have, by Lemma 3.1,
at $(z+ip/2,x)$,

(7.15)
$\frac{1}{\sum_{j=1}^{3}(\lambda_{j+1}-\lambda_{j})e^{-\lambda_{j+2}L}}\cdot\frac{1}{\sum_{j=1}^{3}(\bar{\lambda}_{j+1}-\bar{\lambda}_{j})e^{-\bar{\lambda}_{j+2}L}}\\\\[6.0pt]
=\frac{e^{\lambda_{1}L}e^{\bar{\lambda}_{1}L}}{(\lambda_{3}-\lambda_{2})(\bar{\lambda}_{3}-\bar{\lambda}_{2})}\Big{(}1+O\big{(}e^{-C|z|^{1/3}}\big{)}\Big{)}.$

We next deal with the numerator of $B_{D}(z+ip/2,x)$. Set, for
$(z,x)\in\mathbb{R}\times(0,L)$,

(7.16)
$f(z,x)=\sum_{j=1}^{3}(\lambda_{j}e^{\lambda_{j}L}-\lambda_{j+1}e^{\lambda_{j+1}L})e^{\lambda_{j+2}x},\quad
g(z,x)=\sum_{j=1}^{3}(\bar{\lambda}_{j}e^{\bar{\lambda}_{j}L}-\bar{\lambda}_{j+1}e^{\bar{\lambda}_{j+1}L})e^{\bar{\lambda}_{j+2}x},$

and 888The index $m$ stands the main part.

$f_{m}(z,x)=\lambda_{3}e^{\lambda_{3}L}e^{\lambda_{2}x}-\lambda_{2}e^{\lambda_{2}L}e^{\lambda_{3}x}-\lambda_{3}e^{\lambda_{3}L}e^{\lambda_{1}x},\quad
g_{m}(z,x)=\bar{\lambda}_{3}e^{\bar{\lambda}_{3}L}e^{\bar{\lambda}_{2}x}-\bar{\lambda}_{2}e^{\bar{\lambda}_{2}L}e^{\bar{\lambda}_{3}x}-\bar{\lambda}_{3}e^{\bar{\lambda}_{3}L}e^{\bar{\lambda}_{1}x}.$

We have

$\int_{0}^{L}f(z+ip/2,x)g(z+ip/2,x)\varphi_{x}(x)\,dx\\\\[6.0pt]
=\int_{0}^{L}f_{m}(z+ip/2,x)g_{m}(z+ip/2,x)\varphi_{x}(x)\,dx+\int_{0}^{L}(f-f_{m})(z+ip/2,x)g_{m}(z+ip/2,x)\varphi_{x}(x)\,dx\\\\[6.0pt]
+\int_{0}^{L}f_{m}(z+ip/2,x)(g-g_{m})(z+ip/2,x)\varphi_{x}(x)\,dx+\int_{0}^{L}(f-f_{m})(z+ip/2,x)(g-g_{m})(z+ip/2,x)\varphi_{x}(x)\,dx.$

By Lemma 3.1, we have

(7.17)
$\int_{0}^{L}|(f-f_{m})(z+ip/2,x)g_{m}(z+ip/2,x)\varphi_{x}(x)|\,dx\\\\[6.0pt]
+\int_{0}^{L}|(f-f_{m})(z+ip/2,x)(g-g_{m})(z+ip/2,x)\varphi_{x}(x)|\,dx\\\\[6.0pt]
+\int_{0}^{L}|f_{m}(z+ip/2,x)(g-g_{m})(z+ip/2,x)\varphi_{x}(x)|\,dx\leq
C|e^{(\lambda_{3}+\bar{\lambda}_{3})L}|e^{-C|z|^{1/3}}.$

We next estimate

(7.18) $\int_{0}^{L}f_{m}(z+ip/2,x)g_{m}(z+ip/2,x)\varphi_{x}(x)\\\\[6.0pt]
=\int_{0}^{L}f_{m}(z+ip/2,x)g_{m}(z+ip/2,x)\left(\sum_{j=1}^{3}\eta_{j+2}(\eta_{j+1}-\eta_{j})e^{\eta_{j+2}x}\right)\,dx.$

We first have, by Lemma 3.1, at $z+ip/2$,

(7.19)
$\int_{0}^{L}\Big{(}-\lambda_{3}e^{\lambda_{3}L}e^{\lambda_{2}x}\bar{\lambda}_{2}e^{\bar{\lambda}_{2}L}e^{\bar{\lambda}_{3}x}-\lambda_{2}e^{\lambda_{2}L}e^{\lambda_{3}x}\bar{\lambda}_{3}e^{\bar{\lambda}_{3}L}e^{\bar{\lambda}_{2}x}+\lambda_{2}e^{\lambda_{2}L}e^{\lambda_{3}x}\bar{\lambda}_{2}e^{\bar{\lambda}_{2}L}e^{\bar{\lambda}_{3}x}\Big{)}\\\\[6.0pt]
\times\left(\sum_{j=1}^{3}\eta_{j+2}(\eta_{j+1}-\eta_{j})e^{\eta_{j+2}x}\right)\,dx\mathop{=}^{\eqref{pro-
eta1}}A^{-1}e^{(\lambda_{3}+\bar{\lambda}_{3}+\lambda_{2}+\bar{\lambda}_{2})L}\Big{(}S_{1}(z+ip/2)+O\big{(}e^{-C|z|^{1/3}}\big{)}\Big{)},$

where

(7.20)
$S_{1}(z):=\sum_{j=1}^{3}\eta_{j+2}^{2}(\eta_{j+1}-\eta_{j})\left(\frac{\lambda_{2}\bar{\lambda}_{2}}{\lambda_{3}+\bar{\lambda}_{3}+\eta_{j+2}}-\frac{\lambda_{2}\bar{\lambda}_{3}}{\lambda_{3}+\bar{\lambda}_{2}+\eta_{j+2}}-\frac{\lambda_{3}\bar{\lambda}_{2}}{\lambda_{2}+\bar{\lambda}_{3}+\eta_{j+2}}\right)(z).$

We next obtain, by Lemma 3.1, at $z+ip/2$,

(7.21)
$\int_{0}^{L}\Big{(}\lambda_{3}e^{\lambda_{3}L}e^{\lambda_{1}x}\bar{\lambda}_{3}e^{\bar{\lambda}_{3}L}e^{\bar{\lambda}_{1}x}-\lambda_{3}e^{\lambda_{3}L}e^{\lambda_{1}x}\bar{\lambda}_{3}e^{\bar{\lambda}_{3}L}e^{\bar{\lambda}_{2}x}-\lambda_{3}e^{\lambda_{3}L}e^{\lambda_{2}x}\bar{\lambda}_{3}e^{\bar{\lambda}_{3}L}e^{\bar{\lambda}_{1}x}\Big{)}\\\\[6.0pt]
\times\left(\sum_{j=1}^{3}\eta_{j+2}(\eta_{j+1}-\eta_{j})e^{\eta_{j+2}x}\right)\,dx=e^{(\lambda_{3}+\bar{\lambda}_{3})L}\Big{(}S_{2}(z+ip/2)+O(e^{-C|z|^{1/3}})\Big{)},$

where

(7.22)
$S_{2}(z):=\sum_{j=1}^{3}\eta_{j+2}(\eta_{j+1}-\eta_{j})\left(-\frac{\lambda_{3}\bar{\lambda}_{3}}{\lambda_{1}+\bar{\lambda}_{1}+\eta_{j+2}}+\frac{\lambda_{3}\bar{\lambda}_{3}}{\lambda_{1}+\bar{\lambda}_{2}+\eta_{j+2}}+\frac{\lambda_{3}\bar{\lambda}_{3}}{\lambda_{2}+\bar{\lambda}_{1}+\eta_{j+2}}\right)(z).$

We have, at $z+ip/2$,

(7.23)
$\int_{0}^{L}\lambda_{3}e^{\lambda_{3}L}e^{\lambda_{2}x}\bar{\lambda}_{3}e^{\bar{\lambda}_{3}L}e^{\bar{\lambda}_{2}x}\left(\sum_{j=1}^{3}\eta_{j+2}(\eta_{j+1}-\eta_{j})e^{\eta_{j+2}x}\right)\,dx=e^{(\lambda_{3}+\bar{\lambda}_{3})L}S_{3}(z+ip/2),$

where

(7.24)
$S_{3}(z):=\sum_{j=1}^{3}\frac{\eta_{j+2}(\eta_{j+1}-\eta_{j})\lambda_{3}\bar{\lambda}_{3}\Big{(}e^{\lambda_{2}L+\bar{\lambda}_{2}L+\eta_{j+2}L}-1\Big{)}}{\lambda_{2}+\bar{\lambda}_{2}+\eta_{j+2}}(z).$

We finally get, by Lemma 3.1, at $z+ip/2$,

(7.25)
$\left|\int_{0}^{L}\Big{(}\lambda_{3}e^{\lambda_{3}L}e^{\lambda_{1}x}\bar{\lambda}_{2}e^{\bar{\lambda}_{2}L}e^{\bar{\lambda}_{3}x}+\lambda_{2}e^{\lambda_{2}L}e^{\lambda_{3}x}\bar{\lambda}_{3}e^{\bar{\lambda}_{3}L}e^{\bar{\lambda}_{1}x}\Big{)}\Big{(}\sum_{j=1}^{3}\eta_{j+2}(\eta_{j+1}-\eta_{j})e^{\eta_{j+2}x}\Big{)}\,dx\right|\\\\[6.0pt]
=|e^{(\lambda_{3}+\bar{\lambda}_{3})L}|O(e^{-Cz^{1/3}}).$

By Lemma 3.1, we have, at $z+ip/2$,

(7.26)
$\left\\{\begin{array}[]{c}\lambda_{1}+\bar{\lambda}_{1}+\lambda_{2}+\bar{\lambda}_{2}+\lambda_{3}+\bar{\lambda}_{3}=O(z^{-1/3}),\\\\[6.0pt]
\lambda_{1}+\bar{\lambda}_{1}+\lambda_{3}+\bar{\lambda}_{3}=O(z^{-1/3}),\\\\[6.0pt]
(\lambda_{3}-\lambda_{2})(\bar{\lambda}_{3}-\bar{\lambda}_{2})=3z^{2/3}(1+O(z^{-1/3})).\end{array}\right.$

We claim that

(7.27) $|z|^{-1/3}|S_{1}(z+ip/2)|+|S_{2}(z+ip/2)|+|S_{3}(z+ip/2)|=O(1)\mbox{
for large positive $z$}.$

Admitting this, by combining (7.15), (7.19), (7.21), (7.23), (7.25), and
(7.26), and using (7.27), we obtain

(7.28)
$\int_{0}^{L}B_{D}(z+ip/2,x)\,dx=\frac{A^{-1}S_{1}(z+ip/2)}{3z^{2/3}}+O(|z|^{-2/3}).$

We first derive the the asymptotic behavior of $S_{1}(z+ip/2)$. We have, by
Lemma 3.1, at $z+ip/2$,

(7.29)
$\lambda_{2}\bar{\lambda}_{2}=z^{2/3}+O(1),\quad\lambda_{2}\bar{\lambda}_{3}=z^{2/3}e^{i\pi/3}+O(1),\quad\lambda_{3}\bar{\lambda}_{2}=z^{2/3}e^{-i\pi/3}+O(1),$

and

(7.30)
$\frac{1}{\lambda_{3}+\bar{\lambda}_{3}+\eta_{j+2}}=\frac{1+O(z^{-1/3})}{\sqrt{3}z^{1/3}},\quad\frac{1}{\lambda_{2}+\bar{\lambda}_{3}+\eta_{j+2}}=\frac{1+O(z^{-1/3})}{(e^{i\pi/6}+i)z^{1/3}},\\\\[6.0pt]
\quad\frac{1}{\lambda_{3}+\bar{\lambda}_{2}+\eta_{j+2}}=\frac{1+O(z^{-1/3})}{(e^{-i\pi/6}-i)z^{1/3}}.$

It follows that

(7.31)
$S_{1}(z+ip/2)=\left(\frac{1}{\sqrt{3}}+2\Re\frac{e^{i\pi/3}}{e^{i\pi/6}+i}\right)\sum_{j=1}^{3}\eta_{j+2}^{2}(\eta_{j+1}-\eta_{j})|z|^{1/3}+O(1)\\\\[6.0pt]
=\sqrt{3}\sum_{j=1}^{3}\eta_{j+2}^{2}(\eta_{j+1}-\eta_{j})|z|^{1/3}+O(1).$

We next deal with $S_{2}(z+ip/2)$. Since

$\sum_{j=1}^{3}\eta_{j+2}(\eta_{j+1}-\eta_{j})=0,$

it follows that

$S_{2}(z+ip/2)=\sum_{j=1}^{3}\eta_{j+2}(\eta_{j+1}-\eta_{j})\left(-\frac{\lambda_{3}\bar{\lambda}_{3}}{\lambda_{1}+\bar{\lambda}_{1}+\eta_{j+2}}+\frac{\lambda_{3}\bar{\lambda}_{3}}{\lambda_{1}+\bar{\lambda}_{1}}\right)(z+ip/2)\\\\[6.0pt]
+\sum_{j=1}^{3}\eta_{j+2}(\eta_{j+1}-\eta_{j})\left(\frac{\lambda_{3}\bar{\lambda}_{3}}{\lambda_{1}+\bar{\lambda}_{2}+\eta_{j+2}}-\frac{\lambda_{3}\bar{\lambda}_{3}}{\lambda_{1}+\bar{\lambda}_{2}}\right)(z+ip/2)\\\\[6.0pt]
+\sum_{j=1}^{3}\eta_{j+2}(\eta_{j+1}-\eta_{j})\left(\frac{\lambda_{3}\bar{\lambda}_{3}}{\lambda_{2}+\bar{\lambda}_{1}+\eta_{j+2}}-\frac{\lambda_{3}\bar{\lambda}_{3}}{\lambda_{2}+\bar{\lambda}_{1}}\right)(z+ip/2).$

Using Lemma 3.1, we derive that

(7.32) $S_{2}(z+ip/2)=O(1).$

We next derive the asymptotic behavior of $S_{3}(z+ip/2)$. From Lemma 3.1, we
have, at $z+ip/2$,

(7.33) $\lambda_{2}+\bar{\lambda}_{2}=O(z^{-2/3}).$

Using (7.8), we derive from (7.24) that

$S_{3}(z+ip/2)=z^{2/3}\sum_{j=1}^{3}\frac{(\eta_{j+1}-\eta_{j})\eta_{j+2}}{\eta_{j+2}}\Big{(}A^{-1}\eta_{j+2}-1\Big{)}+O(1)\\\\[6.0pt]
=z^{2/3}\sum_{j=1}^{3}(\eta_{j+1}-\eta_{j})\Big{(}A^{-1}\eta_{j+2}-1\Big{)}+O(1).$

Since $\sum_{j=1}^{3}(\eta_{j+1}-\eta_{j})=0$ and
$\sum_{j=1}^{3}(\eta_{j+1}-\eta_{j})\eta_{j+2}=0$, it follows that

(7.34) $S_{3}(z+ip/2)=O(1).$

Combing (7.28) and (7.31) yields

$\int_{0}^{L}B_{D}(z+ip/2,x)\,dz=E_{D}|z|^{-1/3}+O(z^{-2/3}),$

where

(7.35)
$E_{D}=\frac{1}{\sqrt{3}A}\sum_{j=1}^{3}\eta_{j+2}^{2}(\eta_{j+1}-\eta_{j}).$

The conclusion in the case where $z$ is large and negative can be derived from
the case where $z$ is positive and large as follows. Define, for
$(z,x)\in\mathbb{R}\times(0,L)$, with large $|z|$,

$M_{D}(z,x)=\frac{\sum_{j=1}^{3}(\lambda_{j}e^{\lambda_{j}L}-\lambda_{j+1}e^{\lambda_{j+1}L})e^{\lambda_{j+2}x}}{\sum_{j=1}^{3}(\lambda_{j+1}-\lambda_{j})e^{-\lambda_{j+2}L}}.$

Then

$B_{D}(z+ip/2,x)=|M_{D}(z+ip/2,x)|^{2}\varphi_{x}(x).$

It is clear from the definition of $M_{D}$ that

$M_{D}(-z,x)=\overline{M_{D}(\bar{z},x)}.$

We then have

$B_{D}(-z+ip/2,x)=|M_{D}(-z+ip/2,x)|^{2}\varphi_{x}(x)=|M_{D}(z-ip/2,x)|^{2}\varphi_{x}(x).$

We thus obtain the result in the case where $z$ is negative and large by the
corresponding expression for large positive $z$ in which $p$ is replaced by
$-p$. The conclusion follows. ∎

As a consequence of Lemmas 7.1 and 7.2, we obtain

######  Proposition 7.1.

Let $L\in{\mathcal{N}}_{D}$. Let $u\in H^{1/3}(0,+\infty)$ with compact
support in $[0,+\infty)$, and let $y\in C([0,+\infty);L^{2}(0,L))\cap
L^{2}_{\operatorname{loc}}\big{(}[0,+\infty);H^{1}(0,L)\big{)}$ be the unique
solution of (7.9). Assume that $y(t,\cdot)=0$ for large $t$. We have

(7.36)
$\int_{0}^{+\infty}\int_{0}^{L}|y(t,x)|^{2}\Phi_{x}(t,x)\mathop{}\mkern-3.0mu{d}x\mathop{}\mkern-3.0mu{d}t=\int_{\mathbb{R}}|\hat{u}(z+iq/2)|^{2}\int_{0}^{L}B(z+iq/2,x)\,dx\mathop{}\mkern-3.0mu{d}z,$

where

$B(z,x)=\left|\frac{\sum_{j=1}^{3}\big{(}\lambda_{j}e^{\lambda_{j}L}-\lambda_{j+1}e^{\lambda_{j+1}L}\big{)}e^{\lambda_{j+2}x}}{\sum_{j=1}^{3}(\lambda_{j+1}-\lambda_{j})e^{-\lambda_{j+2}L}}\right|^{2}\phi_{x}(x),$

with $\lambda_{j}=\lambda_{j}(z)$ ($j=1,2,3$) being the three solutions of
$\lambda^{3}+\lambda+iz=0$. Moreover,

(7.37) $\int_{0}^{L}B(z+iq/2,x)\,dx=E|z|^{-1/3}+O(|z|^{-2/3}),$

where

$E=-\frac{be^{2a}}{2\sqrt{3}aL^{2}}(b^{2}+9a^{2})>0.$

Recall that $q$, $a$, $b$, $\phi$, and $\Phi$ are defined in Theorem 1.1.

###### Proof.

We will apply Lemmas 7.1 and 7.2 with $p=q$,

$\eta_{1}=\alpha+i\beta,\quad\eta_{2}=\alpha-i\beta,\quad\mbox{ and
}\quad\eta_{3}=-2\alpha,$

where $\alpha$ and $\beta$ are defined in (1.9). Thus

(7.38) $A=\eta_{3}e^{-\eta_{3}L}=-2\alpha e^{2\alpha L}=\frac{2a}{L}e^{-2a}.$

We have

$\displaystyle\sqrt{3}AE_{D}=$
$\displaystyle\eta_{1}^{2}(\eta_{3}-\eta_{2})+\eta_{2}^{2}(\eta_{1}-\eta_{3})+\eta_{3}^{2}(\eta_{2}-\eta_{1})$
$\displaystyle=$
$\displaystyle(\alpha+i\beta)^{2}(-3\alpha+i\beta)+(\alpha-i\beta)^{2}(3\alpha+i\beta)-8i\alpha^{2}\beta$
$\displaystyle=$
$\displaystyle-12i\alpha^{2}\beta+2i\beta(\alpha^{2}-\beta^{2})-8i\alpha^{2}\beta.$

This implies

(7.39)
$\sqrt{3}AE_{D}=-2i(\beta^{2}+9\alpha^{2})\beta=\frac{2ib}{L^{3}}(b^{2}+9a^{2}).$

We obtain

$E_{D}=\frac{ibe^{2a}}{\sqrt{3}aL^{2}}(b^{2}+9a^{2}).$

Since, by Lemma 4.1,

$\varphi=-2i\phi,$

the conclusion now follows from Lemmas 7.1 and 7.2. ∎

Using Proposition 7.1, we derive the following result which is the key
ingredient for the analysis of the local controllability of the KdV system
(1.1) in small time.

######  Proposition 7.2.

Let $L\in{\mathcal{N}}_{D}$ and $T<1$. Let $u\in H^{1/3}(0,+\infty)$ and let
$y\in C([0,+\infty);L^{2}(0,L))\cap
L^{2}_{\operatorname{loc}}\big{(}[0,+\infty);H^{1}(0,L)\big{)}$ be the unique
solution of (7.9). Assume that $u(t)=0$ for $t>T$, and $y(t,\cdot)=0$ for
large $t$. Then

(7.40)
$\int_{0}^{\infty}\int_{0}^{L}|y(t,x)|^{2}\Phi_{x}(t,x)\mathop{}\mkern-3.0mu{d}x\mathop{}\mkern-3.0mu{d}t=E\|ue^{q\cdot/2}\|_{[H^{1/6}(0,+\infty)]^{*}}^{2}\Big{(}1+O(T^{2/9})\Big{)}.$

###### Proof.

We have, by Proposition 7.1 and Lemma B.2,

$|\hat{u}(z+iq/2)|^{2}\left|\int_{0}^{L}B(z+iq/2,x)\,dx\right|\leq\frac{C|\hat{u}(z+iq/2)|^{2}}{1+|z|^{1/3}}.$

Applying Proposition 7.1 again and using the fact
$\int_{\mathbb{R}}\,dz=\int_{z\in\mathbb{R};|z|<m}\,dz+\int_{z\in\mathbb{R};|z|\geq
m}\,dz$, we derive that

(7.41)
$\left|\int_{\mathbb{R}}|\hat{u}(z+iq/2)|^{2}\int_{0}^{L}B(z+iq/2,x)\mathop{}\mkern-3.0mu{d}x\mathop{}\mkern-3.0mu{d}z-E\int_{\mathbb{R}}\frac{|\hat{u}(z+iq/2)|^{2}}{(1+|z|^{2})^{1/6}}\mathop{}\mkern-3.0mu{d}z\right|\\\\[6.0pt]
\leq
C\int_{|z|<m}\frac{|\hat{u}(z+iq/2)|^{2}}{(1+|z|^{2})^{1/6}}+Cm^{-1/3}\int_{|z|>m}\frac{|\hat{u}(z+iq/2)|^{2}}{(1+|z|^{2})^{1/6}}\mathop{}\mkern-3.0mu{d}z.$

Applying Lemma 7.3 below with $s=1/6$, we have

(7.42) $|\hat{u}(z+iq/2)|\leq
C(|z|T^{4/3}+T^{1/3})\|ue^{q\cdot/2}\|_{[H^{1/6}(0,+\infty)]^{*}}.$

Combining (7.41) and (7.42) yields

(7.43)
$\left|\int_{\mathbb{R}}|\hat{u}(z+iq/2)|^{2}\int_{0}^{L}B(z+iq/2,x)\mathop{}\mkern-3.0mu{d}x\mathop{}\mkern-3.0mu{d}z-E\int_{\mathbb{R}}\frac{|\hat{u}(z+iq/2)|^{2}}{(1+|z|^{2})^{1/6}}\mathop{}\mkern-3.0mu{d}z\right|\\\\[6.0pt]
\leq
C\|ue^{q\cdot/2}\|_{[H^{1/6}(0,+\infty)]^{*}}^{2}(T^{8/3}m^{8/3}+T^{2/3}m^{2/3}+m^{-\frac{1}{3}}).$

By choosing $m=T^{-2/3}$, we obtain

$\left|\int_{\mathbb{R}}|\hat{u}(z+iq/2)|^{2}\int_{0}^{L}B(z+iq/2,x)\mathop{}\mkern-3.0mu{d}x\mathop{}\mkern-3.0mu{d}z-E\int_{\mathbb{R}}\frac{|\hat{u}(z+iq/2)|^{2}}{(1+|z|^{2})^{1/6}}\mathop{}\mkern-3.0mu{d}z\right|\leq
C\|ue^{q\cdot/2}\|_{H^{-1/6}(0,+\infty)}^{2}T^{2/9}.$

The conclusion follows. ∎

In the proof of Proposition 7.2, we have used the following lemma.

######  Lemma 7.3.

Let $0<T<1$, $0<s<1/2$ and let $v\in[H^{s}(0,+\infty)]^{*}$ with
$\operatorname{supp}v\subset[0,T]$. Extend $v$ by $0$ for $t<0$ and still
denote this extension by $v$. There exists a positive constant $C$ independent
of $T$ and $v$ such that, for $z\in\mathbb{R}$,

(7.44) $|\hat{v}(z)|\leq
C(|z|T^{\frac{3}{2}-s}+T^{\frac{1}{2}-s})\|v\|_{[H^{s}(0,+\infty)]^{*}}.$

###### Proof.

Fix $\chi\in C^{1}(\mathbb{R})$ such that $\chi=1$ in $[0,T]$, $\chi=0$ for
$t\geq 2T$, and $0\leq\chi\leq 1$ and $\chi^{\prime}\leq C/T$ in $\mathbb{R}$.
Then

(7.45)
$|\hat{v}(z)|\leq\|v\|_{[H^{s}(0,+\infty)]^{*}}\|e^{-iz\cdot}\chi\|_{H^{s}(0,+\infty)}.$

We have

(7.46)
$C\|e^{-iz\cdot}\chi\|_{H^{s}(0,+\infty)}^{2}\leq\int_{0}^{4T}\int_{0}^{4T}\frac{|\chi(x)|^{2}|e^{-izx}-e^{-izy}|^{2}}{|x-y|^{1+2s}}\,dx\,dy\\\\[6.0pt]
+\int_{0}^{4T}\int_{0}^{4T}\frac{|\chi(x)-\chi(y)|^{2}}{|x-y|^{1+2s}}\,dx\,dy+\int_{0}^{2T}\int_{4T}^{\infty}\frac{1}{|x-y|^{1+2s}}\,dx\,dy.$

Using the fact, for $x\in\mathbb{R}$,

$|e^{ix}-1|^{2}=4\sin^{2}(x/2)\leq x^{2},$

we derive that the RHS of (7.46) is bounded by

(7.47) $Cz^{2}T^{3-2s}+CT^{1-2s}+CT^{1-2s}.$

The conclusion now follows from (7.45) and (7.47). ∎

##  8\. Small time local controllability properties of the KdV system - Proof
of Theorem 1.2

The main result of this section is the following, which implies, in
particular, Theorem 1.2 after applying Proposition 7.2 and Lemma 8.2 at the
end of this section. This result is also the key ingredient in the proof of
Assertion i) of Theorem 1.3.

######  Theorem 8.1.

Let $L\in{\mathcal{N}}_{D}$, $T>0$, and $\alpha>0$. Assume that

(8.1)
$\int_{0}^{T}\int_{0}^{L}|\xi(t,x)|^{2}\Phi(t,x)\,dx\,dt\geq\alpha\|v\|_{[H^{1/6}(0,T)]^{*}}^{2}$

for all $v\in H^{1/3}(0,T)$ such that $\xi(T,\cdot)=0$ in $(0,L)$ where
$\xi\in X_{T}$ is the unique solution of the system

(8.2) $\left\\{\begin{array}[]{cl}\xi_{t}+\xi_{x}+\xi_{xxx}=0&\mbox{ in
}(0,T)\times(0,L),\\\\[6.0pt] \xi(\cdot,0)=\xi_{x}(\cdot,L)=0&\mbox{ in
}(0,T),\\\\[6.0pt] \xi(\cdot,L)=v&\mbox{ in }(0,T),\\\\[6.0pt]
\xi(0,\cdot)=0&\mbox{ in }(0,L).\end{array}\right.$

There exists $\varepsilon_{0}>0$ depending only on $\alpha$, $T$, and $L$ such
that for all $0<\varepsilon<\varepsilon_{0}$, and for all solutions $y\in
X_{T/2}$ of the system

(8.3) $\left\\{\begin{array}[]{cl}y_{t}+y_{x}+y_{xxx}+yy_{x}=0&\mbox{ in
}(0,T/2)\times(0,L),\\\\[6.0pt] y(\cdot,0)=y_{x}(\cdot,L)=0&\mbox{ in
}(0,T/2),\\\\[6.0pt] y(\cdot,L)=u&\mbox{ in }(0,T/2),\\\\[6.0pt]
y(0,\cdot)=\varepsilon\phi&\mbox{ in }(0,L),\end{array}\right.$

with $\|u\|_{H^{1/2}(0,T/2)}<\varepsilon_{0}$, we have

$y(T/2,\cdot)\neq 0.$

Recall that $\phi$ is defined in (1.8).

###### Proof.

Let $\varepsilon_{0}$ be a small positive constant, which depends only on
$\alpha$, $T$, and $L$, and is determined later. We prove Theorem 8.1 by
contradiction. Assume that there exists a solution $y\in
C\big{(}[0,+\infty);L^{2}(0,L)\big{)}\cap
L^{2}_{\operatorname{loc}}\big{(}[0,+\infty);H^{1}(0,L)\big{)}$ of (8.3) with
$y(t,\cdot)=0$ for $t\geq T/2$, for some $0<\varepsilon<\varepsilon_{0}$, for
some $u\in H^{1/3}(0,+\infty)$ with $\operatorname{supp}u\subset[0,T/2]$ and
$\|u\|_{H^{1/2}(0,T)}<\varepsilon_{0}$.

Using the fact $y(t,\cdot)=0$ for $t\geq T/2$, from Proposition 3.4, we have,
for $\varepsilon_{0}$ small,

(8.4) $\|y\|_{L^{2}\big{(}\mathbb{R}_{+};H^{1}(0,L)\big{)}}\leq
C\Big{(}\|y_{0}\|_{L^{2}(0,L)}+\|u\|_{H^{1/3}(\mathbb{R}_{+})}\Big{)},$

which in turn implies, by Proposition 3.1 and Lemma 8.1 below,

(8.5) $\|y\|_{L^{2}\big{(}\mathbb{R}_{+}\times(0,L)\big{)}}\leq
C\Big{(}\|y_{0}\|_{L^{2}(0,L)}+\|u\|_{L^{2}(\mathbb{R}_{+})}\Big{)}.$

Here and in what follows, $C$ denotes a positive constant depending only on
$T$ and $L$ ($C$ thus does not depend on $\alpha$).

Let $y_{1}$ and $y_{2}$ be the solution of the following systems

(8.6) $\left\\{\begin{array}[]{cl}y_{1,t}+y_{1,x}+y_{1,xxx}=0&\mbox{ in
}\mathbb{R}_{+}\times(0,L),\\\\[6.0pt]
y_{1}(\cdot,0)=y_{1,x}(\cdot,L)=0&\mbox{ in }\mathbb{R}_{+},\\\\[6.0pt]
y_{1}(\cdot,L)=u&\mbox{ in }\mathbb{R}_{+},\\\\[6.0pt] y_{1}(0,\cdot)=0&\mbox{
in }(0,L),\end{array}\right.$ (8.7)
$\left\\{\begin{array}[]{cl}y_{2,t}+y_{2,x}+y_{2,xxx}+y_{1}y_{1,x}=0&\mbox{ in
}\mathbb{R}_{+}\times(0,L),\\\\[6.0pt]
y_{2}(\cdot,0)=y_{2,x}(\cdot,L)=y_{2}(\cdot,L)=0&\mbox{ in
}\mathbb{R}_{+},\\\\[6.0pt] y_{2}(0,\cdot)=0&\mbox{ in
}(0,L).\end{array}\right.$

Then, by Proposition 3.1,

(8.8) $\|y_{1}\|_{X_{T}}\leq C\|u\|_{H^{1/3}(\mathbb{R}_{+})}$

and

(8.9) $\|y_{1}\|_{L^{2}\big{(}(0,T)\times(0,L)\big{)}}\leq
C\|u\|_{L^{2}(\mathbb{R}_{+})},$

which in turn imply, by Proposition 3.1 and Lemma 8.1 below,

(8.10) $\|y_{2}\|_{X_{T}}\leq C\|y_{1}y_{1,x}\|_{L^{1}((0,T);L^{2}(0,L))}\leq