C\|u\|_{L^{2}(\mathbb{R}_{+})}^{1/2}\|u\|_{H^{1/3}(\mathbb{R}_{+})}^{3/2},$

and

(8.11) $\|y_{2}\|_{L^{2}\big{(}(0,T)\times(0,L)\big{)}}\leq
C\|y_{1}^{2}\|_{L^{1}((0,T);L^{2}(0,L))}\leq
C\|u\|_{L^{2}(\mathbb{R}_{+})}^{3/2}\|u\|_{H^{1/3}(\mathbb{R}_{+})}^{1/2}.$

Set

$\delta y=y-y_{1}-y_{2}\mbox{ in }\mathbb{R}_{+}\times(0,L).$

We have

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

Applying Proposition 3.1 and using Lemma 8.1 below, we derive that

(8.12) $\|\delta y\|_{X_{T}}\leq
C\Big{(}\|y_{1}y_{2,x}\|_{L^{1}((0,T);L^{2}(0,L))}+\|y_{2}y_{1,x}\|_{L^{1}((0,T);L^{2}(0,L))}+\|y_{2}y_{2,x}\|_{L^{1}((0,T);L^{2}(0,L))}\Big{)}\\\\[6.0pt]
\mathop{\leq}^{\eqref{thm-NL-y1-1}-\eqref{thm-
NL-y2-2}}C\|u\|_{L^{2}(\mathbb{R}_{+})}\|u\|_{H^{1/3}(\mathbb{R}_{+})}^{2}+C\varepsilon,$

and

(8.13) $\|\delta y\|_{L^{2}\big{(}(0,T)\times(0,L)\big{)}}\leq
C\Big{(}\|y\delta y_{x}\|_{L^{1}((0,T);L^{2}(0,L))}+\|(y_{1}+y_{2})\delta
y_{x}\|_{L^{1}((0,T);L^{2}(0,L))}\Big{)}\\\\[6.0pt]
+C\Big{(}\|(y_{1}+y_{2})\delta
y\|_{L^{1}((0,T);L^{2}(0,L))}+\|y_{1}y_{2}\|_{L^{1}((0,T);L^{2}(0,L))}+\|y_{2}^{2}\|_{L^{1}((0,T);L^{2}(0,L))}\Big{)}\\\\[6.0pt]
\mathop{\leq}^{\eqref{thm-NL-yyy}-\eqref{thm-NL-yyy-1},\eqref{thm-
NL-y1-1}-\eqref{thm-NL-
dy-1}}C\|u\|_{L^{2}(\mathbb{R}_{+})}^{3/2}\|u\|_{H^{1/3}(\mathbb{R}_{+})}^{5/2}+C\|u\|_{L^{2}(\mathbb{R}_{+})}^{2}\|u\|_{H^{1/3}(\mathbb{R}_{+})}+C\varepsilon\\\\[6.0pt]
\leq
C\|u\|_{L^{2}(\mathbb{R}_{+})}^{3/2}\|u\|_{H^{1/3}(\mathbb{R}_{+})}^{3/2}+C\varepsilon.$

In (8.12), we absorbed the contribution in the RHS of

$\|y\delta y_{x}\|_{L^{1}((0,T);L^{2}(0,L))}+\|\delta
y(y_{1}+y_{2})_{x}\|_{L^{1}((0,T);L^{2}(0,L))},$

which is bounded by $C\varepsilon_{0}\|\delta y\|_{X_{T}}$. Combining (8.11)
and (8.13) yields

(8.14) $\|y-y_{1}\|_{L^{2}\big{(}(0,T)\times(0,L)\big{)}}\leq
C\|u\|_{L^{2}(\mathbb{R}_{+})}^{3/2}\|u\|_{H^{1/3}(\mathbb{R}_{+})}^{1/2}+C\varepsilon.$

Since $y=0$ for $t\geq T/2$ and $u=0$ for $t\geq T/2$, after using the
regularizing effect of the linear KdV equation, and considering the projection
into $\mathcal{M}_{D}^{\perp}$, we derive that

(8.15) $\|y_{1}(T,\cdot)\|_{L^{2}(0,L)}\leq
C\|u\|_{L^{2}(\mathbb{R}_{+})}^{3/2}\|u\|_{H^{1/3}(\mathbb{R}_{+})}^{1/2}+C\varepsilon.$

Since $y_{1}(T,\cdot)\in\mathcal{M}_{D}^{\perp}$ by Lemma 5.1, it follows from
Proposition 5.1 that there exists $u_{1}\in H^{1/3}(0,T)$ such that

(8.16) $\|u_{1}\|_{H^{1/3}(0,T)}\leq C\|y_{1}(T,\cdot)\|_{L^{2}(0,L)}$

and the solution $\widetilde{y}_{1}\in X_{T}$ of the system

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

satisfies

$\widetilde{y}_{1}(T,\cdot)=-y_{1}(T,\cdot).$

Using (8.15), we derive from (8.16) that

(8.17) $\|u_{1}\|_{H^{1/3}(0,T)}\leq
C\|u\|_{L^{2}(\mathbb{R}_{+})}^{3/2}\|u\|_{H^{1/3}(\mathbb{R}_{+})}^{1/2}+C\varepsilon,$

which in turn implies, by Proposition 3.1,

(8.18) $\|\widetilde{y}_{1}\|_{X_{T}}\leq
C\|u\|_{L^{2}(\mathbb{R}_{+})}^{3/2}\|u\|_{H^{1/3}(\mathbb{R}_{+})}^{1/2}+C\varepsilon.$

Set

$\widetilde{y}=y_{1}+\widetilde{y}_{1}\mbox{ in }(0,T)\times(0,L).$

Then

(8.19)
$\left\\{\begin{array}[]{cl}\widetilde{y}_{t}+\widetilde{y}_{x}+\widetilde{y}_{xxx}=0&\mbox{
in }(0,T)\times(0,L),\\\\[6.0pt]
\widetilde{y}(\cdot,0)=\widetilde{y}_{x}(\cdot,L)=0&\mbox{ in
}(0,T),\\\\[6.0pt] \widetilde{y}(\cdot,L)=u+u_{1}&\mbox{ in }(0,T),\\\\[6.0pt]
\widetilde{y}(0,\cdot)=\widetilde{y}(T,\cdot)=0&\mbox{ in
}(0,L).\end{array}\right.$

We have

(8.20)
$\|y-\widetilde{y}\|_{L^{2}\big{(}(0,T)\times(0,L)\big{)}}\leq\|y-y_{1}\|_{L^{2}\big{(}(0,T)\times(0,L)\big{)}}+\|\widetilde{y}_{1}\|_{L^{2}\big{(}(0,T)\times(0,L)\big{)}}\\\\[6.0pt]
\mathop{\leq}^{\eqref{thm-NL-y1-0},\eqref{thm-NL-
ty1}}C\|u\|_{L^{2}(\mathbb{R}_{+})}^{3/2}\|u\|_{H^{1/3}(\mathbb{R}_{+})}^{1/2}+C\varepsilon.$

Multiplying the equation of $y$ with $\Phi(t,x)$, integrating by parts on
$[0,L]$, and using (1.11), we have

(8.21)
$\frac{d}{dt}\int_{0}^{L}y(t,x)\Phi(t,x)\mathop{}\mkern-3.0mu{d}x-\frac{1}{2}\int_{0}^{L}y^{2}(t,x)\Phi_{x}(t,x)\mathop{}\mkern-3.0mu{d}x=0.$

Integrating (8.21) from 0 to $T$ and using the fact $y(T,\cdot)=0$ yield

(8.22)
$\int_{0}^{L}y_{0}(x)\Phi(0,x)\mathop{}\mkern-3.0mu{d}x+\frac{1}{2}\int_{0}^{T}\int_{0}^{L}y^{2}(t,x)\Phi_{x}(t,x)\mathop{}\mkern-3.0mu{d}x\mathop{}\mkern-3.0mu{d}t=0.$

We have

(8.23)
$\|y^{2}-\widetilde{y}^{2}\|_{L^{1}\big{(}(0,T)\times(0,L)\big{)}}\leq\|y-\widetilde{y}\|_{L^{2}\big{(}(0,T)\times(0,L)\big{)}}\|y+\widetilde{y}\|_{L^{2}\big{(}(0,T)\times(0,L)\big{)}}\\\\[6.0pt]
\mathop{\leq}^{\eqref{thm-NL-yyy-1},\eqref{thm-NL-
yhy}}C\|u\|_{L^{2}(\mathbb{R}_{+})}^{5/2}\|u\|_{H^{1/3}(\mathbb{R}_{+})}^{1/2}+C\varepsilon_{0}\varepsilon.$

Combining (8.22) and (8.23) yields

(8.24)
$\int_{0}^{L}y_{0}(x)\Phi(0,x)\mathop{}\mkern-3.0mu{d}x+\frac{1}{2}\int_{0}^{T}\int_{0}^{L}\widetilde{y}^{2}(t,x)\Phi_{x}(t,x)\mathop{}\mkern-3.0mu{d}x\mathop{}\mkern-3.0mu{d}t\\\
\leq
C\|u\|_{L^{2}(\mathbb{R}_{+})}^{5/2}\|u\|_{H^{1/3}(\mathbb{R}_{+})}^{1/2}+C\varepsilon_{0}\varepsilon.$

Applying (8.1) to $\widetilde{y}$ after noting (8.19), we derive from (8.24)
that, for $\varepsilon_{0}$ sufficiently small,

$\varepsilon\|\phi\|_{L^{2}(0,L)}^{2}+\frac{1}{2}\alpha\|u+u_{1}\|_{[H^{1/6}(\mathbb{R}_{+})]^{*}}^{2}\leq
C\|u\|_{L^{2}(\mathbb{R}_{+})}^{5/2}\|u\|_{H^{1/3}(\mathbb{R}_{+})}^{1/2}+C\varepsilon_{0}\varepsilon.$

Using (8.17), it follows that, for sufficiently small $\varepsilon_{0}$,

$\alpha\|u\|_{[H^{1/6}(\mathbb{R}_{+})]^{*}}^{2}\leq
C\|u\|_{L^{2}(\mathbb{R}_{+})}^{5/2}\|u\|_{H^{1/3}(\mathbb{R}_{+})}^{1/2}.$

We derive from Lemma A.1 that

$\alpha\|u\|_{[H^{1/6}(0,T/2)]^{*}}^{2}\leq
C\|u\|_{L^{2}(0,T/2)}^{5/2}\|u\|_{H^{1/3}(0,T/2)}^{1/2}.$

Note that

$\|u\|_{L^{2}(0,T/2)}\leq
C\|u\|_{[H^{1/6}(0,T/2)]^{*}}^{3/4}\|u\|_{H^{1/2}(0,T/2)}^{1/4}$

and

$\|u\|_{H^{1/3}(0,T/2)}\leq
C\|u\|_{[H^{1/6}(0,T/2)]^{*}}^{1/4}\|u\|_{H^{1/2}(0,T/2)}^{3/4}.$

This yields

$\|u\|_{L^{2}(0,T/2)}^{5/2}\|u\|_{H^{1/3}(0,T/2)}^{1/2}\leq
C\|u\|_{[H^{1/6}(0,T/2)]^{*}}^{2}\|u\|_{H^{1/2}(0,T/2)}.$

So, for fixed sufficiently small $\varepsilon_{0}$,

$u=0.$

Hence $y(t,\cdot)=\varepsilon\Psi(t,\cdot)\not\equiv 0$ for all $t>0$. We have
a contradiction.

The proof is complete. ∎

In the proof of Theorem 8.1, we repeatedly used the following useful result.

######  Lemma 8.1.

Let $L>0$ and $T>0$ and let $f\in L^{2}\big{(}(0,T);H^{1}(0,L)\big{)}$ and
$g\in L^{2}\big{(}(0,T)\times(0,L)\big{)}$. Then

(8.25) $\|fg\|_{L^{1}\big{(}(0,T);L^{2}(0,L)\big{)}}\leq
C\|f\|_{L^{2}\big{(}(0,T)\times(0,L)\big{)}}^{\frac{1}{2}}\|f\|_{L^{2}\big{(}(0,T);H^{1}(0,L)\big{)}}^{\frac{1}{2}}\|g\|_{L^{2}\big{(}(0,T)\times(0,L)\big{)}}$

and

(8.26) $\|(fg)_{x}\|_{L^{1}((0,T);[H^{1}(0,L)]^{*})}\\\\[6.0pt] \leq
C\Big{(}\|fg\|_{L^{1}((0,T);L^{2}(0,T))}+\|(fg)(\cdot,0)\|_{L^{1}(0,T)}+\|(fg)(\cdot,L)\|_{L^{1}(0,T)}\Big{)},$

for some positive constant $C$ depending only on $L$. Consequently, for
$f,g\in L^{2}\big{(}(0,T);H^{1}(0,L)\big{)}$,

(8.27) $\|fg_{x}\|_{L^{1}\big{(}(0,T);L^{2}(0,L)\big{)}}\leq
C\|f\|_{L^{2}\big{(}(0,T)\times(0,L)\big{)}}^{\frac{1}{2}}\|f\|_{L^{2}\big{(}(0,T);H^{1}(0,L)\big{)}}^{\frac{1}{2}}\|g_{x}\|_{L^{2}\big{(}(0,T)\times(0,L)\big{)}}.$

###### Proof.

We first establish (8.25). We have

(8.28)
$\|fg\|_{L^{1}\big{(}(0,T);L_{2}(0,L)\big{)}}^{2}=\int_{0}^{T}\left(\int_{0}^{L}|fg|^{2}(s,x)\,dx\right)^{\frac{1}{2}}\,ds\\\\[6.0pt]
\leq\int_{0}^{T}\sup_{x\in[0,L]}|f(s,x)|\left(\int_{0}^{L}|g|^{2}(s,x)\,dx\right)^{\frac{1}{2}}\,ds.$

Using the fact

(8.29) $\sup_{x\in[0,L]}|f(s,x)|\leq
C\left(\int_{0}^{L}|f(s,x)|^{2}\,dx+\int_{0}^{L}|f(s,x)||f_{x}(s,x)|^{2}\,dx\right)^{\frac{1}{2}}\\\\[6.0pt]
\leq
C\|f(s,\cdot)\|_{L^{2}(0,L)}^{\frac{1}{2}}\|f(s,\cdot)\|_{H^{1}(0,L)}^{\frac{1}{2}},$

we derive from (8.28) that

$\|fg_{x}\|_{L^{1}\big{(}(0,T);L_{2}(0,L)\big{)}}^{2}\leq
C\int_{0}^{T}\|f(s,\cdot)\|_{L^{2}(0,L)}^{\frac{1}{2}}\|f(s,\cdot)\|_{H^{1}(0,L)}^{\frac{1}{2}}\|g(s,\cdot)\|_{L^{2}(0,L)}\,ds\\\\[6.0pt]
\leq
C\|f\|_{L^{2}\big{(}(0,T)\times(0,L)\big{)}}^{\frac{1}{2}}\|f\|_{L^{2}\big{(}(0,T);H^{1}(0,L)\big{)}}^{\frac{1}{2}}\|g\|_{L^{2}\big{(}(0,T)\times(0,L)\big{)}}\quad(\mbox{by
H\"{o}lder's inequality}),$

which is (8.25).

We next derive (8.26). We have, for $0\leq t\leq T$,

$\int_{0}^{L}(fg)_{x}(t,x)\varphi(x)\,dx=-\int_{0}^{L}fg(t,x)\varphi_{x}(x)\,dx+(fg)(t,L)\varphi(L)-(fg)(t,0)\varphi(0)\mbox{
for }\varphi\in H^{1}(0,L).$

This implies, by the trace theory applied to $\varphi\in H^{1}(0,L)$, for
$0\leq t\leq T$,

$\|(fg)_{x}(t,\cdot)\|_{[H^{1}(0,L)]^{*}}\leq
C\Big{(}\|(fg)(t,\cdot)\|_{L^{2}(0,L)}+\|(fg)(t,0)\|_{L^{1}(0,T)}+\|(fg)(t,L)\|_{L^{1}(0,T)}\Big{)}.$

Estimate (8.26) follows.

The proof is complete. ∎

To derive Theorem 1.2 from Theorem 8.1, we also need to use the following
simple lemma.

######  Lemma 8.2.

Let $T>0$, $0<s<1$, $\varphi\in C^{1}([0,+\infty])$. Then

$\|\varphi v\|_{[H^{s}(\mathbb{R}_{+})]^{*}}\leq
C\|v\|_{[H^{s}(\mathbb{R}_{+})]^{*}}\mbox{ for
}v\in[H^{s}(\mathbb{R}_{+})]^{*}\mbox{ with
}\operatorname{supp}v\subset[0,T],$

where $C$ is a positive constant independent of $v$.

###### Proof.

Without loss of generality, one might assume that
$\operatorname{supp}\varphi\subset[0,2T]$. Using the duality between
$[H^{s}(\mathbb{R}_{+})]^{*}$ and $H^{s}(\mathbb{R}_{+})$, it suffices to
prove that

$\|\varphi V\|_{H^{s}(\mathbb{R}_{+})}\leq
C\|V\|_{H^{s}(\mathbb{R}_{+})}\mbox{ for }V\in H^{s}(\mathbb{R}_{+}),$

for some positive constant $C$ independent of $V$. This follows immediately
from the Gagliardo-Nirenberg characterization of the norm of
$H^{s}(\mathbb{R}_{+})$. ∎

##  9\. Local, exact controllability of the KdV equation in a positive time -
Proof of Theorem 1.3

This section consists of two subsections. In the first one, we give the proof
of Assertion $i)$. The proof of Assertion $ii)$ is given in the second
subsection.

###  9.1. Proof of Assertion $i)$ of Theorem 1.3

We first consider the case $\omega=0$. We have

$\left|\sum_{j=1}^{3}\big{(}\lambda_{j}e^{\lambda_{j}L}-\lambda_{j+1}e^{\lambda_{j+1}L}\big{)}e^{\lambda_{j+2}x}\right|^{2}(z)\\\\[6.0pt]
=\sum_{j=1}^{3}\big{(}\lambda_{j}e^{\lambda_{j}L}-\lambda_{j+1}e^{\lambda_{j+1}L}e^{\lambda_{j+2}x}\big{)}(z)\sum_{j=1}^{3}\big{(}\lambda_{j}e^{\lambda_{j}L}-\lambda_{j+1}e^{\lambda_{j+1}L}e^{\lambda_{j+2}x}\big{)}(-\bar{z})$

and similarly,

$|\Xi(z)|^{2}=\Xi(z)\Xi(-\bar{z}),$

where $\Xi$ is defined in (6.4). It follows that

$|\Xi(\cdot+iq/2)|^{2}|\Omega(\cdot+iq/2)|^{2}\mbox{ is an analytic function
in $z\in\mathbb{R}$}.$

There thus exists a finite set $\\{z_{j};1\leq j\leq m\\}\subset\mathbb{R}$
such that

$\Omega(z_{j}+iq/2)=0\mbox{ for }1\leq j\leq m\quad\mbox{ and
}\quad\Omega(z+iq/2)\neq 0\mbox{ for }z\in\mathbb{R}\setminus\\{z_{j};1\leq
j\leq m\\}.$

Since $\Omega(\cdot+iq/2)$ is continuous in $\mathbb{R}$ and
$\Omega(z+iq/2)\to+\infty$ as $|z|\to+\infty$ by (7.37) of Proposition 7.1 and
Lemma 3.1, we derive that for each $\delta>0$ (small), there exists $\rho>0$
(depending on $\delta$) such that

$\Omega(z+iq/2)\geq\rho\mbox{ in
}J_{\delta}:=\mathbb{R}\setminus\cup_{j=1}^{m}(z_{j}-\delta,z_{j}+\delta).$

Let $T>0$ be arbitrary and let $y$ be a solution of (7.9) such that
$y(t,\cdot)=0$ for $t\geq T$. Applying Proposition 7.1 and taking into account
the fact $\omega\geq 0$, we get

(9.1)
$\int_{0}^{+\infty}\int_{0}^{L}y^{2}(t,x)\Phi_{x}(t,x)\mathop{}\mkern-3.0mu{d}x\mathop{}\mkern-3.0mu{d}t=\int_{\mathbb{R}}\int_{0}^{L}|\hat{u}(z+iq/2)|^{2}B(z+iq/2,x)\,dx\,dz,$

where we extend $u$ by $0$ for $t<0$ and $t>T$. This implies

(9.2)
$\int_{0}^{+\infty}\int_{0}^{L}y^{2}(t,x)\Phi_{x}(t,x)\mathop{}\mkern-3.0mu{d}x\mathop{}\mkern-3.0mu{d}t\\\\[6.0pt]
=\int_{J_{\delta}}\int_{0}^{L}|\hat{u}(z+iq/2)|^{2}B(z+iq/2,x)\,dx\,dz+\int_{\mathbb{R}\setminus
J_{\delta}}\int_{0}^{L}|\hat{u}(z+iq/2)|^{2}B(z+iq/2,x)\,dx\,dz\\\\[6.0pt]
\geq
C_{1}\rho\|ue^{q\cdot/2}\|_{[H^{1/6}(0,T)]^{*}}^{2}-C_{2}\rho\int_{J_{\delta}}|\hat{u}(z+iq/2)|^{2}\,dz\\\\[6.0pt]
\mathop{\geq}^{\lx@cref{creftypecap~refnum}{lem-
multiplier}}C_{1}\rho\|u\|_{[H^{1/6}(0,T)]^{*}}^{2}-C_{T}\rho\delta\|u\|_{[H^{1/6}(0,T)]^{*}}^{2}.$

By choosing $\delta$ sufficiently small, we arrive that

(9.3)
$\int_{0}^{\infty}\int_{0}^{+\infty}y^{2}(t,x)\Phi_{x}(t,x)\mathop{}\mkern-3.0mu{d}x\mathop{}\mkern-3.0mu{d}t\geq
C\rho\|u\|_{[H^{1/6}(0,T)]^{*}}^{2}.$

Applying Theorem 8.1, we derive that system (1.1) is not locally null
controllable at time $T/2$. Since $T$ is arbitrary, the conclusion follows.

We next deal with the case $\omega>0$. It is clear that (9.3) holds with
$\rho=\omega$. The conclusion follows by Theorem 8.1 as before since $T$ is
arbitrary. ∎

###  9.2. Proof of Assertion $ii)$ of Theorem 1.3

Set

(9.4) $\sigma(t)=e^{-\frac{1}{1-t^{2}}}\mathds{1}_{(-1,1)},$

where $\mathds{1}_{(-1,1)}$ denotes the characteristic function of the
interval $(-1,1)$. Then $\sigma\in C^{\infty}_{c}(\mathbb{R})$ and

(9.5) $|\hat{\sigma}(z)|\leq c_{1}e^{-c_{2}\sqrt{|z|}}\mbox{ for
}z\in\mathbb{R},$

see e.g., [44]. Let $m\in\mathbb{N}$ be large and denote

$\sigma_{m}(t)=\frac{1}{m}\sigma(t/m)\mbox{ for }t\in\mathbb{R}.$

Let $z_{0}\in\mathbb{R}$ be such that

$\Omega(z_{0}+iq/2)=\omega<0.$

Consider $u_{m}$ defined by

(9.6) $\hat{u}_{m}(z)=e^{imz}\hat{\sigma}_{m}(z-iq/2-z_{0})\det Q(z)\Xi(z).$

Applying Paley-Wiener’s theorem, see e.g., [41], and using (9.5), one can
check that the function $v_{m}$ which is defined by
$\hat{v}_{m}(z)=\hat{\sigma}_{m}(z-iq/2-z_{0})\det Q(z)\Xi(z)$ belongs to
$L^{2}(\mathbb{R})$ with support in $[-m,m]$. We then derive that $u_{m}\in
L^{2}(\mathbb{R})$ with support in $[0,2m]$ since $u_{m}(t)=v_{m}(t+m)$ for
$t\in\mathbb{R}$. Using (9.5), one can also derive that $u_{m}\in
C^{\infty}(\mathbb{R})$.

Let $y=y_{m}$ be the unique solution of (6.12) with $u=u_{m}$. From the
definition of $u_{m}$, we deduce that

$\operatorname{supp}y\subset[0,m]\times[0,L],$

since, by Lemma 6.1,

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

is an entire function with modulus is bounded by $Ce^{(m+\varepsilon)|z|}$ for
all $\varepsilon>0$, and $\hat{y}_{xx}(\cdot,0)\in L^{2}(\mathbb{R})$. It
follows that

$\operatorname{supp}y_{xx}(\cdot,0)\subset[0,m].$

Since $y(m,x)\in\mathcal{M}_{D}^{\perp}$, applying Lemma 5.3, we derive that

$y(t,x)=0\mbox{ for }(t,x)\in[m,+\infty)\times[0,L].$

We thus can apply Proposition 7.1 to $y$.

Since, by Proposition 7.1,

$\int_{0}^{\infty}\int_{0}^{+\infty}y^{2}(t,x)\Phi_{x}(t,x)\mathop{}\mkern-3.0mu{d}x\mathop{}\mkern-3.0mu{d}t=\int_{\mathbb{R}}\frac{|\hat{u}_{m}(z+iq/2)|^{2}}{|H(z+iq/2)|^{2}}\Omega(z+iq/2)\,dz,$

it follows from (9.6) that

(9.7)
$\int_{0}^{\infty}\int_{0}^{+\infty}y^{2}(t,x)\Phi_{x}(t,x)\mathop{}\mkern-3.0mu{d}x\mathop{}\mkern-3.0mu{d}t=e^{-mq}\int_{\mathbb{R}}|\hat{\sigma}_{m}(z-z_{0})|^{2}|\Xi(z+iq/2)|^{2}\Omega(z+iq/2)\,dz.$

We have, by (9.5)

(9.8) $|\hat{\sigma}_{m}(z)|\leq c_{1}e^{-c_{2}\sqrt{m|z|}}\mbox{ for
}z\in\mathbb{R}.$ (9.9) $|\hat{\sigma}_{m}(z)|\geq c_{3}\mbox{ for
}z\in\mathbb{R}\mbox{ with }|z|<c_{4}/m,$ (9.10)
$\inf_{z\in\mathbb{R}}|\Xi(z+iq/2)|>0,\quad|\Xi(z+iq/2)|\leq C(|z|+1)\mbox{
for }z\in\mathbb{R},$ (9.11) $|\Omega(z+iq/2)|\leq c_{5}e^{c_{6}|z|^{1/3}},$

for some positive constant $c_{1}$, …, $c_{6}$, independent of $m$.

Using (9.8), (9.9), (9.10), (9.11), and the fact
$\Omega(z_{0}+iq/2)=\omega<0$, it follows from (9.7) that

$\int_{0}^{\infty}\int_{0}^{+\infty}y^{2}(t,x)\Phi_{x}(t,x)\mathop{}\mkern-3.0mu{d}x\mathop{}\mkern-3.0mu{d}t<0,$

for large $m$. Fix such an $m$ and set

$T=2m.$

We thus have just proved, after scaling, that there exists $U_{1}\in
H^{1/3}(0,T)$ such that

(9.12) $Y_{1}(T,\cdot)=0\quad\mbox{ and
}\quad\mbox{proj}_{\mathcal{M}_{D}}Y_{2}(T,\cdot)=-\phi,$

where $Y_{1},Y_{2}\in X_{T}$ are the unique solution of the systems

(9.13) $\left\\{\begin{array}[]{cl}Y_{1,t}+Y_{1,x}+Y_{1,xxx}=0&\mbox{ in
}(0,T)\times(0,L),\\\\[6.0pt] Y_{1}(\cdot,0)=Y_{1,x}(\cdot,L)=0&\mbox{ in
}(0,T),\\\\[6.0pt] Y_{1}(\cdot,L)=U_{1}&\mbox{ in }(0,T),\\\\[6.0pt]
Y_{1}(0,\cdot)=0&\mbox{ in }(0,L),\end{array}\right.$ (9.14)
$\left\\{\begin{array}[]{cl}Y_{2,t}+Y_{2,x}+Y_{2,xxx}+Y_{1}Y_{1,x}=0&\mbox{ in
}(0,T)\times(0,L),\\\\[6.0pt] Y_{2}(\cdot,0)=Y_{2,x}(\cdot,L)=0&\mbox{ in
}(0,T),\\\\[6.0pt] Y_{2}(\cdot,L)=0&\mbox{ in }(0,T).\end{array}\right.$

We now establish the local controllability for the time $T$ which is $2m$. In
what follows, we consider $T=2m$. Fix $y_{0},y_{T}\in L^{2}(0,L)$ with small
$L^{2}(0,L)$-norms. We first consider the case

$\rho=\|y_{0}\|_{L^{2}(0,L)}>0,$ (9.15) $\int_{0}^{L}y_{0}\phi_{1}\,dx\geq{\bf
c}_{1}\rho\int_{0}^{L}\phi^{2}\,dx,$

and

(9.16) $\|y_{T}\|_{L^{2}(0,L)}\leq\frac{{\bf c}_{1}e^{qT}\rho}{10}.$

for some fixed constant ${\bf c}_{1}$ independent of $\rho$.

For $r>0$ and $y_{0}\in L^{2}(0,L)$, denote $B_{r}(y_{0})$ the open ball
centered at 0 and of radius $r$ in $L^{2}(0,L)$. We also denote
$\overline{B_{r}(y_{0})}$ its closure in $L^{2}(0,L)$.

Let $c$ be a small positive constant determined later. For
$\varphi\in\overline{B_{c\rho}(y_{0})}$, let $u_{1}$ and $u_{2}$ be controls
in $H^{1/3}(0,T)$ for which the solutions $y_{1}$ and $y_{2}$ in $X_{T}$ of
the systems

(9.17) $\left\\{\begin{array}[]{cl}y_{1,t}+y_{1,x}+y_{1,xxx}=0&\mbox{ in
}(0,T)\times(0,L),\\\\[6.0pt] y_{1}(\cdot,0)=y_{1}(\cdot,L)=0&\mbox{ in
}(0,T),\\\\[6.0pt] y_{1,x}(\cdot,L)=u_{1}&\mbox{ in }(0,T),\\\\[6.0pt]
y_{1}(0,\cdot)=0&\mbox{ in }(0,L),\end{array}\right.$ (9.18)
$\left\\{\begin{array}[]{cl}y_{2,t}+y_{2,x}+y_{2,xxx}+y_{1}y_{1,x}=0&\mbox{ in
}(0,T)\times(0,L),\\\\[6.0pt] y_{2}(\cdot,0)=y_{2}(\cdot,L)=0&\mbox{ in
}(0,T),\\\\[6.0pt] y_{2,x}(\cdot,L)=u_{2}&\mbox{ in }(0,T),\\\\[6.0pt]
y_{2}(0,\cdot)=\mbox{proj}_{{\mathcal{M}}_{D}}\varphi/\rho&\mbox{ in
}(0,L),\end{array}\right.$

satisfy

$y_{1}(T,\cdot)=0\quad\mbox{ and }\quad y_{2}(T,\cdot)=y_{T}/\rho.$

Moreover, one can choose $u_{1}$ and $u_{2}$ as Lipschitz functions of
$\mbox{proj}_{{\mathcal{M}}_{D}}\varphi/\rho$ with the Lipschitz constants
bounded by positive constants independent of $\rho$. For example, using Lemma
5.1 and (9.12), one can take 999It is useful to note that
$e^{-qT}\alpha_{0}-\alpha_{T}>0$ by (9.15) and (9.16) if $c$ is sufficiently
small.

$u_{1}=(e^{-qT}\alpha_{0}-\alpha_{T})U_{1},$

where $\alpha_{0}\phi=\mbox{proj}_{\mathcal{M}_{D}}\varphi/\rho$ and
$\alpha_{T}\phi=\mbox{proj}_{\mathcal{M}_{D}}y_{T}/\rho$, and

$u_{2}=\mathcal{L}\big{(}-\mbox{proj}_{\mathcal{M}_{D}^{\perp}}y_{3}(T,\cdot)+\mbox{proj}_{\mathcal{M}_{D}^{\perp}}y_{T}/\rho\big{)}+\hat{\mathcal{L}}(\mbox{proj}_{\mathcal{M}_{D}}\varphi/\rho),$

where $y_{3}\in X_{T}$ is the unique solution of the system

(9.19)
$\left\\{\begin{array}[]{cl}y_{3,t}+y_{3,x}+y_{3,xxx}+y_{1}y_{1,x}=0&\mbox{ in
}(0,T)\times(0,L),\\\\[6.0pt]
y_{3}(\cdot,0)=y_{3}(\cdot,L)=y_{3,x}(\cdot,L)=0&\mbox{ in }(0,T),\\\\[6.0pt]
y_{3}(0,\cdot)=\mbox{proj}_{{\mathcal{M}}_{D}^{\perp}}\varphi/\rho&\mbox{ in
}(0,T).\end{array}\right.$

Here $\mathcal{L}$ and $\hat{\mathcal{L}}$ are the operators given in
Proposition 5.1.

For $\varphi\in\overline{B_{c\rho}(y_{0})}$, let $y\in X_{T}$ be the unique
solution of the backward linear KdV system

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

where $v(t)=\rho^{1/2}y_{1,x}(t,0)+\rho y_{2,x}(t,0)$ with $y_{1}$ and $y_{2}$
being defined by (9.17) and (9.18).

Note that $y\in X_{T}$ is well-defined if $\rho$ is sufficiently small by
Proposition 3.5. We will denote

(9.21) $H_{0}(\varphi)=y(0,\cdot)\mbox{ in }(0,L)\mbox{ and
}H(\varphi)=y\mbox{ in }(0,T)\times(0,L).$

Consider the map

$\begin{array}[]{cccc}\Lambda\colon&\overline{B_{c\rho}(y_{0})}&\to&L^{2}(0,L)\\\\[6.0pt]
&\varphi&\mapsto&\varphi-H_{0}(\varphi)+y_{0}.\end{array}$

We will prove that

(9.22) $\Lambda(\varphi)\in\overline{B_{c\rho}(y_{0})},$

and

(9.23)
$\|\Lambda(\varphi)-\Lambda(\widetilde{\varphi})\|_{L^{2}(0,L)}\leq\lambda\|\varphi-\widetilde{\varphi}\|_{L^{2}(0,L)},$

for some $\lambda\in(0,1)$. Assuming this, one derives from the contraction
mapping theorem that there exists a unique
$\varphi_{0}\in\overline{B_{c\rho}(y_{T})}$ such that
$\Lambda(\varphi_{0})=\varphi_{0}$. As a consequence, with $y=H(\varphi_{0})$,

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

and $y(\cdot,L)$ is hence a required control.

We next establish (9.22) and (9.23). Indeed, assertion (9.22) follows from the
fact

$\|\varphi-H_{0}(\varphi)\|_{L^{2}(0,L)}\leq
C\|\varphi\|_{L^{2}(0,L)}^{3/2}\mbox{ for
}\varphi\in\overline{B_{c\rho}(y_{0})}.$

This can be proved using the approximation via the power series method as
follows. Applying Proposition 3.3 to $y$, we derive from (9.20) that

(9.24) $\|y\|_{X_{T}}\leq C\rho^{1/2}.$

Set 101010The index $a$ stands the approximation.

(9.25) $y_{a}=\rho^{1/2}y_{1}+\rho y_{2}\mbox{ in }(0,T)\times(0,L).$

We have

(9.26)
$(y-y_{a})_{t}+(y-y_{a})_{x}+(y-y_{a})_{xxx}+(y-y_{a})y_{x}+y_{a}(y-y_{a})_{x}=h(t,x),$

where

$h(t,x)=-y_{a}y_{a,x}+\rho
y_{1}y_{1,x}=-\rho^{3/2}(y_{1}y_{2})_{x}-\rho^{2}y_{2}y_{2,x}.$

Applying Proposition 3.3 to $y-y_{a}$, for small $\rho$, one can ignore
(absorb) the contribution from the last two terms in the LHS of (9.26), to
obtain

(9.27) $\|y-y_{a}\|_{X_{T}}\leq
C\|h\|_{L^{1}\big{(}(0,T);L^{2}(0,L)\big{)}}\leq C\rho^{3/2}.$

Assertion (9.22) follows since $H_{0}(\varphi)=y(0,\cdot)$ and
$\varphi=y_{a}(0,\cdot)$.

We next establish (9.23). To this end, we estimate

$\Big{(}\varphi-
H_{0}(\varphi)\Big{)}-\Big{(}\widetilde{\varphi}-H_{0}(\widetilde{\varphi})\Big{)}.$

For $\widetilde{\varphi}\in\overline{B_{c\rho}(y_{0})}$, denote
$\widetilde{u}_{1},\widetilde{u}_{2}$ and
$\widetilde{y}_{1},\widetilde{y}_{2},\widetilde{y}_{a},\widetilde{y}$ the
corresponding functions which are defined in the same way as the functions
$u_{1},u_{2}$, and $y_{1},y_{2},y_{a},y$ considered for $\varphi$.

We have

(9.28)
$(y-\widetilde{y})_{t}+(y-\widetilde{y})_{x}+(y-\widetilde{y})_{xxx}+yy_{x}-\widetilde{y}\widetilde{y}_{x}=0,$
(9.29)
$(y_{a}-\widetilde{y}_{a})_{t}+(y_{a}-\widetilde{y}_{a})_{x}+(y_{a}-\widetilde{y}_{a})_{xxx}+y_{a}y_{a,x}-\widetilde{y}_{a}\widetilde{y}_{a,x}=g(t,x),$

where

(9.30)
$g(t,x)=\rho^{3/2}\Big{(}(y_{1}y_{2})_{x}-(\widetilde{y}_{1}\widetilde{y}_{2})_{x}\Big{)}+\rho^{2}\Big{(}y_{2}y_{2,x}-\widetilde{y}_{2}\widetilde{y}_{2,x}\Big{)}.$

Applying Proposition 3.5 to $y-\widetilde{y}$, for $\rho$ small, one can
ignore (absorb) the contribution of the last two terms in the LHS of (9.28),
to obtain

(9.31) $\|y-\widetilde{y}\|_{X_{T}}\leq
C\|y_{a,x}(\cdot,0)-\widetilde{y}_{a,x}(\cdot,0)\|_{H^{1/3}(0,T)}\leq
C\rho^{-1/2}\|\varphi-\widetilde{\varphi}\|_{L^{2}(0,L)}.$

Using (9.27) for $y-y_{a}$ and similar fact for
$\widetilde{y}-\widetilde{y}_{a}$, we obtain

(9.32) $\|(y-y_{a},\widetilde{y}-\widetilde{y}_{a})\|_{X_{T}}\leq
C\rho^{3/2}.$

From the definition of $g$ in (9.30), we deduce that

(9.33) $\|g\|_{L^{1}\big{(}(0,T);L^{2}(0,L)\big{)}}\leq
C\rho^{1/2}\|\varphi-\widetilde{\varphi}\|_{L^{2}(0,L)}.$

Applying Proposition 3.1 to $y_{a}-\widetilde{y}_{a}$ after ignoring
(absorbing) the contribution of
$y_{a}y_{a,x}-\widetilde{y}_{a}\widetilde{y}_{a,x}$, we derive from (9.29)
that

(9.34) $\|y_{a}-\widetilde{y}_{a}\|_{X_{T}}\leq
C\rho^{-1/2}\|\varphi-\widetilde{\varphi}\|_{L^{2}(0,L)}\leq C\rho^{1/2}.$

Set

$Y=y-y_{a}-(\widetilde{y}-\widetilde{y}_{a})\mbox{ in }(0,T)\times(0,L).$

Using (9.28) and (9.29), we have

(9.35) $\partial_{t}Y+\partial_{x}Y+\partial_{xxx}Y=f(t,x)\mbox{ in
}(0,T)\times(0,L),$

where

$f(t,x)=-g(t,x)-\Big{(}yy_{x}-\widetilde{y}\widetilde{y}_{x}-(y_{a}y_{a,x}-\widetilde{y}_{a}\widetilde{y}_{a,x})\Big{)}.$

From (9.31), (9.32), (9.34), and (9.33), we obtain

(9.36)
$\|yy_{x}-\widetilde{y}\widetilde{y}_{x}-(y_{a}y_{a,x}-\widetilde{y}_{a}\widetilde{y}_{a,x})\|_{L^{1}\big{(}(0,T);L^{2}(0,L)\big{)}}\leq
C\rho^{1/2}\|\varphi-\widetilde{\varphi}\|_{L^{2}(0,L)}.$

Using (9.33) and (9.36), and applying Proposition 3.5 to $Y$, we derive from
(9.35) that

$\|(y-y_{a}-\widetilde{y}+\widetilde{y}_{a})(T,\cdot)\|_{L^{2}(0,L)}\leq
C\rho^{1/2}\|\varphi-\phi\|_{L^{2}(0,L)}.$

Assertion (9.23) follows.

We next consider the general case. One can bring this case to the previous
case as follows. Set

$\rho=\|y_{0}\|_{L^{2}(0,L)}+\|y_{T}\|_{L^{2}(0,L)}.$

Without loss of generality, one might assume that $\rho>0$ since otherwise,
one just takes zero as a control.

Fix $\varepsilon>0$ small. By Proposition 7.1, there exists $v_{1}\in
H^{1/3}(0,\varepsilon)$ such that if $y_{1}\in X_{\varepsilon}$ is the
solution of (9.17) with $y_{1}(\cdot,L)=v_{1}$ and $y_{2}\in X_{\varepsilon}$
is the solution of (9.18) with $y_{2}(\cdot,L)=0$ then

$y_{1}(\varepsilon,\cdot)=0,\quad\mbox{ and
}\quad\int_{0}^{L}y_{2}(\varepsilon,\cdot)\phi\geq c.$

Set

$u=\rho^{1/2}\gamma^{1/2}v_{1}\mbox{ in }(0,\varepsilon).$

Then the unique solution $y\in X_{\varepsilon}$ of the KdV system (1.1)
verifies the properties of the previous case, for large positive $\gamma$. We
are now in the position to apply the previous case with the initial datum
$y(\varepsilon,\cdot)$. The proof is complete. ∎

##  Appendix A Hardy type inequalities

This section is devoted to the following result related to the Hardy
inequality.

######  Lemma A.1.

Let $-\infty<a<b<c\leq+\infty$, and $0<s<1/2$ and let $u\in H^{s}(a,b)$. Set

$v=\left\\{\begin{array}[]{cl}u&\mbox{ in }(a,b),\\\\[6.0pt] 0&\mbox{ in
}(b,c).\end{array}\right.$

Then $v\in H^{s}(a,c)$ and

$\|v\|_{H^{s}(a,c)}\leq C\|u\|_{H^{s}(a,b)},$

for some positive constant $C$ depending only on $a,b,c$, and $s$.

###### Proof.

For notational ease, we assume that $a=-1$ and $b=0$. Without loss of
generality, we then can assume that $c=+\infty$. Let $V\in H^{s}(\mathbb{R})$
be an extension of $u$ such that

$\|V\|_{H^{s}(\mathbb{R})}\leq C\|u\|_{H^{s}(-1,0)}.$

Applying [36, Theorem 1.1] with $\gamma=-s$, $\tau=2$, $p=2$ to $V$, one
obtains

$\||x|^{-s}V\|_{L^{2}(\mathbb{R})}\leq C\|V\|_{H^{s}(\mathbb{R})}.$

The condition $s<1/2$ is required here. This yields

(A.1) $\||x|^{-s}u\|_{L^{2}(-1,0)}\leq C\|u\|_{H^{s}(-1,0)}.$

Using the equivalent Gagliardo-Nirenberg definition of the semi-norm $H^{s}$,
we have

$\|v\|_{H^{s}(-1,+\infty)}^{2}\sim\int_{-1}^{\infty}\int_{-1}^{\infty}\frac{|v(x)-v(y)|^{2}}{|x-y|^{1+2s}}\,dx\,dy+\int_{-1}^{\infty}|v|^{2}\,dx.$

Since

$\int_{-1}^{\infty}\int_{-1}^{\infty}\frac{|v(x)-v(y)|^{2}}{|x-y|^{1+2s}}\,dx\,dy\leq\int_{-1}^{0}\int_{-1}^{0}\frac{|u(x)-u(y)|^{2}}{|x-y|^{1+2s}}\,dx\,dy+C_{s}\int_{-1}^{0}\frac{|u(x)|^{2}}{|x|^{2s}}\,dx,$

and

$\int_{-1}^{\infty}|v|^{2}\,dx=\int_{-1}^{0}|u|^{2}\,dx,$

we derive from (A.1) that

$\|v\|_{H^{s}(-1,+\infty)}\leq C\|u\|_{H^{s}(-1,0)}.$

The proof is complete. ∎

##  Appendix B On the zeros of $\det_{Q}$

We begin this section with

######  Lemma B.1.

Let $L>0$, $z\in\mathbb{C}$, and $\varphi\in C^{\infty}([0,L])$ be such that

$\varphi_{xxx}+\varphi_{x}+iz\varphi=0\mbox{ in }[0,L],$

and

$\varphi(0)=\varphi(L)=\varphi_{x}(L)=0.$

Then

$\Im(z)\geq 0.$

###### Proof.

Set

$\Psi(t,x)=\Re\Big{\\{}e^{izt}\varphi(x)\Big{\\}}\mbox{ in
}\mathbb{R}_{+}\times[0,L].$

Then $\Psi$ is a solution of the linearized KdV system

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

We then derive that

$\frac{d}{dt}\int_{0}^{L}|\Psi(t,x)|^{2}\,dx\leq 0.$

This implies

$\Im(z)\geq 0.$

The proof is complete. ∎

The following result is useful.

######  Lemma B.2.

Let $L\in{\mathcal{N}}_{D}$. Then $\det Q(z+iq/2)\neq 0$ for $z\in\mathbb{R}$.

###### Proof.

We prove the assertion by contradiction. Assume that $\det Q(z+iq/2)=0$ for
some $z\in\mathbb{R}$. Then there exists $\varphi\in C^{\infty}[0,L]$ such
that

$\varphi_{xxx}+\varphi_{x}+i(z+iq/2)\varphi=0\mbox{ in }[0,L],$

and

$\varphi(0)=\varphi(L)=\varphi_{x}(L)=0.$

Applying Lemma B.1, we have

$\Im(z+iq/2)=q/2\leq 0.$

We have a contradiction. ∎

##  Appendix C A lemma related to the moment method

The following lemma from [20] is used in the proof of Proposition 3.1.

######  Lemma C.1.

Let $\varphi$ be an analytic function in $\mathbb{C}$ such that $\varphi$ has
a finite number of zeros on the real line, and

(C.1) $|\varphi(z)|\leq c_{1}e^{c_{2}|z|^{\alpha}}\mbox{ in }\mathbb{C},$

for some $0<\alpha<1$, and $c_{1},c_{2}>0$. Let $T_{1},T_{2}>0$, and $h\in
H^{s}(\mathbb{R})$ for some $s\leq 0$ with support in $(0,T_{1})$. There
exists $g\in C^{\infty}(\mathbb{R})$ with support in $[T_{1},T_{1}+T_{2}]$
such that if $z$ is a real solution of order $m$ of the equation
$\varphi(z)=0$, then $z$ is a also a real solution of order $m$ of the
equation $\hat{h}-\hat{g}=0$, and

(C.2) $\|g\|_{H^{k}(\mathbb{R})}\leq C_{k}\|h\|_{H^{s}(\mathbb{R})}\mbox{ for
}k\in\mathbb{N},$

for some positive constant $C_{k}$ depending only on $k$, $T_{1}$, $T_{2}$,
$s$, and real zeros and their multiplicity of $\varphi$.

###### Proof.

The proof of Lemma C.1 is as in [20], where a special case is considered. The
construction of $g$, inspired by the moment method, see e.g. [44], can be done
as follows. Set $\eta(t)=e^{-1/(t^{2}-(T_{2}/2)^{2})}\mathds{1}_{|t|<T_{2}}$
for $t\in\mathbb{R}$. Assume that $z_{1}$, …, $z_{k}$ are real, distinct
solutions of the equation $\varphi(z)=0$, and $m_{1}$, …, $m_{k}$ are the
corresponding orders. Set, for $z\in\mathbb{C}$,

$\zeta(z)=\sum_{i=1}^{k}\left(\hat{\eta}(z-z_{i})\mathop{\prod_{j=1}^{k}}_{j\neq
i}(z-z_{j})^{m_{j}}\Big{(}\sum_{l=0}^{m_{i}}c_{i,l}(z-z_{i})^{l}\Big{)}\right),$

where $c_{i,l}\in\mathbb{C}$ is chosen such that

$\frac{d^{l}}{dz^{l}}\Big{(}e^{i(T_{1}+T_{2}/2)z}\zeta(z)\Big{)}_{z=z_{i}}=\frac{d^{l}}{dz^{l}}\hat{h}_{3}(z_{i})\mbox{
for }0\leq l\leq m_{i},\;1\leq i\leq k.$

This can be done since $\hat{\eta}(0)\neq 0$. Since

$|\hat{\eta}(z)|\leq Ce^{T_{2}|\Im(z)|/2},$

and, by [44, Lemma 4.3],

$|\hat{\eta}(z)|\leq C_{1}e^{-C_{2}|z|^{1/2}}\mbox{ for }z\in\mathbb{R},$

using (C.1), and applying Paley-Wiener’s theorem, one can prove that $\zeta$
is the Fourier transform of a function $\psi$ of class $C^{1}$; moreover,
$\psi$ has the support in $[-T_{2}/2,T_{2}/2]$. Set, for $z\in\mathbb{C}$,

$g(t)=\psi(t+T_{1}+T_{2}/2).$

Using the fact $\hat{g}(z)=e^{i(T_{1}+T_{2}/2)z}\zeta(z)$, one can check that
$\hat{g}-\hat{h}$ has zeros $z_{1}$, …, $z_{k}$ with the corresponding orders
$m_{1}$, …, $m_{k}$. One can check that

$\|\psi\|_{H^{k}(\mathbb{R})}\leq
C_{T,L,k}\sum_{i=1}^{k}\sum_{l=0}^{m_{i}}\left|\frac{d^{l}}{dz^{l}}\hat{h}(z_{i})\right|,$

which yields

$\|\psi\|_{H^{k}(\mathbb{R})}\leq C_{T,L,k}\|h\|_{H^{s}(\mathbb{R})}.$

The required properties of $g$ follow. ∎

##  Appendix D Scilab program for computing $L_{n}$ and checking the local
controllability property for $n=0,1,2,3$.

Here is the Scilab program which gives the results in Remark 1.3.

⬇

clc

a=0;

b=0;

L=0;

q=0;

alpha=0;

beta=0;

value=0;

function[B] = B(x)

B = (4*x^2*%e^(-6*x) - x^2)^(1/2);

endfunction

function[F] = F(x)

F = B(x) * cos (B(x)) + x* sin (B(x));

endfunction

function[l] = l(x)

l = (B(x)*B(x) - 3*x^2)^(1/2);

endfunction

k=0; j=-0.44; j0=j;

//Here are the outcomes for n=0;

//value=-0.0140641; L=4.5183604; a=-0.5065520; b=4.6027563; q=0.2354919;

//k=1; j=-0.67; j0=j;

// // Here are the outcomes for n=1:

// // value=-0.0061256; L=10.866906; a=-0.6903700; b=10.932497; q=0.1291104;

//k=2; j=-0.78; j0=j;

//// Here are the outcomes for n=2:

//// value=-0.0036196; L=17.177525; a=-0.7947960; b=17.232599; q=0.0933315;

//k=3; j=-0.86; j0=j;

//// Here are the outcomes for n=3:

//// value=-0.0024525; L=23.476776; a=-0.8687610; b=23.524949; q=0.0744156;

while j > -2

j=j - 10^(-6);

if (B(j)> %pi + 2*k*%pi) & (B(j)<3*%pi/2+ 2*k*%pi) & (F(j)<0)

then j0=j0-10^(-6);

else

j=-3;

end

end

a=j0

b=B(a)

L=l(a)

q= - 2*a*(a^2 + b^2)/ L^3

alpha=-a/L

beta=-b/L

function[p] = phix(x)

p = -alpha*beta*%e^(alpha*x)*cos(beta*x)

+ beta^2*%e^(alpha*x)*sin(beta*x)

- 2*alpha*beta*%e^(-2*alpha*x)

+ 3*alpha^2*%e^(alpha*x)*sin(beta*x)

+ 3*alpha*beta*%e^(alpha*x)*cos(beta*x);

endfunction

function[p] = P(x)

p = inv_coeff([-q/2 + %i*x 1 0 1]);

endfunction

function[g] = G(z,x)

lambda= roots(P(z));

g = (lambda(1)*%e^(lambda(1)*L) - lambda(2)*%e^(lambda(2)*L))

*%e^(lambda(3)*x)

+ (lambda(2)*%e^(lambda(2)*L) - lambda(3)*%e^(lambda(3)*L))

*%e^(lambda(1)*x)

+(lambda(3)*%e^(lambda(3)*L) - lambda(1)*%e^(lambda(1)*L))

*%e^(lambda(2)*x);

endfunction

function[h] = H(z,x)

h = abs(G(z,x))*abs(G(z,x))*phix(x);

endfunction

function[omega]=Omega(z)

function[xi] = Xi(x)

xi = abs(G(z,x))*abs(G(z,x))*phix(x);

endfunction

omega = intg(0,L,Xi);

end

value=Omega(0)

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