# Existence of global solutions to the nonlocal mKdV equation on the line

Anran Liu  Engui Fan∗, School of Mathematical Sciences and Key Laboratory of
Mathematics for Nonlinear Science, Fudan University, Shanghai,200433, China
* Corresponding author and e-mail address<EMAIL_ADDRESS>
###### Abstract

In this paper, we address the existence of global solutions to the Cauchy
problem for the integrable nonlocal modified Korteweg-de vries (nonlocal mKdV)
equation with the initial data $u_{0}\in H^{3}(\mathbb{R})\cap
H^{1,1}(\mathbb{R})$ with the $L^{1}(\mathbb{R})$ small-norm assumption. A
Lipschitz $L^{2}$-bijection map between potential and reflection coefficient
is established by using inverse scattering method based on a Riemann-Hilbert
problem associated with the Cauchy problem. The map from initial potential to
reflection coefficient is obtained in direct scattering transform. The inverse
scattering transform goes back to the map from scattering coefficient to
potential by applying the reconstruction formula and Cauchy integral operator.
The bijective relation naturally yields the existence of a global solutions in
a Sobolev space $H^{3}(\mathbb{R})\cap H^{1,1}(\mathbb{R})$ to the Cauchy
problem.

###### keywords:

Nonlocal mKdV equation; Riemann-Hilbert problem; Plemelj projection operator;
Lipschitz continuous; global solutions. Mathematics Subject Classification:
35P25; 35Q51; 35Q15; 35A01; 35G25.

††journal:

###### Contents

  1. 1 Introduction and main results
  2. 2 Direct scattering transforms
    1. 2.1 Lipschitz continuity of the Jost functions
    2. 2.2 Lipschitz continuity of scattering data
  3. 3 Inverse scattering transform
    1. 3.1 Set-up of RH problem
    2. 3.2 Solvability of the RH problem
    3. 3.3 Estimate on solutions to the RH problem
  4. 4 Reconstruction and estimates of the potential
    1. 4.1 Estimates on the negative half-line
    2. 4.2 Estimates on the positive half-line
  5. 5  Global existence and Lipschitz continuity
    1. 5.1 Time evolution of scattering data
    2. 5.2 The proof of main results

##  1 Introduction and main results

In this paper, we establish the global existence of solutions to the Cauchy
problem for the nonlocal mKdV equation

$\displaystyle u_{t}(x,t)+u_{xxx}(x,t)+6\sigma u(x,t)u(-x,-t)u_{x}(x,t)=0,$
(1) $\displaystyle u(x,0)=u_{0}(x),$ (2)

where $u_{0}\in H^{3}(\mathbb{R})\cap H^{1,1}(\mathbb{R})$ and $\sigma=\pm 1$
denote the focusing and defocusing cases, respectively.

The nonlocal mKdV equation (1), introduced in [1, 2], can be regarded as the
integrable nonlocal extension of the classical mKdV equation

$u_{t}(x,t)+u_{xxx}(x,t)+6\sigma u^{2}(x,t)u_{x}(x,t)=0,$ (3)

by replacing $u^{2}(x,t)$ with the _PT_ -symmetric term $u(x,t)u(-x,-t)$ [3].
In physical application, the nonlocal mKdV equation (1) possesses the shifted
parity and delayed time reversal symmetry, and thus it can be related to the
Alice-Bob system [4]. For instance, a special approximate solution of the
nonlocal mKdV was applied to theoretically capture the salient features of two
correlated dipole blocking events in atmospheric dynamical systems [5].

There is much work on the study of various mathematical properties for the
nonlocal mKdV equation (1). The $N$-soliton solutions for the nonlocal mKdV
equation (1) with zero boundary conditions were constructed by using the
Darboux transformation and the inverse scattering transform respectively [7,
8]. Further the Riemann-Hilbert (RH) method was used to construct N-soliton
solutions for the nonlocal mKdV equations (1) with nonzero boundary conditions
[6]. The long-time asymptotics for the nonlocal mKdV equation (1) with
decaying initial data was investigated in [9] via the nonlinear steepest-
descent method developed by Deift and Zhou [10]. Recently, we obtained the
long time asymptotic behavior for the Cauchy problem of the nonlocal mKdV
equation (1) with nonzero initial data in the solitonic regions by using the
$\bar{\partial}$-steepest-descent method [11, 12]. This method, introduced by
McLaughlin and Miller [13, 14], has been extensively implemented in the long-
time asymptotic analysis and the soliton resolution conjecture of some
integrable systems [15, 16, 17, 18, 19]. However, the existence global
solutions to the Cauchy problem (1)-(2) for the nonlocal mKdV equation is
still unknown to our best knowledge. A technical difficulty to apply partial
differential analytical technique for proving global existence of the nonlocal
mKdV equation (1) comes from the fact that the mass and energy conservation
laws to the equation (1) are in the form

$\displaystyle I_{0}=\int_{\mathbb{R}}q(x,t)\overline{q(-x,-t)}dx,$
$\displaystyle
I_{1}=\int_{\mathbb{R}}[q_{x}(x,t)\overline{q_{x}(-x,-t)}-\sigma
q^{2}(-x,t)\overline{q^{2}(-x,-t)}]dx,$

do not preserve any reasonable norm and may be negative. In contrast with
this, the mass and energy conservation laws of the classical mKdV equation (3)
allows to obtain a priory estimates for establishing a unique global solution.

The main purpose in the present paper is to overcome this difficulty and
establish the global existence of solutions to the Cauchy problem (1)-(2) in
an appropriate Sobolev space by applying the inverse scattering theory. Our
principal result is now stated as follows.

######  Theorem 1.

Let the initial data $u_{0}\in H^{3}(\mathbb{R})\cap H^{1,1}(\mathbb{R})$ such
that the spectral problem (5) admits no eigenvalues or resonances. Then

  * 1.

There exists a $L^{2}$-bijection map between the potential $u$ and reflection
coefficients $r_{1,2}$,

$H^{3}(\mathbb{R})\cap H^{1,1}(\mathbb{R})\ni u\mapsto r_{1,2}\in
H^{1,1}(\mathbb{R})\cap L^{2,3}(\mathbb{R}),$ (4)

which is Lipschitz continuous.

  * 2.

There exists a unique global solution $u\in C([0,\infty),H^{3}(\mathbb{R})\cap
H^{1,1}(\mathbb{R}))$ to the Cauchy problem (1)-(2). Furthermore, the map

$H^{3}(\mathbb{R})\cap H^{1,1}(\mathbb{R})\ni u_{0}\mapsto u\in
C\left([0,\infty);H^{3}(\mathbb{R})\cap H^{1,1}(\mathbb{R})\right)$

is Lipschitz continuous.

A key in proving the above result is to establish a Lipschitz
$L^{2}$-bijection (4) between solution and scattering coefficient by using
inverse scattering method [21, 22, 23] . The $L^{2}$-bijection (4) implies
that global well-posdeness of the Cauchy problem (1)-(2) in the space
$H^{3}(\mathbb{R})\cap H^{1,1}(\mathbb{R})$.

The structure of the paper is as follows. In Section 2, we focus on the direct
scattering transform to the Cauchy problem (1)-(2). We especially establish
the Lipschitz continuous maps from the initial data to the Jost function and
the reflection coefficient. In Section 3, we carry out the inverse scattering
transform to set up a RH problem associated with the Cauchy problem (1)-(2),
Further the solvability of the RH problem is shown. In Section 4, we
reconstruct and estimate the potential from the solutions of the RH problem on
positive half line $\mathbb{R}^{+}$ and negative half line $\mathbb{R}^{-}$
respectively. We further establish a Lipschitz continuous mapping from the
reflection coefficients to the potentials. In Section 5, we perform the time
evolution of the reflection coefficients and the RH problem. Then, we prove
that there exists a unique global solution to the initial value problem
(1)-(2) of the nonlocal mKdV equation in the space $H^{3}(\mathbb{R})\cap
H^{1,1}(\mathbb{R})$.

##  2 Direct scattering transforms

In this section, we state some main results on the direct scattering transform
associated with the Cauchy problem (1)-(2). The details can be found in [1,
2].

###  2.1 Lipschitz continuity of the Jost functions

The nonlocal mKdV equation (1) admits the Lax pair

$\displaystyle\psi_{x}-iz\sigma_{3}\psi=Q\psi,$ (5)
$\displaystyle\psi_{t}-4iz^{3}\sigma_{3}\psi=(4z^{2}Q-2iz(Q_{x}-Q^{2})\sigma_{3}+2Q^{3}-Q_{xx})\psi,$
(6)

where

$\displaystyle Q=\begin{pmatrix}0&u(x,t)\\\ -\sigma u(-x,-t)&0\end{pmatrix},\
\ \sigma_{3}=\begin{pmatrix}1&0\\\ 0&-1\end{pmatrix}.$

Define the Jost functions $\psi^{\pm}(x,z)$ to the spectral problem (5) with
the following boundary conditions

$\psi^{\pm}(x,z)\sim e^{izx\sigma_{3}},\ x\to\pm\infty.$

Making a transformation

$m^{\pm}(x,z)=\psi^{\pm}(x,z)e^{-izx\sigma_{3}},$

then

$\lim\limits_{x\to\pm\infty}m^{\pm}(x,z)=I,$

and $m^{\pm}(x,z)$ satisfy the Voterra integral equations

$\displaystyle
m^{\pm}(x,z)=I+\int_{\pm\infty}^{x}e^{-iz(y-x)\operatorname{ad}\sigma_{3}}Qm^{\pm}(y,z)dy,$
(7)

where $e^{\operatorname{ad}\sigma_{3}}A:=e^{\sigma_{3}}Ae^{-\sigma_{3}}$.

Denote $m^{\pm}(x,z)=[m_{1}^{\pm}(x,z),m_{2}^{\pm}(x,z)]$ . From symmetry of
Lax pair, we can get

$m_{1}^{\pm}(x,z)=\sigma\Lambda\overline{m_{2}^{\mp}(-x,-\overline{z}}),\quad
m_{2}^{\pm}(x,z)=\Lambda\overline{m_{1}^{\mp}(-x,-\overline{z}}),$ (8)

where

$\Lambda=\begin{pmatrix}0&\sigma\\\ 1&0\end{pmatrix}.$

It can be further shown that functions $m^{+}_{1}(x,z)$ and $m^{-}_{2}(x,z)$
are analytic in $z\in\mathbb{C}^{+}$, whereas the functions $m^{+}_{2}(x,z)$
and $m^{-}_{1}(x,z)$ are analytic in $z\in\mathbb{C}^{-}$. There is a matrix
$S(z)$ satisfying

$m^{+}(x,z)=m^{-}(x,z)e^{izx\sigma_{3}}S(z),$ (9)

where

$S(z)=\begin{pmatrix}a(z)&c(z)\\\ b(z)&d(z)\end{pmatrix}.$

From (9), we deduce that

$\displaystyle a(z)=\det[m_{1}^{+}(x,z),m_{2}^{-}(x,z)]$ (10) $\displaystyle
d(z)=\det[m_{1}^{-}(x,z),m_{2}^{+}(x,z)]$ (11) $\displaystyle
b(z)=\det[m_{1}^{-}(x,z),m_{1}^{+}(x,z)]e^{-2izx}.$ (12)

It can be shown that $a(z)$ is analytic in $C^{+}$ and $a(z)\to 1$ as
$z\to\infty$ in $\overline{C^{+}}$ while $d(z)$ is analytic in $C^{-}$ and
$d(z)\to 1$ as $z\to\infty$ in $\overline{C^{-}}$.

From $(\ref{symmetry})$ we can get the scattering coefficients satisfy the
following symmetry:

$a(z)=\bar{a}(-\bar{z}),d(z)=\overline{d(-\bar{z})},c(z)=-\sigma\overline{b(-\bar{z})}.$

We define the reflection coefficient:

$r_{1}(z)=b(z)/a(z),r_{2}(z)=c(z)/d(z)=-\sigma\overline{b(-z)}/d(z)\quad
z\in\mathbb{R}.$

The determinant of $S(z)$ is

$a(z)d(z)+\sigma b(z)\overline{b(-\bar{z})}=1.$

In the follows, we prove the existence of $m^{\pm}(x,z)$. For

$f(\cdot,z)=(f_{1}(\cdot,z),f_{2}(\cdot,z))^{\top}\in L^{\infty}(\mathbb{R}),$

define

$(K_{u}f)(x,z):=-\int_{x}^{\infty}{\rm diag}(1,e^{2iz(y-x))}Q(y)f(y,z)dy,$
(13)

then the equation (7) can be written as

$(I-K_{u})m_{1}^{+}(x,z)=e_{1}.$ (14)

Then the operators of $K_{u}$ has the following property.

######  Lemma 2.1.

Let $u\in L^{1}(\mathbb{R})$, for fixed $z\in\overline{\mathbb{C}^{+}}$,
$I-K_{u}$ is an invertible operator in $L^{\infty}(\mathbb{R})$.

###### Proof.

Notice that $z\in\overline{\mathbb{C}^{+}}$ and

$\displaystyle|K_{u}f(x,z)|$
$\displaystyle\leq|(K_{u}f)_{1}(x,z)|+|(K_{u}f)_{2}(x,z)|$
$\displaystyle=|\int_{x}^{\infty}u(-y)f_{1}dy|+|\int_{x}^{\infty}u(y)e^{2iz(y-x)}f_{2}dy|$
$\displaystyle\leq\|u\|_{L^{1}(\mathbb{R})}\|f\|_{L^{\infty}(\mathbb{R})},$

which implies that $K_{u}$ is a bounded operator in $L^{\infty}(\mathbb{R})$
for any fixed $z\in\overline{\mathbb{C}^{+}}$.

Similar to the analysis described above, we have for $n\geq 1$,

$|K_{u}^{n}f|\leq(1/n!)\|u\|_{L^{1}(\mathbb{R})}^{n}\|f\|_{L^{\infty}(\mathbb{R})}.$
(15)

From above analysis we know for any fixed $z\in\overline{\mathbb{C}^{+}}$,
$K_{u}^{n}$ is a bounded operator in $L^{\infty}(\mathbb{R})$ and

$\|K_{u}^{n}\|_{L^{\infty}(\mathbb{R})\to
L^{\infty}(\mathbb{R})}\leq(1/n!)\|u\|_{L^{1}(R)}^{n},$

which yields $1-K_{u}$ is an invertible operator in $L^{\infty}(\mathbb{R})$.
Moreover

$\|(1-K_{u})^{-1}\|_{L^{\infty}(\mathbb{R})\to L^{\infty}(\mathbb{R})}\leq
e^{\|u\|_{L^{1}(\mathbb{R})}}.$

∎

Next, we study the asymptotics of the Jost functions $m^{\pm}(x,z)$.

######  Lemma 2.2.

If $u\in H^{3}(\mathbb{R})\cap H^{1,1}(\mathbb{R})$, then as $|{\rm
Im}z|\rightarrow\infty$, for every $x\in\mathbb{R}$,

$\displaystyle m_{1}^{\pm}(x;z)=$ $\displaystyle
e_{1}+p_{1}^{\pm}(x)(2iz)^{-1}+q_{1}^{\pm}(x)(2iz)^{-2}+g_{1}^{\pm}(x)(2iz)^{-3}+\mathcal{O}(z^{-4}),$
$\displaystyle m_{2}^{\pm}(x;z)=$ $\displaystyle
e_{2}+p_{2}^{\pm}(x)(2iz)^{-1}+q_{2}^{\pm}(x)(2iz)^{-2}+\mathcal{O}(z^{-3}),$

where

$\displaystyle p_{1}^{\pm}(x)=\left[\sigma\int_{x}^{\pm\infty}u(-y)u(y)dy,\
-\sigma u(-x)\right]^{\top},$ (16) $\displaystyle
q_{1}^{\pm}(x)=\begin{bmatrix}\sigma\int_{x}^{\pm\infty}\partial
u(-y)u(y)dy+\int_{x}^{\pm\infty}u(-x_{1})u(x_{1})\int_{x_{2}}^{\pm\infty}u(-x_{2})u(x_{2})\\\
-\sigma\partial u(-x)-u(-x)\int_{x}^{\pm\infty}u(-y)u(y)dy\end{bmatrix},$
$\displaystyle g_{1}^{\pm}(x)=$
$\displaystyle\begin{bmatrix}\sigma\int_{x}^{\pm\infty}u(y)\partial^{2}u(-y)dy+\int_{x}^{\pm\infty}u(x_{1})\partial
u(-x_{1})\int_{x_{1}}^{\pm\infty}u(-x_{2})u(x_{2})+\int_{x}^{\pm\infty}u(-y)^{2}u(y)^{2}dy\\\
+\sigma\int_{x}^{\pm\infty}u(-x_{1})u(x_{1})\int_{x_{1}}^{\pm\infty}u(-x_{2})u(x_{2})\int_{x_{2}}^{\pm\infty}u(-x_{3})u(x_{3})+\int_{x}^{\pm\infty}u(-x_{1})u(x_{1})\int_{x_{1}}^{\pm\infty}u(x_{2})\partial
u(-x_{2})\\\ \\\ -\partial u(-x)\int_{x}^{\pm\infty}u(-y)u(y)dy-\sigma
u(-x)\int_{x}^{\pm\infty}u(-x_{1})u(x_{1})\int_{x_{1}}^{\pm\infty}u(-x_{2})u(x_{2})\\\
-u(-x)\int_{x}^{\pm\infty}\partial
u(-y)u(y)dy-u^{2}(-x)u(x)-\sigma\partial^{2}u(-x)\end{bmatrix},$
$\displaystyle
p_{2}^{\pm}(x)=\left[-u(x),\sigma\int_{x}^{\pm\infty}u(y)u(-y)dy\right]^{\top},$
(17) $\displaystyle q_{2}^{\pm}(x)=\begin{bmatrix}-\sigma\partial
u(-x)-u(-x)\int_{x}^{\pm\infty}u(-y)u(y)dy\\\
\sigma\int_{x}^{\pm\infty}\partial
u(-y)u(y)dy+\int_{x}^{\pm\infty}u(-x_{1})u(x_{1})\int_{x_{2}}^{\pm\infty}u(-x_{2})u(x_{2})\end{bmatrix}.$

###### Proof.

We will only to prove the statement for $m_{1}^{\pm}(x;z)$, while
$m_{2}^{\pm}(x;z)$ is similar to be proved. Rewriting (7) as the component
form:

$m_{11}^{+}(x;z)=1-\int_{x}^{\infty}u(y)m_{21}^{+}(y;z)dy.$
$m_{21}^{+}(x;z)=\sigma\int_{x}^{\infty}e^{2iz(y-x)}u(-y)m_{11}^{+}(y;z)dy.$

We have proved that for every $x\in\mathbb{R}$, $m_{1}^{\pm}(x;z)$ is analytic
in $z\in\mathbb{C}^{+}$. Noticing $u\in H^{3}(\mathbb{R})\cap
H^{1,1}(\mathbb{R})\hookrightarrow L^{\infty}$ which yields (7) is bounded for
every $z\in\mathbb{C}^{+}$ and the integrand converges to $e_{1}$ as
$|Imz|\rightarrow\infty$. Integrating by part and recalling $u\in
H^{3}(\mathbb{R})\cap H^{1,1}(\mathbb{R})$ we can get

$\displaystyle m_{11}^{+}(x;z)=$ $\displaystyle
1+1/(2iz)\sigma\int_{x}^{\infty}u(y)u(-y)m_{11}^{+}dy+1/(2iz)^{2}\sigma\int_{x}^{\infty}u(y)\partial
u(-y)m_{11}^{+}dy$
$\displaystyle+1/(2iz)^{2}\sigma\int_{x}^{\infty}u(y)u(-y)m_{21}^{+}dy+\mathcal{O}(z^{-3})$

and

$m_{21}^{+}(x;z)=-\sigma u(-x)m_{11}^{+}/(2iz)-\sigma(\partial
u(-x)m_{11}^{+}-u(x)u(-x)m_{21}^{+})/(2iz)^{2}+\mathcal{O}(z^{-3}).$

Letting $|z|\rightarrow\infty$ and noticing $m_{1}^{+}\rightarrow e_{1}$ as
$|Imz|\rightarrow\infty$ we get the expanding formula of $m_{1}^{\pm}(x;z)$.

∎

Similar to above analysis, we define $f(x;z)$ as a binary function :

$f(x,z)=(f_{1}(x;z),f_{2}(x;z))^{\top}\in L_{x}^{\infty}(\mathbb{R})\otimes
L_{z}^{2}(\mathbb{R}).$

$K_{u}$ $\partial_{z}K_{u}$ as operators:

$K_{u}f(x;z):=-\int_{x}^{\infty}{\rm diag}(1,e^{2iz(y-x))})Q(y)f(y,z)dy,$
$(\partial_{z}K_{u}f)(x,z):=-\int_{x}^{\infty}{\rm
diag}(0,2i(y-x)e^{2iz(y-x)})Q(y)f(y,z)dy,$

We need to proof the following lemma first.

######  Lemma 2.3.

If $w\in L^{2}(\mathbb{R})$, then

$\sup_{x\in\rm{\mathbb{R}}}\left\|\int_{x}^{\infty}e^{-2iz(x-y)}w(y)dy\right\|_{L^{2}_{z}(\mathbb{R})}\leq\sqrt{\pi}\|w\|_{L^{2}(\mathbb{R})}.$
(18)

If $w\in H^{3}(\mathbb{R})$, then for every $n=1,2,3$, we have

$\sup_{x\in\rm{\mathbb{R}}}\left\|(2iz)^{n}\int_{x}^{\infty}e^{-2iz(x-y)}w(y)dy+\sum_{k=0}^{n-1}(-2iz)^{k+1}\partial_{x}^{k}w(x)\right\|_{L^{2}_{z}(\mathbb{R})}\leq\sqrt{\pi}\|\partial^{n}_{x}w\|_{L^{2}(\mathbb{R})}.$
(19)

If $w\in L^{2,1}(\mathbb{R})$, then for every $x_{0}\in\mathbb{R}^{+}$, we
have

$\sup_{x\in(x_{0},+\infty)}\left\|\langle
x\rangle\int_{x}^{\infty}e^{-2iz(x-y)}w(y)dy\right\|_{L^{2}_{z}(\mathbb{R})}\leq\sqrt{\pi}\|w\|_{L^{2,1}(x_{0},+\infty)}.$
(20)

Furthermore, if $w\in H^{1,1}(\mathbb{R})$, then for every
$x_{0}\in\mathbb{R}^{+}$, we have

$\sup_{x\in(x_{0},+\infty)}\left\|\langle
x\rangle[(2iz)\int_{x}^{\infty}e^{-2iz(x-y)}w(y)dy+w(x)]\right\|_{L^{2}_{z}(\mathbb{R})}\leq\sqrt{\pi}\|w\|_{H^{1,1}(x_{0},+\infty)}$
(21)

and for every $x_{0}\in\mathbb{R}$, we have

$\sup_{x\in\mathbb{R}}\left\|(2iz)\int_{x}^{\infty}(y-x)e^{-2iz(x-y)}w(y)dy\right\|_{L^{2}_{z}(\mathbb{R})}\leq\sqrt{\pi}\|w\|_{H^{1,1}(\mathbb{R})},$
(22)

where $\langle x\rangle=(1+x^{2})^{\frac{1}{2}}$.

###### Proof.

The bounds (18),(19) for $n=1$, and (20) were given in [23]. It remains to
prove the estimate (19) for $n=2,3$, (21) and (22). For every $x\in\mathbb{R}$
and $z\in\mathbb{R}$, define

$f(x;z)=\int_{x}^{\infty}e^{-2iz(x-y)}w(y)dy=\int_{0}^{\infty}e^{2izy}w(x+y)dy.$

Using the Plancherel’s theorem, we have

$\|f(x;z)\|_{L^{2}_{z}}^{2}=\pi\int_{0}^{\infty}|w(x+y)|^{2}dy=\pi\int_{x}^{\infty}|w(y)|^{2}dy.$

Further more, if $x\in\mathbb{R^{+}}$,we have

$\|f(x;z)\|_{L^{2}_{z}}^{2}=\pi\int_{0}^{\infty}|w(x+y)|^{2}dy=\pi\int_{x}^{\infty}|w(y)|^{2}dy\leq\pi\langle
x\rangle^{-2}\int_{x}^{\infty}\langle y\rangle^{2}|w(y)|^{2}dy,$

which yields (20).

Integrate by part, we get

$2izf(x;z)+w(x)=\int_{-\infty}^{x}e^{-2iz(x-y)}\partial_{y}w(y)dy,$ (23)

and

$(2iz)^{2}f(x;z)+2izw(x)+\partial_{x}w(x)=\int_{x}^{\infty}e^{-2iz(x-y)}\partial^{2}_{y}w(y)dy.$

We can easily get

$\left\|2iz\int_{x}^{\infty}e^{-2iz(x-y)}w(y)dy+w(x)\right\|_{L^{2}_{z}}^{2}={\pi}\int_{x}^{\infty}|\partial_{y}w(y)|^{2}dy,$

and

$\begin{split}&\left\|(2iz)^{2}\int_{x}^{\infty}e^{-2iz(x-y)}wdy+2izw(x)+\partial_{x}w(x)\right\|_{L^{2}_{z}}^{2}={\pi}\int_{x}^{\infty}|\partial^{2}_{y}w(y)|^{2}dy.\end{split}$

Then, we get the equation (19) for $n=1,2$. We can get $n=3$ by similar
analysis.

Combine (20) with (23) we can get (21). Replacing $w$ with $(y-x)w$ and
repeating above process, we can get (22). ∎

######  Proposition 2.4.

Let $u\in H^{3}(\mathbb{R})\cap H^{1,1}(\mathbb{R})$, $1-K_{u}$ is an
invertible operator in $L_{x}^{\infty}(\mathbb{R})\otimes
L_{z}^{2}(\mathbb{R})$.

###### Proof.

The conclusion is easy to get from Lemma 2.1. ∎

######  Proposition 2.5.

Let $u\in H^{3}(\mathbb{R})$, then for every $x\in\mathbb{R}$, we have:

$\displaystyle m_{1}^{\pm}-e_{1}\in L_{z}^{2}(\mathbb{R}),$ (24)
$\displaystyle(2iz)(m_{1}^{\pm}-e_{1})-p_{1}^{\pm}\in L_{z}^{2}(\mathbb{R}),$
$\displaystyle(2iz)^{2}(m_{1}^{\pm}-e_{1})-(2iz)p_{1}^{\pm}-q_{1}^{\pm}\in
L_{z}^{2}(\mathbb{R}),$
$\displaystyle(2iz)^{3}(m_{1}^{\pm}-e_{1})-(2iz)^{2}p_{1}^{\pm}-(2iz)q_{1}^{\pm}-g_{1}^{\pm}\in
L_{z}^{2}(\mathbb{R}),$

and

$\displaystyle m_{2}^{\pm}-e_{2}\in L_{z}^{2}(\mathbb{R}),$ (25)
$\displaystyle(2iz)(m_{2}^{\pm}-e_{2})-p_{2}^{\pm}\in L_{z}^{2}(\mathbb{R}),$
$\displaystyle(2iz)^{2}(m_{2}^{\pm}-e_{2})-(2iz)p_{2}^{\pm}-q_{2}^{\pm}\in
L_{z}^{2}(\mathbb{R}).$

If $u\in H^{1,1}(\mathbb{R})$ then for every $x\in\mathbb{R}^{+}$, we have:

$(2iz)\partial_{z}m_{1}^{\pm}\in L_{z}^{2}(\mathbb{R}).$ (26)

###### Proof.

We only prove (24),(26) while the others are similar to prove.

Recalling

$(1-K_{u})[m_{1}^{+}-e_{1}]=K_{u}e_{1},$

and from Proposition 2.4, we know $1-K_{u}$ is an invertible operator in
$L_{x}^{\infty}(\mathbb{R})\otimes L_{z}^{2}(\mathbb{R})$. Therefore
$m_{1}^{+}(x;\cdot)-e_{1}\in L^{2}(\mathbb{R})$ if $K_{u}e_{1}(x;\cdot)\in
L^{2}(\mathbb{R})$. We write $K_{u}e_{1}$ in the following form:

$(K_{u}e_{1})(x;z)=[0,\sigma\int_{x}^{\infty}u(-y)e^{2iz(y-x)dy}]^{\top}.$

Recalling Lemma 2.3, we know

$\sup_{x\in\mathbb{R}}\left\|\int_{x}^{\infty}\mathrm{e}^{2iz(x-y)}\sigma
u(-y)\mathrm{d}y\right\|_{L^{2}}\leq\sqrt{\pi}\|u\|_{L^{2}},$

then

$\displaystyle\|m_{1}^{+}-e_{1}\|_{L_{x}^{\infty}\otimes L_{z}^{2}}$
$\displaystyle\leq\|(1-K_{u})^{-1}\|_{L_{x}^{\infty}\otimes L_{z}^{2}\to
L_{x}^{\infty}\otimes L_{z}^{2}}\|K_{u}e_{1}\|_{L_{x}^{\infty}\otimes
L_{z}^{2}}$ (27) $\displaystyle\leq c\|u\|_{L^{2}}.$

Similar to above analysis, we can get

$(I-K_{u})[(2iz)(m_{1}^{+}-e_{1})-p_{1}^{+}]=(2iz)K_{u}e_{1}-(I-K_{u})p_{1}^{+},$

where

$(I-K_{u})p_{1}^{+}=[0,-u(-x)-\int_{x}^{\infty}e^{2iz(x_{1}-x_{2})}u(-x_{1})\int_{x_{1}}^{\infty}u(-x_{2})u(x_{2})dx_{2}dx_{1}]^{\top}.$

Then, we can get

$\displaystyle(I-K_{u})[(2iz)(m_{1}^{+}-e_{1})-p_{1}^{+}]=2iz\int_{x}^{\infty}e^{-2iz(x-y)}\sigma
u(-y)dy$ (28) $\displaystyle+\sigma
u(-x)+\int_{x}^{\infty}e^{2iz(x_{1}-x)}u(-x_{1})\int_{x_{1}}^{\infty}u(x_{2})u(-x_{2})dx_{2}dx_{1}.$

From Lemma 2.3 again, notice that $u\in H^{2}(\mathbb{R})$, then

$\displaystyle\sup_{x\in\rm{R}}\left\|(2iz)\sigma\int_{x}^{\infty}e^{-2iz(x-y)}u(-y)dy+\sigma
u(-x)\right\|_{L^{2}}\leq c\|\partial_{x}u\|_{L^{2}}$

and

$\displaystyle\sup_{x\in\rm{R}}\|\int_{x}^{\infty}e^{2iz(x_{1}-x)}u(-x_{1})\int_{x_{1}}^{\infty}u(-x_{2})u(x_{2})dx_{2}dx_{1}\|_{L^{2}}\leq
c\|u\|_{L^{2}}^{2}.$

Where $c$ is a constant.

Combing with (28) we have:

$\displaystyle\sup_{x\in\rm{R}}\left\|\int_{x}^{\infty}(2iz)\mathrm{e}^{2iz(x-y)}\sigma
u(-y)\mathrm{d}y-(I-K_{u})p_{1}^{+}e_{2}\right\|_{L^{2}}$ $\displaystyle\leq
c(\|\partial_{x}u\|_{L^{2}}+\|u\|_{L^{2}}^{2}).$

Inserting operator $I-K_{u}$ on $L^{2}_{x}(\mathbb{R})\otimes
L^{2}_{z}(\mathbb{R})$, we get (24) and (25).

We derive on both side of (14),

$(1-K_{u})\partial_{z}m_{1}^{+}=(\partial_{z}K_{u})m_{1}^{+},$

then, we get

$(1-K_{u})\partial_{z}m_{1}^{+}=(\partial_{z}K_{u})[m_{1}^{+}-e_{1}]+\partial_{z}K_{u}e_{1},$
(29)

and

$\displaystyle(I-K_{u})(2iz)\partial_{z}m_{1}^{+}$ (30)
$\displaystyle=\partial_{z}K_{u}[(2iz)(m_{1}^{+}-e_{1})-p_{1}^{+}]+(2iz)\partial_{z}K_{u}e_{1}+\partial_{z}K_{u}p_{1}^{+}.$

Recalling lemma 2.3, we can get for every $x_{0}\in\mathbb{R^{+}}$,

$\displaystyle\sup_{x\in(x_{0},\infty)}\|\langle
x\rangle(m_{1}^{+}-e_{1})\|_{L^{2}}$ (31) $\displaystyle\leq
c\|\int_{x}^{\infty}e^{-2iz(x-y)}\sigma u(-y)dy\|_{L^{2}}\leq
c\|u\|_{L^{2,1}},$ $\displaystyle\sup_{x\in(x_{0},\infty)}\|\langle
x\rangle(2iz)[(m_{1}^{+}-e_{1})-p_{1}^{+}]\|_{L^{2}}$ (32) $\displaystyle\leq
c\|\int_{x}^{\infty}e^{-2iz(x-y)}\partial\sigma u(-y)dy\|_{L^{2}}\leq
c\|u\|_{H^{1,1}}.$

Combine (29) with (31) and recall lemma2.3 we can get:

$\displaystyle\|(I-K_{u})\partial_{z}m_{1}^{+}\|_{L^{2}}\leq\|\int_{x}^{\infty}2i(y-x)\sigma
u(-y)e^{-2iz(x-y)}(m_{11}^{+}-1)dy\|_{L^{2}}$
$\displaystyle+\|\int_{x}^{\infty}2i(y-x)e^{-2iz(x-y)}\sigma
u(-y)dy\|_{L^{2}}$ $\displaystyle\leq
c\|u\|_{L^{1}}\sup_{x\in(x_{0},\infty)}\|\langle
x\rangle(m_{1}^{+}-e_{1})\|_{L^{2}}+c\|u\|_{L^{2,1}}\leq c\|u\|_{H^{1,1}}.$

Combine (32) with (30) and recall lemma2.3 we can get:

$\displaystyle\|(I-K_{u})(2iz)\partial_{z}m_{1}^{+}\|_{L^{2}}\leq\|\int_{x}^{\infty}2i(y-x)e^{-2iz(x-y)}\sigma
u(-y)[(2iz)(m_{11}^{+}-1)-p_{11}^{+}]dy\|_{L^{2}}$
$\displaystyle+\|(2iz)\int_{x}^{\infty}2i(y-x)e^{-2iz(x-y)}\sigma
u(-y)dy\|_{L^{2}}+\|\int_{x}^{\infty}2i(y-x)e^{-2iz(x-y)}\sigma
u(-y)p_{11}^{+}dy\|_{L^{2}}$ $\displaystyle\leq
c\|u\|_{L^{1}}\sup_{x\in(x_{0},\infty)}\|\langle
x\rangle[(2iz)(m_{1}^{+}-e_{1})-p_{1}^{+}]\|_{L^{2}}\leq c\|u\|_{H^{1,1}}.$

Inserting operator $I-K_{u}$ on $L^{2}_{x}(\mathbb{R})\otimes
L^{2}_{z}(\mathbb{R})$, we get (26). We finally improved the lemma. ∎

We has infect constructed the following maps:

$\displaystyle H^{1,1}\ni u\rightarrow z\partial_{z}m_{1}^{\pm}(x;\cdot)\in
L^{2},$ $\displaystyle H^{1,1}\ni u\rightarrow
z\partial_{z}m_{2}^{\pm}(x;\cdot)\in L^{2},$ $\displaystyle H^{3}\ni
u\rightarrow
z^{3}(m_{1}^{\pm}(x;\cdot)-e_{1})-z^{2}p_{1}^{\pm}-zq_{1}^{\pm}-g_{1}^{\pm}\in\L^{2},$
$\displaystyle H^{3}\ni u\rightarrow
z^{3}(m_{2}^{\pm}(x;\cdot)-e_{2})-z^{2}p_{2}^{\pm}-zq_{2}^{\pm}-g_{2}^{\pm}\in\L^{2}.$

Next, we will show this map and remainders of the Jost function in function
space $L^{\infty}_{x}(\mathbb{R})\otimes L^{2}_{z}(\mathbb{R})$ is Lipschitz
continuous.

######  Corollary 2.6.

Let $u,\tilde{u}\in H^{1,1}(\mathbb{R})$ satisfy
$\|u\|_{H^{1,1}(\mathbb{R})}\leq U$ and
$\|\tilde{u}\|_{H^{1,1}(\mathbb{R})}\leq U$ for some $U>0$ then there is a
positive U-dependent constant $C(U)$ such that for every $x\in\mathbb{R}$:

$\left\|\check{m}_{1}^{\pm}(x;\cdot)-\check{\tilde{m}}_{1}^{\pm}(x;\cdot)\right\|_{L^{2}}+\left\|\check{m}_{2}^{\pm}(x;\cdot)-\check{\tilde{m}}_{2}^{\pm}(x;\cdot)\right\|_{L^{2}}\leq
C(U)\|u-\tilde{u}\|_{H^{1,1}},$

where

$\displaystyle\check{m}_{1}^{\pm}(x;\cdot):=(2iz)\partial_{z}m_{1}^{\pm}(x;\cdot),$
$\displaystyle\check{m}_{2}^{\pm}(x;\cdot):=(2iz)\partial_{z}m_{2}^{\pm}(x;\cdot).$

Moreover, if $u,\tilde{u}\in H^{3}(\mathbb{R})$ satisfy
$\|u\|_{H^{3}(\mathbb{R})}\leq U$ and $\|\tilde{u}\|_{H^{3}(\mathbb{R})}\leq
U$ for some $U>0$ then there is a positive U-dependent constant $C(U)$ such
that for every $x\in\mathbb{R}$:

$\left\|\hat{m}_{1}^{\pm}(x;\cdot)-\hat{\tilde{m}}_{1}^{\pm}(x;\cdot)\right\|_{L^{2}}+\left\|\hat{m}_{2}^{\pm}(x;\cdot)-\hat{\tilde{m}}_{2}^{\pm}(x;\cdot)\right\|_{L^{2}}\leq
C(U)\|u-\tilde{u}\|_{H^{3}},$

where

$\displaystyle\hat{m}_{1}^{\pm}(x;\cdot):=(2iz)^{3}(m_{1}^{\pm}(x;\cdot)-e_{1})-(2iz)^{2}p_{1}^{\pm}-(2iz)q_{1}^{\pm}-g_{1}^{\pm},$
$\displaystyle\hat{m}_{2}^{\pm}(x;\cdot):=(2iz)^{3}(m_{2}^{\pm}(x;\cdot)-e_{2})-(2iz)^{2}p_{2}^{\pm}-(2iz)q_{2}^{\pm}-g_{2}^{\pm}.$

###### Proof.

Recalling (14), we can get

$\displaystyle
m_{1}^{+}-\tilde{m}_{1}^{+}=(I-K_{u})^{-1}[K_{u}e_{1}-\tilde{K_{u}}e_{1}]+((I-K_{u})^{-1}-(I-\tilde{K_{u}})^{-1})\tilde{K_{u}}e_{1}$
$\displaystyle=(I-K_{u})^{-1}[K_{u}e_{1}-\tilde{K_{u}}e_{1}]+(I-K_{u})^{-1}(\tilde{K_{u}}-K_{u})(I-\tilde{K_{u}})^{-1}\tilde{K_{u}}e_{1}.$

Where

$\sup_{x\in\mathbb{R}}\|K_{u}e_{1}-\tilde{K_{u}}e_{1}\|_{L^{2}}=\sup_{x\in\mathbb{R}}\|\int_{x}^{\infty}e^{2iz(y-x)}(u-\tilde{u})dy\|_{L^{2}}.$

Using lemma2.3 we can get

$\sup_{x\in\mathbb{R}}\|K_{u}e_{1}-\tilde{K_{u}}e_{1}\|_{L^{2}}\leq
c\sup_{x\in\mathbb{R}}\|u-\tilde{u}\|_{L^{2}}.$

Furthermore, for every $f\in L^{2}_{z}(\mathbb{R})\otimes
L^{\infty}_{x}(\mathbb{R})$,

$\|(K_{u}-\tilde{K_{u}})f\|_{L_{x}^{\infty}\otimes L_{z}^{2}}\leq
C_{2}(U)e^{\|u-\tilde{u}\|_{L^{1}}}\|f\|_{L_{x}^{\infty}\otimes L_{z}^{2}}.$
(33)

Then, we can get

$\sup_{x\in\mathbb{R}}\|m_{1}^{+}-\tilde{m}_{1}^{+}\|_{L^{2}}\leq
c\|u-\tilde{u}\|_{L^{2}}.$

Using lemma2.3 again, we can also get

$\sup_{x\in(x_{0},\infty)}\|\langle
x\rangle(m_{1}^{+}-\tilde{m}_{1}^{+})\|_{L^{2}}\leq
c\|u-\tilde{u}\|_{L^{2,1}}.$ (34)

Direct calculation yields:

$\displaystyle\ \partial_{z}m_{1}^{+}-\partial_{z}\tilde{m}_{1}^{+}$
$\displaystyle=(I-K_{u})^{-1}\partial_{z}K_{u}(m_{1}^{+}-e_{1})-(I-\tilde{K_{u}})^{-1}\partial_{z}\tilde{K_{u}}(\tilde{m}_{1}^{+}-e_{1})$
(35)
$\displaystyle-[(I-K_{u})^{-1}\partial_{z}K_{u}e_{1}-(I-\tilde{K_{u}})^{-1}\partial_{z}\tilde{K_{u}}e_{1}]$
$\displaystyle=(I-K_{u})^{-1}[\partial_{z}K_{u}(m_{1}^{+}-e_{1})-\partial_{z}\tilde{K_{u}}(\tilde{m}_{1}^{+}-e_{1})]$
$\displaystyle+(I-K_{u})^{-1}(\tilde{K_{u}}-K_{u})(I-\tilde{K_{u}})^{-1}\partial_{z}\tilde{K_{u}}\tilde{m}_{1}^{+}$
$\displaystyle-(I-K_{u})^{-1}(\partial_{z}K_{u}e_{1}-\partial_{z}\tilde{K_{u}}e_{1})$
$\displaystyle+(I-K_{u})^{-1}(\tilde{K_{u}}-K_{u})(I-\tilde{K_{u}})^{-1}\partial_{z}\tilde{K_{u}}e_{1}.$

Recalling lemma(2.3) and (34), we get

$\|(2iz)\partial_{z}K_{u}m_{1}^{+}-(2iz)\partial_{z}\tilde{K_{u}}\tilde{m}_{1}^{+}\|_{L^{2}}\leq
C_{1}(U)\|u-\tilde{u}\|_{H^{1,1}},$ (36)

Where $C_{1}(U)$ is an U-dependent positive constant. We notice that $1-K_{u}$
is an invertible operator in $L_{x}^{\infty}(\mathbb{R})\otimes
L_{z}^{2}(\mathbb{R})$ and recall (33) we get the bound (2.6). The other
follow by repeating the same analysis. ∎

###  2.2 Lipschitz continuity of scattering data

The propose of Lemma 2.2 and Proposition 2.5 are to introduce the stand form
of the scattering relations and study properties of that.

######  Lemma 2.7.

If $u\in H^{1,1}(\mathbb{R})$, then the function $a(z)$ is continued
analytically in $\mathbb{C}^{+}$. In addition, we have:

$a(z)-1,d(z)-1,b(z)\in H^{1,1}(\mathbb{R}).$

Moreover, if $u\in H^{3}(\mathbb{R})$, then

$b(z)\in L^{2,3}(\mathbb{R}).$

###### Proof.

By using (2.1), (10) and (12), we get

$b(z)=\sigma\int_{\mathbb{R}}e^{2izy}u(-y)m_{11}^{+}dy,$ (37)
$\partial_{z}b(z)=\sigma\int_{\mathbb{R}}e^{2izy}u(-y)\partial_{z}m_{11}^{+}dy+\sigma\int_{\mathbb{R}}2iye^{2izy}u(-y)m_{11}^{+}dy,$
(38) $\displaystyle a(z)-1=-\int_{\mathbb{R}}u(y)m_{21}^{+}dy,$ (39)
$\displaystyle d(z)-1=\sigma\int_{\mathbb{R}}u(-y)m_{12}^{+}dy,$
$\displaystyle\partial_{z}a(z)=-\int_{\mathbb{R}}u(y)\partial_{z}m_{21}^{+}dy,$
(40)
$\displaystyle\partial_{z}d(z)=\sigma\int_{\mathbb{R}}u(-y)\partial_{z}m_{21}^{+}dy.$

We can easily obtain the following limit for the scattering coefficient $a(z)$
along a contour in $\mathbb{C}^{+}$ extended to $Imz\rightarrow\infty$:

$a(z)\rightarrow 1$

In order to prove $b(z)\in L_{z}^{2,3}(\mathbb{R})$, we rewrite (12) as
following form:

$\displaystyle(2iz)^{3}b(z)=\sigma\int_{\mathbb{R}}e^{2izy}u[(2iz)^{3}(m_{11}^{+}-1)-(2iz)^{2}p_{11}^{+}-(2iz)q_{11}^{+}-g_{11}^{+}]dy$
(41)
$\displaystyle+(2iz)^{3}\sigma\int_{\mathbb{R}}e^{2izy}udy+(2iz)^{2}\sigma\int_{\mathbb{R}}e^{2izy}up_{11}^{+}dy+(2iz)\sigma\int_{\mathbb{R}}e^{2izy}uq_{11}^{+}dy$
$\displaystyle+\sigma\int_{\mathbb{R}}e^{2izy}ug_{11}^{+}dy.$

Using lemma2.3, we get

$\displaystyle\|\sigma\int_{\mathbb{R}}e^{2izy}u[(2iz)^{3}(m_{11}^{+}-1)-(2iz)^{2}p_{11}^{+}-(2iz)q_{11}^{+}-g_{11}^{+}]dy\|_{L^{2}}$
$\displaystyle\leq
c\|u\|_{L^{1}}\sup_{x\in\mathbb{R}}\|(2iz)^{3}(m_{11}^{+}-1)-(2iz)^{2}p_{11}^{+}-(2iz)q_{11}^{+}-g_{11}^{+}\|_{L^{2}}$
$\displaystyle\leq c\|u\|_{L^{1}}\|u\|_{H^{3}}.$

We proved the first term of (41) is bounded in $L^{2}$ space. For the other
terms of (41), using Plancherel’s formula, we can get:

$\displaystyle\|(2iz)^{3}\int_{\mathbb{R}}e^{2izy}udy\|_{L^{2}}\leq
c\|u\|_{H^{3}},$
$\displaystyle\|(2iz)^{2}\int_{\mathbb{R}}e^{2izy}up_{11}^{+}dy\|_{L^{2}}\leq
c\|u\|_{H^{3}},$
$\displaystyle\|(2iz)\int_{\mathbb{R}}e^{2izy}uq_{11}^{+}dy\|_{L^{2}}\|_{L^{2}}\leq
c\|u\|_{H^{3}},$
$\displaystyle\|\int_{\mathbb{R}}e^{2izy}ug_{11}^{+}dy\|_{L^{2}}\|_{L^{2}}\leq
c\|u\|_{H^{3}}.$

Then we get $b(z)\in L^{2,3}(\mathbb{R})$.

In order to prove $\partial_{z}b(z)\in L_{z}^{2,1}(\mathbb{R})$, we rewrite
(38) as following form:

$\displaystyle(2iz)\partial_{z}b(z)$
$\displaystyle=\sigma\int_{\mathbb{R}}e^{2izy}u(2iz)\partial_{z}m_{11}^{+}+\sigma\int_{\mathbb{R}}(2iy)e^{2izy}u[(2iz)(m_{11}^{+}-1)-p_{11}^{+}]dy$
(42)
$\displaystyle+(2iz)\sigma\int_{\mathbb{R}}(2iy)e^{2izy}udy+\sigma\int_{\mathbb{R}}(2iy)e^{2izy}up_{11}^{+}dy.$

Using Lemma 2.3 again, we get

$\displaystyle\|\int_{\mathbb{R}}e^{2izy}u(2iz)\partial_{z}m_{11}^{+}dy\|_{L^{2}}\leq
c\|u\|_{L^{1}}\sup_{x\in\mathbb{R}}\|(2iz)\partial_{z}m_{11}^{+}\|_{L^{2}}$
$\displaystyle\leq c\|u\|_{L^{1}}\|u\|_{H^{1,1}}.$

We proved the first term of (42) is bounded in $L^{2}$ space. For the other
terms of (42), using Plancherel’s formula, we can get:

$\displaystyle\|(2iz)\int_{\mathbb{R}}e^{2izy}(2iy)udy\|_{L^{2}}\leq
c\|u\|_{H^{1,1}},$
$\displaystyle\|\int_{\mathbb{R}}e^{2izy}(2iy)up_{11}^{+}dy\|_{L^{2}}\leq
c\|u\|_{H^{1,1}}.$

We get $\partial_{z}b(z)\in L^{2,1}(\mathbb{R})$. Then we have proved the
bounds of $b(z)$. The conclusion of $a(z),d(z)$ can be proved similar. ∎

It is easy to find that we have constructed the following two maps:

$H^{1,1}(\mathbb{R})\ni u\rightarrow a(z)-1,d(z)-1,b(z)\in
H^{1,1}(\mathbb{R})$

and

$H^{3}(\mathbb{R})\ni u\rightarrow b(z)\in L^{2,3}(\mathbb{R}).$

We will show the two maps are Lipschitz continuous.

######  Corollary 2.8.

Let $u\in H^{1,1}(\mathbb{R})\cap H^{3}(\mathbb{R})$ then,

$H^{1,1}(\mathbb{R})\ni u\rightarrow a(z)-1,d(z)-1,b(z)\in
H_{z}^{1,1}(\mathbb{R})$

and

$H^{3}(\mathbb{R})\ni u\rightarrow b(z)\in L_{z}^{2,3}(\mathbb{R})$

are Lipschitz continuous.

###### Proof.

From the representations (37), (38), (39), (40) and the Lipschitz continuity
of the Jost function $m_{1}^{\pm}$ and $m_{2}^{\pm}$, we can obtain the
Lipschitz continuity of the scattering coefficients. ∎

For the reflection coefficients $r(z)$, we have the following results:

######  Lemma 2.9.

If $u\in H^{1,1}(\mathbb{R})\cap H^{3}(\mathbb{R})$ then we have

$r_{1,2}(z)\in H^{1,1}(\mathbb{R})\cap L^{2,3}(\mathbb{R}).$

As well, the mapping

$H^{1,1}(\mathbb{R})\cap H^{3}(\mathbb{R})\ni u\mapsto r_{1,2}\in
H^{1,1}(\mathbb{R})\cap L^{2,3}(\mathbb{R})$

is Lipschitz continuous.

###### Proof.

Let $r_{1}$ and $\tilde{r}_{1}$ denote the reflection coefficients
corresponding to $u$ and $\tilde{u}$, respectively. Owing to

$r_{1}-\tilde{r}_{1}=\frac{b-\tilde{b}}{a}+\frac{\tilde{b}((\tilde{a}-1)-(a-1))}{a\tilde{a}},$

the Lipschitz continuity of $r_{1}$ follows from the Lipschitz continuity of
$a-1$ and $b$ and we can get similar conclusion of $r_{2}$ . ∎

######  Lemma 2.10.

If $u\in H^{1,1}\cap H^{3}$ with $L^{1}$-norm such that
$\|u\|_{L^{1}}\exp\\{2\|u\|_{L^{1}}\\}<1$, then the spectral problem admits no
eigenvalues or resonances, the scattering coefficients $a(z)$ and $d(z)$ admit
no zeros in $\mathbb{C}^{+}\cup\mathbb{R}$ and $\mathbb{C}^{-}\cup\mathbb{R}$,
respectively.

###### Proof.

The small-norm condition implies that $\|u\|_{L^{1}(\mathbb{R})}<1$. Recall
that $m_{1}^{+}=e_{1}+K_{u}m_{1}^{+}$ in Lemma 2.1 and the operator $I-K_{u}$
is invertible and bounded from $L^{\infty}_{x}$ to $L^{\infty}_{x}$ and we
reach that for every $z\in\mathbb{C}^{+}$,

$\|m_{1}^{+}(\cdot;z)-e_{1}\|_{L^{\infty}}=\|(I-K_{u})^{-1}\|\|K_{u}e_{1}(\cdot;z)\|_{L^{\infty}}\leq\|u\|_{L^{1}}e^{2\|u\|_{L^{1}}}.$

We derive for every $z\in\mathbb{C}^{+}$,

$|a(z)|\geq
1-\left|\int_{\mathbb{R}}u(y)m_{21}^{+}(y;z)dy\right|>1-\|u\|_{L^{1}}e^{2\|u\|_{L^{1}}}.$

Due to the continuity of $a(z)$, we also obtain that

$|a(z)|\geq 1-\|u\|_{L^{1}(\mathbb{R})}e^{2\|u\|_{L^{1}}}>0,\ \
z\in\mathbb{R},$

then $a(z)$ admits no zeros in $\mathbb{C}^{+}\cup\mathbb{R}$. Carrying out a
similar manipulation for $d(z)$, we see that $d(z)$ admits no zeros in
$\mathbb{C}^{-}\cup\mathbb{R}$. ∎

######  Lemma 2.11.

If $u\in L^{2,1}(\mathbb{R})$ with $L^{1}(\mathbb{R})$ small-norm such that

$\displaystyle
1-\|u\|_{L^{1}(\mathbb{R})}(1+2e^{2\|u\|_{L^{1}(\mathbb{R})}})>0,$ (43)

then for every $z\in\mathbb{R}$, we have $|r_{1,2}(z)|<1$.

###### Proof.

Rewrite (37) for $b(z)$ as

$b(z)=\sigma\int_{\mathbb{R}}e^{2izy}u(-y)m_{11}^{+}dy,$ (44)

Under the condition (43).For every $z\in\mathbb{R}$, we obtain

$\displaystyle|b(z)|\leq\|m_{11}^{+}(\cdot;z)-1\|_{L^{\infty}}\|u\|_{L^{1}}+\|u\|_{L^{1}}\leq\|u\|_{L^{1}(\mathbb{R})}+\|u\|_{L^{1}(\mathbb{R})}e^{2\|u\|_{L^{1}(\mathbb{R})}}$
$\displaystyle<1-\|u\|_{L^{1}(\mathbb{R})}e^{2\|u\|_{L^{1}(\mathbb{R})}}\leq
a(z),$

which yields

$|r_{1}(z)|=\frac{|b(z)|}{|a(z)|}<1.$

Similarly, we get $|r_{2}(z)|<1$. ∎

##  3 Inverse scattering transform

In this section, we will set up a RH problem and show the existence and
uniqueness of its solution for the given data $r(z)\in H^{1,1}(\mathbb{R})\cap
L^{2,3}(\mathbb{R})$ .

###  3.1 Set-up of RH problem

Define

$M(x,z)=\begin{cases}\begin{pmatrix}m_{1}^{(+)}&m_{2}^{(-)}\end{pmatrix}\begin{pmatrix}a^{-1}&0\\\
0&1\end{pmatrix}&\operatorname{Im}z>0\\\
\begin{pmatrix}m_{1}^{(-)}&m_{2}^{(+)}\end{pmatrix}\begin{pmatrix}1&0\\\
0&d^{-1}\end{pmatrix},&\operatorname{Im}z<0.\end{cases}$

then $M(x,z)$ satisfy the following RH-problem

######  Problem 1.

Find a matrix function $M(x;z)$ satisfying

(i) $M(x;z)\rightarrow I+\mathcal{O}\left(z^{-1}\right)$ as
$z\rightarrow\infty$.

(ii) For $M(x;z)$ admits the following jump condition

$\displaystyle M_{+}(x;z)$ $\displaystyle=M_{-}(x;z)V_{x}(z),$ (45)

where

$V_{x}(z):=\left(\begin{array}[]{cc}1+\sigma r_{1}r_{2}&\sigma
r_{2}(z)e^{2\mathrm{i}xz}\\\
r_{1}(z)e^{-2\mathrm{i}xz}&1\end{array}\right),z\in\mathbb{R}.$ (46)

The reconstruction formula is given by

$u(x):=2\mathrm{i}\lim_{z\rightarrow\infty}zM_{12}(x,z),u(-x):=2\mathrm{i}\sigma\lim_{z\rightarrow\infty}zM_{21}(x,z).$
(47)

We write (112) in the form

$M_{+}(x;z)-M_{-}(x;z)=M_{-}(x;z)S(x;z),\quad z\in\mathbb{R},$

where

$S(x;z)=\left(\begin{array}[]{cc}\sigma r_{1}r_{2}&\sigma
r_{2}(z)e^{2\mathrm{i}xz}\\\ r_{1}(z)e^{-2\mathrm{i}xz}&0\end{array}\right),$

Introduce a transformation

$\Psi_{\pm}(x;z)=M_{\pm}(x;z)-{I},$

then we obtain a new RH problem for $\Psi(x;z)$

$\displaystyle\Psi_{+}(x;z)-\Psi_{-}(x;z)=\Psi_{-}(x;z)S(x;z)+S(x;z),\quad
z\in\mathbb{R},$ $\displaystyle\Psi_{\pm}(x;z)\rightarrow
0,\quad|z|\rightarrow\infty,\ z\in\mathbb{C}\backslash\mathbb{R}.$

###  3.2 Solvability of the RH problem

We introduce the Cauchy operator

$\mathcal{C}(f)(z)=\frac{1}{2\pi\mathrm{i}}\int_{\mathbb{R}}\frac{f(\zeta)}{\zeta-z}d\zeta,\quad
z\in\mathbb{C}\backslash\mathbb{R}$

and Plemelj projection operator

$\mathcal{P}^{\pm}(f)(x;z):=\lim_{\varepsilon\searrow
0}\frac{1}{2\pi\mathrm{i}}\int_{\mathbb{R}}\frac{f(\zeta)}{\zeta-(z\pm\mathrm{i}\varepsilon)}d\zeta,\quad
z\in\mathbb{R},$ (48)

where $f(z)\in L^{2}(\mathbb{R})$.

######  Proposition 3.1.

( [23]) For every $f\in L^{p}(\mathbb{R})$ with $\ 1\leq p<\infty$, the Cauchy
operator $\mathcal{C}(f)$ is analytic off the real line, decays to zero as
$|z|\rightarrow\infty$, and approaches to $\mathcal{P}^{\pm}(f)$ almost
everywhere when a point $z\in C^{\pm}$ approaches to a point on the real axis
by any non-tangential contour from $\mathbb{C}^{\pm}$. If $1<p<\infty$, then
there exists a positive constant $C_{p}$( $C_{p}=1,p=2$ ) such that

$\|\mathcal{P}^{\pm}(f)\|_{L^{p}}\leq C_{p}\|f\|_{L^{p}},$ (49)

If $f\in L^{1}(\mathbb{R})$, then the Cauchy operator admits the asymptotic

$\lim_{|z|\rightarrow\infty}z\mathcal{C}(f)(z)=-\frac{1}{2\pi
i}\int_{\mathbb{R}}f(s)ds.$ (50)

######  Lemma 3.2.

Let $r_{1,2}(z)\in H^{1}(\mathbb{R})$ , satisfy $|r_{1,2}(z)|\leq 1$ then
there exist positive constants $c_{-}$ and $c_{+}$ such that for every
$x\in\mathbb{R}$ and every column-vector $g\in\mathbb{C}^{2}$, we have

${\rm Re}\ g^{*}(I+S(x;z))g\geq c_{-}g^{*}g,\quad z\in\mathbb{R}$ (51)

and

$\|(I+S(x;z))g\|\leq c_{+}\|g\|,\quad z\in\mathbb{R},$ (52)

where the asterisk denotes the Hermite conjugate.

###### Proof.

The original scattering matrix $S(x;z)$ is not Hermitian due to the fact that
there is no relationship between $a(z)$ and $d(z)$. Therefore, we define
Hermitian part of $S(x;z)$ by

$\begin{split}S_{H}(x;z)&=\frac{1}{2}(S(x;z)+S^{*}(x;z))\\\
&=\left(\begin{array}[]{cc}\sigma\rm
Re(r_{1}r_{2})&\frac{1}{2}(\bar{r}_{1}+\sigma r_{2})e^{2izx}\\\
\frac{1}{2}(r_{1}+\sigma\bar{r}_{2})e^{-2izx}&0\end{array}\right).\end{split}$
(53)

Since $|r_{1,2}(z)|<1$, the 2-order principle minor of the matrix $I+S_{H}$

$1+\sigma{\rm
Re}(r_{1}r_{2})-\frac{1}{4}|r_{1}+\sigma\bar{r}_{2}|^{2}=1-\frac{1}{4}|r_{1}-\sigma\bar{r}_{2}|^{2}>0,$

which indicates that the 1-order principle minor
$1+\sigma\rm{Re}(r_{1}r_{2})>0$. Thus the matrix $I+S_{H}$ is positive
definite.

In view of the algebra theory, for a Hermitian matrix, there exists an unitary
matrix $A$ such that

$A^{*}(I+S_{H})A={\rm diag}(\mu_{+},\mu_{-}),$ (54)

where $\mu_{\pm}$ are the eigenvalues of the matrix $I+S_{H}$

$\mu_{\pm}(z)=\frac{2+\sigma\rm{Re}(r_{1}r_{2})\pm\sqrt{\rm{Re}^{2}(r_{1}r_{2})+|r_{1}+\sigma\bar{r}_{2}|^{2}}}{2}.$

Note $\mu_{+}(z)>\mu_{-}(z)>0$ as $I+S_{H}$ is positive definite. And it
follows from $r_{1,2}(z)\rightarrow 0$ that $\mu_{-}(z)\rightarrow 1$ as
$|z|\rightarrow\infty,\ z\in\mathbb{R}$. Together with $r_{1,2}(k)\in
H^{1}(\mathbb{R})$, there exists a positive constant $c_{-}$ such that
$\mu_{-}>c_{-}$.

Consequently, for every $g\in\mathbb{C}^{2}$, utilizing (54), we have

$c_{-}g^{*}g<\mu_{-}g^{*}g\leq{\rm Re}g^{*}(I+S(x;z))g=g^{*}(I+S_{H})g,$

this completes the proof of the bound (51).

Calculating $(I+S(x;z))g$ componentwise and utilizing $|r_{1,2}(z)|<1$ gives
that

$\begin{split}\|(I+S(x;z))g\|^{2}\leq&2(1+|r_{1}|+|r_{2}|)^{2}\|g\|^{2}\\\
&+2{\rm
Re}\\{((\sigma+r_{1}r_{2})\bar{r}_{2}+r_{1})e^{-2izx}g^{(1)}\overline{g^{(2)}}\\}\\\
\leq&\left((|r_{1}|+1)^{2}+(|r_{2}|+1)^{2}+(|r_{1}|+|r_{2}|)^{2}\right)\|g\|^{2},\end{split}$
(55)

here the norm for a 2-component vector $f$ is
$\|f\|^{2}=|f^{(1)}|^{2}+|f^{(2)}|^{2}$. Therefore, we take

$c_{+}=\sup_{z\in\mathbb{R}}\sqrt{(|r_{1}|+1)^{2}+(|r_{2}|+1)^{2}+(|r_{1}|+|r_{2}|)^{2}}<+\infty,$

then one obtain the bound (52). ∎

######  Lemma 3.3.

Let $r_{1,2}(z)\in H^{1}(\mathbb{R})$ ,satisfy $|r_{1,2}(z)|\leq 1$ then for
every $F(z)\in L^{2}_{z}(\mathbb{R})$, there exists a unique solution
$\Psi(z)\in L_{z}^{2}(\mathbb{R})$ of the equation

$(I-P_{S}^{-})\Psi(z)=F(z),\ z\in\mathbb{R},$ (56)

where $P_{S}^{-}\Psi=P^{-}(\Psi S)$.

###### Proof.

Since $I-P_{S}^{-}$ is a Fredholm operator of the index zero, by Fredholm’s
alternative theorem, there exists a unique solution of the equation (56) if
and only if the homogeneous equation

$(I-P_{S}^{-})g=0,\ z\in\mathbb{R},$ (57)

admits zero solution in $L^{2}_{z}(\mathbb{R})$.

Assume that $g(z)\in L^{2}_{z}(\mathbb{R})$ and $g(z)\neq 0$ is a solution of
equation (57). Define two analytic functions in
$\mathbb{C}\backslash\mathbb{R}$

$g_{1}(z)=\mathcal{C}(gS)(z),\quad g_{2}(z)=\mathcal{C}(gS)^{*}(z).$

The functions $g_{1}(z)$ and $g_{2}(z)$ are well-defined due to $S(z)\in
L_{z}^{2}(\mathbb{R})\cap L_{z}^{\infty}(\mathbb{R})$.

We integrate the function $g_{1}(z)g_{2}(z)$ along the semi-circle of radius R
centered at zero in $\mathbb{C}^{+}$. It follows from Cauchy theorem that

$\oint g_{1}(z)g_{2}(z)dz=0.$

Since $g(z)S(z)\in L_{z}^{1}(\mathbb{R})$, using yields
$g_{1}(z),g_{2}(z)=\mathcal{O}(z^{-1}),\ |z|\rightarrow\infty$. Hence, the
integral on the arc approaches to zero as the radius approaches to infinity.
Therefore, we obtain

$\begin{split}0&=\int_{\mathbb{R}}g_{1}(z)g_{2}(z)dz=\int_{R}\mathcal{P}^{+}(gS)[\mathcal{P}^{-}(gS)]^{*}\rm
dz\\\ &=\int_{\mathbb{R}}[P^{-}(gS)+gS][\mathcal{P}^{-}(gS)]^{*}\rm
dz.\end{split}$ (58)

Utilizing the assumption $\mathcal{P}^{-}(gS)=g$, we have

$\int_{R}g(I+S)g^{*}dz=0.$

We get ${\rm Re}\ g(I+S)g^{*}>c_{-}g^{*}g$ with $c_{-}$ is a positive
constant. Thus the function $g(z)$ has to be zero. This contradicts to the
assumption $g\neq 0$. Therefore, $g=0$ is a unique solution to the equation
$(I-P^{-}_{S})g=0$ in $L^{2}_{z}(\mathbb{R})$. Finally there exists a unique
solution to the equation $(I-P_{S}^{-})\Psi(z)=F(z)$. ∎

######  Lemma 3.4.

Let $r_{1,2}(z)\in H^{1}(\mathbb{R})$ satisfy $|r_{1,2}(z)|\leq 1$ then for
every $x\in\mathbb{R}$, there exist unique solutions $\Psi_{\pm}(x;z)\in
L^{2}_{z}(\mathbb{R})$ satisfying the equation

$\Psi_{+}(x;z)-\Psi_{-}(x;z)=\Psi_{-}S(x;z)+S(x;z),\quad z\in\mathbb{R}.$

###### Proof.

Owing to $S(x;z)\in L^{2}_{z}(\mathbb{R})$, we have $\mathcal{P_{S}}^{-}(z)\in
L^{2}_{z}(\mathbb{R})$ by (49). Then for every $x\in\mathbb{R}$, there exists
a unique solution $\Psi_{-}(x;z)\in L^{2}_{z}(\mathbb{R})$ satisfying the
equation

$\Psi_{-}(x;z)=\mathcal{P}^{-}(\Psi_{-}(x;z)S(x;z)+S(x;z)),\quad
z\in\mathbb{R}.$ (59)

Based on the existence of $\Psi_{-}(x;z)$, we define a function
$\Psi_{+}(x;z)$ by

$\Psi_{+}(x;z)=\mathcal{P}^{+}(\Psi_{-}(x;z)S(x;z)+S(x;z)),\quad
z\in\mathbb{R}.$ (60)

Besides, analytic extensions of $\Psi_{\pm}(x;z)$ to $z\in\mathbb{C}^{\pm}$
are defined by Cauchy operator

$\Psi_{\pm}(x;z)=\mathcal{C}(\Psi_{-}(x;z)S(x;z)+S(x;z)),\quad
z\in\mathbb{C}^{\pm}.$ (61)

Finally we obtain the solution $\Psi_{\pm}(x;z)\in L^{2}_{z}(\mathbb{R})$.
Moreover, given the property of the Cauchy operator and the Plemelj projection
operator, the solutions $\Psi_{\pm}(x;z)$ are analytic functions for
$z\in\mathbb{C}^{\pm}$. ∎

######  Lemma 3.5.

Let $r_{1,2}\in H^{1}(\mathbb{R})$ ,satisfy $|r_{1,2}(z)|\leq 1$ then the
operator $(I-\mathcal{P}_{S}^{-})^{-1}$ is bounded from
$L_{z}^{2}(\mathbb{R})$ to $L_{z}^{2}(\mathbb{R})$, and there exists a
constant $c$ that only depends on $\|r(z)\|_{L_{z}^{\infty}}$ such that

$\|(I-P_{S}^{-})^{-1}f\|_{L_{z}^{2}}\leq c\|f\|_{L_{z}^{2}}$

###### Proof.

For every $f(z)\in L_{z}^{2}(\mathbb{R})$ that there exists a solution
$\Psi(z)\in L_{z}^{2}(\mathbb{R})$ to $(I-\mathcal{P}_{S}^{-})\Psi(z)=f(z)$.
Note that $\mathcal{P}^{+}-\mathcal{P}^{-}=I$, we decompose the function into
$\Psi=\Psi_{+}-\Psi_{-}$ with

$\Psi_{-}-\mathcal{P}^{-}(\Psi_{-}S)=\mathcal{P}^{-}(f),\quad\Psi_{+}-\mathcal{P}^{-}(\Psi_{+}S)=\mathcal{P}^{+}(f).$
(62)

Since $\mathcal{P}^{\pm}(f)\in L_{z}^{2}(\mathbb{R})$, there exist unique
solutions $\Psi_{\pm}(z)\in L_{z}^{2}(\mathbb{R})$ for (59) which implies the
decomposition is unique. Therefore, we only need the estimates of $\Psi_{\pm}$
in $L^{2}_{z}(\mathbb{R})$.

To deal with $\Psi_{-}$, define two analytic functions in
$\mathbb{C}\backslash\mathbb{R}$

$g_{1}(z)=\mathcal{C}(\Psi_{-}S)(z),\quad
g_{2}(z)=\mathcal{C}(\Psi_{-}S+f)^{*}(z),$

Analogous manipulation , we integrate on the semi-circle in the upper half-
plane and have

$\oint g_{1}(z)g_{2}(z)dz=0.$

Since $g_{1}(z)=\mathcal{O}(z^{-1})$ and $g_{2}(z)\rightarrow 0$ as
$|z|\rightarrow\infty$, we have

$\begin{split}0=&\int_{\mathbb{R}}\mathcal{P}^{+}(\Psi_{-}S)[\mathcal{P}^{-}(\Psi_{-}S+f)]^{*}dz\\\
=&\int_{\mathbb{R}}(\mathcal{P}^{-}(\Psi_{-}S)+\Psi_{-}S)[\mathcal{P}^{-}(\Psi_{-}S+f)]^{*}dz\\\
=&\int_{\mathbb{R}}(\Psi_{-}-\mathcal{P}^{-}(f)+\Psi_{-}S)\Psi_{-}^{*}dz.\end{split}$
(63)

Using the bound (51) , (52) and the Hölder inequality, there exists a positive
constant $c_{-}$ such that

$c_{-}\|\Psi_{-}\|_{L^{2}}^{2}\leq{\rm
Re}\int_{\mathbb{R}}\Psi_{-}(I+S)\Psi_{-}^{*}dz={\rm
Re}\int_{\mathbb{R}}\mathcal{P}^{-}(f)\Psi_{-}^{*}dz\leq\|f\|_{L^{2}}\|\Psi_{-}\|_{L^{2}},$

which completes the estimates of $\Psi_{-}$:

$\|(I-\mathcal{P}_{S}^{-})^{-1}P^{-}f\|_{L^{2}_{z}}\leq
c_{-}^{-1}\|f\|_{L^{2}_{z}}.$ (64)

To deal with $\Psi_{+}$, define two functions in
$\mathbb{C}\backslash\mathbb{R}$

$g_{1}(z)=\mathcal{C}(\Psi_{+}S)(z),\quad
g_{2}(z)=\mathcal{C}(\Psi_{+}S+f)^{*}(z).$

Performing the similar procedure leads to

$\begin{split}0=&\oint g_{1}(z)g_{2}(z)dz\\\
=&\int_{\mathbb{R}}\mathcal{P}^{-}(\Psi_{+}S)[\mathcal{P}^{+}(\Psi_{+}S+f)]^{*}dz\\\
=&\int_{\mathbb{R}}[\Psi_{+}-\mathcal{P}^{+}(f)][\Psi_{+}(I+S)]^{*}dz,\end{split}$
(65)

where we have used (59). Using the bounds (51) and (52) , there are positive
constants $c_{-}$ and $c_{+}$ such that

$c_{-}\|\Psi_{+}\|_{L^{2}}^{2}\leq{\rm
Re}\int_{\mathbb{R}}\Psi_{+}(I+S)^{*}\Psi_{+}^{*}dz={\rm
Re}\int_{\mathbb{R}}\mathcal{P}^{+}(f)(I+S)^{*}\Psi_{+}^{*}dz\leq
c_{+}\|f\|_{L^{2}}\|\Psi_{+}\|_{L^{2}},$

which means

$\|(I-P_{S}^{-})^{-1}P^{+}f\|_{L^{2}_{z}}\leq
c_{-}^{-1}c_{+}\|f\|_{L^{2}_{z}}.$ (66)

Combining (64) and (66), we obtain

$\|(I-\mathcal{P}_{S}^{-})^{-1}f\|_{L_{z}^{2}}\leq c\|f\|_{L_{z}^{2}},$

where $c$ is a constant that only depends on $\|r(z)\|_{L_{z}^{\infty}}$. ∎

###  3.3 Estimate on solutions to the RH problem

Next, we see solutions to the RH problem for $M(x;z)$. Denote the functions
$M_{\pm}$ column-wise

$M_{\pm}(x;z)=[\mu_{\pm}(x;z),\nu_{\pm}(x;z)],$

then the functions $\Psi_{\pm}$ can be written as

$\Psi_{\pm}(x;z)=[\mu_{\pm}(x;z)-e_{1},\nu_{\pm}(x;z)-e_{2}].$

We have

$\mu_{\pm}(x;z)-e_{1}=\mathcal{P}^{\pm}(M_{-}S)^{(1)}(x;z),\quad
z\in{\mathbb{R}}$ (67)

and

$\nu_{\pm}(x;z)-e_{2}=\mathcal{P}^{\pm}(M_{-}S)^{(2)}(x;z),\quad
z\in{\mathbb{R}}.$ (68)

Combining (67) with (68), we obtain

$M_{\pm}(x;z)={I}+\mathcal{P}^{\pm}(M_{-}(x;\cdot)S(x;\cdot))(z),\quad
z\in\mathbb{R}.$ (69)

There exist unique solutions $M_{\pm}(x;z)$ to Eq.(69). Further, analytic
extensions of $M_{\pm}(x;z)$ to $z\in\mathbb{C}^{\pm}$ are

$M_{\pm}(x;z)={I}+\mathcal{C}(M_{-}(x;\cdot)S(x;\cdot))(z),\quad
z\in{\mathbb{C}}^{\pm}.$ (70)

######  Lemma 3.6.

Let $r_{1,2}(z)\in H^{1}(\mathbb{R})$, satisfy $|r_{1,2}(z)|\leq 1$ then there
exists a constant $c$ only depending on $\|r(z)\|_{L^{\infty}}$ such that for
every $x\in\mathbb{R}$,

$\|M_{\pm}(x;\cdot)-I\|_{L^{2}}\leq c(\|r_{1}\|_{L^{2}}+\|r_{1}\|_{L^{2}}).$
(71)

###### Proof.

Due to $r_{1,2}\in H^{1}(\mathbb{R})$, we get $r_{1,2}\in
L^{2}(\mathbb{R})\cap L^{\infty}(\mathbb{R})$ and $S(x;\cdot)\in L^{2}_{z}$.
Moreover, there exists a constant $c$ only depending on
$\|r_{1,2}\|_{L^{\infty}}$ such that

$\|S(x;\cdot)\|_{L_{z}^{2}}\leq c(\|r_{1}\|_{L^{2}}+\|r_{1}\|_{L^{2}}),$

We obtain

$\|M_{\pm}-I\|_{L^{2}}=\|\psi_{\pm}\|_{L^{2}}\leq
c(\|r_{1}\|_{L^{2}}+\|r_{1}\|_{L^{2}}),$

where we have used the equation
$(I-\mathcal{P}_{S}^{-})\Psi_{\pm}=\mathcal{P}^{-}S$, and $c$ is another
constant only depending on $\|r_{1,2}(z)\|_{L^{\infty}}$. ∎

######  Proposition 3.7.

For every $x_{0}\in\mathbb{R}^{-}$ and every $r_{1,2}(z)\in
H^{1,1}(\mathbb{R})$, we have

$\displaystyle\sup_{x\in(-\infty,x_{0})}\|\langle
x\rangle\mathcal{P}^{+}(r_{2}e^{2izx})\|_{L^{2}_{z}}\leq c\|r_{2}\|_{H^{1}},$
(72) $\displaystyle\sup_{x\in(-\infty,x_{0})}\|\langle
x\rangle\mathcal{P}^{-}(r_{1}e^{-2izx})\|_{L^{2}_{z}}\leq c\|r_{1}\|_{H^{1}},$
(73) $\displaystyle\sup_{x\in(-\infty,x_{0})}\|\langle
x\rangle\mathcal{P}^{+}(zr_{2}e^{2izx})\|_{L^{2}_{z}}\leq
c\|r_{2}\|_{H^{1,1}},$ $\displaystyle\sup_{x\in(-\infty,x_{0})}\|\langle
x\rangle\mathcal{P}^{-}(zr_{1}e^{-2izx})\|_{L^{2}_{z}}\leq
c\|r_{1}\|_{H^{1,1}},$

where $\langle x\rangle=(1+x^{2})^{1/2}$. Moreover, if $r_{1,2}\in
L^{2,3}(\mathbb{R})$, then we have

$\displaystyle\sup_{x\in\mathbb{R}}\|z^{i}\mathcal{P}^{+}(z^{k}r_{2}e^{2izx})\|_{L^{2}_{z}}\leq\|z^{i+k}r_{2}\|_{L^{2}_{z}},$
(74)
$\displaystyle\sup_{x\in\mathbb{R}}\|z^{i}\mathcal{P}^{-}(z^{k}rr_{1}e^{-2izx})\|_{L^{2}_{z}}\leq\|z^{i+k}r_{1}\|_{L^{2}_{z}},$
(75)

where $i,k=0,1,2,3,i+k\leq 3$ and $c$ is a constant that depends on
$\|r_{1,2}\|_{L^{\infty}}$ .

###### Proof.

By an analogous analysis in [23], we can easy get this proposition. ∎

In order to obtain estimates on the vector columns $\mu_{-}(x;z)-e_{1}$ and
$\nu_{+}(x;z)-e_{2}$ that will be needed in the subsequent section, we rewrite
functions $\mu_{-}(x;z)-e_{1}$ and $\nu_{+}(x;z)-e_{2}$ by (69) as

$\mu_{-}(x;z)-e_{1}=\mathcal{P}^{-}(r_{1}e^{-2izx}\nu_{+}(x;z))(z),z\in\mathbb{R}$
(76)

and

$\nu_{+}(x;z)-e_{2}=\mathcal{P}^{+}({\sigma
r_{2}}e^{2izx}\mu_{-}(x;z))(z),z\in\mathbb{R},$ (77)

where we have used the fact

$\begin{split}M_{-}S=&[\mu_{-},\nu_{-}]\left(\begin{array}[]{cc}\sigma
r_{1}r_{2}&\sigma r_{2}e^{2izx}\\\
r_{1}e^{-2izx}&0\end{array}\right)=[r_{1}e^{-2izx}\nu_{+},\sigma
r_{2}e^{2izx}\mu_{-}].\end{split}$ (78)

Define a function

$N(x;z)=[\mu_{-}(x;z)-e_{1},\nu_{+}(x;z)-e_{2}],$

which satisfies

$\begin{split}N-\mathcal{P}^{+}(NS_{+})-\mathcal{P}^{-}(NS_{-})=F,\end{split}$
(79)

where

$\begin{split}&F(x;z)=[\mathcal{P}^{-}(r_{1}e^{-2izx})e_{2},\mathcal{P}^{+}(\sigma
r_{2}e^{2izx})e_{1}],\\\ &S_{+}(x;z)=\left(\begin{array}[]{cc}0&\sigma
r_{2}e^{2izx}\\\ 0&0\end{array}\right),\quad
S_{-}(x;z)=\left(\begin{array}[]{cc}0&0\\\
r_{1}e^{-2izx}&0\end{array}\right).\end{split}$

######  Lemma 3.8.

Let $r_{1,2}\in H^{1,1}(\mathbb{R})$, then for every $x_{0}\in\mathbb{R}^{-}$,
we have

$\displaystyle\sup_{x\in(-\infty,x_{0})}\|\langle
x\rangle\mu_{-}^{(2)}(x;z)\|_{L^{2}_{z}}\leq c\|r_{1}\|_{H^{1}},$ (80)
$\displaystyle\sup_{x\in(-\infty,x_{0})}\|\langle
x\rangle\nu_{+}^{(1)}(x;z)\|_{L^{2}_{z}}\leq c\|r_{2}\|_{H^{1}},$ (81)
$\displaystyle\sup_{x\in(-\infty,x_{0})}\|\langle
x\rangle\partial_{x}\mu_{-}^{(2)}(x;z)\|_{L^{2}_{z}}\leq c\|r_{1}\|_{H^{1,1}}$
(82) $\displaystyle\sup_{x\in(-\infty,x_{0})}\|\langle
x\rangle\partial_{x}\nu_{+}^{(1)}(x;z)\|_{L^{2}_{z}}\leq
c\|r_{2}\|_{H^{1,1}},$ (83)

where $c$ is a constant that only depends on $\|r_{1,2}\|_{L^{\infty}}$. In
addition, if $r\in L^{2,3}$, then we have

$\displaystyle\sup_{x\in\mathbb{R}}\|z^{k}\partial^{i}_{x}\mu_{-}^{(2)}(x;z)\|_{L^{2}_{z}}\leq
c\|r_{1}\|_{L^{2,i+k}}$ (84)
$\displaystyle\sup_{x\in\mathbb{R}}\|z^{k}\partial^{i}_{x}\nu_{+}^{(1)}(x;z)\|_{L^{2}_{z}}\leq
c\|r_{2}\|_{L^{2,i+k}},$ (85)

where $i,k=0,1,2,3,i+k\leq 3$ and $c$ is a constant that depends on
$\|r_{1,2}\|_{L^{\infty}}$ .

###### Proof.

Note $\mathcal{P}^{+}-\mathcal{P}^{-}=I$ and $S_{+}+S_{-}=(I-S_{+})S$, Eq.(79)
can be rewritten as

$G-\mathcal{P}^{-}(GS)=F$ (86)

with $G=N(I-S_{+})$. And the matrix $G(x;z)$ is written component-wise as

$G(x;z)=\left(\begin{array}[]{cc}\mu_{-}^{(1)}(x;z)-1&\nu_{+}^{(1)}-\sigma
r_{2}e^{2izx}(\mu_{-}^{(1)}(x;z)-1)\\\
\mu_{-}^{(2)}(x;z)&\nu_{+}^{(2)}-1-\sigma
r_{2}e^{2izx}\mu_{-}^{(2)}(x;z)\end{array}\right).$

Comparing the second row of $F(x;z)$ with $G(x;z)$ and utilizing the bound
(73), we have

$\begin{split}&\sup_{x\in(-\infty,x_{0})}\|\langle
x\rangle\mu_{-}^{(2)}\|_{L^{2}_{z}}\leq c\sup_{x\in(-\infty,x_{0})}\|\langle
x\rangle\mathcal{P}^{-}(r_{1}e^{-2izx})\|_{L^{2}_{z}},\\\
&\|\nu_{+}^{(2)}-1-r_{1}e^{-2izx}\mu_{-}^{(2)}(x;z)\|_{L^{2}_{z}}\leq
c\|\mathcal{P}^{-}(r_{1}e^{-2izx})\|_{L^{2}_{z}},\end{split}$ (87)

where $c$ is a constant that depends on $\|r\|_{L^{\infty}}$. Substituting the
bound (73) into (87), we obtain the estimate (80).

Similarly, comparing the first row of $F(x;z)$ and $G(x;z)$ yields

$\begin{split}&\|\mu_{-}^{(1)}(x;z)-1\|_{L^{2}_{z}}\leq
c\|\mathcal{P}^{+}(\sigma r_{2}e^{2izx})\|_{L^{2}_{z}},\\\
&\|\nu_{+}^{(1)}(x;z)-\sigma
r_{2}e^{2izx}(\mu_{-}^{(1)}(x;z)-1)\|_{L^{2}_{z}}\leq
c\|\mathcal{P}^{+}(\sigma r_{2}e^{2izx})\|_{L^{2}_{z}}.\end{split}$ (88)

Taking derivative in x of (79), we obtain

$\partial_{x}N-\mathcal{P}^{+}(\partial_{x}N)S_{+}-\mathcal{P}^{-}(\partial_{x}N)S_{-}=F_{1}$
(89)

with

$\begin{split}F_{1}=&\partial_{x}F+\mathcal{P}^{+}N\partial_{x}S_{+}+\mathcal{P}^{-}N\partial_{x}S_{-}\\\
=&2i[e_{2}\mathcal{P}^{-}(-zr_{1}e^{-2izx}),e_{1}\mathcal{P}^{+}(z\sigma
r_{2}e^{2izx})]\\\
&+2i\left(\begin{array}[]{cc}\mathcal{P}^{-}(-zr_{1}(z)e^{-2izx}\nu_{+}^{(1)}(x;z))&\mathcal{P}^{+}(z\sigma
r_{2}e^{2izx}(\mu_{-}^{(1)}(x;z)-1))\\\
\mathcal{P}^{-}(-zr_{1}e^{-2izx}(\nu_{+}^{(2)}(x;z)-1))&\mathcal{P}^{+}(z\sigma
r_{2}e^{2izx}\mu_{-}^{(2)}(x;z))\end{array}\right).\end{split}$

According to the estimates (74) and (75), together with the triangle equality,
we obtain

$\displaystyle z(\mu_{-}(x;z)-e_{1})\in
L_{x}^{\infty}((-\infty,x_{0});L^{2}_{z}(\mathbb{R})),$ $\displaystyle
z(\nu_{+}(x;z)-e_{2})\in
L_{x}^{\infty}((-\infty,x_{0});L^{2}_{z}(\mathbb{R})).$

On account of the bound (72), (73) and $r_{1,2}(z)\in L^{\infty}(\mathbb{R})$
, we conclude that $F_{1}$ belongs to
$L_{x}^{\infty}((-\infty,x_{0});L^{2}_{z}(\mathbb{R}))$. On account of the
bound (80), (81) and $r_{1,2}(z)\in L^{\infty}(\mathbb{R})$ , we conclude that
$\sup_{x\in(-\infty,x_{0})}\langle x\rangle F_{1}$ belongs to
$L_{x}^{\infty}((-\infty,x_{0});L^{2}_{z}(\mathbb{R}))$. We get (82) and (83).

Taking $j$-order derivative of (79), we obtain

$\partial^{j}_{x}N-\mathcal{P}^{+}(\partial^{j}_{x}N)S_{+}-\mathcal{P}^{-}(\partial^{j}_{x}N)S_{-}=F_{j},$
(90)

with

$F_{j}=\partial_{x}F_{j-1}+\mathcal{P}^{+}\partial^{j-1}_{x}N_{x}\partial_{x}S_{+}+\mathcal{P}^{-}\partial^{j-1}_{x}N_{x}\partial_{x}S_{-}.$

Where $j=2,3$. Repeating the analysis for (86),and using (74), (75) we derive
the estimates (84), (85). ∎

##  4 Reconstruction and estimates of the potential

We shall now recover the potential $u$ from the matrices $M_{\pm}$, which
satisfy the integral equations (70). This will gives us the map

$H^{1,1}(\mathbb{R})\cap L^{2,3}(\mathbb{R})\ni r_{1,2}\mapsto u\in
H^{1,1}(\mathbb{R})\cap H^{3}(\mathbb{R}).$

###  4.1 Estimates on the negative half-line

Recalling (47) which show that

$u(x)=2iz\lim_{z\rightarrow\infty}(M_{\pm})_{12}.$ (91)

We will use (91) to get estimates of $u$ on the negative half-line. It follows
from (70) and (47) that

$u(x)=2i\lim_{|z|\rightarrow\infty}z\mathcal{C}((M_{-}S)_{12}.$ (92)

Since $r\in H^{1,1}(\mathbb{R})\cap L^{2,3}(\mathbb{R})$, we have
$S(x;\cdot)\in L^{1}(\mathbb{R})\cap L^{2}(\mathbb{R})$. Besides, the estimate
(73) implies that $M_{-}(x;\cdot)-I\in L^{2}(\mathbb{R})$. Therefore, we
arrive at $L^{1}(\mathbb{R})$. Subsequently, Applying (50) to (92), we obtain

$\begin{split}u(x)&=\frac{1}{\pi}\int_{\mathbb{R}}\sigma
r_{2}e^{2izx}\mu_{-}^{(1)}(x;z)dz\\\ &=\frac{1}{\pi}\int_{\mathbb{R}}\sigma
r_{2}e^{2izx}P^{-}(re^{-2izx}\nu_{+}^{(1)})\sigma
r_{2}e^{2izx}dz+\frac{1}{\pi}\int_{R}\sigma r_{2}e^{2izx}dz,\end{split}$ (93)

where we have used the identity
$\mu_{-}^{(1)}(x;z)-1=P^{-}(r_{1}e^{-2izx}\nu_{+}^{(1)})$ .

######  Lemma 4.1.

Let $r_{1,2}(z)\in H^{1,1}(\mathbb{R})\cap L^{2,3}(\mathbb{R})$ , then $u\in
H^{1,1}(\mathbb{R})\cap H^{3}(\mathbb{R})$. Moreover, we have

$\|u\|_{H^{1,1}(\mathbb{R}^{-})\cap H^{3}(\mathbb{R}^{-})}\leq
c(\|r_{1}\|_{H^{1,1}(\mathbb{R})\cap
L^{2,3}(\mathbb{R})}+\|r_{2}\|_{H^{1,1}(\mathbb{R})\cap
L^{2,3}(\mathbb{R})}),$ (94)

where $c$ is a constant that depends on $\|r_{1,2}\|_{L^{\infty}}$ and
$\|zr_{1,2}\|_{L^{\infty}}$.

###### Proof.

Recalling (93) and the results from the Fourier theory. For a function
$r_{1,2}(z)\in L^{2}(\mathbb{R})$, by Parseval’s equation, we have

$\|r_{1,2}\|_{L^{2}}=\|\hat{r}_{1,2}\|_{L^{2}},$

where the function $\hat{r}_{1,2}$ denotes the Fourier transform. Since
$L^{2,3}(\mathbb{R})\cap H^{1,1}(\mathbb{R})$, the second term of (93) belongs
to $H^{1,1}(\mathbb{R})\cap H^{3}(\mathbb{R})$ due to the property
$\widehat{\partial_{z}r_{1,2}(z)}=x\hat{r}_{1,2}(x)$.

Let

$I(x)=\int_{\mathbb{R}}\sigma r_{2}e^{2izx}(\mu_{-}^{(1)}(x;z)-1)dz.$ (95)

Substituting (77) into (95) and applying the Fubini’s theorem yields

$\begin{split}I(x)=&\int_{\mathbb{R}}\sigma
r_{2}e^{2izx}\lim_{\epsilon\rightarrow 0}\frac{1}{2\pi
i}\int_{\mathbb{R}}\frac{r_{1}(s)e^{-2isx}\nu_{+}^{(1)}(s)}{s-(z-i\epsilon)}dsdz\\\
=&-\int_{\mathbb{R}}r_{1}(s)e^{-2isx}\nu_{+}^{(1)}(s)\lim_{\epsilon\rightarrow
0}\frac{1}{2\pi i}\int_{\mathbb{R}}\frac{\sigma
r_{2}e^{2izx}}{z-(s+i\epsilon)}dzds\\\
=&-\int_{\mathbb{R}}r_{1}(z)e^{-2izx}\nu_{+}^{(1)}(z)P^{+}(\sigma
r_{2}e^{2isx})(z)dz.\end{split}$

Therefore, for every $x_{0}\in\mathbb{R}^{-}$, utilizing the Hölder’s
inequality and the estimates (72) and (81), we find

$\begin{split}&\sup_{x\in(-\infty,x_{0})}\left|\langle
x\rangle^{2}I(x)\right|\leq\|r_{1}\|_{L^{\infty}}\sup_{x\in(-\infty,x_{0})}\|\langle
x\rangle\nu_{+}^{(1)}\|_{L^{2}}\\\
&\qquad\times\sup_{x\in(-\infty,x_{0})}\|\langle x\rangle P^{+}(\sigma
r_{2}e^{2isx})\|_{L^{2}}\leq c\|r_{1}\|_{H^{1}}\|r_{2}\|_{H^{1}},\end{split}$
(96)

where $c$ is a constant only depends on $\|r\|_{L^{\infty}}$, further, we
obtain

$\begin{split}\|\langle x\rangle I(x)\|_{L^{2}(\mathbb{R}^{-})}\leq
c\|r_{1}\|_{H^{1}}\|r_{2}\|_{H^{1}},\end{split}$

where $c$ is another constant only depends on $\|r\|_{L^{\infty}}$. Combining
the results of the two terms of (96) leads to

$\|u(x)\|_{L^{2,1}(\mathbb{R}^{-})}\leq
c(1+\|r_{1}\|_{H^{1}}+\|r_{2}\|_{H^{1}})(\|r_{1}\|_{H^{1}}+\|r_{2}\|_{H^{1}}).$
(97)

This completes the proof of $u\in L^{2,1}(\mathbb{R}^{-})$. By the Fourier
theory, the derivative of the second term of (93) belongs to
$L^{2}(\mathbb{R})$. For the second term $I(x)$, we differentiate $I(x)$ in
$x$ and obtain

$\begin{split}I^{\prime}(x)=&\partial\int_{\mathbb{R}}\sigma
r_{2}e^{2izx}(\mu_{-}^{(1)}(x;z)-1)dz\\\
=&-2i\int_{\mathbb{R}}r_{1}(z)e^{-2izx}\nu_{+}^{(1)}(x;z)P^{+}(s\sigma
r_{2}e^{2isx})(z)dz\\\
&-2i\int_{\mathbb{R}}zr_{1}(z)e^{-2izx}\nu_{+}^{(1)}(x;z)P^{+}(\sigma
r_{2}e^{2isx})(z)dz\\\
&-\int_{\mathbb{R}}r_{1}(z)e^{-2izx}\partial_{x}\nu_{+}^{(1)}(x;z)P^{+}(\sigma
r_{2}e^{2isx})(z)dz,\end{split}$

where we have used Eq.(76) and the Fubini’s theorem.

Utilizing the estimates (72), (81) we find that for every
$x_{0}\in\mathbb{R}^{-}$,

$\begin{split}&\sup_{x\in(-\infty,x_{0})}|\langle
x\rangle^{2}I^{\prime}(x)|\leq\|r\|_{L^{\infty}}\sup_{x\in(x_{0},+\infty)}\left(2\|\langle
x\rangle\nu_{+}^{(1)}\|_{L^{2}}\|\langle x\rangle
P^{+}(z\overline{r}e^{2izx})\|_{L^{2}}\right.\\\ &\left.+\|\langle
x\rangle\nu_{+}^{(1)}\|_{L^{2}}\|\langle x\rangle P^{+}(z\sigma
r_{2}e^{2izx})\|_{L^{2}}+\|\langle
x\rangle\partial_{x}\nu_{+}^{(1)}\|_{L^{2}}\|\langle x\rangle P^{+}(\sigma
r_{2}e^{2izx})\|_{L^{2}}\right)\\\ &\leq c\|r_{1}\|_{H^{1,1}\cap
L^{2,3}}\|r_{2}\|_{H^{1,1}\cap L^{2,3}},\end{split}$

which implies that

$\begin{split}\|\langle x\rangle
I^{\prime}(x)\|_{L^{2}(\mathbb{R}^{-})}\leq&c\|r_{1}\|_{H^{1,1}\cap
L^{2,3}}\|r_{2}\|_{H^{1,1}\cap L^{2,3}},\end{split}$ (98)

with $c$ is another constant that depends on $\|r\|_{L^{\infty}}$ .
Subsequently, we obtain $I^{\prime}(x)\in L^{2,1}(\mathbb{R}^{-})$. We
conclude that $u\in H^{1,1}(\mathbb{R}^{-})$. The estimate (94) can be
obtained from (97) and (98) with another constant $c$ depending on
$\|r\|_{L^{\infty}}$.

For $I^{\prime\prime}(x)$, we get

$\begin{split}&\sup_{x\in(-\infty,x_{0})}|\langle x\rangle
I^{\prime\prime}(x)|\leq\|r_{1}\|_{L^{\infty}}\sup_{x\in(x_{0},+\infty)}\left(4\|\langle
x\rangle\nu_{+}^{(1)}\|_{L^{2}}\|z^{2}P^{+}(\sigma
r_{2}e^{2izx})\|_{L^{2}}\right.\\\ &+\|\langle
x\rangle\nu_{+}^{(1)}\|_{L^{2}}\|\langle x\rangle P^{+}(z^{2}\sigma
r_{2}e^{2izx})\|_{L^{2}}+\|\partial^{2}_{x}\nu_{+}^{(1)}\|_{L^{2}}\|\langle
x\rangle P^{+}(\sigma r_{2}e^{2izx})\|_{L^{2}}\\\
&+2\|z\nu_{+}^{(1)}\|_{L^{2}}\|\langle x\rangle P^{+}(z\sigma
r_{2}e^{2izx})\|_{L^{2}}+\|\langle
x\rangle\partial_{x}\nu_{+}^{(1)}\|_{L^{2}}\|zP^{+}(\sigma
r_{2}e^{2izx})\|_{L^{2}}\\\
&\left.+2\|\partial_{x}\nu_{+}^{(1)}\|_{L^{2}}\|\langle x\rangle P^{+}(s\sigma
r_{2}e^{2izx})\|_{L^{2}}\right)\leq c\|r_{1}\|_{H^{1,1}\cap
L^{2,3}}\|r_{2}\|_{H^{1,1}\cap L^{2,3}}.\end{split}$

For $I^{\prime\prime\prime}(x)$ similar to above analysis, we get

$\begin{split}&\sup_{x\in(-\infty,x_{0})}|\langle x\rangle
I^{\prime\prime\prime}(x)|\leq\|r_{1}\|_{L^{\infty}}\sup_{x\in(x_{0},+\infty)}\left(8\|\langle
x\rangle\nu_{+}^{(1)}\|_{L^{2}}\|P^{+}(z^{3}\sigma
r_{2}e^{2izx})\|_{L^{2}}\right.\\\ &+\|\langle
x\rangle\nu_{+}^{(1)}\|_{L^{2}}\|z^{3}P^{+}(\sigma
r_{2}e^{2izx})\|_{L^{2}}+\|\partial^{3}_{x}\nu_{+}^{(1)}\|_{L^{2}}\|\langle
x\rangle P^{+}(\sigma r_{2}e^{2izx})\|_{L^{2}}\\\ &+4\|\langle
x\rangle\partial_{x}\nu_{+}^{(1)}\|_{L^{2}}\|P^{+}(s^{2}\sigma
r_{2}e^{2izx})\|_{L^{2}}+2\|\langle
x\rangle\partial^{2}_{x}\nu_{+}^{(1)}\|_{L^{2}}\|P^{+}(z^{2}\sigma
r_{2}e^{2izx})\|_{L^{2}}\\\ &+2\|\langle
x\rangle\nu_{+}^{(1)}\|_{L^{2}}\|z^{2}P^{+}(s\sigma
r_{2}e^{2izx})\|_{L^{2}}+4\|\langle
x\rangle\nu_{+}^{(1)}\|_{L^{2}}\|zP^{+}(z^{2}\sigma
r_{2}e^{2izx})\|_{L^{2}}\\\ &+\|\partial_{x}\nu_{+}^{(1)}\|_{L^{2}}\|\langle
x\rangle s^{2}P^{+}(\sigma
r_{2}e^{2izx})\|_{L^{2}}+\|z\partial^{2}_{x}\nu_{+}^{(1)}\|_{L^{2}}\|\langle
x\rangle P^{+}(\sigma r_{2}e^{2izx})\|_{L^{2}}\\\ &\left.+4\|\langle
x\rangle\nu_{+}^{(1)}\|_{L^{2}_{z}}\|P^{+}(s^{2}\sigma
r_{2}e^{2izx})\|_{L^{2}}+2\|z\partial_{x}\nu_{+}^{(1)}\|_{L^{2}}\|\langle
x\rangle P^{+}(z\overline{r}e^{2izx})\|_{L^{2}}\right)\\\ &\leq
c\|r_{1}\|_{H^{1,1}\cap L^{2,3}}\|r_{2}\|_{H^{1,1}\cap L^{2,3}}.\end{split}$

Then we conclude that $u\in H^{3}(\mathbb{R}^{-})$. Finally we prove the
conclusion. ∎

We actually get the following map through above analysis.

$H^{1,1}(\mathbb{R})\cap L^{2,3}(\mathbb{R})\ni r_{1,2}\mapsto u\in
H^{3}(\mathbb{R}^{-})\cap H^{1,1}(\mathbb{R}^{-}).$

We will prove the map is Lipschitz continuous.

######  Lemma 4.2.

Let $r\in H^{1,1}(\mathbb{R})\cap L^{2,3}(\mathbb{R})$ , then the mapping

$H^{1,1}(\mathbb{R})\cap L^{2,3}(\mathbb{R})\ni r_{1,2}\mapsto u\in
H^{3}(\mathbb{R}^{-})\cap H^{1,1}(\mathbb{R}^{-})$

is Lipschitz continuous.

###### Proof.

Let $r_{1,2},\tilde{r}_{1,2}\in H^{1,1}(\mathbb{R})\cap L^{2,3}(\mathbb{R})$.
Let the functions $u$ and $\tilde{u}$ are the corresponding potentials
respectively. We will show that there exists a constant $c$ that depends on
$\|r_{1,2}\|_{L^{\infty}}$ such that

$\|u-\tilde{u}\|_{H^{3}(\mathbb{R}^{-})\cap H^{1,1}(\mathbb{R}^{-})}\leq
c(\|r_{1}-\tilde{r}_{1}\|_{H^{1,1}(\mathbb{R})\cap
L^{2,3}(\mathbb{R})}+\|r_{2}-\tilde{r}_{2}\|_{H^{1,1}(\mathbb{R})\cap
L^{2,3}(\mathbb{R})}).$ (99)

From (96), we have

$\begin{split}u-\tilde{u}=&\frac{1}{\pi}\int_{\mathbb{R}}(\sigma
r_{2}-\sigma\tilde{r}_{2})e^{2izx}dz+\frac{1}{\pi}\int_{\mathbb{R}}(\sigma
r_{2}-\sigma\tilde{r}_{2})e^{2izx}(\mu_{-}^{(1)}(x;z)-1)dz\\\
&+\frac{1}{\pi}\int_{R}\sigma
r_{2}e^{2izx}(\mu_{-}^{(1)}(x;z)-\tilde{\mu}_{-}^{(1)}(x;z))dz.\end{split}$

Repeating the analysis in the proof of Lemma 4.1, we obtain the Lipschitz
continuity of $u$. ∎

###  4.2 Estimates on the positive half-line

Recalling (47) again, we can get

$u(-x)=2iz\sigma\lim_{z\rightarrow\infty}(M_{\pm})_{21}.$ (100)

Performing the same manipulation for (91) yields

$\begin{split}u(-x)=&-\frac{\sigma}{\pi}\int_{\mathbb{R}}r_{1}(z)e^{-2izx}(\nu_{-}^{(2)}(x;z)+\sigma
r_{2}(z)e^{2ikx}\mu_{-}^{(2)}(x;z))dz\\\
=&-\frac{\sigma}{\pi}\int_{\mathbb{R}}r_{1}(z)e^{-2izx}\nu_{+}^{(2)}(x;z)dz,\end{split}$
(101)

where we have used the identity $\nu_{+}^{(2)}=\sigma
r_{2}(z)e^{2izx}\mu_{-}^{(2)}+\nu_{-}^{(2)}$.

Similar to condition $x_{0}\in\mathbb{R}^{-}$, we summarize the above analysis
as the following lemma:

######  Lemma 4.3.

Let $r_{1,2}(z)\in H^{3}(\mathbb{R})\cap H^{1,1}(\mathbb{R})$ satisfy
$|r_{1,2}(z)|<1$, then $u\in H^{1,1}(\mathbb{R}^{+})\cap
L^{2,3}(\mathbb{R}^{+})$. Moreover, we have

$\|u\|_{H^{1,1}(\mathbb{R}^{+})\cap L^{2,3}(\mathbb{R}^{+})}\leq
c(\|r_{1}\|_{H^{3}(\mathbb{R})\cap
H^{1,1}(\mathbb{R})}+\|r_{2}\|_{H^{3}(\mathbb{R})\cap H^{1,1}(\mathbb{R})}),$
(102)

where c is a constant that depends on $\|r_{1,2}\|_{L^{\infty}}$ and
$\|zr_{1,2}\|_{L^{\infty}}$.

###### Proof.

We rewrite (101) as

$u(-x)=-\frac{\sigma}{\pi}\int_{\mathbb{R}}r_{1}(z)e^{-2izx}dz-\frac{\sigma}{\pi}\int_{\mathbb{R}}r_{1}(z)e^{-2izx}(\nu_{+}^{(2)}(x;z)-1)dz.$
(103)

Let

$\hat{r}_{1}(-x)=\int_{\mathbb{R}}r_{1}(z)e^{-2iz(-x)}dz.$ (104)

According to the Fourier theory, we have
$-x\hat{r}_{1}(-x)=\widehat{\partial_{z}r_{1}(z)}(x)$ and
$\|x\hat{r}_{1}(-x)\|_{L^{2}(\mathbb{R})}=\|\partial_{z}r_{1}(z)\|_{L^{2}(\mathbb{R})}$.
Let

$I_{2}(x)=-\frac{\sigma}{\pi}\int_{\mathbb{R}}r_{1}(z)e^{-2izx}(\nu_{+}^{(2)}(x;z)-1)dz.$

Repeating the analysis in the proof of Lemma 4.1, we obtain $u\in
H^{1,1}(\mathbb{R}^{+})\cap L^{2,3}(\mathbb{R}^{+})$ and

$\|u\|_{H^{1,1}(\mathbb{R}^{+})\cap L^{2,3}(\mathbb{R}^{+}}\leq
c(\|r_{1}\|_{H^{3}(\mathbb{R})\cap
H^{1,1}(\mathbb{R})}+\|r_{2}\|_{H^{3}(\mathbb{R})\cap H^{1,1}(\mathbb{R})}),$
(105)

where c is a constant that depends on $\|r_{1,2}\|_{L^{\infty}}$ and
$\|zr_{1,2}\|_{L^{\infty}}$. ∎

######  Lemma 4.4.

Let $r_{1,2}(z)\in H^{1,1}(\mathbb{R})\cap L^{2,3}(\mathbb{R})$ , then the
following map

$H^{1,1}(\mathbb{R})\cap L^{2,3}(\mathbb{R})\ni r_{1,2}\mapsto u\in
H^{3}(\mathbb{R}^{+})\cap H^{1,1}(\mathbb{R}^{+})$

is Lipschitz continuous.

Summarize the results from Lemma 4.2 to Lemma 4.4, we have the following
proposition:

######  Proposition 4.5.

Let $r_{1,2}(z)\in H^{1,1}(\mathbb{R})\cap L^{2,3}(\mathbb{R})$, then we have
$u\in H^{3}(\mathbb{R})\cap H^{1,1}(\mathbb{R})$ and

$\|u\|_{H^{3}(\mathbb{R})\cap H^{1,1}(\mathbb{R})}\leq
c(\|r_{1}\|_{H^{1,1}(\mathbb{R})\cap
L^{2,3}(\mathbb{R})}+\|r_{2}\|_{H^{1,1}(\mathbb{R})\cap
L^{2,3}(\mathbb{R})}).$

Moreover, the mapping

$H^{1,1}(\mathbb{R})\cap L^{2,3}(\mathbb{R})\ni r_{1,2}\mapsto u\in
H^{3}(\mathbb{R})\cap H^{1,1}(\mathbb{R})$

is Lipschitz continuous.

##  5  Global existence and Lipschitz continuity

###  5.1 Time evolution of scattering data

From Section 2 to Section 4, for the initial data $u(x,0)\in
H^{3}(\mathbb{R})\cap H^{1,1}(\mathbb{R})$, we only consider the spatial
spectral problem (5) and obtain its unique normalized solution

$\displaystyle m_{1}^{\pm}(x,0;z)\rightarrow e_{1},\ \
m_{1}^{\pm}(x,0;z)\rightarrow e_{2},\ \ x\rightarrow\pm\infty,$ (106)

which cannot satisfy the time spectral problem (6) since they are short of a
function about the time $t$. For every $t\in[0,T]$, we define the normalized
Jost functions of the Lax pair (5) and (6):

$\displaystyle m_{1}^{\pm}(x,t;z)=m_{1}^{\pm}(x,0;z)e^{4iz^{3}t},$ (107)
$\displaystyle m_{2}^{\pm}(x,t;z)=m_{2}^{\pm}(x,0;z)e^{-4iz^{3}t},$ (108)

with the potential $u(\cdot,0)\in H^{3}(\mathbb{R})\cap H^{1,1}(\mathbb{R})$.
It follows that for every $t\in[0,T]$, we have

$m_{1}^{\pm}(x,t;z)\rightarrow e^{4iz^{3}t}e_{1}\quad x\rightarrow\pm\infty,$
$m_{2}^{\pm}(x,t;z)\rightarrow e^{-4iz^{3}t}e_{2}\quad x\rightarrow\pm\infty.$

Repeating the analysis as the proof of Lemma2.1, we prove that there exist
unique solutions of the Volttera’s integral equations for Jost functions
$m_{1}^{\pm}(x,t;z)$ and $m_{2}^{\pm}(x,t;z)$, and the Jost functions
$m_{1}^{\pm}(x,t;z)$ and $m_{2}^{\pm}(x,t;z)$ admit the same analytic property
as $m_{1}^{\pm}(x,0;z)$ and $m_{2}^{\pm}(x,0;z)$. As well, for every
$(x,t)\in\mathbb{R}\times\mathbb{R}^{+}$ and every $z\in\mathbb{R}$, the Jost
functions $m_{1}^{\pm}(x,t;z)$ and $m_{2}^{\pm}(x,t;z)$ are supposed to
satisfy the scattering relation

$\begin{split}&m_{1}^{+}(x,t;z)=a(t;z)m_{1}^{-}(t,x;z)+b(t;z)e^{-2izx}m_{2}^{-}(t,x;z),\\\
&m_{2}^{+}(x,t;z)=c(t;z)e^{2izx}m_{1}^{-}(t,x;z+d(t;z)m_{2}^{-}(t,x;z).\end{split}$

By the Crammer’s law and the evolution relation (107)-(108), we obtain the
evolution of the scattering coefficients

$\begin{split}&a(t;z)=W(m_{1}^{+}(0,0;z)e^{4iz^{3}t},m_{2}^{-}(0,0;z)e^{-4iz^{3}t})=a(0;z),\\\
&d(t;z)=W(m_{1}^{-}(0,0;z)e^{4iz^{3}t},m_{2}^{+}(0,0;z)e^{-4iz^{3}t})=d(0;z),\\\
&b(t;z)=W(m_{1}^{-}(0,0;z)e^{-4iz^{3}t},m_{1}^{+}(0,0;z)e^{-4iz^{3}t})=b(0;z)e^{8iz^{3}t}.\end{split}$

Direct calculation shows that the reflection coefficients are given by

$\displaystyle
r_{1}(t;z)=\frac{b(t;z)}{a(t;z)}=\frac{b(0;z)}{a(0;z)}e^{8iz^{3}t}=r_{1}(0;z)e^{8iz^{3}t},$
(109) $\displaystyle
r_{2}(t;z)=\frac{\bar{b}(t;-z)}{d(t;z)}=\frac{\bar{b}(0;-z)}{d(0;z)}e^{8iz^{3}t}=r_{2}(0;z)e^{-8iz^{3}t},$
(110)

where $r_{1,2}(0;z)$ are initial reflection data founded from the initial data
$u(x,0)$.

######  Proposition 5.1.

If $r_{1,2}(0;z)\in H^{1,1}(\mathbb{R})\cap L^{2,3}(\mathbb{R})$, then for any
fixed $T>0$ and $t\in[0,T]$ or $t\in[-T,0]$, we have $r(t;z)\in
H^{1,1}(\mathbb{R})\cap L^{2,3}(\mathbb{R}).$

###### Proof.

By (110), we obtain

$\|r_{1}(t;\cdot)\|_{L^{2,3}(\mathbb{R})}=\|r_{1}(0;\cdot)\|_{L^{2,3}(\mathbb{R})}.$

For every $t\in[0,T]$, we have

$\begin{split}\|z\partial_{z}r_{1}(t;\cdot)\|_{L^{2}(\mathbb{R})}=&\|z\partial_{z}r_{1}(0;\cdot)+24iz^{3}tr_{1}(t;z)\|_{L^{2}(\mathbb{R})}\\\
\leq&\|z\partial_{z}r_{1}(0;\cdot)\|_{L^{2}(\mathbb{R})}+24T\|r_{1}(0;\cdot)\|_{L^{2,3}(\mathbb{R})}.\end{split}$
(111)

Therefore, we infer that $r_{1}(t;\cdot)\in H^{1,1}(\mathbb{R})\cap
L^{2,3}(\mathbb{R})$ for every $t\in[0,T]$ as $r_{1}(0;\cdot)\in
H^{1,1}(\mathbb{R})\cap L^{2,3}(\mathbb{R}).$ We can get similar conclusion
for $t\in[-T,0]$ and $r_{2}$. ∎

Using the time-dependent data $r_{1,2}(t;z)$ we can construct a time-dependent
RH-problem

######  Problem 2.

Find a matrix function $M(x,t;z)$ satisfying

(i) $M(x,t;z)\rightarrow I+\mathcal{O}\left(z^{-1}\right)$ as
$z\rightarrow\infty$.

(ii) For $M(x,t;z)$ admits the following jump condition

$\displaystyle M_{+}(x,t;z)$ $\displaystyle=M_{-}(x,t;z)V_{x,t}(z),$ (112)

where

$V_{x,t}(z):=\left(\begin{array}[]{cc}1+\sigma r_{1}r_{2}&\sigma
r_{2}e^{2\mathrm{i}\theta(x,t;z)}\\\
r_{1}e^{-2\mathrm{i}\theta(x,t;z)}&1\end{array}\right),z\in\mathbb{R}.$ (113)

Where $\theta(x,t;z)=zx+4z^{3}t$.

######  Theorem 5.2.

Assume that

$M(x,t;z)=\begin{cases}m_{+}(x;z,t)&\operatorname{Im}z>0\\\
m_{-}(x;z,t)&\operatorname{Im}z<0.\end{cases}$

is the solution of Problem2. Then $M(x,t;z)$ satisfies the following system od
linear differential equations:

$\displaystyle M_{x}(x,t;z)=iz[\sigma_{3},M]+QM,$ (114) $\displaystyle
M_{t}(x,t;z)=4iz^{3}[\sigma_{3},M]+(4z^{2}Q-2iz(Q_{x}-Q^{2})\sigma_{3}+2Q^{3}-Q_{xx})M,$

where

$Q(x,t)=\begin{bmatrix}0&u(x,t)\\\ -u(-x,-t)&0\end{bmatrix},$ (115)

and

$u(x,t)=2iz\lim_{z\rightarrow\infty}M_{12}(x,t).$ (116)

###### Proof.

Define

$\displaystyle LM=M_{x}(x,t;z)-iz[\sigma_{3},M]-QM,$ (117) $\displaystyle
NM=M_{t}(x,t;z)-4iz^{3}[\sigma_{3},M]-(4z^{2}Q-2iz(Q_{x}-Q^{2})\sigma_{3}+2Q^{3}-Q_{xx})M.$

Where $M(x,t;z)$ satisfies Problem2 and
$u(x,t)=2iz\lim_{z\rightarrow\infty}M_{12}(x,t).$

Then, by direct calculation we know $LM$ and $NM$ satisfy

$(LM)_{+}=(LM)_{-}V_{x,t},\ (NM)_{+}=(NM)_{-}V_{x,t}.$

Using definition of Problem2 we can rewrite $M(x,t;z)$ in the following form

$M(x,t;z)=I+\frac{M_{1}}{z}+\frac{M_{2}}{z^{2}}+\cdot\cdot\cdot+\frac{M_{n}}{z^{n}}+\cdot\cdot\cdot$
(118)

Substitute (118) into (117) and compare the coefficients of $z$ on both side
of equation and noticing (116). We can get

$LM\sim\mathcal{O}(z^{-1}),\ NM\sim\mathcal{O}(z^{-1}).$

In other words, if $M(x,t;z)$ satisfies Problem2 and
$u(x,t)=2iz\lim_{z\rightarrow\infty}M_{12}(x,t)$ then $LM$ and $NM$ satisfy
problem2 and $LM\rightarrow 0,\ NM\rightarrow 0$ as $z\rightarrow 0$. Noticing
Lemma3.4 and proposition5.1 we know Problem2 has unique solution which yields
$LM=NM=0$. Then we get (114). ∎

So under the time evolution of the scattering data $r(x,t;z)$ in (110), the
function reconstructed from the Problem 1 through the reconstruction formula
(93) under time-dependent scattering data $r(t;z)$ is still a solution of the
Mkdv equation (9).

###  5.2 The proof of main results

In this section, we will prove the existence of the local and global solutions
to the Cauchy problem . The scheme behind the proof can be described as below.

######  Lemma 5.3.

Let the initial data $u_{0}(x)\in H^{3}(\mathbb{R})\cap H^{1,1}(\mathbb{R})$ ,
then there exists a unique local solution to the Cauchy problem (1)-(2).

$\displaystyle u\in C([0,T],H^{3}(\mathbb{R})\cap H^{1,1}(\mathbb{R})).$

Furthermore, the map

$H^{3}(\mathbb{R})\cap H^{1,1}(\mathbb{R})\ni u_{0}\mapsto u\in C([0,T],\in
H^{3}(\mathbb{R})\cap H^{1,1}(\mathbb{R}))$

is Lipschitz continuous.

###### Proof.

Performing a similar analysis as Lemma 4.2 and lemma 4.4, we can establish a
RH problem for $r(t;z)$ for every $t\in[0,T]$ and address the existence and
uniqueness of a solution to the RH problem. Further, the potential $u(t,x)$
can be recovered from the reflection coefficients $r(t;z)$. Moreover, the
potential $u(t,\cdot)\in H^{1,1}\cap H^{3}$ for every $t\in[0,T]$ and is
Lipschitz continuous of $r(t;z)$. Thus we have

$\begin{split}\|u(t;\cdot)\|_{H^{1,1}\cap H^{3}}&\leq c_{1}\|r(t;\cdot)\|_{\in
H^{1,1}\cap L^{2,3}}\\\ &\leq c_{2}r(0;\cdot)\|_{H^{1,1}\cap L^{2,3}}\leq
c_{3}\|u_{0}\|_{H^{1,1}\cap H^{3}}\end{split}$ (119)

where the positive constants $c_{1}$,$c_{2}$ and $c_{3}$ depends on
$\|r\|_{L^{\infty}},\ \|zr\|_{L^{\infty}}$ and $(T,\
\|u_{0}\|_{H^{1,1}(\mathbb{R})\cap H^{3}(\mathbb{R})})$, respectively.

Next we show $u(x,t)$ is continuous with respect to $t\in[0,T]$ under the
$H^{1,1}(\mathbb{R})\cap H^{3}(\mathbb{R})$ norm. Let $t\in[0,T]$ and $|\Delta
t|<1$ such that $t+\Delta t\in[0,T]$, then with the Lipschitz continuity from
$u(t,x)$ to $r(t;z)$ in Proposition 4.5, we have

$\displaystyle\|u(t+\Delta t,x)-u(t,x)\|_{H^{1,1}(\mathbb{R})\cap
H^{3}(\mathbb{R})}$ $\displaystyle\leq c(\|r_{1}(t+\Delta
t;z)-r_{1}(t;z)\|_{H^{1,1}(\mathbb{R})\cap
L^{2,3}(\mathbb{R})}+\|r_{2}(t+\Delta
t;z)-r_{2}(t;z)\|_{H^{1,1}(\mathbb{R})\cap L^{2,3}(\mathbb{R})})$
$\displaystyle\leq c|\Delta t|(\|r_{1}(0;z)\|_{H^{1,1}(\mathbb{R})\cap
L^{2,3}(\mathbb{R})}+\|r_{2}(0;z)\|_{H^{1,1}(\mathbb{R})\cap
L^{2,3}(\mathbb{R})})\leq c|\Delta t|\rightarrow 0,\ \Delta t\to 0,$

which together with the estimate (119) implies that there exists a unique
local solution $u(x,t)\in C([0,T],H^{3}(\mathbb{R})\cap H^{1,1}(\mathbb{R}))$
to the Cauchy problem (1)-(2) and the map

$H^{1,1}(\mathbb{R})\cap H^{3}(\mathbb{R})\ni u_{0}(x)\mapsto u(t,x)\in
C([0,T],H^{1,1}(\mathbb{R})\cap H^{3}(\mathbb{R}))$

is Lipschitz continuous. ∎

Finally we give the proof of Theorem 1.

###### Proof of Theorem 1.

Suppose the maximal time in which the local solution in Lemma 5.3 exists is
$T_{max}$.

If $T_{max}=\infty$, then the local solution is global.

If the local solution exists in the closed interval $[0,T_{max}]$, we can use
$u(T_{max},\cdot)\in H^{3}(\mathbb{R})\cap H^{1,1}(\mathbb{R})$ as a new
initial data. By a similar analysis as the previous sections, there exists a
positive constant $T_{1}$ such that the solution $u\in
C([T_{max},T_{max}+T_{1}],H^{1,1}(\mathbb{R})\cap H^{3}(\mathbb{R}))$ exists.
This contradicts with the maximal time assumption.

If the local solution exists in the open interval $[0,T_{max})$. According to
(119), we have

$\|u(t,x)\|_{H^{1,1}(\mathbb{R})\cap H^{3}(\mathbb{R})}\leq
c_{3}(T_{max})\|u_{0}\|_{H^{1,1}(\mathbb{R})\cap H^{3}(\mathbb{R})},\quad
t\in[0,T_{max}).$

Due to the continuity of $u(t,x)$ to the time $t$, the limit of $u(t,x)$ as
$t$ approaches to $T_{max}$ exists. Let $u_{max}(x):=\lim_{t\rightarrow
T_{max}}u(t,x)$. Taking the limit by $t\rightarrow T_{max}$ in (119), we have

$\|u_{max}\|_{H^{1,1}(\mathbb{R})\cap H^{3}(\mathbb{R})}\leq
c_{3}(T_{max})\|u_{0}\|_{H^{1,1}(\mathbb{R})\cap H^{3}(\mathbb{R})},$

which implies that we can extend the local solution $u\in
C([0,T_{max}),H^{1,1}(\mathbb{R})\cap H^{3}(\mathbb{R}))$ to $u\in
C([0,T_{max}],H^{1,1}(\mathbb{R})\cap H^{3}(\mathbb{R}))$, this contradicts
with the premise that $[0,T_{max})$ is the maximal open interval. ∎

Acknowledgements

This work is supported by the National Natural Science Foundation of China
(Grant No. 12271104, 51879045).

Data Availability Statements

The data that supports the findings of this study are available within the
article.

Conflict of Interest

The authors have no conflicts to disclose.

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