11footnotetext:  School of Mathematical Sciences and Key Laboratory for
Nonlinear Science, Fudan University, Shanghai 200433, P.R.
China.22footnotetext:  Department of Mathematics, University of Texas at
Arlington, TX 76019.

#  Existence of global solutions to the Cauchy problem for the modified
Camassa-Holm equation with a linear dispersion term

Yiling YANG1 Engui FAN1 and Yue LIU2 Email address<EMAIL_ADDRESS>Corresponding author and email address<EMAIL_ADDRESS>Email address:
<EMAIL_ADDRESS>
###### Abstract

In this paper, we address the existence of global solutions to the Cauchy
problem of the modified Camassa-Holm (mCH) equation with a linear dispersion
term

$\displaystyle m_{t}+\left(m\left(u^{2}-u_{x}^{2}\right)\right)_{x}+\kappa
u_{x}=0,\quad m=u-u_{xx},$ $\displaystyle m(x,0)=m_{0}(x),$

where $\kappa$ is a positive constant characterizing the effect of the linear
dispersion. The key to prove this result is to establish a bijective maps
between potentials and reflection coefficients with the inverse scattering
transform theory, in which the Volttera integral operator and Cauchy integral
operator play an important role. Based on the spectral analysis of the Lax
pair associated with the mCH equation, the solution of the Cauchy problem is
characterized via the solution of a RH problem in the new scale $(y,t)$. By
using the reconstruction formula and estimates on the solution of the time-
dependent RH problem, we further affirm the existence of a unique global
solution to the Cauchy problem for the mCH equation.
MSC 2020: 35Q51; 35Q15; 37K15; 35C20.

###### Contents

  1. 1 Introduction
    1. 1.1 Main results
    2. 1.2 Some notations
  2. 2 Direct scattering transform
    1. 2.1 Spectral analysis on the Lax pair
    2. 2.2 Scattering data
    3. 2.3 A RH problem associated with initial value problem
  3. 3 The RH problem
    1. 3.1 The Cauchy projection operator
    2. 3.2 Solvability of the RH problem
    3. 3.3 Estimates on the solution of RHPs
  4. 4 Time evolution
  5. 5 Existence of global solution

##  1 Introduction

The present paper is concerned with the existence of global solutions to the
Cauchy problem of the modified Camassa-Holm (mCH) equation with a linear
dispersion term [1, 2, 3]

$\displaystyle m_{t}+\left(m\left(u^{2}-u_{x}^{2}\right)\right)_{x}+\kappa
u_{x}=0,\quad m=u-u_{xx},$ (1.1) $\displaystyle m(x,0)=m_{0}(x),$ (1.2)

where $\kappa$ is a positive constant characterizing the effect of the linear
dispersion. It is known that the mCH equation (1.1) is an integrable
modification to the well-known Camassa-Holm (CH) equation

$\displaystyle m_{t}+(um)_{x}+u_{x}m=0,\quad m=u-u_{xx},$ (1.3)

which was appeared in [4] as a integrable equation proposed by Fuchssteiner
and Fokas and first introduced by Camassa and Holm as a model for the
unidirectional propagation of shallow-water waves [5]. The CH equation (1.3)
has been attracted considerable interest and been studied extensively due to
its rich mathematical structures and remarkable properties, such as Painleve-
type asymptotics, peakon and multi-peakon solutions, bi-Hamiltonian structure,
algebro-geometric solutions, wave-breaking phenomena [6, 7, 8, 9, 10].

In recent years, various modifications and generalizations of the CH equation
have been introduced. Novikov applied the perturbative symmetry approach in
order to classify integrable equations of the form

$(1-\partial_{x}^{2})u_{t}=F(u,u_{x},u_{xx},\cdots),$

two integrable CH-type equations with cubic nonlinearity which admit peakons
have been discovered [11]. One is the mCH equation

$\displaystyle m_{t}+\left(m\left(u^{2}-u_{x}^{2}\right)\right)_{x}=0,\quad
m=u-u_{xx}$ (1.4)

and an other one is the so-called Novikov equation

$\displaystyle m_{t}+\left(m_{x}u+3mu_{x}\right)u=0,\quad m=u-u_{xx}.$ (1.5)

In an equivalent form, the mCH equation was given by Fokas [12], Fuchssteiner
[13], Olver and Rosenau [14] and Qiao [15], where the equation was derived
from the two-dimensional Euler system, the M/W-shape solitons and
peakon/cuspon solutions were presented. So the mCH equation (1.4) is also
referred to as the Fokas-Olver-Rosenau-Qiao equation [16], but is mostly known
as the mCH equation. Adding a linear dispersion term $\kappa u_{x}$ to the
original mCH equation (1.4) leads to the CH equation (1.1) [1, 2, 3]. It can
be shown that the mCH equation (1.4) on a nonzero background or the mCH
equation (1.1) with decaying initial data may support smooth soliton solutions
[3, 17, 18]. The mCH equation (1.1) admits a Lax pair and its smooth dark
soliton solutions were obtained by the method of inverse scattering
transformation method [1, 17]. The wave-breaking and those peakons for the mCH
equation (1.4) with zero background data were also investigated in [19]. With
the aid of reciprocal transformation, the Bäcklund transformation and
nonlinear superposition formula for the mCH equation (1.4) were given [20].
Much interesting work on mCH equation (1.1) and mCH equation (1.4) can be seen
[21, 22, 23, 24, 25, 26, 27, 28, 29]. The long time asymptotic behavior of the
mCH equation (1.1) was established by using $\bar{\partial}$-steepest decadent
analysis in [30]. Recently we obtained the existence of global solutions to
the Cauchy problem of the mCH equation (1.4) with nonzero background
$u(t,x)\to 1$ as $|x|\to\infty$ [31].

In the present paper, we establish the bijectivity between potential and
reflection coefficient for the mCH equation (1.1) and global well-posedness
for the initial value problem of the mCH equation (1.1)-(1.2) with the inverse
scattering method. A crucial approach for proving Theorem 1 is to establish a
$L^{2}$-bijection map to the Cauchy problem (1.1)-(1.2) between potential and
reflection coefficient via the Volttera integral operator and Cauchy integral
operator. The integrability or inverse scattering transform approach has been
successfully applied to show the existence of global solutions to the Cauchy
problem for integrable systems such as the NLS, AKNS, derivative NLS equations
[33, 34, 35, 36, 37, 38].

###  1.1 Main results

A principal results of the present paper is now stated as follows.

######  Theorem 1.

Assume that $m_{0}\in H^{2,1}(\mathbb{R})$ with small norm. Then there exists
a unique global solution $m$ to the Cauchy problem (1.1)-(1.2) with
$m(0)=m_{0}$ such that $m\in C([0,+\infty),H^{2,1}(\mathbb{R})).$

######  Remark 2.

It can be shown that the result presented in Theorem 1 is not inconsistent
with the blow up result in [2].

######  Remark 3.

Compared with [31], we add more requirements on $m_{0}$. The main reason is
that the RH problem of (1.1) loses an explicit symmetry expression between $z$
and $-1/z$. Hence, when we use the solution $M(z)$ of RH problem to
reconstitute the solution $m$ of the Cauchy problem (1.1)-(1.2) in Section 5,
we need the small norm of $M(z)$. And to controlled the norm of $M$, we assume
that the linear equation (2.1) has no any eigenvalue and the small norm of
$r$, which comes from the small norm assumption of initial value $m_{0}$.

This paper is arranged as follows. In section 2, we recall some main results
on the construction process of RH problem [32, 30] and establish the direct
scattering map from the initial data $m_{0}$ to scattering data
$\\{r,\\{z_{k},c_{k}\\}_{k=1}^{N_{0}}\\},\
\\{\tilde{r},\\{z_{k},\tilde{c}_{k}\\}_{k=1}^{N_{0}}\\}$ which are defined in
Definition 3. Under the scattering data
$\\{r,\\{z_{k},c_{k}\\}_{k=1}^{N_{0}}\\},\
\\{\tilde{r},\\{z_{k},\tilde{c}_{k}\\}_{k=1}^{N_{0}}\\}$, we give two RH
problems respectively in Subsection 2.3. Then in Section 3, we give the unique
existence of this two RH problems under any scattering data in
$H^{1,1}(\mathbb{R}_{k})\otimes\mathbb{C}^{N_{0}}\otimes\mathbb{C}^{N_{0}}$.
In section 4, we establish the relationship between the scattering data and
time $t$. Finally, in 5, we give the proof of the following theorem.

###  1.2 Some notations

In this subsection, we fix some notations used this paper. If $I$ is an
interval on the real line $\mathbb{R}$ and $\mathsf{X}$ is a Banach space,
then $C^{0}(I,\mathsf{X})$ denotes the space of continuous functions on $I$
taking values in $\mathsf{X}$. It is equipped with the norm

$\|f\|_{C^{0}(I,\mathsf{X})}=\sup_{x\in I}\|f(x)\|_{\mathsf{X}}.$

Moreover, denote $C^{0}_{B}(\mathsf{X})$ as a space of bounded continuous
functions on $\mathsf{X}$.

If the entries $f_{1}$ and $f_{2}$ are in space $\mathsf{X}$, then we call
vector $\vec{f}=(f_{1},f_{2})^{T}$ is in space $\mathsf{X}$ with the norm
$\parallel\vec{f}\parallel_{\mathsf{X}}\triangleq\parallel
f_{1}\parallel_{\mathsf{X}}+\parallel f_{2}\parallel_{\mathsf{X}}$. Similarly,
if every entries of matrix $A$ are in space $\mathsf{X}$, then we call $A$ is
also in space $\mathsf{X}$.

We introduce the normed spaces:

  * •

A weighted $L^{p}(\mathbb{R})$ space is specified by

$L^{p,s}(\mathbb{R})=\left\\{f(x)\in L^{p}(\mathbb{R})|\hskip
2.84544pt\langle\cdot\rangle^{s}f(x)\in L^{p}(\mathbb{R})\right\\};$

  * •

A weighted Sobolev space is defined by

$H^{k,s}(\mathbb{R})=\left\\{f(x)\in L^{2}(\mathbb{R})|\hskip
2.84544pt\langle\cdot\rangle^{s}\partial^{j}f(x)\in L^{2}(\mathbb{R}),\text{
for }j=1,...,k\right\\}.$

For the simplicity, the norm of $f(x)\in L^{p}(\mathbb{R})$ and $g(x)\in
L^{p,s}(\mathbb{R})$ are abbreviated to $\parallel f\parallel_{p}$, $\parallel
g\parallel_{p,s}$ respectively.

Finally, the letter $C$ will be used to denote universal positive constants
which may vary from line to line. We also use the notation $A\lesssim B$ to
denote the bound of the form $A\leqslant CB$. To emphasize the implied
constant to depend on some parameter $\alpha$, we shall indicate this by
$C(\alpha)$.

##  2 Direct scattering transform

The mCH equation (1.1) is completely integrable and admits the Lax pair [32]

$\Phi_{x}=X\Phi,\hskip 14.22636pt\Phi_{t}=T\Phi,$ (2.1)

where

$X=-\frac{ki}{4}\sigma_{3}+\frac{i\lambda m(x,t)}{2}\sigma_{2},$
$T=\frac{ki}{\lambda^{2}2}\sigma_{3}+\frac{ki}{4}\left(u^{2}-u_{x}^{2}\right)\sigma_{3}-i\left(\frac{2ui-
ku_{x}}{2\lambda}+\frac{\lambda}{2}\left(u^{2}-u_{x}^{2}\right)m\right)\sigma_{2}$

with

$k=k(z)=z-\frac{1}{z},\hskip
8.5359pt\lambda=\lambda(z)=\frac{1}{2}(z+\frac{1}{z}).$

Here, $z\in\mathbb{C}$ is a spectral parameter. Before constructing the direct
scattering transform, the first step is to introduce a new space variable

$y(x,t)=x-\int_{x}^{+\infty}\left(\sqrt{m(s,t)^{2}+1}-1\right)ds.$ (2.2)

Noting that (1.1) is equivalent to the following equation

$\displaystyle(\sqrt{m^{2}+1})_{t}+(\sqrt{m^{2}+1}(u^{2}-u_{x}^{2}))_{x}=0,$

then we adduce that

$\displaystyle dy=\sqrt{m^{2}+1}dx-\sqrt{m^{2}+1}(u^{2}-u_{x}^{2})dt.$ (2.3)

Therefore the Lax pair (2.1) becomes

$\displaystyle\Phi_{y}=Y\Phi,\hskip 14.22636pt\Phi_{t}=Q\Phi$ (2.4)
$\displaystyle Y=\frac{1}{\sqrt{m^{2}+1}}X,\hskip
14.22636ptQ=(u^{2}-u_{x}^{2})X+T.$ (2.5)

In direct scattering transform, we first consider the partial spectral problem
in the Lax pair (2.1) under $t=0$. So in this subsection, the variable $t$ is
omitted as usual. For example, $\Phi(z;0,y)$ is just written as $\Phi(z;y)$ or
even $\Phi$.

###  2.1 Spectral analysis on the Lax pair

We quickly review main results on the spectral analysis of the Lax pair in
[30]. Under $m_{0}\in H^{2,1}$, via (2.2), the map $x\to y(x,0)$ is continuous
differentiable bijection.

Define a transformation

$\Phi(z;y)=F(y)\mu(z;y)e^{-\frac{i}{4}(z-\frac{1}{z})y\sigma_{3}},$ (2.6)

where

$\displaystyle
F(y)=\sqrt{\frac{q+1}{2q}}\left(\begin{array}[]{cc}1&\frac{-im_{0}}{q+1}\\\
\frac{-im_{0}}{q+1}&1\end{array}\right),\hskip
14.22636ptq=\sqrt{m_{0}^{2}+1}.$ (2.9)

It is thereby inferred that

$\displaystyle\mu_{y}=-\frac{i}{4}(z-\frac{1}{z})[\sigma_{3},\mu]+P\mu,$
(2.10)

where

$\displaystyle
P=\frac{im_{0,x}}{2q^{3}}\sigma_{1}+\frac{m_{0}}{2zq^{2}}\left(\begin{array}[]{cc}-im_{0}&1\\\
-1&im_{0}\end{array}\right),$

which leads to two Volterra type integrals

$\mu_{\pm}(y,z)=I+\int^{y}_{\pm\infty}e^{-\frac{i}{4}(z-\frac{1}{z})(y-s)\hat{\sigma}_{3}}P(s,z)\mu_{\pm}(s)ds.$
(2.11)

Denote the matrix in column

$\mu^{\pm}(y,z)=\left(\mu^{\pm}_{1}(y,z),\mu^{\pm}_{2}(y,z)\right),$

where the subscript $1$ and $2$ denote the first and second columns of
$\mu^{\pm}(y,z)$, respectively. And $\mu_{\pm}(z)$ admit two kinds of
reduction conditions

$\mu_{\pm}(z)=\sigma_{2}\overline{\mu_{\pm}(\bar{z})}\sigma_{2}=\sigma_{1}\mu_{\pm}(-z)\sigma_{1}.$
(2.12)

and

$\mu_{\pm}(z)=F^{-2}\sigma_{2}\mu_{\pm}(-z^{-1})\sigma_{2},$ (2.13)

Before analyzing the property of $\mu_{\pm}$, we first give its existence.

######  Proposition 1.

Supposing that $m_{0}\in H^{2,1}$, then $\mu_{\pm}(y,z)$ exist unique such
that $\mu_{\pm}(y,z)-I\in L^{\infty}(\mathbb{R}^{\pm}\times\mathbb{R})\cap
C^{0}(\mathbb{R}^{\pm},L^{2}(\mathbb{R}))\cap
L^{2}(\mathbb{R}^{\pm}\times\mathbb{R})$, $\partial_{z}\mu_{\pm}(y,z),\
k(z)\partial_{z}\mu_{\pm}(y,z)\in C^{0}(\mathbb{R}^{\pm},L^{2}(\mathbb{R}))$.
Furthermore, the non-diagonal terms of $k(z)(\mu_{\pm}(y,z)-I)$ belongs in
$C^{0}(\mathbb{R}^{\pm},L^{2}(\mathbb{R}))$.

###### Proof.

Invoking the definition of variable $y$ in (2.2), $m_{0}\in H^{2,1}_{x}$ is
equivalent to $m_{0}\in H^{2,1}_{y}$. From (2.13), it follows that
$\sup_{|z|\leq 1}|\mu_{\pm}(z)|=\sup_{|z|\geq
1}|F^{-2}\sigma_{2}\mu_{\pm}(-z^{-1})\sigma_{2}|$, which means that

$\displaystyle\inf_{x\in\mathbb{R}}|F(x)^{-2}|\sup_{|z|\geq
1}|\mu_{\pm}(z)|\leq\sup_{|z|\leq
1}|\mu_{\pm}(z)|\leq\|F^{-2}\|_{\infty}\sup_{|z|\geq 1}|\mu_{\pm}(z)|,$
$\displaystyle\int_{|z|\leq 1}|z^{-j}\mu_{\pm}(z)|^{2}dz\leq\int_{|z|\geq
1}\|F^{-2}\|_{\infty}^{2}z^{2j-2}\|mu_{\pm}(z)|^{2}dz,\ j=0,1,2.$

Thus, it is sufficient to analyze the spaces
$L^{\infty}(\mathbb{R}^{\pm}\times\\{|z|\geq 1\\})$,
$C^{0}(\mathbb{R}^{\pm},L^{2}(\\{|z|\geq 1\\}))$ and
$L^{2}(\mathbb{R}^{\pm}\times\mathbb{R})$ which are abbreviated to
$L^{\infty}$, $C^{0}$, $L^{2}$ respectively. We denote

$\displaystyle n(y,z):=\mu_{+,1}(y,z)-e_{1},$ (2.14)

where $e_{1}=(1,0)^{T}$. Introduce the integral operator $\mathcal{T}$

$\displaystyle\mathcal{T}(f)(y,z)=-\int_{y}^{+\infty}K(y,s,z)f(s,z)ds,$ (2.15)

where integral kernel $K(y,s,z)$ is

$\displaystyle K(y,s,z)=$
$\displaystyle\frac{im_{0,x}(s)}{2q^{3}(s)}\left(\begin{array}[]{cc}0&1\\\
e^{\frac{i}{2}(z-1/z)(y-s)}&0\end{array}\right)$ (2.18)
$\displaystyle+\frac{1}{2z}\frac{m_{0}(s)}{q^{2}(s)}\left(\begin{array}[]{cc}-im_{0}(s)&1\\\
-e^{\frac{i}{2}(z-1/z)(y-s)}&e^{\frac{i}{2}(z-1/z)(y-s)}im_{0}(s)\end{array}\right)\
x<s.$ (2.21)

Then the first column of (2.11) is changed to

$\displaystyle n=\mathcal{T}(e_{1})+\mathcal{T}(n),$ (2.22)

where

$\displaystyle\mathcal{T}(e_{1})=-\int_{y}^{+\infty}\left(\begin{array}[]{cc}\frac{-im_{0}^{2}(s)}{2zq^{2}(s)}\\\
e^{\frac{i}{2}(z-1/z)(y-s)}\left(\frac{im_{0,x}(s)}{2q^{3}(s)}-\frac{m_{0}(s)}{2zq^{2}(s)}\right)\end{array}\right)ds,$

with

$\displaystyle|\mathcal{T}(e_{1})|\leq(\|m_{0}\|_{2}^{2}+\|m_{0}\|_{1}+\|m_{0,x}\|_{1})/2,$
$\displaystyle\int_{|z|\geq
1}|\mathcal{T}(e_{1})|^{2}dz\lesssim\int_{y}^{+\infty}|m_{0}|^{2}ds+\int_{y}^{+\infty}|m_{0,x}|^{2}ds,$
$\displaystyle\int_{\mathbb{R}^{+}}\int_{|z|\geq
1}|\mathcal{T}(e_{1})|^{2}dzdy\lesssim\|m_{0}\|_{2,1/2}+\|m_{0,x}\|_{2,1/2}.$

From we shown in [30], it follows that the integral operator $\mathcal{T}$
maps $L^{\infty}\cap C^{0}\cap L^{2}$ to itself with

$\displaystyle\|\mathcal{T}\|\leq C(\|m_{0}\|_{H^{1,1}}).$

In addition, $(I-\mathcal{T})^{-1}$ exists as a bounded operator on
$L^{\infty}\cap C^{0}\cap L^{2}$ also admitting

$\displaystyle\|(I-\mathcal{T})^{-1}\|\leq C(\|m_{0}\|_{H^{1,1}}).$

Thereby we conclude that $n=(I-\mathcal{T})^{-1}\mathcal{T}(e_{1})\in
L^{\infty}\cap C^{0}\cap L^{2}$. On the other hand, by denoting
$\mathcal{T}_{z}$ as a integral operator with the integral kernel

$\displaystyle\partial_{z}K(y,s,z)=$
$\displaystyle\frac{-m_{0}(s)}{2z^{2}q^{2}(s)}\left(\begin{array}[]{cc}-im_{0}(s)&1\\\
0&0\end{array}\right)-\frac{(1+1/z^{2})(y-s)m_{0,x}(s)e^{\frac{i}{2}(z-1/z)(y-s)}}{4q^{3}(s)}\left(\begin{array}[]{cc}0&0\\\
1&0\end{array}\right)$
$\displaystyle-\frac{m_{0}(s)e^{\frac{i}{2}(z-1/z)(y-s)}}{4zq^{2}(s)}(2/z-i(1+1/z^{2})(y-s))\left(\begin{array}[]{cc}0&0\\\
-1&im_{0}(s)\end{array}\right),\ x<s,$

take $z$-derivative in(2.22) and obtain

$\displaystyle\partial_{z}n=\partial_{z}\mathcal{T}(e_{1})+\mathcal{T}_{z}(n)+T(\partial_{z}n),$
(2.23)

where

$\displaystyle\partial_{z}\mathcal{T}(e_{1})=\int_{y}^{+\infty}\left(\begin{array}[]{cc}\frac{-im_{0}^{2}(s)}{2z^{2}q^{2}(s)}\\\
e^{\frac{i}{2}(z-1/z)(y-s)}\left(\frac{m_{0}(s)}{2z^{2}q^{2}(s)}+\frac{i}{2}(1+1/z^{2})(y-s)\left(\frac{im_{0,x}(s)}{2q^{3}(s)}-\frac{m_{0}(s)}{2zq^{2}(s)}\right)\right)\end{array}\right)ds,$

with estimate

$\displaystyle\|\partial_{z}\mathcal{T}(e_{1})\|_{C^{0}}\leq
C(\|m_{0}\|_{H^{1,1}}).$

Noting that for any functions $f(y),\ g(y,z)$, it is found that

$\displaystyle\int_{|z|\geq
1}\big{|}\int_{y}^{+\infty}f(s)g(s,z)ds\big{|}^{2}dz\leq\|f\|_{2}\|g\|_{L^{2}}^{2},$
$\displaystyle\text{or }\int_{|z|\geq
1}\big{|}\int_{y}^{+\infty}f(s)g(s,z)ds\big{|}^{2}dz\leq\|f\|_{1}\|g\|_{C^{0}}.$

Thus,

$\displaystyle\|\mathcal{T}_{z}n\|_{C^{0}}\lesssim
C(\|m_{0}\|_{H^{1,1}})\|n\|_{L^{2}},$

which leads to $\partial_{z}n$ exists in $C^{0}$ with

$\displaystyle\partial_{z}n=(I-\mathcal{T})^{-1}(\mathcal{T}_{z}n+\mathcal{T}(e_{1})).$
(2.24)

Denoting $n=(n_{1},n_{2})^{T}$, by integration by parts, it adduced that

$\displaystyle\frac{i}{2}(z-1/z)n_{2}$
$\displaystyle=\int^{y}_{+\infty}e^{\frac{i}{2}(z-1/z)(y-s)}\partial_{s}\left(\frac{im_{0,x}}{2q^{3}}-\frac{m_{0}}{2zq^{2}}\right)ds$
$\displaystyle+\int^{y}_{+\infty}e^{\frac{i}{2}(z-1/z)(y-s)}\left[\partial_{s}\left(\frac{im_{0,x}}{2q^{3}}-\frac{m_{0}}{2zq^{2}}\right)n_{1}+\partial_{s}\left(\frac{im_{0}^{2}}{2zq^{2}}\right)n_{2}\right]ds$
$\displaystyle+\int^{y}_{+\infty}e^{\frac{i}{2}(z-1/z)(y-s)}\left[\left(\frac{im_{0,x}}{2q^{3}}-\frac{m_{0}}{2zq^{2}}\right)\partial_{s}n_{1}+\frac{im_{0}^{2}}{2zq^{2}}\partial_{s}n_{2}\right]ds.$
(2.25)

The $C^{0}$ norm of the first term of the right hand of above equation is
controlled by
$\|\partial_{s}\left(\frac{im_{0,x}}{2q^{3}}-\frac{m_{0}}{2q^{2}}\right)\|_{2}$,
namely, $\|m_{0}\|_{H^{2}}$. And it follows that

$\displaystyle\|\text{second term of r.h.s \eqref{kn2}}\|_{C^{0}}\leq
C(\|m_{0}\|_{H^{2,1}})\|n\|_{L^{2}}.$

From the definition of $n$ in (2.14) and the equation (2.10), it holds that

$\displaystyle\partial_{s}n=\frac{i}{2}(z-1/z)\left(\begin{array}[]{cc}0\\\
n_{2}\end{array}\right)+\left(\begin{array}[]{cc}\frac{-im_{0}^{2}}{2zq^{2}}(n_{1}+1)+\left(\frac{im_{0,x}}{2q^{3}}+\frac{m_{0}}{2zq^{2}}\right)n_{2}\\\
\left(\frac{im_{0,x}}{2q^{3}}-\frac{m_{0}}{2zq^{2}}\right)(n_{1}+1)+\frac{im_{0}^{2}}{2zq^{2}}n_{2}\end{array}\right).$

Therefore, the third term of r.h.s (2.25) can be rewritten as

$\displaystyle\text{third term of r.h.s
\eqref{kn2}}=\int^{y}_{+\infty}e^{\frac{i}{2}(z-1/z)(y-s)}\left[\frac{-im_{0}^{2}}{4zq^{2}}\left(\frac{im_{0,x}}{q^{3}}-\frac{m_{0}}{zq^{2}}\right)-\frac{m_{0}^{4}}{4z^{2}q^{4}}\right]ds$
$\displaystyle+\int^{y}_{+\infty}e^{\frac{i}{2}(z-1/z)(y-s)}\left(\frac{im_{0,x}}{q^{3}}-\frac{m_{0}}{zq^{2}}\right)\left(\frac{-im_{0}^{2}}{2zq^{2}}n_{1}+\left(\frac{im_{0,x}}{2q^{3}}+\frac{m_{0}}{2zq^{2}}\right)n_{2}\right)ds$
$\displaystyle+\int^{y}_{+\infty}e^{\frac{i}{2}(z-1/z)(y-s)}\frac{im_{0}^{2}}{2q^{2}}\left[\frac{i(1-1/z^{2})}{2}n_{2}+\left(\frac{im_{0,x}}{2zq^{3}}-\frac{m_{0}}{2z^{2}q^{2}}\right)n_{1}+\frac{im_{0}^{2}}{2z^{2}q^{2}}n_{2}\right]ds.$

Thus it is accomplished that

$\displaystyle\|\text{third term of r.h.s \eqref{kn2}}\|_{C^{0}}\leq
C(\|m_{0}\|_{H^{2,1}})\|n\|_{L^{2}}.$

For $\partial_{z}n$, obviously,

$\displaystyle\|k(z)(\partial_{z}\mathcal{T}(e_{1}))_{1}\|_{C^{0}}\leq\|m_{0}\|_{2},$

and

$\displaystyle
k(z)(\partial_{z}\mathcal{T}(e_{1}))_{2}=\int_{y}^{+\infty}e^{\frac{i}{2}(z-1/z)(y-s)}\frac{i}{2}(1-1/z^{2})\frac{m_{0}(s)}{2zq^{2}(s)}ds$
$\displaystyle+\int_{y}^{+\infty}e^{\frac{i}{2}(z-1/z)(y-s)}\frac{(z-1/z)(1+1/z^{2})(y-s)}{4}\left(\frac{im_{0,x}(s)}{2q^{3}(s)}-\frac{m_{0}(s)}{2zq^{2}(s)}\right)ds.$

Similarly, the $C^{0}$ norm of first integral in the right side of above
equation is controlled by $\|m_{0}\|_{1}$, and from integration by parts, the
second integral in the right side of above equation is controlled by
$\|m_{0}\|_{H^{2,1}}$. And

$\displaystyle
k(z)\mathcal{T}_{z}n=\int_{y}^{+\infty}\frac{i}{2}(1-1/z^{2})\frac{-m_{0}(s)}{2zq^{2}(s)}\left(\begin{array}[]{cc}-im_{0}(s)&1\\\
-e^{\frac{i}{2}(z-1/z)(y-s)}&im_{0}(s)e^{\frac{i}{2}(z-1/z)(y-s)}\end{array}\right)n(s,z)ds$
(2.28)
$\displaystyle-\int_{y}^{+\infty}\frac{i}{2}(1-1/z^{2})\frac{m_{0}(s)e^{-i\frac{i}{2}(z-1/z)(y-s)}(1+1/z^{2})(y-s)}{4q^{2}(s)}\left(\begin{array}[]{cc}0&0\\\
-1&im_{0}(s)\end{array}\right)n(s,z)ds$ (2.31)
$\displaystyle+\int_{y}^{+\infty}\frac{i}{2}(z-1/z)\frac{-(1+1/z^{2})(y-s)m_{0,x}(s)e^{\frac{i}{2}(z-1/z)(y-s)}}{4q^{3}(s)}\left(\begin{array}[]{cc}0&0\\\
1&0\end{array}\right)n(s,z)ds.$ (2.34)

The $C^{0}$ norm of first two integral in the right side of above equation is
controlled by $\|m_{0}\|_{2,1}$. And for the last integral, integration by
parts gives that

$\displaystyle\int_{y}^{+\infty}\frac{i}{2}(z-1/z)\frac{-(1+1/z^{2})(y-s)m_{0,x}(s)e^{\frac{i}{2}(z-1/z)(y-s)}}{4q^{3}(s)}n_{1}(s,z)ds$
$\displaystyle=$
$\displaystyle\int_{y}^{+\infty}(1+1/z^{2})e^{\frac{i}{2}(z-1/z)(y-s)}\partial_{s}\left(\frac{(y-s)m_{0,x}(s)}{4q^{3}(s)}\right)n_{1}(s,z)ds$
$\displaystyle+\int_{y}^{+\infty}(1+1/z^{2})e^{\frac{i}{2}(z-1/z)(y-s)}\frac{(y-s)m_{0,x}(s)}{4q^{3}(s)}(\partial_{s}n_{1})(s,z)ds$
$\displaystyle=$
$\displaystyle\int_{y}^{+\infty}(1+1/z^{2})e^{\frac{i}{2}(z-1/z)(y-s)}\partial_{s}\left(\frac{(y-s)m_{0,x}(s)}{4q^{3}(s)}\right)n_{1}(s,z)ds$
$\displaystyle+\int_{y}^{+\infty}(1+1/z^{2})e^{\frac{i}{2}(z-1/z)(y-s)}\frac{(y-s)m_{0,x}(s)}{4q^{3}(s)}\left(\frac{-im_{0}^{2}}{2zq^{2}}(n_{1}+1)+\left(\frac{im_{0,x}}{2q^{3}}+\frac{m_{0}}{2zq^{2}}\right)n_{2}\right)ds.$

Hence, we obtain that

$\displaystyle\|k(z)\mathcal{T}_{z}n\|_{C^{0}}\leq C(\|m_{0}\|_{H^{2,1}}),$

which together with (2.24) gives that

$\displaystyle\|k(z)\partial_{z}n\|_{C^{0}}\leq C(\|m_{0}\|_{H^{2,1}}).$

Thus we accomplish our proof. ∎

For convenience, denote the banach space $\mathsf{Y}^{(\pm)}$ as a space of
$2\times 2$ matrix functions $f(y,z)$ which admits that $f(y,z)\in
L^{\infty}(\mathbb{R}^{\pm}\times\mathbb{R})\cap
C^{0}(\mathbb{R}^{\pm},L^{2}(\mathbb{R}))\cap
L^{2}(\mathbb{R}^{\pm}\times\mathbb{R})$, $\partial_{z}f(y,z),\
k(z)\partial_{z}f(y,z)\in C^{0}(\mathbb{R}^{\pm},L^{2}(\mathbb{R}))$, and the
non-diagonal terms of $k(z)(f(y,z)-I)$ belongs in
$C^{0}(\mathbb{R}^{\pm},L^{2}(\mathbb{R}))$. And its norm is defined by the
sum of the norm listed above.

Note that the operator equations in above proposition are both linear, from
which we obtain that $m_{0}\to\mu^{\pm}-I$ is locally lipshitz continuous from
$H^{2,1}$ to the space $\mathsf{Y}^{(\pm)}$ respectively. And when $m\equiv
0$, it follows that $\mu^{\pm}\equiv I$. So for any neighborhood $G$ of $0$ in
$H^{2,1}$, for any $m_{0}\in G$ and its corresponding jost function
$\mu^{\pm}-I$, there exists a constant $C(G)$ such that

$\displaystyle\|\mu^{\pm}-I\|_{\mathsf{Y}^{(\pm)}}\leq
C(G)\|m_{0}\|_{H^{2,1}}.$

Similar to the proof of the above proposition, it follows that as functions of
spectral parameter $z$, $\mu^{-}_{1}(z)$ and $\mu^{+}_{2}(z)$ are analytic in
$\mathbb{C}^{+}$ while $\mu^{+}_{1}(z)$ and $\mu^{-}_{2}(z)$ are analytic in
$\mathbb{C}^{-}$. In addition, by Lebesgue’s dominated convergence theorem,
$\lim_{z\to\infty}z(\mu_{\pm}(z)-I)$ exists in their analytic region
respectively such that $\mu_{\pm}(z)$ admit the asymptotics in their analytic
region respectively

$\displaystyle\mu_{\pm}(z)=I+\dfrac{D_{1}}{z}+\mathcal{O}(z^{-2}),\hskip
14.22636ptz\rightarrow\infty,$ (2.35)

where

$D_{1}(y)=\dfrac{im_{x}}{(1+m^{2})^{3/2}}\sigma_{2}+\frac{i}{2}\int_{y}^{\pm\infty}\left(\frac{m_{x}^{2}}{q^{6}}+\frac{m^{2}}{q^{2}}\right)ds\sigma_{3}.$
(2.36)

By Abel’s formula, $\det\mu^{\pm}=1$. This asymptotics behavior together with
(2.13) also imply that

$\displaystyle\mu_{\pm}(0;y)=F(y)^{-2}=\frac{1}{q(y)}I+\frac{im_{0}(y)}{q(y)}\sigma_{1}.$
(2.37)

To analyze the property near $z=\pm i$, we let $U(\pm i)$ be a bounded closed
neighborhood of $z=\pm i$ in $\mathbb{C}\setminus\\{0\\}$ and define a new
transformation

$\mu^{0,\pm}(z)=\Phi_{\pm}(z)e^{\frac{k}{2}x\sigma_{3}},$ (2.38)

which leads to two Volterra type integrals

$\mu^{0,\pm}(x,z)=I+\int^{x}_{\pm\infty}e^{-\frac{i}{4}(z-\frac{1}{z})(x-s)\hat{\sigma}_{3}}L_{0}(s,z)\mu^{0,\pm}(s)ds.$
(2.39)

Then $\mu^{0,\pm}$ admits the following proposition.

######  Proposition 2.

Supposing that $m_{0}\in H^{2,1}(\mathbb{R})$, then for every
$x\in\mathbb{R}$, $\mu^{0,-}_{1}(x,\cdot),\ \mu^{0,+}_{2}(x,\cdot),$ exist
unique in $L^{\infty}(U(i))$ while $\mu^{0,+}_{1}(x,\cdot),\
\mu^{0,-}_{2}(x,\cdot),$ exist unique in $L^{\infty}(U(-i))$. And they
satisfies the limits as follows

$\displaystyle\lim_{z\to i\text{ in }U(i)}\left(\mu^{0,-}_{1},\
\mu^{0,+}_{2}\right)=\lim_{z\to-i\text{ in }U(-i)}\left(\mu^{0,+}_{1},\
\mu^{0,-}_{2}\right)=I,$ (2.40) $\displaystyle\lim_{z\to i\text{ in
}U(i)}\frac{\left(\mu^{0,-}_{1},\
\mu^{0,+}_{2}\right)-I}{z-i}=\lim_{z\to-i\text{ in
}U(-i)}\frac{\left(\mu^{0,+}_{1},\ \mu^{0,-}_{2}\right)-I}{z-i}=\mu^{0,(1)},$
(2.41)

where

$\mu^{0,(1)}=\left(\begin{array}[]{cc}0&-\frac{1}{2}(u+u_{x})\\\
-\frac{1}{2}(u-u_{x})&0\end{array}\right).$

###### Proof.

We only consider $\mu^{0,+}_{1}$, which adimts

$\displaystyle\mu^{0,+}_{1}=e_{1}+\mathcal{T}_{0}\mu^{0,+}_{1},$ (2.42)

where $\mathcal{T}_{0}$ is a integral operator on $L^{\infty}(U(i))$ with
integral kernel

$\displaystyle
K_{0}(x,s,z)=\frac{m(s)}{4}(z+1/z)\left(\begin{array}[]{cc}0&1\\\
-e^{\frac{i}{2}(z-\frac{1}{z})(x-s)}&0\end{array}\right),\ x<s.$

Analogous to the proof of Proposition 1, $I-\mathcal{T}_{0}$ is invertible
such that $\|I-\mathcal{T}_{0}\|\leq C(\|m_{0}\|,x,U(i))$. And it immediately
comes that $\lim_{z\to i\text{ in }U(i)}\mu^{0}=I$. Furthermore,

$\displaystyle\frac{\mu^{0,+}_{1}-e_{1}}{z-i}=\frac{z+i}{4z}\int^{x}_{+\infty}m(s)\left(\begin{array}[]{cc}0&1\\\
-e^{\frac{i}{2}(z-\frac{1}{z})(x-s)}&0\end{array}\right)\mu^{0,+}_{1}ds.$

Again by Lebesgue’s dominated convergence theorem,

$\displaystyle\lim_{z\to i\text{ in }U(i)}\frac{\mu^{0,+}_{1}-e_{1}}{z-i}$
$\displaystyle=\frac{1}{2}\int^{x}_{+\infty}m(s)\lim_{z\to i\text{ in
}U(i)}\left(\begin{array}[]{cc}0&1\\\
-e^{\frac{i}{2}(z-\frac{1}{z})(x-s)}&0\end{array}\right)\mu^{0,+}_{1}ds$
$\displaystyle=\frac{1}{2}\int^{x}_{+\infty}m(s)\left(\begin{array}[]{cc}0\\\
-e^{-(x-s)}\end{array}\right)ds=-\frac{1}{2}(u-u_{x}).$

∎

The relations (2.6) and (2.38) leads to

$\mu_{\pm}(z)=F^{-1}(x)\mu^{0,\pm}(z)e^{\frac{i}{4}(z-\frac{1}{z})c_{\pm}(x)\sigma_{3}},$
(2.43)

where

$\displaystyle c_{\pm}(x)=\int_{\pm\infty}^{x}(q-1)dy.$ (2.44)

###  2.2 Scattering data

Since $\Phi_{\pm}(z;y)$ are two fundamental matrix solutions of the Lax pair
(2.1), there exists a linear relation between $\Phi_{+}(z;y)$ and
$\Phi_{-}(z;y)$, namely

$\Phi_{-}(z;y)=\Phi_{+}(z;y)S(z),$ (2.45)

where $S(z)$ is called scattering matrix

$S(z)=\left(\begin{array}[]{cc}a(z)&-\overline{b(\bar{z})}\\\\[4.0pt]
b(z)&\overline{a(\bar{z})}\end{array}\right),\hskip 14.22636pt\det[S(z)]=1.$

Combing with the transformation (2.6), the equation (2.45) is changed to

$\displaystyle\mu_{-}(z)=\mu_{+}(z)e^{-\frac{i}{4}(z-\frac{1}{z})y\hat{\sigma}_{3}}S(z).$
(2.46)

It is shown that $S(z)$ has following symmetry reductions

$S(z)=\overline{S(\bar{z}^{-1})}=\sigma_{3}S\left(-z^{-1}\right)\sigma_{3}.$
(2.47)

Further, taking $z\rightarrow i$ in combining (2.43) and(2.41), we get the
asymptotic of $a(z)$

$\displaystyle
a(z)=e^{\frac{1}{2}\int_{\mathbb{R}}(q-1)dx}\left(1+\mathcal{O}\left((z-i)^{2}\right)\right),\hskip
8.5359pt\text{as }z\to i.$ (2.48)

From (2.46), $a(z)$ and $b(z)$ can be expressed by $\mu_{\pm}(z)$ at $x=0$ as

$\displaystyle
a(z)=\mu_{-}^{11}(0,z)\overline{\mu_{+}^{11}(0,z)}+\mu_{-}^{21}(0,z)\overline{\mu_{+}^{21}(0,z)},$
(2.49)
$\displaystyle\overline{b(z)}=\mu_{-}^{11}(0,z)\mu_{+}^{21}(0,z)-\mu_{-}^{21}(0,z)\mu_{+}^{11}(0,z).$
(2.50)

So $a(z)$ is analytic on $\mathbb{C}^{+}$ and continuous on $\mathbb{R}$. From
(2.50) and (2.35), we obtain the asymptotic of $a(z)$

$\displaystyle a(z)=1+\mathcal{O}(z^{-1}),\hskip
14.22636ptz\rightarrow\infty.$ (2.51)

####  2.2.1 The discrete spectrum

The function $a(z)$ may have zeros on $\mathbb{C}^{+}\cap\mathbb{R}$. If there
exists a $z_{0}$ on $\mathbb{R}$ such that $a(z_{0})=0$, then $z_{0}$ is
called a resonance or a spectral singularity of the spectral problem (2.4)
which is eliminated in our paper. The brings some deficiency to estimate the
Riemann-Hilbert problem, so the research of resonance need another way to
estimate as in [33, 38]. On the other hand, from the analyticity and
asymptotic as $z\to\infty$ in (2.51), it follows that the number of zeros of
$a(z)$ must be finite.

######  Definition 1.

The set of the zeros of $a(z)$ is called the discrete spectrum, which is
finite and denote by $\\{z_{k}\\}_{k=1}^{N_{0}}$.

On the other hand, a point $z_{0}\in\mathbb{C}^{+}$ satisfies $a(z_{0})=0$ if
and only if $z_{0}$ is an eigenvalue of the spatial part of the Lax pair
(2.4). Furthermore, the symmetry of $a(z)$ in (2.47) implies that

$\displaystyle a(z_{0})=a(-\bar{z}_{0})=a(\bar{z}_{0}^{-1})=a(-z_{0}^{-1})=0.$

So the discrete spectrum appears in pairs.

Noting that $a(z)$ is determined by the initial data $m_{0}$, we give the
following definition.

######  Definition 2.

$\mathsf{U}(0)$ is a subset of $H^{2,1}(\mathbb{R})$ which contains all
$m_{0}\in H^{2,1}(\mathbb{R})$ such that the function $a(z)$ determined by
$m_{0}$ has no zeros on $\mathbb{R}$ and only has simple zeros on
$\mathbb{C}^{+}$. Furthermore, denote

$\mathsf{U}_{N}=\\{m_{0}\in H^{2,1}(\mathbb{R});\ \text{corresponding
}a(z)\text{ only has }\text{N simple zeros on }\mathbb{C}^{+}\\}.$

Thus, $\mathsf{U}(0)=\cup_{N=1}^{\infty}\mathsf{U}_{N}$.

######  Remark 4.

From (2.45), it follows that

$\displaystyle a(z)=1+\int_{\mathbb{R}}\left(P\mu_{-}\right)_{11}dx,$ (2.52)

with $P$ defined in (2.9). Thus $\|a-1\|_{\infty}$ is controlled by the
$L^{\infty}$ norm of jost function, so from Proposition 1, $\|a-1\|_{\infty}$
is controlled by $\|m_{0}\|_{H^{1,1}}$. When $\|m_{0}\|_{H^{1,1}}$ is small,
$a$ must has no zero on $\mathbb{C}^{+}\cap\mathbb{R}$. Specially, when
$m_{0}\equiv 0$, $a(z)\equiv 1$, so there exist a neighborhood of $0\in
H^{2,1}(\mathbb{R})$ contained in $\mathsf{U}(0)$.

####  2.2.2 reflection coefficients

We define the reflection coefficients by

$r(z)=\frac{b(z)}{a(z)},\hskip 14.22636pt\tilde{r}(z)=.$ (2.53)

The rest analysis in this section is given by taking $r$ as an example,
because $r$ and $\tilde{r}$ have same property.

$r(z)$ admits symmetry reductions

$r(z)=\overline{r(z^{-1})}=r(-z^{-1}).$

In addition, (2.53), (2.50) and (2.51) lead to $a(0)=1,\ b(0)=0,$ which gives
$r(0)=0$. The above symmetry of $r$ implies that $r(z(k))$ is a well defined
function of $k$ with $k=z-1/z$. Denote

$\displaystyle\rho(k)=r(z(k)).$ (2.54)

For convenience, when $\rho\in H^{1,1}(\mathbb{R})$, we also call that $r\in
H^{1,1}(\mathbb{R}_{k})$ with

$\displaystyle\|r\|_{H^{1,1}(\mathbb{R}_{k})}=\|\rho\|_{H^{1,1}}.$

######  Lemma 1.

If initial data $m_{0}\in\mathsf{U}\subset H^{2,1}(\mathbb{R})$ admits no
eigenvalues or spectral singularity, then the reflection coefficient $r\in
H^{1,1}(\mathbb{R}_{k})$.

###### Proof.

Noting that for any function $f(z)$ satisfying $f(z)=f(-1/z)$, by denoting
$z_{+}(k)=(k+\sqrt{k^{2}+4})/2$, $z_{-}(k)=(k-\sqrt{k^{2}+4})/2$, it is shown
that

$\displaystyle\int_{\mathbb{R}}|f(z)|^{p}dz=\int_{\mathbb{R}}\frac{|f(k)|^{p}}{1+z_{-}(k)^{-2}}dk+\int_{\mathbb{R}}\frac{|f(k)|^{p}}{1+z_{+}(k)^{-2}}dk=\int_{\mathbb{R}}|f(k)|^{p}dk.$

Together with
$|\partial_{k}f|=|\partial_{z}f||\partial_{k}/\partial_{z}|\lesssim|\partial_{z}f|$,
it is sufficient to prove that $k(\cdot)r$, $k(\cdot)r^{\prime}\in
L^{2}(\mathbb{R})$. For $m_{0}\in\mathsf{U}$, $1/a$ is bounded on
$\mathbb{R}$. And Proposition 1 gives that
$\|(\cdot)b\|_{2}\leq\|\mu_{-}^{11}\|{L^{\infty}}\|(\cdot)\mu_{+}^{21}\|_{C^{0}}+\|(\cdot)\mu_{-}^{21}\|_{C^{0}}\|\mu_{+}^{11}\|{L^{\infty}}$,
which leads to $zr\in L^{2}(\mathbb{R})$. And for the $z$-derivative of $r$,
(2.53) also gives that

$r^{\prime}(z)=\frac{a(z)b^{\prime}(z)-a^{\prime}(z)b(z)}{a(z)^{2}},$

where $a$, $b\in L^{\infty}(\mathbb{R})$ and $(\cdot)a^{\prime},\
(\cdot)b^{\prime}\in L^{2}(\mathbb{R})$ via Proposition 1. And combining
$(\cdot)b,\ (\cdot)b^{\prime}\in L^{2}(\mathbb{R})$, it follows that
$(\cdot)b\in L^{\infty}(\mathbb{R})$. Then from $(\cdot)^{2}b^{\prime}\in
L^{2}(\mathbb{R})$, we arrive at that $(\cdot)^{j}r^{\prime}\in
L^{2}(\mathbb{R})$ for $j=0,1$. Finally from
$r(z)=\overline{r(\bar{z}^{-1})}$, it is adduced that $k(\cdot)^{j}r$,
$k(\cdot)^{j}r^{\prime}$ in is $L^{2}(\mathbb{R})$ for $j=0,1$. ∎

The above proposition also shows that the map $m_{0}\to a-1$, $m_{0}\to
a^{\prime}$ and $m_{0}\to b$ is locally lipshitz continuous from
$H^{2,1}(\mathbb{R})$ to $L^{\infty}(\mathbb{R})$, $L^{2,1}(\mathbb{R})$ and
$H^{1,1}(\mathbb{R})$ respectively.

And we arrive at the corollary below.

######  Corollary 1.

If initial data $m_{0}\in H^{2,1}(\mathbb{R})$ admits no eigenvalues or
spectral singularity and $\|m_{0}\|_{H^{2,1}}\leq\rho_{0}$ for some
$\rho_{0}>0$, then the reflection coefficient $r\in H^{1,1}(\mathbb{R}_{k})$
with

$\displaystyle\|r\|_{H^{1,1}(\mathbb{R}_{k})}\leq
C(\rho_{0})\|m_{0}\|_{H^{2,1}}.$ (2.55)

####  2.2.3 The norming constant

For convenience, denote the phase function

$\theta(z)=-\frac{1}{4}\left(z-\frac{1}{z}\right)y.$ (2.56)

As shown in [30], the following residue conditions hold

$\displaystyle\operatorname*{Res}_{z=z_{j}}\left[\frac{\mu_{-,1}(z)}{a(z)}\right]=c_{j}e^{-2i\theta(z_{j})}\mu_{+,2}(z_{j}),\hskip
8.5359pt\operatorname*{Res}_{z=z_{j}}\left[\frac{\mu_{+,2}(z)}{a(z)}\right]=\tilde{c}_{j}e^{-2i\theta(z_{j})}\mu_{-,1}(z_{j}),$
$\displaystyle\operatorname*{Res}_{z=\bar{z}_{j}}\left[\frac{\mu_{-,2}(z)}{\overline{a(\bar{z})}}\right]=-\bar{c}_{j}e^{2i\theta(\bar{z}_{j})}\mu_{+,1}(\bar{z}_{j}),\hskip
8.5359pt\operatorname*{Res}_{z=\bar{z}_{j}}\left[\frac{\mu_{+,1}(z)}{\overline{a(\bar{z})}}\right]=-\bar{\tilde{c}}_{j}e^{2i\theta(\bar{z}_{j})}\mu_{-,2}(\bar{z}_{j}),$

with some constant $c_{j},\ \tilde{c}_{j}$, which is called as norming
constants. Then together with subsubsection 2.2.1, 2.2.2, we give the
definition of scattering data.

######  Definition 3.

The set $\\{r,\\{z_{j},c_{j}\\}_{j=1}^{N_{0}}\\},\
\\{\tilde{r},\\{z_{j},\tilde{c}_{j}\\}_{j=1}^{N_{0}}\\}$ is called scattering
data. Here, $r,\ \tilde{r}$ is defined in (2.53). $z_{j}$ is defined in
Definition 1 and $c_{j},\ \tilde{c}_{j}$ come from above equations.

Therefore, we come to the conclusion in Lemma LABEL:lemmasec2.

###  2.3 A RH problem associated with initial value problem

By the Jost functions $\mu^{\pm}(z;y)$ and the function $a(z)$, two piecewise
analytical matrices are defined as follows,

$\displaystyle
M_{l}(z;y)=\left\\{\begin{array}[]{ll}\left(\frac{\mu^{-}_{1}(z;y)}{a(z)},\mu^{+}_{2}(z;y)\right),&\text{as
}z\in\mathbb{C}^{+},\\\\[12.0pt]
\left(\mu^{+}_{1}(z;y),\frac{\mu^{-}_{2}(z;y)}{a^{*}(z)}\right),&\text{as
}z\in\mathbb{C}^{-},\\\ \end{array}\right.$ (2.59) $\displaystyle
M_{r}(z;y)=\left\\{\begin{array}[]{ll}\left(\mu^{-}_{1}(z;y),\frac{\mu^{+}_{2}(z;y)}{a(z)}\right),&\text{as
}z\in\mathbb{C}^{+},\\\\[12.0pt]
\left(\frac{\mu^{+}_{1}(z;y)}{a^{*}(z)},\mu^{-}_{2}(z;y)\right),&\text{as
}z\in\mathbb{C}^{-}.\\\ \end{array}\right.$ (2.62)

which solves the following RH problems respectively.

######  RHP 1.

Find a matrix-valued function $M_{l}(z):=M_{l}(z;y)$ which satisfies:

$\blacktriangleright$ Analyticity: $M_{l}(z)$ is meromorphic in
$\mathbb{C}\setminus\mathbb{R}$;

$\blacktriangleright$ Symmetry:

$\displaystyle
M_{l}(z)=\sigma_{3}\overline{M_{l}(-\bar{z})}\sigma_{3}=\sigma_{2}\overline{M_{l}(\bar{z})}\sigma_{2}=\sigma_{3}M_{l}(0)^{-1}M_{l}(-1/z)\sigma_{3};$
(2.63)

$\blacktriangleright$ Jump condition: $M_{l}$ has continuous boundary values
$[M_{l}]_{\pm}(z)$ on $\mathbb{R}$ and

$[M_{l}]_{+}(z)=[M_{l}]_{-}(z)\tilde{V}(z),\hskip 14.22636ptz\in\mathbb{R},$
(2.64)

where

$V_{l}(z)=\left(\begin{array}[]{cc}1+|r(z)|^{2}&e^{2\theta(z)}\overline{r(z)}\\\
e^{-2\theta(z)}r(z)&1\end{array}\right);$ (2.65)

$\blacktriangleright$ Asymptotic behaviors:
$M_{l}(z)=I+\mathcal{O}(z^{-1}),\hskip 14.22636ptz\rightarrow\infty.$

$\blacktriangleright$ Residue conditions: $M_{l}(z)$ has simple poles at each
point in $\\{z_{k}\\}_{k=1}^{N_{0}}$ with:

$\displaystyle\operatorname*{Res}_{z=z_{n}}M_{l}(z)=\lim_{z\to
z_{n}}M_{l}(z)\left(\begin{array}[]{cc}0&0\\\
c_{n}e^{-2it\theta_{n}}&0\end{array}\right),$ (2.68)
$\displaystyle\operatorname*{Res}_{z=\bar{z}_{n}}M_{l}(z)=\lim_{z\to\bar{z}_{n}}M_{l}(z)\left(\begin{array}[]{cc}0&-\bar{c}_{n}e^{2it\theta_{n}}\\\
0&0\end{array}\right).$ (2.71)

######  RHP 2.

Find a matrix-valued function $M_{r}(z)=M_{r}(z;y)$ which satisfies:

$\blacktriangleright$ same Analyticity, Symmetry and Asymptotic behaviors as
in RHP 1;

$\blacktriangleright$ Jump condition: $M_{r}(z)$ has continuous boundary
values $[M_{r}]_{\pm}(z)$ on $\mathbb{R}$ and

$[M_{r}]_{+}(z)=[M_{r}]_{-}(z)V(z),\hskip 14.22636ptz\in\mathbb{R},$ (2.72)

where

$V_{r}(z)=\left(\begin{array}[]{cc}1&e^{2\theta(z)}\overline{\tilde{r}(z)}\\\
e^{-2\theta(z)}\tilde{r}(z)&1+|r(z)|^{2}\end{array}\right).$ (2.73)

$\blacktriangleright$ Residue conditions: $M_{r}(z)$ has simple poles at each
point in $\\{z_{k}\\}_{k=1}^{N_{0}}$ with:

$\displaystyle\operatorname*{Res}_{z=z_{n}}M_{r}(z)=\lim_{z\to
z_{n}}M_{r}(z)\left(\begin{array}[]{cc}0&\tilde{c}_{n}e^{2it\theta_{n}}\\\
0&0\end{array}\right),$ (2.76)
$\displaystyle\operatorname*{Res}_{z=\bar{z}_{n}}M_{r}(z)=\lim_{z\to\bar{z}_{n}}M_{r}(z)\left(\begin{array}[]{cc}0&0\\\
-\bar{\tilde{c}}_{n}e^{-2it\theta_{n}}&0\end{array}\right).$ (2.79)

######  Remark 5.

Here, the expression $M_{r}(z;y)$ and $M_{l}(z;y)$ instead of $M_{r}(z;y)$ and
$M_{l}(z;y)$ is used to emphasize that $M_{r}$, $M_{l}$ are the matrix
functions of $z$, and $y$ are only a parameter.

For convenience, we define

$\displaystyle M(z;y)=\left\\{\begin{array}[]{ll}M_{l}(z;y),&\text{as
}y\in[0,+\infty),\\\\[12.0pt] M_{r}(z;y),&\text{as }y\in(-\infty,0],\\\
\end{array}\right.$ (2.82)

Thus from the asymptotic behavior of the functions $\mu_{\pm}$ and (2.48), we
arrive at the proposition as follows.

######  Lemma 2.

The reconstruction formula of $u(y)$ is given by

$q(y)=\frac{1}{M_{11}(0)},$ (2.83)

where

$x(y)=y+c_{+}(x)=y-\ln\left(\frac{M_{12}(i)+M_{22}(i)}{M_{11}(i)+M_{21}(i)}\right).$
(2.84)

And as $z\to\infty$, it is shown from (2.35) that

$\displaystyle\lim_{z\to\infty}z\left(M_{l}-I\right)=\eta\sigma_{2}+\zeta^{(+)}\sigma_{3},\
\lim_{z\to\infty}z\left(M_{r}-I\right)=\eta\sigma_{2}+\zeta^{(-)}\sigma_{3},$
(2.85)

where

$\displaystyle\eta(y)=\dfrac{-im_{0,x}(y)}{(1+m_{0}(y)^{2})^{3/2}},\
\zeta^{(\pm)}(y)=\frac{i}{2}\int_{y}^{\pm\infty}\left(\frac{m_{0,x}(s)^{2}}{q(s)^{6}}+\frac{m_{0}(s)^{2}}{q(s)^{2}}\right)ds.$
(2.86)

On the other hand, the following proposition reveals that the solutions
reconstructed from two RH problems in above formulas are actually same at
$y=0.$

######  Proposition 3.

For $y\in[0,+\infty)$, denote $q_{l}$, $\eta_{l}$, $\zeta_{l}$ as those
recovered from $M_{l}(z;y)$, and for $y\in(-\infty,0]$ denote $q_{r}$,
$\eta_{r}$, $\zeta_{r}$ as those recovered from $M_{r}(z;y)$. Then
$\eta_{l}(0)=\eta_{r}(0)$, $\zeta_{l}(0)=\zeta_{r}(0)$ and
$q_{l}(0)=q_{r}(0)$.

###### Proof.

Denote

$\displaystyle\tilde{a}(z)=\left\\{\begin{array}[]{ll}a(z),&\text{as
}z\in\mathbb{C}^{+},\\\\[12.0pt] a^{*}(z)^{-1},&\text{as
}z\in\mathbb{C}^{-},\\\ \end{array}\right.\hskip 8.5359pt\tilde{a}\to I,\
|z|\to\infty.$ (2.89)

Then $M_{l}(z;y)=M_{r}(z;y)\tilde{a}(z)^{-\sigma_{3}}$. Thus the result is
deduced from (2.48),(2.51), $a(0)=0$ and Lemma 2. ∎

##  3 The RH problem

In this section, we shall provide a rigorous framework to study the the RH
problem of complex analysis.

###  3.1 The Cauchy projection operator

For a given function $f(z)\in L^{p}(\mathbb{R})$, $1\leqslant p<\infty$, the
Cauchy operator $\mathcal{C}$ is defined by

$\displaystyle\mathcal{C}f(z)=\frac{1}{2\pi
i}\int_{\mathbb{R}}\frac{f(s)}{s-z}ds,\ z\in\mathbb{C}\setminus\mathbb{R}.$
(3.1)

The function $\mathcal{C}f$ is analytic off the real line such that
$\mathcal{C}f(\cdot+i\gamma)$ is in $L^{p}(\mathbb{R})$ for each
$0\neq\gamma\in\mathbb{R}$. When $z$ approaches to a point on the real line
transversely from the upper and lower half planes, that is, if $\gamma\to\pm
0$ in the following expression, the Cauchy operator $\mathcal{C}$ becomes the
Cauchy projection operator $\mathcal{C}_{\pm}$:

$\displaystyle\mathcal{C}_{\pm}f(z)=\lim_{\gamma\to 0}\frac{1}{2\pi
i}\int_{\mathbb{R}}\frac{f(s)}{s-(z\pm i\gamma)}ds,\ z\in\mathbb{R},$ (3.2)
$\displaystyle(\mathcal{C}_{\pm}f)^{\wedge}(z)=\pm X_{\pm}(z)\hat{f}(z),$
(3.3)

where $X_{\pm}$ denotes the characteristic function on $\mathbb{R}^{\pm}$.

For any $f\in L^{p}(\mathbb{R})$, $1\leqslant p<\infty$, the Cauchy integral
$\mathcal{C}f$ is analytical off the real line, decays to zero as
$|z|\to\infty$, and approaches to $\mathcal{C}_{\pm}f$ almost everywhere when
a point $z\in\mathbb{C}^{\pm}$ approaches to a point on the real axis by any
non-tangential contour from $\mathbb{C}^{\pm}$[40]. If $1<p<\infty$, then
there exists a positive constant $C_{p}$ such that

$\displaystyle\parallel\mathcal{C}_{\pm}f\parallel_{L^{p}}\leqslant
C_{p}\parallel f\parallel_{L^{p}}.$ (3.4)

Moreover, $C_{2}=1$. When $f\in L^{1}(\mathbb{R})$, as $z\to\infty$,
$\mathcal{C}f(z)=\mathcal{O}(z^{-1})$.

###  3.2 Solvability of the RH problem

In this subsection, we prove solvability of the RH problem 2 about $M_{l}$ for
$y\in[0,+\infty)$ under the given scattering data
$\\{r,\\{z_{k},c_{k}\\}_{k=1}^{N_{0}}\\}\in
H^{1,1}(\mathbb{R}_{k})\otimes\mathbb{C}^{N_{0}}\otimes\mathbb{C}^{N_{0}}$.
For $z\in\mathbb{R}$, denote $w=w_{+}+w_{-}$ with with

$\displaystyle w_{+}(z;y)=\left(\begin{array}[]{cc}0&0\\\
r(z)e^{-2i\theta(z;y)}&0\end{array}\right),\
w_{-}(z;y)=\left(\begin{array}[]{cc}0&\bar{r}(z)e^{2i\theta(z;y)}\\\
0&0\end{array}\right).$ (3.9)

In view of the Beals-Coifman theorem in [8], the RH problem 2 has a solution
given by the Cauchy operator

$\displaystyle M(z;y)=$ $\displaystyle I+\mathcal{C}(\mu w)(z;y)$
$\displaystyle+\sum_{j=1}^{N_{0}}\left(\frac{c_{n}e^{-2i\theta(z_{j})}}{z-z_{n}}\mu_{2}(z_{n};y),\frac{-\bar{c}_{n}e^{2i\theta(\bar{z}_{j})}}{z-\bar{z}_{n}}\mu_{1}(\bar{z}_{n};y)\right),\
z\in\mathbb{C}$ (3.10)

if and only if there is a solution $\mu(z;y)$ of the Fredholm integral
equations

$\displaystyle\mu(z;y)=$ $\displaystyle
I+\mathcal{C}_{w}(I)(z;y)+\mathcal{C}_{w}(\mu-I)(z;y)$
$\displaystyle+\sum_{j=1}^{N_{0}}\left(\frac{c_{j}e^{-2i\theta(z_{j})}}{z-z_{n}}\mu_{2}(z_{j};y),\frac{-\bar{c}_{j}e^{2i\theta(\bar{z}_{j})}}{z-\bar{z}_{n}}\mu_{1}(\bar{z}_{j};y)\right),\
z\in\mathbb{R},$ (3.11) $\displaystyle\mu_{2}(z_{n};y)=$ $\displaystyle
e_{2}+\mathcal{C}(\bar{r}e^{2i\theta}(\mu_{1}-e_{1}))+\sum_{j=1}^{N_{0}}\frac{-\bar{c}_{j}e^{2i\theta(\bar{z}_{j})}}{z_{n}-\bar{z}_{j}}\mu_{1}(\bar{z}_{j};y),$
(3.12) $\displaystyle\mu_{1}(\bar{z}_{n};y)=$ $\displaystyle
e_{1}+\mathcal{C}(re^{-2i\theta}(\mu_{2}-e_{2}))+\sum_{j=1}^{N_{0}}\frac{c_{j}e^{-2i\theta(z_{j})}}{z_{n}-z_{j}}\mu_{2}(z_{j};y),$
(3.13)

with

$\mathcal{C}_{w}f(z;y)=\mathcal{C}_{+}(fw_{-})(z;y)+\mathcal{C}_{-}(fw_{+})(z;y).$

And a simple calculation gives that $\|\mathcal{C}_{w}I\|_{2}\leq
C(\|r\|_{2}).$ The above equations also imply that

$\displaystyle\mu_{2}(z_{n};y)=e_{2}+\mathcal{C}(\mu_{2}-e_{2})(z_{n};y),\
\mu_{1}(\bar{z}_{n};y)=e_{1}+\mathcal{C}(\mu_{1}-e_{1})(\bar{z}_{n};y).$
(3.14)

This inspires us to only analyze the property of $\mu$ on $z\in\mathbb{R}$. We
rewrite (3.11) as

$\displaystyle\mu(z;y)-I=\mathcal{K}(I)(z;y)+\mathcal{K}(\mu-I)(z;y),\
z\in\mathbb{R},$ (3.15)

where $\mathcal{K}$ is a operator on $L^{2}(\mathbb{R})$ such that for any
matrix function $f=(f_{1},f_{2})\in L^{2}(\mathbb{R})$,

$\displaystyle\mathcal{K}(f)(z;y)=((\mathcal{A}_{1}+\mathcal{A}_{2})f_{2},(\mathcal{B}_{1}+\mathcal{B}_{2})f_{1}),$
(3.16)
$\displaystyle\mathcal{A}_{1}(f_{2})(z;y)=\mathcal{C}_{-}(re^{-2i\theta}f_{2}),\hskip
8.5359pt\mathcal{A}_{2}(f_{2})(z;y)=\sum_{j=1}^{N_{0}}\frac{c_{j}e^{-2i\theta(z_{j})}}{z-z_{j}}\mathcal{C}(f_{2})(z_{j};y),$
(3.17)
$\displaystyle\mathcal{B}_{1}(f_{1})(z;y)=\mathcal{C}_{+}(\bar{r}e^{2i\theta}f_{1}),\hskip
8.5359pt\mathcal{B}_{2}(f_{1})(z;y)=\sum_{j=1}^{N_{0}}\frac{-\bar{c}_{j}e^{2i\theta(\bar{z}_{j})}}{z-\bar{z}_{j}}\mathcal{C}(f_{1})(\bar{z}_{j};y).$
(3.18)

$\mathcal{K}$ can also be a operator on the row function in
$L^{2}(\mathbb{R})$. The following proposition is given to show the
invertibility of $I-\mathcal{K}$.

######  Proposition 4.

If $r\in H^{s}(\mathbb{R})$, $s>1/2$ then $I-\mathcal{K}$ is a bounded
Fredholm operator: $L^{2}(\mathbb{R})\to L^{2}(\mathbb{R})$ for
$y\in[0,+\infty)$ with index zero. Moreover, $(I-\mathcal{K})^{-1}$ exists and
is also a bounded linear operator: $L^{2}(\mathbb{R})\to L^{2}(\mathbb{R})$.
And there exists a constant $C$ such that

$\displaystyle\parallel\left(I-\mathcal{K}\right)^{-1}\parallel\leqslant C.$
(3.19)

###### Proof.

We taking the first row of (3.11) for example to give the proof and denote it
as $(\mu_{11}-1,\mu_{12})$. Recall the definition of $\theta$ in (2.56), then

$\displaystyle|e^{-2i\theta(z_{j})}|=e^{-\frac{y}{2}(1+1/|z_{j}|^{2})\text{Im}z_{j}}=|e^{2i\theta(\bar{z}_{j})}|.$
(3.20)

For any $f=(f_{1},f_{2})\in L^{2}(\mathbb{R})$,

$\displaystyle\mathcal{K}^{2}f=\left((\mathcal{A}_{1}+\mathcal{A}_{2})(\mathcal{B}_{1}+\mathcal{B}_{2})f_{1},(\mathcal{B}_{1}+\mathcal{B}_{2})(\mathcal{A}_{1}+\mathcal{A}_{2})f_{2}\right).$

Here, $\mathcal{A}_{2}$ and $\mathcal{B}_{2}$ are finite rank operator.
Therefore, $\mathcal{A}_{1}\mathcal{B}_{2}$, $\mathcal{A}_{2}\mathcal{B}_{1}$,
$\mathcal{B}_{1}\mathcal{A}_{2}$ and $\mathcal{B}_{2}\mathcal{A}_{1}$ compact
operators. As for $\mathcal{B}_{1}\mathcal{A}_{1}$ and
$\mathcal{A}_{1}\mathcal{B}_{1}$, taking $\mathcal{A}_{1}\mathcal{B}_{1}$ as
an example, it is shown that

$\displaystyle(\mathcal{A}_{1}\mathcal{B}_{1}(f_{1}^{\vee}))^{\wedge}(z)={\mathcal{C}_{-}(\mathcal{C}_{+}(f^{\vee}w_{-})w_{+})}^{\wedge}(z)=\int_{\mathbb{R}}K(z,\lambda;y)f_{1}(\lambda)d\lambda,$

where $K(z,\lambda;y)$ is a scalar function with

$\displaystyle
K(z,\lambda;y)=-X_{-}(z)\int_{\mathbb{R}}X_{+}(s){(\bar{r}e^{-2\theta(\cdot;y)})}^{\wedge}(z-s)(re^{2\theta(\cdot;y)})^{\wedge}(s-\lambda)ds.$

Since

$\|K(\cdot,\cdot;y)\|_{L^{2}(\mathbb{R}\otimes\mathbb{R})}\leqslant\|(\bar{r}e^{-2\theta(\cdot;y)})^{\wedge}\|_{L^{2,s}}\|(re^{2\theta(\cdot;y)})^{\wedge}\|_{L^{2,s}}\leqslant
C(y)\|r\|_{H^{s}}^{2}$

for a positive constant $C(y)$ which is finite for every $y\in[0,+\infty)$, it
follows that
$\mathcal{A}_{1}\mathcal{B}_{1}=\mathcal{C}_{-}(\mathcal{C}_{+}(\cdot
w_{-})w_{+})$ is the Hilbert-Schmidt operator. It in turn implies that
$\mathcal{K}^{2}$ is compact. Therefore, it is inferred in [39] that
$I-\mathcal{K}^{2}$ is a Fredholm operator with index zero, and so is
$I-\mathcal{K}^{2}$. Therefore, to prove $I-\mathcal{K}$ is invertible, it is
sufficient to prove that $I-\mathcal{K}$ is injection. If there exists a
matrix function $f\in L^{2}(\mathbb{R})$ such that $(I-\mathcal{K})f=0$, we
denote the first row of $f$ as $(f_{11},f_{12})$. Define

$\displaystyle g_{11}=\mathcal{C}(re^{-2\theta}f_{12}),\
g_{12}=\mathcal{C}(\bar{r}e^{2\theta}f_{11}).$ (3.21)

Thus, it appears that for $z\in\mathbb{R}$,

$\displaystyle[g_{11}]_{-}(z)=f_{11}(z),\hskip
68.28644pt[g_{11}]_{+}(z)=f_{11}(z)+re^{-2\theta}f_{12}(z),$
$\displaystyle[g_{12}]_{-}(z)=f_{12}(z)+\bar{r}e^{2\theta}f_{11}(z),\ \
[g_{12}]_{+}(z)=f_{12}(z).$

And the function $g_{11}g_{11}^{*}+g_{12}g_{12}^{*}$ is a analytic function on
$\mathbb{C}^{+}$. Using the result in the Cauchy-Goursat theorem gives that

$\displaystyle
0=\int_{\mathbb{R}_{+}}g_{11}(z)g_{11}^{*}(z)+g_{12}(z)g_{12}^{*}(z)dz,$

where $\mathbb{R}_{+}$ is used to denote $\mathbb{R}$ when it is viewed as the
boundary of $\mathbb{C}^{+}$. On the other hand

$\displaystyle
0=\int_{\mathbb{R}^{+}}g_{11}(z)g_{11}^{*}(z)+g_{12}(z)g_{12}^{*}(z)ddz$
$\displaystyle=\int_{\mathbb{R}}|f_{11}(z)|^{2}+|f_{12}(z)|^{2}dz,$

which implies that $\|f_{11}\|_{2}=\|f_{12}\|_{2}=0$. And the second row of
$f$ can obtain the same result, which finally implies that $f=0$.

The last step is to prove that $\|(I-\mathcal{K})^{-1}\|$ is uniform bound for
$y\in[0,+\infty)$. Similarly, we only give the estimate of the row function.
For any matrix function $h=(h_{11},h_{12})\in L^{2}(\mathbb{R})$, denote

$\displaystyle
f^{(+)}:=(f^{(+)}_{1},f^{(+)}_{2})=(I-\mathcal{K})^{-1}(\mathcal{C}_{-}h_{11},\mathcal{C}_{+}h_{12}),$
$\displaystyle
f^{(-)}:=(f^{(-)}_{1},f^{(-)}_{2})=(I-\mathcal{K})^{-1}(\mathcal{C}_{+}h_{11},\mathcal{C}_{-}h_{12}).$

Then it holds that $f^{(+)}-f^{(-)}=(I-\mathcal{K})^{-1}(-h_{11},h_{12})$. And
also denote the analytic functions on $\mathbb{C}\setminus\mathbb{R}$ with

$\displaystyle
g_{11}^{(\pm)}=\mathcal{C}(re^{-2\theta}f^{(\pm)}_{2})+\mathcal{A}_{2}f^{(\pm)}_{2},\hskip
14.22636ptg_{12}^{(\pm)}=\mathcal{C}(\bar{r}e^{2\theta}f_{1}^{(\pm)})+\mathcal{B}_{2}f^{(\pm)}_{1}.$
(3.22)

For $f^{(+)}$, the difference is that in this case, for $z\in\mathbb{R}$,

$\displaystyle[g^{({+})}_{11}]_{-}(z)=f^{(+)}_{1}(z)-\mathcal{C}_{-}h_{11}(z),$
$\displaystyle[g^{(+)}_{11}]_{+}(z)=f^{(+)}_{1}(z)-\mathcal{C}_{-}h_{11}(z)+re^{-2\theta}f^{(+)}_{2}(z),$
$\displaystyle[g^{(+)}_{12}]_{-}(z)=f^{(+)}_{2}(z)-\mathcal{C}_{+}h_{12}(z)-\bar{r}e^{2\theta}f^{(+)}_{1}(z),$
$\displaystyle[g^{(+)}_{12}]_{+}(z)=f^{(+)}_{2}(z)-\mathcal{C}_{+}h_{12}(z).$

Performing the same manipulations as above, we obtain that

$\displaystyle 0$
$\displaystyle=\int_{\mathbb{R}_{+}}g^{(+)}_{11}(z)(g^{(+)}_{11}+\mathcal{C}h_{11})^{*}(z)+(g^{(+)}_{12}+\mathcal{C}h_{12})(z)(g^{(+)}_{12})^{*}(z)ddz$
$\displaystyle=\int_{\mathbb{R}}|f^{(+)}_{1}(z)|^{2}+|f^{(+)}_{2}(z)|^{2}-(f^{(+)}_{2}(z)\overline{\mathcal{C}_{+}h_{2}(z)}+\mathcal{C}_{-}h_{1}(z)\overline{f^{(+)}_{1}(z)})dz,$

which leads to that

$\displaystyle\|f^{(+)}\|_{2}^{2}$
$\displaystyle=\int_{\mathbb{R}}|f^{(+)}_{1}(z)|^{2}+|f^{(+)}_{2}(z)|^{2}dz=\big{|}\int_{\mathbb{R}}f^{(+)}_{2}(z)\overline{\mathcal{C}_{+}h_{2}(z)}+\mathcal{C}_{-}h_{1}(z)\overline{f^{(+)}_{1}(z)}dz\big{|}\leq\|f^{(+)}\|_{2}\|h\|_{2}.$

Therefore, it is concluded that $\|f^{(+)}\|_{2}\leq\|h\|_{2}$. And for
$f^{(-)}$, note that for $z\in\mathbb{R}$,

$\displaystyle[g^{({-})}_{11}]_{-}(z)=f^{(-)}_{1}(z)-\mathcal{C}_{+}h_{11}(z),$

and for $z\in\mathbb{C}\setminus\mathbb{R}$,

$\displaystyle\mathcal{C}([g^{({-})}_{11}]_{-})(z)=\left\\{\begin{array}[]{ll}0,&z\in\mathbb{C}^{+},\\\\[12.0pt]
g^{({-})}_{11},&z\in\mathbb{C}^{-},\\\ \end{array}\right.$

which implies that $\mathcal{C}_{-}(f^{(-)}_{1})(z)\equiv 0$ and
$f^{(-)}_{1}(z)=\mathcal{C}_{+}h_{11}(z)$. Thus, $g^{({-})}_{11}=0$.
Similarly, it also follows that $f^{(-)}_{2}(z)=\mathcal{C}_{-}h_{12}(z)$.
Then we arrive at that $\|f^{(-)}\|_{2}\leq\|h\|_{2}$. Therefore, it is
accomplished that

$\displaystyle\|f^{(+)}-f^{(-)}\|_{2}=\|(I-\mathcal{K})^{-1}(-h_{11},h_{12})\|_{2}\leq
2\|h\|_{2},$

which means $\|(I-\mathcal{K})^{-1}\|\leq 2$. ∎

As a consequence of this proposition, the solution of (3.11) exists with

$\displaystyle\mu(z;y)=I+(I-\mathcal{K})^{-1}(\mathcal{K}(I))(z;y),$ (3.23)

and

$\displaystyle\|\mu(z;y)-I\|_{2}\leq 2\|\mathcal{K}(I)\|_{2}\leq 2\|r\|_{2}.$
(3.24)

Therefore, from (3.10), it follows that

$\displaystyle\|M(z;y)-I\|_{2}=\|\mathcal{C}(w)(z;y)+\mathcal{C}((\mu-I)w)(z;y)\|\leq(2\|r\|_{\infty}+1)\|r\|_{2},$
(3.25)

When $y\in\mathbb{R}^{-}$, following the same process as $M_{l}$, $M_{r}$
admits similar property in proposition below.

######  Proposition 5.

If $\tilde{r}\in H^{s}(\mathbb{R})$, $s>1/2$, $y\in\mathbb{R}^{-}$, then
$M_{r}$ exists unique with estimate

$\displaystyle\parallel
M_{r}-I\parallel_{2}\leq\|\tilde{r}\|_{2}(2\|\tilde{r}\|_{\infty}+1).$ (3.26)

###  3.3 Estimates on the solution of RHPs

For further estimates, we require $r\in H^{1,1}(\mathbb{R})$. Consider the
$y$-derivative at both side of the equation (3.11)

$\displaystyle\partial_{y}\mu(z;y)=(\partial_{y}\mathcal{K})(I)(z;y)+(\partial_{y}\mathcal{K})(\mu-I)(z;y)+\mathcal{K}(\partial_{y}\mu)(z;y),$
(3.27)

where for any $f\in L^{2}$

$\displaystyle(\partial_{y}\mathcal{K})f(z;y)=$
$\displaystyle\mathcal{C}_{+}(f\partial_{y}w_{-})(z;y)+\mathcal{C}_{-}(f\partial_{y}w_{+})(z;y)$
$\displaystyle=$
$\displaystyle\frac{1}{2}\mathcal{C}_{+}(ik(\cdot)fw_{-})(z;y)+\frac{1}{2}\mathcal{C}_{-}(ik(\cdot)f\partial_{y}w_{+})(z;y).$

And it holds that

$\displaystyle\|(\partial_{y}\mathcal{K})(I)\|_{2}\leq\|k(\cdot)r\|_{{2}},\hskip
14.22636pt\|(\partial_{y}\mathcal{K})(\mu-I)\|\leq\|\mu-I\|_{2}\|\|k(\cdot)r\|_{{\infty}}.$

It is thereby inferred that $\partial_{y}\mu$ exists with
$\partial_{y}\mu=(I-\mathcal{K})^{-1}((\partial_{y}\mathcal{K})(I)+(\partial_{y}\mathcal{K})(\mu-I))$,
and

$\displaystyle\|\partial_{y}\mu\|_{2}\leq$
$\displaystyle\|\mathcal{C}_{+}(k(\cdot)\bar{r}e^{2\theta})\|_{2}+\|\mathcal{C}_{-}(k(\cdot)re^{-2\theta})\|_{2}$
$\displaystyle+4(\|\mathcal{C}_{+}(\bar{r}e^{2\theta})\|_{2}+\|\mathcal{C}_{-}(re^{-2\theta})\|_{2})\|k(\cdot)r\|_{{\infty}}$
(3.28)

Therefore, we arrive at the $y$-derivative of $M(z;y)$ with

$\displaystyle\partial_{y}M(z;y)=\mathcal{C}((\partial_{y}\mu)w)(z;y)+\frac{i}{2}\mathcal{C}(k(\cdot)\mu
w)(z;y).$ (3.29)

Next, we give a estimate of $M$ and its $y$-derivative at $z=i$ and $z=0$,
which will be used in the reconstruction formula. The following proposition
gives that $M(i)$ and $M(0)$ is uniformly bounded for $y\in\mathbb{R}^{+}$.

######  Proposition 6.

If $r\in H^{1,1}(\mathbb{R}_{k})$ satisfies
$\max_{j=0,1}\|(\cdot)^{j}r\|_{L^{2}(\mathbb{R}_{k})\cap
L^{\infty}(\mathbb{R}_{k})}\leq\rho_{0}$ for some $\rho_{0}>0$, then there
exists a positive constant $C(\rho_{0})$ such that

$\displaystyle|M(i)-I|,\ |M(0)-I|\leq
C(\rho_{0})\|r\|_{L^{2}(\mathbb{R}_{k})\cap L^{\infty}(\mathbb{R}_{k})},$
$\displaystyle|\partial_{y}M(i)|,\ |\partial_{y}M^{i,(1)}(y)|,\leq
C(\rho_{0})\|(\cdot)r\|_{L^{2}(\mathbb{R}_{k})\cap
L^{\infty}(\mathbb{R}_{k})}.$

###### Proof.

The result comes from Hölder inequality immediately. For example,

$\displaystyle|M(i;y)-I|$
$\displaystyle=\Big{|}\frac{1}{2\pi}\int_{\mathbb{R}}\frac{(\mu(s;y)-I)w(s;y)}{s-i}ds+\frac{1}{2\pi}\int_{\mathbb{R}}\frac{w(s;y)}{s-i}ds\Big{|}$
$\displaystyle\leq(\|r\|_{\infty}\|\mu-I\|_{2}+\|r\|_{2})\|1/(\cdot-i)\|_{2}$
$\displaystyle\leq C\|r\|_{2}(2\|r\|_{\infty}+1)\leq
C(\rho_{0})\|r\|_{L^{2}(\mathbb{R}_{k})\cap L^{\infty}(\mathbb{R}_{k})}.$

The others can be estimate in the same way. ∎

######  Corollary 2.

There exists a constant $\epsilon_{0}>0$ such that when
$\max_{j=0,1}\|(\cdot)^{j}r\|_{L^{2}(\mathbb{R}_{k})\cap
L^{\infty}(\mathbb{R}_{k})}<\epsilon_{0}$, it follows that

$\Big{|}\left(\frac{M_{11}(i;y)+M_{21}(i;y)}{M_{12}(i;y)+M_{22}(i;y)}\right)\partial_{y}\left(\frac{M_{12}(i;y)+M_{22}(i;y)}{M_{11}(i;y)+M_{21}(i;y)}\right)\Big{|}<1.$

##  4 Time evolution

According to the time spectral problem in the Lax pair (2.1) and scattering
relation (2.45), it is found that the time evolution scattering data $a(z;t)$
and $b(z;t)$ satisfy the equations

$\displaystyle\partial_{t}a(z;t)=0,\
\partial_{t}b(z;t)=-\dfrac{2iz(z^{2}-1)}{z^{2}+1}b(z;t),\ z\in\mathbb{R},$
(4.1)

which yields

$\displaystyle a(z;t)=a(z;0),\
b(z;t)=e^{-\frac{2iz(z^{2}-1)t}{z^{2}+1}}b(z;0),\ z\in\mathbb{R}.$ (4.2)

Therefore, we define the time-dependent scattering data as in Lemma
LABEL:pror, which proof is similar as it in [31].

Furthermore, we denote $M_{l}(z;t,y)$ as the solution for the RH problem 1
under time-dependent reflection coefficient $r(\cdot;t)$. $M_{r}(z;0,y)$ also
has a similar time evolution denoted by $M_{r}(z;t,y)$ , which is omitted
here. From Propositions 4 and 5, the following proposition holds immediately.

######  Proposition 7.

If $r(\cdot;0)$ and $\tilde{r}(\cdot;0)$ belong in $H^{1,1}(\mathbb{R}_{k})$,
$M_{l}(z;t,y)$ and $M_{r}(z;t,y)$ exists unique for every
$t\in\mathbb{R}^{+}$.

For convenience, we define

$\displaystyle M(z;t,y)=\left\\{\begin{array}[]{ll}M_{l}(z;t,y),&\text{as
}y\in[0,+\infty),\\\\[12.0pt] M_{r}(z;t,y),&\text{as }y\in(-\infty,0],\\\
\end{array}\right.$ (4.5)

The crux of the matter is whether the function $u(t,y)$ reconstituted from the
time-depended RH problem in Lemma 2 is the solution of (1.1). By standard
processing it can be verified that the reconstruction formula remain
accessible for $M_{l}(z;t,y)$ with $y\in\mathbb{R}^{+}$ and for $M_{r}(z;t,y)$
with $y\in\mathbb{R}^{-}$. The first step is to prove the existence of
$\partial_{t}M(z;t,y)$. We present the details when $y\in\mathbb{R}^{+}$. Take
$t$-derivative at the both side of (3.11) and obtain

$\displaystyle\partial_{t}\mu(z;y)=\partial_{t}\mathcal{C}(w)(z;y)+(\partial_{t}\mathcal{K})(\mu-I)(z;y)+\mathcal{K}(\partial_{t}\mu)(z;y),$
(4.6)

where

$\displaystyle\|\partial_{t}\mathcal{C}(w)\|_{2}=\Big{\|}\mathcal{C}\left(\frac{z(z^{2}-1)}{(z^{2}+1)^{2}}w\right)\Big{\|}_{2}\leq
C\|r\|_{2},$
$\displaystyle\|(\partial_{t}\mathcal{K})(\mu-I)\|_{2}=\Big{\|}\mathcal{K}\left(\frac{z(z^{2}-1)}{(z^{2}+1)^{2}}(\mu-I)\right)\Big{\|}_{2}\leq
C\|r\|_{\infty}\|\mu-I\|_{2}.$

So $\partial_{t}\mu(z;y)$ exist unique in $L^{2}(\mathbb{R})$ with

$\displaystyle\partial_{t}\mu(z;y)=(I-\mathcal{K})^{-1}\left(\partial_{t}\mathcal{C}(w)+(\partial_{t}\mathcal{K})(\mu-I)\right)(z;y),$

which gives the existence of $\partial_{t}M(z;t,y)$ with

$\displaystyle\partial_{t}M(z;t,y)=\mathcal{C}(\partial_{t}\mu
w)(z;y)+\mathcal{C}((\mu-I)\partial_{t}w)(z;y)+\mathcal{C}(\partial_{t}w)(z;y).$
(4.7)

Moreover, The following proposition is obtained similar to the proof of
Proposition 6,

######  Proposition 8.

If $r\in H^{1,1}(\mathbb{R}_{k})$ satisfies
$\|r\|_{H^{1,1}(\mathbb{R}_{k})}\leq\rho_{0}$ for some $\rho_{0}>0$,
$y\in\mathbb{R}^{+}$, then there exists a positive constant $C(\rho_{0})$ such
that

$\displaystyle|\partial_{t}M(i)|,\ $
$\displaystyle|\partial_{t}\partial_{y}M(i)|,\ |\partial_{t}M^{i,(1)}(y)|,\
|\partial_{t}\partial_{y}M^{i,(1)}(y)|,$
$\displaystyle|\partial_{t}\partial_{y}M^{i,(1)}(y)|\leq
C(\rho_{0})\|r\|_{H^{1,1}(\mathbb{R}_{k})}.$

The symmetry of $M$ in (2.63) and the fact that det$M$=1 inspire us to denote

$\displaystyle M(0)=\left(\begin{array}[]{cc}\beta_{0}&\eta_{0}\\\
\eta_{0}&\beta_{0}\end{array}\right),\hskip
8.5359ptM(i)=\left(\begin{array}[]{cc}f_{0}&\frac{\eta_{0}}{2f_{0}}\\\
\frac{\beta_{0}-1}{\eta_{0}}f_{0}&\frac{\beta_{0}+1}{f_{0}}\end{array}\right),\hskip
8.5359ptM^{i,(1)}=\left(\begin{array}[]{cc}\frac{\beta_{0}-1}{\eta_{0}}g_{1}&g_{2}\\\
g_{1}&\frac{\beta_{0}-1}{\eta_{0}}g_{2}\end{array}\right).$

Here, when $\eta_{0}=0$, $(\beta_{0}-1)/\eta_{0}=0$. And we recall the symbol
that

$\displaystyle\eta=\lim_{z\to\infty}z(M(z)-I)_{21}.$

In addition, define

$\displaystyle\Psi(z;y,t)=M(z;y,t)e^{-\frac{i}{4}(z-1/z)y\sigma_{3}+2i\frac{z(z^{2}-1)}{(z^{2}+1)^{2}}t\sigma_{3}},$
(4.8)

which admits the proposition below.

######  Proposition 9.

$\Psi$ defined above satisfies the differential equation

$\displaystyle\Psi_{y}=A\Psi,\hskip 14.22636pt\Psi_{t}=B\Psi,$ (4.9)

where

$\displaystyle A=$
$\displaystyle-\frac{iz}{4}\sigma_{3}-2\eta\sigma_{1}+\frac{i}{4z}(\beta_{0}^{2}+\eta_{0}^{2})\sigma_{3}+\frac{1}{2z}\eta_{0}\beta_{0}\sigma_{2},$
$\displaystyle B=$
$\displaystyle\frac{1}{(z+i)^{2}}\sigma_{3}M(0)^{-1}-\frac{1}{(z-i)^{2}}\sigma_{3}M(0)^{-1}$
$\displaystyle+\frac{1}{z-i}\left(\begin{array}[]{cc}2\frac{\beta_{0}-1}{\eta_{0}}f_{0}g_{2}+\frac{g_{1}\eta_{0}}{f_{0}}&-f_{0}g_{2}-(\beta_{0}-1)\frac{g_{1}}{f_{0}}\\\
f_{0}g_{2}+(\beta_{0}+1)\frac{g_{1}}{f_{0}}&-(2\frac{\beta_{0}-1}{\eta_{0}}f_{0}g_{2}+\frac{g_{1}\eta_{0}}{f_{0}})\end{array}\right)+\frac{i}{z-i}\sigma_{3}M(0)^{-1}$
$\displaystyle+\frac{1}{z+i}\left(\begin{array}[]{cc}2\frac{\beta_{0}-1}{\eta_{0}}f_{0}g_{2}+\frac{g_{1}\eta_{0}}{f_{0}}&-f_{0}g_{2}-(\beta_{0}+1)\frac{g_{1}}{f_{0}}\\\
f_{0}g_{2}+(\beta_{0}-1)\frac{g_{1}}{f_{0}}&-(2\frac{\beta_{0}-1}{\eta_{0}}f_{0}g_{2}+\frac{g_{1}\eta_{0}}{f_{0}})\end{array}\right)+\frac{i}{z+i}\sigma_{3}M(0)^{-1}.$

Moreover, via denoting

$\displaystyle\tilde{u}=-\frac{g_{1}}{f_{0}}-2\frac{\beta_{0}-1}{\eta_{0}^{2}}f_{0}g_{2},\hskip
14.22636pt\tilde{u}_{x}=\frac{g_{1}}{f_{0}}-2\frac{\beta_{0}-1}{\eta_{0}^{2}}f_{0}g_{2},$
$\displaystyle\tilde{q}=\frac{1}{\beta_{0}},\hskip
14.22636pt\tilde{m}=\frac{\eta_{0}}{\beta_{0}i},$

the compatibility condition

$\displaystyle A_{t}+AB-B_{y}-BA=0$

yields the mCH equation in the $(y,t)$ variables.

###### Proof.

The definition of $\Psi$ implies that the jump of $\Psi$ is independent of $y$
and $t$. Consequently, $\Psi_{y}\Psi^{-1}$ and $\Psi_{t}\Psi^{-1}$ have no
jump. We analyze $\Psi_{y}\Psi^{-1}$ first. (4.8) gives that

$\displaystyle\Psi_{y}\Psi^{-1}=M_{y}M^{-1}-\frac{i}{4}\left(z-\frac{1}{z}\right)M\sigma_{3}M^{-1},$

which is a meromorphic function, with possible singularities at $z=0$ and
$z=\infty$. As $z\to\infty$,

$\displaystyle\Psi_{y}\Psi^{-1}=-\frac{i}{4}z\sigma_{3}-2\eta\sigma_{1}+\mathcal{O}(1/z),$

while as $z\to 0$,

$\displaystyle\Psi_{y}\Psi^{-1}=\frac{i}{4z}M(0)\sigma_{3}M(0)^{-1}+\mathcal{O}(1).$

Therefore, the function

$\displaystyle\Psi_{y}\Psi^{-1}-\frac{i}{4z}M(0)\sigma_{3}M(0)^{-1}+\frac{i}{4}z\sigma_{3}+2\eta\sigma_{1}$

is holomorphic in $\mathbb{C}$ and vanish at $z\to\infty$. Then, by
Liouville’s theorem, it vanishes identically, which leads to the result
$\Psi_{y}=A\Psi$. Similarly,

$\displaystyle\Psi_{t}\Psi^{-1}=M_{t}M^{-1}+2i\frac{z(z^{2}-1)}{(z^{2}+1)^{2}}M\sigma_{3}M^{-1},$

is a meromorphic function, with possible singularities at $z=\pm i$. From the
decomposition

$\displaystyle
2i\frac{z(z^{2}-1)}{(z^{2}+1)^{2}}=\frac{i}{z+i}+\frac{i}{z-i}+\frac{1}{(z+i)^{2}}-\frac{1}{(z-i)^{2}},$

we obtain that $\Psi_{t}=B\Psi$. Under evaluating the compatibility equation
$A_{t}+AB-B_{y}-BA=0$ at the singular points for $A$ and $B$, we get algebraic
and differential equations among the coefficients of $A$ and $B$, which can be
reduced to the mCH equation in the $(y,t)$ variables. ∎

##  5 Existence of global solution

The main goal of this section is give the proof the Lemma LABEL:lemmaD. In
view of the result in Section 3, we use $M_{l}(z;t,y)$ to recover the solution
$m(t,x(y,t))$ for $y\in\mathbb{R}^{+}$, and $M_{r}(z;t,y)$ to recover the
solution $m(t,x(y,t))$ for $y\in\mathbb{R}^{-}$ through Lemma 2. The first
step is to give the proof of the lemma below.

######  Lemma 3.

Assume that the initial data $m_{0}\in H^{2,1}(\mathbb{R})$ such that
$\|m_{0}\|_{H^{2,1}}$ is small. Then there exists a unique global solution
$m(t,y)$ for every $t\in\mathbb{R}^{+}$ to the Cauchy problem
(1.1)-(LABEL:mch1) for the mCH equation in
$C([0,+\infty);L^{\infty}(\mathbb{R}))$.

###### Proof.

When $m_{0}\in H^{2,1}(\mathbb{R})$ Lemma 1 gives that $r(z;0)\in
H^{1,1}(\mathbb{R}_{k})$ As shown in Proposition 7, for every $t>0$,
$M(z;t,y)$ defined in (4.5) exists unique for every $t\in\mathbb{R}^{+}$,
$y\in\mathbb{R}$.

As shown in Theorem 4.1 [19], there exists a time $T>0$ such that the initial-
value problem (1.1) has unique solution in $m(t,\cdot)\in
C([0,T],H^{2}(\mathbb{R}))$. Hence, $m$ reconstituted from $M(z;t,y)$ in Lemma
2 must unique. It remains to prove $m(t,\cdot)\in
C([0,+\infty);L^{\infty}(\mathbb{R}))$, which is equivalent to
$q(t,\cdot)=\sqrt{1+m(t,\cdot)^{2}}\in C([0,+\infty);L^{\infty}(\mathbb{R}))$.

As shown in Lemma LABEL:pror, for ever $t\in[0,T]$,
$\|r(\cdot;t)\|_{L^{2}(\mathbb{R}_{k})\cap L^{\infty}(\mathbb{R}_{k})}$ is
independent of $t$. The symmetry of $M(z;t,y)$ in (2.63) implies that

$\displaystyle M_{11}(0;t,y)=M_{22}(0;t,y)\in\mathbb{R},\hskip
14.22636ptM_{12}(0;t,y)=M_{21}(0;t,y)i\in\mathbb{R}$

Noting that det$M\equiv 1$, thus $|M_{11}(0;t,y)|\leq 1$. As shown in
Proposition 6, when $\|r(\cdot;0)\|_{L^{2}(\mathbb{R}_{k})\cap
L^{\infty}(\mathbb{R}_{k})}$ is small, it holds that for every $t\in[0,T]$,

$\Big{|}M_{11}(0;t,y)-1\Big{|}\leq C(\|r(\cdot;0)\|_{L^{2}(\mathbb{R}_{k})\cap
L^{\infty}(\mathbb{R}_{k})})<1.$

Therefore, from (2.83) and Proposition 6, it appears that

$\displaystyle 1\leq|q(t,y)|<(1-C(\|r(\cdot;0)\|_{L^{2}(\mathbb{R}_{k})\cap
L^{\infty}(\mathbb{R}_{k})}))^{-1}<\infty.$

If there exists a maximal existence time $T_{max}>0$ such that $q(t,\cdot)$
only exists in $C([0,T_{max}];L^{\infty}(\mathbb{R}))$, then it must appears
that $\lim_{t\to T_{max}-}\|q(t,\cdot)\|_{\infty}=\infty$, from which we
derive a contradiction. This final argument yields the proof of Lemma 3.

∎

Because the estimate in above section is under the scaling transformation of
$y$. The equivalency between the integral norm in $\mathbb{R}_{x}$ and
$\mathbb{R}_{y}$ is first given to arrive at the $L^{2}$-estimate under
variable $x$.

######  Lemma 4.

Assume that $\max_{j=0,1}\|(\cdot)^{j}r(\cdot;0)\|_{L^{2}(\mathbb{R}_{k})\cap
L^{\infty}(\mathbb{R}_{k})}<\epsilon_{0}$. Then for any function
$h(x(\cdot))\in L^{p}(\mathbb{R}_{y})$ with $x(y)$ is given in (2.84),
$1\leqslant p\leqslant\infty$, there has $h\in L^{p}(\mathbb{R}_{x})$.

###### Proof.

As shown in Lemma LABEL:pror, for ever $t\in\mathbb{R}^{+}$, $j=0,1$,
$\|(\cdot)^{j}r(\cdot;t)\|_{L^{2}(\mathbb{R}_{k})\cap
L^{\infty}(\mathbb{R}_{k})}$ is independent of $t$. Recall the symmetry of
$M(z;t,y)$ that

$\displaystyle M(z;t,y)=\sigma_{3}\overline{M(-\bar{z};t,y)}\sigma_{3}.$

It implies that $M_{11}(i)$, $M_{22}(i)\in\mathbb{R}$ while $M_{12}(i)$,
$M_{21}(i)\in i\mathbb{R}$. Noting that det$M\equiv 1$, then the boundedness
of Proposition 6 gives that the functions $|M_{11}(i;t,y)+M_{21}(i;t,y)|$ and
$|M_{12}(i;t,y)+M_{22}(i;t,y)|$ has nonzero infimum for every
$t\in\mathbb{R}^{+},\ y\in\mathbb{R}$. In fact, taking
$|M_{11}(i;t,y)+M_{21}(i;t,y)|$, $y\in\mathbb{R}^{+}$, as an example,

$\displaystyle 1$
$\displaystyle=\big{|}M_{22}(i;t,y)M_{11}(i;t,y)+\text{Im}\left(M_{12}(i;t,y)M_{21}(i;t,y)\right)\big{|}$
$\displaystyle\lesssim|M(i)||M_{11}(i;t,y)+M_{21}(i;t,y)|,$

which leads to that

$\displaystyle|M_{11}(i;t,y)+M_{21}(i;t,y)|\geq
1/C(\|r(\cdot;0)\|_{L^{2}(\mathbb{R}_{k})\cap L^{\infty}(\mathbb{R}_{k})})>0.$

(2.84) gives that

$\partial_{y}x(t,y)=1-\left(\frac{M_{11}(i;t,y)+M_{21}(i;t,y)}{M_{12}(i)+M_{22}(i;t,y)}\right)\partial_{y}\left(\frac{M_{12}(i;t,y)+M_{22}(i;t,y)}{M_{11}(i)+M_{21}(i;t,y)}\right).$

As shown in Corollary 2, when
$\max_{j=0,1}\|(\cdot)^{j}r\|_{L^{2}(\mathbb{R}_{k})\cap
L^{\infty}(\mathbb{R}_{k})}<\epsilon_{0}$, it holds that for every
$t\in\mathbb{R}^{+}$,

$\Big{|}\left(\frac{M_{11}(i)+M_{21}(i)}{M_{12}(i)+M_{22}(i)}\right)\partial_{y}\left(\frac{M_{12}(i;t,y)+M_{22}(i;t,y)}{M_{11}(i;t,y)+M_{21}(i;t,y)}\right)\Big{|}<1,$

which means $0<\partial_{y}x(t,y)<2$. Thus the map $y\to x$ is bijection and
$|\partial_{x}y(t,\cdot)|$ exists in $L^{\infty}(\mathbb{R})$ with

$\displaystyle|\partial_{x}y(t,\cdot)|\leq
C(\max_{j=0,1}\|(\cdot)^{j}r(\cdot;0)\|_{L^{2}(\mathbb{R}_{k})\cap
L^{\infty}(\mathbb{R}_{k})}).$

Thus Lemma 4 comes from the boundedness of $|\partial_{x}y|$ and
$|\partial_{y}x|$ immediately ∎

And we give the following elementary result from Fourier theory.

######  Lemma 5.

If $f(z)$ is a complex valued function defined on $\mathbb{R}$ and admits the
symmetry $f(z)=f(-1/z)$, then $f(z(k))$ is well-defined under the variable
$k=z-1/z$. Denote $\tilde{f}(k)=f(z(k))$, then
(a) if $f\in H^{1,1}(\mathbb{R}_{k})$, then:

$\displaystyle\int_{\mathbb{R}}f(z)e^{-\frac{i}{2}(z-1/z)y}dz=\int_{\mathbb{R}}\tilde{f}(k)e^{-\frac{i}{2}ky}dk,$

which implies that

$\displaystyle\int_{\mathbb{R}}f(z)e^{-\frac{i}{2}(z-1/z)y}dz\in
H^{1,1}(\mathbb{R}_{y}).$

(b) Denote $\mathcal{C}^{k}$ as the Cauchy operator on $k$-plane. Therefore
$\mathcal{C}_{\pm}(f)$ is also well-defined under the variable $k$ and

$\displaystyle\mathcal{C}_{\pm}(f)(z)=\mathcal{C}^{k}_{\pm}(f)(k).$ (5.1)

And
$\|\mathcal{C}_{\pm}(f)\|_{L^{2}(\mathbb{R}_{z})}=\|\mathcal{C}^{k}_{\pm}(f)\|_{L^{2}(\mathbb{R}_{k})}$.
(c) if $g$ also admits that $g(z)=g(-1/z)$ and $\tilde{g}(k)=g(z(k))$ is in
$H^{1}(\mathbb{R}_{k})$, then

$\displaystyle\int_{\mathbb{R}^{+}}\|\mathcal{C}_{+}(fe^{2\theta})\|_{2}^{2}\|\mathcal{C}_{\pm}(ge^{\pm
2\theta})\|_{2}^{2}dy\lesssim\|\tilde{f}\|_{H^{1/4}(\mathbb{R}_{k})}\|\tilde{g}\|_{H^{1/4}(\mathbb{R}_{k})},$
$\displaystyle\int_{\mathbb{R}^{+}}y^{2}\|\mathcal{C}_{+}(fe^{2\theta})\|_{2}^{2}\|\mathcal{C}_{\pm}(ge^{\pm
2\theta})\|_{2}^{2}dy\lesssim\|\tilde{f}\|_{H^{3/4}(\mathbb{R}_{k})}\|\tilde{g}\|_{H^{3/4}(\mathbb{R}_{k})}.$

###### Proof.

Recalling that for $k=z-1/z$, the map $k\to z$ is that

$\displaystyle z_{+}(k)=(k+\sqrt{k^{2}+4})/2:\ \mathbb{R}\to\mathbb{R}^{+},$
$\displaystyle z_{-}(k)=(k-\sqrt{k^{2}+4})/2:\ \mathbb{R}\to\mathbb{R}^{-}.$

Therefore,

$\displaystyle\int_{\mathbb{R}}f(z)e^{-\frac{i}{2}(z-1/z)y}dz$
$\displaystyle=\int_{\mathbb{R}}\tilde{f}(k)e^{-\frac{i}{2}ky}\frac{1}{1+z_{+}(k)^{-2}}dk+\int_{\mathbb{R}}\tilde{f}(k)e^{-\frac{i}{2}ky}\frac{1}{1+z_{-}(k)^{-2}}dk$
$\displaystyle=\int_{\mathbb{R}}\tilde{f}(k)e^{-\frac{i}{2}ky}dk.$

And (b) comes from the fact that

$\displaystyle\frac{1}{s-z}+\dfrac{1}{s^{2}}\cdot\frac{1}{-\frac{1}{s}-z}=\frac{1+\frac{1}{s^{2}}}{s-\frac{1}{s}-z+\frac{1}{z}}.$

For (c), we give the details of the calculation for
$\|\mathcal{C}_{+}(fe^{2\theta})\|_{2}^{2}\|\mathcal{C}_{+}(ge^{2\theta})\|_{2}^{2}$.
It follows that

$\displaystyle\int_{\mathbb{R}^{+}}$
$\displaystyle\|\mathcal{C}_{+}(fe^{2\theta})\|_{2}^{2}\|\mathcal{C}_{+}(ge^{2\theta})\|_{2}^{2}dy=\int_{\mathbb{R}^{+}}\|\mathcal{C}^{k}_{+}(fe^{2\theta})\|_{L^{2}(\mathbb{R}_{k})}^{2}\|\mathcal{C}^{k}_{+}(ge^{2\theta})\|_{L^{2}(\mathbb{R}_{k})}^{2}dy$
$\displaystyle=\int_{\mathbb{R}^{+}}\int_{\mathbb{R}}X_{+}(k)|\hat{\tilde{f}}(k+y/4\pi)|^{2}dk\int_{\mathbb{R}}X_{+}(k)|\hat{\tilde{g}}(k+y/4\pi)|^{2}dkdy$
$\displaystyle=\int_{\mathbb{R}^{+}}\int_{y/4\pi}^{\infty}|\hat{\tilde{f}}(k)|^{2}dk\int_{y/4\pi}^{\infty}|\hat{\tilde{g}}(k)|^{2}dkdy$
$\displaystyle\leq\left(\int_{\mathbb{R}^{+}}(\int_{y/4\pi}^{\infty}|\hat{\tilde{f}}(k)|^{2}dk)^{2}dy\right)^{1/2}\left(\int_{\mathbb{R}^{+}}(\int_{y/4\pi}^{\infty}|\hat{\tilde{g}}(k)|^{2}dk)^{2}dy\right)^{1/2}$
$\displaystyle\leq\int_{\mathbb{R}^{+}}(\int_{0}^{k/4\pi}|\hat{\tilde{f}}(k)|^{4}dy)^{1/2}dk\int_{\mathbb{R}^{+}}(\int_{0}^{k/4\pi}|\hat{\tilde{g}}(k)|^{4}dy)^{1/2}dk$
$\displaystyle=\int_{\mathbb{R}^{+}}|\hat{\tilde{f}}(k)|^{2}(k/4\pi)^{1/2}dk\int_{\mathbb{R}^{+}}|\hat{\tilde{g}}(k)|^{2}(k/4\pi)^{1/2}dk\lesssim\|\hat{\tilde{f}}\|_{L^{2,1/4}(\mathbb{R}_{k})}^{2}\|\hat{\tilde{g}}\|_{L^{2,1/4}(\mathbb{R}_{k})}^{2},$

and

$\displaystyle\int_{\mathbb{R}^{+}}y^{2}$
$\displaystyle\|\mathcal{C}_{+}(fe^{2\theta})\|_{2}^{2}\|\mathcal{C}_{\pm}(ge^{\pm
2\theta})\|_{2}^{2}dy=\int_{\mathbb{R}^{+}}\int_{y/4\pi}^{\infty}y|\hat{\tilde{f}}(k)|^{2}dk\int_{y/4\pi}^{\infty}y|\hat{\tilde{g}}(k)|^{2}dkdy$
$\displaystyle\leq\left(\int_{\mathbb{R}^{+}}(\int_{y/4\pi}^{\infty}y|\hat{\tilde{f}}(k)|^{2}dk)^{2}dy\right)^{1/2}\left(\int_{\mathbb{R}^{+}}(\int_{y/4\pi}^{\infty}y|\hat{\tilde{g}}(k)|^{2}dk)^{2}dy\right)^{1/2}$
$\displaystyle\leq\int_{\mathbb{R}^{+}}(\int_{0}^{k/4\pi}y^{2}|\hat{\tilde{f}}(k)|^{4}dy)^{1/2}dk\int_{\mathbb{R}^{+}}(\int_{0}^{k/4\pi}y^{2}|\hat{\tilde{g}}(k)|^{4}dy)^{1/2}dk$
$\displaystyle=\frac{1}{3}\int_{\mathbb{R}^{+}}|\hat{\tilde{f}}(k)|^{2}(k/4\pi)^{3/2}dk\int_{\mathbb{R}^{+}}|\hat{\tilde{g}}(k)|^{2}(k/4\pi)^{3/2}dk\lesssim\|\hat{\tilde{f}}\|_{L^{2,3/4}(\mathbb{R}_{k})}^{2}\|\hat{\tilde{g}}\|_{L^{2,3/4}(\mathbb{R}_{k})}^{2},$

from which we complete our proof. ∎

We shall now recover $m$ from the RH problem $M$ and give $L^{2}$-estimate of
$m(t,\cdot)$, and further obtain our main result Lemma LABEL:lemmaD.

###### Proof of Lemma LABEL:lemmaD.

We only give the details of the proof of $m(t,\cdot)\in
H^{2,1}(\mathbb{R}^{+}_{y})$. Together with Lemma 4 and the fact that
$|\partial_{x}y|$, $|\partial_{x}^{2}y|$ is bounded shown in Lemma 3, it
follows that $m(t,\cdot)\in H^{2,1}(\mathbb{R}^{+}_{x})$. Recall the
reconstruction formula in Lemma 2:

$\displaystyle\eta=\lim_{z\to\infty}zM_{12}.$ (5.2)

From (3.10), it is adduced that

$\displaystyle\eta=\frac{i}{2\pi}\int_{\mathbb{R}}(\mu_{11}-1)\bar{r}e^{2\theta}dz+\frac{i}{2\pi}\int_{\mathbb{R}}\bar{r}e^{2\theta}dz.$

For convenience, denote

$\displaystyle\eta_{1}=\int_{\mathbb{R}}(\mu_{11}-1)\bar{r}e^{2\theta}dz,\
\eta_{2}=\int_{\mathbb{R}}\bar{r}e^{2\theta}dz.$ (5.3)

Consequently, Lemma 5 gives that $\eta_{2}\in H^{1,1}(\mathbb{R}_{y})$. And
for $\eta_{1}$, invoking (3.11), it holds that

$\displaystyle\eta_{1}=\int_{\mathbb{R}}\mathcal{C}_{-}(re^{-2\theta}\mu_{12})\bar{r}e^{2\theta}dz=-\int_{\mathbb{R}}\mathcal{C}_{+}(\bar{r}e^{2\theta})re^{-2\theta}\mu_{12}dz,$
(5.4)

from which and (3.24) we obtain that

$\displaystyle|\eta_{1}|\leq\|\mathcal{C}_{+}(\bar{r}e^{2\theta})\|_{2}\|\mu_{12}\|_{2}\|r\|_{\infty}\leq
2\|\mathcal{C}_{+}(\bar{r}e^{2\theta})\|_{2}\|r\|_{\infty}(\|\mathcal{C}_{+}(\bar{r}e^{2\theta})\|_{2}+\|\mathcal{C}_{-}(re^{-2\theta})\|_{2}).$

For $\|\mathcal{C}_{+}(\bar{r}e^{2\theta})\|_{2}^{2}$, Lemma 5 gives that

$\displaystyle\int_{\mathbb{R}^{+}}\big{|}\|\mathcal{C}_{+}(\bar{r}e^{2\theta})\|_{2}^{2}\big{|}^{2}dy\lesssim\|\hat{r}\|_{L^{2,1/4}(\mathbb{R}_{k})}^{4},$

and

$\displaystyle\int_{\mathbb{R}^{+}}\big{|}y\|\mathcal{C}_{+}(\bar{r}e^{2\theta})\|_{2}^{2}\big{|}^{2}dy$
$\displaystyle\lesssim\|\hat{r}\|_{L^{2,3/4}(\mathbb{R}_{k})}^{4}.$

Similar calculation of
$\|\mathcal{C}_{+}(\bar{r}e^{2\theta})\|_{2}\|\mathcal{C}_{-}(re^{-2\theta})\|_{2}$
gives that $\eta_{1}\in L^{2,1}(\mathbb{R}^{+}_{y})$. Take $y$-derivative in
(5.4) and obtain

$\displaystyle\partial_{y}\eta_{1}=$
$\displaystyle-\int_{\mathbb{R}}\mathcal{C}_{+}(\frac{i}{2}k(\cdot)\bar{r}e^{2\theta})re^{-2\theta}\mu_{12}dz-\int_{\mathbb{R}}\mathcal{C}_{+}(\bar{r}e^{2\theta})\frac{i}{2}k(z)re^{-2\theta}\mu_{12}dz$
$\displaystyle-\int_{\mathbb{R}}\mathcal{C}_{+}(\bar{r}e^{2\theta})re^{-2\theta}\partial_{y}\mu_{12}dz,$

which implies that

$\displaystyle|\partial_{y}\eta_{1}|\leq$
$\displaystyle\|r\|_{\infty}\left(\|\mathcal{C}_{+}(k(\cdot)\bar{r}e^{2\theta})\|_{2}\|\mu_{12}\|_{2}+\|\mathcal{C}_{+}(\bar{r}e^{2\theta})\|_{2}\|\mu_{12}\|_{2}+\|\mathcal{C}_{+}(\bar{r}e^{2\theta})\|_{2}\|\partial_{y}\mu_{12}\|_{2}\right)$
$\displaystyle\leq$ $\displaystyle
2\|r\|_{\infty}(\|\mathcal{C}_{+}(\bar{r}e^{2\theta})\|_{2}+\|\mathcal{C}_{-}(re^{-2\theta})\|_{2})\left(\|\mathcal{C}_{+}(k(\cdot)\bar{r}e^{2\theta})\|_{2}+\|\mathcal{C}_{+}(\bar{r}e^{2\theta})\|_{2}\right)$
$\displaystyle+2\|r\|_{\infty}\|\mathcal{C}_{+}(\bar{r}e^{2\theta})\|_{2}(\|\mathcal{C}_{+}(k(\cdot)\bar{r}e^{2\theta})\|_{2}+\|\mathcal{C}_{-}(k(\cdot)re^{-2\theta})\|_{2})$
$\displaystyle+8\|r\|_{\infty}\|\mathcal{C}_{+}(\bar{r}e^{2\theta}))\|_{2}(\|\mathcal{C}_{+}(\bar{r}e^{2\theta})\|_{2}+\|\mathcal{C}_{-}(re^{-2\theta})\|_{2})\|k(\cdot)r\|_{{\infty}}$

The last inequality comes from (3.28). We give the details of the estimate of
$\|\mathcal{C}_{+}(\bar{r}e^{2\theta})\|_{2}\|\mathcal{C}_{+}(k(\cdot)\bar{r}e^{2\theta})\|_{2}$.
Similar claims can be applied to the others.

Via Lemma 5, it holds that

$\displaystyle\int_{\mathbb{R}^{+}}\|\mathcal{C}_{+}(\bar{r}e^{2\theta})\|_{2}^{2}\|\mathcal{C}_{+}(k(\cdot)\bar{r}e^{2\theta})\|_{2}^{2}dy\lesssim\|(\cdot)r\|_{H^{1/4}(\mathbb{R}_{k})}^{2}\|r\|_{H^{1/4}(\mathbb{R}_{k})}^{2},$

and

$\displaystyle\int_{\mathbb{R}^{+}}y^{2}\|\mathcal{C}_{+}(\bar{r}e^{2\theta})\|_{2}^{2}\|\mathcal{C}_{+}(k(\cdot)\bar{r}e^{2\theta})\|_{2}^{2}dy\lesssim\|r\|_{H^{3/4}(\mathbb{R}_{k})}^{2}\|(\cdot)r\|_{H^{3/4}(\mathbb{R}_{k})}^{2}.$

Thus it is accomplished that $|\partial_{y}\eta_{1}|\in
L^{2,1}(\mathbb{R}_{y})$.

Therefore, we finally conclude that $\eta\in H^{1,1}(\mathbb{R}^{+}_{y})$.

On the other hand, for

$\displaystyle\zeta^{(+)}=-\lim_{z\to\infty}z(M_{22}-1).$ (5.5)

Combining (3.10) and (3.11), it is adduced that

$\displaystyle\zeta^{(+)}=\frac{1}{2\pi
i}\int_{\mathbb{R}}\mathcal{C}_{-}(re^{-2\theta})\bar{r}e^{2\theta}ds-\frac{1}{2\pi
i}\int_{\mathbb{R}}(\mu_{22}-1)re^{-2\theta}\mathcal{C}_{+}(\bar{r}e^{2\theta})ds.$
(5.6)

An analogical calculation gives that $\zeta^{(+)}\in
H^{1,1}(\mathbb{R}^{+}_{y})$. Recall that

$\displaystyle\eta=\dfrac{-im_{x}}{(1+m^{2})^{3/2}},\
\zeta^{(+)}=\frac{i}{2}\int_{y}^{+\infty}\left(\frac{m_{x}^{2}}{q^{6}}+\frac{m^{2}}{q^{2}}\right)ds.$

Then by the boundedness of $m$ and Sobolev-Gagliardo-Nirenberg inequality, we
finally obtain that $m(t,\cdot)\in H^{2,1}(\mathbb{R}^{+}_{y})$.

This gives the desired result in Lemma LABEL:lemmaD. ∎

Acknowledgements

The work of Fan and Yang is partially supported by the National Science
Foundation of China under grants 12271104, 51879045 and 12247182. The work of
Liu is partially supported by the Simons Foundation under grant 499875.

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