# Existence of global solutions for the modified Camassa-Holm equation with a
nonzero background

Yiling YANG  School of Mathematical Sciences and Key Laboratory for Nonlinear
Science, Fudan University, Shanghai 200433, P. R. China.
<EMAIL_ADDRESS>,  Engui Fan  School of Mathematical Sciences and Key
Laboratory for Nonlinear Science, Fudan University, Shanghai 200433, P. R.
China<EMAIL_ADDRESS>and  Yue Liu  Department of Mathematics, University
of Texas at Arlington, TX 76019<EMAIL_ADDRESS>
###### Abstract.

Consideration in the present paper is the existence of global solutions for
the modified Camassa-Holm (mCH) equation with a nonzero background initial
value. The mCH equation is completely integrable and can be considered as a
model for the unidirectional propagation of shallow-water waves. By applying
the inverse scattering transform with an application of the Cauchy projection
operator, the existence of a unique global solution to the mCH equation in the
line with a nonzero background initial value is established in the weighted
Sobolev space $H^{2,1}(\mathbb{R})\cap H^{1,2}(\mathbb{R})$ based on the
representation of a Riemann-Hilbert (RH) problem associated with the Cauchy
problem to the mCH equation. A crucial technique used is to derive the
boundedness of the solution in the Sobolev space $W^{1,\infty}(\mathbb{R}),$
then reconstruct a new RH problem for the Cauchy projection operator of
reflection coefficients. The regularity of the global solution is achieved by
the refined estimate arguments on those solutions of the corresponding RH
problem.

Keywords: Modified Camassa-Holm equation; global solution; Riemann-Hilbert
problem; Cauchy projection operator.

AMS Subject Classification 2020: 35P25; 35Q15; 35Q35; 35A01.

###### Contents

  1. 1 Introduction
  2. 2 Direct scattering transform
    1. 2.1 The Jost function
    2. 2.2 Scattering map
    3. 2.3 The RH problems
    4. 2.4 Reconstruction of the potential $m$
  3. 3 Inverse scattering transform
    1. 3.1 The Cauchy projection operator
    2. 3.2 Solvability of the RH problem
    3. 3.3 Time evolution
    4. 3.4 Estimates on solutions of the RH problem
  4. 4 Existence of global solutions
  5. 5 Regularity
    1. 5.1 Transitions of the spectral problem
    2. 5.2 A new RH problem
    3. 5.3 Proof of the main theorem

##  1\. Introduction

The present paper is concerned with the existence of global solutions for the
modified Camassa-Holm (mCH) equation [20, 33] , namely,

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

with nonzero background $u(t,x)\to 1$ as $|x|\to\infty.$ It is known that the
mCH equation (1.1) is an integrable modification to the well-known Camassa-
Holm (CH) equation

$\displaystyle m_{t}+(um)_{x}+u_{x}m=0,\quad m=u-u_{xx},$ (1.2)

which was appeared in [21] 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] (see also [13] for a
rigorous justification in shallow-water approximation). The CH equation (1.2)
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 [3, 5, 10, 11, 12, 14,
19, 22, 30, 31].

It is observed that all the nonlinear terms in the CH equation (1.2) are
quadratic. Different from the CH equation (1.2), the mCH equation (1.1)
contains the cubic nonlinearity which was introduced by Fuchssteiner [20],
Olver and Rosenau [33] by applying the general method of tri-Hamiltonian
duality to the bi-Hamiltonian representation of the modified Korteweg-de Vries
equation (see also [35]). Applying the scaling transformation and taking
parameter limit, the mCH equation (1.1) can be reduced a short pulse equation
[37]

$\displaystyle u_{xt}=u+\frac{1}{6}(u^{3})_{xx}.$

More recently, the mCH equation (1.1) was considered as a model for the
unidirectional propagation for shallow-water waves of mild amplitude over a
flat bottom [9], where the solution $u$ is related to the horizontal velocity
in certain level of water.

It was shown that the mCH equation (1.1) admits a Lax pair [35, 38]

$\Phi_{x}=U\Phi,\hskip 14.22636pt\Phi_{t}=V\Phi,$ (1.3)

where

$\displaystyle U=\frac{1}{2}\left(\begin{array}[]{cc}-1&{\lambda}m\\\ -\lambda
m&1\end{array}\right),\ \ \lambda=-\frac{1}{2}(z+z^{-1}),$ (1.6)
$\displaystyle
V=\left(\begin{array}[]{cc}\lambda^{-2}+\frac{1}{2}(u^{2}-u_{x}^{2})&-\lambda^{-1}(u-u_{x})-\frac{1}{2}\lambda(u^{2}-u_{x}^{2})m\\\\[5.0pt]
\lambda^{-1}(u+u_{x})+\frac{1}{2}\lambda(u^{2}-u_{x}^{2})m&-\lambda^{-2}-\frac{1}{2}(u^{2}-u_{x}^{2})\end{array}\right),$
(1.9)

$\Phi$ is a complex matrix valued function of $x,t$ and $\lambda$ with the
spectral parameter $\lambda\in\mathbb{C}$ independent of $x$ and $t$, and the
momentum potential $m$ is the solution of the mCH equation (1.1).

It is noted that the global smooth one-soliton solution of the mCH equation
(1.1) with nonzero background data was obtained by using the RH method [4]. On
the other hand, the soliton of the mCH equation (1.1) with zero background
data is a weak solution in the form of peaked wave, that is,
$\varphi_{c}(x-ct)=\sqrt{\frac{3c}{2}}e^{-|x-ct|},\;c>0.$ In addition, the
quasi-periodic solutions with periodic background data were constructed by
using algebro-geometric method [24].

The wave-breaking and those peakons for the mCH equation (1.1) with zero
background data were also investigated in [23]. With the aid of reciprocal
transformation, the Bäcklund transformation and nonlinear superposition
formula for the mCH equation (1.1) were given [39]. The existence of the
global peakon solutions and the large time asymptotic behaviour of these kind
of non-smooth solitons were investigated in [6]. Recently the long time
asymptotic behavior of the mCH equation (1.1) was established by using
$\bar{\partial}$-steepest decadent analysis in [40].

It is known that the Cauchy problem associated with the mCH equation (1.1) is
the locally well-posed in the Sobolev space $H^{s}(\mathbb{R}),s>5/2$ [23].
However, to the best of our knowledge, the global well-posedness to the mCH
equation (1.1) on the line seems still unknown.

The main purpose in the present paper is to establish the global existence of
solutions to the mCH equation (1.1) with a nonzero background by applying
inverse scattering theory. It is found, however, that directly using this
approach to the mCH equation (1.1) will confront some substantial
difficulties, which need to be overcome and are much different from the NLS,
AKNS, derivative NLS equations [2, 15, 17, 27, 34, 42]. For example,

  * 1)

In the RH problem associated with the mCH equation (1.1), there are two
singularities $z=\pm 1$, which give rise to difficulty in directly using Zhou
vanishing lemma. However, to remove the singularity at $z=\pm 1$ by change of
variable, the function $\eta=1-1/m$ will appear in the condition of the RH
problem. The boundness and space estimate of $\eta$ must be obtained to show
that $m-1\in H^{2,1}(\mathbb{R})\cap H^{1,2}(\mathbb{R})$.

  * 2)

In the estimates of the Volterra integral and the Cauchy projector operators,
there is an oscillating term $e^{i(z-1/z)x}$, which is difficult to directly
apply the Fourier transform in real axis $\mathbb{R}$. This technical
difficulty could be conquered by splitting the original spectral problem into
two new spectral problems by two transformations.

  * 3)

In the jump matrix, the reflection coefficient $|r(z)|=1,\ z=\pm 1$, which
prevents from establishing the Lipschitz continuous for solutions. To deal
with this issue, we can take an advantage of the nice properties of the time
evolution for scattering data to directly obtain estimates of the global
solutions from the time dependent RH problems without a prior estimate on the
potential function $m.$

In what follows, we address our main result on the existence of global
solutions for the Cauchy problem associated with the mCH equation, namely,

$\displaystyle\left\\{\begin{array}[]{ll}m_{t}+\left(m\left(u^{2}-u_{x}^{2}\right)\right)_{x}=0,\quad
m=u-u_{xx},\\\\[4.0pt] m(0,x)=m_{0}(x),\\\ \end{array}\right.$ (1.12)

where $m_{0}=(1-\partial_{x}^{2})u_{0}$ with
$u_{0}=(1-\partial_{x}^{2})^{-1}m_{0}=p\ast m_{0},$
$p(x)=\frac{1}{2}\,e^{-|x|},$ and the symbol ‘$\ast$’ denotes the convolution
product on $\mathbb{R}$, defined by

$(f\ast g)(x)=\int_{\mathbb{R}}f(y)g(x-y)dy.$

The principal result of the present paper is now stated as follows.

######  Theorem 1.1.

Assume that $m_{0}(x)>0,\forall x\in\mathbb{R}$ and $m_{0}-1\in
H^{2,1}(\mathbb{R})\cap H^{1,2}(\mathbb{R})$. Then there exists a unique
global solution $m$ of the Cauchy problem (1.12) such that $m-1\in
C([0,+\infty);H^{2,1}(\mathbb{R})\cap H^{1,2}(\mathbb{R})).$

A crucial approach for proving this theorem is to establish a
$L^{2}$-bijection map between solution $m$ and scattering data by applying the
inverse scattering transform theory to the Cauchy problem (1.12), see Figure
1. Starting from the given initial value $m_{0}-1\in H^{2,1}(\mathbb{R})\cap
H^{1,2}(\mathbb{R})$, the direct scattering transform gives rise to the
scattering data and a RH problem associated with the Cauchy problem (1.12).
Then the inverse scattering transform goes back to the global solution $m(t)$
satisfying $m-1\in C([0,+\infty);H^{2,1}(\mathbb{R})\cap H^{1,2}(\mathbb{R}))$
to the Cauchy problem (1.12) via the RH problem with associated with time-
dependent scattering data.

######  Remark 1.1.

Note that if $m_{0}(x)>0$, $\forall x\in\mathbb{R}$, then the solution $m$ of
(1.12) satisfies $\inf_{x\in\mathbb{R}}m(t,x)>0$, for $t\in[0,T)$ with any
given $T>0$ [23]. Using the blow-up criterion in Theorem 4.2 [23] with the
energy estimates, it follows from Theorem 1.1 that for any $s\geq 2,$ there
exists a unique global solution $m$ of the Cauchy problem (1.12) satisfying
$m-1\in C([0,+\infty);H^{s,1}(\mathbb{R})\cap H^{s-1,2}(\mathbb{R})),$ with
the initial data $m_{0}(x)>0,\forall x\in\mathbb{R}$ and $m_{0}-1\in
H^{s,1}(\mathbb{R})\cap H^{s-1,2}(\mathbb{R}),\,\forall s\geq 2.$

$m_{0}-1\in H^{2,1}(\mathbb{R})\cap
H^{1,2}(\mathbb{R})$$(r(0;z),\\{z_{j},c_{j}(0)\\}_{j=1}^{2N_{0}})\in Y$$m-1\in
C([0,\infty),H^{2,1}(\mathbb{R})\cap
H^{1,2}(\mathbb{R}))$$(r(t;z),\\{z_{j},c_{j}(t)\\}_{j=1}^{2N_{0}})\in Y$Time
evolutionDirect scatteringInverse scattering Figure 1. Bijective relation
between potential $m$ and reflection coefficients

The rest of the paper is organized as follows. In Section 2, we give
construction of the basic RH problems associated with the Cauchy problem
(1.12) and establish a direct inverse transform from the initial value $m_{0}$
to the scattering data at $t=0$. In Section 3, we carry out the inverse
scattering transform from time-dependent scattering data and solvability of
the regular RH problem. In Section 4, we show the existence of a unique global
solution $m-1\in C([0,+\infty);W^{1,\infty}(\mathbb{R}))$ to the Cauchy
problem (1.12). In the last section, Section 5, we reconstruct a new RH
problem to arrive at the regularity of the solution $m-1$. Finally, we
demonstrate the existence of a unique global solution satisfying $m-1\in
C([0,+\infty);H^{2,1}(\mathbb{R})\cap H^{1,2}(\mathbb{R}))$ via the new RH
problem and refined estimates on the solution $m$. Notations. Throughout, the
norm in the classical space $L^{p}(\mathbb{R})$ for any $1\leq p\leq\infty$
will be written as $\|\cdot\|_{L^{p}}.$ We use the normed spaces defined in
the following. A weighted $L^{p}(\mathbb{R})$ space for any $1\leq
p\leq\infty$ is specified by

$L^{p,s}(\mathbb{R})=\left\\{f\in L^{p}(\mathbb{R})\hskip 5.69046pt|\ \hskip
2.84544pt\langle\cdot\rangle^{s}f\in L^{p}(\mathbb{R})\right\\},\quad s\geq 0$

with $\langle x\rangle=(1+x^{2})^{\frac{1}{2}}$ equipped with the norm

$\|f\|_{p,s}=\|\langle\cdot\rangle^{s}f\|_{L^{p}}.\quad$

The Sobolev space $W^{k,p}$ is defined by

$W^{k,p}(\mathbb{R})=\left\\{f\in L^{p}(\mathbb{R})\hskip 5.69046pt|\ \hskip
2.84544pt\partial_{x}^{j}f\in L^{p}(\mathbb{R}),\;j=1,2,...,k\right\\},$

equipped with the norm

$\|f\|_{W^{k,p}}=\left(\sum_{j=0}^{k}\|\partial_{x}^{j}f\|^{p}_{L^{p}}\right)^{1/p},$

and a weighted Sobolev space $H^{k,s}$ is defined by

$H^{k,s}(\mathbb{R})=\left\\{f\in L^{2}(\mathbb{R})\hskip 5.69046pt|\ \hskip
2.84544pt\langle\cdot\rangle^{s}\partial_{x}^{j}f\in
L^{2}(\mathbb{R}),\;j=0,1,...,k\right\\},\;s\geq 0$

equipped with the norm

$\|f\|_{H^{k,s}}=\sum_{j=0}^{k}\|\langle\cdot\rangle^{s}\partial_{x}^{j}f\|_{L^{2}},\;s\geq
0.$

And the Fourier transform is normalized by

$\displaystyle\hat{f}(z)=(\mathcal{F}f)(z)=\int_{\mathbb{R}}f(s)e^{-2\pi
isz}ds,\ \check{g}(z)=(\mathcal{F}^{-1}g)(z)=\int_{\mathbb{R}}g(s)e^{2\pi
isz}ds.$

Let $I$ be an interval on the real line $\mathbb{R}$ and $X$ be a Banach
space. We denote $C_{B}(I;X)$ the space of bounded continuous functions on $I$
taking values in $X$ equipped with the norm

$\|f\|_{C_{B}(I;X)}=\sup_{x\in I}\|f(\cdot,x)\|_{X}.$

Define a Banach space

$\displaystyle
Y=\left\\{\left(r,\\{z_{j},c_{j}\\}_{j=1}^{2{N_{0}}}\right)\in\left(H^{1,2}(\mathbb{R})\cap
H^{2,1}(\mathbb{R})\right)\otimes\mathbb{C}^{2{N_{0}}}\otimes\mathbb{C}^{2{N_{0}}}\right\\},$
(1.13)

equipped with the norm

$\displaystyle\Big{\|}\left(r,\\{z_{j},c_{j}\\}_{j=1}^{2{N_{0}}}\right)\Big{\|}_{Y}=\|r\|_{H^{1,2}\cap
H^{2,1}}+\sum_{j=1}^{2{N_{0}}}|z_{j}|+\sum_{j=1}^{2{N_{0}}}|c_{j}|.$ (1.14)

In addition, for any vector or matrix $A$, the superscript $A^{T}$ denotes the
transpose of $A$. Denote $f^{*}(z)=\overline{f(\bar{z})}$. Finally, the letter
$C$ will be used to denote universal positive constants which may vary from
line to line. We also use the notation $A\mathchar 13358\relax B$ to denote
the bound of the form $A\mathchar 13366\relax CB$. To emphasize the implied
constant to depend on some parameter $\alpha$, we shall indicate this by
$C(\alpha)$.

##  2\. Direct scattering transform

Our attention in this section is to consider the direct scattering transform
for the spatial spectral problem in the Lax pair (1.3) as $t=0$ with the
initial value $m_{0}-1\in H^{1,2}\cap H^{2,1}$. The variable $t$ is omitted as
usual, for example, $\Phi(z;0,x)$ is just written as $\Phi(z;x)$.

###  2.1. The Jost function

Following the approach on the scheme of construction of a basic RH problem [1,
4, 7, 8, 34, 40, 41], we define a transformation for $m_{0}-1\in
L^{1}(\mathbb{R})$

$\displaystyle\mu(z;x)=D(z)\Phi(z;x)e^{p(z;x)\sigma_{3}},$ (2.1)

where

$\displaystyle D(z)=\left(\begin{array}[]{cc}1&-\frac{2\lambda}{2+ik}\\\
-\frac{2\lambda}{2+ik}&1\end{array}\right),\ \ k=z-z^{-1},\
\lambda=-\frac{1}{2}(z+z^{-1}),$ (2.4) $\displaystyle
p(z;x)=\frac{ik}{4}(x+\int_{+\infty}^{x}(m_{0}(s)-1)ds).$

Then the Jost function $\mu(z;x)$ satisfies the following spectral problem

$\displaystyle\mu_{x}(z;x)=-\frac{ik}{4}m_{0}[\sigma_{3},\mu(z;x)]+P\mu(z;x),\
P=\dfrac{m_{0}-1}{ik}\left(\begin{array}[]{cc}1&\lambda\\\
-\lambda&-1\end{array}\right),$ (2.7)

where the Lie bracket $[A,B]$ is defined by $[A,B]=AB-BA.$ We now denote the
matrix in column by
$\mu^{\pm}(z;x)=\left(\mu^{\pm}_{1}(z;x),\mu^{\pm}_{2}(z;x)\right),$ where the
subscripts $1$ and $2$ denote the first and second columns of
$\mu^{\pm}(z;x)$, respectively. Then (2.7) leads to two Volterra type
integrals

$\displaystyle\mu^{\pm}_{1}(z;x)=e_{1}+\int^{x}_{\pm\infty}\text{diag}(1,e^{2(p(z;s)-p(z;x))})P(z;s)\mu^{\pm}_{1}(z;s)ds,$
(2.8)
$\displaystyle\mu^{\pm}_{2}(z;x)=e_{2}+\int^{x}_{\pm\infty}\text{diag}(e^{-2(p(z;s)-p(z;x))},1)P(z;s)\mu_{2}^{\pm}(z;s)ds,$
(2.9)

where $e_{1}=(1,0)^{T}$, $e_{2}=(0,1)^{T}$. Then the existence of $\mu^{\pm}$
and their analytic properties as functions of $z$ are the subject of the
following proposition.

######  Proposition 2.1.

Let $m_{0}-1\in L^{1}(\mathbb{R})$. Then there exists unique solution
$\mu^{\pm}(z;\cdot)\in L^{\infty}(\mathbb{R})$ for every
$z\in\mathbb{R}\setminus\\{\pm 1\\}$. Moreover, for every $x\in\mathbb{R}$,
$z\in\mathbb{C}^{+}$, $\mu^{-}_{1}(z;x)$ and $\mu^{+}_{2}(z;x)$ extend
analytically into solutions of (2.8) and (2.9) respectively, while for every
$x\in\mathbb{R}$, $z\in\mathbb{C}^{-}$, $\mu^{+}_{1}(z;x)$ and
$\mu^{-}_{2}(z;x)$ extend analytically into solutions of (2.8) and (2.9)
respectively.

###### Proof.

It suffices to prove the statement by taking $\mu^{-}_{2}$ as example. Rewrite
(2.9) as

$\displaystyle\mu^{\pm}_{2}(z;x)=e_{2}+\mathcal{T}\mu^{\pm}_{2}(z;x),$

where $\mathcal{T}$ is a linear integral operator defined by

$\displaystyle\mathcal{T}f(z;x)$
$\displaystyle=\int^{x}_{+\infty}K(z;x,s)f(z;s)ds,$ (2.10)

for any $f(z;\cdot)\in L^{\infty}(\mathbb{R})$ with integral kernel
$K(z;x,s)={\rm diag}(e^{\frac{i}{2}(z-1/z)\int^{s}_{x}m_{0}(l)dl},1)P(z;s).$
Then $\mathcal{T}$ is a bounded integral operator on $L^{\infty}$ for every
$z\in\overline{\mathbb{C}^{-}}\setminus\\{\pm 1\\}$ with

$\displaystyle\|\mathcal{T}\|\leq C(z)\|m_{0}-1\|_{L^{1}}.$

For every $z\in\overline{\mathbb{C}^{-}}\setminus\\{\pm 1\\}$, any
$n=2,3,...$, $f(z;\cdot)\in L^{\infty}(\mathbb{R})$

$\displaystyle\mathcal{T}^{n}f(z;x)$
$\displaystyle=\int^{x}_{+\infty}\int^{y_{1}}_{+\infty}\dots\int^{y_{n-1}}_{+\infty}K(z;x,y_{1})\dots
K(z;x,y_{n})f(z;y_{n})dy_{n}\dots dy_{1}$
$\displaystyle:=\int^{x}_{+\infty}K_{n}(z;x,y_{n})f(z;y_{n})dy_{n},$ (2.11)

with

$\displaystyle|K(z;x,y_{n})|\mathchar
13358\relax\frac{\left(\int_{y_{n}}^{+\infty}|m_{0}(\zeta)-1|d\zeta\right)^{n-1}}{(n-1)!}|m_{0}({y_{n}})-1|.$

Then $\mathcal{T}^{n}$ admits

$\displaystyle\|\mathcal{T}^{n}\|\leq C(z)\frac{\|m_{0}-1\|_{L^{1}}^{n}}{n!}.$

Therefore, the series $\sum_{n=1}^{\infty}\mathcal{T}^{n}$ is absolute
convergence such that $\mu^{\pm}_{2}(z;\cdot)$ exists in $L^{\infty}$ with

$\displaystyle\mu^{\pm}_{2}(z;x)=\sum_{n=0}^{\infty}\mathcal{T}^{n}e_{2}(z;x),$

for every $z\in\overline{\mathbb{C}^{-}}\setminus\\{\pm 1\\}$. Analyticity of
$\mu^{\pm}_{2}(x;\cdot)$ in $\mathbb{C}^{-}$ for every $x\in\mathbb{R}$
follows from the absolute and uniform convergence of the Neumann series of
analytic functions in $z$. This completes the proof of Proposition 2.1. ∎

######  Remark 2.1.

Similar to Lemma 1 in [34], although the estimate in (2.51) requires
$\parallel m_{0}-1\parallel_{L^{1}}<1$, real axis $\mathbb{R}$ can be split
into a finite number of subintervals $I_{n}$ such that
$\int_{I_{n}}|m_{0}(s)-1|ds<1$ on each subinterval. Thus, through that
$\mathcal{T}$ is a linear integral operator, Proposition 2.5 can be
accomplished without any restriction on $\|m_{0}-1\|_{L^{1}}<1$.

Applying (2.8) and (2.9) also implies that as $|z|\to\infty$,

$\mu^{\pm}(z;x)=I+\mathcal{O}(1/z),$

with

$\displaystyle\lim_{z\to\infty}z\mu^{\pm}_{12}(z)=\eta,\ \eta=1-1/m_{0}.$
(2.12)

On the other hand, the Jost functions exhibit singular behavior for $z$ near
$\pm 1$. The singular behavior of these solutions at $z=\pm 1$ plays a non-
trivial and unavoidable role in our analysis.

######  Proposition 2.2.

Let $m_{0}-1\in L^{1,1}(\mathbb{R})$. Then $\mu^{\pm}$ have singularities on
$z=\pm 1$ with a special structure

$\displaystyle\mu^{\pm}(z;x)=\frac{i\alpha_{\pm}(x)}{2(z-1)}\left(\begin{array}[]{cc}-1&1\\\
-1&1\end{array}\right)+\mathcal{O}(1),\quad\text{as }z\to 1,$
$\displaystyle\mu^{\pm}(z;x)=\frac{i\alpha_{\pm}(x)}{2(z+1)}\left(\begin{array}[]{cc}-1&-1\\\
1&1\end{array}\right)+\mathcal{O}(1),\quad\text{as }z\to-1,$

where $\alpha_{\pm}(x)\in\mathbb{R}$.

###### Proof.

We take $z$ near $1$ as an example. The property of $z$ near $-1$ can be
obtained from the symmetry immediately. Denote

$\displaystyle\upsilon^{\pm}=(z^{2}-1)D^{-1}\mu^{\pm}$ (2.13)

with $D$ defined in (2.4). Then the integral representations of $\mu^{\pm}$ in
(2.8) and (2.9) give that the two columns of $\upsilon^{\pm}$ admit the
following integral equations,

$\displaystyle\upsilon^{\pm}_{1}(z;x)=\left(\begin{array}[]{cc}(z-i)^{2}/2\\\\[4.0pt]
i(z^{2}+1)/2\end{array}\right)+\int^{x}_{\pm\infty}\frac{m_{0}(s)-1}{2}\Gamma(z,x,s)\upsilon^{\pm}_{1}(z;s)ds,$
(2.16)
$\displaystyle\upsilon^{\pm}_{2}(z;x)=\left(\begin{array}[]{cc}i(z^{2}+1)/2\\\\[4.0pt]
(z-i)^{2}/2\end{array}\right)+\int^{x}_{\pm\infty}\frac{(m_{0}(s)-1)\gamma^{2}(z,x,s)}{2}\Gamma(z,x,s)\upsilon^{\pm}_{2}(z;s)ds,$
(2.19)

where $\Gamma$ is a bounded analytic function with respect to $z$ in a
neighbourhood of $z=\pm 1$,

$\displaystyle\Gamma(z,x,s)$
$\displaystyle=D^{-1}(z)\left(\begin{array}[]{cc}1&0\\\
0&\gamma(z,x,s)\end{array}\right)D(z)\sigma_{3}$ (2.22)
$\displaystyle=\frac{(z-i)^{2}}{2(z^{2}-1)}\left(\begin{array}[]{cc}1-\gamma(z,x,s)d(z)^{2}&-d(z)(1-\gamma(z,x,s))\\\
-d(z)(1-\gamma(z,x,s))&d(z)^{2}-\gamma(z,x,s)\end{array}\right),$ (2.25)

and

$\displaystyle\gamma(z,x,s)=e^{\frac{1}{2}(z-1/z)\int^{s}_{x}m_{0}(\eta)d\eta},\
d(z)=\frac{z^{2}+1}{i(z-i)^{2}}.$ (2.26)

The proof also resembles the analysis of the Jost function in [40] by taking
$\upsilon^{+}_{1}$ as an example. For any matrix function $f(z;x)\in
L^{\infty}((1/2,2)\times\mathbb{R}^{+})$, define a integral operator
$\tilde{\mathcal{T}}$

$\displaystyle\tilde{\mathcal{T}}f(z;x)=\int^{x}_{+\infty}\frac{m_{0}(s)-1}{2}\Gamma(z,x,s)\sigma_{3}f(z;s)ds.$

Noting that

$\displaystyle\Big{|}\frac{1}{z^{2}-1}\left(1-\gamma(z,x,s)\right)\Big{|}=\Big{|}\int_{x}^{s}\frac{e^{\frac{1}{2}(z-1/z)\int^{\xi}_{x}m_{0}(\eta)d\eta}}{m_{0}(\xi)}d\xi\Big{|}\mathchar
13358\relax|s-x|,$

it is ascertained that

$\|(m_{0}(\cdot)-1)\Gamma(z;\cdot)\|_{L^{1}}\leq\|m_{0}-1\|_{L^{1}}+2\|m_{0}-1\|_{L^{1}}^{2}+4\|m_{0}-1\|_{L^{1,1}}.$

Then the similar process as one in Proposition 2.5 implies that
$(I-\tilde{\mathcal{T}})^{-1}$ exists with
$\|(I-\tilde{\mathcal{T}})^{-1}\|\leq\exp\\{\|m_{0}-1\|_{L^{1}}+2\|m_{0}-1\|_{L^{1}}^{2}+4\|m_{0}-1\|_{L^{1,1}}\\}$.
This in turn implies that

$\displaystyle\upsilon^{+}_{1}=(I-\tilde{\mathcal{T}})^{-1}(e_{1}(z-i)^{2}/2+e_{2}i(z^{2}+1)/2).$

Furthermore, when $z=1$, combining (2.13),
$D(1)=\left(\begin{array}[]{cc}1&1\\\ 1&1\end{array}\right)$, and $\det
D(z)=-\frac{z^{2}-2z-1}{2(z^{2}-1)}$, it then follows that

$\displaystyle\mu^{\pm}_{1}(z)=\frac{(\upsilon^{\pm})_{11}(1)+(\upsilon^{\pm})_{21}(1)}{2(z-1)}\left(\begin{array}[]{cc}1\\\
1\end{array}\right)+\mathcal{O}(1).$

Together with the symmetry (2.27), it is concluded that through denoting
$\alpha_{\pm}=i(\upsilon^{\pm})_{11}(1)+i(\upsilon^{\pm})_{21}(1)$, we obtain
the result of Proposition 2.2. ∎

In view of tr$U=0$ in the spectral problem (1.3), it follows from Abel’s
formula that the determinant $\det(\Phi^{\pm}(z;x))=\det(\mu^{\pm}(z;x))=1$.
And from the symmetry of the function $p$ and the matrix function $P$, the
Jost functions $\mu^{\pm}(z;x)$ admit three kinds of reduction conditions

$\mu^{\pm}(z;x)=\sigma_{1}\overline{\mu^{\pm}(\bar{z};x)}\sigma_{1}=\sigma_{2}\mu^{\pm}(-z;x)\sigma_{2}=\sigma_{1}\mu^{\pm}(z^{-1};x)\sigma_{1}.$
(2.27)

On the other hand, $\mu^{\pm}(z;x)$ satisfy a same special asymptotic as $z\to
i$. Based on the approach in [4], if $m_{0}-1\in L^{1}(\mathbb{R})$, then
$\mu^{\pm}(z;x)$ have the following expansions as $z\to i$,

$\displaystyle\mu^{\pm}(z;x)=e^{\frac{1}{2}\int_{x}^{\pm\infty}m_{0}(s)-1ds\sigma_{3}}$
$\displaystyle+(z-i)e^{-\frac{1}{2}\int_{x}^{\pm\infty}m_{0}(s)-1ds\sigma_{3}}\left(\begin{array}[]{cc}0&-\frac{u_{0}(x)+\partial_{x}u_{0}(x)}{2}\\\
\frac{\partial_{x}u_{0}(x)-u_{0}(x)}{2}&0\end{array}\right)+\mathcal{O}\left((z-i)^{2}\right).$
(2.30)

###  2.2. Scattering map

In this subsection, we will derive basic properties on the scattering data of
the Lax pair for the general initial data $m_{0}$ in the weighted Sobolev
spaces with $m_{0}-1\in H^{2,1}(\mathbb{R})\cap H^{1,2}(\mathbb{R})$.

The Jost functions $\mu^{\pm}(z;x)$ satisfy a scattering relation on
$\mathbb{R}$ by

$\mu^{+}(z;x)=\mu^{-}(z;x)e^{p(z;x)\mathaccent 866{\sigma}_{3}}S(z),$ (2.31)

where $S(z)$ is the scattering matrix independent of $x$ and

$S(z)=\left(\begin{array}[]{cc}a(z)&b(z)\\\\[4.0pt]
\overline{b(z)}&\overline{a(z)}\end{array}\right),\hskip
14.22636pt\det(S(z))=|a(z)|^{2}-|b(z)|^{2}=1.$

In addition, the relation formula (2.31) also gives that

$\displaystyle a(z)=\det(\mu_{1}^{+},\mu_{2}^{-}),\
b(z)=e^{-2p}\det(\mu_{2}^{+},\mu_{2}^{-}).$ (2.32)

This thus implies that $a(z)$ is analytical on $\mathbb{C}^{-}$, and
continuous up to $\mathbb{R}$ except at $z=\pm 1$. And $|z|\to\infty$,
$a(z)\to 1$. And $b(z)$ is continuous on $\mathbb{R}$ except at $z=\pm 1$,
where $z=\pm 1$ are the only poles of $a(z)$ and $b(z)$ on $\mathbb{R}$.
Specially, at $z=\pm 1$, in the generic case, $a(z)$ and $b(z)$ have the same
order singularities. While in the non-generic case, since $a(z)$ and $b(z)$
are both continuous at $z=\pm 1$, which can be dealt with in a standard
approach [34], we only consider the generic case in the present paper. The
discrete spectrum. The function $a(z)$ may have zeros on $\mathbb{C}^{-}$. 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 (1.3). This $z_{0}$ is
called the discrete spectrum. In addition, the zeros of $a(z)$ are important
component of scattering data. The following proposition demonstrates that the
zeros of $a(z)$ take a very restricted form.

######  Proposition 2.3.

The zeros of $a(z)$ in $\mathbb{C}^{-}$ are simple and finite, and they are
restricted to the unit circle.

###### Proof.

Since $a(z)$ is analytical and $a(z)\to 1$ as $|z|\to\infty$, then $a(z)$ has
only a finite number of zeros on $\mathbb{C}^{-}$. Rewrite the spatial part in
(1.3) as

$\displaystyle-\frac{i}{m_{0}}\sigma_{2}(2\partial_{x}+\sigma_{3})\Phi=\lambda\Phi,$

which is self-adjoint. So if $z_{0}$ is a zero of $a(z)$, the eigenvalue
$\lambda(z_{0})$ must be real. Together with $|a(z)|\geq 1$ for
$z\in\mathbb{R}$, the zero $z_{0}$ must locate on the unit circle in
$\mathcal{C}^{-}$. Moreover, it is observed from (2.30) and (2.31) that

$\displaystyle a(-i)=\exp\\{\frac{1}{2}\int_{\mathbb{R}}(m_{0}-1)ds\\}\neq 0.$
(2.33)

Thus $z_{0}\neq-i$. It is given by (2.31) that

$\displaystyle a(z)=\det(D)\det(\Phi^{+}_{1},\Phi^{-}_{2}).$ (2.34)

So there exists a constant $b_{0}$ such that

$\displaystyle\Phi^{+}_{1}(z_{0})=b_{0}\Phi^{-}_{2}(z_{0}).$ (2.35)

Differentiating (2.34) with respect to $z$ at $z_{0}$, it appears that

$\displaystyle\partial_{z}a(z_{0})=\det(D(z_{0}))(\det(\partial_{z}\Phi^{+}_{1},\Phi^{-}_{2})(z_{0})+\det(\Phi^{+}_{1},\partial_{z}\Phi^{-}_{2})(z_{0}))\mathchar
13356\relax A_{1}+A_{2}.$

It is inferred from (1.3) that

$\displaystyle\partial_{x}A_{1}=\det(D(z_{0}))\det(\partial_{z}U\cdot\Phi^{+}_{1},\Phi^{-}_{2})(z_{0}),\hskip
8.5359pt\partial_{x}A_{2}=\det(D(z_{0}))\det(\Phi^{+}_{1},\partial_{z}U\cdot\Phi^{-}_{2})(z_{0}),$

where $\partial_{z}U=-\frac{im_{0}}{4}(1-\frac{1}{z^{2}})\sigma_{2}.$ Together
with (2.35), it is deduced that

$\displaystyle
A_{1}=\det(D(z_{0}))(1-z_{0}^{-2})b_{0}\int_{x}^{+\infty}\frac{m_{0}}{4}((\Phi^{-}_{22})^{2}+(\Phi^{-}_{12})^{2})dy,$
$\displaystyle
A_{2}=\det(D(z_{0}))(1-z_{0}^{-2})b_{0}\int^{x}_{-\infty}\frac{m_{0}}{4}((\Phi^{-}_{22})^{2}+(\Phi^{-}_{12})^{2})dy.$

In the above expression, $D(z_{0})$ is a real matrix. Via the symmetry of
$\Psi$, it is ascertained that $\Psi(z_{0}),\ b_{0}\in\mathbb{R}$. It is
thereby inferred that $\Phi(z_{0})\in\mathbb{R}$. Hence,

$\displaystyle\partial_{z}a(z_{0})=\det(D(z_{0}))(1-\frac{1}{z_{0}^{2}})b_{0}\int_{\mathbb{R}}\frac{m_{0}}{4}((\Phi^{-}_{22})^{2}+(\Phi^{-}_{12})^{2})dy\neq
0.$

Simplicity of the zeros of $a$ follows immediately. This completes the proof
of Proposition 2.3. ∎

The zeros of $a(z)$ in the fourth quadrant are denoted as
$z_{1},...,z_{N_{0}}$. Thus we define
$Z_{0}=\\{z_{j},-\bar{z}_{j}\\}_{j=1}^{N_{0}}$ as the set of ${2N_{0}}$ zeros
of $a(z)$ on $\mathbb{C}^{-}$. For convenience, we denote
$z_{j}=-\bar{z}_{j-N_{0}}$, $j={N_{0}}+1,...,2{N_{0}}$. Then
$Z_{0}=\\{z_{j}\\}_{j=1}^{2N_{0}}$.

The reflection coefficient. We now define the reflection coefficient by

$r(z)=\dfrac{b(z)}{a(z)},\ z\in\mathbb{R}.$ (2.36)

From its definition, $r(z)$ is continuous on $\mathbb{R}$ with
$r(1)=-r(-1)=-1$. And $r(z)$ admits symmetries

$\displaystyle r(z)=\overline{r(z^{-1})}=-\overline{r(-z)}.$ (2.37)

This in turn implies from det$(S(z))=1$ that

$\displaystyle|a(z)|^{-2}+|r(z)|^{2}=1,\ z\in\mathbb{R}.$

So $|r(z)|^{2},\ |a(z)|^{-1}\mathchar 13366\relax 1$. Furthermore,
$|a(z)|^{-2}=1-|r(z)|^{2}\neq 0$ for $z\neq\pm 1$.

We analyze the boundedness and integrability of the reflection coefficient
$r(z)$ when $t=0$, which is one of determining factors for the RH problem. The
next lemma gives a property of the direct scattering map: $m_{0}\to r$.

######  Lemma 2.1.

If initial data $m_{0}-1\in H^{2,1}(\mathbb{R})\cap H^{1,2}(\mathbb{R})$, then
the reflection coefficient $r\in H^{2,1}(\mathbb{R})\cap H^{1,2}(\mathbb{R})$.

The proof of Lemma 2.1 is approached via a series of propositions and follows
a similar method to the Section 3 in [40]. By the symmetry of $r$ in (2.37),
it is adduced that

$\displaystyle\int_{\mathbb{R}}|r(z)|^{2}dz=\int_{\mathbb{R}_{-}}|r(z)|^{2}dz+\int_{\mathbb{R}_{+}}|r(z)|^{2}dz=2\int_{\mathbb{R}_{+}}|r(z)|^{2}dz$
$\displaystyle=2\int_{0}^{1}|r(z)|^{2}dz+2\int_{1}^{+\infty}|r(z)|^{2}dz=2\int_{1}^{+\infty}|r(z)|^{2}(1+\frac{1}{z^{2}})dz\mathchar
13366\relax 4\int_{1}^{+\infty}|r(z)|^{2}dz.$

Thus it suffices to consider the property of $r(z)$ at the interval
$[1,+\infty)$. We divide the interval $[1,+\infty)$ into two parts $[1,C]$ and
$(C,\infty)$ for any constant $C>1$. And we first analyze the property of
$r(z)$, namely, the function $\mu^{\pm}(z;x)$ at the interval $z\in(C,\infty)$
for any constant $C>1$. To simplify notation, without loss of generality, we
take $C=2$. In the following demonstration, our attention is now focused on
three function spaces $C_{B}\left(\mathbb{R}^{\pm};L^{2}(2,\infty)\right)$,
$L^{\infty}\left((2,\infty)\times\mathbb{R}^{\pm}\right)$ and
$L^{2}\left((2,\infty)\times\mathbb{R}^{\pm}\right)$ with the norms
$\parallel\cdot\parallel_{C_{B}}$,$\parallel\cdot\parallel_{L^{\infty}}$ and
$\parallel\cdot\parallel_{L^{2}}$ respectively. We start with estimates
obtained with the Fourier theory.

######  Proposition 2.4.

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

$\displaystyle\big{\|}\int_{x}^{+\infty}f(s)e^{2ik\int^{s}_{x}m_{0}(l)dl}ds\big{\|}_{C_{B}}\mathchar
13358\relax\|f\|_{L^{2}}$ (2.38)
$\displaystyle\big{\|}\int_{x}^{+\infty}f(s)e^{2ik\int^{s}_{x}m_{0}(l)dl}ds\big{\|}_{L^{2}\left((C,\infty)\times\mathbb{R}^{\pm}\right)}\mathchar
13358\relax\|f\|_{L^{2,1/2}}.$ (2.39)

###### Proof.

For any $g\in L^{2}$, it transpires that

$\displaystyle\big{|}\int_{\mathbb{R}}\int_{x}^{+\infty}X_{(C,+\infty)}(k)g(k)f(s)e^{2ik\int^{s}_{x}m_{0}(l)dl}dsdk\big{|}=\big{|}\int_{x}^{+\infty}\int_{\mathbb{R}}X_{(C,+\infty)}(k)g(k)f(s)e^{2ik\int^{s}_{x}m_{0}(l)dl}dkds\big{|}$
$\displaystyle=\big{|}\int_{x}^{+\infty}f(s)(X_{(C,+\infty)}g)^{\vee}(\frac{\int^{s}_{x}m_{0}(l)dl}{\pi})ds\big{|}\mathchar
13366\relax(\int_{x}^{+\infty}|f(s)|^{2}ds)^{1/2}\|X_{(C,+\infty)}g\|_{L^{2}},$

where $C$ is a real constant. where $X_{(C,+\infty)}(k)$ is the truncation
function on interval $(C,+\infty)$. Then (2.38) follows directly. It also
gives that

$\displaystyle\int_{\mathbb{R}^{+}}\int_{C}^{+\infty}\big{|}\int_{\mathbb{R}}\int_{x}^{+\infty}f(s)e^{2ik\int^{s}_{x}m_{0}(l)dl}ds\big{|}dkdx\leq\int_{\mathbb{R}^{+}}\int_{x}^{+\infty}|f(s)|^{2}dsdx\mathchar
13358\relax\|f\|_{L^{2,1/2}}^{2}.$

Thus we complete the proof of Proposition 2.4. ∎

######  Proposition 2.5.

If initial data $m_{0}-1\in H^{2,1}(\mathbb{R})\cap H^{1,2}(\mathbb{R})$, then
there exists a unique the Jost function $\mu^{\pm}(z;x)$ with
$\mu^{\pm}(z;x)-I\in C_{B}\left(\mathbb{R}^{\pm};L^{2}(2,\infty)\right)\cap
L^{\infty}\left((2,\infty)\times\mathbb{R}^{\pm}\right)\cap
L^{2}\left((2,\infty)\times\mathbb{R}^{\pm}\right)$ satisfying

$\displaystyle\parallel\mu^{\pm}(z;x)-I\parallel_{L^{\infty}\cap
C_{B}}\mathchar 13366\relax\frac{C(\|m_{0}-1\|_{H^{2,1}\cap
H^{1,2}})}{\underset{x\in\mathbb{R}}{\inf}m_{0}},$
$\displaystyle\parallel\mu^{\pm}(z;x)-I\parallel_{L^{2}}\mathchar 13366\relax
C(\|m_{0}-1\|_{H^{2,1}\cap H^{1,2}}).$

###### Proof.

The proof basically follows the approach in Proposition 2.1. We prove this
proposition only by taking the second column $\mu_{2}^{+}(z;x)$ as an example.
Denote

$\mu_{2}^{+}(z;x)-e_{2}\mathchar 13356\relax
h(z;x)=(h_{1}(z;x),h_{2}(z;x))^{T}.$

The second column of the Volterra integral equation (2.9) can be written as an
operator equation

$\displaystyle h(z;x)=h^{(0)}(z;x)+\mathcal{T}h(z;x),$ (2.40)

where $\mathcal{T}$ is defined in (2.10). And

$\displaystyle h^{(0)}(z;x)$
$\displaystyle=Te_{2}=\int^{x}_{+\infty}i\dfrac{m_{0}(s)-1}{z^{2}-1}\left(\begin{array}[]{cc}\frac{z^{2}+1}{2}e^{\frac{i}{2}(z-1/z)\int^{s}_{x}m_{0}(l)dl}\\\\[4.0pt]
z\end{array}\right)ds\mathchar
13356\relax(h_{1}^{(0)}(z;x),h_{2}^{(0)}(z;x))^{T}.$

It is obvious that $h^{(0)}\in L^{\infty}((2,\infty)\times\mathbb{R}^{+})$.
Considering the first component of $h^{(0)}(z;x)$ and integrating by parts
yields

$\displaystyle h^{(0)}_{1}(z;x)=$
$\displaystyle\dfrac{(z^{2}+1)z}{(z^{2}-1)^{2}}\frac{m_{0}-1}{m_{0}}-\dfrac{(z^{2}+1)z}{(z^{2}-1)^{2}}\int^{x}_{+\infty}e^{\frac{i}{2}(z-1/z)\int^{s}_{x}m_{0}(l)dl}\partial_{s}\left(\frac{m_{0}-1}{m_{0}}\right)ds,$
(2.41)

which implies that

$\displaystyle\sup_{x\in\mathbb{R}^{+}}\int_{\mathbb{R}}|h^{(0)}_{1}(z;x)|^{2}dz\mathchar
13366\relax\frac{\|m_{0}-1\|_{H^{2,1}}}{\underset{x\in\mathbb{R}}{\inf}m_{0}}.$

Obviously, $h^{(0)}_{2}$ has the same estimate. Thus,

$h^{(0)}(z;x)\in C_{B}\left(\mathbb{R}^{+};L^{2}(2,\infty)\right)\cap
L^{\infty}\left((2,\infty)\times\mathbb{R}^{+}\right),$

with $\|h^{(0)}\|_{L^{\infty}\cap C_{B}}\mathchar
13366\relax\frac{2\|m_{0}-1\|_{H^{2,1}}}{\underset{x\in\mathbb{R}}{\inf}m_{0}}$.
Recall the change of variable $k=\frac{1}{4}\left(z-\frac{1}{z}\right)$. When
$z\in(2,+\infty)$, we have $k\in(3/8,+\infty)$. Note that
$\frac{1}{4}|dz|\mathchar 13366\relax|dk|\mathchar
13366\relax\frac{5}{16}|dz|$. Then for any function $f(z)$ well-defined in
$f(z(k))$, it follows that

$f(z)\in L^{2}(2,+\infty)\Longleftrightarrow f(z(k))\in L^{2}(3/8,+\infty).$

Under this change of variable, $h^{(0)}\in
L^{2}\left((2,\infty)\times\mathbb{R}^{+}\right)$ follows from Proposition 2.4
with $\|h^{(0)}\|_{L^{2}}\leq C(\|m_{0}-1\|_{H^{2,1}\cap H^{1,2}})$.

The next step is to prove that $\mathcal{T}$ is a bounded linear operator on
$C_{B}\left(\mathbb{R}^{+};L^{2}(2,\infty)\right)\cap
L^{\infty}\left((2,\infty)\times\mathbb{R}^{+}\right)\cap
L^{2}\left((2,\infty)\times\mathbb{R}^{+}\right)$. In fact, for any
$f=(f_{1},f_{2})^{T}\in
L^{\infty}\left((2,\infty)\times\mathbb{R}^{+}\right)$, it appears that

$\displaystyle|\mathcal{T}f(z;x)|$
$\displaystyle=\Big{|}\int^{x}_{+\infty}\frac{m_{0}(s)-1}{z^{2}-1}\left(\begin{array}[]{c}e^{\frac{i}{2}(z-1/z)\int^{s}_{x}m_{0}(l)dl}(-zf_{1}(z;s)+\frac{z^{2}+1}{2}f_{2}(z;s))\\\\[4.0pt]
zf_{2}(z;s)-\frac{z^{2}+1}{2}f_{1}(z;s)\end{array}\right)ds\Big{|}$ (2.44)
$\displaystyle\mathchar 13366\relax\parallel
f\parallel_{L^{\infty}}\int^{+\infty}_{x}|m_{0}-1|ds\mathchar
13366\relax\parallel m_{0}-1\parallel_{L^{1}}\parallel
f\parallel_{L^{\infty}}.$ (2.45)

For $f\in C_{B}\left(\mathbb{R}^{+};L^{2}(2,\infty)\right)$, it is inferred
that

$\displaystyle\left(\int_{2}^{+\infty}|\mathcal{T}f(z;x)|^{2}dz\right)^{1/2}$
$\displaystyle\mathchar
13366\relax\int^{+\infty}_{x}\left(\int_{2}^{+\infty}\Big{|}\frac{m_{0}(s)-1}{z^{2}-1}\left(\begin{array}[]{c}e^{\frac{i}{2}(z-1/z)\int^{s}_{x}m_{0}(l)dl}(-zf_{1}(z;s)+\frac{z^{2}+1}{2}f_{2}(z;s))\\\\[4.0pt]
zf_{2}(z;s)-\frac{z^{2}+1}{2}f_{1}(z;s)\end{array}\right)\Big{|}^{2}dz\right)^{1/2}ds$
(2.48) $\displaystyle\mathchar
13366\relax\int^{+\infty}_{x}|m_{0}(s)-1|\left(\int_{2}^{+\infty}(|f_{1}(z;s)|+|f_{2}(z;s)|)^{2}dz\right)^{1/2}ds\mathchar
13366\relax\parallel m_{0}-1\parallel_{L^{1}}\parallel f\parallel_{C_{B}}.$
(2.49)

For $f\in L^{2}\left((2,\infty)\times\mathbb{R}^{+}\right)$, it follows that

$\displaystyle\left(\int_{\mathbb{R}^{+}}\int_{2}^{+\infty}|\mathcal{T}f(z;x)|^{2}dzdx\right)^{1/2}$
$\displaystyle\mathchar
13358\relax\int_{\mathbb{R}^{+}}\left(\int^{+\infty}_{x}|m_{0}(s)-1|\int_{2}^{+\infty}(|f_{1}(z;s)|+|f_{2}(z;s)|)^{2}dz)^{1/2}ds\right)dx$
$\displaystyle\mathchar
13366\relax\left(\int_{\mathbb{R}^{+}}(\int^{s}_{0}|m_{0}(s)-1|\int_{2}^{+\infty}(|f_{1}(z;s)|+|f_{2}(z;s)|)^{2}dzdx)^{1/2}ds\right)^{2}$
$\displaystyle\mathchar 13366\relax\parallel
m_{0}-1\parallel_{L^{2,1/2}}\parallel f\parallel_{L^{2}}.$ (2.50)

In general, for $\mathcal{T}^{n}$ defined in (2.11), $n=2,3,...$, it admits

$\displaystyle|\mathcal{T}^{n}f(z;x)|\mathchar 13366\relax\parallel
f\parallel_{L^{\infty}}\int^{+\infty}_{x}\frac{\left(\int_{s}^{+\infty}|m_{0}(\zeta)-1|d\zeta\right)^{n-1}}{(n-1)!}|m_{0}(s)-1|ds\mathchar
13366\relax\frac{\parallel m_{0}-1\parallel_{L^{1}}^{n}}{n!}\parallel
f\parallel_{L^{\infty}},$

and

$\displaystyle\left(\int_{2}^{+\infty}|\mathcal{T}^{n}f(z;x)|^{2}dz\right)^{1/2}\mathchar
13366\relax\int^{+\infty}_{x}\frac{\left(\int_{s}^{+\infty}|m_{0}(\zeta)-1|d\zeta\right)^{n-1}}{(n-1)!}|m_{0}(s)-1|ds\parallel
f\parallel_{C_{B}}\mathchar 13366\relax\frac{\parallel
m_{0}-1\parallel_{L^{1}}^{n}}{n!}\parallel f\parallel_{C_{B}}.$

So the operator $(I-\mathcal{T})^{-1}$ exists on
$C_{B}\left(\mathbb{R}^{\pm};L^{2}(2,\infty)\right)\cap
L^{\infty}\left((2,\infty)\times\mathbb{R}^{\pm}\right)$ and admits

$\displaystyle\parallel(I-\mathcal{T})^{-1}\parallel\mathchar 13366\relax
e^{\parallel m_{0}-1\parallel_{L^{1}}}.$ (2.51)

And for the space $L^{2}\left((2,\infty)\times\mathbb{R}^{\pm}\right)$, a
similar estimate gives that

$\displaystyle\|\mathcal{T}^{n}\|\mathchar 13366\relax\frac{\parallel
m_{0}-1\parallel_{L^{1}}^{n-1}}{(n-1)!}\|m_{0}-1\|{L^{2,1/2}},$ (2.52)

which leads to

$\displaystyle\parallel(I-\mathcal{T})^{-1}\parallel\mathchar 13366\relax
1+e^{\parallel m_{0}-1\parallel_{L^{1}}}\|m_{0}-1\|{L^{2,1/2}}.$

Finally, it follows from the operator equation in (2.40) that

$h(z;x)=(I-\mathcal{T})^{-1}h^{(0)}\ (z;x)\in
C_{B}\left(\mathbb{R}^{+};L^{2}(2,\infty)\right)\cap
L^{\infty}\left((2,\infty)\times\mathbb{R}^{+}\right)\cap
L^{2}\left((2,\infty)\times\mathbb{R}^{+}\right).$

This completes the proof of Proposition 2.5. ∎

Next, in order to obtain the property of $z$-derivative, we establish the
following estimate of the Jost function $\mu^{\pm}(z;x)$ with initial data
$m_{0}(x)-1$.

######  Proposition 2.6.

Let $m_{0}-1\in H^{2,1}(\mathbb{R})\cap H^{1,2}(\mathbb{R})$. Then the
$z$-derivatives of $\mu^{\pm}(z;x)$ admit the estimates

$\displaystyle\big{\|}\partial_{z}\mu^{\pm}(z;x)\big{\|}_{C_{B}},\
\big{\|}\partial_{z}^{2}\mu^{\pm}(z;x)\big{\|}_{C_{B}}\mathchar 13366\relax
C(\|m_{0}-1\|_{H^{2,1}\cap H^{1,2}},\underset{x\in\mathbb{R}}{\inf}m_{0}).$
(2.53)

###### Proof.

By taking derivative in (2.40) with respect to $z$, it is found that

$\displaystyle\partial_{z}h(z;x)=h^{(1)}(z;x)+\mathcal{T}(\partial_{z}h)(z;x),\
\partial_{z}^{2}h(z;x)=h^{(2)}(z;x)+\mathcal{T}(\partial_{z}^{2}h)(z;x)$
(2.54)

where

$\displaystyle
h^{(1)}(z;x)=\partial_{z}h^{(0)}(z;x)+(\partial_{z}\mathcal{T})h(z;x),$
$\displaystyle
h^{(2)}(z;x)=\partial_{z}^{2}h^{(0)}(z;x)+(\partial_{z}\mathcal{T})(\partial_{z}h)(z;x)+(\partial_{z}^{2}\mathcal{T})h(z;x).$

For $h^{(1)}$, $\partial_{z}\mathcal{T}$ is a linear integral operator from
$L^{2}\left((2,\infty)\times\mathbb{R}^{+}\right)$ to
$C_{B}\left(\mathbb{R}^{+};L^{2}(2,\infty)\right)$ defined by

$\displaystyle\partial_{z}\mathcal{T}f(z;x)$
$\displaystyle=\int^{x}_{+\infty}\partial_{z}K(z;x,s)f(z;s)ds,$ (2.55)

where the concrete expression of $\partial_{z}K$ is

$\displaystyle\partial_{z}K(z;x,s)=$
$\displaystyle-\frac{1}{2}(1+1/z^{2})(\int^{s}_{x}m_{0}(l)dl)e^{\frac{i}{2}(z-1/z)\int^{s}_{x}m_{0}(l)dl}(m_{0}(s)-1)\left(\begin{array}[]{cc}-\frac{z}{z^{2}-1}&\frac{z^{2}+1}{2(z^{2}-1)}\\\
0&0\end{array}\right)$ $\displaystyle+i(m_{0}(s)-1){\rm
diag}(e^{\frac{i}{2}(z-1/z)\int^{s}_{x}m_{0}(l)dl},1)(i\partial_{z}(\frac{z^{2}+1}{2(z^{2}-1)})\sigma_{2}-\partial_{z}(\frac{z}{z^{2}-1})\sigma_{3}).$

For $z\in(2,\infty)$, direct calculation gives that

$\displaystyle|\partial_{z}K(z;x,s)|\mathchar
13358\relax|\int^{s}_{x}m_{0}(l)dl||m_{0}(s)-1|+|m_{0}(s)-1|.$

Thus for $f\in L^{2}\left((2,\infty)\times\mathbb{R}^{+}\right)$, it appears
that

$\displaystyle\|\partial_{z}\mathcal{T}f\|_{C_{B}}=\sup_{x\in\mathbb{R}^{+}}(\int_{2}^{+\infty}\big{|}\int^{x}_{+\infty}\partial_{z}K(z;x,s)f(z;s)ds\big{|}^{2}dz)^{1/2}$
$\displaystyle\mathchar
13366\relax\sup_{x\in\mathbb{R}^{+}}\int^{x}_{+\infty}(\int_{2}^{+\infty}|\partial_{z}K(z;x,s)f(z;s)|^{2}dz)^{1/2}ds$
$\displaystyle\mathchar
13358\relax\sup_{x\in\mathbb{R}^{+}}\int^{x}_{+\infty}(|\int^{s}_{x}m_{0}(l)dl||m_{0}(s)-1|+|m_{0}(s)-1|)(\int_{2}^{+\infty}|f(z;s)|^{2}dz)^{1/2}ds$
$\displaystyle\leq(\|m_{0}(s)-1\|_{L^{2,1}}+\|m_{0}(s)-1\|_{L^{2}})\|f\|_{L^{2}},$

which gives the boundedness of $\partial_{z}\mathcal{T}$. Thus it follows that
$h^{(1)}\in C_{B}\left(\mathbb{R}^{+};L^{2}(2,\infty)\right)$. By following
the similar proof of Proposition 2.5, the estimate of $\partial_{z}\mu^{\pm}$
in Proposition 2.6 is obtained. And the proof of the estimate of
$\partial_{z}^{2}\mu^{\pm}$ can be completed in the same way. ∎

By virtue of (2.8) and (2.31), the scattering data $b(z)$ is expressed with

$\displaystyle b(z)=$
$\displaystyle\int_{\mathbb{R}}e^{-\frac{i}{2}(z-1/z)(\int_{s}^{+\infty}(m_{0}(l)-1)dl-s)}\frac{i(m_{0}(s)-1)}{2(z^{2}-1)}\left(2z\mu^{+}_{12}(z;s)-(z^{2}+1)\mu^{+}_{22}(z;s)\right)ds,$
(2.56)

which reveals the following result.

######  Proposition 2.7.

Let $m_{0}(x)-1\in H^{2,1}(\mathbb{R})\cap H^{1,2}(\mathbb{R})$. Then $b(z)\in
H^{1,2}(2,\infty)\cap H^{2,1}(2,\infty)$ and

$\|b\|_{H^{1,2}(2,\infty)\cap H^{2,1}(2,\infty)}\mathchar 13366\relax
C(\|m_{0}-1\|_{H^{2,1}\cap H^{1,2}},\underset{x\in\mathbb{R}}{\inf}m_{0}).$

###### Proof.

(2.56) can be rewritten in the form

$\displaystyle b(z)=$
$\displaystyle\int_{\mathbb{R}}e^{-\frac{i}{2}(z-1/z)(\int_{s}^{+\infty}(m_{0}(l)-1)dl-s)}\frac{i(m_{0}(s)-1)}{2(z^{2}-1)}\left(2z\mu^{+}_{12}(z;s)-(z^{2}+1)(\mu^{+}_{22}(z;s)-1)\right)ds$
$\displaystyle-\int_{\mathbb{R}}e^{-\frac{i}{2}(z-1/z)(\int_{s}^{+\infty}(m_{0}(l)-1)dl-s)}\frac{z^{2}+1}{2(z^{2}-1)}i(m_{0}(s)-1)ds.$
(2.57)

Applying Proposition 2.5 thus yields the $L^{2}$-property of the first
integral. As for the second integral, it suffices to give estimate of the
integral
$\int_{\mathbb{R}}e^{-\frac{i}{2}(z-1/z)(\int_{s}^{+\infty}(m_{0}(l)-1)dl-s)}(m_{0}-1)ds$.

Recall the change of variable $k=\frac{1}{4}\left(z-\frac{1}{z}\right)$.
Consequently, it suffices to show

$\int_{\mathbb{R}}e^{-2i\cdot(\int_{s}^{+\infty}(m_{0}(l)-1)dl-s)}(m_{0}-1)ds\in
L^{2}(3/8,+\infty).$

The proof resemble the one in Proposition 2.4. For any $f(k)\in
L^{2}(3/8,+\infty)$, it transpires that

$\displaystyle\Big{|}\int_{3/8}^{+\infty}f(k)\int_{\mathbb{R}}e^{-2ik(\int_{s}^{+\infty}(m_{0}(l)-1)dl-s)}(m_{0}(s)-1)dsdk\Big{|}$
$\displaystyle=\Big{|}\int_{\mathbb{R}}(m_{0}(s)-1)\mathaccent
866{f}(\frac{1}{\pi}\int_{s}^{+\infty}(m_{0}(l)-1)dl-\frac{s}{\pi})ds\Big{|}$
$\displaystyle\mathchar 13366\relax\parallel
m_{0}-1\parallel_{L^{2}}\Big{\|}\mathaccent
866{f}(\frac{1}{\pi}\int_{s}^{+\infty}(m_{0}(l)-1)dl-\frac{s}{\pi})\Big{\|}_{L^{2}},\mathchar
13358\relax\parallel m_{0}-1\parallel_{L^{2}}\parallel\mathaccent
866{f}\parallel_{L^{2}}^{2}.$

This in turn implies that $b\in L^{2}(2,+\infty)$.

Our attention is now turned to the estimate of weighted integral property of
the function $b$. Applying integration by parts, $zb(z)$ is rewritten as

$\displaystyle zb(z)=$
$\displaystyle-\frac{z^{3}}{(z^{2}-1)^{2}}\int_{\mathbb{R}}e^{-\frac{i}{2}(z-1/z)(\int_{s}^{+\infty}(m_{0}(l)-1)dl-s)}\partial_{s}(\frac{2(m_{0}-1)}{m_{0}}\mu^{+}_{12})ds$
$\displaystyle+\frac{(z^{2}+1)z^{2}}{(z^{2}-1)^{2}}\int_{\mathbb{R}}e^{-\frac{i}{2}(z-1/z)(\int_{s}^{+\infty}(m_{0}(l)-1)dl-s)}\partial_{s}(\frac{m_{0}-1}{m_{0}}(\mu^{+}_{22}-1))ds$
$\displaystyle+\frac{(z^{2}+1)z^{2}}{(z^{2}-1)^{2}}\int_{\mathbb{R}}e^{-\frac{i}{2}(z-1/z)(\int_{s}^{+\infty}(m_{0}(l)-1)dl-s)}\partial_{s}(\frac{m_{0}-1}{m_{0}})ds.$
(2.58)

By substituting (2.7) into the partial derivative in above expression, it is
deduced that

$\displaystyle\partial_{s}(\frac{2(m_{0}-1)}{m_{0}}\mu^{+}_{12})=-2\partial_{s}(\frac{1}{m_{0}})\mu^{+}_{12}+\frac{i(m_{0}-1)}{2}(z-\frac{1}{z})\mu^{+}_{12}+\frac{2i(m_{0}-1)^{2}}{m(z^{2}-1)}(z\mu^{+}_{12}-\frac{z^{2}+1}{2}\mu^{+}_{22}),$
$\displaystyle\partial_{s}(\frac{m_{0}-1}{m_{0}}(\mu^{+}_{22}-1))=\partial_{s}(\frac{m_{0}-1}{m_{0}})(\mu^{+}_{22}-1)+\frac{m_{0}-1}{m_{0}}(P\mu^{+})_{22},$

which follows from Proposition 2.5 that the both expressions in the above
belong to $L^{2}$, where $(P\mu^{+})_{22}$ is the $(2,2)$-entry element of
matrix $P\mu^{+}$. Specially, equations above give that the $\mu^{+}_{12}$
item in the integral representation (2.58) of $zb(z)$ has

$\displaystyle-\frac{3(z^{2}-1)^{2}}{2z^{2}}\int_{\mathbb{R}}e^{-\frac{i}{2}(z-1/z)(\int_{s}^{+\infty}(m_{0}(l)-1)dl-s)}\frac{i(m_{0}-1)z\mu^{+}_{12}}{z^{2}-1}ds$
$\displaystyle=$
$\displaystyle\int_{\mathbb{R}}e^{-\frac{i}{2}(z-1/z)(\int_{s}^{+\infty}(m_{0}(l)-1)dl-s)}(\frac{2i(m_{0}-1)^{2}}{m_{0}(z^{2}-1)}(z\mu^{+}_{12}-\frac{z^{2}+1}{2}\mu^{+}_{22})-2\partial_{s}(\frac{1}{m})\mu^{+}_{12})ds,$

which belongs to $L^{2}(2,\infty)$. Hence in view of (2.58), it is
accomplished that $\langle\cdot\rangle b\in L^{2}(2,\infty)$. Similarly,
continuously using integration by parts and the weight transform to
$x$-derivative of $m_{0}$ again leads to $\langle\cdot\rangle^{2}b\in
L^{2}(\mathbb{R})$. For the $z$-derivative of $b(z)$, the estimate is obtained
in the same way. Here, the essential tool is also that the norm of
$z$-derivative of $b(z)$ can be controlled by the weighted integral norm of
$m_{0}-1$ from integration by parts and Proposition 2.4. In consequence, we
obtain the desired result as indicated in Proposition 2.7. ∎

Similarly, from (2.8) and (2.31), the scattering data $a(z)$ could be
expressed by

$\displaystyle a(z)=$ $\displaystyle
1-\int_{\mathbb{R}}\frac{i(m_{0}(s)-1)}{2(z^{2}-1)}(2z\mu^{+}_{11}(z;s)-(z^{2}+1)\mu^{+}_{21}(z;s))ds.$
(2.59)

According to Propositions 2.5 and 2.6, we have the following result.

######  Proposition 2.8.

Let $m_{0}-1\in H^{2,1}(\mathbb{R})\cap H^{1,2}(\mathbb{R})$. Then $a(z)$ is
continuous up to the real line $\mathbb{R}$, except at $z=\pm 1$. Moreover,
$a\in W^{2,\infty}(2,+\infty)$ with $|a(z)|\geq 1$ on $\mathbb{R}$.

The following proposition gives the property of $r(z)$ near $z=\pm 1$.

######  Proposition 2.9.

Let $m_{0}-1\in H^{2,1}(\mathbb{R})\cap H^{1,2}(\mathbb{R})$. Then $r$,
$\partial_{z}r$ and $\partial_{z}^{2}r$ are bounded near $z=\pm 1$.

###### Proof.

The proof of this proposition resemble as the one in the case $(2,+\infty)$.
In view of the definition of $r$ in (2.36), it is necessary to analyze the
integral property of the function $b$. Since $b$ can be represented by the
Jost function $\mu^{\pm}$, the strategy is to first estimate $\mu^{\pm}$ in
this proof. It is sufficient to prove this proposition near $z=1$. As shown in
Proposition 2.2, $\mu^{\pm}$ have singularities on $z=\pm 1$. Furthermore, by
the symmtry of $\mu^{\pm}$ in (2.27), as $z\to 1$,

$\displaystyle\mu^{\pm}(z;x)=\frac{i\alpha_{\pm}(x)}{2(z-1)}\left(\begin{array}[]{cc}-1&1\\\
-1&1\end{array}\right)+\mu^{\pm,(0)}+\mathcal{O}(z-1),$ (2.62)

with $\mu^{\pm,(0)}_{11}=\mu^{\pm,(0)}_{22}\in\mathbb{R}$,
$\mu^{\pm,(0)}_{21}=\mu^{\pm,(0)}_{12}\in\mathbb{R}$. From (2.32), it appears
that

$\displaystyle a=\mu^{+}_{11}\mu^{-}_{22}-\mu^{+}_{21}\mu^{-}_{12},\ \
b=e^{-2p}(\mu^{+}_{12}\mu^{-}_{22}-\mu^{+}_{22}\mu^{-}_{12}).$

Then in view of (2.62), as $z\to 1$,

$\displaystyle a(z)=\frac{ic^{(0)}}{2(z-1)}+\mathcal{O}(1),\
b(z)=\frac{-ic^{(0)}}{2(z-1)}+\mathcal{O}(1),$

where
$c^{(0)}=\alpha_{-}(\mu^{+,(0)}_{11}-\mu^{+,(0)}_{21})+\alpha_{+}(\mu^{-,(0)}_{12}-\mu^{-,(0)}_{22})$.
Therefore, the definition of $r$ gives that $r$ is bounded at $z=1$. To
analyze the $z$-derivative property of $\mu^{\pm}$ near $z=\pm 1$, the
$z$-derivative of $\upsilon^{\pm}$ needs to be considered. To see this, it is
found from (2.16) that

$\displaystyle\partial_{z}\upsilon_{1}^{\pm}(z;x)=$
$\displaystyle\left(\begin{array}[]{cc}z-i\\\\[4.0pt]
iz\end{array}\right)+\int^{x}_{\pm\infty}\frac{m_{0}(s)-1}{2}\partial_{z}\Gamma(z,x,s)\sigma_{3}\upsilon^{\pm}(z;s)ds+\tilde{\mathcal{T}}(\partial_{z}\upsilon^{\pm})(z;x),$
(2.65)

where the calculation gives that $\partial_{z}\Gamma$ can be divided into two
parts by

$\displaystyle\partial_{z}\Gamma(z,x,s)=\frac{1}{z^{2}-1}\Gamma^{(1)}(z,x,s)+\Gamma^{(0)}(z,x,s).$

$\Gamma^{(1)}$ and $\Gamma^{(0)}$ are both analytic and bounded matrix
functions with respect to $z$ near $z=\pm 1$ whose expressions are omitted for
brevity. Apply a similar approach as $z\to 1$ to get

$\displaystyle(z-1)\partial_{z}\upsilon_{1}^{\pm}=\frac{\upsilon_{z,1}^{\pm,(-1)}(x)}{2}+\mathcal{O}(z-1).$
(2.66)

Therefore, it thereby follows again from the symmetry (2.27) and

$\displaystyle\partial_{z}\mu^{\pm}(z;x)=-\frac{2z}{(z^{2}-1)^{2}}D(z)\upsilon^{\pm}(z;x)+\frac{1}{z^{2}-1}D(z)\partial_{z}\upsilon^{\pm}(z;x)+\frac{1}{z^{2}-1}\partial_{z}D(z)\upsilon^{\pm}(z;x),$

that as $z\to 1$

$\displaystyle\partial_{z}\mu^{\pm}(z;x)=\frac{i\alpha_{\pm}^{(1)}(x)}{4(z-1)^{2}}\left(\begin{array}[]{cc}1&-1\\\
1&-1\end{array}\right)+\frac{\mu_{z}^{\pm,(1)}(x)}{z-1}+\mathcal{O}(1),$
(2.69)

with

$\displaystyle\alpha_{\pm}^{(1)}=-i(2\upsilon^{\pm}_{11}+2\upsilon^{\pm}_{21}+\upsilon_{z,11}^{\pm,(-1)}+\upsilon_{z,21}^{\pm,(-1)})\in\mathbb{R},$
$\displaystyle\mu_{z,11}^{\pm,(-1)}=\mu_{z,22}^{\pm,(-1)}\in i\mathbb{R},\ \
\mu_{z,12}^{\pm,(-1)}=\mu_{z,21}^{\pm,(-1)}\in i\mathbb{R}.$

Taking $z$-derivative of $a$ and $b$ in (2.32), it is deduced that

$\displaystyle\partial_{z}a=\partial_{z}\mu^{+}_{11}\mu^{-}_{22}+\mu^{+}_{11}\partial_{z}\mu^{-}_{22}-\partial_{z}\mu^{+}_{21}\mu^{-}_{12}-\mu^{+}_{21}\partial_{z}\mu^{-}_{12},$
$\displaystyle\partial_{z}b=\partial_{z}e^{-2p}(\mu^{+}_{12}\mu^{-}_{22}-\mu^{+}_{22}\mu^{-}_{12})+e^{-2p}(\partial_{z}\mu^{+}_{12}\mu^{-}_{22}+\mu^{+}_{12}\partial_{z}\mu^{-}_{22}-\partial_{z}\mu^{+}_{22}\mu^{-}_{12}-\mu^{+}_{22}\partial_{z}\mu^{-}_{12}).$

Thus by (2.62) and (2.69), as $z\to 1$, we have

$\displaystyle\partial_{z}a=\frac{ic^{(1)}}{4(z-1)^{2}}+\mathcal{O}\left((z-1)^{-1}\right),\
\partial_{z}b=\frac{-ic^{(1)}}{4(z-1)^{2}}+\mathcal{O}\left((z-1)^{-1}\right),$
(2.70)

where
$c^{(1)}=\alpha_{-}^{(1)}(\mu^{+,(0)}_{11}-\mu^{+,(0)}_{21})+\alpha_{+}^{(1)}(\mu^{-,(0)}_{12}-\mu^{-,(0)}_{22})+2\alpha_{-}(\mu^{+,(-1)}_{11}-\mu^{+,(-1)}_{21})+2\alpha_{+}(\mu^{-,(-1)}_{12}-\mu^{-,(-1)}_{22})$.
Then $a\partial_{z}b-b\partial_{z}a=\mathcal{O}(z-1)^{-2}$ as $z\to 1$. This
also gives the boundedness of

$\displaystyle\partial_{z}r=\frac{1}{a^{2}}(a\partial_{z}b-b\partial_{z}a).$

The analysis of $\partial_{z}^{2}r$ resembles the methods shown above, which
is omitted here. This completes the proof of Proposition 2.9. ∎

###### Proof of Lemma 2.1.

In view of the definition of $r$ in (2.36), the result of Lemma 2.1 is
obtained directly from Proposition 2.8 and Proposition 2.7. ∎

The norming constant. We denote the norming constant $c_{j}$ by

$\displaystyle c_{j}=\frac{1}{b_{j}a^{\prime}(z_{j})},\ j=1,...,{2N_{0}},$
(2.71)

where $b_{j}$ is the ratio of $\mu^{+}_{1}$ and $\mu^{-}_{2}$ such that

$\displaystyle\mu^{+}_{1}(z_{j})=b_{j}e^{-2p(z_{j})}\mu^{-}_{2}(z_{j}).$
(2.72)

The symmetry of $\mu^{\pm}$ implies that $b_{j}\in\mathbb{R}$ and
$\mu^{+}_{1}(-\bar{z}_{j})=b_{j}e^{-2p(-\bar{z}_{j})}\mu^{-}_{2}(-\bar{z}_{j})$.
Thus $c_{j}=c_{j+{N_{0}}}$.

###  2.3. The RH problems

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

$\displaystyle
M_{l}(z;x)=\left\\{\begin{array}[]{ll}\left(\frac{\mu^{-}_{1}(z;x)}{a^{*}(z)},\mu^{+}_{2}(z;x)\right),&\text{as
}z\in\mathbb{C}^{+},\\\\[12.0pt]
\left(\mu^{+}_{1}(z;x),\frac{\mu^{-}_{2}(z;x)}{a(z)}\right),&\text{as
}z\in\mathbb{C}^{-},\\\ \end{array}\right.$ (2.75) $\displaystyle
M_{r}(z;x)=\left\\{\begin{array}[]{ll}\left(\mu^{-}_{1}(z;x),\frac{\mu^{+}_{2}(z;x)}{a^{*}(z)}\right),&\text{as
}z\in\mathbb{C}^{+},\\\\[12.0pt]
\left(\frac{\mu^{+}_{1}(z;x)}{a(z)},\mu^{-}_{2}(z;x)\right),&\text{as
}z\in\mathbb{C}^{-}.\\\ \end{array}\right.$ (2.78)

In order to construct the RH problem only depending explicitly on the
scattering data, we introduce the space variable

$y(x)=x+\int_{+\infty}^{x}(m(s)-1)ds.$ (2.79)

Under this new scaling, when $t=0$, the functions $M_{l}$ and $M_{r}$ denoting
in (2.75) and (2.78) respectively become

$\displaystyle M_{l}(z):=M_{l}(z;y(x)),\ \ M_{r}(z):=M_{r}(z;y(x))$ (2.80)

Obviously,

$M_{l}(z;y)\to I,\ \ y\to\infty,\ \ {\rm and}\ \ M_{r}(z;y)\to I,\ \
y\to-\infty,$

which inspires us to consider $M_{l}(z;y)$ on $y\in[0,+\infty)$ and
$M_{r}(z;y)$ on $y\in(-\infty,0]$. And from the property of $\mu^{\pm}(z;x)$,
(2.31) and (2.72), it then is inferred that $M_{l}(z)$ solves the following RH
problem,

######  The RH problem 1.

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

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

$\bullet$ Symmetry:
$M_{l}(z)=\sigma_{1}\overline{M_{l}(\bar{z})}\sigma_{1}=\sigma_{2}M_{l}(-z)\sigma_{2}=\sigma_{1}M_{l}(z^{-1})\sigma_{1}$;

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

$[M_{l}]_{+}(z)=[M_{l}]_{-}(z)V_{l}(z),\hskip 14.22636ptz\in\mathbb{R},$
(2.81)

where

$V_{l}(z)=\left(\begin{array}[]{cc}1-|r(z)|^{2}&r(z)e^{2\theta(z)}\\\
-\bar{r}(z)e^{-2\theta(z)}&1\end{array}\right),$ (2.82)

with

$\displaystyle\theta(z)=p(z;x(y))=-\frac{1}{4}i\left(z-z^{-1}\right)y;$ (2.83)

$\bullet$ Asymptotic behavior:

$\displaystyle M_{l}(z)=I+\mathcal{O}(z^{-1}),\hskip
14.22636pt|z|\rightarrow\infty;$ (2.84)

$\bullet$ Singularity: $M_{l}(z)$ has singularities at $z=\pm 1$ with

$\displaystyle
M_{l}(z)=\frac{i\alpha_{+}(x)}{2(z-1)}\left(\begin{array}[]{cc}-c&1\\\
-c&1\end{array}\right)+\mathcal{O}(1),\ z\to 1\ \text{ in }\ \mathbb{C}^{+},$
(2.87) $\displaystyle
M_{l}(z)=\frac{i\alpha_{+}(x)}{2(z+1)}\left(\begin{array}[]{cc}-c&-1\\\
c&1\end{array}\right)+\mathcal{O}(1),\ z\to-1\ \text{ in }\ \mathbb{C}^{+},$
(2.90)

where $c=0$ in generic case, while $c\neq 0$ in non-generic case;

$\bullet$ Residue condition: $M_{l}(z)$ has simple poles at each point in
$Z_{0}\cup\bar{Z_{0}}$ with

$\displaystyle\operatorname*{Res}_{z=z_{j}}M_{l}(z)=\lim_{z\to
z_{j}}M_{l}(z)\left(\begin{array}[]{cc}0&c_{j}e^{2\theta(z_{j})}\\\
0&0\end{array}\right),$ (2.93)
$\displaystyle\operatorname*{Res}_{z=\bar{z}_{j}}M_{l}(z)=\lim_{z\to\bar{z}_{j}}M_{l}(z)\left(\begin{array}[]{cc}0&0\\\
\bar{c}_{j}e^{-2\theta(\bar{z}_{j})}&0\end{array}\right),$ (2.96)

where $2\theta(z_{j})=2\theta(\bar{z}_{j})=y$Im$z_{j}$.

The set $(r,\\{z_{j},c_{j}\\}_{j=1}^{2N_{0}})$ as the scattering data is
crucial for construction of the RH problem 1. Analogously, it is also found
that $M_{r}(z)$ solves the following RH problem,

######  The RH problem 2.

Find a matrix-valued function $M_{r}(z)=M_{r}(z;y)$ which satisfies the same
analyticity, symmetry and asymptotic behavior as the RH problem 1, and

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

$[M_{r}]_{+}(z)=[M_{r}]_{-}(z)V_{r}(z),\hskip 14.22636ptz\in\mathbb{R},$
(2.97)

where

$V_{r}(z)=\left(\begin{array}[]{cc}1&\tilde{r}(z)e^{2\theta(z)}\\\
-\bar{\tilde{r}}(z)e^{-2\theta(z)}&1-|\tilde{r}(z)|^{2}\end{array}\right),\ \
\tilde{r}(z)=\dfrac{b(z)}{a^{*}(z)}.$ (2.98)

$\bullet$ Singularity: $M_{r}(z)$ has singularities at $z=\pm 1$ with

$\displaystyle
M_{r}(z)=\frac{i\alpha_{+}(x)}{2(z-1)}\left(\begin{array}[]{cc}-1&c\\\
-1&c\end{array}\right)+\mathcal{O}(1),\ z\to 1\ \text{ in }\ \mathbb{C}^{+},$
(2.101) $\displaystyle
M_{r}(z)=\frac{i\alpha_{+}(x)}{2(z+1)}\left(\begin{array}[]{cc}-1&-c\\\
1&c\end{array}\right)+\mathcal{O}(1),\ z\to-1\ \text{ in }\ \mathbb{C}^{+},$
(2.104)

where $c=0$ in the generic case, while $c\neq 0$ in the non-generic case;

$\bullet$ Residue condition: $M_{r}(z)$ has simple poles at each point in
$Z_{0}\cup\bar{Z_{0}}$ with

$\displaystyle\operatorname*{Res}_{z=z_{j}}M_{r}(z)=\lim_{z\to
z_{j}}M_{r}(z)\left(\begin{array}[]{cc}0&0\\\
\tilde{c}_{j}e^{-2\theta(z_{j})}&0\end{array}\right),$ (2.107)
$\displaystyle\operatorname*{Res}_{z=\bar{z}_{j}}M_{r}(z)=\lim_{z\to\bar{z}_{j}}M_{r}(z)\left(\begin{array}[]{cc}0&\bar{\tilde{c}}_{j}e^{2\theta(\bar{z}_{j})}\\\
0&0\end{array}\right).$ (2.110)

In the residue condition,

$\displaystyle\tilde{c}_{j}=\tilde{c}_{j+{N_{0}}}=\frac{b_{j}}{a^{\prime}(z_{j})},\
j=1,...,{N_{0}},$ (2.111)

where $b_{j}$ is defined in (2.72). Obviously, $\tilde{r}(z)$ has the same
properties as $r(z)$.

If the RH problems 1 and 2 have solutions, it must be unique. In fact, If
$M_{l}$ and $\tilde{M}_{l}$ both are solution of the RH problems 1, then it is
verifies that $\tilde{M}_{l}^{-1}M_{l}$ is a analytic function on $\mathbb{C}$
and tends to $I$ as $|z|\to\infty$. So by Liouvill’s theorem, it follows that
$M_{l}=\tilde{M}_{l}$.

To overcome difficulties from the singularity conditions of the RH problems 1
and 2 on $z=\pm 1$, we consider a matrix function $M_{l}^{(1)}(z)$ which is
the solution of the following RH problem without singularity on $z=\pm 1$.

######  The RH problem 3.

Find a matrix-valued function $M_{l}^{(1)}(z)=M_{l}^{(1)}(z;y)$ having no
singularity on $z=\pm 1$ and satisfying same analyticity, jump condition,
Asymptotic behavior and residue condition as the RH problem 1, and

$\bullet$ Symmetry:
$M_{l}^{(1)}(z)=\sigma_{1}\overline{M_{l}^{(1)}(\bar{z})}\sigma_{1}=\sigma_{2}M_{l}^{(1)}(-z)\sigma_{2}$.

And we will show in Section 3.1 that this RH problem exists a unique solution.

Then it can easily verify that $M_{l}(z;y)$ defined in the following
transformation is the solution of the original RH problem 1:

$\displaystyle\left(I-\frac{\sigma_{1}}{z}\right)M_{l}(z;y)=\left(I-\frac{\sigma_{1}M_{l}^{(1)}(0;y)^{-1}}{z}\right)M_{l}^{(1)}(z;y).$
(2.112)

Similarly, for $y\in(-\infty,0]$, by means of the following transformation

$\displaystyle\left(I-\frac{\sigma_{1}}{z}\right)M_{r}(z;y)=\left(I-\frac{\sigma_{1}M_{r}^{(1)}(0;y)^{-1}}{z}\right)M_{r}^{(1)}(z;y),$
(2.113)

it is easy to check that $M_{r}(z)$ is the solution of the RH problem 2, where
the matrix function $M_{r}^{(1)}(z;y)$ satisfies the above similar type RH
problem 3 under
$\left(\tilde{r},\\{z_{j},\tilde{c}_{j}\\}_{j=1}^{2N_{0}}\right)$ by replacing
$M_{l}^{(1)}(z;y)$ with $M_{r}^{(1)}(z;y)$.

###  2.4. Reconstruction of the potential $m$

The aim in this subsection is to give the reconstruction formula of $m$. By
the construction of $M_{l}(z;y)$ on $y\in[0,+\infty)$ and $M_{r}(z;y)$ on
$y\in(-\infty,0]$ in (2.75) and (2.78) respectively, $m_{0}$ can be
reconstructed from the behavior of the analytic continuations at $z=\infty$
and $z=i$ of $M_{l}(z;y)$ and $M_{r}(z;y)$ from (2.12) and (2.30).

######  Proposition 2.10.

If $M(z;y)$ is the solution of the RH problem 1 for $y\in[0,+\infty)$ or the
RH problem 2 for $y\in(-\infty,0]$, then the potential function $m$ can be
recovered from $M(z;y)$ with

$\displaystyle m=\left(1-\eta\right)^{-1},\ \eta=\lim_{z\to\infty}zM_{12}(z).$
(2.114)

Moreover, as $z\to i$, $M(z;y)$ has the following expansion:

$\displaystyle M(z)=\left(\begin{array}[]{cc}a_{1}&0\\\
0&a_{1}^{-1}\end{array}\right)+\left(\begin{array}[]{cc}0&a_{2}\\\
a_{3}&0\end{array}\right)(z-i)+\mathcal{O}\left((z-i)^{2}\right),$ (2.119)

where $a_{j}$ is a real function independent of $z$. In addition,
$u=(1-\partial_{x}^{2})^{-1}m$ is obtained from $M$ directly, that is

$\displaystyle u(y)=1-M_{11}(i)\lim_{z\to
i}\left(\frac{M_{12}(z)-M_{12}(i)}{z-i}\right)-M_{22}(i)\lim_{z\to
i}\left(\frac{M_{21}(z)-M_{21}(i)}{z-i}\right),$ $\displaystyle
x=y+2\log\left(M_{11}(i)\right).$ (2.120)

On the other hand, the following proposition reveals that the solutions
reconstructed from two RH problems in Proposition 2.10 are actually same at
$y=0.$

######  Proposition 2.11.

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

###### Proof.

Denote

$\displaystyle\tilde{a}(z)=\left\\{\begin{array}[]{ll}a^{*}(z),&\text{as
}z\in\mathbb{C}^{+},\\\\[12.0pt] a^{-1}(z),&\text{as }z\in\mathbb{C}^{-},\\\
\end{array}\right.\hskip 8.5359pt\tilde{a}\to I,\ |z|\to\infty.$ (2.123)

Then $M_{l}(z;y)=M_{r}(z;y)\tilde{a}(z)^{-\sigma_{3}}$ with
$\lim_{z\to\infty}\left(zM_{r,12}(z;y)\right)=\lim_{z\to\infty}\left(zM_{l,12}(z;y)\right)$
for $y=0$. And it is deduced from (2.120) that $u_{l}(y)=u_{r}(y)$. ∎

Therefore, we define

$\displaystyle
M^{(1)}(z;y)=\left\\{\begin{array}[]{ll}M^{(1)}_{l}(z;y),&\text{as
}y\in[0,+\infty),\\\\[12.0pt] M^{(1)}_{r}(z;y),&\text{as }y\in(-\infty,0],\\\
\end{array}\right.$ (2.126)

which is used to recover the solution $m_{0}(x)$ through the following lemma,
which is a directly result from Proposition 2.10 and (3.41).

######  Lemma 2.2.

Suppose that $M^{(1)}(z;y)$ is the solution of the RH problem 3. Then

$\displaystyle M^{(1)}(0;y)=\left(\begin{array}[]{cc}\alpha(y)&i\beta(y)\\\
-i\beta(y)&\alpha(y)\end{array}\right),\ \alpha^{2}-\beta^{2}=1,$

where $\alpha(y)$ and $\beta(y)$ are two real functions. And when $\beta\neq
0$, as $z\to i$, $M^{(1)}(z;y)$ has the expansion

$\displaystyle M^{(1)}(z;y)=$
$\displaystyle\left(\begin{array}[]{cc}f_{1}(y)&\frac{i\beta}{\alpha+1}f_{2}(y)\\\
-\frac{i\beta}{\alpha+1}f_{1}(y)&f_{2}(y)\end{array}\right)$
$\displaystyle+\left(\begin{array}[]{cc}\frac{i\beta}{\alpha+1}g_{1}(y)&g_{2}(y)\\\
g_{1}(y)&-\frac{i\beta}{\alpha+1}g_{2}(y)\end{array}\right)(z-i)+\mathcal{O}((z-i)^{2}),$

where $g_{1}(y)$, $g_{2}(y)$, $f_{1}(y)$ and $f_{2}(y)$ are real functions.
Then the following formula gives $m_{0}(x)=m(x(y))$ with

$\displaystyle u_{0}(y)=$ $\displaystyle
1-\alpha_{2}(y)\alpha_{1}(y)-\alpha_{3}(y)\alpha_{1}(y)^{-1},$ (2.127)
$\displaystyle x(y)=$ $\displaystyle y+2\log\left(\alpha_{1}(y)\right),$
(2.128) $\displaystyle m_{0}(y)=$ $\displaystyle u_{0}(y)-u_{0,xx}(y),$
(2.129)

where

$\displaystyle\alpha_{1}(y)=\left(1-\frac{\beta}{\alpha+1}\right)f_{1},\ \
\alpha_{2}(y)=\frac{\beta}{\alpha+1}f_{2}+\left(1-\frac{\beta}{\alpha+1}\right)g_{2},$
(2.130)
$\displaystyle\alpha_{3}(y)=\frac{-\beta}{\alpha+1}f_{1}+\left(1-\frac{\beta}{\alpha+1}\right)g_{1}.$
(2.131)

######  Remark 2.2.

In the case $\beta=0$, then $\alpha=\pm 1$. When $\alpha=1$, it only needs to
take $\frac{\beta}{\alpha+1}=0$ in the above formula. But when $\alpha=-1$, by
the symmetry of $M^{(1)}(z;y)$, it appears that as $z\to i$, $M^{(1)}(z;y)$
has the expansion

$\displaystyle M^{(1)}(z;y)=$
$\displaystyle\left(\begin{array}[]{cc}0&f_{1}(y)i\\\
f_{1}(y)^{-1}i&0\end{array}\right)+\left(\begin{array}[]{cc}g_{1}(y)i&0\\\
0&g_{2}(y)i\end{array}\right)(z-i)+\mathcal{O}\left((z-i)^{2}\right).$

Under this case, the function $\alpha_{j}$ in the reconstruction formula
become

$\displaystyle\alpha_{1}(y)=f_{1}^{-1}(y),\ \alpha_{2}(y)=f_{1}(y)+g_{2}(y),\
\alpha_{3}(y)=f_{1}^{-1}(y)+g_{1}(y).$

On the other hand, we declare that $\alpha_{1}(y)\neq 0$, which is from
det$(M^{(1)}(z;y))\equiv 1$.

##  3\. Inverse scattering transform

In this section, we shall provide a rigorous framework to study the inverse
scattering transform based on the RH problem of complex analysis.

###  3.1. The Cauchy projection operator

For a given function $f(z)\in L^{p}(\mathbb{R})$, $1\mathchar 13366\relax
p<\infty$, 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)
$\displaystyle(\mathcal{C}_{\pm}f)^{\wedge}(z)=X_{\pm}(z)\hat{f}(z),$ (3.3)

where $X_{\pm}$ denotes the characteristic function on $\mathbb{R}^{\pm}$. The
following proposition summarizes the basic properties of the Cauchy projection
operators [18].

######  Proposition 3.1.

[18] For any $f\in L^{p}(\mathbb{R})$, $1\mathchar 13366\relax p<\infty$, 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 $\mathbb{C}^{\pm}$. If $1<p<\infty$, then
there exists a positive constant $C_{p}$ such that

$\displaystyle\parallel\mathcal{C}_{\pm}f\parallel_{L^{p}}\mathchar
13366\relax C_{p}\parallel f\parallel_{L^{p}}.$ (3.4)

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 3 about
$M^{(1)}_{l}$ for $y\in[0,+\infty)$ under the given scattering data:

$R=\left(r(\cdot),\\{z_{j},c_{j}\\}_{j=1}^{2{N_{0}}}\right)\in Y,$

where $Y$ is a Banach space defined in (1.13), and $r(z)$ admits the
symmetries

$\displaystyle r(z)=\overline{r(z^{-1})}=-\overline{r(-z)},\ r(1)=-r(-1)=-1.$
(3.5)

For $z\in\mathbb{R}$, denote $w=w_{+}+w_{-}$ with

$\displaystyle w_{+}(z;y)=\left(\begin{array}[]{cc}0&0\\\
-\bar{r}(z)e^{-2\theta(z;y)}&0\end{array}\right),\
w_{-}(z;y)=\left(\begin{array}[]{cc}0&r(z)e^{2\theta(z;y)}\\\
0&0\end{array}\right).$ (3.10)

Then in view of the Beals-Coifman theorem in [1], the RH problem 3 has a
solution given by the Cauchy operator

$\displaystyle M^{(1)}_{l}(z;y)=I+\mathcal{C}(\mu
w)(z;y)+\sum_{j=1}^{2{N_{0}}}\left(\frac{\bar{c}_{j}e^{y\text{Im}z_{j}}}{z-\bar{z}_{j}}\mu_{2}(\bar{z}_{j};y),\frac{c_{j}e^{y\text{Im}z_{j}}}{z-z_{j}}\mu_{1}(z_{j};y)\right),$
(3.11)

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

$\displaystyle\mu(z;y)=I+\mathcal{C}_{w}\mu(z;y)+\sum_{j=1}^{2{N_{0}}}\left(\frac{\bar{c}_{j}e^{y\text{Im}z_{j}}}{z-\bar{z}_{j}}\mu_{2}(\bar{z}_{j};y),\frac{c_{j}e^{y\text{Im}z_{j}}}{z-z_{j}}\mu_{1}(z_{j};y)\right).$
(3.12)

And $\mu_{1}(z;y)$ and $\mu_{2}(z;y)$ denote the first and second columns of
$\mu(z;y)$ respectively, and

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

Thus the attention turns to the estimate and solvability of the Fredholm
integral equation (3.12). The proof basically follows the idea in [41, 42],
which was later used in [15, 26] and so on. To this end, we take the first row
of $\mu(z)$ as an instance. A simple calculation gives that
$\|\mathcal{C}_{w}I\|_{L^{2}}\mathchar 13366\relax 2\|R\|_{Y}$ for $\|R\|_{Y}$
in (1.14). We denote a new column vector function by

$\displaystyle N(z;y):=\left(N_{11}(z;y),N_{12}(z;y)\right)^{T},\hskip
8.5359ptN_{11}(z;y)=\mu_{11}(z;y)-1,\hskip 8.5359ptN_{12}(z;y)=\mu_{12}(z;y).$
(3.13)

Via (3.12), it admits that for $z\in\mathbb{R}$:

$\displaystyle
N_{11}(z;y)=\mathcal{C}_{-}(-\bar{r}e^{-2\theta(\cdot;y)}N_{12}(\cdot;y))(z;y)+\sum_{j=1}^{2{N_{0}}}\frac{N_{12}(\bar{z}_{j};y)\bar{c}_{j}e^{y\text{Im}z_{j}}}{z-\bar{z}_{j}},$
(3.14) $\displaystyle
N_{12}(z;y)=N_{0,2}(z;y)+\mathcal{C}_{+}(re^{2\theta(\cdot;y)}N_{11}(\cdot;y))(z;y)+\sum_{j=1}^{2{N_{0}}}\frac{N_{11}(z_{j};y)c_{j}e^{\text{Im}z_{j}y}}{z-z_{j}},$
(3.15)

and for $j=1,...,2N_{0}$,

$\displaystyle
N_{11}(z_{j};y)=\mathcal{C}_{-}(-\bar{r}e^{-2\theta(\cdot;y)}N_{12}(\cdot;y))(z_{j};y)+\sum_{j=1}^{2{N_{0}}}\frac{N_{12}(\bar{z}_{j};y)\bar{c}_{j}e^{y\text{Im}z_{j}}}{z_{j}-\bar{z}_{j}},$
(3.16) $\displaystyle
N_{12}(\bar{z}_{j};y)=N_{0,2}(\bar{z}_{j};y)+\mathcal{C}_{+}(re^{2\theta(\cdot;y)}N_{11}(\cdot;y))(\bar{z}_{j};y)+\sum_{j=1}^{2{N_{0}}}\frac{N_{11}(z_{j};y)c_{j}e^{\text{Im}z_{j}y}}{\bar{z}_{j}-z_{j}},$
(3.17)

where

$\displaystyle
N_{0,2}(z;y)=\mathcal{C}_{+}(re^{2\theta})(z;y)+\sum_{j=1}^{2{N_{0}}}\frac{c_{j}e^{\text{Im}z_{j}y}}{z-z_{j}}.$
(3.18)

Combining above equations, it immediately follows that

$\displaystyle N_{11}(z_{j};y)=\mathcal{C}(N_{11})(z_{j};y),\hskip
8.5359ptN_{12}(\bar{z}_{j};y)=\mathcal{C}(N_{12})(\bar{z}_{j};y),$ (3.19)

which inspires us to only consider $N_{11},\ N_{12}\in L^{2}(\mathbb{R}).$ The
integral equation (3.12) can be rewritten as

$\displaystyle N(z;y)=N_{0}(z;y)+\mathcal{K}N(z;y),$ (3.20) $\displaystyle
N_{0}(z;y)=\left(0,N_{0,2}(z;y)\right)^{T},\hskip
8.5359pt\mathcal{K}=\left(\begin{array}[]{cc}0&\mathcal{K}^{(1)}\\\
\mathcal{K}^{(2)}&0\end{array}\right),$ (3.23)

where $\mathcal{K}^{(1)}$, $\mathcal{K}^{(2)}:L^{2}(\mathbb{R})\to
L^{2}(\mathbb{R})$ are two operators acting on scalar functions defined as
follows. Denote $h$ as a generic function in $L^{2}(\mathbb{R})$. Then

$\displaystyle\mathcal{K}^{(1)}=\mathcal{K}^{(1)}_{1}+\mathcal{K}^{(1)}_{2},\hskip
8.5359pt\mathcal{K}^{(2)}=\mathcal{K}^{(2)}_{1}+\mathcal{K}^{(2)}_{2},$ (3.24)
$\displaystyle\mathcal{K}^{(1)}_{1}h(z;y)=\mathcal{C}_{-}(-\bar{r}e^{-2\theta(\cdot;y)}h)(z;y),\
\mathcal{K}^{(1)}_{2}h(z;y)=\sum_{j=1}^{2{N_{0}}}\frac{\mathcal{C}(h)(\bar{z}_{j};y)\bar{c}_{j}e^{y\text{Im}z_{j}}}{z-\bar{z}_{j}},$
(3.25)
$\displaystyle\mathcal{K}^{(2)}_{1}h(z;y)=\mathcal{C}_{+}(re^{2\theta(\cdot;y)}h)(z;y),\
\mathcal{K}^{(2)}_{2}h(z)=\sum_{j=1}^{2{N_{0}}}\frac{\mathcal{C}(h)(z_{j};y)c_{j}e^{y\text{Im}z_{j}}}{z-z_{j}}.$
(3.26)

It is adduced that $N_{0}\in L^{2}(\mathbb{R})$. On the other hand, on account
of $\|r\|_{L^{\infty}}=1$, there exists a positive constant $C(\|R\|_{Y})$
such that
$\|\mathcal{K}\|=\|\mathcal{K}^{(1)}\|+\|\mathcal{K}^{(2)}\|\mathchar
13366\relax C(\|R\|_{Y})$. Our expectation is to demonstrate $N\in
L^{2}(\mathbb{R})$ through (3.20). Owing to the special structure of
$\mathcal{K}$, we consider the operator

$\displaystyle\mathcal{K}^{2}=\left(\begin{array}[]{cc}\mathcal{K}^{(1)}\mathcal{K}^{(2)}&0\\\
0&\mathcal{K}^{(2)}\mathcal{K}^{(1)}\end{array}\right).$

For simplicity of notations, we drop the dependence of the operator
$\mathcal{K}$, the functions $\mu$, $M^{(1)}_{l}$ and $N$ from the variable of
$y$.

The following proposition is given to show the invertibility of
$I-\mathcal{K}$.

######  Proposition 3.2.

If $R=\left(r,\\{z_{j},c_{j}\\}_{j=1}^{2{N_{0}}}\right)\in Y$, 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 $\|\mathcal{K}^{2}\|\to 0$, as
$y\to\infty$.

###### Proof.

Note that operators $\mathcal{K}^{(1)}_{2}$, $\mathcal{K}^{(2)}_{2}$ are both
finite-rank. So it suffices to analyze the property of
$\mathcal{K}^{(1)}_{1}\mathcal{K}^{(2)}_{1}$ and
$\mathcal{K}^{(2)}_{1}\mathcal{K}^{(1)}_{1}$. Take
$\mathcal{K}^{(2)}_{1}\mathcal{K}^{(1)}_{1}$ as an example. A simple
calculation gives that
$\|\mathcal{K}^{(2)}_{1}\mathcal{K}^{(1)}_{1}\|\mathchar 13366\relax 1$ as
$|r|\mathchar 13366\relax 1$. Recall the definition of $X_{\pm},$ that is,
$X_{\pm}$ is the characteristic function on $\mathbb{R}^{\pm}$. For any $f\in
L^{2}(\mathbb{R})$,

$\displaystyle{(\mathcal{K}^{(2)}_{1}\mathcal{K}^{(1)}_{1}\check{f})}^{\wedge}(z)=\int_{\mathbb{R}}K(z,\lambda;y)f(\lambda)d\lambda,$

where the integral kernel is

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

Since

$\|K(\cdot,\cdot;y)\|_{L^{2}(\mathbb{R}\otimes\mathbb{R})}\mathchar
13366\relax\|(\bar{r}e^{-2\theta(\cdot;y)})^{\wedge}\|_{L^{2,1}}\|(re^{2\theta(\cdot;y)})^{\wedge}\|_{L^{2,1}}\mathchar
13366\relax C(y)\|r\|_{W^{1,2}}^{2}$

for a positive constant $C(y)$ which is finite for every $y\in[0,+\infty)$, it
follows that $\mathcal{K}^{(2)}_{1}\mathcal{K}^{(1)}_{1}$ is the Hilbert-
Schmidt operator. It in turn implies that $\mathcal{K}^{2}$ is compact.
Therefore, it is inferred in [32] that $I-\mathcal{K}^{2}$ is a Fredholm
operator with index zero, and so is $I-\mathcal{K}$. We now only need to prove
that the operator $I-\mathcal{K}:L^{2}\to L^{2}$ is injective. For
convenience, take ${N_{0}}=1$. Suppose that
$\vec{f}=\left(f_{1},f_{2}\right)^{T}\in L^{2}$ is the solution of the
homogeneous equation: $(I-\mathcal{K})\vec{f}=0$. Denote
$\vec{g}(z)=(g_{1}^{*}(z),g_{2}(z))$ with

$\displaystyle
g_{1}(z)=\mathcal{C}(-\bar{r}e^{-2\theta(\cdot;y)}f_{2})(z)+\mathcal{K}^{(1)}_{2}(f_{2})(z),\
g_{2}(z)=\mathcal{C}(re^{2\theta(\cdot;y)}f_{1})(z)+\mathcal{K}^{(2)}_{2}((f_{1})^{T})(z),$

where $g_{1}(z)$ and $g_{2}(z)$ are analytic in $\mathbb{C}^{-}$ and
$\mathbb{C}^{+}$ respectively, and meromorphic in
$\mathbb{C}\setminus\mathbb{R}$. Thus, $[g_{1}]_{-}=f_{1}$,
$[g_{2}]_{+}=f_{2}$. Using the result in the Cauchy-Goursat theorem gives that

$\displaystyle 0=\int_{\mathbb{R}^{+}}\vec{g}(z)[\vec{g}^{*}(z)]^{T}dz.$

On the other hand

$\displaystyle 0=\int_{\mathbb{R}^{+}}\vec{g}(z)[\vec{g}^{*}(z)]^{T}dz$
$\displaystyle=\int_{\mathbb{R}}|f_{1}|^{2}+|f_{2}|^{2}+2\text{Re}\left(f_{1}(z)f_{2}^{*}(z)r(z)e^{2\theta(z;y)}\right)dz$
$\displaystyle\geq\int_{\mathbb{R}}\left(|f_{1}(z)|^{2}+|f_{2}(z)|^{2}\right)(1-|r|)dz.$

Note that $1-|r(z)|>0,z\not=\pm 1$, and $1-|r(z)|=0$ if and only if $z=\pm 1$.
So the above inequality implies that $f_{1}(z)=f_{2}(z)=0$ for $z\neq\pm 1$,
namely, $f(z)=0,\ a.e.$ So $\|f\|_{L^{2}}=0$. The final step is to prove
$\|\mathcal{K}^{2}\|\to 0$ as $y\to\infty$. Because $e^{y\text{Im}z_{j}}$
decays to zero exponentially as $y\to\infty$, it is sufficient to demonstrate
$\|\mathcal{K}^{(1)}_{1}\mathcal{K}^{(2)}_{1}\|$,
$\|\mathcal{K}^{(2)}_{1}\mathcal{K}^{(1)}_{1}\|\to 0$ as $y\to\infty$. Taking
$\mathcal{K}^{(1)}_{1}\mathcal{K}^{(2)}_{1}$ to give details, like in [41],
$\bar{r}$ can be approximated in $L^{\infty}$ norm by a series of functions
$\\{r_{n}\\}_{1}^{\infty}$. Here, $\bar{r}_{n}\in L^{2}$ is a retinal function
with simple poles $\\{\xi^{(n)}_{j}\\}$ on $\mathbb{C}^{+}$.Then
$\mathcal{K}^{(1)}_{1}\mathcal{K}^{(2)}_{1}f$ is approximated in $L^{2}$ norm
by
$\mathcal{C}_{-}(-\bar{r}_{n}e^{-2\theta(\cdot;y)}\mathcal{C}_{+}(re^{2\theta(\cdot;y)}f))$
uniformly for $\|f\|_{L^{2}}\mathchar 13366\relax 1$. We denote the residue of
$r_{n}$ at $\xi^{(n)}_{j}$ by $r_{n,j}$. Note that

$\displaystyle(-(\cdot-\xi^{(n)}_{j})^{-1}e^{-2\theta(\cdot;y)})^{\wedge}(z)=\int_{\mathbb{R}}\frac{e^{-2\pi
izs}e^{i(s-1/s)y/2}}{s-\xi^{(n)}_{j}}ds=X_{-}(z)e^{-2\pi
iz\xi^{(n)}_{j}+i(\xi^{(n)}_{j}-1/\xi^{(n)}_{j})y/2}.$

There appears that

$\displaystyle(-\bar{r}_{n}e^{-2\theta(\cdot;y)})^{\wedge}(z)=\sum_{\xi^{(n)}_{j}\in\mathbb{C}^{+}}X_{-}(z)\bar{r}_{n,j}e^{-2\pi
iz\xi^{(n)}_{j}+i(\xi^{(n)}_{j}-1/\xi^{(n)}_{j})y/2}.$

Therefore,

$\displaystyle\left(\mathcal{C}_{-}(-\bar{r}_{n}e^{-2\theta(\cdot;y)}\mathcal{C}_{+}(re^{2\theta(\cdot;y)}f))\right)^{\wedge}(z)=X_{-}(z)\left(\mathcal{C}_{+}(re^{2\theta(\cdot;y)}f)\right)^{\wedge}*(-\bar{r}_{n}e^{-2\theta(\cdot;y)})^{\wedge}(z)$
$\displaystyle=\sum_{\xi^{(n)}_{j}\in\mathbb{C}^{+}}X_{-}(z)e^{i(\xi^{(n)}_{j}-1/\xi^{(n)}_{j})y/2}\bar{r}_{n,j}\left(\mathcal{C}_{+}(re^{2\theta(\cdot;y)}f)\right)^{\wedge}*(X_{-}e^{-2\pi
i(\cdot)\xi^{(n)}_{j}})(z),$

approaches zero in $L^{2}$ as $y\to\infty$ uniformly for
$\|f\|_{L^{2}}\mathchar 13366\relax 1$. This completes the proof of
Proposition 3.2. ∎

Proposition 3.2 also gives that $I-\mathcal{K}$ is uniformly bounded for
$y\in[0,+\infty),$ and it transpires that there exists a constant
$C(\|R\|_{Y})$ such that

$\displaystyle\parallel\left(I-\mathcal{K}\right)^{-1}\parallel\mathchar
13366\relax C(\|R\|_{Y}).$ (3.27)

As a consequence of this proposition, the solution of (3.20) exsits with

$\displaystyle N(z)=((I-\mathcal{K})^{-1}(N_{0}))(z),$ (3.28)

which leads to the following proposition immediately.

######  Proposition 3.3.

If $R=\left(r,\\{z_{j},c_{j}\\}_{j=1}^{2{N_{0}}}\right)\in Y$, then there
exists a unique solution $M^{(1)}_{l}$ of the RH problem 3 with

$\displaystyle M^{(1)}_{l}(z)=I+\mathcal{C}(\mu
w)(z)+\sum_{j=1}^{2{N_{0}}}\left(\frac{\bar{c}_{j}e^{y\text{Im}z_{j}}}{z-\bar{z}_{j}}\mu_{2}(\bar{z}_{j}),\frac{c_{j}e^{y\text{Im}z_{j}}}{z-z_{j}}\mu_{1}(z_{j})\right),$
(3.29)

where $\mu$ comes from the definition of $N$ in (3.13).

Furthermore, it is adduced from (3.19) that

$\displaystyle|N_{11}(z_{j};y)|=|\mathcal{C}(N_{11})(z_{j};y)|\leq
C(\text{Im}z_{j})\|N_{11}\|_{L^{2}},\hskip
8.5359pt|N_{12}(\bar{z}_{j};y)|=|\mathcal{C}(N_{12})(\bar{z}_{j};y)|\leq
C(\text{Im}z_{j})\|N_{12}\|_{L^{2}}.$

Together with (3.20), it gives that

$\displaystyle\parallel\mu-I\parallel_{L^{2}}\mathchar
13366\relax\parallel\left(I-\mathcal{K}\right)^{-1}\parallel(\parallel
C_{w}I\parallel_{L^{2}}+\sum_{j=1}^{2{N_{0}}}c_{j}e^{y\text{Im}z_{j}}\Big{\|}\frac{1}{z-z_{j}}\Big{\|}_{L^{2}}),$
(3.30)
$\displaystyle\sum_{j=1}^{2{N_{0}}}|\mu_{2}(\bar{z}_{j})|+\sum_{j=1}^{2{N_{0}}}|\mu_{1}(z_{j})-1|\mathchar
13366\relax
C(\text{Im}z_{j})\parallel\left(I-\mathcal{K}\right)^{-1}\parallel(\parallel
C_{w}I\parallel_{L^{2}}+\sum_{j=1}^{2{N_{0}}}c_{j}e^{y\text{Im}z_{j}}\Big{\|}\frac{1}{z-z_{j}}\Big{\|}_{L^{2}}).$
(3.31)

Meanwhile, there exists a positive constant $C(\|R\|_{Y})$ such that

$\displaystyle\sum_{j=1}^{2{N_{0}}}|\mu_{2}(\bar{z}_{j})|+\sum_{j=1}^{2{N_{0}}}|\mu_{1}(z_{j})-1|+\parallel\mu-I\parallel_{L^{2}}\mathchar
13366\relax C(\|R\|_{Y}).$ (3.32)

######  Remark 3.1.

Compared with derivative NLS equation [34], where the corresponding jump
matrix admits positive definite part with reflection coefficient
$\|r\|_{L^{\infty}}<1$. So the norm of their corresponding
$(I-\mathcal{K})^{-1}$ can be directly controlled by
$(1-\|r\|_{L^{\infty}})^{-1}$. However, for the mCH equation, its
corresponding reflection coefficient $r(z)$ satisfies $|r(\pm 1)|=1$. So we
lose the Lipschitz continuity of the solution $M^{(1)}_{l}$ with respect to
reflection coefficient $r(z)$.

When $y\in(-\infty,0)$, following the same process as $M_{l}^{(1)}(z)$,
$M_{r}^{(1)}(z)$ admits similar property in the following proposition.

######  Proposition 3.4.

If $R\in Y$, $y\in(-\infty,0)$, then there exists a positive constant
$C(\|R\|_{Y})$, such that

$\displaystyle\parallel
M_{r}^{(1)}-I\parallel_{L^{2}}+\sum_{j=1}^{2{N_{0}}}|M_{r,2}^{(1)}(\bar{z}_{j})|+\sum_{j=1}^{2{N_{0}}}|M_{r,1}^{(1)}(z_{j})|\mathchar
13366\relax C(\|R\|_{Y}),$ (3.33)

where $M_{r,2}^{(1)}$ denotes the $j$-column of $M_{r}^{(1)}$.

###  3.3. Time evolution

The crucial argument of inverse scattering theory is how to recover the
solution $m(t,x)$ to the initial-value problem (1.12) from scattering data. It
is necessary to give the time evolution of the RH problem from the time
evolution of scattering data. According to the time spectral problem in the
Lax pair (1.3) and scattering relation (2.31), it is found that the time
evolution scattering data $a(z;t)$ and $b(z;t)$ satisfy the equations

$\displaystyle\partial_{t}a(z;t)=0,\
\partial_{t}b(z;t)=-\dfrac{2iz(z^{2}-1)}{z^{2}+1}b(z;t),\ z\in\mathbb{R},$
(3.34)

which yields

$\displaystyle a(z;t)=a(z;0),\
b(z;t)=e^{-\frac{2iz(z^{2}-1)t}{z^{2}+1}}b(z;0),\ z\in\mathbb{R}.$ (3.35)

Therefore, we define the time-dependent scattering data in the following
lemma.

######  Lemma 3.1.

Define time-dependent scattering data
${R}(t)=\left(r(\cdot;t),\\{z_{j}(t),c_{j}(t)\\}_{j=1}^{2{N_{0}}}\right)$ with

$\displaystyle
r(z;t)=\exp\left({-\frac{2iz(z^{2}-1)t}{(z^{2}+1)^{2}}}\right)r(z;0),\
z\in\mathbb{R},\ z_{j}(t)=z_{j}(0),\
c_{j}(t)=\exp{\left(\frac{t\text{Im}z_{j}}{\text{Re}^{2}z_{j}}\right)}c_{j}(0).$

For any $T>0,$ if $R(0)\in Y$, then for every $t\in[0,T)$, $R(t)\in Y$.
Moreover,

$\displaystyle\|R(t)\|_{Y}\mathchar 13366\relax C(T)\|R(0)\|_{Y},$ (3.36)

with a constant $C(T)>0.$

###### Proof.

Since $\big{|}e^{-\frac{2iz(z^{2}-1)t}{(z^{2}+1)^{2}}}\big{|}=1$, for
$z\in\mathbb{R}$, $j=0,1,2$, we have

$\|(\cdot)^{j}r(\cdot;t)\|_{L^{2}}=\|(\cdot)^{j}e^{-\frac{2i(\cdot)((\cdot)^{2}-1)t}{((\cdot)^{2}+1)^{2}}}r(\cdot;0)\|_{L^{2}}=\|r(\cdot;0)\|_{L^{2,j}}.$

Note that for $j=0,1,2$,

$\displaystyle
z^{j}\partial_{z}r(z;t)=2it\frac{z^{j}(z^{4}-6z^{2}+1)}{(z^{2}+1)^{3}}e^{-\frac{2iz(z^{2}-1)t}{(z^{2}+1)^{2}}}r(z;0)+e^{-\frac{2iz(z^{2}-1)t}{(z^{2}+1)^{2}}}\partial_{z}r(z;0).$

It is thereby inferred that

$\displaystyle\|(\cdot)^{j}\partial_{z}r(\cdot;t)\|_{L^{2}}^{2}\mathchar
13366\relax
2t\sup_{z\in\mathbb{R}}\frac{|z^{j}(z^{4}-6z^{2}+1)|}{(z^{2}+1)^{3}}\|r(\cdot;0)\|_{L^{2}}^{2}+\|\partial_{z}r(\cdot;0)\|_{L^{2}}^{2}.$

In the same way for $j=0,1$,

$\displaystyle z^{j}\partial_{z}^{2}r(z;t)=$
$\displaystyle(it\frac{z^{j+1}(z^{4}+2z^{2}-7)}{(z^{2}+1)^{4}}-\frac{t^{2}z^{j}(z^{4}-6z^{2}+1)^{2}}{(z^{2}+1)^{6}})4e^{-\frac{2iz(z^{2}-1)t}{(z^{2}+1)^{2}}}r(z;0)$
$\displaystyle+4it\frac{z^{j}(z^{4}-6z^{2}+1)}{(z^{2}+1)^{3}}e^{-\frac{2iz(z^{2}-1)t}{(z^{2}+1)^{2}}}\partial_{z}r(z;0)+z^{j}e^{-\frac{2iz(z^{2}-1)t}{(z^{2}+1)^{2}}}\partial_{z}^{2}r(z;0).$

Consequently,

$\displaystyle\|(\cdot)^{j}\partial_{z}^{2}r(\cdot;t)\|_{L^{2}}^{2}\mathchar
13366\relax$ $\displaystyle
4(t\sup_{z\in\mathbb{R}}\frac{|z^{j+1}(-z^{4}+2z^{2}+7)|}{(z^{2}+1)^{4}}+t^{2}\sup_{z\in\mathbb{R}}\frac{|z|^{j}(z^{4}-6z^{2}+1)}{(z^{2}+1)^{3}})\|r(\cdot;0)\|_{L^{2}}^{2}$
$\displaystyle+4t\sup_{z\in\mathbb{R}}\frac{|z^{j}(z^{4}-6z^{2}+1)|}{(z^{2}+1)^{3}}\|\partial_{z}r(\cdot;0)\|_{L^{2}}^{2}+\|(\cdot)^{j}\partial_{z}^{2}r(\cdot;0)\|_{L^{2}}^{2}.$

This in turn implies that $r(\cdot;t)\in H^{1,2}(\mathbb{R})\cap
H^{2,1}(\mathbb{R})$. The estimate of others scattering data can be
accomplished in the same way from the fact that $|z_{j}|=1$ with Im$z_{j}<0$
and $\exp{\left(\frac{t\text{Im}z_{j}}{\text{Re}^{2}z_{j}}\right)}<1$, which
implies the desired result as indicated above. ∎

Therefore, invoking of Propositions 3.3 and 3.4, under time-dependent
scattering data the RH problems 3 exists unique solution. We denote
$M^{(1)}(z;t,y)$ as the solution for the RH problem 3 under time-dependent
scattering data $R(t)$. And $M^{(1)}_{r}(z;0,y)$ also has a similar time
evolution denoted by $M^{(1)}_{r}(z;t,y)$ , which is omitted here. Thus the
time evolution of $M^{(1)}$ defined in (2.126) is given by

$\displaystyle
M^{(1)}(z;t,y)=\left\\{\begin{array}[]{ll}M^{(1)}_{l}(z;t,y),&\text{as
}y\in[0,+\infty),\\\\[12.0pt] M^{(1)}_{r}(z;t,y),&\text{as
}y\in(-\infty,0],\\\ \end{array}\right.$ (3.39)

The crux of the matter is whether the function $m(t,y)$ reconstructed from the
time-depended RH problem in (2.114) is the solution of the mCH equation (1.1).
The approach is to demonstrate that $M^{(1)}(z;t,y)$ satisfies some Lax pair
equations with similar structure as (1.3) whose compatibility condition is the
mCH equation (1.1). To achieve this aim, we first give the following
proposition.

######  Proposition 3.5.

$M^{(1)}(z;t,y)$ defined in (3.39) is differentiable with respect to $t>0$.

###### Proof.

We give the details of the proof of the case $y\in[0,+\infty)$. In this case,
under time-dependent scattering data $R(t)$,
$M^{(1)}(z;t,y)=M^{(1)}_{l}(z;t,y)$. From the definition of
$M^{(1)}_{l}(z;t,y)$ in (3.11), the existence of
$\partial_{t}M^{(1)}_{l}(z;t,y)$ is equivalent to the existence of
$\partial_{t}\mu(z;t,y)$. By taking $t$-derivative of the operator equation
(3.20) under time-dependent scattering data $R(t)$, it follows that

$\displaystyle\partial_{t}N(z;t,y)=\left(\partial_{t}N_{0}(z;t,y)+(\partial_{t}\mathcal{K})(N)(z;t,y)\right)+\mathcal{K}(\partial_{t}N)(z;t,y),$
(3.40)

where

$\displaystyle\partial_{t}N(z;t,y)=\left(0,\mathcal{C}_{+}(\partial_{t}r(\cdot;t)e^{2\theta})(z;t,y)+\sum_{j=1}^{2N_{0}}\frac{\partial_{t}c_{j}(t)e^{\text{Im}z_{j}y}}{z-z_{j}}\right)^{T},$
$\displaystyle\partial_{t}\mathcal{K}=\left(\begin{array}[]{cc}0&\partial_{t}\mathcal{K}^{(1)}\\\
\partial_{t}\mathcal{K}^{(2)}&0\end{array}\right).$

For brevity, we omit the concrete expression of operators
$\partial_{t}\mathcal{K}^{(1)}$ and $\partial_{t}\mathcal{K}^{(2)}$. From
Lemma 3.1,

$\displaystyle\partial_{t}r(z;t)=-\frac{2iz(z^{2}-1)}{(z^{2}+1)^{2}}\exp\left({-\frac{2iz(z^{2}-1)t}{(z^{2}+1)^{2}}}\right)r(z;0),\
z\in\mathbb{R}$
$\displaystyle\partial_{t}c_{j}(t)=\frac{\text{Im}z_{j}}{\text{Re}^{2}z_{j}}\exp{\left(\frac{t\text{Im}z_{j}}{\text{Re}^{2}z_{j}}\right)}c_{j}(0),$

with $|\partial_{t}r(z;t)|\mathchar 13358\relax|r(z;0)|$,
$|\partial_{t}c_{j}(t)|\mathchar 13358\relax|c_{j}(0)|$. Therefore,
$\partial_{t}N_{0}(\cdot;t,y)+(\partial_{t}\mathcal{K})(N)(\cdot;t,y)\in
L^{2}(\mathbb{R})$. Then it is inferred from Proposition 3.2 that
$\partial_{t}N(z;t,y)$ exists with

$\displaystyle\partial_{t}N(z;t,y)=\left(I-\mathcal{K}\right)^{-1}\left(\partial_{t}N_{0}(z;t,y)+(\partial_{t}\mathcal{K})(N)(z;t,y)\right)(z;t,y).$

It also implies the existence of $\partial_{t}\mu(z;t,y)$, from which we
obtain the result of this proposition. ∎

Therefore, by standard processing (see for example [4, 16]), the following
proposition can be verified.

######  Proposition 3.6.

Assume that $M^{(1)}(z;t,y)$ is the solution for the RH problem 3 under time-
dependent scattering data $R(t)$. The matrix function $M(z;t,y)$ is defined by

$\displaystyle\left(I-\frac{\sigma_{1}}{z}\right)M(z;t,y)=\left(I-\frac{\sigma_{1}M^{(1)}(t,y;0)^{-1}}{z}\right)M^{(1)}(z;t,y).$
(3.41)

Define

$\displaystyle\breve{M}(z;t,y)=M(z;t,y)e^{-\breve{p}(z;t,y)\sigma_{3}},$
(3.42)

where
$\breve{p}(z;t,y)=-\frac{i(z^{2}-1)}{4z}(-y+\frac{8z^{2}}{(z^{2}+1)^{2}}t)$.
Then $\breve{M}(z;t,y)$ satisfies the system of linear differential equations

$\displaystyle\breve{M}_{y}(z;t,y)=\breve{U}\breve{M},\ \ \
\breve{M}_{t}(z;t,y)=\breve{V}\breve{M},$ (3.43)

where

$\displaystyle\breve{U}=-\frac{i(z^{2}-1)}{4z}\sigma_{3}+\frac{ih_{1}(t,y)}{z-1}\left(\begin{array}[]{cc}1&-1\\\
1&-1\end{array}\right)+\frac{ih_{1}(t,y)}{z+1}\left(\begin{array}[]{cc}1&1\\\
-1&-1\end{array}\right)+ih_{1}(t,y)\left(\begin{array}[]{cc}0&-1\\\
1&0\end{array}\right),$
$\displaystyle\breve{V}=\frac{2i(z^{2}-1)z}{(z^{2}+1)^{2}}\sigma_{3}+\frac{ih_{2}(t,y)}{z-1}\left(\begin{array}[]{cc}1&-1\\\
1&-1\end{array}\right)+\frac{ih_{2}(t,y)}{z+1}\left(\begin{array}[]{cc}1&1\\\
-1&-1\end{array}\right)$
$\displaystyle+\frac{1}{z-1}\left(\begin{array}[]{cc}0&h_{3}(t,y)\\\
h_{4}(t,y)&0\end{array}\right)+\frac{1}{z+1}\left(\begin{array}[]{cc}0&h_{4}(t,y)\\\
h_{3}(t,y)&0\end{array}\right),$ $\displaystyle
h_{1}(t,y)=-\frac{1}{2}\lim_{z\to\infty}zM_{12}(z;t,y),$ $\displaystyle
h_{2}(t,y)=-i\lim_{z\to
1}(z-1)(\breve{M}_{t}(z;t,y)^{-1}\breve{M}(z;t,y))_{11},$ $\displaystyle
h_{3}(t,y)=2M_{11}(i;t,y)\lim_{z\to i}\frac{M_{12}(z;t,y)}{z-i},$
$\displaystyle h_{4}(t,y)=-2M_{22}(i;t,y)\lim_{z\to
i}\frac{M_{21}(z;t,y)}{z-i},$

and the subscript $ij$ used above means the $(i,j)$-entry element of matrix.
The compatibility condition between two equations in the system (3.43),
namely,

$\breve{U}_{t}-\breve{V}_{y}+[\breve{U},\breve{V}]=0$

yields the mCH equation (1.1) with the variables $(t,y)$.

Consequently, under the time evolution of the scattering data in Lemma 3.1, it
can be verified that the reconstruction formula remain accessible for
$M^{(1)}(z;t,y)$.

###  3.4. Estimates on solutions of the RH problem

In this subsection, to estimate the solution of the Cauchy problem (1.12)
through reconstruction formula, a key step is to give the proof of estimates
of $M_{l}^{(1)}(z)$ for $y\in[0,+\infty)$ which is the solution of the RH
problem 3 under scattering data $R(t)$. (3.11) implies that as $z\to i$,
$M_{l}^{(1)}(z)$ has the following expansion.

$\displaystyle
M_{l}^{(1)}(z)=M_{l}^{(1)}(i)+M_{l}^{(1),i}(z-i)+\mathcal{O}\left((z-i)^{2}\right),$
(3.44)

with

$\displaystyle M^{(1)}_{l}(i)=I+\mathcal{C}(\mu
w)(i)+\sum_{j=1}^{2{N_{0}}}\left(\frac{\bar{c}_{j}e^{y\text{Im}z_{j}}}{i-\bar{z}_{j}}\mu_{2}(\bar{z}_{j}),\frac{c_{j}e^{y\text{Im}z_{j}}}{i-z_{j}}\mu_{1}(z_{j})\right),$
$\displaystyle M_{l}^{(1),i}=\frac{1}{2\pi
i}\int_{\mathbb{R}}\dfrac{\mu(s)w(s)}{(s-i)^{2}}ds-\sum_{j=1}^{2{N_{0}}}\left(\frac{\bar{c}_{j}e^{y\text{Im}z_{j}}}{(i-\bar{z}_{j})^{2}}\mu_{2}(\bar{z}_{j}),\frac{c_{j}e^{y\text{Im}z_{j}}}{(i-z_{j})^{2}}\mu_{1}(z_{j})\right).$

######  Proposition 3.7.

If $R\in Y,$ $y\in[0,+\infty)$, there exists a positive constant
$C(\|R\|_{Y})$ such that $M_{l}^{(1)}(i)$, $M_{l}^{(1),i}$ and
$M_{l}^{(1)}(0)$ are estimated by

$\displaystyle|M_{l}^{(1)}(i)-1|,\ |M_{l}^{(1),i}|,\ |M_{l}^{(1)}(0)|\mathchar
13366\relax C(\|R\|_{Y}).$ (3.45)

###### Proof.

From (3.30), for $z=0$,
$\sum_{j=1}^{2{N_{0}}}|\frac{\bar{c}_{j}e^{y\text{Im}z_{j}}}{\bar{z}_{j}}\mu_{2}(\bar{z}_{j})|\;{\rm
and}\;\sum_{j=1}^{2{N_{0}}}|\frac{c_{j}e^{y\text{Im}z_{j}}}{z_{j}}\mu_{1}(z_{j})|$
are bounded. And the integral term becomes

$\displaystyle|\mathcal{C}(\mu w)(0)|$
$\displaystyle=\Big{|}\mathcal{C}((\mu-I)w)+\mathcal{C}w(0)\Big{|}\mathchar
13366\relax\frac{1}{2\pi}\left(\|\mu-I\|_{L^{2}}\|w/z\|_{L^{2}}+\|w/z\|_{L^{1}}\right).$
(3.46)

It remains to analyze $\int_{\mathbb{R}}|s^{-1}w(s)|ds$ and
$\int_{\mathbb{R}}|s^{-1}w(s)|^{2}ds$. In view of the definition of $w$,
$|r|\mathchar 13366\relax 1$ and the symmetry of $r$, it is sufficient to
estimate $\int_{0}^{+\infty}|s^{-1}r(s)|ds$ and
$\int_{0}^{+\infty}|s^{-1}r(s)|^{2}ds$. Note that $r(z)\in
H^{1,2}(\mathbb{R})$. In order to estimate these two integrations near $s=0$,
namely, $\int_{0}^{1}|s^{-1}r(s)|ds$ and $\int_{0}^{1}|s^{-1}r(s)|^{2}ds$, we
use the symmetry of $r$, that is, $r(z)=\overline{r(1/z)},$ and change of the
variable $z\to 1/z$ to obtain

$\displaystyle\int_{0}^{1}|s^{-1}r(s)|ds=\int_{1}^{+\infty}s^{-1}|r(s)|ds\mathchar
13358\relax\parallel r\parallel_{L^{2}},$
$\displaystyle\int_{0}^{1}|s^{-1}r(s)|^{2}ds=\int_{1}^{+\infty}|r(s)|^{2}ds\mathchar
13366\relax\parallel r\parallel_{L^{2}}^{2}.$

Hence the estimate of $M_{l}^{(1)}(0)$ is obtained immediately. For
$M_{l}^{(1)}(i)$ and $M_{l}^{(1),i}$, applying the Hölder inequality and the
estimates in (3.30) gives the result in (3.45). ∎

Now take derivative of the inhomogeneous equation (3.20) in $y$ to obtain

$\displaystyle\partial_{y}N(z)=$
$\displaystyle\partial_{y}N_{0}(z)+\partial_{y}(\mathcal{K})N(z)+\mathcal{K}(\partial_{y}N)(z),$
(3.47) $\displaystyle\partial_{y}^{2}N(z)=$
$\displaystyle\partial_{y}^{2}N_{0}(z)+2\partial_{y}(\mathcal{K})(\partial_{y}N)(z)+\partial_{y}^{2}(\mathcal{K})N(z)+\mathcal{K}(\partial_{y}^{2}N)(z).$
(3.48)

From the definition of $\mathcal{K}$ and $N_{0}$, it transpires that

$\displaystyle\partial_{y}N_{0}(z)=\left(0,\mathcal{C}_{+}(-\frac{i}{2}(s-1/s)re^{2\theta})(z)+\sum_{j=0}^{2N_{0}}\frac{\text{Im}z_{j}c_{j}e^{\text{Im}z_{j}y}}{z-z_{j}}\right)^{T},$
$\displaystyle\partial_{y}^{2}N_{0}(z)=\left(0,\mathcal{C}_{+}(-\frac{1}{4}(s-1/s)^{2}re^{2\theta})(z)+\sum_{j=0}^{2N_{0}}\frac{\text{Im}^{2}z_{j}c_{j}e^{\text{Im}z_{j}y}}{z-z_{j}}\right)^{T},$
$\displaystyle\partial_{y}\mathcal{K}=\left(\begin{array}[]{cc}0&\partial_{y}\mathcal{K}^{(1)}\\\
\partial_{y}\mathcal{K}^{(2)}&0\end{array}\right),\
\partial_{y}^{2}\mathcal{K}=\left(\begin{array}[]{cc}0&\partial_{y}^{2}\mathcal{K}^{(1)}\\\
\partial_{y}^{2}\mathcal{K}^{(2)}&0\end{array}\right).$ (3.53)

For brevity, we omit the concrete expression of operators
$\partial_{y}\mathcal{K}^{(1)}$, $\partial_{y}\mathcal{K}^{(2)}$,
$\partial_{y}^{2}\mathcal{K}^{(1)}$ and $\partial_{y}^{2}\mathcal{K}^{(2)}$.
From $r(z)\in H^{1,2}(\mathbb{R})\cap H^{2,1}(\mathbb{R})$, $\partial_{y}K$
and $\partial_{y}^{2}K$ are also the bounded operators on $L^{2}(\mathbb{R})$.
Thus, $\partial_{y}N_{0}+(\partial_{y}K)N$ belongs to $L^{2}(\mathbb{R})$,
which implies that

$\displaystyle\partial_{y}N(z)=\left(I-\mathcal{K}\right)^{-1}(\partial_{y}N_{0}+(\partial_{y}\mathcal{K})N)(z)\in
L^{2}(\mathbb{R}).$

Namely there exists a positive constant $C(\|R\|_{Y})$ such that

$\displaystyle\|\partial_{y}\mu\|_{L^{2}}\mathchar 13366\relax C(\|R\|_{Y}).$
(3.54)

So
$\partial_{y}^{2}N_{0}+2(\partial_{y}\mathcal{K})(\partial_{y}N)+(\partial_{y}^{2}\mathcal{K})N$
also belongs to $L^{2}(\mathbb{R})$ such that

$\displaystyle\partial_{y}^{2}N=\left(I-\mathcal{K}\right)^{-1}(\partial_{y}^{2}N_{0}+2(\partial_{y}\mathcal{K})(\partial_{y}N)+(\partial_{y}^{2}\mathcal{K})N)\in
L^{2}(\mathbb{R}).$ (3.55)

Therefore there exists a positive constant $C(\|R\|_{Y})$ such that

$\displaystyle\|\partial_{y}^{2}\mu\|_{L^{2}}\mathchar 13366\relax
C(\|R\|_{Y}).$ (3.56)

We now consider the function $M_{l}^{(1)}$. A direct computation reveals that

$\displaystyle\partial_{y}M^{(1)}_{l}(z)=\mathcal{C}(\partial_{y}\mu\
w)(z)+\mathcal{C}(\mu\partial_{y}w)(z)$
$\displaystyle+\sum_{j=1}^{2{N_{0}}}\left(\frac{\bar{c}_{j}e^{y\text{Im}z_{j}}}{z-\bar{z}_{j}}(\text{Im}z_{j}\mu_{2}(\bar{z}_{j})+\partial_{y}\mu_{2}(\bar{z}_{j})),\frac{c_{j}\text{Im}z_{j}e^{y\text{Im}z_{j}}}{z-z_{j}}(\text{Im}z_{j}\mu_{1}(z_{j})+\partial_{y}\mu_{1}(z_{j}))\right),$
$\displaystyle\partial_{y}^{2}M^{(1)}_{l}(z)=\mathcal{C}(\partial_{y}^{2}\mu\
w)(z)+\mathcal{C}(\partial_{y}\mu\partial_{y}w)(z)+\mathcal{C}(\mu\partial_{y}^{2}w)(z)$
$\displaystyle+\sum_{j=1}^{2{N_{0}}}\left(\frac{\bar{c}_{j}e^{y\text{Im}z_{j}}}{z-\bar{z}_{j}}(\text{Im}z_{j}^{2}\mu_{2}(\bar{z}_{j})+2\partial_{y}\text{Im}z_{j}\mu_{2}(\bar{z}_{j})+\partial_{y}^{2}\mu_{2}(\bar{z}_{j})),0\right)$
$\displaystyle+\sum_{j=1}^{2{N_{0}}}\left(0,\frac{c_{j}\text{Im}z_{j}e^{y\text{Im}z_{j}}}{z-z_{j}}(\text{Im}^{2}z_{j}\mu_{1}(z_{j})+2\text{Im}z_{j}\partial_{y}\mu_{1}(z_{j})+\partial_{y}^{2}\mu_{1}(z_{j}))\right).$

As $z\to i$, they have the asymptotic expansions in the following,

$\displaystyle\partial_{y}M^{(1)}_{l}(z)=\partial_{y}M^{(1)}_{l}(i)+\partial_{y}M^{(1),i}_{l}(z-i)+\mathcal{O}\left((z-i)^{2}\right),$
(3.57)
$\displaystyle\partial_{y}^{2}M^{(1)}_{l}(z)=\partial_{y}^{2}M^{(1)}_{l}(i)+\partial_{y}^{2}M^{(1),i}_{l}(z-i)+\mathcal{O}\left((z-i)^{2}\right),$
(3.58)

with

$\displaystyle\partial_{y}M^{(1),i}_{l}=\frac{1}{2\pi
i}\int_{\mathbb{R}}\dfrac{\partial_{y}\mu(s)w(s)}{(s-i)^{2}}ds+\frac{1}{2\pi
i}\int_{\mathbb{R}}\dfrac{\left(\mu(s)-I\right)\partial_{y}w(s)}{(s-i)^{2}}ds+\frac{1}{2\pi
i}\int_{\mathbb{R}}\dfrac{\partial_{y}w(s)}{(s-i)^{2}}ds$
$\displaystyle-\sum_{j=1}^{2{N_{0}}}\left(\frac{\bar{c}_{j}e^{y\text{Im}z_{j}}}{(i-\bar{z}_{j})^{2}}(\text{Im}z_{j}\mu_{2}(\bar{z}_{j})+\partial_{y}\mu_{2}(\bar{z}_{j})),\frac{c_{j}\text{Im}z_{j}e^{y\text{Im}z_{j}}}{(i-z_{j})^{2}}(\text{Im}z_{j}\mu_{1}(z_{j})+\partial_{y}\mu_{1}(z_{j}))\right),$
$\displaystyle\partial_{y}^{2}M^{(1),i}_{l}=\frac{1}{2\pi
i}\int_{\mathbb{R}}\dfrac{\partial_{y}^{2}\mu(s)w(s)}{(s-i)^{2}}ds+\frac{1}{2\pi
i}\int_{\mathbb{R}}\dfrac{\partial_{y}\mu(s)\partial_{y}w(s)}{(s-i)^{2}}ds+\frac{1}{2\pi
i}\int_{\mathbb{R}}\dfrac{\left(\mu(s)-I\right)\partial_{y}^{2}w(s)}{(s-i)^{2}}ds$
$\displaystyle+\frac{1}{2\pi
i}\int_{\mathbb{R}}\dfrac{\partial_{y}^{2}w(s)}{(s-i)^{2}}ds-\sum_{j=1}^{2{N_{0}}}\left(\frac{\bar{c}_{j}e^{y\text{Im}z_{j}}}{(i-\bar{z}_{j})^{2}}(\text{Im}z_{j}^{2}\mu_{2}(\bar{z}_{j})+2\partial_{y}\text{Im}z_{j}\mu_{2}(\bar{z}_{j})+\partial_{y}^{2}\mu_{2}(\bar{z}_{j})),0\right)$
$\displaystyle-\sum_{j=1}^{2{N_{0}}}\left(0,\frac{c_{j}\text{Im}z_{j}e^{y\text{Im}z_{j}}}{(i-z_{j})^{2}}(\text{Im}^{2}z_{j}\mu_{1}(z_{j})+2\text{Im}z_{j}\partial_{y}\mu_{1}(z_{j})+\partial_{y}^{2}\mu_{1}(z_{j}))\right).$

Otherwise, the equation (3.11) gives that for $|z|\to\infty$,

$\displaystyle\lim_{z\to\infty}z(M^{(1)}_{l}(z)-I)=-\frac{1}{2\pi
i}\int_{\mathbb{R}}\left(\mu(s)-I\right)w(s)ds-\frac{1}{2\pi
i}\int_{\mathbb{R}}w(s)ds$
$\displaystyle+\sum_{j=1}^{2{N_{0}}}\left(\bar{c}_{j}e^{y\text{Im}z_{j}}\mu_{2}(\bar{z}_{j}),c_{j}e^{y\text{Im}z_{j}}\mu_{1}(z_{j})\right)$
(3.59)
$\displaystyle\lim_{z\to\infty}z\partial_{y}M^{(1)}_{l}(z)=-\frac{1}{2\pi
i}\int_{\mathbb{R}}\partial_{y}\mu(s)w(s)ds-\frac{1}{2\pi
i}\int_{\mathbb{R}}\left(\mu(s)-I\right)\partial_{y}w(s)ds-\frac{1}{2\pi
i}\int_{\mathbb{R}}\partial_{y}w(s)ds$
$\displaystyle+\sum_{j=1}^{2{N_{0}}}\left(\bar{c}_{j}e^{y\text{Im}z_{j}}(\text{Im}z_{j}\mu_{2}(\bar{z}_{j})+\partial_{y}\mu_{2}(\bar{z}_{j})),c_{j}\text{Im}z_{j}e^{y\text{Im}z_{j}}(\text{Im}z_{j}\mu_{1}(z_{j})+\partial_{y}\mu_{1}(z_{j}))\right).$
(3.60)

Similar to the proof in Lemma 2.2, combining the Hölder inequality, (3.30),
(3.54) and (3.56), it transpires that for $j=1,2,$

$\displaystyle|\partial_{y}^{j}M^{(1)}_{l}(i)|,\
|\partial_{y}^{j}M^{(1),i}_{l}|,\ |\partial_{y}^{j}M^{(1)}_{l}(0)|,\
|\lim_{z\to\infty}zM^{(1)}_{l,12}(z)|,\
|\lim_{z\to\infty}z\partial_{y}M^{(1)}_{l,12}(z)|\mathchar 13366\relax
C\left(\|R\|_{Y}\right),$ (3.61)

where the subscript $12$ means the $(1,2)$-entry element of matrix.

When $y\in(-\infty,0)$, $M_{r}^{(1)}(z)$ has a similar property, which we omit
its proof.

######  Proposition 3.8.

Supposed that $R\in Y,$ $y\in(-\infty,0)$. Then as $z\to i$, $M^{(1)}_{r}$ has
the following asymptotic expansion,

$\displaystyle
M^{(1)}_{r}(z)=M^{(1)}_{r}(i)+M^{(1),i}_{r}(z-i)+\mathcal{O}\left((z-i)^{2}\right),$

and there exists a positive constant $C(\|R\|_{Y})$ such that for $j=0,1,2$,
$\partial_{y}^{j}M^{(1)}_{r}(i)$, $\partial_{y}^{j}M^{(1),i}_{r}$,
$\partial_{y}^{j}M^{(1)}_{r}(0)$, $\lim_{z\to\infty}zM^{(1)}_{r,12}(z)$ and
$\lim_{z\to\infty}z\partial_{y}M^{(1)}_{r,12}(z)$ are bounded, that is,

$\displaystyle|\partial_{y}^{j}M^{(1)}_{r}(i)|,\
|\partial_{y}^{j}M^{(1),i}_{r}|,\
|\partial_{y}^{j}M^{(1)}_{r}(0)|,|\lim_{z\to\infty}zM^{(1)}_{r,12}(z)|,|\lim_{z\to\infty}z\partial_{y}M^{(1)}_{r,12}(z)|\mathchar
13366\relax C(\|R\|_{Y}).$

##  4\. Existence of global solutions

In view of the result in Section 3, we use $M^{(1)}_{l}(z;t,y)$ to recover the
solution $m(t,x(y,t))$ for $y\in[0,+\infty)$, and $M^{(1)}_{r}(z;t,y)$ to
recover the solution $m(t,x(y,t))$ for $y\in(-\infty,0)$ through equations
(2.127)-(2.128). The main aim of this section is to give the proof of the
following lemma.

######  Lemma 4.1.

Assume that the initial data $m_{0}>0$ and $m_{0}-1\in H^{2,1}(\mathbb{R})\cap
H^{1,2}(\mathbb{R})$. Then there exists a unique solution $m(t)\in
C([0,+\infty);W^{1,\infty}(\mathbb{R}))$ for the Cauchy problem (1.12).

The proof of Lemma 4.1 is approached via a series of propositions and lemmas.
The estimates of the functions established in the following lemma will be used
to reconstruct solution in (2.127).

######  Lemma 4.2.

Let $R(0)\in Y.$ For every $t\in[0,T),\ T>0$, the functions $\eta(t,y)$ and
$\alpha_{j}(t,y),j=1,2,3$ defined in (2.114) with the scattering data $R(t)$
in Lemma 3.1, (2.130)-(2.131) respectively satisfy

$\displaystyle\|\alpha_{1}^{-1}(t,\cdot)\|_{L^{\infty}},\
\|\alpha_{1}(t,\cdot)-1\|_{W^{2,\infty}},\
\|\alpha_{j}(t,\cdot)\|_{W^{2,\infty}}\mathchar 13366\relax
C(T,\|R(0)\|_{Y}),\ j=2,3,\quad{\rm and}$
$\displaystyle\|\eta(t,\cdot)\|_{W^{1,\infty}}\mathchar 13366\relax
C(T,\|R(0)\|_{Y}).$

Moreover,

$|u(t,y)-1|\mathchar 13366\relax C(T,\|R(0)\|_{Y}).$

###### Proof.

As shown in Lemma 3.1, for any $t\in[0,T),$ $R(t)\in Y$ with
$\|R(t)\|_{Y}\mathchar 13366\relax C(T,\|R(0)\|_{Y}).$ In view of Propositions
3.2 and 3.4, under the scattering data ${R}(t)$, the RH problem 3 has a unique
solution $M^{(1)}_{l}(z;t,y)$ for every $t\in[0,T)$. Except for
$\|1/\alpha_{1}(t,\cdot)\|_{L^{\infty}}$, the other estimates can be obtained
from Propositions 3.7 and 3.8 and (3.61) immediately. Now consider the
boundedness of $\|1/\alpha_{1}(t,\cdot)\|_{L^{\infty}}$. in fact, when
$\alpha\neq-1$, apply Proposition 3.7 to give the boundedness of $f_{j}(t,y)$
and $\frac{\beta}{\alpha+1}f_{j}(t,y)$, $j=1,2$. This implies that
$\frac{\beta}{\alpha+1}$ is also bounded. On the other hand, det$M^{(1)}\equiv
1$, namely, $\alpha_{1}f_{2}(1+\frac{\beta}{\alpha+1})=1$. Therefore,
$1/\alpha_{1}=f_{2}(1+\frac{\beta}{\alpha+1})$ is bounded. And when
$\alpha=1$, the boundedness of $1/\alpha_{1}$ is a directly result from the
boundedness of $f_{1}$. This completes the proof of Lemma 4.2. ∎

It then follows from Lemma 3.1, Lemma 4.2, and the reconstruction formulas
(2.114), (2.127) and (2.128) that there exist $u(t,y)$ and $\eta(t,y)$ as the
functions of $y$ in $W^{1,\infty}(\mathbb{R})$ for $t\in[0,T),$ with any
$T>0.$ However, the reconstruction formula of $m$ in (2.114) is

$\displaystyle m(t,y)=(1-\eta(t,y))^{-1}.$ (4.1)

So it is hard to obtain the boundedness of $m$ from the reconstruction formula
directly. Therefore, our attention is focused on the definition of $m$ in
(1.1) with $m=u-u_{xx}$. Note that
$\partial_{x}=(1+2(\partial_{y}\alpha_{1})/\alpha_{1})\partial_{y}$, so

$\displaystyle u_{x}=$
$\displaystyle(1+2(\partial_{y}\alpha_{1})/\alpha_{1})\left(-\partial_{y}(\alpha_{2})\alpha_{1}-\alpha_{1}\partial_{y}\alpha_{2}-\partial_{y}(\alpha_{3})\alpha_{1}^{-1}+\partial_{y}(\alpha_{1})\alpha_{3}\alpha_{1}^{-2}\right),$
$\displaystyle u_{xx}=$ $\displaystyle
2(\alpha_{1}\partial_{y}^{2}\alpha_{1}-\partial_{y}\alpha_{1}^{2})\alpha_{1}^{-2}u_{x}+(1+2\partial_{y}(\alpha_{1})\alpha_{1}^{-1})^{2}\left(-\partial_{y}^{2}(\alpha_{2})\alpha_{1}-2\partial_{y}\alpha_{1}\partial_{y}\alpha_{2}-\partial_{y}^{2}(\alpha_{1})\alpha_{2}\right)$
$\displaystyle+\alpha_{1}^{-4}(\alpha_{1}+2\partial_{y}\alpha_{1})^{2}\left(-\alpha_{1}\partial_{y}^{2}\alpha_{3}+2\partial_{y}\alpha_{1}\partial_{y}\alpha_{3}+\partial_{y}^{2}(\alpha_{1})\alpha_{3}-2\partial_{y}(\alpha_{1})^{2}\alpha_{3}\alpha_{1}^{-1}\right).$

Similarly as the above analysis, applying (3.61) and Proposition 3.8 gives the
boundedness of $u_{xx}$. In fact, for every $T\in\mathbb{R}^{+}$, $t\in[0,T)$,
under the scattering data $R(t)$ in Lemma 3.1, $u_{x}(t,y),$ $u_{xx}(t,y)$
defined above are bounded respectively. In addition, their $L^{\infty}$-norm
can be controlled by $C(T,\|R(0)\|_{Y})$ with

$\displaystyle|u_{x}(t,y)|,\ |u_{xx}(t,y)|\mathchar 13366\relax
C(\|R(t)\|_{Y})\mathchar 13366\relax C(T,\|R(0)\|_{Y}),$ (4.2)

Therefore, it is inferred that for every $T>0,$ $t\in[0,T)$, $m(t,y)$ exists
as a bounded function of $y$ with

$\displaystyle|m(t,y)-1|\mathchar 13366\relax C(\|R(t)\|_{Y})\mathchar
13366\relax C(T,\|R(0)\|_{Y}).$ (4.3)

Together with Lemma 4.2, it appears that for every $t\in[0,T)$, $m(t,\cdot)\in
W^{1,\infty}$.

The final step of the proof of Lemma 4.1 is to verify the continuity of
$\eta(t,y)$ with respective to $t$. Analogously, we take the case
$y\in[0,+\infty)$ as an example. The first step is to analyze the continuity
about $t$ of the RH problem.

######  Lemma 4.3.

Both of the operator $\mathcal{K}(t)$ defined in (3.23) corresponding to
scattering data $R(t)$ and $\left(I-\mathcal{K}(t)\right)^{-1}$ are continuous
at $t\in[0,T),$ for any $T>0.$

###### Proof.

From the definition of $\mathcal{K}(t)$ in (3.23), it is sufficient to give
the details of the proof of $\mathcal{K}^{(1)}_{11}(t)$ and
$\mathcal{K}^{(1)}_{12}(t)$ when $y\in[0,\infty)$. The others can be obtained
analogously. For any $t_{0}\in\mathbb{R}^{+}$ and any $f\in
L^{2}(\mathbb{R})$,

$\displaystyle\|\mathcal{K}^{(1)}_{11}(t)f-\mathcal{K}^{(1)}_{11}(t_{0})f\|_{L^{2}}=$
$\displaystyle\Big{\|}\mathcal{C}_{-}(e^{-2\theta}f\bar{r}(z;0)e^{-\frac{2iz(z^{2}-1)t_{0}}{(z^{2}+1)^{2}}}(1-e^{-\frac{2iz(z^{2}-1)(t-t_{0})}{(z^{2}+1)^{2}}}))\Big{\|}_{L^{2}}$
$\displaystyle\mathchar
13366\relax\big{\|}f\big{\|}_{L^{2}}\big{\|}e^{\frac{2iz(z^{2}-1)(t-t_{0})}{(z^{2}+1)^{2}}}-1\big{\|}_{L^{\infty}}\mathchar
13366\relax C\big{\|}f\big{\|}_{L^{2}}|t-t_{0}|.$

So $\mathcal{K}^{(1)}_{11}(t)$ is continuous on $t\in[0,T)$. And for vector
$(f_{1},...,f_{2{N_{0}}})^{T}\in\mathbb{C}^{2{N_{0}}}$,

$\displaystyle\|\mathcal{K}^{(1)}_{12}(t)(f_{1},...,f_{2{N_{0}}})^{T}-\mathcal{K}^{(1)}_{12}(t_{0})(f_{1},...,f_{2{N_{0}}})^{T}\|_{L^{2}}$
$\displaystyle=\Big{\|}\sum_{j=1}^{2{N_{0}}}\frac{f_{j}\bar{c}_{j}e^{y\text{Im}z_{j}}}{z-\bar{z}_{j}}e^{-\frac{2iz(z^{2}-1)t_{0}}{(z^{2}+1)^{2}}}(1-e^{-\frac{2iz(z^{2}-1)(t-t_{0})}{(z^{2}+1)^{2}}})\Big{\|}_{L^{2}}$
$\displaystyle\mathchar
13366\relax\sum_{j=1}^{2{N_{0}}}\Big{\|}\frac{f_{j}\bar{c}_{j}e^{y\text{Im}z_{j}}}{z-\bar{z}_{j}}\Big{\|}_{L^{2}}\Big{\|}1-e^{-\frac{2iz(z^{2}-1)(t-t_{0})}{(z^{2}+1)^{2}}}\Big{\|}_{L^{\infty}}\mathchar
13366\relax C({z_{j}}_{1}^{2{N_{0}}})|t-t_{0}|\sum_{j=1}^{2{N_{0}}}|f_{j}|.$
(4.4)

Since the map: $\mathcal{K}^{(1)}\to(\mathcal{K}^{(1)})^{-1}$ is a
homeomorphism in the Banach algebra, $\left(I-\mathcal{K}(t)\right)^{-1}$ is
also continuous on $t\in[0,T)$. ∎

######  Proposition 4.1.

Suppose that $M^{(1)}(z;t,y)$ is the solution of the RH problem 3 with
scattering data $R(t)\in Y$. Then $M^{(1)}(z;t,y)$ is continuous on
$t\in[0,T),$ for any $T>0$.

###### Proof.

Note that $\left(I-\mathcal{K}(t)\right)^{-1}$ and $N_{0}(t)$ are both
continuous on $t\in[0,T)$. Thus continuity of the function $N(z;t,y)$ on
$t\in[0,T)$ follows from the expression

$\displaystyle N(z;t)=\left(I-\mathcal{K}(t)\right)^{-1}(N_{0}(t))(z).$

From the definition of $N(z;t,y)$ in (3.13) and

$\displaystyle M^{(1)}_{l}(z)=I+\mathcal{C}(\mu
w)(z)+\sum_{j=1}^{2{N_{0}}}\left(\frac{\bar{c}_{j}e^{y\text{Im}z_{j}}}{z-\bar{z}_{j}}\mu_{2}(\bar{z}_{j}),\frac{c_{j}e^{y\text{Im}z_{j}}}{z-z_{j}}\mu_{1}(z_{j})\right),$

continuity of $M^{(1)}_{l}(z;t,y)$ on $t\in[0,T)$ is immediately accomplished.
This completes the proof of Proposition 4.1. ∎

We are now in the position to prove Lemma 4.1.

###### Proof of Lemma 4.1.

Since the initial data $m_{0}>0\ $ with $m_{0}-1\in H^{2,1}(\mathbb{R})\cap
H^{1,2}(\mathbb{R})$, applying Lemma 2.1 gives that the scattering data
$R(0)\in Y$. Thus Lemma 3.1 also reveals that after time evolution, the
scattering data $R(t)\in Y$. It is then adduced from Lemma 4.2 and the
estimate in (4.3) that $m(t,\cdot)\in W^{1,\infty},$ for any
$t\in[0,T),\,T>0$. Moreover, it is deduced from Proposition 4.1 that
$M^{(1)}(z;t,y)$ is continuous on $t\in[0,T)$. In consequence, it follows from
the reconstruction formula in (2.114) that $\eta(t)$ and $m(t)\in
W^{1,\infty}(\mathbb{R})$ are continuous on $t\in[0,T)$. This completes the
proof of Lemma 4.1. ∎

##  5\. Regularity

The aim of this section is to derive the regularity of the solution $m$ and
show that $m-1\in C([0,+\infty);H^{2,1}(\mathbb{R})\cap H^{1,2}(\mathbb{R}))$.

###  5.1. Transitions of the spectral problem

The first step is to establish a new RH problem by using the Lax pair (1.3).
When $t=0$, consider two column function vectors

$\displaystyle\tilde{\Phi}_{1}=\left(\begin{array}[]{cc}1&0\\\\[4.0pt]
-\frac{m_{0}-1}{m_{0}}&-2\lambda\end{array}\right)\Psi_{1},\hskip
22.76228pt\tilde{\Phi}_{2}=\left(\begin{array}[]{cc}-2\lambda&-\frac{m_{0}-1}{m_{0}}\\\\[4.0pt]
0&1\end{array}\right)\Psi_{2}.$ (5.5)

Then a straightforward computation shows that they satisfy two new spectral
problems

$\displaystyle\tilde{\Phi}_{j,x}=-\frac{1}{4}ik\,m_{0}\,\sigma_{3}\tilde{\Phi}_{j}+P_{j}\tilde{\Phi}_{j},\
j=1,2,$ (5.6)

with

$\displaystyle
P_{1}=\frac{i(m_{0}-1)}{2k}\left(\begin{array}[]{cc}\eta-2&1\\\\[4.0pt]
-\eta^{2}-\frac{4}{m_{0}}&-\eta+2\end{array}\right)-\left(\begin{array}[]{cc}0&0\\\\[4.0pt]
\eta_{x}&0\end{array}\right),$ $\displaystyle
P_{2}=\frac{i(m_{0}-1)}{2k}\left(\begin{array}[]{cc}\eta-2&\eta^{2}+\frac{4}{m_{0}}\\\\[4.0pt]
-1&-\eta+2\end{array}\right)-\left(\begin{array}[]{cc}0&\eta_{x}\\\\[4.0pt]
0&0\end{array}\right),$

where $\eta(0,y)=\frac{m_{0}-1}{m_{0}}$. Furthermore, define the functions
$\nu_{j}^{\pm}(x;k),j=1,2$ by the transformations

$\displaystyle\nu_{j}^{\pm}(x;k)=e^{(-1)^{j-1}p(x;k)}\tilde{\Phi}_{j}(x;k),$
(5.7)

which satisfy the following Volterra integral equations,

$\displaystyle\nu_{1}^{\pm}(x;k)=e_{1}+\int_{\pm\infty}^{x}\left(\begin{array}[]{cc}1&0\\\\[4.0pt]
0&e^{2(p(x;k)-p(s;k))}\end{array}\right)P_{1}(s,k)\nu_{1}^{\pm}(s,k)ds,$
(5.10)
$\displaystyle\nu_{2}^{\pm}(x,k)=e_{2}+\int_{\pm\infty}^{x}\left(\begin{array}[]{cc}e^{-2(p(x;k)-p(s;k))}&0\\\\[4.0pt]
0&1\end{array}\right)P_{2}(s;k)\nu_{2}^{\pm}(s,k)ds.$ (5.13)

It then transpires from a direct calculation that

######  Proposition 5.1.

If initial data $m_{0}-1\in H^{2,1}(\mathbb{R})\cap H^{1,2}(\mathbb{R})$, then
the Jost functions $\nu_{j}^{\pm}(x;k),j=1,2$ admit the following asymptotics

$\displaystyle\nu_{1}^{\pm}(x;k)=e_{1}+\mathcal{O}(k^{-1}),\quad|k|\to\infty,$
(5.14)
$\displaystyle\nu_{2}^{\pm}(x;k)=e_{2}+\left(\begin{array}[]{cc}2i\frac{\partial_{x}m_{0}}{m_{0}^{3}}\\\\[4.0pt]
b_{\pm}(x)\end{array}\right)k^{-1}+\mathcal{O}(k^{-2}),\quad|k|\to\infty,$
(5.17)

with

$b_{\pm}(x)=\frac{i}{2}\int_{\pm\infty}^{x}\frac{m_{0}(s)^{2}-1}{m_{0}(s)}ds,$

and they satisfy the following limits along a contour in the domains of their
analyticity

$\displaystyle\nu_{1}^{\pm}(x,k)=-\frac{i\alpha_{\pm}}{k}\left(\begin{array}[]{cc}1\\\\[4.0pt]
2-\eta\end{array}\right)+\mathcal{O}(1),\quad k\to 0,$ (5.20)
$\displaystyle\nu_{2}^{\pm}(x,k)=\frac{i\alpha_{\pm}}{k}\left(\begin{array}[]{cc}2-\eta\\\\[4.0pt]
1\end{array}\right)+\mathcal{O}(1),\quad k\to 0.$ (5.23)

###  5.2. A new RH problem

We now define

$\displaystyle\frac{\nu^{-}_{1}}{a^{*}(z)}-\nu^{+}_{1}=r_{2}e^{iky/2}p^{l}_{+},\hskip
8.5359pt\frac{p^{l}_{-}}{a}-p^{l}_{+}=r_{1}e^{-iky/2}\nu^{+}_{2},$ (5.24)
$\displaystyle\nu^{-}_{1}-\frac{\nu^{+}_{1}}{a^{*}(z)}=\tilde{r}_{2}e^{iky/2}p^{r}_{-},\hskip
8.5359pt\frac{p^{r}_{+}}{a}=\frac{\nu^{+}_{1}}{a^{*}(z)}r_{1}e^{-iky/2}\nu^{+}_{2}+(1-r_{1}r_{2})p^{r}_{-},$
(5.25)

where

$\displaystyle r_{1}(k)=\frac{z}{z^{2}+1}r(z),\
r_{2}(k)=\frac{z^{2}+1}{z}\overline{r(z)},\ r_{1}(k)r_{2}(k)=|r(z)|^{2},$
(5.26) $\displaystyle\tilde{r}_{1}(k)=\frac{z}{z^{2}+1}\tilde{r}(z),\
\tilde{r}_{2}(k)=\frac{z^{2}+1}{z}\overline{\tilde{r}(z)},\
\tilde{r}_{1}(k)\tilde{r}_{2}(k)=|r(z)|^{2}.$ (5.27)

Obviously, $p_{\pm}(k;y)$ are both well defined under variable $k=z-1/z$. By
using these two column vectors, we construct two functions by

$\displaystyle
M^{(2)}_{l}(k;y)=\left\\{\begin{array}[]{ll}\left(\begin{array}[]{cc}\eta-2,&1\end{array}\right)\left(\begin{array}[]{cc}\frac{\nu_{1}^{1}(k)}{a^{*}(k)},&p^{l}_{+}(k)\end{array}\right),&\text{as
}k\in\mathbb{C}^{+},\\\\[12.0pt]
\left(\begin{array}[]{cc}\eta-2,&1\end{array}\right)\left(\begin{array}[]{cc}\nu_{1}^{+}(k),&\frac{p^{l}_{-}(k)}{a(k)}\end{array}\right),&\text{as
}k\in\mathbb{C}^{-},\\\ \end{array}\right.\ y\in\mathbb{R}^{+},$ (5.34)
$\displaystyle
M^{(2)}_{r}(k;y)=\left\\{\begin{array}[]{ll}\left(\begin{array}[]{cc}\eta-2,&1\end{array}\right)\left(\begin{array}[]{cc}\nu_{1}^{1}(k),&\frac{p_{+}^{r}(k)}{a^{*}(k)}\end{array}\right),&\text{as
}k\in\mathbb{C}^{+},\\\\[12.0pt]
\left(\begin{array}[]{cc}\eta-2,&1\end{array}\right)\left(\begin{array}[]{cc}\frac{\nu_{1}^{+}(k)}{a(k)},&p_{-}^{r}(k)\end{array}\right),&\text{as
}k\in\mathbb{C}^{-},\\\ \end{array}\right.\ y\in\mathbb{R}^{-}.$ (5.41)

It is easy to see that these two functions satisfy the following two RH
problems, respectively.

######  The RH problem 4.

Find a vector-valued function $M_{l}^{(2)}(k)=M^{(2)}_{l}(k;y)$ which
satisfies

$\bullet$ Analyticity: $M_{l}^{(2)}(k)$ is meromorphic in
$\mathbb{C}\setminus\mathbb{R}$;

$\bullet$ Jump condition: $M_{l}^{(2)}(k)$ has continuous boundary values
$[M_{l}^{(2)}]_{\pm}(k)$ on $\mathbb{R}$ and

$[M^{(2)}_{l}]_{+}(k)=[M^{(2)}_{l}]_{-}(k)\mathaccent 869{V}_{l}(k),\hskip
14.22636pt\mathaccent
869{V}_{l}(k)=\left(\begin{array}[]{cc}1-|r|^{2}&r_{1}e^{-2\theta}\\\
-r_{2}e^{2\theta}&1\end{array}\right);$

$\bullet$ Asymptotic behavior:

$\displaystyle M^{(2)}_{l}(k)=(\eta-2,1)+\mathcal{O}(k^{-1}),\hskip
14.22636pt|k|\rightarrow\infty\ in\ \mathbb{C};$ (5.42)

$\bullet$ Residue condition: $M^{(2)}_{l}(k)$ has simple poles at each point
in $\\{k_{j},\bar{k}_{j}\\}_{j=1}^{{N_{0}}}$ with

$\displaystyle\operatorname*{Res}_{k=k_{j}}M^{(2)}_{l}(k)=\lim_{k\to
k_{j}}M^{(2)}_{l}(k)\left(\begin{array}[]{cc}0&d_{j}e^{2y\text{Im}z_{j}}\\\
0&0\end{array}\right),$ (5.45)
$\displaystyle\operatorname*{Res}_{k=\bar{k}_{j}}M^{(2)}_{l}(k)=\lim_{k\to\bar{k}_{j}}M^{(2)}_{l}(k)\left(\begin{array}[]{cc}0&0\\\
\tilde{d}_{j}e^{2y\text{Im}z_{j}}&0\end{array}\right).$ (5.48)

Here, $k_{j}=z_{j}-1/z_{j}$, $d_{j}=-\bar{z}_{j}c_{j}$ and
$\tilde{d}_{j}=(2\text{Re}z_{j})^{2}\bar{d}_{j}$ for $j=1,...,{N_{0}}$. For
$j=1,...,{N_{0}}$, the zeros $z_{j}$ and $z_{j+{N_{0}}}$ of $a$ in the
$z$-plane condense to one point $k_{j}$ in the $k$-plane.

######  The RH problem 5.

Find a vector-valued function $M_{r}^{(2)}(k)=M^{(2)}_{r}(k;y)$ which
satisfies

$\bullet$ Analyticity: $M_{r}^{(2)}(k)$ is meromorphic in
$\mathbb{C}\setminus\mathbb{R}$;

$\bullet$ Jump condition: $M_{r}^{(2)}(k)$ has continuous boundary values
$[M_{r}^{(2)}]_{\pm}(k)$ on $\mathbb{R}$ and

$[M^{(2)}_{r}]_{+}(k)=[M^{(2)}_{r}]_{-}(k)\mathaccent 869{V}_{r}(k),\hskip
14.22636pt\mathaccent
869{V}_{r}(k)=\left(\begin{array}[]{cc}1&\tilde{r}_{1}e^{-2\theta}\\\
-\tilde{r}_{2}e^{2\theta}&1-|r|^{2}\end{array}\right);$ (5.49)

$\bullet$ Asymptotic behavior:

$\displaystyle M^{(2)}_{r}(k)=(\eta-2,1)+\mathcal{O}(k^{-1}),\hskip
14.22636pt|k|\rightarrow\infty\ in\ \mathbb{C};$ (5.50)

$\bullet$ Residue condition: $M^{(2)}_{r}(k)$ has simple poles at each point
in $\\{k_{j},\bar{k}_{j}\\}_{j=1}^{{N_{0}}}$ with

$\displaystyle\operatorname*{Res}_{k=k_{j}}M^{(2)}_{r}(k)=\lim_{k\to
k_{j}}M^{(2)}_{r}(k)\left(\begin{array}[]{cc}0&0\\\
\bar{d}_{j}e^{-2y\text{Im}z_{j}}&0\end{array}\right),$ (5.53)
$\displaystyle\operatorname*{Res}_{k=\bar{k}_{j}}M^{(2)}_{r}(k)=\lim_{k\to\bar{k}_{j}}M^{(2)}_{r}(k)\left(\begin{array}[]{cc}0&\tilde{\bar{d}}_{j}e^{-2y\text{Im}z_{j}}\\\
0&0\end{array}\right).$ (5.56)

In view of the asymptotic expression of the Jost function in Proposition 5.1,
construction of $M^{(2)}_{l}(k)$ and $M^{(2)}_{r}(k)$ in (5.34)-(5.41), it is
thereby inferred that

$\displaystyle\frac{2im_{x}}{m^{3}}=\lim_{k\to\infty}k(M^{(2)}_{1}-\eta+2),\hskip
8.5359ptb_{+}=\frac{i}{2}\int_{+\infty}^{x}\frac{m(s)^{2}-1}{m(s)}ds=\lim_{k\to\infty}k(M^{(2)}_{2}-1),$
(5.57)

where $M^{(2)}=M_{l}^{(2)}$ for $y\in\mathbb{R}^{+}$ and $M^{(2)}=M_{r}^{(2)}$
for $y\in\mathbb{R}^{-}$ with $M^{(2)}_{j}$ as the $j$-element of $M^{(2)}$.
Similar to Section 3, when $y=0$, it is found that

$\displaystyle\lim_{k\to\infty}[M^{(2)}_{l}(k;0)-(\eta(0)-2,1)]=\lim_{k\to\infty}[M^{(2)}_{r}(k;0)-(\eta(0)-2,1)].$
(5.58)

It is our purpose here to analyze the property of $M^{(2)}$. We only give the
proof of case $y\in\mathbb{R}^{+}$, since the estimate of $y\in\mathbb{R}^{-}$
could be obtained using the same method. To see this, we have the following
lemma derived from $r\in H^{1,2}(\mathbb{R})\cap H^{2,1}(\mathbb{R})$ and the
symmetry of $r$.

######  Lemma 5.1.

If $r\in H^{1,2}(\mathbb{R})\cap H^{2,1}(\mathbb{R})$, then $r_{1}(k)\in
H^{1,3}(\mathbb{R})\cap H^{2,2}(\mathbb{R})$, $r_{2}(k)\in
W^{2,2}(\mathbb{R})\cap H^{1,1}(\mathbb{R})$, and there exists a positive
constant $C$ such that

$\displaystyle\parallel r_{1}\parallel_{H^{1,3}\cap H^{2,2}},\ \parallel
r_{2}\parallel_{W^{2,2}\cap H^{1,1}}\mathchar 13366\relax C\parallel
r\parallel_{H^{1,2}\cap H^{2,1}}.$ (5.59)

For $k\in\mathbb{R}$, denote

$\displaystyle\tilde{w}=\tilde{w}_{+}+\tilde{w}_{-},\
\tilde{w}_{+}(k)=\left(\begin{array}[]{cc}0&0\\\
-r_{2}(k)e^{2\theta}&0\end{array}\right),\
\tilde{w}_{-}(k)=\left(\begin{array}[]{cc}0&r_{1}(k)e^{-2\theta}\\\
0&0\end{array}\right),$ (5.64)

and $\mathcal{C}^{k}$ is the Cauchy operator on the $k$-plane. Then the RH
problem 4 has a solution given by

$\displaystyle
M^{(2)}_{l}(k)=(\eta-2,1)+\mathcal{C}^{k}(\tilde{\mu}\tilde{w})(k)+\sum_{j=1}^{{N_{0}}}\left(\frac{\tilde{d}_{j}e^{y\text{Im}z_{j}}}{k-\bar{k}_{j}}\tilde{\mu}_{2}(\bar{k}_{j}),\frac{d_{j}e^{y\text{Im}z_{j}}}{k-k_{j}}\tilde{\mu}_{1}(k_{j})\right),$
(5.65)

if and only if there is a solution $\tilde{\mu}$ of the Fredholm integral
equation in the $k$-plane

$\displaystyle\tilde{\mu}(k)=(\eta-2,1)+\mathcal{C}^{k}_{\tilde{w}}\tilde{\mu}(k)+\sum_{j=1}^{{N_{0}}}\left(\frac{\tilde{d}_{j}e^{y\text{Im}z_{j}}}{k-\bar{k}_{j}}\tilde{\mu}_{2}(\bar{k}_{j}),\frac{d_{j}e^{y\text{Im}z_{j}}}{k-k_{j}}\tilde{\mu}_{1}(k_{j})\right),$
(5.66)

with

$\mathcal{C}^{k}_{\tilde{w}}f(z)=\mathcal{C}^{k}_{+}(f\tilde{w}_{-})(z)+\mathcal{C}_{-}^{k}(f\tilde{w}_{+})(z).$

Introduce a column vector function

$\displaystyle\tilde{N}(k)=$
$\displaystyle\left(\tilde{\mu}_{1}(k)-(\eta-2),\tilde{\mu}_{2}(k)-1\right)^{T},$

which satisfies the following equation,

$\displaystyle\tilde{N}(k)=$
$\displaystyle\tilde{N}_{0}(k)+\tilde{\mathcal{K}}\tilde{N}(k),\hskip
8.5359pt\tilde{\mathcal{K}}=\left(\begin{array}[]{cc}0&\tilde{\mathcal{K}}^{(1)}\\\
\tilde{\mathcal{K}}^{(2)}&0\end{array}\right),$ (5.69)
$\displaystyle\tilde{N}_{0}(k)=$
$\displaystyle\left((\eta-2)\mathcal{C}_{-}^{k}(-r_{2}e^{-2\theta})(z)+\sum_{j=1}^{N_{0}}(\eta-2)\frac{\tilde{d}_{j}e^{\text{Im}z_{j}y}}{k-\bar{k}_{j}},\mathcal{C}_{+}^{k}(r_{1}e^{2\theta})(z)+\sum_{j=1}^{N_{0}}\frac{d_{j}e^{\text{Im}z_{j}y}}{k-k_{j}}\right)^{T}.$
(5.70)

where the definitions of $\tilde{\mathcal{K}^{(1)}}$ and
$\tilde{\mathcal{K}^{(2)}}$ are similar to (3.25)-(3.26). We now verify that
$\tilde{N}\in L^{2}(\mathbb{R})$. In fact, it is shown in Proposition 3.2 that
$I-\mathcal{K}$ and $(I-\mathcal{K})^{-1}$ are two bounded linear operators
on: $L^{2}(\mathbb{R})\to L^{2}(\mathbb{R})$ of the $z$-plane. It suffices to
derive the estimate of bound for the integral operators on the $k$-plane.

In the following proposition, the superscript for the $z$-plane and the
$k$-plane is used on two Cauchy projection operators with
$\mathcal{C}_{\pm}^{k}$, $\mathcal{C}^{k}$ and $\mathcal{C}_{\pm}^{z}$,
$\mathcal{C}^{z}$.

######  Proposition 5.2.

If $f\in L^{p}(\mathbb{R})$, $1\mathchar 13366\relax p<\infty$ and
$f(z)=f(-\frac{1}{z})$, then

$\displaystyle\mathcal{C}_{\pm}^{z,(e)}f(z)=\mathcal{C}_{\pm}^{k}f(k),$ (5.71)

where

$\displaystyle\mathcal{C}_{\pm}^{z,(e)}f(z)=\frac{1}{2\pi
i}\left(\int_{-\infty}^{-1}+\int_{-1}^{0}+\int_{0}^{1}+\int_{1}^{\infty}\right)\frac{f(s)ds}{s-(z\pm
0i)}.$ (5.72)

If $g\in L^{p}(\mathbb{R})$, $1\mathchar 13366\relax p<\infty$ and
$g(z)=-g(-\frac{1}{z})$, then

$\displaystyle\mathcal{C}_{\pm}^{z,(o)}g(z)=\mathcal{C}_{\pm}^{k}g(k),$ (5.73)

where

$\displaystyle\mathcal{C}_{\pm}^{z,(o)}g(z)=\frac{1}{2\pi
i}\left(\int_{-1}^{-\infty}+\int_{-1}^{0}+\int_{0}^{1}+\int_{\infty}^{1}\right)\frac{g(s)ds}{s-(z\pm
0i)}.$ (5.74)

In addition,

$\displaystyle\left(z+\frac{1}{z}\right)\mathcal{C}_{\pm}^{z,(e)}f(z)=\mathcal{C}_{\pm}^{z,(e)}\left((z+\frac{1}{z})f\right)(z),$
(5.75)
$\displaystyle\left(z+\frac{1}{z}\right)\mathcal{C}_{\pm}^{z,(o)}g(z)=\mathcal{C}_{\pm}^{z,(o)}\left((z+\frac{1}{z})g\right)(z).$
(5.76)

This proposition is a directly consequence of

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

The following proposition about $\tilde{\mathcal{K}}$ is deduced in the same
way as Proposition 3.2.

######  Proposition 5.3.

The operator $I-\tilde{\mathcal{K}}$ is invertible on $L^{2}(\mathbb{R})$ and
$(I-\tilde{\mathcal{K}})^{-1}$ is also a bounded operator. Moreover, there
exists a positive constant $C(\|R(0)\|_{Y})$ such that

$\displaystyle\|(I-\tilde{\mathcal{K}})^{-1}\|\mathchar 13366\relax
C(\|R(0)\|_{Y}).$ (5.77)

Therefore, $\tilde{N}$ is solvable by

$\displaystyle\tilde{N}(k)=(I-\tilde{\mathcal{K}})^{-1}(\tilde{N}_{0})(k),$
(5.78)

with

$\displaystyle\|\tilde{N}\|_{L^{2}}\mathchar 13366\relax
C(\|R(0)\|_{Y}\|\tilde{N}_{0}\|_{L^{2}}.$ (5.79)

###### Proof.

Same as in Proposition 3.2, $\tilde{\mathcal{K}}^{2}$ is a compact operator,
and $I-\tilde{\mathcal{K}}$ is injective. In fact, if a vector function

$\tilde{f}(k)=(f_{1}(k),f_{2}(k))$

is a solution of equation $\tilde{\mathcal{K}}f=f$, it then follows from
Proposition 5.2 that

$f(z)=((z+1/z)^{-1}f_{1}(z-1/z),f_{2}(z-1/z))$

is a solution of $\mathcal{K}f=f$. Thus the kernel of $\tilde{\mathcal{K}}$
must be trivial. So it is inferred that $(I-\tilde{\mathcal{K}})^{-1}$ exists
as a bounded operator. To obtain $\|\tilde{\mathcal{K}}^{2}\|\to 0$ as
$y\to\infty$, similarly it only needs to analyze the integral kernel of
$\mathcal{F}\tilde{K}^{(2)}_{1}\tilde{K}_{1}^{(1)}\mathcal{F}^{-1}$: for $f\in
L^{2}(\mathbb{R})$,

$\displaystyle(\tilde{K}^{(2)}_{1}\tilde{K}_{1}^{(1)}\check{f})^{\wedge}(z)=\int_{\mathbb{R}}\tilde{K}(z,\lambda)f(\lambda)d\lambda,$

where

$\displaystyle\tilde{K}(z,\lambda)=X_{+}(z)\int_{-y}^{-\infty}\hat{r}_{2}(z-s)\hat{r}_{1}(s-\lambda)ds.$

Then it is concluded that as $y\to\infty$,

$\displaystyle\|\tilde{K}\|_{L^{2}\times L^{2}}\mathchar 13358\relax\langle
y\rangle^{-1}\|\hat{r}_{2}\|_{L^{2}}\|\hat{r}_{1}\|_{L^{2,2}}\to 0.$

This completes the proof of Proposition 5.3. ∎

In view of Proposition 5.3, it also gives solvability of $\tilde{\mu}$ for the
RH problem 4.

######  Lemma 5.2.

For $R\in Y$, $y\in\mathbb{R}^{+}$, there exists a unique solution for the RH
problem 4 with

$\displaystyle
M^{(2)}_{l}(k)=(\eta-2,1)+\mathcal{C}^{k}(\tilde{\mu}\tilde{w})(k)+\sum_{j=1}^{{N_{0}}}\left(\frac{\tilde{d}_{j}e^{y\text{Im}z_{j}}}{k-\bar{k}_{j}}\tilde{\mu}_{2}(\bar{k}_{j}),\frac{d_{j}e^{y\text{Im}z_{j}}}{k-k_{j}}\tilde{\mu}_{1}(k_{j})\right).$
(5.80)

Moreover,

$\displaystyle\parallel\tilde{\mu}-(\eta-2,1)\parallel_{L^{2}}+\sum_{j=1}^{{N_{0}}}\big{|}\tilde{\mu}_{1}(k_{j})-(\eta-2)\big{|}+\sum_{j=1}^{{N_{0}}}\big{|}\tilde{\mu}_{2}(\bar{k}_{j})-1\big{|}\mathchar
13366\relax C(\|R\|_{Y})\|\tilde{N}_{0}\|_{L^{2}}.$

The estimates in the following lemma are decisively arriving at the
$y$-integrability.

######  Lemma 5.3.

For any $f,g\in W^{1,2}(\mathbb{R})$, it appears that

$\displaystyle\sup_{y\in\mathbb{R}^{+}}\parallel\mathcal{C}_{\pm}(fe^{\mp
i\cdot y/2})(\cdot)\parallel_{L^{\infty}}\mathchar 13366\relax C\parallel
f\parallel_{W^{1,2}},$ (5.81)
$\displaystyle\sup_{y\in\mathbb{R}^{+}}\parallel\langle
y\rangle\mathcal{C}_{\pm}(fe^{\mp i\cdot
y/2})(\cdot)\parallel_{L^{2}}\mathchar 13366\relax C\parallel
f\parallel_{W^{1,2}},$ (5.82) $\displaystyle\sup_{y\in\mathbb{R}^{+}}\langle
y\rangle^{2}\Big{|}\int_{\mathbb{R}}f(s)e^{\pm
isy/2}\mathcal{C}_{\pm}(-ge^{\mp iky/2})(s)ds\Big{|}\mathchar 13366\relax
C\parallel g\parallel_{W^{1,2}}\parallel f\parallel_{W^{1,2}},$ (5.83)

where $\langle y\rangle=(1+y^{2})^{1/2}$ and $C$ is a positive constant.
Moreover, when $f,g\in W^{2,2}$

$\displaystyle\sup_{y\in\mathbb{R}^{+}}\parallel\langle
y\rangle^{2}\mathcal{C}_{\pm}(fe^{i\cdot
y/2})(\cdot)\parallel_{L^{2}}\mathchar 13366\relax C\parallel
f\parallel_{W^{2,2}},$ (5.84) $\displaystyle\sup_{y\in\mathbb{R}^{+}}\langle
y\rangle^{4}\Big{|}\int_{\mathbb{R}^{+}}g(s)e^{-isy/2}\mathcal{C}_{+}(-fe^{iky/2})(s)ds\Big{|}\mathchar
13366\relax C\parallel f\parallel_{W^{2,2}}\parallel g\parallel_{W^{2,2}},$
(5.85)

###### Proof.

Take $\mathcal{C}_{-}$ as an example. It is noted that

$\displaystyle\mathcal{C}_{-}(fe^{iky/2})(k)=-(X_{-}(fe^{i\cdot
y/2})^{\wedge})^{\vee}(k).$

So

$\displaystyle\parallel\mathcal{C}_{\pm}(fe^{\mp i\cdot
y/2})(\cdot)\parallel_{L^{\infty}}\mathchar 13366\relax\parallel
X_{\pm}(fe^{i\cdot y/2})^{\wedge}\parallel_{L^{1}}\mathchar
13366\relax\parallel\hat{f}\parallel_{L^{2,1}}.$

On the other hand,

$\displaystyle\parallel\mathcal{C}_{\pm}(fe^{\mp i\cdot
y/2})(\cdot)\parallel_{L^{2}}=\|X_{\pm}(fe^{i\cdot
y/2})^{\wedge}\parallel_{L^{2}}=\left(\int_{-\infty}^{0}\big{|}\hat{f}(z-\frac{y}{4\pi})\big{|}^{2}dz\right)^{1/2}$
$\displaystyle=\left(\int_{-\infty}^{-\frac{y}{4\pi}}\big{|}\hat{f}(z)\big{|}^{2}dz\right)^{1/2}\mathchar
13358\relax\langle
y\rangle^{-1}\left(\int_{-\infty}^{-\frac{y}{4\pi}}\big{|}\langle
z\rangle\hat{f}(z)\big{|}^{2}dz\right)^{1/2}\mathchar 13366\relax\langle
y\rangle^{-1}\parallel f\parallel_{W^{1,2}}.$

The last step follows from that $\langle y\rangle\mathchar 13358\relax\langle
s\rangle$ for $y\geq 0$. And for (5.83), we have

$\displaystyle\int_{\mathbb{R}}f(s)e^{isy/2}\mathcal{C}_{+}(-ge^{-iky/2})(s)ds=-\int_{\mathbb{R}}\int_{\mathbb{R}}f(s)e^{isy/2}X_{+}(k)\hat{g}(k+\frac{y}{4\pi})e^{2\pi
iks}dkds$
$\displaystyle=-\int_{\frac{y}{4\pi}}^{+\infty}\check{f}(k)\hat{g}(k)ds.$

Then (5.83) is accomplished in the same calculation. And when $f\in W^{2,2}$,
the above inequality becomes

$\displaystyle\parallel\mathcal{C}_{\pm}(fe^{\mp i\cdot
y/2})(\cdot)\parallel_{L^{2}}\mathchar 13358\relax\langle
y\rangle^{-2}\left(\int_{-\infty}^{-\frac{y}{4\pi}}\big{|}\langle
z\rangle^{2}\hat{f}(z)\big{|}^{2}dz\right)^{1/2}\mathchar 13366\relax\langle
y\rangle^{-2}\parallel f\parallel_{W^{2,2}},$
$\displaystyle\big{|}\int_{\mathbb{R}}f(s)e^{isy/2}\mathcal{C}_{+}(-ge^{-iky/2})(s)ds\big{|}\mathchar
13366\relax\langle y\rangle^{-4}\parallel f\parallel_{W^{2,2}}\parallel
g\parallel_{W^{2,2}},$

thereby concluding the proof of Lemma 5.3. ∎

Applying Lemma 5.3 finally leads to for $j=0,1,2$,

$\displaystyle\langle y\rangle^{j}\|\tilde{N}_{0}\|_{L^{2}}\mathchar
13366\relax C(\|R\|_{Y}).$

Combining with Lemma 5.2, it is deduced that

$\displaystyle\langle
y\rangle^{j}\left[\parallel\tilde{\mu}-(\eta-2,1)\parallel_{L^{2}}+\sum_{j=1}^{{N_{0}}}\big{|}\tilde{\mu}_{1}(k_{j})-(\eta-2)\big{|}+\sum_{j=1}^{{N_{0}}}\big{|}\tilde{\mu}_{2}(\bar{k}_{j})-1\big{|}\right]\mathchar
13366\relax C(\|R\|_{Y}).$ (5.86)

Consider the $y$-derivative of $M^{(2)}_{l}$. Invoking Lemma 5.2, it is
adduced that

$\displaystyle\partial_{y}M^{(2)}_{l}(k)=$
$\displaystyle(\partial_{y}\eta,0)+\mathcal{C}^{k}(\partial_{y}\tilde{\mu}\tilde{w})(k)+\mathcal{C}^{k}(\tilde{\mu}\partial_{y}\tilde{w})(k)+\sum_{j=1}^{{N_{0}}}\left(\frac{\tilde{d}_{j}e^{y\text{Im}z_{j}}}{k-\bar{k}_{j}}\partial_{y}\tilde{\mu}_{2}(\bar{k}_{j}),\frac{d_{j}e^{y\text{Im}z_{j}}}{k-k_{j}}\partial_{y}\tilde{\mu}_{1}(k_{j})\right)$
$\displaystyle+\sum_{j=1}^{{N_{0}}}\left(\frac{\text{Im}z_{j}\tilde{d}_{j}e^{y\text{Im}z_{j}}}{k-\bar{k}_{j}}\tilde{\mu}_{2}(\bar{k}_{j}),\frac{\text{Im}z_{j}d_{j}e^{y\text{Im}z_{j}}}{k-k_{j}}\tilde{\mu}_{1}(k_{j})\right),$

where $\partial_{y}\tilde{\mu}$ satisfies that

$\displaystyle\partial_{y}\tilde{N}(k)=$
$\displaystyle\left(\partial_{y}\tilde{\mu}_{1}(k)-\partial_{y}\eta,\partial_{y}\tilde{\mu}_{1}(k_{1})-\partial_{y}\eta,...,\partial_{y}\tilde{\mu}_{1}(k_{{N_{0}}})-\partial_{y}\eta\right)^{T}$
$\displaystyle\otimes\left(\partial_{y}\tilde{\mu}_{2}(k),\partial_{y}\tilde{\mu}_{2}(\bar{k}_{1}),...,\partial_{y}\tilde{\mu}_{2}(\bar{k}_{{N_{0}}})\right)^{T},$
$\displaystyle\partial_{y}\tilde{N}(k)=$
$\displaystyle\partial_{y}\tilde{N}_{0}(k)+(\partial_{y}\tilde{K})\tilde{N}(k)+\tilde{K}(\partial_{y}\tilde{N})(k),\hskip
8.5359pt\partial_{y}\tilde{K}=\left(\begin{array}[]{cc}0&\partial_{y}\tilde{A}\\\
\partial_{y}\tilde{B}&0\end{array}\right),$
$\displaystyle\partial_{y}\tilde{N}_{0}(k)=$
$\displaystyle\mathcal{C}_{-}^{k}(-(\frac{ik(\eta-2)}{2}+\partial_{y}\eta)r_{2}e^{-2\theta})(z)$
$\displaystyle\otimes\left(((\eta-2)\text{Im}z_{1}+\partial_{y}\eta)\frac{\tilde{d}_{1}e^{\text{Im}z_{1}y}}{k-\bar{k}_{1}},...,((\eta-2)\text{Im}z_{{N_{0}}}+\partial_{y}\eta)\frac{\tilde{d}_{{N_{0}}}e^{\text{Im}z_{{N_{0}}}y}}{k-\bar{k}_{{N_{0}}}}\right)^{T}$
$\displaystyle\otimes\left(\mathcal{C}_{+}^{k}(-ikr_{1}e^{2\theta})(z),\frac{\text{Im}z_{1}d_{1}e^{\text{Im}z_{1}y}}{k-k_{1}},...,\frac{\text{Im}z_{{N_{0}}}d_{{N_{0}}}e^{\text{Im}z_{{N_{0}}}y}}{k-k_{{N_{0}}}}\right)^{T}.$

It in turn implies from $\|\partial_{y}\tilde{K}\|\mathchar
13358\relax\|R(0)\|_{Y}$ that
$\partial_{y}\tilde{N}_{0}+(\partial_{y}\tilde{K})\tilde{N}\in L^{2}$ and