# The asymptotic stability of solitons in the focusing Hirota equation on the
line

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

In this paper, the $\overline{\partial}$-steepest descent method and Bäcklund
transformation are used to study the asymptotic stability of solitons for the
following Cauchy problem of focusing Hirota equation

$\displaystyle
iq_{t}+\alpha(2|q|^{2}q+q_{xx})+i\beta(q_{xxx}+6|q|^{2}q_{x})=0,$
$\displaystyle q(x,0)=q_{0}(x),$

where $q_{0}\in
H^{1}(\mathbb{R})\,\cap\,L^{2,s}(\mathbb{R}),s\in(\frac{1}{2},1].$ We first
express the solution of the Cauchy problem in term of the solution of a
Riemann-Hilbert (RH) problem. Then the RH problem is further decomposed into
pure radiation solution and solitons solution, which are solved by using
$\overline{\partial}$-techniques and Bäcklund transformation respectively. As
a directly consequence, we obtain the asymptotic stability of solitons for the
Hirota equation.

###### keywords:

Hirota equation; Riemann-Hilbert problem; $\overline{\partial}$-steepest
descent method, Bäcklund transformation; asymptotic stability. Mathematics
Subject Classification: 35P25; 35Q51; 35Q15; 35A01; 35G25.

††journal:

###### Contents

  1. 1 Introduction and main results
  2. 2 Direct and inverse scattering transforms
    1. 2.1 Notations
    2. 2.2 Jost functions and scattering data
    3. 2.3 A Riemann-Hilbert problem
  3. 3 Dispersion for pure radiation solutions
    1. 3.1 A regular RH problem
    2. 3.2 A mixed RH Problem and its decomposition
    3. 3.3 A solvable model near $z_{1}$ and $z_{2}$
    4. 3.4 The $\overline{\partial}$ argument
  4. 4 The asymptotic stability of the solitons
    1. 4.1 The Bäcklund transformation
    2. 4.2 The proof of main result
  5. A A parabolic cylinder model
  6. B RH problem under BT

##  1 Introduction and main results

In this paper, we apply $\overline{\partial}$-steepest descent method and
Bäcklund transformation (BT) to obtain the asymptotic stability of solitons
for the Cauchy problem of the Hirota equation,

$\displaystyle
iq_{t}+\alpha(2|q|^{2}q+q_{xx})+i\beta(q_{xxx}+6|q|^{2}q_{x})=0,$ (1.1)
$\displaystyle q(x,0)=q_{0}(x),\ \ (x,t)\in\mathbb{R}\times\mathbb{R}^{+},$
(1.2)

where $q_{0}\in
H^{1}(\mathbb{R})\,\cap\,L^{2,s}(\mathbb{R}),s\in(\frac{1}{2},1]$, the real
constants $\alpha$ and $\beta$ stand for the second-order and third-order
dispersions. The Hirota equation (1.1) is a typically mathematical physical
model, which encompasses the well-known NLS equation and derivative NLS
equation [1, 2]. It is a more accurate approximation than the NLS equation in
describing wave propagation in the ocean and optical fiber [3, 4, 5].

In recent years, much work has been done to study the various mathematical
properties of the Hirota equation. For example, exact solutions such as
multisoliton solutions, breather solutions, rational solutions and rogue wave
solutions for the Hirota equation was widely studied [6, 7, 8, 9]. The
$N$-soliton solutions for the Hirota equation with non-zero boundary condition
were constructed by using the Riemann-Hilbert (RH) method [10]. The initial
boundary value problem for the Hirota Equation on the half line was analyzed
by using the Fokas unified method [11]. The long time asymptotics for the
Hirota equation with decaying initial data was investigated via the nonlinear
steepest descent method [12]. We further obtained the Painlevé asymptotics for
the Hirota equation with Schwartz Cauchy data in the transition region [13].
It was shown that the Cauchy problem for the Hirota equation is globally well-
posed in the space $H^{s}(\mathbb{R}),s\geq 1$ [14], and admits the
conservation law $||q(t)||_{L^{2}}=||q_{0}||_{L^{2}},\ t\in\mathbb{R}$.
Further the initial-boundary vale problem for the Hirota equation is well-
posed in the space $H^{s}(\mathbb{R}),s>1/2$ [15]. The orbital stability of
solitons for the Hirota equation in Sobolev space $H^{1}(\mathbb{R})$ was
shown [16].

In this paper, we use the $\overline{\partial}$-steepest descent method and
Bäcklund transformation to acquire the asymptotic stability of solitons for
the Hirota equation in Sobolev space
$H^{1}(\mathbb{R})\,\cap\,L^{2,s}(\mathbb{R})$. The asymptotic analysis of
solitons for focusing Hirota equation is necessarily more detailed than the
defocusing case due to the presence of solitons which correspond to discrete
spectrum of the non self-adjoint operator associated with (1.1).

We are interested in asymptotic stability of solitons in the Hirota equation
(1.1) given by the explicit expressions [17]

$\displaystyle q_{(\eta,\xi,\gamma,\beta,\alpha)}(x,t)$ $\displaystyle=2\eta
e^{2i(-\xi
x-4\beta\xi^{3}t-2\alpha\xi^{2}t+12\beta\xi\eta^{2}t+2\alpha\eta^{2}t)+i\gamma}$
$\displaystyle\times\mbox{sech}(-2\eta
x-24\beta\eta\xi^{2}t+8\beta\eta^{3}t-8\alpha\eta\xi t),$ (1.3)

where $\lambda=\xi+i\eta$ is a complex nonzero parameter that determines by
solitons. We consider here the question of their asymptotic stability that is
when $q_{0}$ is close to $q_{(\eta,\xi,\gamma)}$ for a particular
$(\eta,\xi,\gamma)$. Our principal result is now stated as follows.

######  Theorem 1.1.

Let $q_{0}\in H^{1}(\mathbb{R})\,\cap\,L^{2,s}(\mathbb{R})$ for fixed
$s\in(1/2,1]$ and a soliton $q_{(\eta,\xi,\gamma)}(0,x)$ of the Hirota
equation. Then, there exist positive constants
$\epsilon_{0}=\epsilon_{0}(\eta_{0},\xi_{0})$, $T=T(\eta_{0},\xi_{0})$,
$C=C(\eta_{0},\xi_{0})$ and if

$\epsilon:=||q_{(\eta_{0},\xi_{0},\gamma_{0})}(\cdot,0)-q_{0}||_{H^{1}(\mathbb{R})\,\cap\,L^{2,s}(\mathbb{R})}<\epsilon_{0}.$
(1.4)

then there exist a state $q_{(\eta_{1},\xi_{1},\gamma_{1})}(x,t)$ such that
for the solution of the Cauchy problem (1.1)-(1.2), we have

$|(\eta_{1},\xi_{1},\gamma_{1},x)-(\eta_{0},\xi_{0},\gamma_{0},,x_{0})|<C\epsilon,$
(1.5)

and for all $|t|\geq T$,

$||q(\cdot,t)-q_{(\eta_{1},\xi_{1},\gamma_{1})}(\cdot,t)||_{L^{\infty}}<C\epsilon|t|^{-\frac{1}{2}}.$
(1.6)

Organization of the paper is as follows. In Section 2, we describe the inverse
scattering transform and the solution to the Cauchy problem (1.1)-(1.2) is
characterized in term of the solution of a RH problem 2.1 (see below). In
Section 3, we introduce the $\overline{\partial}$ analysis to define
continuous extensions of the jump matrix and we construct a global model
solution which captures the leading order asymptotic behavior of the solution.
Removing this component of the solution results $\overline{\partial}$ problem
which we analyzed at last. In Section 4 Bäcklund transformation establish the
relation between soliton and soliton free solutions. We estimates the norm of
the transformation in Lemma 4.6, and further give the proof of Theorem 1.1.

##  2 Direct and inverse scattering transforms

In this section, we will analyze the spectral problem of the focusing Hirota
equation (1.1) to obtain Jost functions and scattering matrix related to
initial value $q_{0}$.

###  2.1 Notations

With regard to complex variables, given a variable $z$ or a function $f(z)$,
we denote by $z^{*}$ and $f^{*}(z)$ their respective complex conjugates; The
symbol $\overline{\partial}$ denoted the derivative with respect to $z^{*}$,
i.e. if $z=x+iy$, then

$\overline{\partial}f=\frac{1}{2}(f_{x}+if_{y}).$

We introduce the Japanese bracket $\langle x\rangle=\sqrt{1+x^{2}}$ and the
following normed spaces: A weighted space $L^{p,s}(\mathbb{R})$ defined with
the norm

$\|q\|_{L^{p,s}(\mathbb{R})}:=\|\langle x\rangle^{s}q\|_{L^{p}(\mathbb{R})};$

A Sobolev space $W^{k,p}(\mathbb{R})$ defined with the norm
$\|q\|_{W^{k,p}(\mathbb{R})}:=\sum_{j=0}^{k}\|\partial^{j}q\|_{L^{p}(\mathbb{R})},$
where $\partial^{j}q$ is the $j^{th}$ weak derivative of $q$;

A Sobolev space $H^{k}(\mathbb{R})$ defined with the norm
$\|q\|_{H^{k}(\mathbb{R})}:=\|\langle
x\rangle^{k}\mathcal{F}(q)\|_{L^{2}(\mathbb{R})},$ where $\mathcal{F}(q)$ is
the Fourier transform of $q$;

Recall that $L^{2,s}(\mathbb{R})$ is embedded into $L^{1}(\mathbb{R})$ for any
$s>\frac{1}{2}$. Based on this fact we consider the potential function
$q(x)\in L^{1}(\mathbb{R})$ for simplicity.

###  2.2 Jost functions and scattering data

The focusing Hirota equation (1.1) is just a compatibility condition of the
system of linear partial differential equation

$\displaystyle\Psi_{x}=M(x,t;z)\Psi,\quad M(x,t;z)=-iz\sigma_{3}+Q,$ (2.1)
$\displaystyle\Psi_{t}=N(x,t;z)\Psi,\quad\,N(x,t;z)=-i(4\beta z^{3}+2\alpha
z^{2})\sigma_{3}+V,$ (2.2)

where $\Psi=\Psi(x,t;z)$ is a 2$\times$2 matrix-valued eigenfunction,
$z\in\mathbb{C}$ is a spectral parameter, and

$Q=\begin{pmatrix}0&q(x,t)\\\ -q^{*}(x,t)&0\end{pmatrix},\ \
\sigma_{3}=\begin{pmatrix}1&0\\\ 0&-1\end{pmatrix},$ $V=4\beta
z^{2}Q+zV_{1}+V_{0},\quad V_{1}=2\alpha Q-2i\beta(Q_{x}+Q^{2})\sigma_{3},$
$V_{0}=-i\alpha(Q_{x}+Q^{2}+q_{0}^{2})\sigma_{3}+\beta[Q_{x},Q]+\beta(2Q^{3}-Q_{xx}).$

The Lax pair (2.1)-(2.2) admits a Jost function with asymptotics

$\Psi(x,t;z)\thicksim e^{-it\theta(z)\sigma_{3}},\quad|x|\to\infty,$ (2.3)

where

$\theta=\theta(x,t;z)=z\frac{x}{t}+2\alpha z^{2}+4\beta z^{3}.$ (2.4)

Define the eigenfunctions $\Phi(x,t;z)$

$\Phi(x,t;z)=\Psi(x,t;z)e^{it\theta(z)\sigma_{3}},$ (2.5)

then $\Phi(x,t;z)\to\mathnormal{I},\ \ |x|\to\infty$, and we obtain a
equivalent Lax pair

$\displaystyle\Phi_{x}+iz[\sigma_{3},\Phi]=Q\Phi,$ (2.6)
$\displaystyle\Phi_{t}+i(2\alpha z^{2}+4\beta z^{3})[\sigma_{3},\Phi]=V\Phi,$
(2.7)

which can be written in full derivative form

$d(e^{it\theta(z)\hat{\sigma}_{3}}\Phi)=e^{it\theta(z)\hat{\sigma}_{3}}[(Qdx+Vdt)\Phi].$
(2.8)

These solutions can be expressed as Volterra type integrals

$\displaystyle\Phi_{-}(x,t;z)=\mathnormal{I}+\int_{-\infty}^{x}e^{-iz(x-y)\hat{\sigma_{3}}}Q(y,t)\Phi_{-}(y,t;z)dy,$
(2.9)
$\displaystyle\Phi_{+}(x,t;z)=\mathnormal{I}-\int_{x}^{+\infty}e^{-iz(x-y)\hat{\sigma_{3}}}Q(y,t)\Phi_{+}(y,t;z)dy,$
(2.10)

where $e^{\hat{\sigma}_{3}}A=e^{\sigma_{3}}Ae^{-\sigma_{3}}$ with $A$ is
$2\times 2$ matrix.

Let $\Phi_{\pm,m}(x,t;z)(m=1,2)$ denote the $m$-th column of
$\Phi_{\pm}(x,t;z)$, then it can be shown that $\Phi_{-,1}(x,t;z)$ and
$\Phi_{+,2}(x,t;z)$ are analytic in $C^{+}=\\{z\,|\,\mbox{Im}z>0\\}$ and
continuous in $C^{+}\cup\,\mathbb{R}=\\{z\,|\,\mbox{Im}z\geq 0\\}$,
$\Phi_{+,1}(x,t;z)$ and $\Phi_{-,2}(x,t;z)$ are analytic in
$C^{-}=\\{z\,|\,\mbox{Im}z<0\\}$ and continuous in
$C^{-}\cup\mathbb{R}=\\{z\,|\,\mbox{Im}z\leq 0\\}$.

Since $\Phi_{+}(x,t;z)e^{-it\theta(z)\sigma_{3}}$ and
$\Phi_{-}(x,t;z)e^{-it\theta(z)\sigma_{3}}$ are fundamental matrix solution of
the Lax pair (2.1)-(2.2), there exists a scattering matrix
$S(z)=(s_{ij}(z))_{2\times 2}$ satisfying

$\Phi_{-}(x,t;z)=\Phi_{+}(x,t;z)e^{-it\theta(z)\hat{\sigma}_{3}}S(z),\quad
z\in\mathbb{R}.$ (2.11)

The scattering coefficients can be expressed by using Wronskians

$\displaystyle s_{11}(z)=\det(\Phi_{-,1}(x,t;z)\quad\Phi_{+,2}(x,t;z)),$
$\displaystyle
s_{12}(z)=e^{-2it\theta(z)}\det(\Phi_{+,2}(x,t;z)\quad\Phi_{-,2}(x,t;z)),$
$\displaystyle
s_{21}(z)=e^{2it\theta(z}\det(\Phi_{+,1}(x,t;z)\quad\Phi_{-,1}(x,t;z)),$
$\displaystyle s_{22}(z)=\det(\Phi_{-,2}(x,t;z)\quad\Phi_{+,1}(x,t;z)),$

where scattering data $s_{11}(z)$ is analytic in $C^{+}$ and continuous on
$C^{+}\cup\mathbb{R}$ and $s_{22}(z)$ is analytic in $C^{-}$ and continuous on
$C^{-}\cup\mathbb{R}$. In addition, $s_{12}(z)$ and $s_{21}(z)$ are continuous
in $\mathbb{R}$. It follows the symmetric relation

$\Phi(x,t;z)=\sigma_{2}\Phi^{*}(x,t;z^{*})\sigma_{2},\quad
S(z)=\sigma_{2}S^{*}(z^{*})\sigma_{2}.$

We can obtain $s_{11}(z)=s_{22}^{*}(z^{*})$ and
$s_{12}(z)=-s_{21}^{*}(z^{*})$. Define reflection coefficient

$r(z):=\frac{s_{21}(z)}{s_{11}(z)},\,\,z\in\mathbb{R},$ (2.12)

then it can be shown that [18, 19].

######  Lemma 2.1.

There exists an open dense set $\mathcal{G}\subset L^{1}(\mathbb{R})$ such
that, for $q\in\mathcal{G}$, the scattering function $s_{11}(z)$ has at most a
finite number of zeros forming a set $\mathcal{Z}_{+}=\\{z_{1},\dots,z_{N}\\}$
in $\mathbb{C}_{+}$, with $s_{11}(z)\neq 0$ for all $z\in\mathbb{R}$ and
$s_{11}^{\prime}(z_{k})\neq 0$ for all $k$. Denote
$\mathcal{Z}_{-}=\\{z^{*}_{1},\dots,z^{*}_{N}\\}$ and the cardinality
$q\to\sharp\mathcal{Z}$ is locally constant near $q$ in $\mathcal{G}$ and the
map $\mathcal{G}\ni q\to(z_{1},\dots,z_{N})\in\mathbb{C}_{+}^{N}$ is locally
Lipschitz.

We denote by $\mathcal{G}_{N}$ the open subset of $\mathcal{G}$ formed by the
elements such that $\sharp\mathcal{Z}=N$. We call the potential in
$\mathcal{G}$ _generic_. $\mathcal{G}_{1}$ simply contains the soliton (1).
From [20] show that small $L^{1}$-perturbations to the solitons are in
$\mathcal{G}_{1}$.

######  Lemma 2.2.

Let $s\in(\frac{1}{2},1]$. For $q\in
H^{1}(\mathbb{R})\,\cap\,L^{2,s}(\mathbb{R})\cap\,\mathcal{G}$ we have $r\in
H^{s}(\mathbb{R})\cap L^{2,1}(\mathbb{R})$. Furthermore, the map
$H^{1}(\mathbb{R})\,\cap\,L^{2,s}(\mathbb{R})\cap\mathcal{G}\ni q\to r\in
H^{s}(\mathbb{R})\cap L^{2,1}(\mathbb{R})$ is locally Lipschitz.

###### Proof.

For any fixed $\kappa_{0}>0$ there exists a positive constant $C$ such that if
$||q||_{H^{1}(\mathbb{R})\,\cap\,L^{2,s}(\mathbb{R})}\leq\kappa_{0}$
satisfying

$\displaystyle||\Phi_{-,j}(x,\cdot)-e_{j}||_{H^{s}_{z}(\mathbb{R})}\leq
C||q||_{L^{2,s}(\mathbb{R})},\quad\forall\,\,x\leq 0,j=1,2,$
$\displaystyle||\Phi_{+,j}(x,\cdot)-e_{j}||_{H^{s}_{z}(\mathbb{R})}\leq
C||q||_{L^{2,s}(\mathbb{R})},\quad\forall\,\,x\geq 0,j=1,2,$
$\displaystyle||\Phi_{-,j}(x,\cdot)-e_{j}||_{L_{z}^{2,1}(\mathbb{R})}\leq
C||q||_{H^{1}(\mathbb{R})},\quad\forall\,\,x\leq 0,j=1,2,$
$\displaystyle||\Phi_{+,j}(x,\cdot)-e_{j}||_{L_{z}^{2,1}(\mathbb{R})}\leq
C||q||_{H^{1}(\mathbb{R})},\quad\forall\,\,x\geq 0,j=1,2.$ (2.13)

We now consider the case $j=1$ and the minus sign only to prove (2.13). We
based on the fact that if there is $s\in(0,1]$ such that for an $f\in
L^{2}(\mathbb{R})$ we have

$||f(\cdot+h)-f(\cdot)||_{L^{2}(\mathbb{R})}\leq C|h|^{s},\qquad\forall
h\in\mathbb{R},$ (2.14)

then $f\in H^{s}(\mathbb{R})=\dot{H}^{s}(\mathbb{R})\cap L^{2}({\mathbb{R})}$
and there is a positive constant $c$ independent of $f$ such that
$||f||_{\dot{H}^{s}(\mathbb{R})}\leq cC,$ where $\dot{H}^{s}(\mathbb{R})$
defined with
$||q||_{\dot{H}^{s}(\mathbb{R})}:=|||x|^{s}\mathcal{F}{(q)}||_{L^{2}(\mathbb{R})}$.
Define

$Kf(x,z):=\int_{-\infty}^{x}\begin{pmatrix}e^{-2i(x-y)z}&0\\\
0&e^{2i(x-y)z}\end{pmatrix}Q(q(y))f(y,z)dy.$ (2.15)

According to the [21, 22] we obtain

$||Ke_{2}||_{L_{x}^{\infty}(\mathbb{R},L^{2}_{z}(\mathbb{R}))}\leq||q||_{L^{2}(\mathbb{R})},$

hence we acquire that

$||(1-K)^{-1}||_{L_{x}^{\infty}((-\infty,x_{0}),L_{z}^{2}(\mathbb{R})\to
L_{x}^{\infty}(-\infty,x_{0}),L_{z}^{2}(\mathbb{R}))}\leq
e^{||q||_{L^{1}}},\quad\forall x_{0}\leq+\infty.$

We consider the case for $x\leq 0$ we obtain

$\displaystyle(1-K)(\Phi_{-,1}(x,z)-e_{1})=Ke_{1}=\int_{-\infty}^{x}\begin{pmatrix}0\\\
-q^{*}e^{2i(x-y)z}\end{pmatrix}dy,$

and

$\displaystyle||\Phi_{-,1}(x,z)-e_{1}||_{L^{2}_{z}(\mathbb{R})}$
$\displaystyle\leq
e^{||q||_{L^{1}}}\Big{|}\Big{|}\int_{-\infty}^{x}e^{-2i(x-y)z}q^{*}(y)dy\Big{|}\Big{|}_{L_{z}^{2}(\mathbb{R})}$
$\displaystyle\leq e^{||q||_{L^{1}}}\Big{(}\int_{-\infty}^{x}\langle
y\rangle^{2s}|q(y)|^{2}dy\Big{)}^{1/2}\langle x\rangle^{-s}$
$\displaystyle\leq e^{||q||_{L^{1}}}||q||_{L^{2,s}(\mathbb{R})}\langle
x\rangle^{-s},$ (2.16)

and

$\displaystyle||\Phi_{-,1}(x,z)-e_{1}||_{L_{z}^{2,1}(\mathbb{R})}$
$\displaystyle\leq
e^{||q||_{L^{1}}}\Big{|}\Big{|}\int_{-\infty}^{x}ze^{2i(x-y)z}q^{*}(y)dy\Big{|}\Big{|}_{L_{z}^{2}(\mathbb{R})}$
$\displaystyle\leq
e^{||q||_{L^{1}}}\Big{(}\int_{-\infty}^{x}ze^{2i(x-y)z}q^{*}(y)+q^{*}(y)dy\Big{|}\Big{|}_{L_{z}^{2}(\mathbb{R})}$
$\displaystyle\leq e^{||q||_{L^{1}}}\sqrt{\pi}||q||_{H^{1}(\mathbb{R})}.$
(2.17)

Next we define $\mathcal{N}(x,z):=\Phi_{-,1}(x,z+h)-\Phi_{-,1}(x,z)$ for
$h\in\mathbb{R}$ to estimate the form (2.14) by calculating the inequality
$C\lesssim||q||_{L^{2,s}(\mathbb{R})}$, we have

$\displaystyle(1-K)$
$\displaystyle\mathcal{N}(x,z)=\int_{-\infty}^{x}\begin{pmatrix}e^{-2i(x-y)(z+h)}-e^{-2i(x-y)z}&0\\\
0&e^{2i(x-y)(z+h)}-e^{2i(x-y)z}\end{pmatrix}$ $\displaystyle
Q(q(y))(\Phi_{-,1}(y,z)-e_{1})dy+\int_{-\infty}^{x}\begin{pmatrix}0\\\
(e^{2i(x-y)(z+h)}-e^{2i(x-y)z})q^{*}(y)\end{pmatrix}dy.$ (2.18)

Using the Fourier transform $\mathcal{F}$ we have for $x\leq 0$ the second
term r.h.s (2.18)

$\displaystyle\bigg{|}\bigg{|}\int_{-\infty}^{x}$
$\displaystyle(e^{2i(x-y)(z+h)}-e^{2i(x-y)z})q^{*}(y)dy\bigg{|}\bigg{|}_{L_{z}^{2}}$
$\displaystyle=||\mathcal{F}^{*}[q(\cdot+x)\chi_{\mathbb{R}_{-}}](z+h)-\mathcal{F}^{*}[q(\cdot+x)\chi_{\mathbb{R}_{-}}](z)||_{L_{z}^{2}}$
$\displaystyle\leq
C||\mathcal{F}^{*}[q(\cdot+x)\chi_{\mathbb{R}_{-}}](z)||_{H_{z}^{s}(\mathbb{R})}|h|^{s}$
$\displaystyle=||q(y+x)||_{L_{y}^{2,s}(\mathbb{R}_{-})}|h|^{s}\leq||q||_{L^{2,s}(\mathbb{R})}|h|^{s},$
(2.19)

and the first term r.h.s (2.18) is

$\displaystyle\int_{-\infty}^{x}$
$\displaystyle\begin{pmatrix}e^{-2i(x-y)(z+h)}-e^{-2i(x-y)z}&0\\\
0&e^{2i(x-y)(z+h)}-e^{2i(x-y)z}\end{pmatrix}$ $\displaystyle
Q(q(y))(\Phi_{-,1}(y,z)-e_{1})dy$ $\displaystyle\leq
2^{(1-s)}|h|^{s}\int_{-\infty}^{x}|y|^{s}|q(y)||\Phi_{-,1}(y,z)-e_{1}||_{L_{z}^{2}(\mathbb{R})}dy$
$\displaystyle\leq
2^{(1-s)}|h|^{s}e^{||q||_{L^{1}}}||q||_{L^{2,s}(\mathbb{R})}\int_{-\infty}^{x}|y|^{s}\langle
y\rangle^{-s}|q(y)|dy$ $\displaystyle\leq
2^{(1-s)}|h|^{s}e^{||q||_{L_{1}}}||q||_{L^{2,s}(\mathbb{R})}||q||_{L^{1}}.$
(2.20)

Then, we obtain

$||\Phi_{-,1}(x,z+h)-\Phi_{-,1}(x,z)||_{L_{z}^{2}(\mathbb{R})}\leq
C|h|^{s}||q||_{L^{2,s}(\mathbb{R})},\ x\leq 0,$

where $C$ is a fixed constant for
$||q||_{H^{1}(\mathbb{R})\,\cap\,L^{2,s}(\mathbb{R})}\leq\kappa_{0}$, for a
preassigned bound $\kappa_{0}$. This implies that for all $x\leq 0$ we have

$||m_{-,1}(x,z)-e_{1}||_{\dot{H}_{z}^{s}(\mathbb{R})}\leq
C||q||_{L^{2,s}(\mathbb{R})},$

for some positive constant $C$ and above the assume have been proved. The
other cases similar. Now we conclude that (2.13) be established. Then
$s_{12}(z)\in H^{s}(\mathbb{R})\,\cap\,L^{2,1}(\mathbb{R})$ because

$\displaystyle s_{12}(z)$
$\displaystyle=\det(\Phi_{+,2}(0,z),\Phi_{-,2}(0,z))$
$\displaystyle=\det(\Phi_{+,2}(0,z)-e_{2},\Phi_{-,2}(0,z)-e_{2})$
$\displaystyle+\det(\Phi_{+,2}(0,z)-e_{2},e_{2})+\det(e_{2},\Phi_{-,2}(0,z)-e_{2}),$
(2.21)

and we notice that $H^{s}(\mathbb{R})$ is a Banach algebra with respect to
pointwise multipilication for any $s>\frac{1}{2}$. Similarly, $(s_{11}-1)\in
H^{s}(\mathbb{R})\,\cap\,L^{2,1}(\mathbb{R})$ because

$\displaystyle s_{11}(z)$
$\displaystyle=\det(\Phi_{-,1}(0,z),\Phi_{+,2}(0,z))$
$\displaystyle=\det(\Phi_{-,1}(0,z)-e_{1},\Phi_{+,2}(0,z))+\det(e_{1},\Phi_{+,2}(0,z)-e_{2})+\det(e_{1},e_{2})$
$\displaystyle=1+\det(\Phi_{-,1}(0,z)-e_{1},\Phi_{+,2}-e_{2})$
$\displaystyle+\det(e_{1},\Phi_{+,2}-e_{2})+\det(\Phi_{-,1}(0,z)-e_{1},e_{2}).$
(2.22)

We conclude that if $q\in
H^{1}(\mathbb{R})\,\cap\,L^{2,s}(\mathbb{R})\cap\,\mathcal{G}$ then $r\in
H^{s}(\mathbb{R})\cap L^{2,1}(\mathbb{R})$ and this shows that we accquire a
map $H^{1}(\mathbb{R})\,\cap\,L^{2,s}(\mathbb{R})\cap\,\mathcal{G}\ni q\to
r\in H^{s}(\mathbb{R})\cap L^{2,1}(\mathbb{R})$.The locally Lipschitz of the
map $H^{1}(\mathbb{R})\,\cap\,L^{2,s}(\mathbb{R})\cap\,\mathcal{G}\ni q\to
r\in H^{s}(\mathbb{R})\cap L^{2,1}(\mathbb{R})$ we skip.
∎

###  2.3 A Riemann-Hilbert problem

Define the collection what we need in the space

$\displaystyle\mathcal{S}(s,n):=$ $\displaystyle\\{r(z)\in
H^{s}(\mathbb{R})\cap
L^{2,1}(\mathbb{R}),\quad(z_{1},\dots,z_{n})\in\mathbb{C}_{+}^{n},$
$\displaystyle(c_{1},\dots,c_{n})\in\mathbb{C}_{*}^{n}:=\mathbb{C}^{n}\setminus\\{0\\}\\},$
(2.23)

is called the scattering data for for initial data $q_{0}$ and the map
$\mathcal{P}:q_{0}\mapsto\mathcal{S}$ is called the (forward) scattering map.

######  Proposition 2.3.

If $r(z)\in H^{s}(\mathbb{R})\,\cap\,L^{2,1}(\mathbb{R})$, then for every
$t\in\mathbb{R}$, we have $r(z,t)=r(z)e^{i(4\beta z^{3}+2\alpha z^{2})t}\in
H^{s}(\mathbb{R})\cap L^{2,1}(\mathbb{R})$.

###### Proof.

By (2.12), we obtain

$\displaystyle||r(\cdot,t)||_{L^{2,1}(\mathbb{R})}=|||z|r(z)e^{i(4\beta
z^{3}+2\alpha z^{2})t}||_{L^{2}(\mathbb{R})}$
$\displaystyle\qquad\qquad\qquad=|||z|r(z)||_{L^{2}(\mathbb{R})}=||r(z)||_{L^{2,1}(\mathbb{R})},$
$\displaystyle and$
$\displaystyle||r(\cdot,t)||_{H^{s}(\mathbb{R})}=|||z|^{s}\mathcal{F}(r(z)e^{i(4\beta
z^{3}+2\alpha z^{2})t})||_{L^{2}(\mathbb{R})}$
$\displaystyle\qquad\qquad\qquad=(2\pi)^{-1}|||z|^{s}\mathcal{F}(r(z))*\mathcal{F}(e^{i(4\beta
z^{3}+2\alpha z^{2})t})||_{L^{2}(\mathbb{R})}$
$\displaystyle\qquad\qquad\qquad=(2\pi)^{-1}||r(z)||_{H^{s}(\mathbb{R})}.$

Therefore, we completed the proof of Proposition 2.3. ∎

The essential fact of integrability is that if the potential $q_{0}$ evolves
according to (1.1) the the time evolution of the scattering data $\mathcal{D}$
is trivial and the collection

$\displaystyle\mathcal{D}(t):=$ $\displaystyle\\{r(z)e^{i(4\beta z^{3}+2\alpha
z^{2})t}\in H^{s}(\mathbb{R})\cap
L^{2,1}(\mathbb{R}),\quad(z_{1},\dots,z_{n})\in\mathbb{C}_{+}^{n},$
$\displaystyle(c_{1}e^{i(4\beta z_{1}^{3}+2\alpha
z_{1}^{2})t},\dots,c_{n}e^{i(4\beta z_{n}^{3}+2\alpha
z_{n}^{2})t})\in\mathbb{C}^{n}\setminus\\{0\\}\\}.$ (2.24)

The inverse scattering map $\mathcal{P}^{-1}:\mathcal{D}(r)\mapsto q(x,t)$
seeks to recover the solution of (1.1) from its scattering data.

We construct the function

$M(z):=\begin{cases}\big{(}\frac{\Phi_{-,1}(x,t;z)}{s_{11}(z)},\Phi_{+,2}(x,t;z)\big{)},\quad
z\in\mathbb{C}^{+},\\\ \\\
(\Phi_{+,1}(x,t;z),\frac{\Phi_{-,2}(x,t;z)}{s_{22}(z)}\big{)},\quad\,z\in\mathbb{C}^{-},\end{cases}$
(2.25)

which satisfies the following RH problem.

######  RHP 2.1.

Find an analytic function
$M(z):\mathbb{C}\setminus(\mathbb{R}\,\cup\,\mathcal{Z}\,\cup\,\mathcal{Z}^{*})\to
SL_{2}(\mathbb{C})$ with the following properties

  1. 1.

$M(z)=I+\mathcal{O}(z^{-1})$ as $z\to\infty$.

  2. 2.

$M(z)$ takes continuous boundary values $M_{\pm}(z)$ which satisfy the jump
relation

$M_{+}(z)=M_{-}(z)V(z),$

where

$V(z)=\begin{pmatrix}1+|r(z)|^{2}&r^{*}(z)e^{-2it\theta(z)}\\\
r(z)e^{2it\theta(z)}&1\end{pmatrix}.$ (2.26)

  3. 3.

$M(z)$ has simple poles at each
$\mathcal{Z}:=\mathcal{Z}_{+}\cup\mathcal{Z}_{-}$ at which

$\displaystyle\underset{z=z_{k}}{\mbox{Res}}M(z)=\underset{z\to
z_{k}}{\mbox{lim}}M(z)\begin{pmatrix}0&0\\\
c_{k}e^{-2it\theta(z)}&0\end{pmatrix},$ (2.27)
$\displaystyle\underset{z=z_{k}^{*}}{\mbox{Res}}M(z)=\underset{z\to
z_{k}^{*}}{\mbox{lim}}M(z)\begin{pmatrix}0&-c_{k}^{*}e^{2it\theta(z^{*})}\\\
0&0\end{pmatrix}.$ (2.28)

It’s a simple consequence of Liouville’s theorem that if a solution exist, it
is unique. The existence of solutions of RHP 2.1 for any
$(x,t)\in\mathbb{R}\times\mathbb{R}$ follows by means of Zhou’s vanishing
lemma argument [23]. Expanding this solution as

$M(z)=I+\frac{M^{(1)}(x,t)}{z}+\mathcal{O}(z^{-2}),\ z\to\infty,$

and one can find the solution

$M(x,z)=I-\underset{\zeta\in\mathcal{Z}}{\sum}\frac{M_{x}(\zeta)V(\zeta)}{\zeta-z}+\frac{1}{2\pi
i}\int_{\mathbb{R}}\frac{M_{x}(V(\zeta)-I)}{\zeta-z}d\zeta,$ (2.29)

where $M_{x}(z)$ is defined for $z\in\mathbb{R}\cup\mathcal{Z}$ in the space
$M_{2\times 2}(\mathbb{C})$ of complex $2\times 2$ matrices and satisfies
system (2.31) and (2.32) written below and it follows that the solution of
(1.1) is given by

$q(x,t)=2i\underset{z\to\infty}{\mbox{lim}}(zM(x,t;z))_{12}=2i(M^{(1)})_{12}.$
(2.30)

######  Lemma 2.4.

Fix $s\in(\frac{1}{2},1]$ and suppose that $r\in H^{s}(\mathbb{R})\cap
L^{2,1}(\mathbb{R})$. Then, for any $x\in\mathbb{R}$ there exists and unique a
solution $M_{x}:\mathbb{R}\cup\mathcal{Z}\to\mbox{M}_{2\times 2}(\mathbb{C})$
of the following system of integral and algebraic equation:

$M_{x}(z)=I-\underset{\zeta\in\mathcal{Z}}{\sum}\frac{M_{x}(\zeta)V(\zeta)}{\zeta-z}+\underset{\epsilon\to
0}{\mbox{lim}}\int_{\mathbb{R}}\frac{M_{x}(\zeta)(V(\zeta)-I)}{\zeta-(z-i\epsilon)}d\zeta,\quad
z\in\mathbb{R},$ (2.31)

and

$M_{x}(z)=I-\underset{\zeta\in\mathcal{Z}\setminus\\{z\\}}{\sum}\frac{M_{x}(\zeta)V(\zeta)}{\zeta-z}+\frac{1}{2\pi
i}\int_{\mathbb{R}}\frac{M_{x}(\zeta)(V(\zeta)-I)}{\zeta-z}d\zeta,\quad
z\in\mathcal{Z},$ (2.32)

such that $(M_{x}(z)-I)\in L_{z}^{2}(\mathbb{R})$.

###### Proof.

At first we introduce the Cauchy operator $C_{\mathbb{R}}$ acting on functions
$h(z)\in L^{2}(\mathbb{R})$

$(C_{\mathbb{R}}h)(z)=\frac{1}{2\pi
i}\int_{\mathbb{R}}\frac{h(\zeta)}{\zeta-z}d\zeta,\quad
z\in\mathbb{C}\setminus\mathbb{R},$ (2.33)

with the boundary

$(C_{\mathbb{R}}^{\pm}h)(z)=\underset{\epsilon\to 0}{\mbox{lim}}\frac{1}{2\pi
i}\int_{\mathbb{R}}\frac{h(\zeta)}{\zeta-(z\pm i\epsilon)}d\zeta,\quad
z\in\mathbb{R}.$ (2.34)

Then we define the operator $C_{V_{x}}$ by $C_{V_{x}}h:=C^{-}(h(V_{x}-I))$
where $h(z)\in L^{2}(\mathbb{R})$ the system of integral and algebraic
equation (2.33) and (2.34) reduces to

$(I-C_{V_{x}})(M_{x}-I)+\underset{\zeta\in\mathcal{Z}}{\sum}\frac{M_{x}(\zeta)V(\zeta)}{\zeta-z}=C_{V_{x}}I,\quad
z\in\mathbb{R},$ (2.35)

and

$\displaystyle M_{x}(z)$
$\displaystyle+\underset{\zeta\in\mathcal{Z}\setminus\\{z\\}}{\sum}\frac{M_{x}(\zeta)V(\zeta)}{\zeta-z}-\frac{1}{2\pi
i}\int_{\mathbb{R}}\frac{M_{x}(\zeta)(V(\zeta)-I)}{\zeta-z}d\zeta$
$\displaystyle=I+\frac{1}{2\pi
i}\int_{\mathbb{R}}\frac{(V(\zeta)-I)}{\zeta-z}d\zeta,\quad z\in\mathcal{Z}.$
(2.36)

According to the [25], for there exists a fixed $c$ such that for
$||r||_{L^{\infty}(\mathbb{R})}=\rho$ the operator $I-C_{V_{x}}$ is invertible
in $L^{2}(\mathbb{R})$ and $||(I-C_{V_{x}})^{-1}||_{L^{2}\to L^{2}}\leq
c\langle\rho\rangle^{2}$. We solve the inhomogeneous system (2.35) and (2.36)
using the Fredholm alternative Theorem by admitting exactly one solution if
and only of $f=0$ is the only solution $f:\mathbb{R}\,\cup\mathcal{Z}\to
M_{2\times 2}(\mathbb{C})$ with $f\,\big{|}_{\mathbb{R}}\in L^{2}(\mathbb{R})$
of

$\displaystyle(I-C_{V_{x}})f+\underset{\zeta\in\mathcal{Z}}{\sum}\frac{f(\zeta)V(\zeta)}{\zeta-z}=0,\quad
z\in\mathbb{R},$ $\displaystyle
f(z)+\underset{\zeta\in\mathcal{Z}\setminus\\{z\\}}{\sum}\frac{f(\zeta)V(\zeta)}{\zeta-z}-\frac{1}{2\pi
i}\int_{\mathbb{R}}\frac{f(\zeta)(V(\zeta)-I)}{\zeta-z}d\zeta=0,\quad
z\in\mathcal{Z}.$ (2.37)

For $z\in\mathbb{C}\setminus(\mathcal{Z}\cup\mathbb{R})$ we set

$M(z):=-\underset{\zeta\in\mathcal{Z}}{\sum}\frac{f(\zeta)(V(\zeta)}{\zeta-z}+\frac{1}{2\pi
i}\int_{\mathbb{R}}\frac{f(\zeta)(V(\zeta)-I)}{\zeta-z}d\zeta.$ (2.38)

Notice that

$M_{-}(z)=-\underset{\zeta\in\mathcal{Z}}{\sum}\frac{f(\zeta)V(\zeta)}{\zeta-z}+C^{-}(f(V-I)).$
(2.39)

Using the operator $C^{-}(f(V-I))=C_{V_{x}}f$ and (2.37) and obtain
$M_{-}(z)=f(z)$ for $z\in\mathbb{R}$. Then we have

$\displaystyle M_{+}$
$\displaystyle=-\underset{\zeta\in\mathcal{Z}}{\sum}\frac{f(\zeta)V(\zeta)}{\zeta-z}+C^{+}(f(V-I))$
$\displaystyle=-\underset{\zeta\in\mathcal{Z}}{\sum}\frac{f(\zeta)V(\zeta)}{\zeta-z}+C^{-}(f(V-I))(z)+f(z)(V(z)-I)$
$\displaystyle=M_{-}(z)+M_{-}(z)(V-I)=M_{-}(z)V(z).$

We obtain

$\displaystyle 0$
$\displaystyle=\int_{\mathbb{R}}C^{+}(f(V-I))(C^{-}(f(f(V-I)))^{*}$
$\displaystyle=\int_{\mathbb{R}}\Big{(}M_{+}(z)+\underset{\zeta\in\mathcal{Z}}{\sum}\frac{f(\zeta)V(\zeta)}{\zeta-z}\Big{)}\Big{(}M_{-}(z)+\underset{\zeta\in\mathcal{Z}}{\sum}\frac{f(\zeta)V(\zeta)}{\zeta-z}\Big{)}dz$
$\displaystyle=\int_{\mathbb{R}}M_{+}M_{-}^{*}dz+\underset{\zeta\in\mathcal{Z}}{\sum}\int_{\mathbb{R}}\frac{M_{+}(z)dz}{\zeta^{*}-z}V^{*}(\zeta)f^{*}(\zeta)+\bigg{(}\underset{\zeta\in\mathcal{Z}}{\sum}\int_{\mathbb{R}}\frac{M_{-}(z)dz}{\zeta^{*}-z}V^{*}(\zeta)f^{*}(\zeta)\bigg{)}^{*}$
$\displaystyle+\underset{A\in\\{+,-\\}}{\sum}\underset{\zeta,\xi\in\mathcal{Z}_{A}}{\sum}f(\zeta)V(\zeta)V^{*}(\xi)f^{*}(\xi)\int_{\mathbb{R}}\frac{dz}{(\zeta-z)(\xi^{*}-z)}.$
(2.40)

The $A=+$ term in the last line cancels with the following:

$\displaystyle\underset{\zeta\in\mathcal{Z}_{+}}{\sum}$
$\displaystyle\int_{\mathbb{R}}\frac{M_{+}(z)dz}{\zeta^{*}-z}V^{*}(\zeta)f^{*}(\zeta)=\underset{\zeta\in\mathcal{Z}_{+}}{\sum}\int_{\mathbb{R}}\frac{C^{+}(f(V-I))(z)dz}{\zeta^{*}-z}V^{*}(\zeta)f^{*}(\zeta)$
$\displaystyle-\underset{\zeta\in\mathcal{Z}_{+}}{\sum}\underset{\xi\in\mathcal{Z}}{\sum}\int_{\mathbb{R}}\frac{dz}{(\zeta^{*}-z)(\xi-z)}f(\xi)V(\xi)V^{*}(\zeta)f^{*}(\zeta)$
$\displaystyle=-\underset{\zeta,\xi\in\mathcal{Z}_{+}}{\sum}\int_{\mathbb{R}}\frac{dz}{(\zeta^{*}-z)(\xi-z)}f(\xi)V(\xi)V^{*}(\zeta)f^{*}(\zeta).$

Using the relationship

$\mbox{Res}(M,\zeta^{*})V^{*}(\zeta)=f(\zeta^{*})V(\zeta^{*})V^{*}(\zeta)=-f(\zeta^{*})V^{*}(\zeta)\\\
V^{*}(\zeta)=0,$

and

$\displaystyle\underset{\zeta\mathcal{Z}_{-}}{\sum}$
$\displaystyle\int_{\mathbb{R}}\frac{M_{+}(z)dz}{\zeta^{*}-z}V^{*}(\zeta)f^{*}(\zeta)=\underset{\zeta\in\mathcal{Z}_{-}}{\sum}\int_{\mathbb{R}}\Big{(}M_{+}(z)-\frac{\mbox{Res}(m,\zeta^{*})}{z-\zeta^{*}}\Big{)}\frac{V^{*}(\zeta)f^{(}\zeta)}{\zeta^{*}-z}dz$
$\displaystyle=-2\pi
i\underset{\zeta\in\mathcal{Z}_{-}}{\sum}f(\zeta^{*})V^{*}(\zeta)f^{*}(\zeta)+\underset{\zeta\in\mathcal{Z}_{-}}{\sum}\underset{\xi\in\mathcal{Z}_{+}\setminus\\{\xi^{*}\\}}{\sum}f(\xi)V(\xi)V^{*}(\zeta)f^{*}(\zeta)\frac{2\pi
i}{\zeta^{*}-\xi}$ $\displaystyle=-2\pi
i\underset{\zeta\in\mathcal{Z}_{-}}{\sum}f(\zeta^{*})V^{*}(\zeta)f^{*}(\zeta).$
(2.41)

Notice (2.37) and (2.38) and for all $\zeta\in\mathcal{Z}$ we obtain

$\underset{z\to\zeta}{\mbox{lim}}\Big{(}M(z)-\frac{\mbox{Res}(M,\zeta)}{z-\zeta}\Big{)}=C(f(V-I))(\zeta)-\underset{\zeta\in\mathcal{Z}\setminus\\{\zeta\\}}{\sum}\frac{f(\zeta^{\prime})V(\zeta^{\prime})}{\zeta^{\prime}-\zeta}=f(\zeta).$

The following term cancels with the $A=-$ term in the last line of (2.40)

$\displaystyle\Big{(}\underset{\zeta\in\mathcal{Z}_{-}}{\sum}$
$\displaystyle\int_{\mathbb{R}}\frac{M_{-}(z)dz}{\zeta^{*}-z}V^{*}(\zeta)f^{*}(\zeta)\Big{)}^{*}=\underset{\zeta\in\mathcal{Z}_{-}}{\sum}f(\zeta)V(\zeta)\Big{[}\int_{\mathbb{R}}\frac{(C^{-}(f(V-I)))^{*}}{\zeta-z}dz$
$\displaystyle-\underset{\xi\mathcal{Z}}{\sum}\int_{\mathbb{R}}\frac{dz}{(\zeta-z)(\xi^{*}-z)}f(\zeta)V(\zeta)V^{*}(\xi)f^{*}(\xi)\Big{]}$
$\displaystyle=-\underset{\zeta,\xi\in\mathcal{Z}_{-}}{\sum}f(\zeta)V(\zeta)V^{*}(\xi)f^{*}(\xi)\int_{\mathbb{R}}\frac{dz}{(\zeta-z)(\xi^{*}-z)}.$
(2.42)

We get

$\displaystyle\Big{(}\underset{\zeta\in\mathcal{Z}_{+}}{\sum}$
$\displaystyle\int_{\mathbb{R}}\frac{M_{-}(z)dz}{\zeta^{*}-z}V^{*}(\zeta)f^{*}(\zeta)\Big{)}^{*}$
$\displaystyle=\Big{(}\underset{\zeta\in\mathcal{Z}_{+}}{\sum}\int_{\mathbb{R}}\Big{(}M_{-}(z)-\frac{\mbox{Res}(m,\zeta^{*})}{z-\zeta^{*}}\Big{)}\frac{V^{*}(\zeta)f^{*}(\zeta)}{\zeta^{*}-z}dz\Big{)}^{*}$
$\displaystyle=\Big{(}2\pi
i\underset{\zeta\in\mathcal{Z}_{+}}{\sum}f(\zeta^{*})V^{*}(\zeta)f^{*}(\zeta)\Big{)}^{*}=-2\pi
i\underset{\zeta\in\mathcal{Z}_{+}}{\sum}f(\zeta)V(\zeta)f^{*}(\zeta^{*}).$
(2.43)

The terms from (2.42) and (2.43) cancel out in (2.40) because of
$V(\xi^{*})=-V^{*}(\zeta)$. Then, by $M_{+}=M_{-}V$ yields

$0=\int_{\mathbb{R}}M_{-}(z)V(z)M_{-}^{*}(z)dz.$ (2.44)

We obtain the relationship $M_{-}(z)=f(z)=0$ for $z\in\mathbb{R}$ since $V(z)$
is strictly positive. And according to (2.39) we acquire that
$f(\zeta)V(\zeta)=0$ for $\zeta\in\mathcal{Z}$. Then we completed the proof
that if $f$ solves (2.37) then $f=0$. ∎

The result of the Lemma 2.4 implies that the map $\mathcal{G}_{N}\cap
H^{1}(\mathbb{R})\,\cap\,L^{2,s}(\mathbb{R})\to\mathcal{S}(s,n)$ is one-to-
one. The result is due to Zhou [24], we just the completeness of it. Now we
just consider the case of pure radiation solutions of the Hirota equation with
$N=0$. And the Lemma 2.5 establish the fact that the map $\mathcal{G}_{0}\cap
H^{1}(\mathbb{R})\,\cap\,L^{2,s}(\mathbb{R})\to\mathcal{S}(s,0)$ is only one-
to-one but also onto.

######  Lemma 2.5.

Let $r\in H^{s}(\mathbb{R})\cap L^{2,1}(\mathbb{R})$,
$\mathcal{Z}=\varnothing$ and consider the potential $q$ defined by the
reconstructing formula (2.30). Then $q\in
H^{1}(\mathbb{R})\,\cap\,L^{2,s}(\mathbb{R})$. Furthermore, for any positive
$\kappa_{0}$, there is a constant $C$ such that for
$||r||_{L^{\infty}(\mathbb{R})}\leq\kappa_{0}$, we have

$||q||_{H^{1}(\mathbb{R})\,\cap\,L^{2,s}(\mathbb{R})}\leq
C||r||_{H^{s}(\mathbb{R})\,\cap\,L^{2,1}(\mathbb{R})}.$

###### Proof.

We now construct the inverse scattering transform from
$H^{s}(\mathbb{R})\,\cap\,L^{2,1}(\mathbb{R})$ to
$H^{1}(\mathbb{R})\,\cap\,L^{2,s}(\mathbb{R})$. Let $r\in
H^{s}(\mathbb{R})\,\cap\,L^{2,1}(\mathbb{R})$ be given. The inverse problem is
formulated as RH Problem 2.1.

We factorize the jump matrix $V(z)$ on the line
$z\in(-\infty,z_{1})\cup(z_{2},+\infty)$ where

$V(z)=V_{-}^{-1}(z)V_{+}(z)=\begin{pmatrix}1&r^{*}(z)e^{-2it\theta(z)}\\\
0&1\end{pmatrix}\begin{pmatrix}1&0\\\ r(z)e^{2it\theta}&1\end{pmatrix}.\quad$
(2.45)

This problem is adapted for studying the decay behavior of $q$ as
$x\to-\infty$. Next, we prove the decay behavior of $q$ as $x\to-\infty$. The
decay behavior of $q$ as $x\to\infty$ can be obtained in a similar manner.

Let

$C_{w_{x}}h:=C^{+}(hw_{x-})+C^{-}(hw_{x+}),$

where $w_{x\pm}:=\pm(V_{\pm}-I).$ Then we consider a function $\mu\in
I+L^{2}(\mathbb{R})$ such that

$(I-C_{w})(\mu_{x})(z)=I.$ (2.46)

We acquire the potential $q(x,t)$ in (2.30) by
$w(\zeta):=V_{+}(\zeta)-V_{-}(\zeta)$ in the case of $c_{j}=0$ which can be
expressed also as

$M(x,z)=I+C^{\pm}(\mu_{x}(\zeta)w(\zeta))=I+\frac{1}{2\pi
i}\int_{\mathbb{R}}\frac{\mu_{x}(\zeta)w(\zeta)}{\zeta-z}d\zeta.$ (2.47)

Then differentiating the jump relation $M_{+}(z)=M_{-}(z)V(z)$, we obtain

$\frac{d}{dx}M_{+}+iz[M_{+},\sigma_{3}]=(\frac{d}{dx}M_{-}+iz[M_{-},\sigma_{3}])V.$

A simple calculation shows that

$iz[M_{\pm},\sigma_{3}]=Q+C^{\pm}(i[\mu_{x}w,\sigma_{3}]).$

Thus $M_{\pm}$ solve the differential equation
$\frac{d}{dx}M_{\pm}=iz[\sigma_{3},M_{\pm}]+QM_{\pm}$. Set $\mathcal{I}(r)=q$.
The following results show that $\mathcal{I}$ maps
$H^{s}(\mathbb{R})\,\cap\,L^{2,1}(\mathbb{R})$ to
$H^{1}(\mathbb{R})\,\cap\,L^{2,s}(\mathbb{R})$. For a fixed $c_{s}$ and $x\leq
0$, using the argument in Lemma 3.4 [21] we obtain

$||C^{\pm}w(z)||_{L_{z}^{2}(\mathbb{R})}\leq c_{s}\langle
x\rangle^{-s}||r||_{H^{s}(\mathbb{R})\,\cap\,L^{2,1}(\mathbb{R})},$ (2.48)

which can be induced by the different definition of the operator in (2.1) from
which we get

$||C_{w}I||_{L^{2}(\mathbb{R})}\leq 2c_{s}\langle
x\rangle^{-s}||r||_{H^{s}(\mathbb{R})\,\cap\,L^{2,s}(\mathbb{R})}.$ (2.49)

Then we take into account the inequality

$\mu_{x}-I=(I-C_{w})^{-1}C_{w}I,$ (2.50)

and correspondingly

$||\mu_{x}-I||_{L_{z}^{2}}\leq||(I-C_{w})^{-1}||_{L_{z}^{2}\to
L_{z}^{2}}||C_{w}I||_{L_{z}^{2}}.$ (2.51)

For fixed $c$ and from the Lemma 5.1 [25] we obtain
$||(I-C_{w})^{-1}||_{L_{z}^{2}\to L_{z}^{2}}\leq C\langle\rho\rangle^{2}$
where $\rho:=||r||_{L^{\infty}(\mathbb{R})}$. We conclude that for $x\leq 0$
and for any $\kappa_{0}$ there is a constant $C$ such that

$||\mu_{x}-I||_{L_{z}^{2}}\leq C\langle
x\rangle^{-s}||r||_{H^{s}(\mathbb{R})\,\cap\,L^{2,s}(\mathbb{R})},$ (2.52)

for $\rho\leq\kappa_{0}$. As above, we write

$M-I=\int_{\mathbb{R}}((I-C_{w})^{-1}(V_{+}-V_{-}))dz=\int_{1}+\int_{2}+\int_{3},$

where

$\int_{1}=\int(V_{+}-V_{-}),\qquad\qquad\int_{2}=\int(C_{w}I)(V_{+}-V_{-}),$
$\int_{3}=\int(C_{w})(I-C_{w})^{-1}C_{w}I)(V_{+}-V_{-})=\int(C_{w})(\mu-I)(V_{+}-V_{-}).$

We remark that for calculating $q$, the estimate of $\int_{2}$ is not needed
because it is diagonal and $\hat{\sigma_{3}}\int_{2}=0$. But the estimate is
useful for other problems. Clearly, $\int(V_{+}-V_{-})\in
H^{s}(\mathbb{R})\cap L^{2,1}(\mathbb{R})$ by the Fourier transform. Using the
triangularity of $V_{\pm}$, the fact that $C^{+}-C^{-}=1$ Cauchy’s theorem,
using Lemma 3.4 [21] we obtain for some $c>0$

$\displaystyle|\int_{2}|$
$\displaystyle=|\int[(C^{+}(I-V_{-}))(V_{+}-I)+(C^{-}(V_{+}-I)(I-V_{-}))]$
$\displaystyle=|\int[C^{+}(I-V_{-}))(C^{-}(I-V_{+}))+(C^{-}(V_{+}-I))(C^{+}(I-V_{-}))]$
$\displaystyle\leq c(1+x^{2})^{-1},$

and

$||\mu-I||_{L^{2}}=||(I-C_{w})^{-1}C_{w}I||_{L^{2}}\leq c(1+x^{2})^{-1/2},$

we have

$\displaystyle|\int_{3}|$
$\displaystyle=|\int[(C^{+}(\mu-I)(I-V_{-})(V_{+}-I))]+(C^{-}(\mu-I)(V_{+}-I)(I-V_{-}))$
$\displaystyle\leq c(1+x^{2})^{-1}.$

According to Theorem 3.6 [21], we have

$q(x)\in(H^{1}\,\cap\,L^{2,s})(-\infty,z_{1})\,\cup\,(z_{2},+\infty),$

moreover,

$||q||_{H^{1}(\mathbb{R})\,\cap\,L^{2,s}(\mathbb{R})}\leq
C||r||_{H^{s}(\mathbb{R})\,\cap\,L^{2,1}(\mathbb{R})}.$

For the case of $q(x)\in(H^{1}\,\cap\,L^{2,s})(z_{1},z_{2})$ we consider the
instead the decompsition

$V=U_{L}U_{0}U_{R}=\begin{pmatrix}1&0\\\
\frac{r(z)e^{2it\theta}}{1+|r(z)|^{2}}&1\end{pmatrix}\begin{pmatrix}1+|r(z)|^{2}&0\\\
0&\frac{1}{1+|r(z)|^{2}}\end{pmatrix}\begin{pmatrix}1&\frac{r^{*}(z)e^{-2it\theta}}{1+|r(z)|^{2}}\\\
0&1\end{pmatrix},$ (2.53)

which we consider jump matrix
$\tilde{V}:=\delta_{-}^{\sigma_{3}}V\delta_{+}^{-\sigma_{3}}$ define in the RH
problem 3.1 and correspondingly we acquire estimates

$||\tilde{q}||_{H^{1}(\mathbb{R})\,\cap\,L^{2,s}(\mathbb{R})}\leq
C||\tilde{r}||_{H^{s}(\mathbb{R})\cap L^{2,1}(\mathbb{R})}\\\ \leq
c||r||_{H^{s}(\mathbb{R})\cap L^{2,1}(\mathbb{R})}$

for the function $\tilde{q}$ associated to $\tilde{r}:=r\delta_{+}\delta_{-}$
and for fixed $c$ when $\rho\leq\kappa_{0}$ by proceeding as above. Finally,
$\tilde{q}=q$ and more details see [21]. ∎

##  3 Dispersion for pure radiation solutions

In this section, we consider the elements of $\mathcal{G}$ such that
$\mathcal{Z}=\varnothing$ generate pure radiation solutions of the Hirota
equation.

###  3.1 A regular RH problem

From the function (2.4), we get two stationary points

$\displaystyle z_{1}=\frac{-\alpha-\sqrt{\alpha^{2}-3\beta x/t}}{6\beta},$
(3.1) $\displaystyle z_{2}=\frac{-\alpha+\sqrt{\alpha^{2}-3\beta
x/t}}{6\beta}.$ (3.2)

The signature table of Reit$\theta(z)$ is given in Figure 3.1.

Rez$-$$-$$+$$+$$-$$+$$z_{2}$$z_{1}$ Figure 3.1: The function $e^{2it\theta}$
decay in yellow domains and increase in white domains.

First we consider the scalar factorization problem

$\begin{cases}\delta_{+}(z)=\delta_{-}(z)(1+|r(z)|^{2}),\quad
z\in(z_{1},z_{2}),\\\
\delta_{+}(z)=\delta_{-}(z),\quad\quad\qquad\qquad\,\,\mathbb{R}\setminus(z_{1},z_{2}),\\\
\delta(z)\to 1,\qquad\qquad\qquad\,\,\,\,\qquad z\to\infty,\end{cases}$ (3.3)

which admits a solution by the Plemelj formula

$\displaystyle\delta(z)=\exp\big{[}i\int_{z_{1}}^{z_{2}}\frac{\nu(s)}{s-z}ds\big{]},$
(3.4)

where

$\displaystyle\nu(s)=-\frac{1}{2\pi}\log(1+|r(s)|^{2}).$ (3.5)

Define $z_{0}=\frac{|z_{2}-z_{1}|}{2}$ then function $\delta(z)$ has an
expansion

$\displaystyle\delta(z)$
$\displaystyle=\exp\big{(}i\int_{z_{1}}^{z_{2}}\frac{\nu(s)-\chi(s)\nu(z_{2})(s-z_{2}+1)}{s-z}ds+i\nu(z_{2})\int_{z_{0}}^{z_{2}}\frac{s-z_{2}+1}{s-z}ds$
$\displaystyle=\exp\\{i\beta(z,z_{2})+i\nu(z_{2})+i\nu(z_{2})[(z-z_{2})\log(z-z_{2})$
$\displaystyle-(z-z_{2}+1)\log(z-z_{2}+1)+i\nu(z_{2})\log(z-z_{2})\\}$
$\displaystyle=e^{i\nu(z_{2})+i\beta(z,z_{2})}(z-z_{2})^{i\nu(z_{2})}e^{i\nu(z_{2})[(z-z_{2})\log(z-z_{2})-(z-z_{2}+1)\log(z-z_{2}+1)]},$

where $\chi(s)$ is characteristic function of $(z_{0},z_{2})$

$\beta(z,z_{2})=\int_{z_{1}}^{z_{2}}\frac{\nu(s)-\nu(z_{2})\chi(s)(s-z_{2}+1)}{s-z}ds.$

Let $\rho:=||r||_{L^{\infty}}(\mathbb{R})$, the function $\delta(z)$
satisfying the properties following:

  * 1.

when $z\notin[z_{1},z_{2}]$, we obtain $\delta(z)\delta^{*}(z^{*})=1$ and
$\langle\rho\rangle^{-1}\leq|\delta(z)|\leq\langle\rho\rangle$;

  * 2.

For $\mp\mbox{Im}z>0$ we have $|\delta^{\pm}(z)|\leq 1$.

Next we just consider the stationary point $z=z_{2}$, and the point $z=z_{1}$
is similar.

######  Lemma 3.1.

Define
$L_{\phi}=\tilde{z}+e^{-i\phi}\mathbb{R}=\\{z=\tilde{z}+e^{-i\phi}q:q\in\mathbb{R}\\}$.
For $s\in(\frac{1}{2},1]$ then there is a fixed $C(\rho,s)$ such that that for
any $\tilde{z}\in\mathbb{R}$ and any $\phi\in(0,\pi)$

$\displaystyle||\beta(e^{-i\phi}\cdot,\tilde{z})||_{H^{s}(\mathbb{R})}\leq
C(\rho,s)||r||_{H^{s}(\mathbb{R})},$ (3.6)
$\displaystyle|\beta(z,\tilde{z})-\beta(\tilde{z},\tilde{z})|\leq
C(\rho,s)||r||_{H^{s}(\mathbb{R})\,\cap\,L^{2,1}(\mathbb{R})}|z-\tilde{z}|^{s-\frac{1}{2}},\,\,z\,\in
L_{\phi}.$ (3.7)

###### Proof.

When $s=1$ there have

$||C_{\mathbb{R}}f||_{H^{\tau}(L_{\phi})}\leq
C_{\tau}||f||_{H^{\tau}(\mathbb{R})},\ \tau=0,1$

which are proved in Lemma 23.3 [27]. By interpolation calculation we get the
case $\tau=s$ where $s\in(0,1)$. And the (3.7) can be obtained from (3.6) of
the following elementary estimate when $s\in(1/2,1]$

$|f(x)-f(y)|\leq C_{s}||f||_{H^{s}(\mathbb{R})}|x-y|^{s-\frac{1}{2}},$ (3.8)

for all $x,y\in\mathbb{R}$ and $f\in H^{s}(\mathbb{R})$ for a fixed $C_{s}$.
Noting that

$f(x+h)-f(x)=\frac{1}{\sqrt{2\pi}}\int
e^{ix\xi}(e^{ih\xi}-1)\hat{f}(\xi)d\xi,$

Then for any $\kappa>0$ we have for a fixed $C_{s}$

$\displaystyle|f(x+h)-f(x)|$
$\displaystyle\leq\frac{||f||_{H^{s}}}{\sqrt{2\pi}}\Big{[}\big{(}|h|\int_{|\xi|\leq\kappa}|\xi|^{2-2s}d\xi\big{)}^{\frac{1}{2}}+\big{(}\int_{|\xi|\geq\kappa}|\xi|^{-2s}d\xi\big{)}^{\frac{1}{2}}\Big{]}$
$\displaystyle\leq
C_{s}(|h|\kappa^{\frac{3-2s}{2}}+\kappa^{\frac{1-2s}{2}})||f||_{H^{s}}.$

By Plancherel formula we have

$\displaystyle|f(x+h)-f(x)|$ $\displaystyle\leq
C_{s}\frac{||f||_{L^{2,1}}}{\sqrt{2\pi}}\Big{[}\big{(}|h|\int_{|\xi|\leq\kappa}|\xi|^{1-2s}d\xi\big{)}^{\frac{1}{2}}+\big{(}\int_{|\xi|\geq\kappa}|\xi|^{-2s}d\xi\big{)}^{\frac{1}{2}}\Big{]}$
$\displaystyle\leq C_{s}\kappa^{\frac{1-2s}{2}}||f||_{L^{2,1}},$ (3.9)

which equals $2C_{s}|h|^{s-\frac{1}{2}}||f||_{H^{s}}$ for $\kappa=|h|^{-1}$.∎

We define a new unknown function

$\displaystyle M^{(1)}(z)=M(z)\delta^{-\sigma_{3}}(z),$ (3.10)

which satisfies the following RH problem.

######  RHP 3.1.

Find an analytic function $M^{(1)}(z):\mathbb{C}\setminus\mathbb{R}\to
SL_{2}(\mathbb{C})$ with the following properties

  1. 1.

$M^{(1)}(z)=I+\mathcal{O}(z^{-1})$ as $z\to\infty$.

  2. 2.

$M^{(1)}(z)$ takes continuous boundary values $M^{(1)}_{\pm}(z)$ and satisfy
the jump relation $M_{+}^{(1)}(z)=M_{-}^{(1)}(z)V^{(1)}(z)$, where

$\displaystyle V^{(1)}(z)=\delta_{-}^{\sigma_{3}}V(z)\delta_{+}^{-\sigma_{3}}$
$\displaystyle=\begin{cases}\begin{pmatrix}1&r^{*}(z)\delta^{2}(z)e^{-2it\theta(z)}\\\
0&1\end{pmatrix}\begin{pmatrix}1&0\\\
r(z)\delta^{-2}(z)e^{2it\theta}&1\end{pmatrix},\quad
z\in(-\infty,z_{1})\cup(z_{2},\infty),\\\ \\\ \begin{pmatrix}1&0\\\
\frac{r(z)\delta^{-2}e^{2it\theta}}{1+|r(z)|^{2}}&1\end{pmatrix}\begin{pmatrix}1&\frac{r^{*}(z)\delta^{-2}e^{-2it\theta}}{1+|r(z)|^{2}}\\\
0&1\end{pmatrix},\,\,\qquad\qquad\quad z\in(z_{1},z_{2}).\end{cases}$

###  3.2 A mixed RH Problem and its decomposition

In this section we construct a mixed RH Problem by follow closely the argument
of [26, 21, 29].

Fix a smooth cut-off function of compact support, with $\chi(x)\geq 0$ for any
$x$ and $\int\chi dx=1$. Let
$\chi_{\epsilon}(x)=\epsilon^{-1}\chi(\epsilon^{-1}x)$ for $\epsilon\neq 0$.
We define $\mathbf{r}(z)$ as follows

$\mathbf{r}(z)=\begin{cases}r(\mbox{Re}z),\,\quad\quad\,\,\,\,\,\,\,\,\mbox{for}\,\,\mbox{Im}z=0,\\\
\chi_{\mbox{Imz}}*r(\mbox{Re}z),\,\,\mbox{for}\,\,\mbox{Im}z\neq
0.\end{cases}$ (3.11)

For convenience of expression, for $j=1,2$ we define rays

$\displaystyle
L_{j}=\\{z_{j}+\mathbb{R}^{+}e^{i\frac{\pi}{4}}\\}\cup\\{z_{j}+\mathbb{R}^{+}e^{i\frac{5\pi}{4}}\\}=\Sigma_{j2}\cup\Sigma_{j3},$
(3.12)
$\displaystyle\overline{L}_{j}=\\{z_{j}+\mathbb{R}^{+}e^{-i\frac{\pi}{4}}\\}\cup\\{z_{j}+\mathbb{R}^{+}e^{i\frac{3\pi}{4}}\\}=\Sigma_{j1}\cup\Sigma_{j4},$
(3.13) $\displaystyle I_{12}=(z_{1},0),\quad
I_{22}=(0,z_{2}),\,\,\,\,I_{11}=(-\infty,z_{1}),\,\,\,\,I_{21}=(z_{2},\infty).$

These rays divide the complex plane $\mathbb{C}$ into ten domains
$\Omega_{ij},i=1,2;j=1,3,4,6;\Omega_{2},\Omega_{5}$. See Figure 3.2.

Rez$\Omega_{21}$$\Omega_{26}$$\Omega_{2}$$z_{3}$$z_{4}$$z_{2}$$\Omega_{23}$$\Omega_{24}$$\Omega_{11}$$\Omega_{16}$$\Omega_{5}$$z_{1}$$\Omega_{13}$$\Omega_{14}$$\Sigma_{21}$$\Sigma_{11}$$\Sigma_{22}$$\Sigma_{12}$$\Sigma_{23}$$\Sigma_{13}$$\Sigma_{24}$$\Sigma_{14}$$l_{1}$$l_{2}$
Figure 3.2: The continuous extension domains $\Omega_{nj},\ n=1,2;j=1,3,4,6$
and contour $\Sigma^{(2)}$ for mixed RH problem $M^{(2)}(z)$

We define functions
$R_{ij}:\overline{\Omega}_{ij}\to\mathbb{C},(i=1,2;j=1,3,4,6)$ and a constant
$c$ with boundary values satisfying

$\displaystyle
R_{j1}(z)=\begin{cases}r(z),\qquad\qquad\qquad\qquad\quad\qquad\quad\qquad\qquad\,\,\,z\in
I_{j1},\\\
\hat{r}_{0}(z-z_{j})^{-2i\nu(z_{j})}\delta^{2},\quad\quad\quad\,\,\,\,\,\,\qquad\qquad\qquad
z\in\Sigma_{j1},\end{cases}$ (3.14) $\displaystyle
R_{j3}(z)=\begin{cases}\frac{r^{*}(z)}{1+|r(z)|^{2}},\qquad\qquad\qquad\,\,\qquad\qquad\qquad\,\,\quad\,\,\,\,\,\,\,z\in
I_{j2},\\\
\frac{\hat{r}_{0}^{*}(z)}{1+|r(z_{j})|^{2}}(z-z_{j})^{2i\nu(z_{j})}\delta^{-2},\,\,\,\,\,\,\,\,\quad\quad\qquad\qquad\,\,z\in\Sigma_{j3},\end{cases}$
(3.15) $\displaystyle
R_{j4}(z)=\begin{cases}\frac{r(z)}{1+|r(z)|^{2}},\qquad\qquad\qquad\,\qquad\qquad\qquad\,\,\quad\,\,\,\,\,\,\,\,\,z\in
I_{j4},\\\
\frac{\hat{r}_{0}(z)}{1+|r(z_{j})|^{2}}(z-z_{j})^{-2i\nu(z_{j})}\delta^{2},\,\,\,\,\,\,\,\,\,\qquad\qquad\qquad
z\in\Sigma_{j4},\end{cases}$ (3.16) $\displaystyle
R_{j6}(z)=\begin{cases}r^{*}(z),\qquad\qquad\qquad\,\,\,\,\qquad\quad\qquad\qquad\qquad\,\,\,\,z\in
I_{j6},\\\
\hat{r}_{0}^{*}(z-z_{j})^{2i\nu(z_{j})}\delta^{-2},\quad\qquad\,\,\,\,\,\,\,\,\qquad\qquad\qquad
z\in\Sigma_{j6},\end{cases}$ (3.17)

where $\hat{r}_{0}=r(z_{j})e^{-2i\nu(z_{j})-2\beta(z_{j},z_{j})}$ .

######  Proposition 3.2.

Fix $\lambda_{0}>0$ and assume
$||r||_{H^{s}(\mathbb{R})\,\cap\,L^{2,1}(\mathbb{R})}<\lambda$ a preassigned
$s\in(1/2,1]$. Then for $j=1,2;j=1,2,3,4$ there exist functions
$R_{ji}:\overline{\Omega_{ji}}\to\mathbb{C}$ such that forall
$z\in\Omega_{ji}+z_{j}$ and $\psi(x)=-\chi(x)-x\chi^{\prime}(x)$, we have for
a fixed $c$

$\displaystyle|\overline{\partial}R_{ji}(z)|$ $\displaystyle\leq
c||r||_{H^{s}(\mathbb{R})\,\cap L^{2,1}(\mathbb{R})}|z-z_{j}|^{s-\frac{3}{2}}$
$\displaystyle+c|\partial_{\tiny\rm\mbox{Re}z}\mathbf{r}(z)|+c|(\mbox{Im
z})^{-1}\psi_{\tiny\rm\mbox{Im}z}*r(\mbox{Re z})|.$ (3.18)

###### Proof.

We consider the stationary point at $z_{2}$, the case at $z_{1}$ is similar.
For $\zeta\in\Omega_{13}$ we use $0<\arg(\zeta-z_{1})<\pi/4$ and set
$G(\zeta):=g(\arg(\zeta-z_{1}))$ where the function $g:[0,\pi/4]\to[0,1]$ is a
smooth function such that $g(\psi)=1$ for all $\psi\in[0,\pi/6]$ and
$g(\pi/4)=0$. the function $G$ is continuous in
$\Omega_{13}\setminus\\{z_{1}\\}$ and fulfills

$G(\zeta)=\begin{cases}1,\quad\mbox{for}\,\zeta\in I_{12},\\\
0,\quad\mbox{for}\,\zeta\in\Sigma_{13},\\\ \end{cases}$ (3.19)

Moreover, we can find a constant $c$ such that

$|\overline{\partial}G(\zeta)\leq c|\zeta-z_{1}|^{-1}$

Consider the function of $R_{2i}(z)$ which can be defined explicitly by
setting for $z-z_{2}=u+iv$. For $i=1,3$ we obtain

$\displaystyle R_{21}(z)=G(\zeta)\mathbf{r}(z)+(1-G(\zeta)))f_{21}(u+iv),$
$\displaystyle
R_{23}(z)=\cos(2(\arg(z-z_{0})-\pi))\frac{\mathbf{r}^{*}(z)}{1+|\mathbf{r}(z)|^{2}}$
$\displaystyle\quad\qquad\,\,+(1-\cos(2\arg(z-z_{0})-\pi))f_{23}(u+iv).$
(3.20)

For $i=4,6$ which can be defined similarly. Now we consider the boundary
values and consider case $i=1$ only. We have

$\overline{\partial}R_{21}=(\mathbf{r}-f_{21})\overline{\partial}b+\frac{b}{2}(\chi_{\mbox{Im}z}*r^{\prime}(\mbox{Re}z)+i(\mbox{Im}z)^{-1}\psi_{\mbox{Im}z}*r(\mbox{Re}z)),$

with $\psi(x)=-\chi(x)-x\chi^{\prime}(x)$ and
$f_{21}\frac{\hat{r}_{0}^{*}(z)}{1+|r(z_{1})|^{2}}(z-z_{1})^{2i\nu(z_{j})}\delta^{-2}(1-\Xi(\zeta))$.
Notice that $\hat{\psi}(0)=0$. Then we have the bound

$\displaystyle|\overline{\partial}R_{21}|$
$\displaystyle\leq|\chi_{\mbox{Im}z}*r^{\prime}(\mbox{Re}z)|+|(\mbox{Im}z)^{-1}\psi_{\mbox{Im}z}*r(\mbox{Re}z)|$
$\displaystyle+\frac{c}{|z-z_{2}|}|(r(z)-r(z_{2})|+|f_{21}(z)-r(z_{2})|).$
(3.21)

By (3.8) we acquire that

$|r(z)-r(z_{2})|\leq
C|z-z_{2}|^{s-\frac{1}{2}}||r||_{H^{s}(\mathbb{R})\,\cap\,L^{2.s}(\mathbb{R})}$

to get the desired estimate for $|\overline{\partial}R_{21}|$ we need to bound
the last line. Then, we have

$\displaystyle
f_{21}(z)-r(z_{2})=r(z_{2})\times[\exp(2i\mu(z_{2})((z-z_{2})\log(z-z_{2})$
$\displaystyle-(z-z_{2}+1)\log(z-z_{2}+1))+2(\beta(z,z_{2})-\beta(z_{2},z_{2})))-1].$
(3.22)

According to Lemma 3.1 we have

$\beta(z,z_{2})-\beta(z_{2},z_{2})=C(\rho,s)||r||_{H^{s}(\mathbb{R})\cap
L^{2,1}(\mathbb{R})}|z-z_{2}|^{s-\frac{1}{2}},$

and notice that both $(z-z_{2})\log(z-z_{2})$ and $(z-z_{2}+1)\log(z-z_{2}+1)$
are $\mathcal{O}(|z-z_{2}|^{s-\frac{1}{2}})$ when $z\to z_{2}$. We get the
desired estimate for $|\overline{\partial}R_{21}|$. ∎

Based on Proposition 3.2, we define a new unknown function

$\displaystyle\mathcal{R}^{(2)}=\begin{cases}\begin{pmatrix}1&0\\\
R_{j1}e^{2it\theta}\delta^{-2}&1\end{pmatrix}^{-1},\qquad
z\in\Omega_{j1},\\\\[8.0pt] \begin{pmatrix}1&R_{j3}e^{-2it\theta}\delta^{2}\\\
0&1\end{pmatrix}^{-1},\quad\,\,\,z\in\Omega_{j3},\\\\[8.0pt]
\begin{pmatrix}1&0\\\
R_{j4}e^{2it\theta}\delta^{-2}&1\end{pmatrix},\qquad\quad\,z\in\Omega_{j4},\\\\[8.0pt]
\begin{pmatrix}1&R_{j6}e^{-2it\theta}\delta^{2}\\\
0&1\end{pmatrix},\qquad\quad\,\,z\in\Omega_{j6},\\\\[5.0pt]
\begin{pmatrix}1&0\\\
0&1\end{pmatrix},\qquad\qquad\qquad\quad\,\,\,\,\,z\in\Omega_{2}\cup\Omega_{5}.\end{cases}$
(3.23)

We introduce jump contour

$\Sigma^{(2)}=L_{1}\cup\overline{L}_{1}\cup L_{2}\cup\overline{L}_{2}\cup
l_{1}\cup l_{2},$

then it can be shown that

$\displaystyle M^{(2)}(z)=M^{(1)}(z)\mathcal{R}^{(2)}$ (3.24)

satisfies the following RH problem.

######  RHP 3.2.

Find a continuous function $M^{(2)}(z):\mathbb{C}\setminus\Sigma^{(2)}\to
SL_{2}(\mathbb{C})$ with the following properties

  1. 1.

$M^{(2)}(z)=I+\mathcal{O}(z^{-1})$ as $z\to\infty$.

  2. 2.

For $z\in\Sigma^{(2)}$, the boundary values satisfy the jump relation
$M^{(2)}_{+}(z)=M^{(2)}_{-}(z)V^{(2)}(z)$, where

$\displaystyle V^{(2)}(z)=\begin{cases}\begin{pmatrix}1&0\\\
R_{j1}(z)\delta^{-2}(z)e^{2it\theta}&1\end{pmatrix},\quad\,\,z\in\Sigma_{j1},\\\
\\\ \begin{pmatrix}1&-R_{j3}(z)\delta^{2}e^{-2it\theta}\\\
0&1\end{pmatrix},\quad\quad z\in\Sigma_{j3},\\\ \\\ \begin{pmatrix}1&0\\\
R_{j4}(z)\delta^{-2}e^{2it\theta}&1\end{pmatrix},\quad\,\quad\,\,\,\,\,z\in\Sigma_{j4},\\\
\\\ \begin{pmatrix}1&-R_{j6}(z)\delta^{2}e^{-2it\theta}\\\
0&1\end{pmatrix},\quad\,\,\quad z\in\Sigma_{j6},\\\ \\\
\begin{pmatrix}1&(R_{23}-R_{13})e^{-2it\theta}\delta^{2}\\\
0&1\end{pmatrix},\qquad z\in l_{1},\\\ \\\ \begin{pmatrix}1&0\\\
(R_{14}-R_{24})e^{2it\theta}\delta^{-2}&1\end{pmatrix},\qquad z\in l_{2}.\\\
\end{cases}$ (3.25)

  3. 3.

For $\mathbb{C}\setminus\Sigma^{(2)}$ we have
$\overline{\partial}M^{(2)}(z)=M^{(2)}(z)\overline{\partial}\mathcal{R}^{(2)}(z)$
where

$\overline{\partial}\mathcal{R}^{(2)}=\begin{cases}\begin{pmatrix}1&0\\\
-\overline{\partial}R_{j1}(z)e^{2it\theta}\delta^{-2}&1\end{pmatrix},\qquad\,\,z\in\Omega_{j1},\\\
\\\ \begin{pmatrix}1&-\overline{\partial}R_{j3}(z)e^{-2it\theta}\delta^{2}\\\
0&1\end{pmatrix},\quad\,\,\,\,\,\,\,\,z\in\Omega_{j3},\\\ \\\
\begin{pmatrix}1&0\\\
\overline{\partial}R_{j4}(z)e^{2it\theta}\delta^{-2}&1\end{pmatrix},\qquad\quad\,z\in\Omega_{j4},\\\
\\\ \begin{pmatrix}1&\overline{\partial}R_{j6}(z)e^{-2it\theta}\delta^{2}\\\
0&1\end{pmatrix},\qquad\quad\,\,z\in\Omega_{j6},\\\ \\\
0,\qquad\,\,\,\qquad\qquad\qquad\quad\qquad\quad
z\in\Omega_{j2}\cup\Omega_{j5}.\end{cases}$ (3.26)

Next we decompose $M^{(2)}$ in the form

$\displaystyle M^{(2)}(z)=M^{(3)}(z)M^{rhp}(z),$ (3.27)

where $M^{rhp}(z)$ as a pure RH problem resulting from $M^{(2)}(z)$ by setting
$\overline{\partial}\mathcal{R}^{(2)}\equiv 0,\
z\in\mathbb{C}\setminus\Sigma^{(2)}$, which satisfies the following RH problem

######  RHP 3.3.

Find a analytical function $M^{rhp}(z):\mathbb{C}\setminus\Sigma^{(2)}\to
SL_{2}(\mathbb{C})$ with the following properties

  1. 1.

$M^{rhp}(z)=I+\mathcal{O}(z^{-1})$ as $z\to\infty$.

  2. 2.

For $z\in\Sigma^{(2)}$, the boundary values satisfy the jump relation
$M^{rhp}_{+}(z)=MM^{rhp}_{-}(z)V^{(2)}(z)$, where $V^{(2)}(z)$ is given by
(3.25).

While the $M^{(3)}(z)$ defined by (3.27) satisfies the following RH Problem

######  RHP 3.4.

Find a continuous matrix-valued function $M^{(3)}(z):\mathbb{C}\to
SL_{2}(\mathbb{C})$ with the following properties:

  1. 1.

$M^{(3)}(z)=I+\mathcal{O}(z^{-1})$ as $z\to\infty$.

  2. 2.

For $z\in\mathbb{C}$, we have

$\overline{\partial}M^{(3)}(z)=M^{(3)}(z)W^{(3)}(z),$

where
$W^{(3)}(z):=M^{rhp}(z)\overline{\partial}\mathcal{R}^{(2)}(z)M^{rhp}(z)^{-1}$
and $\overline{\partial}\mathcal{R}^{(2)}(z)$ is defined by (3.26).

###  3.3 A solvable model near $z_{1}$ and $z_{2}$

We introduce jump contour

$\Sigma^{(3)}=L_{1}\cup\overline{L}_{1}\cup L_{2}\cup\overline{L}_{2},$

and consider the following RH problem

######  RHP 3.5.

Find a analytical function $M^{loc}(z):\mathbb{C}\setminus\Sigma^{(3)}\to
SL_{2}(\mathbb{C})$ with the following properties

  1. 1.

$M^{loc}(z)=I+\mathcal{O}(z^{-1})$ as $z\to\infty$.

  2. 2.

For $z\in\Sigma^{(3)}$, the boundary values satisfy the jump relation
$M^{loc}_{+}(z)=M^{loc}_{-}(z)V^{(2)}(z)$, where $V^{(2)}(z)$ is given by
(3.25).

This RH problem is solvable and its solution is given by

$\displaystyle M^{loc}(z)=M^{pc}(\zeta_{1})+M^{pc}(\zeta_{2}),$

where $M^{pc}(\zeta_{j})$ is the solution of the parabolic cylinder model in
A, and $\zeta_{j}=\sqrt{(-1)8t(\alpha+6\beta z_{j})}(z-z_{j}),\ j=1,2$.

Direct calculation shows that there is constant $c>0$ such that

$\|V_{l}-I\|_{L^{\infty}(l_{1}\cup l_{2})}\lesssim e^{-ct},\quad t\to\infty.$

Further by using small RH problem, it can be shown that

######  Proposition 3.3.

$\displaystyle M^{rhp}(z)=M^{loc}(z)(I+O(e^{-ct})).$

###  3.4 The $\overline{\partial}$ argument

It is well understood that the solution to the $\overline{\partial}$ problem
3.4 can be given with the following Cauchy integral

$M^{(3)}(z)=I+\frac{1}{2\pi
i}\int_{\mathbb{C}}\frac{M^{(3)}(z)W^{(3)}(z)}{\varsigma-z}d\varsigma.$ (3.28)

Let $||r||_{H^{s}(\mathbb{R})\,\cap\,L^{2,s}(\mathbb{R})}\leq\lambda_{0}$,
define the following operator

$JH(z):=\frac{1}{\pi}\int_{\mathbb{C}}\frac{H(\varsigma)W(\varsigma)}{\varsigma-z}dA(\varsigma),$
(3.29)

then it can be shown that

######  Lemma 3.4.

The operator $J:L^{\infty}(\mathbb{C})\to L^{\infty}(\mathbb{C})\cap
C^{0}(\mathbb{C})$, and there exist a $C=C(\lambda_{0})$ such that

$||J||_{L^{\infty}(\mathbb{C})\to L^{\infty}(\mathbb{C})}\leq
Ct^{\frac{1-2s}{4}},\,\,\,t>0.$ (3.30)

###### Proof.

For convenience of expression the proof below only consider the case when
$z=z_{2}$ and from the defination of the operator of $J$ we suppose $H\in
L^{\infty}(\Omega_{21})$ then

$|JH(z)|\leq||H||_{L^{\infty}(\Omega_{21}})||\delta^{-2}||_{L^{\infty}(\Omega_{21})}\int_{\Omega_{21}}\frac{|\overline{\partial}R_{21}(\varsigma)e^{2it\theta}|}{|\varsigma-z|}dA(\varsigma),$
(3.31)

and from the properties of $\delta(z)$ we acquire that
$||\delta^{-2}||_{L^{\infty}(\Omega_{21})}\leq 1$. For $j=1,2,3$ we have the
bound of $I_{j}$ from (3.18) following

$\displaystyle
I_{j}=\int_{\Omega_{21}}\frac{|X_{j}(\zeta)e^{2it\theta}|}{|\varsigma-z|}dA(\varsigma),\qquad
X_{1}(z):=\partial_{\mbox{Re}z}\mathbf{r}(z),$ $\displaystyle
X_{2}(z):=||r||_{H^{s}(\mathbb{R})}|z-z_{2}|^{s-\frac{3}{2}},\quad
X_{3}(z):=(\mbox{Im}z)^{-1}\psi_{\mbox{Im}z}*r(\mbox{Re}z).$ (3.32)

Following the prove in Section 2.4 [29] and recall the expression of $z_{2}$
we set $\varsigma-z_{2}=u+iv$ and $z-z_{2}=\alpha+i\beta$, the region
$\Omega_{21}$ corresponds to $u\geq v\geq 0$ we have

$\displaystyle I_{1}$
$\displaystyle=\int_{\Omega_{21}}\frac{|\partial_{u}\mathbf{r}(\varsigma)|e^{-8t(\alpha+6\beta
z_{2})uv}}{|\varsigma-z|}dudv\lesssim\int_{0}^{\infty}dv\int_{v}^{\infty}\frac{|\partial_{u}\mathbf{r}(\varsigma)|e^{-8tuv}e^{-tz_{2}uv}}{|\varsigma-z|}du$
$\displaystyle\lesssim\int_{0}^{\infty}dve^{-8tv^{2}}e^{-tz_{2}v^{2}}||\partial_{u}\mathbf{r}(u,v)||_{L_{u}^{2}(v,\infty)}||((u-\alpha)^{2}+(v-\beta)^{2}))^{-1}||_{L_{u}^{2}(v,\infty)}.$
(3.33)

It is easy to check the relationship
$||((u-\alpha)^{2}+(v-\beta)^{2})^{-1}||_{L_{u}^{2}(v,\infty)}\leq
C|v-\beta|^{-\frac{1}{2}}$ is achieved. For fixed $C$ by using the Plancherel
we obtain

$\displaystyle||\partial_{u}\mathbf{r}(u,v)||_{L_{u}^{2}}$
$\displaystyle=||\partial_{u}\int_{\mathbb{R}}v^{-1}\chi(v^{-1}(u-t))r(t)dt||_{L_{u}^{2}}=||\xi\hat{\chi}(v\xi)\hat{r}(\xi)||_{L^{2}}$
$\displaystyle\leq
v^{s-1}||\xi^{1-s}\hat{\chi}(\xi)||_{L^{\infty}}||r||_{H^{s}}\leq
Cv^{s-1}||r||_{H^{s}}.$ (3.34)

Direct calculation yields

$\displaystyle I_{1}$
$\displaystyle\lesssim||r||_{H^{s}}\int_{\mathbb{R}}dve^{-8tv^{2}}e^{-tz_{2}v^{2}}|v|^{s-1}|v-\beta|^{-\frac{1}{2}}$
$\displaystyle\lesssim||r||_{H^{s}}\int_{\mathbb{R}}dve^{-t(1+z_{2})v^{2}}|v|^{s-1}|v-\beta|^{-\frac{1}{2}}$
$\displaystyle\lesssim((1+z_{2})t)^{\frac{1-2s}{4}}||r||_{H^{s}}\int_{\mathbb{R}}dve^{-v^{2}}(|v|^{s-\frac{3}{2}}+|v-\sqrt{t}\beta|^{s-\frac{3}{2}})$
$\displaystyle\lesssim(\int_{\mathbb{R}}e^{-v^{2}}|v|^{s-\frac{3}{2}}dv)||r||_{H^{s}}t^{\frac{1-2s}{4}}.$
(3.35)

For the last inequality we used the fact that for any $c\in\mathbb{R}$

$\displaystyle\int_{\mathbb{R}}e^{-v^{2}}|v-c|^{s-\frac{3}{2}}dv\leq\int_{|v|\leq|v-c|}e^{-v^{2}}|v|^{s-\frac{3}{2}}dv$
$\displaystyle+\int_{|v|\geq|v-c|}e^{-(v-c)^{2}}|v-c|^{s-\frac{3}{2}}dv\leq
2\int_{\mathbb{R}}e^{-v^{2}}|v|^{s-\frac{3}{2}}dv.$ (3.36)

The estimate for $I_{3}$ is similar after replacing (3.34) with

$\displaystyle||v^{-2}\int$
$\displaystyle\psi(v^{-1}(u-t))r(t)dt||_{L_{u}^{2}}=||v^{-1}\xi^{-s}\hat{\psi}(v\xi)\xi^{s}\hat{r}(\xi)||_{L^{2}}$
$\displaystyle\leq
v^{s-1}||\xi^{-s}\hat{\psi}(\xi)||_{L^{\infty}}||r||_{H^{s}}\leq
Cv^{s-1}||r||_{H^{s}},$ (3.37)

the schwartz function $\hat{\psi}$ with $\hat{\psi}(0)=0$ lead to the latter
bound in above inequality. In the similar method we estimate the $I_{2}$ as

$\displaystyle
I_{2}\lesssim\int_{0}^{\infty}e^{-8tv^{2}}e^{-tz_{2}v^{2}}dv|||\varsigma-
z_{2}|^{s-\frac{3}{2}}||_{L^{p}(v,\infty)}|||\varsigma-z|^{-1}||_{L^{q}(v,\infty)},$
(3.38)

where$\frac{1}{p}+\frac{1}{q}=1$. By [16] we get

$|||\varsigma-z|^{-1}||_{L^{q}(v,\infty)}\leq C|v-\beta|^{\frac{1}{q}-1},$
(3.39)

and

$\displaystyle|||\varsigma-
z_{2}|^{s-\frac{3}{2}}||_{L^{p}(v,\infty)}=\big{(}\int_{v}^{\infty}|u+iv|^{p(s-\frac{3}{2})}du\big{)}^{\frac{1}{p}}$
$\displaystyle=\big{(}\int_{v}^{\infty}(u^{2}+v^{2})^{p\frac{2s-3}{4}}du\big{)}^{\frac{1}{p}}=v^{\frac{2s-3}{2}+\frac{1}{p}}\big{(}\int_{v}^{\infty}(u^{2}+1)^{p\frac{2s-3}{4}}du\big{)}^{\frac{1}{p}}.$
(3.40)

So by(3.38) and using again (3.36), we obtain

$\displaystyle I_{2}$
$\displaystyle\lesssim\int_{0}^{\infty}e^{-8tv^{2}}e^{-tz_{2}v^{2}}v^{\frac{2s-3}{2}+\frac{1}{p}}|v-\beta|^{\frac{1}{q}-1}dv$
$\displaystyle\lesssim\int_{0}^{\infty}e^{-8tv^{2}}e^{-tz_{2}v^{2}}v^{\frac{2s-3}{2}}dv\leq
Ct^{\frac{1-2s}{4}}.$ (3.41)

And by above estimates using standard facts, like dominated convergence we can
proof the $J(L^{\infty})\subset C^{0}$ and the readers can prove it for
themselves. ∎

This lemma implies that the integral equation (3.28) has an unique solution,
and we further obtain the following result.

######  Proposition 3.5.

There exists $\epsilon_{0}>0$ such that for $||r||_{H^{s}(\mathbb{R})\cap
L^{2,s}(\mathbb{R})}<\epsilon$ there exist constants $T$ and $c$ such that for
$t\geq T$ and for $z\in\Omega_{j2}\cup\Omega_{j5}$

$\displaystyle M^{(3)}(z)=I+M^{(3)}_{1}z^{-1}+\mathcal{O}(z^{-2}),$ (3.42)

where

$\displaystyle|M^{(3)}_{1}|\leq c||q_{0}||_{H^{1}(\mathbb{R})\cap
L^{2,s}(\mathbb{R})}t^{-\frac{2s+1}{4}},\quad\mbox{for}\quad t\geq T.$ (3.43)

###### Proof.

In general we just consider the case when $j=2,i=1$. According to
$M^{(3)}_{1}=\frac{1}{\pi}\int_{\mathbb{C}}M^{(3)}WdA$ we acquire that

$|M^{(3)}_{1}|\leq\frac{||M^{(3)}||_{\infty}}{\pi}\int_{\Omega_{21}}|W|dA.$

By using

$\displaystyle||e^{-8t(\alpha+6\beta z_{2})uv}||_{L_{u}^{q}(v,\infty)}$
$\displaystyle=(8qt(\alpha+6\beta z_{2})v)^{-\frac{1}{q}}e^{-8qt(\alpha+6\beta
z_{2})v^{2}},$

We obtain

$\displaystyle\int$
${}_{\Omega_{21}}|X_{1}(\zeta)e^{2it\theta}|dA\leq||r||_{H^{s}\cap
L^{2,1}}\int_{0}^{\infty}v^{s-1}||e^{-8t(\alpha+6\beta
z_{2})uv}||_{L_{u}^{2}(v,\infty)}dv$ $\displaystyle\lesssim
t^{-\frac{1}{2}}\int_{0}^{\infty}v^{s-\frac{3}{2}}e^{-t(\alpha+6\beta
z_{2})v^{2}}dv||r||_{H^{s}\cap
L^{2,1}}=C_{s}t^{-\frac{2s+1}{4}}||r||_{H^{s}\cap L^{2,1}}.$

For $l=2$, we get

$\displaystyle\int_{\Omega_{21}}|X_{2}e^{2it\theta}|dA\leq||r||_{H^{s}\cap
L^{2,1}}\int_{0}^{\infty}|||\varsigma-
z_{2}|^{s-\frac{3}{2}}||_{L^{p}(v,\infty)}||e^{-8t(\alpha+6\beta
z_{2})uv}||_{L^{q}_{u}(v,\infty)}dv$ $\displaystyle\leq C||r||_{H^{s}\cap
L^{2,1}}t^{-\frac{1}{q}}\int_{0}^{\infty}v^{\frac{2s-3}{2}+\frac{1}{p}-\frac{1}{q}}e^{-t(\alpha+6\beta
z_{2})v^{2}}dv\leq C_{s}t^{-\frac{2s+1}{4}}||r||_{H^{s}\cap L^{2,1}}.$

Now we acquire (3.42) by
$||r||_{H^{s}(\mathbb{R})\,\cap\,L^{2,1}(\mathbb{R})}\leq
C||q_{0}||_{H^{1}(\mathbb{R})\,\cap\,L^{2,s}(\mathbb{R})}$ for a fixed $C$ by
the Lipschitz continuous of Lemma 2.2. ∎

######  Theorem 3.6.

Fix $s\in(\frac{1}{2},1]$ and let $q_{0}\in H^{1}(\mathbb{R})\cap
L^{2,s}(\mathbb{R})\cap\mathcal{G}_{0}$. Then there exist constants
$C(q_{0})>0$ and $T(q_{0})>0$ such that the solution of the Hirota equation
(1.1) satisfies

$||q(t,\cdot)||_{L^{\infty}(\mathbb{R})}\leq
C(q_{0})t^{-\frac{1}{2}},\,\,\mbox{for all}\,\,|t|\geq T(q_{0}).$ (3.44)

There are further constants $C_{0}>0,T_{0}>0$ and small $\epsilon>0$ such that
for $||q_{0}||_{H^{1}(\mathbb{R})\cap L^{2,s}(\mathbb{R})}<\epsilon$, we can
take $C(q_{0})=C_{0}||q_{0}||_{H^{1}(\mathbb{R})\cap L^{2,s}(\mathbb{R})}$ and
$T(u_{0})=T_{0}$.

###### Proof.

Recalling a series of transformations (3.10), (3.24) and (3.27), we have

$M(z)=M^{(3)}(z)M^{rhp}(z)R^{(2)}(z)^{-1}\delta^{\sigma_{3}}.$

Taking the limit $z\to\infty$ along $R^{(2)}(z)=I$ leads to

$M_{1}=M^{(3)}_{1}+\sum_{j=1}^{2}\frac{M^{pc}_{1}}{\sqrt{(-1)^{j}8t(\alpha+6\beta
z_{j})}}+\delta^{\sigma_{3}}.$

By using reconstruction formula (2.30), (3.43) for $t\geq T(s,\lambda_{0})$
and a fixed $C=C(s,\lambda_{0})$, we have

$|q(t,x)|\leq Ct^{-\frac{1}{2}},$

which proves Theorem 3.6 for $q_{0}\in H^{1}(\mathbb{R})\cap
L^{2,s}(\mathbb{R})\cap\mathcal{G}_{0}$. ∎

##  4 The asymptotic stability of the solitons

###  4.1 The Bäcklund transformation

In this section we consider scattering data $\\{r\equiv
0,\\{(z_{k},c_{k})_{k=1}^{N}\\}$ for which the reflection coefficient vanishes
identically correspond to $N$-soliton solution of (1.1). Especially when
$N=1$, the single soliton is given by (1), which is a localized pulse with
speed $v=-(12\beta\xi^{2}-4\beta\eta^{2}+4\alpha\eta^{2})$ and maximum
amplitude $2\eta$. Since $\mathcal{G}_{1}$ is an open subset of
$L^{1}(\mathbb{R})$ and the soliton (1) belong to it. If the value of
$\epsilon_{0}>0$ in the bound (1.4) is small enough, then the initial datum
$q_{0}$ belong to $\mathcal{G}_{1}$. Notice also that the positive constant
$\epsilon_{0}$ can be taken independent of $(\eta_{0},x_{0})$.

In this section we consider the initial datum $q_{0}$ satisfying the bound
$(\ref{eq:1.4})$ and the scattering datum associated with it belongs to the
space $\mathcal{S}(1,1)$ defined in (2.23) is close to those of the soliton
$q_{(\eta_{0},\xi_{0},\gamma_{0}))}(0,x)$ by Lemma 2.1 and Lemma 2.2, we also
obtain that when $q_{0}\in H^{1}(\mathbb{R})\cap L^{2,s}(\mathbb{R})$ implies
$r\in H^{s}(\mathbb{R})\cap L^{2,1}(\mathbb{R})$. Furthermore, by the
Lipschitz continuity of $q_{0}\to r$ and the fact that the soliton has
$r\equiv 0$, we have $||r||_{H^{s}(\mathbb{R})\cap L^{2,1}(\mathbb{R})}\leq
C\epsilon$ with $C=C(\eta_{0},\xi_{0},\gamma_{0})$ and the value of $\epsilon$
is given in (1.4).

We define now a map

$\mathcal{G}_{1}\times\mathbb{C}_{+}\times\mathbb{C}_{*}\ni(q_{0},z_{s},c_{1})\mapsto\tilde{q}_{0}\in\mathcal{G}_{0},$
(4.1)

via the transformation

$\tilde{r}(z):=r(z)\frac{z-z_{s}}{z-z^{*}_{s}}.$ (4.2)

From its definition, we acquire that $\tilde{r}\in H^{s}(\mathbb{R})\cap
L^{2,1}(\mathbb{R})$ if $r\in H^{s}(\mathbb{R})\cap L^{2,1}(\mathbb{R})$ and
there exist a constant $C>0$ such that $||\tilde{r}||_{H^{s}(\mathbb{R})\cap
L^{2,1}(\mathbb{R})}\leq C||r||_{H^{s}(\mathbb{R})\cap L^{2,1}(\mathbb{R})}$.
Then we define
$\tilde{q}\in\mathcal{G}_{0}\,\cap\,H^{1}\,\cap\,L^{2,s}(\mathbb{R})$ by the
reconstruction formula (2.30) with the corresponding RH problem is solved for
the scattering datum in $\mathcal{S}(1,0)=\\{\tilde{r}\in
H^{s}(\mathbb{R})\cap L^{2,1}(\mathbb{R})\\}$. By Lemma 2.5 we know that
$\tilde{q_{0}}\in\mathcal{G}_{0}\,\cap H^{s}(\mathbb{R})\cap
L^{2,1}(\mathbb{R})$ with norm $||\tilde{q}_{0}||_{H^{s}(\mathbb{R})\cap
L^{2,1}(\mathbb{R})}\leq C||\tilde{r}||_{H^{s}(\mathbb{R})\cap
L^{2,1}(\mathbb{R})}\leq C\epsilon$.

The BT technique is a method that can derive multisoliton solution starting
from a trivial seed solution in a purely algebraic procedure for a integrable
equation. Main feature of BT is that the Lax pair associated with the
integrable equation remains covariant under the gauge transformation.

######  Lemma 4.1.

Define the BT that generates a new solution of the Hirota equation as

$\tilde{q}=q-\mathbf{B},\quad\mathbf{B}:=2i(z_{s}-z_{s}^{*})\frac{f_{1}f^{*}_{2}}{|f_{1}|^{2}+|f_{2}|^{2}},$
(4.3)

where

$f_{1}:=e^{-ixz_{s}}m_{11}(t,x,z_{s})-\frac{c_{1}m_{12}(t,x;z_{s})e^{ixz_{s}+i(4\beta
z_{s}^{3}+2\alpha z_{s}^{2})t}}{2i\mbox{Im}(z_{s})},$
$f_{2}:=e^{-ixz_{s}}m_{21}(t,x,z_{s})-\frac{c_{1}m_{22}(t,x,z_{s})e^{ixz_{s}+i(4\beta
z_{s}^{3}+2\alpha z_{s}^{2})t}}{2i\mbox{Im}(z_{s})}.$

The proof will be given in the B.

######  Remark 4.2.

The soliton (1) can be recovered with the BT (4.3) by taking

$\displaystyle r=0,\quad z_{s}=\xi+i\eta,\quad\tilde{q}=0,$ $\displaystyle
f_{1}=e^{-ixz_{s}},\quad\mbox{and}\quad
f_{2}=-\frac{c_{1}}{2i\eta}e^{ixz_{s}+i(4\beta z_{s}^{3}+2\alpha
z_{s}^{2})t}.$

######  Lemma 4.3.

Let $z_{s}=\alpha_{1}+i\beta_{1}$ with $\beta_{1}>0$. There is $\epsilon_{0}$
sufficiently small such that for
$||q_{0}||_{H^{1}(\mathbb{R})\,\cap\,L^{2,s}(\mathbb{R})}<\epsilon_{0}$, there
is a constant $C$ such that

$\displaystyle|I-V_{+}(z_{s})|\leq
Ce^{-t8\beta_{1}^{2}}||q_{0}||_{H^{1,s}(\mathbb{R})},\,\,\mbox{if}\,\,z_{s}\in\Omega_{j1}+z_{j},$
$\displaystyle|I-U^{-1}_{R}(z_{s})|\leq
Ce^{-t8\beta_{1}^{2}}||q_{0}||_{H^{1,s}(\mathbb{R})},\,\,\mbox{if}\,\,z_{s}\in\Omega_{j3}+z_{j}.$
(4.4)

###### Proof.

For $j=2,i=1,3$ from (3.20) we obtain that

$||R_{i}||_{L^{\infty}(\Omega_{2i}+z_{2})}\leq
C^{\prime}||r||_{H^{s}(\mathbb{R})\cap L^{2,2}(\mathbb{R})}\leq
C||q_{0}||_{L^{2,s}(\mathbb{R})\cap H^{2}(\mathbb{R})}.$

If $z_{s}\in\Omega_{21}+z_{2}$ we have $\alpha_{1}-z_{2}\geq\beta_{1}$ and so

$|e^{-2it\theta}|\lesssim e^{-8t(\alpha-z_{2})\beta_{1}}\lesssim
e^{-t8\beta_{1}^{2}}.$

If $z_{s}\in\Omega_{23}+z_{2},$ we have similarly $|e^{2it\theta}|\leq
e^{-t8\beta_{1}^{2}}$ which yield (4.4).∎

By Theorem 3.6, we know that there exist constants $C_{0}$ and $T>0$ such that
for all $t>T$, we have

$||\tilde{q}(t,\cdot)||_{L^{\infty}(\mathbb{R})}\leq
C_{0}\epsilon|t|^{-\frac{1}{2}}.$

Since there is a constant $C>0$ such that
$||\tilde{q}_{0}||_{H^{1}(\mathbb{R})\,\cap\,L^{2,s}(\mathbb{R})}\leq
C\epsilon$. In order to prove Theorem 1.1 we need to focus only on
$\mathbf{B}$, we focusing on $t\gg 1$ we know that

######  Lemma 4.4.

Suppose that
$||q_{0}||_{H^{1}(\mathbb{R})\,\cap\,L^{2,s}(\mathbb{R})}<\epsilon_{0}$ and
$z_{s}\in\mathbb{C}_{+}$, then there are a $\epsilon_{0}>0$, a $c>0$ and a
$T>0$ such that

$|I-M^{(3)}(z_{s})|\leq ct^{-\frac{2s+1}{4}}||q_{0}||_{H^{1}(\mathbb{R})\cap
L^{2,s}(\mathbb{R})},\quad\mbox{for}\quad t\geq T.$ (4.5)

###### Proof.

Notice the function $M^{(3)}(z)$ satisfying the inequality with

$|M^{(3)}-I|\leq\frac{||M^{(3)}||_{\infty}}{\pi}\sum_{i}\int_{\Omega_{2i}}\frac{|W|}{|\zeta-
z_{s}|}dA,$

and similarly as in Proposition 3.5 we just consider the case when $j=2,i=1$.
We set $z_{s}=\alpha_{1}+i\beta_{1}$ and for $l=1,3$ and just like the case in
the (3.32) and (3.33) we obtain

$\displaystyle\int_{\Omega_{21}}\frac{|X_{l}(\varsigma)e^{2it\theta}|}{|\varsigma-
z_{2}|}dA\leq||r||_{H^{s}(\mathbb{R})}[A_{1}+A_{2}],$ (4.6)

where

$A_{1}:=\int_{0}^{\frac{\rho_{1}}{2}}v^{s-1}\Big{|}\Big{|}\frac{e^{-8t(\alpha+6\beta
z_{2})uv}}{|\varsigma-z_{2}|}\Big{|}\Big{|}_{L_{u}^{2}(v,\infty)}dv,$

and

$A_{2}:=\int_{\frac{\rho_{1}}{2}}^{\infty}v^{s-1}\Big{|}\Big{|}\frac{e^{-8t(\alpha+6\beta
z_{2})uv}}{|\varsigma-z_{2}|}\Big{|}\Big{|}_{L_{u}^{2}(v,\infty)}dv.$

We get

$\displaystyle A_{1}$
$\displaystyle=\int_{0}^{\frac{\rho_{1}}{2}}v^{s-1}\Big{|}\Big{|}\frac{e^{-8t(\alpha+6\beta
z_{2})uv}}{\sqrt{(u+z_{2}-\alpha_{1})^{2}+(v-\beta_{1})^{2}}}\Big{|}\Big{|}_{L_{u}^{2}(v,\infty)}dv$
$\displaystyle\leq
C^{\prime}(\beta_{1})\int_{0}^{\frac{\rho_{1}}{2}}v^{s-1}\Big{|}\Big{|}e^{-8t(\alpha+6\beta
z_{2})uv}\Big{|}\Big{|}_{L_{u}^{2}(v,\infty)}dv\leq
C(s,\beta_{1})t^{-\frac{4s+1}{8}}.$ (4.7)

By using (4.4) and consider the case when $t\geq 1$ and $e^{-8tv^{2}}\leq
e^{-t\gamma_{1}^{2}}e^{-4v^{2}}$ for $v\geq\frac{\beta_{1}}{2}$ we acquire the
bound for $A_{2}$ is

$\displaystyle A_{2}$
$\displaystyle\leq\int_{\frac{\rho_{1}}{2}}^{\infty}e^{-8t(\alpha+6\beta
z_{2})v^{2}}v^{s-1}|||\varsigma-z_{2}|^{-1}||_{L_{u}^{2}(v,\infty)}dv$
$\displaystyle\leq C\int_{\frac{\rho_{1}}{2}}^{\infty}e^{-8t(\alpha+6\beta
z_{2})v^{2}}v^{s-1}|v-\beta_{1}|^{-\frac{1}{2}}dv$ $\displaystyle\leq
Ce^{-t\beta_{1}^{2}}\int_{0}^{\infty}e^{-4(\alpha+6\beta
z_{2})v^{2}}v^{s-1}|v-\beta_{1}|^{-\frac{1}{2}}dv\leq
C^{\prime}e^{-t\beta_{1}^{2}}.$ (4.8)

For $l=2$ we similarly have

$\displaystyle\int_{\Omega_{21}}\frac{|X_{2}(\varsigma)e^{2it\theta}|}{|\varsigma-
z_{2}|}dA\leq||r||_{H^{s}(\mathbb{R})}[B_{1}+B_{2}],$ (4.9)

where

$B_{1}:=\int_{0}^{\frac{\rho_{1}}{2}}\int_{v}^{\infty}|\varsigma-
z_{2}|^{s-\frac{3}{2}}\frac{e^{-8t(\alpha+6\beta z_{2})uv}}{|\varsigma-
z_{2}|}dv,$

and

$B_{2}:=\int_{\frac{\rho_{1}}{2}}^{\infty}\int_{v}^{\infty}|\varsigma-
z_{2}|^{s-\frac{3}{2}}\frac{e^{-8t(\alpha+6\beta z_{2})uv}}{|\varsigma-
z_{2}|}dv.$

Then we obtain $B_{1}\leq C(\beta,s)t^{-\frac{2s+1}{8}}$ and $|\varsigma-
z_{2}|\geq\beta_{1}/2$. And when $t\geq 1$ we have

$\displaystyle B_{2}$
$\displaystyle\leq\int_{\frac{\rho_{1}}{2}}^{\infty}e^{-8t(\alpha+6\beta
z_{2})v^{2}}|||\varsigma-
z_{2}|^{s-\frac{3}{2}}||_{L^{p}(v,\infty)}|||\varsigma-
z_{2}|^{-1}||_{L^{q}(v,\infty)}dv$ $\displaystyle\leq
Ce^{-t\beta_{1}^{2}/2}\int_{0}^{\infty}e^{-4(\alpha+6\beta
z_{2})v^{2}}v^{\frac{2s-3}{2}+\frac{1}{p}}|v-\beta_{1}|^{\frac{1}{q}-1}dv\leq
C_{s}e^{-ty_{1}^{2}/2}.$ (4.10)

∎

######  Lemma 4.5.

Fix $\lambda_{0}>0$. Then there is a $C>0$ and a $T>0$ such that for
$||\tilde{r}||_{H^{s}(\mathbb{R})}<\lambda_{0}$ we have for $t\geq T$

$\displaystyle\big{|}m_{11}(t,x,z_{s})-\delta(z_{s})\big{|}+\big{|}m_{22}(t,x,z_{s})-\delta^{-1}(z_{s})\big{|}$
$\displaystyle\qquad\qquad\qquad\qquad\leq
C||\tilde{r}||_{H^{s}(\mathbb{R})}t^{-\frac{1}{2}}(||\tilde{r}||_{H^{s}(\mathbb{R})}+t^{-\frac{2s-1}{8}}),$
(4.11)
$\displaystyle\big{|}m_{12}(t,x,z_{s})-\frac{\delta^{-1}(z_{s})\beta_{12}}{\sqrt{(-1)^{j}8t(\alpha+6\beta
z_{j})}(z_{s}-z_{j})}\big{|}$
$\displaystyle\qquad\qquad\qquad\qquad+|m_{22}(t,x,z_{s})-\frac{\delta(z_{s})\beta_{21}}{\sqrt{(-1)^{j}8t(\alpha+6\beta
z_{j})}(z_{s}-z_{j})}\big{|}$ $\displaystyle\qquad\qquad\qquad\qquad\leq
C(||\tilde{r}||_{H^{s}(\mathbb{R})}t^{-\frac{2s+1}{8}}).$ (4.12)

###### Proof.

We focus on $j=2$ and recall the Lemma 4.4 we have the expansion of function
$M^{(3)}=I+\mathcal{O}(||\tilde{r}||_{H^{s}(\mathbb{R})}t^{-\frac{2s+1}{8}})$
and Lemma 4.3 we get similar expansions for $V_{+}(z_{s})$ and $U_{R}(z_{s})$.
We furthermore know by the property of $|\delta^{\pm}|\leq 1+\rho^{2}$ for
$\rho=||\tilde{r}||_{L^{\infty}(\mathbb{R})}$ and we acquire
$|\beta_{12}|+|\beta_{21}|<C||\tilde{r}||_{L^{\infty}(\mathbb{R})}$. From A we
have the expansion

$M^{pc}(\zeta_{2})=I+\frac{M^{pc}_{1}}{\sqrt{8t(\alpha+6\beta
z_{2})}(z_{s}-z_{2})}+\mathcal{O}(||\tilde{r}||_{H^{s}(\mathbb{R})}t^{-1}).$

These observations yield Lemma 4.5.∎

###  4.2 The proof of main result

Now we start to analyze the term $\mathbf{B}$ in (4.3). Consider the following
inequalities:

$\displaystyle\big{|}e^{-ixz_{s}}m_{11}(t,x,z_{s})\big{|}>10\big{|}\frac{c_{1}m_{12}(t,x;z_{s})e^{ixz_{s}+i(2\alpha
z_{s}^{2}+4\beta z_{s}^{3})t}}{2i\mbox{Im}(z_{s})}\big{|},$ (4.13)
$\displaystyle
10\big{|}e^{-ixz_{1}}m_{21}(t,x;z_{s})\big{|}<\big{|}\frac{c_{1}m_{22}(t,x,z_{s})e^{ixz_{s}+i(2\alpha
z_{s}^{2}+4\beta z_{s}^{3})t}}{2i\mbox{Im}(z_{s})}\big{|}.$ (4.14)

######  Lemma 4.6.

Given $\epsilon>0$ small, there exist $T(\epsilon_{0})>0$ and $C>0$ such that
if $||\tilde{r}||_{H^{s}(\mathbb{R})\,\cap\,L^{2,s}(\mathbb{R})}<\epsilon_{0}$
and if $(t,x)$ is such that at least one of (4.13) and (4.14) is false, then
we have $|\mathbf{B}|<Ct^{-\frac{1}{2}}\epsilon$ for $t\geq T(\epsilon_{0})$.

###### Proof.

For the $\epsilon$ of (1.4) and $\rho=||\tilde{r}||_{L^{\infty}(\mathbb{R})}$.
We are only consider the case when $t$ is large by assuming that for $(t,x)$
inequality(4.13) is false, Lemma 4.5 implies for $t\geq T$

$\displaystyle|m_{12}|$
$\displaystyle\leq(1+\rho^{2})|k_{1}|t^{-\frac{1}{2}}+C\epsilon
t^{-\frac{2s+1}{8}}$ $\displaystyle\leq t^{-\frac{1}{2}}\epsilon
K\big{(}\frac{1}{2}(1+\rho^{2})^{-1}-Ct^{\frac{1-2s}{8}}-C\epsilon\big{)}$
$\displaystyle\leq t^{-\frac{1}{2}}\epsilon K|m_{22}|,$ (4.15)

for a fixed and sufficiently large constant $K$. Then, if (4.13) is false and
$t\geq T$, both terms in (4.13) are bounded from above by

$\Big{|}\frac{c_{1}m_{22}(t,x;z_{s})e^{ixz_{s}+i(2\alpha z_{s}^{2}+4\beta
z_{s}^{3})t}}{2i\mbox{Im}(z_{s})}\Big{|}.$

For $t\geq T$ by the same argument of (4.15) we have also

$\big{|}e^{-ixz_{1}}m_{21}(x,t;z_{s})\big{|}\leq t^{-\frac{1}{2}}\epsilon
K|e^{-ixz_{s}}m_{11}(x,t;z_{s})\big{|}.$ (4.16)

We conclude that for $t\geq T$ and if $(t,x)$ is in the domain where (4.15) is
false, we have for some fixed $K$

$|\mathbf{B}|\leq K\frac{\big{|}m_{12}e^{ixz_{s}-i(2\alpha z_{s}^{2}+4\beta
z_{s}^{3})t}m^{*}_{22}e^{-ixz^{*}_{s}-i(2\alpha z^{*2}_{s}+4\beta
z_{s}^{*3})t}\big{|}}{|m_{22}e^{ixz_{s}+i(2\alpha z_{s}^{2}+4\beta
z_{s}^{3})t}|^{2}}=K\frac{|m_{12}|}{|m_{22}|}\leq\frac{CK}{\sqrt{t}}\epsilon.$
(4.17)

Now we assume that $(t,x)$ is such that (4.13) is true. Notice that by (4.13)
and (4.16) we have for a fixed $K$

$\frac{|f_{1}e^{ixz_{s}^{*}}m_{21}^{*}|}{||f||^{2}}\leq
K\frac{\big{|}e^{-ixz_{s}}{m_{11}e^{-ixz_{s}^{*}}}m_{21}^{*}|}{|e^{-ixz_{s}}m_{11}|^{2}}=K\frac{|m_{21}|}{|m_{11}|}\leq\frac{CK}{\sqrt{t}}\epsilon.$
(4.18)

Consider the suppose that $(t,x)$ is such that (4.13) and (4.14) are not both
true, we assume now that (4.13) is true and (4.14) is false. Then by (4.18),
for a fixed $K$

$|\mathbf{B}|\leq 4|\mbox{Im}z_{s}|\frac{f_{1}f_{2}}{||f||^{2}}\leq
K\frac{|f_{1}e^{ixz_{s}^{*}}m_{21}^{*}|}{||f||^{2}}\leq\frac{CK}{\sqrt{t}}\epsilon.$
(4.19)

This prove the Lemma 4.5 for values of $(t,x)$ for which (4.13) and (4.14) are
true. ∎

######  Lemma 4.7.

Fix $\rho_{0}>0$ and let $\rho:=||r||_{L^{\infty}(\mathbb{R})}$ and assume
$\rho<\rho_{0}$, then for fix $z_{s}=\alpha_{1}+i\beta_{1}$ with $\beta_{1}>0$
there exists a constant $C$ independent from $z_{j}$ such that

$\displaystyle|\delta(z_{s})-\nu(z_{s})|\leq C||r||_{L^{2}}^{2},$
$\displaystyle\mbox{where}\,\,\,\,\,\nu(z_{s}):=\exp\Big{(}\frac{1}{2\pi
i}\int_{-\infty}^{\alpha_{1}}\frac{\log(1+|r(\varsigma)|^{2}}{\varsigma-
z_{s}}d\varsigma\Big{)},$ (4.20)

and fix $K>0$ then for $|z_{j}-\alpha_{1}|\leq K/\sqrt{t}$ there exists a
constant $C$ such that

$|\delta(z_{s})-\nu(z_{s})|\leq\frac{C}{\sqrt{t}\beta_{1}}\log(1+\rho^{2}).$
(4.21)

###### Proof.

According to the Proposition 3.5 and for a fixed $c$ and for

$\gamma(z):=\frac{1}{2\pi
i}\int_{z_{1}}^{z_{2}}\frac{\log(1+|r(s)|^{2})}{s-z}ds,$

rewrite as $\delta(z)=\delta(z)^{\gamma(z)}$ and we obtain

$\Big{|}\gamma(z_{s})-\frac{1}{2\pi
i}\int_{z_{1}}^{\alpha_{1}}\frac{\log(1+|r(s)|^{2})}{s-z_{s}}ds\Big{|}=\frac{1}{2\pi}\Big{|}\int_{\alpha_{1}}^{z_{2}}\frac{\log(1+r(s)|^{2}}{s-\alpha_{1}-i\beta_{1}}ds\Big{|}\leq\frac{c}{\beta_{1}}||r||_{L^{2}}^{2}.$
(4.22)

This yield (4.20) since the bound $|\delta(z)|\leq(1+\rho^{2})$ is independent
from $z_{2}$. Similarly (4.21) follows from

$\Big{|}\gamma(z_{s})-\frac{1}{2\pi
i}\int_{z_{1}}^{\alpha_{1}}\frac{\log(1+r(s)|^{2})}{s-z_{s}}ds\Big{|}=\frac{1}{2\pi}\Big{|}\int_{\alpha_{1}}^{z_{2}}\frac{\log(1+|r(s)|^{2})}{s-\alpha_{1}-i\beta_{1}}ds\Big{|}.$
(4.23)

This yield Lemma 4.7.∎

Then, We assume that (4.13) and by (4.14) are true, from the inequality (4.17)
and (4.18), up to terms bounded by $Ct^{-\frac{1}{2}}\epsilon$, what is left
is the analysis of

$-2i\frac{e^{-ixz_{s}}m_{11}c_{1}^{*}m^{*}_{22}e^{-ixz_{s}^{*}-(2\alpha
z_{s}^{*2}+4\beta z_{s}^{*3})t}}{||f||^{2}}.$ (4.24)

Set now

$b^{2}=\big{|}e^{-ixz_{s}}m_{11}\big{|}^{2}+\big{|}\frac{c_{1}m_{22}e^{ixz_{s}+(2\alpha
z_{s}^{2}+4\beta z_{s}^{3})t}}{2i\mbox{Im}(z_{s})}\big{|}^{2},$

and expand

$||f||^{2}=b^{2}\big{(}1+\mathcal{O}(b^{-1}\big{|}c_{1}m_{12}e^{ixz_{s}+(2\alpha
z_{s}^{2}+4\beta
z_{s}^{3})t}\big{|})+\mathcal{O}(b^{-1}|m_{21}(t,x;z_{1})e^{-ixz_{1}})\big{)}.$

Then the quantity in (4.14) is of the form

$\displaystyle-2i$ $\displaystyle
e^{-ixz_{s}}m_{11}\frac{c_{1}^{*}m_{22}^{*}e^{-ixz_{s}^{*}-(2\alpha
z_{s}^{*2}+4\beta z_{s}^{*3})t}}{b^{2}}$
$\displaystyle\times\big{(}1+\mathcal{O}(b^{-1}\big{|}m_{12}e^{ixz_{s}+(2\alpha
z_{s}^{2}+4\beta
z_{s}^{3})t}\big{|})+\mathcal{O}(b^{-1}|m_{21}e^{-ixz_{s}})\big{)}.$ (4.25)

We claim that the quality in (4.24) equals

$-2i\frac{e^{-ixz_{1}}\delta(z_{s})(\delta(z_{s}))^{-1}c_{1}^{*}e^{-ixz_{s}^{*}-(2\alpha
z_{s}^{*2}+4\beta z_{s}^{*3})t}}{b^{2}}(1+\mathcal{O}(\epsilon
t^{-\frac{1}{2}})).$ (4.26)

To prove this claim, since
$m_{ii}=\delta^{-(-1)^{i}}(z_{1})+\mathcal{O}(\epsilon^{-\frac{1}{2})}$ and
$|\delta^{\pm 1}(z_{1})|\geq\langle\rho\rangle^{-2}$, we have

$b^{2}=(\big{|}\delta(z_{s})e^{-ixz_{s}}\big{|}^{2}+\big{|}\frac{c_{1}\delta(z_{s})^{-1}e^{ixz_{s}+(2\alpha
z_{s}^{2}+4\beta
z_{s}^{3})t}}{2i\mbox{Im}(z_{s})}\big{|}^{2})(1+\mathcal{O}(\epsilon
t^{-\frac{1}{2}})).$

We have $\mathcal{O}(b^{-1}\big{|}c_{1}m_{12}e^{ixz_{s}+(2\alpha
z_{s}^{2}+4\beta z_{s}^{3})t}\big{|})=\mathcal{O}(\epsilon t^{-\frac{1}{2}})$
by

$b^{-1}\big{|}c_{1}m_{12}e^{ixz_{s}+(2\alpha z_{s}^{2}+4\beta
z_{s}^{3})t}\big{|}\leq\frac{|m_{12}e^{ixz_{s}+(2\alpha z_{s}^{2}+4\beta
z_{s}^{3})t}|}{|m_{22}e^{ixz_{s}+(2\alpha z_{s}^{2}+4\beta
z_{s}^{3})t}|}=\frac{|m_{12}|}{|m_{22}|}\leq C\epsilon t^{-\frac{1}{2}}.$

We have
$\mathcal{O}(b^{-1}\big{|}m_{21}e^{ixz_{s}^{*}}\big{|})=\mathcal{O}(\epsilon
t^{-\frac{1}{2}})$ by

$b^{-1}\big{|}m_{21}e^{ixz_{s}}\big{|}\leq\frac{|m_{12}e^{-ixz_{s}}|}{|m_{11}e^{-ixz_{s}}|}=\frac{|m_{21}|}{|m_{11}|}\leq
C\epsilon t^{-\frac{1}{2}}.$

hence (4.26) is proved. Consider the term in (4.26), now for $z_{s}=\xi+i\eta$
and $\nu(z_{s})$ defined in (3.5) and inserting trivial factors $\nu/\nu=1$
and $\nu^{*}/\nu^{*}=1$, the expression in (4.26) equals

$\frac{4\eta e^{-ixz_{s}-i(2\alpha z_{s}^{2}+4\beta
z_{s}^{3})t}e^{-ixz_{s}^{*}-i(2\alpha z_{s}^{*2}+4\beta
z_{s}^{*3})t}\frac{\delta(z_{s})}{\nu(z_{s})}\frac{\nu^{*}(z_{s})}{\delta^{*}(z_{s})}\frac{\nu(z_{s})}{\nu(z_{s})}}{\tilde{b}^{2}},$
(4.27)

where

$\displaystyle\tilde{b}^{2}=\big{|}$ $\displaystyle e^{-2\eta
x-24\beta\xi^{2}t+8\beta\eta^{3}t-8\alpha\eta\xi
t}\big{|}^{2}|\frac{\delta(z_{s})}{\nu(z_{s})}||\nu(z_{s})|$
$\displaystyle\qquad\qquad\qquad+\big{|}e^{2\eta
x+24\beta\xi^{2}t-8\beta\eta^{3}t+8\alpha\eta\xi
t}\big{|}^{2}|\frac{\nu(z_{s})}{\delta(z_{s})}||\nu(z_{s})|^{-1}.$

Now fix a constant $\kappa>0$,Then (4.27) differs from the soliton solution

$\displaystyle 2$ $\displaystyle\eta e^{2i(-\xi
x-4\beta\xi^{3}t-2\alpha\xi^{2}t+12\beta\xi\eta^{2}t+2\alpha\eta^{2}t)+i\gamma}$
$\displaystyle\qquad\qquad\times\mbox{sech}(-2\eta
x-24\beta\xi^{2}t+8\beta\eta^{3}t-8\alpha\eta\xi t+\log(|\nu(z_{s})|)),$
(4.28)

by less than $c\kappa t^{-\frac{1}{2}}\epsilon$ by the sum of the following
two error terms: notice the (4.27) and (4.28) can be bounded, up to a constant
factor $C=C(\xi,\eta,\alpha,\beta)$, by the sum of the following two error
terms:

$\frac{\big{|}\frac{\delta(z_{s})}{\nu(z_{s})}\frac{\nu^{*}(z_{s})}{\delta^{*}(z_{s})}-1\big{|}}{e^{8(3\beta|z_{2}-\xi|^{2}t-\beta\eta^{3}t+\alpha\eta|z_{2}-\xi|t)}(1+||\tilde{r}||_{L^{\infty}(\mathbb{R})}^{2})^{-1}},$
(4.29)

and

$\displaystyle\big{|}$
$\displaystyle\mbox{sech}(-8(3\beta|z_{2}-\xi|^{2}t-\beta\eta^{3}t+\alpha\eta|z_{2}-\xi|t+\log(|\nu(z_{s})|))$
$\displaystyle-\mbox{sech}(-8(3\beta|z_{2}-\xi|^{2}t-\beta\eta^{3}t+\alpha\eta|z_{2}-\xi|t+\log(|\nu(z_{s})|)+\log\big{(}\frac{|\delta(z_{s})|}{|\nu(z_{s})|}\big{)})\big{|}.$
(4.30)

At first we bounded (4.29) and for $|z_{2}-\xi|\geq\kappa t^{-\frac{1}{2}}$
formula (4.29) is bounded by
$Ce^{8(3\beta|\sqrt{t}\kappa}e^{-\beta\eta^{3}t+\alpha\eta\kappa\sqrt{t})}.$
For $|z_{2}-\xi|\leq\kappa t^{-\frac{1}{2}}$ we bounded (4.29) by (4.21)

$(1+||\tilde{r}||_{L^{\infty}(\mathbb{R})}^{2})\Big{|}\frac{\delta(z_{s})}{\nu(z_{s})}\frac{\nu^{*}(z_{s})}{\delta^{*}(x_{s})}-1\Big{|}\leq
4\frac{C}{\sqrt{t}}(1+||\tilde{r}||_{L_{\infty}^{2}(\mathbb{R})})\leq
Kt^{-\frac{1}{2}}\epsilon^{2}.$

According to Lagrange Theorem, (4.30) is bounded

$\displaystyle\mbox{sech}(-8(3\beta|z_{2}-\xi|^{2}t$
$\displaystyle-\beta\eta^{3}t+\alpha\eta|z_{2}-\xi|t+\log(|\nu(z_{s})|)$
$\displaystyle+\log\big{(}\frac{|\delta(z_{s})|}{|\nu(z_{s})|}\big{)})\Big{|}\log\Big{(}\frac{|\delta(z_{s})|}{|\nu(z_{s})|}\Big{)}\Big{|},$

for some $c\in(0,1)$. This satisfies bounds similar to those satisfied by
(4.29).

We consider initial potential $q_{0}\in
H^{1}(\mathbb{R})\,\cap\,L^{2,s}(\mathbb{R})$ we need to show that when one of
(4.13) and (4.14) is false to complete the proof of Theorem 1.1, then the
function in (4.28) is $\mathcal{O}(\epsilon t^{-\frac{1}{2}})$. By Lemma 4.5
the fact that (4.13) and (4.14) false means that for a fixed $C=C(\rho_{0})>0$
we have

$|e^{-2ixz_{s}-2i(\alpha z_{s}^{*2}+2\beta z_{s}^{*3})t}|\leq C\epsilon
t^{-\frac{1}{2}},$

and

$|e^{2ixz^{*}_{s}+2i(\alpha z_{s}^{2}+2\beta z_{s}^{3})t}|\leq C\epsilon
t^{-\frac{1}{2}}.$

Any of these yield our claim that the function in (4.28) is
$\mathcal{O}(\epsilon t^{-\frac{1}{2})}$ Then we complete the proof of Theorem
1.1 for $q_{0}\in H^{1}(\mathbb{R})\,\cap\,L^{2,s}(\mathbb{R})$ and we notice
that when $|t|\geq T(\epsilon_{0})$ the soliton in formula (1) is given by
formula (4.28).

In the end of this paper of we will explain the ground states
$q_{\xi,\eta,\gamma_{\pm},\alpha,\beta}(t,x-x_{\pm})$ in the statement of
Theorem 1.1 are in general distinct. The + ground state has been computed
explicitly in (4.28).

######  Lemma 4.8.

Considering $t<-T(\lambda_{0})$ ground state is given by formula (4.30) but
with $\nu(z_{s})$ replaced by

$\Lambda(z_{s})=\exp\Big{(}\frac{1}{2\pi
i}\int_{\alpha_{1}}^{\infty}\frac{(\log(1+|r(s)|^{2})}{s-z_{s}}ds)\Big{)}.$
(4.31)

###### Proof.

Notice that if $q(t,x)$ solves (1.1) then $u(t,x):=q^{*}(-t,x)$ solves the
Hirota equation with initial value $q^{*}_{0}(x)$, and if $(r(z),z_{s},c_{1})$
are the spectral data of $q_{0}\in\mathcal{G}_{1}$, then we have
$q^{*}_{0}\in\mathcal{G}_{1}$ with spectral data
$(r^{*}(-z),-z_{s}^{*},-c_{1}^{*})$. According to (4.28) we obtain when
$t\to-\infty$

$\displaystyle u(-t,x)$ $\displaystyle\sim-2\eta e^{2i(\xi
x+4\beta\xi^{3}t+2\alpha\xi^{2}t-12\beta\xi\eta^{2}t-2\alpha\eta^{2}t)-i\gamma}$
$\displaystyle\times\mbox{sech}(-2\eta
x-24\beta\xi^{2}t+8\beta\eta^{3}t-8\alpha\eta\xi t+\log(|\Lambda(z_{s})|)).$

And the complex conjugate of $\Lambda(z_{s})$ can be written as

$\Lambda^{*}(z_{s})=\exp\Big{(}\frac{1}{2\pi
i}\int_{-\infty}^{\xi}\frac{(\log(1+|r(s)|^{2})}{s+\xi-i\eta}ds)\Big{)}.$

Then (4.31) is true and using $q(t,x)=u^{*}(-t,x)$ and so taking the complex
conjugate of the above formula, we obtain for $t\to-\infty$

$\displaystyle q(t,x)$ $\displaystyle\sim 2\eta e^{2i(-\xi
x-4\beta\xi^{3}t-2\alpha\xi^{2}t+12\beta\xi\eta^{2}t+2\alpha\eta^{2}t)+i\gamma}$
$\displaystyle\times\mbox{sech}(-2\eta
x-24\beta\xi^{2}t+8\beta\eta^{3}t-8\alpha\eta\xi t+\log(|\Lambda(z_{s})|)).$

thus completing the proof of Lemma 4.8. ∎

##  Appendix A A parabolic cylinder model

Here we describe the solution of the parabolic cylinder model problem. Let
$\Sigma^{pc}=\cup_{n=1}^{4}\Sigma_{j}$, where $\Sigma_{j}$ denotes the complex
contour

$\Sigma_{j}=\\{\zeta\in\mathbb{C}|\arg\zeta=\frac{2j-1}{4}\pi\\},\quad
j=1,\dots,4.$ (A.1)

Denote $\Omega_{j},j=1,\dots,6$ be the six maximally connected open sectors in
$\mathbb{C}\setminus(\Sigma^{pc}\cup\mathbb{R})$ where its labelled
sequentially as one encircles the origin in a counterclockwise fashion.
Finally, fix $r_{0}\in\mathbb{C}$ and let

$\kappa:=-\frac{1}{2\pi}\log(1+|r_{0}|^{2}).$ (A.2)

Then consider the following RH problem.

######  RHP A.1.

Fix $r_{0}\in\mathbb{C}$, find an analytic function
$M^{pc}(\cdot):\mathbb{C}\setminus\Sigma^{pc}\to SL_{2}(\mathbb{C})$

  1. 1.

$M^{pc}(\zeta)=I+\zeta^{-1}M^{pc^{(1)}}+\mathcal{O}(\zeta^{-2})$ uniformly as
$\zeta\to\infty$.

  2. 2.

For $\zeta\in\Sigma^{pc}$, the continuous values $M^{pc}_{\pm}(\zeta)$ satisfy
the jump relation

$M^{pc}_{+}(\zeta)=M^{pc}_{-}(\zeta)V^{pc}(\zeta),$

where

$V^{pc}(\zeta)=\begin{cases}\begin{pmatrix}1&0\\\
r_{0}\zeta^{-2i\kappa}e^{i\zeta^{2}/2}&1\end{pmatrix},\quad\,\,\,\,\quad\qquad\mbox{arg}\zeta=\frac{\pi}{4},\\\
\\\ \begin{pmatrix}1&r_{0}^{*}\zeta^{2i\kappa}e^{-i\zeta^{2}/2}\\\
0&1\end{pmatrix},\quad\,\,\quad\,\qquad\mbox{arg}\zeta=-\frac{\pi}{4},\\\ \\\
\begin{pmatrix}1&\frac{r_{0}^{*}}{1+|r_{0}|^{2}}\zeta^{2i\kappa}e^{-\zeta^{2}/2}\\\
0&1\end{pmatrix},\quad\qquad\mbox{arg}\zeta=\frac{3\pi}{4},\\\ \\\
\begin{pmatrix}1&0\\\
\frac{r_{0}}{1+|r_{0}|^{2}}\zeta^{-2i\kappa}e^{i\zeta^{2}/2}&1\end{pmatrix},\quad\qquad\mbox{arg}\zeta=-\frac{\pi}{4},\end{cases}$
(A.3)

See Figure A.1.

Re
$\zeta$$\Sigma_{21}$$\Sigma_{24}$$\Sigma_{23}$$\Sigma_{22}$$0$$\Omega_{1}$$\Omega_{2}$$\Omega_{3}$$\Omega_{4}$$\Omega_{5}$$\Omega_{6}$
Figure A.1: The contours $\Sigma_{j}$ and sectors $\Omega_{j}$ in the
$\zeta$-plane defining RHP A.1

The solution of the RH problem A.1 can be given by

$M^{pc}(\zeta,r)=I+\frac{1}{\zeta}\begin{pmatrix}0&-i\beta_{12}\\\
i\beta_{21}&0\end{pmatrix}+\mathcal{O}(\zeta^{-2}),$ (A.4)

where $\beta_{12}$ and $\beta_{21}$ are the complex constants

$\displaystyle\beta_{12}=\frac{-i\sqrt{2\pi}e^{i\pi/4}e^{-\pi\kappa/2}}{r_{0}\Gamma(-i\kappa)},$
$\displaystyle\beta_{21}=\frac{i\sqrt{2\pi}e^{-i\pi/4}e^{-\pi\kappa/2}}{r^{*}_{0}\Gamma(i\kappa)}=\frac{\kappa}{\beta_{12}}.$
(A.5)

It can be shown that

######  Lemma A.1.

Let $\rho=||r||_{L^{\infty}(\mathbb{R})}$, there exists a $C$ such that

$\displaystyle|M^{pc}(\zeta)|\leq C,\quad\mbox{for
all}\,\,\zeta\notin\mathbb{R},$ (A.6)
$\displaystyle\Big{|}M^{pc}(\zeta)-I-\zeta^{-1}{M^{pc}}^{(1)}\Big{|}\leq
C\rho|\zeta|^{-2}\quad\mbox{for }\,\,|\zeta|\geq 1.$ (A.7)

##  Appendix B RH problem under BT

The following calculations are standard and can be found in [25]. Let $q(x)\in
H^{1,1}(\mathbb{R})$ be given and consider the associated ZS-AKNS operator and
its reflection coefficient function $r(z)\in H^{1,1}(\mathbb{R})$. Support
that for each $x\in\mathbb{R}$, $2\times 2$ matrix
$\Phi(x,t;z)=\Psi(x,t;z)e^{it\theta(z)\sigma_{3}}$ solves the corresponding RH
problem with a finite number of simple bound states at

$\mathcal{Z}=\\{z\,|\,z=z_{1},\dots,z_{n}\in\mathbb{C}^{+}\\},\
\mathcal{Z}^{*}=\\{z|z=z^{*}_{1},\dots,z^{*}_{n}\in\mathbb{C}_{-}\\}.$

######  RHP B.1.

Find an analytic function
$\Psi:\mathbb{C}\setminus(\mathbb{R}\cup\mathcal{Z}\cup\mathcal{Z}^{*})\to
SL_{2}(\mathbb{C})$ with the following properties

  1. 1.

$\Psi(x,z)e^{it\theta\sigma_{3}}=I+\mathcal{O}(z^{-1})$ as $z\to\infty$.

  2. 2.

$\Psi(x,z)e^{it\theta\sigma_{3}}$ takes continuous boundary values
$\Psi_{\pm}(x,z)e^{it\theta\sigma_{3}}$ which satisfy the jump relation
$\Psi_{+}(x,z)e^{it\theta\sigma_{3}}=\Psi_{-}(x,z)e^{it\theta\sigma_{3}}V(z)$
where

$V(z)=\begin{pmatrix}1+|r(z)|^{2}&r^{*}(z)\\\ r(z)&1\end{pmatrix}.$ (B.1)

  3. 3.

$\Psi(z)$ has simple poles at each $z_{k}\in\mathcal{Z}$ and
$z_{k}^{*}\in\mathcal{Z^{*}}$ ($1\leq k\leq n$) at which

$\underset{z=z_{k}}{\mbox{Res}}\Psi(x,z)e^{it\theta\sigma_{3}}=\underset{z\to
z_{k}}{\mbox{lim}}\Psi(x,z)e^{it\theta\sigma_{3}}\begin{pmatrix}0&0\\\
c_{k}&0\end{pmatrix},$ (B.2)
$\underset{z=z_{k}^{*}}{\mbox{Res}}\Psi(x,z)e^{it\theta\sigma_{3}}=\underset{z\to
z_{k}^{*}}{\mbox{lim}}\Psi(x,z)e^{it\theta\sigma_{3}}\begin{pmatrix}0&-c_{k}^{*}\\\
0&0\end{pmatrix}.$ (B.3)

The goal is to add in another simple bound state at
$z=\xi\in\mathbb{C}^{+}\setminus\\{z_{1},\dots,z_{n}\\}$ and simultaneously at
$z=\xi^{*}\in\mathbb{C}^{-}\setminus\\{z^{*}_{1},\dots,z^{*}_{n}\\}.$ We use a
Darboux transformation
$(z+P)(\partial_{x}-L)=(\partial_{x}-\overline{L})(z+P)$ where $P$ can chosen
in the form $P=\mathfrak{b}(x)P_{0}\mathfrak{b}^{-1}(x)$ and $P_{0}$ is a
constant matrix and $\mathfrak{b}=\mathfrak{b}(x)$ solves the equation
$b^{\prime}=Qb-i\frac{\sigma_{3}}{2}bP_{0}$, here the approprate choice here
is $P_{0}=-\begin{pmatrix}\xi&0\\\ 0&\overline{\xi}\end{pmatrix}$, where
$\mathfrak{b}$ is determined below. Set

$\tilde{\Psi}(x,z)=\mathfrak{b}(x)\mu(z)\mathfrak{b}^{-1}(x)\Psi(x,z)\mu^{-1}(z),$
(B.4)

where $\mu(z)=z+P_{0}=\begin{pmatrix}z-\xi&0\\\ 0&z-\xi^{*}\end{pmatrix}$.
Note that $\tilde{\Psi}(x,z)e^{it\theta\sigma_{3}}\to I$ as $z\to\infty$. Let
$\tilde{c}(\xi)$ be any nonzero constant. We want to choose $\mathfrak{b}(x)$
so that $\tilde{\Psi}$ has a simple pole in the first column at $z=\xi$ and a
simple pole in the second column at $z=\xi^{*}$ such that for
$x\in\mathbb{R}$,

$\underset{z=\xi}{\mbox{Res}}\tilde{\Psi}e^{it\theta\sigma_{3}}=\underset{z\to\xi}{\mbox{lim}}\tilde{\Psi}e^{it\theta\sigma_{3}}\begin{pmatrix}0&0\\\
\tilde{c}(\xi)&0\end{pmatrix},$
$\underset{z=\xi^{*}}{\mbox{Res}}\tilde{\Psi}e^{it\theta\sigma_{3}}=\underset{z\to\xi^{*}}{\mbox{lim}}\tilde{\Psi}e^{it\theta\sigma_{3}}\begin{pmatrix}0&-\tilde{c^{*}(\xi)}\\\
0&0\end{pmatrix}.$

Since

$\mathfrak{b}^{-1}\tilde{\Psi}=\begin{pmatrix}(\mathfrak{b}^{-1}\Psi)_{11}&(\mathfrak{b}^{-1}\Psi)_{12}\frac{z-\xi}{z-\xi^{*}}\\\
\\\
(\mathfrak{b}^{-1}\Psi)_{21}\frac{z-\xi^{*}}{z-\xi}&(\mathfrak{b}^{-1}\Psi)_{21}\end{pmatrix},$

we have

$\underset{z_{\xi}}{\mbox{Res}}\,\mathfrak{b}^{-1}\tilde{\Psi}=\begin{pmatrix}0&0\\\
(\mathfrak{b}^{-1}\Psi)_{21}(x,\xi)(\xi-\xi^{*})&0\end{pmatrix},$

but

$\underset{z\to\xi}{\mbox{lim}}\,\mathfrak{b}^{-1}\tilde{\Psi}\begin{pmatrix}0&0\\\
\tilde{c}(\xi)&0\end{pmatrix}=\begin{pmatrix}0&0\\\
\tilde{c}(\xi)(\mathfrak{b}^{-1}\Psi)_{22}(x,\xi)&0\end{pmatrix},$

and hence we must have

$(\xi-\xi^{*})(e_{2},\mathfrak{b}^{-1}\Psi(x,\xi)e_{1})=\tilde{c}(\xi)(e_{2},\mathfrak{b}^{-1}\Psi(x,\xi)e_{2}).$

Therefore, it follows necessarily that

$\mathfrak{b}(x)e_{1}=c_{1}(x)\big{(}\Psi(x,\xi)e_{1}-\frac{\tilde{c}(\xi)}{\xi-\xi^{*}}\Psi(x,\xi)e_{2}\big{)},$

for some nonzero function $c_{1}(x)$. Similarly for $z=\xi^{*}$, we see that

$\mathfrak{b}(x)e_{2}=c_{2}(x)\big{(}-\frac{\tilde{c}^{*}(\xi)}{\xi-\xi^{*}}\Psi(x,\xi^{*})e_{1}+\Psi(x,\xi^{*})e_{2}\big{)},$

for some nonzero function $c_{2}(x)$, Observe that $c_{1}(x)$ and $c_{2}(x)$
factor out in the formula (B.1) for $\tilde{\Psi}$. Set

$\mathfrak{b}=\big{(}\Psi(x,\xi)\begin{pmatrix}1\\\
-\frac{\tilde{c}(\xi)}{\xi-\xi^{*}}\\\
\end{pmatrix}\quad\Psi(x,\xi^{*})\begin{pmatrix}-\frac{\tilde{c}^{*}(\xi)}{\xi-\xi^{*}}\\\
1\end{pmatrix}\big{)}.$ (B.5)

From the symmetry we see that $\mathfrak{b}_{2}=\begin{pmatrix}0&1\\\
-1&0\end{pmatrix}\mathfrak{b}_{1}^{*}$, where
$\mathfrak{b}=(\mathfrak{b}_{1},\mathfrak{b}_{2})$. Thus,
$\mbox{det}\mathfrak{b}(x)=|\mathfrak{b}_{1})_{1}(x)|^{2}+|(\mathfrak{b}_{1})_{2}(x)|^{2}>0$
and hence $\mathfrak{b}(x)$ is invertible for all $x\in\mathbb{R}$. The jump
matrix $\tilde{V}$ for $\Psi(x,z)$ is given by

$\displaystyle\tilde{V}(z)$
$\displaystyle=\tilde{\Psi}_{-}^{-1}(x,z)\tilde{\Psi}_{+}(x,z)=\mu(z)V(z)\mu^{-1}(z)$
(B.6) $\displaystyle=\begin{pmatrix}1+|\tilde{r}(z)|^{2}&\tilde{r}^{*}(z)\\\
\tilde{r}(z)&1\end{pmatrix},\qquad z\in\mathbb{R},$ (B.7)

where

$\tilde{r}(z)=r(z)\frac{z-\xi}{z-\xi^{*}}.$

A straightforward calculation shows that for $1\leq k\leq n$,

$\displaystyle\underset{z_{k}}{\mbox{Res}}\tilde{\Psi}=\underset{z\to
z_{k}}{\mbox{lim}}\tilde{\Psi}\begin{pmatrix}0&0\\\
\tilde{c}_{k}(z_{k})&0\end{pmatrix},$
$\displaystyle\underset{z_{k}^{*}}{\mbox{Res}}\tilde{\Psi}=\underset{z\to
z_{k}^{*}}{\mbox{lim}}\tilde{\Psi}\begin{pmatrix}0&-\tilde{c}_{k}(z_{k})^{*}\\\
0&0\end{pmatrix},$ (B.8)

where

$\tilde{c}(z_{k})=c(z_{k})\frac{z_{k}-\xi^{*}}{z_{k}-\xi}.$

The above calculations show that
$\tilde{M}(x,z)=\tilde{\Psi}(x,z)e^{-it\theta\sigma_{3}}$ solves the RH
Problem 2.1. Note that

$\tilde{a}(z)=\frac{z-\xi}{z-\xi^{*}}a(z),$

where $z(z)$ and $\tilde{a}(z)$ are the scattering functions for $\Psi(x,z)$
and $\tilde{\Psi}(x,z)$, respectively. From the fact that
$\Psi_{x}+iz[\sigma_{3},\Psi]=Q\Psi,Q=\begin{pmatrix}0&q\\\
-q^{*}&0\end{pmatrix}$, we have

$Q=-i[\sigma_{3},\Phi_{1}],\quad\Psi=I+\frac{\Phi_{1}}{z}+\mathcal{O}(z^{-1}),$

as $z\to\infty$ in any cone $|\mbox{Im}z|>c|\mbox{Re}z|$, and $c>0$. Let
$\mu_{1}\begin{pmatrix}\xi&0\\\ 0&\xi^{*}\end{pmatrix}.$ For
$\tilde{\Psi}(x,z)=\tilde{\Phi}(x,z)e^{-it\theta\sigma_{3}}$

$\displaystyle\tilde{\Phi}$
$\displaystyle=\mathfrak{b}\big{(}I-\frac{\mu_{1}}{z}\big{)}\mathfrak{b}^{-1}\big{(}I+\frac{\Phi_{1}}{z}+\mathcal{O}(z^{-1})\big{)}\big{(}I-\frac{\mu_{1}}{z}\big{)}^{-1}$
(B.9)
$\displaystyle=I+\frac{\Psi_{1}-\mathfrak{b}\mu_{1}\mathfrak{b}^{-1}+\mu_{1}}{z}+\mathcal{O}(z^{-1}),$
(B.10)

and hence

$\displaystyle\tilde{q}(x)$
$\displaystyle=-i[\sigma_{2},\Phi_{1}-\mathfrak{b}\mu_{1}\mathfrak{b}^{-1}+\mu_{1}]_{12}$
(B.11)
$\displaystyle=q(x)+i(\xi-\xi^{*})\frac{(\mathfrak{b}_{1})_{1}(\mathfrak{b}_{1})_{2}^{*}}{|(\mathfrak{b}_{1})_{1}|^{2}+|(\mathfrak{b}_{1})_{2}|^{2}}.$
(B.12)

One can also Darboux transformations similar to (B.1) to remove eigenvalues.
We do not provide any further details, except to not that at each step, if the
poles at $z=z_{k},z_{k}^{*}$ are removed, then
$r(z)\to\tilde{r}(z)\frac{z-z_{k}^{*}}{z-z_{k}}$, etc.

Acknowledgements

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

Data Availability Statements

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

Conflict of Interest

The authors have no conflicts to disclose.

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