11footnotetext:  School of Mathematical Sciences, Fudan University, Shanghai
200433, P.R. China.

# Long time asymptotics for the nonlocal mKdV equation with finite density
initial data

Xuan Zhoua, Engui Fana Corresponding author, with e-mail address as
<EMAIL_ADDRESS>
###### Abstract

In this paper, we consider the Cauchy problem for an integrable real nonlocal
(also called reverse-space-time) mKdV equation with nonzero boundary
conditions

$\displaystyle q_{t}(x,t)-6\sigma q(x,t)q(-x,-t)q_{x}(x,t)+q_{xxx}(x,t)=0,$
$\displaystyle q(x,0)=q_{0}(x),\lim_{x\to\pm\infty}q_{0}(x)=q_{\pm},$

where $|q_{\pm}|=1$ and $q_{+}=\delta q_{-}$, $\sigma\delta=-1$. Based on the
spectral analysis of the Lax pair, we express the solution of the Cauchy
problem of the nonlocal mKdV equation in terms of a Riemann-Hilbert problem.
In a fixed space-time solitonic region $-6<x/t<6$, we apply
$\bar{\partial}$-steepest descent method to analyze the long-time asymptotic
behavior of the solution $q(x,t)$. We find that the long time asymptotic
behavior of $q(x,t)$ can be characterized with an $N(\Lambda)$-soliton on
discrete spectrum and leading order term $\mathcal{O}(t^{-1/2})$ on continuous
spectrum up to an residual error order $\mathcal{O}(t^{-1})$.

Keywords: Nonlocal mKdV equation, Riemann-Hilbert problem,
$\bar{\partial}$-steepest descent method, Long time asymptotics, Soliton
resolution.
Mathematics Subject Classification: 35Q51; 35Q15; 35C20; 37K15.

###### Contents

  1. 1 Introduction
  2. 2 The spectral analysis and the RH problem
    1. 2.1 Notations
    2. 2.2 The Lax pair and spectral analysis
    3. 2.3 A RH problem
  3. 3 Distribution of Saddle Points and Signature Table
  4. 4 Deformation of the RH Problem
    1. 4.1 Jump matrix factorizations
    2. 4.2 Characteristic lines and estimates for ${\rm Im}\theta(z)$
    3. 4.3 Mixed $\bar{\partial}$-RH problem
  5. 5 Decomposition of the mixed $\bar{\partial}$-RH problem
    1. 5.1 Analysis on the pure RH Problem
      1. 5.1.1 Outer model RH problem
      2. 5.1.2 A local solvable RH model near phase points
      3. 5.1.3 The small norm RH problem for error function
    2. 5.2 Analysis on the pure $\bar{\partial}$-Problem
  6. 6 Long time asymptotics for nonlocal mKdV equation

##  1 Introduction

Since the inverse scattering transform (IST) technique, one of the most
powerful tool to investigate solitons of nonlinear integrable models, was
firstly presented by Gardner, Greene, Kruskal and Miurra [1]. The dressing
method and Riemann-Hilbert (RH) approach are modern versions of IST [3, 2].
The RH approach allows getting the solution $q(x,t)$ of the original nonlinear
problem by solving a RH problem with specific jump matrix on a given curve.
Generally, the Cauchy problems of nonlinear integrable systems can be solved
by suing IST or RH method only in the case of refectioness potentials. So a
natural idea is to study the asymptotic behavior of solutions in general case.

The analysis of the long time behavior of the solution of the initial value
problem for a nonlinear integrable equation, the so-called nonlinear version
of the steepest-decent method has proved to be extremely efficient. Being
developed by many researchers, the method was first carried out with IST
method by Manakov in 1974 [4]. Being developed by many researchers, the method
has finally been put into a rigorous shape by Deift and Zhou in 1993 [5]. They
developed rigorous analytic method to present the long time asymptotic
representation of the solution for defocusing mKdV equation by deforming
contours to reduce the original RH problem to a model whose solution can be
derived in terms of parabolic cylinder functions. Then this method has been
widely applied to the focusing NLS equation, KdV equation, Camassa-Holm
equation, Degasperis-Procesi, Fokas-Lenells equation etc [6, 7, 8, 9, 10, 11].

Recently, McLaughlin and Miller have developed a nonlinear steepest descent-
method for the asymptotic analysis of RH problems based on the analysis of
$\bar{\partial}$-problems, rather than the asymptotic analysis of singular
integrals on contours [12, 13]. When it is applied to integrable systems, the
$\bar{\partial}$-steepest descent method also has displayed some advantages,
such as avoiding delicate estimates involving $L^{p}$ estimates of Cauchy
projection operators, and leading the non-analyticity in the RH problem
reductions to a $\bar{\partial}$–problem in some sectors of the complex plane.
Dieng and McLaughlin obtained sharp asymptotics of solutions of the defocusing
nonlinear NLS equation[14], based on $\bar{\partial}$-method and under
essentially minimal regularity assumptions on initial data. Cussagna and
Jenkins study the defocusing NLS equation with finite density initial data
[15]. Borghese, Jenkins and McLaughlin have studied the Cauchy problem for the
focusing NLS equation using the $\bar{\partial}$-generalization of the
nonlinear steepest descent method, which has been conjectured for a long time
[16]; Jenkins and Liu studied the derivative NLS equation for generic initial
data in a weighted Sobolev space[17].

In this paper, we investigate the long-time asymptotic behavior for the Cauchy
problem of an integrable real nonlocal mKdV equation under nonzero boundary
conditions

$\displaystyle q_{t}(x,t)-6\sigma q(x,t)q(-x,-t)q_{x}(x,t)+q_{xxx}(x,t)=0,$
(1.1) $\displaystyle q(x,0)=q_{0}(x),\lim_{x\to\pm\infty}q_{0}(x)=q_{\pm},$
(1.2)

where $|q_{\pm}|=1$ and $q_{+}=\delta q_{-}$, $\sigma\delta=-1$. (1.1) was
introduced in [18, 19], where $\sigma=\pm 1$ denote the defocusing and
focusing cases, respectively, $q(x,t)$ is a real function. (1.1) can be
regarded as the integrable nonlocal extension of the local mKdV equation

$q_{t}(x,t)-6\sigma q^{2}(x,t)q_{x}(x,t)+q_{xxx}(x,t)=0,$ (1.3)

which appears in various of physical fields [20]. From the perspective of
their structure, (1.3) can be translated to (1.1) by replacing $q^{2}(x,t)$
with the _PT_ -symmetric term $q(x,t)q(-x,-t)$ (In the real field, the
definition of _PT_ symmetry is given by _P_ : $x\rightarrow-x$ and _T_ :
$t\rightarrow-t$) [21]. The general form of the nonlocal mKdV equation was
shown to appear in the nonlinear oceanic and atmospheric dynamical system
[22]. Particularly, (i) for the _PT_ -symmetric case $q(-x,-t)=q(x,t)$, (1.1)
reduces to the usual mKdV equation (1.3); (ii) for the anti-_PT_ -symmetric
case $q(-x,-t)=-q(x,t)$, the focusing (defocusing) nonlocal mKdV equation
(1.1) reduces to the defocusing (focusing) mKdV equation; (iii) for
$q(-x,-t)=-q(-x,-t)$, the focusing (defocusing) nonlocal mKdV equation (1.1)
reduces to the defocusing (focusing) nonlocal mKdV equation [25].

The Darboux transformation was used to seek for soliton solutions of the
focusing (1.1) [23], and the IST for the focusing (1.1) with ZBC was presented
[24]. In [25], Zhang and Yan rigorously analyzed dynamical behaviors of
solitons and their interactions for four distinct cases of the reflectionless
potentials for both focusing and defocusing nonlocal mKdV equations with
NZBCs. Moreover, the long-time asymptotics for the nonlocal defocusing mKdV
equation with decaying initial data were investigated via Deift-Zhou steepest-
descent method [30]

In the present paper, we study the long-time behavior of the Cauchy problem
for nonlocal mKdV equation (1.1) with finite density type initial data in the
case of $\sigma\delta=-1$. The rest of the paper is organized as follows. In
Section 2, we get down to the spectral analysis on the Lax pair. Based on Jost
solutions and scattering data, the RH problem for $m(z)$ is established. In
Section 3, we give the distribution of saddle points for different $\xi$ and
signature table. In Section 4,we conduct several deformations to obtain a
mixed $\bar{\partial}$-RH problem for $m^{(2)}(z)$. In Section 5, we decompose
$m^{(2)}(z)$ into a pure RH problem for $m^{(2)}_{RHP}$ and a pure
$\bar{\partial}$-Problem for $m^{(3)}(z)$. Furthermore, The $m^{(2)}_{RHP}$
can be obtained via an modified reflectionless RH problem
$m^{\Lambda}(z|\sigma_{d}^{\Lambda})$ and an inner model $m^{(lo)}(z)$ for the
stationary phase point $\zeta_{k}$ which are approximated by parabolic
cylinder model. In Section 6, we derive the long-time behavior of nonlocal
mKdV equation (1.1) in the soliton region.

##  2 The spectral analysis and the RH problem

###  2.1 Notations

We recall some notations. $\sigma_{1},\sigma_{2},\sigma_{3}$ are classical
Pauli matrices as follows

$\sigma_{1}=\begin{bmatrix}0&1\\\
1&0\end{bmatrix},\quad\sigma_{2}=\begin{bmatrix}0&-i\\\
i&0\end{bmatrix},\quad\sigma_{3}=\begin{bmatrix}1&0\\\ 0&-1\end{bmatrix}.$

We introduce the Japanese bracket $\langle x\rangle:=\sqrt{1+|x|^{2}}$ and the
normed spaces:

  * •

A weighted $L^{p,s}(\mathbb{R})$ is defined by

$L^{p,s}(\mathbb{R})=\\{u\in L^{p}(\mathbb{R}):\langle x\rangle^{s}u(x)\in
L^{p}(\mathbb{R})\\}.$

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

  * •

A Sobolev space is defined by

$W^{m,p}(\mathbb{R})=\\{u\in L^{p}(\mathbb{R}):\partial^{j}u(x)\in
L^{p}(\mathbb{R})\quad{\rm for}\quad j=0,1,2,\dots,m\\}.$

And
$\|u\|_{W^{m,p}(\mathbb{R})}:=\sum_{j=0}^{m}\|\partial^{j}u\|_{L^{p}(\mathbb{R})}$.
Additionally, we are used to expressing
$H^{m}(\mathbb{R}):=W^{m,2}(\mathbb{R})$.

  * •

A weighted Sobolev space is defined by

$H^{m,s}(\mathbb{R}):=L^{2,s}(\mathbb{R})\cap H^{m}(\mathbb{R}).$

In this paper, we use $a\lesssim b$ to express $\exists c=c(\xi)>0$, s.t.
$a\leqslant cb$.

###  2.2 The Lax pair and spectral analysis

The nonlocal mKdV equation (1.1) posses the nonlocal Lax pair

$\Phi_{x}=X\Phi,\quad\Phi_{t}=T\Phi,$ (2.1)

where

$\begin{split}&X=ik\sigma_{3}+Q,\\\ &T=[4k^{2}+2\sigma
q(x,t)q(-x,-t)]X-2ik\sigma_{3}Q_{x}+[Q_{x},Q]-Q_{xx},\\\
&Q=\begin{bmatrix}0&q(x,t)\\\ \sigma q(-x,-t)&0\end{bmatrix},\sigma=\pm
1,\end{split}$ (2.2)

and $\Phi=\Phi(x,t;k)$ is a matrix eigenfunction, $k$ is an iso-spectral
parameter.

Taking $q(x,t)=q_{\pm}$ in the Lax pair (2.1), we get the spectral problems

$\phi_{x}=X_{\pm}\phi,\quad\phi_{t}=T_{\pm}\phi,$ (2.3)

with $Q_{\pm}=\lim_{x\to\pm\infty}Q(x,t)=\begin{bmatrix}0&q_{\pm}\\\
-q_{\pm}&0\end{bmatrix}$, we have the fundamental matrix solution of Eq. (2.3)
as

$\displaystyle\phi_{\pm}(x,t;k)=\left\\{\begin{aligned}
&E_{\pm}(k)e^{it\theta(x,t;k)\sigma_{3}},\quad k\neq\pm i,\\\
&I+[x+(4k^{2}-2)t]X_{\pm}(k),\quad k=\pm i,\end{aligned}\right.$ (2.4)

where

$E_{\pm}(k)=\begin{bmatrix}1&\frac{iq_{\pm}}{k+\lambda}\\\
\frac{iq_{\pm}}{k+\lambda}&1\end{bmatrix},\hskip
14.22636pt\lambda^{2}-k^{2}=1,\hskip
14.22636pt\theta(x,t;k)=\lambda[x+(4k^{2}-2)t].$ (2.5)

To avoid multi-valued case of eigenvalue $\lambda$, we introduce a
uniformization variable

$z=k+\lambda,$ (2.6)

and obtain two single-valued functions

$k=\frac{1}{2}(z-\frac{1}{z}),\hskip
14.22636pt\lambda=\frac{1}{2}(z+\frac{1}{z}).$ (2.7)

We can define two domains $D_{+}$, $D_{-}$ and their boundary $\Sigma$ on
$z$-plane by

$\displaystyle D_{+}=\\{z\in\mathbb{C}:(|z|-1)Imz>0\\},$ $\displaystyle
D_{-}=\\{z\in\mathbb{C}:(|z|-1)Imz<0\\},$
$\displaystyle\Sigma=\mathbb{R}\cup\\{z\in\mathbb{C}:|z|=1\\}.$

which are shown in the Figure 1.

RezImz$z_{k}$$-\overline{z}_{k}$$\mathrm{i}\omega_{k}$$\frac{\mathrm{i}}{\omega_{k}}$$\overline{z}_{k}^{-1}$$-z_{k}^{-1}$
Figure 1: The complex $z$-plane showing the discrete spectrums [zeros of
scattering data $s_{11}(z)$ (red) in yellow region and those of scattering
data $s_{22}(z)$ (blue) in white region]. The yellow and white regions stand
for $D_{+}$ and $D_{-}$, respectively.

We know that the Jost solutions $\Phi_{\pm}(x,t,z)$ satisfy

$\Phi_{\pm}(x,t,z)\sim E_{\pm}(z)e^{\mathrm{i}t\theta(x,t,z)\sigma_{3}}.$
(2.8)

For convenience, we introduce the modified Jost solutions $\mu_{\pm}(x,t,z)$
by eliminating the exponential oscillations

$\mu_{\pm}(x,t,z)=\Phi_{\pm}(x,t,z)e^{-\mathrm{i}t\theta(x,t,z)\sigma_{3}},$
(2.9)

such that

$\lim_{x\to\pm\infty}\mu_{\pm}(x,t,z)=E_{\pm}(z),$ (2.10)

and $\mu_{\pm}$ admit the Volterra type integral equations

$\displaystyle\mu_{\pm}(x,t;z)=E_{\pm}(z)+\left\\{\begin{aligned}
&\int_{\pm\infty}^{x}E_{\pm}(z)e^{\mathrm{i}\lambda(x-y)\sigma_{3}}[E_{\pm}^{-1}(z)\Delta
Q_{\pm}(y,t)\mu_{\pm}(y,t,z)]dy,\quad k\neq\pm i,\\\
&\int_{\pm\infty}^{x}[I+(x-y)X_{\pm}]\Delta
Q_{\pm}(y,t)\mu_{\pm}(y,t,z)dy,\quad k=\pm i,\end{aligned}\right.$ (2.11)

where $\Delta Q_{\pm}(x,t)=Q(x,t)-Q_{\pm}$. For convenience, we let
$\Sigma_{0}=\Sigma/{\pm\mathrm{i}}$. Similarly to [15], we have the following
propositions:

######  Proposition 2.1.

Given $n\in\mathbb{N}_{0}$, let $q\mp q_{\pm}\in$ $L^{1,n+1}(\mathbb{R})$,
$q^{\prime}\in$ $W^{1,1}(\mathbb{R})$.

  * •

$\mu_{+,1}$ and $\mu_{-,2}$ can be analytically extended to $D_{+}$ and
continuously extended to $D_{+}\cup\Sigma_{0}$, $\mu_{-,1}$ and $\mu_{+,2}$
can be analytically extended to $D_{-}$ and continuously extended to
$D_{-}\cup\Sigma_{0}$;

  * •

The map $q\rightarrow$ $\frac{\partial^{n}}{\partial z^{n}}\mu_{\pm,i}(z)$
$(i=1,2,n\geq 0)$ are Lipschitz continuous, specifically, for any
$x_{0}\in\mathbb{R}$, $\mu_{-,1}(z)$ and $\mu_{+,2}(z)$ are continuously
differentiable mappings:

$\displaystyle\partial_{z}^{n}\mu_{+,1}:\bar{D}_{+}\setminus\\{0,\pm\mathrm{i}\\}\rightarrow
L^{\infty}_{loc}\\{\bar{D}_{+}\setminus\\{0,\pm\mathrm{i}\\},C^{1}([x_{0},\infty),\mathbb{C}^{2})\cap
W^{1,\infty}([x_{0},\infty),\mathbb{C}^{2})\\},$ (2.12)
$\displaystyle\partial_{z}^{n}\mu_{-,2}:\bar{D}_{+}\setminus\\{0,\pm\mathrm{i}\\}\rightarrow
L^{\infty}_{loc}\\{\bar{D}_{+}\setminus\\{0,\pm\mathrm{i}\\},C^{1}((-\infty,x_{0}],\mathbb{C}^{2})\cap
W^{1,\infty}((-\infty,x_{0}],\mathbb{C}^{2})\\},$ (2.13)

$\mu_{+,1}(z)$ and $\mu_{-,2}(z)$ are continuously differentiable mappings:

$\displaystyle\partial_{z}^{n}\mu_{-,1}:\bar{D}_{-}\setminus\\{0,\pm\mathrm{i}\\}\rightarrow
L^{\infty}_{loc}\\{\bar{D}_{-}\setminus\\{0,\pm\mathrm{i}\\},C^{1}((-\infty,x_{0}],\mathbb{C}^{2})\cap
W^{1,\infty}((-\infty,x_{0}],\mathbb{C}^{2})\\},$ (2.14)
$\displaystyle\partial_{z}^{n}\mu_{+,2}:\bar{D}_{-}\setminus\\{0,\pm\mathrm{i}\\}\rightarrow
L^{\infty}_{loc}\\{\bar{D}_{-}\setminus\\{0,\pm\mathrm{i}\\},C^{1}([x_{0},\infty),\mathbb{C}^{2})\cap
W^{1,\infty}([x_{0},\infty),\mathbb{C}^{2})\\}.$ (2.15)

  * •

Let $K$ be a compact neighborhood of $\\{-\mathrm{i},\mathrm{i}\\}$ in
$\bar{D}_{+}\setminus\\{0\\}$. Set $x^{\pm}=\max\\{\pm x,0\\}$, then there
exists a constant $C$ such that for $z\in K$ we have

$|\mu_{+,1}(z)-(1,\mathrm{i}z^{-1})^{\rm T}|\leq C\langle x^{-}\rangle
e^{C\int_{x}^{\infty}\langle
y-x\rangle|q-1|dy}\|q-\tilde{q}\|_{L^{1,1}(x,\infty)},$ (2.16)

i.e., the map $z\rightarrow\mu_{+,1}(z)$ extends as a continuous map to the
points $\pm\mathrm{i}$ with values in $C^{1}([x_{0},\infty),\mathbb{C})\cap
W^{1,n}([x_{0},\infty),\mathbb{C})$ for any preassigned $x_{0}\in\mathbb{R}$.
Moreover, the map $q\rightarrow\mu_{1}^{+}(z)$ is locally Lipschitz continuous
from:

$L^{1,1}(\mathbb{R})\rightarrow
L^{\infty}(\bar{D}_{+}\setminus\\{0\\},C^{1}([x_{0},\infty),\mathbb{C})\cap
W^{1,\infty}([x_{0},\infty),\mathbb{C}).$ (2.17)

Analogous statements hold for $\mu_{+,2}$ and for $\mu_{-,j}$ $(j=1,2)$.
Furthermore, the maps $z\rightarrow\partial_{z}^{n}\mu_{+,1}(z)$ and
$q\rightarrow\partial_{z}^{n}\mu_{+,1}(z)$ also satisfy:

$|\partial_{z}^{n}\mu_{+,1}(z)|\leq
F_{n}\left[(1+|x|)^{n+1}\|q-1\|_{L^{1,n+1}(x,\infty)}\right],\hskip
14.22636ptz\in K.$ (2.18)

The asymptotic behavior of $\mu_{\pm,j}$, $j=1,2$ could be described by
following proposition.

######  Proposition 2.2.

Suppose that $q\mp q_{\pm}\in$ $L^{1,n+1}(\mathbb{R})$ and $q^{\prime}\in$
$W^{1,1}(\mathbb{R})$. Then as $z\rightarrow\infty$, we have

$\displaystyle\mu_{\pm,1}(z)=e_{1}-\frac{\mathrm{i}}{z}\left(\begin{array}[]{c}\int^{x}_{\pm\infty}[\sigma
q(y,t)q(-y,-t)+1]dy\\\ \sigma q(-x,-t)\end{array}\right)+\mathcal{O}(z^{-2}),$
(2.21)
$\displaystyle\mu_{\pm,2}(z)=e_{2}+\frac{\mathrm{i}}{z}\left(\begin{array}[]{c}q(x,t)\\\
\int^{x}_{\pm\infty}[\sigma
q(y,t)q(-y,-t)+1]dy\end{array}\right)+\mathcal{O}(z^{-2}),$ (2.24)

and as $z\rightarrow 0$, we have

$\displaystyle\mu_{\pm,1}(z)=\mathrm{i}\frac{q_{\pm}}{z}e_{2}+\mathcal{O}(1),$
(2.25)
$\displaystyle\mu_{\pm,2}(z)=\mathrm{i}\frac{q_{\pm}}{z}e_{1}+\mathcal{O}(1),$
(2.26)

where $e_{1}=(1,0)^{\rm T}$, $e_{2}=(0,1)^{\rm T}$.

It follows that $\Phi_{\pm}(x,t;z)$ are fundamental solutions of Lax pair
(2.1) as $z\in\Sigma_{0}$, thus there exists a constant scattering matrix
$S(z)=(s_{ij}(z))_{2\times 2}$ (independence of $x$ and $t$) such that

$\Phi{+}(x,t,z)=\Phi_{-}(x,t,z)S(z),\quad z\in\Sigma_{0},$ (2.27)

where $s_{ij}(z)$ can be expressed as

$\begin{split}&s_{11}(z)=\frac{{\rm
det}(\Phi_{+,1},\Phi_{-,2})}{1+z^{-2}},\quad s_{12}(z)=\frac{{\rm
det}(\Phi_{+,2},\Phi_{-,2})}{1+z^{-2}},\\\ &s_{21}(z)=\frac{{\rm
det}(\Phi_{-,1},\Phi_{+,1})}{1+z^{-2}},\quad s_{22}(z)=\frac{{\rm
det}(\Phi_{-,1},\Phi_{+,2})}{1+z^{-2}}.\end{split}$ (2.28)

By calculating, we get the symmetry relations of the Jost solutions and
scattering matrix for variables and isospectral parameter.

######  Proposition 2.3.

The symmetries for Jost solutions $\Phi_{\pm}(x,t,z)$ and scattering matrix
$S(z)$ in $z\in\Sigma$ are listed as follows:

  * •

The first symmtry

$\Phi_{\pm}(x,t,z)=\sigma_{4}\overline{\Phi_{\mp}(-x,-t,-\bar{z})}\sigma_{4},\quad
S(z)=\sigma_{4}[\overline{S(-\bar{z})}]^{-1}\sigma_{4},$ (2.29)

where $\sigma_{4}$ is defined as

$\displaystyle\sigma_{4}=\left\\{\begin{aligned}
&\sigma_{1},\quad\sigma=-1,\\\ &\sigma_{2},\quad\sigma=1.\end{aligned}\right.$
(2.30)

  * •

The second symmetry

$\Phi_{\pm}(x,t,z)=\overline{\Phi_{\pm}(-x,-t,-\bar{z})},\quad
S(z)=\overline{S(-\bar{z})}.$ (2.31)

  * •

The third symmetry

$\Phi_{\pm}(x,t,z)=\frac{\mathrm{i}}{z}\Phi_{\pm}(x,t,-z^{-1})\sigma_{3}Q_{\pm},\quad
S(z)=(\sigma_{3}Q_{-})^{-1}S(-z^{-1})(\sigma_{3}Q_{+}).$ (2.32)

We use the scattering coefficients $s_{ij}(z),i,j=1,2$ to define reflection
coefficients

$\rho(z)=\frac{s_{21}(z)}{s_{11}(z)},\quad\tilde{\rho}(z)=\frac{s_{12}(z)}{s_{22}(z)},\quad
z\in\Sigma.$ (2.33)

The following proposition provides some essential properties for
$s_{ij}(z),i,j=1,2$ and $\rho(z),\tilde{\rho}(z)$.

######  Proposition 2.4.

Let $q\mp q_{\pm}\in$ $L^{1,n+1}(\mathbb{R})$ and $q^{\prime}\in$
$W^{1,1}(\mathbb{R})$, then

  * 1)

For $z\in\left\\{z\in\mathbb{C}:|z|=1\right\\}/\\{\pm\mathrm{i}\\}$,

$|\rho(z)\tilde{\rho}(z)|\leq 1-|s_{11}(z)|^{-2}<1.$ (2.34)

  * 2)

$s_{ij}(z),i,j=1,2$ and the reflection coefficient $\rho(z),\tilde{\rho}(z)$
satisfy the symmetries

$s_{11}(z)=-\sigma s_{22}(-z^{-1}),\quad s_{12}(z)=-\sigma
s_{21}(-z^{-1}),\quad\tilde{\rho}(z)=\rho(-z^{-1}).$ (2.35)

  * 3)

The scattering data have the asymptotics

$\displaystyle\lim_{z\rightarrow\infty}(s_{11}(z)-1)z=i\int_{\mathbb{R}}\left[\sigma
q(y,t)q(-y,-t)-1\right]dy,$ (2.36) $\displaystyle\lim_{z\rightarrow
0}s_{11}(z)=-\sigma,$ (2.37)
$\displaystyle|s_{21}(z)|=\mathcal{O}(|z|^{-2}),\hskip 14.22636pt\text{as
}|z|\rightarrow\infty,$ (2.38)
$\displaystyle|s_{21}(z)|=\mathcal{O}(|z|^{2}),\hskip 14.22636pt\text{as
}|z|\rightarrow 0,$ (2.39)

So that

$\displaystyle\rho(z),\tilde{\rho}(z)\sim z^{-2},\hskip
5.69046pt|z|\rightarrow\infty;\hskip 14.22636pt\rho(z),\tilde{\rho}(z)\sim
0,\hskip 5.69046pt|z|\rightarrow 0.$ (2.40)

  * 4)

Although $s_{11}(z)$ and $s_{21}(z)$ have singularities at points
$\pm\mathrm{i}$, we can claim that the reflection coefficient
$\rho(z),\tilde{\rho}(z)$ remain bounded at $z=\pm\mathrm{i}$ and
$|\rho(\pm\mathrm{i})|=1,|\tilde{\rho}(\pm\mathrm{i})|=1$. In fact, by direct
calculation, we obtain

$s_{11}(z)=\frac{\mp s^{\pm}}{z\mp\mathrm{i}}+\mathcal{O}(1),\quad
s_{21}(z)=\frac{-\sigma s^{\pm}}{z\mp\mathrm{i}}+\mathcal{O}(1),$ (2.41)
$\lim_{z\to\pm\mathrm{i}}\rho(z)=\lim_{z\to\pm\mathrm{i}}\tilde{\rho}(z)=\pm\sigma,$
(2.42)

where $s^{\pm}=\frac{1}{2\mathrm{i}}{\rm
det}(\Phi_{+,1}(\pm\mathrm{i}),\Phi_{-2}(\pm\mathrm{i}))$.

Since one cannot exclude the possibilities of zeros for $s_{11}(z)$ and
$s_{22}(z)$ along $\Sigma$. To solve the Riemann-Hilbert problem in the
inverse process, we only consider the potentials without spectral
singularities, i.e., $s_{11}(z)\neq 0$, $s_{22}(z)\neq 0$ for $z\in\Sigma$.

The next proposition shows that, given data $q_{0}(x)$ with sufficient
smoothness and decay properties, the reflection coefficients will also be
smooth and decaying.

######  Proposition 2.5.

For given $q\mp q_{\pm}\in L^{1,2}(\mathbb{R})$ and $q^{\prime}\in$
$W^{1,1}(\mathbb{R})$, we then have $\rho(z),\tilde{\rho}(z)\in
H^{1}(\Gamma)$, where $\Gamma$ is defined in (4.2).

###### Proof.

Proposition 2.1 and (2.28) indicate that $s_{11}(z)$ and $s_{21}(z)$ are
continuous to $\Sigma/\\{0,\pm\mathrm{i}\\}$. Noticing $s_{11}(z)\neq 0$,
$s_{22}(z)\neq 0$ for $z\in\Sigma$, then $\rho(z)$ is continuous to
$\Sigma/\\{0,\pm\mathrm{i}\\}$. From (LABEL:rhob1) and (2.42), we know that
$\rho(z)$ is bounded in the small neighborhood of $\\{0,\pm\mathrm{i}\\}$ and
$\rho(z)\in L^{1}(\Gamma)\cap L^{2}(\Gamma)$. Next, we just need to prove that
$\rho^{\prime}(z)\in L^{2}(\Gamma)$. For $\delta_{0}>0$ sufficiently small,
from Proposition 2.1, the maps

$q\rightarrow{\rm det}(\Phi_{+,1},\Phi_{-,2}),\quad q\rightarrow{\rm
det}(\Phi_{-,1},\Phi_{+,1})$ (2.43)

are locally Lipschitz maps from

$\\{q:q\in L^{1,n+1}(\mathbb{R}),q^{\prime}\in
W^{1,1}(\mathbb{R})\\}\rightarrow
W^{n,\infty}(\mathbb{R}/(-\delta_{0},\delta_{0})),n\geq 0.$ (2.44)

Actually, $q\rightarrow\Phi_{+,1}(z,0)$ is, by Proposition 2.1, a locally
Lipschitz map with values in
$W^{n,\infty}(\bar{D}_{+}/(0,\delta_{0},\mathbb{C}^{2}))$. For
$q\rightarrow\Phi_{-,2}(z,0)$ and $q\rightarrow\Phi_{-,1}(z,0)$ the same is
true. These and (2.36) imply that $q\rightarrow\rho(z)$ is a locally Lipschitz
map from the domain in (2.44) into

$W^{n,\infty}(I_{\delta_{0}})\cap H^{n}(I_{\delta_{0}}),$ (2.45)

where $I_{\delta_{0}}=\Gamma/(-\delta_{0},\delta_{0})\cup
dist(\pm\mathrm{i};\delta_{0})$. Now fix $\delta_{0}$ sufficiently small such
that the $3$ intervals $dist(z;\pm\mathrm{i})\leq\delta_{0}$ and
$|z-0|\leq\delta_{0}$ have no intersection. In the complement of their union

$|\partial^{j}_{z}\rho(z)|\leq C_{\delta_{0}}\langle z\rangle^{-1},\quad
j=0,1.$ (2.46)

Let $|z-\mathrm{i}|<\delta_{0}$, then using the definition of $s_{+}$ we have

$\rho(z)=\frac{s_{21}(z)}{s_{11}(z)}=\frac{{\rm
det}[\Phi_{-,1},\Phi_{+,1}]}{{\rm
det}[\Phi_{+,1},\Phi_{-,2}]}=\frac{\int_{\mathrm{i}}^{z}F(s)ds-\sigma
s^{+}}{\int_{\mathrm{i}}^{z}G(s)ds-s^{+}},$ (2.47)

where

$F(s)=\partial_{s}{\rm det}[\Phi_{-,1}(s)\Phi_{+,1}(s)],\quad
G(s)=\partial_{s}{\rm det}[\Phi_{+,1},\Phi_{-,2}].$ (2.48)

If $s_{+}\neq 0$ then it is clear from the above formula that
$\rho^{\prime}(z)$ exist and is bounded near $\mathrm{i}$.

If $s_{+}=0$, then $z=\mathrm{i}$ is not the pole of $s_{11}(z)$ and
$s_{21}(z)$, so that $s_{11}(z)$ and $s_{21}(z)$ are continuous at
$z=\mathrm{i}$, then

$\rho(z)=\frac{\int_{\mathrm{i}}^{z}F(s)ds}{\int_{\mathrm{i}}^{z}G(s)ds}.$
(2.49)

From (2.28), we have

$(1+z^{2})s_{11}(z)=z^{2}{\rm det}[\Phi_{+,1},\Phi_{-,2}].$ (2.50)

Since $s_{+}=0$ implies that ${\rm
det}[\Phi_{+,1}(\mathrm{i}),\Phi_{-,2}(\mathrm{i})]=0$, differentiating (2.50)
at $z=\mathrm{i}$ we get

$2\mathrm{i}s_{11}(\mathrm{i})=-\partial_{z}{\rm
det}[\Phi_{+,1}(z),\Phi_{-,2}(z)]|_{z=\mathrm{i}}=-\mathrm{i}G(\mathrm{i}).$
(2.51)

With $|s_{11}(\mathrm{i})|^{2}=1+|s_{21}(\mathrm{i})|^{2}>1$, we have
$G(\mathrm{i})\neq 0$. It follows that the derivative $\rho^{\prime}(z)$ is
bounded around $\mathrm{i}$. The same proof holds at $z=-\mathrm{i}$ and
$z=0$. It follows that $\rho^{\prime}(z)\in L^{2}(\Gamma)$.

According to the symmetry $\tilde{\rho}(z)=\rho(-z^{-1})$, we have
$\tilde{\rho}(z)\in H^{1}(\Gamma)$. ∎

######  Corollary 2.1.

For given $q\mp q_{\pm}\in L^{1,2}(\mathbb{R})$, $q^{\prime}\in
W^{1,1}(\mathbb{R})$, we then have $r(z)\in H^{1,1}(\Gamma)$.

###### Proof.

Since $H^{1,1}(\mathbb{R})=L^{2,1}(\mathbb{R})\cap H^{1}(\mathbb{R})$, what we
need to prove is $r\in L^{2,1}(\mathbb{R})$. With (2.40), we can see that

$|z|^{2}r^{2}(z)\sim|z|^{-2},\quad|z|\rightarrow\infty$ (2.52)

Thus

$\int_{\Gamma}|\langle z\rangle r(z)|^{2}<\infty,$ (2.53)

which help us obtain the result. ∎

Suppose that $s_{11}(z)$ has $N_{1}$ and $N_{2}$ simple zeros, respectively,
in $D_{+}\cap\\{z\in\mathbb{C}:Rez>0\\}$ denoted by $z_{k}$,
$k=1,2,\cdot\cdot\cdot,N_{1}$ and in $D_{+}\cap\\{z\in\mathbb{C}:Rez=0\\}$
denoted by $\mathrm{i}\omega_{k}$, $k=1,2,\cdot\cdot\cdot,N_{2}$. It follows
from the symmetry relations of the scattering coefficients that

$\begin{split}&s_{11}(z_{k})=s_{11}(-\bar{z}_{k})=s_{22}(-z_{k}^{-1})=s_{22}(\bar{z}_{k}^{-1}),\quad
k=1,2,\cdot\cdot\cdot,N_{1},\\\
&s_{11}({\mathrm{i}\omega_{k}})=s_{22}(\mathrm{i}\omega_{k}^{-1}),\quad
k=1,2,\cdot\cdot\cdot,N_{2}.\end{split}$ (2.54)

It is convenient to define that

$\displaystyle\eta_{k}=\left\\{\begin{aligned} &z_{k},\quad
k=1,2,\cdot\cdot\cdot,N_{1},\\\ &-\bar{z}_{k-N_{1}},\quad
k=N_{1}+1,N_{1}+2,\cdot\cdot\cdot,2N_{1}.\\\
&\mathrm{i}\omega_{k-2N_{1}},\quad
k=2N_{1}+1,2N_{1}+2,\cdot\cdot\cdot,2N_{1}+N_{2},\end{aligned}\right.$ (2.55)

and

$\hat{\eta}_{k}=-\eta_{k}^{-1},\quad k=1,2,\cdot\cdot\cdot,2N_{1}+N_{2}.$
(2.56)

Thus, the discrete spectrum is given by

$\mathcal{Z}\cup\hat{\mathcal{Z}}=\\{\eta_{k},\hat{\eta}_{k}\\}_{k=1}^{2N_{1}+N_{2}},$
(2.57)

and the distribution of $\mathcal{Z}\cup\hat{\mathcal{Z}}$ on the $z$-plane is
shown in Figure 1.

Given $z_{0}\in\mathcal{Z}$, it follows from the Wronskian representations and
$s_{11}(z_{0})=0$ that $\Phi_{+,1}(x,t,z_{0})$ and $\Phi_{-,2}(x,t,z_{0})$ are
linearly dependent; Given $z_{0}\in\hat{\mathcal{Z}}$, it follows from the
Wronskian representations $s_{22}(z_{0})=0$ and that $\Phi_{+,2}(x,t,z_{0})$
and $\Phi_{-,1}(x,t,z_{0})$ are linearly dependent. For convenience, we denote
the proportional coefficient in the following definition of $b[z_{0}]$ by
$\frac{\Phi_{+,1}(x,t,z_{0})}{\Phi_{-,2}(x,t,z_{0})}$ or
$\frac{\Phi_{+,2}(x,t,z_{0})}{\Phi_{-,1}(x,t,z_{0})}$ according to the region
$z_{0}$ belongs to. Let

$\displaystyle b[z_{0}]=\left\\{\begin{aligned}
&\frac{\Phi_{+,1}(x,t,z_{0})}{\Phi_{-,2}(x,t,z_{0})},\quad
z_{0}\in\mathcal{Z},\\\
&\frac{\Phi_{+,2}(x,t,z_{0})}{\Phi_{-,1}(x,t,z_{0})},\quad
z_{0}\in\hat{\mathcal{Z}},\end{aligned}\right.\quad\left\\{\begin{aligned}
&A[z_{0}]=\frac{b[z_{0}]}{s^{{}^{\prime}}_{11}(z_{0})},\quad
z_{0}\in\mathcal{Z},\\\
&A[z_{0}]=\frac{b[z_{0}]}{s^{{}^{\prime}}_{22}(z_{0})},\quad
z_{0}\in\hat{\mathcal{Z}}.\end{aligned}\right.$ (2.58)

######  Proposition 2.6.

For the given $z_{0}\in\mathcal{Z}\cup\hat{\mathcal{Z}}$, there exist three
relations for $b[z_{0}]$, $s^{\prime}_{11}(z_{0})$ and
$s^{\prime}_{22}(z_{0})$:

  * •

The first relation

$b[z_{0}]=-\frac{\sigma}{\overline{b[-\bar{z}_{0}]}},\quad
s^{\prime}_{11}(z_{0})=-\overline{s^{\prime}_{11}(-\bar{z}_{0})},\quad
s^{\prime}_{22}(z_{0})=-\overline{s^{\prime}_{22}(-\bar{z}_{0})}.$ (2.59)

  * •

The second relation

$b[z_{0}]=-\overline{b[-\bar{z}_{0}]},\quad
s^{\prime}_{11}(z_{0})=-\overline{s^{\prime}_{11}(-\bar{z}_{0})},\quad
s^{\prime}_{22}(z_{0})=-\overline{s^{\prime}_{22}(-\bar{z}_{0})}.$ (2.60)

  * •

The third relation

$b[z_{0}]=-\sigma b[-z_{0}^{-1}],\quad s^{\prime}_{11}(z_{0})=-\sigma
z_{0}^{-2}s^{\prime}_{22}(-z_{0}^{-1}).$ (2.61)

From the first relation, one concludes that imaginary discrete spectrum
$\mathrm{i}\omega_{k}$ exists if and only if $\sigma=-1$. That is to say that
as $\sigma=1$, one has $N_{2}=0$. We give the following relation among
discrete spectrum $\mathcal{Z}\cup\hat{\mathcal{Z}}$.

######  Proposition 2.7.

The relations for $b[\cdot]$ and $A[\cdot]$ in
$\mathcal{Z}\cup\hat{\mathcal{Z}}$ are given by

$\begin{split}&b[\eta_{k}]=\overline{b[-\bar{\eta}_{k}]}=-\sigma
b[\hat{\eta}_{k}]=-\sigma b[\bar{\eta}_{k}^{-1}],\quad b^{2}[\eta_{k}]=1\\\
&s^{\prime}_{11}(\eta_{k})=-\overline{s^{\prime}_{11}(-\bar{\eta}_{k})}=-\sigma\hat{\eta}_{k}^{2}s^{\prime}_{22}(\hat{\eta}_{k})=\sigma\hat{\eta}_{k}^{2}s^{\prime}_{22}(\bar{\eta}_{k}^{-1}).\end{split}$
(2.62)

As $\sigma=-1$, one has

$b[\mathrm{i}\omega_{k}]=b[\mathrm{i}\omega_{k}^{-1}],\quad
s^{\prime}_{11}(\mathrm{i}\omega_{k})=-\omega_{k}^{2}s^{\prime}_{22}(\mathrm{i}\omega_{k}^{-1}).$
(2.63)

Then,

$A[\hat{\eta}_{k}]=\hat{\eta}_{k}^{2}A[\eta_{k}].$ (2.64)

###  2.3 A RH problem

Define a sectionally meromorphic matrix as follows

$m(z)=m(x,t;z):=\left\\{\begin{aligned}
\left(\frac{\mu_{+,1}(x,t;z)}{s_{11}(z)},\mu_{-,2}(x,t;z)\right),\quad z\in
D_{+},\\\
\left(\mu_{-,1}(x,t;z),\frac{\mu_{+,2}(x,t;z)}{s_{22}(z)}\right),\quad z\in
D_{-},\end{aligned}\right.$ (2.65)

and

$m_{\pm}(x,t,z)=\lim_{z^{\prime}\to z\atop z\in D_{\pm}}m(x,t,z^{\prime})\quad
z\in\Sigma,$ (2.66)

the multiplicative matrix Riemann-Hilbert problem can be proposed as follows.

######  RHP 2.1.

Find a $2\times 2$ matrix-valued function $m(x,t;z)$ such that

  * *

Analyticity: $m(z)$ is analytical in
$\mathbb{C}\backslash(\Sigma\cup\mathcal{Z}\cup\hat{\mathcal{Z}})$ and has
simple poles in
$\mathcal{Z}\cup\hat{\mathcal{Z}}=\\{\eta_{k},\bar{\eta}_{k}\\}_{k=1}^{2N_{1}+N_{2}}$.

  * *

Jump relation: $m_{+}(x,t,z)=m_{-}(x,t,z)v(z)$, where

$v(z)=\begin{bmatrix}1-\rho(z)\tilde{\rho}(z)&-\tilde{\rho}(z)e^{2\mathrm{i}t\theta(x,t,z)}\\\
\rho(z)e^{-2\mathrm{i}t\theta(x,t,z)}&1\end{bmatrix},\quad z\in\Sigma.$ (2.67)

  * *

Asymptotic behavior:

$\displaystyle m(x,t;z)=I+\mathcal{O}(z^{-1}),\quad z\rightarrow\infty,$
(2.68) $\displaystyle
m(x,t;z)=\frac{\mathrm{i}}{z}\sigma_{3}Q_{-}+\mathcal{O}(1),\quad z\rightarrow
0.$ (2.69)

  * *

Residue conditions

$\displaystyle\underset{z=\eta_{k}}{\rm
Res}m(z)=\lim_{z\rightarrow\eta_{k}}m(z)\begin{bmatrix}0&0\\\
A[\eta_{k}]e^{-2\mathrm{i}t\theta(x,t,\eta_{k})}&0\end{bmatrix},$ (2.70)
$\displaystyle\underset{z=\hat{\eta}_{k}}{\rm
Res}m(z)=\lim_{z\rightarrow\hat{\eta}_{k}}m(z)\begin{bmatrix}0&A[\hat{\eta}_{k}]e^{2\mathrm{i}t\theta(,x,t,\hat{\eta}_{k})}\\\
0&0\end{bmatrix},$ (2.71)

where $\theta(x,t,z)=\lambda(z)[\frac{x}{t}+(4k^{2}(z)-2)]$.

The potential $q(x,t)$ is found by the reconstruction formula

$q(x,t)=-i(m_{1})_{12}=-i\lim_{z\rightarrow\infty}(zm)_{12},$ (2.72)

where $m_{1}$ appears in the expansion of $m=I+z^{-1}m_{1}+O(z^{-2})$ as
$z\rightarrow\infty$.

##  3 Distribution of Saddle Points and Signature Table

We notice that the long-time asymptotic behavior of RHP 2.1 is influenced by
the growth and decay of the exponential function

$e^{\pm
2it\theta},\quad\theta(z)=\frac{1}{2}\left(z+\frac{1}{z}\right)\left[\frac{x}{t}-2+\left(z-\frac{1}{z}\right)^{2}\right],$
(3.1)

which not only appear in jump matrix $v(z)$ but also in the residue condition.
Based on this observation, we shall make analysis for the real part of $\pm
2it\theta$ to ensure the exponential decaying property. Let $\xi=\frac{x}{t}$,
we consider the stationary phase points and the real part of
$2\mathrm{i}t\theta$:

$\theta^{\prime}(z)=-\frac{(1-z^{2})(3z^{4}+\xi z^{2}+3)}{2z^{4}},$ (3.2)
${\rm Re}[2\mathrm{i}t\theta]=-2t{\rm Im}\theta=-t{\rm
Im}z(1-|z|^{-2})[\xi-3+(1+|z|^{-2}+|z|^{-4})(3{\rm Re}^{2}z-{\rm Im}^{2}z)].$
(3.3)

From Eq. (3.2), we find six stationary phase points of $\theta(z)$:

$\pm 1,\quad\pm\frac{\sqrt{-\xi\pm\sqrt{-36+\xi^{2}}}}{\sqrt{6}}.$ (3.4)

Except to $z=\pm 1$, there are also four stationary phase points, whose
distribution depends on different $\xi$ is as follows:

  * i.

For $\xi<-6$, the four phase points are located on real axis $\mathbb{R}$
corresponding to Figure 2. Among them, two are inside the unit circle and the
other two are outside the unit circle;

  * ii.

For $-6<\xi<6$, the four phase points are all located on the unit circle and
they are symmetrical to each other, which is corresponded to Figure 2. We will
mainly discuss this case in the present paper;

  * iii.

For $\xi>6$, the four phase points are located on $\mathbb{R}$ the imaginary
axis $\mathrm{i}\mathbb{R}$ corresponding to Figure 2. The two of them are
inside the unit circle and the other two are outside the unit circle;.

Moreover, the decaying regions of ${\rm Re}(2\mathrm{i}t\theta)$ are shown in
Figure 3.

Figure 2: Plots of the distributions for saddle points: (a) $\xi<-6$, (b)
$-6<\xi<6$, (c) $\xi>6$. The black dotted curve is an unit circle. The red
curve shows the ${\rm Re}\theta^{\prime}(z)=0$, and the green curve shows the
${\rm Im}\theta^{\prime}(z)=0$. The intersection points are the saddle points
which express $\theta^{\prime}(z)=0$.

Figure 3: Signature table of ${\rm Re}(2it\theta)$ with different $\xi$: (a)
$\xi<-6$, (b) $-6<\xi<6$, (c) $\xi>6$. ${\rm Re}(2it\theta)<0$ in the green
region and ${\rm Re}(2it\theta)>0$ in the white region. In other words,
$|e^{2it\theta}|\rightarrow 0$ as $t\rightarrow\infty$ in the green region and
$|e^{-2it\theta}|\rightarrow 0$ as $t\rightarrow\infty$ in the white region.
Moreover, ${\rm Re}(2it\theta)=0$ on the green curve.

######  Remark 3.1.

According to (3.1), $\theta(z)$ allows the following symmetry:

$\theta(-z^{-1})=-\theta(z),\quad\theta(-\bar{z})=-\overline{\theta(z)},\quad\theta(-\bar{z}^{-1})=-\overline{\theta(z)}.$
(3.5)

##  4 Deformation of the RH Problem

###  4.1 Jump matrix factorizations

Now we use factorizations of the jump matrix along $\Sigma$ to deform the
contours onto those on which the oscillatory jump is traded for exponential
decay. Record the six stationary phase points as
$\zeta_{j},j=1,2,\cdot\cdot\cdot,6$, and $\alpha_{1,2}=\pm\mathrm{i}$, see in
Figure 4. Then by the well known factorizations of $v(z)$:

$\displaystyle v(x,t;z)=\left\\{\begin{aligned}
&\begin{bmatrix}1&-\tilde{\rho}e^{2\mathrm{i}t\theta}\\\
0&1\end{bmatrix}\begin{bmatrix}1&0\\\ \rho
e^{-2\mathrm{i}t\theta}&0\end{bmatrix},\quad z\in\tilde{\Gamma}\\\
&\begin{bmatrix}1&0\\\
\frac{\rho}{1-\rho\tilde{\rho}}e^{-2\mathrm{i}t\theta}&1\end{bmatrix}\begin{bmatrix}1-\rho\tilde{\rho}&0\\\
0&\frac{1}{1-\rho\tilde{\rho}}\end{bmatrix}\begin{bmatrix}1&-\frac{\tilde{\rho}}{1-\rho\tilde{\rho}}e^{2\mathrm{i}t\theta}\\\
0&1\end{bmatrix},\quad z\in\Gamma,\end{aligned}\right.$ (4.1)

where

$\tilde{\Gamma}=\widehat{\zeta_{2}\zeta_{3}}\cup\widehat{\zeta_{5}\zeta_{6}},\quad\Gamma=\Sigma/\tilde{\Gamma}.$
(4.2)

We will utilize these factorizations to deform the jump contours so that the
oscillating factor $e^{\pm 2\mathrm{i}\theta}$ are decaying in corresponding
region respectively. For this purpose, we introduce the following scalar RH
problem

######  RHP 4.1.

Find a scalar function $\delta(z)\triangleq\delta(z;\xi)$, which is defined by
the following properties:

  * *

Analyticity: $\delta(z)$ is analytical in $\mathbb{C}\backslash\Gamma$.

  * *

Jump relation:

$\begin{split}&\delta{+}(z)=\delta{-}(z)(1-\rho(z)\tilde{\rho}(z)),\quad
z\in{\Gamma};\\\ &\delta{+}(z)=\delta{-}(z),\quad
z\in\tilde{\Gamma}.\end{split}$ (4.3)

  * *

Asymptotic behavior:

$\delta(z)\rightarrow 1,z\rightarrow\infty$ (4.4)

Utilizing the Plemelj’s formula, we are arriving

$\delta(z)={\rm exp}\left[-\frac{1}{2\pi i}\int_{\Gamma}{\rm
log}\left(1-\rho(s)\tilde{\rho}(s)\right)\frac{1}{s-z}ds\right].$ (4.5)

Taking $\nu(z)=-\frac{1}{2\pi}{\rm log}(1-\rho(z)\tilde{\rho}(z))$, then we
can express

$\delta(z)={\rm exp}\left(i\int_{\Gamma}\frac{\nu(s)}{s-z}ds\right).$ (4.6)

######  Remark 4.1.

From the symmetries of $S(z)$ in Proposition 2.3, we can derive the symmetry
relation between $\rho(z)$ and $\tilde{\rho}(z)$, i.e.
$\tilde{\rho}(z)=\overline{\rho(z)}$ when $|z|=1$, then for
$z\in\\{z\in\mathbb{C}:|z|=1\\}\cap\Gamma$:

$\nu(z)=-\frac{1}{2\pi}{\rm log}(1-|\rho(z)|^{2})$ (4.7)

is a positive real function.

For brevity, we denote $\mathcal{N}=\\{1,2,\cdot\cdot\cdot,2N_{1}+N_{2}\\}$.
Moreover, we introduce a small positive constant $\varrho$:

$\varrho=\frac{1}{2}{\rm min}\left\\{\underset{k\in\mathcal{N}}{\rm
min}\left\\{|{\rm Im}(\eta_{k})|,|{\rm
Im}(\hat{\eta}_{k})|\right\\},{\underset{\lambda,\mu\in\mathcal{Z}\cup\hat{\mathcal{Z}},\lambda\neq\mu}{\rm
min}|\lambda-\mu|},{\underset{\lambda\in\mathcal{Z}\cup\hat{\mathcal{Z}},i=1,2,3,4}{\rm
min}|\lambda-\zeta_{i}|}\right\\}.$ (4.8)

Taking $\delta_{0}<\varrho$, we define $\triangle,\nabla$ and $\Lambda$ of
$\mathcal{N}$ as follows:

$\begin{split}&\triangle=\\{k\in\mathcal{N}:{\rm
Re}(2\mathrm{i}\theta(\eta_{k}))<0\\},\quad\nabla=\\{k\in\mathcal{N}:Re(2\mathrm{i}\theta(\eta_{k}))>0\\},\\\
&\Lambda=\\{k\in\mathcal{N}:|{\rm
Re}(2\mathrm{i}\theta(\eta_{k}))|<\delta_{0}\\}.\end{split}$ (4.9)

To distinguish different type of zeros, we further give

$\begin{split}&\triangle_{1}=\\{k\in\\{1,2,\cdot\cdot\cdot,N_{1}\\}:{\rm
Re}(2\mathrm{i}\theta(z_{k}))<0\\},\quad\nabla_{1}=\\{k\in\\{1,2,\cdot\cdot\cdot,N_{1}\\}:{\rm
Re}(2\mathrm{i}\theta(z_{k}))>0\\},\\\
&\triangle_{2}=\\{k\in\\{1,2,\cdot\cdot\cdot,N_{1}\\}:{\rm
Re}(2\mathrm{i}\theta(\mathrm{i}\omega_{k}))<0\\},\quad\nabla_{2}=\\{k\in\\{1,2,\cdot\cdot\cdot,N_{1}\\}:{\rm
Re}(2\mathrm{i}\theta(\mathrm{i}\omega_{k})))>0\\},\\\
&\Lambda_{1}=\\{k\in\\{1,2,\cdot\cdot\cdot,N_{1}\\}:|{\rm
Re}(2\mathrm{i}\theta(z_{k}))|<\delta_{0}\\},\quad\Lambda_{2}=\\{k\in\\{1,2,\cdot\cdot\cdot,N_{1}\\}:|{\rm
Re}(2\mathrm{i}\theta(\mathrm{i}\omega_{k}))|<\delta_{0}\\}.\end{split}$
(4.10)

Define the function

$T(z):=T(z;\xi)=\prod_{k\in\triangle_{1}}\prod_{l\in\triangle_{2}}\frac{(z+z_{k}^{-1})(z-\bar{z}_{k}^{-1})(z-\mathrm{i}\omega_{l}^{-1})}{(zz_{k}^{-1}-1)(z\bar{z}_{k}^{-1}+1)(\mathrm{i}\omega_{l}^{-1}z+1)}{\rm
exp}[\mathrm{i}\int_{\Gamma}\nu(s)(\frac{1}{s-z}-\frac{1}{2s})ds].$ (4.11)

######  Proposition 4.1.

The function defined by Eq. (4.11) has following properties:

  * (a)

$T(z)$ is meromorphic in $\mathbb{C}\backslash\Gamma$, and for each
$k\in\triangle_{1}$, $l\in\triangle_{2}$, $z_{k}$, $-\bar{z}_{k}$,
$\mathrm{i}\omega_{l}$ are simple poles and $-z_{k}^{-1}$, $\bar{z}_{k}^{-1}$,
$\mathrm{i}\omega_{l}^{-1}$ are c simple zeros of $T(z)$.

  * (b)

$T(z)=-[T(-z^{-1})]^{-1}$.

  * (c)

For $z\in\Gamma$,

$\frac{T_{+}(z)}{T_{-}(z)}=1-\rho(z)\tilde{\rho}(z).$ (4.12)

Particularly, for $z\in\\{z:|z|=1\\}\cap\Gamma$,

$\frac{T_{+}(z)}{T_{-}(z)}=1-|\rho(z)|^{2}.$ (4.13)

  * (d)

Define $\mathbb{R}_{i+}=[{\rm Re}\zeta_{i},+\infty)$, $i=1,4$,
$\mathbb{R}_{i-}=(-\infty,{\rm Re}\zeta_{i}],\ i=2,3$. Then, as
$z\rightarrow\zeta_{i}$ along any ray
$\zeta_{i}+e^{\mathrm{i}\phi}\mathbb{R}_{i\pm}$ with $|\phi|<\pi$,

$|T(z)-T_{i}(\zeta_{i})(z-\zeta_{i})^{-\mathrm{i}\nu(\zeta_{i})}|\leq
c\left\|{\rm
log}(1-\rho(z)\tilde{\rho}(z))\right\|_{H^{1}(\Gamma)}|z-\zeta_{i}|^{1/2},$
(4.14)

where

$\begin{split}&T_{i}(\xi)=\prod_{k\in\triangle_{1}}\prod_{l\in\triangle_{2}}\frac{(\zeta_{i}+z_{k}^{-1})(\zeta_{i}-\bar{z}_{k}^{-1})(\zeta_{i}-\mathrm{i}\omega_{l}^{-1})}{(\zeta_{i}z_{k}^{-1}-1)(\zeta_{i}\bar{z}_{k}^{-1}+1)(\mathrm{i}\omega_{l}^{-1}\zeta_{i}+1)}{\rm
exp}[\mathrm{i}\beta_{i}(\zeta_{i},\xi)],\\\
&\beta_{i}(z,\xi)=\nu(\zeta_{i}){\rm
log}(z-\alpha_{l})+\int_{\widehat{\zeta_{i}\zeta_{k}}}\frac{\nu(s)-\nu(\zeta_{i})}{s-z}ds+\int_{\Gamma\backslash\widehat{\zeta_{i}\zeta_{k}}}\frac{\nu(s)}{s-z}ds-\int_{\Gamma}\frac{\nu(s)}{2s}ds,\end{split}$
(4.15)

$i=1,2,3,4$, where $k=5$ when $i=1,4$ and $k=6$ when $i=2,3$, $l=1$ when
$i=1,2$ and $l=2$ when $i=3,4$.

###### Proof.

Properties $(a)$, $(b)$ and $(c)$ can be obtain by simple calculation from the
definition of $T(z)$ in (4.11). And for $(d)$, analogously to , rewrite

${\rm
exp}[\mathrm{i}\int_{\Gamma}\nu(s)(\frac{1}{s-z}-\frac{1}{2s})ds]=(z-\zeta_{i})^{-\mathrm{i}\nu(\zeta_{i})}{\rm
exp}[\mathrm{i}\beta_{i}(z,\xi)],$ (4.16)

and note the fact that

$|(z-\zeta_{i})^{-\mathrm{i}\nu(\zeta_{i})}|\leq
e^{\pi\nu(\zeta_{i})}=(1-|\rho(\zeta_{i})|^{2})^{-1/2},$ (4.17)

and

$|\beta_{i}(z,\xi)-\beta_{i}(\zeta_{i},\xi)|\leq c\left\|{\rm
log}(1-\rho(z)\tilde{\rho}(z))\right\|_{H^{1}(\Gamma)}|z-\zeta_{i}|^{1/2}.$
(4.18)

The result then follows promptly. ∎

By using $T(z)$, the new matrix-valued function $m^{(1)}(z)$ is defined as

$m^{(1)}(z)=T(\infty)^{\sigma_{3}}m(z)T(z)^{-\sigma_{3}},$ (4.19)

which satisfies the following RH problem.

######  RHP 4.2.

Find a $2\times 2$ matrix-valued function $m^{(1)}(x,t;z)$ such that

  * *

$m^{(1)}(z)$ is meromorphic in $\mathbb{C}\backslash\Sigma$.

  * *

The non-tangential limits $m^{(1)}_{\pm}(z)$ exist for any
$z\in\Sigma=\tilde{\Gamma}\cup\Gamma$ and satisfy the jump relation
$m_{+}^{(1)}(z)=m_{-}^{(1)}(z)v^{(1)}(z)$, where

$\displaystyle v(x,t;z)=\left\\{\begin{aligned}
&\begin{bmatrix}1&-\tilde{\rho}T^{2}e^{2\mathrm{i}t\theta}\\\
0&1\end{bmatrix}\begin{bmatrix}1&0\\\ \rho
T^{-2}e^{-2\mathrm{i}t\theta}&0\end{bmatrix},\quad z\in\tilde{\Gamma}\\\
&\begin{bmatrix}1&0\\\
\frac{\rho}{1-\rho\tilde{\rho}}T_{-}^{-2}e^{-2\mathrm{i}t\theta}&1\end{bmatrix}\begin{bmatrix}1&-\frac{\tilde{\rho}}{1-\rho\tilde{\rho}}T_{+}^{2}e^{2\mathrm{i}t\theta}\\\
0&1\end{bmatrix},\quad z\in\Gamma,\end{aligned}\right.$ (4.20)

  * *

Asymptotic behavior

$m^{(1)}(x,t;z)=I+\mathcal{O}(z^{-1}),\quad z\rightarrow\infty.$ (4.21)

  * *

Residue conditions

$\displaystyle\underset{z=\eta_{k}}{\rm Res}m^{(1)}(z)=\left\\{\begin{aligned}
&\lim_{z\rightarrow\eta_{k}}m^{(1)}(z)\begin{bmatrix}0&0\\\
A[\eta_{k}]T^{-2}(\eta_{k})e^{-2it\theta(\eta_{k})}&0\end{bmatrix},\quad
k\in\triangle,\\\
&\lim_{z\rightarrow\eta_{k}}m^{(1)}(z)\begin{bmatrix}0&\frac{1}{A[\eta_{k}]}[(\frac{1}{T})^{\prime}(\eta_{k})]^{-2}e^{2it\theta(\eta_{k})}\\\
0&0\end{bmatrix},\quad k\in\nabla,\end{aligned}\right.$ (4.22)
$\displaystyle\underset{z=\hat{\eta}_{k}}{\rm
Res}m^{(1)}(z)=\left\\{\begin{aligned}
&\lim_{z\rightarrow\hat{\eta}_{k}}m^{(1)}(z)\begin{bmatrix}0&A[\hat{\eta}_{k}]T^{2}(\hat{\eta}_{k})e^{2it\theta(\hat{\eta}_{k})}\\\
0&0\end{bmatrix},\quad k\in\triangle,\\\
&\lim_{z\rightarrow\hat{\eta}_{k}}m^{(1)}(z)\begin{bmatrix}0&0\\\
\frac{1}{A[\hat{\eta}_{k}]}\frac{1}{[T^{\prime}(\hat{\eta}_{k})]^{2}}e^{-2it\theta(\hat{\eta}_{k})}&0\end{bmatrix},\quad
k\in\nabla.\end{aligned}\right.$ (4.23)

###### Proof.

The analyticity of $m^{(1)}(z)$ directly follows from its definition (4.2). By
simple computation, we can obtain the jump relation and residue condition from
(4.2), (4.20), (LABEL:rhp1resa) and (4.64)as well as the jump relation of RHP
2.1. As for asymptotic behaviors, we notice that
$T(\infty)^{\sigma_{3}}m(z)T(z)^{-\sigma_{3}}\rightarrow
I,z\rightarrow\infty$, thus the asymptotic behaviors of $m^{(1)}(z)$ is
obtained. ∎

###  4.2 Characteristic lines and estimates for ${\rm Im}\theta(z)$

In this section, we construct a new matrix function $m^{(2)}$ for deforming
the contour $\Sigma$ into a contour $\Sigma_{2}$ such that: First, $m^{(2)}$
has no jump on $\Sigma$. For this purpose, we choose the boundary values of
$R^{(2)}$ through the factorization of $v^{(1)}$ where the new jumps on
$\Sigma_{2}$ match a well known model RH problem; Second, we need to control
the norm of $\mathcal{R}^{(2)}$, so that the $\partial$-contribution to the
long-time asymptotics of $q(x,t)$ can be ignored; Third, the residues are
unaffected by the transformation.

For this purpose, we first introduce some new contours, see in Figure 4:

${\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}\rm
Re}z$Imz$\alpha_{1}$$\alpha_{2}$$\zeta_{5}$$\zeta_{1}$$\zeta_{2}$$\zeta_{6}$$\zeta_{3}$$\zeta_{4}$$O$$\Sigma_{01}$$\Sigma_{02}$$\Sigma_{03}$$\Sigma_{04}$$\Sigma_{52}$$\Sigma_{54}$$\Sigma_{51}$$\Sigma_{53}$$\Sigma_{62}$$\Sigma_{64}$$\Sigma_{61}$$\Sigma_{63}$$\Sigma_{01}^{{}^{\prime}}$$\Sigma_{02}^{{}^{\prime}}$$\Sigma_{03}^{{}^{\prime}}$$\Sigma_{04}^{{}^{\prime}}$$\Sigma_{14}$$\Sigma_{12}$$\Sigma_{11}$$\Sigma_{13}$$\Sigma_{24}$$\Sigma_{22}$$\Sigma_{21}$$\Sigma_{23}$$\Sigma_{31}$$\Sigma_{33}$$\Sigma_{34}$$\Sigma_{32}$$\Sigma_{41}$$\Sigma_{43}$$\Sigma_{44}$$\Sigma_{42}$

Figure 4: The green curves are the origin jump contours
$\Sigma=\mathbb{R}\cup\\{Z\in\mathbb{C}:|z|=1\\}$, the blue curves are the
opening contours in region
$\\{z\in\mathbb{C}:|e^{-2\mathrm{i}t\theta(z)}|\rightarrow 0\\}$ while the red
curves are the opening contours in region
$\\{z\in\mathbb{C}:|e^{2\mathrm{i}t\theta(z)}|\rightarrow 0\\}$. These arrows
represent directions of jump contours.

.

  * 1.

To open the lens at $(0,0)$, we fix an angle $\theta_{0}>0$ sufficiently small
such that the set $\\{z\in\mathbb{C}:\left|\frac{{\rm Re}z}{z}\right|\\}$ does
not intersect any of the disks $|z-\eta_{k}|<\varrho$ or
$|z-\hat{\eta}_{k}|<\varrho$. Let

$\phi(\xi)={\rm min}\left\\{\theta_{0},{\rm
arccos}\sqrt{\frac{|\xi|+6}{12}}\right\\}$ (4.24)

and define

$\begin{split}&\Sigma_{01}=l_{01}e^{\mathrm{i}\phi(\xi)},\quad\Sigma_{02}=l_{02}e^{\mathrm{i}(\pi-\phi(\xi)}),\\\
&\Sigma_{03}=l_{03}e^{\mathrm{i}(\pi+\phi(\xi))},\quad\Sigma_{04}=l_{04}e^{\mathrm{i}(2\pi-\phi(\xi)}),\end{split}$
(4.25)

where

$l_{01}=l_{04}=(0,\frac{1}{2}),\quad l_{02}=l_{03}=(-\frac{1}{2},0).$ (4.26)

In the same way, we can define $\Sigma_{5j}$, $\Sigma_{6j}$ and
$\Sigma_{0j}^{{}^{\prime}}$, $j=1,2,3,4.$

  * 2.

Let

$\begin{split}&l_{11}=(0,{\rm Re}\zeta_{1}),\quad l_{12}=(1,{\rm
Re}\zeta_{1}),\\\ &l_{13}=({\rm Re}\zeta_{1},3),\quad l_{14}=({\rm
Re}\zeta_{1},2.4).\end{split}$ (4.27)

Then define

$\begin{split}&\Sigma_{11}=l_{11}e^{\frac{\pi}{4}},\quad\Sigma_{12}=l_{12}e^{-\frac{\pi}{4}},\\\
&\Sigma_{13}=l_{13}e^{-\frac{\pi}{4}},\end{split}$ (4.28)

and the straight part of $\Sigma_{14}$ as $l_{14}e^{\frac{\pi}{4}}$. Moreover,
$\Sigma_{ij}$ can be defined similarly, where $i=2,3,4,\ j=1,2,3,4$.

Inspired by the idea of Cuccagna and Burgers in [16, 15], we open the jump
contour $\Sigma$ at $(0,0)$ by a small angle and at six stationary phase
points by $\frac{\pi}{4}$, see the blue and red curves in Figure 4. It is
worth noting that there are two opened jump contours of opposite directions on
each $\Sigma_{ij}$, $i=5,6,\ j=1,2,3,4$. Thus, they can cancel each other out
and the contours $\Sigma_{ij}$ are disappeared, where $i=5,6,\ j=1,2,3,4$. In
a word, the boundary of $\mathcal{R}^{(2)}$ can be defined as:

$\begin{split}&\widetilde{\Sigma}^{(2)}=L_{0}\cup
L^{{}^{\prime}}_{0}\cup(\underset{i=1,2,3,4}{\cup}L_{i}),\\\
&L_{0}=\underset{j=1,2,3,4}{\cup}\Sigma_{0j},\quad
L^{{}^{\prime}}_{0}=\underset{j=1,2,3,4}{\cup}\Sigma^{{}^{\prime}}_{0j},\\\
&L_{i}=\underset{j=1,2,3,4}{\cup}\Sigma_{ij},\quad
i=1,2,\cdot\cdot\cdot,6.\end{split}$ (4.29)

See Figure 5. Additionally, domains of $\mathcal{R}^{(2)}$ are presented in
Figure 6.

${\rm
Re}z$$\zeta_{5}$$\zeta_{1}$$\zeta_{2}$$\zeta_{6}$$\zeta_{3}$$\zeta_{4}$$\Sigma_{01}$$\Sigma_{02}$$\Sigma_{03}$$\Sigma_{04}$$\Sigma_{01}^{{}^{\prime}}$$\Sigma_{02}^{{}^{\prime}}$$\Sigma_{03}^{{}^{\prime}}$$\Sigma_{04}^{{}^{\prime}}$$\Sigma_{14}$$\Sigma_{12}$$\Sigma_{11}$$\Sigma_{13}$$\Sigma_{24}$$\Sigma_{22}$$\Sigma_{21}$$\Sigma_{23}$$\Sigma_{31}$$\Sigma_{33}$$\Sigma_{34}$$\Sigma_{32}$$\Sigma_{41}$$\Sigma_{43}$$\Sigma_{44}$$\Sigma_{42}$
Figure 5: Boundary of $\mathcal{R}^{(2)}$.

${\rm Re}z$${\rm
Im}z$$\zeta_{1}$$\zeta_{2}$$\zeta_{3}$$\zeta_{4}$$\Omega_{01}$$\Omega_{02}$$\Omega_{03}$$\Omega_{04}$$\Omega_{01}^{{}^{\prime}}$$\Omega_{02}^{{}^{\prime}}$$\Omega_{03}^{{}^{\prime}}$$\Omega_{04}^{{}^{\prime}}$$\Omega_{14}$$\Omega_{12}$$\Omega_{11}$$\Omega_{13}$$\Omega_{24}$$\Omega_{22}$$\Omega_{21}$$\Omega_{23}$$\Omega_{31}$$\Omega_{33}$$\Omega_{34}$$\Omega_{32}$$\Omega_{41}$$\Omega_{43}$$\Omega_{44}$$\Omega_{42}$

Figure 6: Domains of $\mathcal{R}^{(2)}$.

.

######  Proposition 4.2.

For $z=|z|e^{\mathrm{i}w}$,

$\begin{split}&{\rm Re}[2\mathrm{i}\theta(z)]>-(1-|z|^{-2})|z|{\rm
sin}wg(z)>0,\quad
z\in\Omega_{01}\cup\Omega_{02}\cup\Omega^{{}^{\prime}}_{03}\cup\Omega^{{}^{\prime}}_{04},\\\
&{\rm Re}[2\mathrm{i}\theta(z)]<-(1-|z|^{-2})|z|{\rm sin}wg(z)<0,\quad
z\in\Omega_{02}\cup\Omega_{03}\cup\Omega^{{}^{\prime}}_{01}\cup\Omega^{{}^{\prime}}_{02},\end{split}$
(4.30)

where

$g(z)=3(|z|+|z|^{-1})^{2}{\rm cos}^{2}w+\xi-4>0.$ (4.31)

###### Proof.

We give a proof for $z\in\Omega_{01}$, the others are similar. For
$z=|z|e^{\mathrm{i}w}\in\Omega_{01}$,

${\rm Re}[2\mathrm{i}\theta(z)]=-(1-|z|^{-2})|z|{\rm
sin}w[3(|z|+|z|^{-1})^{2}{\rm cos}^{2}w+\xi-4],$ (4.32)

let

$g(z)=3(|z|+|z|^{-1})^{2}{\rm cos}^{2}w+\xi-4.$ (4.33)

Noticing $w\in(0,\phi(\xi))$, thus $\sqrt{\frac{|\xi|+6}{12}}<{\rm cos}w<1$.
Then observing that $|z|+|z|^{-1}\geq 2$, we can easy obtain $g(z)>0$. As a
result,

${\rm Re}[2\mathrm{i}\theta(z)]>-(1-|z|^{-2})|z|{\rm
sin}wg(z)>0,z\in\Omega_{01}.$ (4.34)

∎

###  4.3 Mixed $\bar{\partial}$-RH problem

We choose $\mathcal{R}^{(2)}(z):=\mathcal{R}^{(2)}(z;\xi)$ as:

$\displaystyle\mathcal{R}^{(2)}(z)=\left\\{\begin{aligned}
&\begin{bmatrix}1&0\\\ R_{0j}e^{-2\mathrm{i}t\theta}&1\end{bmatrix},\quad
z\in\Omega_{0j},\quad j=1,2,\\\
&\begin{bmatrix}1&-R_{0j}e^{2\mathrm{i}t\theta}\\\ 0&1\end{bmatrix}^{-1},\quad
z\in\Omega_{0j},\quad j=3,4,\\\
&\begin{bmatrix}1&-R_{0j}^{{}^{\prime}}e^{2\mathrm{i}t\theta}\\\
0&1\end{bmatrix}^{-1},\quad z\in\Omega_{0j}^{{}^{\prime}},\quad j=1,2,\\\
&\begin{bmatrix}1&0\\\
R_{0j}^{{}^{\prime}}e^{-2\mathrm{i}t\theta}&1\end{bmatrix},\quad
z\in\Omega_{0j}^{{}^{\prime}},\quad j=3,4,\\\
&\begin{bmatrix}1&-R_{i1}e^{2\mathrm{i}t\theta}\\\ 0&1\end{bmatrix}^{-1},\quad
z\in\Omega_{i1},\quad i=1,2,3,4.\\\ &\begin{bmatrix}1&0\\\
R_{i2}e^{-2\mathrm{i}t\theta}&1\end{bmatrix},\quad z\in\Omega_{i2},\quad
i=1,2,3,4.\\\ &\begin{bmatrix}1&-R_{i3}e^{2\mathrm{i}t\theta}\\\
0&1\end{bmatrix}^{-1},\quad z\in\Omega_{i3},\quad i=1,2,3,4.\\\
&\begin{bmatrix}1&0\\\ R_{i4}e^{-2\mathrm{i}t\theta}&1\end{bmatrix},\quad
z\in\Omega_{i4},\quad i=1,2,3,4,\\\ &I,\quad elsewhere,\end{aligned}\right.$
(4.35)

where the functions $R_{0j}$, $R_{0j}^{{}^{\prime}}$ and $R_{ij}$ are defined
as the following two propositions.

######  Proposition 4.3 (Opening lens at a small tangle).

$R_{0j}:\overline{\Omega}_{0j}\rightarrow\mathbb{C}$ and
$R_{0j}^{{}^{\prime}}:\overline{\Omega}_{0j}^{{}^{\prime}}\rightarrow\mathbb{C}$,
$j=1,2,3,4$ are continuous on
$\overline{\Omega}_{0j},\overline{\Omega}_{0j}^{{}^{\prime}}$ respectively,
$j=1,2,3,4$. Their boundary values are as follows:

$\displaystyle R_{0j}(z)=$ $\displaystyle\left\\{\begin{aligned}
&\frac{\rho(z)}{1-\rho(z)\tilde{\rho}(z)}T_{-}^{-2}(z),\quad z\in l_{0j},\quad
j=1,2,\\\ &0,\quad\quad z\in\Sigma_{0j},\quad j=1,2,\end{aligned}\right.$
(4.36) $\displaystyle R_{0j}(z)=$ $\displaystyle\left\\{\begin{aligned}
&\frac{\tilde{\rho}(z)}{1-\rho(z)\tilde{\rho}(z)}T_{+}^{2}(z),\quad z\in
l_{0j},\quad j=3,4,\\\ &0,\quad\quad z\in\Sigma_{0j},\quad
j=3,4.\end{aligned}\right.$ (4.37) $\displaystyle R_{0j}^{{}^{\prime}}(z)=$
$\displaystyle\left\\{\begin{aligned}
&\frac{\tilde{\rho}(z)}{1-\rho(z)\tilde{\rho}(z)}T_{+}^{2}(z),\quad z\in
l_{0j}^{{}^{\prime}},\quad j=1,2,\\\ &0,\quad\quad
z\in\Sigma_{0j}^{{}^{\prime}},\quad j=1,2,\end{aligned}\right.$ (4.38)
$\displaystyle R_{0j}^{{}^{\prime}}(z)=$ $\displaystyle\left\\{\begin{aligned}
&\frac{\rho(z)}{1-\rho(z)\tilde{\rho}(z)}T_{-}^{-2}(z),\quad z\in
l_{0j}{{}^{\prime}},\quad j=3,4,\\\ &0,\quad\quad
z\in\Sigma_{0j}{{}^{\prime}},\quad j=3,4.\end{aligned}\right.$ (4.39)

Moreover, $R_{0j}$ and $R_{0j}^{{}^{\prime}}$ have following property: for j =
1; 2; 3; 4;

$\begin{split}&|\bar{\partial}R_{0j}(z)|\lesssim|z|^{-\frac{1}{2}}+\left|\left(\frac{\rho}{1-\rho\tilde{\rho}}\right)^{{}^{\prime}}({\rm
Re}z)\right|,\quad j=1,2,\\\
&|\bar{\partial}R_{0j}^{{}^{\prime}}(z)|\lesssim|z|^{-\frac{1}{2}}+\left|\left(\frac{\rho}{1-\rho\tilde{\rho}}\right)^{{}^{\prime}}({\rm
Re}z)\right|\quad j=3,4,\\\
&|\bar{\partial}R_{0j}(z)|\lesssim|z|^{-\frac{1}{2}}+\left|\left(\frac{\tilde{\rho}}{1-\rho\tilde{\rho}}\right)^{{}^{\prime}}({\rm
Re}z)\right|,\quad j=3,4,\\\
&|\bar{\partial}R_{0j}^{{}^{\prime}}(z)|\lesssim|z|^{-\frac{1}{2}}+\left|\left(\frac{\tilde{\rho}}{1-\rho\tilde{\rho}}\right)^{{}^{\prime}}({\rm
Re}z)\right|\quad j=1,2.\\\ \end{split}$ (4.40)

###### Proof.

Taking $R_{01}(z)$ as an example, its extensions can be constructed by:

$R_{01}(z)=\frac{\rho({\rm Re}z)}{1-\rho({\rm Re}z)\tilde{\rho}({\rm
Re}z)}T_{-}^{-2}(z){\rm cos}(\frac{\pi}{2\phi(\xi)}\varphi),$ (4.41)

where $z=re^{\mathrm{i}\varphi}$, ${\rm Re}z=r{\rm cos}\varphi$. Utilizing

$\bar{\partial}=\frac{1}{2}e^{\mathrm{i}\varphi}(\partial
r+\mathrm{i}r^{-1}\partial{\varphi}),$ (4.42)

we have

$\bar{\partial}R_{01}(z)=-\frac{\pi}{4\phi(\xi)}\mathrm{i}r^{-1}e^{\mathrm{i}\varphi}{\rm
sin}(\frac{\pi}{2\phi(\xi)}\varphi)\left(\frac{\rho}{1-\rho\tilde{\rho}}\right)({\rm
Re}z)+\frac{1}{2}T_{-}^{-2}(z){\rm
cos}(\frac{\pi}{2\phi(\xi)}\varphi)\left(\frac{\rho}{1-\rho\tilde{\rho}}\right)^{{}^{\prime}}({\rm
Re}z).$ (4.43)

Consequently,

$\begin{split}&|\bar{\partial}R_{01}(z)|\lesssim\frac{1}{|z|}\left|\left(\frac{\rho}{1-\rho\tilde{\rho}}\right)({\rm
Re}z)\right|+\left|\left(\frac{\rho}{1-\rho\tilde{\rho}}\right)^{{}^{\prime}}({\rm
Re}z)\right|,\\\ &\quad\quad\quad\quad\lesssim
c_{1}|z|^{-\frac{1}{2}}+c_{2}\left|\left(\frac{\rho}{1-\rho\tilde{\rho}}\right)^{{}^{\prime}}({\rm
Re}z)\right|,\end{split}$ (4.44)

where

$\begin{split}&\left|\left(\frac{\rho}{1-\rho\tilde{\rho}}\right)({\rm
Re}z)\right|=\left|\int_{0}^{{\rm
Re}z}\left(\frac{\rho}{1-\rho\tilde{\rho}}\right)^{{}^{\prime}}(s)ds\right|,\\\
&\leq\left\|\left(\frac{\rho}{1-\rho\tilde{\rho}}\right)^{{}^{\prime}}(s)\right\|_{L^{2}(0,{\rm
Re}z)}|z|^{\frac{1}{2}}\end{split}$ (4.45)

∎

######  Proposition 4.4 (Opening lens at stationary phase points).

$R_{ij}:\overline{\Omega}_{ij}\rightarrow\mathbb{C}$, $i,j=1,2,3,4$ are
continuous on $\overline{\Omega}_{ij}$ with boundary values:

$\displaystyle R_{i1}(z)=\left\\{\begin{aligned}
&\tilde{\rho}(z)T^{2}(z),\quad z\in\widehat{\zeta_{i}\alpha_{k}}\\\
&\tilde{\rho}(\zeta_{i})T^{2}(\zeta_{i})(z-\zeta_{i})^{-2\mathrm{i}\nu(\zeta_{i})}(1-\chi_{\mathcal{Z}}(z)),\quad
z\in\Sigma_{i1}\end{aligned}\right.$ (4.46)

$\displaystyle R_{i2}(z)=\left\\{\begin{aligned}
&\frac{\rho(z)}{1-\rho(z)\tilde{\rho}(z)}T_{-}^{-2}(z),\quad
z\in\widehat{\zeta_{i}\zeta_{l}}\\\
&\tilde{\rho}(\zeta_{i})T_{-}^{-2}(\zeta_{i})(z-\zeta_{i})^{2\mathrm{i}\nu(\zeta_{i})}(1-\chi_{\mathcal{Z}}(z)),\quad
z\in\Sigma_{i2}\end{aligned}\right.$ (4.47)

$\displaystyle R_{i3}(z)=\left\\{\begin{aligned}
&\frac{\tilde{\rho}(z)}{1-\rho(z)\tilde{\rho}(z)}T_{+}^{2}(z),\quad
z\in\widehat{\zeta_{i}\zeta_{l}}\\\
&\tilde{\rho}(\zeta_{i})T_{+}^{2}(\zeta_{i})(z-\zeta_{i})^{-2\mathrm{i}\nu(\zeta_{i})}(1-\chi_{\mathcal{Z}}(z)),\quad
z\in\Sigma_{i3}\end{aligned}\right.$ (4.48)

$\displaystyle R_{i4}(z)=\left\\{\begin{aligned} &\rho(z)T^{-2}(z),\quad
z\in\widehat{\zeta_{i}\alpha_{k}}\\\
&\rho(\zeta_{i})T^{-2}(\zeta_{i})(z-\zeta_{i})^{2\mathrm{i}\nu(\zeta_{i})}(1-\chi_{\mathcal{Z}}(z)),\quad
z\in\Sigma_{i4}\end{aligned}\right.$ (4.49)

$i=1,2,3,4$, $k=1$ when $i=1,2$ and $k=2$ when $i=3,4$, $l=5$ when $i=1,2$ and
$l=6$ when $i=3,4$, and $\chi_{\mathcal{Z}}(z)\in C_{0}^{\infty}$ is defined
as

$\displaystyle\chi_{\mathcal{Z}}(z)=\left\\{\begin{aligned} &1,\quad{\rm
dist(z,\mathcal{Z}\cup\mathcal{\hat{Z}})}<\varrho/3\\\ &0,\quad{\rm
dist(z,\mathcal{Z}\cup\mathcal{\hat{Z}})}>2\varrho/3.\end{aligned}\right.$
(4.50)

Moreover, $R_{ij}(z),i,j=1,2,3,4$ have following properties:

$\begin{split}&|R_{ij}(z)|\lesssim(1+|z|^{2})^{-\frac{1}{4}}+c,\\\
&|\bar{\partial}R_{ij}(z)|=\mathcal{O}(1),\quad i,j=1,2,3,4.\end{split}$
(4.51)

###### Proof.

We give the details for $R_{11}(z)$ only. The other cases are easily inferred.
The continuous extension of $R_{11}(z)$ on $\Omega_{11}$ can be constructed by

$R_{11}(z)=[\gamma\tilde{\rho}(z)+(1-\gamma)g_{11}(z)]T^{2}(z)(1-\chi_{\mathcal{Z}}(z)),\quad\gamma\in[0,1],$
(4.52)

where $g_{11}(z)$ is defined as

$g_{11}(z)=\tilde{\rho}(\zeta_{1})T^{2}(\zeta_{1})T^{-2}(z)(z-\zeta_{1})^{-2\mathrm{i}\nu(\zeta_{1})}.$
(4.53)

Firstly, we have

$\begin{split}|\tilde{\rho}(\zeta_{1})(z-\zeta_{1})^{-2\mathrm{i}\nu(\zeta_{1})}|&=|\tilde{\rho}(\zeta_{1})|e^{2\nu(\zeta_{1}){\rm
arg}(z-\zeta_{1})}\\\
&\leq|\tilde{\rho}(\zeta_{1})|(1-|\tilde{\rho}(\zeta_{1})|^{2})^{-1}.\end{split}$
(4.54)

Recall $\tilde{\rho}(z)\in H^{1}(\Gamma)$, we obtain

$|\tilde{\rho}(z)|\lesssim\left|(1+z^{2})^{-\frac{1}{4}}\right|\lesssim(1+|z|^{2})^{-\frac{1}{4}}.$
(4.55)

Thus,

$\begin{split}&|R_{11}(z)|\lesssim(1+|z|^{2})^{-\frac{1}{4}}+c,\\\
&|\bar{\partial}R_{11}(z)|\leq|[\gamma\tilde{\rho}(z)+(1-\gamma)g_{11}(z)]T^{2}(z)\bar{\partial}\chi_{\mathcal{Z}}(z)|=\mathcal{O}(1).\end{split}$
(4.56)

∎

${\rm Re}z$${\rm
Im}z$$\zeta_{1}$$\zeta_{2}$$\zeta_{3}$$\zeta_{4}$$\Sigma_{14}$$\Sigma_{12}$$\Sigma_{11}$$\Sigma_{13}$$\Sigma_{24}$$\Sigma_{22}$$\Sigma_{21}$$\Sigma_{23}$$\Sigma_{31}$$\Sigma_{33}$$\Sigma_{34}$$\Sigma_{32}$$\Sigma_{41}$$\Sigma_{43}$$\Sigma_{44}$$\Sigma_{42}$
Figure 7: The jump contours $\Sigma^{(2)}$ of $m^{(2)}$.

Define $\Sigma^{(2)}=\underset{i=1,2,3,4}{\cup}L_{i}$, which can be referred
in the following Figure 7. We now use $\mathcal{R}^{(2)}(z)$ to define a new
transformation

$m^{(2)}(z)=m^{(1)}(z)\mathcal{R}^{(2)}(z),$ (4.57)

which satisfies the following mixed $\bar{\partial}$-RH problem.

######  RHP 4.3.

Find a $2\times 2$ matrix-valued function $m^{(2)}(x,t;z)$ such that

  * *

$m^{(2)}(z)$ is continuous in
$\mathbb{C}\backslash(\Sigma^{(2)}\cup\mathcal{Z}\cup\hat{\mathcal{Z}})$.

  * *

Jump relation: $m^{(2)}_{+}(z)=m^{(2)}_{-}(z)v^{(2)}(z)$, $z\in\Sigma_{(2)}$,
where

$v^{(2)}(z)=[\mathcal{R}^{(2)}_{-}]^{-1}v^{(1)}\mathcal{R}^{(2)}_{+}=I+(1-\chi_{\mathcal{Z}}(z))\delta
v^{(2)}(z),$ (4.58) $\displaystyle\delta v^{(2)}(z)=\left\\{\begin{aligned}
&\begin{bmatrix}0&-\tilde{\rho}(\zeta_{i})T^{2}(\zeta_{i})(z-\zeta_{i})^{2\mathrm{i}\nu(\zeta_{i})}e^{2\mathrm{i}t\theta(z)}\\\
0&0\end{bmatrix},\quad z\in\Sigma_{i1},i=1,2,3,4,\\\ &\begin{bmatrix}0&0\\\
\frac{\rho(\zeta_{i})}{1-\rho(\zeta_{i})\tilde{\rho}(\zeta_{i})}T_{-}^{-2}(\zeta_{i})(z-\zeta_{i})^{-2\mathrm{i}\nu(\zeta_{i})}e^{-2\mathrm{i}t\theta(z)}&0\end{bmatrix},\quad
z\in\Sigma_{i2},i=1,2,3,4,\\\
&\begin{bmatrix}0&-\frac{\tilde{\rho}(\zeta_{i})}{1-\rho(\zeta_{i})\tilde{\rho}(\zeta_{i})}T_{+}^{-2}(\zeta_{i})(z-\zeta_{i})^{2\mathrm{i}\nu(\zeta_{i})}e^{2\mathrm{i}t\theta(z)}\\\
0&0\end{bmatrix},\quad z\in\Sigma_{i3},i=1,2,3,4,\\\ &\begin{bmatrix}0&0\\\
-\rho(\zeta_{i})T^{-2}(\zeta_{i})(z-\zeta_{i})^{-2\mathrm{i}\nu(\zeta_{i})}e^{-2\mathrm{i}t\theta(z)}&0\end{bmatrix},\quad
z\in\Sigma_{i4},i=1,2,3,4,\\\ &\begin{bmatrix}0&0\\\ 0&0\end{bmatrix},\quad
elsewhere.\end{aligned}\right.$ (4.59)

  * *

For
$z\in\mathbb{C}\backslash(\Sigma^{(2)}\cup\mathcal{Z}\cup\hat{\mathcal{Z}})$,

$\bar{\partial}m^{(2)}=m^{(2)}\bar{\partial}\mathcal{R}^{(2)},$ (4.60)

where

$\displaystyle\bar{\partial}\mathcal{R}^{(2)}=\left\\{\begin{aligned}
&\begin{bmatrix}0&-\bar{\partial}R_{i1}e^{2\mathrm{i}t\theta}\\\
0&0\end{bmatrix},\quad z\in\Omega_{i1},i=1,2,3,4,\\\ &\begin{bmatrix}0&0\\\
\bar{\partial}R_{i2}e^{-2\mathrm{i}t\theta}&0\end{bmatrix},\quad
z\in\Omega_{i2},i=1,2,3,4,\\\
&\begin{bmatrix}0&\bar{\partial}R_{i3}e^{2\mathrm{i}t\theta}\\\
0&0\end{bmatrix},\quad z\in\Omega_{i3},i=1,2,3,4,\\\ &\begin{bmatrix}0&0\\\
\bar{\partial}R_{i4}e^{-2\mathrm{i}t\theta}&0\end{bmatrix},\quad
z\in\Omega_{i4},i=1,2,3,4,\\\ &\begin{bmatrix}0&0\\\ 0&0\end{bmatrix},\quad
elsewhere.\\\ \end{aligned}\right.$ (4.61)

  * *

Asymptotic behavior

$m^{(2)}(x,t;z)=I+\mathcal{O}(z^{-1}),\quad z\rightarrow\infty.$ (4.62)

  * *

Residue conditions

$\displaystyle\underset{z=\eta_{k}}{\rm Res}m^{(2)}(z)=\left\\{\begin{aligned}
&\lim_{z\rightarrow\eta_{k}}m^{(2)}(z)\begin{bmatrix}0&0\\\
A[\eta_{k}]T^{-2}(\eta_{k})e^{-2it\theta(\eta_{k})}&0\end{bmatrix},\quad
k\in\triangle,\\\
&\lim_{z\rightarrow\eta_{k}}m^{(2)}(z)\begin{bmatrix}0&\frac{1}{A[\eta_{k}]}[(\frac{1}{T})^{\prime}(\eta_{k})]^{-2}e^{2it\theta(\eta_{k})}\\\
0&0\end{bmatrix},\quad k\in\nabla,\end{aligned}\right.$ (4.63)
$\displaystyle\underset{z=\hat{\eta}_{k}}{\rm
Res}m^{(2)}(z)=\left\\{\begin{aligned}
&\lim_{z\rightarrow\hat{\eta}_{k}}m^{(2)}(z)\begin{bmatrix}0&A[\hat{\eta}_{k}]T^{2}(\hat{\eta}_{k})e^{2it\theta(\hat{\eta}_{k})}\\\
0&0\end{bmatrix},\quad k\in\triangle,\\\
&\lim_{z\rightarrow\hat{\eta}_{k}}m^{(2)}(z)\begin{bmatrix}0&0\\\
\frac{1}{A[\hat{\eta}_{k}]}\frac{1}{[T^{\prime}(\hat{\eta}_{k})]^{2}}e^{-2it\theta(\hat{\eta}_{k})}&0\end{bmatrix},\quad
k\in\nabla.\end{aligned}\right.$ (4.64)

${\rm Re}z$${\rm
Im}z$$\zeta_{1}$$\zeta_{2}$$\zeta_{3}$$\zeta_{4}$$\Omega_{14}$$\Omega_{12}$$\Omega_{11}$$\Omega_{13}$$\Omega_{24}$$\Omega_{22}$$\Omega_{21}$$\Omega_{23}$$\Omega_{31}$$\Omega_{33}$$\Omega_{34}$$\Omega_{32}$$\Omega_{41}$$\Omega_{43}$$\Omega_{44}$$\Omega_{42}$

Figure 8: Domains of $m^{(2)}$.

.

##  5 Decomposition of the mixed $\bar{\partial}$-RH problem

To solve RHP 4.3, we decompose $m^{(2)}$ into a pure RH problem for
$m^{(2)}_{RHP}$ with $\bar{\partial}\mathcal{R}^{(2)}\equiv 0$ and a pure
$\bar{\partial}$-problem with nonzero $\bar{\partial}$-derivatives, which can
be shown as the following structure

$\displaystyle m^{(2)}=m^{(3)}m^{(2)}_{RHP}\left\\{\begin{aligned}
&\bar{\partial}\mathcal{R}^{(2)}\equiv 0\longrightarrow m^{(2)}_{RHP},\\\
&\bar{\partial}\mathcal{R}^{(2)}\neq 0\longrightarrow
m^{(3)}=m^{(2)}[m^{(2)}_{RHP}]^{-1}.\end{aligned}\right.$ (5.1)

$m^{(2)}_{RHP}$ satisfies the following RH problem.

######  RHP 5.1.

Find a $2\times 2$ matrix-valued function $m^{(2)}_{RHP}(x,t;z)$ such that

  * *

$m^{(2)}_{RHP}(z)$ is analytic in
$\mathbb{C}\backslash(\Sigma^{(2)}\cup\mathcal{Z}\cup\hat{\mathcal{Z}})$.

  * *

Jump relation: $m^{(2)}_{PHP+}(z)=m^{(2)}_{RHP-}(z)v^{(2)}(z)$,
$z\in\Sigma^{(2)}$, where

  * *

For
$z\in\mathbb{C}\backslash(\Sigma^{(2)}\cup\mathcal{Z}\cup\hat{\mathcal{Z}})$,

$\bar{\partial}m^{(2)}\equiv 0.$ (5.2)

  * *

Asymptotic behavior

$m^{(2)}_{RHP}(x,t;z)=I+\mathcal{O}(z^{-1}),\quad z\rightarrow\infty.$ (5.3)

  * *

$m^{(2)}_{RHP}$ has same jump matrix and residue conditions as $m^{(2)}$.

Define $U_{\zeta_{i}}$ as

$U_{\zeta_{i}}=\left\\{z:|z-\zeta_{i}|<\varrho\right\\},i=1,2,3,4.$ (5.4)

Then the jump matrix in RHP 5.1 admits the following estimates.

######  Proposition 5.1.

As $t\rightarrow\infty$, there exist positive constants $K_{i},i=1,2,3,4$ that
the jump matrix $v^{(2)}(z)$ admits the following estimate

$\begin{split}&\|v^{(2)}(z)-I\|_{L^{\infty}(\Sigma_{ij}\backslash
U_{\zeta_{i}})}=\mathcal{O}(e^{-K_{i}|z-\zeta_{i}|t}),\\\
&\|v^{(2)}(z)-I\|_{L^{\infty}(\Sigma_{ij}\cap
U_{\zeta_{i}})}=\mathcal{O}(|z-\zeta_{i}|^{-1/2}t^{-1/2}),\quad
i,j=1,2,3,4.\end{split}$ (5.5)

###### Proof.

We prove the case of $z\in\Sigma_{11}\backslash U_{\zeta_{1}}$ and
$z\in\Sigma_{11}\cap U_{\zeta_{1}}$, then the another case can be proved in
similar way. Denote $\zeta_{1}=a_{1}+\mathrm{i}b_{1}$, we can obtain

$\xi=-3(\zeta_{1}^{2}+\zeta_{1}^{-2}).$ (5.6)

For $z\in\Sigma_{11}$,
$z=\zeta_{1}+|z-\zeta_{1}|e^{\mathrm{i}\frac{3}{4}\pi}$, we have

${\rm
Re}[2\mathrm{i}t\theta(z)]=-t(1-|z|^{-2})\left(b_{1}+\frac{\sqrt{2}}{2}|z-\zeta_{1}|\right)f(z),$
(5.7)

where

$f(z)=\left[12b_{1}^{2}-9+(1+|z|^{-2}+|z|^{-4})(3-4b_{1}^{2}-\sqrt{2}(3\sqrt{1-b_{1}^{2}}+b_{1})|z-\zeta_{1}|+|z-\zeta_{1}|^{2})\right].$
(5.8)

Observe that

$\begin{split}f(z)&\lesssim|z-\zeta_{1}|(1+|z|^{-2}+|z|^{-4})\left(|z-\zeta_{1}|-\sqrt{2}(3\sqrt{1-b_{1}^{2}}+b_{1})\right)\\\
&\leq-\sqrt{2}|z-\zeta_{1}|\left(1+\frac{1}{2}(1+b_{1})^{2}+\frac{1}{4}(1+b_{1})^{4}\right)\left(2\sqrt{1-b_{1}^{2}}+b_{1}\right)\\\
&\lesssim-|z-\zeta_{1}|,\end{split}$ (5.9)

we obtain

$\|v^{(2)}(z)-I\|_{L^{\infty}(\Sigma_{11})}=\left|-\tilde{\rho}(\zeta_{1})T^{2}(\zeta_{1})(z-\zeta_{1})^{2\mathrm{i}\nu(\zeta_{1})}e^{2\mathrm{i}t\theta(z)}\right|\lesssim
e^{-K_{1}|z-\zeta_{1}|t}.$ (5.10)

Thus, for $z\in\Sigma_{11}\backslash U_{\zeta_{1}}$,

$\|v^{(2)}(z)-I\|_{L^{\infty}(\Sigma_{11}\backslash U_{\zeta_{1}})}\lesssim
e^{-K_{1}|z-\zeta_{1}|t},$ (5.11)

while for $z\in\Sigma_{11}\cap U_{\zeta_{1}}$,

$\|v^{(2)}(z)-I\|_{L^{\infty}(\Sigma_{11}\cap U_{\zeta_{1}})}\lesssim
e^{-K_{1}|z-\zeta_{1}|t}\lesssim|z-\zeta_{1}|^{-1/2}t^{-1/2}.$ (5.12)

∎

This proposition implies that the jump matrix $v^{(2)}$ uniformly goes to $I$
in terms of exponentially small error outside $U_{\zeta_{i}},i=1,2,3,4$. So we
can ignore the jump relation of $m^{(2)}_{RHP}$ outside the
$U_{\zeta_{i}},i=1,2,3,4$ and decompose $m^{(2)}_{RHP}$ in the following form

$\displaystyle m^{(2)}_{RHP}(z)=\left\\{\begin{aligned}
&E(z)m^{(out)}(z),\quad z\in
D\backslash(\underset{i=1,2,3,4}{\cup}U_{\zeta_{i}}),\\\
&E_{(}z)m^{(out)}(z)m^{(lo)}(z),\quad z\in U_{\zeta_{i}},\quad
i=1,2,3,4.\end{aligned}\right.$ (5.13)

In this decomposition, $m^{(out)}$ solves the pure RHP obtained by ignoring
the jump conditions of RHP 5.1, which will be solved in next Section 5.1.1;
$m^{(lo)}$ uses parabolic cylinder functions to build a matrix to match jumps
of $m^{(2)}_{RHP}$ in a neighborhood $\zeta_{i},i=1,2,3,4$ which is shown in
Section 5.1.2. $E(z)$ is an error function and a solution of a small norm
Riemann-Hilbert problem which is shown in Section 5.1.3.

###  5.1 Analysis on the pure RH Problem

####  5.1.1 Outer model RH problem

In this subsection, we build a reflectionless case of RHP 2.1 to show that its
solution can approximated with $m^{(out)}$. As the main contribution to
$m^{(out)}$ comes from its scattering data

$\sigma_{d}=\left\\{\left(\eta_{k},{c}_{k}\right)\right\\}_{k=1}^{2N_{1}+N_{2}},$
(5.14) $\displaystyle\tilde{c}_{k}=\left\\{\begin{aligned}
&A[\eta_{k}]T^{-2}(\eta_{k}),\quad z\in\triangle\\\
&\frac{1}{A[\hat{\eta}_{k}]}\frac{1}{[T^{\prime}(\hat{\eta}_{k})]^{2}},\quad
z\in\nabla,\end{aligned}\right.$ (5.15)

thus $m^{(out)}(z)=m^{(out)}(z|\sigma_{d}^{out})$, and
$m^{(out)}(z|\sigma_{d}^{out})$ can be construct as

######  RHP 5.2.

Find a $2\times 2$ matrix-valued function $m^{(out)}(z|\sigma_{d}^{out})$ such
that

  * *

$m^{(out)}(z|\sigma_{d}^{out})$ is analytic in
$\mathbb{C}\backslash\left(\mathcal{Z}\cup\hat{\mathcal{Z}}\right)$.

  * *

$m^{(out)}(z|\sigma_{d}^{out})=I+\mathcal{O}(z^{-1}),\quad
z\rightarrow\infty$.

  * *

Residue conditions: for $\eta_{k}\in\mathcal{Z}$ and
$\hat{\eta}_{k}\in\hat{\mathcal{Z}}$,

$\displaystyle\underset{z=\eta_{k}}{\rm
Res}m^{(out)}(z|\sigma_{d}^{out})=\left\\{\begin{aligned}
&\lim_{z\rightarrow\eta_{k}}m^{(out)}(z|\sigma_{d}^{out})\begin{bmatrix}0&0\\\
A[\eta_{k}]T^{-2}(\eta_{k})e^{-2it\theta(\eta_{k})}&0\end{bmatrix},\quad
k\in\triangle,\\\
&\lim_{z\rightarrow\eta_{k}}m^{(out)}(z|\sigma_{d}^{out})\begin{bmatrix}0&\frac{1}{A[\eta_{k}]}[(\frac{1}{T})^{\prime}(\eta_{k})]^{-2}e^{2it\theta(\eta_{k})}\\\
0&0\end{bmatrix},\quad k\in\nabla,\end{aligned}\right.$ (5.16)
$\displaystyle\underset{z=\hat{\eta}_{k}}{\rm
Res}m^{(out)}(z|\sigma_{d}^{out})=\left\\{\begin{aligned}
&\lim_{z\rightarrow\hat{\eta}_{k}}m^{(out)}(z|\sigma_{d}^{out})\begin{bmatrix}0&A[\hat{\eta}_{k}]T^{2}(\hat{\eta}_{k})e^{2it\theta(\hat{\eta}_{k})}\\\
0&0\end{bmatrix},\quad k\in\triangle,\\\
&\lim_{z\rightarrow\hat{\eta}_{k}}m^{(out)}(z|\sigma_{d}^{out})\begin{bmatrix}0&0\\\
\frac{1}{A[\hat{\eta}_{k}]}\frac{1}{[T^{\prime}(\hat{\eta}_{k})]^{2}}e^{-2it\theta(\hat{\eta}_{k})}&0\end{bmatrix},\quad
k\in\nabla.\end{aligned}\right.$ (5.17)

Furthermore, we set a new RH problem $m(z|\sigma_{d})$, which is a
reflectionless case $\rho(z)=\tilde{\rho}(z)=0$ of original RHP 2.1, and
satisfies

######  RHP 5.3.

Find a $2\times 2$ matrix-valued function $m(z|\sigma_{d})$ such that

  * *

$m(z|\sigma_{d})$ is analytic in
$\mathbb{C}\backslash\left(\mathcal{Z}\cup\hat{\mathcal{Z}}\right)$.

  * *

$m(z|\sigma_{d})=I+\mathcal{O}(z^{-1}),\quad z\rightarrow\infty$.

  * *

Residue conditions: for $\eta_{k}\in\mathcal{Z}$ and
$\hat{\eta}_{k}\in\hat{\mathcal{Z}}$,

$\begin{split}&\underset{z=\eta_{k}}{\rm
Res}m(z|\sigma_{d})=\lim_{z\rightarrow\eta_{k}}m(z|\sigma_{d})N_{k},\\\
&\underset{z=\hat{\eta}_{k}}{\rm
Res}m(z|\sigma_{d})=-\hat{\eta}_{k}^{2}\lim_{z\rightarrow\hat{\eta}_{k}}m(z|\sigma_{d})\sigma_{2}N_{k}\sigma_{2},\end{split}$
(5.18)

where

$\sigma_{d}=\left\\{\left(\eta_{k},A[\eta_{k}]\right)\right\\}_{k=1}^{2N_{1}+N_{2}},$
(5.19)

is the corresponding scattering data.

$N_{k}=\begin{bmatrix}0&0\\\
\gamma_{k}(x,t)&0\end{bmatrix},\quad\gamma_{k}(x,t)=A[\eta_{k}]e^{-2\mathrm{i}\theta(x,t,\eta_{k})}$
(5.20)

######  Proposition 5.2.

For giving scattering data

$\sigma_{d}=\left\\{\left(\eta_{k},A[\eta_{k}]\right)\right\\}_{k=1}^{2N_{1}+N_{2}}.$
(5.21)

RHP 5.3 has an unique solution.

###### Proof.

The uniqueness of the solution can be deduced from the Liouville’s theorem
directly. By Plemelj’s formulae, we can construct $m(z|\sigma_{d})$ as the
following form

$m(z|\sigma_{d})=I+\frac{\mathrm{i}}{z}\sigma_{3}Q_{-}+\sum_{k=1}^{2N_{1}+N_{2}}\frac{\underset{z=\eta_{k}}{\rm
Res}m(z|\sigma_{d})}{z-\eta_{k}}+\sum_{k=1}^{2N_{1}+N_{2}}\frac{\underset{z=\hat{\eta}_{k}}{\rm
Res}m(z|\sigma_{d})}{z-\hat{\eta}_{k}},$ (5.22)

where

$\underset{z=\eta_{k}}{\rm
Res}m(z|\sigma_{d})=a(\eta_{k})N_{k}=\begin{bmatrix}a_{12}(\eta_{k})\gamma_{k}&0\\\
a_{22}(\eta_{k})\gamma_{k}&0\end{bmatrix}\triangleq\begin{bmatrix}\alpha_{k}&0\\\
\beta_{k}&0\end{bmatrix}.$ (5.23)

Take advantage of the symmetry:
$m(-z^{-1}|\sigma_{d})=-\mathrm{i}zm(z|\sigma_{d})\begin{bmatrix}0&\sigma\\\
-1&0\end{bmatrix},\sigma=\pm 1$, we have

$\underset{z=\hat{\eta}_{k}}{\rm
Res}m(z|\sigma_{d})=\begin{bmatrix}0&\mathrm{i}\hat{\eta}_{k}\alpha_{k}\\\
0&\mathrm{i}\hat{\eta}_{k}\beta_{k}\end{bmatrix}.$ (5.24)

Thus (5.22) can be written as

$m(z|\sigma_{d})=I+\frac{\mathrm{i}}{z}\sigma_{3}Q_{-}+\sum_{k=1}^{2N_{1}+N_{2}}\frac{1}{z-\eta_{k}}\begin{bmatrix}\alpha_{k}&0\\\
\beta_{k}&0\end{bmatrix}+\sum_{k=1}^{2N_{1}+N_{2}}\frac{1}{z-\hat{\eta}_{k}}\begin{bmatrix}0&\mathrm{i}\hat{\eta}_{k}\alpha_{k}\\\
0&\mathrm{i}\hat{\eta}_{k}\beta_{k}\end{bmatrix}.$ (5.25)

Substituting this formulae into

$\underset{z=\eta_{j}}{\rm
Res}(z|\sigma_{d})=\lim_{z\rightarrow\eta_{j}}m(z|\sigma_{d})N_{j},$ (5.26)

we can derive following equations

$\alpha_{j}-\mathrm{i}\sum_{k=1}^{2N_{1}+N_{2}}\frac{\hat{\eta}_{k}\gamma_{j}}{\eta_{j}-\hat{\eta}_{k}}=-\sigma\mathrm{i}\frac{\gamma_{j}}{\eta_{j}},$
(5.27)
$\beta_{j}-\mathrm{i}\sum_{k=1}^{2N_{1}+N_{2}}\frac{\hat{\eta}_{k}\gamma_{j}}{\eta_{j}-\hat{\eta}_{k}}=\gamma_{j}.$
(5.28)

Denote the coefficient matrix of (5.27) and (5.28) as $A$ and $B$,
respectively. It is easy to prove that $|A|\neq 0$ and $|B|\neq 0$, so by
Cramer’s Rule, both (5.27) and (5.28) have an unique solution. ∎

######  Corollary 5.1.

Denote $q_{sol}(x,t,\sigma_{d})$ the soliton solution with scattering data

$\sigma_{d}=\left\\{\left(\eta_{k},A[\eta_{k}]\right)\right\\}_{k=1}^{2N_{1}+N_{2}}.$
(5.29)

By reconstruction formula formulae (2.72), the soliton solution is given by

$q_{sol}(x,t,\sigma_{d})=-\mathrm{i}\lim_{z\rightarrow\infty}\left[zm(z|\sigma_{d})\right]_{12}=-\sigma+\sum_{k=1}^{2N_{1}+N_{2}}\hat{\eta_{k}}\alpha_{k}.$
(5.30)

In reflectionless case, the transmission coefficient admits following trace
formula

$a(z)=\prod_{k=1}^{2N_{1}+N_{2}}\frac{z-\eta_{k}}{z-\hat{\eta}_{k}},$ (5.31)

whose poles can be split into two parts. Let
$\diamondsuit\subseteq\left\\{1,2,\cdot\cdot\cdot,2N_{1}+N_{2}\right\\}$ and
define

$a_{\diamondsuit}(z)=\prod_{k\in\diamondsuit}\frac{z-\eta_{k}}{z-\hat{\eta}_{k}}.$
(5.32)

We make a renormalization transformation

$m^{\diamondsuit}(z|\sigma_{d}^{\lozenge})=m(z|\sigma_{d})a_{\lozenge}(z)^{\sigma_{3}},$
(5.33)

which satisfies the following RH problem.

######  RHP 5.4.

Find a $2\times 2$ matrix-valued function
$m^{\lozenge}(z|\sigma_{d}^{\lozenge})$ such that

  * *

$m^{\diamondsuit}(z|\sigma_{d}^{\diamondsuit})$ is analytic in
$\mathbb{C}\backslash\left(\mathcal{Z}\cup\hat{\mathcal{Z}}\right)$.

  * *

$m^{\diamondsuit}(z|\sigma_{d}^{\diamondsuit})=I+\mathcal{O}(z^{-1}),\quad
z\rightarrow\infty$.

  * *

Residue conditions: for $\eta_{k}\in\mathcal{Z}$ and
$\hat{\eta}_{k}\in\hat{\mathcal{Z}}$,

$\begin{split}&\underset{z=\eta_{k}}{\rm
Res}m^{\diamondsuit}(z|\sigma_{d}^{\diamondsuit})=\lim_{z\rightarrow\eta_{k}}m^{\diamondsuit}(z|\sigma_{d}^{\diamondsuit})N^{\diamondsuit}_{k},\\\
&\underset{z=\hat{\eta}_{k}}{\rm
Res}m^{\diamondsuit}(z|\sigma_{d}^{\diamondsuit})=-\hat{\eta}_{k}^{2}\lim_{z\rightarrow\hat{\eta}_{k}}m^{\diamondsuit}(z|\sigma_{d}^{\diamondsuit})\sigma_{2}N^{\diamondsuit}_{k}\sigma_{2},\end{split}$
(5.34)

where

$\sigma^{\diamondsuit}_{d}=\left\\{\left(\eta_{k},\hat{c}_{k}\right)\right\\}_{k=1}^{2N_{1}+N_{2}},$
(5.35) $\displaystyle\hat{c}_{k}=\left\\{\begin{aligned}
&\frac{1}{A[\eta_{k}]}[a_{\diamondsuit}^{{}^{\prime}}(\eta_{k})]^{-2},\quad
k\in\diamondsuit,\\\ &A[\eta_{k}][a_{\diamondsuit}(\eta_{k})]^{2},\quad
k\in\mathcal{N}\backslash\diamondsuit\end{aligned}\right.$ (5.36)

is the corresponding scattering data, and

$\displaystyle N_{k}^{\diamondsuit}=\left\\{\begin{aligned}
&\begin{bmatrix}0&\gamma_{k}^{\diamondsuit}\\\ 0&0\end{bmatrix},\quad
k\in\diamondsuit\\\ &\begin{bmatrix}0&0\\\
\gamma_{k}^{\diamondsuit}&0\end{bmatrix},\quad
k\in\mathcal{N}\backslash\diamondsuit,\end{aligned}\right.$ (5.37)
$\displaystyle\gamma^{\diamondsuit}_{k}=\left\\{\begin{aligned}
&\frac{1}{A[\eta_{k}]}[a_{\diamondsuit}^{{}^{\prime}}(\eta_{k})]^{-2}e^{2\mathrm{i}t\theta(\eta_{k})},\quad
k\in\diamondsuit,\\\
&A[\eta_{k}][a_{\diamondsuit}(\eta_{k})]^{2}e^{-2\mathrm{i}t\theta(\eta_{k})},\quad
k\in\mathcal{N}\backslash\diamondsuit.\end{aligned}\right.$ (5.38)

######  Proposition 5.3.

For giving scattering data
$\sigma^{\diamondsuit}_{d}=\left\\{\left(\eta_{k},\hat{c}_{k}\right)\right\\}_{k=1}^{2N_{1}+N_{2}}$
without reflection, RHP 5.4 has an unique solution and

$q_{sol}(x,t;\sigma_{d}^{\diamondsuit})=-\mathrm{i}\lim_{z\to\infty}\left[zm^{\diamondsuit}(z|\sigma_{d}^{\diamondsuit})\right]_{12}=-\mathrm{i}\lim_{z\to\infty}\left[zm(z|\sigma_{d})\right]_{12}=q_{sol}(x,t;\sigma_{d}).$
(5.39)

###### Proof.

Since the transform (5.33) is explicit, we obtain the existence and uniqueness
of the solution of the RHP 5.4 by Proposition 5.2. And from reconstruction
formula, (5.39) can be derived. ∎

We observe that $m^{(out)}(z|\sigma_{d}^{out})$ has reflection, and its
reflection mainly from $T(\eta_{k})$. A nature idea is to connect
$m^{(out)}(z|\sigma_{d}^{out})$ with reflectionless scattering data
$\sigma^{\triangle}_{d}=\left\\{\left(\eta_{k},\hat{c}_{k}\right)\right\\}_{k=1}^{2N_{1}+N_{2}}$.
For this purpose, we take $a_{\diamondsuit}(z)$ as

$a_{\triangle}(z)=\prod_{k=1}^{2N_{1}}\prod_{l=1}^{N_{2}}\frac{(zz_{k}-1)(z\bar{z}_{k}^{-1}+1)(\mathrm{i}\omega_{l}^{-1}z+1)}{(z+z_{k}^{-1})(z-\bar{z}_{k}^{-1})(z-\mathrm{i}\omega_{l}^{-1})}$
(5.40)

in (5.32). Then $T(z)=a_{\triangle}^{-1}(z)\tilde{\delta}(z)$, where
$\tilde{\delta}(z)=\delta(z){\rm
exp}\left[-\mathrm{i}\int_{\Gamma}\frac{\nu(s)}{2s}ds\right]$. Therefore, the
scattering data (5.14) in RHP 5.2 can be rewritten as

$\sigma_{d}^{out}=\left\\{\eta_{k},\tilde{c}_{k}\right\\}_{k=1}^{2N_{1}+N_{2}},$
(5.41) $\displaystyle\tilde{c}_{k}=\left\\{\begin{aligned}
&A[\eta_{k}]a_{\triangle}^{2}(\eta_{k})\tilde{\delta}^{-2}(\eta_{k}),\quad
k\in\triangle\\\
&\frac{1}{A[\eta_{k}]}[a_{\triangle}^{{}^{\prime}}(\eta_{k})]^{-2}\tilde{\delta}^{2}(\eta_{k}),\quad
k\in\nabla.\end{aligned}\right.$ (5.42)

Under this new scattering data, RHP 5.2 becomes

######  RHP 5.5.

Find a $2\times 2$ matrix-valued function $m^{(out)}(z|\sigma^{out}_{d})$ such
that

  * *

$m^{(out)}(z|\sigma^{out}_{d})$ is analytic in
$\mathbb{C}\backslash\left(\mathcal{Z}\cup\hat{\mathcal{Z}}\right)$.

  * *

$m^{(out)}(z|\sigma^{out}_{d})=I+\mathcal{O}(z^{-1}),\quad
z\rightarrow\infty$.

  * *

Residue conditions: for $\eta_{k}\in\mathcal{Z}$ and
$\hat{\eta}_{k}\in\hat{\mathcal{Z}}$,

$\begin{split}&\underset{z=\eta_{k}}{\rm
Res}m^{(out)}(z|\sigma^{out}_{d})=\lim_{z\rightarrow\eta_{k}}m^{(out)}(z|\sigma^{out}_{d})N^{out}_{k},\\\
&\underset{z=\hat{\eta}_{k}}{\rm
Res}m^{(out)}(z|\sigma^{out}_{d})=-\hat{\eta}_{k}^{2}\lim_{z\rightarrow\hat{\eta}_{k}}m^{(out)}(z|\sigma^{out}_{d})\sigma_{2}N^{out}_{k}\sigma_{2},\end{split}$
(5.43)

where

$\displaystyle N_{k}^{out}=\left\\{\begin{aligned} &\begin{bmatrix}0&0\\\
A[\eta_{k}]a_{\triangle}^{2}(\eta_{k})\tilde{\delta}^{-2}(\eta_{k})&0\end{bmatrix},\quad
k\in\triangle\\\
&\begin{bmatrix}0&\frac{1}{A[\eta_{k}]}[a_{\triangle}^{{}^{\prime}}(\eta_{k})]^{-2}\tilde{\delta}^{2}(\eta_{k})\\\
0&0\end{bmatrix},\quad k\in\nabla.\end{aligned}\right.$ (5.44)

It is easy to verify that the solution of RHP 5.5 is given by

$m^{(out)}(z|\sigma^{out}_{d})=m^{\triangle}(z|\sigma_{d}^{\triangle})\tilde{\delta}(z)^{\sigma_{3}}.$
(5.45)

######  Proposition 5.4.

RHP 5.5 has an unique solution. Moreover, the relation between $N$-soliton
solution with reflection data and $N$-soliton solution without reflection data
as follows

$q_{sol}(x,t;\sigma_{d}^{out})=cq_{sol}(x,t;\sigma_{d}^{\triangle}),$ (5.46)

where $c={\rm exp}\left[-\mathrm{i}\int_{\Gamma}\frac{\nu(s)}{2s}ds\right]$.

###### Proof.

Since the transform (5.45) is explicit, we obtain the existence and uniqueness
of the solution of the RHP 5.5 by Proposition 5.3. And

$\begin{split}q_{sol}(x,t;\sigma_{d}^{out})&=-\mathrm{i}\lim_{z\to\infty}\left[zm^{(out)}(z|\sigma_{d}^{out})\right]_{12}=-\mathrm{i}\lim_{z\to\infty}\left[zm^{\triangle}(z|\sigma_{d}^{\triangle})\tilde{\delta}(z)^{\sigma_{3}}\right]_{12}\\\
&=-\mathrm{i}c\lim_{z\to\infty}\left[zm^{\triangle}(z|\sigma_{d}^{\triangle})\right]_{12}=cq_{sol}(x,t;\sigma_{d}^{\triangle}).\end{split}$
(5.47)

∎

Next, we consider the asymptotic behavior of
$m^{\triangle}(z|\sigma_{d}^{\triangle})$. Recall

$\Lambda=\\{k\in\mathcal{N}:|Re(2\mathrm{i}\theta(\eta_{k}))|<\delta_{0}\\},$
(5.48)

and further define

$\begin{split}&\Lambda_{+}=\\{k\in\mathcal{N}:0\leq
Re(2\mathrm{i}\theta(\eta_{k}))<\delta_{0}\\},\\\
&\Lambda_{-}=\\{k\in\mathcal{N}:-\delta_{0}<Re(2\mathrm{i}\theta(\eta_{k}))\leq
0\\}.\end{split}$ (5.49)

######  Proposition 5.5.

For giving scattering data
$\sigma^{\triangle}_{d}=\left\\{\left(\eta_{k},\hat{c}_{k}\right)\right\\}_{k=1}^{2N_{1}+N_{2}}$,
we have

$m^{\triangle}(z|\sigma_{d}^{\triangle})=\left(I+\mathcal{O}(e^{-\delta_{0}t})\right)m^{\Lambda}(z|\sigma_{d}^{\Lambda}),$
(5.50)

where $m^{\Lambda}(z|\sigma_{d}^{\Lambda})$ is a solution for RH problem
defined by scattering data $\sigma^{\Lambda}_{d}$,
$\sigma^{\Lambda}_{d}=\\{(\eta_{k},\hat{c}_{k}),k\in\Lambda\\}$.

###### Proof.

For $k\in\triangle$,

$N_{k}^{\triangle}=\begin{bmatrix}0&\gamma_{k}^{\triangle}\\\
0&0\end{bmatrix},\quad\gamma_{k}^{\triangle}=\frac{1}{A[\eta_{k}]}[a_{\triangle}^{{}^{\prime}}(\eta_{k})]^{-2}e^{2\mathrm{i}t\theta(\eta_{k})}.$
(5.51)

Therefore, for $k\in\triangle\cap\Lambda_{-}$,
$|\gamma_{k}^{\triangle}|=\mathcal{O}(1)$ while for
$k\in\triangle\backslash\Lambda_{-}$,

$|\gamma_{k}^{\triangle}|\lesssim
e^{-\delta_{0}t}=\mathcal{O}(e^{-\delta_{0}t});$ (5.52)

As for $k\in\nabla$, it has same estimate as above. In a word, as
$t\rightarrow\infty$,

$\displaystyle\|N_{k}^{\triangle}\|=\left\\{\begin{aligned}
&\mathcal{O}(1),\quad k\in\Lambda\\\ &\mathcal{O}(e^{-\delta_{0}t}),\quad
k\in\mathcal{N}\backslash\Lambda,\end{aligned}\right.$ (5.53)

Make discs with sufficiently small radius for each discrete spectrum
$\eta_{k},\hat{\eta}_{k},k\in\mathcal{N}\backslash\Lambda$ so that they do not
intersect each other. Define

$\displaystyle\Phi=\left\\{\begin{aligned}
&I-\frac{1}{z-\eta_{k}}N_{k}^{\triangle},\quad z\in D_{k},\\\
&I+\frac{\hat{\eta_{k}}^{2}}{z-\hat{\eta}_{k}}\sigma_{2}N_{k}^{\triangle},\quad
z\in\hat{D}_{k}\sigma_{2},\\\ &I,\quad elsewhere.\end{aligned}\right.$ (5.54)

Make a transformation

$\widehat{m}^{\triangle}(z|\sigma_{d}^{\triangle})=m^{\triangle}(z|\sigma_{d})\Phi(z).$
(5.55)

By (5.54), we have

$\widehat{m}^{\triangle}_{+}(z|\sigma_{d}^{\triangle})=\widehat{m}^{\triangle}_{-}(z|\sigma_{d}^{\triangle})\widehat{v}(z),\quad
z\in\widehat{\Sigma}=\cup_{k\in\mathcal{N}\backslash\Lambda}\left(\partial
D_{k}\cup\partial\hat{D}_{k}\right),$ (5.56)

where $\widehat{v}(z)=\Phi(z)$, which satisfies the following estimate

$\|\widehat{v}-I\|_{L^{\infty}(\widehat{\Sigma})}=\mathcal{(}e^{-\delta_{0}t}).$
(5.57)

Take $\diamondsuit=\Lambda$, then $m^{\Lambda}(z|\sigma_{d}^{\Lambda})$ and
$\widehat{m}^{\triangle}_{+}(z|\sigma_{d}^{\triangle})$ have the same poles
and residue conditions in $\Lambda$. Therefore,

$\varepsilon(z)=\widehat{m}^{\triangle}_{+}(z|\sigma_{d}^{\triangle})[m^{\Lambda}(z|\sigma_{d}^{\Lambda})]^{-1}$
(5.58)

have no poles, and allows the following jump relation

$\varepsilon_{+}(z)=\varepsilon_{-}(z)v_{\epsilon}(z),$ (5.59)

where
$v_{\epsilon}=m^{\Lambda}\widehat{v}[m^{\Lambda}]^{-1}\sim\widehat{v},z\rightarrow\infty$,
and

$\|v_{\epsilon}-I\|_{L^{\infty}(\widehat{\Sigma})}=\mathcal{(}e^{-\delta_{0}t}).$
(5.60)

According to the properties of Small norm RH problem, we know that
$\varepsilon(z)$ exists and

$\varepsilon(z)=I+\mathcal{(}-\delta_{0}t).$ (5.61)

Finally, combine (5.55) and (5.58), we obtain

$m^{\triangle}(z|\sigma_{d}^{\triangle})=\left(I+\mathcal{O}(e^{-\delta_{0}t})\right)m^{\Lambda}(z|\sigma_{d}^{\Lambda}).$
(5.62)

∎

######  Corollary 5.2.

Denote $q_{sol}(x,t;\sigma_{d}^{\triangle})$ as the corresponding $N$-soliton
solution under scattering data
$\sigma^{\triangle}_{d}=\left\\{\left(\eta_{k},\hat{c}_{k}\right)\right\\}_{k=1}^{2N_{1}+N_{2}}$,
then

$q_{sol}(x,t;\sigma_{d}^{out})=cq_{sol}(x,t;\sigma_{d}^{\triangle})=cq_{sol}(x,t;\sigma_{d}^{\Lambda})+\mathcal{O}(e^{-\delta_{0}t})\quad
t\rightarrow\infty.$ (5.63)

####  5.1.2 A local solvable RH model near phase points

From the Proposition 5.1, we find that $v^{(2)}-I$ does not have a uniformly
small jump for $t\rightarrow\infty$ in the neighborhood $U_{\xi_{i}}$ of
$\xi_{i},i=1,2,3,4$ , therefore we establish a local model $m^{(lo)}$ which
exactly matches the jumps of $m^{(2)}_{RHP}$ on $\Sigma^{(lo)}$ for function
$E(z)$ and then it has a uniform estimate on the decay of the jump, where
$\Sigma^{(lo)}$ is defined as:

$\begin{split}&\Sigma^{(lo)}=\underset{i=1,2,3,4}{\cup}\left(L_{i}\cap
U_{\zeta_{i}}\right),\\\ &\Sigma_{i}^{(lo)}=L_{i}\cap
U_{\zeta_{i}},i=1,2,3,4,\end{split}$ (5.64)

see in Figure 9.

######  RHP 5.6.

Find a $2\times 2$ matrix-valued function $m^{(lo)}(z)$ such that

  * *

Analyticity: $m^{(lo)}(z)$ is analytic in $\mathbb{C}\backslash\Sigma^{(lo)}$.

  * *

Jump relation: $m_{+}^{(lo)}(z)=m_{-}^{(lo)}(z)v^{(2)}(z),\quad
z\in\Sigma^{(lo)}$.

  * *

Asymptotic behaviors: $m^{(lo)}(z)=I+\mathcal{O}(z^{-1}),\quad
z\rightarrow\infty$.

$\zeta_{1}$$\zeta_{2}$$\zeta_{3}$$\zeta_{4}$ Figure 9: The jump contours
$\Sigma^{(lo)}$ of $m^{(lo)}$.

This RHP exist jump relation but no poles and the analysis of it can be solved
by the so-called Beals-Coifman operator theory. Now we use the Beals-Coifman
theory to establish the relationship between $m^{(lo)}$ and
$\underset{i=1,2,3,4}{\sum}m_{i}^{(lo)}$, where $m_{i}^{(lo)}$ can be
constructed by parabolic cylinder equation.

On $\Sigma^{(lo)}_{ij},i,j=1,2,3,4$, $v^{(2)}$ allows a factorization

$\left(I-w_{ij}^{-}\right)^{-1}\left(I+w_{ij}^{+}\right),$ (5.65)
$w_{ij}^{-}=I-[v^{(2)}]^{-1}=v^{(2)}-I,\quad w_{ij}^{+}=0,$ (5.66)

and the superscript $\pm$ indicate the analyticity in the positive/negative
neighborhood of the contour.

Recall the Cauchy projection operator $C_{\pm}$ on $\Sigma^{(lo)}_{ij}$,
$i,j=1,2,3,4$

$C_{\pm}f(z)=\lim_{z\leftarrow s\in\Sigma^{(lo)}_{ij,\pm}}\frac{1}{2\pi
i}\int_{\Sigma^{(lo)}_{ij}}\frac{f(s)}{s-z}ds,$ (5.67)

we can define the Beals-Coifman operator on $\Sigma^{(lo)}_{ij}$,
$i,j=1,2,3,4$ as follows

$C_{w_{ij}}(f):=C_{+}(fw_{ij}^{-})+C_{-}(fw_{ij}^{+}).$ (5.68)

Then we define

$w_{i}=\sum_{j=1}^{4}w_{ij},\quad
w=\sum_{i=1}^{4}w_{i}=\sum_{i,j=1}^{4}w_{ij},$ (5.69)

then we obtain $C_{w}=\sum_{i=1}^{4}C_{w_{i}}=\sum_{i,j=1}^{4}C_{w_{ij}}$. Now
we introduce the following theorem, which plays a vital role in the steepest
method

######  Theorem 5.1.

If $\mu\in I+L^{2}(\Sigma)$ is the solution of the singular integral equation

$\mu=I+C_{w}(\mu),$ (5.70)

Then there exists unique solution to the RHP for $m^{(lo)}$ written as

$m^{(lo)}=I+C(\mu w).$ (5.71)

Based on the above discussions, we now try to construct the Beals-Cofiman
solution of $m^{(lo)}$. We start with the following lemma

######  Lemma 5.1.

The matrix functions $w_{ij}$ defined in (5.66) admit the following estimation

$\|w_{ij}\|_{L^{2}(\Sigma^{(lo)}_{ij})}=\mathcal{O}(t^{-1/2}).$ (5.72)

This lemma implies that $1-C_{w}$, $1-C_{w_{i}}$ and $1-C_{w_{ij}}$ exist.
Moreover, with the Theorem 5.1, the Beals-Cofiman solution for $m^{(lo)}$
exist unique as

$m^{(lo)}=I+\frac{1}{2\pi
i}\int_{\Sigma^{(lo)}}\frac{(1-C_{w})^{-1}Iw}{s-z}ds.$ (5.73)

However, the integral $I+\frac{1}{2\pi
i}\int_{\Sigma^{(lo)}}\frac{(1-C_{w})^{-1}Iw}{s-z}ds$ is still hard to
compute. Follow the standard procedure of Deift-Zhou [6], we can separate the
contributions from each saddle point. Before executing this procedure, we need
the following lemma.

######  Lemma 5.2.

As $t\rightarrow+\infty$, for $i\neq j$

$\|C_{w_{i}}C_{w_{i}}\|_{L^{2}(\Sigma^{(lo)})}=\mathcal{O}(t^{-1}),\quad\|C_{w_{i}}C_{w_{j}}\|_{L^{\infty}(\Sigma^{(lo)})\rightarrow
L^{2}(\Sigma^{(lo)})}=\mathcal{O}(t^{-1})$ (5.74)

###### Proof.

Thanks to the observation of Varzugin [32], we have

$\displaystyle 1-\sum_{i\neq
j}C_{w_{i}}C_{w_{j}}\left(1-C_{w_{j}}\right)^{-1}=\left(1-C_{w}\right)\left(1+\sum_{i=1}^{4}C_{w_{i}}\left(1-C_{w_{i}}\right)^{-1}\right),$
(5.75) $\displaystyle 1-\sum_{i\neq
j}\left(1-C_{w_{i}}\right)^{-1}C_{w_{i}}C_{w_{j}}=\left(1+\sum_{i=1}^{4}C_{w_{i}}\left(1-C_{w_{i}}\right)^{-1}\right)\left(1-C_{w}\right).$
(5.76)

Take use of Lemma 5.1, we prove this lemma. ∎

Now we can separate the contribution of Beals-Cofiman solution for $m^{(lo)}$
from each stationary phase point, which is expressed by the following
proposition.

######  Proposition 5.6.

As $t\rightarrow+\infty$

$\int_{\Sigma^{(lo)}}\frac{\left(1-C_{w}\right)^{-1}Iw}{s-z}=\sum_{i=1}^{4}\int_{\Sigma^{(lo)}_{i}}\frac{\left(1-C_{w_{i}}\right)^{-1}Iw_{i}}{s-z}+O(t^{-1}).$
(5.77)

###### Proof.

Firstly, we can decompose the resolvent $(1-C_{w})^{-1}I$ as

$(1-C_{w})^{-1}I=I+\sum_{i=1}^{4}C_{w_{j}}(1-C_{w_{i}})^{-1}I+QPRI,$ (5.78)

where

$\displaystyle Q:=1+\sum_{i=1}^{4}C_{w_{i}}(1-C_{w_{i}})^{-1},$ (5.79)
$\displaystyle P:=\left(1-\sum_{i\neq
j}C_{w_{i}}C_{w_{k}}\left(1-C_{w_{j}}\right)^{-1}\right)^{-1}$ (5.80)
$\displaystyle R:=\sum_{i\neq j}C_{w_{i}}C_{w_{j}}(1-C_{w_{j}})^{-1}$ (5.81)

By Cauchy-Schwarz inequality

$|\int
QPRIw|\leqslant\|Q\|_{L^{2}(\Sigma^{(lo)}_{i})}\|P\|_{L^{2}(\Sigma^{(lo)}_{i})}\|R\|_{L^{2}(\Sigma^{(lo)}_{i})}\|w\|_{L^{2}}\lesssim
t^{-1}.$ (5.82)

The rest of the proof is trivial. ∎

Through the above conclusion, we consider to reduce above RHP 5.6 to a model
RHP whose solution can be given explicitly in terms of parabolic cylinder
functions on every contour $\Sigma^{(lo)}_{i}$ respectively. For briefly, we
only give the details of $\Sigma_{1}^{(lo)}$ , the model of other critical
point can be constructed similar.

######  RHP 5.7.

Find a $2\times 2$ matrix-valued function $m_{1}^{(lo)}(z)$ such that

  * *

Analyticity: $m_{1}^{(lo)}(z)$ is analytic in
$\mathbb{C}\backslash\Sigma_{1}^{(lo)}$.

  * *

Jump relation: $m_{1+}^{(lo)}(z)=m_{1-}^{(lo)}(z)v_{1}^{(lo)}(z),\quad
z\in\Sigma_{1}^{(lo)}$, where

$\displaystyle v_{1}^{(lo)}(z)=\left\\{\begin{aligned}
&\begin{bmatrix}1&-\tilde{\rho}(\zeta_{1})T^{2}(\zeta_{1})(z-\zeta_{1})^{2\mathrm{i}\nu(\zeta_{1})}e^{2\mathrm{i}t\theta(z)}\\\
0&1\end{bmatrix},\quad z\in\Sigma^{(lo)}_{11},\\\ &\begin{bmatrix}1&0\\\
\frac{\rho(\zeta_{1})}{1-\rho(\zeta_{1})\tilde{\rho}(\zeta_{1})}T_{-}^{-2}(\zeta_{1})(z-\zeta_{1})^{-2\mathrm{i}\nu(\zeta_{1})}e^{-2\mathrm{i}t\theta(z)}&1\end{bmatrix},\quad
z\in\Sigma^{(lo)}_{12},\\\
&\begin{bmatrix}1&-\frac{\tilde{\rho}(\zeta_{1})}{1-\rho(\zeta_{1})\tilde{\rho}(\zeta_{1})}T_{+}^{-2}(\zeta_{1})(z-\zeta_{1})^{2\mathrm{i}\nu(\zeta_{1})}e^{2\mathrm{i}t\theta(z)}\\\
0&1\end{bmatrix},\quad z\in\Sigma^{(lo)}_{13},\\\ &\begin{bmatrix}1&0\\\
-\rho(\zeta_{1})T^{-2}(\zeta_{1})(z-\zeta_{1})^{-2\mathrm{i}\nu(\zeta_{1})}e^{-2\mathrm{i}t\theta(z)}&1\end{bmatrix},\quad
z\in\Sigma^{(lo)}_{14}.\end{aligned}\right.$ (5.83)

  * *

Asymptotic behaviors: $m^{(lo)}_{1}(z)=I+\mathcal{O}(z^{-1}),\quad
z\rightarrow\infty$.

$\Sigma^{lo,1}_{4}$$\Sigma^{lo,1}_{11}$$\Sigma^{lo,1}_{12}$$\Sigma^{lo,1}_{13}$${\rm
Re}z$$\left(\begin{array}[]{cc}1&0\\\
-\rho(\zeta_{1})T^{-2}(\zeta_{1})(z-\zeta_{1})^{-2\mathrm{i}\nu(\zeta_{1})}e^{-2\mathrm{i}t\theta(z)}&1\end{array}\right)$$\left(\begin{array}[]{cc}1&-\frac{\tilde{\rho}(\zeta_{1})}{1-\rho(\zeta_{1})\tilde{\rho}(\zeta_{1})}T_{+}^{-2}(\zeta_{1})(z-\zeta_{1})^{2\mathrm{i}\nu(\zeta_{1})}e^{2\mathrm{i}t\theta(z)}\\\
0&1\end{array}\right)$$\left(\begin{array}[]{cc}1&-\tilde{\rho}(\zeta_{1})T^{2}(\zeta_{1})(z-\zeta_{1})^{2\mathrm{i}\nu(\zeta_{1})}e^{2\mathrm{i}t\theta(z)}\\\
0&1\end{array}\right)$$\left(\begin{array}[]{cc}1&0\\\
\frac{\rho(\zeta_{1})}{1-\rho(\zeta_{1})\tilde{\rho}(\zeta_{1})}T_{-}^{-2}(\zeta_{1})(z-\zeta_{1})^{-2\mathrm{i}\nu(\zeta_{1})}e^{-2\mathrm{i}t\theta(z)}&1\end{array}\right)$$\zeta_{1}$
Figure 10: The contour $\Sigma^{(lo)}_{1}$ and the jump matrix on it.

For $z$ near $\zeta_{1}$, we have

$\theta(z)=\theta(\zeta_{1})+\theta^{{}^{\prime\prime}}(\zeta_{1})(z-\zeta_{1})^{2}+o((z-\zeta_{1})^{2}),$
(5.84)

where $\theta^{{}^{\prime\prime}}(\zeta_{1})$ is complex value and non-zero.
To match RHP 5.7 with a parabolic cylinder model, we introduce the following
scaling transformation

$u_{1}=2t^{\frac{1}{2}}\sqrt{\theta^{{}^{\prime\prime}}(\zeta_{1})}(z-\zeta_{1}).$
(5.85)

Under this scaling transformation, we define

$\tilde{\rho}_{\zeta_{1}}=\tilde{\rho}(\zeta_{1})T^{2}(\zeta_{1})e^{2\mathrm{i}t\theta(\zeta_{1})-\mathrm{i}\nu(\zeta_{1}){\rm
log}(4t\theta^{{}^{\prime\prime}}(\zeta_{1}))}.$ (5.86)

Similarly, we can define $\rho_{\zeta_{1}}$,
$\frac{\rho_{\zeta_{1}}}{1-\rho_{\zeta_{1}}\tilde{\rho}_{\zeta_{1}}}$ and
$\frac{\tilde{\rho}_{\zeta_{1}}}{1-\rho_{\zeta_{1}}\tilde{\rho}_{\zeta_{1}}}$.
Moreover, the jump matrix $v_{1}^{(lo)}(z)$ approximates to the jump of a
parabolic cylinder model problem as follow:

######  RHP 5.8.

Find a $2\times 2$ matrix-valued function $m_{1}^{(pc)}(u_{1},\xi)$ such that

  * *

Analyticity: $m_{1}^{(pc)}(u_{1},\xi)$ is analytic in
$\mathbb{C}\backslash\Sigma_{1}^{(pc)}$.

  * *

Jump relation:
$m_{1+}^{(pc)}(u_{1})=m_{1-}^{(pc)}(u_{1})v_{1}^{(pc)}(u_{1}),\quad
u_{1}\in\Sigma_{1}^{(pc)}$, where

$\displaystyle v_{1}^{(pc)}(u_{1})=\left\\{\begin{aligned}
&\begin{bmatrix}1&-\tilde{\rho}_{\zeta_{1}}u_{1}^{-2\mathrm{i}\nu(\zeta_{1})}e^{\frac{\mathrm{i}}{2}u_{1}^{2}}\\\
0&1\end{bmatrix},\quad u_{1}\in\Sigma^{(pc)}_{11},\\\ &\begin{bmatrix}1&0\\\
\frac{\rho_{\zeta_{1}}}{1-\rho_{\zeta_{1}}\tilde{\rho}_{\zeta_{1}}}u_{1}^{2\mathrm{i}\nu(\zeta_{1})}e^{-\frac{\mathrm{i}}{2}u_{1}^{2}}&1\end{bmatrix},\quad
u_{1}\in\Sigma^{(pc)}_{12},\\\
&\begin{bmatrix}1&-\frac{\tilde{\rho}_{\zeta_{1}}}{1-\rho_{\zeta_{1}}\tilde{\rho}_{\zeta_{1}}}u_{1}^{-2\mathrm{i}\nu(\zeta_{1})}e^{\frac{\mathrm{i}}{2}u_{1}^{2}}\\\
0&1\end{bmatrix},\quad u_{1}\in\Sigma^{(pc)}_{13},\\\ &\begin{bmatrix}1&0\\\
-\rho_{\zeta_{1}}u_{1}^{2\mathrm{i}\nu(\zeta_{1})}e^{-\frac{\mathrm{i}}{2}u_{1}^{2}}&1\end{bmatrix},\quad
u_{1}\in\Sigma^{(pc)}_{14}.\end{aligned}\right.$ (5.87)

  * *

Asymptotic behaviors:
$m_{1}^{(pc)}(u_{1})=I+m_{1,1}^{(pc)}(\zeta_{1})u_{1}^{-1}+\mathcal{O}(u_{1}^{-2}),\quad
z\rightarrow\infty$.

$\Sigma^{lo,1}_{4}$$\Sigma^{lo,1}_{11}$$\Sigma^{lo,1}_{12}$$\Sigma^{lo,1}_{13}$${\rm
Re}u_{1}$$\left(\begin{array}[]{cc}1&0\\\
-\tilde{\rho}_{\zeta_{1}}u_{1}^{-2\mathrm{i}\nu(\zeta_{1})}e^{\frac{\mathrm{i}}{2}u_{1}^{2}}&1\end{array}\right)$$\left(\begin{array}[]{cc}1&\frac{\rho_{\zeta_{1}}}{1-\rho_{\zeta_{1}}\tilde{\rho}_{\zeta_{1}}}u_{1}^{2\mathrm{i}\nu(\zeta_{1})}e^{-\frac{\mathrm{i}}{2}u_{1}^{2}}\\\
0&1\end{array}\right)$$\left(\begin{array}[]{cc}1&-\frac{\tilde{\rho}_{\zeta_{1}}}{1-\rho_{\zeta_{1}}\tilde{\rho}_{\zeta_{1}}}u_{1}^{-2\mathrm{i}\nu(\zeta_{1})}e^{\frac{\mathrm{i}}{2}u_{1}^{2}}\\\
0&1\end{array}\right)$$\left(\begin{array}[]{cc}1&0\\\
-\rho_{\zeta_{1}}u_{1}^{2\mathrm{i}\nu(\zeta_{1})}e^{-\frac{\mathrm{i}}{2}u_{1}^{2}}&1\end{array}\right)$$O$
Figure 11: The contour $\Sigma^{(pc)}_{1}$ and the jump matrix on it.

The solution of this RHP can be referred in as follows

$m_{1}^{(pc)}(u_{1})=I+m_{1,1}^{(pc)}(\zeta_{1})u_{1}^{-1}+\mathcal{O}(u_{1}^{-2}),$
(5.88)

where

$m_{1,1}^{(pc)}(\zeta_{1})=\begin{bmatrix}0&\beta^{\zeta_{1}}_{12}\\\
-\beta^{\zeta_{1}}_{21}&0\end{bmatrix},$ (5.89)

and

$\displaystyle\left\\{\begin{aligned}
&\beta^{\zeta_{1}}_{12}=-\frac{\sqrt{2\pi}e^{\frac{\pi}{4}\mathrm{i}}e^{-\frac{\pi}{2}}\nu}{\rho_{\zeta_{1}}\Gamma(a)},\\\
&\beta^{\zeta_{1}}_{21}=-\frac{\sqrt{2\pi}e^{-\frac{\pi}{4}\mathrm{i}}e^{-\frac{\pi}{2}}\nu(1-\rho_{\zeta_{1}}\tilde{\rho}_{\zeta_{1}})}{\rho_{\zeta_{1}}\Gamma(a)}.\end{aligned}\right.$
(5.90)

Substitute (5.85) into above consequence, we get

$m_{1}^{(lo)}(z)=I+\frac{1}{2}\frac{t^{-\frac{1}{2}}}{\sqrt{\theta^{{}^{\prime\prime}}(\zeta_{1})}(z-\zeta_{1})}m_{1,1}^{(pc)}(\zeta_{1})+\mathcal{O}(t^{-1}).$
(5.91)

Moreover, the local RHP around the other stationary phase points can be solved
by the same way:

$m_{i}^{(lo)}(z)=I+\frac{1}{2}\frac{t^{-\frac{1}{2}}}{\sqrt{\theta^{{}^{\prime\prime}}(\zeta_{i})}(z-\zeta_{i})}m_{i,1}^{(pc)}(\zeta_{i})+\mathcal{O}(t^{-1}),$
(5.92)

where

$m_{i,1}^{(pc)}(\zeta_{i})=\begin{bmatrix}0&\beta^{\zeta_{i}}_{12}\\\
-\beta^{\zeta_{i}}_{21}&0\end{bmatrix},$ (5.93)

and

$\displaystyle\left\\{\begin{aligned}
&\beta^{\zeta_{i}}_{12}=-\frac{\sqrt{2\pi}e^{\frac{\pi}{4}\mathrm{i}}e^{-\frac{\pi}{2}}\nu}{\rho_{\zeta_{i}}\Gamma(a)},\\\
&\beta^{\zeta_{i}}_{21}=-\frac{\sqrt{2\pi}e^{-\frac{\pi}{4}\mathrm{i}}e^{-\frac{\pi}{2}}\nu(1-\rho_{\zeta_{i}}\tilde{\rho}_{\zeta_{i}})}{\rho_{\zeta_{i}}\Gamma(a)},\end{aligned}\right.$
(5.94)

$i=1,2,3,4$.

Thus the solution $m^{(lo)}$ can be expressed as the following proposition as
$t\rightarrow\infty$.

######  Proposition 5.7.

As $t\rightarrow\infty$,

$m^{(lo)}(z)=I+\frac{1}{2}t^{-\frac{1}{2}}\sum_{i=1}^{4}\frac{m_{i,1}^{(pc)}(\zeta_{i})}{\sqrt{\theta^{{}^{\prime\prime}}(\zeta_{i})}(z-\zeta_{i})}+\mathcal{O}(t^{-1}).$
(5.95)

####  5.1.3 The small norm RH problem for error function

In this section, we consider the error matrix-function $E(z)$. Define

$U(\xi)=\underset{i=1,2,3,4}{\cup}U_{\zeta_{i}},$ (5.96)

the $E(z)$ allows the following RHP

######  RHP 5.9.

Find a $2\times 2$ matrix-valued function $E(z)$ such that

  * *

$E(z)$ is analytical in $\mathbb{C}\backslash\Sigma^{(E)}$, where

$\Sigma^{(E)}=\partial U(\xi)\cup\left(\Sigma^{(2)}\backslash U(\xi)\right);$
(5.97)

  * *

$E(z)$ takes continuous boundary values $E_{\pm}(z)$ on $\Sigma^{(E)}$ and

$E_{+}(z)=E_{-}(z)v^{(E)}(z),$ (5.98)

where

$v^{(E)}(z)=\left\\{\begin{aligned}
&m^{(out)}(z)v^{(2)}(z)\left[{m^{(out)}(z)}\right]^{-1},\quad
z\in\Sigma^{(2)}\backslash U(\xi),\\\
&m^{(out)}(z)m^{(lo)}(z)\left[{m^{(out)}(z)}\right]^{-1}.\quad z\in\partial
U(\xi).\end{aligned}\right.$ (5.99)

  * *

asymptotic behavior

$E(z)=I+\mathcal{O}(z^{-1}),\quad z\rightarrow\infty.$ (5.100)

${\rm Re}z$${\rm Im}z$$\zeta_{1}$$\zeta_{2}$$\zeta_{3}$$\zeta_{4}$ Figure 12:
The jump contours $\Sigma^{(E)}$ of $m^{(E)}$.

By Proposition 5.1, we have the following estimates of $v^{(E)}$:

$\displaystyle\left|v^{(E)}-I\right|=\left\\{\begin{aligned}
&\mathcal{O}(e^{-\delta_{0}t}),\quad z\in\Sigma_{2}\backslash U(\xi),\\\
&\mathcal{O}(t^{-\frac{1}{2}}),\quad z\in\partial U(\xi).\end{aligned}\right.$
(5.101)

According to Beals-Cofiman theory, we consider the trivial decomposition of
$v^{(E)}$

$v^{(E)}=b_{-}^{-1}b_{+},\quad b_{-}=I,\quad b_{+}=v^{(E)},$ (5.102)

thus,

$\begin{split}&(w_{E})_{-}=I-b_{-}=0,\quad(w_{E})_{+}=b_{+}-I=v^{(E)}-I,\\\
&w_{E}=(w_{E})_{-}+(w_{E})_{+}=v^{(E)}-I,\\\
&C_{w_{E}}f=C_{-}\left(f(w_{E})_{+}\right)+C_{+}\left(f(w_{E})_{-}\right)=C_{-}\left(f(v^{(E)}-I)\right),\end{split}$
(5.103)

where $C_{-}$ is the Cauchy projection operator

$C_{-}f(z)=\lim_{z^{{}^{\prime}}\rightarrow
z\in\Sigma^{(E)}}\frac{1}{2\pi\mathrm{i}}\int_{\Sigma^{(E)}}\frac{f(s)}{s-z^{{}^{\prime}}}ds,$
(5.104)

and $\|C_{-}\|_{L^{2}(\Sigma^{(E)})}$ is bounded. As a result, $E(z)$ in RHP
5.9 can be given by

$E(z)=I+\frac{1}{2\pi\mathrm{i}}\int_{\Sigma^{(E)}}\frac{\mu_{E}(v^{(E)}(s)-I)}{s-z^{{}^{\prime}}}ds,$
(5.105)

where $\mu_{E}\in L^{2}(\Sigma^{(E)})$, and satisfies

$(I-C_{w_{E}})\mu_{E}=I.$ (5.106)

######  Proposition 5.8.

RHP 5.9 has an unique solution $E(z)$.

###### Proof.

We can obtain the following estimate by the definition of $C_{w}$:

$\|C_{w_{E}}\|_{L^{2}(\Sigma^{(E)})}\leq\|C_{-}\|_{L^{2}(\Sigma^{(E)})}\|v^{(E)}-I\|_{l^{\infty}(\Sigma^{(E)})}\lesssim\mathcal{O}(t^{-1/2}),$
(5.107)

which implies $I-C_{w_{E}}$ is invertible for sufficiently large $t$.
Furthermore, the existence and uniqueness for $\mu$ and $E(z)$ cab be proved.
∎

In order to reconstruct the solution $q(x,t)$ of (1.1), we need the asymptotic
behavior of $E(z)$ as $z\rightarrow\infty$.

######  Proposition 5.9.

As $z\rightarrow\infty$, we have

$E(z)=I+\frac{E_{1}}{z}+\mathcal{O}(z^{-2}),$ (5.108)

where

$E_{1}=\sum_{i=1}^{4}\frac{t^{-1/2}}{2\mathrm{i}\sqrt{\theta^{{}^{\prime\prime}}(\zeta_{i})}}m^{(out)}(\zeta_{i})m^{(pc)}_{i1}[m^{(out)}(\zeta_{i})]^{-1}+\mathcal{O}(t^{-1}).$
(5.109)

###### Proof.

Recall (5.105), we have

$E_{1}=-\frac{1}{2\pi\mathrm{i}}\int_{\Sigma^{(E)}}\mu_{E}(s)(v^{(E)}(s)-I)ds=I_{1}+I_{2}+I_{3},$
(5.110)

where

$\displaystyle I_{1}=-\frac{1}{2\pi i}\oint_{\partial
U(\xi)}\mu_{E}(s)\left(v^{(E)}(s)-I\right)ds,$ (5.111) $\displaystyle
I_{2}=-\frac{1}{2\pi i}\int_{\Sigma^{(E)}\backslash
U(\xi)}\mu_{E}(s)\left(v^{(E)}(s)-I\right)ds,$ (5.112) $\displaystyle
I_{3}=-\frac{1}{2\pi i}\int_{\Sigma^{(E)}}(\mu_{E}(s)-I)({v^{(E)}}(s)-I)ds.$
(5.113)

Take use of Proposition 5.1, we can obtain $|I_{2}|=\mathcal{O}(t^{-1})$. As
for $I_{3}$, we have

$|I_{3}|\leq
c\|\mu_{E}(s)-I\|_{L^{2}(\Sigma^{(E)})}\|v^{(E)}-I\|_{L^{2}(\Sigma^{(E)})}=\mathcal{O}(t^{-1}).$
(5.114)

Finally,

$\displaystyle I_{1}$ $\displaystyle=-\frac{1}{2\pi
i}\sum_{i=1}^{4}\oint_{\partial
U_{\zeta_{i}}}m^{(out)}(s)\left(m_{i}^{(lo)}(s)-I\right){m^{(out)}(s)}^{-1}ds$
$\displaystyle=-\frac{1}{2\pi i}\sum_{i=1}^{4}\oint_{\partial
U_{\zeta_{i}}}\frac{t^{-1/2}}{2\mathrm{i}\sqrt{\theta^{{}^{\prime\prime}}(\zeta_{i})}(z-\zeta_{i})}m^{(out)}(s)m_{i,1}^{(pc)}{m^{(out)}(s)}^{-1}ds$
$\displaystyle=\sum_{i=1}^{4}\frac{t^{-1/2}}{2\mathrm{i}\sqrt{\theta^{{}^{\prime\prime}}(\zeta_{i})}}m^{(out)}(\zeta_{i})m_{i,1}^{(pc)}{m^{(out)}(\zeta_{i})}^{-1}.$
(5.115)

∎

###  5.2 Analysis on the pure $\bar{\partial}$-Problem

Now we define the function

$m^{(3)}(z)=m^{(2)}(z)(m^{(2)}_{RHP}(z))^{-1}.$ (5.116)

Then $m^{(3)}$ satisfies the following $\bar{\partial}$-Problem.

######  $\bar{\partial}$-Problem 5.1.

Find a $2\times 2$ matrix-valued function $m^{(3)}(z)$ such that

  * *

$m^{(3)}(z)$ is continuous in $\mathbb{C}$ and analytic in
$\mathbb{C}\backslash\left(\underset{i,j=1,2,3,4}{\cup}\Omega_{ij}\right)$;

  * *

asymptotic behavior

$m^{(3)}(z)=I+\mathcal{O}(z^{-1}),\quad z\rightarrow\infty;$ (5.117)

  * *

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

$\bar{\partial}m^{(3)}(z)=m^{(3)}(z)W^{(3)}(z);$ (5.118)

where
$W^{(3)}=m^{(2)}_{RHP}(z)\bar{\partial}R^{(2)}(z)(m^{(2)}_{RHP}(z))^{-1}$,

$\displaystyle W^{(3)}(z)=\left\\{\begin{aligned}
&m^{(2)}_{RHP}(z)\begin{bmatrix}0&-\bar{\partial}R_{i1}e^{2\mathrm{i}\theta}\\\
0&0\end{bmatrix}(m^{(2)}_{RHP}(z))^{-1},\quad z\in\Omega_{i1},i=1,2,3,4,\\\
&m^{(2)}_{RHP}(z)\begin{bmatrix}0&0\\\
\bar{\partial}R_{i2}e^{-2\mathrm{i}\theta}&0\end{bmatrix}(m^{(2)}_{RHP}(z))^{-1},\quad
z\in\Omega_{i2},i=1,2,3,4,\\\
&m^{(2)}_{RHP}(z)\begin{bmatrix}0&\bar{\partial}R_{i3}e^{2\mathrm{i}\theta}\\\
0&0\end{bmatrix}(m^{(2)}_{RHP}(z))^{-1},\quad z\in\Omega_{i3},i=1,2,3,4,\\\
&m^{(2)}_{RHP}(z)\begin{bmatrix}0&0\\\
\bar{\partial}R_{i4}e^{-2\mathrm{i}\theta}&0\end{bmatrix}(m^{(2)}_{RHP}(z))^{-1},\quad
z\in\Omega_{i4},i=1,2,3,4,\\\ &\begin{bmatrix}0&0\\\ 0&0\end{bmatrix},\quad
elsewhere.\end{aligned}\right.$ (5.119)

###### Proof.

The normalization condition and $\bar{\partial}$-derivative of $m^{(3)}(z)$
follow immediately from the properties of $m^{(2)}(z)$ and $m^{(2)}_{RHP}(z)$.
Then we prove the following claims.

  * Claim 1:

$m^{(3)}$ has no jumps;
Since $m^{(2)}$ and $m^{(2)}_{RHP}$ take the same jump matrix, we have

$\begin{split}[m^{(3)}_{-}]^{-1}m^{(3)}_{+}&=m^{(2)}_{RHP-}[m^{(2)}_{-}]^{-1}m^{(2)}_{+}\left[m^{(2)}_{RHP+}\right]^{-1}\\\
&=m^{(2)}_{RHP-}v^{(2)}\left[m^{(2)}_{RHP+}\right]^{-1}=I,\end{split}$ (5.120)

  * Claim 2:

$m^{(3)}$ has no singularity at $z=0$.
Near $z=0$, we have

$\left[m^{(2)}_{RHP}\right]^{-1}=\left(I+\frac{i}{z}\sigma_{3}Q_{-}\right)^{-1}=\frac{\sigma_{2}\left[m^{(2)}_{RHP}\right]^{\textrm{T}}\sigma_{2}}{1+z^{-2}}.$
(5.121)

Thus

$\lim_{z\rightarrow 0}m^{(3)}=\lim_{z\rightarrow
0}\frac{m^{(2)}\sigma_{2}\left[m^{(2)}_{RHP}\right]^{\textrm{T}}\sigma_{2}}{1-z^{-2}}=(\mathrm{i}\sigma_{3}Q_{-}\sigma_{2})^{2}=I=\mathcal{O}(1).$
(5.122)

  * Claim 3:

$m^{(3)}$ has no singularities at $z=\pm\mathrm{i}$.
This follows from observing that the symmetries of RH problem applied to the
local expansion of $m^{(2)}(z)$ and $m^{(2)}_{RHP}(z)$ imply that

$\begin{split}&m^{(2)}(z)=\begin{bmatrix}c_{1\pm}&\pm\sigma c_{1\pm}\\\
c_{2\pm}&\pm\sigma c_{2\pm}\end{bmatrix}+\mathcal{O}(z\mp\mathrm{i}),\\\
&\left[m^{(2)}_{RHP}(z)\right]^{-1}=\frac{\pm
1}{2\mathrm{i}(z\mp\mathrm{i})}\sigma_{2}\begin{bmatrix}\gamma_{1\pm}&\pm\sigma\gamma_{1\pm}\\\
\gamma_{2\pm}&\pm\sigma\gamma_{2\pm}\end{bmatrix}^{\textrm{T}}\sigma_{2}+\mathcal{O}(1),\end{split}$
(5.123)

where
$c_{j\pm}=m^{(2)}_{j1}(\pm\mathrm{i}),\gamma_{j\pm}=m^{(2)}_{RHPj1}(\pm\mathrm{i}),j=1,2$.
Taking the product it’s immediately clear the singular part of $m^{(3)}(z)$
vanishes at $z=\pm\mathrm{i}$.

  * Claim 4:

$m^{(3)}$ has no singularities at $\mathcal{Z}\cup\hat{\mathcal{Z}}$.
For $\eta\in\mathcal{Z}\cup\hat{\mathcal{Z}}$, let $\mathcal{N}_{\eta}$ denote
the nilpotent matrix which appears in the residue condition of $m^{(2)}$ and
$m^{(2)}_{RHP}$, we have Laurent expansions at $z=\eta$

$\displaystyle
m^{(2)}(z)=a(\eta)\left(\frac{\mathcal{N}_{\eta}}{z-\eta}+I\right)+\mathcal{O}\left(\left(z-\eta\right)\right),$
(5.124) $\displaystyle
m^{(2)}_{RHP}(z)=A(\eta)\left(\frac{\mathcal{N}_{\eta}}{z-\eta}+I\right)+\mathcal{O}\left(\left(z-\eta\right)\right),$
(5.125)

then, we have

$m^{(3)}=m^{(2)}\left[m^{(2)}_{RHP}\right]^{-1}=\left[a(\eta)\left(\frac{\mathcal{N}_{\eta}}{z-\eta}+I\right)\right]\left[\left(-\frac{\mathcal{N}_{\eta}}{z-\eta}+I\right)\sigma_{2}A^{\rm
T}(\eta)\sigma_{2}\right]=\mathcal{O}(1).$ (5.126)

∎

Now we consider the long time asymptotic behavior of $m^{(3)}$. The solution
of $\bar{\partial}$-Problem 5.1 can be solved by the following integral
equation

$m^{(3)}(z)=I-\frac{1}{\pi}\iint_{\mathbb{C}}\frac{m^{(3)}(s)W^{(3)}(s)}{s-z}dA(s),$
(5.127)

where $A(s)$ is the Lebesgue measure on $\mathbb{C}$. Denote $S$ as the
Cauchy-Green integral operator

$S[f](z)=-\frac{1}{\pi}\iint\frac{f(s)W^{(3)}(s)}{s-z}dA(s),$ (5.128)

then (5.127) can be written as the following operator equation

$(1-S)m^{(3)}(z)=I.$ (5.129)

To prove the existence of the operator at large time, we present the following
lemma.

######  Lemma 5.3.

The norm of the integral operator $S$ decay to zero as $t\rightarrow\infty$,
and

$\|S\|_{L^{\infty}\rightarrow L^{\infty}}\leq ct^{-1/4}.$ (5.130)

###### Proof.

Taking $z\in\Omega_{11}$ for example, the other cases are similar. Denote

$s=u+\mathrm{i}v-\zeta_{1}=(u-a_{1})+\mathrm{i}(v-b_{1}),\quad
z=\alpha+\mathrm{i}\eta,$ (5.131)

then$\left|e^{(2\mathrm{i}t\theta)}\right|=e^{{\rm
Re}(2\mathrm{i}t\theta)}\leq e^{-t(v-b_{1})}$ and

$\begin{split}\left\|\frac{1}{s-z}\right\|^{2}_{L^{2}(a_{1}+b_{1}-v,2a_{1})}&=\int_{a_{1}+b_{1}-v}^{2a_{1}}\frac{1}{|s-z|^{2}}du\leq\int_{0}^{+\infty}\frac{1}{|s-z|^{2}}\\\
&=\int_{0}^{+\infty}\frac{1}{(u-a_{1}-\alpha)^{2}+(v-b-\eta)^{2}}du\\\
&=\frac{1}{|v-b_{1}-\eta|}\int_{0}^{+\infty}\frac{1}{1+y^{2}}dy\\\
&=\frac{\pi}{2|v-b_{1}-\eta|},\end{split}$ (5.132)

where $y=\frac{u-a_{1}-\alpha}{v-b_{1}-\eta}$. Therefore, we obtain

$\begin{split}\left|S(f)\right|&\leq\frac{1}{\pi}\iint_{\Omega_{11}}\frac{\left|fm^{(2)}_{RHP}\bar{\partial}\mathcal{R}^{(2)}[m^{(2)}_{RHP}]^{-1}\right|}{|s-z|}dA(s)\\\
&\lesssim\iint_{\Omega_{11}}\frac{\bar{\partial}\chi_{\mathcal{Z}}(z)e^{2\mathrm{i}t\theta}}{|s-z|}dA(s)\\\
&\lesssim\int_{b_{1}}^{+\infty}\left\|\frac{1}{s-z}\right\|^{2}_{L^{2}(0,+\infty)}e^{-t(v-b_{1})}dv\\\
&\lesssim\int_{b_{1}}^{+\infty}\frac{e^{-t(v-b_{1})}}{\sqrt{v-b_{1}-\eta}}dv\\\
&=\int_{0}^{\eta}\frac{e^{-t\kappa}}{\sqrt{\kappa-\eta}}d\kappa+\int_{\eta}^{+\infty}\frac{e^{-t\kappa}}{\sqrt{\kappa-\eta}}dv,\end{split}$
(5.133)

where

$\int_{0}^{\eta}\frac{e^{-t\kappa}}{\sqrt{\kappa-\eta}}d\kappa=\int_{0}^{1}\frac{\sqrt{\eta}e^{-tw\eta}}{\sqrt{w-1}}dw\lesssim
t^{-1/4}\int_{0}^{1}\frac{1}{\sqrt{\sqrt{w}(w-1)}}dw\lesssim t^{-1/4},$
(5.134)
$\int_{\eta}^{+\infty}\frac{e^{-t\kappa}}{\sqrt{\kappa-\eta}}d\kappa\lesssim\int_{0}^{+\infty}\frac{e^{-tw}}{\sqrt{w}}dw=t^{-1/2}\int_{0}^{+\infty}\frac{e^{-\lambda}}{\lambda}d\lambda\lesssim
t^{-1/2}.$ (5.135)

∎

Based on the above discussion, we have the following proposition.

######  Proposition 5.10.

As $t\rightarrow\infty$, $(I-S)^{-1}$ exists, which implies $\bar{\partial}$
Problem 5.1 has an unique solution.

Aim at the asymptotic behavior of $m^{(3)}$, we make the asymptotic expansion

$m^{(3)}=I+z^{-1}m^{(3)}_{1}(x,t)+\mathcal{O}(z^{-2}),\quad{\rm as}\quad
z\rightarrow\infty$ (5.136)

where

$m^{(3)}_{1}(x,t)=\frac{1}{\pi}\iint_{\mathbb{C}}m^{(3)}(s)W^{(3)}(s)dA(s).$
(5.137)

To recover the solution of (1.1), we shall discuss the asymptotic behavior of
$m^{(3)}_{1}(x,t)$, thus we have the following proposition.

######  Proposition 5.11.

As $t\rightarrow\infty$,

$|m^{(3)}_{1}(x,t)|\lesssim t^{-1}.$ (5.138)

###### Proof.

$\begin{split}\left|m_{1}^{(3)}\right|&\leq\frac{1}{\pi}\iint_{\Omega_{11}}\left|m^{(2)}_{RHP}\bar{\partial}\mathcal{R}^{(2)}[m^{(2)}_{RHP}]^{-1}\right|dA(s)\\\
&\lesssim\int_{2a_{1}}^{b_{1}+1}\int_{a_{1}+b_{1}-v}^{2a_{1}}|\bar{\partial}\chi_{\mathcal{Z}}(z)|e^{-t(v-b_{1})}dudv\\\
&\lesssim t^{-1}.\end{split}$ (5.139)

∎

##  6 Long time asymptotics for nonlocal mKdV equation

######  Proposition 6.1.

For giving reflectionless scattering data

$\sigma_{d}^{\Lambda}=\\{(\eta_{k},\hat{c}_{k}),k\in\Lambda\\},$ (6.1)

its corresponding $N(\Lambda)$-solution for nonlocal mKdV equation (1.1)
allows the following long time asymptotics

$q(x,t)=cq_{sol}(x,t;\sigma_{d}^{\Lambda})-t^{-1/2}f+\mathcal{O}(t^{-1}),\quad
t\rightarrow\infty,$ (6.2)

where

$f=\sum_{i=1}^{4}\frac{1}{2\sqrt{\theta^{{}^{\prime\prime}}(\zeta_{i})}}\left(m_{11}^{2}(\zeta_{i})\beta_{12}^{\zeta_{i}}+m_{12}^{2}(\zeta_{i})\beta_{21}^{\zeta_{i}}\right).$
(6.3)

###### Proof.

Recall all the transformations for $m(x,t;z)$, we obtain

$m(z)=T(\infty)^{-\sigma_{3}}m^{(3)}(z)E(z)m^{(out)}(z)\mathcal{R}^{(2)}(z)^{-1}T(z)^{\sigma_{3}},$
(6.4)

then for $z\rightarrow\infty$ out
$\underset{j=1,2,3,4}{\cup}\overline{\Omega}_{ij}$, $i=1,2,\cdot\cdot\cdot,6,$

$m=\left(I+\frac{m_{1}^{(3)}}{z}+\cdot\cdot\cdot\right)\left(I+\frac{E_{1}}{z}+\cdot\cdot\cdot\right)\left(I+\frac{m_{1}^{(out)}}{z}+\cdot\cdot\cdot\right),$
(6.5)

obviously,

$m_{1}=m_{1}^{(out)}+m_{1}^{(3)}+E_{1}.$ (6.6)

Take use the potential recovering formulae (2.72), we have

$\begin{split}q(x,t)&=-\mathrm{i}\left[m_{1}^{(out)}+m_{1}^{(3)}+E_{1}\right]_{12}\\\
&=cq_{sol}(x,t;\sigma_{d}^{\Lambda})-t^{-1/2}f+\mathcal{O}(t^{-1}).\end{split}$
(6.7)

∎

Acknowledgements

This work is supported by the National Natural Science Foundation of China
(Grant No. 11671095, 51879045).

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