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

#  Long time asymptotic behavior for the derivative Schr$\ddot{o}$dinger
equation with nonzero boundary conditions

Yiling YANG1, Qiaoyuan CHENG1 and Engui FAN1 Corresponding author and email
address<EMAIL_ADDRESS>
###### Abstract

In this paper, we apply $\overline{\partial}$ steepest descent method to study
the Cauchy problem for the derivative nonlinear Schr$\ddot{o}$dinger equation
with nonzero boundary conditions

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

where $\lim_{x\to\pm\infty}q_{0}(x)=q_{\pm},\ \ |q_{\pm}|=1$. Based on the
spectral analysis of the Lax pair, we express the solution of the derivative
Schr$\ddot{o}$dinger equation in terms of solutions of a Riemann-Hilbert
problem. In a fixed space-time solitonic region $-3<x/t<-1$, we compute the
long time asymptotic expansion of the solution $q(x,t)$, which implies soliton
resolution conjecture and can be characterized with an $N(\Lambda)$-soliton
whose parameters are modulated by a sum of localized soliton-soliton
interactions as one moves through the region; the residual error order
$\mathcal{O}(t^{-3/4})$ from a $\overline{\partial}$ equation.
Keywords: Derivative Schr$\ddot{o}$dinger equation, Riemann-Hilbert problem,
$\overline{\partial}$ steepest descent method, soliton resolution, asymptotic
stability.
AMS: 35Q51; 35Q15; 37K15; 35C20.

###### Contents

  1. 1 Introduction
  2. 2 The spectral analysis and a RH problem
  3. 3 Deformation to a mixed $\bar{\partial}$-RH problem
  4. 4 Mixed $\bar{\partial}$-RH Problem
  5. 5  Decomposition of the mixed $\bar{\partial}$-RH problem
  6. 6  Asymptotic of $\mathcal{N}(\Lambda)$-soliton solutions
  7. 7 The small norm RH problem for error function
  8. 8 Analysis on the pure $\bar{\partial}$-Problem
  9. 9 Asymptotic for the DNLS equation

##  1 Introduction

The study on the long-time behavior of nonlinear wave equations which is
solvable by the inverse scattering method was first carried out by Manakov in
1974 [1]. By using this method, Zakharov and Manakov give the first result for
large-time asymptotic of solutions for the NLS equation with decaying initial
data [2]. The inverse scattering method also worked for long-time behavior of
integrable systems such as KdV, Landau-Lifshitz and the reduced Maxwell-Bloch
system [3, 4, 5]. In 1993, Deift and Zhou developed a nonlinear steepest
descent method to rigorously obtain the long-time asymptotics behavior of the
solution for the MKdV equation by deforming contours to reduce the original
Riemann-Hilbert (RH) problem to a model one whose solution is calculated in
terms of parabolic cylinder functions [6]. Since then this method has been
widely applied to the focusing NLS equation, KdV equation, Fokas-Lenells
equation, short-pulse equation and Camassa-Holm equation etc. [7, 8, 9, 10,
11, 12].

In recent years, McLaughlin and Miller further presented a $\bar{\partial}$
steepest descent method which combine steepest descent with
$\bar{\partial}$-problem rather than the asymptotic analysis of singular
integrals on contours to analyze asymptotic of orthogonal polynomials with
non-analytical weights [13, 14]. 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
which can be solved by being recast into an integral equation and by using
Neumann series. Dieng and McLaughin used it to study the defocusing NLS
equation under essentially minimal regularity assumptions on finite mass
initial data [15]; This $\bar{\partial}$ steepest descent method was also
successfully applied to prove asymptotic stability of N-soliton solutions to
focusing NLS equation [16]; Jenkins et.al studied soliton resolution for the
derivative nonlinear NLS equation for generic initial data in a weighted
Sobolev space [17]. Their work provided the soliton resolution property for
derivative NLS equation, which decomposes the solution into the sum of a
finite number of separated solitons and a radiative parts when $t\to\infty$.
And the dispersive part contains two components, one coming from the
continuous spectrum and another from the interaction of the discrete and
continuous spectrum. For finite density initial data, Cussagna and Jenkins
studied the defocusing NLS equation [18].

In this paper, we study the long time asymptotic behavior for the derivative
nonlinear Schr$\ddot{o}$dinger (DNLS) equation with nonzero boundary
conditions

$\displaystyle iq_{t}+q_{xx}+i\sigma(|q|^{2}q)_{x}=0,$ (1.1) $\displaystyle
q(x,0)=q_{0}(x),\hskip 14.22636pt$ (1.2)

where $\lim_{x\to\pm\infty}q_{0}(x)=q_{\pm},\ \ |q_{\pm}|=1$. Since the
solution space of the equation (1.1) with $\sigma=1$ and $\sigma=-1$ by the
simple mapping $q(x,t)\to q(-x,t)$, we only need to consider the case
$\sigma=-1$ in our paper. The DNLS equation as a completely integrable system
was first proposed by Kaup and Newell [19].

The DNLS equation is often used to describe various nonlinear waves. For
instance, DNLS equation governs the evolution of small but finite amplitude
nonlinear Alfv$\acute{e}$n waves which propagates quasi-parallel to the
magnetic field in space plasma physics [20, 21, 22, 23, 24], sub-picosecond
pulses in single mode optical fibers [26, 27, 25]. Moreover, DNLS equation
also describe weak nonlinear electromagnetic waves in ferromagnetic [28],
dielectric [29] and anti-ferromagnetic systems under external magnetic fields
[30]. Either zero boundary conditions or nonzero boundary conditions for the
DNLS equation have well physically significant. For problems of nonlinear
Alfve waves, weak nonlinear electromagnetic waves in magnetic and dielectric
media, waves propagating strictly parallel to the ambient magnetic fields are
modeled by zero boundary conditions, while those oblique waves are modeled by
the nonzero boundary conditions. In optical fibers, pulses under bright
background waves are modeled by the zero boundary conditions. Much work on the
DNLS equation were also developed in [31, 32, 33, 34, 35].

Zhang and Yan presented the inverse scattering transform of the DNLS equation
(1.1) for both zero/nonzero boundary conditions in terms of the matrix RH
problems [36]. For Schwartz initial value $q_{0}\in\mathcal{S}(\mathbb{R})$,
Xu and Fan derived the long-time asymptotic for (1.1) without soliton [37]

$q(x,t)=t^{-\frac{1}{2}}\alpha(\lambda_{0})e^{\frac{ix^{2}}{4t}-i\nu(\lambda_{0})\log
t}+\mathcal{O}(t^{-1}\log t).$ (1.3)

The long-time asymptotic for (1.1) with step-like initial data was further
investigated [38]. Recently for generic initial data in $H^{2,2}(\mathbb{R})$,
applying $\bar{\partial}$ steepest descent method, Jenkins et al obtained the
following asymptotics for the equation (1.1) [17]

$q(x,t)=q_{sol}(x,t;D_{I})+t^{-\frac{1}{2}}f(x,t)+\mathcal{O}(t^{-\frac{3}{4}}),$
(1.4)

where $q_{sol}(x,t;D_{I})$ is the soliton solutions of the equation (1.1) with
modulating reflectionless scattering data. In our paper, for finite density
initial data $q_{0}-q_{\pm}\in H^{1,1}(\mathbb{R})$, we apply $\bar{\partial}$
steepest descent method to obtain the following long-time asymptotic of the
DNLS equation (1.1)

$\displaystyle
q(x,t)=exp\left\\{\frac{i}{2}\int_{-\infty}^{x}(|q^{r}_{\Lambda}(x,t)|^{2}-1)dy\right\\}T(\infty)^{-2}q^{r}_{\Lambda}(x,t)+\mathcal{O}(t^{-3/4}).$
(1.5)

This paper is arranged as follows. In section 2, we recall some main results
on the construction process of the RH problem with respect to the initial
problem of the DNLS equation (1.1) obtained in [33, 36], which will be used to
analyze long-time asymptotics of the DNLS equation in our paper. In section 3,
we introduce a function $T(z)$ to define a new RH problem for $M^{(1)}(z)$,
which admits a regular discrete spectrum and two triangular decompositions of
the jump matrix. In section 4, by introducing a matrix-valued function $R(z)$,
we obtain a mixed $\bar{\partial}$-RH problem for $M^{(2)}(z)$ by continuous
extension of $M^{(1)}(z)$. In section 5, we decompose $M^{(2)}(z)$ into a
model RH problem for $M^{(r)}(z)$ and a pure $\bar{\partial}$ Problem for
$M^{(3)}(z)$. The $M^{r}(z)$ can be obtained via an modified reflectionless RH
problem $M^{(r)}_{\Lambda}(z)$ for the soliton components which is solved in
Section 6. In section 7, the error function $E(z)$ between $M^{(r)}(z)$ and
$M^{(r)}_{\Lambda}(z)$ can be computed with a small-norm RH problem. In
Section 8, we analyze the $\bar{\partial}$-problem for $M^{(3)}(z)$. Finally,
in Section 9, based on the result obtained above, a relation formula is found

$\displaystyle
M(z)=T(\infty)^{\sigma_{3}}M^{(3)}(z)E(z)M^{(r)}_{\Lambda}(z)R^{(2)}(z)^{-1}T(z)^{-\sigma_{3}},$

from which we then obtain the long-time asymptotic behavior for the DNLS
equation (1.1) via a reconstruction formula.

##  2 The spectral analysis and a RH problem

The DNLS equation (1.1) is completely integrable and admits the Lax pair [19]

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

while

$X=ik^{2}\sigma_{3}+kQ,$ $T=-\left(2k^{2}+Q^{2}\right)X-ikQ_{x}\sigma_{3},$

where $k\in\mathbb{C}$ is a spectral parameter and

$Q=\left(\begin{array}[]{cc}0&q\\\ -\bar{q}&0\end{array}\right),\ \
\sigma_{3}=\left(\begin{array}[]{cc}1&0\\\ 0&-1\end{array}\right).$ (2.2)

By using the boundary condition (1.1), the Lax pair (2.1) becomes

$\Phi_{\pm,x}\sim X_{\pm}\Phi_{\pm},\hskip 14.22636pt\Phi_{\pm,t}\sim
T_{\pm}\Phi_{\pm},\ \ \ \ x\rightarrow\pm\infty,$ (2.3)

where

$X_{\pm}=ik^{2}\sigma_{3}+kQ_{\pm},\hskip
14.22636ptT_{\pm}=-\left(2k^{2}-1\right)X_{\pm},$ (2.4)

and

$Q_{\pm}=\left(\begin{array}[]{cc}0&q_{\pm}\\\
-\bar{q}_{\pm}&0\end{array}\right).$

The eigenvalues of the matrix $X_{\pm}$ are $\pm ik\lambda$, which satisfy

$\lambda^{2}=k^{2}+1.$ (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

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

Define two domains $D^{+}$, $D^{-}$ and their boundary $\Sigma$ on $z$-plane
by

$\displaystyle D^{-}=\\{z:{\rm Re}z{\rm Im}z<0\\},\hskip
14.22636ptD^{+}=\\{z:{\rm Re}z{\rm Im}z>0\\},$ $\displaystyle\Sigma=\\{z:{\rm
Re}z{\rm Im}z=0\\}=\mathbb{R}\cup i\mathbb{R}\backslash\\{0\\},$

which are shown in Figure 1.

$\mathbb{R}$$i\mathbb{R}$$D^{+}$$D_{-}$$D_{-}$$D^{+}$$0$ Figure 1:  The
domains $D^{-}$, $D^{+}$ and boundary $\Sigma=\mathbb{R}\cup
i\mathbb{R}\backslash\\{0\\}$.

We derive the solution of the asymptotic spectral problem (2.3)

$\Phi_{\pm}\sim Y_{\pm}e^{ik(z)\lambda(z)x\sigma_{3}},$ (2.8)

where

$Y_{\pm}=\left(\begin{array}[]{cc}1&\frac{iq_{\pm}}{z}\\\
\frac{i\bar{q}_{\pm}}{z}&1\end{array}\right).$

By making transformation

$\mu_{\pm}=\Phi_{\pm}e^{-ik\lambda x\sigma_{3}},$ (2.9)

then we have

$\displaystyle\mu_{\pm}\sim Y_{\pm},\hskip 14.22636ptx\rightarrow\pm\infty,$
$\displaystyle\det[\Phi_{\pm}]=\det[\mu_{\pm}]=\det[Y_{\pm}]=1+z^{-2},$

and $\mu_{\pm}$ satisfy the Volterra integral equations

$\displaystyle\mu_{\pm}(z)=Y_{\pm}+\int_{\pm\infty}^{x}Y_{\pm}e^{ik\lambda(x-y){\hat{\sigma}}_{3}}[Y_{\pm}^{-1}\Delta
X_{\pm}\mu_{\pm}(z)]dy,\ \ \ z\not=\pm i,\ \ $ (2.10)
$\displaystyle\mu_{\pm}(z)=Y_{\pm}+\int_{\pm\infty}^{x}\left[I+(x-y)X_{\pm}(z)\right]\Delta
X_{\pm}\mu_{\pm}(z)dy,\ \ \ z=\pm i,$ (2.11)

where $\Delta X_{\pm}=k\left(Q-Q_{\pm}\right)$.

It can be shown that the eigenfunction $\mu_{\pm}$ admit symmetry [36].

######  Proposition 1.

Jost functions admit two reduction conditions on the $z$-plane:

The first symmetry reduction:

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

The second symmetry reduction:

$\mu_{\pm}(z)=\frac{i}{z}\mu_{\pm}(-z^{-1})\sigma_{3}Q_{\pm}.$ (2.13)

And for $z\in\Sigma^{0}=\Sigma\setminus\left\\{\pm i\right\\}$, there exist
scattering matrix which is a linear relation between $\Phi_{+}$ and $\Phi_{-}$

$\Phi_{+}(x,t,z)=\Phi_{-}(x,t,z)S(z),$ (2.14)

where

$S(z)=\left(\begin{array}[]{cc}a(z)&-\overline{b(\bar{z})}\\\\[4.0pt]
b(z)&\overline{a(\bar{z})}\end{array}\right),\hskip 14.22636pt\det[S(z)]=1$
(2.15)

with symmetry reduction:

$S(z)=\sigma_{1}\overline{S(-\bar{z})}\sigma_{1}=(\sigma_{3}Q_{-})^{-1}S\left(-z^{-1}\right)\sigma_{3}Q_{+}.$
(2.16)

And the reflection coefficients are defined by

$\rho(z)=\frac{b(z)}{a(z)},\hskip
14.22636pt\tilde{\rho}(z)=-\overline{\rho(\bar{z})},$ (2.17)

with symmetry reduction:

$\rho(z)=\overline{\tilde{\rho}(-\bar{z})}=\frac{\bar{q}_{-}}{q_{-}}\tilde{\rho}(-z^{-1}).$
(2.18)

Then

$\displaystyle a(z)=\frac{{\rm
Wr}(\Phi_{+}^{1},\Phi_{-}^{2})}{1+z^{-2}},\hskip 14.22636ptb(z)=\frac{{\rm
Wr}(\Phi_{-}^{1},\Phi_{+}^{1})}{1+z^{-2}}.$ (2.19)

Although $a(z)$ and $b(z)$ has singularities at points $\pm i$, $|\rho(\pm
i)|=1$. The uniqueness and existences of Lax pair from [36]:

######  Proposition 2.

If $q-q_{\pm}\in L^{1,1}(\mathbb{R}_{\pm})$, the fundamental eigenfunctions
$\mu_{\pm}$ defined by (2.10) and (2.11) exist and is the unique. Define
$\mu_{\pm}=(\mu_{\pm}^{1},\mu_{\pm}^{2})$ with $\mu_{\pm}^{1}$ and
$\mu_{\pm}^{2}$ denoting the first and second column of $\mu_{\pm}$
respectively. Then $\mu_{+}^{1}$and $\mu_{-}^{2}$ are analytical on the
$D^{+}$ , and continuous in $\overline{D^{+}}$; $\mu_{-}^{1}$ and
$\mu_{+}^{2}$ are analytical on the $D^{-}$, and continuous in
$\overline{D^{-}}$. Moreover, form (2.19), $a(z)$ is analytical on the $D^{+}$
, and continuous in $\overline{D^{+}}\setminus\left\\{\pm i\right\\}$.
Further, $\lambda a(z)$ is analytical on the $D^{+}$ , and continuous in
$\overline{D^{+}}$. $b(z)$ and $\lambda b(z)$ are continuous in $\Sigma^{0}$
and $\Sigma$ respectively.

The zeros of $a(z)$ on $\Sigma$ are known to occur and they correspond to
spectral singularities. They are excluded from our analysis in the this paper.
To deal with our following work, we assume our initial data satisfy this
assumption.

######  Assumption 1.

The initial data $q-q_{\pm}(x)\in L^{1,1}(\mathbb{R}^{\pm})$ and it generates
generic scattering data which satisfy that

1\. a(z) has no zeros on $\Sigma$.

2\. a(z) only has finite number of simple zeros.

3\.  $\rho(z)$ and $\tilde{\rho}(z)$ belong to $W^{2,\infty}(\Sigma)\cap
W^{1,2}(\Sigma)$.

Suppose that $a(z)$ has $N_{1}$ simple zeros $z_{1},...,z_{N_{1}}$ on
$D^{+}\cap\\{z\in\mathbb{C}:{\rm Im}z>0,|z|>1\\}$, and $N_{2}$ simple zeros
$w_{1},...,w_{m}$ on the circle
$\\{z=e^{i\varphi}:0<\varphi<\frac{\pi}{2}\\}$. The symmetries (2.16) imply
that

$\displaystyle a(\pm
z_{n})=0\Longleftrightarrow\overline{a(\pm\bar{z}_{n})}=0\Longleftrightarrow\overline{a\left(\pm\frac{1}{z_{n}}\right)}=0$
$\displaystyle\Longleftrightarrow a\left(\pm\frac{1}{\bar{z}}\right)=0,\hskip
14.22636ptn=1,...,N_{1},$

and on the circle

$a(\pm w_{m})=0\Longleftrightarrow\overline{a(\pm\bar{w}_{m})}=0,\hskip
14.22636ptm=1,...,N_{2}.$

So the zeros of $a(z)$ come in pairs. It is convenient to define
$\zeta_{n}=z_{n}$, $\zeta_{n+N_{1}}=-z_{n}$,
$\zeta_{n+2N_{1}}=\bar{z}_{n}^{-1}$ and $\zeta_{n+3N_{1}}=-\bar{z}_{n}^{-1}$
for $n=1,\cdot\cdot\cdot,N_{1}$; $\zeta_{m+4N_{1}}=w_{m}$ and
$\zeta_{m+4N_{1}+N_{2}}=-w_{m}$ for $m=1,\cdot\cdot\cdot,N_{2}$. Therefore,
the discrete spectrum is

$\mathcal{Z}=\left\\{\zeta_{n},\
\bar{\zeta}_{n}\right\\}_{n=1}^{4N_{1}+2N_{2}},$ (2.20)

with $\zeta_{n}\in D^{+}$ and $\bar{\zeta}_{n}\in D^{-}$. And the distribution
of $\mathcal{Z}$ on the $z$-plane is shown in Figure 2.

Re$z$Im$z$$z_{n}$$\bar{z}_{n}$$-\frac{1}{z_{n}}$$-\frac{1}{\bar{z}_{n}}$$\frac{1}{\bar{z}_{n}}$$\frac{1}{z_{n}}$$-\bar{z}_{n}$$-z_{n}$$w_{m}$$\bar{w}_{m}$$-\bar{w}_{m}$$-w_{m}$
Figure 2: Distribution of the discrete spectrum $\mathcal{Z}$. The red one is
unit circle.

As shown in [33], denote norming constant $c_{n}=b_{n}/a^{\prime}(z_{n})$.
Then we have residue conditions as

$\displaystyle\operatorname*{Res}_{z=\pm
z_{n}}\left[\frac{\mu^{1}_{+}(z)}{a(z)}\right]=c_{n}e^{-2ik(\pm
z_{n})\lambda(\pm z_{n})x}\mu^{2}_{-}(\pm z_{n}),$ (2.21)
$\displaystyle\operatorname*{Res}_{z=\pm\bar{z}_{n}^{-1}}\left[\frac{\mu^{1}_{+}(z)}{a(z)}\right]=\pm\frac{\bar{q}_{-}}{q_{-}}\bar{z}_{n}^{-2}\bar{c}_{n}e^{-2ik(\pm\bar{z}_{n}^{-1})\lambda(\pm\bar{z}_{n}^{-1})x}\mu^{2}_{-}(\pm\bar{z}_{n}^{-1}),$
(2.22)
$\displaystyle\operatorname*{Res}_{z=\pm\bar{z}_{n}}\left[\frac{\mu^{2}_{+}(z)}{\overline{a(\bar{z})}}\right]=-\bar{c}_{n}e^{2ik(\pm\bar{z}_{n})\lambda(\pm\bar{z}_{n})x}\mu^{1}_{-}(\pm\bar{z}_{n}),$
(2.23) $\displaystyle\operatorname*{Res}_{z=\pm
z_{n}^{-1}}\left[\frac{\mu^{2}_{+}(z)}{\overline{a(\bar{z})}}\right]=\pm\frac{q_{-}}{\bar{q}_{-}}z_{n}^{-2}c_{n}e^{-2ik(\pm\bar{z}_{n}^{-1})\lambda(\pm
z_{n}^{-1})x}\mu^{2}_{-}(\pm z_{n}^{-1}).$ (2.24)

For $m=1,...,N_{2}$, there also have
$c_{N_{1}+m}=b_{N_{1}+m}/a^{\prime}(w_{m})$ and

$\displaystyle\operatorname*{Res}_{z=\pm
w_{m}}\left[\frac{\mu^{1}_{+}(z)}{a(z)}\right]=c_{N_{1}+m}e^{-2ik(\pm
w_{m})\lambda(\pm w_{m})x}\mu^{2}_{-}(\pm w_{m}),$ (2.25)
$\displaystyle\operatorname*{Res}_{z=\pm\bar{w}_{m}}\left[\frac{\mu^{2}_{+}(z)}{\overline{a(\bar{z})}}\right]=-\bar{c}_{N_{1}+m}e^{2ik(\pm\bar{w}_{m})\lambda(\pm\bar{w}_{m})x}\mu^{1}_{-}(\pm\bar{w}_{m}).$
(2.26)

For brevity, we introduce a new constant $C_{n}$ as: for $n=1,...,N_{1}$,
$C_{n}=C_{n+N_{1}}=c_{n}$,
$C_{n+2N_{1}}=-C_{n+3N_{1}}=\frac{\bar{q}_{-}}{q_{-}}\bar{z}_{n}^{-2}\bar{c}_{n}$;
for $m=1,...,N_{2}$, $C_{m+4N_{1}}=C_{m+4N_{1}+N_{2}}=c_{m+N_{1}}$, and the
collection $\sigma_{d}=\left\\{\zeta_{n},C_{n}\right\\}^{4N_{1}+2N_{2}}_{n=1}$
is called the _scattering data_.

Now we are going to take into account the time evolution of scattering data.
If $q$ also depends on t (i.e. $q$ = $q(x,t))$, we can obtain the functions
$a$ and $b$ as above for all times $t\in R$. Taking account of the t-part in
(2.1), the t- derivative of $a$ and $b$ comes to

$\displaystyle a_{t}(z;t)=0,\hskip 14.22636ptb_{t}(z;t)=-(2k^{2}-1)k\lambda
b(z;t).$ (2.27)

Then we can obtain time dependence of scattering data which can be expressed
as the following replacement

$\displaystyle C(\zeta_{n})\rightarrow
C(t,\zeta_{n})=c(0,\zeta_{n})e^{-(2k(\zeta_{n})^{2}-1)k(\zeta_{n})\lambda(\zeta_{n})t},$
(2.28) $\displaystyle r(z)\rightarrow r(t,z)=r(0,z)e^{-(2k^{2}-1)k\lambda t}$
(2.29)

In particular, if at time $t=0$ the initial function $q(x,0)$ produces
$4N_{1}+2N_{2}$ simple zeros $\zeta_{1}$,…,$\zeta_{4N_{1}+2N_{2}}$ of $a(z;0)$
and if $q$ evolves accordingly to the (1.1), then $q(x,t)$ will produce
exactly the same N simple zeros at any other time $t\in R$. The scattering
data with time $t$ is given by

$\left\\{e^{-(2k^{2}-1)k\lambda
t}r(z),\left\\{\zeta_{n},e^{-(2k(\zeta_{n})^{2}-1)k(\zeta_{n})\lambda(\zeta_{n})t}C_{n}\right\\}^{4N_{1}+2N_{2}}_{n=1}\right\\},$

where
$\left\\{r(z),\left\\{\zeta_{n},C_{n}\right\\}^{4N_{1}+2N_{2}}_{n=1}\right\\}$
are obtained from the initial data $q(x,0)=q_{0}(x)$. Denote the phase
function

$\theta(z)=k(z)\lambda(z)\left[x/t-(2k(z)^{2}-1)\right],$ (2.30)

and for convenience we denote $\theta_{n}=\theta(\zeta_{n})$.

To propose and solve the matrix RH problem in the following inverse problem,
we finally give the asymptotic behaviors of the modified Jost solutions and
scattering matrix as $z\to\infty$ and $z\to 0$.

######  Proposition 3.

The Jost solutions posses the following asymptotic behaviors

$\displaystyle\mu_{\pm}(x,t,z)=e^{i\nu_{\pm}(x,t;q)\sigma_{3}}+O(z^{-1}),\hskip
14.22636ptz\rightarrow\infty,$ (2.31)
$\displaystyle\mu_{\pm}(x,t,z)=\frac{i}{z}e^{i\nu_{\pm}(x,t;q)\sigma_{3}}\sigma_{3}Q_{\pm}+O(1),\hskip
14.22636ptz\rightarrow 0,$ (2.32)

where

$\nu_{\pm}(x,t;q)=\frac{1}{2}\int_{\pm\infty}^{x}(|q|^{2}-1)dy.$ (2.33)

The scattering matrices admit asymptotic behaviors

$\displaystyle S(z)=e^{-i\nu_{0}(t;q)\sigma_{3}}+O(z^{-1}),\hskip
14.22636ptz\rightarrow\infty,$ (2.34) $\displaystyle S(z)={\rm
diag}\left(\frac{q_{-}}{q_{+}},\frac{q_{+}}{q_{-}}\right)e^{i\nu_{0}(t;q)\sigma_{3}}+O(z),\hskip
14.22636ptz\rightarrow 0,$ (2.35)

where

$\nu_{0}(t;q)=\frac{1}{2}\int_{-\infty}^{+\infty}(|q|^{2}-1)dy.$ (2.36)

Further we have $\rho(0)=\tilde{\rho}(0)=0$.

Moreover, from trace formulae we have

$a(z)=\prod_{j=1}^{4N_{1}+2N_{2}}\frac{z-\zeta_{j}}{z-\bar{\zeta}_{j}}\exp\left\\{-\frac{1}{2\pi
i}\int_{\Sigma}\frac{\log(1-\rho(s)\tilde{\rho}(s))}{s-z}ds\right\\}.$ (2.37)

Then by taking $z\to 0$, theta condition is obtained:

$\displaystyle\arg\frac{q_{-}}{q_{+}}+2\nu_{0}=8\sum_{n=1}^{N_{1}}\arg(z_{n})+4\sum_{m=1}^{N_{2}}\arg(w_{m})+\frac{1}{2\pi}\int_{\Sigma}\frac{\log(1-\rho(s)\tilde{\rho}(s))}{s}ds+2j\pi,\hskip
14.22636pt$ (2.38)

where $j$ is a integer.

Define a sectionally meromorphic matrix

$M(z;x,t)=\left\\{\begin{array}[]{ll}\left(a(z)^{-1}\mu_{+}^{1},\mu_{-}^{2}\right),&\text{as
}z\in D^{+},\\\\[12.0pt]
\left(\mu_{-}^{1},\overline{a(\bar{z})}^{-1}\mu_{+}^{2}\right),&\text{as }z\in
D^{-},\\\ \end{array}\right.$ (2.39)

which solves the following (time-dependent) RHP.

RHP0. Find a matrix-valued function $M(z)$ which satisfies:

$\blacktriangleright$ Analyticity: $M(z)$ is meromorphic in
$\mathbb{C}\setminus\Sigma$ and has single poles $\mathcal{Z}$;

$\blacktriangleright$ Symmetry:
$M(z)=\sigma_{2}\overline{M(\bar{z})}\sigma_{2}$=$\sigma_{1}\overline{M(-\bar{z})}\sigma_{1}=\frac{i}{z}M(-1/z)\sigma_{3}Q_{-}$;

$\blacktriangleright$ Jump condition: $M(z)$ has continuous boundary values
$M_{\pm}(z)$ on $\Sigma$ and

$M_{+}(z)=M_{-}(z)V(z),\hskip 14.22636ptz\in\Sigma,$ (2.40)

where

$V(z)=\left(\begin{array}[]{cc}1-\tilde{\rho}(z)\rho(z)&-e^{2it\theta}\tilde{\rho}(z)\\\
e^{-2it\theta}\rho(z)&1\end{array}\right);$ (2.41)

$\blacktriangleright$ Asymptotic behaviors:

$\displaystyle M(z)=e^{i\nu_{-}(x,t;q)\sigma_{3}}+\mathcal{O}(z^{-1}),\hskip
14.22636ptz\rightarrow\infty,$ (2.42) $\displaystyle
M(z)=\frac{i}{z}e^{i\nu_{-}(x,t;q)\sigma_{3}}\sigma_{3}Q_{-}+\mathcal{O}(1),\hskip
14.22636ptz\rightarrow 0;$ (2.43)

$\blacktriangleright$ Residue conditions: $M$ has simple poles at each point
in $\mathcal{Z}\cup\bar{\mathcal{Z}}$ with:

$\displaystyle\operatorname*{Res}_{z=\zeta_{n}}M(z)=\lim_{z\to\zeta_{n}}M(z)\left(\begin{array}[]{cc}0&0\\\
C_{n}e^{-2it\theta_{n}}&0\end{array}\right),$ (2.46)
$\displaystyle\operatorname*{Res}_{z=\bar{\zeta}_{n}}M(z)=\lim_{z\to\bar{\zeta}_{n}}M(z)\left(\begin{array}[]{cc}0&-\bar{C}_{n}e^{2it\bar{\theta}_{n}}\\\
0&0\end{array}\right).$ (2.49)

From the asymptotic behavior in Proposition 2, the reconstruction formula of
$q(x,t)$ is given by

$q(x,t)=\exp\left\\{\frac{i}{2}\int_{-\infty}^{x}(|q(x,t)|^{2}-1)dy\right\\}m(x,t),$
(2.50)

where

$\displaystyle m(x,t)=\lim_{z\to\infty}\left[zM\right]_{12}.$ (2.51)

Take modulus on both sides of (2.50) yields

$\displaystyle|q(x,t)|=|m(x,t)|,$

which is substituted back into (2.50) leads to

$q(x,t)=\exp\left\\{\frac{i}{2}\int_{-\infty}^{x}(|m(x,t)|^{2}-1)dy\right\\}m(x,t).$
(2.52)

##  3 Deformation to a mixed $\bar{\partial}$-RH problem

We find that the long-time asymptotic of RHP0 is affected by the growth and
decay of the exponential function $e^{\pm 2it\theta}$ appearing in both the
jump relation and the residue conditions. Therefore, in this section, we
introduce a new transform $M(z)\to M^{(1)}(z)$, which make that the
$M^{(1)}(z)$ is well behaved as $t\to\infty$ along any characteristic line.

Let $\xi=\frac{x}{t}$, to obtain asymptotic behavior of $e^{2it\theta}$ as
$t\to\infty$, we consider the real part of $2it\theta$:

$\text{Re}(2it\theta)=-t\text{Im}z\text{Re}z\left[\left(\xi+2\right)\left(1+|z|^{-4}\right)-\left(\text{Re}^{2}z-\text{Im}^{2}z\right)\left(1+|z|^{-8}\right)\right].$
(3.1)

The signature of $\text{Im}\theta$ are shown in Figure 3.

Figure 3: In these figure we take $\xi=-4,-3,-2.6,-1.5,-1,0$ respectively to
show all type of $\text{Im}\theta$. The green curve is unit circle. In the red
region, $\text{Im}\theta>0$ while $\text{Im}\theta=0$ on the red curve. And
$\text{Im}\theta<0$ in the white region.

In our paper, we only consider the case $-3<\xi<-1$ which is corresponding to
Figure 3 (c), because of its well property.

For brevity, we introduce some notations with respect to subscripts

$\displaystyle\mathcal{N}\triangleq\left\\{1,...,4N_{1}+2N_{2}\right\\},\ \
\nabla=\left\\{n\in\mathcal{N}|\text{Im}\theta_{n}\leq 0\right\\},$
$\displaystyle\Delta=\left\\{n\in\mathcal{N}|\text{Im}\theta_{n}>0\right\\},\Lambda=\left\\{n\in\mathcal{N}|\text{Im}\theta_{n}=0\right\\}.$
(3.2)

For $n\in\Delta$, the residue of $M(z)$ at $\zeta_{n}$ in (2.46) are unbounded
as $t\to\infty$. Similarly, for $n\in\nabla$, the residue at $\zeta_{n}$
approach to be zero as $t\to\infty$. Define

$\rho_{0}=\min_{n\in\Delta\cup\nabla\setminus\Lambda}|\text{Im}\theta_{n}|\neq
0.$ (3.3)

To distinguish different type of zeros, we further give

$\displaystyle\nabla_{1}=\left\\{j\in\left\\{1,...,N_{1}\right\\}|\text{Im}\theta(z_{j})\leq
0\right\\},\Delta_{1}=\left\\{j\in\left\\{1,...,N_{1}\right\\}|\text{Im}\theta(z_{j})>0\right\\},$
$\displaystyle\nabla_{2}=\left\\{i\in\left\\{1,...,N_{2}\right\\}|\text{Im}\theta(w_{i})\leq
0\right\\},\Delta_{2}=\left\\{i\in\left\\{1,...,N_{2}\right\\}|\text{Im}\theta(w_{i})>0\right\\},$
$\displaystyle\Lambda_{1}=\left\\{j_{0}\in\left\\{1,...,N_{1}\right\\}|\text{Im}\theta(z_{j_{0}})=0\right\\},\Lambda_{2}=\left\\{i_{0}\in\left\\{1,...,N_{2}\right\\}|\text{Im}\theta(w_{i_{0})}=0\right\\}.$

For the poles $\zeta_{n}$ with $n\notin\Lambda$, we want to trap them for
jumps along small closed circles enclosing themselves respectively. The jump
matrix in (2.41) also needs to be restricted. Recall the well known
factorizations of $V(z)$:

$\displaystyle V(z)$
$\displaystyle=\left(\begin{array}[]{cc}1&-\tilde{\rho}e^{2it\theta}\\\
0&1\end{array}\right)\left(\begin{array}[]{cc}1&0\\\ \rho
e^{-2it\theta}&1\end{array}\right)$ (3.8)
$\displaystyle=\left(\begin{array}[]{cc}1&\\\ \frac{\rho
e^{-2it\theta}}{1-\rho\tilde{\rho}}&1\end{array}\right)(1-\rho\tilde{\rho})^{\sigma_{3}}\left(\begin{array}[]{cc}1&-\frac{\tilde{\rho}e^{2it\theta}}{1-\rho\tilde{\rho}}\\\
0&1\end{array}\right).$ (3.13)

We will use these factorizations to deform the jump contours so that
exponentials $e^{\pm 2it\theta}$ are decaying in corresponding regions
respectively. Define functions

$\displaystyle\delta(z)$ $\displaystyle=\exp\left(-\frac{1}{2\pi
i}\int_{i\mathbb{R}}\left(\dfrac{1}{s-z}-\frac{1}{2s}\right)\log(1-\rho(s)\tilde{\rho}(s))ds\right);$
(3.14) $\displaystyle T(z)$
$\displaystyle=T(z,\xi)=\prod_{n\in\Delta}\dfrac{z-\zeta_{n}}{\bar{\zeta}_{n}^{-1}z-1}\delta(z)$
$\displaystyle=\prod_{j\in\Delta_{1}}\dfrac{z^{2}-z_{j}^{2}}{\bar{z}_{j}^{-2}z^{2}-1}\dfrac{z^{2}-\bar{z}_{j}^{-2}}{z_{j}^{2}z^{2}-1}\prod_{i\in\Delta_{2}}\dfrac{z^{2}-w_{i}^{2}}{w_{i}^{2}z^{2}-1}\delta(z).$
(3.15)

In the above formulas, we choose the principal branch of power and logarithm
functions.

######  Proposition 4.

The function defined by (3.15) has following properties:
(a) $T$ is meromorphic in $\mathbb{C}\setminus\mathbb{R}$, and for each
$n\in\Delta$, $T(z)$ has simple zeros $\zeta_{n}$ and simple poles
$\bar{\zeta}_{n}$;
(b) $T(z)=\overline{T^{-1}(\bar{z})}=T^{-1}(-z^{-1})$;
(c) For $z\in\mathbb{R}$, as z approaching the real axis from above and below,
$T$ has boundary values $T_{\pm}$, which satisfy:

$T_{+}(z)=(1-\rho(z)\tilde{\rho}(z))T_{-}(z),\hskip 14.22636ptz\in
i\mathbb{R};$ (3.16)

(d) $\lim_{z\to\infty}T(z)\triangleq T(\infty)$, where

$T(\infty)=\prod_{j\in\Delta_{1}}\bar{z}_{j}^{2}z_{j}^{-2}\prod_{i\in\Delta_{2}}\bar{w}_{i}^{2}\exp\left(\frac{1}{4\pi
i}\int_{i\mathbb{R}}s^{-1}\log(1-\rho(s)\tilde{\rho}(s))ds\right),$ (3.17)

with $|T(\infty)|=1$;
(e) As $|z|\to\infty$ with $|arg(z)|\leq c<\pi$,

$T(z)=T(\infty)\left(1+z^{-1}\frac{1}{2\pi
i}\int_{i\mathbb{R}}\log(1-\rho(s)\tilde{\rho}(s))ds+\mathcal{O}(z^{-2})\right);$
(3.18)

(f) $T(z)$ is continuous at $z=0$, and

$\lim_{z\to 0}T(z)=T(0)=T(\infty)^{-1};$ (3.19)

(g) $\frac{a(z)}{T(z)}$ is holomorphic in $D^{+}$. And its absolute value is
bounded in $D^{+}\cap\left\\{z\in\mathbb{C}|\text{Re}z>0\right\\}$.
Additionally, the ratio extends as a continuous function on $i\mathbb{R}$.

###### Proof.

Properties (a), (b), (d) and (f) can be obtain by simple calculation. And (c)
follows from the Plemelj formula. By the Laurent expansion (e) immediately.
For brevity, we omit calculation. For (g), from (2.37) we have

$\frac{a(z)}{T(z)}=T(\infty)^{-1}\prod_{j\in\nabla_{1}}\dfrac{z^{2}-z_{j}^{2}}{\bar{z}_{j}^{-2}z^{2}-1}\dfrac{z^{2}-\bar{z}_{j}^{-2}}{z_{j}^{2}z^{2}-1}\prod_{i\in\nabla_{2}}\dfrac{z^{2}-w_{i}^{2}}{w_{i}^{2}z^{2}-1}\exp\left\\{-\frac{1}{2\pi
i}\int_{\mathbb{R}}\frac{\log(1-\rho(s)\tilde{\rho}(s))}{s-z}\right\\}.$

So $\frac{a(z)}{T(z)}$ is holomorphic in $D^{+}$. And in above expression, all
factors except the last integral is bounded for $z\in D^{+}$. From (2.17),
$1-\rho(s)\tilde{\rho}(s)=1+|\rho(s)|^{2}$. Let $z=x+yi$, then the real part
of the exponential is
$-\frac{y}{2\pi}\int_{\mathbb{R}}\frac{\log(1+|\rho(s)|^{2})}{|s-z|^{2}}ds$
which can be bounded as follows:

$\displaystyle|\frac{y}{2\pi}\int_{\mathbb{R}}\frac{\log(1+|\rho(s)|^{2})}{|s-z|^{2}}ds|$
$\displaystyle\leq\frac{1}{2\pi}\parallel\log(1+|\rho(s)|^{2})\parallel_{L^{\infty}(\mathbb{R})}\parallel\frac{y}{(s-x)^{2}+y^{2}}\parallel_{L^{1}(\mathbb{R})}$
$\displaystyle\lesssim\parallel\rho(s)\parallel_{L^{\infty}(\mathbb{R})}.$

∎

Additionally, let $\varrho$ be a positive constant stratifying

$\varrho=\frac{1}{2}\min\left\\{\min_{j\neq
i\in\mathcal{N}}|\zeta_{i}-\zeta_{j}|,\min_{j\in\mathcal{N}}\left\\{|\text{Im}\zeta_{j}|,|\text{Re}\zeta_{j}|\right\\},\min_{j\in\mathcal{N}\setminus\Lambda,\text{Im}\theta(z)=0}|\zeta_{j}-z|\right\\}.$
(3.20)

By above definition, for every $n\in\mathcal{N}$, we define disks
$\mathbb{D}(\zeta_{n},\varrho)$, such that they pairwise disjoint, also
disjoint with $\left\\{z\in\mathbb{C}|\text{Im}\theta(z)=0\right\\}$ and
$\Sigma$. Introduce a piecewise matrix function

$G(z)=\left\\{\begin{array}[]{ll}\left(\begin{array}[]{cc}1&0\\\
-C_{n}(z-\zeta_{n})^{-1}e^{-2it\theta_{n}}&1\end{array}\right),&\text{as
}z\in\mathbb{D}(\zeta_{n},\varrho),n\in\nabla\setminus\Lambda;\\\\[12.0pt]
\left(\begin{array}[]{cc}1&-C_{n}^{-1}(z-\zeta_{n})e^{2it\theta_{n}}\\\
0&1\end{array}\right),&\text{as
}z\in\mathbb{D}(\zeta_{n},\varrho),n\in\Delta;\\\
\left(\begin{array}[]{cc}1&\bar{C}_{n}(z-\bar{\zeta}_{n})^{-1}e^{2it\bar{\theta}_{n}}\\\
0&1\end{array}\right),&\text{as
}z\in\mathbb{D}(\bar{\zeta}_{n},\varrho),n\in\nabla\setminus\Lambda;\\\
\left(\begin{array}[]{cc}1&0\\\
\bar{C}_{n}^{-1}(z-\bar{\zeta}_{n})e^{-2it\bar{\theta}_{n}}&1\end{array}\right),&\text{as
}z\in\mathbb{D}(\bar{\zeta}_{n},\varrho),n\in\Delta;\\\ I&\text{as }z\text{ in
elsewhere};\end{array}\right..$ (3.21)

Now we use $T(z)$ and $G(z)$ to define a new matrix-valued function
$M^{(1)}(z)$.

$M^{(1)}(z)=T(\infty)^{-\sigma_{3}}M(z)G(z)T(z)^{\sigma_{3}},$ (3.22)

which then satisfies the following RH problem.

RHP1. Find a matrix-valued function $M^{(1)}(z)$ which satisfies:

$\blacktriangleright$ Analyticity: $M^{(1)}(z)$ is meromorphic in
$\mathbb{C}\setminus\Sigma^{(1)}$, where

$\Sigma^{(1)}=\mathbb{R}\cup
i\mathbb{R}\cup\left[\cup_{n\in\mathcal{N}\setminus\Lambda}\left(\partial\mathbb{D}(\bar{\zeta}_{n},\varrho)\cup\partial\mathbb{D}(\zeta_{n},\varrho)\right)\right],$
(3.23)

is shown in Figure 4;

$\blacktriangleright$ Symmetry:
$M^{(1)}(z)=\sigma_{2}\overline{M^{(1)}(\bar{z})}\sigma_{2}$=$\sigma_{1}\overline{M^{(1)}(-\bar{z})}\sigma_{1}=\frac{i}{z}M^{(1)}(-1/z)\sigma_{3}Q_{-}$;

$\blacktriangleright$ Jump condition: $M^{(1)}$ has continuous boundary values
$M^{(1)}_{\pm}$ on $\Sigma^{(1)}$ and

$M^{(1)}_{+}(z)=M^{(1)}_{-}(z)V^{(1)}(z),\hskip 14.22636ptz\in\Sigma^{(1)},$
(3.24)

where

$V^{(1)}(z)=\left\\{\begin{array}[]{ll}\left(\begin{array}[]{cc}1&-e^{2it\theta}\tilde{\rho}(z)T^{-2}(z)\\\
0&1\end{array}\right)\left(\begin{array}[]{cc}1&0\\\
e^{-2it\theta}\rho(z)T^{2}(z)&1\end{array}\right),&\text{as
}z\in\mathbb{R};\\\\[12.0pt] \left(\begin{array}[]{cc}1&0\\\
\frac{e^{-2it\theta}\tilde{\rho}(z)T_{+}^{2}(z)}{1-\tilde{\rho}(z)\rho(z)}&1\end{array}\right)\left(\begin{array}[]{cc}1&-\frac{e^{2it\theta}\tilde{\rho}(z)T_{-}^{-2}(z)}{1-\tilde{\rho}(z)\rho(z)}\\\
0&1\end{array}\right),&\text{as }z\in i\mathbb{R};\\\\[12.0pt]
\left(\begin{array}[]{cc}1&0\\\
-C_{n}(z-\zeta_{n})^{-1}T^{2}(z)e^{-2it\theta_{n}}&1\end{array}\right),&\text{as
}z\in\partial\mathbb{D}(\zeta_{n},\varrho),n\in\nabla\setminus\Lambda;\\\\[12.0pt]
\left(\begin{array}[]{cc}1&-C_{n}^{-1}(z-\zeta_{n})T^{-2}(z)e^{2it\theta_{n}}\\\
0&1\end{array}\right),&\text{as
}z\in\partial\mathbb{D}(\zeta_{n},\varrho),n\in\Delta;\\\
\left(\begin{array}[]{cc}1&\bar{C}_{n}(z-\bar{\zeta}_{n})^{-1}T^{-2}(z)e^{2it\bar{\theta}_{n}}\\\
0&1\end{array}\right),&\text{as
}z\in\partial\mathbb{D}(\bar{\zeta}_{n},\varrho),n\in\nabla\setminus\Lambda;\\\
\left(\begin{array}[]{cc}1&0\\\
\bar{C}_{n}^{-1}(z-\bar{\zeta}_{n})e^{-2it\bar{\theta}_{n}}T^{2}(z)&1\end{array}\right),&\text{as
}z\in\partial\mathbb{D}(\bar{\zeta}_{n},\varrho),n\in\Delta;\\\
\end{array}\right.;$ (3.25)

$\blacktriangleright$ Asymptotic behaviors:

$\displaystyle
M^{(1)}(z)=e^{i\nu_{-}(x,t;q)\sigma_{3}}+\mathcal{O}(z^{-1}),\hskip
14.22636ptz\rightarrow\infty,$ (3.26) $\displaystyle
M^{(1)}(z)=\frac{i}{z}e^{i\nu_{-}(x,t;q)\sigma_{3}}\sigma_{3}Q_{-}+\mathcal{O}(1),\hskip
14.22636ptz\rightarrow 0;$ (3.27)

$\blacktriangleright$ Residue conditions: $M^{(1)}$ has simple poles at each
point $\zeta_{n}$ and $\bar{\zeta}_{n}$ for $n\in\Lambda$ with:

$\displaystyle\operatorname*{Res}_{z=\zeta_{n}}M^{(1)}(z)=\lim_{z\to\zeta_{n}}M^{(1)}(z)\left(\begin{array}[]{cc}0&0\\\
C_{n}e^{-2it\theta_{n}}T^{2}(\zeta_{n})&0\end{array}\right),$ (3.30)
$\displaystyle\operatorname*{Res}_{z=\bar{\zeta}_{n}}M^{(1)}(z)=\lim_{z\to\bar{\zeta}_{n}}M^{(1)}(z)\left(\begin{array}[]{cc}0&-\bar{C}_{n}T^{-2}(\bar{\zeta}_{n})e^{2it\bar{\theta}_{n}}\\\
0&0\end{array}\right).$ (3.33)

###### Proof.

Note that the triangular factors (3.21) trades poles $\zeta_{n}$ and
$\bar{\zeta}_{n}$ to jumps on the disk boundaries
$\partial\mathbb{D}(\zeta_{n},\varrho)$ and
$\partial\mathbb{D}(\bar{\zeta}_{n},\varrho)$ respectively for
$n\in\mathcal{N}\setminus\Lambda$. Then by simple calculation we can obtain
the residues condition and jump condition from (2.46), (2.49) (2.41), (3.21)
and (3.22). The analyticity and symmetry of $M^{(1)}(z)$ is directly from its
definition, the Proposition 4, (3.21) and the properties of $M$. As for
asymptotic behaviors, from $\lim_{z\to 0}G(z)=\lim_{z\to\infty}G(z)=I$ and
Proposition 4 (f), we obtain that $M^{(1)}(z)$ has same asymptotic behaviors
as $M(z)$. ∎

Figure 4: The blue curve, including $\mathbb{R}$, $i\mathbb{R}$ and the small
circles constitute $\Sigma^{(1)}$. Because Im$\theta(w_{m})=0$, it remain the
pole of $M^{(1)}$. And Im$\theta(z_{n})\neq 0$, so we change it to jump on
$\partial\mathbb{D}(\zeta_{n},\varrho)$.

##  4 Mixed $\bar{\partial}$-RH Problem

In this section, we make continuous extension for the jump matrix $V^{(1)}$ to
remove the jump from $\Sigma$. Besides, the new problem is hoped to takes
advantage of the decay/growth of $e^{2it\theta(z)}$ for $z\notin\Sigma$. For
this purpose, we introduce new eight regions:

$\displaystyle\Omega_{2n+1}=\left\\{z\in\mathbb{C}|n\pi/2\leq\arg z\leq
n\pi/2+\varphi\right\\},$ (4.1)
$\displaystyle\Omega_{2n+2}=\left\\{z\in\mathbb{C}|(n+1)\pi/2-\varphi\leq\arg
z\leq(n+1)\pi/2\right\\},$ (4.2)

where $n=0,1,2,3$ and $\varphi>0$ is an fixed sufficiently small angle
achieving following conditions:
1\. $\frac{2|\xi+2|}{|\xi+2|+1}<\cos 2\varphi<1$;
2\. each $\Omega_{i}$ doesn’t intersect any of $\mathbb{D}(\zeta_{n},\varrho)$
or $\mathbb{D}(\bar{\zeta}_{n},\varrho)$.
Define new contours as follow:

$\displaystyle\Sigma_{k}=e^{(k-1)i\pi/4+\varphi}R_{+},\hskip
14.22636ptk=1,3,5,7,$ (4.3)
$\displaystyle\Sigma_{k}=e^{ki\pi/4-\varphi}R_{+},\hskip 14.22636ptk=2,4,6,8,$
(4.4) $\displaystyle\tilde{\Sigma}=\Sigma_{1}\cup\Sigma_{2}...\cup\Sigma_{8},$
(4.5)

which is the boundary of $\Omega_{k}$ respectively. In addition, let

$\displaystyle\Omega=\Omega_{1}\cup...\cup\Omega_{8}.$ (4.6)
$\displaystyle\Sigma^{(2)}=\cup_{n\in\mathcal{N}\setminus\Lambda}\left(\partial\mathbb{D}(\bar{\zeta}_{n},\varrho)\cup\partial\mathbb{D}(\zeta_{n},\varrho)\right),$
(4.7)

which are shown in Figure 5.

$\Sigma_{1}$$\Sigma_{2}$$\Sigma_{3}$$\Sigma_{4}$$\Sigma_{5}$$\Sigma_{6}$$\Sigma_{7}$$\Sigma_{8}$Re$z$Im$z$$\Omega_{3}$$\Omega_{2}$$\Omega_{1}$$\Omega_{6}$$\Omega_{5}$$\Omega_{4}$$\Omega_{7}$$\Omega_{8}$$0$$z_{n}$$\bar{z}_{n}$$-\frac{1}{z_{n}}$$-\frac{1}{\bar{z}_{n}}$$\frac{1}{\bar{z}_{n}}$$\frac{1}{z_{n}}$$-\bar{z}_{n}$$-z_{n}$$1$$w_{m}$$\bar{w}_{m}$$-\bar{w}_{m}$$-w_{m}$
Figure 5: The yellow region is $\Omega$. The blue circle constitute
$\Sigma^{(2)}$ together.

######  Lemma 1.

Let $\xi=\frac{x}{t}\in(-3,-1)$, and $F(r)=r^{2}+\frac{1}{r^{2}}$ is a real-
valued function. Then for $z=re^{i\phi}$, the imaginary part of phase function
(3.1) satisfies

$\displaystyle\text{Im }\theta(z)\leq\frac{1}{16}|\sin
2\phi|(|\xi+2|-1)F(r)^{2},\hskip 14.22636pt\text{as
}z\in\Omega_{1},\Omega_{3},\Omega_{5},\Omega_{7};$ (4.8)
$\displaystyle\text{Im }\theta(z)\geq\frac{1}{16}|\sin
2\phi|(1-|\xi+2|)F(r)^{2},\hskip 14.22636pt\text{as
}z\in\Omega_{2},\Omega_{4},\Omega_{6},\Omega_{8}.$ (4.9)

###### Proof.

We only prove the case $z\in\Omega_{1},$ and the other regions are similarly.
From (3.1) we have

$\displaystyle\text{Im }\theta(z)$
$\displaystyle=\frac{1}{2}\text{Im}z\text{Re}z\left[\left(\xi+2\right)\left(1+|z|^{-4}\right)-\left(\text{Re}^{2}z-\text{Im}^{2}z\right)\left(1+|z|^{-8}\right)\right]$
$\displaystyle=\frac{1}{4}r^{2}\sin
2\phi\left[\left(\xi+2\right)\left(1+r^{-4}\right)-r^{2}\cos
2\phi\left(1+r^{-8}\right)\right]$ $\displaystyle=\frac{1}{4}\sin
2\phi\left[\left(\xi+2\right)F(r)-\cos 2\phi\left(F(r)^{2}-2\right)\right].$
(4.10)

$F(r)\geq 2$ leads to $2\leq\frac{F(r)^{2}}{2}$. For $z\in\Omega_{1}$,
$\frac{2|\xi+2|}{|\xi+2|+1}<\cos 2\varphi<\cos 2\phi$, then we have

$\frac{|\xi+2|}{\cos 2\phi}F(r)\leq\frac{|\xi+2|+1}{4}F(r)^{2}.$ (4.11)

Substitute above inequality into (4.10) we obtain the consequence immediately.
∎

Introduce a small enough constant $1>\epsilon_{0}>0$ with
$(1-\epsilon_{0})\cos\varphi>\frac{1}{2}$. Let $X_{1}\in
C_{0}^{\infty}\left(\mathbb{R},[0,1]\right)$, which is support in
$(1-\epsilon_{0},1+\epsilon_{0})$. And $X_{0}$ has support in
$(-\epsilon_{0},\epsilon_{0})$ with $X_{0}(z)=X_{1}(1+z)$. In addition, we
denote following functions for brief:

$\displaystyle p_{1}(z)=p_{5}(z)=\rho(z),\hskip
14.22636ptp_{2}(z)=p_{6}(z)=\dfrac{\tilde{\rho}(z)}{1-\rho(z)\tilde{\rho}(z)},$
(4.12) $\displaystyle
p_{3}(z)=p_{7}(z)=\dfrac{\rho(z)}{1-\rho(z)\tilde{\rho}(z)},\hskip
14.22636ptp_{4}(z)=p_{8}(z)=\tilde{\rho}(z).$ (4.13)

Then the next step is to construct a matrix function $R^{(2)}$. We need to
remove jump on $\mathbb{R}$ and $i\mathbb{R}$, and have some mild control on
$\bar{\partial}R^{(2)}$ sufficient to ensure that the
$\bar{\partial}$-contribution to the long-time asymptotics of $q(x,t)$ is
negligible. So we choose $R^{(2)}(z)$ as

$R^{(2)}(z)=\left\\{\begin{array}[]{lll}\left(\begin{array}[]{cc}1&R_{j}(z)e^{2it\theta}\\\
0&1\end{array}\right),&z\in\Omega_{j},j=2,4,6,8;\\\ \\\
\left(\begin{array}[]{cc}1&0\\\
R_{j}(z)e^{-2it\theta}&1\end{array}\right),&z\in\Omega_{j},j=1,3,5,7;\\\ \\\
I,&elsewhere;\\\ \end{array}\right.$ (4.14)

where the functions $R_{j}$, $j=1,2,..,8$, is defined in following
Proposition.

######  Proposition 5.

$R_{j}$: $\bar{\Omega}_{j}\to C$, $j=1,2,..,8$ have boundary values as follow:

$\displaystyle
R_{1}(z)=\Bigg{\\{}\begin{array}[]{ll}-\rho(z)T(z)^{2}&z\in\mathbb{R}^{+},\\\
0&z\in\Sigma_{1},\\\ \end{array},\hskip
17.07182ptR_{2}(z)=\Bigg{\\{}\begin{array}[]{ll}0&z\in\Sigma_{2},\\\
\dfrac{\tilde{\rho}(z)T_{+}(z)^{2}}{1-\rho(z)\tilde{\rho}(z)}&z\in
i\mathbb{R}^{+},\\\ \end{array}$ (4.19) $\displaystyle
R_{3}(z)=\Bigg{\\{}\begin{array}[]{ll}\dfrac{\rho(z)T_{-}(z)^{2}}{1-\rho(z)\tilde{\rho}(z)}&z\in
i\mathbb{R}^{+},\\\ 0&z\in\Sigma_{3},\\\
\end{array},R_{4}(z)=\Bigg{\\{}\begin{array}[]{ll}0&z\in\Sigma_{4},\\\
-\tilde{\rho}(z)T(z)^{-2}&z\in\mathbb{R}^{-},\\\ \end{array}$ (4.24)
$\displaystyle
R_{5}(z)=\Bigg{\\{}\begin{array}[]{ll}-\rho(z)T(z)^{2}&z\in\mathbb{R}^{-},\\\
0&z\in\Sigma_{5},\end{array},\hskip
14.22636ptR_{6}(z)=\Bigg{\\{}\begin{array}[]{ll}0&z\in\Sigma_{6},\\\
\dfrac{\tilde{\rho}(z)T_{+}(z)^{2}}{1-\rho(z)\tilde{\rho}(z)}&z\in
i\mathbb{R}^{-},\\\ \end{array}$ (4.29) $\displaystyle
R_{7}(z)=\Bigg{\\{}\begin{array}[]{ll}\dfrac{\rho(z)T_{-}(z)^{2}}{1-\rho(z)\tilde{\rho}(z)}&z\in
i\mathbb{R}^{-},\\\ 0&z\in\Sigma_{7},\end{array},\hskip
17.07182ptR_{8}(z)=\Bigg{\\{}\begin{array}[]{ll}0&z\in\Sigma_{8},\\\
-\tilde{\rho}(z)T(z)^{-2}&z\in\mathbb{R}^{+}.\\\ \end{array}$ (4.34)

$R_{j}$ have following property: for $j=1,5,4,8,$

$\displaystyle|\bar{\partial}R_{j}(z)|\lesssim|p_{j}^{\prime}(|z|)|+|z|^{-1/2},\text{for
all $z\in\Omega_{j}$;}$ (4.35)

and for $j=2,3,6,7,$

$\displaystyle|\bar{\partial}R_{j}(z)|\lesssim|z\mp i|,\text{for all
$z\in\Omega_{j}$ in a small fixed neighborhood of $\pm i$},$ (4.36)
$\displaystyle|\bar{\partial}R_{j}(z)|\lesssim|p_{j}^{\prime}(i|z|)|+|z|^{-1/2}+|\bar{\partial}X_{1}(|z|)|,\text{for
all $z\in\Omega_{j}$}.$ (4.37)

And

$\bar{\partial}R_{j}(z)=0,\hskip 14.22636pt\text{if }z\in elsewhere.$ (4.38)

###### Proof.

Case I: $z\in\bar{\Omega}_{j}$, $j=1,5,4,8$.
Take $R_{1}(z)$ as an example with extensions

$R_{1}(z)=p_{1}(|z|)T^{2}(z)\cos(k_{0}\arg z),\hskip
14.22636ptk_{0}=\frac{2\pi}{\varphi}.$ (4.39)

The other cases are easily inferred. $p_{1}(|z|)=\rho(|z|)$ is bounded. Denote
$z=re^{i\phi}$, then we have
$\bar{\partial}=\frac{e^{i\phi}}{2}\left(\partial_{r}+\frac{i}{r}\partial_{\phi}\right)$.
So

$\displaystyle\bar{\partial}R_{1}(z)=\frac{e^{i\phi}}{2}T^{2}(z)\left(p_{1}^{\prime}(r)\cos(k_{0}\phi)-\frac{i}{r}p_{1}(r)k_{0}\sin(k_{0}\phi)\right).$
(4.40)

To bound second term we use Cauchy-Schwarz inequality and obtain

$|p_{1}(r)|=|\rho(r)|=|\rho(r)-\rho(0)|=|\int_{0}^{r}\rho^{\prime}(s)ds|\leq\parallel\rho^{\prime}(s)\parallel_{L^{2}}r^{1/2}.$
(4.41)

And note that $T(z)$ is a bounded function in $\bar{\Omega}_{1}$. Then the
boundedness of (4.35) follows immediately.
Case II: $z\in\bar{\Omega}_{j}$, $j=2,3,6,7.$
The details of the proof are only given for $R_{2}$. Unlike the vanishing
boundary condition case in [16], the determinant of $M(z)$ is $1+z^{-2}$. So
to bound the $\bar{\partial}$-derivative construct by $R^{(2)}$ in following
section, the property of $R^{(2)}$ at $\pm i$ needs to be control. For this
purpose, we make small adjustments to the extensions of $R_{2}$ as

$R_{2}(z)=R_{21}(z)+R_{22}(z),$ (4.42)

with a constant $\delta_{0}$ stratifying $\varphi>\delta_{0}\epsilon_{0}$ and

$\displaystyle R_{21}(z)$
$\displaystyle=[1-X_{1}(|z|)]p_{2}(i|z|)T^{-2}(z)\cos[k_{0}(\frac{\pi}{2}-\arg
z)],$ (4.43) $\displaystyle R_{22}(z)$
$\displaystyle=f(|z|)g(z)\cos[k_{0}(\frac{\pi}{2}-\arg z)]$
$\displaystyle-\frac{i|z|}{k_{0}}X_{0}(\frac{\arg
z}{\delta_{0}})f^{\prime}(|z|)g(z)\sin[k_{0}(\frac{\pi}{2}-\arg z)].$ (4.44)

Among above function,

$\displaystyle f(z)=X_{1}(z)\frac{\bar{b}(z)}{a(z)},\hskip
14.22636ptg(z)=\left(\frac{a(z)}{T(z)}\right)^{2}.$ (4.45)

Then $f(z)\in W{2,\infty}$. Obviously, $R_{21}(z)\equiv 0$ with $|z|$ in the
support of $X_{1}$ and $R_{22}(z)\equiv 0$ out the support of $X_{1}$. Note
that

$\displaystyle|p_{2}(z)|=|\dfrac{\tilde{\rho}(z)}{1-\rho(z)\tilde{\rho}(z)}|=|\dfrac{\tilde{\rho}(z)}{1-|\rho(z)|^{2}}|\lesssim|\rho(z)|,\hskip
14.22636pt\text{for $z$ out of supp}(X_{1}).$ (4.46)

Similarly in case I, $R_{21}(z)$ can be bounded as

$|\bar{\partial}R_{21}(z)|\lesssim(1-X_{1}(|z|))\left(|p_{2}^{\prime}(i|z|)|+|z|^{-1/2}\right)+|\bar{\partial}X_{1}(|z|)|.$
(4.47)

As for $R_{22}(z)$, $z=re^{i\phi}$,

$\displaystyle\bar{\partial}R_{22}(z)=$
$\displaystyle\frac{e^{i\phi}}{2}g(z)\cos[k_{0}(\frac{\pi}{2}-\varphi)]f^{\prime}(ir)\left(1-X_{0}(\frac{\varphi}{\delta_{0}})\right)$
$\displaystyle+\sin[k_{0}(\frac{\pi}{2}-\varphi)]\left[\frac{ik_{0}}{r}f(ir)+\frac{1}{\delta_{0}k_{0}}X_{0}^{\prime}(\frac{\varphi}{\delta_{0}})f^{\prime}(ir)\right]$
$\displaystyle-\frac{i}{k_{0}}\sin[k_{0}(\frac{\pi}{2}-\varphi)]X_{0}(\frac{\arg
z}{\delta_{0}})(rf^{\prime}(ir))^{\prime}.$ (4.48)

So $|\bar{\partial}R_{22}(z)|$ is bounded, and we can write
$|\bar{\partial}R_{22}(z)|\lesssim X_{1}(z)|z|^{-1/2}$. So (4.36) is obtained.
In addition, for $z\sim i$,

$|\bar{\partial}R_{22}(z)|\lesssim|\sin[k_{0}(\frac{\pi}{2}-\varphi)]|+|1-X_{0}(\frac{\varphi}{\delta_{0}})|=\mathcal{O}(\varphi),$
(4.49)

from which (4.36) follows immediately. ∎

In addition, from Proposition 1, $R^{(2)}$ achieve the symmetry:

$R^{(2)}(z)=\sigma_{2}\overline{R^{(2)}(\bar{z})}\sigma_{2}=\sigma_{1}\overline{R^{(2)}(-\bar{z})}\sigma_{1}=\sigma_{3}Q_{-}R^{(2)}(-1/z)\sigma_{3}Q_{-}.$
(4.50)

We now use $R^{(2)}$ to define the new transformation

$M^{(2)}(z)=M^{(1)}(z)R^{(2)}(z),$ (4.51)

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

RHP2. Find a matrix valued function $M^{(2)}(z;x,t)$ with following
properties:

$\blacktriangleright$ Analyticity: $M^{(2)}(z;x,t)$ is continuous in
$\mathbb{C}$, sectionally continuous first partial derivatives in
$\mathbb{C}\setminus\left(\Sigma^{(2)}\cup\left\\{\zeta_{n},\bar{\zeta}_{n}\right\\}_{n\in\Lambda}\right)$
and meromorphic out $\bar{\Omega}$;

$\blacktriangleright$ Symmetry:
$M^{(2)}(z)=\sigma_{2}\overline{M^{(2)}(\bar{z})}\sigma_{2}$=$\sigma_{1}\overline{M^{(2)}(-\bar{z})}\sigma_{1}=\frac{i}{z}M^{(2)}(-1/z)\sigma_{3}Q_{-}$;

$\blacktriangleright$ Jump condition: $M^{(2)}$ has continuous boundary values
$M^{(2)}_{\pm}$ on $\Sigma^{(2)}$ and

$M^{(2)}_{+}(z;x,t)=M^{(2)}_{-}(z;x,t)V^{(2)}(z),\hskip
14.22636ptz\in\Sigma^{(2)},$ (4.52)

where

$V^{(2)}(z)=\left\\{\begin{array}[]{ll}\left(\begin{array}[]{cc}1&0\\\
-C_{n}(z-\zeta_{n})^{-1}T^{2}(z)e^{-2it\theta_{n}}&1\end{array}\right),&\text{as
}z\in\partial\mathbb{D}(\zeta_{n},\varrho),n\in\nabla\setminus\Lambda;\\\\[12.0pt]
\left(\begin{array}[]{cc}1&-C_{n}^{-1}(z-\zeta_{n})T^{-2}(z)e^{2it\theta_{n}}\\\
0&1\end{array}\right),&\text{as
}z\in\partial\mathbb{D}(\zeta_{n},\varrho),n\in\Delta;\\\
\left(\begin{array}[]{cc}1&\bar{C}_{n}(z-\bar{\zeta}_{n})^{-1}T^{-2}(z)e^{2it\bar{\theta}_{n}}\\\
0&1\end{array}\right),&\text{as
}z\in\partial\mathbb{D}(\bar{\zeta}_{n},\varrho),n\in\nabla\setminus\Lambda;\\\
\left(\begin{array}[]{cc}1&0\\\
\bar{C}_{n}^{-1}(z-\bar{\zeta}_{n})e^{-2it\bar{\theta}_{n}}T^{2}(z)&1\end{array}\right),&\text{as
}z\in\partial\mathbb{D}(\bar{\zeta}_{n},\varrho),n\in\Delta;\\\
\end{array}\right.;$ (4.53)

$\blacktriangleright$ Asymptotic behaviors:

$\displaystyle
M^{(2)}(z)=e^{i\nu_{-}(x,t;q)\sigma_{3}}+\mathcal{O}(z^{-1}),\hskip
14.22636ptz\rightarrow\infty,$ (4.54) $\displaystyle
M^{(2)}(z)=\frac{i}{z}e^{i\nu_{-}(x,t;q)\sigma_{3}}\sigma_{3}Q_{-}+\mathcal{O}(1),\hskip
14.22636ptz\rightarrow 0;$ (4.55)

$\blacktriangleright$ $\bar{\partial}$-Derivative: For $z\in\mathbb{C}$ we
have

$\displaystyle\bar{\partial}M^{(2)}=M^{(2)}\bar{\partial}R^{(2)},$ (4.56)

where

$\bar{\partial}R^{(2)}=\left\\{\begin{array}[]{lll}\left(\begin{array}[]{cc}0&\bar{\partial}R_{j}(z)e^{2it\theta}\\\
0&0\end{array}\right),&z\in\Omega_{j},j=1,3,5,7,\\\ \\\
\left(\begin{array}[]{cc}0&0\\\
\bar{\partial}R_{j}(z)e^{-2it\theta}&0\end{array}\right),&z\in\Omega_{j},j=2,4,6,8,\\\
\\\ 0&elsewhere;\\\ \end{array}\right.$ (4.57)

$\blacktriangleright$ Residue conditions: $M^{(2)}$ has simple poles at each
point $\zeta_{n}$ and $\bar{\zeta}_{n}$ for $n\in\Lambda$ with:

$\displaystyle\operatorname*{Res}_{z=\zeta_{n}}M^{(2)}(z)=\lim_{z\to\zeta_{n}}M^{(2)}(z)\left(\begin{array}[]{cc}0&0\\\
C_{n}e^{-2it\theta_{n}}T^{2}(\zeta_{n})&0\end{array}\right),$ (4.60)
$\displaystyle\operatorname*{Res}_{z=\bar{\zeta}_{n}}M^{(2)}(z)=\lim_{z\to\bar{\zeta}_{n}}M^{(2)}(z)\left(\begin{array}[]{cc}0&-\bar{C}_{n}T^{-2}(\bar{\zeta}_{n})e^{2it\bar{\theta}_{n}}\\\
0&0\end{array}\right).$ (4.63)

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

To solve RHP2, we decompose it into a model RH problem for $M^{(r)}(z)$ with
$\bar{\partial}R^{(2)}\equiv 0$ and a pure $\bar{\partial}$-Problem with
nonzero $\bar{\partial}$-derivatives. For the first step, we establish a RH
problem for the $M^{(r)}(z)$ as follows.

RHP3. Find a matrix-valued function $M^{(r)}(z)$ with following properties:

$\blacktriangleright$ Analyticity: $M^{(r)}(z)$ is meromorphic in
$\mathbb{C}\setminus\Sigma^{(2)}$;

$\blacktriangleright$ Jump condition: $M^{r}$ has continuous boundary values
$M^{(r)}_{\pm}$ on $\Sigma^{(2)}$ and

$M^{(r)}_{+}(z)=M^{(r)}_{-}(z)V^{2}(z),\hskip 14.22636ptz\in\Sigma^{(2)};$
(5.1)

$\blacktriangleright$ Symmetry:
$M^{(r)}(z)=\sigma_{2}\overline{M^{(r)}(\bar{z})}\sigma_{2}$=$\sigma_{1}\overline{M^{(r)}(-\bar{z})}\sigma_{1}=\frac{i}{z}M^{(r)}(-1/z)\sigma_{3}Q_{-}$;

$\blacktriangleright$ $\bar{\partial}$-Derivative: $\bar{\partial}R^{(2)}=0$,
for $z\in\mathbb{C}$;

$\blacktriangleright$ Asymptotic behaviors:

$\displaystyle
M^{(r)}(z)=e^{i\nu_{-}(x,t;q)\sigma_{3}}+\mathcal{O}(z^{-1}),\hskip
14.22636ptz\rightarrow\infty,$ (5.2) $\displaystyle
M^{(r)}(z)=\frac{i}{z}e^{i\nu_{-}(x,t;q)\sigma_{3}}\sigma_{3}Q_{-}+\mathcal{O}(1),\hskip
14.22636ptz\rightarrow 0;$ (5.3)

$\blacktriangleright$ Residue conditions: $M^{(r)}$ has simple poles at each
point $\zeta_{n}$ and $\bar{\zeta}_{n}$ for $n\in\Lambda$ with:

$\displaystyle\operatorname*{Res}_{z=\zeta_{n}}M^{(r)}(z)=\lim_{z\to\zeta_{n}}M^{(r)}(z)\left(\begin{array}[]{cc}0&0\\\
C_{n}e^{-2it\theta_{n}}T^{2}(\zeta_{n})&0\end{array}\right),$ (5.6)
$\displaystyle\operatorname*{Res}_{z=\bar{\zeta}_{n}}M^{(r)}(z)=\lim_{z\to\bar{\zeta}_{n}}M^{(r)}(z)\left(\begin{array}[]{cc}0&-\bar{C}_{n}T^{-2}(\bar{\zeta}_{n})e^{2it\bar{\theta}_{n}}\\\
0&0\end{array}\right).$ (5.9)

The unique existence and asymptotic of $M^{(r)}(z)$ will shown in section 6.

We now use $M^{(r)}(z)$ to construct a new matrix function

$M^{(3)}(z)=M^{(2)}(z)M^{(r)}(z)^{-1},$ (5.10)

which removes analytical component $M^{(r)}(z)$ to get a pure
$\bar{\partial}$-problem.

$\bar{\partial}$-problem4. Find a matrix-valued function $M^{(3)}(z)$ with
following properties:

$\blacktriangleright$ Analyticity: $M^{(3)}(z)$ is continuous and has
sectionally continuous first partial derivatives in $\mathbb{C}$.

$\blacktriangleright$ Asymptotic behavior:

$\displaystyle M^{(3)}(z)\sim I+\mathcal{O}(z^{-1}),\hskip
14.22636ptz\rightarrow\infty;$ (5.11)

$\blacktriangleright$ $\bar{\partial}$-Derivative: We have

$\bar{\partial}M^{(3)}(z)=M^{(3)}(z)W,\ \ z\in\mathbb{C},$

where

$W=M^{(r)}(z)\bar{\partial}R^{(2)}(z)M^{(r)}(z)^{-1}.$ (5.12)

###### Proof.

By using properties of the solutions $M^{(2)}(z)$ and $M^{(r)}(z)$ for RHP3
and $\bar{\partial}$-problem 4, the analyticity is obtained immediately. And
for its Asymptotic behavior, from
$M^{(r)}(z)^{-1}=(1+z^{-2})\sigma_{2}M^{(r)}(z)^{T}\sigma_{2}$ we have

$\displaystyle\lim_{z\to 0}M^{(3)}(z)=$ $\displaystyle\lim_{z\to
0}\dfrac{(zM^{(2)}(z))\sigma_{2}(zM^{(r)}(z)^{T})\sigma_{2}}{1+z^{2}}$
$\displaystyle=$ $\displaystyle
ie^{i\nu_{-}(x,t;q)\sigma_{3}}\sigma_{3}Q_{-}\sigma_{2}(ie^{i\nu_{-}(x,t;q)\sigma_{3}}\sigma_{3}Q_{-})^{T}\sigma_{2}=I.$
(5.13)

Since $M^{(2)}(z)$ and $M^{(r)}(z)$ achieve same jump matrix, we have

$\displaystyle M_{-}^{(3)}(z)^{-1}M_{+}^{(3)}(z)$
$\displaystyle=M_{-}^{(2)}(z)^{-1}M_{-}^{(r)}(z)M_{+}^{(r)}(z)^{-1}M_{+}^{(2)}(z)$
$\displaystyle=M_{-}^{(2)}(z)^{-1}V^{(2)}(z)^{-1}M_{+}^{(2)}(z)=I,$

which implies $M^{(3)}(z)$ has no jumps and is everywhere continuous. We also
can show that $M^{(3)}(z)$ has no pole. For
$\lambda\in\left\\{\zeta_{n},\bar{\zeta}_{n}\right\\}_{n\in\Lambda}$, let
$\mathcal{N}_{\lambda}$ denote the nilpotent matrix which appears in the left
side of the corresponding residue condition of RHP4 and RHP5, we have the
Laurent expansions in $z-\lambda$

$\displaystyle
M^{(2)}(z)=a(\lambda)\left[\dfrac{\mathcal{N}_{\lambda}}{z-\lambda}+I\right]+\mathcal{O}(z-\lambda),$
$\displaystyle
M^{(r)}(z)=A(\lambda)\left[\dfrac{\mathcal{N}_{\lambda}}{z-\lambda}+I\right]+\mathcal{O}(z-\lambda),$

where $a(\lambda)$ and $A(\lambda)$ are the constant matrix in their
respective expansions. Then

$\displaystyle M^{3}(z)$
$\displaystyle=\left\\{a(\lambda)\left[\dfrac{\mathcal{N}_{\lambda}}{z-\lambda}+I\right]\right\\}\left\\{\left[\dfrac{-\mathcal{N}_{\lambda}}{z-\lambda}+I\right]\sigma_{2}A(\lambda)^{T}\sigma_{2}\right\\}+\mathcal{O}(z-\lambda)$
$\displaystyle=\mathcal{O}(1),$ (5.14)

which implies that $M^{(3)}(z)$ has removable singularities at $\lambda$. And
the $\bar{\partial}$-derivative of $M^{(3)}(z)$ come from $M^{(3)}(z)$ due to
analyticity of $M^{(r)}(z)$. In addition, unlike the zero boundary case, we
must check its property at $\pm i$. The symmetries of $M^{(2)}(z)$ and
$M^{(r)}(z)$ imply that

$\displaystyle M^{(2)}(z)=$ $\displaystyle\left(\begin{array}[]{cc}\gamma&\pm
q_{-}\gamma\\\
\pm\bar{q}_{-}\bar{\gamma}&\bar{\gamma}\end{array}\right)+\mathcal{O}(z\mp
i),$ (5.17) $\displaystyle M^{(r)}(z)=$ $\displaystyle\dfrac{\pm i}{2(z\mp
i)}\left(\begin{array}[]{cc}\bar{\iota}&\mp q_{-}\iota\\\
\mp\bar{q}_{-}\bar{\iota}&\iota\end{array}\right)+\mathcal{O}(1),$ (5.20)

for two constants $\gamma$ and $\iota$. Then the singular part of $M^{(3)}(z)$
vanishes at $z=\pm i$ by simple calculation immediately. ∎

The unique existence and asymptotic of $M^{(3)}(z)$ will shown in section 7.

##  6  Asymptotic of $\mathcal{N}(\Lambda)$-soliton solutions

In this section, we build a reflectionless RH problem and show that its
solution can approximated with $M^{(r)}$.

First we show the existence and uniqueness of solution of the above RHP3 which
is related with original RH problem 0.

######  Proposition 6.

The solution $M^{(r)}(z)$ of the RH problem 3 with scattering data
$\left\\{r(z),\left\\{\zeta_{n},C_{n}\right\\}_{n\in\Lambda}\right\\}$ exists
and is unique. By an explicit transformation, $M^{(r)}(z)$ is equivalent to a
reflectionless solution of the original RHP0 with modified scattering data
$\left\\{0,\left\\{\zeta_{n},\mathring{c}_{n}\right\\}_{n\in\Lambda}\right\\}$,
where

$\mathring{c}_{n}(x,t)=C_{n}\exp\left\\{-\frac{1}{i\pi}\int_{\mathbb{R}}\log(1-|\rho(s)|^{2})\left(\frac{1}{s-\zeta_{n}}-\frac{1}{2s}\right)\right\\}.$
(6.1)

###### Proof.

To transform $M^{(r)}(z)$ to the soliton-solution of RHP0, the jumps and poles
need to be restored. We reverses the triangularity effected in (3.22) and
(4.51):

$N(z)=\left(\prod_{n\in\Delta}\zeta_{n}\right)^{-\sigma_{3}}M^{(r)}(z)T^{-\hat{\sigma}_{3}}G^{-1}(z)\left(\prod_{n\in\Delta}\dfrac{z-\zeta_{n}}{\bar{\zeta}_{n}^{-1}z-1}\right)^{-\sigma_{3}},$
(6.2)

with $G(z)$ defined in (3.21). First we verify $N(z)$ satisfying RHP0. This
transformation to $N(z)$ preserves the normalization conditions at the origin
and infinity obviously. And comparing with (3.22), this transformation restore
the jump on $\mathbb{D}(\bar{\zeta}_{n},\varrho)$ and
$\mathbb{D}(\zeta_{n},\varrho)$ to residue for $n\notin\Lambda$. As for
$n\in\Lambda$, take $\zeta_{n}$ as an example. Substitute (5.9) into the
transformation:

$\displaystyle\operatorname*{Res}_{z=\zeta_{n}}N(z)=$
$\displaystyle\left(\prod_{n\in\Delta}\zeta_{n}\right)^{-\sigma_{3}}\operatorname*{Res}_{z=\zeta_{n}}M^{(r)}(z)T^{-\hat{\sigma}_{3}}G(z)^{-1}\left(\prod_{n\in\Delta}\dfrac{z-\zeta_{n}}{\bar{\zeta}_{n}^{-1}z-1}\right)^{-\sigma_{3}}$
$\displaystyle=$
$\displaystyle\lim_{z\to\zeta_{n}}-\left(\prod_{n\in\Delta}\zeta_{n}\right)^{-\sigma_{3}}M^{(r)}(z)\left(\begin{array}[]{cc}0&0\\\
C_{n}e^{-2it\theta_{n}}T^{2}(\zeta_{n})&0\end{array}\right)\left(\prod_{n\in\Delta}\dfrac{z-\zeta_{n}}{\bar{\zeta}_{n}^{-1}z-1}\right)^{-\sigma_{3}}$
(6.5) $\displaystyle=$
$\displaystyle\lim_{z\to\zeta_{n}}N(z)\left(\begin{array}[]{cc}0&0\\\
\mathring{c}_{n}e^{-2it\theta_{n}}&0\end{array}\right).$ (6.8)

Its analyticity and symmetry follow from the Proposition of $M^{(r)}(z)$,
$T(z)$ and $G(z)$ immediately. So $N(z)$ is solution of RHP0 with absence of
reflection, whose unique exact solution exists and can be obtained as
described similarly in [18]. So $M^{(r)}(z)$ unique exists. ∎

Although $M^{(r)}(z)$ admits uniqueness and existence, we can’t give its
explicit expression. The jump matrix is uniformly near identity and doesn’t
contribute to the asymptotic behavior of the solution.

######  Lemma 2.

The jump matrix $V^{(2)}(z)$ in (4.53) satisfies

$\displaystyle\parallel
V^{(2)}-I\parallel_{L^{\infty}(\Sigma^{(2)})}=\mathcal{\mathcal{O}}(e^{-2\rho_{0}t}),$
(6.9)

where $\rho_{0}$ is defined by (3.3).

###### Proof.

Take $z\in\partial\mathbb{D}(\zeta_{n},\varrho),$ $n\in\nabla\setminus\Lambda$
as an example.

$\displaystyle\parallel
V^{(2)}-I\parallel_{L^{\infty}(\partial\mathbb{D}(\zeta_{n},\varrho))}$
$\displaystyle=|C_{n}(z-\zeta_{n})^{-1}T^{2}(z)e^{-2it\theta_{n}}|$
$\displaystyle\lesssim\varrho^{-1}e^{-\text{Re}(2it\theta_{n})}\lesssim
e^{2t\text{Im}(\theta_{n})}\leq e^{-2\rho_{0}t}.$ (6.10)

The last step follows from that for $n\in\nabla\setminus\Lambda$,
$\text{Im}\theta_{n}<0$. ∎

######  Corollary 1.

For $1\leq p\leq+\infty$, the jump matrix $V^{(2)}(z)$ satisfies

$\parallel V^{(2)}-I\parallel_{L^{p}(\Sigma^{(2)})}\leq K_{p}e^{-2\rho_{0}t},$
(6.11)

for some constant $K_{p}\geq 0$ depending on $p$.

This estimation of $V^{(2)}$ inspires us to consider to completely ignore the
jump condition on $M^{(r)}(z)$, because there is only exponentially small
error (in t). We decompose $M^{(r)}(z)$ as

$M^{(r)}(z)=E(z)M^{(r)}_{\Lambda}(z),$ (6.12)

where $E(z)$ is a error function, which is a solution of a small-norm RH
problem and we discuss it in Section 8. $M^{(r)}_{\Lambda}(z)$ solves RHP3
with $V^{(2)}\equiv 0$.

RHP5. Find a matrix-valued function $M^{(r)}_{\Lambda}(z;x,t)$ with following
properties:

$\blacktriangleright$ Analyticity: $M^{(r)}_{\Lambda}(z;x,t)$ is analytical in
$\mathbb{C}\setminus\left\\{\zeta_{n},\bar{\zeta}_{n}\right\\}_{n\in\Lambda}$;

$\blacktriangleright$ Symmetry:
$M^{(r)}_{\Lambda}(z)=\sigma_{2}\overline{M^{(r)}_{\Lambda}(\bar{z})}\sigma_{2}$=$\sigma_{1}\overline{M^{(r)}_{\Lambda}(-\bar{z})}\sigma_{1}=\frac{i}{z}M^{(r)}_{\Lambda}(-1/z)\sigma_{3}Q_{-}$;

$\blacktriangleright$ Asymptotic behaviors:

$\displaystyle
M^{(r)}_{\Lambda}(z;x,t)=e^{i\nu_{-}(x,t;q)\sigma_{3}}+\mathcal{O}(z^{-1}),\hskip
14.22636ptz\rightarrow\infty,$ (6.13) $\displaystyle
M^{(r)}_{\Lambda}(z;x,t)=\frac{i}{z}e^{i\nu_{-}(x,t;q)\sigma_{3}}\sigma_{3}Q_{-}+\mathcal{O}(1),\hskip
14.22636ptz\rightarrow 0;$ (6.14)

$\blacktriangleright$ Residue conditions: $M^{(r)}_{\Lambda}$ has simple poles
at each point $\zeta_{n}$ and $\bar{\zeta}_{n}$ for $n\in\Lambda$ with:

$\displaystyle\operatorname*{Res}_{z=\zeta_{n}}M^{(r)}_{\Lambda}(z)=\lim_{z\to\zeta_{n}}M^{(r)}_{\Lambda}(z)\left(\begin{array}[]{cc}0&0\\\
C_{n}e^{-2it\theta_{n}}T^{2}(\zeta_{n})&0\end{array}\right),$ (6.17)
$\displaystyle\operatorname*{Res}_{z=\bar{\zeta}_{n}}M^{(r)}_{\Lambda}(z)=\lim_{z\to\bar{\zeta}_{n}}M^{(r)}_{\Lambda}(z)\left(\begin{array}[]{cc}0&-\bar{C}_{n}T^{-2}(\bar{\zeta}_{n})e^{2it\bar{\theta}_{n}}\\\
0&0\end{array}\right).$ (6.20)

######  Proposition 7.

The RHP5 exists an unique solution. Moreover, $M^{(r)}_{\Lambda}(z)$ is
equivalent to a reflectionless solution of the original RHP0 with modified
scattering data
$\left\\{0,\left\\{\zeta_{n},\mathring{c}_{n}\right\\}_{n\in\Lambda}\right\\}$
as follows:
Case I: if $\Lambda=\varnothing$, then

$M^{(r)}_{\Lambda}(z)=e^{i\nu_{-}(x,t;q^{r}_{\Lambda})\sigma_{3}}+\frac{i}{z}e^{i\nu_{-}(x,t;q^{r}_{\Lambda})\sigma_{3}}\sigma_{3}Q_{-};$
(6.21)

Case I: if $\Lambda\neq\varnothing$ with
$\Lambda_{1}=\left\\{z_{j_{k}}\right\\}_{k=1}^{n_{1}}$ and
$\Lambda_{2}=\left\\{w_{i_{s}}\right\\}_{s=1}^{n_{2}}$, then

$\displaystyle M^{(r)}_{\Lambda}(z)$
$\displaystyle=e^{i\nu_{-}(x,t;q^{r}_{\Lambda})\sigma_{3}}+\frac{i}{z}e^{i\nu_{-}(x,t;q^{r}_{\Lambda})\sigma_{3}}\sigma_{3}Q_{-}$
$\displaystyle+\sum_{s=1}^{n_{2}}\left[\left(\begin{array}[]{cc}\frac{\alpha_{s}}{z-w_{i_{s}}}&\frac{\overline{\kappa_{s}}}{z-\bar{w}_{i_{s}}}\\\
\frac{\kappa_{s}}{z-w_{i_{s}}}&\frac{\overline{\alpha_{s}}}{z-\bar{w}_{i_{s}}}\end{array}\right)+\left(\begin{array}[]{cc}-\frac{\alpha_{s}}{z+w_{i_{s}}}&\frac{\overline{\kappa_{s}}}{z+\bar{w}_{i_{s}}}\\\
\frac{\kappa_{s}}{z+w_{i_{s}}}&-\frac{\overline{\alpha_{s}}}{z+\bar{w}_{i_{s}}}\end{array}\right)\right]$
(6.26)
$\displaystyle+\sum_{k=1}^{n_{1}}\left[\left(\begin{array}[]{cc}\frac{\beta_{k}}{z-z_{j_{k}}}&\frac{\overline{\varsigma_{k}}}{z-\bar{z}_{j_{k}}}\\\
\frac{\varsigma_{k}}{z-z_{j_{k}}}&\frac{\overline{\beta_{k}}}{z-\bar{z}_{j_{k}}}\end{array}\right)+\left(\begin{array}[]{cc}-\frac{\beta_{k}}{z+z_{j_{k}}}&\frac{\overline{\varsigma_{k}}}{z+\bar{z}_{j_{k}}}\\\
\frac{\varsigma_{k}}{z+z_{j_{k}}}&-\frac{\overline{\beta_{k}}}{z+\bar{z}_{j_{k}}}\end{array}\right)\right]$
(6.31)
$\displaystyle+\sum_{k=1}^{n_{1}}i\left[\left(\begin{array}[]{cc}\frac{-\overline{q_{-}\beta_{k}}}{\bar{z}_{j_{k}}z-1}&\frac{-q_{-}\varsigma_{k}}{z_{j_{k}}z-1}\\\
\frac{-\overline{q_{-}\varsigma_{k}}}{\bar{z}_{j_{k}}z-1}&\frac{q_{-}\beta_{k}}{z_{j_{k}}z-1}\end{array}\right)+\left(\begin{array}[]{cc}\frac{\overline{q_{-}\beta_{k}}}{\bar{z}_{j_{k}}z+1}&\frac{-q_{-}\varsigma_{k}}{z_{j_{k}}z+1}\\\
\frac{-\overline{q_{-}\varsigma_{k}}}{\bar{z}_{j_{k}}z+1}&\frac{-q_{-}\beta_{k}}{z_{j_{k}}z+1}\end{array}\right)\right],$
(6.36)

where $\beta_{k}=\beta_{k}(x,t)$, $\varsigma_{k}=\varsigma_{k}(x,t)$,
$\alpha_{s}=\alpha_{s}(x,t)$ and $\kappa_{s}=\kappa_{s}(x,t)$ with linearly
dependant equations:

$\displaystyle
c_{j_{k}}^{-1}T(z_{j_{k}})^{-2}e^{-2i\theta(z_{j_{k}})t}\beta_{k}$
$\displaystyle=\frac{i}{z_{j_{k}}}e^{i\nu_{-}(x,t;q^{r}_{\Lambda})}q_{-}+\sum_{h=1}^{n_{2}}\left(\frac{\overline{\kappa_{h}}}{z_{j_{k}}-\bar{w}_{i_{h}}}+\frac{\overline{\kappa_{h}}}{z_{j_{k}}+\bar{w}_{i_{h}}}\right)$
$\displaystyle+\sum_{l=1}^{n_{1}}\left(\frac{\overline{\varsigma_{l}}}{z_{j_{k}}-\bar{z}_{j_{l}}}+\frac{\overline{\varsigma_{l}}}{z_{j_{k}}+\bar{z}_{j_{l}}}-\frac{iq_{-}\varsigma_{l}}{z_{j_{l}}z_{j_{k}}-1}-\frac{iq_{-}\varsigma_{l}}{z_{j_{l}}z_{j_{k}}+1}\right),$
(6.37) $\displaystyle
c_{j_{k}}^{-1}T(z_{j_{k}})^{-2}e^{-2i\theta(z_{j_{k}})t}\varsigma_{k}$
$\displaystyle=\frac{i}{z_{j_{k}}}e^{-i\nu_{-}(x,t;q^{r}_{\Lambda})}\bar{q}_{-}+\sum_{h=1}^{n_{2}}\left(\frac{\overline{\alpha_{h}}}{z_{j_{k}}-\bar{w}_{i_{h}}}-\frac{\overline{\alpha_{h}}}{z_{j_{k}}+\bar{w}_{i_{h}}}\right)$
$\displaystyle+\sum_{l=1}^{n_{1}}\left(\frac{\overline{\beta_{l}}}{z_{j_{k}}-\bar{z}_{j_{l}}}-\frac{\overline{\beta_{l}}}{z_{j_{k}}+\bar{z}_{j_{l}}}+\frac{iq_{-}\beta_{l}}{z_{j_{l}}z_{j_{k}}-1}-\frac{iq_{-}\beta_{l}}{z_{j_{l}}z_{j_{k}}+1}\right),$
(6.38)

and

$\displaystyle
c_{i_{s}+N_{1}}^{-1}T(w_{i_{s}})^{-2}e^{-2i\theta(w_{i_{s}})t}\alpha_{k}$
$\displaystyle=\frac{i}{w_{i_{s}}}e^{i\nu_{-}(x,t;q^{r}_{\Lambda})}q_{-}+\sum_{h=1}^{n_{2}}\left(\frac{\overline{\kappa_{h}}}{w_{i_{s}}-\bar{w}_{i_{h}}}+\frac{\overline{\kappa_{h}}}{w_{i_{s}}+\bar{w}_{i_{h}}}\right)$
$\displaystyle+\sum_{l=1}^{n_{1}}\left(\frac{\overline{\varsigma_{l}}}{w_{i_{s}}-\bar{z}_{j_{l}}}+\frac{\overline{\varsigma_{l}}}{w_{i_{s}}+\bar{z}_{j_{l}}}-\frac{iq_{-}\varsigma_{l}}{z_{j_{l}}w_{i_{s}}-1}-\frac{iq_{-}\varsigma_{l}}{z_{j_{l}}w_{i_{s}}+1}\right),$
(6.39) $\displaystyle
c_{i_{s}+N_{1}}^{-1}T(w_{i_{s}})^{-2}e^{-2i\theta(w_{i_{s}})t}\kappa_{k}$
$\displaystyle=\frac{i}{w_{i_{s}}}e^{-i\nu_{-}(x,t;q^{r}_{\Lambda})}\bar{q}_{-}+\sum_{h=1}^{n_{2}}\left(\frac{\overline{\alpha_{h}}}{w_{i_{s}}-\bar{w}_{i_{h}}}-\frac{\overline{\alpha_{h}}}{w_{i_{s}}+\bar{w}_{i_{h}}}\right)$
$\displaystyle+\sum_{l=1}^{n_{1}}\left(\frac{\overline{\beta_{l}}}{w_{i_{s}}-\bar{z}_{j_{l}}}-\frac{\overline{\beta_{l}}}{w_{i_{s}}+\bar{z}_{j_{l}}}+\frac{iq_{-}\beta_{l}}{z_{j_{l}}w_{i_{s}}-1}-\frac{iq_{-}\beta_{l}}{z_{j_{l}}w_{i_{s}}+1}\right),$
(6.40)

for $k=1,...,n_{1}$, $s=1,...,n_{2}$ respectively.

###### Proof.

The uniqueness of solution follows from the Liouville’s theorem. Case I can be
simple obtain. As for Case II, the symmetries of $M^{(r)}_{\Lambda}(z)$ means
that it admits a partial fraction expansion of following form as above. And to
obtain $\beta_{k}$, $\varsigma_{k}$,$\alpha_{s}$ and $\kappa_{s}$, we
substitute (6.36) into (6.20) and obtain four linearly dependant equations set
above. ∎

######  Corollary 2.

When $\rho(s)\equiv 0$, the scattering matrices $S(z)\equiv I$, which means
$q_{-}=q_{+}$. Denote $q^{r}_{\Lambda}(x,t)$ is the
$\mathcal{N}(\Lambda)$-soliton with scattering data
$\left\\{0,\left\\{\zeta_{n},\mathring{c}_{n}\right\\}_{n\in\Lambda}\right\\}$.
By the reconstruction formula (2.51) and (2.52), the solution
$q^{r}_{\Lambda}(x,t)$ of (1.1) with scattering data
$\left\\{0,\left\\{\zeta_{n},\mathring{c}_{n}\right\\}_{n\in\Lambda}\right\\}$
is given by:

$\displaystyle
q^{r}_{\Lambda}(x,t)=e^{i\nu_{-}(x,t;q^{r}_{\Lambda})}\lim_{z\to\infty}z\left[M^{(r)}_{\Lambda}\right]_{12}.$
(6.41)

Then in case I,

$u^{r}_{\Lambda}(x,t)=\lim_{z\to\infty}z|\left[M^{r}_{\Lambda}\right]_{12}|=1.$
(6.42)

So $\nu_{-}(x,t;q^{r}_{\Lambda})=0$ and

$\displaystyle q^{r}_{\Lambda}(x,t)=q_{-}.$ (6.43)

And in case II,

$\displaystyle u^{r}_{\Lambda}(x,t)$
$\displaystyle=\lim_{z\to\infty}z|\left[M^{(r)}_{\Lambda}\right]_{12}|$
$\displaystyle=|ie^{i\nu_{-}(x,t;q^{r}_{\Lambda})}q_{-}+2\sum_{s=1}^{n_{2}}\bar{\kappa}_{k}+2\sum_{k=1}^{n_{1}}(\bar{\varsigma}_{k}-iq_{-}\varsigma_{k})|,$
(6.44)

which leads to
$\nu_{-}(x,t;q^{r}_{\Lambda})=\frac{1}{2}\int_{-\infty}^{x}(|u^{r}_{\Lambda}(y,t)|^{2}-1)dy$
and

$\displaystyle q^{r}_{\Lambda}(x,t)$
$\displaystyle=\lim_{z\to\infty}e^{i\nu_{-}(x,t;q^{r}_{\Lambda})}z\left[M^{(r)}_{\Lambda}\right]_{12}$
$\displaystyle=e^{2i\nu_{-}(x,t;q^{r}_{\Lambda})}\left(ie^{i\nu_{-}(x,t;q^{r}_{\Lambda})}q_{-}+2\sum_{s=1}^{n_{2}}\bar{\kappa}_{k}+2\sum_{k=1}^{n_{1}}(\bar{\varsigma}_{k}-iq_{-}\varsigma_{k})\right).$
(6.45)

##  7 The small norm RH problem for error function

In this section, we consider the error matrix-function $E(z)$ and show that
for large times, the error function $E(z)$ solves a small norm RH problem
which can be expanded asymptotically. From the definition (6.12), we can
obtain a RH problem for the matrix function $E(z)$.

RHP6 Find a matrix-valued function $E(z)$ with following properties:

$\blacktriangleright$ Analyticity: $E(z)$ is analytical in
$\mathbb{C}\setminus\Sigma^{(2)}$;

$\blacktriangleright$ Asymptotic behaviors:

$\displaystyle E(z)\sim I+\mathcal{O}(z^{-1}),\hskip
14.22636pt|z|\rightarrow\infty;$ (7.1)

$\blacktriangleright$ Jump condition: $E$ has continuous boundary values
$E_{\pm}$ on $\Sigma^{(2)}$ satisfying

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

where the jump matrix $V^{E}$ is given by

$V^{E}(z)=M^{(r)}_{\Lambda}(z)V^{(2)}(z)M^{(r)}_{\Lambda}(z)^{-1}.$ (7.2)

Proposition 7 implies that $M^{(r)}_{\Lambda}(z)$ is bound on $\Sigma^{(2)}$.
By using Lemma 2 and Corollary 1, we have the following estimates

$\parallel V^{E}-I\parallel_{L^{p}}\lesssim\parallel
V^{(2)}-I\parallel_{L^{p}}=\mathcal{O}(e^{-2\rho_{0}t}),$ (7.3)

for $1\leq p\leq+\infty$. This uniformly vanishing bound $\parallel
V^{E}-I\parallel$ establishes RHP6 as a small-norm RH problem. Therefore, the
existence and uniqueness of the RHP6 can shown by using a small-norm RH
problem

$E(z)=I+\frac{1}{2\pi
i}\int_{\Sigma^{(2)}}\dfrac{\left(I+\eta(s)\right)(V^{E}-I)}{s-z}ds,$ (7.4)

where the $\eta\in L^{2}(\Sigma^{(2)})$ is the unique solution of following
equation

$(1-C_{E})\eta=C_{E}\left(I\right),$ (7.5)

here $C_{E}$:$L^{2}(\Sigma^{(2)})\to L^{2}(\Sigma^{(2)})$ is a integral
operator defined by

$C_{E}(f)(z)=C_{-}\left(f(V^{E}-I)\right).$ (7.6)

The Cauchy projection operator $C_{-}$ on $\Sigma^{(2)}$ is

$C_{-}(f)(s)=\lim_{z\to\Sigma^{(2)}_{-}}\frac{1}{2\pi
i}\int_{\Sigma^{(2)}}\dfrac{f(s)}{s-z}ds.$ (7.7)

Then by (7.2) we have

$\parallel C_{E}\parallel\leq\parallel C_{-}\parallel\parallel
V^{E}-I\parallel_{L^{\infty}}\lesssim\mathcal{O}(e^{-2\rho_{0}t}),$ (7.8)

which means $\parallel C_{E}\parallel<1$ for sufficiently large t, therefore
$1-C_{E}$ is invertible, and $\eta$ exists and is unique. Moreover,

$\parallel\eta\parallel_{L^{2}(\Sigma^{(2)})}\lesssim\dfrac{\parallel
C_{E}\parallel}{1-\parallel
C_{E}\parallel}\lesssim\mathcal{O}(e^{-2\rho_{0}t}).$ (7.9)

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

######  Proposition 8.

For $E(z)$ defined in (7.4), it stratifies

$|E(z)-I|\lesssim\mathcal{O}(e^{-2\rho_{0}t}).$ (7.10)

As $z\to\infty$, the large $z$ expansion of $E$ is

$\displaystyle E(z)=I+E_{1}z^{-1}+\mathcal{O}(z^{-2}),$ (7.11)

where

$E_{1}=-\frac{1}{2\pi i}\int_{\Sigma^{(2)}}\left(I+\eta(s)\right)(V^{E}-I)ds,$
(7.12)

satisfying long time asymptotic behavior condition

$E_{1}\lesssim\mathcal{O}(e^{-2\rho_{0}t}).$ (7.13)

###### Proof.

By combining (7.9) and (7.3), we obtain

$|E(z)-I|\leq|(1-C_{E})(\eta)|+|C_{E}(\eta)|\lesssim\mathcal{O}(e^{-2\rho_{0}t}).$
(7.14)

As $z\to\infty$,geometrically expanding $(s-z)^{-1}$ for $z$ large in (7.4)
leads to (7.11). Finally for $E_{1}$,

$\displaystyle|E_{1}|\lesssim\parallel
V^{E}-I\parallel_{L^{1}}+\parallel\eta\parallel_{L^{2}}2\parallel
V^{E}-I\parallel_{L^{2}}\lesssim\mathcal{O}(e^{-2\rho_{0}t}).$ (7.15)

∎

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

Now we consider the asymptotics behavior of $M^{(3)}(z)$. The
$\bar{\partial}$-problem 4 of $M^{(3)}(z)$ is equivalent to the integral
equation

$M^{(3)}(z)=I+\frac{1}{\pi}\int_{\mathbb{C}}\dfrac{M^{(3)}(s)W^{(3)}(s)}{z-s}dm(s),$
(8.1)

where $m(s)$ is the Lebesgue measure on the $\mathbb{C}$. Denote $C_{z}$ as
the left Cauchy-Green integral operator defined by

$fC_{z}(z)=\frac{1}{\pi}\int_{C}\dfrac{f(s)W^{(3)}(s)}{z-s}dm(s).$

Then above equation (8.1) can be rewritten as

$M^{(3)}(z)=I\cdot\left(I-C_{z}\right)^{-1}.$ (8.2)

The existence of operator $\left(I-C_{z}\right)^{-1}$ is given by the
following Lemma.

######  Lemma 3.

The norm of the integral operator $C_{z}$ decay to zero as $t\to\infty$:

$\parallel C_{z}\parallel_{L^{\infty}\to L^{\infty}}\lesssim t^{-1/2},$ (8.3)

which implies that $\left(I-C_{z}\right)^{-1}$ exists.

###### Proof.

For any $f\in L^{\infty}$,

$\displaystyle\parallel fC_{z}\parallel_{L^{\infty}}$
$\displaystyle\leq\parallel
f\parallel_{L^{\infty}}\frac{1}{\pi}\int_{C}\dfrac{|W^{(3)}(s)|}{|z-s|}dm(s),$

where $W(s)=M^{(r)}(z)\bar{\partial}R^{(2)}(z)M^{(r)}(z)^{-1}$. So we only
need to estimate the integral

$\frac{1}{\pi}\int_{C}\dfrac{|W(s)|}{|z-s|}dm(s).$

Since $W(s)\equiv 0$ out of $\bar{\Omega}$, we only need to focus on the
estimate

$\frac{1}{\pi}\int_{\Omega}\dfrac{|W(s)|}{|z-s|}dm(s).$

Unlike the zero boundary case in [16], here $\det M^{(r)}(z)=1+z^{-2}$, and
Proposition 8 implies that $|M^{(r)}(z)|\lesssim\sqrt{1+|z|^{-2}}$. So

$\frac{1}{\pi}\int_{\Omega}\dfrac{|W(s)|}{|z-s|}dm(s)\lesssim\frac{1}{\pi}\int_{\Omega}\dfrac{|\bar{\partial}R^{(2)}(s)|}{|z-s|}\frac{1+|s|^{-2}}{|1+s^{-2}|}dm(s).$
(8.4)

Then for $j=1,4,5,8$, $|M^{(r)}(z)|$ is bounded in $\Omega_{j}$. But when
$z\in\Omega_{j}$ for $j=2,3,6,7$, the singularity at $z=\pm i$ need to be
treat more carefully. So in following calculation, we take $\Omega_{2}$ in the
second case as an example, because it is more elaborate than $\Omega_{j}$ for
$j=1,4,5,8$. Denote three sub-region of $\Omega_{2}$ as

$\displaystyle D_{1}=\mathbb{D}(0,1-\epsilon_{0})\cap\Omega_{2}$
$\displaystyle,\hskip
8.5359ptD_{2}=\mathbb{D}(0,1+\epsilon_{0})\setminus\mathbb{D}(0,1-\epsilon_{0})\cap\Omega_{2},$
$\displaystyle D_{3}=\Omega_{2}\setminus\mathbb{D}(0,1+\epsilon_{0}).$ (8.5)

Then the integral $\int_{\Omega_{2}}\dfrac{|W(s)|}{|z-s|}dm(s)$ is divide to
three part:

$\displaystyle
I_{i}=\int_{D_{i}}\dfrac{|\bar{\partial}R^{(2)}(s)|}{|z-s|}\frac{1+|s|^{-2}}{|1+s^{-2}|}dm(s),\text{
for }i=1,2,3.$ (8.6)

Let $s=u+vi=re^{i\vartheta}$, $z=x+yi$. In the following calculation, we will
use the inequality

$\parallel|s-z|^{-1}\parallel_{L^{q}(\mathbb{R}^{+})}=\left\\{\int^{+\infty}_{0}\left[\left(\frac{v-y}{u-x}\right)^{2}+1\right]^{-\frac{q}{2}}d\left(\frac{v-y}{|u-x|}\right)\right\\}^{\frac{1}{q}}|u-x|^{-\frac{1}{p}}\lesssim|u-x|^{-\frac{1}{p}},$
(8.7)

with $1\leq q<+\infty$ and $\frac{1}{p}+\frac{1}{q}=1$. For $s\in D_{3}$,
$|s|>1+\epsilon_{0}$, then

$\frac{1+|s|^{-2}}{|1+s^{-2}|}<\frac{1+|s|^{2}}{|s|^{2}-1}<1+\frac{2}{\epsilon_{0}^{2}+2\epsilon_{0}}<\infty.$
(8.8)

Then together with (4.14), we have

$\displaystyle
I_{3}\lesssim\int_{\Omega_{2}}\dfrac{|\bar{\partial}R^{(2)}(s)|}{|z-s|}dm(s)=\int_{\Omega_{2}}\dfrac{|\bar{\partial}R_{2}(s)e^{2it\theta}|}{|z-s|}dm(s).$
(8.9)

Moreover, by Lemma 1,

$|e^{2it\theta}|\leq e^{-c\sin 2\vartheta F(r)^{2}}\leq e^{-2cuv}\leq
e^{-2cu},$ (8.10)

where $c$ is a positive constant, and the last step follows form

$v\geq\max\left\\{1+\epsilon_{0},\frac{u}{\tan\varphi}\right\\}\geq
1+\epsilon_{0}>1.$

Substitute (4.35) and above inequality into (8.9) and obtain:

$\displaystyle
I_{3}\lesssim\int_{0}^{+\infty}\int_{\frac{u}{\tan\varphi}}^{+\infty}\dfrac{|p_{2}^{\prime}(ir)|e^{-4cut}}{|z-s|}dvdu+\int_{0}^{+\infty}\int_{\frac{u}{\tan\varphi}}^{+\infty}\dfrac{|r|^{-1/2}e^{-2cut}}{|z-s|}dvdu$
$\displaystyle+\int_{0}^{+\infty}\int_{\frac{u}{\tan\varphi}}^{+\infty}\dfrac{|\bar{\partial}X_{1}(r)|e^{-4cut}}{|z-s|}dvdu.$

By Cauchy-Schwarz inequality, the first item have

$\displaystyle\int_{0}^{+\infty}\int_{\frac{u}{\tan\varphi}}^{+\infty}\dfrac{|p_{2}^{\prime}(ir)|e^{-4cut}}{|z-s|}dvdu$
$\displaystyle\leq\int_{0}^{+\infty}\parallel\tilde{\rho}^{\prime}\parallel_{L^{2}(i\mathbb{R})}\parallel|s-z|^{-1}\parallel_{L^{2}(\mathbb{R}^{+})}e^{-2cut}du$
$\displaystyle\leq\int_{0}^{+\infty}e^{-2cut}|u-x|^{-\frac{1}{2}}du\lesssim
t^{-\frac{1}{2}}.$ (8.11)

So does the last item. Before we estimating the second item, we consider for
$p>2$,

$\displaystyle\left(\int_{\frac{u}{\tan\varphi}}^{+\infty}|\sqrt{u^{2}+V^{(2)}}|^{-\frac{p}{2}}dv\right)^{\frac{1}{p}}$
$\displaystyle=\left(\int_{\frac{u}{\sin\varphi}}^{+\infty}|r|^{-\frac{p}{2}+1}v^{-1}dr\right)^{\frac{1}{p}}\lesssim
u^{-\frac{1}{2}+\frac{1}{p}}.$ (8.12)

Then

$\displaystyle\int_{0}^{+\infty}\int_{\frac{u}{\tan\varphi}}^{+\infty}\dfrac{|r|^{-1/2}e^{-2cut}}{|z-s|}dvdu$
$\displaystyle\leq\int_{0}^{+\infty}\parallel|r|\parallel_{L_{v}^{p}(\frac{u}{\tan\varphi},+\infty)}\parallel|s-z|^{-1}\parallel_{L^{q}(\mathbb{R}^{+})}e^{-2cut}du$
$\displaystyle\leq\int_{0}^{+\infty}e^{-2cut}|u-x|^{-\frac{1}{p}}u^{-\frac{1}{2}+\frac{1}{p}}du\lesssim
t^{-\frac{1}{2}}.$ (8.13)

Combing above inequality we final have $I_{3}\lesssim t^{-\frac{1}{2}}$. As
for $I_{2}$, the singularity at $i$ can be balanced by (4.36), and recall that
$1>\epsilon_{0}>0$ with $(1-\epsilon_{0})\cos\varphi>\frac{1}{2}$

$\displaystyle I_{2}$
$\displaystyle\lesssim\int_{0}^{2}\int_{1/2}^{2}\dfrac{e^{-2cut}}{|z-s|}\frac{1+|s|^{2}}{|s+i|}dvdu\lesssim\int_{0}^{2}\int_{1/2}^{2}\dfrac{e^{-2cut}}{|z-s|}dvdu$
$\displaystyle\lesssim\int_{0}^{2}|u-x|^{-1/2}e^{-2cut}du\lesssim|t|^{-1/2}.$
(8.14)

Finally, consider $I_{1}$, similarly we have

$\displaystyle
I_{1}\lesssim\int_{0}^{1-\epsilon_{0}}\int_{u}^{1-\epsilon_{0}}\left(|p_{2}^{\prime}(ir)|+|r|^{-1/2}+|\bar{\partial}X_{1}(r)|\right)\dfrac{e^{-2cut}}{|z-s|}dvdu,$
(8.15)

which can be estimated same as $I_{3}$. So the proof is completed. ∎

As $z\to\infty$, $M^{(3)}(z)$ has asymptotic expansion:

$M^{(3)}(z)=I-M^{(3)}_{1}(x,t)z+\mathcal{O}(z^{-2}),$ (8.16)

where $M^{(3)}_{1}$ is a $z$-independent coefficient. The asymptotic behavior
of $M^{(3)}_{1}$ given by following Proposition.

######  Proposition 9.

As $z\to\infty$, the expansion above holds with

$M^{(3)}_{1}(x,t)=\frac{1}{\pi}\int_{C}M^{(3)}(s)W^{(3)}(s)dm(s).$ (8.17)

There exist constants $T_{1}$, such that for all $t>T_{1}$, $M^{(3)}_{1}(x,t)$
satisfies

$|M^{(3)}_{1}(x,t)|\lesssim t^{-3/4}.$ (8.18)

###### Proof.

Lemma 3 and (8.2) implies that for large $t$, $\parallel
M^{(3)}(z)\parallel_{L^{\infty}}\lesssim 1$. The proof proceeds along the same
lines as the proof of above Proposition. For same reason, we only estimate the
integral on $\Omega_{2}$. Like in the above Proposition,

$\frac{1}{\pi}\int_{\Omega_{2}}M^{(3)}(s)W^{(3)}(s)dm(s)\lesssim\frac{1}{\pi}\int_{\Omega_{2}}|\bar{\partial}R_{2}(s)e^{2it\theta}|\frac{1+|s|^{-2}}{|1+s^{-2}|}dm(s).$
(8.19)

Let $s=u+vi=re^{i\vartheta}$. And we also divide right integral of above
inequality to three part

$I_{i+3}=\frac{1}{\pi}\int_{D_{i}}|\bar{\partial}R_{2}(s)e^{2it\theta}|\frac{1+|s|^{-2}}{|1+s^{-2}|}dm(s).$
(8.20)

For $I_{4}$, $\frac{1+|s|^{-2}}{|1+s^{-2}|}<\infty$, so

$\displaystyle I_{4}\lesssim$
$\displaystyle\int_{0}^{+\infty}\int_{\frac{u}{\tan\varphi}}^{+\infty}|p_{2}^{\prime}(ir)|e^{-2cuvt}dvdu+\int_{0}^{+\infty}\int_{\frac{u}{\tan\varphi}}^{+\infty}|r|^{-\frac{1}{2}}e^{-2cuvt}dvdu$
$\displaystyle+\int_{0}^{+\infty}\int_{\frac{u}{\tan\varphi}}^{+\infty}|\bar{\partial}X_{1}(r)|e^{-2cuvt}dvdu.$
(8.21)

Note that

$\displaystyle\left(\int_{\frac{u}{\tan\varphi}}^{+\infty}e^{-2cuvtq}dv\right)^{\frac{1}{q}}$
$\displaystyle=\left(\int_{\frac{u}{\tan\varphi}}^{+\infty}e^{-2cuvtq}d(2cuvtq)\right)^{\frac{1}{q}}(2cutq)^{-\frac{1}{q}}$
$\displaystyle\lesssim e^{-c^{\prime}u^{2}t}(ut)^{-\frac{1}{q}},$ (8.22)

where $c^{\prime}$ is a positive constant. Then the first integral in (8.21)
have

$\displaystyle\int_{0}^{+\infty}\int_{\frac{u}{\tan\varphi}}^{+\infty}|p_{2}^{\prime}(ir)|e^{-2cuvt}dvdu$
$\displaystyle\lesssim
t^{-\frac{1}{2}}\int_{0}^{+\infty}\parallel\tilde{\rho}^{\prime}\parallel_{L^{2}(i\mathbb{R})}u^{-\frac{1}{q}}e^{-c^{\prime}u^{2}t}du\lesssim
t^{-\frac{3}{4}}.$

The last integral can be bounded in same way. To estimate the second item, we
also use Cauchy-Schwarz inequality for $4>p>2$ and $\frac{1}{q}+\frac{1}{p}=1$

$\displaystyle\int_{0}^{+\infty}\int_{\frac{u}{\tan\varphi}}^{+\infty}|r|^{-\frac{1}{2}}e^{-2cuvt}dvdu\lesssim
t^{-\frac{1}{q}}\int_{0}^{+\infty}u^{\frac{2}{p}-\frac{3}{2}}e^{-c^{\prime}u^{2}t}du\lesssim
t^{-\frac{3}{4}}.$ (8.23)

The bound for $I_{4}$ follows in the same manner as for $I_{6}$. Turning to
$I_{5}$, we also use $|\bar{\partial}R_{2}(z)|\lesssim|z-i|$ and obtain

$\displaystyle I_{5}$
$\displaystyle\lesssim\int_{D_{2}}\dfrac{e^{-2cut}}{|z-s|}\frac{1+|s|^{2}}{|s+i|}dm(s)\lesssim\int_{1/2}^{2}\int_{u}^{2}e^{-2cuvt}dvdu$
$\displaystyle=\int_{1/2}^{2}(cut)^{-1}\left(e^{-2cu^{2}t}-e^{-4cut}\right)du\lesssim
t^{-1}.$ (8.24)

This estimate is strong enough to obtain the result. ∎

##  9 Asymptotic for the DNLS equation

Now we begin to construct the long time asymptotics of the DNLS equation
(1.1). Inverting the sequence of transformations (3.22), (4.51), (5.10) and
(6.12), we have

$\displaystyle M(z)=$ $\displaystyle
T(\infty)^{\sigma_{3}}M^{(3)}(z)E(z)M^{(r)}_{\Lambda}(z)R^{(2)}(z)^{-1}T(z)^{-\sigma_{3}}.$
(9.1)

To reconstruct the solution $q(x,t)$ by using (2.50), we take $z\to\infty$ out
of $\bar{\Omega}$. In this case, $R^{(2)}(z)=I$. Further using Propositions 4,
8 and 9, we can obtain that

$\displaystyle M(z)=$ $\displaystyle
T(\infty)^{\sigma_{3}}\left(I+M^{(3)}_{1}(z)z^{-1}\right)E(z)M^{(r)}_{\Lambda}(z)$
$\displaystyle T(\infty)^{-\sigma_{3}}\left(1+z^{-1}\frac{1}{2\pi
i}\int_{\mathbb{R}}\log(1-\rho(s)\tilde{\rho}(s))ds\right)^{-\sigma_{3}}+\mathcal{O}(z^{-2}),$
(9.2)

whose admits long time asymptotics

$\displaystyle M(z)=$ $\displaystyle
T(\infty)^{\sigma_{3}}M^{(r)}_{\Lambda}(z)T(\infty)^{-\sigma_{3}}\left(1+z^{-1}\frac{1}{2\pi
i}\int_{\mathbb{R}}\log(1-\rho(s)\tilde{\rho}(s))ds\right)^{-\sigma_{3}}$
$\displaystyle+\mathcal{O}(z^{-2})+\mathcal{O}(t^{-3/4}).$

From (2.51),

$\displaystyle|m(x,t)|$
$\displaystyle=|\lim_{z\to\infty}z\left[M(z)\right]_{12}|=|T(\infty)^{-2}||\lim_{z\to\infty}z\left[M^{(r)}_{\Lambda}(z)\right]_{12}$
$\displaystyle+\lim_{z\to\infty}\left[M^{(r)}_{\Lambda}(z)\right]_{12}\frac{1}{2\pi
i}\int_{\mathbb{R}}\log(1-\rho(s)\tilde{\rho}(s))ds|+\mathcal{O}(t^{-3/4})$
$\displaystyle=|q^{r}_{\Lambda}(x,t)|+\mathcal{O}(t^{-3/4}),$ (9.3)

where $q^{r}_{\Lambda}(x,t)$ is given in Corollary 2. Then from (2.52),

$\displaystyle q(x,t)$
$\displaystyle=exp\left\\{\frac{i}{2}\int_{-\infty}^{x}(|m(x,t)|^{2}-1)dy\right\\}m(x,t)$
$\displaystyle=exp\left\\{\frac{i}{2}\int_{-\infty}^{x}(|q^{r}_{\Lambda}(x,t)|^{2}-1)dy\right\\}T(\infty)^{-2}q^{r}_{\Lambda}(x,t)+\mathcal{O}(t^{-3/4}).$
(9.4)

Therefore, we achieve main result of this paper.

######  Theorem 1.

Let $q(x,t)$ be the solution for the initial-value problem (1.1) with generic
data $u_{0}(x)\in H^{1,1}(\mathbb{R})$ and scatting data
$\left\\{r(z),\left\\{\zeta_{n},C_{n}\right\\}^{4N_{1}+2N_{2}}_{n=1}\right\\}$.
Let $\xi=\frac{x}{t}$ with $-3<\xi<-1$. Denote $q^{r}_{\Lambda}(x,t)$ be the
$\mathcal{N}(\Lambda)$-soliton solution corresponding to scattering data
$\left\\{0,\left\\{\zeta_{n},\mathring{c}_{n}\right\\}_{n\in\Lambda}\right\\}$
shown in Corollary 2. And $\Lambda$ is defined in (3.2). There exist a large
constant $T_{1}=T_{1}(\xi)$, for all $T_{1}<t\to\infty$,

$\displaystyle
q(x,t)=exp\left\\{\frac{i}{2}\int_{-\infty}^{x}(|q^{r}_{\Lambda}(x,t)|^{2}-1)dy\right\\}T(\infty)^{-2}q^{r}_{\Lambda}(x,t)+\mathcal{O}(t^{-3/4}),$
(9.5)

where $u^{r}_{\Lambda}(x,t)$ and $T(z)$ are show in Propositions 4 and
Corollary 2 respectively.

The long time asymptotic expansion (9.5) shows the soliton resolution of for
the initial value problem of the the derivative nonlinear schr$\ddot{o}$dinger
equation, which can be characterized with an $\mathcal{N}(\Lambda)$-soliton
whose parameters are modulated by a sum of localized soliton-soliton
interactions . Our results also show that the poles on curve soliton solutions
of short-pulse equation has dominant contribution to the solution as
$t\to\infty$.

Acknowledgements

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

## References

  * [1] S.V. Manakov, Nonlinear Fraunhofer diffraction, Sov. Phys.-JETP 38(1974), 693-696.
  * [2] V. E. Zakharov, S. V. Manakov,  Asymptotic behavior of nonlinear wave systems integrated by the inverse scattering method,  Soviet Physics JETP, 44(1976), 106-112.
  * [3] P. C. Schuur,  Asymptotic analysis of soliton products,  Lecture Notes in Mathematics, 1232, 1986.
  * [4] R. F. Bikbaev,  Asymptotic-behavior as t-infinity of the solution to the cauchy-problem for the landau-lifshitz equation,  Theor. Math. Phys, 77(1988), 1117-1123.
  * [5] R. F. Bikbaev,  Soliton generation for initial-boundary-value problems,  Phys. Rev. Lett., 68(1992), 3117-3120.
  * [6] X. Zhou, P. Deift,  A steepest descent method for oscillatory Riemann-Hilbert problems.  Ann. Math., 137(1993), 295-368.
  * [7] X. Zhou, P. Deift,  Long-time behavior of the non-focusing nonlinear Schr$\ddot{o}$dinger equation–a case study,  Lectures in Mathematical Sciences, Graduate School of Mathematical Sciences, University of Tokyo, 1994.
  * [8] P. Deift, X. Zhou,  Long-time asymptotics for solutions of the NLS equation with initial data in a weighted Sobolev space,  Comm. Pure Appl. Math., 56(2003), 1029-1077.
  * [9] K. Grunert, G. Teschl,  Long-time asymptotics for the Korteweg de Vries equation via noninear steepest descent.  Math. Phys. Anal. Geom., 12(2009), 287-324.
  * [10] A. B. de Monvel, A. Kostenko, D. Shepelsky, G. Teschl,  Long-time asymptotics for the Camassa-Holm equation.,  SIAM J. Math. Anal, 41(2009), 1559-1588.
  * [11] J. Xu, E. G. Fan,  Long-time asymptotics for the Fokas-Lenells equation with decaying initial value problem: Without solitons,  J. Differential Equations, 259(2015), 1098-1148.
  * [12] J. Xu,  Long-time asymptotics for the short pulse equation,  J. Differential Equations, 265(2018), 3494-3532.
  * [13] K. T. R. McLaughlin, P. D. Miller,  The $\bar{\partial}$ steepest descent method and the asymptotic behavior of polynomials orthogonal on the unit circle with fixed and exponentially varying non-analytic weights,  Int. Math. Res. Not., (2006), Art. ID 48673.
  * [14] K. T. R. McLaughlin, P. D. Miller,  The $\bar{\partial}$ steepest descent method for orthogonal polynomials on the real line with varying weights,  Int. Math. Res. Not., (2008), Art. ID 075.
  * [15] M. Dieng, K. D. T. McLaughlin,  Dispersive asymptotics for linear and integrable equations by the Dbar steepest descent method,  Nonlinear dispersive partial differential equations and inverse scattering, 253-291, Fields Inst. Commun., 83, Springer, New York, 2019
  * [16] M. Borghese, R. Jenkins, K. T. R. McLaughlin, Miller P,  Long-time asymptotic behavior of the focusing nonlinear Schr$\ddot{o}$dinger equation,  Ann. I. H. Poincar$\acute{e}$ Anal, 35(2018), 887-920.
  * [17] R. Jenkins, J. Liu, P. Perry, C. Sulem,  Soliton resolution for the derivative nonlinear Schr$\ddot{o}$dinger equation,  Commun. Math. Phys., 363(2018), 1003-1049.
  * [18] S. Cuccagna, R. Jenkins,  On asymptotic stability of N-solitons of the defocusing nonlinear Schr$\ddot{o}$dinger equation,  Comm. Math. Phys, 343(2016), 921-969.
  * [19] D. J. Kaup and A. C. Newell,  An exact solution for a Derivative Nonlinear Schr$\ddot{o}$dinger equation,  J. Math. Phys.,  19(1978), 798-801.
  * [20] A. Rogister,  Parallel propagation of nonlinear low-frequency waves in high-$\beta$ plasma,  Phys. Fluids., 14(1971), 2733-2739.
  * [21] E. Mj$\phi$lhus,  On the modulational instability of hydromagnetic waves parallel to the magnetic field,  J. Plasma Phys., 16(1976), 321-334.
  * [22] K. Mio, T. Ogino, K. Minami, S. Takeda,  Modified nonlinear Schr$\ddot{o}$dinger equation for Alfv$\acute{e}$n waves propagating along the magnetic field in cold plasmas,  J. Phys. Soc. Jpn., 41(1976), 265-271.
  * [23] E. Mj$\phi$lhus,  Nonlinear Alfv$\acute{e}$n waves and the DNLS equation: oblique aspects,  Phys. Scr., 40(1989), 227-237.
  * [24] E. Mj$\phi$lhus, T. Hada, in: T. Hada, H. Matsumoto (Eds.),  Nonlinear Waves and Chaos in Space Plasmas,  Terrapub, Tokio, (1997).
  * [25] G. P. Agrawal,  Nonlinear Fiber Optics.  Academic Press, Boston, 1989.
  * [26] D. Anderson and M. Lisak,  Nonlinear Asymmetric Self-phase modulation and self-steepening of pulses in long Optical Waveguides,  Phys. Rev. A., 27(1983), 1393-1398.
  * [27] N.Tzoar and M.Jain,  Self-phase Modulation in long-geometry optical waveguide,  Phys. Rev. A., 23(1981), 1266-1270.
  * [28] I. Nakata,  Weak nonlinear electromagnetic waves in a ferromagnet propagating parallel to an external magnetic field,  J. Phys. Soc. Jpn., 60(1991), 3976-3977.
  * [29] I. Nakata, H. One and M. Yosida,  Solitons in a dielectric medium under an external magnetic field,  Prog. Theor. Phys., 90(1993), 739-742.
  * [30] M. Daniel, V. Veerakumar,  Propagation of electromagnetic soliton in antiferromagnetic medium,  Phys. Lett. A, 302(2002), 77-86.
  * [31] G. Q. Zhou, N. N. Huang  An N-soliton solution to the DNLS equation based on revised inverse scattering transform,  J. Phys. A: Math. Theor.,  40(2007), 13607.
  * [32] T. Kawata, H. Inoue,  Exact solutions of the derivative nonlinear Schr$\ddot{o}$dinger equation under the nonvanishing conditions,  J. Phys. Soc. Jpn.,  44(1978), 1968-1976.
  * [33] X. J. Chen, W. K. Lam,  Inverse scattering transform for the derivative nonlinear Schr$\ddot{o}$dinger equation with nonvanishing boundary conditions,  Phys. Rev. E,  69(2004), 066604.
  * [34] X. J. Chen, J. Yang, W. K. Lam,  N-soliton solution for the derivative nonlinear Schr$\ddot{o}$dinger equation with nonvanishing boundary conditions,  J. Phys. A,  39(2006), 3263.
  * [35] V. Lashkin,  N-soliton solution and perturbation theory for the derivative nonlinear Schr$\ddot{o}$dinger equation with nonvanishing boundary conditions,  J. Phys. A, 40(2007), 6119.
  * [36] G. Zhang, Z. Yan,  The Derivative nonlinear Schr$\ddot{o}$dinger equation with zero/Nonzero Bboundary conditions: inverse scattering transforms and N-double-pole solutions,  J. Non. Sci., 2(2002).
  * [37] J. Xu and E. G. Fan, Inverse scattering for the derivative nonlinear Schrodinger equation: A Riemann-Hilbert approach, arXiv:1209.4245v1.
  * [38] J. Xu, E. Fan, Y. Chen,  Long-time asymptotic for for the derivative nonlinear Schr$\ddot{o}$dinger equation with step-like initial value,  Math. Phys. Anal. Geometry, 16(2013), 253-288.