# Soliton resolution for a coupled generalized nonlinear Schrödinger equations
with weighted Sobolev initial data 111Corresponding author.
 _E-mail addresses_<EMAIL_ADDRESS><EMAIL_ADDRESS>(S. F. Tian)

Zhi-Qiang Li, Shou-Fu Tian∗ and Jin-Jie Yang  School of Mathematics, China
University of Mining and Technology, Xuzhou 221116, People’s Republic of China

###### Abstract

In this work, we employ the $\bar{\partial}$ steepest descent method in order
to study the Cauchy problem of the cgNLS equations with initial conditions in
weighted Sobolev space $H^{1,1}(\mathbb{R})=\\{f\in
L^{2}(\mathbb{R}):f^{\prime},xf\in L^{2}(\mathbb{R})\\}$. The large time
asymptotic behavior of the solution $u(x,t)$ and $v(x,t)$ are derived in a
fixed space-time cone
$S(x_{1},x_{2},v_{1},v_{2})=\\{(x,t)\in\mathbb{R}^{2}:x=x_{0}+vt,~{}x_{0}\in[x_{1},x_{2}],~{}v\in[v_{1},v_{2}]\\}$.
Based on the resulting asymptotic behavior, we prove the solution resolution
conjecture of the cgNLS equations which contains the soliton term confirmed by
$|\mathcal{Z}(\mathcal{I})|$-soliton on discrete spectrum and the
$t^{-\frac{1}{2}}$ order term on continuous spectrum with residual error up to
$O(t^{-\frac{3}{4}})$.

###### keywords:

Integrable system , The coupled generalized nonlinear Schrödinger equations ,
Infinite conservation laws , Riemann-Hilbert problem , $\bar{\partial}$
steepest descent method , Soliton resolution.

††journal: Journal of LaTeX Templates

##  1 Introduction

The classical nonlinear Schrödinger(NLS) equation, i.e.,

$\displaystyle iu_{t}\pm u_{xx}+2|u|^{2}u=0,$

is a fundamental physical model and has been applied in various fields such as
deep water waves[1], plasma physics[2, 3], nonlinear optical fibers[4, 5],
etc. Since it plays a significant role in the field of nonlinear science, lots
of researchers and scholars do many works on the NLS equation and its
extensions[6]-[11]. It is worth noting that in nonlinear science such as
optical fibers, the NLS equation only describes the propagation of optical
solitons in scalar fields of mono-mode fibers. However, because the change of
nonlinear phase stems from the cross-phase modulation in multi-mode fibers or
birefringent fibers, we need take the interaction of some field components at
different frequencies or polarisations into consideration. The coupled
nonlinear Schrödinger system came into being to describe the characteristics
of such soliton. For example, the well-known Manakov system[12], i.e.

$\displaystyle iu_{t}\pm\frac{1}{2}u_{xx}+(|u|^{2}+|v|^{2})u=0,$
$\displaystyle iv_{t}\pm\frac{1}{2}v_{xx}+(|u|^{2}+|v|^{2})v=0,$

can be used to describe the intense electromagnetic pulse propagation in
birefringent fiber.

In this work, considering that there are more effects of nonlinear factors in
practise, we derive a novel integrable coupled generalized nonlinear
Schrödinger(cgNLS) equation that reads

$\displaystyle\begin{split}iu_{t}-i\alpha
u_{x}+u_{xx}-2u^{2}v+4\beta^{2}u^{3}v^{2}+4i\beta(uv)_{x}u+\gamma u=0,\\\
iv_{t}-i\alpha
v_{x}-v_{xx}+2uv^{2}-4\beta^{2}u^{2}v^{3}+4i\beta(uv)_{x}v-\gamma
v=0,\end{split}$ (1.1)

where $u(x,t)$ and $v(x,t)$ are complex function of variables $x$, $t$, and
$\alpha$, $\beta$, $\gamma$ are arbitrary constants. Additionally, we will
derive the Lax pair, infinite conservation laws and Hamiltonian function
associated with the cgNLS equation. Furthermore, we investigate the soliton
resolution of the cgNLS equation with the initial value condition

$\displaystyle u(x,0)=u_{0}(x),v(x,0)=v_{0}(x)\in H^{1,1}(\mathbb{R}),$ (1.2)

where $H^{1,1}(\mathbb{R})$ is a weighted Sobolev space, i.e.,

$\displaystyle H^{1,1}(\mathbb{R})=\\{f\in L^{2}(\mathbb{R}):f^{\prime},xf\in
L^{2}(\mathbb{R})\\}.$ (1.3)

In particular, some unique cases of (1.1) can be obtained by fixing the
coefficients $\alpha$, $\beta$ and $\gamma$.

  * 1.

When $\alpha=\gamma=0$ and $\beta\neq 0$, Eq.(1.1) degenerates into the
coupled nonlinear Kundu-Eckhaus equation, i.e.,

$\displaystyle iu_{t}+u_{xx}-2u^{2}v+4\beta^{2}u^{3}v^{2}+4i\beta(uv)_{x}u=0,$
$\displaystyle iv_{t}-v_{xx}+2uv^{2}-4\beta^{2}u^{2}v^{3}+4i\beta(uv)_{x}v=0,$

which is a completely integrable system and possesses a Lax representation,
the Hamiltonian structure and some other properties[13]-[15].

  * 2.

When $\alpha\neq 0$, $\gamma\neq 0$, $\beta=0$, and $v=-u^{*}$, Eq.(1.1)
degenerates into the modified Landau-Lifshitz(mLL) equation, i.e.,

$\displaystyle iu_{t}-i\alpha u_{x}+u_{xx}+2|u|^{2}u+\gamma u=0,$

which can be employed to depict the dynamic behavior of local magnetization in
electromagnetics. Also, a research for the mLL equation reveals the
accumulation of energy plays an important role in the generation of magnetic
rogue waves[16]. Furthermore, the long time asymptotic behavior of the mLL
equation with nonzero boundary condition have been studied[17].

  * 3.

When $\alpha=\gamma=\beta=0$, and $v=-u^{*}$, Eq.(1.1) degenerates into the
classical focusing nonlinear Schrödinger(NLS) equation, i.e.,

$\displaystyle iu_{t}+u_{xx}+2|u|^{2}u=0,$

which has played an important part in nonlinear science.

It is not hard to check that the cgNLS equation is not self-adjoint, so that
the soliton solution will emerge for zero boundary conditions. Therefore, it
is significant to consider the question that how to obtain the long time
asymptotic behavior of the cgNLS equation in the domain of solution solutions.

In order to obtain the long time behavior of nonlinear evolution equations,
researchers have done a series work. In 1974, using the inverse scattering
method, the long time behavior of nonlinear wave equation was first carried
out by Manakov [18]. In 1976, the long time asymptotic solutions of nonlinear
Schrödinger(NLS) equation with decaying initial value was obtained by Zakharov
and Manakov [19]. In 1993, a nonlinear steepest descent method was developed
to study the modified Korteweg-de Vries(MKdV) equation and obtained the
solution rigorously by Defit and Zhou [20]. Then, this approach has been
applied widespread. However, it has been shown in literature [21, 22] that if
the initial value is smooth and decay fast enough then the error term is
$o(\frac{\log t}{t})$. Then, after the hard work of many researchers [23], the
error term becomes $O(t^{-(\frac{1}{2}+\iota)})$ for any $0<\iota<\frac{1}{4}$
with the condition that the initial value belongs to the weighted Sobolev
space (1.3).

In recent years, McLaughlin and Miller have made contributions to develop the
method for the long time asymptotic analysis of Riemann-Hilbert problem(RHP).
Combining steepest descent with $\bar{\partial}$-problem, they presented a
$\bar{\partial}$ steepest descent method which has been used to analyse the
long time asymptotic behavior of series equations [24, 25]. Since then, lots
of significant works have been done by applying this method. For example, with
the assumption of finite mass initial data, Dieng and McLaughlin studied the
defocusing NLS equation [26]; with finite density initial data, the defocusing
NLS equation was investigated by Cuccagna and Jenkins [27]; Borghese and
Jenkins studied the Cauchy problem for the focusing NLS equation via applying
this method [28]. Of course, there are a series of great work about this
$\bar{\partial}$ steepest descent method [29]-[33]. Compared with the
nonlinear steepest descent method, it is not necessary to study the delicate
estimates involving $L^{p}$ estimates of Cauchy projection operators by
applying the $\bar{\partial}$ steepest descent method. Also, it improves the
error estimates, i.e., if the initial value belongs to the weighted Sobolev
space (1.3), the error becomes $o(t^{-\frac{3}{4}})$ which has been shown in
[26].

In this work, we are going to use the $\bar{\partial}$ steepest descent method
to study the long time asymptotic behavior of the cgNLS equation with the
initial value $u_{0}(x)$ and $v_{0}(x)$ which belong to the weighted Sobolev
space (1.3).

The outline of this work is as follows. In section 2, based on the Lax pair of
the cgNLS equation, we introduce two kinds of eigenfunction and scattering
matrix. Also, the analytical, symmetries and asymptotic properties are
analyzed. Then the Riemann-Hilbert problem is constructed for the cgNLS
equation with initial problem. In section 3, we introduce the matrix function
$T(z)$ to separate the jump matric which is defined in (2.2) near the phrase
point $z_{0}=-\frac{1}{4}\left(\frac{x}{t}+\alpha\right)$. In section 4, we
make the continuous extension of the jump matrix off the real axis by
introducing a matrix function $R^{(2)}(z)$ and get a mixed
$\bar{\partial}$-Riemann-Hilbert(RH) problem. In section 5, we decompose the
mixed $\bar{\partial}$-RH problem into two parts which are a model RH problem
with $\bar{\partial}R^{(2)}=0$ and a pure $\bar{\partial}$-RH problem with
$\bar{\partial}R^{(2)}\neq 0$, i.e., $M_{RHP}$ and $M^{(3)}$. In section 6, we
solve the model RH problem $M_{RHP}$ via an outer model $M^{(out)}(z)$ for the
soliton part and an inner model near the phase point $z_{0}$ which can be
solved by a parabolic cylinder model problem. Also, the error function $E(z)$
with a small-norm RH problem is obtained. In section 7, the pure
$\bar{\partial}$-RH problem is studied. Finally, in section 8, we obtain the
soliton resolution and long time asymptotic behavior of the cgNLS equation.

##  2 The spectral analysis of cgNLS equation

The Lax pair of the equation (1.1) reads

$\displaystyle\begin{split}\psi_{x}=U\psi,\quad\psi_{t}=V\psi,\\\
\psi=(\psi_{1},\psi_{2})^{T},\end{split}$ (2.1)

where $\psi_{i},(i=1,2)$ are eigenfunctions, and

$\displaystyle\begin{split}U&=-iz\sigma_{3}-i\beta uv\sigma_{3}+U_{0},\\\
V&=-2iz^{2}\sigma_{3}+a_{1}\sigma_{3}+2zU_{0}-i(U_{0})_{x}\sigma_{3}-2\beta
U_{0}^{3}+\alpha U_{0}\end{split}$ (2.2)

with $U_{0}=\left(\begin{array}[]{cc}0&u\\\ v&0\end{array}\right)$,
$\sigma_{3}=\left(\begin{array}[]{cc}1&0\\\ 0&-1\end{array}\right)$ and
$a_{1}=\beta(u_{x}v-uv_{x})+4i\beta^{2}u^{2}v^{2}-i(1-\alpha\beta)uv$. We can
check that $U,V$ satisfy the zero curvature equation $U_{t}-V_{x}+[U,V]=0$,
which is the compatibility condition of (1.1).

Considering the initial conditions that $u_{0}(x),v_{0}(x)\in
H^{1,1}(\mathbb{R})$, letting $x\rightarrow\pm\infty$, we can get an
asymptotic scattering problem and construct the two Jost solutions, i.e.,
$\psi_{\pm}\sim e^{-i(zx+(2z^{2}+\alpha z-\frac{1}{2}\gamma)t)\sigma_{3}}$.

Then, we introduce $m_{\pm}(x,t;z)$ which satisfies that

$\displaystyle m_{\pm}(x,t;z)=\psi_{\pm}(x,t;z)e^{i(zx+(2z^{2}+\alpha
z-\frac{1}{2}\gamma)t)\sigma_{3}}.$ (2.3)

The equivalent Lax pair of Eq.(2.1) can be derived as

$\displaystyle\begin{split}m_{x}&+iz[\sigma_{3},m]=(-i\beta
uv\sigma_{3}+U_{0})m\triangleq U_{1}m,\\\ m_{t}&+i(2z^{2}+\alpha
z-\frac{1}{2}\gamma)[\sigma_{3},m]=(a_{1}\sigma_{3}+2zU_{0}-i(U_{0})_{x}\sigma_{3}-2\beta
U_{0}^{3}+\alpha U_{0})m\triangleq V_{1}m,\end{split}$ (2.4)

which can be written in full derivative form, i.e.,

$\displaystyle d(e^{i(zx+(2z^{2}+\alpha
z-\frac{1}{2}\gamma)t)\sigma_{3}}m)=e^{i(zx+(2z^{2}+\alpha
z-\frac{1}{2}\gamma)t)\sigma_{3}}(U_{1}dx+V_{1}dt)m.$ (2.5)

Evaluating a solution of Eq.(2.5) with the form

$\displaystyle
m=\mathcal{X}+\frac{m_{1}}{z}+\frac{m_{2}}{z^{2}}+o(z^{-3}),~{}~{}z\rightarrow\infty,$
(2.6)

where $\mathcal{X}$, $m_{1}$ and $m_{2}$ are independent of variable $z$.
Substituting Eq.(2.6) into Eq.(2.4), and comparing the coefficients of the
same power of $z$, we show that

$\displaystyle\begin{split}\mathcal{X}_{x}&=(-i\beta
uv\sigma_{3})\mathcal{X},\\\
\mathcal{X}_{t}&=(\beta(u_{x}v-uv_{x})+4i\beta^{2}u^{2}v^{2}-i\alpha\beta
uv)\sigma_{3}\mathcal{X}.\end{split}$ (2.7)

We note that Eq.(1.1) possesses the conservation law

$\displaystyle(-i\beta
uv)_{t}=(\beta(u_{x}v-uv_{x})+4i\beta^{2}u^{2}v^{2}-i\alpha\beta uv)_{x}.$
(2.8)

Therefore, if we define

$\displaystyle\mathcal{X}(x,t)=e^{i\int_{(-\infty,\infty)}^{(x,t)}\Delta\sigma_{3}},$
(2.9)

where

$\displaystyle\Delta(x,t)=-\beta
uvdx+(-i\beta(u_{x}v-uv_{x})+4\beta^{2}u^{2}v^{2}-\alpha\beta uv)dt,$ (2.10)

the two equations in (2.7) for $\mathcal{X}$ are consistent and are both
satisfied. Now, introducing a new function $\mu$

$\displaystyle
m(x,t,z)=e^{i\int_{(-\infty,\infty)}^{(x,t)}\Delta\sigma_{3}}\mu(x,t,z).$
(2.11)

It is obvious that

$\displaystyle\mu=I+O(\frac{1}{z}),\quad z\rightarrow\infty.$ (2.12)

Then, from Eq.(2.5), we can derive that

$\displaystyle d(e^{i(zx+(2z^{2}+\alpha
z-\frac{1}{2}\gamma)t)\sigma_{3}}\mu)=e^{i(zx+(2z^{2}+\alpha
z-\frac{1}{2}\gamma)t)\sigma_{3}}(U_{2}dx+V_{2}dt)\mu.$ (2.13)

where

$\displaystyle
U_{2}=e^{-i\int_{(-\infty,\infty)}^{(x,t)}\Delta\hat{\sigma}_{3}}\left(\begin{array}[]{cc}0&u\\\
v&0\end{array}\right),$ (2.16) $\displaystyle
V_{2}=e^{-i\int_{(-\infty,\infty)}^{(x,t)}\Delta\hat{\sigma}_{3}}\left(\begin{array}[]{cc}iuv&2zu+iu_{x}-2\beta
u^{2}v+\alpha u\\\ 2zv-iv_{x}-2\beta uv^{2}+\alpha v&-iuv\end{array}\right).$
(2.19)

Also, the Lax pair (2.4) can be transformed into

$\displaystyle\mu_{x}+iz[\sigma_{3},\mu]=U_{2}\mu,$ (2.20)
$\displaystyle\mu_{t}+i(2z^{2}+\alpha
z-\frac{1}{2}\gamma)[\sigma_{3},\mu]=V_{2}\mu.$ (2.21)

Based on the Eq.(2.13), we can select two special integration paths i.e.,
$(-\infty,t)\rightarrow(x,t)$ and $(+\infty,t)\rightarrow(x,t)$, and obtain
the following Volterra type integrals

$\displaystyle\begin{matrix}\mu_{\pm}(x,t;z)=\mathbb{I}+\int_{\pm\infty}^{x}e^{-iz(x-y)\hat{\sigma}_{3}}[U_{2}(y,t;z)\mu_{-}(y,t;z)]\,dy,\end{matrix}$
(2.22)

from which we can derive the analytical properties of $\mu_{\pm}$.

  * 1.

It is assumed that $u(x)-u_{0},v(x)-v_{0}\in H^{1,1}(\mathbb{R})$. Then,
$\mu_{-,1},\mu_{+,2}$ are analytic in $C_{+}$ and $\mu_{-,2},\mu_{+,1}$ are
analytic in $C_{-}$ and they can be recorded as
$\mu^{+}_{-,1},\mu^{+}_{+,2},\mu^{-}_{-,2},\mu^{-}_{+,1}$, respectively. The
$\mu_{\pm,j}(j=1,2)$ mean the $j$-th column of $\mu_{\pm}$.

For $z\in\mathbb{R}$ both $\psi_{+}$ and $\psi_{-}$ are two fundamental matrix
solutions of the scattering problem. Therefore,

$\displaystyle\psi_{+}(x,t;z)=\psi_{-}(x,t;z)S(z),\quad
x,t\in\mathbb{R},~{}~{}~{}z\in\mathbb{R},$ (2.23)

where $S(z)$ is a $2\times 2$ matrix and independent of the variable $x$ and
$t$. According to the Abel’s theorem [35] and the Lax pair (2.1), we know that
$(\det\psi)_{x}=(\det\psi)_{t}=0$. Furthermore, we have
$\det\psi_{\pm}=\det\mu_{\pm}=1$.Then, we can derive that $\psi_{\pm}$ are
reversible. Therefore, we obtain that

$\displaystyle\begin{split}&s_{11}(z)=Wr\left(\psi_{+,1},\psi_{-,2}\right),\quad
s_{22}(z)=Wr\left(\psi_{-,1},\psi_{+,2}\right),\\\
&s_{12}(z)=Wr\left(\psi_{+,2},\psi_{-,2}\right),\quad
s_{21}(z)=Wr\left(\psi_{-,1},\psi_{+,1}\right),\end{split}$ (2.24)

where $\psi_{\pm,j}$ mean the $j$-column of $\psi_{\pm}$, respectively, and
$Wr$ infers to the Wronskians determinant. Then, according to the analytical
properties of $\mu_{\pm}$ and the relationship between $\psi_{\pm}$ and
$\mu_{\pm}$, we obtain that $s_{11}$ is analytic in $\mathbb{C}^{+}$, and
$s_{22}$ is analytic in $\mathbb{C}^{-}$.

It is easy to check that if $\mu(x,t;z)$ is the solution of Eq.(2.20) then
$-\sigma\mu^{*}(x,z^{*})\sigma$ also is the solution of Eq.(2.20) and it
follows

$\displaystyle\mu_{\pm}(z)=-\sigma\mu^{*}_{\pm}(z^{*})\sigma.$ (2.25)

Then, it is obvious that

$\displaystyle S(z)=-\sigma S^{*}(z^{*})\sigma.$ (2.26)

Based on Eq.(2.12) and Eq.(2.23), we can derive that
$S(z)\rightarrow\mathbb{I}$, $z\rightarrow\infty$.

######  Assumption 2.1.

In the following analysis, we make the assumption to avoid the many
pathologies possible, i.e.,

  * 1.

For $z\in\mathbb{R}$, no spectral singularities exist, i.e, $s_{11}(z)\neq 0$;

  * 2.

Suppose that $s_{11}(z)$ has $N$ zero points, denoted as
$\mathcal{Z}=\\{(z_{j},Imz_{j}>0)^{N}_{j=1}\\}$. So that $s_{22}(\lambda)$ has
$N$ zero points $\mathcal{Z}^{*}=\\{(z^{*}_{j},Imz^{*}_{j}<0)^{N}_{j=1}\\}$.

  * 3.

The discrete spectrum is simple, i.e., if $z_{0}$ is the zero of $s_{11}(z)$,
then $s^{\prime}_{11}(z_{0})\neq 0$.

Furthermore, based on Eq.(2.24) and Eq.(2.26), there exists norming constants
$c_{j}$ such that

$\mu_{-,1}(z_{j})=c_{j}e^{2it\theta(z_{j})}\mu_{+,2}(z_{k});~{}\mu_{-,2}(z^{*}_{j})=-c^{*}_{j}e^{-2it\theta(z^{*}_{j})}\mu_{+,1}(z^{*}_{j}),$

where $\theta(z)=(z\frac{x}{t}+2z^{2}+\alpha z-\frac{1}{2}\gamma)$.

Now, we introduce a sectionally meromorphic matrices

$\displaystyle M(x,t;z)=\left\\{\begin{aligned}
&M^{+}(x,t;z)=\left(\frac{\mu_{-,1}(x,t;z)}{s_{11}(z)},\mu_{+,2}(x,t;z)\right),\quad
z\in\mathbb{C}^{+},\\\
&M^{-}(x,t;z)=\left(\mu_{+,1}(x,t;z),\frac{\mu_{-,2}(x,t;z)}{s_{22}(z)}\right),\quad
z\in\mathbb{C}^{-},\end{aligned}\right.$ (2.27)

where $M^{\pm}(x,t;z)=\lim\limits_{\varepsilon\rightarrow 0^{+}}M(x,t;z\pm
i\varepsilon),~{}\varepsilon\in\mathbb{R}$, and reflection coefficients

$\displaystyle
r(z)=\frac{s_{21}(z)}{s_{11}(z)},~{}~{}\frac{s_{12}(z)}{s_{22}(z)}=-\frac{s^{*}_{21}(z^{*})}{s^{*}_{11}(z^{*})}=-r^{*}(z^{*})=-r^{*}(z),~{}~{}z\in\mathbb{R}.$
(2.28)

Based on the above analysis, the matrix function $M(x,t;z)$ admits the
following matrix RHP.

######  Riemann-Hilbert Problem 2.2.

Find an analysis function $M(x,t;z)$ with the following properties:

  * 1.

$M(x,t;z)$ is meromorphic in $C\setminus\mathbb{R}$.

  * 2.

$M^{+}(x,t;z)=M^{-}(x,t;z)G(x,t;z)$, $z\in\mathbb{R}$, where

$\displaystyle
G(x,t;z)=\left(\begin{array}[]{cc}1+|r(z)|^{2}&r^{*}(z)e^{-2it\theta(z)}\\\
r(z)e^{2it\theta(z)}&1\end{array}\right);$ (2.31)

  * 3.

$M(z)=I+O(z^{-1})$ as $z\rightarrow\infty$;

  * 4.

At each $z_{j}\in\mathcal{Z}$ and $z^{*}_{j}\in\mathcal{Z}^{*}$, $M(z)$
satisfies the residue condition, i.e.,

$\displaystyle\mathop{Res}_{z=z_{j}}M=\lim_{z\rightarrow
z_{j}}M\left(\begin{array}[]{cc}0&0\\\
c_{j}e^{2it\theta}&0\end{array}\right),\mathop{Res}_{z=z^{*}_{j}}M=\lim_{z\rightarrow
z^{*}_{j}}M\left(\begin{array}[]{cc}0&-c^{*}_{j}e^{-2it\theta}\\\
0&0\end{array}\right).$ (2.36)

######  Remark 2.3.

By referring to the Zhou’s vanishing lemma, the existence of the solutions of
RHP 2.2 for $(x,t)\in\mathbb{R}^{2}$ is guaranteed. According to a consequence
of Liouville’s theorem, we know that if a solution exists, it is unique.

Then, expanding this solution as $z\rightarrow\infty$ and combining with
Eq.(2.20), we can reconstruct the solution of cgNLS equation as

$\displaystyle\begin{split}u(x,t)=2i\lim_{z\rightarrow\infty}(zM)_{12}e^{2i\int_{(-\infty,\infty)}^{(x,t)}\Delta},~{}~{}v(x,t)=2i\lim_{z\rightarrow\infty}(zM)_{21}e^{-2i\int_{(-\infty,\infty)}^{(x,t)}\Delta}.\end{split}$
(2.37)

##  3 Conjugation

In this section, in order to renormalize the Riemann-Hilbert problem(2.2), we
introduce a function to establish a transformation $M\mapsto M^{(1)}$.

In jump matrix (2.31), the oscillation term is $e^{2it\theta(z)}$, from which
we find a phase point

$\displaystyle z_{0}=-\frac{1}{4}\left(\frac{x}{t}+\alpha\right).$

Consequently, $\theta(z)$ can be written as

$\displaystyle\theta(z)=2z^{2}-4z_{0}z-\frac{1}{2}\gamma=2(z-z_{0})^{2}-2z^{2}_{0}-\frac{1}{2}\gamma.$
(3.1)

And we have $Re(2i\theta)=-8(Rez-z_{0})Imz$. Then, we derive the decaying
domains of the oscillation term.

$Rez$$Imz$$|e^{2i\theta(z)}|\rightarrow 0$$|e^{2i\theta(z)}|\rightarrow
0$$|e^{-2i\theta(z)}|\rightarrow 0$$|e^{-2i\theta(z)}|\rightarrow 0$

Figure 1. Exponential decaying domains.

To make the following analysis more convenient, we introduce some notations.

$\displaystyle\begin{aligned}
&\triangle^{-}_{z_{0}}=\\{k\in\\{1,\cdots,N\\}|Re(z_{k})<z_{0}\\},\\\
&\triangle^{+}_{z_{0}}=\\{k\in\\{1,\cdots,N\\}|Re(z_{k})>z_{0}\\}.\end{aligned}$
(3.2)

For $\mathcal{I}=[a,b]$, define

$\displaystyle\mathcal{Z}(\mathcal{I})=\\{z_{k}\in\mathcal{Z}:Rez_{k}\in\mathcal{I}\\},$
$\displaystyle\mathcal{Z}^{-}(\mathcal{I})=\\{z_{k}\in\mathcal{Z}:Rez_{k}<a\\},$
$\displaystyle\mathcal{Z}^{+}(\mathcal{I})=\\{z_{k}\in\mathcal{Z}:Rez_{k}>b\\}.$

For $z_{0}\in\mathcal{I}$, define

$\displaystyle\triangle^{-}_{z_{0}}(\mathcal{I})=\\{k\in\\{1,\cdots,N\\}:a\leq
Rez_{k}<z_{0}\\},$
$\displaystyle\triangle^{+}_{z_{0}}(\mathcal{I})=\\{k\in\\{1,\cdots,N\\}:a<Rez_{k}\leq
z_{0}\\}.$

Next, we introduce the function

$\displaystyle\delta(z)=\exp[i\int_{-\infty}^{z_{0}}\frac{\nu(s)}{s-z}ds],~{}~{}\nu(s)=-\frac{1}{2\pi}\log(1+|r(s)|^{2}).$

and

$\displaystyle
T(z)=\prod_{k\in\Delta_{z_{0}}^{-}}\frac{z-z^{*}_{k}}{z-z_{k}}\delta(z),$
(3.3)

which has the following properties.

######  Proposition 3.4.

$T(z)$ admits that
($a$) $T$ is meromorphic in $C\setminus(-\infty,z_{0}]$;
($b$) For $z\in C\setminus(-\infty,z_{0}]$, $T^{*}(z^{*})=\frac{1}{T(z)}$;
($c$) For $z\in(-\infty,z_{0}]$, the boundary values $T_{\pm}$ satisfy

$\displaystyle T_{+}(z)/T_{-}(z)=1+|r(z)|^{2},z\in(-\infty,z_{0}];$ (3.4)

($d$) As $|z|\rightarrow\infty$ with $|arg(z)|\leq c<\pi$,

$\displaystyle
T(z)=1+\frac{i}{z}[2\sum_{k\in\Delta_{z_{0}}^{-}}Imz_{k}-\int_{-\infty}^{z_{0}}\nu(s)ds]+O(z^{-2});$
(3.5)

($e$) As $z\rightarrow z_{0}$ along any ray $z_{0}+e^{i\phi}R_{+}$ with
$|\phi|\leq c<\pi$

$\displaystyle|T(z-z_{0})-T_{0}(z_{0})(z-z_{0})^{i\nu(z_{0})}|\leq C\parallel
r\parallel_{H^{1}(R)}|z-z_{0}|^{\frac{1}{2}},$ (3.6)

where $T_{0}(z_{0})$ is the complex unit

$\displaystyle\begin{split}&T_{0}(z_{0})=\prod_{k\in\Delta_{z_{0}}^{-}}(\frac{z_{0}-z^{*}_{k}}{z_{0}-z_{k}})e^{i\beta(z_{0},z_{0})},\\\
&\beta(z,z_{0})=-\nu(z_{0})\log(z-z_{0}+1)+\int_{-\infty}^{z_{0}}\frac{\nu(s)-\chi(s)\nu(z_{0})}{s-z}ds,\end{split}$
(3.7)

with $\chi(s)=1$ as $s\in(z_{0}-1,z_{0})$, and $\chi(s)=0$ as
$s\in(-\infty,z_{0}-1]$.

###### Proof.

For part $(a)$, $(b)$ and $(c)$, it is obvious to obtain their results from
the definition of $T(z)$ shown in (3.3). For part $(d)$, we expand the
$\prod_{k\in\Delta_{z_{0}}^{-}}\frac{z-z^{*}_{k}}{z-z_{k}}$ and $\delta(z)$
for large $z$. Then, a direct calculation can show the result. For part $(e)$,
we rewrite (3.3) as

$\displaystyle T(z)=T(z,z_{0})$
$\displaystyle=\prod_{k\in\Delta_{z_{0}}^{-}}\left(\frac{z-z^{*}_{k}}{z-z_{k}}\right)\exp\left(i\int_{z_{0}-1}^{z_{0}}\frac{\nu(z_{0})}{s-z}ds+i\int_{-\infty}^{z_{0}}\frac{\nu(s)-\chi(s)\nu(z_{0})}{s-z}ds\right)$
$\displaystyle=\prod_{k\in\Delta_{z_{0}}^{-}}\left(\frac{z-z^{*}_{k}}{z-z_{k}}\right)(z-z_{0})^{i\nu(z_{0})}\exp(i\beta(z,z_{0}).$

Then, according to the fact that

$\displaystyle|(z-z_{0})^{i\nu(z_{0})}|\leq|e^{-\pi\nu(z_{0})}|=\sqrt{1+|r(z_{0})|^{2}},$

and the Lemma shown in [34], i.e.,

$\displaystyle|\beta(z,z_{0})-\beta(z_{0},z_{0})|\leq
c||r||_{H^{1}(\mathbb{R})}|z-z_{0}|^{\frac{1}{2}},$

the result of $(e)$ can be obtained directly. ∎

Now, using the function $T(z)$, we establish a transformation

$\displaystyle M^{(1)}(z)=M(z)T(z)^{-\sigma_{3}},$ (3.8)

which admits the following matrix RHP.

######  Riemann-Hilbert Problem 3.5.

Find an analysis function $M^{(1)}$ with the following properties:

  * 1.

$M^{(1)}$ is meromorphic on $C\setminus R$;

  * 2.

$M^{(1)}(z)=I+O(z^{-1})$ as $z\rightarrow\infty$;

  * 3.

$M^{(1)}_{\pm}(z)$ satisfy the jump relationship
$M^{(1)}_{+}(z)=M^{(1)}_{-}(z)V^{(1)}(z)$, where

$\displaystyle V^{(1)}=\left\\{\begin{aligned}
\left(\begin{array}[]{cc}1&r^{*}(z)T(z)^{2}e^{-2it\theta}\\\ 0&1\\\
\end{array}\right)\left(\begin{array}[]{cc}1&0\\\
r(z)T(z)^{-2}e^{2it\theta}&1\\\ \end{array}\right)\triangleq
W_{L}W_{R},z\in(z_{0},\infty),\\\ \left(\begin{array}[]{cc}1&0\\\
\frac{r(z)T_{-}(z)^{-2}}{1+|r(z)|^{2}}e^{2it\theta}&1\\\
\end{array}\right)\left(\begin{array}[]{cc}1&\frac{r^{*}(z)T_{+}(z)^{2}}{1+|r(z)|^{2}}e^{-2it\theta}\\\
0&1\\\ \end{array}\right)\triangleq
U_{L}U_{R},z\in(-\infty,z_{0}).\end{aligned}\right.$ (3.9)

  * 4.

$M^{(1)}(z)$ has simple poles at each $z_{k}\in Z$ and $z_{k}\in Z$ at which

$\displaystyle\begin{split}\mathop{Res}\limits_{z_{k}}M^{(1)}=\left\\{\begin{aligned}
&\lim_{z\rightarrow
z_{k}}M^{(1)}\left(\begin{array}[]{cc}0&c_{k}^{-1}\left((\frac{1}{T})^{\prime}(z_{k})\right)^{-2}e^{-2it\theta}\\\
0&0\\\ \end{array}\right),k\in\Delta_{z_{0}}^{-}\\\ &\lim_{z\rightarrow
z_{k}}M^{(1)}\left(\begin{array}[]{cc}0&0\\\
c_{k}^{-1}T^{-2}(z_{k})e^{2it\theta}&0\\\
\end{array}\right),k\in\Delta_{z_{0}}^{+}\end{aligned}\right.\\\
\mathop{Res}\limits_{z^{*}_{k}}M^{(1)}=\left\\{\begin{aligned}
&\lim_{z\rightarrow z^{*}_{k}}M^{(1)}\left(\begin{array}[]{cc}0&0\\\
-\bar{c}_{k}^{-1}(T^{\prime}(z^{*}_{k}))^{-2}e^{2it\theta}&0\\\
\end{array}\right),k\in\Delta_{z_{0}}^{-}\\\ &\lim_{z\rightarrow
z^{*}_{k}}M^{(1)}\left(\begin{array}[]{cc}0&-c^{*}_{k}(T(z^{*}_{k}))^{2}e^{-2it\theta}\\\
0&0\\\
\end{array}\right),k\in\Delta_{z_{0}}^{+}\end{aligned}\right.\end{split}$
(3.10)

###### Proof.

Firstly, according to the definition of $M^{(1)}$, Proposition 3.4 and the
properties of $M(x,t;z)$, we can easily obtain the analyticity, jump matrix
and asymptotic behavior of $M^{(1)}$. Then, we consider the residue
conditions. When $k\in\triangle^{+}_{z_{0}}$, $T(z)$ is analytic at the points
$z_{k}$, $z^{*}_{k}$. Thus, at these points, the residue conditions can be
obtained directly from (2.36) and (3.8). When $k\in\triangle^{-}_{z_{0}}$,
$z_{k}$ is the pole of $T(z)$ and $T^{-1}(z)$ is analytic at $z_{k}$.
Consequently, we have

$\displaystyle\mathop{Res}\limits_{z=z_{k}}M^{(1)}_{1}$
$\displaystyle=\mathop{Res}\limits_{z=z_{k}}(M_{1}T^{-1})=0,$ $\displaystyle
M^{(1)}_{1}(z_{k})$ $\displaystyle=\lim_{z\rightarrow
z_{k}}(M_{1}(z)T^{-1}(z))=c_{k}e^{2it\theta(z_{k})}M_{2}(z_{k})(\frac{1}{T})^{\prime}(z_{k}),$
$\displaystyle\mathop{Res}\limits_{z=z_{k}}M^{(1)}_{2}$
$\displaystyle=\mathop{Res}\limits_{z=z_{k}}(M_{2}T)=M_{2}(z_{k})\lim_{z\rightarrow
z_{k}}T(z)(z-z_{k})$
$\displaystyle=c^{-1}_{k}e^{-2it\theta(z_{k})}M^{(1)}_{1}(z_{k})((\frac{1}{T})^{\prime}(z_{k}))^{-1},$

from which the first formula shown in (3.10) can be obtained. Similarly, it is
easy to derive the residue condition at $z^{*}_{k}$ with
$k\in\triangle^{-}_{z_{0}}$.

∎

##  4 Continuous extension to a mixed $\bar{\partial}$-RH problem

In this section, we are going to make the continuous extension of the jump
matrix off the real axis. It should be pointed that the extension is not
necessary and the oscillation term along the new contours are decaying. To
achieve this, we define the contours

$\displaystyle\begin{split}\Sigma_{j}=&-\frac{1}{4}(\frac{x}{t}+\alpha)+e^{i(2j-1)\pi/4}\mathbb{R}_{+},j=1,2,3,4,\\\
\Sigma_{R}=&R\cup\Sigma_{1}\cup\Sigma_{2}\cup\Sigma_{3}\cup\Sigma_{4}.\end{split}$
(4.1)

Moreover, define

$\displaystyle\rho=\frac{1}{2}\min_{\lambda,\zeta\in\mathcal{Z}\cup\mathcal{Z}^{*}\lambda\neq\mu}|\lambda-\zeta|,$
(4.2)

and $\chi_{Z}\in C_{0}^{\infty}(C,[0,1])$ which is supported near the discrete
spectrum such that

$\displaystyle\chi_{Z}(z)=\left\\{\begin{aligned}
&1,~{}~{}dist(z,\mathcal{Z}\cup\mathcal{Z}^{*})<\rho/3,\\\
&0,~{}~{}dist(z,\mathcal{Z}\cup\mathcal{Z}^{*})>2\rho/3.\end{aligned}\right.$
(4.3)

It is easy to check that
$dist(\mathcal{Z}\cup\mathcal{Z}^{*},R)>\rho,k=1,2,\cdots,N.$

Next, we need to extend the jump matrix onto the new contours along which
oscillation term are decaying. Therefore, we introduce a transformation

$\displaystyle M^{(2)}=M^{(1)}R^{(2)},$ (4.4)

where $R^{(2)}$ is selected to admit some conditions.

  * 1.

The purpose of the transformation is to deform the contour $\mathbb{R}$ to the
contour $\Sigma^{(2)}$ where
$\Sigma^{(2)}=\Sigma_{1}\cup\Sigma_{2}\cup\Sigma_{3}\cup\Sigma_{4}$. So on the
real axis, $M^{(2)}$ must have no jump.

  * 2.

To guarantee that the $\bar{\partial}$-contribution to the large-time
asymptotic solution of $u(x,t)$ and $v(x,t)$, the norm of $R^{(2)}$ should be
controlled.

  * 3.

The introduced transformation need to have no impact on the residue condition.

Thus, we define $R^{(2)}$ as

$\displaystyle R^{(2)}=\left\\{\begin{aligned}
&\left(\begin{array}[]{cc}1&0\\\ R_{1}e^{2it\theta}&1\\\
\end{array}\right)^{-1}\triangleq W_{R}^{-1},~{}&z\in\Omega_{1},\\\
&\left(\begin{array}[]{cc}1&R_{3}e^{-2it\theta}\\\ 0&1\\\
\end{array}\right)^{-1}\triangleq U_{R}^{-1},~{}&z\in\Omega_{3},\\\
&\left(\begin{array}[]{cc}1&0\\\ R_{4}e^{2it\theta}&1\\\
\end{array}\right)\triangleq U_{L},~{}&z\in\Omega_{4},\\\
&\left(\begin{array}[]{cc}1&R_{6}e^{-2it\theta}\\\ 0&1\\\
\end{array}\right)\triangleq W_{L},~{}&z\in\Omega_{6},\\\
&\left(\begin{array}[]{cc}1&0\\\ 0&1\\\
\end{array}\right),~{}&z\in\Omega_{2}\cup\Omega_{5},\end{aligned}\right.$
(4.5)

where $R_{j},j=1,3,4,6$ are defined in the following proposition.

######  Proposition 4.6.

There exists functions $R_{j}:\Omega_{j}\rightarrow C,j=1,3,4,6$ such that

$\displaystyle R_{1}(z)=\left\\{\begin{aligned}
&r(z)T^{-2}(z),~{}~{}~{}~{}z\in(z_{0},\infty),\\\
&f_{1}=r(z_{0})T_{0}^{-2}(z_{0})(z-z_{0})^{-2i\nu(z_{0})}(1-\chi_{Z}(z)),z\in\Sigma_{1},\end{aligned}\right.$
$\displaystyle R_{3}(z)=\left\\{\begin{aligned}
&\frac{r^{*}(z)}{1+|r(z)|^{2}}T_{+}^{2}(z),~{}~{}~{}~{}z\in(-\infty,z_{0}),\\\
&f_{3}=\frac{r^{*}(z_{0})}{1+|r(z_{0})|^{2}}T_{0}^{2}(z_{0})(z-z_{0})^{2i\nu(z_{0})}(1-\chi_{Z}(z)),z\in\Sigma_{2},\end{aligned}\right.$
$\displaystyle R_{4}(z)=\left\\{\begin{aligned}
&\frac{r(z)}{1+|r(z)|^{2}}T_{-}^{2}(z),~{}~{}~{}~{}z\in(-\infty,z_{0}),\\\
&f_{4}=\frac{r(z_{0})}{1+|r(z_{0})|^{2}}T_{0}^{-2}(z_{0})(z-z_{0})^{-2i\nu(z_{0})}(1-\chi_{Z}(z)),z\in\Sigma_{3},\end{aligned}\right.$
$\displaystyle R_{6}(z)=\left\\{\begin{aligned}
&r^{*}(z)T^{2}(z),~{}~{}~{}~{}z\in(z_{0},\infty),\\\
&f_{6}=r^{*}(z_{0})T_{0}^{2}(z_{0})(z-z_{0})^{2i\nu(z_{0})}(1-\chi_{Z}(z)),z\in\Sigma_{4}.\end{aligned}\right.$

And $R_{j}$ admit that

$\displaystyle|R_{j}(z)|\leq c_{1}\sin^{2}(\arg
z)+c_{2}\left<Rez\right>^{-1/2},$ (4.6)
$\displaystyle|\bar{\partial}R_{j}(z)|\leq
c_{1}\bar{\partial}\chi_{Z}(z)+c_{2}|z-z_{0}|^{-1/2}+c_{3}|r^{\prime}(Rez)|,$
(4.7)
$\displaystyle\bar{\partial}R_{j}(z)=0,z\in\Omega_{2}\cup\Omega_{5},or~{}dist(z,\mathcal{Z}\cup\mathcal{Z}^{*})\leq\rho/3.$
(4.8)

where $\left<Rez\right>=\sqrt{1+(Rez)^{2}}$.

$Rez$$z_{0}$$\Omega_{5}$$\Omega_{2}$$\Omega_{6}$$\Omega_{1}$$\Omega_{3}$$\Omega_{4}$$\mathcal{R}^{(2)}=U_{R}^{-1}$$\mathcal{R}^{(2)}=W_{R}^{-1}$$\mathcal{R}^{(2)}=W_{L}$$\mathcal{R}^{(2)}=U_{L}$${z}_{k}$${\bar{z}}_{k}$$\Sigma_{1}$$\Sigma_{4}$$\Sigma_{2}$$\Sigma_{3}$

Figure 2. Definition of $R^{(2)}$ in different domains.

Then, using the transformation(4.4) and the definition of $R^{(2)}$, we derive
that $M^{(2)}$ admits the following mixed $\bar{\partial}$-RH problem.

######  Riemann-Hilbert Problem 4.7.

Find a matrix value function $M^{(2)}$, admitting

  * 1.

$M^{(2)}(x,t,z)$ is continuous in
$\mathbb{C}\backslash(\Sigma^{(2)}\cup\mathcal{Z}\cup\mathcal{Z}^{*})$.

  * 2.

$M_{+}^{(2)}(x,t,z)=M_{-}^{(2)}(x,t,z)V^{(2)}(x,t,z),$ $z\in\Sigma^{(2)}$,
where the jump matrix $V^{(2)}(x,t,z)$ satisfies

$\displaystyle
V^{(2)}(z)=(R_{-}^{(2)})^{-1}V^{(1)}R_{+}^{(2)}=I+(1-\chi_{\mathcal{Z}}(z))\bar{V}^{(2)},$
(4.9)

with

$\displaystyle\bar{V}^{(2)}(z)=\left\\{\begin{aligned} &\begin{pmatrix}0&0\\\
r(z_{0})T_{0}(z_{0})^{-2}(z-z_{0})^{-2i\nu(z_{0})}e^{2it\theta}&0\end{pmatrix}&z\in\Sigma_{1},\\\
&\begin{pmatrix}0&\frac{r^{*}(z_{0})T_{0}(z_{0})^{2}}{1+|{r(z_{0}})|^{2}}(z-z_{0})^{2i\nu(z_{0})}e^{-2it\theta}\\\
0&0\end{pmatrix}&z\in\Sigma_{2},\\\ &\begin{pmatrix}0&0\\\
\frac{r(z_{0})T_{0}(z_{0})^{-2}}{1+|{r(z_{0}})|^{2}}(z-z_{0})^{-2i\nu(z_{0})}e^{2it\theta}&0\end{pmatrix}&z\in\Sigma_{3},\\\
&\begin{pmatrix}0&r^{*}(z_{0})T_{0}(z_{0})^{2}(z-z_{0})^{2i\nu(z_{0})}e^{-2it\theta}\\\
0&0\end{pmatrix}&z\in\Sigma_{4}.\end{aligned}\right.$ (4.10)

  * 3.

$M^{(2)}(x,t,z)\rightarrow I,$ $z\rightarrow\infty$.

  * 4.

For $\mathbb{C}\backslash(\Sigma^{(2)}\cup\mathcal{Z}\cup\mathcal{Z}^{*})$,
$\bar{\partial}M^{(2)}=M^{(2)}\bar{\partial}\mathcal{R}^{(2)}(z),$ where

$\displaystyle\bar{\partial}\mathcal{R}^{(2)}=\left\\{\begin{aligned}
&\begin{pmatrix}0&0\\\
-\bar{\partial}R_{1}e^{2it\theta}&0\end{pmatrix},&z\in\Omega_{1},\\\
&\begin{pmatrix}0&-\bar{\partial}R_{3}e^{-2it\theta}\\\
0&0\end{pmatrix},&z\in\Omega_{3},\\\ &\begin{pmatrix}0&0\\\
\bar{\partial}R_{4}e^{2it\theta}&0\end{pmatrix},&z\in\Omega_{4},\\\
&\begin{pmatrix}0&-\bar{\partial}R_{6}e^{-2it\theta}\\\
0&0\end{pmatrix},&z\in\Omega_{6},\\\ &\begin{pmatrix}0&0\\\
0&0\end{pmatrix},&z\in\Omega_{2}\cup\Omega_{5}.\end{aligned}\right.$ (4.11)

  * 5.

$M^{(2)}$ admits the residue conditions at poles $z_{k}\in\mathcal{Z}$ and
$z^{*}_{k}\in\mathcal{Z}^{*}$, i.e.,

$\displaystyle\mathop{Res}_{z=z_{k}}M^{(2)}=\left\\{\begin{aligned}
&\mathop{lim}_{z\rightarrow
z_{k}}M^{(2)}\begin{pmatrix}0&c_{k}^{-1}(\frac{1}{T})^{{}^{\prime}}(z_{k})^{-2}e^{-2it\theta(z_{k})}\\\
0&0\end{pmatrix},&&k\in\triangle_{z_{0}}^{-}\\\ &\mathop{lim}_{z\rightarrow
z_{k}}M^{(2)}\begin{pmatrix}0&0\\\
c_{k}^{-1}T(z_{k})^{-2}e^{2it\theta(z_{k})}&0\end{pmatrix},&&k\in\triangle_{z_{0}}^{+}\end{aligned}\right.$
$\displaystyle\mathop{Res}_{z=z^{*}_{k}}M^{(2)}=\left\\{\begin{aligned}
&\mathop{lim}_{z\rightarrow z^{*}_{k}}M^{(2)}\begin{pmatrix}0&0\\\
-(c^{*}_{k})^{-1}T^{{}^{\prime}}(\bar{z}_{k})^{-2}e^{-2it\theta(\bar{z}_{k})}&0\end{pmatrix},&&k\in\triangle_{z_{0}}^{-}\\\
&\mathop{lim}_{z\rightarrow
z^{*}_{k}}M^{(2)}\begin{pmatrix}0&-c^{*}_{k}T(\bar{z}_{k})^{2}e^{-2it\theta(\bar{z}_{k})}\\\
0&0\end{pmatrix},&&k\in\triangle_{z_{0}}^{+}\end{aligned}\right.$

Then, from Eq.(2.37) and the transformation (3.8) and (4.4), we can
reconstruct the solution of cgNLS equation as

$\displaystyle\begin{split}u(x,t)=2i\lim_{z\rightarrow\infty}(zM^{(2)})_{12}e^{2i\int_{(-\infty,\infty)}^{(x,t)}\Delta},~{}~{}v(x,t)=2i\lim_{z\rightarrow\infty}(zM^{(2)})_{21}e^{-2i\int_{(-\infty,\infty)}^{(x,t)}\Delta}.\end{split}$
(4.12)

$Rez$$z_{0}$$\Omega_{5}$$\Omega_{2}$$\Omega_{6}$$\Omega_{1}$$\Omega_{3}$$\Omega_{4}$$\mathcal{R}^{(2)}=U_{R}^{-1}$$\mathcal{R}^{(2)}=W_{R}^{-1}$$\mathcal{R}^{(2)}=W_{L}$$\mathcal{R}^{(2)}=U_{L}$${z}_{k}$${\bar{z}}_{k}$$\Sigma_{1}$$\Sigma_{4}$$\Sigma_{2}$$\Sigma_{3}$

Figure 3. Jump matrix $V^{(2)}$, yellow parts support $\bar{\partial}$
derivative: $\bar{\partial}R^{(2)}\neq 0$. White parts do not support
$\bar{\partial}$ derivative: $\bar{\partial}R^{(2)}=0$.

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

In this section, we are going to decompose the mixed $\bar{\partial}$-RH
problem, i.e., RHP 4.7 into two parts which include a model RH problem with
$\bar{\partial}R^{(2)}=0$ and a pure $\bar{\partial}$-RH problem. We denote
$M_{RHP}$ as the solution when $\bar{\partial}R^{(2)}=0$ in the mixed
$\bar{\partial}$-RH problem 4.7. Then, if the existence of the solution of
$M_{RHP}$ can be proved and its asymptotic expansion for large $t$ can be
constructed, the mixed $\bar{\partial}$-RH problem 4.7 can be reduced to a
pure $\bar{\partial}$-RH problem. Now, we first show the pure
$\bar{\partial}$-RH problem. The proof of the existence and asymptotic of
$M_{RHP}$ will be shown in the following analysis.

######  Riemann-Hilbert Problem 5.8.

Find a matrix value function $M_{RHP}$, admitting

  * 1.

$M_{RHP}$ is analytical in
$\mathbb{C}\backslash(\Sigma^{(2)}\cup\mathcal{Z}\cup\mathcal{Z}^{*})$;

  * 2.

$M_{RHP,+}(x,t,z)=M_{RHP,-}(x,t,z)V^{(2)}(x,t,z),$ $z\in\Sigma^{(2)}$, where
$V^{(2)}(x,t,z)$ is the same with the jump matrix appears in RHP 4.7;

  * 3.

As $z\rightarrow\infty$, $M_{RHP}(x,t,z)=I+o(z^{-1})$;

  * 4.

$M_{RHP}$ possesses the same residue condition with $M^{(2)}$.

Now, using the $M_{RHP}$, we construct a transformation

$\displaystyle M^{(3)}(z)=M^{(2)}(z)M_{RHP}(z)^{-1}.$ (5.1)

Then, we can get the following pure $\bar{\partial}$-RH problem.

######  Riemann-Hilbert Problem 5.9.

Find a matrix value function $M^{(3)}$, admitting

  * 1.

$M^{(3)}$ is continuous with sectionally continuous first partial derivatives
in $\mathbb{C}\backslash(\Sigma^{(2)}\cup\mathcal{Z}\cup\mathcal{Z}^{*})$;

  * 2.

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

$\displaystyle
W^{(3)}=M_{RHP}^{(2)}(z)\bar{\partial}R^{(2)}M_{RHP}^{(2)}(z)^{-1};$ (5.2)

  * 3.

As $z\rightarrow\infty$,

$\displaystyle M^{(3)}(z)=I+o(z^{-1}).$ (5.3)

###### Proof.

Based on the properties of the $M_{RHP}$ and $M^{(2)}$ that has been shown in
RHP 5.8 and (4.7), we can derive the analyticity and asymptotic properties of
$M^{(3)}$ easily. According to the construction of the $M_{RHP}$, we know that
$M_{RHP}$ possesses the same jump matrix with $M^{(2)}$. Consequently, we
obtain that

$\displaystyle M^{(3)}_{-}(z)^{-1}M^{(3)}_{+}(z)$
$\displaystyle=M_{RHP,-}(z)M^{(2)}_{-}(z)^{-1}M^{(2)}_{+}(z)M_{RHP,+}(z)^{-1}$
$\displaystyle=M_{RHP,-}(z)V^{2}(z)(M_{RHP,-}(z)V^{2}(z))^{-1}=\emph{I},$

from which we know that $M^{(3)}$ has no jump. Next, we futher explain that
there exists no pole in $M^{(3)}$. We use $N_{k}$ to denote nilpotent matrix
which appears in the left side of the residue condition of RHP 4.7 and RHP
5.8. Then, we obtain the Laurent expansions

$\displaystyle
M^{(2)}(z)=C(z_{k})\left[\frac{N_{k}}{z-z_{k}}+\emph{I}\right]+o(z-z_{k}),$
$\displaystyle
M_{RHP}(z)=\hat{C}(z_{k})\left[\frac{N_{k}}{z-z_{k}}+\emph{I}\right]+o(z-z_{k}),$

where $C(z_{k})$ and $\hat{C}(z_{k})$ are constant terms. Then, we can derive
that

$\displaystyle M^{(2)}(z)M_{RHP}(z)^{-1}=o(1),$

which means has removable singularities at $z_{k}$. Finally, based on the
definition of $M^{(3)}$, we have

$\displaystyle\bar{\partial}M^{(3)}(z)$
$\displaystyle=\bar{\partial}(M^{(2)}(z)M_{RHP}(z)^{-1})=\bar{\partial}M^{(2)}(z)M_{RHP}(z)^{-1}=M^{(2)}(z)\bar{\partial}R^{(2)}(z)M_{RHP}(z)^{-1}$
$\displaystyle=M^{(2)}(z)M_{RHP}(z)^{-1}(M_{RHP}(z)\bar{\partial}R^{(2)}(z)M_{RHP}(z)^{-1})=M^{(3)}(z)W^{(3)}(z),$

i.e., the second condition of RHP 5.9. ∎

##  6 The pure RH problem

In this section, we construct the solution $M_{RHP}$ of RHP 5.8. Define

$\displaystyle\mathcal{U}_{z_{0}}=\\{z:|z-z_{0}|<\rho/2\\},$

and we decompose $M_{RHP}$ into two parts

$\displaystyle M_{RHP}(z)=\left\\{\begin{aligned}
&E(z)M^{(out)}(z),&&z\in\mathbb{C}\backslash\mathcal{U}_{z_{0}},\\\
&E(z)M^{(out)}(z)M^{(pc)}(z_{0},r_{0}),&&z\in\mathcal{U}_{z_{0}},\end{aligned}\right.$
(6.1)

where $M^{(out)}$ solve a model RHP, $M^{(pc)}$ is a known parabolic cylinder
model and $E(z)$ is an error function which is a solution of a small-norm
Riemann-Hilbert problem.

###  6.1 Outer model RH problem: $M^{(out)}$

Considering the construction of $M_{RHP}$, we know that $M_{RHP}$ admits a
pure Riemann-Hilbert problem. Additionally, for the jump matrix $V^{(2)}$, we
have the following estimate.

$\displaystyle||V^{(2)}-\emph{I}||_{L^{\infty}(\Sigma^{(2)})}=\left\\{\begin{aligned}
&o(e^{-4t|z-z_{0}|^{2}}),&&z\in\Sigma^{(2)}\backslash\mathcal{U}_{z_{0}},\\\
&o|z-z_{0}|^{-1}t^{-\frac{1}{2}},&&z\in\Sigma^{(2)}\cap\mathcal{U}_{z_{0}}.\end{aligned}\right.$
(6.2)

Recall that $z_{0}=-\frac{1}{4}\left(\frac{x}{t}+\alpha\right)$. Due to
$|z-z_{0}|\geqslant\rho/2$ outside $\mathcal{U}_{z_{0}}$, if we omit the jump
condition of $M_{RHP}$, the estimate infers to that exponentially small error
exists. Next, we establish a model RH problem and prove that its solution can
be approximated by finite sum of soliton solutions.

######  Riemann-Hilbert Problem 6.10.

Find a matrix value function $M^{(out)}$, admitting

  * 1.

$M^{(out)}(x,t;z)$ is analytical in
$\mathbb{C}\backslash(\Sigma^{(2)}\cup\mathcal{Z}\cup\mathcal{Z}^{*})$;

  * 2.

As $z\rightarrow\infty$,

$\displaystyle M^{(out)}(x,t;z)=I+o(z^{-1});$ (6.3)

  * 3.

$M^{(out)}(x,t;z)$ has simple poles at each point in
$\mathcal{Z}\cup\mathcal{Z}^{*}$ admitting the same residue condition in RHP
4.7 with $M^{(out)}(x,t;z)$ replacing $M^{(2)}(x,t;z)$.

To obtain the solution of $M^{(out)}(x,t;z)$, we first study RHP 2.2 for the
case of reflectionless. Under this condition, we know that $M$ has no jump and
get the following Riemann-Hilbert problem from RHP 2.2.

######  Riemann-Hilbert Problem 6.11.

Find a matrix value function $M(x,t;z|\sigma_{d})$, admitting

  * 1.

$M(x,t;z|\sigma_{d})$ is analytical in
$\mathbb{C}\setminus(\mathcal{Z}\bigcup\mathcal{Z}^{*})$;

  * 2.

$M(x,t;z|\sigma_{d})=I+O(z^{-1})$, $z\rightarrow\infty$;

  * 3.

$M(x,t;z|\sigma_{d})$ satisfies the following residue conditions at simple
poles $z_{k}\in\mathcal{Z}$ and $z_{k}^{*}\in\mathcal{Z}^{*}$

$\displaystyle\begin{aligned}
&\mathop{Res}_{z=z_{k}}M(x,t;z|\sigma_{d})=\mathop{lim}_{z\rightarrow
z_{k}}M(x,t;z|\sigma_{d})N_{k},\\\
&\mathop{Res}_{z=z_{k}^{*}}M(x,t;z|\sigma_{d})=\mathop{lim}_{z\rightarrow
z_{k}^{*}}M(x,t;z|\sigma_{d})\sigma_{2}N^{*}_{k}\sigma_{2},\end{aligned}$
(6.4)

where $\sigma_{d}=\\{(z_{k},c_{k}),z_{k}\in\mathcal{Z}\\}^{N}_{k=1}$, which
satisfies $z_{k}\neq z_{j}$ for $k\neq j$, are scattering data , and

$\displaystyle N_{k}=\left(\begin{aligned} \begin{array}[]{cc}0&0\\\
\gamma_{k}(x,t)&0\end{array}\end{aligned}\right),~{}\gamma_{k}(x,t)=c_{k}e^{2it\theta(z_{k})},$
(6.5) $\displaystyle\theta(z_{k})=2z^{2}+\alpha z-\gamma+\frac{x}{t}z.$ (6.6)

Then, according to the Liouville’s theorem, the uniqueness of the solution is
obvious. Based on the symmetry that shown in Eq.(2.25), we obtain
$M(x,t;z|\sigma_{d})=-\sigma M^{*}(x,t;z^{*}|\sigma_{d})\sigma$, from which we
can derive the following expansion, i.e.,

$\displaystyle
M(x,t;z|\sigma_{d})=\emph{I}+\sum_{k=1}^{N}\left[\frac{1}{z-z_{k}}\left(\begin{aligned}
\begin{array}[]{cc}\zeta_{k}(x,t)&0\\\
\eta_{k}(x,t)&0\end{array}\end{aligned}\right)+\frac{1}{z-z^{*}_{k}}\left(\begin{aligned}
\begin{array}[]{cc}0&-\eta^{*}_{k}(x,t)\\\
0&\zeta^{*}_{k}(x,t)\end{array}\end{aligned}\right)\right],$ (6.7)

where $\zeta_{k}(x,t)$ and $\eta_{k}(x,t)$ are unknown coefficients to be
determined. Next, using the similar way that has been shown in literature[28],
we obtain the following proposition.

######  Corollary 6.12.

For the given scattering data $\sigma_{d}$, the RHP 6.11 exists the unique
solution.

So, for given scattering data $\sigma_{d}$, the unique solution of RHP(6.11)
can be shown that

$\displaystyle
u_{sol}(x,t;\sigma_{d})=2i\mathop{lim}_{z\rightarrow\infty}(zM(z|\sigma_{d}))_{12}e^{2i\int_{(-\infty,\infty)}^{(x,t)}\Delta},$
(6.8) $\displaystyle
v_{sol}(x,t;\sigma_{d})=2i\mathop{lim}_{z\rightarrow\infty}(zM(z|\sigma_{d}))_{21}e^{-2i\int_{(-\infty,\infty)}^{(x,t)}\Delta}.$
(6.9)

Now, we need to establish the relation between $M(x,t;z|\sigma_{d})$ and
$M^{(out)}(x,t;z)$. We first give the following proposition, and further
explained it in the following two subsections.

######  Corollary 6.13.

There exist unique solution $M^{(out)}$ of RHP 6.10. Particularly,

$\displaystyle
M^{(out)}(z)=M^{\vartriangle_{z_{0}}^{-}}(z)\delta(z)^{\sigma_{3}}=M^{\vartriangle_{z_{0}}^{-}}(z|\sigma_{d}^{out})$
(6.10)

where $M^{\vartriangle_{z_{0}}^{-}}(z)$ is the solution of RHP 6.14 with
$\vartriangle=\vartriangle_{z_{0}}^{-}$ and
$\sigma_{d}^{out}=\\{(z_{k},\widetilde{c}_{k}(z_{0}))\\}_{k=1}^{N}$ with

$\displaystyle\widetilde{c}_{k}(z_{0})=c_{k}e^{\frac{i}{\pi}\int_{-\infty}^{z_{0}}\frac{log(1+|r(s)|^{2})}{s-z_{k}}ds}.$
(6.11)

In addition,

$\displaystyle
u_{sol}(x,t;\sigma_{d}^{out})=\lim_{z\rightarrow\infty}2izM_{12}^{(out)}(x,t;z)e^{2i\int_{(-\infty,\infty)}^{(x,t)}\Delta},$
(6.12) $\displaystyle
v_{sol}(x,t;\sigma_{d}^{out})=\lim_{z\rightarrow\infty}2izM_{21}^{(out)}(x,t;z)e^{-2i\int_{(-\infty,\infty)}^{(x,t)}\Delta},$
(6.13)

where $u_{sol}(x,t;\sigma_{d}^{out})$ and $v_{sol}(x,t;\sigma_{d}^{out})$ are
the $N$-soliton solution of Eq.(1.1) corresponding the scattering data
$\sigma_{d}^{out}$.

####  6.1.1 Renormalization of the Riemann-Hilbert problem for reflectionless
case

For the case of reflectionless, recall that

$\displaystyle
M(x,t;z|\sigma_{d})=\left(\frac{\mu_{-,1}(x,t;z)}{s_{11}(z)},\mu_{+,2}(x,t;z)\right),~{}~{}z\in\mathbb{C}^{+},~{}~{}s_{11}(z)=\prod_{k=1}^{N}\left(\frac{z-z_{k}}{z-z^{*}_{k}}\right).$
(6.14)

Let $\vartriangle\subseteq\\{1,2,\cdots,N\\}$,
$\bigtriangledown\subseteq\\{1,2,\cdots,N\\}\setminus\Delta$, and define

$\displaystyle
s_{11,\vartriangle}=\prod_{k\in\vartriangle}\frac{z-z_{k}}{z-z^{*}_{k}},\quad
s_{11,\triangledown}=\frac{s_{11}}{s_{11,\vartriangle}}=\prod_{k\in\triangledown}\frac{z-z_{k}}{z-z^{*}_{k}}.$
(6.15)

Then, we introduce the normalization transformation

$\displaystyle
M^{\vartriangle}(x,t;z|\sigma_{d}^{\vartriangle})=M(x,t;z|\sigma_{d})s_{11,\vartriangle}(z)^{\sigma_{3}},$
(6.16)

which splits the poles between the columns of $M(x,t;z|\sigma_{d})$ by
selecting different $\vartriangle$. Then, we can get the modified RIemann-
Hilbert problem.

######  Riemann-Hilbert Problem 6.14.

Given scattering data $\sigma_{d}=\\{(z_{k},c_{k})\\}^{N}_{k=1}$ and
$\vartriangle\subseteq\\{1,2,\cdots,N\\}$. Find a matrix value function
$M^{\vartriangle}$, admitting

  * 1.

$M^{\vartriangle}(x,t;z|\sigma_{d})$ is analytical in
$\mathbb{C}\setminus(\mathcal{Z}\bigcup\mathcal{Z}^{*})$;

  * 2.

$M^{\vartriangle}(x,t;z|\sigma_{d})=I+O(z^{-1})$, $z\rightarrow\infty$;

  * 3.

$M^{\vartriangle}(x,t;z|\sigma_{d})$ satisfies the following residue
conditions at simple poles $z_{k}\in\mathcal{Z}$ and
$z_{k}^{*}\in\mathcal{Z}^{*}$

$\displaystyle\begin{aligned}
&\mathop{Res}_{z=z_{k}}M^{\vartriangle}(x,t;z|\sigma_{d})=\mathop{lim}_{z\rightarrow
z_{k}}M^{\vartriangle}(x,t;z|\sigma_{d})N^{\vartriangle}_{k},\\\
&\mathop{Res}_{z=z_{k}^{*}}M^{\vartriangle}(x,t;z|\sigma_{d})=\mathop{lim}_{z\rightarrow
z_{k}^{*}}M^{\vartriangle}(x,t;z|\sigma_{d})\sigma_{2}(N^{\vartriangle}_{k})^{*}\sigma_{2},\end{aligned}$
(6.17)

where

$\displaystyle N_{k}^{\vartriangle}=\left\\{\begin{aligned}
\left(\begin{array}[]{cc}0&\gamma_{k}^{\vartriangle}\\\ 0&0\\\
\end{array}\right),\quad k\in\vartriangle,\\\ \left(\begin{array}[]{cc}0&0\\\
\gamma_{k}^{\vartriangle}&0\\\ \end{array}\right),\quad
k\in\vartriangle,\end{aligned}\right.~{}~{}\gamma_{k}^{\vartriangle}=\left\\{\begin{aligned}
&c_{k}^{-1}(s_{11,\vartriangle}^{{}^{\prime}}(z_{k}))^{-2}e^{-2it\theta(z_{k})}\quad
k\in\vartriangle,\\\
&c_{k}(s_{11,\vartriangle}(z_{k}))^{2}e^{2it\theta(z_{k})}\quad
k\in\vartriangle,\end{aligned}\right.$ (6.18)
$\displaystyle\theta(z_{k})=2z^{2}+\alpha z-\gamma+\frac{x}{t}z.$

Since $M^{\vartriangle}(x,t;z|\sigma_{d})$ is directly transformed from
$M(x,t;z|\sigma_{d})$, it is obvious that RHP 6.14 has unique solution.

####  6.1.2 Long-time behavior of soliton solutions

When $N=1$, the scattering data $\sigma_{d}=\\{(z_{1}=\xi+i\eta,c_{1})\\}$, we
derive that

$\displaystyle u(x,t)=2\eta
sech(2\eta\Omega(x,t))e^{-2i((2(\xi-\eta)^{2}-\alpha\xi+\gamma)t+\xi
x)}e^{-i(\frac{\pi}{2}+arg(c_{1}))}e^{2i\int_{(-\infty,\infty)}^{(x,t)}\Delta},$
(6.19) $\displaystyle v(x,t)=2\eta
sech(2\eta\Omega(x,t))e^{2i((2(\xi-\eta)^{2}-\alpha\xi+\gamma)t+\xi
x)}e^{i(\frac{\pi}{2}+arg(c_{1}))}e^{-2i\int_{(-\infty,\infty)}^{(x,t)}\Delta},$
(6.20)

where $\Omega(x,t)=4\xi t+\alpha
t-x-\frac{1}{2\eta}log(\frac{|c_{1}|}{2\eta})$. In fact, after the elastic
collisions, the $N$-soliton asymptotically separate into $N$ single-soliton
solutions as $t\rightarrow\infty$, of course, except the non-generic case
which we do not discuss here such as two points of scattering data lie on a
vertical line.

Define a distance

$\displaystyle\mu(I)=\min_{z_{k}\in Z\setminus
Z(I)}\left\\{Im(z_{k})dist(Rez_{k},I)\right\\},$ (6.21)

and a space-time cone

$\displaystyle
S(x_{1},x_{2},v_{1},v_{2})=\\{(x,t),x=x_{0}+vt~{}with~{}x_{0}\in[x_{1},x_{2}],v\in[v_{1},v_{2}]\\}.$
(6.22)

$x$$t$$x_{2}$$x_{1}$$x=v_{2}t+x_{2}$$x=v_{1}t+x_{2}$$x=v_{2}t+x_{1}$$x=v_{2}t+x_{1}$$S$

Figure 4. Space-time $S(x_{1},x_{2},v_{1},v_{2})$.

######  Proposition 6.15.

For given scattering data
$\sigma_{d}^{\vartriangle_{z_{0}}^{-}}=\\{(z_{k},\hat{c}_{k})\\},$ for any
fixed $x_{1},x_{2},v_{1},v_{2}\in R$ and $x_{1}<x_{2}$, $v_{1}<v_{2}$. Let
$\mathcal{I}=\left[-\frac{v_{2}}{2},-\frac{v_{1}}{2}\right]$, then
$t\rightarrow\infty$ and $(x,t)\in S(x_{1},x_{2},v_{1},v_{2})$, we have

$\displaystyle
M^{\vartriangle_{z_{0}}^{\mp}}(z|\sigma_{d}^{\vartriangle_{z_{0}}^{\pm}})=(I+O(e^{-8\mu
t}))M^{\vartriangle_{z_{0}}^{\pm}(\mathcal{I})}(z|\hat{\sigma}_{d}(\mathcal{I})),$
(6.23)

where
$M^{\vartriangle_{z_{0}}^{\mp}(\mathcal{I})}(z|\hat{\sigma}_{d}(\mathcal{I}))$
is $N(I)=|\mathcal{Z}(\mathcal{I})|$-soliton solutions corresponding to
scattering data

$\displaystyle\hat{\sigma}_{d}(\mathcal{I})=\\{(z_{k},c_{k}(\mathcal{I})),z_{k}\in\mathcal{Z}(\mathcal{I})\\},~{}c_{k}(\mathcal{I})=c_{k}\prod_{z_{j}\in\mathcal{Z}\setminus\mathcal{Z}(\mathcal{I})}\left(\frac{z_{k}-z_{j}}{z_{k}-z^{*}_{j}}\right)^{2}.$
(6.24)

$Rez$$-\frac{v_{1}}{2}$$-\frac{v_{2}}{2}$$z_{5}$$\bar{z}_{5}$$z_{2}$$\bar{z}_{2}$$z_{1}$$\bar{z}_{1}$$z_{3}$$\bar{z}_{3}$$z_{4}$$\bar{z}_{4}$

Figure 5. For fixed $v_{1}<v_{2}$,
$I=\left[-\frac{v_{2}}{2},-\frac{v_{1}}{2}\right]$.

From RHP 6.14, if we chose $\vartriangle=\vartriangle_{z_{0}}^{\mp}$, it is
easy to check that

$\displaystyle||N_{k}^{\vartriangle_{z_{0}}^{\mp}}||=\left\\{\begin{aligned}
&o(1)\quad k\in\mathcal{Z}(\mathcal{I}),\\\ &o(e^{-8\mu(\mathcal{I})t})\quad
k\in\mathcal{Z}\setminus\mathcal{Z}(\mathcal{I}),\end{aligned}\right.~{}~{}t\rightarrow\pm\infty,$
(6.25)

which infers to the residues with
$z_{k}\in\mathcal{Z}\setminus\mathcal{Z}(\mathcal{I})$ have little
contribution to the solution $M^{\vartriangle_{z_{0}}^{\pm}}$.

Next, for each discrete spectrum point
$z_{k}\in\mathcal{Z}\setminus\mathcal{Z}(\mathcal{I})$, we make a small disk
$D_{k}$ whose radius is smaller than $\mu(\mathcal{I})$. Denoting that
$\partial D_{k}$ is the boundary of $D_{k}$. Then, we introduce that

$\displaystyle\phi(z)=\left\\{\begin{aligned}
&I-\frac{N_{k}^{\vartriangle^{-}(\mathcal{I})}}{z-z_{k}}\quad z\in D_{k},\\\
&I+\frac{\sigma(N^{*}_{k})^{\vartriangle^{-}(\mathcal{I})\sigma}}{z-z_{k}}\quad
z\in D^{*}_{k},\\\ &I,\quad elsewhere,\end{aligned}\right.$ (6.26)

where

$\displaystyle\vartriangle^{-}(\mathcal{I})=\\{k:Rez_{k}<-\frac{v_{2}}{2}\\},~{}~{}\vartriangle^{+}(\mathcal{I})=\\{k:Rez_{k}>-\frac{v_{1}}{2}\\}.$

Making the transformation
$\hat{M}^{\vartriangle_{z_{0}}^{\mp}}(z)=M^{\vartriangle_{z_{0}}^{\mp}}(z)\phi(z)$,
we can derive that $\hat{M}^{\vartriangle_{z_{0}}^{\pm}}(z)$ has new jump in
each boundary of the disk which can be denoted as $\hat{V}$, i.e.,

$\displaystyle\hat{M}^{\vartriangle_{z_{0}}^{\pm}}_{+}(z)=\hat{M}^{\vartriangle_{z_{0}}^{\pm}}_{-}(z)\hat{V},~{}~{}z\in\partial
D_{k}\cup D^{*}_{k}.$ (6.27)

By using the estimate (6.25), we have

$\displaystyle||\hat{V}-I||=o(e^{-8\mu(\mathcal{I})t}),~{}~{}z\in\partial
D_{k}\cup D^{*}_{k},~{}~{}t\rightarrow\pm\infty.$ (6.28)

It is obvious that $\hat{M}^{\vartriangle_{z_{0}}^{\pm}}(z|\sigma_{d})$ and
$M^{\vartriangle_{z_{0}}^{\mp}}(z|\hat{\sigma}_{d})$ the same poles and
residue conditions. Thus,

$\displaystyle\varepsilon(z)=\hat{M}^{\vartriangle_{z_{0}}^{\pm}}(z|\sigma_{d})[M^{\vartriangle_{z_{0}}^{\mp}}(z|\hat{\sigma}_{d})]^{-1}$
(6.29)

has no poles. However, there exists jump for
$z\in\cup_{z_{k}\in\mathcal{Z}\setminus\mathcal{Z}(\mathcal{I})}(\partial
D_{k}\cup\partial D^{*}_{k})$,

$\displaystyle\varepsilon_{+}(z)=\varepsilon_{-}(z)V_{\varepsilon},$

where
$V_{\varepsilon}=M^{\vartriangle_{z_{0}}^{\mp}}\hat{V}(^{\vartriangle_{z_{0}}^{\mp}})^{-1}$
which implies that

$\displaystyle||V_{\varepsilon}-I||=o(e^{-8\mu(\mathcal{I})t}),~{}~{}t\rightarrow\pm\infty.$
(6.30)

Then, using the theory of small-norm Riemann-Hilbert problems, one can easily
derive that

$\displaystyle\varepsilon(z)=I+o(e^{-8\mu(\mathcal{I})t}),~{}~{}t\rightarrow\pm\infty,$

which together with
$\hat{M}^{\vartriangle_{z_{0}}^{\mp}}(z)=M^{\vartriangle_{z_{0}}^{\mp}}(z)\phi(z)$
gives the formula (6.23).

###  6.2 Local solvable model near phase point
$z=z_{0}=-\frac{1}{4}(\frac{x}{t}+\alpha)$

From (6.2), we can easily find that $V^{(2)}-I$ does not have a uniform
estimate for large time. Therefore, we use a local model
$M^{(out)}(z)M^{(pc)}(z_{0},r_{0})$ to match the the jumps of $M_{RHP}$ on
$\Sigma^{(2)}\cap\mathcal{U}_{z_{0}}$ to make the jump uniformly for the
function $E(z)$. Recall the definition of $\theta(z)$ (3.1) and introduce the
transformation

$\displaystyle\lambda=\lambda(z)=2\sqrt{2t}(z-z_{0}).$ (6.31)

Then, we can derive that

$\displaystyle 2t\theta=\frac{1}{2}\lambda^{2}-4tz_{0}^{2}-2\gamma t,$ (6.32)

which means $\mathcal{U}_{z_{0}}$ can be mapped into an expanding neighborhood
of $\lambda=0$. If we let

$\displaystyle
r_{0}(z_{0})=r(z_{0})T_{0}(z_{0})^{-2}e^{2i(\nu(z_{0})log(2\sqrt{2t}))}e^{-4itz_{0}^{2}-2it\gamma}$
(6.33)

and consider the fact that $1-\chi_{\mathcal{Z}}=1$ as
$z\in\mathcal{U}_{z_{0}}$, the jump of $M_{RHP}$ in $\mathcal{U}_{z_{0}}$ can
be expressed as

$\displaystyle
V^{(2)}(z)\mid_{z\in\mathcal{U}_{z_{0}}}=\left\\{\begin{aligned}
\lambda(z)^{i\nu\hat{\sigma}_{3}e^{-\frac{i\lambda(z)^{2}}{4}\hat{\sigma}_{3}}}\left(\begin{array}[]{cc}1&0\\\
r_{0}(z_{0})&1\\\ \end{array}\right),\quad z\in\Sigma_{1},\\\
\lambda(z)^{i\nu\hat{\sigma}_{3}e^{-\frac{i\lambda(z)^{2}}{4}\hat{\sigma}_{3}}}\left(\begin{array}[]{cc}1&\frac{r^{*}_{0}(z_{0})}{1+|r_{0}(z_{0})|^{2}}\\\
0&1\\\ \end{array}\right),\quad z\in\Sigma_{2},\\\
\lambda(z)^{i\nu\hat{\sigma}_{3}e^{-\frac{i\lambda(z)^{2}}{4}\hat{\sigma}_{3}}}\left(\begin{array}[]{cc}1&0\\\
\frac{r_{0}(z_{0})}{1+|r_{0}(z_{0})|^{2}}&1\\\ \end{array}\right),\quad
z\in\Sigma_{3},\\\
\lambda(z)^{i\nu\hat{\sigma}_{3}e^{-\frac{i\lambda(z)^{2}}{4}\hat{\sigma}_{3}}}\left(\begin{array}[]{cc}1&r^{*}_{0}(z_{0})\\\
0&1\\\ \end{array}\right),\quad z\in\Sigma_{4}.\end{aligned}\right.$ (6.34)

Obviously, the above jump $V^{(2)}(z)\mid_{z\in\mathcal{U}_{z_{0}}}$ is
equivalent to the jump of the parabolic cylinder model problem(A.5) which we
have obtain the solutions in appendix $A$. Then, since $M^{(out)}(z)$ an
analytic and bounded function in the $\mathcal{U}_{z_{0}}$,
$M^{(out)}(z)M^{(pc)}(z_{0},r_{0})$ which has been defined in (6.1) admits the
jump $V^{(2)}(z)$ of $M_{RHP}(z)$.

###  6.3 The small-norm RHP for $E(z)$

Recall the transform (6.1), we can derive that

$\displaystyle E(z)=\left\\{\begin{aligned}
&M_{RHP}(z)M^{(out)}(z)^{-1},&&z\in\mathbb{C}\backslash\mathcal{U}_{z_{0}},\\\
&M_{RHP}(z)M^{(pc)}(z_{0},r_{0})^{-1}M^{(out)}(z)^{-1},&&z\in\mathcal{U}_{z_{0}},\end{aligned}\right.$
(6.35)

which is analytic in $\mathbb{C}\setminus\Sigma^{(E)}$ where

$\displaystyle\Sigma^{(E)}=\partial\mathcal{U}_{z_{0}}\bigcup(\Sigma^{(2)}\backslash\mathcal{U}_{z_{0}}),$
(6.36)

with clockwise direction for $\partial\mathcal{U}_{z_{0}}$. Then we can check
that $E(z)$ satisfies the Riemann-Hilbert problem.

######  Riemann-Hilbert Problem 6.16.

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

  * 1.

$E$ is analytical in $\mathbb{C}\backslash\Sigma^{(E)}$;

  * 2.

$E(z)=I+O(z^{-1})$, $z\rightarrow\infty$;

  * 3.

$E_{+}(z)=E_{-}(z)V^{(E)}(z)$, $z\in\Sigma^{(E)}$, where

$\displaystyle V^{(E)}(z)=\left\\{\begin{aligned}
&M^{(out)}(z)V^{(2)}(z)M^{(out)}(z)^{-1},&&z\in\Sigma^{(2)}\backslash\mathcal{U}_{z_{0}},\\\
&M^{(out)}(z)M^{(pc)}(\xi,r_{0})M^{(out)}(z)^{-1},&&z\in\partial\mathcal{U}_{z_{0}}.\end{aligned}\right.$
(6.37)

$\partial\mathcal{U}_{z_{0}}$$\Sigma^{(2)}\setminus\mathcal{U}_{z_{0}}$

Figure 6. Jump contour
$\Sigma^{(E)}=\partial\mathcal{U}_{z_{0}}\cup(\Sigma^{(2)}\setminus\mathcal{U}_{z_{0}})$.

Base on the estimate $V^{(2)}-I$ shown in (6.2) and the boundedness of
$M^{(out)}$, it is easy to check that

$\displaystyle|V^{(E)}(z)-I|=\left\\{\begin{aligned}
&\mathcal{O}(e^{-2t|z-z_{0}|^{2}})&&z\in\Sigma^{(2)}\backslash\mathcal{U}_{z_{0}},\\\
&\mathcal{O}(t^{-1/2})&&z\in\partial\mathcal{U}_{z_{0}},\end{aligned}\right.$
(6.38)

then, we can derive that

$\displaystyle||(z-z_{0})^{k}(V^{(E)}-I)||_{L^{p}(\Sigma^{(E)})}=o(t^{-1/2}),~{}~{}p\in[1,+\infty],~{}k\geq
0.$ (6.39)

As a small-norm Riemann-Hilbert problem, the existence and uniqueness of RHP
6.16 are guaranteed and we obtain that

$\displaystyle E(z)=I+\frac{1}{2\pi
i}\int_{\Sigma_{E}}\frac{\mu_{E}(s)(V^{(E)}(s)-I)}{s-z}ds$ (6.40)

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

$\displaystyle(1-C_{\omega_{E}})\mu_{E}=I,$ (6.41)

where $C_{\omega_{E}}$ is an integral operator which is defined by

$\displaystyle C_{\omega_{E}}f=C_{-}(f(V^{(E)}-I)),$ $\displaystyle
C_{-}f(z)\lim_{z\rightarrow\Sigma_{-}^{(E)}}\int_{\Sigma_{E}}\frac{f(s)}{s-z}ds,$

where $C_{-}$ is the Cauchy projection operator. Then, according to the
properties of the Cauchy projection operator $C_{-}$, and the estimate of
$V^{(E)}-I$ shown in (6.38), we can derive that

$\displaystyle\|C_{\omega_{E}}\|_{L^{2}(\Sigma^{(E)})}\lesssim\|C_{-}\|_{L^{2}(\Sigma^{(E)})}\|V^{(E)}-I\|_{L^{\infty}(\Sigma^{(E)})}\lesssim\mathcal{O}(t^{-1/2}).$
(6.42)

which implies that $1-C_{\omega_{E}}$ is invertible and thus the existence and
uniqueness of $\mu_{E}$ are guaranteed, and thus of $E(z)$. Now, it can be
explained that the definition of $M_{RHP}$ is reasonable, and in turn we can
solve (5.1) to the unknown $M^{(3)}$ which admits the pure
$\bar{\partial}$-Problem(5.9).

Then, to reconstruct the solutions of $u(x,t)$ and $v(x,t)$, the asymptotic
behavior of $E(z)$ as $z\rightarrow\infty$ is needed. According to the
expression (6.40) of $E(z)$, we obtain that

$\displaystyle E(z)=I+z^{-1}E_{1}+\mathcal{O}(z^{-2}),\quad
z\rightarrow\infty,$ (6.43)

where

$\displaystyle
E_{1}(x,t)=-\frac{1}{2i\pi}\int_{\Sigma_{E}}\mu_{E}(s)(V^{(E)}(s)-I)ds.$
(6.44)

Then, by applying (6.41) and the estimate (6.38), (6.39), we obtain that

$\displaystyle
E_{1}(x,t)=-\frac{1}{2i\pi}\oint_{\partial\mathcal{U}_{z_{0}}}(V^{(E)}(s)-I)ds+o(t^{-1}).$
(6.45)

Finally, applying (6.37), (6.31) and (A.6) shown in Appendix $A$, we derive
that

$\displaystyle\begin{split}E_{1}(x,t)&=\frac{1}{2i\sqrt{2t}}M^{(out)}(z_{0})M_{1}^{(pc)}(z_{0})M^{(out)}(z_{0})^{-1}+\mathcal{O}(t^{-1})\\\
&=\frac{1}{2i\sqrt{2t}}M^{(out)}(z_{0})\begin{pmatrix}0&\beta_{12}\\\
-\beta_{21}&0\end{pmatrix}M^{(out)}(z_{0})^{-1}+\mathcal{O}(t^{-1}),\end{split}$
(6.46)

with

$\displaystyle\beta_{12}(z_{0})=\beta_{21}^{*}(z_{0})=\tau(z_{0},+)e^{ix^{2}/(4t)+i\alpha^{2}t/(4)-2\alpha
x-i\nu(z_{0})\log|8t|}.$ (6.47)

Here, $|\tau(z_{0},+)|^{2}=|\nu(z_{0})^{2}|$ and

$\displaystyle\arg\tau(z_{0},+)=\frac{\pi}{4}+\arg\Gamma(i\nu(z_{0}))-\arg
r(z_{0})-2\int^{z_{0}}_{-\infty}\ln|z-z_{0}|\mathrm{d}\nu(s).$

##  7 Pure $\bar{\partial}$-Problem

In this section, we analysis the remaining $\bar{\partial}$-Problem. The
$\bar{\partial}$-RH problem 5.9 for $M^{(3)}(z)$ is equivalent to the
following integral equation

$\displaystyle
M^{(3)}(z)=\mathrm{I}-\frac{1}{\pi}\iint_{\mathbb{C}}\frac{M^{(3)}W^{(3)}}{s-z}\mathrm{d}A(s),$
(7.1)

where $\mathrm{d}A(s)$ is Lebesgue measure. Further, we write the equation
(7.1) in operator form

$\displaystyle(\mathrm{I}-\mathrm{S})M^{(3)}(z)=\mathrm{I},$ (7.2)

where $\mathrm{S}$ is Cauchy operator

$\displaystyle\mathrm{S}[f](z)=-\frac{1}{\pi}\iint_{\mathbb{C}}\frac{f(s)W^{(3)}(s)}{s-z}\mathrm{d}A(s).$
(7.3)

We need to prove that the inverse operator $(\mathrm{I}-\mathrm{S})^{-1}$
exists, so that the solution $M^{(3)}(z)$ exists.

######  Lemma 7.17.

For large $t$, there exists a constant $c$ that makes the operator (7.3)
admits the following relation

$\displaystyle||\mathrm{S}||_{L^{\infty}\rightarrow L^{\infty}}\leq
ct^{-1/4}.$ (7.4)

###### Proof.

We mainly prove the case that the matrix function supported in the region
$\Omega_{1}$, the other case can be proved similarly. Denoted that $f\in
L^{\infty}(\Omega_{1})$ and $s=p+iq$, then based on (4.11) and (5.2), we can
obtain the following inequality

$\displaystyle|S[f](z)|$
$\displaystyle\leq\frac{1}{\pi}\iint_{\Omega_{1}}\frac{|f(s)M_{RHP}(s)\bar{\partial}R_{1}(s)M_{RHP}(s)^{-1}|}{|s-z|}df(s)$
(7.5)
$\displaystyle\leq\big{|}\big{|}f\big{|}\big{|}_{L^{\infty}(\Omega_{1})}\big{|}\big{|}M_{RHP}\big{|}\big{|}_{L^{\infty}(\hat{\Omega}_{1})}\big{|}\big{|}M_{RHP}^{-1}\big{|}\big{|}_{L^{\infty}(\hat{\Omega}_{1})}\frac{1}{\pi}\iint_{\Omega_{1}}\frac{|\bar{\partial}R_{1}(s)||e^{-8tq(p-z_{0})}|}{|s-z|}df(s),$
(7.6)

where $\hat{\Omega}_{1}\triangleq\Omega_{1}\cap supp(1-\chi_{\mathcal{Z}})$ is
bounded far from the scattering data $z_{k}$ of $M_{RHP}$, and thus
$||M_{RHP}^{\pm 1}||_{L^{\infty}(\hat{\Omega}_{1})}||$ are finite.

Based on (4.6) and the estimates shown in Appendix $B$, from (7.5), we obtain
that

$\displaystyle||\mathrm{S}||_{L^{\infty}\rightarrow L^{\infty}}\leq
c(I_{1}+I_{2}+I_{3})\leq ct^{-1/4},$ (7.7)

where

$\displaystyle
I_{1}=\iint_{\Omega_{1}}\frac{|\bar{\partial}\chi_{\mathcal{Z}}(s)||e^{-8tq(p-z_{0})}|}{|s-z|}df(s),~{}~{}I_{2}=\iint_{\Omega_{1}}\frac{|r^{\prime}(p)||e^{-8tq(p-z_{0})}|}{|s-z|}df(s),$
(7.8)

and

$\displaystyle
I_{3}=\iint_{\Omega_{1}}\frac{|s-z_{0}|^{-\frac{1}{2}}|e^{-8tq(p-z_{0})}|}{|s-z|}df(s).$
(7.9)

∎

Next, our purpose is to recover the large time asymptotic behaviors of
$u(x,t)$ and $v(x,t)$. According to (2.37), we need determine the coefficient
of the $z^{-1}$ term of $M^{(3)}$ in the Laurent expansion at infinity. Base
on (7.1), we can derive that

$\displaystyle
M^{(3)}(z)=\mathrm{I}-\frac{1}{\pi}\int\int_{\mathbb{C}}\frac{M^{(3)}(s)W^{(3)}(3)}{s-z}\mathrm{d}A(s)=\mathrm{I}+\frac{M^{(3)}_{1}}{z}+o(z^{-2}),$

where

$\displaystyle
M^{(3)}_{1}=\frac{1}{\pi}\int\int_{\mathbb{C}}M^{(3)}(s)W^{(3)}(s)\mathrm{d}A(s).$

In what follows, the the coefficient of the $z^{-1}$ term of $M^{(3)}$ in the
Laurent expansion at infinity, i.e., $M^{(3)}_{1}$ satisfies the following
property.

######  Lemma 7.18.

For large $t$, there exists a constant $c$ that makes $M^{(3)}_{1}$ admits the
following inequality

$\displaystyle M^{(3)}_{1}\leq ct^{-3/4}.$ (7.10)

The proof of this Lemma is similar to the process that shown in Appendix $B$.

##  8 Soliton resolution for the cgNLS equation

Now, we are going to construct the long time asymptotic of the cgNLS equation
1.1). Recall a series of transformation process including (3.8), (4.4), (5.1)
and (6.1), i.e.,

$\displaystyle M(z)\leftrightarrows M^{(1)}(z)\leftrightarrows
M^{(2)}(z)\leftrightarrows M^{(3)}(z)\leftrightarrows E(z),$

we then obtain

$\displaystyle
M(z)=M^{(3)}(z)E(z)M^{(out)}(z)R^{(2)^{-1}}(z)T^{\sigma_{3}}(z),~{}~{}z\in\mathbb{C}\setminus\mathcal{U}_{z_{0}}.$

In order to recover the solution $u(x,t)$ and $v(x,t)$, we take
$z\rightarrow\infty$ vertically which implies $z\in\Omega_{2}$ or
$z\in\Omega_{5}$, thus $R^{(2)}(z)=I$. Then, we have

$\displaystyle
M=\left(\mathrm{I}+\frac{M^{(3)}_{1}}{z}+\cdots\right)\left(\mathrm{I}+\frac{E_{1}}{z}+\cdots\right)\left(\mathrm{I}+\frac{M_{1}^{(out)}}{z}+\cdots\right)\left(\mathrm{I}+\frac{T_{1}\sigma_{3}}{z}+\cdots\right),$

from which we can derive that

$\displaystyle M_{1}=M_{1}^{(out)}+E_{1}+M_{1}^{(3)}+T_{1}\sigma_{3}.$

Then, according to the reconstruction formula (2.37) and the Lemma 7.18, we
obtain that

$\displaystyle
u(x,t)e^{-2i\int_{(-\infty,\infty)}^{(x,t)}\Delta}=2i(M_{1}^{(out)})_{12}+2i(E_{1})_{12}+O(t^{-3/4}),$
(8.1) $\displaystyle
v(x,t)e^{2i\int_{(-\infty,\infty)}^{(x,t)}\Delta}=2i(M_{1}^{(out)})_{21}+2i(E_{1})_{21}+O(t^{-3/4}).$
(8.2)

Based on the Corollary 6.13 and the formulae (6.46), (6.47), we obtain the
soliton resolution

$\displaystyle
u(x,t)e^{-2i\int_{(-\infty,\infty)}^{(x,t)}\Delta}=u_{sol}(x,t;\sigma_{d}^{out})e^{-2i\int_{(-\infty,\infty)}^{(x,t)}\Delta}+t^{-\frac{1}{2}}f^{+}_{1}+O(t^{-\frac{3}{4}}),$
$\displaystyle
v(x,t)e^{2i\int_{(-\infty,\infty)}^{(x,t)}\Delta}=v_{sol}(x,t;\sigma_{d}^{out})e^{-2i\int_{(-\infty,\infty)}^{(x,t)}\Delta}+t^{-\frac{1}{2}}f^{+}_{2}+O(t^{-\frac{3}{4}}).$

where

$\displaystyle
f^{+}_{1}=\frac{1}{\sqrt{2}}m_{11}^{2}\tau(z_{0},+)e^{\theta_{1}(x,t)}+\frac{1}{2\sqrt{2}}m_{12}^{2}\tau^{*}(z_{0},+)e^{\theta_{2}(x,t)},$
$\displaystyle
f^{+}_{2}=-\frac{1}{\sqrt{2}}m_{22}^{2}\tau^{*}(z_{0},+)e^{\theta_{2}(x,t)}-\frac{1}{2\sqrt{2}}m_{21}^{2}\tau(z_{0},+)e^{\theta_{1}(x,t)},$

with $z_{0}=-\frac{1}{4}(\frac{x}{t}-\alpha)$,

$\displaystyle\theta_{1}(x,t)=\frac{ix^{2}}{4t}+i\frac{\alpha^{2}t}{4}-2\alpha
x-i\nu(z_{0})\log|8t|,$
$\displaystyle\theta_{2}(x,t)=-\frac{ix^{2}}{4t}-i\frac{\alpha^{2}t}{4}-2\alpha
x+i\nu(z_{0})\log|8t|,$

and $m_{11}$, $m_{12}$, $m_{21}$ and $m_{22}$ come from the elements of the
matrix solution of RHP 6.11. Via using Proposition 6.15, there exists
exponential errors which are absorbed into the $O(t^{-3/4})$ term when we
replace $u_{sol}(x,t;\sigma_{d}^{out})$ and $v_{sol}(x,t;\sigma_{d}^{out})$
with $u_{sol}(x,t;\hat{\sigma}_{d}(I))$ and
$v_{sol}(x,t;\hat{\sigma}_{d}(I))$, respectively. Then, we consider the term
$e^{-2i\int_{(-\infty,\infty)}^{(x,t)}\Delta}$. As $t\rightarrow\infty$, we
have

$\displaystyle e^{-2i\int_{(-\infty,\infty)}^{(x,t)}\Delta}=$ $\displaystyle
e^{-2i\beta\int_{-\infty}^{x}u(s,t)v(s,t)ds}$ $\displaystyle=$ $\displaystyle
e^{-2i\beta\int_{(-\infty,\infty)}^{(x,t)}\mathcal{M}(s,t)ds}+O(t^{-\frac{3}{2}}),$

where

$\displaystyle\mathcal{M}^{+}(x,t)=|u_{sol}(x,t;\hat{\sigma}_{d}(I))v_{sol}(x,t;\hat{\sigma}_{d}(I))+t^{-1}f^{+}_{1}f^{+}_{2}|.$
(8.3)

Finally, we can give the more general solution of $u(x,t)$ and $v(x,t)$ and
summarise as the following theorem.

######  Theorem 8.19.

With the initial value conditions that $u_{0}(x),v_{0}(x)\in
H^{1,1}(\mathbb{R})$ and the assumption that the initial values
$u_{0}(x),v_{0}(x)$ satisfy the Assumption (2.1), let $u(x,t)$ and $v(x,t)$ be
the solution of cgNLS equations (1.1). The scattering data denoted as
$\\{r,\\{z_{k},c_{k}\\}_{k=1}^{N}\\}$ which generated from the initial values
$u_{0}(x),v_{0}(x)$. For fixed $x_{1},x_{2},v_{1},v_{2}\in\mathbb{R}$ with
$x_{1}<x_{2}$, $v_{1}<v_{2}$, and
$\mathcal{I}=\left[-\frac{v_{2}}{2},-\frac{v_{1}}{2}\right]$,
$z_{0}=-\frac{1}{4}(\frac{x}{t}+\alpha)$, then as $t\rightarrow\infty$ and
$(x,t)\in S(x_{1},x_{2},v_{1},v_{2})$ which is defined in (6.22), the solution
$u(x,t)$ and $v(x,t)$ can be expressed as

$\displaystyle\begin{split}u(x,t)=(u_{sol}(x,t;\hat{\sigma}_{d}(I))+t^{-\frac{1}{2}}f^{\pm}_{1})e^{2i\beta\int_{(-\infty,\infty)}^{(x,t)}\mathcal{M}(s,t)ds}+O(t^{-\frac{3}{4}}),\\\
v(x,t)=(v_{sol}(x,t;\hat{\sigma}_{d}(I))+t^{-\frac{1}{2}}f^{\pm}_{2})e^{-2i\beta\int_{(-\infty,\infty)}^{(x,t)}\mathcal{M}(s,t)ds}+O(t^{-\frac{3}{4}}).\end{split}$
(8.4)

Here, $u_{sol}(x,t;\hat{\sigma}_{d}(I))$ and
$v_{sol}(x,t;\hat{\sigma}_{d}(I))$ are the
$N(\mathcal{I})=|\mathcal{Z}(\mathcal{I})|$ soliton, and

$\displaystyle\mathcal{M}^{\pm}(x,t)=|u_{sol}(x,t;\hat{\sigma}_{d}(I))v_{sol}(x,t;\hat{\sigma}_{d}(I))+t^{-1}f^{\pm}_{1}f^{\pm}_{2}|,$
$\displaystyle
f^{\pm}_{1}=\frac{1}{\sqrt{2}}m_{11}^{2}\tau(z_{0},\pm)e^{\theta^{\mp}_{1}(x,t)}+\frac{1}{\sqrt{2}}m_{12}^{2}\tau^{*}(z_{0},\pm)e^{\theta^{\pm}_{2}(x,t)},$
$\displaystyle
f^{\pm}_{2}=-\frac{1}{\sqrt{2}}m_{22}^{2}\tau^{*}(z_{0},\pm)e^{\theta^{\pm}_{2}(x,t)}-\frac{1}{\sqrt{2}}m_{21}^{2}\tau(z_{0},\pm)e^{\theta^{\mp}_{1}(x,t)},$

with $|\tau(z_{0},\pm)|^{2}=|\nu(z_{0})^{2}|$,

$\displaystyle\arg\tau(z_{0},\pm)=\frac{\pi}{4}+\arg\Gamma(i\nu(z_{0}))-\arg
r(z_{0})\mp 2\int^{z_{0}}_{-\infty}\ln|z-z_{0}|\mathrm{d}\nu(s),$
$\displaystyle\theta^{\mp}_{1}(x,t)=\frac{ix^{2}}{4t}+i\frac{\alpha^{2}t}{4}-2\alpha
x\mp i\nu(z_{0})\log|8t|,$
$\displaystyle\theta^{\pm}_{2}(x,t)=-\frac{ix^{2}}{4t}-i\frac{\alpha^{2}t}{4}-2\alpha
x\pm i\nu(z_{0})\log|8t|,$

and $m_{11}$, $m_{12}$, $m_{21}$ and $m_{22}$ come from the elements of the
matrix solution of RHP6.11.

######  Remark 8.20.

The results shown in Theorem 8.19 need the condition that the initial values
$u_{0}(x),v_{0}(x)\in H^{1,1}(\mathbb{R})$, then the inverse scattering
transform possesses well mapping properties, and it guarantees the condition
that there exists no discrete spectrum on the real axis. It is noted that the
asymptotic results only depend on the $H^{1}(\mathbb{R})$ norm of $r$,
therefore, for any $u_{0}(x),v_{0}(x)\in H^{1,1}(\mathbb{R})$ admitting the
Assumption (2.1), the process of the large-time analysis and calculations
shown in this work is unchanged.

## Acknowledgements

This work was supported by the National Natural Science Foundation of China
under Grant No. 11975306, the Natural Science Foundation of Jiangsu Province
under Grant No. BK20181351, the Six Talent Peaks Project in Jiangsu Province
under Grant No. JY-059, and the Fundamental Research Fund for the Central
Universities under the Grant Nos. 2019ZDPY07 and 2019QNA35.

## Appendix A: The parabolic cylinder model problem

Here, we describe the solution of parabolic cylinder model problem[36, 37].
Define the contour $\Sigma^{pc}=\cup_{j=1}^{4}\Sigma_{j}^{pc}$ where

$\displaystyle\Sigma_{j}^{pc}=\left\\{\lambda\in\mathbb{C}|\arg\lambda=\frac{2j-1}{4}\pi\right\\}.$
(A.1)

For $r_{0}\in\mathbb{C}$, let $\nu(r)=-\frac{1}{2\pi}\log(1+|r_{0}|^{2})$,
consider the following parabolic cylinder model Riemann-Hilbert problem.

######  Riemann-Hilbert Problem 8.21.

Find a matrix-valued function $M^{(pc)}(\lambda)$ such that

$\displaystyle\bullet\quad M^{(pc)}(\lambda)~{}\text{is analytic
in}~{}\mathbb{C}\setminus\Sigma^{pc},$ (A.2) $\displaystyle\bullet\quad
M_{+}^{(pc)}(\lambda)=M_{-}^{(pc)}(\lambda)V^{(pc)}(\lambda),\quad\lambda\in\Sigma^{pc},$
(A.3) $\displaystyle\bullet\quad
M^{(pc)}(\lambda)=I+\frac{M_{1}}{\lambda}+O(\lambda^{2}),\quad\lambda\rightarrow\infty.$
(A.4)

where

$\displaystyle V^{(pc)}(\lambda)=\left\\{\begin{aligned}
\lambda^{i\nu\hat{\sigma}_{3}e^{-\frac{i\lambda^{2}}{4}\hat{\sigma}_{3}}}\left(\begin{array}[]{cc}1&0\\\
r_{0}&1\\\ \end{array}\right),\quad\lambda\in\Sigma_{1}^{pc},\\\
\lambda^{i\nu\hat{\sigma}_{3}e^{-\frac{i\lambda^{2}}{4}\hat{\sigma}_{3}}}\left(\begin{array}[]{cc}1&\frac{r^{*}_{0}}{1+|r_{0}|^{2}}\\\
0&1\\\ \end{array}\right),\quad\lambda\in\Sigma_{2}^{pc},\\\
\lambda^{i\nu\hat{\sigma}_{3}e^{-\frac{i\lambda^{2}}{4}\hat{\sigma}_{3}}}\left(\begin{array}[]{cc}1&0\\\
\frac{r_{0}}{1+|r_{0}|^{2}}&1\\\
\end{array}\right),\quad\lambda\in\Sigma_{3}^{pc},\\\
\lambda^{i\nu\hat{\sigma}_{3}e^{-\frac{i\lambda^{2}}{4}\hat{\sigma}_{3}}}\left(\begin{array}[]{cc}1&r^{*}_{0}\\\
0&1\\\ \end{array}\right),\quad\lambda\in\Sigma_{4}^{pc},\end{aligned}\right.$
(A.5)

$\Sigma_{1}^{pc}$$\Sigma_{4}^{pc}$$\Sigma_{2}^{pc}$$\Sigma_{3}^{pc}$$0$$\Omega_{6}$$\Omega_{1}$$\Omega_{5}$$\Omega_{2}$$\Omega_{4}$$\Omega_{3}$$\lambda^{i\nu\hat{\sigma}_{3}}e^{-\frac{i\lambda^{2}}{4}\hat{\sigma}_{3}}\left(\begin{array}[]{cc}1&0\\\
r_{0}&1\\\
\end{array}\right)$$\lambda^{i\nu\hat{\sigma}_{3}}e^{-\frac{i\lambda^{2}}{4}\hat{\sigma}_{3}}\left(\begin{array}[]{cc}1&r^{*}_{0}\\\
0&1\\\
\end{array}\right)$$\lambda^{i\nu\hat{\sigma}_{3}}e^{-\frac{i\lambda^{2}}{4}\hat{\sigma}_{3}}\left(\begin{array}[]{cc}1&\frac{r^{*}_{0}}{1+|r_{0}|^{2}}\\\
0&1\\\
\end{array}\right)$$\lambda^{i\nu\hat{\sigma}_{3}}e^{-\frac{i\lambda^{2}}{4}\hat{\sigma}_{3}}\left(\begin{array}[]{cc}1&0\\\
\frac{r_{0}}{1+|r_{0}|^{2}}&1\\\ \end{array}\right)$

Figure 7. Jump matrix $V^{(pc)}$.

We know that the parabolic cylinder equation can be expressed as [38]

$\displaystyle\left(\frac{\partial^{2}}{\partial
z}+(\frac{1}{2}-\frac{z^{2}}{2}+a)\right)D_{a}=0.$

As shown in the literature[20, 39], we obtain the explicit solution
$M^{(pc)}(\lambda,r_{0})$:

$\displaystyle
M^{(pc)}(\lambda,r_{0})=\Phi(\lambda,r_{0})\mathcal{P}(\lambda,r_{0})e^{\frac{i}{4}\lambda^{2}\sigma_{3}}\lambda^{-i\nu\sigma_{3}},$

where

$\displaystyle\mathcal{P}(\lambda,r_{0})=\left\\{\begin{aligned}
&\left(\begin{array}[]{cc}1&0\\\ -r_{0}&1\\\
\end{array}\right),\quad&\lambda\in\Omega_{1},\\\
&\left(\begin{array}[]{cc}1&-\frac{r^{*}_{0}}{1+|r_{0}|^{2}}\\\ 0&1\\\
\end{array}\right),\quad&\lambda\in\Omega_{3},\\\
&\left(\begin{array}[]{cc}1&0\\\ \frac{r_{0}}{1+|r_{0}|^{2}}&1\\\
\end{array}\right),\quad&\lambda\in\Omega_{4},\\\
&\left(\begin{array}[]{cc}1&r^{*}_{0}\\\ 0&1\\\
\end{array}\right),\quad&\lambda\in\Omega_{6},\\\
&~{}~{}~{}\emph{I},\quad&\lambda\in\Omega_{2}\cup\Omega_{5},\end{aligned}\right.$

and

$\displaystyle\Phi(\lambda,r_{0})=\left\\{\begin{aligned}
\left(\begin{array}[]{cc}e^{-\frac{3\pi\nu}{4}}D_{i\nu}\left(e^{-\frac{3i\pi}{4}}\lambda\right)&-i\beta_{12}e^{-\frac{\pi}{4}(\nu-i)}D_{-i\nu-1}\left(e^{-\frac{i\pi}{4}}\lambda\right)\\\
i\beta_{21}e^{-\frac{3\pi(\nu+i)}{4}}D_{i\nu-1}\left(e^{-\frac{3i\pi}{4}}\lambda\right)&e^{\frac{\pi\nu}{4}}D_{-i\nu}\left(e^{-\frac{i\pi}{4}}\lambda\right)\\\
\end{array}\right),\quad\lambda\in\mathbb{C}^{+},\\\
\left(\begin{array}[]{cc}e^{\frac{\pi\nu}{4}}D_{i\nu}\left(e^{\frac{i\pi}{4}}\lambda\right)&-i\beta_{12}e^{-\frac{3\pi(\nu-i)}{4}}D_{-i\nu-1}\left(e^{\frac{3i\pi}{4}}\lambda\right)\\\
i\beta_{21}e^{\frac{\pi}{4}(\nu+i)}D_{i\nu-1}\left(e^{\frac{i\pi}{4}}\lambda\right)&e^{-\frac{3\pi\nu}{4}}D_{-i\nu}\left(e^{\frac{3i\pi}{4}}\lambda\right)\\\
\end{array}\right),\quad\lambda\in\mathbb{C}^{-},\end{aligned}\right.$

with

$\displaystyle\beta_{12}=\frac{\sqrt{2\pi}e^{i\pi/4}e^{-\pi\nu/2}}{r_{0}\Gamma(-i\nu)},\quad\beta_{21}=\frac{-\sqrt{2\pi}e^{-i\pi/4}e^{-\pi\nu/2}}{r_{0}^{*}\Gamma(i\nu)}=\frac{\nu}{\beta_{12}}.$

Then, it is not hard to obtain the asymptotic behavior of the solution by
using the well known asymptotic behavior of $D_{a}(z)$,

$\displaystyle
M^{(pc)}(r_{0},\lambda)=I+\frac{M_{1}^{(pc)}}{i\lambda}+O(\lambda^{-2}),$
(A.6)

where

$\displaystyle M_{1}^{(pc)}=\begin{pmatrix}0&\beta_{12}\\\
-\beta_{21}&0\end{pmatrix}.$

## Appendix B: Detailed calculations for the pure $\bar{\partial}$-Problem

######  Proposition 8.22.

For $t>0$, there exists constants $c_{j}(j=1,2,3)$ such that $I_{j}(j=1,2,3)$
which defined in (7.8) and (7.9) possess the following estimate

$\displaystyle I_{j}\leq c_{j}t^{-\frac{1}{4}},~{}~{}j=1,2,3.$ (B.1)

###### Proof.

Let $s=p+iq$ and $z=\xi+i\eta$. Using the fact that

$\displaystyle\Big{|}\Big{|}\frac{1}{s-z}\Big{|}\Big{|}_{L^{2}}(q+z_{0})=(\int_{q+z_{0}}^{\infty}\frac{1}{|s-z|^{2}}dp)^{\frac{1}{2}}\leq\frac{\pi}{q-\eta},$

we can derive that

$\displaystyle\begin{split}|I_{1}|&\leq\int_{0}^{+\infty}\int_{q+z_{0}}^{+\infty}\frac{|\bar{\partial}\chi_{\mathcal{Z}}(s)|e^{-8tq(p-z_{0})}}{|s-z|}dpdq\\\
&\leq\int_{0}^{+\infty}e^{-8tq^{2}}\big{|}\big{|}\bar{\partial}\chi_{\mathcal{Z}}(s)\big{|}\big{|}_{L^{2}}(q+z_{0})\Big{|}\Big{|}\frac{1}{s-z}\Big{|}\Big{|}_{L^{2}}(q+z_{0})dq\\\
&\leq c_{1}\int_{0}^{+\infty}\frac{e^{-8tq^{2}}}{\sqrt{|q-\eta|}}dq\leq
c_{1}t^{-\frac{1}{4}}.\end{split}$ (B.2)

Similarly, considering that $r\in H^{1,1}(\mathbb{R})$, we obtain the estimate

$\displaystyle|I_{2}|\leq\int_{0}^{+\infty}\int_{q+z_{0}}^{+\infty}\frac{|r^{\prime}(p)|e^{-8tq^{2}}}{|s-z|}dpdq\leq
c_{2}t^{-\frac{1}{4}}.$ (B.3)

To obtain the estimate of $I_{3}$, we consider the following $L^{k}(k>2)$ norm

$\displaystyle\Big{|}\Big{|}\frac{1}{\sqrt{|s-z_{0}|}}\Big{|}\Big{|}_{L^{k}}\leq\left(\int_{q+z_{0}}^{+\infty}\frac{1}{|p-z_{0}+iq|^{\frac{k}{2}}}dp\right)^{\frac{1}{k}}\leq
cq^{\frac{1}{k}-\frac{1}{2}}.$ (B.4)

Similarly, we can derive that

$\displaystyle\Big{|}\Big{|}\frac{1}{|s-z|}\Big{|}\Big{|}_{L^{k}}\leq
c|q-\eta|^{\frac{1}{k}-\frac{1}{2}}.$ (B.5)

By applying (B.4) and (B.4), it is not hard to check that

$\displaystyle\begin{split}|I_{3}|&\leq\int_{0}^{+\infty}\int_{q}^{+\infty}\frac{|z-z_{0}|^{-\frac{1}{2}}e^{-8tq(p-z_{0})}}{|s-z|}dpdq\\\
&\leq\int_{0}^{+\infty}e^{-8tq^{2}}\Big{|}\Big{|}\frac{1}{\sqrt{|s-z_{0}|}}\Big{|}\Big{|}_{L^{k}}\Big{|}\Big{|}\frac{1}{|s-z|}\Big{|}\Big{|}_{L^{k}}dq\leq
c_{3}t^{-\frac{1}{4}}.\end{split}$ (B.6)

Now, we complete the estimates of $I_{j}(j=1,2,3)$. ∎

## References

  * [1] V.E. Zakharov, Stability of periodicwaves of finite amplitude on the surface of a deep fluid, Sov. Phys. J. Appl. Mech. Tech. Phys. 4 (1968) 190-194.
  * [2] V.E. Zakharov, Collapse of Langmuir waves, Sov. Phys. JETP, 35 (1972) 908-914.
  * [3] V.E. Zakharov , A.B. Shabat, Exact theory of two-dimensional self-focusing and one-dimensional self-modulation of waves in nonlinear media, Sov. Phys. JETP, 34 (1972) 62-69.
  * [4] G.P. Agrawal, Nonlinear Fiber Optics Academic, San Diego, 1989.
  * [5] A. Hasegawa and Y. Kodama, Solitons in Optical Communications Clarendon, Oxford, 1995.
  * [6] S.F. Tian, T.T. Zhang, Long-time asymptotic behavior for the Gerdjikov-Ivanov type of derivative nonlinear Schrödinger equation with time-periodic boundary condition, Proc. Am. Math. Soc. 146 (2018) 1713-1729.
  * [7] S.F. Tian, Initial-boundary value problems for the general coupled nonlinear Schrödinger equation on the interval via the Fokas method, J. Differ. Equ. 262 (2017) 506-558.
  * [8] S.F. Tian, The mixed coupled nonlinear Schrödinger equation on the half-line via the Fokas method, Proc. R. Soc. Lond. A, 472(2195) (2016) 20160588.
  * [9] Z.Y. Yan, An initial-boundary value problem for the integrable spin-1 Gross-Pitaevskii equations with a $4\times 4$ Lax pair on the half-line, Chaos, 27(5) (2017) 053117.
  * [10] B. Guo, N. Liu, Y. Wang, A Riemann-Hilbert approach for a new type coupled nonlinear Schrödinger equations, J. Math. Anal. Appl. 459(1) (2018) 145-158.
  * [11] D.S. Wang, B. Guo, X. Wang, Long-time asymptotics of the focusing Kundu-Eckhaus equation with nonzero boundary conditions, J. Differ. Equ. 266(9) (2019) 5209-5253.
  * [12] S.V. Manakov. On the theory of two-dimensional stationary selffocussing of electromagnetic waves. Sov. Phys. JETP, 65 (1973) 505-516.
  * [13] X. Geng, H. W. Tam, Darboux Transformation and Soliton Solutions for Generalized Nonlinear Schrödinger Equations. J. Phys. Soc. Jpn. 68(5) (1999), 1508-1512.
  * [14] G.Z. Tu, Liouville Integrability of Zero Curvature Equations, (Springer, Berlin) 1990.
  * [15] S. Kakei, N. Sasa, J. Satsuma, Bilinearization of a Generalized Derivative Nonlinear Schrödinger Equation, J. Phys. Soc. Jpn. 64 (1995) 1519.
  * [16] F. Zhao, Z. D. Li, Q. Y. Li, L. Wen, G. S. Fu, W. M. Liu, Magnetic rogue wave in a perpendicular anisotropic ferromagnetic nanowire with spin-transfer torque, Ann. Phys. 327(9) (2012) 2085-2095.
  * [17] W.Q. Peng, S.F. Tian, Long-time asymptotics in the modified Landau-Lifshitz equation with nonzero boundary conditions, arXiv:1912.00542.
  * [18] S.V. Manakov, Nonlinear Fraunhofer diffraction, Sov. Phys. JETP 38 (1974) 693-696.
  * [19] V. E. Zakharov, S. V. Manakov, Asymptotic behavior of nonlinear wave systems integrated by the inverse scattering method, Sov. Phys. JETP, 44 (1976) 106-112.
  * [20] P. Deift, X. Zhou, A steepest descent method for oscillatory Riemann¨CHilbert problems. Asymptotics for the MKdV equation, Ann. Math. 137(2) (1993) 295-368.
  * [21] P. Deift, X. Zhou, Long-time asymptotics for integrable systems. Higher order theory, Comment. Phys.-Math. 165(1) (1994) 175-191.
  * [22] P. Deift, X. Zhou, Long-Time Behavior of the Non-Focusing Nonlinear Schrödinger Equation, a Case Study, Lectures in Mathematical Sciences, New Ser., vol. 5, Graduate School of Mathematical Sciences, University of Tokyo, 1994.
  * [23] P. Deift, X. Zhou, Long-time asymptotics for solutions of the NLS equation with initial data in a weighted Sobolev space, Commun. Pure Appl. Math. 56(8) (2003) 1029-1077.
  * [24] 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.
  * [25] 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. IMRN (2008), Art. ID 075.
  * [26] M. Dieng, K. D. T. McLaughlin, Long-time Asymptotics for the NLS equation via dbar methods, arXiv: 0805.2807.
  * [27] S. Cuccagna, R. Jenkins, On asymptotic stability of $N$-solitons of the defocusing nonlinear Schrödinger equation, Comm. Math. Phys. 343 (2016) 921-969.
  * [28] M. Borghese, R. Jenkins, K. T. R. McLaughlin, Long-time asymptotic behavior of the focusing nonlinear Schrödinger equation, Ann. I. H. Poincaré Anal, 35 (2018) 887-920.
  * [29] R. Jenkins, J. Liu, P. Perry, C. Sulem, Soliton Resolution for the derivative nonlinear Schrödinger equation, Commun. Math. Phys. 363 (2018) 1003-1049.
  * [30] R. Jenkins, J. Liu, P. Perry, C. Sulem, Global well-posedness for the derivative nonlinear Schrödinger equation, Commun. Part. Diff. Equ. 43(8) (2018) 1151-1195.
  * [31] Y. L. Yang, E.G. Fan, Soliton Resolution for the Short-pluse Equation, arXiv:2005.12208.
  * [32] Q. Y. Cheng, E.G. Fan, Soliton resolution for the focusing Fokas-Lenells equation with weighted Sobolev initial data, arXiv:2010.08714.
  * [33] R. H. Ma, E.G. Fan, Longtime asymptotic behavior of the focusing nonlinear Kundu-Eckhaus equation, arXiv:1912.01425.
  * [34] R. Beals, R.R. Coifman, Linear spectral problems, non-linear equations and the $\bar{\partial}$-method, Inverse Probl. 5 (1989) 87-130.
  * [35] Z.Q. Li, S.F. Tian, J.J. Yang, Riemann-Hilbert approach and soliton solutions for the higher-order dispersive nonlinear Schrödinger equation with nonzero boundary conditions, arXiv:1911.01624.
  * [36] A. Its, Asymptotic behavior of the solutions to the nonlinear Schrödinger equation, and isomonodromic deformations of systems of linear differential equations, Dokl. Akad. Nauk SSSR (ISSN 0002-3264) 261(1) (1981) 14-18.
  * [37] J. Liu, P. Perry, C. Sulem, Long-time behavior of solutions to the derivative nonlinear Schrödinger equation for soliton-free initial data, Ann. I. H. Poincaré, Anal. Non Linéaire, 35 (2018) 217-265.
  * [38] F.W.J. Olver, A.B. Olde Daalhuis, D.W. Lozier, B.I. Schneider, R.F. Boisvert, C.W. Clark, B.R. Miller, B.V. Saunders, NIST Digital Library of Mathematical Functions, (2016). http://dlmf.nist.gov/.
  * [39] R. Jenkins, K. McLaughlin, Semiclassical limit of focusing NLS for a family of square barrier initial data, Commun. Pure Appl. Math. 67 (2) (2014) 246-320.