# Long-time asymptotic behavior of the Hunter-Saxton equation

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

With $\bar{\partial}$-generalization of the Deift-Zhou steepest descent
method, we investigate the long-time asymptotics of the solution to the Cauchy
problem for the Hunter-Saxton (HS) equation

$\displaystyle u_{txx}-2\omega u_{x}+2u_{x}u_{xx}+uu_{xxx}=0,\quad
x\in\mathbb{R},\ t>0,$ (0.1) $\displaystyle u(x,0)=u_{0}(x),$ (0.2)

where $u_{0}\in H^{3,4}(\mathbb{R})$ and $\omega>0$ is a constant. Using the
new scale $(y,t)$ and a series of deformations to a Riemann-Hilbert problem
associated with the Cauchy problem, we obtain the long-time asymptotic
approximations of the solution $u(x,t)$ in two space-time regions: The
solution of the HS equation decays as the speed of $\mathcal{O}(t^{-1/2})$ in
the region $y/t>0$; While in the region $y/t<0$, the solution of the HS
equation is depicted by a parabolic cylinder model with an residual error
order $\mathcal{O}(t^{-1+\frac{1}{2p}})$ with $p>2$.

###### keywords:

Hunter-Saxton equation, Riemann-Hilbert problem, $\bar{\partial}$-steepest
descent method, long-time asymptotics.
Mathematics Subject Classification: 35Q51; 35Q15; 35C20; 35P25.

###### Contents

  1. 1 Introduction
  2. 2 The IST under lower regular initial data
    1. 2.1 The Lax pair and Jost solutions
    2. 2.2 Scattering map from $u_{0}(x)$ to $r(k)$
    3. 2.3 The Riemann-Hilbert problem of the HS equation
  3. 3 Asymptotic analysis in the region $y/t<0$
    1. 3.1 Triangular decomposition
    2. 3.2 A mixed $\bar{\partial}$-RH problem and its decomposition
    3. 3.3 Analysis on pure RH problem
      1. 3.3.1 A local solvable RH model near stationary points
      2. 3.3.2 Error function with small norm
    4. 3.4 Analysis on the pure $\bar{\partial}$-problem
    5. 3.5 Long-time asymptotics for the HS equation
  4. 4 Asymptotic analysis in the region $y/t>0$
    1. 4.1 A mixed $\bar{\partial}$-RH problem and its decomposition
    2. 4.2 Analysis on the pure RH problem
    3. 4.3 Analysis on the pure $\bar{\partial}$-problem
    4. 4.4 Long-time asymptotics for the HS equation

##  1 Introduction

This paper is concerned with the long time asymptotic behavior to the solution
of the Cauchy problem for the Hunter-Saxton (HS) equation [1]

$\displaystyle u_{txx}-2\omega u_{x}+2u_{x}u_{xx}+uu_{xxx}=0,\quad
x\in\mathbb{R},\ t>0,$ (1.1) $\displaystyle u(x,0)=u_{0}(x),$ (1.2)

where $u_{0}\in H^{3,4}(\mathbb{R})$ and $\omega>0$ is a constant. The HS
equation (1.1) was derived by Hunter and Saxton as a short-wave limit of the
Camassa-Holm (CH) equation [2]

$u_{t}-u_{txx}-2\omega u_{x}+3uu_{x}=2u_{x}u_{xx}+uu_{xxx},$ (1.3)

which was discovered as a model for unidirectional propagation of small
amplitude shallow water waves by Camassa and Holm [11]. The CH equation also
appears in the study of the propagation of axially symmetric waves in
hyperelastic rods [4, 5]. Constantin and Lannes further found that the CH
equation arises in the modeling of the propagation of shallow water waves over
a flat bed [6]. Due to its astonishing amounts of structures and relating
interesting physical phenomena, the CH equation has received extensive
attention. For example, Constantin et al. proved the existence and uniqueness
of global weak solutions for the CH equation (1.3) [7], and discussed the
scatting problem of the CH equation [8]. Besides, Constantin et al showed the
orbital stability of the peakons and the solitary wave solution for the Cauchy
problem of the CH equation (1.3) [9, 10]. Boutet de Monvel et al. first
investigated the related Riemann-Hilbert (RH) problem for the CH equation [11,
12]. They further obtained long-time asymptotics and Painlevé-type asymptotics
for CH equation on the line and half-line by using the nonlinear steepest
descent method [11, 12, 13, 14]. This method developed by Deift and Zhou [15]
is an effective tool to rigorously analyze the long-time asymptotics behavior
of integrable systems [16, 17, 18, 19, 20, 21].

As a short-wave limit of the CH equation, the HS equation has also received
considerable attention. For instance, the HS equation (1.1) was found to be a
model for short capillary waves propagating under the action of gravity [22,
23]. Alber et.al studied the weak piecewise solutions of (1.1) by algebraic
geometric methods [24]. The local existence of strong solutions of the
periodic Hunter-Saxton equation has been proved by Yin [25]. The multi-cusp
solutions of (1.1) were obtained by Matsuno et.al [26]. Lenells et al.
established Liouville transformation between the HS hierarchies and the KdV
hierarchies [27]. Zuo proposed a two-component generalization of the
generalized HS equation and established its bi-Hamiltonian Euler equation
[28]. Zhao constructs the conservation laws of the HS equation for liquid
crystal [29]. Lenells provides a rigorous foundation for the geometric
interpretation of the HS equation, and yields explicit expressions for the
spatially periodic solutions of the related Cauchy problem [30]. Holden et.al
developed an existence theory for the Cauchy problem to the stochastic HS
equation and proved several properties of the blow-up of its solutions [31].
For initial data in Sobolev spaces $H^{s}(\mathbb{R})$, $s>3/2$, Tiǧlay proved
local well-posedness of the periodic Cauchy problem for the modified HS
equation [32]. Recently Monvel et.al developed the inverse scattering
transform and obtained long-time asymptotics for the HS equation (1.1) with
Schwartz initial data by using steepest descent method [33].

The aim of our paper is to derive the long time asymptotic behavior of the
solution for the HS equations (1.1) with a lower regularity weighted Sobolev
initial data

$u_{0}\in H^{3,4}(\mathbb{R})=\\{u\in
L^{2}(\mathbb{R})|\langle\cdot\rangle^{4}\partial^{j}u\in L^{2}(\mathbb{R}),\
j=0,1,\cdots,3\\}.$ (1.4)

The key tool to proved this result is the $\bar{\partial}$-generalization of
the steepest descent method, which was first proposed by McLaughlin and Miller
[34, 35]. In recent years, this method shows some advantages and has been
successfully used to study long-time asymptotic analysis, soliton resolution
and asymptotic stability of $N$-soliton solutions to integrable systems in a
weighted Sobolev space [34, 35, 36, 37, 38, 39, 40, 41, 42, 44, 45, 43, 46,
47, 48, 49].

In order to extend the result [33] to the solution with weighted Sobolev
initial data $u_{0}\in H^{3,4}(\mathbb{R})$ using the
$\bar{\partial}$-steepest descent method, there two essential difficulties
need to be overcome. Firstly, we need to establish a scattering map between
initial data and reflection coefficient

$H^{3,4}(\mathbb{R})\ni u_{0}\to r(k)\in H^{4}(\mathbb{R})\cap
H^{1,1}(\mathbb{R}),$

which is given in Proposition 1. This require us to make a series of refined
estimates on eigenfunctions and scattering data starting from initial data
$u_{0}\in H^{3,4}(\mathbb{R})$. Secondly, we need to carefully deal with the
effect of spectral singularity $k=0$. In general, for the Schrödinger equation
and mKdV equation, their solutions can be recovered from the
$k^{-1}$-coefficient of the solution of their related RH problem as
$k\to\infty$ [15, 40, 45]. However, the solution of the HS equation need to be
recovered from the $k^{2}$-coefficient of the solution of a RH problem as
$k\to 0$. This case results in a higher singular term of $k^{-3}$ appearing in
the integral to estimate a pure $\bar{\partial}$-problem for $M^{(3)}(k)$. To
avoid this singularity, we construct a kind of new extension functions
$R_{jl}(k)$ such that $|\bar{\partial}R_{jl}|\lesssim|k|^{2}$ near $k=0$,
which is given in Proposition 3.

Without loss of generality, we only discuss the HS equation (1.1) as
$\omega=1$, because under the scaling transformation

$x=\omega^{-2}\tilde{x},\ t=\omega^{-1}\tilde{t},\
u(x,t)=\omega^{-3}\tilde{u}(\tilde{x},\tilde{t}),\ $ (1.5)

the HS equation (1.1) becomes

$\tilde{u}_{\tilde{t}\tilde{x}\tilde{x}}-2\tilde{u}_{\tilde{x}}+2\tilde{u}_{\tilde{x}}\tilde{u}_{\tilde{x}\tilde{x}}+\tilde{u}\tilde{u}_{\tilde{x}\tilde{x}\tilde{x}}=0.$

Now we address our main result as follows.

######  Theorem 1.

Let $u_{0}\in H^{3,4}(\mathbb{R})$ with $-u_{0xx}+1>0,\ for\ all\
x\in\mathbb{R}$. Then the solution to the Cauchy problem (1.1)-(1.2) has the
following asymptotic approximation as $t\to\infty$:

  * •

when $\xi=\frac{y}{t}>0$, the solution $u(x,t)$ decays fast to $0$,

$u(x,t)=\hat{u}(y(x,t),t)=\mathcal{O}(t^{-1/2}),\quad
x(y,t)=y+\mathcal{O}(t^{-1/2});$ (1.6)

  * •

when $\xi=\frac{y}{t}<0$, the solution $u(x,t)$ has the asymptotic expansion

$u(x,t)=\hat{f}t^{-1/2}+\mathcal{O}(t^{-1+\frac{1}{2p}}),$ (1.7)

and

$x(y,t)=y-\frac{1}{2}\delta_{1}+\frac{\rho_{0}^{3/2}}{2}f_{11}t^{-\frac{1}{2}}+\mathcal{O}(t^{-1+\frac{1}{2p}}).$
(1.8)

where $p>2$, $\hat{f},f_{11}$ and $\delta_{1}$ are given in Section 3. Here
the variable $y$ is defined as

$y=y(x,t)=x-\int_{x}^{\infty}(\sqrt{m(s,t)+1}-1)\mathrm{d}s.$

This paper is arranged as follows. In Section 2, with the condition $u_{0}\in
H^{3,4}(\mathbb{R})$, we first make the direct scatting transform. To improve
the strictness of the integral estimates in $\bar{\partial}$-method, we then
establish a scattering map between the initial data and the reflection
coefficient and prove it detailed. Further we construct a RH problem
associated with the Cauchy problem (1.1)-(1.2). In Section 3, we deal with the
RH problem in the region $\xi<0$ with two stationary points. With a series of
transformations, the original RH problem is deformed into a mixed
$\bar{\partial}$-RH problem that can be decomposed into a pure RH problem and
a $\bar{\partial}$-problem, which are asymptotically analyzed in Subsection
3.3 and Subsection 3.4. In this part, to deal with the spectral singularity,
we construct a new function with special construction near $k=0$ in
Proposition 3. Thus, we obtain the long-time asymptotic result for the
solution $u(x,t)$ of the HS equation (1.1) via the reconstruction formula.
Finally, in Section 4, we show the long-time asymptotics for $u(x,t)$ in the
region $\xi>0$ in the similar way.

##  2 The IST under lower regular initial data

In this section, based on the matrix Lax pair for the HS equation, we
establish a scattering map between the initial data $u_{0}\in
H^{3,4}(\mathbb{R})$ and the reflection coefficient $r\in
H^{4}(\mathbb{R})\cap H^{1,1}(\mathbb{R})$. Further a RH problem associated
with the Cauchy problem (1.1)-(1.2) is constructed.

###  2.1 The Lax pair and Jost solutions

The HS equation (1.1) admits following Lax pair [33]

$\displaystyle\tilde{\Phi}_{x}+\mathrm{i}k\sqrt{m+1}\sigma_{3}\tilde{\Phi}=U\tilde{\Phi},\
$ (2.1)
$\displaystyle\tilde{\Phi}_{t}-\mathrm{i}k\left\\{\frac{1}{2k^{2}}+u\sqrt{m+1}\right\\}\sigma_{3}\tilde{\Phi}=V\tilde{\Phi},$
(2.2)

where $\tilde{\Phi}\equiv\tilde{\Phi}(x,t,k)$ is a matrix function,
$m:=-u_{xx}$, and

$\displaystyle U=\frac{m_{x}}{4(m+1)}\sigma_{1},$ $\displaystyle
V=-\frac{um_{x}}{4(m+1)}\sigma_{1}-\frac{1}{4\mathrm{i}k}\left[\sqrt{m+1}+\frac{1}{\sqrt{m+1}}-2\right]\sigma_{3}.$

Here $\sigma_{1},\sigma_{2},\sigma_{3}$ are the Pauli matrices

$\sigma_{1}=\left(\begin{array}[]{ll}0&1\\\
1&0\end{array}\right),\quad\sigma_{2}=\left(\begin{array}[]{cc}0&-\mathrm{i}\\\
\mathrm{i}&0\end{array}\right),\quad\sigma_{3}=\left(\begin{array}[]{cc}1&0\\\
0&-1\end{array}\right).$

Considering the conservation law of the HS equation, we introduce

$p(x,t,k)=x-\int_{x}^{\infty}(\sqrt{m(s,t)+1}-1)\mathrm{d}s-\frac{t}{2k^{2}}$
(2.3)

with

$p_{x}=\sqrt{m+1},\quad p_{t}=-\frac{1}{2k^{2}}-u\sqrt{m+1}.$

Then the function $\tilde{\Phi}(x,t,k)$ satisfies the following asymptotics

$\tilde{\Phi}(x,t,k)\sim e^{-ikp(x,t,k)\sigma_{3}},\quad x\to\pm\infty.$

Define a new matrix valued function

$\Phi\equiv\Phi(x,t,k)=\tilde{\Phi}e^{ikp(x,t,k)\sigma_{3}}$ (2.4)

satisfying the boundary condition

$\Phi\sim I,\quad x\to\pm\infty.$ (2.5)

Following from (2.1)-(2.2), $\Phi$ admits the new Lax pair

$\displaystyle\Phi_{x}+ikp_{x}[\sigma_{3},\Phi]=U\Phi,$ (2.6)
$\displaystyle\Phi_{t}+ikp_{t}[\sigma_{3},\Phi]=V\Phi,$ (2.7)

which allows $\Phi$ to be written as the total differential

$\mathrm{d}(e^{ikp(x,t,k)\hat{\sigma}_{3}}\Phi)=e^{ikp(x,t,k)\hat{\sigma}_{3}}[(U\mathrm{d}x+V\mathrm{d}t)\Phi].$
(2.8)

For any fixed $t>0$, integrating (2.8) from $\pm\infty$ to $x$, we get

$\Phi_{\pm}=I+\int^{x}_{\pm\infty}e^{-ik\int_{y}^{x}\sqrt{m(\tilde{x},t)+1}\mathrm{d}\tilde{x}\hat{\sigma}_{3}}(U\Phi_{\pm})\mathrm{d}y.$
(2.9)

On the other hand, the HS equation (1.1) also admits another matrix Lax pair

$\displaystyle(\tilde{\Phi}_{0})_{x}+ik\sigma\tilde{\Phi}_{0}=U_{0}\tilde{\Phi}_{0},$
(2.10)
$\displaystyle(\tilde{\Phi}_{0})_{t}+\frac{1}{2ik}\sigma_{3}\tilde{\Phi}_{0}=V_{0}\tilde{\Phi}_{0},$
(2.11)

where $\tilde{\Phi}_{0}\equiv\tilde{\Phi}_{0}(x,t,k)$ is a matrix function and

$\displaystyle U_{0}=-\frac{ik}{2}m(i\sigma_{2}+\sigma_{3}),$ $\displaystyle
V_{0}=\frac{u_{x}}{2}\sigma_{1}+iku[\sigma_{3}+\frac{m}{2}(i\sigma_{2}+\sigma_{3})].$

When $x\to\pm\infty$, $\tilde{\Phi}_{0}\sim
e^{(-ikx-\frac{1}{2ik})\sigma_{3}}$. Take the transformation on
$\tilde{\Phi}_{0}$ with

$\Phi_{0}\equiv\Phi_{0}(x,t,k)=\tilde{\Phi}_{0}e^{(ikx+\frac{t}{2ik})\sigma_{3}},$
(2.12)

then $\Phi_{0}\sim I,\ x\to\pm\infty$. And $\Phi_{0}$ admits the following Lax
pair

$\displaystyle(\Phi_{0})_{x}+ik[\sigma_{3},\Phi_{0}]=U_{0}\Phi_{0},$ (2.13)
$\displaystyle(\Phi_{0})_{t}+\frac{1}{2ik}[\sigma_{3},\Phi_{0}]=V_{0}\Phi_{0}.$
(2.14)

$\Phi_{0}$ admits the following asymptotic result at $k=0$:

$\Phi_{0}=I+\frac{ik}{2}u_{x}(i\sigma_{2}+\sigma_{3})-(ik)^{2}u\sigma_{1}+O(k^{3}),\quad
k\to 0.$ (2.15)

Considering the linear dependence of $\tilde{\Phi}$ and $\tilde{\Phi}_{0}$
[33], together with equations (2.4) and (2.12), we get the following relation:

$\Phi_{\pm}=Q(x,t)\Phi_{0\pm}e^{(-ikx-\frac{t}{2ik})\sigma_{3}}C_{\pm}(k)e^{ikp(x,t,k)\sigma_{3}},$
(2.16)

where

$Q(x,t)=\frac{1}{2}\left(\begin{array}[]{cc}q+q^{-1}&q-q^{-1}\\\
q-q^{-1}&q+q^{-1}\end{array}\right),\quad q=q(x,t)=(m+1)^{\frac{1}{4}},$

and

$C_{+}(k)\equiv I,\quad C_{-}(k)=e^{ikc\sigma_{3}},\quad
c=\int_{-\infty}^{+\infty}(\sqrt{m+1}-1)\mathrm{d}x.$ (2.17)

Denote $\Phi_{\pm}$ as

$\Phi_{\pm}=(\Phi_{\pm}^{(1)}\quad\Phi_{\pm}^{(2)}),$

where $\Phi_{\pm}^{(j)},\ j=1,2$ are columns. By the linear dependence of
$\tilde{\Phi}_{\pm}$, we have

$\Phi_{+}=\Phi_{-}e^{-ikp(x,t,k)\hat{\sigma}_{3}}S(k),$ (2.18)

where

$S(k)=\left(\begin{array}[]{cc}s_{11}(k)&s_{12}(k)\\\
s_{21}(k)&s_{22}(k)\end{array}\right)$

is independent with $x$ and $t$.

Considering the structure of the Lax pair (2.6)-(2.7), $\Phi$ has the
following symmetry:

$\sigma_{1}\overline{\Phi(x,t,\bar{k})}\sigma_{1}=\Phi(x,t,k),\quad\Phi(x,t,-k)=\overline{\Phi(x,t,k)},$
(2.19)

thus

$s_{11}(k)=\overline{s_{22}(\bar{k})},\quad
s_{21}(k)=\overline{s_{12}(\bar{k})}.$

Therefore, we rewrite $S(k)$ as

$S(k)=\left(\begin{array}[]{cc}\overline{a(\bar{k})}&b(k)\\\
\overline{b(\bar{k})}&a(k)\end{array}\right),$ (2.20)

and by (2.18), $a(k)$ and $b(k)$ are represented as

$a(k)=\mid\Phi_{-}^{(1)}\ \Phi_{+}^{(2)}\mid,\quad
b(k)=e^{2ikp(x,t,k)}\mid\Phi_{+}^{(2)}\ \Phi_{-}^{(2)}\mid.$ (2.21)

Based on the asymptotic formula (2.5), we find $|S(k)|\equiv 1$ and for
$k\in\mathbb{R}$,

$|a(k)|^{2}=1+|b(k)|^{2}\geqslant 1.$

Combining (2.15), (2.16) and (2.21), we deduce the expansion of $a(k)$ at
$k=0$:

$a(k)=1+ikc+(ik)^{2}\frac{c^{2}}{2}+O(k^{3}),$ (2.22)

where $c$ is given by (2.17). Besides, $a(k)$ has no zero in $\mathbb{C}$,
which is found from the structure of relating Dirac equation [33].

Define the reflection coefficient $r(k)$ as

$r(k)=-\frac{\bar{b}(k)}{a(k)},\quad k\in\mathbb{R},$ (2.23)

then

$|r(k)|^{2}=1-|a(k)|^{-2}<1,$

and it has been proved that for $u_{0}(x)\in H^{3,4}(\mathbb{R})$,
$1-|r(k)|^{2}>c_{0}>0$, where $c_{0}$ is a constant [33]. Based on the
symmetry (2.19), we can also derive that $|r(k)|=|r(-k)|$.

With the method of iteration and the Neumann series, we proved that
$\Phi_{-}^{(1)},\ \Phi_{+}^{(2)},\ a(k)$ are analytic in the half plane
$\mathbb{C}^{+}:=\\{k\in\mathbb{C}|\mathrm{Im}k>0\\}$; $\Phi_{-}^{(2)},\
\Phi_{+}^{(1)},\ \overline{a(\bar{k})}$ are analytic in
$\mathbb{C}^{-}:=\\{k\in\mathbb{C}|\mathrm{Im}k<0\\}$.

###  2.2 Scattering map from $u_{0}(x)$ to $r(k)$

In this part, we discuss the relation between the initial data $u_{0}$ and the
reflection coefficient $r(k)$. Now we give the main conclusion.

######  Proposition 1.

Suppose the initial data $u_{0}(x)\in H^{3,4}(\mathbb{R})$ satisfying
$-u_{0xx}+1\geq\epsilon_{0}>0$, then the map $u_{0}\to r(k)$ is Lipschitz
continuous from $H^{3,4}(\mathbb{R})$ to $H^{4}(\mathbb{R})\cap
H^{1,1}(\mathbb{R})$.

Denote $\Phi_{\pm}(x,k)=\left(\phi_{jl}^{\pm}(x,k)\right)$ as the solutions of
(2.9) for $t=0$. Considering (2.19), (2.21) as well as $a(k),\ b(k)$ are
independent with $x$ and $t$, taking $x=t=0$, we have

$\displaystyle
a(k)=\phi_{11}^{-}(0,k)\overline{\phi_{11}^{+}(0,k)}-\phi_{21}^{-}(0,k)\overline{\phi_{21}^{+}(0,k)},$
$\displaystyle
e^{2ikc_{0}}\overline{b(k)}=\phi_{11}^{-}(0,k)\phi^{+}_{21}(0,k)-\phi_{21}^{-}(0,k)\phi^{+}_{11}(0,k),$

where $c_{0}=\int_{0}^{\infty}(\sqrt{m+1}-1)\mathrm{d}x$ is real, and thus
$\|\overline{b(k)}\|_{L^{2}(\mathbb{R})}=\|e^{2ikc_{0}}\overline{b(k)}\|_{L^{2}(\mathbb{R})}$.

Define
$\textbf{n}^{\pm}(x,k)=(n^{\pm}_{11}(x,k),n^{\pm}_{21}(x,k))^{T}=(\phi_{11}^{\pm}(x,k)-1,\phi_{21}^{\pm}(x,k))^{T}$,
then $a(k),\ b(k)$ are denoted as

$\displaystyle
a(k)-1=n_{11}^{-}(0,k)\overline{n_{11}^{+}(0,k)}-n_{21}^{-}(0,k)\overline{n_{21}^{+}(0,k)}+n_{11}^{-}(0,k)+\overline{n_{11}^{+}(0,k)},\qquad\quad$
(2.24) $\displaystyle
e^{2ikc_{0}}\overline{b(k)}=n_{11}^{-}(0,k)n^{+}_{21}(0,k)-n_{21}^{-}(0,k)n^{+}_{11}(0,k)+n^{+}_{21}(0,k)-n_{21}^{-}(0,k).$
(2.25)

Therefore, to testify that $r(k)\in H^{4}(\mathbb{R})$, we only have to prove
$\textbf{n}^{\pm}(0,k)\in H^{4}(\mathbb{R})$, that is

######  Proposition 2.

The maps

$u_{0}(x)\to n_{11}^{\pm}(0,k),\quad u_{0}(x)\to n_{21}^{\pm}(0,k)$

are Lipschitz continuous from $H^{3,4}(\mathbb{R})$ to $H^{4}(\mathbb{R})$.

The proof of this Proposition will be detailed later. Now we prove Proposition
1 based on the results in Proposition 2.

###### Proof.

(_Proposition 1_)

As $\textbf{n}^{\pm}(0,k)\in H^{4}(\mathbb{R})$, by (2.24) and (2.25), it’s
obvious that $a(k)$ is bounded and $a^{\prime}(k),\ a^{\prime\prime}(k),$
$a^{\prime\prime\prime}(k),\ a^{\prime\prime\prime\prime}(k)\in
L^{2}(\mathbb{R}),\ b(k)\in H^{4}(\mathbb{R})$. Thus $r(k)\in
H^{4}(\mathbb{R})$.

We next prove $r(k)\in H^{1,1}(\mathbb{R})$, which equals to prove that
$k\overline{b(k)},\ k\overline{b^{\prime}(k)}\in L^{2}(\mathbb{R})$. Based on
(2.9), we find

$\displaystyle kn^{\pm}_{21}(0,k)$
$\displaystyle=k\int_{\pm\infty}^{0}e^{2i\int_{x}^{0}\sqrt{m+1}\mathrm{d}y}\frac{m_{x}}{4\sqrt{m+1}}\mathrm{d}x+k\int_{\pm\infty}^{0}e^{2ik\int_{x}^{0}\sqrt{m+1}\mathrm{d}y}\frac{m_{x}}{4\sqrt{m+1}}n_{11}^{\pm}(x,k)\mathrm{d}x$
$\displaystyle=-\int_{\pm\infty}^{0}\frac{1}{8i}\frac{m_{x}}{m+1}\mathrm{d}e^{2ik\int_{x}^{0}\sqrt{m+1}\mathrm{d}y}-\int_{\pm\infty}^{0}\frac{1}{8i}\frac{m_{x}}{m+1}n_{11}^{\pm}(x,k)\mathrm{d}e^{2ik\int_{x}^{0}\sqrt{m+1}\mathrm{d}y}$
$\displaystyle=-\frac{1}{8i}\frac{m_{x}(0)}{m(0)+1}+I_{1}^{\pm}+I_{2}^{\pm},$

where

$\displaystyle
I_{1}^{\pm}=-\frac{1}{8i}\frac{m_{x}(0)}{m(0)+1}n_{11}^{\pm}(0,k),$
$\displaystyle
I_{2}^{\pm}=\int_{\pm\infty}^{0}\frac{1}{8i}[\frac{m_{x}}{m+1}(1+n_{11}^{\pm}(x,k))]_{x}e^{2ik\int_{x}^{0}\sqrt{m+1}\mathrm{d}y}\mathrm{d}x$

belong to $L^{2}(\mathbb{R})$. Therefore, by (2.25), we have

$\displaystyle e^{2ikc_{0}}k\overline{b(k)}=$
$\displaystyle-\frac{1}{8i}\frac{m_{x}(0)}{m(0)+1}(n^{-}_{11}(0,k)-n^{+}_{11}(0,k))+(I_{1}^{+}+I_{2}^{+})n^{-}_{11}(0,k)$
$\displaystyle-(I_{1}^{-}+I_{2}^{-})n^{+}_{11}(0,k)+(I_{1}^{+}+I_{2}^{+})-(I_{1}^{-}+I_{2}^{-}).$

Thus we conclude that $k\overline{b(k)}\in L^{2}(\mathbb{R})$, and the proof
of $k\overline{b^{\prime}(k)}\in L^{2}(\mathbb{R})$ is similar. ∎

Therefore, we only have to prove Proposition 2, which needs some preparation
work with several lemmas. Take $\textbf{n}^{+}(x,k)$ for example and to
convenient, we replace it by $\textbf{n}(x,k)$. By (2.9), we have

$\textbf{n}(x,k)=\textbf{n}_{0}(x,k)+T\textbf{n}(x,k),$ (2.26)

where $T$ is an integral operator defined by

$T\textbf{f}(x,k)=\int_{x}^{+\infty}K(x,y,k)\textbf{f}(y,k)\mathrm{d}y,$
(2.27)

with the kernel

$K(x,y,k)=\left(\begin{array}[]{cc}0&-\frac{m_{y}}{4(m+1)}\\\
-\frac{m_{y}}{4(m+1)}e^{2ik(h(y)-h(x))}&0\end{array}\right),$ (2.28)

and

$\textbf{n}_{0}(x,k)=T\textbf{e}_{1}=\left(\begin{array}[]{c}0\\\
-\int_{x}^{+\infty}\frac{m_{y}}{4(m+1)}e^{2ik(h(y)-h(x))}\mathrm{d}y\end{array}\right).$
(2.29)

Here the function $h(x)$ is defined as

$h(x)=\int_{x}^{\infty}\sqrt{m+1}\mathrm{d}\zeta,$

and thus

$h(x)-h(y)=\int_{x}^{y}\sqrt{m+1}\mathrm{d}\zeta.$

Taking the partial derivatives of $k$ for (2.26), we get

$\displaystyle(\textbf{n})_{k}=\textbf{n}_{1}+T(\textbf{n})_{k},\quad\textbf{n}_{1}=(\textbf{n}_{0})_{k}+(T)_{k}\textbf{n},$
(2.30)
$\displaystyle(\textbf{n})_{kk}=\textbf{n}_{2}+T(\textbf{n})_{kk},\quad\textbf{n}_{2}=(\textbf{n}_{0})_{kk}+(T)_{kk}\textbf{n}+2(T)_{k}(\textbf{n})_{k},$
(2.31) $\displaystyle(\textbf{n})_{kkk}=\textbf{n}_{3}+T(\textbf{n})_{kkk},\
\textbf{n}_{3}=(\textbf{n}_{0})_{kkk}+(T)_{kkk}\textbf{n}+3(T)_{kk}(\textbf{n})_{k}+3(T)_{k}(\textbf{n})_{kk},\qquad$
(2.32) $\displaystyle(\textbf{n})_{kkkk}=\textbf{n}_{4}+T(\textbf{n})_{kkkk},\
\textbf{n}_{4}=(\textbf{n}_{0})_{kkkk}+(T)_{kkkk}\textbf{n}+4(T)_{kkk}(\textbf{n})_{k}+6(T)_{kk}(\textbf{n})_{kk}+4(T)_{k}(\textbf{n})_{kkk}.\quad\qquad$
(2.33)

To find the solutions of the differential equations (2.26), (2.30), (2.31),
(2.32) and (2.33), we need several lemmas as follows:

######  Lemma 1.

For $u_{0}\in H^{3,4}(\mathbb{R})$, the following estimates hold:

$\|\textbf{n}_{0}\|_{C^{0}(\mathbb{R}^{+},L^{2}(\mathbb{R}))}\lesssim\|m_{x}\|_{L^{2}},\quad\|\textbf{n}_{0}\|_{L^{2}(\mathbb{R}^{+}\times\mathbb{R})}\lesssim\|m_{x}\|_{L^{2,\frac{1}{2}}};$
(2.34)
$\displaystyle\|(\textbf{n}_{0})_{k}\|_{C^{0}(\mathbb{R}^{+},L^{2}(\mathbb{R}))}\lesssim\|m_{x}\|_{L^{2,1}}+\|m\|_{L^{1}}\|m_{x}\|_{L^{2}},$
(2.35)
$\displaystyle\|(\textbf{n}_{0})_{k}\|_{L^{2}(\mathbb{R}^{+}\times\mathbb{R})}\lesssim\|m_{x}\|_{L^{2,\frac{3}{2}}}+\|m\|_{L^{1}}\|m_{x}\|_{L^{2,\frac{1}{2}}};$
$\displaystyle\|(\textbf{n}_{0})_{kk}\|_{C^{0}(\mathbb{R}^{+},L^{2}(\mathbb{R}))}\lesssim\|m_{x}\|_{L^{2,2}}+\|m\|_{L^{1}}\|m_{x}\|_{L^{2,1}}+\|m\|^{2}_{L^{1}}\|m_{x}\|_{L^{2}},$
(2.36)
$\displaystyle\|(\textbf{n}_{0})_{kk}\|_{L^{2}(\mathbb{R}^{+}\times\mathbb{R})}\lesssim\|m_{x}\|_{L^{2,\frac{5}{2}}}+\|m\|_{L^{1}}\|m_{x}\|_{L^{2,\frac{3}{2}}}+\|m\|^{2}_{L^{1}}\|m_{x}\|_{L^{2,\frac{1}{2}}};$
$\displaystyle\|(\textbf{n}_{0})_{kkk}\|_{C^{0}(\mathbb{R}^{+},L^{2}(\mathbb{R}))}\lesssim\|m_{x}\|_{L^{2,3}}+\|m\|_{L^{1}}\|m_{x}\|_{L^{2,2}}+\|m\|^{2}_{L^{1}}\|m_{x}\|_{L^{2,1}}+\|m\|^{3}_{L^{1}}\|m_{x}\|_{L^{2}},$
(2.37)
$\displaystyle\|(\textbf{n}_{0})_{kkk}\|_{L^{2}(\mathbb{R}^{+}\times\mathbb{R})}\lesssim\|m_{x}\|_{L^{2,\frac{7}{2}}}+\|m\|_{L^{1}}\|m_{x}\|_{L^{2,\frac{5}{2}}}+\|m\|^{2}_{L^{1}}\|m_{x}\|_{L^{2,\frac{3}{2}}}+\|m\|^{3}_{L^{1}}\|m_{x}\|_{L^{2,\frac{1}{2}}};$
$\displaystyle\|(\textbf{n}_{0})_{kkkk}\|_{C^{0}(\mathbb{R}^{+},L^{2}(\mathbb{R}))}$
$\displaystyle\lesssim\|m_{x}\|_{L^{2,4}}+\|m\|_{L^{1}}\|m_{x}\|_{L^{2,3}}+\|m\|^{2}_{L^{1}}\|m_{x}\|_{L^{2,2}}+\|m\|^{3}_{L^{1}}\|m_{x}\|_{L^{2,1}}$
(2.38) $\displaystyle+\|m\|^{4}_{L^{1}}\|m_{x}\|_{L^{2}}.$

###### Proof.

We take the proof of (2.35) for example, and the rest are deduced similarly.

Take the derivative with respect to $\textbf{n}_{0}(x,k)$ for $k$, we get

$(\textbf{n}_{0})_{k}(x,k)=\left(\begin{array}[]{c}0\\\
-2i(h(y)-h(x))\int_{x}^{+\infty}\frac{m_{y}}{4(m+1)}e^{2ik(h(y)-h(x))}\mathrm{d}y\end{array}\right).$

Considering that for $y>x$,

$h(x)-h(y)=\int_{x}^{y}\sqrt{m+1}\mathrm{d}\zeta\lesssim(y-x)+\|m\|_{L^{1}},$

we deduce that for any function $\varphi(k)\in L^{2}(\mathbb{R})$ satisfying
$\|\varphi\|_{L^{2}}=1$,

$\displaystyle\|(\textbf{n}_{0})_{k}\|_{L^{2}(\mathbb{R})}=\sup_{\varphi\in
L^{2}(\mathbb{R})}\int_{0}^{\infty}2i(h(x)-h(y))\int_{x}^{+\infty}\frac{m_{y}}{4(m+1)}e^{2ik(h(y)-h(x))}\varphi(k)\mathrm{d}y\mathrm{d}k$
$\displaystyle\lesssim$ $\displaystyle\sup_{\varphi\in
L^{2}(\mathbb{R})}\left(\int_{x}^{+\infty}\frac{(y-x)m_{y}}{4(m+1)}\widehat{\varphi}(h(x)-h(y))\mathrm{d}y+\|m\|_{L^{1}}\int_{x}^{+\infty}\frac{m_{y}}{4(m+1)}\widehat{\varphi}(h(x)-h(y))\mathrm{d}y\right)$
$\displaystyle\lesssim$
$\displaystyle\left(\int_{x}^{+\infty}|ym_{y}|^{2}\mathrm{d}y\right)^{1/2}+\|m\|_{L^{1}}\left(\int_{x}^{+\infty}|m_{y}|^{2}\mathrm{d}y\right)^{1/2}.$

Therefore,

$\|(\textbf{n}_{0})_{k}\|_{C^{0}(\mathbb{R}^{+},L^{2}(\mathbb{R}))}=\sup_{x\geqslant
0}\|(\textbf{n}_{0})_{k}\|_{L^{2}(\mathbb{R})}\lesssim\|m_{x}\|_{L^{2,1}}+\|m\|_{L^{1}}\|m_{x}\|_{L^{2}},$

and

$\displaystyle\|(\textbf{n}_{0})_{k}\|_{L^{2}(\mathbb{R}^{+}\times\mathbb{R})}$
$\displaystyle\lesssim\left(\int_{0}^{+\infty}\int_{x}^{+\infty}|ym_{y}|^{2}\mathrm{d}y\mathrm{d}x+\|m\|_{L^{1}}\int_{0}^{+\infty}\int_{x}^{+\infty}|m_{y}|^{2}\mathrm{d}y\mathrm{d}x\right)^{1/2}$
$\displaystyle\lesssim\left(\int_{0}^{+\infty}\int_{0}^{y}|ym_{y}|^{2}\mathrm{d}x\mathrm{d}y\right)^{1/2}+\|m\|_{L^{1}}\left(\int_{0}^{+\infty}\int_{0}^{y}|m_{y}|^{2}\mathrm{d}x\mathrm{d}y\right)^{1/2}$
$\displaystyle\lesssim\|m_{x}\|_{L^{2,\frac{3}{2}}}+\|m\|_{L^{1}}\|m_{x}\|_{L^{2,\frac{1}{2}}}.$

∎

Next, we deal with the operators $(T)_{k},\ (T)_{kk}$ and $(T)_{kkk}$, which
have the integral kernel $(K)_{k},\ (K)_{kk}$ and $(K)_{kkk}$ separately,
where

$(K)_{k}(x,y,k)=\left(\begin{array}[]{cc}0&0\\\
2i(h(x)-h(y))\frac{m_{y}}{4(m+1)}e^{2ik(h(y)-h(x))}&0\end{array}\right).$
(2.39)

$(K)_{kk}$, $(K)_{kkk}$ and $(K)_{kkkk}$ have the totally same form with
$2i(h(x)-h(y))$ replaced by $(2i(h(x)-h(y)))^{2}$, $(2i(h(x)-h(y)))^{3}$ and
$(2i(h(x)-h(y)))^{4}$. These operators admit following estimates:

######  Lemma 2.

For $u_{0}\in H^{3,4}(\mathbb{R})$, the following operator bounds hold
uniformly, and the operators are Lipschitz continuous of $u_{0}(x)$.

$\displaystyle\|(T)_{k}\|_{L^{2}(\mathbb{R}^{+}\times\mathbb{R})\to
C^{0}(\mathbb{R}^{+},L^{2}(\mathbb{R}))}\lesssim\|m_{x}\|_{L^{2,1}}+\|m\|_{L^{1}}\|m_{x}\|_{L^{2}},$
$\displaystyle\|(T)_{k}\|_{L^{2}(\mathbb{R}^{+}\times\mathbb{R})\to
L^{2}(\mathbb{R}^{+}\times\mathbb{R})}\lesssim\|m_{x}\|_{L^{2,\frac{3}{2}}}+\|m\|_{L^{1}}\|m\|_{L^{2,\frac{1}{2}}};$
$\displaystyle\|(T)_{kk}\|_{L^{2}(\mathbb{R}^{+}\times\mathbb{R})\to
C^{0}(\mathbb{R}^{+},L^{2}(\mathbb{R}))}\lesssim\|m_{x}\|_{L^{2,2}}+\|m\|_{L^{1}}\|m_{x}\|_{L^{2,1}}+\|m\|^{2}_{L^{1}}\|m_{x}\|_{L^{2}},$
$\displaystyle\|(T)_{kk}\|_{L^{2}(\mathbb{R}^{+}\times\mathbb{R})\to
L^{2}(\mathbb{R}^{+}\times\mathbb{R})}\lesssim\|m_{x}\|_{L^{2,\frac{5}{2}}}+\|m\|_{L^{1}}\|m\|_{L^{2,\frac{3}{2}}}+\|m\|^{2}_{L^{1}}\|m_{x}\|_{L^{2,\frac{1}{2}}};$
$\displaystyle\|(T)_{kkk}\|_{L^{2}(\mathbb{R}^{+}\times\mathbb{R})\to
C^{0}(\mathbb{R}^{+},L^{2}(\mathbb{R}))}\lesssim\|m_{x}\|_{L^{2,3}}+\|m\|_{L^{1}}\|m_{x}\|_{L^{2,2}}+\|m\|^{2}_{L^{1}}\|m_{x}\|_{L^{2,1}}+\|m\|^{3}_{L^{1}}\|m_{x}\|_{L^{2}},$
$\displaystyle\|(T)_{kkk}\|_{L^{2}(\mathbb{R}^{+}\times\mathbb{R})\to
L^{2}(\mathbb{R}^{+}\times\mathbb{R})}\lesssim\|m_{x}\|_{L^{2,\frac{7}{2}}}+\|m\|_{L^{1}}\|m\|_{L^{2,\frac{5}{2}}}+\|m\|^{2}_{L^{1}}\|m_{x}\|_{L^{2,\frac{3}{2}}}+\|m\|^{3}_{L^{1}}\|m_{x}\|_{L^{2,\frac{1}{2}}};$
$\displaystyle\|(T)_{kkkk}\|_{L^{2}(\mathbb{R}^{+}\times\mathbb{R})\to
C^{0}(\mathbb{R}^{+},L^{2}(\mathbb{R}))}\lesssim\|m_{x}\|_{L^{2,4}}+\|m\|_{L^{1}}\|m_{x}\|_{L^{2,3}}+\|m\|^{2}_{L^{1}}\|m_{x}\|_{L^{2,2}}$
$\displaystyle+\|m\|^{3}_{L^{1}}\|m_{x}\|_{L^{2,1}}+\|m\|^{4}_{L^{1}}\|m_{x}\|_{L^{2}}.$

The proof of this lemma is the same as Lemma 1.

To solve the equations (2.26), (2.30), (2.31), (2.32) and (2.33), we finally
discuss the existence of the operator $(I-T)^{-1}$. Denote
$f^{*}(x)=\sup_{y\geqslant x}\|f(y,\cdot)\|_{L^{2}(\mathbb{R})}$, then by
(2.28), we find $K(x,y,k)\leqslant g(y)$ and

$(Tf)^{*}(x)\leqslant\int_{x}^{\infty}g(y)f^{*}(y)\mathrm{d}y,$ (2.40)

where

$g(y)=\frac{m_{y}}{2(m+1)}.$

Therefore, the resolvent $(I-T)^{-1}$ exists with following lemma:

######  Lemma 3.

For each $k\in\mathbb{R}$ and $u_{0}\in H^{3,4}(\mathbb{R})$, $(I-T)^{-1}$
exists as a bounded operator from $C^{0}(\mathbb{R}^{+})$ to itself. What’s
more, $\hat{L}:=(I-T)^{-1}-I$ is an integral operator with continuous integral
kernel $L(x,y,k)$ satisfying

$|L(x,y,k)|\leqslant\exp(\|g\|_{L^{1}})g(y).$ (2.41)

###### Proof.

By (2.26), it’s obvious that $T$ is a Volterra operator, and together with
(2.40), we can deduce that $(I-T)^{-1}$ exists unique as a bounded operator on
$C^{0}(\mathbb{R}^{+})$. For the operator $\hat{L}$, the integral kernel
$L(x,y,k)$ is given by

$L(x,y,k)=\left\\{\begin{array}[]{ll}\sum_{n=1}^{\infty}K_{n}(x,y,k),&x\leqslant
y,\\\ 0,&x>y,\end{array}\right.$

where

$K_{n}(x,y,k)=\int_{x\leqslant y_{1}\leqslant\cdots\leqslant
y_{n-1}}K_{(}x,y_{1},k)K(y_{1},y_{2},k)\cdots
K(y_{n-1},y,k)\mathrm{d}y_{n-1}\cdots\mathrm{d}y_{1}.$

By the estimate $|K(x,y,k)|\leqslant g(y)$, we get

$|K_{n}(x,y,k)|\leqslant\frac{1}{(n-1)!}\left(\int_{x}^{\infty}g(t)\mathrm{d}t\right)^{n-1}g(y),$

and then (2.41) follows. ∎

By (2.40), we find that $T$ is a bounded operator as $T:L^{2}\to C^{0},\
T:C^{0}\to L^{2},$ and $T:L^{2}\to T^{2}$. Therefore, by the formula

$\hat{L}=(I-T)^{-1}-I=T+T(I-T)^{-1}T,$

we deduce that $\hat{L}$ is a bounded operator as
$\hat{L}:C^{0}(\mathbb{R}^{+},L^{2}(\mathbb{R}))\to
C^{0}(\mathbb{R}^{+},L^{2}(\mathbb{R}))$ and
$\hat{L}:L^{2}(\mathbb{R}^{+}\times\mathbb{R})\to
L^{2}(\mathbb{R}^{+}\times\mathbb{R})$.

Based on above results, we now turn back to give the proof of Proposition 2.

###### Proof.

(_Proposition 2_)

By (2.26), we find

$\textbf{n}(x,k)=((I-T)^{-1}-I)\textbf{n}_{0}(x,k)+\textbf{n}_{0}(x,k).$
(2.42)

By (2.34) in Lemma 1, $\textbf{n}_{0}(x,k)\in
C^{0}(\mathbb{R}^{+},L^{2}(\mathbb{R}))\cap
L^{2}(\mathbb{R}^{+}\times\mathbb{R})$, and then Lemma 3 guarantees that there
exists unique solution $\textbf{n}(x,k)$ of (2.42) with $\textbf{n}(x,k)\in
C^{0}(\mathbb{R}^{+},L^{2}(\mathbb{R}))\cap
L^{2}(\mathbb{R}^{+}\times\mathbb{R})$. Similarly, together with Lemma 2, we
have

$\displaystyle\textbf{n}_{k}(x,k)\in
C^{0}(\mathbb{R}^{+},L^{2}(\mathbb{R}))\cap
L^{2}(\mathbb{R}^{+}\times\mathbb{R}),\quad\textbf{n}_{kk}(x,k)\in
C^{0}(\mathbb{R}^{+},L^{2}(\mathbb{R}))\cap
L^{2}(\mathbb{R}^{+}\times\mathbb{R}),$ $\displaystyle\textbf{n}_{kkk}(x,k)\in
C^{0}(\mathbb{R}^{+},L^{2}(\mathbb{R}))\cap
L^{2}(\mathbb{R}^{+}\times\mathbb{R}),\quad\textbf{n}_{kkkk}(x,k)\in
C^{0}(\mathbb{R}^{+},L^{2}(\mathbb{R})).$

Taking $x=0$ in all above, we get $\textbf{n}(0,k)\in H^{4}(\mathbb{R})$. ∎

At the end of this part, we give a remark as a supplement of Proposition 1. It
plays an important role in solving the singularity at $k=0$ in following
sections.

######  Remark 1.

If $r(k)\in H^{4}(\mathbb{R})$, then $r(k)\in C^{3}(\mathbb{R})$ by the
Sobolev embedding theorem.

###  2.3 The Riemann-Hilbert problem of the HS equation

Define a new matrix-valued function $M(x,t,k)\equiv M(k)$ as follows:

$M(k)=\left\\{\begin{array}[]{lr}\left(\frac{\Phi_{-}^{(1)}}{a\left(k\right)}\quad\Phi_{+}^{(2)}\right),\quad\mathrm{Im}k>0,\\\
\left(\Phi_{+}^{(1)}\quad\frac{\Phi_{-}^{(2)}}{\overline{a(\bar{k})}}\right),\quad\mathrm{Im}k<0,\end{array}\right.$
(2.43)

which solves the following RH problem

######  RH Problem 1.

Find a $2\times 2$ matrix-valued function $M(k)$ satisfying

  * •

Analyticity: $M(k)$ is analytic in $\mathbb{C}\setminus\mathbb{R};$

  * •

Symmetry: $M(k)=\sigma_{1}\overline{M(\bar{k})}\sigma_{1}$,
$M(-k)=\overline{M(k)};$

  * •

Jump condition: $M(k)$ has continuous boundary values $M_{\pm}(k)$ on
$\mathbb{R}$ and

$M_{+}(k)=M_{-}(k)V(x,t,k),\quad k\in\mathbb{R},$ (2.44)

where

$V(x,t,k)=e^{-ikp(x,t,k)\hat{\sigma}_{3}}V_{0}(k),$ (2.45)

and

$V_{0}(k)=\left(\begin{array}[]{cc}1-|r(k)|^{2}&-\bar{r}(k)\\\
r(k)&1\end{array}\right);$ (2.46)

  * •

Asymptotic behaviors:

$M(k)=I+\mathcal{O}(k^{-1}),\quad k\to\infty.$ (2.47)

Denoting $M(k)=(M_{lj}(k)),\ M_{j}(k)=M_{1j}(k)+M_{2j}(k)$, from the expansion
of $M_{j}(k)$ at $k=0$:

$M_{j}(k)\triangleq M_{j}(0)+M_{j,1}k+M_{j,2}k^{2}+\mathcal{O}(k^{3}),\
j=1,2,$

we find

$\displaystyle M_{1}(k)M_{2}(k)=\sqrt{m+1}(1+2k^{2}u+\mathcal{O}(k^{3})).$
(2.48)

Introduce a new scale $(y,t)$ by

$y\equiv y(x,t)=x-\int_{x}^{\infty}(\sqrt{m(\zeta,t)+1}-1)\mathrm{d}\zeta,$
(2.49)

then we give a new matrix-valued function as

$\widehat{M}(k)\equiv\widehat{M}(y(x,t),t,k):=M(x,t,k).$ (2.50)

By the coordinate transformation (2.49), $\widehat{M}(k)$ obeys the jump
condition

$\widehat{M}_{+}(k)=\widehat{M}_{-}(k)\widehat{V}(y,t,k),$ (2.51)

where

$\quad\widehat{V}(y(x,t),t,k)=V(x,t,k)=e^{-it\theta(\xi,k)\hat{\sigma}_{3}}V_{0}(k),$
(2.52)

and

$\theta(\xi,k)=k\xi-\frac{1}{2k},\quad\xi=\frac{y}{t}.$ (2.53)

Combining (2.48) and (2.50), we reconstruct the formula of $u(x,t)$ as follows

$\displaystyle u(x,t)$ $\displaystyle=\hat{u}(y(x,t),t)=\lim_{k\rightarrow
0}\frac{1}{2k^{2}}\left(\frac{\widehat{M}_{1}(y,t,k)\widehat{M}_{2}(y,t,k)}{\widehat{M}_{1}(y,t,0)\widehat{M}_{2}(y,t,0)}-1\right)$
(2.54)
$\displaystyle=\widehat{M}_{1,1}\widehat{M}_{2,1}(y(x,t),t)+\widehat{M}_{1,2}(y(x,t),t)+\widehat{M}_{2,2}(y(x,t),t),$

where

$x(y,t)=y-\lim_{k\rightarrow
0}\frac{1}{2ik}\left(\frac{\widehat{M}_{1}(y,t,k)}{\widehat{M}_{1}(y,t,0)}-1\right),$
(2.55)

and

$\widehat{M}_{j,l}(y(x,t),t):=M_{j,l}(x,t),\ j,l=1,2.$

By (2.53),

$\mathrm{Im}\theta=\mathrm{Im}k(\xi+\frac{1}{2|k|^{2}}),$ (2.56)

and thus we have the following lemma from direct calculation:

######  Lemma 4.

For $\xi\neq 0$, $\mathrm{Im}\theta(k)>\xi\mathrm{Im}k,\ as\
\mathrm{Im}k>0;\quad\mathrm{Im}\theta(k)<\xi\mathrm{Im}k,\ as\
\mathrm{Im}k<0.$

Before the asymptotic analysis, we first divide the $(y,t)$ half-plane into
two regions as shown in Figure 1 :

$y$$t$$0$$\mathrm{Region}\ \mathrm{I}:\ \xi<0$$\mathrm{Region}\ \mathrm{II}:\
\xi>0$ Figure 1: The different regions of the $(y,t)$ half-plane,
$\xi=\frac{y}{t}$.

  1. 1.

Region I: For $\xi<0$, the sign of $\mathrm{Im}\theta$ is shown in Figure 2,
where $k_{j}$ are the two stationary points:

$k_{1}=-\sqrt{-\frac{1}{2\xi}},\quad k_{2}=\sqrt{-\frac{1}{2\xi}}.$ (2.57)

For convenience, we denote $\rho_{0}=|k_{j}|=\sqrt{-\frac{1}{2\xi}}$;

  2. 2.

Region II: For $\xi>0$, $\mathrm{Im}\theta$ and $\mathrm{Im}k$ have the same
sign, and there is no stationary point.

In following sections, we discuss the asymptotic behavior for the two regions
respectively.

##  3 Asymptotic analysis in the region $y/t<0$

In this section, we investigate the long-time asymptotic behavior in the
region $y/t<0$, which includes two stationary points.

###  3.1 Triangular decomposition

First of all, based on the decay regions as shown in Figure 2, we need to
decompose the jump matrix $\widehat{V}(k)$ into upper and lower triangular
matrix. We therefore introduce a scalar function $\delta(k)$ satisfying the RH
problem as follows.

######  RH Problem 2.

Find a scalar function $\delta(k)$ with the following properties:

  * •

Analyticity: $\delta(k)$ is analytical in $\mathbb{C}\setminus\mathbb{R};$

  * •

Jump condition: $\delta(k)$ has continuous boundary values $\delta_{\pm}$ and

$\left\\{\begin{array}[]{ll}\delta_{+}(k)=\delta_{-}(k)(1-|r|^{2}),&k\in\Gamma_{1},\\\
\delta_{+}(k)=\delta_{-}(k),&k\in(k_{1},k_{2}),\end{array}\right.$ (3.1)

where $\Gamma_{1}=(-\infty,k_{1})\cup(k_{2},\infty);$

  * •

Asymptotic behavior:

$\delta(k)\to 1,\quad k\to\infty.$ (3.2)

$k_{1}$$k_{2}$$\mathrm{Re}k$$\mathrm{Im}k$$0$$+$$+$$+$$-$$-$$-$ Figure 2: The
classification for sign of $\mathrm{Im}\theta$ in the case $\xi<0$. In the
shaded regions, $\mathrm{Im}\theta<0$, which implies $e^{-it\theta}\to 0$ as
$t\to\infty$; while in the white regions, $\mathrm{Im}\theta>0$ and thus
$e^{it\theta}\to 0$ as $t\to\infty.$

By the Plemelj formula, there exists the unique solution for RH problem 2 with

$\delta(k)=\mathrm{exp}\left[i\int_{\Gamma_{1}}\frac{\nu(s)}{s-k}\mathrm{d}s\right],$
(3.3)

where

$\nu(s)=-\frac{1}{2\pi}\log(1-|r(s)|^{2}).$ (3.4)

Therefore, $\delta(k)$ has the following asymptotic expansion at $k=0$:

$\delta(k)=1+\delta_{1}(ik)+\frac{\delta_{1}^{2}}{2}(ik)^{2}+\mathcal{O}(k^{3}),\quad
k\to 0,$ (3.5)

where

$\delta_{1}=\int_{\Gamma_{1}}\frac{\nu(s)}{s^{2}}\mathrm{d}s.$ (3.6)

Define

$\beta_{j}(k)=\int_{\Gamma_{1}}\frac{\nu(s)}{s-k}\mathrm{d}s-\log(k-k_{j})\nu(k_{j}),\
j=1,2,$ (3.7)

then $\delta(k)$ is rewritten as

$\delta(k)=(k-k_{j})^{i\nu(k_{j})}e^{i\beta_{j}(k)}.$ (3.8)

For $r(k)\in H^{1}(\mathbb{R})$,

$|\beta_{j}(k)-\beta_{j}(k_{j})|\lesssim\|r\|_{H^{1}(\mathbb{R})}|k-k_{j}|^{1/2}.$
(3.9)

Now we use $\delta(k)$ to define a new matrix function

$M^{(1)}(k)\equiv M^{(1)}(y,t,k)=\widehat{M}(k)\delta(k)^{-\sigma_{3}},$
(3.10)

which is a solution for the following Riemann-Hilbert problem.

######  RH Problem 3.

Find a $2\times 2$ matrix-valued function $M^{(1)}(k)$ with the following
properties:

  * •

Analyticity: $M^{(1)}(k)$ is analytical in $\mathbb{C}\setminus\mathbb{R};$

  * •

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

$M^{(1)}_{+}(k)=M^{(1)}_{-}(k)V^{(1)}(k),$ (3.11)

where

$V^{(1)}(k)=\left\\{\begin{array}[]{ll}\left(\begin{array}[]{cc}1&0\\\
\frac{r}{1-|r|^{2}}e^{2it\theta}\delta_{-}^{-2}&1\end{array}\right)\left(\begin{array}[]{cc}1&\frac{-\bar{r}}{1-|r|^{2}}e^{-2it\theta}\delta_{+}^{2}\\\
0&1\end{array}\right),&k\in\Gamma_{1},\\\
\left(\begin{array}[]{cc}1&-\bar{r}e^{-2it\theta}\delta_{-}^{2}\\\
0&1\end{array}\right)\left(\begin{array}[]{cc}1&0\\\
re^{2it\theta}\delta_{+}^{-2}&1\end{array}\right),&k\in\mathbb{R}\setminus\Gamma_{1};\end{array}\right.$
(3.12)

  * •

Asymptotic behaviors: $\ M^{(1)}(k)=I+\mathcal{O}(k^{-1}),\ as\ k\to\infty.$

###  3.2 A mixed $\bar{\partial}$-RH problem and its decomposition

Through the triangular decomposition, we make a continuous extension to deal
with the jump condition.

$\Sigma_{11}$$\Sigma_{12}$$\Sigma_{13}$$\Sigma_{14}$$\Sigma_{22}$$\Sigma_{21}$$\Sigma_{24}$$\Sigma_{23}$$k_{1}$$k_{2}$$\Omega_{11}$$\Omega_{12}$$\Omega_{13}$$\Omega_{14}$$\Omega_{22}$$\Omega_{21}$$\Omega_{24}$$\Omega_{23}$$\mathrm{Re}k$$\mathrm{Im}k$$\Sigma_{01}$$\Sigma_{02}$$\Sigma_{03}$$\Sigma_{04}$$0$$+$$+$$+$$-$$-$$-$
Figure 3: The jump lines $\Sigma_{jl}$ and regions $\Omega_{jl}$ for
$\xi<-\epsilon$.

Denote arcs $\Sigma_{0l}$, lines $\Sigma_{jl},$ and domains $\Omega_{jl},\
j=1,2,\ l=1,2,3,4$ as shown in Figure 3. Let

$\Sigma_{j}=\bigcup_{l=1}^{4}\Sigma_{jl},\
\Sigma=\bigcup_{j=0}^{2}\Sigma_{j},\ \Omega_{j}=\bigcup_{l=1}^{4}\Omega_{jl},\
\Omega=\bigcup_{j=1}^{2}\Omega_{j}.$

Now we introduce a new function

$R^{(2)}(k):=\left\\{\begin{array}[]{ll}\left(\begin{array}[]{cc}1&0\\\
R_{jl}e^{2it\theta}&1\end{array}\right),&k\in\Omega_{jl},\ j=1,2,\ l=1,3,\\\
\left(\begin{array}[]{cc}1&R_{jl}e^{-2it\theta}\\\
0&1\end{array}\right),&k\in\Omega_{jl},\ j=1,2,\ l=2,4,\\\
\mathrm{I},&elsewhere,\end{array}\right.$ (3.13)

where $R_{jl}\equiv R_{jl}(k),\ j=1,2,\ l=1,2,3,4$ are functions in
$\overline{\Omega}_{jl}$ admitting the properties in the following
Proposition:

######  Proposition 3.

The functions $R_{jl}:\overline{\Omega}_{jl}\to\mathbb{C},\ j=1,2,\ l=1,2,3,4$
have boundary values as follows:

$\displaystyle
R_{j1}=\left\\{\begin{array}[]{lr}-r(k_{j})(k-k_{j})^{-2i\nu(k_{j})}e^{-2i\beta_{j}(k_{j})},\
k\in\Sigma_{j1},\\\ -r(k)\delta_{+}^{-2},\quad
k\in\Gamma_{j1},\end{array}\right.$ (3.16) $\displaystyle
R_{j2}=\left\\{\begin{array}[]{lr}\frac{\bar{r}(k_{j})}{1-|r(k_{j})|^{2}}(k-k_{j})^{2i\nu(k_{j})}e^{2i\beta_{j}(k_{j})},\
k\in\Sigma_{j2},\\\ \frac{\bar{r}(k)}{1-|r(k)|^{2}}\delta_{+}^{2},\quad
k\in\Gamma_{j2},\end{array}\right.$ (3.19) $\displaystyle
R_{j3}=\left\\{\begin{array}[]{lr}\frac{r(k_{j})}{1-|r(k_{j})|^{2}}(k-k_{j})^{-2i\nu(k_{j})}e^{-2i\beta_{j}(k_{j})},\
k\in\Sigma_{j3},\\\ \frac{r(k)}{1-|r(k)|^{2}}\delta_{-}^{-2},\quad
k\in\Gamma_{j2},\end{array}\right.$ (3.22) $\displaystyle
R_{j4}=\left\\{\begin{array}[]{lr}-\bar{r}(k_{j})(k-k_{j})^{2i\nu(k_{j})}e^{2i\beta_{j}(k_{j})},\
k\in\Sigma_{j4},\\\ -\bar{r}(k)\delta_{-}^{2},\quad
k\in\Gamma_{j1},\end{array}\right.$ (3.25)

where $\Sigma_{jl}$ are shown in Figure 3 and $\Gamma_{ij}$ are intervals with

$\Gamma_{11}=(k_{1},0),\ \Gamma_{21}=(0,k_{2}),\ \Gamma_{12}=(-\infty,k_{1}),\
\Gamma_{22}=(k_{2},\infty).$

For all $j=1,2$, $R_{jl}$ is bounded as $k\in\Omega_{jl}$ and satisfies the
estimates:

$\displaystyle|\bar{\partial}R_{jl}|\lesssim\varphi(\mathrm{Re}k)+|k-k_{j}|^{-1/2}+|r^{\prime}(\mathrm{Re}k)|,\
for\ all\ k\in\Omega_{jl},\ l=1,4;\quad\qquad$ (3.26)
$\displaystyle|\bar{\partial}R_{jl}|\lesssim|k-k_{j}|^{-1/2}+|r^{\prime}(\mathrm{Re}k)|,\quad
for\ all\ k\in\Omega_{jl},\ l=2,3;$ (3.27)
$\displaystyle|\bar{\partial}R_{jl}|\lesssim|k|^{2},\quad for\ all\
k\in\Omega_{jl},\ l=1,4,\ as\ k\to 0,$ (3.28)

where $\varphi\in C^{\infty}_{0}(\mathbb{R},[0,1])$ is a fixed cut-off
function with support near 0.

###### Proof.

Since the functions $R_{jl},\ j=1,2,\ l=1,4$ have the same construction, we
take $R_{11}$ for example.

Denote $k=k_{1}+\rho e^{i\alpha}$, for $k\in\Omega_{11}$,
$\rho=|k-k_{1}|\in[0,\rho_{0}]$, $\alpha\in[0,\pi/4]$. Under the
$(\rho,\alpha)$-coordinate, the $\bar{\partial}$-derivative has the following
representation

$\bar{\partial}=\frac{1}{2}e^{i\alpha}(\partial_{\rho}+i\rho^{-1}\partial_{\alpha}).$
(3.29)

To construct the function $R_{11}$, we introduce a cut-off function
$\chi_{0}(x)\in\ C^{\infty}_{0}([0,1]),$

$\chi_{0}(x)=\left\\{\begin{array}[]{ll}1,&|x|\leqslant\min\\{1,\rho_{0}\\}/8,\\\
0,&|x|\geqslant\min\\{1,\rho_{0}\\}/4.\end{array}\right.$ (3.30)

Define the function $R_{11}$ as

$R_{11}=R_{11,1}+R_{11,2},$

where

$\displaystyle R_{11,1}=$
$\displaystyle-(1-\chi_{0}(\mathrm{Re}k))r(\mathrm{Re}k)\delta_{+}^{-2}cos^{2}(2\alpha)+\tilde{g}_{11}(1-cos^{2}(2\alpha)),$
(3.31) $\displaystyle R_{11,2}=$
$\displaystyle\tilde{f}(\mathrm{Re}k)cos^{2}(2\alpha)+\frac{i}{2}\rho
e^{-i\alpha}sin(2\alpha)cos(2\alpha)\chi_{0}(\alpha)\tilde{f}^{\prime}(\mathrm{Re}k)+sin^{2}(2\alpha)\chi_{0}(\alpha)\tilde{f}(\mathrm{Re}k)$
$\displaystyle-\frac{1}{4}\rho
e^{-i\alpha}sin^{2}(2\alpha)\tilde{f}^{\prime}(\mathrm{Re}k)-\frac{1}{8}\rho^{2}e^{-2i\alpha}sin^{2}(2\alpha)\tilde{f}^{\prime\prime}(\mathrm{Re}k),$

and

$\displaystyle\tilde{g}_{11}=-r(k_{1})(k-k_{1})^{-2i\nu(k_{1})}e^{-2i\beta_{1}(k_{1})},$
$\displaystyle\tilde{f}(\mathrm{Re}k)=-\chi_{0}(\mathrm{Re}k)r(\mathrm{Re}k)\delta_{+}^{-2}.$

Here the function $R_{11,2}$ is built to implement the estimate near $k=0$.

We first deal with $R_{11,1}$. By (3.29), we have

$\displaystyle\bar{\partial}R_{11,1}=$
$\displaystyle\frac{1}{2}\chi_{0}^{\prime}(\mathrm{Re}k)r(\mathrm{Re}k)\delta_{+}^{-2}cos(2\alpha)-\frac{1}{2}[1-\chi_{0}(\mathrm{Re}k)]r^{\prime}(\mathrm{Re}k)\delta_{+}^{-2}cos(2\alpha)$
(3.32)
$\displaystyle+2i\rho^{-1}e^{i\alpha}[1-\chi_{0}(\mathrm{Re}k)]r(\mathrm{Re}k)\delta_{+}^{-2}sin(2\alpha)cos(2\alpha)+2i\rho^{-1}e^{i\alpha}\tilde{g}_{11}sin(2\alpha)cos(2\alpha),$

where $\delta^{-2}$, $\rho^{-1},\ r(\mathrm{Re}k)$ and
$r^{\prime}(\mathrm{Re}k)$ are all bounded in the support of
$\chi_{0}(\mathrm{Re}k)$, thus (3.32) is estimated as

$|\bar{\partial}R_{11,1}|\lesssim\varphi(\mathrm{Re}k)+|r^{\prime}(\mathrm{Re}k)|+\rho^{-1}|\tilde{g}_{11}+r(\mathrm{Re}k)\delta_{+}^{-2}|.$
(3.33)

The last item of the right is rewritten as

$\rho^{-1}|[r(\mathrm{Re}k)-r(k_{1})]\delta_{+}^{-2}+r(k_{1})(k-k_{1})^{-2i\nu(k_{1})}e^{-2i\beta_{1}(k_{1})}[e^{-2i(\beta_{1}(k)-\beta_{1}(k_{1}))}-1]|$

Based on the estimate $|r(\mathrm{Re}k)-r(k_{1})|\lesssim|k-k_{1}|^{1/2}$ and
(3.9), we finally derive that

$\rho^{-1}|\tilde{g}_{11}+r(\mathrm{Re}k)\delta_{+}^{-2}|\lesssim|k-k_{1}|^{-1/2}.$
(3.34)

For $R_{11,2}$, we have

$\displaystyle\bar{\partial}R_{11,2}=$ (3.35)
$\displaystyle\left\\{\frac{1}{2}\tilde{f}^{\prime}(\mathrm{Re}k)cos^{2}(2\alpha)+\frac{i}{4}sin(2\alpha)cos(2\alpha)[8\rho^{-1}e^{i\alpha}\tilde{f}(\mathrm{Re}k)+2\tilde{f}^{\prime}(\mathrm{Re}k)+\rho
e^{-i\alpha}\tilde{f}^{\prime\prime}(\mathrm{Re}k)]\right\\}[\chi_{0}(\alpha)-1]$
$\displaystyle+\left\\{\frac{i}{2}\rho^{-1}e^{i\alpha}\chi_{0}^{\prime}(\alpha)\tilde{f}(\mathrm{Re}k)+[\chi_{0}(\alpha)-\frac{1}{4}]\tilde{f}^{\prime}(\mathrm{Re}k)-\frac{3}{8}\rho
e^{-i\alpha}\tilde{f}^{\prime\prime}(\mathrm{Re}k)-\frac{1}{16}\rho^{2}e^{-2i\alpha}\tilde{f}^{\prime\prime\prime}(\mathrm{Re}k)\right\\}sin^{2}(\alpha)$
$\displaystyle-\frac{1}{4}\rho
e^{-i\alpha}sin(2\alpha)cos(2\alpha)\chi_{0}^{\prime}(\alpha)\tilde{f}^{\prime}(\mathrm{Re}k)$

Obviously, each item of the right of (3.35) is bounded in the support of
$\chi_{0}(\mathrm{Re}k)$, so

$|\bar{\partial}R_{11,2}|\lesssim\varphi(\mathrm{Re}k).$ (3.36)

As $k\to 0$, we have $\alpha\to 0$ and within a small neighborhood of $0$,
$\chi_{0}(\alpha)\equiv 1,\ \chi_{0}^{\prime}(\alpha)\equiv 0$, thus

$\displaystyle|\bar{\partial}R_{11,2}|$
$\displaystyle\lesssim|\tilde{f}(\mathrm{Re}k)+\tilde{f}^{\prime}(\mathrm{Re}k)+\tilde{f}^{\prime\prime}(\mathrm{Re}k)+\tilde{f}^{\prime\prime\prime}(\mathrm{Re}k)|sin^{2}(2\alpha)+|\tilde{f}^{\prime}(\mathrm{Re}k)\chi_{0}^{\prime}(\alpha)sin(2\alpha)|$
(3.37) $\displaystyle\lesssim|k|^{2},$

the last inequality is deduced by Remark 1, which shows that $r(k),\
r^{\prime}(k),\ r^{\prime\prime}(k)$ and $r^{\prime\prime\prime}(k)$ are all
bounded as $k\to 0$. Together (3.33), (3.34), (3.36) with (3.37), we obtain
(3.26) and (3.28).

Meanwhile, $R_{jl},\ j=1,2,\ l=2,3$ are constructed similarly as $R_{11,1}$
without the cut-off function. Taking $R_{12}$ for example:

$R_{12}=\frac{\bar{r}(\mathrm{Re}k)}{1-|r(\mathrm{Re}k)|^{2}}\delta_{+}^{2}cos(2\alpha)+\tilde{g}_{12}(1-cos(2\alpha)),$

where

$\tilde{g}_{12}=\frac{\bar{r}(k_{1})}{1-|r(k_{1})|^{2}}(k-k_{1})^{2i\nu(k_{1})}e^{2i\beta_{1}(k_{1})}.$

Therefore, (3.27) is proved by the same method used in the estimate of
$R_{11,1}$. ∎

We now use $R^{(2)}(k)$ to define a new unknown function:

$M^{(2)}(k)\equiv M^{(2)}(y,t,k)=M^{(1)}(k)R^{(2)}(k).$ (3.38)

Then $M^{(2)}(k)$ satisfies the following mixed $\bar{\partial}$-Riemann-
Hilbert problem.

######  RH Problem 4.

Find a $2\times 2$ matrix-valued function $M^{(2)}(k)$ admitting properties as
follows:

  * •

Analyticity: $M^{(2)}(k)$ is continuous in $\mathbb{C}\setminus\Sigma;$

  * •

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

$M^{(2)}_{+}(k)=M^{(2)}_{-}(k)V^{(2)}(k),$ (3.39)

where

$V^{(2)}(k)=\left\\{\begin{array}[]{ll}(R^{(2)}(k))^{-1},&k\in\Sigma_{jl},\
j=0,1,2,\ l=1,2,\\\ R^{(2)}(k),&k\in\Sigma_{jl},\ j=0,1,2,\
l=3,4;\end{array}\right.$ (3.40)

  * •

Asymptotic behavior:

$M^{(2)}(k)=I+\mathcal{O}(k^{-1}),\ as\ k\to\infty;$ (3.41)

  * •

$\bar{\partial}$-Derivative: For $k\in\mathbb{C}$, we have

$\bar{\partial}M^{(2)}(k)=M^{(2)}(k)\bar{\partial}R^{(2)}(k),$ (3.42)

where

$\bar{\partial}R^{(2)}(k)=\left\\{\begin{array}[]{ll}\left(\begin{array}[]{cc}0&0\\\
\bar{\partial}R_{jl}e^{2it\theta}&0\end{array}\right),&k\in\Omega_{jl},\
j=1,2,\ l=1,3,\\\
\left(\begin{array}[]{cc}0&\bar{\partial}R_{jl}e^{-2it\theta}\\\
0&0\end{array}\right),&k\in\Omega_{jl},\ j=1,2,\ l=2,4,\\\
0,&elsewhere.\end{array}\right.$ (3.43)

Denote

$U_{j}=\left\\{k\|k-k_{j}|\leqslant\frac{1}{4}\rho_{0}\right\\},\ j=1,2.$

and $U=U_{1}\cup U_{2}.$ The jump matrix $V^{(2)}(k)$ has the following
estimates.

######  Proposition 4.

As $t\to\infty$, for $1\leq p_{1}\leq+\infty$, there exists a positive
constant $c_{1}(p)$ relied on $p$ such that

$\|V^{(2)}(k)-I\|_{L^{p}(\Sigma_{jl}\setminus
U_{j})}=\mathcal{O}(e^{-c_{1}(p_{1})t}),$ (3.44)

for $j=1,2$ and $l=1,2,3,4$. And there exists another positive constant
$c_{2}(p)$ relied on $p$ such that

$\|V^{(2)}(k)-I\|_{L^{p}(\Sigma_{0l})}=\mathcal{O}(e^{-c_{2}(p)t}),\
l=1,2,3,4.$ (3.45)

###### Proof.

For $\xi<0$, by (2.56), we find that

$\mathrm{Im}\theta=\mathrm{Im}k\frac{2|k|^{2}\xi+1}{2|k|^{2}},$

and $\frac{2|k|^{2}\xi+1}{2|k|^{2}}$ has nonzero boundary on
$\Sigma_{jl}\setminus U_{j}$. We take the case as $k\in\Sigma_{11}\setminus
U_{1}$ for example: denote $k=k_{1}+le^{\frac{i\pi}{4}},\
l\in(\frac{\rho_{0}}{4},+\infty)$, then for $1\leq p<+\infty,$

$\displaystyle\|V^{(2)}(k)-I\|^{p}_{L^{p}(\Sigma_{11}\setminus U_{1})}$
$\displaystyle=\|-\tilde{g}_{11}e^{2it\theta}\|^{p}_{L^{p}(\Sigma_{11}\setminus
U_{1})}\lesssim\|e^{2it\theta}\|^{p}_{L^{p}(\Sigma_{11}\setminus U_{1})}$
$\displaystyle\lesssim\int_{\frac{\rho_{0}}{4}}^{+\infty}e^{-pc^{\prime}tl}\mathrm{d}l\lesssim
t^{-1}e^{-pc^{\prime}t\rho_{0}/4}.$

For $p=+\infty$, the estimate is obvious.

As for $k\in\Sigma_{01},$ $|V^{(2)}(k)-I|=|R_{11}e^{2it\theta}|$, so (3.45) is
deduced by the same way. ∎

To solve RH problem 4, we decompose $M^{(2)}(k)$ into a pure RH problem
$M^{R}(k)\equiv M^{R}(y,t,k)$ with $\bar{\partial}M^{R}(k)\equiv 0$ and a pure
$\bar{\partial}$-problem with nonzero $\bar{\partial}$-derivatives.

We first give a RH problem for the $M^{R}(k)$:

######  RH Problem 5.

Find a $2\times 2$ matrix-valued function $M^{R}(k)$ with the following
properties:

  * •

Analyticity: $M^{R}(k)$ is analytic in $\mathbb{C}\setminus\Sigma;$

  * •

Jump condition: $M^{R}(k)$ has continuous boundry values $M^{R}_{\pm}(k)$ on
$\Sigma$ and

$M^{R}_{+}(k)=M^{R}_{-}(k)V^{(2)}(k),\quad k\in\Sigma;$ (3.46)

  * •

Asymptotic behavior: $M^{R}(k)=I+\mathcal{O}(k^{-1}),\ as\ k\to\infty;$

  * •

$\bar{\partial}$-Derivative: $\bar{\partial}R^{(2)}=0,$ for all
$k\in\mathbb{C}$.

Next, We use $M^{R}(k)$ to define a new matrix function

$M^{(3)}(k)\equiv M^{(3)}(y,t,k)=M^{(2)}(k)(M^{R}(k))^{-1},$ (3.47)

which is a pure $\bar{\partial}$-problem by removing $M^{R}(k)$ from
$M^{(2)}(k)$.

######  $\bar{\partial}$-problem 1.

Find a $2\times 2$ matrix-valued function $M^{(3)}(k)$ satisfying the
following properties:

  * •

Continuity: $M^{(3)}(k)$ is continuous in $\mathbb{C};$

  * •

Asymptotic behavior: $M^{(3)}(k)\thicksim I+\mathcal{O}(k^{-1}),\quad
k\to\infty;$

  * •

$\bar{\partial}$-Derivative: For $k\in\mathbb{C}$, we have

$\bar{\partial}M^{(3)}(k)=M^{(3)}(k)W(k),$ (3.48)

where

$W(k)=M^{R}(k)\bar{\partial}R^{(2)}(k)(M^{R}(k))^{-1}.$ (3.49)

###  3.3 Analysis on pure RH problem

####  3.3.1 A local solvable RH model near stationary points

$k_{1}$$k_{2}$$0$$\mathrm{Re}k$ Figure 4: The jump lines $\Sigma^{lo}$.

To solve the RH problem $M^{R}(k)$, we use the PC model to estimate the
behavior near stationary points, which will be defined as a new local RH
problem $M^{lo}(k)\equiv M^{lo}(y,t,k)$. And then we find the error function
between $M^{R}(k)$ and $M^{lo}(k)$ to deduce the asymptotic representation of
$M^{R}(k)$.

To introduce the local function, we denote several lines as shown in Figure 4

$\Sigma_{jl}^{lo}=\Sigma_{jl}\cap U_{j},\ j=1,2,\ l=1,2,3,4,$

and

$\Sigma^{lo}=\Sigma^{lo}_{1}\cup\Sigma^{lo}_{2},\quad\Sigma^{lo}_{j}=\cup_{l=1}^{4}\Sigma_{jl}^{lo}.$

Now we consider the following RH problem

######  RH Problem 6.

Find a $2\times 2$ matrix-valued function $M^{lo}(k)$ admitting the properties
as follows:

  * •

Analyticity: $M^{lo}(k)$ is analytical in $\mathbb{C}\setminus\Sigma^{lo};$

  * •

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

$M^{lo}_{+}(k)=M^{lo}_{-}(k)V^{(2)}(k),\quad k\in\Sigma^{lo};$ (3.50)

  * •

Asymptotic behavior: $M^{lo}(k)=I+\mathcal{O}(k^{-1}),\quad k\to\infty.$

For $j=1,2$ and $l=1,2,3,4$, denote

$\omega_{jl}(k)=\left\\{\begin{array}[]{ll}\left(\begin{array}[]{cc}0&0\\\
(-1)^{\frac{l-1}{2}}R_{jl}e^{2it\theta}&0\end{array}\right),&k\in\Sigma_{jl},\
j=1,2,\ l=1,3,\\\
\left(\begin{array}[]{cc}0&(-1)^{\frac{l-2}{2}}R_{jl}e^{-2it\theta}\\\
0&0\end{array}\right),&k\in\Sigma_{jl},\ j=1,2,\ l=2,4,\end{array}\right.$

then $V^{(2)}(k)=I-\omega_{jl}(k)$ for $k\in\Sigma_{jl}$. Let

$\omega_{j}(k)=\sum_{l=1}^{4}\omega_{jl}(k),\
\omega(k)=\omega_{1}+\omega_{2},$

and

$\omega_{jl}^{\pm}(k)=\omega_{jl}(k)|_{\mathbb{C}^{\pm}},\
\omega_{j}^{\pm}(k)=\omega_{j}(k)|_{\mathbb{C}^{\pm}},\
\omega^{\pm}(k)=\omega(k)|_{\mathbb{C}^{\pm}}.$

Considering the Cauchy projection operator $\mathcal{C}_{\pm}$ on $\Sigma$, we
define the following operator:

$\mathcal{C}_{\omega}(f)=\mathcal{C}_{+}(f\omega^{-})+\mathcal{C}_{-}(f\omega^{+}),\
\mathcal{C}_{\omega_{j}}(f)=\mathcal{C}_{+}(f\omega_{j}^{-})+\mathcal{C}_{-}(f\omega_{j}^{+}).$
(3.51)

Thus $\mathcal{C}_{\omega}=\mathcal{C}_{\omega_{1}}+\mathcal{C}_{\omega_{2}}$.

######  Lemma 5.

For the function $\omega_{jl}(k),\ j=1,2,\ l=1,2,3,4$, we have following
estimate:

$\|\omega_{jl}(k)\|_{L^{2}}\lesssim t^{-1/2},$ (3.52)

This Lemma can be directly proved by calculation. And thus
$I-\mathcal{C}_{\omega}$ and $I-\mathcal{C}_{\omega_{j}}$ are reversible.
Therefore, there exists unique solution of RH problem 6, and it is denoted as

$M^{lo}(k)=I+\frac{1}{2\pi
i}\int_{\Sigma^{lo}}\frac{(I-\mathcal{C}_{\omega})^{-1}I\omega}{s-k}\mathrm{d}s.$
(3.53)

Besides, by (3.51) and Lemma 5, we deduce following corollary:

######  Corollary 1.

As $t\to\infty$,

$\|\mathcal{C}_{\omega_{j}}\mathcal{C}_{\omega_{l}}\|_{B(L^{2}(\Sigma^{lo}))}\lesssim
t^{-1},\quad\|\mathcal{C}_{\omega_{j}}\mathcal{C}_{\omega_{l}}\|_{(L^{\infty}(\Sigma^{lo}\to
L^{2}(\Sigma^{lo})))}\lesssim t^{-1}.$

Therefore, by above results, we finally find that the contributions of every
crosses $\Sigma^{lo}_{j},\ j=1,2$ are separated out.

The RH problems near the two stationary points are estimated by PC model
respectively. Here we take the estimate near $k_{1}$ for example. Consider the
following RH problem:

######  RH Problem 7.

Find a $2\times 2$ matrix-valued function $M^{lo,1}(k)\equiv M^{lo,1}(y,t,k)$
satisfying the following properties:

  * •

Analyticity: $M^{lo,1}(k)$ is analytical in
$\mathbb{C}\setminus\Sigma^{lo}_{1};$

  * •

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

$M^{lo,1}_{+}(k)=M^{lo,1}_{-}(k)V^{lo,1}(k),\quad k\in\Sigma_{1}^{lo},$ (3.54)

where

$V^{lo,1}(k)=\left\\{\begin{array}[]{ll}\left(\begin{array}[]{cc}1&0\\\
r(k_{1})(k-k_{1})^{-2i\nu(k_{1})}e^{-2i\beta_{1}(k_{1})}e^{2it\theta}&1\end{array}\right),&k\in\Sigma_{11}^{lo},\\\
\left(\begin{array}[]{cc}1&-\frac{\overline{r(k_{1})}}{1-|r(k_{1})|^{2}}(k-k_{1})^{2i\nu(k_{1})}e^{2i\beta_{1}(k_{1})}e^{-2it\theta}\\\
0&1\end{array}\right),&k\in\Sigma_{12}^{lo},\\\
\left(\begin{array}[]{cc}1&0\\\
\frac{r(k_{1})}{1-|r(k_{1})|^{2}}(k-k_{1})^{-2i\nu(k_{1})}e^{-2i\beta_{1}(k_{1})}e^{2it\theta}&1\end{array}\right),&k\in\Sigma_{13}^{lo},\\\
\left(\begin{array}[]{cc}1&-\overline{r(k_{1})}(k-k_{1})^{2i\nu(k_{1})}e^{2i\beta_{1}(k_{1})}e^{-2it\theta}\\\
0&1\end{array}\right),&k\in\Sigma_{14}^{lo};\end{array}\right.$ (3.55)

  * •

Asymptotic behavior: $M^{lo,1}(k)=I+\mathcal{O}(k^{-1}),\quad k\to\infty.$

Taking Cauchy expansion at $k=k_{1}$ for $\theta$, we get

$\theta=-\frac{1}{k_{1}}-\frac{1}{k_{1}^{3}}(k-k_{1})^{2}+\mathcal{O}((k-k_{1})^{3}).$
(3.56)

Let $\zeta=\zeta(k)$ defined as follows:

$\zeta=t^{1/2}\sqrt{-\frac{4}{k_{1}^{3}}}(k-k_{1}),$ (3.57)

and denote a parameter $\widetilde{r}_{1}$ as

$\widetilde{r}_{1}=r(k_{1})e^{-2i(\beta_{1}(k_{1})+tk_{1}^{-1})}\mathrm{exp}\\{i\nu(k_{1})\mathrm{log}(-4tk_{1}^{-3})\\}.$
(3.58)

Now $M^{lo,1}$ can match the following model RH problem:

######  RH Problem 8.

Find a $2\times 2$ matrix-valued function $M^{pc}(\zeta)$ with the following
properties:

  * •

Asymptotic: $M^{pc}(\zeta)$ is analytical in $\mathbb{C}\setminus\Sigma^{pc}$
with $\Sigma^{pc}={\mathbb{R}e^{\varphi
i}}\cup{\mathbb{R}e^{(\pi-\varphi)i}};$

  * •

Jump condition: $M^{pc}(\zeta)$ has continuous boundary values $M_{\pm}^{pc}$
on $\Sigma^{pc}$ and

$M^{pc}_{+}(\zeta)=M^{pc}_{-}(\zeta)V^{pc}(\zeta),\quad\zeta\in\Sigma^{pc},$
(3.59)

where

$V^{pc}(\zeta)=\left\\{\begin{array}[]{ll}\left(\begin{array}[]{cc}1&0\\\
\widetilde{r}_{1}\zeta^{-2i\nu(k_{1})}e^{\frac{i}{2}\zeta^{2}}&1\end{array}\right),&\zeta\in\mathbb{R}^{+}e^{\varphi
i},\\\
\left(\begin{array}[]{cc}1&-\frac{\overline{\widetilde{r}_{1}}}{1-|\widetilde{r}_{1}|^{2}}\zeta^{2i\nu(k_{1})}e^{-\frac{i}{2}\zeta^{2}}\\\
0&1\end{array}\right),&\zeta\in\mathbb{R}^{+}e^{(\pi-\varphi)i},\\\
\left(\begin{array}[]{cc}1&0\\\
\frac{\widetilde{r}_{1}}{1-|\widetilde{r}_{1}|^{2}}\zeta^{-2i\nu(k_{1})}e^{\frac{i}{2}\zeta^{2}}&1\end{array}\right),&\zeta\in\mathbb{R}^{+}e^{(\pi+\varphi)i},\\\
\left(\begin{array}[]{cc}1&-\overline{\widetilde{r}_{1}}\zeta^{2i\nu(k_{1})}e^{-\frac{i}{2}\zeta^{2}}\\\
0&1\end{array}\right),&\zeta\in\mathbb{R}^{+}e^{-\varphi i};\\\
\end{array}\right.$ (3.60)

  * •

Asymptotic behavior:

$M^{pc}(\zeta)=I+\frac{M_{1}^{pc}}{\zeta}+\mathcal{O}(\zeta^{-2}),\quad\zeta\to\infty.$
(3.61)

There exists the unique solution of RH problem 8 with the asymptotic behavior
as (3.61), where

$M_{1}^{pc}=\left(\begin{array}[]{cc}0&-i\beta^{1}_{12}\\\
i\beta^{1}_{21}&0\end{array}\right),$

and

$\beta^{1}_{12}=\frac{\sqrt{2\pi}e^{-\frac{\pi}{2}\nu(k_{1})}e^{\frac{\pi}{4}i}}{\widetilde{r}_{1}\Gamma(-i\nu(k_{1}))},\quad\beta^{1}_{21}=\frac{\nu(k_{1})}{\beta^{1}_{12}}.$
(3.62)

As proved in A1-A6 of [43], $M^{lo,1}(k)$ is estimated as:

$M^{lo,1}(k)=I+t^{-\frac{1}{2}}\frac{\rho_{0}^{3/2}}{k-k_{1}}\frac{i}{2}\left(\begin{array}[]{cc}0&-\beta^{1}_{12}\\\
\beta^{1}_{21}&0\end{array}\right)+\mathcal{O}(t^{-1}).$

Similarly,

$M^{lo,2}(k)=I+t^{-\frac{1}{2}}\frac{\rho_{0}^{3/2}}{k-k_{2}}\frac{i}{2}\left(\begin{array}[]{cc}0&-\beta^{2}_{12}\\\
\beta^{2}_{21}&0\end{array}\right)+\mathcal{O}(t^{-1}),$

where

$\beta^{2}_{12}=\frac{\sqrt{2\pi}e^{\frac{3\pi}{2}\nu(k_{2})}e^{\frac{3\pi}{4}i}}{\widetilde{r}_{2}\Gamma(i\nu(k_{2}))},\quad\beta^{2}_{21}=\frac{\nu(k_{2})}{\beta^{2}_{12}},$
(3.63)

and

$\widetilde{r}_{2}=r(k_{2})e^{-2i(\beta_{2}(k_{2})+tk_{2}^{-1})}\mathrm{exp}\\{i\nu(k_{2})\mathrm{log}(-4tk_{2}^{-3})\\}.$
(3.64)

Therefore, we have

$M^{lo}(k)=I+t^{\frac{1}{2}}\frac{\rho_{0}^{3/2}i}{2}\sum_{j=1}^{2}\frac{B_{j}}{k-k_{j}}+\mathcal{O}(t^{-1}),$
(3.65)

where

$B_{j}=\left(\begin{array}[]{cc}0&-\beta_{12}^{j}\\\
\beta_{21}^{j}&0\end{array}\right),\quad j=1,2.$ (3.66)

####  3.3.2 Error function with small norm

$k_{1}$$k_{2}$$0$$\mathrm{Re}k$ Figure 5: The jump lines $\Sigma^{(E)}$ for
$E(k)$.

Considering the error between $M^{R}(k)$ and $M^{lo}(k)$, we define a new
function as follows:

$E(k)\equiv
E(y,t,k)=\left\\{\begin{array}[]{ll}M^{R}(k),&k\in\mathbb{C}\setminus U,\\\
M^{R}(k)(M^{lo}(k))^{-1},&k\in U.\end{array}\right.$ (3.67)

Then $E(k)$ is a solution of the new RH problem:

######  RH Problem 9.

Find a $2\times 2$ matrix-valued function $E(k)$ with following properties:

  * •

Analyticity: $E(k)$ is analytical in $\mathbb{C}\setminus\Sigma^{(E)}$, where
$\Sigma^{(E)}=\partial U\cup(\Sigma\setminus U)$ as shown in Figure 5;

  * •

Jump condition: $E(k)$ has continuous boundary values $E_{\pm}(k)$ on
$\Sigma^{(E)}$, and

$E_{+}(k)=E_{-}(k)V^{(E)}(k),$ (3.68)

where

$V^{(E)}(k)=\left\\{\begin{array}[]{ll}V^{(2)},&k\in\Sigma\setminus U,\\\
M^{lo}(k),&k\in\partial U;\end{array}\right.$ (3.69)

  * •

Asymptotic behavior: $E(k)\sim I+\mathcal{O}(k^{-1}),\quad k\to\infty.$

Using Proposition 4, we get the estimates of $V^{(E)}(k)$ as follows:

$\|V^{(E)}(k)-I\|_{p}\lesssim\left\\{\begin{array}[]{ll}e^{-c_{1}(p)t},&k\in\Sigma_{jl}\setminus
U,\ j=1,2,\\\ e^{-c_{2}(p)t},&k\in\Sigma_{0l},\ l=1,2,3,4.\end{array}\right.$
(3.70)

For $k\in\partial U$, by (3.65) and (3.69), we have

$|V^{(E)}(k)-I|=|M^{lo}(k)-I|=\mathcal{O}(t^{-1/2}).$ (3.71)

Therefore, based on the Beals-Coifman theorem, by the same method in Section
3, the solution of the RH problem 9 is represented by

$E(k)=I+\frac{1}{2\pi
i}\int_{\Sigma_{E}}\frac{\mu_{E}(s)(V^{(E)}(s)-I)}{s-k}\mathrm{d}s,$ (3.72)

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

$(1-\mathcal{C}_{\omega_{E}})\mu_{E}=I,$

and the operator $\mathcal{C}_{\omega_{E}}$ is defined by

$\mathcal{C}_{\omega_{E}}f=\mathcal{C}_{-}(f(V^{(E)}-I))$

with the estimate:

$\|\mathcal{C}_{\omega_{E}}\|_{L^{2}(\Sigma^{(E)})}\leqslant\|\mathcal{C}_{-}\|_{L^{2}(\Sigma^{(E)})}\|V^{(E)}-I\|_{L^{\infty}(\Sigma^{(E)})}\leqslant\mathcal{O}(t^{-1/2}),$

which guarantees the existence and uniqueness of $\mu_{E}$ and $M^{(E)}$. We
finally give the asymptotic expansion of $E(k)$ at $k=0$:

######  Proposition 5.

As $k\to 0$, we have

$E(k)=E_{0}+E_{1}k+E_{2}k^{2}+\mathcal{O}(k^{3}),$ (3.73)

where

$\displaystyle E_{0}=E(0)=I+\frac{1}{2\pi
i}\int_{\Sigma^{(E)}}\frac{\mu_{E}(s)(V^{(E)}(s)-I)}{s}\mathrm{d}s,$
$\displaystyle E_{1}=-\frac{1}{2\pi
i}\int_{\Sigma^{(E)}}\frac{\mu_{E}(s)(V^{(E)}(s)-I)}{s^{2}}\mathrm{d}s,\quad
E_{2}=\frac{1}{2\pi
i}\int_{\Sigma^{(E)}}\frac{\mu_{E}(s)(V^{(E)}(s)-I)}{s^{3}}\mathrm{d}s.$

Besides, $E_{j},\ j=0,1,2$ satisfies following long-time asymptotic behavior:

$\displaystyle E_{0}-I=t^{-\frac{1}{2}}H_{0}+\mathcal{O}(t^{-1}),$ (3.74)
$\displaystyle E_{j}=t^{-\frac{1}{2}}H_{j}+\mathcal{O}(t^{-1}),\ j=1,2,$
(3.75)

and by the residue theorem, we write them out as:

$H_{j}=(-1)^{j}\frac{\rho_{0}^{3/2}i}{2}\sum_{l=1}^{2}\frac{B_{l}}{k_{l}^{j+1}}.$
(3.76)

###### Proof.

The proof of (3.74) and (3.75) are similar, so we only give the details for
the first one. Notice that

$\frac{1}{s-k}=\frac{1}{s}\sum_{n=0}^{\infty}(\frac{k}{s})^{n},$

thus

$\displaystyle E_{0}-I=\frac{1}{2\pi
i}\int_{\Sigma^{(E)}}\frac{\mu_{E}(V^{(E)}-I)}{s}\mathrm{d}s$ $\displaystyle=$
$\displaystyle\frac{1}{2\pi
i}\int_{\Sigma^{(E)}}\frac{(\mu_{E}-I)(V^{(E)}-I)}{s}\mathrm{d}s+\frac{1}{2\pi
i}\int_{\Sigma^{(E)}\setminus U}\frac{V^{(E)}-I}{s}\mathrm{d}s+\frac{1}{2\pi
i}\int_{\partial U}\frac{V^{(E)}-I}{s}\mathrm{d}s$ $\displaystyle=$
$\displaystyle\frac{1}{2\pi i}\sum_{j=1}^{2}\int_{\partial
U_{j}}t^{-\frac{1}{2}}\frac{i\rho_{0}^{\frac{3}{2}}B_{j}}{2(s-k_{j})}s^{-1}\mathrm{d}s+\mathcal{O}(t^{-1})=\frac{i\rho_{0}^{\frac{3}{2}}}{2}\sum_{j=1}^{2}\frac{B_{j}}{k_{j}}t^{-\frac{1}{2}}+\mathcal{O}(t^{-1})$
$\displaystyle=$ $\displaystyle H_{0}t^{-\frac{1}{2}}+\mathcal{O}(t^{-1}).$

∎

###  3.4 Analysis on the pure $\bar{\partial}$-problem

The solution $M^{(3)}(k)$ of the $\bar{\partial}$-problem 1 is given by the
integral equation:

$M^{(3)}(k)=I+\frac{1}{\pi}\iint_{\mathbb{C}}\frac{M^{(3)}(s)W^{(3)}(s)}{s-k}\mathrm{d}A(s).$
(3.77)

Denote the integral operator $S^{(3)}$ as

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

Then, (3.77) equals to

$(I-S^{(3)})M^{(3)}(k)=I.$ (3.78)

To prove the existence of $(I-S^{(3)})^{-1}$, we need the following
proposition.

######  Proposition 6.

As $t\to\infty$, the norm of the integral operator $S^{(3)}$ satisfies the
estimate:

$\|S^{(3)}\|_{L^{\infty}\to L^{\infty}}\lesssim t^{-1/2}.$ (3.79)

###### Proof.

We only give the proof of $k\in\Omega_{11}$. Define a new function $f(k)\in
L^{\infty}(\Omega_{11})$, then

$\displaystyle|S(f)|$
$\displaystyle\leqslant\frac{1}{\pi}\iint_{\Omega_{11}}\frac{|f(s)M^{R}(s)\bar{\partial}R^{(2)}(s)(M^{R}(s))^{-1}|}{|s-k|}\mathrm{d}A(s)$
$\displaystyle\leqslant\|f\|_{L^{\infty}}\frac{1}{\pi}\iint_{\Omega_{11}}\frac{|\bar{\partial}R_{12}e^{-2it\theta}|}{|s-k|}\mathrm{d}A(s)\lesssim
I_{1}+I_{2}+I_{3},$

where

$\displaystyle
I_{1}=\iint_{\Omega_{11}}\frac{|\varphi(\mathrm{Re}s)e^{-2it\theta}|}{|s-k|}\mathrm{d}A(s),\
I_{2}=\iint_{\Omega_{11}}\frac{|r^{\prime}(\mathrm{Re}s)e^{-2it\theta}|}{|s-k|}\mathrm{d}A(s),$
$\displaystyle
I_{3}=\iint_{\Omega_{11}}\frac{|s-k_{1}|^{-\frac{1}{2}}|e^{-2it\theta}|}{|s-k|}\mathrm{d}A(s).$

Denote $s=k_{1}+u+iv,\ k=\alpha+i\eta$, by Lemma 4 and $\varphi\in
C_{0}^{\infty}$, we have

$I_{1}\leqslant\int_{0}^{\infty}\int_{v}^{\infty}\frac{|\varphi(u)|}{|s-k|}e^{2\xi
tv}\mathrm{d}u\mathrm{d}v\leqslant\int_{0}^{\infty}e^{-ctv}\|\varphi\|_{L^{2}}\|(s-k)^{-1}\|_{L^{2}}\mathrm{d}v,$

where

$\displaystyle\|(s-k)^{-1}\|_{L^{2}(v,\infty)}$
$\displaystyle\leqslant\left(\int_{-\infty}^{\infty}\frac{1}{(k_{1}+u-\alpha)^{2}+(v-\eta)^{2}}\mathrm{d}u\right)^{1/2}$
(3.80)
$\displaystyle=\left(\frac{1}{|v-\eta|}\int_{-\infty}^{\infty}\frac{1}{1+y^{2}}\mathrm{d}y\right)^{1/2}=\left(\frac{\pi}{|v-\eta|}\right)^{1/2}$

with $y=\frac{k_{1}+u-\alpha}{v-\eta}$. Thus

$I_{1}\lesssim\int_{0}^{\infty}\frac{e^{-ctv}}{\sqrt{|v-\eta|}}\mathrm{d}v\lesssim
t^{-1/2}.$

The estimate of $I_{2}$ is just the same as $I_{1}$ because $r^{\prime}(k)\in
L^{2}(\mathbb{R})$, so we get $I_{2}\lesssim t^{-1/2}.$

Finally, we deal with $I_{3}$. We first give the proof of the estimates as
follows: for $2<p<\infty$ and $q$ satisfying $\frac{1}{p}+\frac{1}{q}=1$, we
have

$\|\frac{1}{\sqrt{|s-k_{1}|}}\|_{L^{p}(v,\infty)}=\left(\int_{v}^{\infty}\frac{1}{|u+iv|^{p/2}}\mathrm{d}u\right)^{1/p}=v^{\frac{1}{p}-\frac{1}{2}}\left(\int_{1}^{\infty}\frac{1}{(1+y^{2})^{p/4}}\mathrm{d}y\right)^{1/p}\lesssim
v^{\frac{1}{p}-\frac{1}{2}},$ (3.81)

and by the same method as (3.80), we find

$\|(s-k)^{-1}\|_{L^{q}(v,\infty)}\lesssim|v-\eta|^{1/q-1}.$ (3.82)

So

$I_{3}\leqslant\int_{0}^{+\infty}e^{-ctv}\int^{\infty}_{v}\frac{|s-1|^{-1/2}}{|s-k|}\mathrm{d}u\mathrm{d}v\lesssim\int_{0}^{+\infty}e^{-ctv}v^{\frac{1}{p}-\frac{1}{2}}|v-\eta|^{\frac{1}{q}-1}\mathrm{d}v\lesssim
t^{-1/2}.$ (3.83)

∎

Therefore, the solution of $\bar{\partial}$-problem 1 is unique and obeys the
formula (3.77). Then we have the following asymptotic estimation of
$M^{(3)}(k)$.

######  Proposition 7.

As $k\rightarrow 0$, $M^{(3)}(k)$ has the asymptotic expansion:

$M^{(3)}(k)=M^{(3)}(0)+M^{(3)}_{1}k+M^{(3)}_{2}k^{2}+O(k^{3}),$ (3.84)

where

$|M^{(3)}(0)-I|=|\frac{1}{\pi}\iint_{\mathbb{C}}\frac{M^{(3)}(s)W^{(3)}(s)}{s}\mathrm{d}A(s)|\lesssim
t^{-1},$ (3.85)

and

$M^{(3)}_{1}=-\frac{1}{\pi}\iint_{\mathbb{C}}\frac{M^{(2)}(s)W^{(3)}(s)}{s^{2}}\mathrm{d}A(s),\quad
M^{(3)}_{2}=-\frac{1}{\pi}\iint_{\mathbb{C}}\frac{M^{(2)}(s)W^{(3)}(s)}{s^{3}}\mathrm{d}A(s)$
(3.86)

satisfying

$|M^{(3)}_{1}|\lesssim t^{-1},\ as\ t\rightarrow\infty,$ (3.87)

and for $2<p<\infty$,

$|M^{(3)}_{2}|\lesssim t^{-1+\frac{1}{2p}},\ as\ t\rightarrow\infty.$ (3.88)

###### Proof.

We take the case as $s\in\Omega_{11}$ for example. Considering the estimate
(3.28) in Proposition 3, we divide the integral region into $\Omega_{11}\cap
B(0)$, $\Omega_{11}\cap B_{k_{1}}$ and
$\Omega^{\prime}_{11}:=\Omega_{11}\setminus(B(0)\cap B_{k_{1}})$, where
$B(0)=\\{k\mid|k|<\epsilon^{\prime}_{0}<\rho_{0}/4\\}$ satisfying

$|\bar{\partial}R_{11}|\lesssim|k|^{2},\ for\ all\ k\in B(0),$

and $B_{k_{1}}=\\{k\mid|k-k_{1}|<\rho_{0}/4\\}$. Then

$|M^{(3)}(0)-I|_{\Omega_{11}}\lesssim(\iint_{\Omega_{11}\cap
B(0)}+\iint_{\Omega_{11}\cap
B_{k_{1}}}+\iint_{\Omega^{\prime}_{11}})\frac{|\bar{\partial}R_{11}|e^{-ct\mathrm{Im}\theta}}{|s|}\mathrm{d}A(s),$
(3.89)

where the first part

$\iint_{\Omega_{11}\cap
B(0)}\frac{|\bar{\partial}R_{11}|e^{-ctv}}{|s|}\mathrm{d}A(s)\lesssim\iint_{\Omega_{2}\cap
B(0)}e^{-ctv}\mathrm{d}A(s)\lesssim t^{-1}.$ (3.90)

For the second part, $|s|>\rho_{0}/2$, so we have

$\displaystyle\iint_{\Omega_{11}\cap
B_{k_{1}}}\frac{|\bar{\partial}R_{11}|e^{-ctv}}{|s|}\mathrm{d}A(s)$
$\displaystyle\lesssim$ $\displaystyle\iint_{\Omega_{11}\cap
B_{k_{1}}}(|\varphi(u)|+|r^{\prime}(u)|)e^{-ctv}\mathrm{d}A(s)+\iint_{\Omega_{11}\cap
B_{k_{1}}}|s-k_{1}|^{-1/2}e^{ct\mathrm{Im}\theta}\mathrm{d}A(s):=I^{(0)}_{1}+I^{(0)}_{2}.$

Notice that $\varphi,\ r^{\prime}(k)\in L^{1}(\mathbb{R})$, thus

$I^{(0)}_{1}\lesssim\iint_{\Omega_{11}}(|\varphi(u)|+|r^{\prime}(u)|)e^{-ctv}\mathrm{d}A(s)\lesssim\int_{0}^{+\infty}(\|\varphi\|_{L^{1}}+\|r^{\prime}\|_{L^{1}})e^{-ctv}\mathrm{d}v\lesssim
t^{-1}.$ (3.91)

As for $I^{(0)}_{2}$, based on the definition of the region $B_{k_{1}}$ and
(2.56), we deduce that

$\mathrm{Im}\theta=uv\left(\frac{\xi}{u}+\frac{1}{2u[(u+k_{1})^{2}+v^{2}]}\right)\leqslant
c_{\xi}uv$ (3.92)

where $c_{\xi}$ is a constant independent with $u,v$. Therefore, by (3.81), we
have

$\displaystyle I^{(0)}_{2}$
$\displaystyle\lesssim\int_{0}^{+\infty}\||s-k_{1}|^{-1/2}\|_{L^{q}}\|e^{-c_{\xi}tuv}\|_{L^{p}}\mathrm{d}v\lesssim
t^{-\frac{1}{p}}\int_{0}^{+\infty}v^{\frac{2}{q}-\frac{1}{2}}e^{-ctv^{2}}\mathrm{d}v$
$\displaystyle=t^{-\frac{1}{p}}(\int_{0}^{1}+\int_{1}^{+\infty})v^{\frac{2}{q}-\frac{1}{2}}e^{-ctv^{2}}\mathrm{d}v=t^{-\frac{1}{p}}(I^{(0)}_{21}+I^{(0)}_{22})$

with

$I^{(0)}_{21}\lesssim
t^{-\frac{1}{4}-\frac{1}{q}}\int_{0}^{\infty}y^{\frac{1}{q}-\frac{3}{4}}e^{-y}\mathrm{d}y\lesssim
t^{-\frac{1}{4}-\frac{1}{q}},$

and

$I^{(0)}_{22}\lesssim\int_{1}^{+\infty}v^{\frac{2}{q}-\frac{1}{2}}e^{-ctv}\mathrm{d}v\lesssim
t^{-\frac{1}{2}-\frac{2}{q}}.$

Therefore,

$I^{(0)}_{2}\lesssim t^{-\frac{5}{4}}.$ (3.93)

For the last part of (3.89), we have

$\displaystyle\iint_{\Omega^{\prime}_{11}}\frac{|\bar{\partial}R_{11}|e^{-ct\mathrm{Im}\theta}}{|s|}\mathrm{d}A(s)$
$\displaystyle\lesssim\int_{0}^{\infty}\int_{v}^{\infty}(|\varphi(u)|+|r^{\prime}(u)|)e^{-ctv}\mathrm{d}u\mathrm{d}v+\int_{\frac{\rho_{0}}{4}}^{\infty}\int_{v}^{\infty}\frac{|s-k_{1}|^{-\frac{1}{2}}}{|s|}e^{-ctv}\mathrm{d}u\mathrm{d}v$
(3.94)
$\displaystyle\lesssim\int_{0}^{\infty}e^{-ctv}(\|\varphi(u)\|_{L^{1}}+\|r^{\prime}(u)\|_{L^{1}})\mathrm{d}v+\int_{\frac{\rho_{0}}{4}}^{\infty}e^{-ctv}v^{-1/2}\mathrm{d}v\lesssim
t^{-1}.$

Summing up all above, we proved (3.85). The estimate (3.87) is deduced
similarly.

Finally, we give the proof of (3.88):

$|M^{(3)}_{2}|\leqslant\iint_{\Omega_{11}\cap
B(0)}\frac{e^{-ct\mathrm{Im}\theta}}{|s|}\mathrm{d}A(s)+(\iint_{\Omega_{11}\cap
B_{k_{1}}}+\iint_{\Omega^{\prime}_{11}})\frac{|\bar{\partial}R_{11}|e^{-ct\mathrm{Im}\theta}}{|s|}\mathrm{d}A(s),$
(3.95)

where the last two parts have been estimated to decay by the speed of at least
$t^{-1}$ as $t\to\infty$. Now we only need to deal with the first item:

As for $|s|<\rho_{0}/4,$ there holds $|s-k_{1}|\leqslant\rho_{0}/4+|k_{1}|$.
For $s\in\Omega_{11}\cap B(0)$, (3.92) holds for some constant
$c^{\prime}_{\xi}$. Based on above, taking $p>2,\ k=0$ in (3.82), we find

$\iint_{\Omega_{11}\cap
B(0)}\frac{e^{-ct\mathrm{Im}\theta}}{|s|}\mathrm{d}A(s)\lesssim\int_{0}^{\rho_{0}/4}\|s^{-1}\|_{L^{p}}\|e^{-c^{\prime}_{\xi}tuv}\|_{L^{q}}\mathrm{d}v\lesssim
t^{-\frac{1}{q}}\int_{0}^{\rho_{0}/4}|v|^{\frac{1}{p}-1}e^{-c^{\prime}_{\xi}tv^{2}}\mathrm{d}v\lesssim
t^{-1+\frac{1}{2p}}.$ (3.96)

Thus, we conclude the proof. ∎

###  3.5 Long-time asymptotics for the HS equation

Finally, we construct the long-time asymptotic approximation for the solution
of the HS equation (1.1). Inverting the transformations (3.10), (3.38), (3.47)
and (3.67), we have

$\widehat{M}(k)=M^{(3)}(k)E(k)(R^{(2)})^{-1}(k)\delta^{\sigma_{3}}(k).$ (3.97)

We take $k\to 0$ out of $\Omega$ so that $R^{(2)}(k)=I$. Then by the results
of Proposition 5 and Proposition 7, we obtain the asymptotic expansion of
$\widehat{M}(k)$ as follows:

$\displaystyle\widehat{M}(k)=$ $\displaystyle
M^{(3)}(k)E(k)\delta^{\sigma_{3}}=(I+\mathcal{O}(t^{-1+\frac{1}{2p}}))(E_{0}+E_{1}k+E_{2}k^{2})(I+i\delta_{1}\sigma_{3}k-\frac{\delta_{1}^{2}}{2}Ik^{2})+\mathcal{O}(k^{3})$
(3.98) $\displaystyle=$
$\displaystyle(I+t^{-\frac{1}{2}}(H_{0}+H_{1}k+H_{2}k^{2}))(I+i\delta_{1}\sigma_{3}k-\frac{\delta_{1}^{2}}{2}Ik^{2})+\mathcal{O}(t^{-1+\frac{1}{2p}})+\mathcal{O}(k^{3}),$

which implies that

$\displaystyle\widehat{M}_{1,1}(y,t)=i\delta_{1}-\frac{t^{-\frac{1}{2}}}{2}\rho_{0}^{3/2}f_{11}+\mathcal{O}(t^{-1}),\
\qquad\widehat{M}_{2,1}(y,t)=-i\delta_{1}-\frac{t^{-\frac{1}{2}}}{2}\rho_{0}^{3/2}f_{12}+\mathcal{O}(t^{-1}),\qquad\qquad$
$\displaystyle\widehat{M}_{1,2}(y,t)=-\frac{1}{2}\delta_{1}^{2}-\frac{t^{-\frac{1}{2}}}{2}\rho_{0}^{3/2}f_{21}+\mathcal{O}(t^{-1}),\quad\widehat{M}_{2,2}(y,t)=-\frac{1}{2}\delta_{1}^{2}-\frac{t^{-\frac{1}{2}}}{2}\rho_{0}^{3/2}f_{22}+\mathcal{O}(t^{-1}),$

where

$\displaystyle
f_{11}=\delta_{1}\sum_{j=1}^{2}\frac{\beta_{21}^{j}}{k_{j}}+i\sum_{j=1}^{2}\frac{\beta_{21}^{j}}{k_{j}^{2}},$
(3.99) $\displaystyle
f_{12}=\delta_{1}\sum_{j=1}^{2}\frac{\beta_{12}^{j}}{k_{j}}-i\sum_{j=1}^{2}\frac{\beta_{12}^{j}}{k_{j}^{2}},$
(3.100) $\displaystyle
f_{21}=\frac{i}{2}\delta_{1}^{2}\sum_{j=1}^{2}\frac{\beta_{21}^{j}}{k_{j}}-\delta_{1}\sum_{j=1}^{2}\frac{\beta_{21}^{j}}{k_{j}^{2}}-i\sum_{j=1}^{2}\frac{\beta_{21}^{j}}{k_{j}^{3}},$
(3.101) $\displaystyle
f_{21}=\frac{i}{2}\delta_{1}^{2}\sum_{j=1}^{2}\frac{\beta_{12}^{j}}{k_{j}}-\delta_{1}\sum_{j=1}^{2}\frac{\beta_{12}^{j}}{k_{j}^{2}}+i\sum_{j=1}^{2}\frac{\beta_{12}^{j}}{k_{j}^{3}}.$
(3.102)

Therefore, by the reconstruction formula (2.54), we conclude that

$\displaystyle u(x,t)$
$\displaystyle=\hat{u}(y(x,t),t)=\widehat{M}_{1,1}\widehat{M}_{2,1}(y(x,t),t)+\widehat{M}_{1,2}(y(x,t),t)+\widehat{M}_{2,2}(y(x,t),t)$
(3.103) $\displaystyle=\hat{f}t^{-1/2}+\mathcal{O}(t^{-1+\frac{1}{2p}}),$

where

$\hat{f}=-\frac{\rho_{0}^{\frac{3}{2}}}{2}\left(i\delta_{1}(f_{12}-f_{11})+f_{21}+f_{22}\right),$
(3.104)

and

$x(y,t)=y-\frac{1}{2i}\widehat{M}_{1,1}(y,t)=y-\frac{1}{2}\delta_{1}+\frac{\rho_{0}^{3/2}}{2}f_{11}t^{-\frac{1}{2}}+\mathcal{O}(t^{-1+\frac{1}{2p}}).$
(3.105)

##  4 Asymptotic analysis in the region $y/t>0$

$\mathrm{Re}k$$\mathrm{Im}k$$0$$1$$\Sigma_{1}$$\Sigma_{2}$$\Sigma_{3}$$\Sigma_{4}$$\Omega_{1}$$\Omega_{4}$$\Omega_{3}$$\Omega_{2}$$+$$-$
Figure 6: The signature of $\mathrm{Im}\theta$ in the case $\xi>0$ and the
jump lines $\Sigma_{j}$

In this section, we investigate the long-time asymptotic behavior in the
region $y/t>0$. In this case, there is no stationary point, and the jump
matrix $\widehat{V}(k)$ takes the following decomposition:

$\widehat{V}(k)=\left(\begin{array}[]{cc}1&-\bar{r}(k)e^{-2it\theta}\\\
0&1\end{array}\right)\left(\begin{array}[]{cc}1&0\\\
r(k)e^{2it\theta}&1\end{array}\right).$ (4.1)

Considering the reconstruction formula of $u$, we choose to open the jump line
at $k=1$. Define several regions and lines as shown in Figure 6:

$\displaystyle\Omega_{2n+1}=\\{k\in\mathbb{C}|n\pi\leq\arg(k-1)\leq
n\pi+\frac{n\pi}{4}\\},$ (4.2)
$\displaystyle\Omega_{2n+2}=\\{k\in\mathbb{C}|(n+1)\pi-\frac{n\pi}{4}\leq\arg(k-1)\leq(n+1)\pi\\},$
(4.3)

where $n=0,1$;

$\Sigma_{j}=e^{\frac{(j-1)i\pi}{2}+\frac{i\pi}{4}}(\Gamma_{+}),\quad\Gamma_{+}=\\{k\in\mathbb{R}|k\geq
1\\},\quad j=1,2,3,4,$ (4.4)

and denote $\Sigma^{(1)}:=\bigcup_{j=1}^{4}\Sigma_{j}$.

###  4.1 A mixed $\bar{\partial}$-RH problem and its decomposition

In this part, we open the jump line at $k=1$ to make a continuous extension,
and the first step is to introduce several new functions:

######  Proposition 8.

Define functions $R_{j}\equiv
R_{j}(k):\bar{\Omega}_{j}\rightarrow\mathbb{C},j=1,2,3,4$ with the following
boundary values:

$\displaystyle R_{1}=\left\\{\begin{array}[]{ll}-r(k),&k\in\Gamma_{+},\\\
-r(1),&k\in\Sigma_{1},\end{array}\right.,\qquad
R_{2}=\left\\{\begin{array}[]{ll}-r(k),&k\in\mathbb{R}\setminus\Gamma_{+},\\\
-r(1),&k\in\Sigma_{2},\end{array}\right.$ (4.9) $\displaystyle
R_{3}=\left\\{\begin{array}[]{ll}-\bar{r}(k),&k\in\mathbb{R}\setminus\Gamma_{+},\\\
-\bar{r}(1),&k\in\Sigma_{3},\end{array}\right.,\quad
R_{4}=\left\\{\begin{array}[]{ll}-\bar{r}(k),&k\in\Gamma_{+},\\\
-\bar{r}(1),&k\in\Sigma_{4},\end{array}\right.$ (4.14)

And $R_{j}$ satisfies the following estimates:

$|\bar{\partial}R_{j}|\lesssim|r^{\prime}(\mathrm{Re}k)|+|k-1|^{-1/2},\ for\
k\in\Omega_{j},\ j=1,4;$ (4.15)
$|\bar{\partial}R_{j}|\lesssim|r^{\prime}(\mathrm{Re}k)|+|k-1|^{-1/2}+\varphi(\mathrm{Re}k),for\
k\in\Omega_{j},\ j=2,3,$ (4.16)

where $\varphi\in C^{\infty}_{0}(\mathbb{R},[0,1])$ is a fixed cut-off
function with support near 0.

In a small fixed neighborhood of 0, $for\ k\in\Omega_{j},\ j=2,3,\ $we have

$|\bar{\partial}R_{j}|\lesssim|k|^{2}\rightarrow 0,\ as\ k\rightarrow 0.$
(4.17)

###### Proof.

We first give the proof of (4.16) and take $R_{2}(k)$ for example. Here we
take the cut-off function $\chi_{0}(x)\in\ C^{\infty}_{0}([0,1])$ as
$\rho_{0}=1$:

$\chi_{0}(x)=\left\\{\begin{array}[]{ll}1,&|x|\leqslant 1/8,\\\
0,&|x|\geqslant 1/4.\end{array}\right.$ (4.18)

Define the function $R_{2}$ as follows:

$R_{2}=R_{21}+R_{22},$ (4.19)

where

$\displaystyle R_{21}=$
$\displaystyle\left(r(1)-[1-\chi_{0}(\mathrm{Re}k)]r(\mathrm{Re}k)\right)cos^{2}(2\alpha)-r(1),$
(4.20) $\displaystyle R_{22}=$
$\displaystyle\tilde{f}(\mathrm{Re}k)cos^{2}(2\alpha)+\frac{i}{2}\rho
e^{-i\alpha}sin(2\alpha)cos(2\alpha)\chi_{0}(\alpha)\tilde{f}^{\prime}(\mathrm{Re}k)+sin^{2}(2\alpha)\chi_{0}(\alpha)\tilde{f}(\mathrm{Re}k)$
$\displaystyle-\frac{1}{4}\rho
e^{-i\alpha}sin^{2}(2\alpha)\tilde{f}^{\prime}(\mathrm{Re}k)-\frac{1}{8}\rho^{2}e^{-2i\alpha}sin^{2}(2\alpha)\tilde{f}^{\prime\prime}(\mathrm{Re}k),$

and

$\tilde{f}(\mathrm{Re}k)=-\chi_{0}(\mathrm{Re}k)r(\mathrm{Re}k).$ (4.21)

Denote $k=1+\rho e^{i\alpha}$, then $\rho=|k-1|$ and
$\bar{\partial}=\frac{1}{2}e^{i\alpha}(\partial_{\rho}+i\rho^{-1}\partial_{\alpha}).$
So we have

$\displaystyle\bar{\partial}R_{21}=$
$\displaystyle\frac{1}{2}\chi_{0}^{\prime}(\mathrm{Re}k)r(\mathrm{Re}k)cos^{2}(2\alpha)-2ie^{i\alpha}\rho^{-1}\chi_{0}(\mathrm{Re}k)r(\mathrm{Re}k)sin(2\alpha)cos(2\alpha)$
(4.22)
$\displaystyle-\frac{1}{2}(1-\chi_{0}(\mathrm{Re}k))r^{\prime}(\mathrm{Re}k)cos^{2}(2\alpha)+2ie^{i\alpha}\rho^{-1}(r(\mathrm{Re}k)-r(1))sin(2\alpha)cos(2\alpha).$

Considering that $r(\mathrm{Re}k),\ \rho^{-1}$ is bounded in the support of
$\chi_{0}(\mathrm{Re}k)$, the first two terms of (4.22) are controlled by a
function $\varphi\in C^{\infty}_{0}(\mathbb{R},[0,1])$. The last term of
(4.22) is estimated based on the inequality:

$|r(\mathrm{Re}k)-r(1)|\lesssim|\mathrm{Re}k-1|^{1/2}\leqslant\rho^{1/2},$
(4.23)

and then

$|\bar{\partial}R_{21}|\lesssim\varphi(\mathrm{Re}k)+|r^{\prime}(\mathrm{Re}k)|+|k-1|^{-1/2}.$
(4.24)

As for $R_{22}$, it’s obvious that it has the same formula as $R_{11,2}$ in
the proof of Proposition 3. We can similarly get the following results:

$|\bar{\partial}R_{11,2}|\lesssim\varphi(\mathrm{Re}k),\
k\in\Omega_{11},\quad|\bar{\partial}R_{11,2}|\lesssim|k|^{2},\quad as\ k\to
0.$ (4.25)

Combining all above, we derive (4.16) and (4.17).

Finally, we give the proof of (4.15). Define $R_{1}$ by

$R_{1}(k)=\left(r(1)-r(\mathrm{Re}k)\right)cos(2\alpha)-r(1),$ (4.26)

and $\bar{\partial}R_{1}$ is estimated by the similar way as
$\bar{\partial}R_{21}$. ∎

Now we introduce a matrix function by the following transformation

$M^{(1)}(k)\equiv M^{(1)}(y,t,k)=\widehat{M}(k)R^{(1)}(k),$ (4.27)

where

$R^{(1)}(k)=\left\\{\begin{array}[]{ll}\left(\begin{array}[]{cc}1&0\\\
R_{j}e^{2it\theta}&1\end{array}\right),&k\in\Omega_{1}\cup\Omega_{2};\\\
\left(\begin{array}[]{cc}1&R_{j}e^{-2it\theta}\\\
0&1\end{array}\right),&k\in\Omega_{3}\cup\Omega_{4};\\\
I,&elsewhere.\end{array}\right.$ (4.28)

Then $M^{(1)}(k)$ is the solution of a mixed $\bar{\partial}$-RH problem as
follows:

######  RH Problem 10.

Find a $2\times 2$ matrix-valued function $M^{(1)}(k)$ admitting the following
properties:

  * •

Analyticity: $M^{(1)}(k)$ is continuous in $\mathbb{C}\setminus\Sigma^{(1)}$;

  * •

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

$M^{(1)}_{+}(k)=M^{(1)}_{-}(k)V^{(1)}(k),\quad k\in\Sigma^{(1)},$ (4.29)

where

$V^{(1)}(k)=\left\\{\begin{array}[]{ll}(R^{(1)}(k))^{-1},&k\in\Sigma_{1}\cup\Sigma_{2},\\\
R^{(1)}(k),&k\in\Sigma_{3}\cup\Sigma_{3};\end{array}\right.$ (4.30)

  * •

Asymptotic behavior: $M^{(1)}(k)\sim I+\mathcal{O}(k^{-1}),\quad as\
k\to\infty;$

  * •

$\bar{\partial}$\- derivative:

$\bar{\partial}M^{(1)}(k)=M^{(1)}(k)\bar{\partial}R^{(1)}(k).$ (4.31)

To solve $M^{(1)}(k)$, we decompose it into a model RH problem for
$M^{R}(k)\equiv M^{R}(y,t,k)$ with $\bar{\partial}M^{R}(k)\equiv 0$ and a pure
$\bar{\partial}$\- Problem. First we give a RH problem for $M^{R}(k)$:

######  RH Problem 11.

Find a $2\times 2$ matrix-valued function $M^{R}(k)$ satisfying the properties
as follows:

  * •

Analyticity: $M^{R}(k)$ is analytic in $\mathbb{C}\setminus\Sigma^{(1)}$;

  * •

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

$M^{R}_{+}(k)=M^{R}_{-}(k)-V^{(1)}(k),\quad k\in\Sigma^{(1)};$ (4.32)

  * •

Asymptotic behavior: $M^{R}(k)\sim I+\mathcal{O}(k^{-1}),\quad as\
k\to\infty.$

Then, we define $M^{(2)}(k)\equiv M^{(2)}(y,t,k)$ as

$M^{(2)}(k)=M^{(1)}(k)M^{R}(k)^{-1}.$ (4.33)

$M^{(2)}(k)$ is solution of a new $\bar{\partial}$-problem as follows:

######  $\bar{\partial}$-problem 2.

Find a $2\times 2$ matrix-valued function $M^{(2)}(k)$ with the following
properties:

  * •

Analyticity: $M^{(2)}(k)$ is continuous in $\mathbb{C}$.

  * •

Asymptotic behavior: $M^{(2)}(k)\rightarrow I,\quad k\rightarrow\infty;$

  * •

$\bar{\partial}$-Derivative:

$\bar{\partial}M^{(2)}(k)=M^{(2)}(k)W(k),\quad k\in\mathbb{C},$ (4.34)

where

$W(k)=M^{R}(k)\bar{\partial}R^{(1)}(k)(M^{R}(k))^{-1}.$ (4.35)

###  4.2 Analysis on the pure RH problem

We first give an estimate for the jump matrix of $M^{R}(k)$ as follows:

######  Proposition 9.

As $t\rightarrow\infty$, we have

$\|V^{(1)}(k)-I\|_{L^{2}(\Sigma^{(1)})}=O(t^{-1/2}).$ (4.36)

###### Proof.

Take the case in $\Sigma_{1}$ for example:

$\|V^{(1)}(k)-I\|_{L^{2}(\Sigma_{1})}=\left(\int_{\Sigma_{1}}|r(1)e^{2it\theta}|^{2}\mathrm{d}s\right)^{1/2}\lesssim\left(\int_{0}^{+\infty}(e^{-ctl})^{2}\mathrm{d}l\right)^{1/2}\lesssim
t^{-1/2},$ (4.37)

where $s=1+le^{\frac{3i}{4}},\ \mathrm{Im}s=\frac{\sqrt{2}}{2}l.$ ∎

To solve the RH problem $M^{R}(k)$, we introduce an operator
$\mathcal{C}_{\omega}:\ L^{2}(\Sigma^{(1)})\to L^{2}(\Sigma^{(1)})$ as:

$\mathcal{C}_{\omega}f=\mathcal{C}_{-}(f(V^{(1)}-I)),$ (4.38)

where $\mathcal{C}_{-}$ is the Cauchy projection operator on $\Sigma^{(1)}$.
Based on Proposition 9, we have

$\|\mathcal{C}_{\omega_{e}}\|_{L^{2}}\leqslant\|\mathcal{C}_{-}\|_{L^{2}}\|V^{(1)}-I\|_{L^{2}}\lesssim
t^{-1/2},$ (4.39)

thus there exists $\mu(k)$ satisfying

$(I-\mathcal{C}_{\omega_{e}})\mu(k)=I.$ (4.40)

Therefore, there exists a unique solution $M^{R}(k)$ for RH problem 11
represented by

$M^{R}(k)=I+\frac{1}{2\pi
i}\int_{\Sigma^{(1)}}\frac{\mu(s)(V^{(1)}(s)-I)}{s-k}\mathrm{d}s$ (4.41)

with the following property:

######  Proposition 10.

$M^{R}(k)$ has the asymptotic expansion as $k\to 0$:

$M^{R}(k)=M^{R}(0)+M^{R}_{1}k+M^{R}_{2}k^{2}+\mathcal{O}(k^{3}),$ (4.42)

where

$M^{R}_{1}=\frac{1}{2\pi
i}\int_{\Sigma^{(1)}}\frac{\mu(s)(V^{(1)}-I)}{s^{2}}\mathrm{d}s,\quad
M^{R}_{2}=\frac{1}{2\pi
i}\int_{\Sigma^{(1)}}\frac{\mu(s)(V^{(1)}-I)}{s^{3}}\mathrm{d}s.$ (4.43)

Besides, they satisfy the estimates as follows:

$|M^{R}(0)-I|\lesssim t^{-1/2},\ |M^{R}_{j}|\lesssim t^{-1/2},\ j=1,2.$ (4.44)

###### Proof.

Considering that $s^{-1}$ is bounded on $\Sigma^{(1)}$, the results are
obvious, so we omit the proof. ∎

###  4.3 Analysis on the pure $\bar{\partial}$-problem

In this part, we deal with the pure $\bar{\partial}$-problem $M^{(2)}(k)$,
which is presented by

$M^{(2)}(k)=I+\frac{1}{\pi}\iint_{\mathbb{C}}\frac{M^{(2)}(s)W(s)}{s-k}\mathrm{d}A(s),$
(4.45)

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

$S[f](k)=\frac{1}{\pi}\iint_{\mathbb{C}}\frac{f(s)W(s)}{s-k}\mathrm{d}A(s),$
(4.46)

then $(I-S)M^{(2)}(k)=I.$ We have an estimate of $S$ as follows:

######  Proposition 11.

As $t\rightarrow\infty$, the norm of the integral operator S decays to zero:

$\|S\|_{L^{\infty}\to L^{\infty}}\lesssim t^{-1/2},$ (4.47)

which implies that $(I-S)^{-1}$ exists.

###### Proof.

The proof is similar as Proposition 6. ∎

Therefore, $M^{(2)}(k)$ exists uniquely and satisfies the following
Proposition:

######  Proposition 12.

As $k\rightarrow 0$, $M^{(2)}(k)$ has the following asymptotic expansion:

$M^{(2)}(k)=M^{(2)}(0)+M^{(2)}_{1}k+M^{(2)}_{2}k^{2}+O(k^{3}),$ (4.48)

where

$|M^{(2)}(0)-I|=|\frac{1}{\pi}\iint_{\mathbb{C}}\frac{M^{(2)}(s)W(s)}{s}\mathrm{d}A(s)|\lesssim
t^{-1/2},$ (4.49)

and

$\displaystyle
M^{(2)}_{1}=-\frac{1}{\pi}\iint_{\mathbb{C}}\frac{M^{(2)}(s)W(s)}{s^{2}}\mathrm{d}A(s),$
(4.50) $\displaystyle
M^{(2)}_{2}=-\frac{1}{\pi}\iint_{\mathbb{C}}\frac{M^{(2)}(s)W(s)}{s^{3}}\mathrm{d}A(s)$
(4.51)

satisfying

$|M^{(2)}_{j}|\lesssim t^{-1/2},\ as\ t\rightarrow\infty,\ for\ j=1,2.$ (4.52)

###### Proof.

The proof is similar as Proposition 7. ∎

###  4.4 Long-time asymptotics for the HS equation

Now we give the asymptotic expression of the solution $\widehat{M}(k)$ in the
case $\xi>0$: as $k\to 0$,

$\displaystyle\widehat{M}(k)=$ $\displaystyle
M^{(2)}(k)M^{R}(k)(R^{(1)})^{-1}(M^{(2)}(0)+M^{(2)}_{1}k+M^{(2)}_{2}k^{2})(M^{R}(0)+M^{R}_{1}k+M^{R}_{2}k^{2})+\mathcal{O}(k^{3})$
(4.53) $\displaystyle=$
$\displaystyle(I+\mathcal{O}(t^{-1/2}))(I+\mathcal{O}(t^{-1/2}))+\mathcal{O}(k^{3})=I+\mathcal{O}(t^{-1/2})+\mathcal{O}(k^{3}),$

which together with the reconstruction formula (2.54) gives the asymptotic
behavior of the solution $u(x,t)$ of the HS equation by

$u(x,t)=u(y(x,t),t)=\mathcal{O}(t^{-1/2}),\quad t\to\infty,$ (4.54)

and

$x(y,t)=y-\frac{1}{2i}\widehat{M}_{1,1}(y,t)=y+\mathcal{O}(t^{-1/2}),\quad
t\to\infty.$ (4.55)

In conclusion, we get the main result of this paper as shown in Theorem 1.

Acknowledgements

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

Data Availability Statements

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

Conflict of Interest

The authors have no conflicts to disclose.

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