# Long-time asymptotic behavior for an extended modified Korteweg-de Vries
equation

Nan Liua,111Corresponding author., Boling Guoa, Deng-Shan Wangb, Yufeng Wangc
aInstitute of Applied Physics and Computational Mathematics, Beijing 100088,
P.R. China
bSchool of Applied Science, Beijing Information Science and Technology
University, Beijing 100192, P.R. China
cCollege of Science, Minzu University of China, Beijing 100081, P.R. China
000E-mail address<EMAIL_ADDRESS>(N. Liu).

> Abstract. We investigate an integrable extended modified Korteweg-de Vries
> equation on the line with the initial value belonging to the Schwartz space.
> By performing the nonlinear steepest descent analysis of an associated
> matrix Riemann–Hilbert problem, we obtain the explicit leading-order
> asymptotics of the solution of this initial value problem as time $t$ goes
> to infinity. For a special case $\alpha=0$, we present the asymptotic
> formula of the solution to the extended modified Korteweg-de Vries equation
> in region $\mathcal{P}=\\{(x,t)\in{\mathbb{R}}^{2}|0<x\leq
> Mt^{\frac{1}{5}},t\geq 3\\}$ in terms of the solution of a fourth order
> Painlevé II equation.
>
> Keywords: Extended modified Korteweg-de Vries equation; Riemann–Hilbert
> problem; Nonlinear steepest descent method; Long-time asymptotics.

##  1 Introduction

It is a well-known fact that the modified Korteweg-de Vries (mKdV) equation is
a fundamental completely integrable model in solitary waves theory, and is
given in canonical form as

$u_{t}+6\sigma u^{2}u_{x}+u_{xxx}=0,$ (1.1)

where $\sigma=\pm 1$, $u=u(x,t)$ is a real function with evolution variable
$t$ and transverse variable $x$. This equation gives rise to multiple soliton
solutions and multiple singular soliton solutions for $\sigma=+1$ and
$\sigma=-1$, respectively. Moreover, the mKdV equation has significant
applications in various physical contexts such as the generation of
supercontinuum in optical fibres, acoustic waves in certain anharmonic
lattices, nonlinear Alfvén waves propagating in plasma and fluid dynamics.

In this paper, we investigate an extended modified Korteweg-de Vries (emKdV)
equation [26], which takes the form

$u_{t}+\alpha(6u^{2}u_{x}+u_{xxx})+\beta(30u^{4}u_{x}+10u_{x}^{3}+40uu_{x}u_{xx}+10u^{2}u_{xxx}+u_{xxxxx})=0,$
(1.2)

where $\alpha>0$ and $\beta>0$ stand for the third- and fifth-order dispersion
coefficients matching with the relevant nonlinear terms, respectively.
Moreover, (1.2) also has certain application for the description of nonlinear
internal waves in a fluid stratified by both density and current [16, 25].
Equation (1.2) is integrable, the infinitely many conservation laws have been
constructed based on the Lax pair, meanwhile, periodic and rational solutions
have been also obtained by means of the $N$-fold Darboux transformation in a
recent paper [26]. The Painlevé test and multi-soliton solutions via the
simplified Hirota direct method for equation (1.2) have been recently studied
in [28]. However, it is noted that the long-time asymptotics for the emKdV
equation (1.2) on the line were not analyzed to the best of our knowledge.

In particular, the purpose of present paper is to consider the initial-value
problem (IVP) for the emKdV equation (1.2) on the line by a Riemann–Hilbert
(RH) approach. Assuming that the initial data $u(x,0)=u_{0}(x)$ are smooth and
decay sufficiently fast as $|x|\rightarrow\infty$, that is,
$u_{0}(x)\in\mathcal{S}({\mathbb{R}})$, one then can show that the solution
$u(x,t)$ of the IVP for (1.2) can be represented in terms of the solution of a
$2\times 2$ matrix RH problem formulated in the complex $k$-plane with the
jump matrices given in terms of two spectral functions $a(k)$, $b(k)$ obtained
from the initial value $u_{0}(x)$. Then, this representation obtained allows
us to apply the nonlinear steepest descent method for the associated RH
problem and to obtain a detailed description for the leading term of the
asymptotics of the solution for the Cauchy problem.

The nonlinear steepest descent method was first introduced in 1993 by Deift
and Zhou [10], where they derived the long-time asymptotics for the IVP for
the mKdV equation (1.1) with $\sigma=-1$. It then turns out to be very
successful for analyzing the long-time asymptotics of IVPs for a large range
of nonlinear integrable evolution equations in a rigorous and transparent
form. Numerous new significant results about the asymptotics theory of
initial-value and initial-boundary value problems for different completely
integrable nonlinear equations were obtained based on the analysis of the
corresponding RH problems [2, 7, 6, 8, 29, 11, 30, 19, 31, 21, 18, 17, 24, 15,
3, 1, 27].

Developing and extending the methods used in [10, 22], our goal here is to
explore the long-time asymptotics of the solution $u(x,t)$ for the emKdV
equation (1.2) on the line. Compared with other integrable equations, the
long-time asymptotic analysis for (1.2) presents some distinctive features.
For example, the spectral curve of emKdV equation (1.2) is more involved and
it possesses four stationary points, which is different from that of mKdV
equation and Hirota equation considered in [10, 19] where the phase function
has only two critical points. We note that in the case of the Camassa–Holm
equation [5, 4], there is a sector $-\frac{1}{4}+C<c=\frac{x}{\aleph t}<0$
where the corresponding phase function also has four stationary points.
Moreover, in the case of the Degasperis–Procesi equation [6], depending on the
range of $x/t$, one can also have four stationary points. However, our main
asymptotic analysis still presents many particular pictures different from
these literatures (see Sections 3 and 4). Therefore, the study of the long-
time asymptotics for the IVP for (1.2) on the line is more interesting. Our
main results of this paper are summarized by the following theorems.

###### Theorem 1.1

Suppose that $u_{0}(x)$ lie in the Schwartz space $S({\mathbb{R}})$ and be
such that no discrete spectrum is present. Then, for any positive constant
$\varepsilon>0$, as $t\rightarrow\infty$, the solution $u(x,t)$ of the Cauchy
problem for emKdV equation (1.2) on the line satisfies the following
asymptotic formula

$u(x,t)=-\frac{u_{as}(x,t)}{\sqrt{t}}+O\bigg{(}\frac{\ln t}{t}\bigg{)},\quad
t\rightarrow\infty,~{}\xi=\frac{x}{t}\in\bigg{(}-\frac{9\alpha^{2}}{20\beta}+\varepsilon,-\varepsilon\bigg{)},$
(1.3)

where the error term is uniform with respect to $x$ in the given range, and
the leading-order coefficient $u_{as}(x,t)$ is given by

$\displaystyle u_{as}(x,t)$
$\displaystyle=\sqrt{\frac{\nu(k_{1})}{k_{1}(3\alpha-40\beta
k_{1}^{2})}}\cos\bigg{(}16tk_{1}^{3}(8\beta k_{1}^{2}-\alpha)$ (1.4)
$\displaystyle-\nu(k_{1})\ln(16t(k_{2}-k_{1})^{2}(3\alpha k_{1}-40\beta
k_{1}^{3}))+\phi_{a}(\xi)\bigg{)}$
$\displaystyle+\sqrt{\frac{\nu(k_{2})}{k_{2}(40\beta
k_{2}^{2}-3\alpha)}}\cos\bigg{(}16tk_{2}^{3}(8\beta k_{2}^{2}-\alpha)$
$\displaystyle+\nu(k_{2})\ln(16t(k_{2}-k_{1})^{2}(40\beta k_{2}^{3}-3\alpha
k_{2}))+\phi_{b}(\xi)\bigg{)},$

where

$\displaystyle\phi_{a}(\xi)$ $\displaystyle=$
$\displaystyle-\frac{\pi}{4}-\arg
r(k_{1})+\arg\Gamma(\text{i}\nu(k_{1}))+2\nu(k_{1})\ln\bigg{(}\frac{k_{1}+k_{2}}{2k_{1}}\bigg{)}$
$\displaystyle-\frac{1}{\pi}\int_{k_{1}}^{k_{2}}\ln\bigg{(}\frac{1+|r(s)|^{2}}{1+|r(k_{1})|^{2}}\bigg{)}\bigg{(}\frac{1}{s-k_{1}}-\frac{1}{s+k_{1}}\bigg{)}\text{d}s,$
$\displaystyle\phi_{b}(\xi)$ $\displaystyle=$ $\displaystyle\frac{\pi}{4}-\arg
r(k_{2})-\arg\Gamma(\text{i}\nu(k_{2}))+2\nu(k_{2})\ln\bigg{(}\frac{2k_{2}}{k_{1}+k_{2}}\bigg{)}$
$\displaystyle-\frac{1}{\pi}\int_{k_{1}}^{k_{2}}\ln\bigg{(}\frac{1+|r(s)|^{2}}{1+|r(k_{2})|^{2}}\bigg{)}\bigg{(}\frac{1}{s-k_{2}}-\frac{1}{s+k_{2}}\bigg{)}\text{d}s,$

and $k_{1},~{}k_{2}$, $\nu(k_{1}),~{}\nu(k_{2})$ are defined by (3.6), (3.7),
(3.43) and (3.45), respectively.

###### Remark 1.1

For $\xi<-\frac{9\alpha^{2}}{20\beta}$, there are no real critical points for
the phase function $\Phi(k)$. Thus, it is easy to proof that the solution
$u(x,t)$ of emKdV equation (1.2) is rapidly decreasing as
$t\rightarrow\infty$. However, for $\xi>0$, there are two different real
stationary points $\pm
k_{0}=\pm\sqrt{\frac{3\alpha}{40\beta}\bigg{(}1+\sqrt{1+\frac{20\beta\xi}{9\alpha^{2}}}\bigg{)}}$,
this implies that it is possible to deform the RH problem through a series of
transformations in exactly the same way as in the similarity region for the
mKdV equation to find the asymptotics (one also can follow the strategy used
in Section 3).

###### Theorem 1.2

Under the assumptions of Theorem 1.1, the solution $u(x,t)$ of equation (4.1),
i.e., $\alpha=0$ in emKdV equation (1.2), satisfies the following asymptotic
formula as $t\rightarrow\infty$:

$u(x,t)=\bigg{(}\frac{8}{5\beta
t}\bigg{)}^{\frac{1}{5}}u_{p}\bigg{(}\frac{-x}{(20\beta
t)^{\frac{1}{5}}}\bigg{)}+O(t^{-\frac{2}{5}}),\quad 0<x\leq Mt^{\frac{1}{5}},$
(1.5)

where the formula holds uniformly with respect to $x$ in the given range for
any fixed $M>1$ and the function $u_{p}(y)$ denotes the solution of the fourth
order Painlevé II equation (A.5).

The organization of this paper is as follows. In Section 2, we show how the
solution of emKdV equation (1.2) can be expressed in terms of the solution of
a $2\times 2$ matrix RH problem and give an auxiliary theorem which is useful
for determining the long-time asymptotics. In Section 3, we derive the long-
time asymptotic behavior of the solution of the emKdV equation (1.2) to prove
our first main Theorem 1.1 in physically interesting region. In Section 4, we
present the asymptotic formula of the solution to a particular case $\alpha=0$
of emKdV equation (1.2) in region $0<x\leq Mt^{\frac{1}{5}}$. A few facts
related to the RH problem associated with the fourth order Painlevé II
equation are collected in Appendix.

##  2 Preliminaries

###  2.1 Riemann–Hilbert formalism

The Lax pair of equation (1.2) is [26]

$\displaystyle\Psi_{x}$ $\displaystyle=X\Psi,\quad X=\text{i}k\sigma_{3}+U,$
(2.1) $\displaystyle\Psi_{t}$ $\displaystyle=T\Psi,~{}\quad
T=(-16\text{i}\beta k^{5}+4\text{i}\alpha k^{3})\sigma_{3}+V,$

(namely, equation (1.2) is the compatibility condition $X_{t}-T_{x}+[X,T]=0$
of equation (2.1)), where $\Psi(x,t;k)$ is a $2\times 2$ matrix-valued
function, $k\in{\mathbb{C}}$ is the spectral parameter and

$\displaystyle\sigma_{3}={\left(\begin{array}[]{cc}1&0\\\\[4.0pt] 0&-1\\\
\end{array}\right)},\quad U={\left(\begin{array}[]{cc}0&u\\\\[4.0pt] -u&0\\\
\end{array}\right)},\quad V={\left(\begin{array}[]{cc}A&B\\\\[4.0pt] C&-A\\\
\end{array}\right)},$ (2.8) $\displaystyle A=8\text{i}\beta
k^{3}u^{2}-\text{i}(6\beta u^{4}+2\alpha u^{2}+4\beta uu_{xx}-2\beta
u_{x}^{2})k,$ $\displaystyle B=-16\beta k^{4}u+8\text{i}\beta
k^{3}u_{x}+(8\beta u^{3}+4\alpha u+4\beta u_{xx})k^{2}-\text{i}(12\beta
u^{2}u_{x}+2\beta u_{xxx}+2\alpha u_{x})k$ $\displaystyle\qquad-6\beta
u^{5}-2\alpha u^{3}-10\beta u^{2}u_{xx}-10\beta uu_{x}^{2}-\beta
u_{xxxx}-\alpha u_{xx},$ $\displaystyle C=-B+16\text{i}\beta
k^{3}u_{x}-\text{i}(24\beta u^{2}u_{x}+4\alpha u_{x}+4\beta u_{xxx})k.$

Introducing a new eigenfunction $\mu(x,t;k)$ by

$\mu(x,t;k)=\Psi(x,t;k)\text{e}^{-\text{i}[kx+(-16\beta k^{5}+4\alpha
k^{3})t]\sigma_{3}},$ (2.9)

we obtain the equivalent Lax pair

$\displaystyle\mu_{x}-\text{i}k[\sigma_{3},\mu]=U\mu,$ (2.10)
$\displaystyle\mu_{t}+\text{i}(16\beta k^{5}-4\alpha
k^{3})[\sigma_{3},\mu]=V\mu.$

We now consider the spectral analysis of the $x$-part of (2.10). Define two
solutions $\mu_{1}$ and $\mu_{2}$ of the $x$-part of (2.10) by the following
Volterra integral equations

$\displaystyle\mu_{1}(x,t;k)$ $\displaystyle=$ $\displaystyle
I+\int_{-\infty}^{x}\text{e}^{\text{i}k(x-x^{\prime})\hat{\sigma}_{3}}[U(x^{\prime},t)\mu_{1}(x^{\prime},t;k)]\text{d}x^{\prime},$
(2.11) $\displaystyle\mu_{2}(x,t;k)$ $\displaystyle=$ $\displaystyle
I-\int^{\infty}_{x}\text{e}^{\text{i}k(x-x^{\prime})\hat{\sigma}_{3}}[U(x^{\prime},t)\mu_{2}(x^{\prime},t;k)]\text{d}x^{\prime},$
(2.12)

where $\hat{\sigma}_{3}$ acts on a $2\times 2$ matrix $X$ by
$\hat{\sigma}_{3}X=[\sigma_{3},X]$, and
$\text{e}^{\hat{\sigma}_{3}}=\text{e}^{\sigma_{3}}X\text{e}^{-\sigma_{3}}$. We
denote by $\mu^{(1)}$ and $\mu^{(2)}$ the columns of a $2\times 2$ matrix
$\mu=(\mu^{(1)}~{}\mu^{(2)})$. Then it follows from (2.11)-(2.12) that for all
$(x,t)$:

(i) $\det\mu_{j}=1$, $j=1,2$.

(ii) $\mu^{(1)}_{2}$ and $\mu^{(2)}_{1}$ are analytic and bounded in
$\\{k\in{\mathbb{C}}|\mbox{Im}k>0\\}$, and
$(\mu^{(1)}_{2}~{}\mu^{(2)}_{1})\rightarrow I$ as $k\rightarrow\infty.$

(iii) $\mu^{(1)}_{1}$ and $\mu^{(2)}_{2}$ are analytic and bounded in
$\\{k\in{\mathbb{C}}|\mbox{Im}k<0\\}$, and
$(\mu^{(1)}_{1}~{}\mu^{(2)}_{2})\rightarrow I$ as $k\rightarrow\infty.$

(iv) $\\{\mu_{j}\\}_{1}^{2}$ are continuous up to the real axis.

(v) Symmetry:

$\overline{\mu_{j}(x,t;\bar{k})}=\mu_{j}(x,t;-k)=\sigma_{2}\mu_{j}(x,t;k)\sigma_{2},$
(2.13)

where $\sigma_{2}$ is the second Pauli matrix,

$\displaystyle\sigma_{2}=\begin{pmatrix}0&-\text{i}\\\ \text{i}&0\\\
\end{pmatrix}.$

The symmetry relation (2.13) can be proved easily due to the symmetries of the
matrix $X$:

$\overline{X(x,t;\bar{k})}=X(x,t;-k)=\sigma_{2}X(x,t;k)\sigma_{2}.$

The solutions of the system of differential equation (2.10) must be related by
a matrix independent of $x$ and $t$, therefore,

$\displaystyle\mu_{1}(x,t;k)=\mu_{2}(x,t;k)\text{e}^{\text{i}[kx+(-16\beta
k^{5}+4\alpha k^{3})t]\hat{\sigma}_{3}}s(k),~{}\det
s(k)=1,~{}k\in{\mathbb{R}}.$ (2.14)

Evaluation at $x\rightarrow\infty,t=0$ gives

$s(k)=\lim_{x\rightarrow\infty}\text{e}^{-\text{i}kx\hat{\sigma}_{3}}\mu_{1}(x,0;k),$
(2.15)

that is,

$s(k)=I+\int_{-\infty}^{\infty}\text{e}^{-\text{i}kx\hat{\sigma}_{3}}[U(x,0)\mu_{1}(x,0;k)]\text{d}x.$
(2.16)

Due to the symmetry (2.13), the matrix-valued spectral function $s(k)$ can be
defined in terms of two scalar spectral functions $a(k)$ and $b(k)$ by

$s(k)=\begin{pmatrix}\bar{a}(k)&b(k)\\\\[4.0pt] -\bar{b}(k)&a(k)\\\
\end{pmatrix},$ (2.17)

where $\bar{a}(k)=\overline{a(\bar{k})}$ and
$\bar{b}(k)=\overline{b(\bar{k})}$ indicate the Schwartz conjugates. The
spectral functions $a(k)$ and $b(k)$ can be determined by $u_{0}(x)$ through
the solution of equation (2.16). On the other hand, $a(k)$ is analytic in the
half-plane $\\{k\in{\mathbb{C}}|\mbox{Im}k>0\\}$ and continuous in
$\\{k\in{\mathbb{C}}|\mbox{Im}k\geq 0\\}$, and $a(k)\rightarrow 1$ as
$k\rightarrow\infty$. Furthermore, $|a(k)|^{2}+|b(k)|^{2}=1$ for
$k\in{\mathbb{R}}$. Finally, $a(-k)=\bar{a}(k),~{}b(-k)=\bar{b}(k)$.

Assuming $u(x,t)$ be a solution of equation (1.2), the analytic properties of
$\mu_{j}(x,t;k)$ stated above allow us to define a piecewise meromorphic,
$2\times 2$ matrix-valued function $M(x,t;k)$ by

$M(x,t;k)=\left\\{\begin{aligned}
&\bigg{(}\frac{\mu_{2}^{(1)}(x,t;k)}{a(k)}~{}\mu_{1}^{(2)}(x,t;k)\bigg{)},\quad\text{Im}k>0,\\\
&\bigg{(}\mu_{1}^{(1)}(x,t;k)~{}\frac{\mu_{2}^{(2)}(x,t;k)}{\bar{a}(k)}\bigg{)},\quad\text{Im}k<0.\end{aligned}\right.$
(2.18)

Then, for each $x\in{\mathbb{R}}$ and $t\geq 0$, the boundary values
$M_{\pm}(x,t;k)$ of $M$ as $k$ approaches ${\mathbb{R}}$ from the sides
$\pm\text{Im}k>0$ are related as follows:

$M_{+}(x,t;k)=M_{-}(x,t;k)J(x,t;k),\quad k\in{\mathbb{R}},$ (2.19)

with

$\displaystyle
J(x,t;k)=\begin{pmatrix}1+|r(k)|^{2}~{}&\bar{r}(k)\text{e}^{-t\Phi(k)}\\\\[4.0pt]
r(k)\text{e}^{t\Phi(k)}~{}&1\end{pmatrix},$ (2.20) $\displaystyle
r(k)=\frac{\bar{b}(k)}{a(k)},\quad\Phi(k)=2\text{i}(-k\frac{x}{t}+16\beta
k^{5}-4\alpha k^{3}).$

In view of the properties of $\mu_{j}(x,t;k)$ and $s(k)$, $M(x,t;k)$ also
satisfies the following properties:
(i) Behavior at $k=\infty$:

$M(x,t;k)\rightarrow I~{}\text{as}~{}k\rightarrow\infty.$ (2.21)

(ii) Symmetry:

$\overline{M(x,t;\bar{k})}=M(x,t;-k)=\sigma_{2}M(x,t;k)\sigma_{2}.$ (2.22)

(iii) Residue conditions: Let $\\{k_{j}\\}_{1}^{N}$ be the set of zeros of
$a(k)$. We assume these zeros are finite in number, simple and no zero is
real, then $M(x,t;k)$ satisfies the following residue conditions:

$\displaystyle\text{Res}_{k=k_{j}}M^{(1)}(x,t;k)=\frac{\text{e}^{t\Phi(k_{j})}}{\dot{a}(k_{j})b(k_{j})}M^{(2)}(x,t;k_{j})=\text{i}\chi_{j}\text{e}^{t\Phi(k_{j})}M^{(2)}(x,t;k_{j}),$
(2.23)
$\displaystyle\text{Res}_{k=\bar{k}_{j}}M^{(2)}(x,t;k)=-\frac{\text{e}^{-t\Phi(\bar{k}_{j})}}{\overline{\dot{a}(k_{j})}\overline{b(k_{j})}}M^{(1)}(x,t;\bar{k}_{j})=\text{i}\bar{\chi}_{j}\text{e}^{-t\Phi(\bar{k}_{j})}M^{(1)}(x,t;\bar{k}_{j}).$

###### Theorem 2.1

Let $\\{r(k),\\{k_{j},\chi_{j}\\}_{1}^{N}\\}$ be the spectral data determined
by $u_{0}(x)$, and define $M(x,t;k)$ as the solution of the associated RH
(2.19) with the jump matrix (2.20), the normalization condition (2.21) and the
residue conditions (2.23). Then, $M(x,t;k)$ exists and is unique. Define
$u(x,t)$ in terms of $M(x,t;k)$ by

$\displaystyle u(x,t)=-2\text{i}\lim_{k\rightarrow\infty}(kM(x,t;k))_{12}.$
(2.24)

Then $u(x,t)$ solves the emKdV equation (1.2). Furthermore, $u(x,0)=u_{0}(x)$.

Proof.In the case when $a(k)$ has no zeros, the existence and uniqueness for
the solution of above RH problem is a consequence of a ‘vanishing lemma’ for
the associated RH problem with the vanishing condition at infinity
$M(k)=O(1/k)$, $k\rightarrow\infty$ (see [23] since
$J^{\dagger}(\bar{k})=J(k)$). If $a(k)$ has zeros, the singular RH problem can
be mapped to a regular one following the approach of [14]. Moreover, it
follows from standard arguments using the dressing method [13] that if $M$
solves the above RH problem and $u(x,t)$ is defined by (2.24), then $u(x,t)$
solves the emKdV equation (1.2). One observes that for $t=0$, the RH problem
reduces to that associated with $u_{0}(x)$, which yields $u(x,0)=u_{0}(x)$,
owing to the uniqueness of the solution of the RH problem.

###  2.2 A model RH problem

After the formulation of the main RH problem, the main idea of analysis of the
long-time behavior is to reduce the original RH problem to a model RH problem
which can be solved exactly. The following theorem is turned out suitable for
determining asymptotics of a class of RH problems which arise in the study of
long-time asymptotics.

Fig. 1: The contour $X=X_{1}\cup X_{2}\cup X_{3}\cup X_{4}$.

Let $X=X_{1}\cup X_{2}\cup X_{3}\cup X_{4}\subset{\mathbb{C}}$ be the cross
defined by

$\displaystyle X_{1}$
$\displaystyle=\\{l\text{e}^{\frac{\text{i}\pi}{4}}|0\leq
l<\infty\\},~{}~{}~{}X_{2}=\\{l\text{e}^{\frac{3\text{i}\pi}{4}}|0\leq
l<\infty\\},$ (2.25) $\displaystyle X_{3}$
$\displaystyle=\\{l\text{e}^{-\frac{3\text{i}\pi}{4}}|0\leq
l<\infty\\},~{}X_{4}=\\{l\text{e}^{-\frac{\text{i}\pi}{4}}|0\leq l<\infty\\},$

and oriented as in Fig. 1. Define the function
$\nu:{\mathbb{C}}\rightarrow(0,\infty)$ by
$\nu(q)=\frac{1}{2\pi}\ln(1+|q|^{2})$. We consider the following RH problems
parametrized by $q\in{\mathbb{C}}$:

$\left\\{\begin{aligned} &M^{X}_{+}(q,z)=M^{X}_{-}(q,z)J^{X}(q,z),~{}~{}z\in
X,\\\ &M^{X}(q,z)\rightarrow
I,~{}~{}~{}\qquad\qquad\qquad~{}z\rightarrow\infty,\end{aligned}\right.$
(2.26)

where the jump matrix $J^{X}(q,z)$ is defined by

$J^{X}(q,z)=\left\\{\begin{aligned} &\begin{pmatrix}1~{}&0\\\\[4.0pt]
q\text{e}^{\frac{\text{i}z^{2}}{2}}z^{2\text{i}\nu(q)}~{}&1\end{pmatrix},~{}~{}~{}\quad\qquad
z\in X_{1},\\\
&\begin{pmatrix}1~{}&-\frac{\bar{q}}{1+|q|^{2}}\text{e}^{-\frac{\text{i}z^{2}}{2}}z^{-2\text{i}\nu(q)}\\\\[4.0pt]
0~{}&1\end{pmatrix},~{}~{}z\in X_{2},\\\ &\begin{pmatrix}1~{}&0\\\\[4.0pt]
-\frac{q}{1+|q|^{2}}\text{e}^{\frac{\text{i}z^{2}}{2}}z^{2\text{i}\nu(q)}~{}&1\end{pmatrix},~{}~{}\quad
z\in X_{3},\\\
&\begin{pmatrix}1~{}&\bar{q}\text{e}^{-\frac{\text{i}z^{2}}{2}}z^{-2\text{i}\nu(q)}\\\\[4.0pt]
0~{}&1\end{pmatrix},~{}~{}~{}\qquad~{}z\in X_{4}.\end{aligned}\right.$ (2.27)

Then we have the following theorem.

###### Theorem 2.2

The RH problem (2.26) has a unique solution $M^{X}(q,z)$ for each
$q\in{\mathbb{C}}$. This solution satisfies

$M^{X}(q,z)=I-\frac{\text{i}}{z}\begin{pmatrix}0~{}&\beta^{X}(q)\\\\[4.0pt]
\overline{\beta^{X}(q)}~{}&0\end{pmatrix}+O\bigg{(}\frac{q}{z^{2}}\bigg{)},\quad
z\rightarrow\infty,~{}q\in{\mathbb{C}},$ (2.28)

where the error term is uniform with respect to $\arg z\in[0,2\pi]$ and the
function $\beta^{X}(q)$ is given by

$\beta^{X}(q)=\sqrt{\nu(q)}\text{e}^{\text{i}\big{(}\frac{\pi}{4}-\arg
q-\arg\Gamma(\text{i}\nu(q))\big{)}},\quad q\in{\mathbb{C}},$ (2.29)

where $\Gamma(\cdot)$ denotes the standard Gamma function. Moreover, for each
compact subset $\mathcal{D}$ of ${\mathbb{C}}$,

$\sup_{q\in\mathcal{D}}\sup_{z\in{\mathbb{C}}\setminus X}|M^{X}(q,z)|<\infty$
(2.30)

and

$\sup_{q\in\mathcal{D}}\sup_{z\in{\mathbb{C}}\setminus
X}\frac{|M^{X}(q,z)-I|}{|q|}<\infty.$ (2.31)

Proof.For the proof of this theorem, we refer the readers to see [10, 11, 1].

##  3 Long-time asymptotics

In this section, we aim to transform the associated original RH problem (2.21)
to a solvable RH problem and then find the explicitly asymptotic formula for
the emKdV equation (1.2). In the following analysis, we suppose that $a(k)\neq
0$ for $\\{k\in{\mathbb{C}}|\text{Im}k\geq 0\\}$ so that no discrete spectrum
is present. Namely, we consider the following RH problem

$\left\\{\begin{aligned}
&M_{+}(x,t;k)=M_{-}(x,t;k)J(x,t;k),~{}k\in{\mathbb{R}},\\\
&M(x,t;k)\rightarrow
I,~{}\qquad\qquad\qquad\qquad~{}k\rightarrow\infty,\end{aligned}\right.$ (3.1)

where the jump matrix $J(x,t;k)$ is defined by

$\displaystyle
J(x,t;k)=\begin{pmatrix}1+|r(k)|^{2}~{}&\bar{r}(k)\text{e}^{-t\Phi(k)}\\\\[4.0pt]
r(k)\text{e}^{t\Phi(k)}~{}&1\end{pmatrix},$ (3.2) $\displaystyle
r(k)=\frac{\bar{b}(k)}{a(k)},~{}\Phi(k)=2\text{i}(-k\xi+16\beta k^{5}-4\alpha
k^{3}),~{}\xi=\frac{x}{t}.$

In view of the symmetry relation in (2.13), we conclude that

$r(-k)=\overline{r(\bar{k})},\quad k\in{\mathbb{R}}.$ (3.3)

Moreover, the relation between the solution $u(x,t)$ of the emKdV equation
(1.2) and $M(x,t;k)$ is

$u(x,t)=-2\text{i}\lim_{k\rightarrow\infty}(kM(x,t;k))_{12}.$ (3.4)

The jump matrix $J$ defined in (3.2) involves the exponentials $\text{e}^{\pm
t\Phi}$, therefore, the sign structure of the quantity Re$\Phi(k)$ plays an
important role in the following analysis. In particular, we suppose

$-\frac{9\alpha^{2}}{20\beta}<\xi<0.$ (3.5)

It follows that there are four different real stationary points located at the
points where $\frac{\partial\Phi}{\partial k}=0$, namely, at

$\displaystyle\pm k_{1}$ $\displaystyle=$
$\displaystyle\pm\sqrt{\frac{3\alpha}{40\beta}\bigg{(}1-\sqrt{1+\frac{20\beta\xi}{9\alpha^{2}}}\bigg{)}},$
(3.6) $\displaystyle\pm k_{2}$ $\displaystyle=$
$\displaystyle\pm\sqrt{\frac{3\alpha}{40\beta}\bigg{(}1+\sqrt{1+\frac{20\beta\xi}{9\alpha^{2}}}\bigg{)}}.$
(3.7)

The signature table for Re$\Phi(k)$ is shown in Fig. 2.

Fig. 2: The signature table for Re$\Phi(k)$ in the complex $k$-plane.

Let $\varepsilon>0$ be given constant. We restrict our attention here to the
physically interesting region
$\xi\in\mathcal{I}=\big{(}-\frac{9\alpha^{2}}{20\beta}+\varepsilon,-\varepsilon\big{)}$.

###  3.1 Transformations of the RH problem

One goes from the original RH problem (3.1) for $M$ to the equivalent RH
problem for the new function $M^{(1)}$ defined by

$M^{(1)}(x,t;k)=M(x,t;k)\delta^{-\sigma_{3}}(k),$ (3.8)

where the complex-valued function $\delta(k)$ is given by

$\delta(k)=\exp\bigg{\\{}\frac{1}{2\pi\text{i}}\bigg{(}\int_{-k_{2}}^{-k_{1}}+\int^{k_{2}}_{k_{1}}\bigg{)}\frac{\ln(1+|r(s)|^{2})}{s-k}\text{d}s\bigg{\\}},~{}k\in{\mathbb{C}}\setminus([-k_{2},-k_{1}]\cup[k_{1},k_{2}]).$
(3.9)

###### Lemma 3.1

The function $\delta(k)$ has the following properties:

(i) $\delta(k)$ satisfies the following jump condition across the real axis
oriented from $-\infty$ to $\infty$:

$\displaystyle\delta_{+}(k)=\delta_{-}(k)(1+|r(k)|^{2}),~{}~{}k\in(-k_{2},-k_{1})\cup(k_{1},k_{2}).$

(ii) As $k\rightarrow\infty$, $\delta(k)$ satisfies the asymptotic formula

$\delta(k)=1+O(k^{-1}),\quad k\rightarrow\infty.$ (3.10)

(iii) $\delta(k)$ and $\delta^{-1}(k)$ are bounded and analytic functions of
$k\in{\mathbb{C}}\setminus([-k_{2},-k_{1}]\cup[k_{1},k_{2}])$ with continuous
boundary values on $(-k_{2},-k_{1})\cup(k_{1},k_{2})$.

(iv) $\delta(k)$ obeys the symmetry

$\delta(k)=\overline{\delta(\bar{k})}^{-1},\quad
k\in{\mathbb{C}}\setminus([-k_{2},-k_{1}]\cup[k_{1},k_{2}]).$

Then $M^{(1)}(x,t;k)$ satisfies the following RH problem

$M^{(1)}_{+}(x,t;k)=M^{(1)}_{-}(x,t;k)J^{(1)}(x,t;k),\quad k\in{\mathbb{R}},$
(3.11)

with the jump matrix
$J^{(1)}=\delta_{-}^{\sigma_{3}}J\delta_{+}^{-\sigma_{3}}$, namely,

$\displaystyle J^{(1)}(x,t;k)=\left\\{\begin{aligned}
&\begin{pmatrix}1~{}&r_{4}(k)\delta^{2}(k)\text{e}^{-t\Phi(k)}\\\\[4.0pt]
0~{}&1\\\ \end{pmatrix}\begin{pmatrix}1~{}&0\\\\[4.0pt]
r_{1}(k)\delta^{-2}(k)\text{e}^{t\Phi(k)}~{}&1\\\
\end{pmatrix},\quad|k|>k_{2},~{}|k|<k_{1},\\\
&\begin{pmatrix}1~{}&0\\\\[4.0pt]
r_{3}(k)\delta_{-}^{-2}(k)\text{e}^{t\Phi}(k)~{}&1\\\
\end{pmatrix}\begin{pmatrix}1~{}&r_{2}(k)\delta_{+}^{2}(k)\text{e}^{-t\Phi(k)}\\\\[4.0pt]
0~{}&1\\\ \end{pmatrix},\quad k_{1}<|k|<k_{2},\end{aligned}\right.$ (3.12)

where we define $\\{r_{j}(k)\\}_{1}^{4}$ by

$\displaystyle r_{1}(k)$
$\displaystyle=r(k),\qquad\qquad~{}~{}r_{2}(k)=\frac{\bar{r}(k)}{1+r(k)\bar{r}(k)},$
(3.13) $\displaystyle r_{3}(k)$
$\displaystyle=\frac{r(k)}{1+r(k)\bar{r}(k)},\quad r_{4}(k)=\bar{r}(k).$

Fig. 3: The open sets $\\{\Omega_{j}\\}_{1}^{4}$ in the complex $k$-plane.

Before processing the next deformation, we first introduce analytic
approximations of $\\{r_{j}(k)\\}_{1}^{4}$ following the idea of [22]. We
define the open subsets $\\{\Omega_{j}\\}_{1}^{4}$, as displayed in Fig. 3
such that

$\displaystyle\Omega_{1}\cup\Omega_{3}$ $\displaystyle=$
$\displaystyle\\{k\in{\mathbb{C}}|\text{Re}\Phi(k)<0\\},$
$\displaystyle\Omega_{2}\cup\Omega_{4}$ $\displaystyle=$
$\displaystyle\\{k\in{\mathbb{C}}|\text{Re}\Phi(k)>0\\}.$

###### Lemma 3.2

There exist decompositions

$r_{j}(k)=\left\\{\begin{aligned}
&r_{j,a}(x,t,k)+r_{j,r}(x,t,k),\quad|k|>k_{2},~{}|k|<k_{1},~{}k\in{\mathbb{R}},~{}j=1,4,\\\
&r_{j,a}(x,t,k)+r_{j,r}(x,t,k),\quad
k_{1}<|k|<k_{2},~{}k\in{\mathbb{R}},~{}j=2,3,\end{aligned}\right.$ (3.14)

where the functions $\\{r_{j,a},r_{j,r}\\}^{4}_{1}$ have the following
properties:

(1) For $\xi\in\mathcal{I}$ and each $t>0$, $r_{j,a}(x,t,k)$ is defined and
continuous for $k\in\bar{\Omega}_{j}$ and analytic for $\Omega_{j}$,
$j=1,2,3,4$.

(2) The functions $r_{1,a}$ and $r_{4,a}$ satisfy, for
$\xi\in\mathcal{I},~{}t>0$,

$|r_{j,a}(x,t,k)|\leq\frac{C}{1+|k|^{2}}\text{e}^{\frac{t}{4}|\text{Re}\Phi(k)|},~{}k\in\bar{\Omega}_{j}\cap\\{k\in{\mathbb{C}}||Rek|>k_{2}\\},~{}j=1,4,$
(3.15)

where the constant $C$ is independent of $\xi,k,t$.

(3) The $L^{1},L^{2}$ and $L^{\infty}$ norms of the functions
$r_{1,r}(x,t,\cdot)$ and $r_{4,r}(x,t,\cdot)$ on
$(-\infty,-k_{2})\cup(k_{2},\infty)\cup(-k_{1},k_{1})$ are $O(t^{-3/2})$ as
$t\rightarrow\infty$ uniformly with respect to $\xi\in\mathcal{I}$.

(4) The $L^{1},L^{2}$ and $L^{\infty}$ norms of the functions
$r_{2,r}(x,t,\cdot)$ and $r_{3,r}(x,t,\cdot)$ on
$(-k_{2},-k_{1})\cup(k_{1},k_{2})$ are $O(t^{-3/2})$ as $t\rightarrow\infty$
uniformly with respect to $\xi\in\mathcal{I}$.

(5) For $j=1,2,3,4$, the following symmetries hold:

$r_{j,a}(x,t,k)=\overline{r_{j,a}(x,t,-\bar{k})},~{}r_{j,r}(x,t,k)=\overline{r_{j,r}(x,t,-\bar{k})}.$
(3.16)

Proof.We first consider the decomposition of $r_{1}(k)$. Denote
$\Omega_{1}=\Omega_{1}^{1}\cup\Omega_{1}^{2}\cup\Omega_{1}^{3}$, where
$\Omega_{1}^{1}$, $\Omega_{1}^{2}$ and $\Omega_{1}^{3}$ denote the parts of
$\Omega_{1}$ in $\\{k\in{\mathbb{C}}|\text{Re}k>k_{2}\\}$,
$\\{k\in{\mathbb{C}}|\text{Re}k<-k_{2}\\}$ and the remaining part,
respectively. We first derive a decomposition of $r_{1}(k)$ in
$\Omega_{1}^{1}$, and then extend it to $\Omega_{1}^{2}$ by symmetry. Then, we
derive a decomposition of $r_{1}(k)$ in $\Omega_{1}^{3}$. Since
$u_{0}(x)\in\mathcal{S}({\mathbb{R}})$, this implies that
$r_{1}(k)=r(k)\in\mathcal{S}({\mathbb{R}})$. Then for $n=0,1,2$, we have

$\displaystyle r_{1}^{(n)}(k)$ $\displaystyle=$
$\displaystyle\frac{\text{d}^{n}}{\text{d}k^{n}}\bigg{(}\sum_{j=0}^{6}\frac{r_{1}^{(j)}(k_{2})}{j!}(k-k_{2})^{j}\bigg{)}+O((k-k_{2})^{7-n}),~{}~{}k\rightarrow
k_{2},$ (3.17)

Let

$f_{0}(k)=\sum_{j=5}^{11}\frac{a_{j}}{(k-\text{i})^{j}},$ (3.18)

where $\\{a_{j}\\}_{5}^{11}$ are complex constants such that

$f_{0}(k)=\sum_{j=0}^{6}\frac{r_{1}^{(j)}(k_{2})}{j!}(k-k_{2})^{j}+O((k-k_{2})^{7}),~{}~{}k\rightarrow
k_{2}.$ (3.19)

It is easy to verify that (3.19) imposes seven linearly independent conditions
on the $a_{j}$, hence the coefficients $a_{j}$ exist and are unique. Letting
$f=r_{1}-f_{0}$, it follows that
(i) $f_{0}(k)$ is a rational function of $k\in{\mathbb{C}}$ with no poles in
$\Omega_{1}^{1}$;
(ii) $f_{0}(k)$ coincides with $r_{1}(k)$ to six order at $k_{2}$, more
precisely,

$\frac{\text{d}^{n}}{\text{d}k^{n}}f(k)=\left\\{\begin{aligned}
&O((k-k_{2})^{7-n}),~{}k\rightarrow k_{2},\\\
&O(k^{-5-n}),~{}~{}~{}~{}~{}\quad k\rightarrow\infty,\end{aligned}\quad
k\in{\mathbb{R}},~{}n=0,1,2.\right.$ (3.20)

Fig. 4: The graph of the function $\phi(k)$ defined in (3.21).

The decomposition of $r_{1}(k)$ can be derived as follows. The map
$k\mapsto\phi=\phi(k)$ defined by

$\phi(k)=-\text{i}\Phi(k)=2(16\beta k^{5}-4\alpha k^{3}-\xi k)$ (3.21)

is a bijection $(k_{2},\infty)\mapsto(-128\beta k_{2}^{5}+16\alpha
k_{2}^{3},\infty)$ (see Fig. 4), so we may define a function $F$ by

$F(\phi)=\left\\{\begin{aligned}
&\frac{(k-\text{i})^{3}}{k-k_{2}}f(k),~{}~{}\phi>-128\beta k_{2}^{5}+16\alpha
k_{2}^{3},\\\ &0,\qquad\qquad~{}~{}~{}~{}~{}\phi\leq-128\beta
k_{2}^{5}+16\alpha k_{2}^{3}.\end{aligned}\right.$ (3.22)

Then,

$\displaystyle
F^{(n)}(\phi)=\bigg{(}\frac{1}{160\beta(k^{2}-k_{2}^{2})(k^{2}+k_{2}^{2}-\frac{3\alpha}{20\beta})}\frac{\partial}{\partial
k}\bigg{)}^{n}\bigg{(}\frac{(k-\text{i})^{3}}{k-k_{2}}f(k)\bigg{)},~{}~{}\phi>-128\beta
k_{2}^{5}+16\alpha k_{2}^{3}.$

By (3.20), $F^{(n)}(\phi)=O(|\phi|^{-3/5})$ as $|\phi|\rightarrow\infty$ for
$n=0,1,2$. In particular,

$\bigg{\|}\frac{\text{d}^{n}F}{\text{d}\phi^{n}}\bigg{\|}_{L^{2}({\mathbb{R}})}<\infty,\quad
n=0,1,2,$ (3.23)

that is, $F$ belongs to $H^{2}({\mathbb{R}})$. By the Fourier transform
$\hat{F}(s)$ defined by

$\displaystyle\hat{F}(s)=\frac{1}{2\pi}\int_{\mathbb{R}}F(\phi)\text{e}^{-\text{i}\phi
s}\text{d}\phi$

where

$F(\phi)=\int_{\mathbb{R}}\hat{F}(s)\text{e}^{\text{i}\phi s}\text{d}s,$
(3.24)

it follows from Plancherel theorem that
$\|s^{2}\hat{F}(s)\|_{L^{2}({\mathbb{R}})}<\infty$. Equations (3.22) and
(3.24) imply

$f(k)=\frac{k-k_{2}}{(k-\text{i})^{3}}\int_{\mathbb{R}}\hat{F}(s)\text{e}^{\text{i}\phi
s}\text{d}s,\quad k>k_{2}.$ (3.25)

Writing

$f(k)=f_{a}(x,t,k)+f_{r}(x,t,k),\quad t>0,~{}k>k_{2},$

where the functions $f_{a}$ and $f_{r}$ are defined by

$\displaystyle f_{a}(x,t,k)$ $\displaystyle=$
$\displaystyle\frac{k-k_{2}}{(k-\text{i})^{3}}\int_{-\frac{t}{4}}^{\infty}\hat{F}(s)\text{e}^{s\Phi(k)}\text{d}s,~{}\quad
t>0,~{}k\in\Omega^{1}_{1},$ $\displaystyle f_{r}(x,t,k)$ $\displaystyle=$
$\displaystyle\frac{k-k_{2}}{(k-\text{i})^{3}}\int^{-\frac{t}{4}}_{-\infty}\hat{F}(s)\text{e}^{s\Phi(k)}\text{d}s,\quad
t>0,~{}k>k_{2},$

we infer that $f_{a}(x,t,\cdot)$ is continuous in $\bar{\Omega}^{1}_{1}$ and
analytic in $\Omega^{1}_{1}$. Moreover, we can get

$\displaystyle|f_{a}(x,t,k)|$ $\displaystyle\leq$
$\displaystyle\frac{|k-k_{2}|}{|k-\text{i}|^{3}}\|\hat{F}(s)\|_{L^{1}({\mathbb{R}})}\sup_{s\geq-\frac{t}{4}}\text{e}^{s\text{Re}\Phi(k)}$
(3.26) $\displaystyle\leq$
$\displaystyle\frac{C|k-k_{2}|}{|k-\text{i}|^{3}}\text{e}^{\frac{t}{4}|\text{Re}\Phi(k)|},\quad
t>0,~{}k\in\bar{\Omega}^{1}_{1},~{}\xi\in\mathcal{I}.$

Furthermore, we have

$\displaystyle|f_{r}(x,t,k)|$ $\displaystyle\leq$
$\displaystyle\frac{|k-k_{2}|}{|k-\text{i}|^{3}}\int_{-\infty}^{-\frac{t}{4}}s^{2}|\hat{F}(s)|s^{-2}\text{d}s$
$\displaystyle\leq$
$\displaystyle\frac{C}{1+|k|^{2}}\|s^{2}\hat{F}(s)\|_{L^{2}({\mathbb{R}})}\sqrt{\int_{-\infty}^{-\frac{t}{4}}s^{-4}\text{d}s},$
$\displaystyle\leq$ $\displaystyle\frac{C}{1+|k|^{2}}t^{-3/2},\quad
t>0,~{}k>k_{2},~{}\xi\in\mathcal{I}.$

Hence, the $L^{1},L^{2}$ and $L^{\infty}$ norms of $f_{r}$ on $(k_{2},\infty)$
are $O(t^{-3/2})$. Letting

$\displaystyle\begin{aligned} r_{1,a}(x,t,k)&=f_{0}(k)+f_{a}(x,t,k),~{}\quad
t>0,~{}k\in\bar{\Omega}_{1}^{1},\\\
r_{1,r}(x,t,k)&=f_{r}(x,t,k),~{}~{}\qquad\qquad t>0,~{}k>k_{2}.\end{aligned}$
(3.28)

For $k<-k_{2}$, we use the symmetry (3.16) to extend this decomposition.

We next derive the decomposition of $r_{1}(k)$ for $-k_{1}<k<k_{1}$. Following
[10, 22], we split $r_{1}(k)$ into even and odd parts as follows:

$r_{1}(k)=r_{+}(k^{2})+kr_{-}(k^{2}),\quad k\in{\mathbb{R}},$ (3.29)

where $r_{\pm}:[0,\infty)\rightarrow{\mathbb{C}}$ are defined by

$\displaystyle
r_{+}(s)=\frac{r_{1}(\sqrt{s})+r_{1}(-\sqrt{s})}{2},~{}r_{-}(s)=\frac{r_{1}(\sqrt{s})-r_{1}(-\sqrt{s})}{2\sqrt{s}},~{}s\geq
0.$

We write $r_{1}(k)$ as the following form of Taylor series

$r_{1}(k)=\sum_{j=0}^{10}q_{j}k^{j}+\frac{1}{10!}\int_{0}^{k}r_{1}^{(11)}(t)(k-t)^{10}\text{d}t,~{}q_{j}=\frac{r_{1}^{(j)}(0)}{j!}.$
(3.30)

It then follows that

$\displaystyle\begin{aligned}
r_{+}(s)&=\sum_{j=0}^{5}q_{2j}s^{j}+\frac{1}{2\times
10!}\int_{0}^{\sqrt{s}}(r_{1}^{(11)}(t)-r_{1}^{(11)}(-t))(\sqrt{s}-t)^{10}\text{d}t,\\\
r_{-}(s)&=\sum_{j=0}^{4}q_{2j+1}s^{j}+\frac{1}{2\times
10!\sqrt{s}}\int_{0}^{\sqrt{s}}(r_{1}^{(11)}(t)+r_{1}^{(11)}(-t))(\sqrt{s}-t)^{10}\text{d}t.\end{aligned}$
(3.31)

Letting $\\{p_{j}^{\pm}\\}_{0}^{4}$ denote the coefficients of the Taylor
series representations

$\displaystyle
r_{\pm}(k^{2})=\sum_{j=0}^{4}p_{j}^{\pm}(k^{2}-k_{1}^{2})^{j}+\frac{1}{4!}\int_{k_{1}^{2}}^{k^{2}}r_{\pm}^{(5)}(t)(k^{2}-t)^{4}\text{d}t,$

we infer that the function $f_{0}(k)$ defined by

$f_{0}(k)=\sum_{j=0}^{4}p_{j}^{+}(k^{2}-k_{1}^{2})^{j}+k\sum_{j=0}^{4}p_{j}^{-}(k^{2}-k_{1}^{2})^{j}$
(3.32)

has the following properties:
(i) $f_{0}(k)$ is a polynomial in $k\in{\mathbb{C}}$ whose coefficients are
bounded.
(ii) The difference $f(k)=r_{1}(k)-f_{0}(k)$, which satisfies

$\frac{\text{d}^{n}}{\text{d}k^{n}}f(k)\leq
C|k^{2}-k_{1}^{2}|^{5-n},~{}-k_{1}<k<k_{1},~{}\xi\in\mathcal{I},~{}n=0,1,2,$
(3.33)

where $C$ is independent of $\xi,k$. The decomposition of $r_{1}(k)$ for
$-k_{1}<k<k_{1}$ can now be derived as follows. Since the function
$k\mapsto\phi$ defined in (3.21) is a bijection
$(-k_{1},k_{1})\rightarrow(128\beta k_{1}^{5}-16\alpha k_{1}^{3},-128\beta
k_{1}^{5}+16\alpha k_{1}^{3})$ (see Fig. 4), we may define a function
$F(\phi)$ by

$F(\phi)=\left\\{\begin{aligned}
&\frac{1}{k^{2}-k_{1}^{2}}f(k),~{}|\phi|<-128\beta k_{1}^{5}+16\alpha
k_{1}^{3},\\\ &0,\qquad\qquad\quad|\phi|\geq-128\beta k_{1}^{5}+16\alpha
k_{1}^{3}.\end{aligned}\right.$ (3.34)

Thus, we have

$F^{(n)}(\phi)=\bigg{(}\frac{1}{160\beta(k^{2}-k_{1}^{2})(k^{2}+k_{1}^{2}-\frac{3\alpha}{20\beta})}\frac{\partial}{\partial
k}\bigg{)}^{n}\frac{f(k)}{k^{2}-k_{1}^{2}},~{}|\phi|<-128\beta
k_{1}^{5}+16\alpha k_{1}^{3}.$ (3.35)

Equations (3.33) and (3.35) imply that

$\displaystyle\bigg{|}\frac{\text{d}^{n}}{\text{d}\phi^{n}}F(\phi)\bigg{|}\leq
C,~{}|\phi|<-128\beta k_{1}^{5}+16\alpha k_{1}^{3},~{}n=0,1,2.$

Therefore, $F(\phi)$ satisfies (3.23). On the other hand, (3.24) and (3.34)
imply

$(k^{2}-k_{1}^{2})\int_{\mathbb{R}}\hat{F}(s)\text{e}^{s\Phi(k)}\text{d}s=\left\\{\begin{aligned}
&f(k),\quad|k|<k_{1},\\\ &0,\qquad~{}|k|\geq k_{1}.\end{aligned}\right.$
(3.36)

Letting

$f(k)=f_{a}(x,t,k)+f_{r}(x,t,k),\quad t>0,~{}|k|k_{1},~{}\xi\in\mathcal{I},$

where the functions $f_{a}$ and $f_{r}$ are defined by

$\displaystyle f_{a}(x,t,k)$ $\displaystyle=$
$\displaystyle(k^{2}-k^{2}_{1})\int_{-\frac{t}{4}}^{\infty}\hat{F}(s)\text{e}^{s\Phi(k)}\text{d}s,~{}\quad
t>0,~{}k\in\Omega^{3}_{1},$ $\displaystyle f_{r}(x,t,k)$ $\displaystyle=$
$\displaystyle(k^{2}-k^{2}_{1})\int^{-\frac{t}{4}}_{-\infty}\hat{F}(s)\text{e}^{s\Phi(k)}\text{d}s,\quad
t>0,~{}|k|<k_{1},$

we infer that $f_{a}(x,t,\cdot)$ is continuous in $\bar{\Omega}^{3}_{1}$ and
analytic in $\Omega^{3}_{1}$. Moreover, we can get from (3.26) and (3.1) that

$\displaystyle|f_{a}(x,t,k)|$ $\displaystyle\leq$ $\displaystyle
C|k^{2}-k^{2}_{1}|\text{e}^{\frac{t}{4}|\text{Re}\Phi(k)|},\quad
t>0,~{}k\in\bar{\Omega}^{3}_{1},~{}\xi\in\mathcal{I},$
$\displaystyle|f_{r}(x,t,k)|$ $\displaystyle\leq$ $\displaystyle
Ct^{-3/2},\qquad\qquad\qquad~{}t>0,~{}|k|<k_{1},~{}\xi\in\mathcal{I}.$

It follows from

$\displaystyle r_{1,a}(x,t,k)=f_{0}(k)+f_{a}(x,t,k),\quad
r_{1,r}(x,t,k)=f_{r}(x,t,k)$

that one get a decomposition of $r_{1}(k)$ for $|k|<k_{1}$ with the properties
listed in the statement of the lemma. Thus, we find a decomposition of
$r_{1}(k)$ for $|k|>k_{2}$ and $|k|<k_{1}$ with the properties listed in the
statement of the lemma. The decomposition of the function $r_{3}(k)$ can be
obtained by a similar procedure as the decomposition of $r_{1}(k)$ for
$|k|<k_{1}$.

Finally, the decompositions of $r_{2}(k)$ and $r_{4}(k)$ can be obtained from
$r_{3}(k)$ and $r_{1}(k)$ by Schwartz conjugation.

Fig. 5: The oriented contour $\Sigma$ and the open sets $\\{D_{j}\\}_{1}^{6}$
in the complex $k$-plane.

The purpose of the next transformation is to deform the contour so that the
jump matrix involves the exponential factor $\text{e}^{-t\Phi}$ on the parts
of the contour where Re$\Phi$ is positive and the factor $\text{e}^{t\Phi}$ on
the parts where Re$\Phi$ is negative. More precisely, we put

$M^{(2)}(x,t;k)=M^{(1)}(x,t;k)G(k),$ (3.37)

where

$G(k)=\left\\{\begin{aligned} &\begin{pmatrix}1&~{}0\\\\[4.0pt]
-r_{1,a}\delta^{-2}\text{e}^{t\Phi}~{}&1\end{pmatrix},~{}~{}\quad k\in
D_{1},\\\ &\begin{pmatrix}1~{}&-r_{2,a}\delta^{2}\text{e}^{-t\Phi}\\\\[4.0pt]
0~{}&1\end{pmatrix},\qquad~{}k\in D_{2},\\\ &\begin{pmatrix}1~{}&0\\\\[4.0pt]
r_{3,a}\delta^{-2}\text{e}^{t\Phi}~{}&1\end{pmatrix},\qquad~{}~{}~{}k\in
D_{3},\\\ &\begin{pmatrix}1~{}&r_{4,a}\delta^{2}\text{e}^{-t\Phi}\\\\[4.0pt]
0~{}&1\end{pmatrix},~{}~{}~{}\qquad k\in D_{4},\\\
&I,~{}~{}~{}~{}~{}~{}~{}~{}\qquad\qquad\qquad\quad k\in D_{5}\cup
D_{6}.\end{aligned}\right.$ (3.38)

Then the matrix $M^{(2)}(x,t;k)$ satisfies the following RH problem

$M_{+}^{(2)}(x,t;k)=M_{-}^{(2)}(x,t;k)J^{(2)}(x,t;k),\quad k\in\Sigma,$ (3.39)

with the jump matrix $J^{(2)}=G_{-}^{-1}(k)J^{(1)}G_{+}(k)$ is given by

$\displaystyle J^{(2)}_{1}$ $\displaystyle=$
$\displaystyle\begin{pmatrix}1&0\\\\[4.0pt]
r_{1,a}\delta^{-2}\text{e}^{t\Phi}&1\end{pmatrix},~{}\qquad
J^{(2)}_{2}=\begin{pmatrix}1&~{}-r_{2,a}\delta^{2}\text{e}^{-t\Phi}\\\
0~{}&1\end{pmatrix},$ $\displaystyle J^{(2)}_{3}$ $\displaystyle=$
$\displaystyle\begin{pmatrix}1&~{}0\\\
-r_{3,a}\delta^{-2}\text{e}^{t\Phi}~{}&1\end{pmatrix},\quad
J^{(2)}_{4}=\begin{pmatrix}1&~{}r_{4,a}\delta^{2}\text{e}^{-t\Phi}\\\
0~{}&1\end{pmatrix},$ (3.40) $\displaystyle J^{(2)}_{5}$ $\displaystyle=$
$\displaystyle\begin{pmatrix}1~{}&r_{4,r}\delta^{2}\text{e}^{-t\Phi}\\\\[4.0pt]
0~{}&1\\\ \end{pmatrix}\begin{pmatrix}1~{}&0\\\\[4.0pt]
r_{1,r}\delta^{-2}\text{e}^{t\Phi}~{}&1\\\ \end{pmatrix},\quad
J^{(2)}_{6}=\begin{pmatrix}1~{}&0\\\\[4.0pt]
r_{3,r}\delta_{-}^{-2}\text{e}^{t\Phi}~{}&1\end{pmatrix}\begin{pmatrix}1~{}&r_{2,r}\delta_{+}^{2}\text{e}^{-t\Phi}\\\\[4.0pt]
0~{}&1\\\ \end{pmatrix},$

with $J^{(2)}_{i}$ denoting the restriction of $J^{(2)}$ to the contour
$\Sigma$ labeled by $i$ in Fig. 5. It is easy to see that the jump matrix
$J^{(2)}$ decays to identity matrix $I$ as $t\rightarrow\infty$ everywhere
except near the critical points $\pm k_{1}$ and $\pm k_{2}$. This implies that
we only need to consider a neighborhood of the critical points $\pm k_{1}$ and
$\pm k_{2}$ when we study the long-time asymptotics of $M^{(2)}(x,t;k)$ in
terms of the corresponding RH problem.

###  3.2 Local models near the critical points $\pm k_{1}$ and $\pm k_{2}$

We introduce the following scaling operators

$\displaystyle\begin{aligned} &S_{-k_{2}}:~{}k\mapsto\frac{z}{4\sqrt{t(40\beta
k_{2}^{3}-3\alpha k_{2})}}-k_{2},\\\
&S_{-k_{1}}:~{}k\mapsto\frac{z}{4\sqrt{t(3\alpha k_{1}-40\beta
k_{1}^{3})}}-k_{1},\\\ &S_{k_{1}}:~{}k\mapsto\frac{z}{4\sqrt{t(3\alpha
k_{1}-40\beta k_{1}^{3})}}+k_{1},\\\
&S_{k_{2}}:~{}k\mapsto\frac{z}{4\sqrt{t(40\beta k_{2}^{3}-3\alpha
k_{2})}}+k_{2}.\end{aligned}$ (3.41)

Integrating by parts in formula (3.9) yields,

$\displaystyle\delta(k)$
$\displaystyle=\bigg{(}\frac{k-k_{2}}{k-k_{1}}\bigg{)}^{-\text{i}\nu(k_{1})}\bigg{(}\frac{k+k_{1}}{k+k_{2}}\bigg{)}^{-\text{i}\nu(k_{1})}\text{e}^{\chi_{1}(k)}$
(3.42)
$\displaystyle=\bigg{(}\frac{k-k_{2}}{k-k_{1}}\bigg{)}^{-\text{i}\nu(k_{2})}\bigg{(}\frac{k+k_{1}}{k+k_{2}}\bigg{)}^{-\text{i}\nu(k_{2})}\text{e}^{\chi_{2}(k)},$

where

$\displaystyle\nu(k_{1})$ $\displaystyle=$
$\displaystyle\frac{1}{2\pi}\ln(1+|r(k_{1})|^{2})>0,$ (3.43)
$\displaystyle\chi_{1}(k)$ $\displaystyle=$
$\displaystyle\frac{1}{2\pi\text{i}}\int_{k_{1}}^{k_{2}}\ln\bigg{(}\frac{1+|r(s)|^{2}}{1+|r(k_{1})|^{2}}\bigg{)}\bigg{(}\frac{1}{s-k}-\frac{1}{s+k}\bigg{)}\text{d}s,$
(3.44) $\displaystyle\nu(k_{2})$ $\displaystyle=$
$\displaystyle\frac{1}{2\pi}\ln(1+|r(k_{2})|^{2})>0,$ (3.45)
$\displaystyle\chi_{2}(k)$ $\displaystyle=$
$\displaystyle\frac{1}{2\pi\text{i}}\int_{k_{1}}^{k_{2}}\ln\bigg{(}\frac{1+|r(s)|^{2}}{1+|r(k_{2})|^{2}}\bigg{)}\bigg{(}\frac{1}{s-k}-\frac{1}{s+k}\bigg{)}\text{d}s.$
(3.46)

Hence, we have

$\displaystyle S_{-k_{2}}(\delta(k)\text{e}^{-\frac{t\Phi(k)}{2}})$
$\displaystyle=$ $\displaystyle\delta_{-k_{2}}^{0}(z)\delta_{-k_{2}}^{1}(z),$
$\displaystyle S_{-k_{1}}(\delta(k)\text{e}^{-\frac{t\Phi(k)}{2}})$
$\displaystyle=$ $\displaystyle\delta_{-k_{1}}^{0}(z)\delta_{-k_{1}}^{1}(z),$
$\displaystyle S_{k_{1}}(\delta(k)\text{e}^{-\frac{t\Phi(k)}{2}})$
$\displaystyle=$ $\displaystyle\delta_{k_{1}}^{0}(z)\delta_{k_{1}}^{1}(z),$
$\displaystyle S_{k_{2}}(\delta(k)\text{e}^{-\frac{t\Phi(k)}{2}})$
$\displaystyle=$ $\displaystyle\delta_{k_{2}}^{0}(z)\delta_{k_{2}}^{1}(z),$

with

$\displaystyle\delta_{-k_{2}}^{0}(z)$ $\displaystyle=$
$\displaystyle\bigg{(}16t(k_{2}-k_{1})^{2}(40\beta k_{2}^{3}-3\alpha
k_{2})\bigg{)}^{-\frac{\text{i}\nu(k_{2})}{2}}\bigg{(}\frac{2k_{2}}{k_{1}+k_{2}}\bigg{)}^{-\text{i}\nu(k_{2})}$
(3.47)
$\displaystyle\times\text{e}^{\chi_{2}(-k_{2})}\text{e}^{-8\text{i}k_{2}^{3}t(8\beta
k_{2}^{2}-\alpha)},$ $\displaystyle\delta_{-k_{2}}^{1}(z)$ $\displaystyle=$
$\displaystyle(-z)^{\text{i}\nu(k_{2})}\text{e}^{(\chi_{2}(z/\sqrt{16tk_{2}(40\beta
k_{2}^{2}-3\alpha)}-k_{2})-\chi_{2}(-k_{2}))}$
$\displaystyle\times\bigg{(}\frac{-z/\sqrt{16tk_{2}(40\beta
k_{2}^{2}-3\alpha)}-k_{1}+k_{2}}{k_{2}-k_{1}}\bigg{)}^{-\text{i}\nu(k_{2})}$
$\displaystyle\times\bigg{(}\frac{k_{1}+k_{2}}{-z/\sqrt{16tk_{2}(40\beta
k_{2}^{2}-3\alpha)}+k_{1}+k_{2}}\bigg{)}^{-\text{i}\nu(k_{2})}$
$\displaystyle\times\bigg{(}\frac{-z/\sqrt{16tk_{2}(40\beta
k_{2}^{2}-3\alpha)}+2k_{2}}{2k_{2}}\bigg{)}^{-\text{i}\nu(k_{2})}$
$\displaystyle\times\exp\bigg{(}\frac{\text{i}z^{2}}{4}\bigg{[}1-\frac{(40\beta
k_{2}^{2}-\alpha)z}{4\sqrt{t}(40\beta k_{2}^{3}-3\alpha
k_{2})^{3/2}}+\frac{5\beta z^{2}}{4tk_{2}(40\beta k_{2}^{2}-3\alpha)^{2}}$
$\displaystyle\qquad\qquad-\frac{\beta z^{3}}{16t^{3/2}(40\beta
k_{2}^{3}-3\alpha k_{2})^{5/2}}\bigg{]}\bigg{)},$
$\displaystyle\delta_{-k_{1}}^{0}(z)$ $\displaystyle=$
$\displaystyle\bigg{(}16t(k_{2}-k_{1})^{2}(3\alpha k_{1}-40\beta
k_{1}^{3})\bigg{)}^{\frac{\text{i}\nu(k_{1})}{2}}\bigg{(}\frac{k_{1}+k_{2}}{2k_{1}}\bigg{)}^{-\text{i}\nu(k_{1})}$
(3.49)
$\displaystyle\times\text{e}^{\chi_{1}(-k_{1})}\text{e}^{-8\text{i}k_{1}^{3}t(8\beta
k_{1}^{2}-\alpha)},$ $\displaystyle\delta_{-k_{1}}^{1}(z)$ $\displaystyle=$
$\displaystyle
z^{-\text{i}\nu(k_{1})}\text{e}^{(\chi_{1}(z/\sqrt{16tk_{1}(3\alpha-40\beta
k_{1}^{2})}-k_{1})-\chi_{1}(-k_{1}))}$
$\displaystyle\times\bigg{(}\frac{k_{2}-k_{1}}{z/\sqrt{16tk_{1}(3\alpha-40\beta
k_{1}^{2})}-k_{1}+k_{2}}\bigg{)}^{-\text{i}\nu(k_{1})}$
$\displaystyle\times\bigg{(}\frac{-z/\sqrt{16tk_{1}(3\alpha-40\beta
k_{1}^{2})}+k_{1}+k_{2}}{k_{1}+k_{2}}\bigg{)}^{-\text{i}\nu(k_{1})}$
$\displaystyle\times\bigg{(}\frac{2k_{1}}{-z/\sqrt{16tk_{1}(3\alpha-40\beta
k_{1}^{2})}+2k_{1}}\bigg{)}^{-\text{i}\nu(k_{1})}$
$\displaystyle\times\exp\bigg{(}-\frac{\text{i}z^{2}}{4}\bigg{[}1+\frac{(40\beta
k_{1}^{2}-\alpha)z}{4\sqrt{t}(3\alpha k_{1}-40\beta
k_{1}^{3})^{3/2}}-\frac{5\beta z^{2}}{4tk_{1}(3\alpha-40\beta k_{1}^{2})^{2}}$
$\displaystyle\qquad\qquad+\frac{\beta z^{3}}{16t^{3/2}(3\alpha k_{1}-40\beta
k_{1}^{3})^{5/2}}\bigg{]}\bigg{)},$ $\displaystyle\delta_{k_{1}}^{0}(z)$
$\displaystyle=$ $\displaystyle\bigg{(}16t(k_{2}-k_{1})^{2}(3\alpha
k_{1}-40\beta
k_{1}^{3})\bigg{)}^{-\frac{\text{i}\nu(k_{1})}{2}}\bigg{(}\frac{2k_{1}}{k_{1}+k_{2}}\bigg{)}^{-\text{i}\nu(k_{1})}$
(3.51)
$\displaystyle\times\text{e}^{\chi_{1}(k_{1})}\text{e}^{8\text{i}k_{1}^{3}t(8\beta
k_{1}^{2}-\alpha)},$ $\displaystyle\delta_{k_{1}}^{1}(z)$ $\displaystyle=$
$\displaystyle(-z)^{\text{i}\nu(k_{1})}\text{e}^{(\chi_{1}(z/\sqrt{16tk_{1}(3\alpha-40\beta
k_{1}^{2})}+k_{1})-\chi_{1}(k_{1}))}$
$\displaystyle\times\bigg{(}\frac{-z/\sqrt{16tk_{1}(3\alpha-40\beta
k_{1}^{2})}-k_{1}+k_{2}}{k_{2}-k_{1}}\bigg{)}^{-\text{i}\nu(k_{1})}$
$\displaystyle\times\bigg{(}\frac{k_{1}+k_{2}}{z/\sqrt{16tk_{1}(3\alpha-40\beta
k_{1}^{2})}+k_{1}+k_{2}}\bigg{)}^{-\text{i}\nu(k_{1})}$
$\displaystyle\times\bigg{(}\frac{z/\sqrt{16tk_{1}(3\alpha-40\beta
k_{1}^{2})}+2k_{1}}{2k_{1}}\bigg{)}^{-\text{i}\nu(k_{1})}$
$\displaystyle\times\exp\bigg{(}\frac{\text{i}z^{2}}{4}\bigg{[}1-\frac{(40\beta
k_{1}^{2}-\alpha)z}{4\sqrt{t}(3\alpha k_{1}-40\beta
k_{1}^{3})^{3/2}}-\frac{5\beta z^{2}}{4tk_{1}(3\alpha-40\beta k_{1}^{2})^{2}}$
$\displaystyle\qquad\qquad-\frac{\beta z^{3}}{16t^{3/2}(3\alpha k_{1}-40\beta
k_{1}^{3})^{5/2}}\bigg{]}\bigg{)},$ $\displaystyle\delta_{k_{2}}^{0}(z)$
$\displaystyle=$ $\displaystyle\bigg{(}16t(k_{2}-k_{1})^{2}(40\beta
k_{2}^{3}-3\alpha
k_{2})\bigg{)}^{\frac{\text{i}\nu(k_{2})}{2}}\bigg{(}\frac{k_{1}+k_{2}}{2k_{2}}\bigg{)}^{-\text{i}\nu(k_{2})}$
(3.53)
$\displaystyle\times\text{e}^{\chi_{2}(k_{2})}\text{e}^{8\text{i}k_{2}^{3}t(8\beta
k_{2}^{2}-\alpha)},$ $\displaystyle\delta_{k_{2}}^{1}(z)$ $\displaystyle=$
$\displaystyle
z^{-\text{i}\nu(k_{2})}\text{e}^{(\chi_{2}(z/\sqrt{16tk_{2}(40\beta
k_{2}^{2}-3\alpha)}+k_{2})-\chi_{2}(k_{2}))}$
$\displaystyle\times\bigg{(}\frac{k_{2}-k_{1}}{z/\sqrt{16tk_{2}(40\beta
k_{2}^{2}-3\alpha)}-k_{1}+k_{2}}\bigg{)}^{-\text{i}\nu(k_{2})}$
$\displaystyle\times\bigg{(}\frac{2k_{2}}{z/\sqrt{16tk_{2}(40\beta
k_{2}^{2}-3\alpha)}+2k_{2}}\bigg{)}^{-\text{i}\nu(k_{2})}$
$\displaystyle\times\bigg{(}\frac{z/\sqrt{16tk_{2}(40\beta
k_{2}^{2}-3\alpha)}+k_{1}+k_{2}}{k_{1}+k_{2}}\bigg{)}^{-\text{i}\nu(k_{2})}$
$\displaystyle\times\exp\bigg{(}-\frac{\text{i}z^{2}}{4}\bigg{[}1+\frac{(40\beta
k_{2}^{2}-\alpha)z}{4\sqrt{t}(40\beta k_{2}^{3}-3\alpha
k_{2})^{3/2}}+\frac{5\beta z^{2}}{4tk_{2}(40\beta k_{2}^{2}-3\alpha)^{2}}$
$\displaystyle\qquad\qquad+\frac{\beta z^{3}}{16t^{3/2}(40\beta
k_{2}^{3}-3\alpha k_{2})^{5/2}}\bigg{]}\bigg{)}.$

For $j=1,2$, let $D_{\varepsilon}(\pm k_{j})$ denote the open disk of radius
$\varepsilon$ centered at $\pm k_{j}$ for a small $\varepsilon>0$. Now we
define

$\displaystyle\tilde{\tilde{M}}(x,t;z)$ $\displaystyle=$ $\displaystyle
M^{(2)}(x,t;k)(\delta_{-k_{2}}^{0})^{\sigma_{3}}(z),\quad k\in
D_{\varepsilon}(-k_{2})\setminus\Sigma,$
$\displaystyle\check{\check{M}}(x,t;z)$ $\displaystyle=$ $\displaystyle
M^{(2)}(x,t;k)(\delta_{-k_{1}}^{0})^{\sigma_{3}}(z),\quad k\in
D_{\varepsilon}(-k_{1})\setminus\Sigma,$ $\displaystyle\check{M}(x,t;z)$
$\displaystyle=$ $\displaystyle
M^{(2)}(x,t;k)(\delta_{k_{1}}^{0})^{\sigma_{3}}(z),\quad~{}~{}k\in
D_{\varepsilon}(k_{1})\setminus\Sigma,$ $\displaystyle\tilde{M}(x,t;z)$
$\displaystyle=$ $\displaystyle
M^{(2)}(x,t;k)(\delta_{k_{2}}^{0})^{\sigma_{3}}(z),\quad~{}~{}k\in
D_{\varepsilon}(k_{2})\setminus\Sigma.$

Then $\check{\check{M}}$, $\check{M}$, $\tilde{\tilde{M}}$ and $\tilde{M}$ are
the sectionally analytic functions of $z$ which satisfy

$\displaystyle\tilde{\tilde{M}}_{+}(x,t;z)$ $\displaystyle=$
$\displaystyle\tilde{\tilde{M}}_{-}(x,t;z)\tilde{\tilde{J}}(x,t;z),\quad
k\in\mathcal{X}_{-k_{2}}^{\varepsilon},$
$\displaystyle\check{\check{M}}_{+}(x,t;z)$ $\displaystyle=$
$\displaystyle\check{\check{M}}_{-}(x,t;z)\check{\check{J}}(x,t;z),\quad
k\in\mathcal{X}_{-k_{1}}^{\varepsilon},$ $\displaystyle\check{M}_{+}(x,t;z)$
$\displaystyle=$ $\displaystyle\check{M}_{-}(x,t;z)\check{J}(x,t;z),\quad
k\in\mathcal{X}_{k_{1}}^{\varepsilon},$ $\displaystyle\tilde{M}_{+}(x,t;z)$
$\displaystyle=$ $\displaystyle\tilde{M}_{-}(x,t;z)\tilde{J}(x,t;z),\quad
k\in\mathcal{X}_{k_{2}}^{\varepsilon},$

where $\mathcal{X}_{\pm k_{j}}=X\pm k_{j}$ denote the cross $X$ defined by
(2.25) centered at $\pm k_{j}$ and $\mathcal{X}_{\pm
k_{j}}^{\varepsilon}=\mathcal{X}_{\pm k_{j}}\cap D_{\varepsilon}(\pm k_{j})$
for $j=1,2$. Moreover, the corresponding jump matrices are given by

$\tilde{\tilde{J}}(x,t;z)=\left\\{\begin{aligned}
&\begin{pmatrix}1~{}&r_{2,a}(\delta_{-k_{2}}^{1})^{2}\\\\[4.0pt]
0~{}&1\end{pmatrix},~{}\qquad k\in(\mathcal{X}_{-k_{2}}^{\varepsilon})_{1},\\\
&\begin{pmatrix}1~{}&0\\\\[4.0pt]
-r_{1,a}(\delta_{-k_{2}}^{1})^{-2}~{}&1\end{pmatrix},\quad
k\in(\mathcal{X}_{-k_{2}}^{\varepsilon})_{2},\\\
&\begin{pmatrix}1~{}&-r_{4,a}(\delta_{-k_{2}}^{1})^{2}\\\\[4.0pt]
0~{}&1\end{pmatrix},~{}~{}\quad
k\in(\mathcal{X}_{-k_{2}}^{\varepsilon})_{3},\\\
&\begin{pmatrix}1~{}&0\\\\[4.0pt]
r_{3,a}(\delta_{-k_{2}}^{1})^{-2}~{}&1\end{pmatrix},\qquad
k\in(\mathcal{X}_{-k_{2}}^{\varepsilon})_{4},\\\ \end{aligned}\right.$ (3.55)
$\check{\check{J}}(x,t;z)=\left\\{\begin{aligned}
&\begin{pmatrix}1~{}&0\\\\[4.0pt]
r_{1,a}(\delta_{-k_{1}}^{1})^{-2}~{}&1\end{pmatrix},~{}\quad
k\in(\mathcal{X}_{-k_{1}}^{\varepsilon})_{1},\\\
&\begin{pmatrix}1~{}&-r_{2,a}(\delta_{-k_{1}}^{1})^{2}\\\\[4.0pt]
0~{}&1\end{pmatrix},~{}~{}~{}~{}k\in(\mathcal{X}_{-k_{1}}^{\varepsilon})_{2},\\\
&\begin{pmatrix}1~{}&0\\\\[4.0pt]
-r_{3,a}(\delta_{-k_{1}}^{1})^{-2}~{}&1\end{pmatrix},~{}~{}k\in(\mathcal{X}_{-k_{1}}^{\varepsilon})_{3},\\\
&\begin{pmatrix}1~{}&r_{4,a}(\delta_{-k_{1}}^{1})^{2}\\\\[4.0pt]
0~{}&1\end{pmatrix},\quad\quad
k\in(\mathcal{X}_{-k_{1}}^{\varepsilon})_{4},\\\ \end{aligned}\right.$ (3.56)
$\check{J}(x,t;z)=\left\\{\begin{aligned}
&\begin{pmatrix}1~{}&r_{2,a}(\delta_{k_{1}}^{1})^{2}\\\\[4.0pt]
0~{}&1\end{pmatrix},~{}\qquad k\in(\mathcal{X}_{k_{1}}^{\varepsilon})_{1},\\\
&\begin{pmatrix}1~{}&0\\\\[4.0pt]
-r_{1,a}(\delta_{k_{1}}^{1})^{-2}~{}&1\end{pmatrix},\quad
k\in(\mathcal{X}_{k_{1}}^{\varepsilon})_{2},\\\
&\begin{pmatrix}1~{}&-r_{4,a}(\delta_{k_{1}}^{1})^{2}\\\\[4.0pt]
0~{}&1\end{pmatrix},~{}~{}\quad
k\in(\mathcal{X}_{k_{1}}^{\varepsilon})_{3},\\\
&\begin{pmatrix}1~{}&0\\\\[4.0pt]
r_{3,a}(\delta_{k_{1}}^{1})^{-2}~{}&1\end{pmatrix},\qquad
k\in(\mathcal{X}_{k_{1}}^{\varepsilon})_{4},\\\ \end{aligned}\right.$ (3.57)

and

$\tilde{J}(x,t;z)=\left\\{\begin{aligned} &\begin{pmatrix}1~{}&0\\\\[4.0pt]
r_{1,a}(\delta_{k_{2}}^{1})^{-2}~{}&1\end{pmatrix},~{}\quad
k\in(\mathcal{X}_{k_{2}}^{\varepsilon})_{1},\\\
&\begin{pmatrix}1~{}&-r_{2,a}(\delta_{k_{2}}^{1})^{2}\\\\[4.0pt]
0~{}&1\end{pmatrix},~{}~{}~{}~{}k\in(\mathcal{X}_{k_{2}}^{\varepsilon})_{2},\\\
&\begin{pmatrix}1~{}&0\\\\[4.0pt]
-r_{3,a}(\delta_{k_{2}}^{1})^{-2}~{}&1\end{pmatrix},~{}~{}k\in(\mathcal{X}_{k_{2}}^{\varepsilon})_{3},\\\
&\begin{pmatrix}1~{}&r_{4,a}(\delta_{k_{2}}^{1})^{2}\\\\[4.0pt]
0~{}&1\end{pmatrix},\quad\quad k\in(\mathcal{X}_{k_{2}}^{\varepsilon})_{4}.\\\
\end{aligned}\right.$ (3.58)

For the jump matrix $\tilde{J}(x,t;z)$, define

$q=r(k_{2}),$

then for any fixed $z\in X$, we have $k(z)\rightarrow k_{2}$ as
$t\rightarrow\infty$. Hence,

$r_{1,a}(k)\rightarrow q,\quad
r_{2,a}(k)\rightarrow\frac{\bar{q}}{1+|q|^{2}},\quad\delta_{k_{2}}^{1}\rightarrow\text{e}^{-\frac{\text{i}z^{2}}{4}}z^{-\text{i}\nu(q)}.$

This implies that the jump matrix $\tilde{J}$ tend to the matrix $J^{X}$
defined in (2.27) for large $t$. In other words, the jumps of $M^{(2)}$ for
$k$ near $k_{2}$ approach those of the function
$M^{X}(\delta_{k_{2}}^{0})^{-\sigma_{3}}$ as $t\rightarrow\infty$. Therefore,
we can approximate $M^{(2)}$ in the neighborhood $D_{\varepsilon}(k_{2})$ of
$k_{2}$ by

$M^{(k_{2})}(x,t;k)=(\delta_{k_{2}}^{0})^{\sigma_{3}}M^{X}(q,z)(\delta_{k_{2}}^{0})^{-\sigma_{3}},$
(3.59)

where $M^{X}(q,z)$ is given by (2.28). On the other hand, according to the
symmetry property (3.3), one can deduce that

$\displaystyle\tilde{\tilde{J}}(x,t;z)\rightarrow\overline{J^{X}(q,-\bar{z})},\quad\text{as}~{}t\rightarrow\infty.$

Thus, by uniqueness of the RH problem, we can approximate $M^{(2)}$ in the
neighborhood $D_{\varepsilon}(-k_{2})$ of $-k_{2}$ by

$M^{(-k_{2})}(x,t;k)=(\delta_{-k_{2}}^{0})^{\sigma_{3}}\overline{M^{X}(q,-\bar{z})}(\delta_{-k_{2}}^{0})^{-\sigma_{3}}.$
(3.60)

For the case of $\check{J}(x,t;z)$, as $t\rightarrow\infty$, we find

$\displaystyle r_{1,a}(k)\rightarrow r(k_{1}),\quad
r_{2,a}(k)\rightarrow\frac{\overline{r(k_{1})}}{1+|r(k_{1})|^{2}},\quad\delta_{k_{1}}^{1}\rightarrow\text{e}^{\frac{\text{i}z^{2}}{4}}(-z)^{\text{i}\nu(k_{1})}.$

This implies as $t\rightarrow\infty$ that

$\displaystyle\check{J}(x,t;z)\rightarrow J^{Y}(p,z)=\left\\{\begin{aligned}
&\begin{pmatrix}1~{}&\frac{\bar{p}}{1+|p|^{2}}\text{e}^{\frac{\text{i}z^{2}}{2}}(-z)^{2\text{i}\nu(p)}\\\\[4.0pt]
0~{}&1\end{pmatrix},~{}~{}~{}\quad z\in X_{1},\\\
&\begin{pmatrix}1~{}&0\\\\[4.0pt]
-p\text{e}^{-\frac{\text{i}z^{2}}{2}}(-z)^{-2\text{i}\nu(p)}~{}&1\end{pmatrix},~{}\quad\quad
z\in X_{2},\\\
&\begin{pmatrix}1~{}&-\bar{p}\text{e}^{\frac{\text{i}z^{2}}{2}}(-z)^{2\text{i}\nu(p)}\\\\[4.0pt]
0~{}&1\end{pmatrix},~{}~{}~{}~{}\qquad z\in X_{3},\\\
&\begin{pmatrix}1~{}&0\\\\[4.0pt]
\frac{p}{1+|p|^{2}}\text{e}^{-\frac{\text{i}z^{2}}{2}}(-z)^{-2\text{i}\nu(p)}~{}&1\end{pmatrix},~{}~{}~{}z\in
X_{4},\\\ \end{aligned}\right.$

if we set

$p=r(k_{1}).$

It is easy to verify that

$\displaystyle J^{Y}(p,z)=\overline{J^{X}(\bar{p},-\bar{z})},$

which in turn implies that

$M^{Y}(p,z)=\overline{M^{X}(\bar{p},-\bar{z})},$ (3.61)

where $M^{Y}(p,z)$ is the unique solution of the following RH problem

$\displaystyle\left\\{\begin{aligned}
&M^{Y}_{+}(p,z)=M^{Y}_{-}(p,z)J^{Y}(p,z),~{}\text{for~{}almost~{}every}~{}z\in
X,\\\ &M^{Y}(p,z)\rightarrow
I,~{}~{}~{}\qquad\qquad\qquad\text{as}~{}z\rightarrow\infty.\end{aligned}\right.$

Therefore, we find that

$\displaystyle
M^{Y}(p,z)=I-\frac{\text{i}}{z}\begin{pmatrix}0~{}&\beta^{Y}(p)\\\\[4.0pt]
\overline{\beta^{Y}(p)}&~{}0\end{pmatrix}+O\bigg{(}\frac{p}{z^{2}}\bigg{)},$

where

$\displaystyle\beta^{Y}(p)=\sqrt{\nu(p)}\text{e}^{-\text{i}(\frac{\pi}{4}+\arg
p+\arg\Gamma(-\text{i}\nu(p)))}.$

As a consequence, we can approximate $M^{(2)}(x,t;k)$ in the neighborhood
$D_{\varepsilon}(k_{1})$ of $k_{1}$ by

$M^{(k_{1})}(x,t;k)=(\delta_{k_{1}}^{0})^{\sigma_{3}}M^{Y}(p,z)(\delta_{k_{1}}^{0})^{-\sigma_{3}}.$
(3.62)

Using again (3.3), we can use

$M^{(-k_{1})}(x,t;k)=(\delta_{-k_{1}}^{0})^{\sigma_{3}}\overline{M^{Y}(p,-\bar{z})}(\delta_{-k_{1}}^{0})^{-\sigma_{3}}$
(3.63)

to approximate $M^{(2)}(x,t;k)$ in the neighborhood $D_{\varepsilon}(-k_{1})$
of $-k_{1}$.

###### Lemma 3.3

For each $t>0$, $\xi\in\mathcal{I}$ and $j=1,2$, the functions $M^{(\pm
k_{j})}(x,t;k)$ defined in (3.62), (3.63), (3.59) and (3.60) are analytic
functions of $k\in D_{\varepsilon}(\pm k_{j})\setminus\mathcal{X}_{\pm
k_{j}}^{\varepsilon}$. Furthermore,

$|M^{(\pm k_{j})}(x,t;k)-I|\leq C,\quad
t>3,~{}\xi\in\mathcal{I},~{}k\in\overline{D_{\varepsilon}(\pm
k_{j})}\setminus\mathcal{X}_{\pm k_{j}}^{\varepsilon},~{}j=1,2.$ (3.64)

Across $\mathcal{X}_{\pm k_{j}}^{\varepsilon}$, $M^{(\pm k_{j})}(x,t;k)$
satisfied the jump condition $M_{+}^{(\pm k_{j})}=M_{-}^{(\pm k_{j})}J^{(\pm
k_{j})}$, where the jump matrix $J^{(\pm k_{j})}$ satisfy the following
estimates for $1\leq n\leq\infty$:

$\|J^{(2)}-J^{(\pm k_{j})}\|_{L^{n}(\mathcal{X}_{\pm
k_{j}}^{\varepsilon})}\leq Ct^{-\frac{1}{2}-\frac{1}{2n}}\ln t,\quad
t>3,~{}\xi\in\mathcal{I},~{}j=1,2,$ (3.65)

where $C>0$ is a constant independent of $t,\xi,k$. Moreover, as
$t\rightarrow\infty$,

$\|(M^{(\pm k_{j})})^{-1}(x,t;k)-I\|_{L^{\infty}(\partial D_{\varepsilon}(\pm
k_{j}))}=O(t^{-1/2}),$ (3.66)

and

$\displaystyle\frac{1}{2\pi\text{i}}\int_{\partial
D_{\varepsilon}(k_{1})}((M^{(k_{1})})^{-1}(x,t;k)-I)\text{d}k$
$\displaystyle=$
$\displaystyle-\frac{(\delta_{k_{1}}^{0})^{\hat{\sigma}_{3}}M^{Y}_{1}(\xi)}{4\sqrt{tk_{1}(3\alpha-40\beta
k_{1}^{2})}}+O(t^{-1}),$ (3.67)
$\displaystyle\frac{1}{2\pi\text{i}}\int_{\partial
D_{\varepsilon}(-k_{1})}((M^{(-k_{1})})^{-1}(x,t;k)-I)\text{d}k$
$\displaystyle=$
$\displaystyle\frac{(\delta_{-k_{1}}^{0})^{\hat{\sigma}_{3}}\overline{M^{Y}_{1}(\xi)}}{4\sqrt{tk_{1}(3\alpha-40\beta
k_{1}^{2})}}+O(t^{-1}),$ (3.68)
$\displaystyle\frac{1}{2\pi\text{i}}\int_{\partial
D_{\varepsilon}(k_{2})}((M^{(k_{2})})^{-1}(x,t;k)-I)\text{d}k$
$\displaystyle=$
$\displaystyle-\frac{(\delta_{k_{2}}^{0})^{\hat{\sigma}_{3}}M^{X}_{1}(\xi)}{4\sqrt{tk_{2}(40\beta
k_{2}^{2}-3\alpha)}}+O(t^{-1}),$ (3.69)
$\displaystyle\frac{1}{2\pi\text{i}}\int_{\partial
D_{\varepsilon}(-k_{2})}((M^{(-k_{2})})^{-1}(x,t;k)-I)\text{d}k$
$\displaystyle=$
$\displaystyle\frac{(\delta_{-k_{2}}^{0})^{\hat{\sigma}_{3}}\overline{M^{X}_{1}(\xi)}}{4\sqrt{tk_{2}(40\beta
k_{2}^{2}-3\alpha)}}+O(t^{-1}),$ (3.70)

where $M^{X}_{1}(\xi)$ and $M^{Y}_{1}(\xi)$ are given by

$M^{X}_{1}(\xi)=-\text{i}\begin{pmatrix}0~{}&\beta^{X}(q)\\\\[4.0pt]
\overline{\beta^{X}(q)}~{}&0\end{pmatrix},\quad
M^{Y}_{1}(\xi)=-\text{i}\begin{pmatrix}0~{}&\beta^{Y}(p)\\\\[4.0pt]
\overline{\beta^{Y}(p)}~{}&0\end{pmatrix}.$ (3.71)

Proof.We just consider the proof for the function $M^{(k_{2})}(x,t;k)$, the
others accordingly follow.

The analyticity of $M^{(k_{2})}$ is obvious. Since
$|\delta_{k_{2}}^{0}(z)|=1$, thus, the estimate (3.64) for $M^{(k_{2})}$
follows from the definition of $M^{(k_{2})}$ in (3.59) and the estimate
(3.25). On the other hand, we have

$\displaystyle
J^{(2)}-J^{(k_{2})}=(\delta_{k_{2}}^{0})^{\hat{\sigma}_{3}}(\tilde{J}-J^{X}),\quad
k\in\mathcal{X}_{k_{2}}^{\varepsilon}.$

However, proceeding the similar calculation as the Lemma 3.35 in [10] (also
can see [9, 29]), we have

$\|\tilde{J}-J^{X}\|_{L^{\infty}((\mathcal{X}_{k_{2}}^{\varepsilon})_{1})}\leq
C|\text{e}^{\frac{\text{i}\gamma}{2}z^{2}}|t^{-1/2}\ln t,\quad
0<\gamma<\frac{1}{2},~{}t>3,~{}\xi\in\mathcal{I},$ (3.72)

for $k\in(\mathcal{X}_{k_{2}}^{\varepsilon})_{1}$, that is,
$z=4\sqrt{tk_{2}(40\beta
k_{2}^{2}-3\alpha)}s\text{e}^{\frac{\text{i}\pi}{4}}$, $0\leq
s\leq\varepsilon$. Thus,

$\|\tilde{J}-J^{X}\|_{L^{1}((\mathcal{X}_{k_{2}}^{\varepsilon})_{1})}\leq
Ct^{-1}\ln t,\quad t>3,~{}\xi\in\mathcal{I}.$ (3.73)

By the general inequality
$\|f\|_{L^{n}}\leq\|f\|^{1-1/n}_{L^{\infty}}\|f\|_{L^{1}}^{1/n}$, we find

$\|\tilde{J}-J^{X}\|_{L^{n}((\mathcal{X}_{k_{2}}^{\varepsilon})_{1})}\leq
Ct^{-1/2-1/2n}\ln t,\quad t>3,~{}\xi\in\mathcal{I}.$ (3.74)

The norms on $(\mathcal{X}_{k_{2}}^{\varepsilon})_{j}$, $j=2,3,4$, are
estimated in a similar way. Therefore, (3.65) follows.

If $k\in\partial D_{\varepsilon}(k_{2})$, the variable
$z=4\sqrt{tk_{2}(40\beta k_{2}^{2}-3\alpha)}(k-k_{2})$ tends to infinity as
$t\rightarrow\infty$. It follows from (2.28) that

$\displaystyle M^{X}(q,z)=I+\frac{M^{X}_{1}(\xi)}{4\sqrt{tk_{2}(40\beta
k_{2}^{2}-3\alpha)}(k-k_{2})}+O\bigg{(}\frac{q}{t}\bigg{)},\quad
t\rightarrow\infty,~{}k\in\partial D_{\varepsilon}(k_{2}),$

where $M^{X}_{1}(\xi)$ is defined by (3.71). Since

$M^{(k_{2})}(x,t;k)=(\delta_{k_{2}}^{0})^{\hat{\sigma}_{3}}M^{X}(q,z),$

thus we have

$(M^{(k_{2})})^{-1}(x,t;k)-I=-\frac{(\delta_{k_{2}}^{0})^{\hat{\sigma}_{3}}M^{X}_{1}(\xi)}{4\sqrt{tk_{2}(40\beta
k_{2}^{2}-3\alpha)}(k-k_{2})}+O\bigg{(}\frac{q}{t}\bigg{)},~{}t\rightarrow\infty,~{}k\in\partial
D_{\varepsilon}(k_{2}).$ (3.75)

The estimate (3.66) immediately follows from (3.75) and $|M_{1}^{X}|\leq C$.
By Cauchy’s formula and (3.75), we derive (3.69).

###  3.3 The final step

We now begin to establish the explicit long-time asymptotic formula for the
emKdV equation (1.2) on the line. Define the approximate solution
$M^{(app)}(x,t;k)$ by

$M^{(app)}=\left\\{\begin{aligned} &M^{(-k_{2})},\quad k\in
D_{\varepsilon}(-k_{2}),\\\ &M^{(-k_{1})},\quad k\in
D_{\varepsilon}(-k_{1}),\\\ &M^{(k_{1})},~{}~{}\quad k\in
D_{\varepsilon}(k_{1}),\\\ &M^{(k_{2})},~{}~{}\quad k\in
D_{\varepsilon}(k_{2}),\\\
&I,\qquad~{}~{}~{}~{}~{}{\text{e}lsewhere}.\end{aligned}\right.$ (3.76)

Let $\hat{M}(x,t;k)$ be

$\hat{M}=M^{(2)}(M^{(app)})^{-1},$ (3.77)

then $\hat{M}(x,t;k)$ satisfies the following RH problem

$\hat{M}_{+}(x,t;k)=\hat{M}_{-}(x,t;k)\hat{J}(x,t;k),\quad k\in\hat{\Sigma},$
(3.78)

where the jump contour $\hat{\Sigma}=\Sigma\cup\partial
D_{\varepsilon}(-k_{2})\cup\partial D_{\varepsilon}(-k_{1})\cup\partial
D_{\varepsilon}(k_{1})\cup\partial D_{\varepsilon}(k_{2})$ is depicted in Fig.
6, and the jump matrix $\hat{J}(x,t;k)$ is given by

$\hat{J}=\left\\{\begin{aligned}
&M^{(app)}_{-}J^{(2)}(M^{(app)}_{+})^{-1},~{}~{}k\in\hat{\Sigma}\cap(D_{\varepsilon}(-k_{2})\cup
D_{\varepsilon}(-k_{1})\cup D_{\varepsilon}(k_{1})\cup
D_{\varepsilon}(k_{2})),\\\ &(M^{(app)})^{-1},\qquad\qquad\quad k\in(\partial
D_{\varepsilon}(-k_{2})\cup\partial D_{\varepsilon}(-k_{1})\cup\partial
D_{\varepsilon}(k_{1})\cup\partial D_{\varepsilon}(k_{2})),\\\
&J^{(2)},\qquad\qquad\qquad\quad~{}~{}~{}k\in\hat{\Sigma}\setminus(\overline{D_{\varepsilon}(-k_{2})}\cup\overline{D_{\varepsilon}(-k_{1})}\cup\overline{D_{\varepsilon}(k_{1})}\cup\overline{D_{\varepsilon}(k_{2})}).\end{aligned}\right.$
(3.79)

Fig. 6: The contour $\hat{\Sigma}$.

For convenience, we rewrite $\hat{\Sigma}$ as follows:

$\hat{\Sigma}=\hat{\Sigma}_{1}\cup\hat{\Sigma}_{2}\cup\hat{\Sigma}_{3}\cup\hat{\Sigma}_{4},$

where

$\displaystyle\hat{\Sigma}_{1}$ $\displaystyle=$
$\displaystyle\bigcup_{1}^{4}\Sigma_{j}\setminus(D_{\varepsilon}(-k_{2})\cup
D_{\varepsilon}(-k_{1})\cup D_{\varepsilon}(k_{1})\cup
D_{\varepsilon}(k_{2})),~{}\hat{\Sigma}_{2}=\bigcup_{5}^{6}\Sigma_{j},$
$\displaystyle\hat{\Sigma}_{3}$ $\displaystyle=$ $\displaystyle\partial
D_{\varepsilon}(-k_{2})\cup\partial D_{\varepsilon}(-k_{1})\cup\partial
D_{\varepsilon}(k_{1})\cup\partial D_{\varepsilon}(k_{2}),$
$\displaystyle\hat{\Sigma}_{4}$ $\displaystyle=$
$\displaystyle\mathcal{X}_{-k_{2}}^{\varepsilon}\cup\mathcal{X}_{-k_{1}}^{\varepsilon}\cup\mathcal{X}_{k_{1}}^{\varepsilon}\cup\mathcal{X}_{k_{2}}^{\varepsilon}.$

and $\\{\Sigma_{j}\\}_{1}^{6}$ denoting the restriction of $\Sigma$ to the
contour labeled by $j$ in Fig. 5. Then we have the following lemma if we let
$\hat{w}=\hat{J}-I$.

###### Lemma 3.4

For $1\leq n\leq\infty$, $t>3$ and $\xi\in\mathcal{I}$, the following
estimates hold:

$\displaystyle\|\hat{w}\|_{L^{n}(\hat{\Sigma}_{1})}$ $\displaystyle\leq$
$\displaystyle C\text{e}^{-ct},$ (3.80)
$\displaystyle\|\hat{w}\|_{L^{n}(\hat{\Sigma}_{2})}$ $\displaystyle\leq$
$\displaystyle Ct^{-3/2},$ (3.81)
$\displaystyle\|\hat{w}\|_{L^{n}(\hat{\Sigma}_{3})}$ $\displaystyle\leq$
$\displaystyle Ct^{-1/2},$ (3.82)
$\displaystyle\|\hat{w}\|_{L^{n}(\hat{\Sigma}_{4})}$ $\displaystyle\leq$
$\displaystyle Ct^{-\frac{1}{2}-\frac{1}{2n}}\ln t.$ (3.83)

Proof.For
$k\in\Omega_{1}\cap\\{k\in{\mathbb{C}}|\text{Re}k>k_{2}\\}\cap\hat{\Sigma}_{1}$,
we have $-|\text{Re}\Phi(k)|\leq-c\varepsilon^{2}.$ Since $\hat{w}$ only has a
nonzero $r_{1,a}\delta^{-2}\text{e}^{t\Phi}$ in $(21)$ entry, hence, for
$t\geq 1$, by (3.15), we get

$\displaystyle|\hat{w}_{21}|$ $\displaystyle=$
$\displaystyle|r_{1,a}\delta^{-2}\text{e}^{t\Phi}|\leq\frac{C}{1+|k|^{2}}\text{e}^{-\frac{3t}{4}|\text{Re}\Phi|}\leq
C\text{e}^{-c\varepsilon^{2}t}.$

In a similar way, the other estimates on $\hat{\Sigma}_{1}$ hold. This proves
(3.80). Since the matrix $\hat{w}$ on $\hat{\Sigma}_{2}$ only involves the
small remainders $r_{j,r}$ for $j=1,\cdots,4$, by Lemma 3.2, the estimate
(3.81) follows. The inequality (3.82) is a consequence of (3.66), (3.76) and
(3.79). For $k\in\mathcal{X}_{\pm k_{j}}^{\varepsilon}$, we find

$\hat{w}=M^{(\pm k_{j})}_{-}(J^{(2)}-J^{(\pm k_{j})})(M^{(\pm
k_{j})}_{+})^{-1},\quad j=1,2.$

Therefore, it follows from (3.64) and (3.65) that the estimate (3.83) holds.

The estimates in Lemma 3.4 imply that

$\begin{aligned} \|\hat{w}\|_{(L^{1}\cap L^{2})(\hat{\Sigma})}&\leq
Ct^{-1/2},\\\ \|\hat{w}\|_{L^{\infty}(\hat{\Sigma})}&\leq Ct^{-1/2}\ln
t,\end{aligned}\quad t>3,~{}\xi\in\mathcal{I}.$ (3.84)

Let $\hat{C}$ denote the Cauchy operator associated with $\hat{\Sigma}$:

$\displaystyle(\hat{C}f)(k)=\int_{\hat{\Sigma}}\frac{f(\zeta)}{\zeta-k}\frac{\text{d}\zeta}{2\pi\text{i}},\quad
k\in{\mathbb{C}}\setminus\hat{\Sigma},~{}f\in L^{2}(\hat{\Sigma}).$

We denote the boundary values of $\hat{C}f$ from the left and right sides of
$\hat{\Sigma}$ by $\hat{C}_{+}f$ and $\hat{C}_{-}f$, respectively. Define the
operator $\hat{C}_{\hat{w}}$:
$L^{2}(\hat{\Sigma})+L^{\infty}(\hat{\Sigma})\rightarrow L^{2}(\hat{\Sigma})$
by $\hat{C}_{\hat{w}}f=\hat{C}_{-}(f\hat{w}),$ that is, $\hat{C}_{\hat{w}}$ is
defined by
$\hat{C}_{\hat{w}}(f)=\hat{C}_{+}(f\hat{w}_{-})+\hat{C}_{-}(f\hat{w}_{+})$
where we have chosen, for simplicity, $\hat{w}_{+}=\hat{w}$ and
$\hat{w}_{-}=0$. Then, by (3.84), we find

$\|\hat{C}_{\hat{w}}\|_{B(L^{2}(\hat{\Sigma}))}\leq
C\|\hat{w}\|_{L^{\infty}(\hat{\Sigma})}\leq Ct^{-1/2}\ln t,$ (3.85)

where $B(L^{2}(\hat{\Sigma}))$ denotes the Banach space of bounded linear
operators $L^{2}(\hat{\Sigma})\rightarrow L^{2}(\hat{\Sigma})$. Therefore,
there exists a $T>0$ such that $I-\hat{C}_{\hat{w}}\in B(L^{2}(\hat{\Sigma}))$
is invertible for all $\xi\in\mathcal{I},$ $t>T$. Following this, we may
define the $2\times 2$ matrix-valued function $\hat{\mu}(x,t;k)$ whenever
$t>T$ by

$\hat{\mu}=I+\hat{C}_{\hat{w}}\hat{\mu}.$ (3.86)

Then

$\hat{M}(x,t;k)=I+\frac{1}{2\pi\text{i}}\int_{\hat{\Sigma}}\frac{(\hat{\mu}\hat{w})(x,t;\zeta)}{\zeta-k}\text{d}\zeta,\quad
k\in{\mathbb{C}}\setminus\hat{\Sigma}$ (3.87)

is the unique solution of the RH problem (3.78) for $t>T$. Moreover, using the
Neumann series (see [22]), the function $\hat{\mu}(x,t;k)$ satisfies

$\|\hat{\mu}(x,t;\cdot)-I\|_{L^{2}(\hat{\Sigma})}=O(t^{-1/2}),\quad
t\rightarrow\infty,~{}\xi\in\mathcal{I}.$ (3.88)

It follows from (3.87) that

$\lim_{k\rightarrow\infty}k(\hat{M}(x,t;k)-I)=-\frac{1}{2\pi\text{i}}\int_{\hat{\Sigma}}(\hat{\mu}\hat{w})(x,t;k)\text{d}k.$
(3.89)

Using (3.80) and (3.88), we have

$\displaystyle\int_{\hat{\Sigma}_{1}}(\hat{\mu}\hat{w})(x,t;k)\text{d}k$
$\displaystyle=$
$\displaystyle\int_{\hat{\Sigma}_{1}}\hat{w}(x,t;k)\text{d}k+\int_{\hat{\Sigma}_{1}}(\hat{\mu}(x,t;k)-I)\hat{w}(x,t;k)\text{d}k$
$\displaystyle\leq$
$\displaystyle\|\hat{w}\|_{L^{1}(\hat{\Sigma}_{1})}+\|\hat{\mu}-I\|_{L^{2}(\hat{\Sigma}_{1})}\|\hat{w}\|_{L^{2}(\hat{\Sigma}_{1})}$
$\displaystyle\leq$ $\displaystyle C\text{e}^{-ct},\quad t\rightarrow\infty.$

By (3.81) and (3.88), the contribution from $\hat{\Sigma}_{2}$ to the right-
hand side of (3.89) is

$\displaystyle
O(\|\hat{w}\|_{L^{1}(\hat{\Sigma}_{2})}+\|\hat{\mu}-I\|_{L^{2}(\hat{\Sigma}_{2})}\|\hat{w}\|_{L^{2}(\hat{\Sigma}_{2})})=O(t^{-3/2}),\quad
t\rightarrow\infty.$

Similarly, by (3.83) and (3.88), the contribution from $\hat{\Sigma}_{4}$ to
the right-hand side of (3.89) is

$O(\|\hat{w}\|_{L^{1}(\hat{\Sigma}_{4})}+\|\hat{\mu}-I\|_{L^{2}(\hat{\Sigma}_{4})}\|\hat{w}\|_{L^{2}(\hat{\Sigma}_{4})})=O(t^{-1}\ln
t),\quad t\rightarrow\infty.$

Finally, by (3.67)-(3.70), (3.82) and (3.88), we can get

$\displaystyle~{}~{}-\frac{1}{2\pi\text{i}}\int_{\hat{\Sigma}_{3}}(\hat{\mu}\hat{w})(x,t;k)\text{d}k$
$\displaystyle\quad=-\frac{1}{2\pi\text{i}}\int_{\hat{\Sigma}_{3}}\hat{w}(x,t;k)\text{d}k-\frac{1}{2\pi\text{i}}\int_{\hat{\Sigma}_{3}}(\hat{\mu}(x,t;k)-I)\hat{w}(x,t;k)\text{d}k$
$\displaystyle\quad=-\frac{1}{2\pi\text{i}}\int_{\partial
D_{\varepsilon}(k_{1})}\bigg{(}(M^{(k_{1})})^{-1}(x,t;k)-I\bigg{)}\text{d}k-\frac{1}{2\pi\text{i}}\int_{\partial
D_{\varepsilon}(-k_{1})}\bigg{(}(M^{(-k_{1})})^{-1}(x,t;k)-I\bigg{)}\text{d}k$
$\displaystyle\qquad-\frac{1}{2\pi\text{i}}\int_{\partial
D_{\varepsilon}(k_{2})}\bigg{(}(M^{(k_{2})})^{-1}(x,t;k)-I\bigg{)}\text{d}k-\frac{1}{2\pi\text{i}}\int_{\partial
D_{\varepsilon}(-k_{2})}\bigg{(}(M^{(-k_{2})})^{-1}(x,t;k)-I\bigg{)}\text{d}k$
$\displaystyle\qquad+O(\|\hat{\mu}-I\|_{L^{2}(\hat{\Sigma}_{3})}\|\hat{w}\|_{L^{2}(\hat{\Sigma}_{3})})$
$\displaystyle\quad=\frac{(\delta_{k_{1}}^{0})^{\hat{\sigma}_{3}}M^{Y}_{1}(\xi)}{4\sqrt{tk_{1}(3\alpha-40\beta
k_{1}^{2})}}-\frac{(\delta_{-k_{1}}^{0})^{\hat{\sigma}_{3}}\overline{M^{Y}_{1}(\xi)}}{4\sqrt{tk_{1}(3\alpha-40\beta
k_{1}^{2})}}$
$\displaystyle~{}\qquad+\frac{(\delta_{k_{2}}^{0})^{\hat{\sigma}_{3}}M^{X}_{1}(\xi)}{4\sqrt{tk_{2}(40\beta
k_{2}^{2}-3\alpha)}}-\frac{(\delta_{-k_{2}}^{0})^{\hat{\sigma}_{3}}\overline{M^{X}_{1}(\xi)}}{4\sqrt{tk_{2}(40\beta
k_{2}^{2}-3\alpha)}}+O(t^{-1}),\quad t\rightarrow\infty.$

Thus, we obtain the following important relation

$\displaystyle\begin{aligned}
\lim_{k\rightarrow\infty}k(\hat{M}(x,t;k)-I)&=\frac{(\delta_{k_{1}}^{0})^{\hat{\sigma}_{3}}M^{Y}_{1}(\xi)}{4\sqrt{tk_{1}(3\alpha-40\beta
k_{1}^{2})}}-\frac{(\delta_{-k_{1}}^{0})^{\hat{\sigma}_{3}}\overline{M^{Y}_{1}(\xi)}}{4\sqrt{tk_{1}(3\alpha-40\beta
k_{1}^{2})}}\\\
&+\frac{(\delta_{k_{2}}^{0})^{\hat{\sigma}_{3}}M^{X}_{1}(\xi)}{4\sqrt{tk_{2}(40\beta
k_{2}^{2}-3\alpha)}}-\frac{(\delta_{-k_{2}}^{0})^{\hat{\sigma}_{3}}\overline{M^{X}_{1}(\xi)}}{4\sqrt{tk_{2}(40\beta
k_{2}^{2}-3\alpha)}}+O(t^{-1}\ln t),\quad t\rightarrow\infty.\end{aligned}$
(3.90)

Taking into account that (3.4), (3.8), (3.37) and (3.77), for sufficient large
$k\in{\mathbb{C}}\setminus\hat{\Sigma}$, we get

$\displaystyle u(x,t)$ $\displaystyle=$
$\displaystyle-2\text{i}\lim_{k\rightarrow\infty}(kM(x,t;k))_{12}$
$\displaystyle=$
$\displaystyle-2\text{i}\lim_{k\rightarrow\infty}k(\hat{M}(x,t;k)-I)_{12}$
$\displaystyle=$
$\displaystyle-\bigg{(}\frac{\beta^{Y}(p)(\delta_{k_{1}}^{0})^{2}+\overline{\beta^{Y}(p)}(\delta_{-k_{1}}^{0})^{2}}{2\sqrt{tk_{1}(3\alpha-40\beta
k_{1}^{2})}}+\frac{\beta^{X}(q)(\delta_{k_{2}}^{0})^{2}+\overline{\beta^{X}(q)}(\delta_{-k_{2}}^{0})^{2}}{2\sqrt{tk_{2}(40\beta
k_{2}^{2}-3\alpha)}}\bigg{)}+O\bigg{(}\frac{\ln t}{t}\bigg{)}.$

Using

$\delta_{-k_{1}}^{0}=\overline{\delta_{k_{1}}^{0}},\quad\delta_{-k_{2}}^{0}=\overline{\delta_{k_{2}}^{0}}$
(3.91)

as $\chi_{j}(-k_{j})=-\chi_{j}(k_{j})=\overline{\chi_{j}(k_{j})},~{}j=1,2$,
and collecting the above computations, we obtain our main results as stated in
the Theorem 1.1.

##  4 Asymptotics for a special case $\alpha=0$

In this section, we consider the long-time asymptotics to the solutions for a
particular case, namely, $\alpha=0$ of emKdV equation (1.2),

$u_{t}+\beta(30u^{4}u_{x}+10u_{x}^{3}+40uu_{x}u_{xx}+10u^{2}u_{xxx}+u_{xxxxx})=0.$
(4.1)

As we all know, the Hirota equation can be reduced to the complex-valued mKdV
equation under the Galilean transformation. Thus, if we rewrite equation (1.2)
as a complex-valued form

$u_{t}+\alpha(6|u|^{2}u_{x}+u_{xxx})+\beta[30|u|^{4}u_{x}+10(u|u_{x}|^{2}+|u|^{2}u_{xx})_{x}+u_{xxxxx}]=0,$
(4.2)

similarly, the Galilean transformation can reduce (4.2) into a complex-valued
form of (4.1), however, where we take $u$ is a real-valued function. In fact,
the fifth order KdV equation indeed can be excluded the third derivative term,
(see [12]), where its prolongation structure was considered.

As in Section 2, under the condition that $a(k)\neq 0$ for
$\\{k\in{\mathbb{C}}|\text{Im}k\geq 0\\}$, the corresponding RH problem
associated with (4.1) is

$\left\\{\begin{aligned}
&N_{+}(x,t;k)=N_{-}(x,t;k)v(x,t;k),~{}k\in{\mathbb{R}},\\\
&N(x,t;k)\rightarrow
I,~{}\qquad\qquad\qquad\qquad~{}k\rightarrow\infty,\end{aligned}\right.$ (4.3)

where the jump matrix $v(x,t;k)$ is defined by

$\displaystyle
v(x,t;k)=\begin{pmatrix}1+|r(k)|^{2}~{}&\bar{r}(k)\text{e}^{-t\theta(k)}\\\\[4.0pt]
r(k)\text{e}^{t\theta(k)}~{}&1\end{pmatrix},$ (4.4) $\displaystyle
r(k)=\frac{\bar{b}(k)}{a(k)},~{}\theta(k)=2\text{i}(16\beta
k^{5}-k\xi),~{}\xi=\frac{x}{t}.$

Also we have

$r(-k)=\overline{r(\bar{k})},\quad k\in{\mathbb{R}}.$ (4.5)

Moreover, the relation between the solution $u(x,t)$ of the equation (4.1) and
$N(x,t;k)$ is

$u(x,t)=-2\text{i}\lim_{k\rightarrow\infty}(kN(x,t;k))_{12}.$ (4.6)

For this case, there are two real and pure imaginary critical points of
$\theta(k)$ located at the points $\pm k_{0}$ and $\pm\text{i}k_{0}$, where

$k_{0}=\sqrt[4]{\frac{\xi}{80\beta}}.$ (4.7)

Our aim in this section is to find the asymptotics of solution $u(x,t)$ to the
equation (4.1) in region $0<x\leq Mt^{\frac{1}{5}}$, where $M>1$ is a
constant. We see that as $t\rightarrow\infty$, the critical points $\pm k_{0}$
approach 0 at least as fast as $t^{-\frac{1}{5}}$, i.e.,
$k_{0}\leq\big{(}\frac{M}{80\beta}\big{)}^{\frac{1}{4}}t^{-\frac{1}{5}}$. We
will show that the asymptotics of the solution $u(x,t)$ in this region is
given in terms of the solution of a fourth order Painlevé II equation.

Let $\Gamma\subset{\mathbb{C}}$ denote the contour
$\Gamma={\mathbb{R}}\cup\Gamma_{1}\cup\Gamma_{2}$, where

$\displaystyle\Gamma_{1}$
$\displaystyle=\\{k_{0}+l\text{e}^{\frac{\pi\text{i}}{6}}|l\geq
0\\}\cup\\{-k_{0}+l\text{e}^{\frac{5\pi\text{i}}{6}}|l\geq 0\\},$ (4.8)
$\displaystyle\Gamma_{2}$
$\displaystyle=\\{k_{0}+l\text{e}^{-\frac{\pi\text{i}}{6}}|l\geq
0\\}\cup\\{-k_{0}+l\text{e}^{-\frac{5\pi\text{i}}{6}}|l\geq 0\\},$

and we orient $\Gamma$ to the right. Let $\mathcal{V}$ and $\mathcal{V}^{*}$
denote the triangular domains shown in Fig. 7.

Fig. 7: The oriented contour $\Gamma$ and the sets $\mathcal{V}$ and
$\mathcal{V}^{*}$.

Denote

$\mathcal{P}=\\{(x,t)\in{\mathbb{R}}^{2}|0<x\leq Mt^{\frac{1}{5}},t\geq 3\\}.$

By Lemma 3.2, we also have the following analytic approximation lemma for
$r(k)$.

###### Lemma 4.1

There exists a decomposition

$r(k)=r_{a}(x,t,k)+r_{r}(x,t,k),\quad k\in(-\infty,-k_{0})\cup(k_{0},\infty),$
(4.9)

where the functions $r_{a}$ and $r_{r}$ satisfy the following properties:
(i) For $(x,t)\in\mathcal{P}$, $r_{a}(x,t,k)$ is defined and continuous for
$k\in\bar{\mathcal{V}}$ and analytic for $\mathcal{V}$.
(ii) The function $r_{a}$ satisfies

$|r_{a}(x,t,k)|\leq\frac{C}{1+|k|^{2}}\text{e}^{\frac{t}{4}|\text{Re}\theta(k)|},~{}k\in\bar{\mathcal{V}},$
(4.10)

and

$|r_{a}(x,t,k)-r(k_{0})|\leq
C|k-k_{0}|\text{e}^{\frac{t}{4}|\text{Re}\theta(k)|},~{}k\in\bar{\mathcal{V}}.$
(4.11)

(iii) The $L^{1},L^{2}$ and $L^{\infty}$ norms of the function
$r_{r}(x,t,\cdot)$ on $(-\infty,-k_{0})\cup(k_{0},\infty)$ are $O(t^{-3/2})$
as $t\rightarrow\infty$ uniformly with respect to $(x,t)\in\mathcal{P}$.
(iv) The following symmetries hold:

$r_{a}(x,t,k)=\overline{r_{a}(x,t,-\bar{k})},~{}r_{r}(x,t,k)=\overline{r_{r}(x,t,-\bar{k})}.$
(4.12)

The first transform is as follows:

$N^{(1)}(x,t;k)=N(x,t;k)\times\left\\{\begin{aligned}
&\begin{pmatrix}1&0\\\\[4.0pt]
-r_{a}(x,t,k)\text{e}^{t\theta(k)}&1\end{pmatrix},~{}~{}~{}~{}k\in\mathcal{V},\\\
&\begin{pmatrix}1&~{}\overline{r_{a}(x,t,\bar{k})}\text{e}^{-t\theta(k)}\\\\[4.0pt]
0&1\end{pmatrix},~{}~{}~{}~{}k\in\mathcal{V}^{*},\\\
&I,\qquad\qquad\qquad\qquad~{}~{}~{}~{}~{}~{}~{}~{}\text{elsewhere}.\end{aligned}\right.$
(4.13)

Then we obtain the RH problem

$N^{(1)}_{+}(x,t;k)=N^{(1)}_{-}(x,t;k)v^{(1)}(x,t,k)$ (4.14)

on the contour $\Gamma$ depicted in Fig. 7. The jump matrix $v^{(1)}(x,t,k)$
is given by

$\displaystyle v^{(1)}_{1}$ $\displaystyle=$
$\displaystyle\begin{pmatrix}1&~{}0\\\\[4.0pt]
r_{a}\text{e}^{t\theta}&~{}1\end{pmatrix},\qquad\qquad\qquad
v^{(1)}_{2}=\begin{pmatrix}1&~{}\bar{r}_{a}\text{e}^{-t\theta}\\\\[4.0pt]
0&~{}1\end{pmatrix},$ $\displaystyle v^{(1)}_{3}$ $\displaystyle=$
$\displaystyle\begin{pmatrix}1&~{}\bar{r}\text{e}^{-t\theta}\\\\[4.0pt]
0&~{}1\end{pmatrix}\begin{pmatrix}1&~{}0\\\\[4.0pt]
r\text{e}^{t\theta}&~{}1\end{pmatrix},~{}~{}~{}~{}v^{(1)}_{4}=\begin{pmatrix}1~{}&\bar{r}_{r}\text{e}^{-t\theta}\\\\[4.0pt]
0~{}&1\\\ \end{pmatrix}\begin{pmatrix}1&~{}0\\\\[4.0pt]
r_{r}\text{e}^{t\theta}&~{}1\end{pmatrix},$

where $v^{(1)}_{i}$ denotes the restriction of $v^{(1)}$ to the contour
labeled by $i$ in Fig. 7.

Let us introduce the new variables $y$ and $z$ by

$y=\frac{-x}{(20\beta t)^{\frac{1}{5}}},\quad z=(20\beta t)^{\frac{1}{5}}k,$
(4.15)

such that

$t\theta(k)=2\text{i}\bigg{(}\frac{4}{5}z^{5}+yz\bigg{)}.$ (4.16)

We now have $-C\leq y<0$. Fix $\varepsilon>0$ and let
$D_{\varepsilon}(0)=\\{k\in{\mathbb{C}}||k|<\varepsilon\\}$ denote the open
disk of radius $\varepsilon$ centered at the origin. Let
$\Gamma^{\varepsilon}=(\Gamma\cap
D_{\varepsilon}(0))\setminus((-\infty,-k_{0})\cup(k_{0},\infty))$. Let $Z$
denote the contour defined in (B.1) with $z_{0}=(20\beta
t)^{\frac{1}{5}}k_{0}=\sqrt[4]{-y}/\sqrt{2}$. The map $k\mapsto z$ maps
$\Gamma^{\varepsilon}$ onto $Z\cap\\{z\in{\mathbb{C}}||z|<(20\beta
t)^{\frac{1}{5}}\varepsilon\\}$. We write
$\Gamma^{\varepsilon}=\cup_{j=1}^{3}\Gamma_{j}^{\varepsilon}$, where
$\Gamma_{j}^{\varepsilon}$ denotes the inverse image of
$Z_{j}\cap\\{z\in{\mathbb{C}}||z|<(20\beta t)^{\frac{1}{5}}\varepsilon\\}$
under this map, see Fig. 8.

Fig. 8: The oriented contour $\hat{\Gamma}$ and $\Gamma^{\varepsilon}$.

For large $t$ and fixed $z$, the jump matrices $\\{v^{(1)}_{j}\\}_{1}^{4}$
tend to the matrix $v^{Z}$ defined in (B.3) if we set $s=r(0)$. Thus we expect
that $N^{(1)}$ approaches the solution $N^{0}(x,t,k)$ defined by

$N^{0}(x,t;k):=N^{Z}(s,y,z,z_{0})$ (4.17)

for large $t$, where $N^{Z}(s,y,z,z_{0})$ is the solution of the model RH
problem for (B.2) with $z_{0}=\sqrt[4]{-y}/\sqrt{2}$. Moreover, if
$(x,t)\in\mathcal{P}$, then $(y,t,z_{0})\in\mathbb{P}$, where $\mathbb{P}$ is
the parameter subset defined in (B.4). Thus, Lemma B.1 ensures that $N^{0}$ is
well-defined by (4.17).

###### Lemma 4.2

For each $(x,t)\in\mathcal{P}$, the function $N^{0}(x,t;k)$ defined in (4.17)
is an analytic function of $k\in
D_{\varepsilon}(0)\setminus\Gamma^{\varepsilon}$ such that

$|N^{0}(x,t;k)|\leq C,\quad(x,t)\in\mathcal{P},~{}k\in
D_{\varepsilon}(0)\setminus\Gamma^{\varepsilon}.$ (4.18)

Across $\Gamma^{\varepsilon}$, $N^{0}$ obeys the jump condition
$N_{+}^{0}=N_{-}^{0}v^{0}$, where the jump matrix $v^{0}$ satisfies, for
$1\leq n\leq\infty$,

$\|v^{(1)}-v^{0}\|_{L^{n}(\Gamma^{\varepsilon})}\leq
Ct^{-\frac{1}{5}(1+\frac{1}{n})},\quad(x,t)\in\mathcal{P}.$ (4.19)

Furthermore, as $t\rightarrow\infty$, we have

$\|N^{0}(x,t;k)^{-1}-I\|_{L^{\infty}(\partial
D_{\varepsilon}(0))}=O(t^{-\frac{1}{5}}),$ (4.20)

and

$\frac{1}{2\pi\text{i}}\int_{\partial
D_{\varepsilon}(0)}(N^{0}(x,t;k)^{-1}-I)\text{d}k=-\frac{\text{i}N_{1}^{0}(y)}{(20\beta
t)^{\frac{1}{5}}},$ (4.21)

where

$N_{1}^{0}(y)=\begin{pmatrix}-2\int^{y}_{-\infty}u_{p}^{2}(y^{\prime})\text{d}y^{\prime}~{}&u_{p}(y)\\\
u_{p}(y)~{}&2\int^{y}_{-\infty}u_{p}^{2}(y^{\prime})\text{d}y^{\prime}\end{pmatrix}.$
(4.22)

Proof.The analyticity and boundedness of $N^{0}$ are a consequence of Lemma
B.1. Moreover,

$v^{(1)}-v^{0}=\left\\{\begin{aligned} &\begin{pmatrix}0&~{}0\\\
(r_{a}(x,t,k)-r(0))\text{e}^{t\theta}&~{}0\end{pmatrix},\qquad\qquad\quad
k\in\Gamma^{\varepsilon}_{1},\\\
&\begin{pmatrix}0&~{}(\overline{r_{a}(x,t,\bar{k}})-\overline{r(0)})\text{e}^{-t\theta}\\\
0&~{}0\end{pmatrix},~{}~{}\quad\qquad\quad k\in\Gamma^{\varepsilon}_{2},\\\
&\begin{pmatrix}|r(k)|^{2}-|r(0)|^{2}&~{}(\overline{r(k)}-\overline{r(0)})\text{e}^{-t\theta}\\\
(r(k)-r(0))\text{e}^{t\theta}&~{}0\end{pmatrix},~{}~{}k\in\Gamma^{\varepsilon}_{3}.\end{aligned}\right.$
(4.23)

For $k=k_{0}+l\text{e}^{\frac{\pi\text{i}}{6}}$, $0\leq l\leq\varepsilon$, we
obtain

$\displaystyle\text{Re}\theta(k)=-16\beta
l^{2}(l^{3}+5\sqrt{3}k_{0}l^{2}+20k_{0}^{2}l+10\sqrt{3}k_{0}^{3})\leq-16\beta|k-k_{0}|^{5}.$

On the other hand, if $|k-k_{0}|\geq k_{0}$, then $|k-k_{0}|\geq|k|/2$, and
hence

$\displaystyle\text{e}^{-12\beta
t|k-k_{0}|^{5}}\leq\text{e}^{-\frac{3}{8}\beta t|k|^{5}}.$

If $|k-k_{0}|<k_{0}$, then $|k|\leq Ct^{-\frac{1}{5}}$, and so

$\displaystyle\text{e}^{-12\beta t|k-k_{0}|^{5}}\leq 1\leq
C\text{e}^{-\frac{3}{8}\beta t|k|^{5}}.$

Thus, for $k=k_{0}+l\text{e}^{\frac{\pi\text{i}}{6}}$, $0\leq
l\leq\varepsilon$, we find

$\text{e}^{-\frac{3}{4}t|\text{Re}\theta|}\leq\text{e}^{-12\beta
t|k-k_{0}|^{5}}\leq C\text{e}^{-\frac{3}{8}\beta t|k|^{5}}\leq
C\text{e}^{-\frac{3}{160}|z|^{5}}.$ (4.24)

As a consequence, we have

$\displaystyle|v^{(1)}-v^{0}|$ $\displaystyle\leq
C|r_{a}(x,t,k)-r(k_{0})|\text{e}^{t\text{Re}\theta}+C|r(k_{0})-r(0)|\text{e}^{t\text{Re}\theta}$
(4.25) $\displaystyle\leq
C|k-k_{0}|\text{e}^{-\frac{3}{4}t|\text{Re}\theta|}+Ck_{0}\text{e}^{-t|\text{Re}\theta|}\leq
C|zt^{-\frac{1}{5}}|\text{e}^{-\frac{3}{160}|z|^{5}}.$

A similar computation shows that (4.25) also holds for
$k=-k_{0}+l\text{e}^{\frac{5\pi\text{i}}{6}}$, $0\leq l\leq\varepsilon$.
Consequently, writing $l=|z|$,

$\|v^{(1)}-v^{0}\|_{L^{\infty}(\Gamma^{\varepsilon}_{1})}\leq
Ct^{-\frac{1}{5}},$ (4.26)

and

$\|v^{(1)}-v^{0}\|_{L^{1}(\Gamma^{\varepsilon}_{1})}\leq
C\int_{0}^{\infty}lt^{-\frac{1}{5}}\text{e}^{-\frac{3}{160}l^{5}}\frac{\text{d}l}{t^{\frac{1}{5}}}\leq
Ct^{-\frac{2}{5}}.$ (4.27)

Using the general inequality
$\|f\|_{L^{n}}\leq\|f\|^{1-1/n}_{L^{\infty}}\|f\|_{L^{1}}^{1/n}$, (4.19) holds
for $k\in\Gamma^{\varepsilon}_{1}$. Similar estimates applying to
$\Gamma^{\varepsilon}_{j}$, $j=2,3$ show that (4.19) holds.

The variable $z=(20\beta t)^{\frac{1}{5}}k$ satisfies $|z|=(20\beta
t)^{\frac{1}{5}}\varepsilon$ if $|k|=\varepsilon$. Thus, equation (B.5) yields

$N^{0}(x,t;k)=I+\frac{\text{i}N^{0}_{1}(y)}{(20\beta
t)^{\frac{1}{5}}k}+O(t^{-\frac{2}{5}}),\quad k\in\partial
D_{\varepsilon}(0),~{}t\rightarrow\infty.$ (4.28)

Thus, (4.20) and (4.21) follow from (4.28) and Cauchy’s formula.

Let $\hat{\Gamma}=\Gamma\cup\partial D_{\varepsilon}(0)$ and assume that the
boundary of $D_{\varepsilon}(0)$ is oriented counterclockwise, see Fig. 8.
Define $\hat{N}(x,t;k)$ by

$\hat{N}(x,t;k)=\left\\{\begin{aligned} &N^{(1)}(x,t;k)N^{0}(x,t;k)^{-1},\quad
k\in D_{\varepsilon}(0),\\\ &N^{(1)}(x,t;k),~{}\qquad\qquad\qquad
k\in{\mathbb{C}}\setminus D_{\varepsilon}(0),\end{aligned}\right.$ (4.29)

then $\hat{N}(x,t;k)$ satisfies the following RH problem

$\hat{N}_{+}(x,t;k)=\hat{N}_{-}(x,t;k)\hat{v}(x,t;k),\quad k\in\hat{\Gamma},$
(4.30)

where the jump contour $\hat{\Gamma}=\Gamma^{\varepsilon}\cup\partial
D_{\varepsilon}(0)\cup({\mathbb{R}}\setminus[-k_{0},k_{0}])\cup\hat{\Gamma}^{\prime}$,
$\hat{\Gamma}^{\prime}=\Gamma\setminus({\mathbb{R}}\cup\overline{D_{\varepsilon}(0)})$
is depicted in Fig. 8, and the jump matrix $\hat{v}(x,t;k)$ is given by

$\hat{v}=\left\\{\begin{aligned}
&N^{0}_{-}v^{(1)}(N^{0}_{+})^{-1},~{}~{}~{}k\in\hat{\Gamma}\cap
D_{\varepsilon}(0),\\\ &(N^{0})^{-1},\qquad\qquad k\in\partial
D_{\varepsilon}(0),\\\
&v^{(1)},\qquad\qquad\quad~{}~{}k\in\hat{\Gamma}\setminus\overline{D_{\varepsilon}(0)}.\end{aligned}\right.$
(4.31)

###### Lemma 4.3

Let $\hat{\omega}=\hat{v}-I$. For each $1\leq n\leq\infty$, the following
estimates hold:

$\displaystyle\|\hat{\omega}\|_{L^{n}(\partial D_{\varepsilon}(0))}\leq
Ct^{-\frac{1}{5}},$ (4.32)
$\displaystyle\|\hat{\omega}\|_{L^{n}(\Gamma^{\varepsilon})}\leq
Ct^{-\frac{1}{5}(1+\frac{1}{n})},$ (4.33)
$\displaystyle\|\hat{\omega}\|_{L^{n}({\mathbb{R}}\setminus[-k_{0},k_{0}])}\leq
Ct^{-\frac{3}{2}},$ (4.34)
$\displaystyle\|\hat{\omega}\|_{L^{n}(\hat{\Gamma}^{\prime})}\leq
C\text{e}^{-ct}.$ (4.35)

Proof.The estimate (4.32) follows from (4.20). For $k\in\Gamma^{\varepsilon}$,
we have

$\displaystyle\hat{\omega}=N^{0}_{-}(v^{(1)}-v^{0})(N^{0}_{+})^{-1},$

as a consequence, (4.18) and (4.19) imply (4.33). On
${\mathbb{R}}\setminus[-k_{0},k_{0}]$, the jump matrix $\hat{\omega}$ only
involves the small remainder $r_{r}$, so the estimate (4.34) holds as a
consequence of Lemma 4.1 and (4.18). Finally, (4.35) follows from
$\text{e}^{-t|\text{Re}\theta|}\leq C\text{e}^{-ct}$ uniformly on
$\hat{\Gamma}^{\prime}$.

As the discussion in Subsection 3.3, the estimates in Lemma 4.3 show that the
RH problem (4.30) for $\hat{N}$ has a unique solution given by

$\hat{N}(x,t;k)=I+\frac{1}{2\pi\text{i}}\int_{\hat{\Gamma}}(\hat{\mu}\hat{\omega})(x,t;s)\frac{\text{d}s}{s-k},$
(4.36)

where $\hat{\mu}=I+(I-\hat{C}_{\hat{w}})^{-1}\hat{C}_{\hat{w}}I,$ and
$\hat{C}_{\hat{w}}f=\hat{C}_{-}(f\hat{w})$, $\hat{C}$ denote the Cauchy
operator associated with $\hat{\Gamma}$. Moreover, the function
$\hat{\mu}(x,t;k)$ satisfies

$\|\hat{\mu}(x,t;\cdot)-I\|_{L^{2}(\hat{\Sigma})}=O(t^{-\frac{1}{5}}),\quad
t\rightarrow\infty,~{}(x,t)\in\mathcal{P}.$ (4.37)

It follows from (4.36) that

$\lim_{k\rightarrow\infty}k(\hat{N}(x,t;k)-I)=-\frac{1}{2\pi\text{i}}\int_{\hat{\Gamma}}(\hat{\mu}\hat{\omega})(x,t;k)\text{d}k.$
(4.38)

By (4.21), (4.32) and (4.37), we can get

$\displaystyle\quad-\frac{1}{2\pi\text{i}}\int_{\partial
D_{\varepsilon}(0)}(\hat{\mu}\hat{\omega})(x,t;k)\text{d}k$
$\displaystyle=-\frac{1}{2\pi\text{i}}\int_{\partial
D_{\varepsilon}(0)}\hat{\omega}(x,t;k)\text{d}k-\frac{1}{2\pi\text{i}}\int_{\partial
D_{\varepsilon}(0)}(\hat{\mu}(x,t;k)-I)\hat{\omega}(x,t;k)\text{d}k$
$\displaystyle=-\frac{1}{2\pi\text{i}}\int_{\partial
D_{\varepsilon}(0)}\bigg{(}(N^{0})^{-1}(x,t;k)-I\bigg{)}\text{d}k+O(\|\hat{\mu}-I\|_{L^{2}(\partial
D_{\varepsilon}(0))}\|\hat{w}\|_{L^{2}(\partial D_{\varepsilon}(0))})$
$\displaystyle=\frac{\text{i}N_{1}^{0}(y)}{(20\beta
t)^{\frac{1}{5}}}+O(t^{-\frac{2}{5}}),\quad t\rightarrow\infty.$

Using (4.33) and (4.37), we have

$\displaystyle\int_{\Gamma^{\varepsilon}}(\hat{\mu}\hat{\omega})(x,t;k)\text{d}k$
$\displaystyle=$
$\displaystyle\int_{\Gamma^{\varepsilon}}\hat{\omega}(x,t;k)\text{d}k+\int_{\Gamma^{\varepsilon}}(\hat{\mu}(x,t;k)-I)\hat{\omega}(x,t;k)\text{d}k$
$\displaystyle\leq$
$\displaystyle\|\hat{\omega}\|_{L^{1}(\Gamma^{\varepsilon})}+\|\hat{\mu}-I\|_{L^{2}(\Gamma^{\varepsilon})}\|\hat{\omega}\|_{L^{2}(\Gamma^{\varepsilon})}$
$\displaystyle\leq$ $\displaystyle Ct^{-\frac{2}{5}},\quad
t\rightarrow\infty.$

By (3.34) and (3.37), the contribution from
${\mathbb{R}}\setminus[-k_{0},k_{0}]$ to the right-hand side of (4.38) is
$O(t^{-\frac{3}{2}})$, and similarly, by (3.35) and (3.37), the contribution
from $\hat{\Gamma}^{\prime}$ to the right-hand side of (4.38) is
$O(\text{e}^{-ct}),$ as $t\rightarrow\infty.$ Thus, we obtain the following
important relation

$\displaystyle\begin{aligned}
\lim_{k\rightarrow\infty}k(\hat{M}(x,t;k)-I)&=\frac{\text{i}N_{1}^{0}(y)}{(20\beta
t)^{\frac{1}{5}}}+O(t^{-\frac{2}{5}}),\quad t\rightarrow\infty.\end{aligned}$
(4.39)

Recalling the definition of $N_{1}^{0}(y)$ in (4.22) and the relation (4.6),
we obtain our another main results stated in the Theorem 1.2.

###### Remark 4.2

We did not directly consider the asymptotic behavior of the solution to
equation (1.2) in region $\mathcal{P}$ because there is no suitable scale
transformation to eliminate $t$ from the coefficients of $k^{3}$ and $k^{5}$
terms of the phase function $\Phi(k)$ given by (3.2) at the same time.

Acknowledgments.

N. Liu was supported by the China Postdoctoral Science Foundation under Grant
no. 2019TQ0041.

Appendix A. Fourth order Painlevé II RH problem

Fig. 9: The oriented contour $Y$.

Let $Y$ denote the contour $Y=Y_{1}\cup Y_{2}$ oriented as in Fig. 9, where

$\displaystyle Y_{1}=\\{l\text{e}^{\frac{\pi\text{i}}{6}}|l\geq
0\\}\cup\\{l\text{e}^{\frac{5\pi\text{i}}{6}}|l\geq 0\\},\quad
Y_{2}=\\{l\text{e}^{-\frac{\pi\text{i}}{6}}|l\geq
0\\}\cup\\{l\text{e}^{-\frac{5\pi\text{i}}{6}}|l\geq 0\\}.$

Lemma A.1 (Fourth order Painlevé II RH problem) Let $s\in{\mathbb{C}}$ be a
complex number. Then the following RH problems parametrized by
$y\in{\mathbb{R}},s\in{\mathbb{C}}$:

$\left\\{\begin{aligned} &N^{Y}_{+}(y,z)=N^{Y}_{-}(y,z)v^{Y}(y,z),~{}~{}z\in
Y,\\\ &N^{Y}(y,z)\rightarrow
I,\qquad\qquad\quad\qquad~{}z\rightarrow\infty,\end{aligned}\right.$ (A.1)

where the jump matrix $v^{Y}(y,z)$ is defined by

$v^{Y}(y,z)=\left\\{\begin{aligned} &\begin{pmatrix}1&~{}0\\\
s\text{e}^{2\text{i}(\frac{4}{5}z^{5}+yz)}&~{}1\end{pmatrix},\qquad z\in
Y_{1},\\\
&\begin{pmatrix}1&~{}\bar{s}\text{e}^{-2\text{i}(\frac{4}{5}z^{5}+yz)}\\\
0&~{}1\end{pmatrix},~{}~{}\quad z\in Y_{2}.\end{aligned}\right.$ (A.2)

has a unique solution $N^{Y}(y,z)$ for each $y\in{\mathbb{R}}$. Moreover,
there exists smooth functions $\\{N_{j}^{Y}(y)\\}^{4}$ of $y\in{\mathbb{R}}$
with decay as $y\rightarrow\infty$ such that

$N^{Y}(y,z)=I+\sum_{j=1}^{4}\frac{N_{j}^{Y}(y)}{z^{j}}+O(z^{-5}),\quad
z\rightarrow\infty,$ (A.3)

uniformly for $y$ in compact subsets of ${\mathbb{R}}$ and for $\arg
z\in[0,2\pi]$. The leading coefficient $N_{1}^{Y}$ is given by

$N_{1}^{Y}(y)=\text{i}\begin{pmatrix}-2\int^{y}_{\infty}u_{p}^{2}(y^{\prime})\text{d}y^{\prime}~{}&u_{p}(y)\\\
u_{p}(y)~{}&2\int^{y}_{\infty}u_{p}^{2}(y^{\prime})\text{d}y^{\prime}\end{pmatrix},$
(A.4)

where the real-valued function $u_{p}(y)$ satisfies the following fourth order
Painlevé II equation (see [20])

$u_{p}^{{}^{\prime\prime\prime\prime}}(y)+40u_{p}^{2}(y)u_{p}^{{}^{\prime\prime}}(y)+40u_{p}(y)u_{p}^{\prime
2}(y)+96u_{p}^{5}(y)+4yu_{p}(y)=0.$ (A.5)

Proof.The jump matrix $v^{Y}$ admits the symmetries

$v^{Y}(y,z)=(v^{Y})^{\dagger}(y,\bar{z})=\overline{v^{Y}(y,-\bar{z})}.$ (A.6)

We infer from the first of these symmetries that the RH problem for
$N^{Y}(y,z)$ admits a vanishing lemma, as a consequence, there exists a unique
solution $N^{Y}(y,z)$ which admits an expansion of the form (A.3). Assume that

$(N^{Y})^{-1}(y,z)=I+\sum_{j=1}^{4}\frac{\varphi_{j}(y)}{z^{j}}+O(z^{-5}),$
(A.7)

a direct calculation shows that

$\displaystyle\varphi_{1}$
$\displaystyle=-N_{1}^{Y},~{}\varphi_{2}=(N_{1}^{Y})^{2}-N_{2}^{Y},~{}\varphi_{3}=N_{1}^{Y}N_{2}^{Y}+N_{2}^{Y}N_{1}^{Y}-(N_{1}^{Y})^{3}-N_{3}^{Y},$
(A.8) $\displaystyle\varphi_{4}$
$\displaystyle=(N_{1}^{Y})^{4}+N_{1}^{Y}N_{3}^{Y}+N_{3}^{Y}N_{1}^{Y}-(N_{1}^{Y})^{2}N_{2}^{Y}-N_{1}^{Y}N_{2}^{Y}N_{1}^{Y}-N_{2}^{Y}(N_{1}^{Y})^{2}+(N_{2}^{Y})^{2}-N_{4}^{Y}.$

Let $\phi(y,z)=N^{Y}(y,z)\text{e}^{-\text{i}(\frac{4}{5}z^{5}+yz)\sigma_{3}}$.
Then the function $\mathcal{Y}(y,z)$ defined by

$\mathcal{Y}=\phi_{y}\phi^{-1}=(N_{y}^{Y}-\text{i}zN^{Y}\sigma_{3})(N^{Y})^{-1}$
(A.9)

is an entire function of $z$, hence according to (A.3), (A.7) and (A.8), we
have

$\mathcal{Y}(y,z)=-\text{i}z\sigma_{3}+\text{i}[\sigma_{3},N_{1}^{Y}].$ (A.10)

Thus, we find

$N_{y}^{Y}-\text{i}zN^{Y}\sigma_{3}=\mathcal{Y}N^{Y}.$ (A.11)

Substituting the expansion (A.3) into (A.11) and collecting terms with
$O(z^{-n})$, one can get

$\displaystyle N_{1y}^{Y}+\text{i}[\sigma_{3},N_{2}^{Y}]$
$\displaystyle=\text{i}[\sigma_{3},N_{1}^{Y}]N_{1}^{Y},$ (A.12) $\displaystyle
N_{2y}^{Y}+\text{i}[\sigma_{3},N_{3}^{Y}]$
$\displaystyle=\text{i}[\sigma_{3},N_{1}^{Y}]N_{2}^{Y},$ $\displaystyle
N_{3y}^{Y}+\text{i}[\sigma_{3},N_{4}^{Y}]$
$\displaystyle=\text{i}[\sigma_{3},N_{1}^{Y}]N_{3}^{Y}.$

Accordingly, since

$\mathcal{Z}=\phi_{z}\phi^{-1}=\bigg{(}N_{z}^{Y}-\text{i}(4z^{4}+y)N^{Y}\sigma_{3}\bigg{)}(N^{Y})^{-1}$
(A.13)

is entire, and thus we get

$\mathcal{Z}=\mathcal{Z}_{0}+\mathcal{Z}_{1}z+\mathcal{Z}_{2}z^{2}+\mathcal{Z}_{3}z^{3}+\mathcal{Z}_{4}z^{4}.$
(A.14)

Substituting the expansion (A.3) and (A.7) into (A.13), it follows from (A.8)
and (A.12) that

$\displaystyle\mathcal{Z}_{4}=-4\text{i}\sigma_{3},~{}\mathcal{Z}_{3}=4\text{i}[\sigma_{3},N_{1}^{Y}],~{}\mathcal{Z}_{2}=-4N_{1y}^{Y},~{}\mathcal{Z}_{1}=-4N_{2y}^{Y}+4N_{1y}^{Y}N_{1}^{Y},$
(A.15)
$\displaystyle\mathcal{Z}_{0}=-4N_{3y}^{Y}-4N_{1y}^{Y}(N_{1}^{Y})^{2}+4N_{1y}^{Y}N_{2}^{Y}+4N_{2y}^{Y}N_{1}^{Y}-\text{i}y\sigma_{3}.$

We have shown that $\phi$ obeys the Lax pair equations

$\left\\{\begin{aligned} \phi_{y}=&\mathcal{Y}\phi,\\\
\phi_{z}=&\mathcal{Z}\phi,\end{aligned}\right.$ (A.16)

where $\mathcal{Y}$ and $\mathcal{Z}$ are given by (A.10) and (A.14),
respectively.

The symmetries (A.6) of the jump matrix $v^{Y}(y,z)$ implies that $N^{Y}(y,z)$
satisfies the symmetries

$N^{Y}(y,z)=(N^{Y})^{\dagger}(y,\bar{z})^{-1}=\sigma_{2}N^{Y}(y,-z)\sigma_{2}.$
(A.17)

In particular, the coefficients $N_{1}^{Y}(y)$, $N_{2}^{Y}(y)$ and
$N^{Y}_{3}(y)$ satisfy

$\displaystyle N_{1}^{Y}=$
$\displaystyle-(N_{1}^{Y})^{\dagger}=-\sigma_{2}N_{1}^{Y}\sigma_{2},$ (A.18)
$\displaystyle N_{2}^{Y}=$
$\displaystyle\sigma_{2}N_{2}^{Y}\sigma_{2},~{}N_{3}^{Y}=-\sigma_{2}N_{3}^{Y}\sigma_{2}.$

Therefore, we can write

$\displaystyle N_{1}^{Y}(y)$
$\displaystyle=\begin{pmatrix}\psi_{1}(y)~{}&\psi_{2}(y)\\\
\psi_{2}(y)~{}&-\psi_{1}(y)\end{pmatrix},$ $\displaystyle N_{2}^{Y}(y)$
$\displaystyle=\begin{pmatrix}f_{1}(y)~{}&f_{2}(y)\\\
-f_{2}(y)~{}&f_{1}(y)\end{pmatrix},$ $\displaystyle N_{3}^{Y}(y)$
$\displaystyle=\begin{pmatrix}g_{1}(y)~{}&g_{2}(y)\\\
g_{2}(y)~{}&-g_{1}(y)\end{pmatrix},$

where $\\{\psi_{j}(y),f_{j}(y),g_{j}(y)\\}_{1}^{2}$ are complex-valued
functions and $\psi_{1}(y),\psi_{2}(y)\in\text{i}{\mathbb{R}}$. Then the
compatibility condition

$\mathcal{Y}_{z}-\mathcal{Z}_{y}+\mathcal{Y}\mathcal{Z}-\mathcal{Z}\mathcal{Y}=0$
(A.20)

of the Lax pair (A.16) can then rewrite as

$-\text{i}\sigma_{3}-\mathcal{Z}_{0y}+\text{i}[\sigma_{3},N_{1}^{Y}]\mathcal{Z}_{0}-\text{i}\mathcal{Z}_{0}[\sigma_{3},N_{1}^{Y}]=0,$
(A.21)

since one can directly calculate that the coefficients of $z,z^{2},z^{3}$ and
$z^{4}$ in (A.20) vanish identically. On the other hand, substituting (A.19)
into (A.12), we find

$\left\\{\begin{aligned} \psi_{1}^{\prime}&=2\text{i}\psi_{2}^{2},\\\
\psi_{2}^{\prime}+2\text{i}f_{2}&=-2\text{i}\psi_{1}\psi_{2},\\\
f_{1}^{\prime}&=-2\text{i}\psi_{2}f_{2},\\\
f_{2}^{\prime}+2\text{i}g_{2}&=2\text{i}\psi_{2}f_{1},\\\
g_{1}^{\prime}&=2\text{i}\psi_{2}g_{2}.\end{aligned}\right.$ (A.22)

Substituting (A.19) into (A.21) and using above relations, it follows from
$(12)$-entry of (A.21) that

$\psi_{2}^{\prime\prime\prime\prime}-40\psi_{2}^{2}\psi_{2}^{\prime\prime}-40\psi_{2}\psi_{2}^{\prime
2}+96\psi_{2}^{5}+4y\psi_{2}=0,$ (A.23)

however, the $(11)$-entry of (A.21) vanish identically. If we set
$\psi_{2}(y)=\text{i}u_{p}(y)$, then $u_{p}(y)$ satisfies the fourth order
Painlevé II equation (A.5). The lemma follows.

Appendix B. Model RH problem for sector $\mathcal{P}$

Fig. 10: The oriented contour $Z$.

Given a number $z_{0}\geq 0$, let $Z$ denote the contour $Z=Z_{1}\cup
Z_{1}\cup Z_{3}$, where the line segments

$\displaystyle Z_{1}$
$\displaystyle=\\{z_{0}+l\text{e}^{\frac{\pi\text{i}}{6}}|l\geq
0\\}\cup\\{-z_{0}+l\text{e}^{\frac{5\pi\text{i}}{6}}|l\geq 0\\},$ (B.1)
$\displaystyle Z_{2}$
$\displaystyle=\\{z_{0}+l\text{e}^{-\frac{\pi\text{i}}{6}}|l\geq
0\\}\cup\\{-z_{0}+l\text{e}^{-\frac{5\pi\text{i}}{6}}|l\geq 0\\},$
$\displaystyle Z_{3}$ $\displaystyle=\\{l|-z_{0}\leq l\leq z_{0}\\}$

are oriented as in Fig. 10. It turns out that the long time asymptotics in the
sector $\mathcal{P}$ is related to the solution $N^{Z}$ of the following
family of RH problems parametrized by $y<0,s\in{\mathbb{C}},z_{0}\geq 0$:

$\left\\{\begin{aligned}
&N^{Z}_{+}(s,y,z,z_{0})=N^{Z}_{-}(s,y,z,z_{0})v^{Z}(s,y,z,z_{0}),~{}z\in Z,\\\
&N^{Z}(s,y,z,z_{0})\rightarrow
I,\qquad\qquad\qquad\qquad\quad\qquad~{}z\rightarrow\infty,\end{aligned}\right.$
(B.2)

where the jump matrix $v^{Z}(s,y,z,z_{0})$ is defined by

$v^{Z}(s,y,z,z_{0})=\left\\{\begin{aligned} &\begin{pmatrix}1&~{}0\\\
s\text{e}^{2\text{i}(\frac{4}{5}z^{5}+yz)}&~{}1\end{pmatrix},\qquad\qquad\qquad\qquad~{}~{}z\in
Z_{1},\\\
&\begin{pmatrix}1&~{}\bar{s}\text{e}^{-2\text{i}(\frac{4}{5}z^{5}+yz)}\\\
0&~{}1\end{pmatrix},~{}~{}\quad\qquad\qquad\qquad~{}~{}z\in Z_{2},\\\
&\begin{pmatrix}1&~{}\bar{s}\text{e}^{-2\text{i}(\frac{4}{5}z^{5}+yz)}\\\
0&~{}1\end{pmatrix}\begin{pmatrix}1&~{}0\\\
s\text{e}^{2\text{i}(\frac{4}{5}z^{5}+yz)}&~{}1\end{pmatrix},\quad z\in
Z_{3}.\end{aligned}\right.$ (B.3)

Lemma B.1 (Model RH problem for sector $\mathcal{P}$) Define the parameter
subset

$\mathbb{P}=\\{(y,t,z_{0})\in{\mathbb{R}}^{3}|-C_{1}\leq y<0,t\geq
3,\sqrt[4]{-y}/\sqrt{2}\leq z_{0}\leq C_{2}\\},$ (B.4)

where $C_{1},C_{2}>0$ are constants. Then for $(y,t,z_{0})\in\mathbb{P}$, the
RH problem (B.2) has a unique solution $N^{Z}(s,y,z,z_{0})$ which satisfies

$N^{Z}(s,y,z,z_{0})=I+\frac{\text{i}}{z}\begin{pmatrix}-2\int^{y}_{\infty}u_{p}^{2}(y^{\prime})\text{d}y^{\prime}~{}&u_{p}(y)\\\
u_{p}(y)~{}&2\int^{y}_{\infty}u_{p}^{2}(y^{\prime})\text{d}y^{\prime}\end{pmatrix}+O\bigg{(}\frac{1}{z^{2}}\bigg{)},\quad
z\rightarrow\infty,$ (B.5)

where $u_{p}(y)$ denotes the solution of the fourth order Painlevé II equation
(A.5) and $N^{Z}(s,y,z,z_{0})$ is uniformly bounded for
$z\in{\mathbb{C}}\setminus Z$. Furthermore, $N^{Z}$ obeys the symmetries

$N^{Z}(s,y,z,z_{0})=(N^{Z})^{\dagger}(s,y,\bar{z},z_{0})^{-1}=\sigma_{2}N^{Z}(s,y,-z,z_{0})\sigma_{2}.$
(B.6)

Proof.Note that

$\displaystyle\text{Re}\bigg{(}2\text{i}\bigg{(}\frac{4}{5}z^{5}+yz\bigg{)}\bigg{)}\leq
l^{2}\bigg{(}-\frac{4}{5}l^{3}-4\sqrt{3}z_{0}l^{2}-16z_{0}^{2}l-8\sqrt{3}z_{0}^{3}\bigg{)}$

for all $z=z_{0}+l\text{e}^{\frac{\pi\text{i}}{6}}$ and
$z=-z_{0}+l\text{e}^{\frac{5\pi\text{i}}{6}}$ with $l\geq 0,z_{0}\geq 0$ and
$-4z_{0}^{4}\leq y<0$. Thus, we have

$\displaystyle|\text{e}^{2\text{i}(\frac{4}{5}z^{5}+yz)}|\leq
C\text{e}^{-|z\pm z_{0}|^{2}(\frac{4}{5}|z\pm z_{0}|^{3}+4\sqrt{3}z_{0}|z\pm
z_{0}|^{2}+16z_{0}^{2}|z\pm z_{0}|+8\sqrt{3}z_{0}^{3})},\quad z\in Z_{1}.$

Analogous estimates hold for $z\in Z_{2}$. However, $|\text{e}^{\pm
2\text{i}(\frac{4}{5}z^{5}+yz)}|=1$ for $z\in Z_{3}$, this shows that
$v^{Z}-I$ exponentially fast as $z\rightarrow\infty$.

The jump matrix $v^{Z}$ obeys the same symmetries (A.6) as $v^{Y}$. In
particular, $v^{Z}$ is Hermitian and positive definite on $Z\cap{\mathbb{R}}$
and satisfies $v^{Z}(s,y,z,z_{0})=(v^{Z})^{\dagger}(s,y,z,z_{0})$ on
$Z\setminus{\mathbb{R}}$. This implies the existence of a vanishing lemma from
which we deduce the unique existence of the solution $N^{Z}$. The symmetries
(B.6) follow from the symmetries of $v^{Z}$. Moreover, the RH problem (B.2)
for $N^{Z}(s,y,z,z_{0})$ can be transformed into the RH problem (A.1) for
$N^{Y}(y,z)$ up to a trivial contour deformation. Thus (B.5) follows from
(A.3) and (A.4).

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