# The Cauchy problem for the Novikov equation under a nonzero background:
Painlevé asymptotics in two transition zones

Zhaoyu Wang  Xuan Zhou  Engui Fan∗, School of Mathematical Sciences and Key
Laboratory of Mathematics for Nonlinear Science, Fudan University,
Shanghai,200433, China  College of Mathematics and Systems Science, Shandong
University of Science and Technology, Qingdao, 266061, China
* Corresponding author and e-mail address<EMAIL_ADDRESS>
###### Abstract

In this paper, we investigate the Painlevé asymptotics in two transition zones
for the solutions to the Cauchy problem of the Novikov equation under a
nonzero background

$\displaystyle u_{t}-u_{txx}+4u_{x}=3uu_{x}u_{xx}+u^{2}u_{xxx},$
$\displaystyle u(x,0)=u_{0}(x),$

where $u_{0}(x)\rightarrow\kappa>0,\ x\rightarrow\pm\infty$ and
$u_{0}(x)-\kappa$ is assumed in the Schwarz space. This result is established
by performing the $\overline{\partial}$-nonlinear steepest descent analysis to
a Riemann-Hilbert problem associated with the the Cauchy problem in a new
scale $(y,t)$ with

$y=x-\int_{x}^{\infty}\left((u-u_{xx}+1)^{2/3}-1\right)ds,$

for large times in two transition zones $y/t\approx-1/8$ and $y/t\approx 1$
respectively. It shown that the leading order term of the asymptotic
approximation comes from the contribution of solitons and singular points,
while the sub-leading term is related to the solution of the Painlevé II
equation.

###### keywords:

Novikov equation , Riemann-Hilbert problem , $\overline{\partial}$-nonlinear
steepest descent method , Painlevé transcendents , large time asymptotics.
Mathematics Subject Classification: 35Q51; 35Q15; 35C20; 37K15.

###### Contents

  1. 1 Introduction
  2. 2 Inverse Scattering and RH Problem
    1. 2.1 The Lax pair and spectral analysis
    2. 2.2 An RH characterization
  3. 3 Interpolation and Conjugation
  4. 4 Painlevé Asymptotics in $|\xi+\frac{1}{8}|t^{2/3}<C$
    1. 4.1 Opening $\bar{\partial}$-lenses
    2. 4.2 A hybrid $\bar{\partial}$-RH problem and its decomposition
    3. 4.3 Contribution from discrete spectrum
    4. 4.4 Contribution from jump contours
      1. 4.4.1 Local model near critical points
      2. 4.4.2 Small norm RH problem for residual error
    5. 4.5 Contribution from $\bar{\partial}$-components
    6. 4.6 Proof of Theorem 1.1-(i)
  5. 5 Painlevé Asymptotics in $|\xi-1|t^{2/3}<C$
    1. 5.1 Opening $\bar{\partial}$-lenses
    2. 5.2 A hybrid $\bar{\partial}$-RH problem and its decomposition
    3. 5.3 Contribution from discrete spectrum
    4. 5.4 Contribution from jump contours
      1. 5.4.1 Local model near critical points
      2. 5.4.2 Small norm RH problem for residual error
    5. 5.5 Contribution from $\bar{\partial}$-components
    6. 5.6 Proof of Theorem 1.1-(ii)
  6. A Modified Painlevé II RH Problem
  7. B Model RH Problem for Transition Zones

##  1 Introduction

In this paper, we are concerned with the Painlevé asymptotics of solutions to
the Cauchy problem for the Novikov equation under a nonzero background

$\displaystyle u_{t}-u_{txx}+4u_{x}=3uu_{x}u_{xx}+u^{2}u_{xxx},$ (1.1)
$\displaystyle u(x,0)=u_{0}(x),\quad x\in\mathbb{R},\ t>0,$ (1.2)
$\displaystyle u_{0}(x)\to\kappa>0,\ x\to\pm\infty,$ (1.3)

where $u=u(x,t)$ is a real-valued function of $x$ and $t$. By introducing the
momentum variable $m=u-u_{xx}$, the Novikov equation (1.1) can be rewritten as

$\displaystyle m_{t}+\left(m_{x}u+3mu_{x}\right)u=0,$ (1.4)

or equivalently,

$\displaystyle(m^{2/3})_{t}+\left(u^{2}m^{2/3}\right)_{x}=0.$ (1.5)

The Novikov equation (1.1) as a new integrable system was derived in the
search for a classification of integrable generalized Camassa-Holm equations
of the form

$\displaystyle(1-\partial_{x}^{2})u_{t}=F(u,u_{x},u_{xx},...)$ (1.6)

possessing infinite hierarchies of higher symmetries.

The Novikov equation, being an integrable equation, possesses a Lax pair. A
scalar Lax pair involving the third order derivative with respect to $x$ has
been provided [1, 2]. Furthermore, by employing the prolongation algebra
method, Hone and Wang introduced a $3\times 3$ matrix Lax pair and established
a bi-Hamiltonian structure for the Novikov equation (1.1) [3]. This Lax pair
was used to explicitly construct peakon solutions on a zero background,
replicating a feature characterizing the waves of great height-waves of
largest amplitude that were exact solutions of the governing equations for
water waves [3, 4, 5, 6, 7]. Hone et al. further derived the explicit formulas
for multipeakon solutions of the Novikov equation (1.1) [8]. Matsuno, using
the Hirota bilinear method, presented parametric representations of smooth
multisoliton solutions for the Novikov equation (1.1) on a nonzero constant
background [9]. He also demonstrated that a smooth soliton converges to a
peakon in the limit where the constant background approaches zero while
maintaining a fixed soliton velocity. Wu et al. obtained $N$-soliton solutions
for the Novikov equation through Darboux transformations [10]. Recently, Chang
et al. applied Pfaffian technique to investigate multipeakons of the Novikov
equation, establishing a connection between the Novikov peakons and the finite
Toda lattice of BKP type, as well as employing Hermite-Pade approximation to
address the Novikov peakon problem [11, 12]. Boutet de Monvel et al. developed
the inverse scattering theory to the Novikov equation (1.1) with a nonzero
constant background [13]. Under a transformation

$\displaystyle
u(x,t)\rightarrow\kappa\tilde{u}(x-\kappa^{2}t,\kappa^{2}t)+\kappa,$ (1.7)

the Cauchy problem of the Novikov equation (1.1)–(1.3) reduces to the
following new Cauchy problem on zero background

$\displaystyle(\tilde{m}^{2/3})_{t}+\left(\tilde{m}^{2/3}\left(u^{2}+2u\right)\right)_{x}=0,\
\tilde{m}=u-u_{xx}+1,$ (1.8) $\displaystyle u(x,0)=u_{0}(x),$ (1.9)

where $u_{0}(x)$ satisfies the sign condition

$u_{0}(x)-u_{0,xx}(x)+1>0.$

Then there exists a unique global solution $u(x,t)$ of the Novikov equation
(1.1), such that $u(x,t)\to 0$ as $x\to\pm\infty$ for all $t>0$ [14]. Building
upon the above characteristic, a Riemann-Hilbert (RH) formalism for the Cauchy
problem (1.8)-(1.9) has been developed in [13] .

The Novikov equation and DP equation possess numerous common characteristics
in their RH problem and face some difficulties. One of the difficulties is the
Lax pair associated with (1.8) has six spectral singularities at
$\varkappa_{n}=e^{\frac{n\pi i}{3}}$ for $n=1,...,6$, which means that the
corresponding RH problem also exhibits spectral singularities. During the
large time asymptotic analysis for the DP equation, Boutet de Monvel et al.
considered a row vector RH problem without singularities to avoid the impact
of singularities [15, 16]. On the contrary, in the case of the Novikov
equation, the situation is different. The solution of the similar row vector
RH problem cannot be directly used to recover $e^{x(y,t)-y}$ [13]. Recently
this difficulty was overcome by the establishment of a small-norm RH problem
near the singular points [17]. Further they obtained the different large time
asymptotic expansions for the solution of the Cauchy problem (1.8)-(1.9) of
the Novikov equation in four different space-time regions

${\rm I.}\ \xi<-1/8;\ {\rm II.}\ -1/8<\xi<0;\ {\rm III.}\ 0<\xi<1,\ \ {\rm
IV.}\ \xi>1,$

with $\xi=y/t$ (The definition of $y$ is in (2.33) and we use the parameter
$y$ to avoid the dependence of the jump matrix relating the two sides of the
RH problem on the parameter $x$ and $t$ and the multivalued nature of the
residue conditions with poles [15]). See Figure 1.1. The main tool is the
$\bar{\partial}$ nonlinear steepest approach introduced by McLaughlin-Miller
[18, 19], which has applied to analyze the large time asymptotics of
integrable systems [20, 21, 22, 23, 24, 25]. The remaining question is: How to
describe the asymptotics of the solution to the Cauchy problem (1.8)-(1.9) in
the transition zones ${\rm V.}\ \xi\approx-1/8$ and ${\rm VI.}\ \xi\approx 1$
as illustrated in Figure 1.1 ?

It is the aim of the present work to fulfill this gap. We present the large
time asymptotic formulas of the Novikov equation in two transition zones ${\rm
V.}\ \xi\approx-1/8$ and ${\rm VI.}\ \xi\approx 1$. The leading term of the
asymptotic expansion is influenced by the discrete spectrum and the
singularities and the sub-leading term is in terms of the solutions of the
Painlevé II equation. Our main result is stated as follows.

$0$yt$II$$III$$I$$VI$$V$$VI$$\xi=-1/8$$\xi=1$ Figure 1.1: The different
asymptotic regions of the Novikov equation in the $(y,t)$-half plane, where
$\xi=y/t$.

######  Theorem 1.1.

Let $u(x,t)=u(y(x,t),t)$ be the solution for the Novikov equation (1.8) with
generic initial data $u_{0}\in\mathcal{S}(\mathbb{R})$ and its associated with
scatting data
$\left\\{r(z),\left\\{\zeta_{n},c_{n}\right\\}^{6N_{0}}_{n=1}\right\\}$. Let
$\xi=y/t$ and $u^{\lozenge}(y,t)$ be the $\mathcal{N}(\lozenge)$-soliton
solution corresponding to the scattering data
$\tilde{\mathcal{D}_{\lozenge}}=\left\\{0,\left\\{\zeta_{n},C_{n}T^{2}(\zeta_{n})\right\\}_{n\in\lozenge}\right\\}$
shown in Corollary 4.1 and Corollary 5.1. Then there exists a large constant
$T_{1}$ such that for all $t>T_{1}$, we have

  * (i)

In the first transition zone $|\xi+\frac{1}{8}|t^{2/3}<C$,

$\displaystyle u(y,t)$
$\displaystyle=u^{\lozenge}(y,t;\tilde{\mathcal{D}}_{\lozenge})\left(T_{1}(e^{\frac{\pi
i}{6}})T_{3}(e^{\frac{\pi i}{6}})\right)^{-1/2}-1$
$\displaystyle+\frac{1}{2}\left(T_{1}(e^{\frac{\pi i}{6}})T_{3}(e^{\frac{\pi
i}{6}})\right)^{-1/2}f_{11}t^{-1/3}+\mathcal{O}(t^{-2/3+2\delta_{1}}),$ (1.10)
$\displaystyle x(y,t)$
$\displaystyle=x^{\lozenge}(y,t;\tilde{\mathcal{D}}_{\lozenge})+\frac{1}{2}\ln
T_{13}(e^{\frac{\pi
i}{6}})+\frac{1}{2}f_{12}t^{-1/3}+\mathcal{O}(t^{-2/3+2\delta_{1}}),$ (1.11)

where $u^{\lozenge}(y,t;\tilde{\mathcal{D}}_{\lozenge})$ and
$x^{\lozenge}(y,t;\tilde{\mathcal{D}}_{\lozenge})$ are defined in Corollary
4.1, $T_{13}$, $f_{11}$, and $f_{12}$ are given by (3.65), (4.206), and
(4.207) respectively. More importantly,
$u^{\lozenge}(y,t;\tilde{\mathcal{D}}_{\lozenge})$ and
$x^{\lozenge}(y,t;\tilde{\mathcal{D}}_{\lozenge})$ include contributions from
solitons and singularities, while functions $f_{11}$ and $f_{12}$ are
constructed by the Painlevé II equation

$u_{ss}=2u^{3}+su,\quad s\in\mathbb{R},$ (1.12)

characterized by

$u(s)\sim\left|r\left(\frac{\sqrt{7}+\sqrt{3}}{2}\right)\right|{\rm
Ai}(s),\quad s\to-\infty.$ (1.13)

  * (ii)

In the second transition zone $|\xi-1|t^{2/3}<C$,

$\displaystyle u(y,t)$
$\displaystyle=u^{\lozenge}(y,t;\tilde{\mathcal{D}}_{\lozenge})\left(T_{1}(e^{\frac{\pi
i}{6}})T_{3}(e^{\frac{\pi
i}{6}})\right)^{-1/2}-1+\frac{1}{2}\left(T_{1}(e^{\frac{\pi
i}{6}})T_{3}(e^{\frac{\pi i}{6}})\right)^{-1/2}f_{21}t^{-1/3}$
$\displaystyle+\frac{1}{2}\left(T_{1}(e^{\frac{\pi i}{6}})T_{3}(e^{\frac{\pi
i}{6}})\right)^{-1/2}f_{23}t^{-2/3}+\mathcal{O}(t^{-1/3-3\delta_{2}}),$ (1.14)
$\displaystyle x(y,t)$
$\displaystyle=x^{\lozenge}(y,t;\tilde{\mathcal{D}}_{\lozenge})+\frac{1}{2}\ln
T_{13}(e^{\frac{\pi
i}{6}})+\frac{1}{2}f_{22}t^{-1/3}+\frac{1}{4}f_{24}t^{-2/3}+\mathcal{O}(t^{-2/3+2\delta_{1}}),$
(1.15)

where $u^{\lozenge}(y,t;\tilde{\mathcal{D}}_{\lozenge})$ and
$x^{\lozenge}(y,t;\tilde{\mathcal{D}}_{\lozenge})$ are defined in Corollary
5.1, $T_{13}$ has the same definition with the above result (i), and $f_{2j}$,
$j=1,\cdots,4$, are defined by (5.149)-(5.151), respectively, which are
constructed by the Painlevé II equation (1.12) with

$u(s)\sim-\left|R_{a}\right|{\rm Ai}(s),\quad s\to-\infty,$ (1.16)

and $R_{a}$ is defined by (5.96).

######  Remark 1.2.

Our main tools are the $\bar{\partial}$ nonlinear steepest descent method and
double technique used in our precious paper [25]. To deal with singularities
in the RH problem, we employ a more simple and effective technique by
transferring the singularities into the transformation, which is different
from that in [17].

The organization of our paper is as follows:

In Section 2, we quickly review some basic results, especially the
construction of a RH problem for $M(z)$ related to the Cauchy problem
(1.1)-(1.3), which will be used to analyze the large time asymptotics of the
Novikov equation. For details, refer to [13, 17].

In Section 3, we convert the original RH problem to a RH problem for
$M^{(2)}(z)$, whose jump contour can be opened with different factorizations
of the jump matrix (2.72) according to the sign of the phase functions (2.68).

In Section 4, we focus on the large time asymptotic analysis in the first
transition zone $|\xi+1/8|t^{2/3}<C$ with the following steps. First of all,
introducing a matrix-valued function $\mathcal{R}^{(3)}(z)$ to make a
continuous extension of $M^{(2)}(z)$ into a hybrid $\bar{\partial}$-RH problem
for $M^{(3)}(z)$, which can be decomposed into a pure RH problem for
$M^{R}(z)$ and a pure $\bar{\partial}$-problem for $M^{(4)}(z)$. We observe
that the contribution to $M^{R}(z)$ arises from two distinct components. The
first contribution originates from the discrete spectrum, where a modified
reflectionless RH problem $M^{O}(z)$ is solved in Subsection 4.3. The second
contribution comes from jump contours, which are approximated by local
solvable models $M^{L}(z)$ near the phase points after colliding in Subsection
4.4. The residual error function results from a small norm RH problem outside
the neighborhood of phase points. The contribution associated with the
$\bar{\partial}$-problem for $M^{(4)}(z)$ will be analyzed in Subsection 4.5.
Finally in Subsection 4.6, based on the result obtained above,we give the
proof of Theorem 1.1-(i).

In Section 5, we investigate the Painlevé asymptotics in the second transition
zone $|\xi-1|t^{2/3}<C$ and complete the proof of Theorem 1.1-(ii) with a
similar way to Section 4.

##  2 Inverse Scattering and RH Problem

In this section, we state some main results on the inverse scattering
transform and the RH problem associated with the Cauchy problem (1.1)–(1.3).
The details can be found in [13, 17].

###  2.1 The Lax pair and spectral analysis

The Novikov equation (1.8) admits the Lax pair [3, 13]

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

where $k$ is a spectral parameter, $\breve{\Phi}_{x}(k)$ is a $3\times 3$
matrix valued eigenfunction, the matrices $\breve{X}$ and $\breve{T}$ are
defined by

$\displaystyle\breve{X}=\left(\begin{array}[]{ccc}0&k\tilde{m}&1\\\
0&0&k\tilde{m}\\\ 1&0&0\end{array}\right),$ (2.5)
$\displaystyle\breve{T}=\left(\begin{array}[]{ccc}-(u+1)u_{x}+\frac{1}{3k^{2}}&\frac{u_{x}}{k}-(u^{2}+2u)k\tilde{m}&u_{x}^{2}+1\\\
\frac{u+1}{k}&-\frac{2}{3k^{2}}&-\frac{u_{x}}{k}-(u^{2}+2u)k\tilde{m}\\\
-u^{2}-2u&\frac{u+1}{k}&(u+1)u_{x}+\frac{1}{3k^{2}}\end{array}\right).$ (2.9)

Let $\omega=e^{\frac{2i\pi}{3}}$. Define $\lambda_{j}(z)$, $j=1,2,3$ are three
roots of the following algebraic equation

$\displaystyle\lambda^{3}(z)-\lambda(z)=k^{2}(z),$ (2.10)

where

$\displaystyle
k(z)^{2}=\frac{1}{3\sqrt{3}}\left(z^{3}+\frac{1}{z^{3}}\right),\quad\lambda_{j}(z)=\frac{1}{\sqrt{3}}\left(\omega^{j}z+\frac{1}{\omega^{j}z}\right).$
(2.11)

Moreover, $k(\kappa_{n})=0$, for
$\kappa_{n}=e^{\frac{i\pi}{6}+\frac{i\pi(n-1)}{3}}$, $n=1,...,6$.

Let

$\displaystyle D(x,t):=\text{diag}\\{q(x,t),q^{-1}(x,t),1\\},\quad
q=q(x,t):=\tilde{m}^{1/3}(x,t),$ $\displaystyle
P(z):=\left(\begin{array}[]{ccc}\lambda_{1}^{2}(z)&\lambda_{2}^{2}(z)&\lambda_{3}^{2}(z)\\\
k(z)&k(z)&k(z)\\\
\lambda_{1}(z)&\lambda_{2}(z)&\lambda_{3}(z)\end{array}\right),$ (2.15)
$\displaystyle
P^{-1}(z):=\left(\begin{array}[]{ccc}\frac{1}{3\lambda_{1}^{2}(z)-1}&0&0\\\
0&\frac{1}{3\lambda_{2}^{2}(z)-1}&0\\\
0&0&\frac{1}{3\lambda_{3}^{2}(z)-1}\end{array}\right)\left(\begin{array}[]{ccc}1&\frac{z}{\lambda_{1}(z)}&\lambda_{1}(z)\\\
1&\frac{z}{\lambda_{2}(z)}&\lambda_{2}(z)\\\
1&\frac{z}{\lambda_{3}(z)}&\lambda_{3}(z)\end{array}\right).$ (2.22)

In order to control the large $k$ behavior of the solutions of above Lax pair,
we introduce a new transformation

$\displaystyle\Phi(z):=P^{-1}(z)D^{-1}(x,t)\breve{\Phi}(k),$ (2.23)

where satisfies the Lax pair

$\displaystyle\Phi_{x}-q^{2}\Lambda(z)\Phi=U\Phi,$ (2.24)
$\displaystyle\Phi_{t}+\left[(u^{2}+2u)q^{2}\Lambda(z)-A(z)\right]\Phi=V\Phi,$
(2.25)

where

$\displaystyle\Lambda(z):=\text{diag}\\{\lambda_{1}(z),\lambda_{2}(z),\lambda_{3}(z)\\},\
\ A(z):=\frac{1}{3k^{2}}+\Lambda(z)^{-1},$ $\displaystyle U:=U_{1}U_{2},\ \
V:=U_{1}(V_{1}+V_{2}\Lambda),$ (2.26) $\displaystyle
U_{1}:=\text{diag}\left\\{\frac{1}{3\lambda_{1}^{2}-1},\frac{1}{3\lambda_{2}^{2}-1},\frac{1}{3\lambda_{3}^{2}-1}\right\\},$
$\displaystyle
U_{2}:=\left(\begin{array}[]{ccc}c_{2}\lambda_{1}&c_{1}(\lambda_{1}\lambda_{2}-\lambda_{2}^{2})+c_{2}\lambda_{2}&c_{1}(\lambda_{1}\lambda_{3}-\lambda_{3}^{2})+c_{2}\lambda_{3}\\\
c_{1}(\lambda_{1}\lambda_{2}-\lambda_{1}^{2})+c_{2}\lambda_{1}&c_{2}\lambda_{2}&c_{1}(\lambda_{3}\lambda_{2}-\lambda_{3}^{2})+c_{2}\lambda_{3}\\\
c_{1}(\lambda_{1}\lambda_{3}-\lambda_{3}^{2})+c_{2}\lambda_{1}&c_{1}(\lambda_{3}\lambda_{2}-\lambda_{2}^{2})+c_{2}\lambda_{2}&c_{2}\lambda_{3}\end{array}\right),$
(2.30)

with $c_{1}=\frac{q_{x}}{q}$ and $c_{2}=q^{-2}-q^{2}$. $V_{1}$ has same form
of $U_{1}$ with $c_{1}$ and $c_{2}$ replaced by
$c_{3}=-(u^{2}+2u)\frac{q_{x}}{q}$ and
$c_{4}=(u^{2}+2u)q^{2}+\frac{u_{x}^{2}+1}{q^{2}}-1$ , respectively. And we
denote $c_{5}=\frac{u_{x}}{q}$, $c_{6}=(u+1)q-1$, then $V_{2}$ can be written
as

$\displaystyle V_{2}:=[V_{2}^{(jl)}]_{3\times 3},\quad
V_{2}^{(jl)}:=c_{5}\left(\frac{1}{\lambda_{l}}-\frac{1}{\lambda_{j}}\right)+c_{6}\left(\frac{\lambda_{j}}{\lambda_{l}}+\frac{\lambda_{l}}{\lambda_{j}}\right).$
(2.31)

Introduce a new eigenfunction $\mu(z;x,t)$ satisfying

$\displaystyle\Phi(z;x,t)=\mu(z;x,t)e^{Q},$ (2.32)

where $Q(z;x,t)$ is a $3\times 3$ diagonal function

$Q(z;x,t)=y(x,t)\Lambda(z)+tA(z),$

where

$\displaystyle Q_{x}=q^{2}\Lambda,\quad
Q_{t}=-\left(u^{2}+2u\right)q^{2}\Lambda-A,$ $\displaystyle
y(x,t)=x-\int_{x}^{\infty}\left(q^{2}(s,t)-1\right)ds.$ (2.33)

Then the Lax pair (2.25) is changed to

$\displaystyle\mu_{x}-[Q_{x},\mu]=U\mu,\quad\mu_{t}-[Q_{t},\mu]=V\mu,$ (2.34)

whose solutions satisfy the Fredholm integral equations

$\mu^{\pm}(z;x,t)=I-\int_{x}^{\pm\infty}e^{-\hat{\Lambda}(z)\int_{x}^{s}q^{2}(v,t)dv}[U\mu_{\pm}(s,t;z)]ds.$
(2.35)

We define six rays at $z=0$

$\Sigma=\cup_{n=1}^{6}L_{n},\ \ \
L_{n}=e^{\frac{\pi(n-1)i}{3}}\mathbb{R}^{+},\ n=1,\cdots,6,$

which divide the complex plane $\mathbb{C}$ into six open cones

$S_{n}=\\{z\in\mathbb{C};\arg z\in({(n-1)\pi}/{3},{n\pi}/{3})\\},\
n=1,\cdots,6,$

see Figure 2.1. Denote the matrix

$\mu^{\pm}=\left(\mu^{\pm}_{1},\mu^{\pm}_{2},\mu^{\pm}_{3}\right),$

where the scripts $1,2$ and $3$ denote the first, second and third column of
$\mu^{\pm}(z)$ respectively. Then from (2.35), we can show that
$\left(\mu^{+}_{1},\mu^{+}_{2},\mu^{+}_{3}\right)$ is analytical in the
domains

$\left(\bar{S}_{1}\cup\bar{S}_{2},\ \bar{S}_{5}\cup\bar{S}_{6},\
\bar{S}_{3}\cup\bar{S}_{4}\right),$

while $\left(\mu^{-}_{1},\mu^{-}_{2},\mu^{-}_{3}\right)$ is analytical in the
domains

$\left(\bar{S}_{4}\cup\bar{S}_{5},\ \bar{S}_{3}\cup\bar{S}_{2},\
\bar{S}_{1}\cup\bar{S}_{6}\right).$

Here, $\bar{S}_{j}$ denotes the closure of $S_{j}$ for $j=1,...,6$,
respectively. The initial points of integration $\infty_{j,l}$ are specified
as follows for each matrix entry $(j,l)$ for $j,l=1,2,3$:

$\displaystyle\infty_{j,l}=\left\\{\begin{array}[]{ll}+\infty,&\text{if
Re}\lambda_{j}\geq\text{Re}\lambda_{l},\\\\[12.0pt] -\infty,&\text{if
Re}\lambda_{j}<\text{Re}\lambda_{l}.\\\ \end{array}\right.$ (2.38)

Then (2.35) can be rewritten as a system of Fredholm integral equations

$\mu_{jl}(z;x,t)=I_{jl}-\int_{x}^{\infty_{jl}}e^{-\int_{x}^{s}q^{2}(v,t)dv(\lambda_{j}(z)-\lambda_{l}(z))}[U\mu(s,t;z)]_{jl}ds.$
(2.39)

$L_{1}$$L_{2}$$L_{3}$$L_{4}$$L_{6}$$L_{5}$$0$$\varkappa_{2}$$\varkappa_{6}$$\varkappa_{3}$$\varkappa_{5}$$\varkappa_{1}$$\varkappa_{4}$$\kappa_{3}$$\kappa_{4}$$\kappa_{1}$$\kappa_{6}$$\kappa_{2}$$\kappa_{5}$$S_{3}$$S_{4}$$S_{1}$$S_{6}$$S_{2}$$S_{5}$
Figure 2.1: The $z$-plane is divided into six analytical domains
$S_{n},n=1,...,6$ by six rays $L_{n}$. $\varkappa_{n}$ and $\kappa_{n}$ are
spectral singularities.

It was shown that the eigenfunction $\mu(z)$ defined by (2.32) has the
following properties [13].

######  Proposition 2.1.

The equations (2.35) uniquely define a $3\times 3$-matrix valued solutions
$\mu(z):=\mu(z;x,t)$ of (2.34) with the following properties:

  * (1)

det $\mu(z)$=1.

  * (2)

$\mu(z)$ is piecewise meromorphic with respect to $\mathbb{C}\setminus\Sigma$,
as function of the spectral parameter $z$.

  * (3)

$\mu(z)$ obeys the symmetries

$\displaystyle\mu(z)=\Gamma_{1}\overline{\mu(\bar{z})}\Gamma_{1}=\Gamma_{2}\overline{\mu(\omega^{2}\bar{z})}\Gamma_{2}=\Gamma_{3}\overline{\mu(\omega\bar{z})}\Gamma_{3}=\overline{\mu(\bar{z}^{-1})},$

where

$\displaystyle\Gamma_{1}=\left(\begin{array}[]{ccc}0&1&0\\\ 1&0&0\\\
0&0&1\end{array}\right),\quad\Gamma_{2}=\left(\begin{array}[]{ccc}0&0&1\\\
0&1&0\\\
1&0&0\end{array}\right),\quad\Gamma_{3}=\left(\begin{array}[]{ccc}1&0&0\\\
0&0&1\\\ 0&1&0\end{array}\right).$ (2.49)

  * (4)

$\mu(z)$ has pole singularities at $\varkappa_{n}=e^{\frac{n\pi i}{3}}$ with
$\mu=\mathcal{O}(\frac{1}{z-\varkappa_{n}})$ as $z\to\varkappa_{n}$.

  * (5)

$\mu(z)\to I$ as $z\to\infty$, and for $z\in\mathbb{C}\setminus\Sigma$,
$\mu(z)$ is bounded as $x\to-\infty$ and $\mu(z)\to I$ as $x\to+\infty$.

######  Remark 2.1.

From the symmetries in Proposition 2.1(3), it follows that the values of $\mu$
at $z$ and at $\omega z$ are related by

$\mu(\omega z)=C^{-1}\mu(z)C,\ \text{where}\
C=\left(\begin{array}[]{ccc}0&1&0\\\ 1&0&0\\\ 0&0&1\end{array}\right).$ (2.50)

Denote $\mu_{\pm}(z;x,t)$ as the limiting values of $\mu(z^{\prime};x,t)$ as
$z^{\prime}\to z$ from the positive or negative side of $L_{n}$, then they are
related as follows

$\displaystyle\mu_{+}(z)=\mu_{-}(z)e^{Q}V_{n}(z)e^{-Q},\quad z\in L_{n},\
n=1,\cdots,6,$ (2.51)

where $V_{n}(z)$ only depends on $z$ and is completely determined by $u(x,0)$,
i.e., by the initial data for the Cauchy problem (1.8). Take
$L_{1}=\mathbb{R}^{+}$ and $L_{4}=\mathbb{R}^{-}$ as an example, $v_{n}$ for
$n=1,4$ has a special matrix structure

$\displaystyle V_{n}(z)=\left(\begin{array}[]{ccc}1&-r_{\pm}(z)&0\\\
\bar{r}_{\pm}(z)&1-|r_{\pm}(z)|^{2}&0\\\ 0&0&1\end{array}\right),\quad
z\in\mathbb{R}^{\pm}.$ (2.55)

$r_{\pm}(z)$ are single scalar functions, with $r_{\pm}(z)\in
L^{\infty}(\mathbb{R}^{\pm})$, and $r_{\pm}(z)=\mathcal{O}(z^{-1})$ as
$z\to\pm\infty$. The symmetry of $\mu(z)$ gives that
$r_{\pm}(z)=\overline{r_{\pm}(z^{-1})}$, therefore it also has $r_{\pm}(z)\in
L^{2}(\mathbb{R}^{\pm})$ and $\lim_{z\to 0^{\pm}}r_{\pm}(z)=0$. Moreover, the
singularities at $\pm 1$ give that $r_{\pm}(\pm 1)=0$. So we define the
_reflection coefficient_

$\displaystyle
r(z)=\left\\{\begin{array}[]{ll}r_{\pm}(z),&z\in\mathbb{R}^{\pm},\\\\[12.0pt]
0,&z=0.\\\ \end{array}\right.$ (2.58)

Then $r\in L^{\infty}(\mathbb{R})\cap L^{2}(\mathbb{R})$ and
$r(z)=\mathcal{O}(z^{-1})$ as $z\to\infty$. In references [13, 26], it was
shown that there exist at most a finite number of simple poles $z_{n}$ of
$\mu(z)$ lying in $S_{1}\cap\\{z\in\mathbb{C};\ |z|>1\\}$ and $w_{m}$ lying in
$S_{1}\cap\\{z\in\mathbb{C};\ |z|=1\\}$. And there are no poles except $\pm
1$, $\pm\omega$ and $\pm\omega^{2}$ on the contour $\Sigma$.

To differentiate this two types of poles, we denote them as $z_{n}$,
$z_{n}^{A}$ and $w_{m}$, $w_{m}^{A}$, respectively. Denote $N_{1}$,
$N_{1}^{A}$, $N_{2}$ and $N_{2}^{A}$ as the number of $z_{n}$, $z_{n}^{A}$,
$w_{m}$, and $w_{m}^{A}$, respectively. The symmetries of $\mu(z)$ imply
$\bar{z}_{n}^{-1}$ and $\frac{1}{\bar{z}_{n}^{A}}$ are also the poles of
$\mu(z)$ in $S_{1}$. It is convenient to define $\zeta_{n}=z_{n}$, and
$\zeta_{n+N_{1}}=\bar{z}_{n}^{-1}$ for $n=1,\cdot\cdot\cdot,N_{1}$;
$\zeta_{m+2N_{1}}=w_{m}$ for $m=1,\cdot\cdot\cdot,N_{2}$;
$\zeta_{j+2N_{1}+N_{2}}=z_{j}^{A}$, and
$\zeta_{j+2N_{1}+N_{2}+N_{1}^{A}}=\frac{1}{\bar{z}_{n}^{A}}$ for
$j=1,\cdot\cdot\cdot,N_{1}^{A}$; $\zeta_{m+2N_{1}+2N_{1}^{A}+N_{2}}=w_{m}^{A}$
for $m=1,\cdot\cdot\cdot,N_{2}^{A}$. For the sake of brevity, let

$\displaystyle N_{0}=2N_{1}+2N_{1}^{A}+N_{2}+N_{2}^{A}.$

Moreover, $\omega\zeta_{n}$, $\omega^{2}\zeta_{n}$, $\bar{\zeta}_{n}$,
$\omega\bar{\zeta}_{n}$, $\omega^{2}\bar{\zeta}_{n}$ are also poles of
$\mu(z)$ in $S_{j}$, $j=2,\cdots,6$. For convenience, let
$\zeta_{n+N_{0}}=\omega\bar{\zeta}_{n}$, $\zeta_{n+2N_{0}}=\omega\zeta_{n},$
$\zeta_{n+3N_{0}}=\omega^{2}\bar{\zeta}_{n}$,
$\zeta_{n+4N_{0}}=\omega^{2}\zeta_{n}$ and $\zeta_{n+5N_{0}}=\bar{\zeta}_{n}$
for $n=1,...,N_{0}$. Therefore, define the discrete spectrum $\mathcal{Z}$ as

$\mathcal{Z}=\left\\{\zeta_{n}\right\\}_{n=1}^{6N_{0}},$ (2.59)

with $\zeta_{n}\in S_{1}$ and $\bar{\zeta}_{n}\in S_{6}$ whose distribution on
the $z$-plane is shown in Figure 2.2.

$L_{1}$$L_{2}$$L_{3}$$L_{4}$$L_{6}$$L_{5}$$\omega\bar{w}_{m}$$\omega^{2}w_{m}$$\frac{\omega}{z_{n}}$$\frac{\omega^{2}}{\bar{z}_{n}}$$\frac{\omega}{\bar{z}_{n}}$$\frac{\omega^{2}}{z_{n}}$$z_{n}$$\bar{z}_{n}$$\omega\bar{z}_{n}$$\omega^{2}z_{n}$$\frac{1}{\bar{z}_{n}}$$\frac{1}{z_{n}}$$\omega
z_{n}$$\omega^{2}\bar{z}_{n}$$w_{m}$$\bar{w}_{m}$$\omega
w_{m}$$\omega^{2}\bar{w}_{m}$$S_{3}$$S_{4}$$S_{1}$$S_{6}$$S_{2}$$S_{5}$ Figure
2.2: Distribution of the discrete spectrum $\mathcal{Z}$ in the $z$-plane.

As shown in [13], denote the _norming constant_ $c_{n}$ and residue conditions
as

$\displaystyle\operatorname*{Res}_{z=\zeta_{n}}\mu(z)=\lim_{z\to\zeta_{n}}\mu(z)e^{Q}\left(\begin{array}[]{ccc}0&-c_{n}&0\\\
0&0&0\\\ 0&0&0\end{array}\right)e^{-Q},$ (2.63)

for $n=1,...,2N_{1}+N_{2}$ and

$\displaystyle\operatorname*{Res}_{z=\zeta_{j}}\mu(z)=\lim_{z\to\zeta_{j}}\mu(z)e^{Q}\left(\begin{array}[]{ccc}0&0&0\\\
0&0&-c_{j+2N_{1}+N_{2}}\\\ 0&0&0\end{array}\right)e^{-Q},$ (2.67)

for $j=1+2N_{1}+N_{2},...,N_{0}$. In addition, the collection
$\sigma_{d}=\left\\{\zeta_{n},C_{n}\right\\}^{6N_{0}}_{n=1}$ is called the
_scattering data_ with $C_{n}=c_{n}$, $C_{n+N_{0}}=\omega\bar{c}_{n}$,
$C_{n+2N_{0}}=\omega c_{n},$ $C_{n+3N_{0}}=\omega^{2}\bar{c}_{n}$,
$C_{n+4N_{0}}=\omega^{2}c_{n}$ and $C_{n+5N_{0}}=\bar{c}_{n}$ for
$n=1,...,N_{0}$.

To deal with our following work, we assume our initial data satisfies that
$u_{0}\in\mathcal{S}(\mathbb{R})$ to generate generic scattering data such
that $\mu(z)$ has no the poles on $L_{n}\setminus\\{\varkappa_{n}\\}$,
$n=1,...,6$ and at the point $z=e^{\frac{\pi i}{6}}$. For the reflection
coefficient $r(z)$, we have the following proposition.

######  Proposition 2.2.

If the initial data $u_{0}(x)\in\mathcal{S}(\mathbb{R})$, then $r(z)$ belongs
to $\mathcal{S}(\mathbb{R})$. There exist fixed constants $C_{1,r}>0$ and
$C_{2,r}$ such that if $u_{0}-u_{0,xx}>C_{2,r}>-1$, $\parallel
u_{0}-u_{0,xx}\parallel_{L^{1}}<C_{1,r}$ and $\parallel
u_{0}\parallel_{W^{3,1}\cup W^{3,\infty}}<C_{1,r}$, then $|r(z)|<1$ for
$z\in\mathbb{R}$. Especially, $|r(\pm 1)|=0$.

###  2.2 An RH characterization

We replace the pair of variables $(x,t)$ by $(y,t)$ with $y$ defined in
(2.33). The price to pay for this is that the solution of the initial problem
can be given only implicitly, or parametrically. It will be given in terms of
functions in the new scale, whereas the original scale will also be given in
terms of functions in the new scale. Let $\xi=\frac{y}{t}$. By the definition
of the new scale $y(x,t)$, we denote the phase function

$\theta_{jl}(z)=-i\left[\xi(\lambda_{j}(z)-\lambda_{l}(z))+\left(\frac{1}{\lambda_{j}(z)}-\frac{1}{\lambda_{l}(z)}\right)\right].$
(2.68)

Especially,

$\theta_{12}(z)=\sqrt{3}\left(z-\frac{1}{z}\right)\left[\xi-\frac{1}{z^{2}-1+z^{-2}}\right],\quad$
(2.69)

with $\theta_{23}(z)=\theta_{12}(\omega z),\text{ and
}\theta_{31}(z)=\theta_{12}(\omega^{2}z)$.

$M(z;y,t):=\mu(z;x(y,t),t).$ (2.70)

which solves the following RH problem.

######  RH problem 2.1.

Find a matrix-valued function $M(z):=M(z;y,t)$ which satisfies:

  * 1.

$M(z)$ is meromorphic in $\mathbb{C}\setminus\Sigma$ and has finite single
poles.

  * 2.

$M(z)=\Gamma_{1}\overline{M(\bar{z})}\Gamma_{1}=\Gamma_{2}\overline{M(\omega^{2}\bar{z})}\Gamma_{2}=\Gamma_{3}\overline{M(\omega\bar{z})}\Gamma_{3}=\overline{M(\bar{z}^{-1})}.$

  * 3.

$M(z)$ has continuous boundary values $M_{\pm}(z)$ on $\Sigma$

$M_{+}(z)=M_{-}(z)V(z),\hskip 8.5359ptz\in\Sigma,$ (2.71)

where

$\displaystyle
V(z)=\begin{cases}\left(\begin{array}[]{ccc}1&-r(z)e^{it\theta_{12}}&0\\\
\bar{r}(z)e^{-it\theta_{12}}&1-|r(z)|^{2}&0\\\ 0&0&1\end{array}\right),\ z\in
L_{1},\\\ \left(\begin{array}[]{ccc}1&0&0\\\ 0&1&-r(\omega
z)e^{it\theta_{23}}\\\ 0&\bar{r}(\omega z)e^{-it\theta_{23}}&1-|r(\omega
z)|^{2}\end{array}\right),\ z\in L_{2},\\\
\left(\begin{array}[]{ccc}1-|r(\omega^{2}z)|^{2}&0&\bar{r}(\omega^{2}z)e^{it\theta_{13}}\\\
0&1&0\\\ -r(\omega^{2}z)e^{-it\theta_{13}}&0&1\end{array}\right),\ z\in
L_{3},\\\ \left(\begin{array}[]{ccc}1&-r(z)e^{it\theta_{12}}&0\\\
\bar{r}(z)e^{-it\theta_{12}}&1-|r(z)|^{2}&0\\\ 0&0&1\end{array}\right),\ z\in
L_{4},\\\ \left(\begin{array}[]{ccc}1&0&0\\\ 0&1&-r(\omega
z)e^{it\theta_{23}}\\\ 0&\bar{r}(\omega z)e^{-it\theta_{23}}&1-|r(\omega
z)|^{2}\end{array}\right),\ z\in L_{5},\\\
\left(\begin{array}[]{ccc}1-|r(\omega^{2}z)|^{2}&0&\bar{r}(\omega^{2}z)e^{it\theta_{13}}\\\
0&1&0\\\ -r(\omega^{2}z)e^{-it\theta_{13}}&0&1\end{array}\right),\ z\in
L_{6}.\end{cases}$ (2.72)

  * 4.

$M(z)=I+\mathcal{O}(z^{-1}),\hskip 14.22636ptz\rightarrow\infty$.

  * 5.

As $z\to\varkappa_{n}=e^{\frac{i\pi(n-1)}{3}}$, $n=1,...,6$, the limit of
$M(z)$ has pole singularities

$\displaystyle M(z)=\frac{1}{z\mp
1}\left(\begin{array}[]{ccc}\alpha_{\pm}&\alpha_{\pm}&\beta_{\pm}\\\
-\alpha_{\pm}&-\alpha_{\pm}&-\beta_{\pm}\\\
0&0&0\end{array}\right)+\mathcal{O}(1),\ z\to\pm 1,$ (2.76) $\displaystyle
M(z)=\frac{\pm\omega^{2}}{z\mp\omega^{2}}\left(\begin{array}[]{ccc}0&0&0\\\
\beta_{\pm}&\alpha_{\pm}&\alpha_{\pm}\\\
-\beta_{\pm}&-\alpha_{\pm}&-\alpha_{\pm}\end{array}\right)+\mathcal{O}(1),\
z\to\pm\omega^{2},$ (2.80) $\displaystyle
M(z)=\frac{\pm\omega}{z\mp\omega}\left(\begin{array}[]{ccc}-\alpha_{\pm}&-\beta_{\pm}&-\alpha_{\pm}\\\
0&0&0\\\
\alpha_{\pm}&\beta_{\pm}&\alpha_{\pm}\end{array}\right)+\mathcal{O}(1),\
z\to\pm\omega,$ (2.84)

with $\alpha_{\pm}=\alpha_{\pm}(y,t)=-\bar{\alpha}_{\pm}$,
$\beta_{\pm}=\beta_{\pm}(y,t)=-\bar{\beta}_{\pm}$ and $M^{-1}(z)$ has same
specific matrix structure with $\alpha_{\pm}$, $\beta_{\pm}$ replaced by
$\tilde{\alpha}_{\pm}$, $\tilde{\beta}_{\pm}$. Moreover, $\left(\alpha_{\pm},\
\beta_{\pm}\right)\neq 0$ iff $\left(\tilde{\alpha}_{\pm},\
\tilde{\beta}_{\pm}\right)\neq 0$.

  * 6.

$M(z)$ has simple poles at each point in $\mathcal{Z}\cup\bar{\mathcal{Z}}$
with

$\displaystyle\operatorname*{Res}\limits_{k=\zeta_{n}}M(k)=\lim_{k\rightarrow\zeta_{n}}M(k)B_{n},$
(2.85)
$\displaystyle\operatorname*{Res}\limits_{k=\omega\bar{\zeta}_{n}}M(k)=\lim_{k\rightarrow\omega\bar{\zeta}_{n}}M(k)\Gamma_{3}(\omega\bar{B}_{n})\Gamma_{3}:=\lim_{k\rightarrow\zeta_{n}}M(k)B_{n+N},$
$\displaystyle\operatorname*{Res}\limits_{k=\omega\zeta_{n}}M(k)=\lim_{k\rightarrow\omega\zeta_{n}}M(k)C^{2}(\omega^{2}B_{n})C^{-2}:=\lim_{k\rightarrow\zeta_{n}}M(k)B_{n+2N},$
$\displaystyle\operatorname*{Res}\limits_{k=\omega^{2}\bar{\zeta}_{n}}M(k)=\lim_{k\rightarrow\omega^{2}\bar{\zeta}_{n}}M(k)\Gamma_{2}(\omega^{2}\bar{B}_{n})\Gamma_{2}:=\lim_{k\rightarrow\zeta_{n}}M(k)B_{n+3N},$
$\displaystyle\operatorname*{Res}\limits_{k=\omega^{2}\zeta_{n}}M(k)=\lim_{k\rightarrow\omega^{2}\zeta_{n}}M(k)C(\omega^{2}B_{n})C^{-1}:=\lim_{k\rightarrow\zeta_{n}}M(k)B_{n+4N},$
$\displaystyle\operatorname*{Res}\limits_{k=\bar{\zeta}_{n}}M(k)=\lim_{k\rightarrow\bar{\zeta}_{n}}M(k)\Gamma_{1}\bar{B}_{n}\Gamma_{1}:=\lim_{k\rightarrow\zeta_{n}}M(k)B_{n+5N},$

where

$\displaystyle
B_{n}=\begin{cases}\left(\begin{array}[]{ccc}0&-c_{n}e^{\mathrm{i}t\theta_{12}(\zeta_{n})}&0\\\
0&0&0\\\ 0&0&0\end{array}\right),\quad n=1,\cdots,2N_{1}+N_{2},\\\
\left(\begin{array}[]{ccc}0&0&0\\\
0&0&-c_{n}e^{\mathrm{i}t\theta_{23}(\zeta_{n})}\\\
0&0&0\end{array}\right),\quad
n=2N_{1}+N_{2}+1,\cdots,2N_{1}+N_{2}+2N_{1}^{A}+N_{2}^{A}.\\\ \end{cases}$
(2.86)

Denote $M(z;y,t)=(M_{jl}(z;y,t))_{jl=1}^{3}$, then the solution of Novikov
equation (1.8) can be obtained by the following reconstruction formula

$\displaystyle u(x,t)=u(y(x,t),t)=$
$\displaystyle\frac{1}{2}\tilde{m}_{1}(y,t)\left(\frac{M_{33}(e^{\frac{i\pi}{6}};y,t)}{M_{11}(e^{\frac{i\pi}{6}};y,t)}\right)^{1/2}$
$\displaystyle+\frac{1}{2}\tilde{m}_{3}(y,t)\left(\frac{M_{33}(e^{\frac{i\pi}{6}};y,t)}{M_{11}(e^{\frac{i\pi}{6}};y,t)}\right)^{-1/2}-1,$
(2.87)

where

$x(y,t)=y+\frac{1}{2}\ln\frac{M_{33}(e^{\frac{i\pi}{6}};y,t)}{M_{11}(e^{\frac{i\pi}{6}};y,t)},$
(2.88)

and

$\displaystyle\tilde{m}_{l}:=\sum_{j=1}^{3}M_{jl}(e^{\frac{i\pi}{6}};y,t),\
l=1,2,3.$

##  3 Interpolation and Conjugation

In this section, we aim to convert the original RH problem to a regular RH
problem which satisfies the following conditions:

  * 1.

It is well behaved as $t\to\infty$ with $\xi$ fixed.

  * 2.

Different factorizations of the jump matrix (2.72) should be taken for
different transition sectors.

For convenience, we denote

$\mathcal{N}:=\left\\{1,...,N_{0}\right\\},\
\tilde{\mathcal{N}}:=\left\\{1,...,2N_{1}+N_{2}\right\\},\
\tilde{\mathcal{N}}^{A}:=\left\\{1+2N_{1}+N_{2},...,N_{0}\right\\}.$

Further to distinguish different type of zeros, we introduce a small positive
constant $\delta_{0}$ to give the partitions $\Delta,\nabla$ and $\lozenge$ of
$\mathcal{N}$ as follow

$\displaystyle\nabla_{1}=\left\\{j\in\tilde{\mathcal{N}};\text{Im}\theta_{12}(\zeta_{j})>\delta_{0}\right\\},\Delta_{1}=\left\\{j\in\tilde{\mathcal{N}};\text{Im}\theta_{12}(\zeta_{j})<0\right\\},$
$\displaystyle\nabla_{2}=\left\\{i\in\tilde{\mathcal{N}}^{A};\text{Im}\theta_{23}(\zeta_{i})>\delta_{0}\right\\},\Delta_{2}=\left\\{i\in\tilde{\mathcal{N}}^{A};\text{Im}\theta_{23}(\zeta_{i})<0\right\\},$
(3.1)
$\displaystyle\nabla=\nabla_{1}\cup\nabla_{2},\Delta=\Delta_{1}\cup\Delta_{2},$
$\displaystyle\lozenge_{1}=\left\\{j_{0}\in\tilde{\mathcal{N}};0\leq\text{Im}\theta_{12}(\zeta_{j_{0}})\leq\delta_{0}\right\\},$
$\displaystyle\lozenge_{2}=\left\\{i_{0}\in\tilde{\mathcal{N}}^{A};0\leq\text{Im}\theta_{23}(\zeta_{i_{0}})\leq\delta_{0}\right\\},$
(3.2) $\displaystyle\lozenge=\\{n\in\mathcal{N};\exists
k\in\\{1,...,5\\},n-kN_{0}\in\lozenge_{1}\cup\lozenge_{2}\\}.$ (3.3)

For $\zeta_{n}$ with $n\in\Delta$, the residue of $M(z)$ at $\zeta_{n}$ in RH
problem 2.1 grows without bound as $t\to\infty$. However, for $\zeta_{n}$ with
$n\in\nabla$, the residue decays to $0$. Denote two constants
$\mathcal{N}(\lozenge)=|\lozenge|$ and

$\rho_{0}=\min_{n\in\mathcal{N}\setminus\lozenge}\left\\{|\text{Im}\theta_{12}(\zeta_{n})|,\
|\text{Im}\theta_{23}(\zeta_{n})|\right\\}>0.$ (3.4)

For the poles $\zeta_{n}$ with $n\in\mathcal{N}\setminus\lozenge$, we want to
convert the residue of these poles into the jumps along small closed loops
enclosing themselves, respectively. The jump matrix $V(z)$ on $\Sigma$ in
(2.71) has the following factorizations:
On the contour $\mathbb{R}$,

$\displaystyle V(z)=\left(\begin{array}[]{ccc}1&0&0\\\
\bar{r}e^{-it\theta_{12}}&1&0\\\
0&0&1\end{array}\right)\left(\begin{array}[]{ccc}1&-re^{it\theta_{12}}&0\\\
0&1&0\\\ 0&0&1\end{array}\right)$ (3.11)
$\displaystyle=\left(\begin{array}[]{ccc}1&\frac{-re^{it\theta_{12}}}{1-|r|^{2}}&0\\\
0&1&0\\\
0&0&1\end{array}\right)\left(\begin{array}[]{ccc}\frac{1}{1-|r|^{2}}&0&0\\\
0&1-|r|^{2}&0\\\ 0&0&1\end{array}\right)\left(\begin{array}[]{ccc}1&0&0\\\
\frac{\bar{r}e^{-it\theta_{12}}}{1-|r|^{2}}&1&0\\\ 0&0&1\end{array}\right);$
(3.21)

On the contour $\omega^{2}\mathbb{R}$,

$\displaystyle V(z)=\left(\begin{array}[]{ccc}1&0&0\\\ 0&1&0\\\
0&\bar{r}(\omega
z)e^{-it\theta_{23}}&1\end{array}\right)\left(\begin{array}[]{ccc}1&0&0\\\
0&1&-r(\omega z)e^{it\theta_{23}}\\\ 0&0&1\end{array}\right)$ (3.28)
$\displaystyle=\left(\begin{array}[]{ccc}1&0&0\\\ 0&1&\frac{-r(\omega
z)e^{it\theta_{23}}}{1-|r(\omega z)|^{2}}\\\
0&0&1\end{array}\right)\left(\begin{array}[]{ccc}1&0&0\\\
0&\frac{1}{1-|r(\omega z)|^{2}}&0\\\ 0&0&1-|r(\omega
z)|^{2}\end{array}\right)\left(\begin{array}[]{ccc}1&0&0\\\ 0&1&0\\\
0&\frac{\bar{r}(\omega z)e^{-it\theta_{23}}}{1-|r(\omega
z)|^{2}}&1\end{array}\right);$ (3.38)

On the contour $\omega\mathbb{R}$,

$\displaystyle
V(z)=\left(\begin{array}[]{ccc}1&0&\bar{r}(\omega^{2}z)e^{it\theta_{13}}\\\
0&1&0\\\ 0&0&1\end{array}\right)\left(\begin{array}[]{ccc}1&0&0\\\ 0&1&0\\\
-r(\omega^{2}z)e^{-it\theta_{13}}&0&1\end{array}\right)$ (3.45)
$\displaystyle=\left(\begin{array}[]{ccc}1&0&0\\\ 0&1&0\\\
\frac{-r(\omega^{2}z)e^{-it\theta_{13}}}{1-|r(\omega^{2}z)|^{2}}&0&1\end{array}\right)\left(\begin{array}[]{ccc}1-|r(\omega^{2}z)|^{2}&0&0\\\
0&1&0\\\
0&0&\frac{1}{1-|r(\omega^{2}z)|^{2}}\end{array}\right)\left(\begin{array}[]{ccc}1&0&\frac{\bar{r}(\omega^{2}z)e^{it\theta_{13}}}{1-|r(\omega^{2}z)|^{2}}\\\
0&1&0\\\ 0&0&1\end{array}\right).$ (3.55)

To achieve our goal, we utilize these factorizations to deform the jump
contours so that the oscillating factor $e^{\pm it\theta_{12}}$, which is
determined by the sign of $\operatorname*{Im}\theta_{12}$, are decaying in the
corresponding regions, respectively,. We consider the signature table of
$\operatorname*{Im}\theta_{12}$

$\displaystyle\operatorname*{Im}\theta_{12}=\sqrt{3}\text{Im}z\left(1+|z|^{-2}\right)\xi-$
$\displaystyle\dfrac{\sqrt{3}\text{Im}z\left(1+|z|^{-2}\right)\left(-|z|^{6}-|z|^{4}+4\text{Re}^{2}z|z|^{2}-|z|^{2}\right)}{|z|^{8}+1+2[(\text{Re}^{2}z-\text{Im}^{2}z)^{2}-4\text{Re}^{2}z\text{Im}^{2}z]-2(1+|z|^{4})(\text{Re}^{2}z-\text{Im}^{2}z)+|z|^{4}}.$
(3.56)

which are depicted in Figure 3.1.

(a) $\xi<-\frac{1}{8}$

(b) $\xi=-\frac{1}{8}$

(c) $-\frac{1}{8}<\xi<0$

(d) $0\leq\xi<1$

(e) $\xi=1$

(f) $\xi>1$

Figure 3.1: The signature table of $\operatorname*{Im}\theta_{12}$ with
different $\xi$: (a) $\xi<-\frac{1}{8}$, (b) $\xi=-\frac{1}{8}$, (c)
$-\frac{1}{8}<\xi<0$, (d) $0\leq\xi<1$, (e) $\xi=1$, (f) $\xi>1$. In the blue
region, $\operatorname*{Im}\theta_{12}<0$ and then
$|e^{-\mathrm{i}t\theta_{12}}|\rightarrow 0$ as $t\rightarrow\infty$; In the
white region, $\operatorname*{Im}\theta_{12}>0$ and then
$|e^{\mathrm{i}t\theta_{12}}|\rightarrow 0$ as $t\rightarrow\infty$. Moreover,
on the blue curve, $\operatorname*{Im}\theta_{12}=0$.

Introduce the notations on the contour $\mathbb{R}$,

$\displaystyle I(\xi)=\left\\{\begin{array}[]{ll}\mathbb{R},&\text{if
}|\xi+\frac{1}{8}|t^{2/3}<C,\\\\[4.0pt] \emptyset,&\text{if
}|\xi-1|t^{2/3}<C,\end{array}\right.$ (3.59)

and on the contours $\omega\mathbb{R}$ and $\omega^{2}\mathbb{R}$,

$I^{\omega}(\xi)=\\{\omega z:z\in I(\xi)\\},\ \
I^{\omega^{2}}(\xi)=\\{\omega^{2}z:z\in I(\xi)\\}.$

To proceed, we introduce the following scalar RH problem.

######  RH problem 3.1.

Find a scalar function $\delta_{1}(z)$ with

  * 1.

$\delta_{1}(z)$ is analytic in $\mathbb{C}\setminus\mathbb{R}$.

  * 2.

$\delta_{1}(z)$ has jump relation:

$\displaystyle\delta_{1,+}(z)=\delta_{1,-}(z)(1-|r(z)|^{2}),\ z\in I(\xi),$
$\displaystyle\delta_{1,+}(z)=\delta_{1,-}(z),\ z\in\mathbb{R}\setminus
I(\xi).$

  * 3.

$\delta_{1}(z)\to 1$ as $z\to\infty$.

By the Plemelj formula, this RH problem admits an unique solution

$\displaystyle\delta_{1}(z)$
$\displaystyle=\delta_{1}(z,\xi)=\exp\left(-i\int_{I(\xi)}\dfrac{\nu(h)dh}{h-z}\right),$
(3.60)

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

Moreover, define

$\displaystyle H(z)$
$\displaystyle=\prod_{n\in\Delta_{1}}\dfrac{z-\zeta_{n}}{z-\bar{\zeta}_{n}}\prod_{m\in\Delta_{2}}\dfrac{z-\omega\zeta_{m}}{z-\omega^{2}\bar{\zeta}_{m}}\delta_{1}(z,\xi)^{-1};$
(3.61) $\displaystyle T_{1}(z)$
$\displaystyle=T_{1}(z,\xi)=\frac{H(\omega^{2}z)}{H(z)};$ (3.62)
$\displaystyle T_{2}(z)$ $\displaystyle=T_{2}(z,\xi)=T_{1}(\omega
z)=\frac{H(z)}{H(\omega z)};$ (3.63) $\displaystyle T_{3}(z)$
$\displaystyle=T_{3}(z,\xi)=T_{1}(\omega^{2}z)=\frac{H(\omega
z)}{H(\omega^{2}z)};$ (3.64) $\displaystyle T_{ij}(z)$
$\displaystyle=T_{ij}(z,\xi)=\frac{T_{i}(z)}{T_{j}(z)},\ i,j=1,2,3.$ (3.65)

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

######  Proposition 3.1.

The function defined by (3.62) and (3.65) has the following properties

  * (a)

$T_{1}$ is meromorphic in $\mathbb{C}\setminus\mathbb{R}$, and for each
$n\in\Delta_{1}$, $T_{1}(z)$ has a simple pole at $\zeta_{n}$ and a simple
zero at $\bar{\zeta}_{n}$, for each $m\in\Delta_{2}$, $T_{1}(z)$ has a simple
pole at $\omega\zeta_{m}$ and a simple zero at $\omega\bar{\zeta}_{m}$.

  * (b)

$\overline{T_{1}(\bar{z})}=T_{1}(\omega z)=T_{1}(z^{-1})$.

  * (c)

$T_{1}(z)$ satisfies the jump condition

$\displaystyle T_{1,-}(z)=(1-|r(z)|^{2})T_{1,+}(z),\hskip 14.22636ptz\in
I(\xi),$ (3.66) $\displaystyle
T_{1,+}(z)=(1-|r(\omega^{2}z)|^{2})T_{1,-}(z),\hskip 14.22636ptz\in
I^{\omega}(\xi).$ (3.67)

  * (d)

$\lim_{z\to\infty}T_{1}(z):=T_{1}(\infty)$ with $T_{1}(\infty)=1$.

  * (e)

$T_{1}(e^{\frac{i\pi}{6}})$ exists as a constants.

  * (f)

$T_{1}(z)$ is continuous at $z=0$ with $T_{1}(0)=1$.

For

$\displaystyle\varrho:=\frac{1}{4}\min$
$\displaystyle\left\\{\min_{j\in\mathcal{N}}|\text{Im}\zeta_{j}|,\min_{j\in\mathcal{N},\arg(z)=\frac{\pi
i}{3}}|\zeta_{j}-z|,\min_{j\in\mathcal{N}\setminus\lozenge,\text{Im}\theta_{ik}(z)=0}|\zeta_{j}-z|,\right.$
$\displaystyle\left.\min_{j\in\mathcal{N}}|\zeta_{j}-e^{\frac{i\pi}{6}}|,\min_{j\neq
k\in\mathcal{N}}|\zeta_{j}-\zeta_{k}|\right\\},$ (3.68)

we also define

$\mathbb{D}_{n}:=\mathbb{D}(\zeta_{n},\varrho)=\\{z:|z-\zeta_{n}|\leq\varrho\\},\
n\in\mathcal{N},$

to be small disks, which are pairwise disjoint, also disjoint with critical
lines $\left\\{z\in\mathbb{C};\text{Im}\theta(z)=0\right\\}$, as well as the
contours $\mathbb{R}$, $\omega\mathbb{R}$ and $\omega^{2}\mathbb{R}$. Besides,
$e^{\frac{i\pi}{6}}\notin\mathbb{D}_{n}$. By the above definition and the
symmetry of poles and $\theta_{jk}(z)$, for every $n$, there exists a
$k\in\\{0,...,5\\}$ such that $n-kN_{0}\in\mathcal{N}\setminus\lozenge$.
Denote a piecewise matrix function

$G(z)=\left\\{\begin{array}[]{ll}I-\frac{B_{n}}{z-\zeta_{n}},&z\in\mathbb{D}_{n},n-kN_{0}\in\nabla,k\in\\{0,...,5\\},\\\\[12.0pt]
\left(\begin{array}[]{ccc}1&0&0\\\
-\frac{z-\zeta_{n}}{C_{n}e^{it\theta_{12}(\zeta_{n})}}&1&0\\\
0&0&1\end{array}\right),&z\in\mathbb{D}_{n},n\in\Delta_{1}\text{ or
}n-2N_{0}\in\Delta_{2},\\\\[12.0pt] \left(\begin{array}[]{ccc}1&0&0\\\
0&1&0\\\
0&-\frac{z-\zeta_{n}}{C_{n}e^{it\theta_{23}(\zeta_{n})}}&1\end{array}\right),&z\in\mathbb{D}_{n},n-N_{0}\in\Delta_{1}\text{
or }n-5N_{0}\in\Delta_{2},\\\\[12.0pt]
\left(\begin{array}[]{ccc}1&0&-\frac{z-\zeta_{n}}{C_{n}e^{-it\theta_{13}(\zeta_{n})}}\\\
0&1&0\\\
0&0&1\end{array}\right),&z\in\mathbb{D}_{n},n-2N_{0}\in\Delta_{1}\text{ or
}n-4N_{0}\in\Delta_{2},\\\\[12.0pt] \left(\begin{array}[]{ccc}1&0&0\\\
0&1&-\frac{z-\zeta_{n}}{C_{n}e^{-it\theta_{23}(\zeta_{n})}}\\\
0&0&1\end{array}\right),&z\in\mathbb{D}_{n},n-3N_{0}\in\Delta_{1}\text{ or
}n-N_{0}\in\Delta_{2},\\\\[12.0pt] \left(\begin{array}[]{ccc}1&0&0\\\ 0&1&0\\\
0&-\frac{z-\zeta_{n}}{C_{n}e^{it\theta_{23}(\zeta_{n})}}&1\end{array}\right),&z\in\mathbb{D}_{n},n-4N_{0}\in\Delta_{1}\text{
or }n\in\Delta_{2},\\\\[12.0pt]
\left(\begin{array}[]{ccc}1&-\frac{z-\zeta_{n}}{C_{n}e^{-it\theta_{12}(\zeta_{n})}}&0\\\
0&1&0\\\
0&0&1\end{array}\right),&z\in\mathbb{D}_{n},n-5N_{0}\in\Delta_{1}\text{ or
}n-3N_{0}\in\Delta_{2},\\\\[12.0pt] I,&z\text{ in
elsewhere},\end{array}\right.$ (3.69)

where $B_{n}$ are defined in (2.86), and define

$T(z)=\text{diag}\\{T_{1}(z),T_{2}(z),T_{3}(z)\\}.$

Now we introduce the following transformation to construct a regular RH
problem.

$M^{(1)}(z;y,t):=M^{(1)}(z)=M(z)G(z)T(z),$ (3.70)

which then satisfies the following RH problem.

######  RH problem 3.2.

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

  * 1.

$M^{(1)}(z)$ is meromorphic in $\mathbb{C}\setminus\Sigma^{(1)}$, where
$\Sigma^{(1)}=\Sigma\cup\left(\underset{n\in\mathcal{N}\setminus\lozenge,k=0,..,5}{\cup}\partial\mathbb{D}_{n+kN_{0}}\right)$.

  * 2.

$M^{(1)}(z)=\Gamma_{1}\overline{M^{(1)}(\bar{z})}\Gamma_{1}=\Gamma_{2}\overline{M^{(1)}(\omega^{2}\bar{z})}\Gamma_{2}=\Gamma_{3}\overline{M^{(1)}(\omega\bar{z})}\Gamma_{3}=\overline{M^{(1)}(\bar{z}^{-1})}$.

  * 3.

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

$V^{(1)}(z)=\left\\{\begin{array}[]{ll}\left(\begin{array}[]{ccc}1&0&0\\\
\bar{r}T_{12}e^{-it\theta_{12}}&1&0\\\
0&0&1\end{array}\right)\left(\begin{array}[]{ccc}1&-rT_{21}e^{it\theta_{12}}&0\\\
0&1&0\\\ 0&0&1\end{array}\right),&z\in\mathbb{R}\setminus I(\xi),\\\\[12.0pt]
\left(\begin{array}[]{ccc}1&-\frac{rT_{21}e^{it\theta_{12}}}{1-|r|^{2}}&0\\\
0&1&0\\\ 0&0&1\end{array}\right)\left(\begin{array}[]{ccc}1&0&0\\\
\frac{\bar{r}T_{12}e^{-it\theta_{12}}}{1-|r|^{2}}&1&0\\\
0&0&1\end{array}\right),&z\in I(\xi),\\\\[12.0pt]
\left(\begin{array}[]{ccc}1&0&0\\\ 0&1&0\\\ 0&\bar{r}(\omega
z)e^{-it\theta_{23}}T_{23}&1\end{array}\right)\left(\begin{array}[]{ccc}1&0&0\\\
0&1&-r(\omega z)T_{32}e^{it\theta_{23}}\\\
0&0&1\end{array}\right),&z\in\omega^{2}\mathbb{R}\setminus
I^{\omega}(\xi),\\\\[12.0pt] \left(\begin{array}[]{ccc}1&0&0\\\
0&1&\frac{-r(\omega z)T_{32}e^{it\theta_{23}}}{1-|r(\omega z)|^{2}}\\\
0&0&1\end{array}\right)\left(\begin{array}[]{ccc}1&0&0\\\ 0&1&0\\\
0&\frac{\bar{r}(\omega z)e^{-it\theta_{23}}T_{23}}{1-|r(\omega
z)|^{2}}&1\end{array}\right),&z\in I^{\omega}(\xi),\\\\[12.0pt]
\left(\begin{array}[]{ccc}1&0&\bar{r}(\omega^{2}z)T_{31}e^{it\theta_{13}}\\\
0&1&0\\\ 0&0&1\end{array}\right)\left(\begin{array}[]{ccc}1&0&0\\\ 0&1&0\\\
-r(\omega^{2}z)T_{13}e^{-it\theta_{13}}&0&1\end{array}\right),&z\in\omega\mathbb{R}\setminus
I^{\omega^{2}}(\xi),\\\\[12.0pt] \left(\begin{array}[]{ccc}1&0&0\\\ 0&1&0\\\
\frac{-r(\omega^{2}z)T_{13}e^{-it\theta_{13}}}{1-|r(\omega^{2}z)|^{2}}&0&1\end{array}\right)\left(\begin{array}[]{ccc}1&0&\frac{\bar{r}(\omega^{2}z)T_{31}e^{it\theta_{13}}}{1-|r(\omega^{2}z)|^{2}}\\\
0&1&0\\\ 0&0&1\end{array}\right),&z\in I^{\omega^{2}}(\xi),\\\\[12.0pt]
T^{-1}(z)G(z)T(z),&z\in\partial\mathbb{D}_{n}\cap(\underset{k=1}{\cup^{3}}S_{2k}),\\\\[12.0pt]
T^{-1}(z)G^{-1}(z)T(z),&z\in\partial\mathbb{D}_{n}\cap(\underset{k=1}{\cup^{3}}S_{2k-1}).\\\
\end{array}\right.$ (3.71)

  * 4.

$M^{(1)}(z;y,t)=I+\mathcal{O}(z^{-1}),\hskip 14.22636ptz\rightarrow\infty$.

  * 5.

As $z\to\varkappa_{l}=e^{\frac{i\pi(l-1)}{3}}$, $l=1,...,6$, the limit of
$M^{(1)}(z)$ has the pole singularities

$\displaystyle M^{(1)}(z)=\frac{1}{z\mp
1}\left(\begin{array}[]{ccc}\alpha_{\pm}&\alpha_{\pm}&\beta_{\pm}\\\
-\alpha_{\pm}&-\alpha_{\pm}&-\beta_{\pm}\\\ 0&0&0\end{array}\right)T(\pm
1)+\mathcal{O}(1),\ z\to\pm 1,$ (3.75) $\displaystyle
M^{(1)}(z)=\frac{1}{z\mp\omega^{2}}\left(\begin{array}[]{ccc}0&0&0\\\
\beta_{\pm}&\alpha_{\pm}&\alpha_{\pm}\\\
-\beta_{\pm}&-\alpha_{\pm}&-\alpha_{\pm}\end{array}\right)T(\pm\omega^{2})+\mathcal{O}(1),\
z\to\pm\omega^{2},$ (3.79) $\displaystyle
M^{(1)}(z)=\frac{1}{z\mp\omega}\left(\begin{array}[]{ccc}-\alpha_{\pm}&-\beta_{\pm}&-\alpha_{\pm}\\\
0&0&0\\\
\alpha_{\pm}&\beta_{\pm}&\alpha_{\pm}\end{array}\right)T(\pm\omega)+\mathcal{O}(1),\
z\to\pm\omega,$ (3.83)

with $\alpha_{\pm}=\alpha_{\pm}(y,t)=-\bar{\alpha}_{\pm}$,
$\beta_{\pm}=\beta_{\pm}(y,t)=-\bar{\beta}_{\pm}$ and $M^{(1)}(z)^{-1}$ has
same specific matrix structure with $\alpha_{\pm}$, $\beta_{\pm}$ replaced by
$\tilde{\alpha}_{\pm}$, $\tilde{\beta}_{\pm}$.

  * 6.

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

$\displaystyle\operatorname*{Res}_{z=\zeta_{n}}M^{(1)}(z)=\lim_{z\to\zeta_{n}}M^{(1)}(z)\left[T^{-1}(z)B_{n}T(z)\right].$
(3.84)

###### Proof.

The above properties of RH problem 3.2 can be directly obtained by the
properties of RH problem 2.1, Proposition 3.1, and (3.69)- (3.70). ∎

Since the jump matrices on the disks
$\mathbb{D}_{n},n\in\mathcal{N}\setminus\lozenge$ decay exponentially to the
identity matrix as $t\to\infty$, it follows that the RH problem is
asymptotically equivalent to the following RH problem.

######  RH problem 3.3.

Find a matrix-valued function $M^{(2)}(z)$ which satisfies:

  * 1.

$M^{(2)}(z)$ is meromorphic in $\mathbb{C}\setminus\Sigma^{(2)}$, where
$\Sigma^{(2)}=\Sigma$.

  * 2.

$M^{(2)}(z)$ obeys the symmetries

$M^{(2)}(z)=\Gamma_{1}\overline{M^{(2)}(\bar{z})}\Gamma_{1}=\Gamma_{2}\overline{M^{(2)}(\omega^{2}\bar{z})}\Gamma_{2}=\Gamma_{3}\overline{M^{(2)}(\omega\bar{z})}\Gamma_{3}=\overline{M^{(2)}(\bar{z}^{-1})}.$
(3.85)

  * 3.

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

$V^{(2)}(z)=\left\\{\begin{array}[]{ll}\left(\begin{array}[]{ccc}1&-\frac{rT_{21}e^{it\theta_{12}}}{1-|r|^{2}}&0\\\
0&1&0\\\ 0&0&1\end{array}\right)\left(\begin{array}[]{ccc}1&0&0\\\
\frac{\bar{r}T_{12}e^{-it\theta_{12}}}{1-|r|^{2}}&1&0\\\
0&0&1\end{array}\right),&z\in I(\xi),\\\\[12.0pt]
\left(\begin{array}[]{ccc}1&0&0\\\ 0&1&\frac{-r(\omega
z)T_{32}e^{it\theta_{23}}}{1-|r(\omega z)|^{2}}\\\
0&0&1\end{array}\right)\left(\begin{array}[]{ccc}1&0&0\\\ 0&1&0\\\
0&\frac{\bar{r}(\omega z)e^{-it\theta_{23}}T_{23}}{1-|r(\omega
z)|^{2}}&1\end{array}\right),&z\in I^{\omega}(\xi),\\\\[12.0pt]
\left(\begin{array}[]{ccc}1&0&0\\\ 0&1&0\\\
\frac{-r(\omega^{2}z)T_{13}e^{-it\theta_{13}}}{1-|r(\omega^{2}z)|^{2}}&0&1\end{array}\right)\left(\begin{array}[]{ccc}1&0&\frac{\bar{r}(\omega^{2}z)T_{31}e^{it\theta_{13}}}{1-|r(\omega^{2}z)|^{2}}\\\
0&1&0\\\ 0&0&1\end{array}\right),&z\in I^{\omega^{2}}(\xi),\\\
\end{array}\right.$ (3.86)

  * 4.

$M^{(2)}(z)=I+\mathcal{O}(z^{-1}),\hskip 14.22636ptz\rightarrow\infty$.

  * 5.

As $z\to\varkappa_{l}=e^{\frac{i\pi(l-1)}{3}}$, $l=1,...,6$,

$\displaystyle M^{(2)}(z)=\frac{1}{z\mp
1}\left(\begin{array}[]{ccc}\alpha_{\pm}&\alpha_{\pm}&\beta_{\pm}\\\
-\alpha_{\pm}&-\alpha_{\pm}&-\beta_{\pm}\\\ 0&0&0\end{array}\right)T(\pm
1)+\mathcal{O}(1),\ z\to\pm 1,$ (3.90) $\displaystyle
M^{(2)}(z)=\frac{1}{z\mp\omega^{2}}\left(\begin{array}[]{ccc}0&0&0\\\
\beta_{\pm}&\alpha_{\pm}&\alpha_{\pm}\\\
-\beta_{\pm}&-\alpha_{\pm}&-\alpha_{\pm}\end{array}\right)T(\pm\omega^{2})+\mathcal{O}(1),\
z\to\pm\omega^{2},$ (3.94) $\displaystyle
M^{(2)}(z)=\frac{1}{z\mp\omega}\left(\begin{array}[]{ccc}-\alpha_{\pm}&-\beta_{\pm}&-\alpha_{\pm}\\\
0&0&0\\\
\alpha_{\pm}&\beta_{\pm}&\alpha_{\pm}\end{array}\right)T(\pm\omega)+\mathcal{O}(1),\
z\to\pm\omega,$ (3.98)

with $\alpha_{\pm}=\alpha_{\pm}(y,t)=-\bar{\alpha}_{\pm}$,
$\beta_{\pm}=\beta_{\pm}(y,t)=-\bar{\beta}_{\pm}$ and $M^{(1)}(z)^{-1}$ has
same specific matrix structure with $\alpha_{\pm}$, $\beta_{\pm}$ replaced by
$\tilde{\alpha}_{\pm}$, $\tilde{\beta}_{\pm}$.

  * 6.

$M^{(2)}(z)$ has simple poles at each point $\zeta_{n}$ and $\bar{\zeta}_{n}$
for $n\in\lozenge$ with

$\displaystyle\operatorname*{Res}_{z=\zeta_{n}}M^{(2)}(z)=\lim_{z\to\zeta_{n}}M^{(2)}(z)\left[T^{-1}(z)B_{n}T(z)\right].$
(3.99)

######  Proposition 3.2.

The solution of RH problem 3.2 can be approximated by the solution of RH
problem 3.3

$M^{(1)}(z)=M^{(2)}(z)(I+\mathcal{O}(e^{-ct})),$ (3.100)

where $c$ is a positive constant.

###### Proof.

The result is derived from the theorem of Beals-Coifman and the corresponding
norm estimates. ∎

##  4 Painlevé Asymptotics in $|\xi+\frac{1}{8}|t^{2/3}<C$

In this section, we consider the asymptotics in $0<(\xi+\frac{1}{8})t^{2/3}<C$
for any positive constant $C$, as the discussions for the other half region is
similar.

From (2.69), the phase function $\theta_{12}(z)$ has eight saddle points
$\xi_{i}$, $i=1,\cdots,8$, such that $\theta^{\prime}_{12}(\xi_{i})=0$.
Introducing

$\tilde{k}=z-\frac{1}{z},$

then one gets

$\theta^{\prime}_{12}(z)=\sqrt{3}\left(\xi-\frac{1-\tilde{k}^{2}}{(\tilde{k}^{2}+1)^{2}}\right)\left(1+\frac{1}{z^{2}}\right)$
(4.1)

Thus, in the $\tilde{k}$-plane the real critical points
$\tilde{k}_{1}\in\mathbb{R}$ are determined by the equation

$\xi=\frac{1-\tilde{k}^{2}}{(\tilde{k}^{2}+1)^{2}},$ (4.2)

or, equivalently, in terms of $\varpi=\tilde{k}^{2}+1\geq 1$, by

$\xi\varpi^{2}+\varpi-2=0.$ (4.3)

In $0<(\xi+\frac{1}{8})t^{2/3}<C$, this equation has two solutions $\geq 1$,
which gives in the $\tilde{k}$-plane four real saddle points
$\pm\kappa_{0},\pm\kappa_{1}$:

$\displaystyle\kappa_{0}(\xi)=\left(\frac{\sqrt{1+8\xi}-1-2\xi}{2\xi}\right)^{\frac{1}{2}},$
(4.4)
$\displaystyle\kappa_{1}(\xi)=\left(-\frac{\sqrt{1+8\xi}+1+2\xi}{2\xi}\right)^{\frac{1}{2}}.$
(4.5)

Applying the inverse map $\tilde{k}\leadsto z$ to obtain the corresponding
saddle points in the $z$-plane we get $\pm p_{0},\pm\frac{1}{p_{0}},\pm
p_{1},\pm\frac{1}{p_{1}}$:

$\displaystyle p_{0}(\xi)=\frac{\sqrt{\kappa_{0}^{2}+4}-\kappa_{0}}{2},$ (4.6)
$\displaystyle p_{1}(\xi)=\frac{\sqrt{\kappa_{1}^{2}+4}-\kappa_{1}}{2}.$ (4.7)

As $\xi\to-\frac{1}{8}^{+}$, $\kappa_{0},\kappa_{1}\to\sqrt{3}$. Thus, in the
$z$-plane pairs of saddle points collide:

$p_{0},p_{1}\to\frac{\sqrt{7}-\sqrt{3}}{2},\quad\frac{1}{p_{0}},\frac{1}{p_{1}}\to\frac{\sqrt{7}+\sqrt{3}}{2}.$

By (3.59) and (3.71), the jump matrix for $M^{(1)}$ on $\mathbb{R}$ is

$V^{(2)}(z)=\left(\begin{array}[]{ccc}1&-\overline{\tilde{r}(z)}e^{it\theta_{12}(z)}&0\\\
0&1&0\\\ 0&0&1\end{array}\right)\left(\begin{array}[]{ccc}1&0&0\\\
\tilde{r}(z)e^{-it\theta_{12}(z)}&1&0\\\ 0&0&1\end{array}\right),$ (4.8)

where

$\tilde{r}(z):=\frac{\overline{r(z)}T_{12}(z)}{1-|r(z)|^{2}}.$ (4.9)

The jump matrix on $\omega\mathbb{R}$ and $\omega^{2}\mathbb{R}$ can be
obtained by the symmetries (3.85).

###  4.1 Opening $\bar{\partial}$-lenses

According to the signature table of $\operatorname*{Im}\theta_{12}$ for
$0<\left(\xi+\frac{1}{8}\right)<C$ illustrated in Figure 1(c), we now want to
remove the jump from the intervals
$(-\infty,-\frac{1}{p_{1}})\cup(-\frac{1}{p_{0}},-p_{0})\cup(-p_{1},p_{1})\cup(p_{0},\frac{1}{p_{0}})$
in such a way that the new problem takes advantage of the decay/growth of
$e^{\pm it\theta_{12}(z)}$. Additionally, we want to "open the lens" in such a
way that the lenses are bounded away from the disks introduced previously to
remove the poles from the RH problem.

For $l=0,1,2$, define

$\displaystyle\Omega_{1}^{l}:=\\{z\in\mathbb{C}:0\leq\arg(z-\omega^{l}\frac{1}{p_{1}})\leq\varphi_{0}\\},$
(4.10)
$\displaystyle\Omega_{2}^{l}:=\\{z\in\mathbb{C}:\pi-\varphi_{0}\leq\arg(z-\omega^{l}\frac{1}{p_{0}})\leq\pi,\
|\operatorname*{Re}(z-\omega^{l}\frac{1}{p_{0}})|\leq\frac{1-p_{0}^{2}}{2p_{0}}\\},$
(4.11)
$\displaystyle\Omega_{3}^{l}:=\\{z\in\mathbb{C}:0\leq\arg(z-\omega^{l}p_{0})\leq\varphi_{0},\
|\operatorname*{Re}(z-\omega^{l}p_{0})|\leq\frac{1-p_{0}^{2}}{2p_{0}}\\},$
(4.12)
$\displaystyle\Omega_{4}^{l}:=\\{z\in\mathbb{C}:\pi-\varphi_{0}\leq\arg(z-\omega^{l}p_{1})\leq\pi,\
|\operatorname*{Re}(z-\omega^{l}p_{1})|\leq\frac{p_{1}}{2}\\},$ (4.13)

where $0<\varphi_{0}<\frac{\pi}{8}$ is a sufficiently small angle such that
each $\Omega_{j}^{l}$ doesn’t intersect the set
$\\{z\in\mathbb{C}:\operatorname*{Im}\theta_{12}(z)=0\\}$ and any small disks
$\mathbb{D}_{n},\ n\in\mathcal{N}$. Denote $\Omega_{j}^{l},\ j=5,6,7,8,$ be
the regions of $\Omega_{j}^{l},\ j=1,2,3,4,$ symmetric about the imaginary
axis.

Moreover, we use $\Sigma_{j}^{l},\ j=1,\cdots,8,\ l=0,1,2,$ to denote the
boundary of $\Omega_{j}^{l},\ j=1,\cdots,8,\ l=0,1,2,$ in the upper half plane
and set

$\displaystyle\Sigma^{l}_{2,3}=\left\\{z\in\mathbb{C}:\frac{z}{\omega^{l}}=\frac{1+p_{0}^{2}}{2p_{0}}+i\rho,\
\rho\in(0,\frac{1-p_{0}^{2}}{2p_{0}\tan\varphi_{0}})\right\\},$ (4.14)
$\displaystyle\Sigma^{l}_{6,7}=\left\\{z\in\mathbb{C}:\frac{z}{\omega^{l}}=-\frac{1+p_{0}^{2}}{2p_{0}}+i\rho,\
\rho\in(0,\frac{1-p_{0}^{2}}{2p_{0}\tan\varphi_{0}})\right\\},$ (4.15)
$\displaystyle I_{1}^{l}:=\omega^{l}(\frac{1}{p_{0}},\frac{1}{p_{1}}),\
I^{l}_{2}:=\omega^{l}(p_{1},p_{0}),\ I^{l}_{3}:=\omega^{l}(-p_{0},-p_{1}),\
I_{4}^{l}=\omega^{l}(-\frac{1}{p_{1}},-\frac{1}{p_{0}}).$ (4.16)

$l_{1}$$l_{4}$$l_{2}$$l_{5}$$l_{3}$$l_{6}$$1$$p_{1}$$p_{0}$$\frac{1}{p_{0}}$$\frac{1}{p_{1}}$$-\omega^{2}$$-\omega^{2}p_{1}$$-\omega^{2}p_{0}$$-\omega^{2}\frac{1}{p_{0}}$$-\omega^{2}\frac{1}{p_{1}}$$\omega$$\omega
p_{1}$$\omega
p_{0}$$\omega\frac{1}{p_{0}}$$\omega\frac{1}{p_{1}}$$-1$$-p_{1}$$-p_{0}$$-\frac{1}{p_{0}}$$\frac{1}{p_{1}}$$\omega^{2}$$\omega^{2}p_{1}$$\omega^{2}p_{0}$$\omega^{2}\frac{1}{p_{0}}$$\omega^{2}\frac{1}{p_{1}}$$-\omega$$-\omega
p_{1}$$-\omega p_{0}$$-\omega\frac{1}{p_{0}}$$-\omega\frac{1}{p_{1}}$$0$
Figure 4.1: The contour $\Sigma^{(3)}$.

Using these rays defined above, we define new contours obtained when opening
jump contours $\omega^{l}\mathbb{R}\setminus\cup_{j=1}^{4}I^{l}_{j},\
l=0,1,2$:

$\displaystyle\Sigma^{l}_{p}=\Sigma^{l}_{2,3}\cup\Sigma^{l}_{6,7},\quad
I^{l}=\cup_{j=1}^{4}I^{l}_{j},$
$\displaystyle\tilde{\Sigma}^{l}=({\underset{j=1,\cdots,8}{\cup}}(\Sigma_{j}^{l}\cup(\Sigma_{j}^{l})^{*}))\cup\Sigma^{l}_{p}\cup(\Sigma^{l}_{p})^{*},$
$\displaystyle\Sigma^{(3)}=\cup_{l=0,1,2}(\tilde{\Sigma}^{l}\cup I^{l}),$

as shown in Figure 4.1. Further we also define opened domains along jump
contours $\omega^{l}\mathbb{R}\setminus I^{l},\ l=0,1,2$:

$\displaystyle\Omega=\underset{l=0,1,2}{\underset{j=1,\cdots,8}{\cup}}\Omega_{j}^{l}\cup(\Omega_{j}^{l})^{*},$

From Figure 1(c), we open the contours $\mathbb{R}\setminus I^{0}$ via
continuous extensions of the jump matrix $V^{(3)}(z)$ by defining appropriate
functions and other contours can be opened by the symmetries. We can construct
a matrix function $\mathcal{R}^{(3)}$ like in [21]. The difference is here we
have extra singularity on the boundary. Hence, to deal with the singularity at
$\varkappa_{k}$, $k=1,...,6$, we need to introduce a fixed cutoff function
$\mathcal{X}(z)$ in $C^{\infty}_{0}(\mathbb{R},[0,1])$ with support near $1$
with

$\displaystyle\mathcal{X}(z)=\left\\{\begin{array}[]{llll}0,&|z-1|>2\varepsilon,\\\\[4.0pt]
1,&|z-1|<\varepsilon,\end{array}\right.$ (4.19)

where $\varepsilon$ is a small enough positive constant satisfying the support
of $\mathcal{X}(z)$ doesn’t contain any of phase points with
$\varepsilon<\frac{1-p_{0}^{2}}{16p_{0}}$.

We now define the continuous extension functions in this case: for
$j=1,\cdots,8$,

$\mathcal{R}^{(3)}(z)=\left\\{\begin{array}[]{lll}\left(\begin{array}[]{ccc}1&0&0\\\
-R_{j}(z)e^{-it\theta_{12}}&1&0\\\
0&0&1\end{array}\right),&z\in\Omega_{j}^{0},\\\
\left(\begin{array}[]{ccc}1&-R_{j}^{*}(z)e^{it\theta_{12}}&0\\\ 0&1&0\\\
0&0&1\end{array}\right),&z\in\Omega_{j}^{0*},\\\
\left(\begin{array}[]{ccc}1&0&0\\\ 0&1&0\\\
-R_{j}(\omega^{2}z)e^{-it\theta_{13}}&0&1\end{array}\right),&z\in\Omega^{1}_{j},\\\
\left(\begin{array}[]{ccc}1&0&-R_{j}^{*}(\omega^{2}z)e^{it\theta_{13}}\\\
0&1&0\\\ 0&0&1\end{array}\right),&z\in\Omega^{1*}_{j},\\\
\left(\begin{array}[]{ccc}1&0&0\\\ 0&1&-R_{j}(\omega z)e^{it\theta_{23}}\\\
0&0&1\end{array}\right),&z\in\Omega^{2}_{j},\\\
\left(\begin{array}[]{ccc}1&0&0\\\ 0&1&0\\\ 0&-R_{j}^{*}(\omega
z)e^{-it\theta_{23}}&1\end{array}\right),&z\in\Omega^{2*}_{j},\\\
I,&elsewhere,\\\ \end{array}\right.$ (4.20)

where the functions $R_{j}(z)$, $j=1,\cdots,8$ are given by the following
proposition.

######  Proposition 4.1.

Define functions $R_{j}$: $\overline{\Omega}_{j}^{0}\to\mathbb{C}$,
$j=1,\cdots,8$, continuous on $\overline{\Omega}_{j}^{0}$, with continuous
first partials on $\Omega_{j}^{0}$, and boundary values

$\displaystyle R_{j}(z)=\begin{cases}\tilde{r}^{*}(z),&z\in\mathbb{R},\\\
\tilde{r}^{*}(\frac{1}{p_{1}}),&z\in\Sigma_{1}^{0},\\\
\tilde{r}^{*}(\frac{1}{p_{0}}),&z\in\Sigma_{2}^{0},\\\
\tilde{r}^{*}(p_{0}),&z\in\Sigma_{3}^{0},\\\
\tilde{r}^{*}(p_{1}),&z\in\Sigma_{4}^{0},\\\
\tilde{r}^{*}(-p_{1}),&z\in\Sigma_{5}^{0},\\\
\tilde{r}^{*}(-p_{0}),&z\in\Sigma_{6}^{0},\\\
\tilde{r}^{*}(-\frac{1}{p_{0}}),&z\in\Sigma_{7}^{0},\\\
\tilde{r}^{*}(-\frac{1}{p_{1}}),&z\in\Sigma_{8}^{0},\end{cases}$ (4.21)

such that

$\displaystyle|\bar{\partial}R_{j}(z)|\lesssim|\tilde{r}^{\prime}(\operatorname*{Re}z)|+|\mathcal{X}^{\prime}(\operatorname*{Re}z)|+|\operatorname*{Re}z-\xi_{j}|^{-1/2},\text{for
all $z\in\Omega_{j}^{0}$},$ (4.22)
$\displaystyle|\bar{\partial}R_{j}(z)|\lesssim|\tilde{r}^{\prime}(\operatorname*{Re}z)|+|\mathcal{X}^{\prime}(\operatorname*{Re}z)|,\
\text{for all $z\in\Omega_{j}^{0}$},$ (4.23) $\displaystyle
R_{j}(z)=\bar{\partial}R_{j}(z)=0,\ \text{ for all }z\in\Omega_{j}^{0}\text{
with }|\operatorname*{Re}z\pm 1|<\varepsilon,$ (4.24)
$\displaystyle\bar{\partial}R_{j}(z)=0,\hskip 14.22636pt\text{ elsewhere},$
(4.25)

where $\xi_{j},j=1,\cdots,8$ stands for the saddle points in the corresponding
regions $\Omega_{j}^{0},j=1,\cdots,8$. Setting $R:\Omega\to\mathbb{C}$ by
$R(z)|_{z\in\Omega_{j}^{0}}=R_{j}(z)$, the extension can preserve the symmetry
$R(z)=\overline{R(\bar{z}^{-1})}$.

###### Proof.

Without loss of generality, we give the details for $R_{1}(z)$ and the other
cases are easily inferred. The extension of $R_{1}(z)$ can be constructed by

$R_{1}(z)=(1-\mathcal{X}(\operatorname*{Re}z))\left(\tilde{r}^{*}(\operatorname*{Re}z)-\tilde{r}^{*}(\frac{1}{p_{1}})\right)\cos\left(\frac{\pi}{2\varphi_{0}}\arg(z-\frac{1}{p_{1}})\right)+\tilde{r}^{*}(\frac{1}{p_{1}}).$
(4.26)

Note that for $z-\frac{1}{p_{1}}=le^{i\varphi}=u+vi$,
$l,\varphi,u,v\in\mathbb{R}$, we have
$\bar{\partial}=\frac{1}{2}\left(\partial_{u}+i\partial_{v}\right)=\frac{e^{i\varphi}}{2}\left(\partial_{l}+il^{-1}\partial_{\varphi}\right)$.
Applying $\bar{\partial}$ operator to (4.26), it is readily seen that

$\displaystyle\bar{\partial}R_{1}(z)$
$\displaystyle=\frac{1}{2}\tilde{r}^{*^{\prime}}(u)\cos\left(\frac{\pi\varphi}{2\varphi_{0}}\right)-\frac{1}{2}\mathcal{X}^{\prime}(u)\left(\tilde{r}^{*}(u)-\tilde{r}^{*}(\frac{1}{p_{1}})\right)\cos\left(\frac{\pi\varphi}{2\varphi_{0}}\right)$
$\displaystyle+\frac{1}{2}(1-\mathcal{X}(u))\left(\tilde{r}^{*}(u)-\tilde{r}^{*}(\frac{1}{p_{1}})\right)\bar{\partial}\cos\left(\frac{\pi\varphi}{2\varphi_{0}}\right).$
(4.27)

From the definition (4.9) and Hölder’s inequality, we obtain

$\displaystyle|\tilde{r}^{*}(u)-\tilde{r}^{*}(\frac{1}{p_{1}})|=|\int_{\frac{1}{p_{1}}}^{u}\tilde{r}^{*^{\prime}}(s)ds|\leq\parallel\tilde{r}^{*^{\prime}}\parallel_{L^{2}(\mathbb{R})}|u-\frac{1}{p_{1}}|^{1/2},$
(4.28)

which yields (4.22). Moreover, (4.23) is obtained from the boundedness of
$\tilde{r}^{\prime}(z)$.

∎

In addition, $\mathcal{R}^{(3)}$ achieves the symmetry:

$\mathcal{R}^{(3)}(z)=\Gamma_{1}\overline{\mathcal{R}^{(3)}(\bar{z})}\Gamma_{1}=\Gamma_{2}\overline{\mathcal{R}^{(3)}(\omega^{2}\bar{z})}\Gamma_{2}=\Gamma_{3}\overline{\mathcal{R}^{(3)}(\omega\bar{z})}\Gamma_{3}=\overline{\mathcal{R}^{(3)}(\bar{z}^{-1})}.$
(4.29)

###  4.2 A hybrid $\bar{\partial}$-RH problem and its decomposition

Now we use $\mathcal{R}^{(3)}$ to define a new transformation

$M^{(3)}(z):=M^{(3)}(z;y,t)=M^{(2)}(z)\mathcal{R}^{(3)}(z),$ (4.30)

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

######  RH problem 4.1.

Find a matrix valued function $M^{(3)}(z)$ with following properties:

  * 1.

$M^{(3)}(z)$ has sectionally continuous first partial derivatives in
$\mathbb{C}\setminus\left(\Sigma^{(3)}\cup\left\\{\zeta_{n},\bar{\zeta}_{n}\right\\}_{n\in\lozenge}\right)$,
and is meromorphic out $\bar{\Omega}$.

  * 2.

$M^{(3)}(z)=\Gamma_{1}\overline{M^{(3)}(\bar{z})}\Gamma_{1}=\Gamma_{2}\overline{M^{(3)}(\omega^{2}\bar{z})}\Gamma_{2}=\Gamma_{3}\overline{M^{(3)}(\omega\bar{z})}\Gamma_{3}=\overline{M^{(3)}(\bar{z}^{-1})}$.

  * 3.

$M^{(3)}(z)$ has continuous boundary values $M^{(3)}_{\pm}(z)$ on
$\Sigma^{(3)}$ with

$M^{(3)}_{+}(z)=M^{(3)}_{-}(z)V^{(3)}(z),\hskip 14.22636ptz\in\Sigma^{(3)},$

where

$V^{(3)}(z)=\left\\{\begin{array}[]{ll}\left(\begin{array}[]{ccc}1&-\overline{\tilde{r}(z)}e^{it\theta_{12}(z)}&0\\\
0&1&0\\\ 0&0&1\end{array}\right)\left(\begin{array}[]{ccc}1&0&0\\\
\tilde{r}(z)e^{-it\theta_{12}(z)}&1&0\\\ 0&0&1\end{array}\right),&z\in
I^{0},\\\\[12.0pt] \left(\begin{array}[]{ccc}1&0&0\\\
0&1&-\overline{\tilde{r}(\omega z)}e^{it\theta_{23}(z)}\\\
0&0&1\end{array}\right)\left(\begin{array}[]{ccc}1&0&0\\\ 0&1&0\\\
0&\tilde{r}(\omega z)e^{-it\theta_{23}(z)}&1\end{array}\right),&z\in
I^{2},\\\\[12.0pt] \left(\begin{array}[]{ccc}1&0&0\\\ 0&1&0\\\
-\overline{\tilde{r}(\omega^{2}z)}e^{-it\theta_{13}(z)}&0&1\end{array}\right)\left(\begin{array}[]{ccc}1&0&\tilde{r}(\omega^{2}z)we^{it\theta_{13}(z)}\\\
0&1&0\\\ 0&0&1\end{array}\right),&z\in I^{1},\\\
\mathcal{R}^{(3)}(z)|_{z\in\Omega_{j+1}}-\mathcal{R}^{(2)}(z)|_{z\in\Omega_{j}},&z\in\Sigma_{p}^{l},l=0,1,2,\\\
\mathcal{R}^{(3)}(z)|_{z\in\Omega^{*}_{j+1}}-\mathcal{R}^{(2)}(z)|_{z\in\Omega^{*}_{j}},&z\in\Sigma_{p}^{l*},l=0,1,2\\\
\mathcal{R}^{(3)}(z),&z\in\tilde{\Sigma}^{l},l=0,1,2\\\
\mathcal{R}^{(3)}(z)^{-1},&z\in\tilde{\Sigma}^{l*},l=0,1,2.\end{array}\right.$
(4.31)

  * 4.

$M^{(3)}(z)=I+\mathcal{O}(z^{-1}),\hskip 14.22636ptz\rightarrow\infty$.

  * 5.

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

$\displaystyle\bar{\partial}M^{(3)}(z)=M^{(3)}(z)\bar{\partial}\mathcal{R}^{(3)}(z),$
(4.32)

where

$\mathcal{R}^{(3)}(z)=\left\\{\begin{array}[]{lll}\left(\begin{array}[]{ccc}0&0&0\\\
-\bar{\partial}R_{j}(z)e^{-it\theta_{12}}&0&0\\\
0&0&0\end{array}\right),&z\in\Omega_{j}^{0},\\\
\left(\begin{array}[]{ccc}0&-\bar{\partial}R_{j}^{*}(z)e^{it\theta_{12}}&0\\\
0&0&0\\\ 0&0&0\end{array}\right),&z\in\Omega_{j}^{0*},\\\
\left(\begin{array}[]{ccc}0&0&0\\\ 0&0&0\\\
-\bar{\partial}R_{j}(\omega^{2}z)e^{-it\theta_{13}}&0&1\end{array}\right),&z\in\Omega^{1}_{j},\\\
\left(\begin{array}[]{ccc}0&0&-\bar{\partial}R_{j}^{*}(\omega^{2}z)e^{it\theta_{13}}\\\
0&0&0\\\ 0&0&0\end{array}\right),&z\in\Omega^{1*}_{j},\\\
\left(\begin{array}[]{ccc}0&0&0\\\ 0&0&-R_{j}(\omega z)e^{it\theta_{23}}\\\
0&0&0\end{array}\right),&z\in\Omega^{2}_{j},\\\
\left(\begin{array}[]{ccc}0&0&0\\\ 0&0&0\\\ 0&-\bar{\partial}R_{j}^{*}(\omega
z)e^{-it\theta_{23}}&0\end{array}\right),&z\in\Omega^{2*}_{j},\\\
I,&elsewhere,\\\ \end{array}\right.$ (4.33)

  * 6.

It has the same singularities with the RH problem 3.2.

  * 7.

$M^{(3)}(z)$ has simple poles at each point $\zeta_{n}$ and $\bar{\zeta}_{n}$
for $n\in\lozenge$ with

$\displaystyle\operatorname*{Res}_{z=\zeta_{n}}M^{(3)}(z)=\lim_{z\to\zeta_{n}}M^{(3)}(z)\left[T^{-1}(z)B_{n}T(z)\right].$
(4.34)

To solve the RH problem 4.1, we decompose it into a pure RH problem for
$M^{R}(z):=M^{R}(z;y,t)$ with $\bar{\partial}\mathcal{R}^{(3)}\equiv 0$ and a
pure $\bar{\partial}$-problem with nonzero $\bar{\partial}$-derivatives.

By omitting the $\bar{\partial}$-derivative part of RH problem 4.1, we obtain
the pure RH problem for $M^{R}(z)$ as follows.

######  RH problem 4.2.

  * Find a matrix-valued function $M^{R}(z)$ with following properties:

  * 1.

$M^{R}(z)$ is holomorphic in $\mathbb{C}\setminus\Sigma^{(3)}$.

  * 2.

$M^{R}(z)$ has continuous boundary values $M^{R}_{\pm}(z)$ on $\Sigma^{(3)}$
and

$M^{R}_{+}(z)=M^{R}_{-}(z)V^{(3)}(z),\hskip 14.22636ptz\in\Sigma^{(3)},$

where $V^{(3)}(z)$ is defined by (4.31).

  * 3.

$M^{R}(z)=\Gamma_{1}\overline{M^{R}(\bar{z})}\Gamma_{1}=\Gamma_{2}\overline{M^{R}(\omega^{2}\bar{z})}\Gamma_{2}=\Gamma_{3}\overline{M^{R}(\omega\bar{z})}\Gamma_{3}=\overline{M^{R}(\bar{z}^{-1})}$.

  * 4.

$\bar{\partial}\mathcal{R}^{(3)}(z)=0$, for $z\in\mathbb{C}$.

  * 5.

$M^{R}(z)=I+\mathcal{O}(z^{-1}),\hskip 14.22636ptz\rightarrow\infty$.

  * 6.

As $z\rightarrow\varkappa_{l}=e^{\frac{i\pi(l-1)}{3}},l=1,...,6$, the limit of
$M^{R}(z)$ have the pole singularities

$\displaystyle M^{R}(z)=\frac{1}{z\mp
1}\left(\begin{array}[]{ccc}\alpha^{(2)}_{\pm}&\alpha^{(2)}_{\pm}&\beta^{(2)}_{\pm}\\\
-\alpha^{(2)}_{\pm}&-\alpha^{(2)}_{\pm}&-\beta^{(2)}_{\pm}\\\
0&0&0\end{array}\right)+\mathcal{O}(1),\ z\to\pm 1,$ $\displaystyle
M^{R}(z)=\frac{1}{z\mp\omega^{2}}\left(\begin{array}[]{ccc}0&0&0\\\
\beta^{(2)}_{\pm}&\alpha^{(2)}_{\pm}&\alpha^{(2)}_{\pm}\\\
-\beta^{(2)}_{\pm}&-\alpha^{(2)}_{\pm}&-\alpha^{(2)}_{\pm}\end{array}\right)+\mathcal{O}(1),\
z\to\pm\omega^{2},$ $\displaystyle
M^{R}(z)=\frac{1}{z\mp\omega}\left(\begin{array}[]{ccc}-\alpha^{(2)}_{\pm}&-\beta^{(2)}_{\pm}&-\alpha^{(2)}_{\pm}\\\
0&0&0\\\
\alpha^{(2)}_{\pm}&\beta^{(2)}_{\pm}&\alpha^{(2)}_{\pm}\end{array}\right)+\mathcal{O}(1),\
z\to\pm\omega,$

with $\alpha^{(2)}_{\pm}=\alpha^{(2)}_{\pm}(y,t)=-\bar{\alpha}^{(2)}_{\pm}$,
$\beta^{(2)}_{\pm}=\beta^{(2)}_{\pm}(y,t)=-\bar{\beta}^{(2)}_{\pm}$ and
$M^{R}(z)^{-1}$ has same specific matrix structure with $\alpha^{(2)}_{\pm}$,
$\beta^{(2)}_{\pm}$ replaced by $\tilde{\alpha}^{(2)}_{\pm}$,
$\tilde{\beta}^{(2)}_{\pm}$.

  * 7.

$M^{R}(z)$ has the simple poles at each point $\zeta_{n}$ and
$\bar{\zeta}_{n}$ for $n\in\lozenge$ with

$\displaystyle\operatorname*{Res}_{z=\zeta_{n}}M^{R}(z)=\lim_{z\to\zeta_{n}}M^{R}(z)\left[T^{-1}(z)B_{n}T(z)\right].$

For convenience, let

$\xi_{a}=\frac{\sqrt{7}+\sqrt{3}}{2},\ \xi_{b}=\frac{\sqrt{7}-\sqrt{3}}{2},\
\xi_{c}=-\xi_{b},\ \xi_{d}=-\xi_{a}.$ (4.35)

Then denote

$\displaystyle
U^{l}_{a}=\left\\{z\in\mathbb{C}:\left|z-\omega^{l}\xi_{a}\right|\leq\varrho^{0}\right\\},\
U^{l}_{b}=\left\\{z\in\mathbb{C}:\left|z-\omega^{l}\xi_{b}\right|\leq\varrho^{0}\right\\},$
$\displaystyle
U^{l}_{c}=\left\\{z\in\mathbb{C}:\left|z-\omega^{l}\xi_{c}\right|\leq\varrho^{0}\right\\},\
U^{l}_{d}=\left\\{z\in\mathbb{C}:\left|z-\omega^{l}\xi_{d}\right|\leq\varrho^{0}\right\\},$

where

$\varrho^{0}:=\min\left\\{\frac{p_{1}}{2},2\left(p_{i}-\frac{\sqrt{7}-\sqrt{3}}{2}\right)t^{\delta_{1}},2\left(\frac{1}{p_{i}}-\frac{\sqrt{7}+\sqrt{3}}{2}\right)t^{\delta_{1}},i=0,1\right\\},\
\frac{1}{9}<\delta_{1}<\frac{1}{6}.$ (4.36)

Denote $U:=\underset{l=0,1,2}{\cup}\left(U^{l}_{a}\cup U^{l}_{b}\cup
U^{l}_{c}\cup U^{l}_{d}\right)$ as the union set of $U^{l}_{j}$ with
$j=a,b,c,d$ and $l=0,1,2$. For $t$ large enough, we have
$\omega^{l}\frac{1}{p_{0}},\omega^{l}\frac{1}{p_{1}}\in U^{l}_{a}$,
$\omega^{l}p_{1},\omega^{l}p_{0}\in U^{l}_{b}$,
$-\omega^{l}p_{0},-\omega^{l}p_{1}\in U^{l}_{c}$ and
$-\omega^{l}\frac{1}{p_{1}},-\omega^{l}\frac{1}{p_{0}}\in U^{l}_{d}$.

From (4.31), we know the jump matrix $V^{(3)}(z)$ uniformly goes to $I$ on
$\Sigma^{(3)}$ except in $U$ , which enlightens us to construct the solution
$M^{R}(z)$ as follows:

$M^{R}(z)=\left\\{\begin{array}[]{ll}E(z)M^{O}(z),&z\notin U,\\\
E(z)M^{O}(z)M^{L}(z),&z\in U.\end{array}\right.$ (4.37)

where $M^{O}(z)$ is an outer model including the influence of solitons,
$M^{L}(z)$ is a local model which can be well approximated by the Painlevé II
RH model, and $E(z)$ is the error function which we will prove exists and
bound it asymptotically. Then we use $M^{R}(z)$ to construct a new matrix
function

$M^{(4)}(z):=M^{(4)}(z;y,t)=M^{(3)}(z)M^{R}(z)^{-1}.$ (4.38)

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

$\bar{\partial}$-problem. Find a matrix-valued function
$M^{(4)}(z):=M^{(4)}(z;y,t)$ such that

  * 1.

$M^{(4)}(z)$ has sectionally continuous first partial derivatives in
$\mathbb{C}$.

  * 2.

$M^{(4)}(z)=I+\mathcal{O}(z^{-1}),\hskip 14.22636ptz\rightarrow\infty$.

  * 3.

$M^{(4)}(z)$ satisfies the $\bar{\partial}$-equation

$\bar{\partial}M^{(4)}(z)=M^{(4)}(z)W^{(4)}(z),\ \ z\in\mathbb{C},$

where

$W^{(4)}(z)=M^{R}(z)\bar{\partial}\mathcal{R}^{(3)}(z)M^{R}(z)^{-1}.$ (4.39)

###### Proof.

The proof of this pure $\bar{\partial}$-problem can be given similarly as
shown in Section 4.3 of [17].

∎

The existence and asymptotics of the above pure $\bar{\partial}$-problem for
$M^{(4)}(z)$ will be shown in Section 4.4.2.

###  4.3 Contribution from discrete spectrum

Now we construct a model solution outside $U$ which ignores the jumps
completely. The outer model $M^{O}(z)$ satisfies the following RH problem.

######  RH problem 4.3.

Find a matrix-valued function $M^{O}(z)$ with the following properties:

  * 1.

$M^{O}(z)$ is analytical in
$\mathbb{C}\setminus\left\\{\zeta_{n},\bar{\zeta}_{n}\right\\}_{n\in\lozenge}$.

  * 2.

$M^{O}(z)=\Gamma_{1}\overline{M^{r}(\bar{z})}\Gamma_{1}=\Gamma_{2}\overline{M^{r}(\omega^{2}\bar{z})}\Gamma_{2}=\Gamma_{3}\overline{M^{r}(\omega\bar{z})}\Gamma_{3}=\overline{M^{r}(\bar{z}^{-1})}$.

  * 3.

$M^{O}(z)=I+\mathcal{O}(z^{-1}),\hskip 14.22636ptz\rightarrow\infty$.

  * 4.

The singularity conditions of $M^{O}(z)$ are the same as $M^{R}(z)$ in RH
problem 4.2.

  * 5.

$M^{O}(z)$ has simple poles at each point $\zeta_{n}$ and $\bar{\zeta}_{n}$
for $n\in\lozenge$ with

$\displaystyle\underset{z=\zeta_{n}}{\operatorname{Res}}M^{O}(z)=\lim_{z\to\zeta_{n}}M^{O}(z)\left[T^{-1}(z)B_{n}T(z)\right].$
(4.40)

######  Proposition 4.2.

RH problem has the unique solution

  * 1.

if $\lozenge=\emptyset$, then

$M^{O}(z)=I+\sum_{l=0}^{2}\left(\frac{M_{+\omega^{l}}}{k-\omega^{l}}+\frac{M_{-\omega^{l}}}{k+\omega^{l}}\right).$
(4.41)

  * 2.

if $\lozenge\neq\emptyset$, then

$M^{O}(z)=I+\sum_{l=0}^{2}\left(\frac{M_{+\omega^{l}}}{k-\omega^{l}}+\frac{M_{-\omega^{l}}}{k+\omega^{l}}\right)+\sum_{n\in\lozenge}\left(\frac{N_{n}(\zeta_{n})}{z-\zeta_{n}}+\frac{\bar{N}_{n}(\bar{\zeta}_{n})}{z-\bar{\zeta}_{n}}\right),$
(4.42)

where

$\underset{z=\zeta_{n}}{\rm Res}M^{\mathrm{O}}(z)=N_{n}(\zeta_{n})$ (4.43)

with $n\in\lozenge$.

###### Proof.

The detailed proof for the existence and uniqueness of solution of the above
RH problem 4.3 can be found in Proposition 4.3 of [17]. From this, we know RH
problem 4.3 has an unique solution $M^{O}(z)$ with modified scattering data
$\tilde{D}_{\lozenge}=\left\\{0,\\{\zeta_{n},C_{n}^{\lozenge}\\}_{n\in\lozenge}\\}\right\\}$,
where $C_{n}^{\lozenge}=C_{n}T^{2}(\zeta_{n})$. Then we prove that
(4.41)-(4.42) solves RH problem 4.3. The conditions in RH problem 4.3 implied
that $M^{O}(z)$ is meromorphic with simple poles at $z=\pm\omega^{l}$ and if
$\lozenge\neq\emptyset$, at $z=\zeta_{n}$. If $\lozenge=\emptyset$, then
(4.41) is an immediate consequence of RH problem 4.3 and Liouville’s theorem.
∎

When $r(z)\equiv 0$, denote $u^{\lozenge}(y,t;\tilde{\mathcal{D}})$ as the
$\mathcal{N}(\lozenge)$-soliton with scattering data
$\tilde{\mathcal{D}}_{\lozenge}$. By the reconstruction formula (2.87) and
(2.88), we then have

######  Corollary 4.1.

The soliton solution of the Novikov equation (1.8) is given by

$\displaystyle u^{\lozenge}(y,t;\tilde{\mathcal{D}}_{\lozenge})=$
$\displaystyle\frac{1}{2}\tilde{m}^{\lozenge}_{1}(y,t)\left(\frac{M^{O}_{33}(e^{\frac{i\pi}{6}};y,t)}{M^{O}_{11}(e^{\frac{i\pi}{6}};y,t)}\right)^{1/2}$
$\displaystyle+\frac{1}{2}\tilde{m}^{\lozenge}_{3}(y,t)\left(\frac{M^{O}_{33}(e^{\frac{i\pi}{6}};y,t)}{M^{O}_{11}(e^{\frac{i\pi}{6}};y,t)}\right)^{-1/2}-1,$
(4.44)

in which

$\displaystyle
x^{\lozenge}(y,t;\tilde{\mathcal{D}}_{\lozenge})=y+\frac{1}{2}\ln\frac{M^{O}_{33}(e^{\frac{i\pi}{6}};y,t)}{M^{O}_{11}(e^{\frac{i\pi}{6}};y,t)},\
\tilde{m}^{\lozenge}_{l}:=\sum_{j=1}^{3}M^{O}_{jl}(e^{\frac{i\pi}{6}};y,t),\
l=1,2,3.$ (4.45)

###  4.4 Contribution from jump contours

Since we have ignored the jump conditions completely in the above section, now
we construct the solutions which match with the jump conditions. Again by the
property

$\|V^{(3)}(z)-I\|_{L^{q}(\Sigma^{(3)}\setminus
U)}=\mathcal{O}(e^{-K_{q}t}),\quad t\to\infty,$ (4.46)

where $K_{q}$ is a positive constant and $1\leq q\leq+\infty$, the jump matrix
is exponentially close to the identity outside of $U$ and hence we need to
investigate the local properties near the saddle points.

####  4.4.1 Local model near critical points

We define a new local contour

$\Sigma^{L}={\Sigma}^{L,0}\cup{\Sigma}^{L,1}\cup{\Sigma}^{L,2},$

where ${\Sigma}^{L,l},\ l=0,1,2$ are the local contours on jump contours
$\omega^{n}\mathbb{R}$, respectively

$\displaystyle{\Sigma}^{L,l}=({\underset{j=a,b,c,d}{\cup}}(\Sigma_{j}^{{}^{l}}\cup\Sigma_{j}^{l*})\cup
I^{l})\cap U,\ l=0,1,2.$

Since there are $12$ colliding points $\omega^{l}\xi_{j},l=0,1,2,j=a,b,c,d$,
so the entire local jump contour $\Sigma^{L}$ consists $12$ separate local
jumps in this case. See Figure 4.2. Further denote the local jump for each
colliding point $\omega^{l}\xi_{j}$

$\displaystyle\Sigma^{L,l}_{j}=(\Sigma_{j}^{l}\cup\Sigma_{j}^{l*}\cup
I^{l}_{j})\cap U^{l}_{j},\ l=0,1,2,\ j=a,b,c,d.$

We consider the following local RH problem.

$l_{1}$$l_{4}$$0$$\xi_{c}$$\xi_{d}$$\xi_{b}$$\xi_{a}$$l_{2}$$l_{6}$$l_{3}$$l_{5}$$0$$\xi_{8}$$\xi_{7}$$\xi_{5}$$\xi_{6}$$\xi_{1}$$\xi_{2}$$\xi_{4}$$\xi_{3}$$\omega^{2}\xi_{5}$$\omega\xi_{5}$$\omega\xi_{4}$$\omega^{2}\xi_{4}$$\omega^{2}\xi_{7}$$\omega\xi_{7}$$\omega\xi_{2}$$\omega^{2}\xi_{2}$$\omega^{2}\xi_{6}$$\omega^{2}\xi_{c}$$\omega^{2}\xi_{b}$$\omega\xi_{c}$$\omega\xi_{b}$$\omega\xi_{6}$$\omega\xi_{3}$$\omega^{2}\xi_{3}$$\omega^{2}\xi_{8}$$\omega\xi_{8}$$\omega\xi_{1}$$\omega^{2}\xi_{1}$$\omega^{2}\xi_{d}$$\omega^{2}\xi_{a}$$\omega\xi_{d}$$\omega\xi_{a}$
Figure 4.2: The local jump contour $\Sigma^{L}$ for
$0<(\xi+\frac{3}{8})t^{2/3}<C$.

######  RH problem 4.4.

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

  * 1.

$M^{L}(z)$ is analytical in $\mathbb{C}\setminus\Sigma^{L}$.

  * 2.

$M^{L}(z)=\Gamma_{1}\overline{M^{L}(\bar{z})}\Gamma_{1}=\Gamma_{2}\overline{M^{L}(\omega^{2}\bar{z})}\Gamma_{2}=\Gamma_{3}\overline{M^{L}(\omega\bar{z})}\Gamma_{3}=\overline{M^{L}(\bar{z}^{-1})}$.

  * 3.

$M^{L}(z)$ has continuous boundary values $M^{L}_{\pm}$ on $\Sigma^{L}$ and

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

where $V^{L}(z)=V^{(3)}(z)\big{|}_{z\in\Sigma^{L}}$.

  * 4.

$M^{L}(z)=I+\mathcal{O}(z^{-1}),\hskip 14.22636ptz\rightarrow\infty.$

This local RH problem, which consists of $12$ local models on
$\Sigma^{L,l}_{j}$ about phase point $\omega^{l}\xi_{j},\ j=a,b,c,d,\
l=0,1,2$, has the jump condition and no poles.

######  RH problem 4.5.

Find a matrix-valued function $M^{L,l}_{j}(z)$ with following properties:

  * 1.

$M^{L,l}_{j}(z)$ is analytical in $\mathbb{C}\setminus\Sigma^{L,l}_{j}$.

  * 2.

$M^{L,l}_{j}(z)$ has continuous boundary values $M^{L,l}_{j,\pm}$ and

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

where $V^{L,l}_{j}(z)=V^{L}(z)|_{z\in\Sigma^{L,l}_{j}}$

  * 3.

$M^{L,l}_{j}(z)=I+\mathcal{O}(z^{-1}),\hskip 14.22636ptz\rightarrow\infty$.

According to the theorem of Beals-Coifman, we know as $t\to\infty$, the
solution $M^{L}(z)$ of the RH problem 4.4 is approximated by the sum of the
separate local model $M^{L,l}_{j}(z)$ of the RH problem 4.5 in the
neighborhoods of $\omega^{l}\xi_{j}$, $j=a,b,c,d,\ =0,1,2$ respectively. As
illustrative example, we only consider a local model at the point $\xi_{a}$ on
$\mathbb{R}$, whose jump contours denotes (red lines in Figure 4.2)

$\Sigma^{L,0}_{a}=(\Sigma_{a}^{0}\cup\Sigma_{a}^{0*}\cup I^{0}_{1})\cap
U_{a}^{0},$

which corresponds to the following local RH problem.

######  RH problem 4.6.

Find a matrix-valued function $M^{L,0}_{a}(z)$ with following properties:

  * 1.

$M^{L,0}_{a}(z)$ is analytical in $\mathbb{C}\setminus\Sigma^{L,0}_{a}$.

  * 2.

$M^{L,0}_{a}(z)$ has continuous boundary values $M^{L,0}_{a\pm}(z)$ and
$M^{L,0}_{a+}(z)=M^{L,0}_{a-}(z)V^{L,0}_{a}(z),\ z\in\Sigma^{L,0}_{a}$, where

$\displaystyle
V^{L,0}_{a}(z)=\left\\{\begin{array}[]{ll}\left(\begin{array}[]{ccc}1&0&0\\\
\tilde{r}(1/p_{1})e^{-it\theta_{12}(z)}&1&0\\\
0&0&1\end{array}\right),&z\in\Sigma_{1}^{0}\cap U_{a}^{0},\\\\[10.0pt]
\left(\begin{array}[]{ccc}1&-\overline{\tilde{r}(1/p_{1})}e^{it\theta_{12}(z)}&0\\\
0&1&0\\\ 0&0&1\end{array}\right),&z\in\Sigma_{1}^{0*}\cap
U_{a}^{0},\\\\[10.0pt] \left(\begin{array}[]{ccc}1&0&0\\\
\tilde{r}(1/p_{0})e^{-it\theta_{12}(z)}&1&0\\\
0&0&1\end{array}\right),&z\in\Sigma_{2}^{0}\cap U_{a}^{0},\\\\[10.0pt]
\left(\begin{array}[]{ccc}1&-\overline{\tilde{r}(1/p_{0})}e^{it\theta_{12}(z)}&0\\\
0&1&0\\\ 0&0&1\end{array}\right),&z\in\Sigma_{2}^{0*}\cap
U_{a}^{0},\\\\[10.0pt]
\left(\begin{array}[]{ccc}1&-\overline{\tilde{r}(z)}e^{it\theta_{12}(z)}&0\\\
0&1&0\\\ 0&0&1\end{array}\right)\left(\begin{array}[]{ccc}1&0&0\\\
\tilde{r}(z)e^{-it\theta_{12}(z)}&1&0\\\
0&0&1\end{array}\right),&z\in(1/p_{0},1/p_{1}).\end{array}\right.$ (4.70)

  * 3.

$M^{L,0}_{a}(z)=I+\mathcal{O}(z^{-1}),\hskip 14.22636ptz\rightarrow\infty$.

To solve the RH problem for $M^{L,0}_{a}(z)$, we observe that for $z\in
U_{a}^{0}$ and $t$ large enough,

$\displaystyle-t\theta_{12}(z)=-t\theta_{12}(\xi_{a})+\frac{8}{3}\hat{z}^{3}+2\tilde{s}\hat{z}+\mathcal{O}(t^{-1/3}\hat{z}^{2}).$
(4.71)

Here, we define the scaled spectral parameter $\hat{z}$ as

$\displaystyle\hat{z}=\left(c_{a}t\right)^{1/3}(z-\xi_{a}),$ (4.72)

where the constant $c_{a}$ is given by

$\displaystyle c_{a}=\frac{21}{256}(14\sqrt{3}-9\sqrt{7}).$ (4.73)

Additionally, we introduce the parameter $\tilde{s}$ as

$\displaystyle\tilde{s}=s_{1}(1+8\xi)t^{2/3}$ (4.74)

with

$\displaystyle
s_{1}=\frac{3^{1/6}(-7+\sqrt{21})}{2^{7/3}7^{1/3}(14\sqrt{3}-9\sqrt{7})}<0,$
(4.75)

which parametrizes the space-time region. Next we show that after scaling, RH
problem 4.6 can be well-approximated by the model RH problem B.1 in B, which
is associated with the Painlevé II equation.

Through this change of variable (4.72), it is directly inferred that
$M^{L,0}_{a}(\hat{z})=M^{L,0}_{a}(z(\hat{z}))$ is holomorphic for
$\hat{z}\in\mathbb{C}\setminus\hat{\Sigma}_{a}^{L,0}$ where

$\displaystyle\hat{\Sigma}_{a}^{L,0}=\cup_{j=1,2}\left(\hat{\Sigma}_{j}^{0}\cup\hat{\Sigma}_{j}^{0*}\right)\cup(\hat{z}_{1},\hat{z}_{2}),$
(4.76)

with $\hat{\Sigma}_{j}^{0},j=1,2$ be the corresponding contours of
$\Sigma_{j}^{0},j=1,2$ after scaling, and

$\displaystyle\hat{z}_{1}=(c_{a}t)^{1/3}(1/p_{1}-\xi_{a}),\
\hat{z}_{2}=(c_{a}t)^{1/3}(1/p_{0}-\xi_{a}).$ (4.77)

In addition, the jump matrix $V^{L,0}_{a}(\hat{z})$ satisfies

$\displaystyle
V^{L,0}_{a}(\hat{z})=\left\\{\begin{array}[]{ll}\left(\begin{array}[]{ccc}1&0&0\\\
\tilde{r}(1/p_{1})e^{-it\theta_{12}((c_{a}t)^{-1/3}\hat{z}+\xi_{a})}&1&0\\\
0&0&1\end{array}\right),&\hat{z}\in\hat{\Sigma}_{1}^{0},\\\\[10.0pt]
\left(\begin{array}[]{ccc}1&-\overline{\tilde{r}(1/p_{1})}e^{it\theta_{12}((c_{a}t)^{-1/3}\hat{z}+\xi_{a})}&0\\\
0&1&0\\\ 0&0&1\end{array}\right),&\hat{z}\in\hat{\Sigma}_{1}^{0*},\\\\[10.0pt]
\left(\begin{array}[]{ccc}1&0&0\\\
\tilde{r}(1/p_{0})e^{-it\theta_{12}((c_{a}t)^{-1/3}\hat{z}+\xi_{a})}&1&0\\\
0&0&1\end{array}\right),&\hat{z}\in\hat{\Sigma}_{2}^{0},\\\\[10.0pt]
\left(\begin{array}[]{ccc}1&-\overline{\tilde{r}(1/p_{0})}e^{it\theta_{12}((c_{a}t)^{-1/3}\hat{z}+\xi_{a})}&0\\\
0&1&0\\\ 0&0&1\end{array}\right),&\hat{z}\in\hat{\Sigma}_{2}^{0*},\\\\[10.0pt]
\mathcal{A}_{1}\left(\begin{array}[]{ccc}1&-\overline{\tilde{r}((c_{a}t)^{-1/3}\hat{z}+\xi_{a})}&0\\\
0&1&0\\\ 0&0&1\end{array}\right)\left(\begin{array}[]{ccc}1&0&0\\\
\tilde{r}((c_{a}t)^{-1/3}\hat{z}+\xi_{a})&1&0\\\
0&0&1\end{array}\right)\mathcal{A}_{1}^{-1},&\hat{z}\in(\hat{z}_{1},\hat{z}_{2}),\end{array}\right.$
(4.101)

where

$\displaystyle\mathcal{A}_{1}=\left(\begin{array}[]{ccc}e^{it\theta_{12}((c_{a}t)^{-1/3}\hat{z}+\xi_{a})/2}&0&0\\\
0&e^{-it\theta_{12}((c_{a}t)^{-1/3}\hat{z}+\xi_{a})/2}&0\\\
0&0&1\end{array}\right).$ (4.105)

Since (4.71) holds, we will show the RH problem for $M^{L,0}_{a}(\hat{z})$ in
the $\hat{z}$-plane can be explicitly approximated by the model RH problem for
$N^{P}(\hat{z})$ in Appendix B. To do this, we give the following basic
estimates.

######  Proposition 4.3.

As $t\to+\infty$,

$\displaystyle|\tilde{r}((c_{a}t)^{-1/3}\hat{z}+\xi_{a})e^{-it\theta_{12}((c_{a}t)^{-1/3}\hat{z}+\xi_{a})}-R_{a}e^{i(8\hat{z}^{3}/3+2\tilde{s}\hat{z})}|\lesssim
t^{-1/3+2\delta_{1}},\ \hat{z}\in(\hat{z}_{1},\hat{z}_{2}),$ (4.106)
$\displaystyle|\tilde{r}(1/p_{1})e^{-it\theta_{12}((c_{a}t)^{-1/3}\hat{z}+\xi_{a})}-R_{a}e^{i(8\hat{z}^{3}/3+2\tilde{s}\hat{z})}|\lesssim
t^{-1/3+2\delta_{1}},\ \hat{z}\in\hat{\Sigma}_{1}^{0},$ (4.107)
$\displaystyle|\tilde{r}(1/p_{0})e^{-it\theta_{12}((c_{a}t)^{-1/3}\hat{z}+\xi_{a})}-R_{a}e^{i(8\hat{z}^{3}/3+2\tilde{s}\hat{z})}|\lesssim
t^{-1/3+2\delta_{1}},\ \hat{z}\in\hat{\Sigma}_{2}^{0}.$ (4.108)

where

$\displaystyle R_{a}:=\tilde{r}(\xi_{a})e^{-it\theta_{12}(\xi_{a})}$ (4.109)

###### Proof.

For $\hat{z}\in(\hat{z}_{1},\hat{z}_{2})$, we have

$\displaystyle|e^{-it\theta_{12}((c_{a}t)^{-1/3}\hat{z}+\xi_{a})}|=|e^{i(8\hat{z}^{3}/3+2\tilde{s}\hat{z})}|=1,$

and

$\displaystyle\left|\tilde{r}((c_{a}t)^{-1/3}\hat{z}+\xi_{a})-\tilde{r}(\xi_{a})\right|=\left|\int_{\xi_{a}}^{(c_{a}t)^{-1/3}\hat{z}+\xi_{a}}\tilde{r}^{\prime}(\eta)d\eta\right|\leq\|\tilde{r}^{\prime}\|_{L^{\infty}(\mathbb{R})}\left|(c_{a}t)^{-1/3}\hat{z}\right|\lesssim
t^{-1/3}.$

For $\hat{z}\in\hat{\Sigma}_{1}^{0}$, since
$\operatorname*{Re}(i(8\hat{z}^{3}/3+2\tilde{s}\hat{z}))<0$, it follows that
$|e^{i(8\hat{z}^{3}/3+2\tilde{s}\hat{z})}|$ is bounded and

$\displaystyle\left|e^{-it\theta_{12}((c_{a}t)^{-1/3}\hat{z}+\xi_{a})}-e^{-it\theta_{12}(\xi_{a})}e^{i(8\hat{z}^{3}/3+2\tilde{s}\hat{z})}\right|=\left|e^{\mathcal{O}(t^{-1/3}\hat{z}^{3})}-1\right|\lesssim
t^{-1/3+2\delta_{1}}.$ (4.110)

On the other hand,

$\displaystyle\left|\tilde{r}(1/p_{1})-\tilde{r}(\xi_{a})\right|=\left|\int_{\xi_{a}}^{1/p_{1}}\tilde{r}^{\prime}(\eta)d\eta\right|\leq\|\tilde{r}^{\prime}\|_{L^{\infty}(\mathbb{R})}\left|1/p_{1}-\xi_{a}\right|\lesssim
t^{-1/3}.$ (4.111)

Then (4.107) can be derived from (4.110) and (4.111). For
$\hat{z}\in\hat{\Sigma}_{2}^{0}$, we can use a similar method to prove
(4.108).

∎

Then we obtain the following proposition as a direct corollary of Proposition
4.3.

######  Proposition 4.4.

As $t\to+\infty$,

$\displaystyle
M_{a}^{L,0}(\hat{z})=\hat{M}_{a}^{L,0}(\hat{z})+\mathcal{O}(t^{-1/3}+2\delta_{1}),$
(4.112)

where $\hat{M}_{a}^{L,0}(\hat{z})$ satisfies the following RH problem.

######  RH problem 4.7.

Find a matrix-valued function $\hat{M}_{a}^{L,0}(\hat{z})$ with properties:

  * 1.

$\hat{M}^{L,0}_{a}(\hat{z})$ is analytical in
$\mathbb{C}\setminus\hat{\Sigma}^{L,0}_{a}$.

  * 2.

$\hat{M}^{L,0}_{a+}(\hat{z})=\hat{M}^{L,0}_{a-}(\hat{z})\hat{V}^{L,0}_{a}(\hat{z}),\
\hat{z}\in\hat{\Sigma}^{L,0}_{a}$, where

$\displaystyle\hat{V}^{L,0}_{a}(\hat{z})=\left\\{\begin{array}[]{ll}\left(\begin{array}[]{ccc}1&0&0\\\
R_{a}e^{i(8\hat{z}^{3}/3+2\tilde{s}\hat{z})}&1&0\\\
0&0&1\end{array}\right),&\hat{z}\in\hat{\Sigma}_{j}^{0},\ j=1,2,\\\\[10.0pt]
\left(\begin{array}[]{ccc}1&-\overline{R_{a}}e^{-i(8\hat{z}^{3}/3+2\tilde{s}\hat{z})}&0\\\
0&1&0\\\ 0&0&1\end{array}\right),&\hat{z}\in\hat{\Sigma}_{j}^{0*},\
j=1,2,\\\\[10.0pt]
\left(\begin{array}[]{ccc}1&-\overline{R_{a}}e^{-i(8\hat{z}^{3}/3+2\tilde{s}\hat{z})}&0\\\
0&1&0\\\ 0&0&1\end{array}\right)\left(\begin{array}[]{ccc}1&0&0\\\
R_{a}e^{i(8\hat{z}^{3}/3+2\tilde{s}\hat{z})}&1&0\\\
0&0&1\end{array}\right),&\hat{z}\in(\hat{z}_{1},\hat{z}_{2}),\end{array}\right.$
(4.128)

  * 3.

$\hat{M}^{L,0}_{a}(\hat{z})=I+\mathcal{O}(\hat{z}^{-1}),\hskip
14.22636pt\hat{z}\rightarrow\infty$.

To match RH problem 4.7 with the model RH problem, we need to convert the
coefficients in front of the exponential terms in the jump matrix into the
pure imaginary number. Rewriting $R_{a}$ as

$\displaystyle R_{a}=|R_{a}|e^{i\varphi_{a}}$ (4.129)

with

$\displaystyle|R_{a}|=|r(\xi_{a})|,\ \varphi_{a}=\arg
R_{a}=\arg\tilde{r}(\xi_{a})-t\theta_{12}(\xi_{a}),$ (4.130)

we make the following transformation

$\displaystyle\tilde{M}^{L,0}_{a}(\hat{z})=\mathcal{B}_{1}\hat{M}^{L,0}_{a}(\hat{z})\mathcal{B}_{1}^{-1},$
(4.131)

where

$\displaystyle\mathcal{B}_{1}=\left(\begin{array}[]{ccc}e^{i(\varphi_{a}/2-\pi/4)}&0&0\\\
0&e^{-i(\varphi_{a}/2-\pi/4)}&0\\\ 0&0&1\end{array}\right).$ (4.135)

Then $\tilde{M}^{L,0}_{a}(\hat{z})$ satisfies the following RH problem.

######  RH problem 4.8.

Find a matrix-valued function $\tilde{M}_{a}^{L,0}(\hat{z})$ with properties:

  * 1.

$\tilde{M}^{L,0}_{a}(\hat{z})$ is analytical in
$\mathbb{C}\setminus\hat{\Sigma}^{L,0}_{a}$.

  * 2.

$\tilde{M}^{L,0}_{a+}(\hat{z})=\tilde{M}^{L,0}_{a-}(\hat{z})\tilde{V}^{L,0}_{a}(\hat{z}),\
\hat{z}\in\hat{\Sigma}^{L,0}_{a}$, where

$\displaystyle\tilde{V}^{L,0}_{a}(\hat{z})=\left\\{\begin{array}[]{ll}\left(\begin{array}[]{ccc}1&0&0\\\
i|r(\xi_{a})|e^{i(8\hat{z}^{3}/3+2\tilde{s}\hat{z})}&1&0\\\
0&0&1\end{array}\right),&\hat{z}\in\hat{\Sigma}_{j}^{0},\ j=1,2,\\\\[10.0pt]
\left(\begin{array}[]{ccc}1&i|r(\xi_{a})|e^{-i(8\hat{z}^{3}/3+2\tilde{s}\hat{z})}&0\\\
0&1&0\\\ 0&0&1\end{array}\right),&\hat{z}\in\hat{\Sigma}_{j}^{0*},\
j=1,2,\\\\[10.0pt]
\left(\begin{array}[]{ccc}1&i|r(\xi_{a})|e^{-i(8\hat{z}^{3}/3+2\tilde{s}\hat{z})}&0\\\
0&1&0\\\ 0&0&1\end{array}\right)\left(\begin{array}[]{ccc}1&0&0\\\
i|r(\xi_{a})|e^{i(8\hat{z}^{3}/3+2\tilde{s}\hat{z})}&1&0\\\
0&0&1\end{array}\right),&\hat{z}\in(\hat{z}_{1},\hat{z}_{2}),\end{array}\right.$
(4.151)

  * 3.

$\tilde{M}^{L,0}_{a}(\hat{z})=I+\mathcal{O}(\hat{z}^{-1}),\hskip
14.22636pt\hat{z}\rightarrow\infty$.

Observing the jump matrix (4.151), the RH problem 4.8 could be explicitly
solved by using the model RH problem in Appendix B.

######  Proposition 4.5.

Let $p=i|r(\xi_{a})|$ and $p_{j}=0,\ j=1,\cdots,n$ in (B.2), then we obtain
$\tilde{M}^{L,0}_{a}(\hat{z})=N^{P}(\hat{z})$.

###### Proof.

Note that $N^{P}(\hat{z})$ is invertible, we define

$\displaystyle\Xi(\hat{z}):=\tilde{M}^{L,0}_{a}(\hat{z})N^{P}(\hat{z})^{-1},$
(4.152)

which satisfies the following RH problem.

######  RH problem 4.9.

Find a matrix-valued function $\Xi(\hat{z})$ with properties:

  * 1.

$\Xi(\hat{z})$ is analytical in $\mathbb{C}\setminus\hat{\Sigma}^{L,0}_{a}$.

  * 2.

$\Xi_{+}(\hat{z})=\Xi_{-}(\hat{z})V^{\Xi}(\hat{z}),\
\hat{z}\in\hat{\Sigma}^{L,0}_{a}$, where

$\displaystyle
V^{\Xi}(\hat{z})=N^{P}_{-}(\hat{z})\tilde{V}^{L,0}_{a}(\hat{z})V^{P}(\hat{z})^{-1}N^{P}_{+}(\hat{z})^{-1}.s$
(4.153)

  * 3.

$\Xi(\hat{z})=I+\mathcal{O}(\hat{z}^{-1}),\hskip
14.22636pt\hat{z}\rightarrow\infty$.

Because of the boundedness of $N^{P}(\hat{z})$ (see (B.4) below), it is
sufficient to estimate the error between the jump matrices
$\tilde{V}^{L,0}_{a}(\hat{z})$ and $V^{P}(\hat{z})$.

$\displaystyle\tilde{V}^{L,0}_{a}(\hat{z})-V^{P}(\hat{z})=$
$\displaystyle\begin{cases}\begin{pmatrix}1&0&0\\\
(i|r(\xi_{a})|-p)e^{i(8\hat{z}^{2}/3+2\tilde{s}\hat{z})}&1&0\\\
0&0&1\end{pmatrix},\quad\hat{z}\in\hat{\Sigma}^{0}_{j},\ j=1,2,\\\
\begin{pmatrix}1&(i|r(\xi_{a})|+p^{*})e^{-i(8\hat{z}^{2}/3+2\tilde{s}\hat{z})}&0\\\
0&1&0\\\ 0&0&1\end{pmatrix},\quad\hat{z}\in\hat{\Sigma}^{0*}_{j},\ j=1,2,\\\
\begin{pmatrix}-|r(\xi_{a})|^{2}+|p|^{2}&(i|r(\xi_{a})|+p^{*})e^{-i(8\hat{z}^{2}/3+2\tilde{s}\hat{z})}&0\\\
(ir(\xi_{a})-p)e^{i(8\hat{z}^{2}/3+2\tilde{s}\hat{z})}&0&0\\\
0&0&1\end{pmatrix},\quad\hat{k}\in(\hat{z}_{1},\hat{z}_{2}).\end{cases}$

Then using Proposition 4.3, we have

$\displaystyle\|V^{\Xi}(\hat{z})-I\|_{L^{1}\cap L^{2}\cap
L^{\infty}(\hat{\Sigma}^{L,0}_{a})}\lesssim t^{-1/3+2\delta_{1}}.$ (4.154)

Thus, the existence and uniqueness of $\Xi(\hat{z})$ are valid by a small norm
RH problem arguments, which also yields

$\displaystyle\Xi(\hat{z})=I+\mathcal{O}(t^{-1/3+2\delta_{1}}),\quad\text{as}\
t\to+\infty.$ (4.155)

∎

Finally, the following proposition follows directly from propositions 4.3-4.5.

######  Proposition 4.6.

As $t\to+\infty$,

$\displaystyle
M_{a}^{L,0}(\hat{z})=\mathcal{B}_{1}^{-1}N^{P}(\hat{z})\mathcal{B}_{1}+\mathcal{O}(t^{-1/3+2\delta_{1}}).$
(4.156)

######  Proposition 4.7.

Moreover, as $\hat{z}\to\infty$,

$\displaystyle
M_{a}^{L,0}(\hat{z})=I+\frac{M_{a1}^{L,0}}{\hat{z}}+\mathcal{O}(\hat{z}^{-2}),$
(4.157)

where $M_{a1}^{L,0}$ is the coefficient of the term $1/\hat{z}$ in the
large-$\hat{z}$ expansion of $M_{a}^{L,0}(\hat{z})$. As $t\to+\infty$,

$\displaystyle
M_{a1}^{L,0}=\frac{i}{2}\begin{pmatrix}-\int_{\tilde{s}}^{\infty}u(\eta)^{2}d\eta&u(\tilde{s})e^{-i\varphi_{a}}&0\\\
-u(\tilde{s})e^{i\varphi_{a}}&\int_{\tilde{s}}^{\infty}u(\eta)^{2}d\eta&0\\\
0&0&0\end{pmatrix}+\mathcal{O}(t^{-1/3+2\delta_{1}}).$ (4.158)

The RH problem for $M_{b}^{L,0}$, $M_{c}^{L,0}$ and $M_{d}^{L,0}$ can be
solved in a similar manner. For $z\in U_{b}^{0}$ and $t$ large enough, we have

$\displaystyle-t\theta_{12}(z)=-t\theta_{12}(\xi_{b})+\frac{8}{3}\check{z}^{3}+2\tilde{s}\check{z}+\mathcal{O}(t^{-1/3}\check{z}^{2}),$
(4.159)

where $\tilde{s}$ is defined by (4.74) and

$\displaystyle\check{z}=(c_{b}t)^{\frac{1}{3}}(z-\xi_{b}),$ (4.160)

is the scaled spectral parameter in this case with

$\displaystyle c_{b}=\frac{21}{256}(14\sqrt{3}+9\sqrt{7}).$ (4.161)

Through the analysis, we obtain the result of the large-$\check{z}$ expansion
of $M_{b}^{L,0}(\check{z})$,

$\displaystyle
M_{b}^{L,0}(\check{z})=I+\frac{M_{b1}^{L,0}}{\check{z}}+\mathcal{O}(\check{z}^{-2}),$
(4.162)

where as $t\to+\infty$,

$\displaystyle
M_{b1}^{L,0}=\frac{i}{2}\begin{pmatrix}-\int_{\tilde{s}}^{\infty}u(\eta)^{2}d\eta&u(\tilde{s})e^{-i\varphi_{b}}&0\\\
-u(\tilde{s})e^{i\varphi_{b}}&\int_{\tilde{s}}^{\infty}u(\eta)^{2}d\eta&0\\\
0&0&0\end{pmatrix}+\mathcal{O}(t^{-1/3+2\delta_{1}}),$ (4.163)

with

$\displaystyle\varphi_{b}=\arg\tilde{r}(\xi_{b})-t\theta_{12}\xi_{b}.$ (4.164)

For $z\in U_{c}^{0}$, and for sufficiently large $t$, the following
relationship holds:

$\displaystyle-t\theta_{12}(z)=-t\theta_{12}(\xi_{c})+\frac{8}{3}\tilde{z}^{3}+2\tilde{s}\tilde{z}+\mathcal{O}(t^{-1/3}\tilde{z}^{2}).$
(4.165)

Here, $\tilde{s}$ is defined in accordance with (4.74), and the scaled
spectral parameter $\tilde{z}$ is given by:

$\displaystyle\tilde{z}=(c_{c}t)^{\frac{1}{3}}(z-\xi_{c}),$ (4.166)

where

$\displaystyle c_{c}=c_{b},$ (4.167)

which is as specified in (4.161). Moreover, as $\tilde{z}\to\infty$,

$\displaystyle
M_{c}^{L,0}(\check{z})=I+\frac{M_{c1}^{L,0}}{\check{z}}+\mathcal{O}(\tilde{z}^{-2}),$
(4.168)

where as $t\to+\infty$,

$\displaystyle
M_{c1}^{L,0}=\frac{i}{2}\begin{pmatrix}-\int_{\tilde{s}}^{\infty}u(\eta)^{2}d\eta&-u(\tilde{s})e^{i\varphi_{b}}&0\\\
u(\tilde{s})e^{-i\varphi_{b}}&\int_{\tilde{s}}^{\infty}u(\eta)^{2}d\eta&0\\\
0&0&0\end{pmatrix}+\mathcal{O}(t^{-1/3+2\delta_{1}}).$ (4.169)

Here we use the symmetry $|r(z)|=|r(-z)|$ whose detailed proof could be given
by the similar method in the proof of Proposition 4.2 in [15].

For $z\in U_{d}^{0}$, and for sufficiently large $t$, the following
relationship holds:

$\displaystyle-t\theta_{12}(z)=-t\theta_{12}(\xi_{d})+\frac{8}{3}\breve{z}^{3}+2\tilde{s}\breve{z}+\mathcal{O}(t^{-1/3}\breve{z}^{2}).$
(4.170)

Here, $\tilde{s}$ is defined in accordance with (4.74), and the scaled
spectral parameter $\breve{z}$ is given by:

$\displaystyle\breve{z}=(c_{d}t)^{\frac{1}{3}}(z-\xi_{d}),$ (4.171)

where

$\displaystyle c_{d}=c_{a},$ (4.172)

which is as specified in (4.73). Moreover, as $\breve{z}\to\infty$,

$\displaystyle
M_{d}^{L,0}(\breve{z})=I+\frac{M_{d1}^{L,0}}{\breve{z}}+\mathcal{O}(\breve{z}^{-2}),$
(4.173)

where as $t\to+\infty$,

$\displaystyle
M_{d1}^{L,0}=\frac{i}{2}\begin{pmatrix}-\int_{\tilde{s}}^{\infty}u(\eta)^{2}d\eta&-u(\tilde{s})e^{i\varphi_{a}}&0\\\
u(\tilde{s})e^{-i\varphi_{a}}&\int_{\tilde{s}}^{\infty}u(\eta)^{2}d\eta&0\\\
0&0&0\end{pmatrix}+\mathcal{O}(t^{-1/3+2\delta_{1}}).$ (4.174)

According to the symmetries of RH problem 2.1, we have the following
proposition.

######  Proposition 4.8.

As $z\to\infty$, the coefficients $M_{j1}^{L,l},\ j=a,b,c,d,\ l=0,1,2$ of the
term of $z^{-1}$ have the following relationships:

$\displaystyle
M_{j1}^{L,1}=\omega\Gamma_{3}\overline{M_{j1}^{L,0}}\Gamma_{3},\quad
M_{j1}^{L,2}=\omega^{2}\Gamma_{2}\overline{M_{j1}^{L,0}}\Gamma_{2}.$ (4.175)

Combining the results above, finally we obtain the following proposition.

######  Proposition 4.9.

As $t\to+\infty$,

$\displaystyle
M^{L}(z)=I+t^{-1/3}\sum_{j=a,b,c,d}c_{j}^{-1/3}\left(\frac{M_{j1}^{L,0}}{z-\xi_{j}}+\frac{\omega\Gamma_{3}\overline{M_{j1}^{L,0}}\Gamma_{3}}{z-\omega\xi_{j}}+\frac{\omega^{2}\Gamma_{2}\overline{M_{j1}^{L,0}}\Gamma_{2}}{z-\omega^{2}\xi_{j}}\right)+\mathcal{O}(t^{-1}),$
(4.176)

where $c_{j},\ j=a,b,c,d$ are defined by (4.73), (4.161), (4.167), and
(4.172), and $M_{j1}^{L,0},\ j=a,b,c,d$ are given by (4.158), (4.163),
(4.169), and (4.174).

####  4.4.2 Small norm RH problem for residual error

In this section, we consider the error matrix-function $E(z)$. From the
definition (4.37), we can obtain the RH problem for $E(z)$:

######  RH problem 4.10.

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

  * 1.

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

$\displaystyle\Sigma^{E}=\partial U\cup(\Sigma^{(3)}\setminus U).$

  * 2.

$E(z)$ has continuous boundary values $E_{\pm}(z)$ satisfying

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

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

$V^{E}(z)=\left\\{\begin{array}[]{llll}M^{O}(z)V^{(3)}(z)M^{O}(z)^{-1},&z\in\Sigma^{(3)}\setminus
U,\\\\[4.0pt] M^{O}(z)M^{L,j}_{j}(z)^{-1}M^{O}(z)^{-1},&z\in\partial
U_{j}^{l},\end{array}\right.$ (4.177)

with $j=a,b,c,d,\ l=0,1,2$. see Figure 4.3.

  * 3.

$E(z)$ has the following asymptotic behavior:

$\displaystyle E(z)=I+\mathcal{O}(z^{-1}),\hskip
14.22636ptz\rightarrow\infty.$

$l_{1}$$l_{4}$$l_{2}$$l_{5}$$l_{3}$$l_{6}$$\xi_{b}$$\xi_{a}$$\xi_{c}$$\xi_{d}$$\omega^{2}\xi_{d}$$\omega^{2}\xi_{c}$$\omega\xi_{b}$$\omega\xi_{a}$$\omega^{2}\xi_{b}$$\omega^{2}\xi_{a}$$\omega\xi_{c}$$\omega\xi_{d}$$0$
Figure 4.3: The jump contour $\Sigma^{E}$. The red circles represents $U$.

We will show that the RH problem 4.10 for $E(z)$ is solvable for large $t$ as
a small norm RH problem. By Proposition 4.9, the jump matrix $V^{E}(z)$ admits
the following estimates.

$\parallel
V^{E}(z)-I\parallel_{L^{p}(\Sigma^{E})}\lesssim\left\\{\begin{array}[]{llll}e^{-ct^{3\delta_{1}}},&z\in\Sigma^{E}\setminus
U,\\\\[4.0pt] t^{-K_{p}},&z\in\partial U,\end{array}\right.$ (4.178)

for some positive $c$ with $K_{\infty}=\delta_{1}$ and $K_{2}=1/6+\delta_{1}$.
Therefore, the existence and uniqueness of RH problem 4.10 can be shown by
using a small-norm RH problem [RN9, RN10]. Moreover, its solution can be given
by

$E(z)=I+\frac{1}{2\pi
i}\int_{\Sigma^{E}}\dfrac{\varpi(\eta)(V^{E}(\eta)-I)}{\eta-z}ds,$ (4.179)

where $\varpi\in I+L^{2}(\Sigma^{E})$ is the unique solution of the following
equation

$\varpi=I+\mathcal{C}_{E}\varpi.$ (4.180)

Here $\mathcal{C}_{E}$ is a integral operator defined by

$\mathcal{C}_{E}(f)(z)=\mathcal{C}^{-}\left(f(V^{E}(z)-I)\right),$ (4.181)

and $\mathcal{C}^{-}$ is the Cauchy projection operator on $\Sigma^{E}$. By
(4.178), we have

$\parallel\mathcal{C}_{E}\parallel_{L^{2}(\Sigma^{E})}\leq\parallel\mathcal{C}^{-}\parallel_{L^{2}(\Sigma^{E})}\parallel
V^{E}(z)-I\parallel_{L^{\infty}(\Sigma^{E})}\lesssim t^{-\delta_{1}},$ (4.182)

which implies that $1-\mathcal{C}_{E}$ is invertible for sufficiently large
$t$. So $\varpi$ exists and is unique with

$\displaystyle\varpi=I+(1-\mathcal{C}_{E})^{-1}(\mathcal{C}_{E}I).$ (4.183)

Moreover, the following estimates hold:

$\parallel\mathcal{C}_{E}I\parallel_{L^{2}(\Sigma^{E})}\lesssim
t^{-1/6-\delta_{1}/2},\quad\parallel\varpi-I\parallel_{L^{2}(\Sigma^{E})}\lesssim
t^{-1/6-\delta_{1}/2}.$ (4.184)

In order to reconstruct the solution $u(y,t)$ of (1.8), it is necessary to
consider the long time asymptotic behavior of $E(e^{\frac{i\pi}{6}})$. Note
that when we estimate its asymptotic behavior, from (4.179) and (4.178) we
only need to consider the calculation on $\partial U$ because it approaches
zero exponentially on the other boundary.

######  Proposition 4.10.

As $z=e^{\frac{i\pi}{6}}$, we have

$\displaystyle E(e^{\frac{i\pi}{6}})=I+\frac{1}{2\pi
i}\int_{\Sigma^{E}}\dfrac{\varpi(\eta)(V^{E}(\eta)-I)}{\eta-e^{\frac{i\pi}{6}}}d\eta,$

with the long time asymptotic behavior

$E(e^{\frac{i\pi}{6}})=I+t^{-1/3}E^{(0)}+\mathcal{O}(t^{-2/3+2\delta_{1}}),$
(4.185)

where $E^{(0)}$ is explicitly computed by

$\displaystyle E^{(0)}$
$\displaystyle=-\sum_{l=0}^{2}\sum_{j\in\\{a,b,c,d\\}}c_{j}^{-1/3}\frac{M^{O}(\omega^{l}\xi_{j})M^{L,j}_{j1}M^{O}(\omega^{l}\xi_{j})^{-1}}{\omega^{l}\xi_{j}-e^{\frac{i\pi}{6}}}$
$\displaystyle=-\sum_{j\in\\{a,b,c,d\\}}c_{j}^{-1/3}\Bigl{(}\frac{M^{O}(\xi_{j})M^{L,0}_{j1}M^{O}(\xi_{j})^{-1}}{\xi_{j}-e^{\frac{i\pi}{6}}}-\frac{\omega
M^{O}(\omega\xi_{j})\Gamma_{3}\overline{M^{L,0}_{j1}}\Gamma_{3}M^{O}(\omega\xi_{j})^{-1}}{\omega\xi_{j}-e^{\frac{i\pi}{6}}}$
$\displaystyle-\frac{\omega^{2}M^{O}(\omega^{2}\xi_{j})\Gamma_{2}\overline{M^{L,0}_{j1}}\Gamma_{2}M^{O}(\omega^{2}\xi_{j})^{-1}}{\omega^{2}\xi_{j}-e^{\frac{i\pi}{6}}}\Bigr{)}.$
(4.186)

###### Proof.

From (4.176) and (4.177), it follows that

$\displaystyle E(e^{\frac{i\pi}{6}})$ $\displaystyle=I+\frac{1}{2\pi
i}\sum_{l=0}^{2}\sum_{j\in\\{a,b,c,d\\}}\oint_{\partial
U^{l}_{j}}\frac{M^{O}(\eta)(M_{j}^{l}(\eta)^{-1}-I)M^{O}(\eta)^{-1}}{\eta-e^{\frac{i\pi}{6}}}d\eta+\mathcal{O}(t^{-1/3-\delta_{1}})$
$\displaystyle=I-\frac{1}{2\pi
i}\sum_{j\in\\{a,b,c,d\\}}(c_{j}t)^{-1/3}\oint_{\partial
U^{0}_{j}}\frac{M^{O}(\eta)M^{L,0}_{j1}M^{O}(\eta)^{-1}}{(\eta-e^{\frac{i\pi}{6}})(\eta-\xi_{j})}d\eta$
$\displaystyle-\frac{1}{2\pi
i}\sum_{j\in\\{a,b,c,d\\}}(c_{j}t)^{-1/3}\oint_{\partial
U^{1}_{j}}\omega\frac{M^{O}(\eta)\Gamma_{3}\overline{M^{L,0}_{j1}}\Gamma_{3}M^{O}(\eta)^{-1}}{(\eta-e^{\frac{i\pi}{6}})(\eta-\omega\xi_{j})}d\eta$
$\displaystyle-\frac{1}{2\pi
i}\sum_{j\in\\{a,b,c,d\\}}(c_{j}t)^{-1/3}\oint_{\partial
U^{2}_{j}}\omega^{2}\frac{M^{O}(\eta)\Gamma_{2}\overline{M^{L,0}_{j1}}\Gamma_{2}M^{O}(\eta)^{-1}}{(\eta-e^{\frac{i\pi}{6}})(\eta-\omega^{2}\xi_{j})}d\eta$
$\displaystyle+\mathcal{O}(t^{-2/3+2\delta_{1}})$
$\displaystyle=I-\sum_{j\in\\{a,b,c,d\\}}(c_{j}t)^{-1/3}M^{O}(e^{\frac{i\pi}{6}})\left(\frac{M^{L,0}_{j1}}{\xi_{j}-e^{\frac{i\pi}{6}}}+\frac{\omega\Gamma_{3}\overline{M^{L,0}_{j1}}\Gamma_{3}}{\omega\xi_{j}-e^{\frac{i\pi}{6}}}+\frac{\omega^{2}\Gamma_{2}\overline{M^{L,0}_{j1}}\Gamma_{2}}{\omega^{2}\xi_{j}-e^{\frac{i\pi}{6}}}\right)M^{O}(e^{\frac{i\pi}{6}})^{-1}$
$\displaystyle+\mathcal{O}(t^{-2/3+2\delta_{1}}).$

∎

###  4.5 Contribution from $\bar{\partial}$-components

Now we consider the asymptotics of $M^{(4)}$ of the $\bar{\partial}$-problem
4.1, whose solution can be given by an integral equation

$M^{(4)}(z)=I+\frac{1}{\pi}\iint_{\mathbb{C}}\dfrac{M^{(4)}(\eta)W^{(4)}(\eta)}{\eta-z}dm(\eta),$
(4.187)

where $m(\eta)$ is the Lebesgue measure on the $\mathbb{C}$. Define the left
Cauchy-Green integral operator,

$fC_{z}(z)=\frac{1}{\pi}\iint_{\mathbb{C}}\dfrac{f(\eta)W^{(4)}(\eta)}{\eta-z}dm(\eta),$

then the above equation (4.187) can be rewritten as

$\left(I-C_{z}\right)M^{(4)}(z)=I.$ (4.188)

Aiming at estimating $M^{(4)}(z)$, we need to evaluate the norm of the
integral operator $\left(I-{C}_{z}\right)^{-1}$ in this transition zone. To
finish this goal, we first give the following lemma.

######  Lemma 4.1.

Let $\Omega_{j}^{l},\ j=1,\cdots,8,\ l=0,1,2$, denote the sectors defined in
(4.10)- (4.13). The following estimates for the imaginary part of the phase
function $\theta_{12}(z)$ in the transition zone $0<(\xi+1/8)t^{2/3}<C$, as
defined in (2.69), are valid. Similar estimates can also be provided for the
imaginary parts of the phase functions $\theta_{23}(z)$ and $\theta_{31}(z)$.

  * 1.

If $|\operatorname*{Re}z(1-|z|^{-2})|\leq 2$, then

$\displaystyle\operatorname*{Im}\theta_{12}(z)\leq-\frac{\sqrt{3}(7-3\tilde{k}_{1}^{2})}{(\tilde{k}_{1}+1)^{2}}\operatorname*{Im}z\left(\operatorname*{Re}z-\xi_{j}\right)^{2},\quad
z\in\Omega_{j}^{0},\ j=1,\cdots,8,$ (4.189a)
$\displaystyle\operatorname*{Im}\theta_{12}(z)\geq\frac{\sqrt{3}(7-3\tilde{k}_{1}^{2})}{(\tilde{k}_{1}+1)^{2}}\operatorname*{Im}z\left(\operatorname*{Re}z-\xi_{j}\right)^{2},\quad
z\in\Omega_{j}^{0*},\ j=1,\cdots,8,$ (4.189b)

  * 2.

If $|\operatorname*{Re}z(1-|z|^{-2})|>2$, then

$\displaystyle\operatorname*{Im}\theta_{12}(z)\leq-\frac{\sqrt{3}}{8}\operatorname*{Im}z,\quad
z\in\Omega_{j}^{0},\ j=1,\cdots,8,$ (4.190a)
$\displaystyle\operatorname*{Im}\theta_{12}(z)\geq\frac{\sqrt{3}}{8}\operatorname*{Im}z,\quad
z\in\Omega_{j}^{0*},\ j=1,\cdots,8,$ (4.190b)

###### Proof.

We give the proof on the sector $\Omega_{1}$ and the proof on other sectors
can be obtained similarly.

For $z\in\Omega_{1}$, denote

$\displaystyle z-1/z:=u+vi,\ u,v\in\mathbb{R},\ \text{and}\
\tilde{k}_{1}:=\xi_{1}-1/\xi_{1}.$

Then we obtain $u=\operatorname*{Re}z(1-1/|z|^{2})$,
$v=\operatorname*{Im}z(1+1/|z|^{2})$. From (2.69),

$\displaystyle\operatorname*{Im}\theta_{12}=\sqrt{3}v\left(\xi+F(u,v)\right),\
F(u,v):=\frac{u^{2}+v^{2}-1}{(u^{2}-v^{2}+1)^{2}+4u^{2}v^{2}}.$ (4.191)

It is evident that

$\displaystyle F(u,v)\leq\begin{cases}&F(u,0),\quad u^{2}\leq 4,\\\ &0,\quad
u^{2}>4.\end{cases}$ (4.192)

Moreover, from (4.2),we have

$\displaystyle\xi+F(u,0)=\frac{1-\tilde{k}_{1}^{2}}{(1+\tilde{k}_{1}^{2})^{2}}+\frac{u^{2}-1}{(u^{2}+1)^{2}}=(u^{2}-\tilde{k}_{1}^{2})\frac{3+u^{2}-\tilde{k}_{1}^{2}(u^{2}-1)}{(1+\tilde{k}_{1}^{2})^{2}(u^{2}+1)^{2}}$
(4.193)

Inserting (4.192) and (4.193) into (4.191), we obtain (4.189a) and (4.190a)
from the ranges of $u$ and $\tilde{k}_{1}$.

∎

By the above lemma 4.1, we can show that for sufficiently large $t$ the
operator $\mathcal{C}_{z}$ is small-norm, so that the resolvent operator
$(I-\mathcal{C}_{z})^{-1}$ exists and can be expressed as a Neumann series.

######  Proposition 4.11.

The norm of the integral operator $C_{z}$ satisfies the inequality

$\parallel C_{z}\parallel_{L^{\infty}\to L^{\infty}}\lesssim t^{-1/3},\ \
t\to\infty.$ (4.194)

###### Proof.

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

$\displaystyle\parallel fC_{z}\parallel_{L^{\infty}}$
$\displaystyle\leq\parallel
f\parallel_{L^{\infty}}\frac{1}{\pi}\iint_{\mathbb{C}}\dfrac{|W^{(4)}(s)|}{|z-s|}dm(s).$

We detail the case for matrix functions having support in the region
$\Omega_{1}^{0}$, the case for the other regions follows similarly. Recall the
definition of
$W^{(4)}(z)=M^{R}(z)\bar{\partial}\mathcal{R}^{(3)}(z)M^{R}(z)^{-1}$. Note
that $W^{(4)}(z)\equiv 0$ out of $\overline{\Omega}$. Proposition 4.9 and 4.10
implies the boundedness of $M^{R}(z)$ and $M^{R}(z)^{-1}$ for
$z\in\overline{\Omega}_{1}$. By (4.33) and (4.39), it follows that

$\frac{1}{\pi}\iint_{\Omega_{1}}\dfrac{|W^{(4)}(\eta)|}{|z-\eta|}dm(\eta)\lesssim\frac{1}{\pi}\iint_{\Omega_{1}}\dfrac{|\bar{\partial}R_{1}(\eta)e^{-it\theta_{12}}|}{|z-\eta|}dm(\eta).$

From Lemma 4.1, we divide $\Omega_{1}^{0}$ in two regions:

  * 1.

$\\{z\in\Omega_{1}^{0}:\operatorname*{Re}z(1-|z|^{-2})\leq
2\\}\subseteq\Omega_{A}:=\\{z\in\Omega_{1}^{0}:\operatorname*{Re}z\leq 3\\}$,

  * 2.

$\\{z\in\Omega_{1}^{0}:\operatorname*{Re}z(1-|z|^{-2})>2\\}\subseteq\Omega_{B}:=\\{z\in\Omega_{1}^{0}:\operatorname*{Re}z\geq
2\\}$.

Set $\eta=u+1/p_{1}+vi$, $z=z_{R}+iz_{I}$, $u,v,z_{R},z_{I}\in\mathbb{R}$, and
$1/q+1/p=1$ with $p>2$.

Referring to (4.22) and (4.23) in Proposition 4.1, then the following integral
can be divided into three part:

$\displaystyle\iint_{\Omega_{1}^{0}}\dfrac{|\bar{\partial}R_{1}(\eta)|e^{t\text{Im}\theta_{12}}}{|z-\eta|}dm(\eta)\lesssim\hat{I}_{1}+\hat{I}_{2}+\hat{I}_{3},$

with

$\displaystyle\hat{I}_{1}:=\iint_{\Omega_{1}^{0}}\dfrac{\left(|\tilde{r}^{\prime}(u+1/p_{1})|+|\mathcal{X}^{\prime}(u+1/p1)|\right)e^{t\text{Im}\theta_{12}}}{|z-\eta|}dm(\eta),$
$\displaystyle\hat{I}_{2}:=\iint_{\Omega_{B}}\dfrac{|u|^{-1/2}e^{t\text{Im}\theta_{12}}}{|z-\eta|}dm(\eta).$

Our task now is to estimate the above integrals $\hat{I}_{i}$, $i=1,2,3$,
respectively.

$\displaystyle\hat{I}_{1}\leq$
$\displaystyle\iint_{\Omega_{A}}\dfrac{\left(|\tilde{r}^{\prime}(u+1/p_{1})|+|\mathcal{X}^{\prime}(u+1/p_{1})|\right)e^{t\text{Im}\theta_{12}}}{|z-\eta|}dm(\eta)$
$\displaystyle+$
$\displaystyle\iint_{\Omega_{B}}\dfrac{\left(|\tilde{r}^{\prime}(u+1/p_{1})|+|\mathcal{X}^{\prime}(u+1/p_{1})|\right)e^{t\text{Im}\theta_{12}}}{|z-\eta|}dm(\eta)$
$\displaystyle\leq$
$\displaystyle\int_{0}^{(3-1/p_{1})\tan\varphi_{0}}\int_{v}^{3}\dfrac{e^{-t\frac{\sqrt{3}(7-3\tilde{k}_{1}^{2})}{(\tilde{k}_{1}+1)^{2}}vu^{2}}}{|z-\eta|}dudv+\int_{2}^{+\infty}\int_{u\tan(\varphi_{0}/3)}^{u\tan\varphi_{0}}\dfrac{e^{-t\frac{\sqrt{3}}{8}v}}{|z-\eta|}dvdu.$

Note that the following basic inequalities hold

$\displaystyle\||z-\eta|^{-1}\|_{L^{q}_{u}(v,\infty)}\lesssim|v-y|^{-1+1/q},\quad\|e^{-tvu^{2}}\|_{L^{p}_{u}(v,\infty)}\lesssim(tv)^{-1/(2p)},\quad\|e^{-tv}\|_{L^{p}_{u}(v,\infty)}\lesssim(pt)^{-1/p}.$
(4.195)

Then using Cauchy-Schwartz’s inequality, we have

$\displaystyle\int_{0}^{(3-1/p_{1})\tan\varphi_{0}}\int_{v}^{3}\dfrac{e^{-t\frac{\sqrt{3}(7-3\tilde{k}_{1}^{2})}{(\tilde{k}_{1}+1)^{2}}vu^{2}}}{|z-\eta|}dudv$
$\displaystyle\lesssim
t^{-1/4}\int_{0}^{(3-1/p_{1})\tan\varphi_{0}}|v-z_{I}|^{-1/2}v^{-1/4}e^{-t\frac{\sqrt{3}(7-3\tilde{k}_{1}^{2})}{(\tilde{k}_{1}+1)^{2}}v^{3}}dv\lesssim
t^{-1/3}.$ (4.196)

Using Hölder’s inequality, we obtain

$\displaystyle\int_{2}^{+\infty}\int_{u\tan(\varphi_{0}/3)}^{u\tan\varphi_{0}}\dfrac{e^{-t\frac{\sqrt{3}}{8}v}}{|z-\eta|}dvdu\lesssim
t^{-1/p}\int_{2}^{\infty}e^{-t\frac{\sqrt{3}}{8}u\tan(\varphi_{0}/3)}|u-z_{R}|^{-1+1/q}du\lesssim
t^{-1}.$ (4.197)

Together (4.196) with (4.197) gives us that

$\displaystyle\hat{I}_{1}\lesssim t^{-1/3}.$ (4.198)

To estimate $\hat{I}_{2}$, it follows from Hölder’s inequality again that

$\displaystyle\hat{I}_{2}\leq$
$\displaystyle\iint_{\Omega_{B}}\dfrac{|u|^{-1/2}e^{t\text{Im}\theta_{12}}}{|z-\eta|}dm(\eta)\leq\int_{2}^{+\infty}\int_{u\tan(\varphi_{0}/3)}^{u\tan\varphi_{0}}\dfrac{|u|^{-1/2}e^{-t\frac{\sqrt{3}}{8}v}}{|z-\eta|}dvdu$
$\displaystyle\lesssim$
$\displaystyle\int_{2}^{+\infty}u^{1/p-1/2}|u-z_{R}|^{-1+1/q}e^{-t\frac{\sqrt{3}}{8}u\tan(\varphi_{0}/3)}du\lesssim
t^{-1/2}.$

Then combining the estimates for $\hat{I}_{1}$ and $\hat{I}_{2}$ yields the
estimate

$\displaystyle\frac{1}{\pi}\iint_{\Omega_{1}}\dfrac{|W^{(4)}(\eta)|}{|z-\eta|}dm(\eta)\lesssim
t^{-1/3}.$

The integral over other regions can be estimated in similar manners, which
finally confirms (4.194).

∎

Then from (4.188), we immediately arrive at the existence and uniqueness of
$M^{(4)}(z)$ for $z\in\mathbb{C}$. Take $z=e^{\frac{i\pi}{6}}$ in (4.187),
then

$M^{(4)}(e^{\frac{\pi
i}{6}})=I+\frac{1}{\pi}\iint_{\mathbb{C}}\dfrac{M^{(4)}(\eta)W^{(4)}(\eta)}{\eta-e^{\frac{\pi
i}{6}}}dm(\eta).$ (4.199)

To reconstruct the solution of (1.8), we need the local behaviors of (4.199)
as $t\to\infty$.

######  Proposition 4.12.

There exists a constant $T_{1}$, such that for all $t>T_{1}$, the solution
$M^{(4)}(z)$ of the $\bar{\partial}$-problem admits the following estimate:

$\displaystyle|M^{(4)}(e^{\frac{\pi i}{6}})-I|\lesssim t^{-2/3}.$ (4.200)

###### Proof.

Similar to the proof of Proposition 4.11, we only give the proof for the
integral over $\Omega_{1}^{0}$ and the integral on other regions can be
obtained in the same way. Proposition 4.11 and (4.188) imply that the
boundedness of $M^{(4)}(z)$ for $z\in\Omega_{1}^{0}$ as $t\to\infty$. Then it
is sufficient to consider the integral

$\displaystyle\iint_{\mathbb{C}}\dfrac{|M^{(4)}(\eta)W^{(4)}(\eta)|}{\eta-e^{\frac{\pi
i}{6}}}dm(\eta)\lesssim\iint_{\mathbb{C}}\dfrac{|W^{(4)}(\eta)|}{\eta-e^{\frac{\pi
i}{6}}}dm(\eta).$

Referring (4.23) in Proposition 4.1, and note that
$|\mathcal{X}^{\prime}(\text{Re}z)|=0$ in $\overline{\Omega_{1}^{0}}$, this
integral can be divided into two parts

$\displaystyle\iint_{\Omega_{1}^{0}}\dfrac{|W^{(4)}(\eta)|}{\eta-e^{\frac{\pi
i}{6}}}dm(\eta)\leq\left(\iint_{\Omega_{A}}+\iint_{\Omega_{B}}\right)\dfrac{|\bar{\partial}R_{1}(\eta)|e^{2t\text{Im}\theta_{12}}}{\eta-e^{\frac{\pi
i}{6}}}dm(\eta).$ (4.201)

Let $\eta=u+1/p_{1}+vi$ with $u,v\in\mathbb{R}$. From (4.23) and
$|\eta-e^{\frac{\pi i}{6}}|^{-1}$ is bounded for $z\in\Omega_{A}$, we have

$\displaystyle\iint_{\Omega_{A}}\dfrac{|\bar{\partial}R_{1}(\eta)|e^{2t\text{Im}\theta_{12}}}{\eta-e^{\frac{\pi
i}{6}}}dm(\eta)$
$\displaystyle\leq\int_{0}^{(3-1/p_{1})\tan\varphi_{0}}\int_{v}^{3}|\tilde{r}^{\prime}(u+1/p_{1})|e^{-t\frac{\sqrt{3}(7-3\tilde{k}_{1}^{2})}{(\tilde{k}_{1}+1)^{2}}vu^{2}}dudv$
$\displaystyle\lesssim\int_{0}^{\infty}\int_{v}^{\infty}e^{-t\frac{\sqrt{3}(7-3\tilde{k}_{1}^{2})}{(\tilde{k}_{1}+1)^{2}}vu^{2}}dudv\lesssim
t^{-2/3}.$

To estimate the integral over $\Omega_{B}$, we use the fact
$|\eta-e^{\frac{\pi i}{6}}|^{-1}\leq|\eta|^{-1}$ and the estimate (4.23) to
obtain that

$\displaystyle\iint_{\Omega_{B}}\dfrac{|\bar{\partial}R_{1}(\eta)|e^{2t\text{Im}\theta_{12}}}{\eta-e^{\frac{\pi
i}{6}}}dm(\eta)\leq\int_{2}^{+\infty}\int_{0}^{u}(u^{2}+v^{2})^{-1/2}e^{-t\frac{\sqrt{3}}{8}v}dvdu\lesssim
t^{-1}.$ (4.202)

Combining the above two estimates, we arrive at the desired result (4.200). ∎

###  4.6 Proof of Theorem 1.1-(i)

Now we begin to construct the large time asymptotics of the Novikov equation
(1.8) in the first transition zone $|\xi+\frac{1}{8}|t^{2/3}<C$. Inverting the
sequence of transformations (3.70), (3.100), (4.30), (4.37) and (4.38), we
have for $z\in\mathbb{C}\setminus U$,

$\displaystyle M(z)=$ $\displaystyle
M^{(4)}(z)E(z)M^{O}(z)\mathcal{R}^{(3)}(z)^{-1}T(z)^{-1}G(z)^{-1}+\mathcal{O}(e^{-ct}).$
(4.203)

To reconstruct the solution $u(x,t)$ by using (2.87), we take $z=e^{\frac{\pi
i}{6}}$. In this case, $\mathcal{R}^{(3)}(z)=G(z)=I$, and then we can obtain
that

$\displaystyle M(e^{\frac{\pi i}{6}})=$ $\displaystyle M^{(4)}(e^{\frac{\pi
i}{6}})E(e^{\frac{\pi i}{6}})M^{O}(e^{\frac{\pi i}{6}})T(e^{\frac{\pi
i}{6}})^{-1}+\mathcal{O}(e^{-ct}).$ (4.204)

Using Proposition 4.12, (4.204) comes down to

$\displaystyle M(e^{\frac{\pi i}{6}})=E(e^{\frac{\pi i}{6}})M^{O}(e^{\frac{\pi
i}{6}})T(e^{\frac{\pi i}{6}})^{-1}+\mathcal{O}(t^{-2/3}).$ (4.205)

Substitute the above estimates into (2.87) and (2.88), and obtain

$\displaystyle u(y,t)$
$\displaystyle=u^{\lozenge}(y,t;\tilde{\mathcal{D}}_{\lozenge})\left(T_{1}(e^{\frac{\pi
i}{6}})T_{3}(e^{\frac{\pi i}{6}})\right)^{-1/2}-1$
$\displaystyle+\frac{1}{2}\left(T_{1}(e^{\frac{\pi i}{6}})T_{3}(e^{\frac{\pi
i}{6}})\right)^{-1/2}f_{11}t^{-1/3}+\mathcal{O}(t^{-2/3+2\delta_{1}}),$
$\displaystyle x(y,t)$
$\displaystyle=y+\frac{1}{2}\ln\frac{M^{O}_{33}(e^{\frac{i\pi}{6}};y,t)}{M^{O}_{11}(e^{\frac{i\pi}{6}};y,t)}+\frac{1}{2}\ln\left(\frac{T_{1}(e^{\frac{\pi
i}{6}})}{T_{3}(e^{\frac{\pi
i}{6}})}\right)+\frac{1}{2}f_{12}t^{-1/3}+\mathcal{O}(t^{-2/3+2\delta_{1}}),$
$\displaystyle=x^{\lozenge}(y,t;\tilde{\mathcal{D}}_{\lozenge})+\frac{1}{2}\ln\left(\frac{T_{1}(e^{\frac{\pi
i}{6}})}{T_{3}(e^{\frac{\pi
i}{6}})}\right)+\frac{1}{2}f_{12}t^{-1/3}+\mathcal{O}(t^{-2/3+2\delta_{1}}),$

where

$\displaystyle f_{11}$
$\displaystyle=\frac{1}{2}\left(\tilde{m}^{\lozenge}_{1}(y,t)\left(\frac{M_{33}^{O}(e^{\frac{\pi
i}{6}})}{M_{11}^{O}(e^{\frac{\pi
i}{6}})}\right)^{1/2}-\tilde{m}^{\lozenge}_{3}(y,t)\left(\frac{M_{33}^{O}(e^{\frac{\pi
i}{6}})}{M_{11}^{O}(e^{\frac{\pi i}{6}})}\right)^{-1/2}\right)f_{12}$
$\displaystyle+(E^{(0)}M^{O}(e^{\frac{\pi
i}{6}}))_{1}\left(\frac{M_{33}^{O}(e^{\frac{\pi
i}{6}})}{M_{11}^{O}(e^{\frac{\pi
i}{6}})}\right)^{1/2}+(E^{(0)}M^{O}(e^{\frac{\pi
i}{6}}))_{3}\left(\frac{M_{33}^{O}(e^{\frac{\pi
i}{6}})}{M_{11}^{O}(e^{\frac{\pi i}{6}})}\right)^{-1/2},$ (4.206)
$\displaystyle f_{12}$ $\displaystyle=(E^{(0)}M^{O}(e^{\frac{\pi
i}{6}}))_{33}M^{O}_{33}(e^{\frac{\pi i}{6}})^{-1}-(E^{(0)}M^{O}(e^{\frac{\pi
i}{6}}))_{11}M^{O}_{11}(e^{\frac{\pi i}{6}})^{-1},$ (4.207)

$(E^{(0)}M^{O})_{ij}$ represents the element in the $i$-th row and $j$-th
column of the matrix $E^{(0)}M^{O}$,
$(E^{(0)}M^{O})_{j}=\sum_{i=1}^{3}(E^{(0)}M^{O})_{ij}$, and
$u^{\lozenge}(y,t;\tilde{\mathcal{D}}_{\lozenge})$,
$x^{\lozenge}(y,t;\tilde{\mathcal{D}}_{\lozenge})$, and
$\tilde{m}^{\lozenge}_{j}(y,t)$ are defined in Corollary 4.1. Bring (3.65)
into the above formulas, we obtain the final result.

##  5 Painlevé Asymptotics in $|\xi-1|t^{2/3}<C$

As we have done in the previous section, the analysis here is performed for
$-C<(\xi-1)t^{2/3}<0$. In this case, the saddle point equation (4.3) has
exactly one solution $\geq 1$, which gives in the $\tilde{k}$-plane two real
saddle points $\pm\kappa_{0}$:

$\kappa_{0}(\xi)=\left(\frac{\sqrt{1+8\xi}-1-2\xi}{2\xi}\right)^{\frac{1}{2}}.$
(5.1)

Applying the inverse map $\tilde{k}\leadsto z$ we get in the $z$-plane four
real saddle points $\pm p_{0},\pm\frac{1}{p_{0}}$:

$p_{0}(\xi)=\frac{\kappa_{0}-\sqrt{\kappa_{0}^{2}+4}}{2}.$ (5.2)

As $\xi\to 1^{-}$, $\kappa_{0}\to 0$ and pairs of saddle points collide:

$p_{0},\frac{1}{p_{0}}\to 1.$

By (3.86), the jump matrix $V^{(2)}(z)$ on $\mathbb{R}$ satisfies

$\displaystyle V^{(2)}(z)=\left(\begin{array}[]{ccc}1&0&0\\\
\tilde{r}(z)e^{-it\theta_{12}(z)}&1&0\\\
0&0&1\end{array}\right)\left(\begin{array}[]{ccc}1&-\overline{\tilde{r}(z)}e^{it\theta_{12}(z)}&0\\\
0&1&0\\\ 0&0&1\end{array}\right),\ $ $\displaystyle z\in\mathbb{R},$ (5.9)

where

$\displaystyle\tilde{r}(z):=\bar{r}(z)T_{12}(z).$ (5.10)

The jump matrix on $\omega\mathbb{R}$ and $\omega^{2}\mathbb{R}$ can be
obtained by the symmetries (3.85). The signature table of
$\operatorname*{Im}\theta_{12}$ for $-C<(\xi-1)t^{2/3}<0$ is depcited in
Figure 1(d), which implies opening lenses around
$(-\infty,-p_{0})\cup(-1/p_{0},1/p_{0})\cup(p_{0},\infty)$ in what follows.

###  5.1 Opening $\bar{\partial}$-lenses

For $l=0,1,2$, denote

$\displaystyle\Omega_{1}^{l}:=\\{z\in\mathbb{C}:0\leq\arg(z-\omega^{l}p_{0})\leq\varphi_{0}\\},$
$\displaystyle\Omega_{2}^{l}:=\\{z\in\mathbb{C}:\pi-\varphi_{0}\leq\arg(z-\omega^{l}/p_{0})\leq\pi,|\operatorname*{Re}(z-\omega^{l}/p_{0})|\leq|1/p_{0}|\\},$
$\displaystyle\Omega_{3}^{l}:=\\{z\in\mathbb{C}:-z\in\Omega_{2}^{l}\\},\quad\Omega_{4}^{l}:=\\{z\in\mathbb{C}:-z\in\Omega_{1}^{l}\\},$

where $0<\varphi_{0}<\frac{\pi}{8}$ is a fixed angle such that each
$\Omega_{j}^{l}$ doesn’t intersect the set
$\\{z\in\mathbb{C}:\operatorname*{Im}\theta_{12}(z)=0\\}$ and any small disks
$\mathbb{D}_{n},\ n\in\mathcal{N}$. Moreover, we use $\Sigma_{j}^{l}$ to
denote the boundary of $\Omega_{j}^{l}$ in the upper half plane. Let

$\displaystyle\tilde{\Sigma}^{l}=\cup_{j=1,\cdots,4}\left(\Sigma_{j}^{l}\cup\Sigma_{j}^{l*}\right)\cup(\omega^{l}1/p_{0},\omega^{l}p_{0})\cup(-\omega^{l}p_{0},-\omega^{l}1/p_{0}),$
(5.11)
$\displaystyle\Sigma^{(3)}=\cup_{l=0,1,2}\tilde{\Sigma}^{l},\quad\Omega=\underset{l=0,1,2}{\underset{j=1,\cdots,4}{\cup}}\left(\Omega_{j}^{l}\cup\Omega_{j}^{l*}\right)$
(5.12)

See Figure 5.1.

$l_{1}$$l_{4}$$l_{2}$$l_{3}$$l_{6}$$l_{5}$$-p_{0}$$-\frac{1}{p_{0}}$$p_{0}$$\frac{1}{p_{0}}$$-\omega^{2}\frac{1}{p_{0}}$$-\omega\frac{1}{p_{0}}$$\omega\frac{1}{p_{0}}$$\omega^{2}\frac{1}{p_{0}}$$-\omega^{2}p_{0}$$-\omega
p_{0}$$\omega
p_{0}$$\omega^{2}p_{0}$$1$$-1$$\varkappa_{2}$$\varkappa_{3}$$\varkappa_{6}$$\varkappa_{5}$
Figure 5.1: The contour $\Sigma^{(3)}$.

Following the same steps as in Section 4.1, we first define the continuous
extension functions in this case. For $j=1,\cdots,4$,

$\mathcal{R}^{(3)}(z)=\left\\{\begin{array}[]{lll}\left(\begin{array}[]{ccc}1&R_{j}^{*}(z)e^{it\theta_{12}}&0\\\
0&1&0\\\ 0&0&1\end{array}\right),&z\in\Omega_{j}^{0},\\\
\left(\begin{array}[]{ccc}1&0&0\\\ R_{j}(z)e^{-it\theta_{12}}&1&0\\\
0&0&1\end{array}\right),&z\in\Omega_{j}^{0*},\\\
\left(\begin{array}[]{ccc}1&0&R_{j}^{*}(\omega^{2}z)e^{it\theta_{13}}\\\
0&1&0\\\ 0&0&1\end{array}\right),&z\in\Omega_{j}^{1},\\\
\left(\begin{array}[]{ccc}1&0&0\\\ 0&1&0\\\
R_{j}(\omega^{2}z)e^{-it\theta_{13}}&0&1\end{array}\right),&z\in\Omega_{j}^{1*},\\\
\left(\begin{array}[]{ccc}1&0&0\\\ 0&1&0\\\ 0&R_{j}^{*}(\omega
z)e^{-it\theta_{23}}&1\end{array}\right),&z\in\Omega_{j}^{2},\\\
\left(\begin{array}[]{ccc}1&0&0\\\ 0&1&R_{j}(\omega z)e^{it\theta_{23}}\\\
0&0&1\end{array}\right),&z\in\Omega_{j}^{2*},\\\ I,&elsewhere,\\\
\end{array}\right.$ (5.13)

where the functions $R_{j}(z)$, $j=1,\cdots,4$ are given by the following
proposition.

######  Proposition 5.1.

It is possible to define functions
$R_{j}:\overline{\Omega_{j}^{0}}\to\mathbb{C}$, $j=1,\cdots,4$, continuous on
$\overline{\Omega_{j}^{0}}$, with continuous first partials on
$\Omega_{j}^{0}$, and boundary values

$\displaystyle R_{j}(z)=\begin{cases}\tilde{r}(z),&z\in\mathbb{R},\\\
\tilde{r}(p_{0}),&z\in\Sigma_{1}^{0},\\\
\tilde{r}(\frac{1}{p_{0}}),&z\in\Sigma_{2}^{0},\\\
\tilde{r}(-\frac{1}{p_{0}}),&z\in\Sigma_{3}^{0},\\\
\tilde{r}(-p_{0}),&z\in\Sigma_{4}^{0},\end{cases}$ (5.14)

such that

$\displaystyle|\bar{\partial}R_{j}(z)|\lesssim|\tilde{r}^{\prime}(\operatorname*{Re}z)|+|\mathcal{X}^{\prime}(\operatorname*{Re}z)|+|\operatorname*{Re}z\mp
p_{0}|^{-1/2},\text{for all $z\in\Omega_{1}^{0}$ for $-$, $\Omega_{4}^{0}$ for
$+$},$ (5.15)
$\displaystyle|\bar{\partial}R_{j}(z)|\lesssim|\tilde{r}^{\prime}(\operatorname*{Re}z)|+|\mathcal{X}^{\prime}(\operatorname*{Re}z)|+|\operatorname*{Re}z\mp\frac{1}{p_{0}}|^{-1/2},\text{for
all $z\in\Omega_{2}^{0}$ for $-$, $\Omega_{3}^{0}$ for $+$},$ (5.16)
$\displaystyle|\bar{\partial}R_{j}(z)|\lesssim|\tilde{r}^{\prime}(\operatorname*{Re}z)|+|\mathcal{X}^{\prime}(\operatorname*{Re}z)|,\
\text{for all $z\in\Omega_{j}^{0}$},$ (5.17) $\displaystyle
R_{j}(z)=\bar{\partial}R_{j}(z)=0,\ \text{ for all }z\in\Omega_{j}^{0}\text{
with }|\operatorname*{Re}z\pm 1|<\varepsilon,$ (5.18)
$\displaystyle\bar{\partial}R_{j}(z)=0,\hskip 14.22636pt\text{ elsewhere},$
(5.19)

Setting $R:\Omega\to\mathbb{C}$ by $R(z)|_{z\in\Omega_{j}^{0}}=R_{j}(z)$, the
extension can preserve the symmetry $R(z)=\overline{R(\bar{z}^{-1})}$.

###### Proof.

The proof is similar to that in Proposition 4.1. Here we omit it. ∎

###  5.2 A hybrid $\bar{\partial}$-RH problem and its decomposition

As in the previous section, we define a transformation

$M^{(3)}(z)=M^{(2)}(z)\mathcal{R}^{(3)}(z),$ (5.20)

where $\mathcal{R}^{(3)}(z)$ is defined by (5.13). Then $M^{(3)}(z)$ satisfies
the following $\bar{\partial}$-RH problem.

######  RH problem 5.1.

Find a matrix valued function $M^{(3)}(z)$ with following properties:

  * 1.

$M^{(3)}(z)$ is continuous in
$\mathbb{C}\setminus\left(\Sigma^{(3)}\cup\left\\{\zeta_{n},\bar{\zeta}_{n}\right\\}_{n\in\lozenge}\right)$.

  * 2.

$M^{(3)}(z)=\Gamma_{1}\overline{M^{(3)}(\bar{z})}\Gamma_{1}=\Gamma_{2}\overline{M^{(3)}(\omega^{2}\bar{z})}\Gamma_{2}=\Gamma_{3}\overline{M^{(3)}(\omega\bar{z})}\Gamma_{3}=\overline{M^{(3)}(\bar{z}^{-1})}$.

  * 3.

$M^{(3)}(z)$ takes continuous boundary values $M^{(3)}_{\pm}(z)$ on
$\Sigma^{(3)}$ with

$M^{(3)}_{+}(z)=M^{(3)}_{-}(z)V^{(3)}(z),\hskip 14.22636ptz\in\Sigma^{(3)},$

where

$V^{(3)}(z)=\left\\{\begin{array}[]{ll}\left(\begin{array}[]{ccc}1&0&0\\\
\tilde{r}(z)e^{-it\theta_{12}(z)}&1&0\\\
0&0&1\end{array}\right)\left(\begin{array}[]{ccc}1&-\overline{\tilde{r}(z)}e^{it\theta_{12}(z)}&0\\\
0&1&0\\\ 0&0&1\end{array}\right),&z\in I^{0},\\\\[12.0pt]
\left(\begin{array}[]{ccc}1&0&0\\\ 0&1&0\\\ 0&\tilde{r}(\omega
z)e^{-it\theta_{23}(z)}&1\end{array}\right)\left(\begin{array}[]{ccc}1&0&0\\\
0&1&-\overline{\tilde{r}(\omega z)}e^{it\theta_{23}(z)}\\\
0&0&1\end{array}\right),&z\in I^{2},\\\\[12.0pt]
\left(\begin{array}[]{ccc}1&0&\tilde{r}(\omega^{2}z)we^{it\theta_{13}(z)}\\\
0&1&0\\\ 0&0&1\end{array}\right)\left(\begin{array}[]{ccc}1&0&0\\\ 0&1&0\\\
-\overline{\tilde{r}(\omega^{2}z)}e^{-it\theta_{13}(z)}&0&1\end{array}\right),&z\in
I^{1},\\\ \mathcal{R}^{(3)}(z),&z\in\tilde{\Sigma}^{l},\\\
\mathcal{R}^{(3)}(z)^{-1},&z\in\tilde{\Sigma}^{l*},\end{array}\right.$ (5.21)

where

$\displaystyle
I^{l}=\omega^{l}(\frac{1}{p_{0}},p_{0})\cup(-p_{0},-\frac{1}{p_{0}}),\
l=0,1,2.$ (5.22)

  * 4.

$M^{(3)}(z)=I+\mathcal{O}(z^{-1}),\hskip 14.22636ptz\rightarrow\infty$.

  * 5.

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

$\displaystyle\bar{\partial}M^{(3)}(z)=M^{(3)}(z)\bar{\partial}\mathcal{R}^{(3)}(z),$
(5.23)

where

$\mathcal{R}^{(3)}(z)=\left\\{\begin{array}[]{lll}\left(\begin{array}[]{ccc}0&\bar{\partial}R_{j}^{*}(z)e^{it\theta_{12}}&0\\\
0&0&0\\\ 0&0&0\end{array}\right),&z\in\Omega_{j}^{0},\\\
\left(\begin{array}[]{ccc}0&0&0\\\
\bar{\partial}R_{j}(z)e^{-it\theta_{12}}&0&0\\\
0&0&0\end{array}\right),&z\in\Omega_{j}^{0*},\\\
\left(\begin{array}[]{ccc}0&0&\bar{\partial}R_{j}^{*}(\omega^{2}z)e^{it\theta_{13}}\\\
0&0&0\\\ 0&0&0\end{array}\right),&z\in\Omega_{j}^{1},\\\
\left(\begin{array}[]{ccc}0&0&0\\\ 0&0&0\\\
\bar{\partial}R_{j}(\omega^{2}z)e^{-it\theta_{13}}&0&0\end{array}\right),&z\in\Omega_{j}^{1*},\\\
\left(\begin{array}[]{ccc}0&0&0\\\ 0&0&0\\\ 0&\bar{\partial}R_{j}^{*}(\omega
z)e^{-it\theta_{23}}&0\end{array}\right),&z\in\Omega_{j}^{2},\\\
\left(\begin{array}[]{ccc}0&0&0\\\ 0&0&\bar{\partial}R_{j}(\omega
z)e^{it\theta_{23}}\\\ 0&0&0\end{array}\right),&z\in\Omega_{j}^{2*},\\\
I,&elsewhere,\\\ \end{array}\right.$ (5.24)

  * 6.

It has the same form of singularity conditions as RH problem 3.2.

  * 7.

$M^{(3)}(z)$ has simple poles at each point $\zeta_{n}$ and $\bar{\zeta}_{n}$
for $n\in\lozenge$ with

$\displaystyle\operatorname*{Res}_{z=\zeta_{n}}M^{(3)}(z)=\lim_{z\to\zeta_{n}}M^{(3)}(z)\left[T^{-1}(z)B_{n}T(z)\right].$
(5.25)

This $\bar{\partial}$-RH problem can be decomposed into a pure RH problem with
$\bar{\partial}\mathcal{R}^{(3)}=0$, which includes the contribution from the
discrete spectrum and jump contours, and a pure $\bar{\partial}$-problem with
$\bar{\partial}\mathcal{R}^{(3)}\neq 0$, which has the influence of
$\bar{\partial}$-components. The next sections are then devoted to the
asymptotic analysis of these three different contributions separately. First,
we consider the pure RH problem, which is obtained from the above
$\bar{\partial}$-RH problem by omitting the $\bar{\partial}$-derivative part,
as follows.

######  RH problem 5.2.

  * Find a matrix-valued function $M^{R}(z)$ with following properties:

  * 1.

$M^{R}(z)$ is holomorphic for $z\in\mathbb{C}\setminus\Sigma^{(3)}$, where
$\Sigma^{(3)}$ is defined in (5.11).

  * 2.

$M^{R}(z)$ has continuous boundary values $M^{R}_{\pm}(z)$ on $\Sigma^{(3)}$
and

$M^{R}_{+}(z)=M^{R}_{-}(z)V^{(3)}(z),\hskip 14.22636ptz\in\Sigma^{(3)},$

where $V^{(3)}(z)$ is defined by (4.31).

  * 3.

$M^{R}(z)=\Gamma_{1}\overline{M^{R}(\bar{z})}\Gamma_{1}=\Gamma_{2}\overline{M^{R}(\omega^{2}\bar{z})}\Gamma_{2}=\Gamma_{3}\overline{M^{R}(\omega\bar{z})}\Gamma_{3}=\overline{M^{R}(\bar{z}^{-1})}$.

  * 4.

$\bar{\partial}\mathcal{R}^{(3)}(z)=0$, for $z\in\mathbb{C}$.

  * 5.

$M^{R}(z)=I+\mathcal{O}(z^{-1}),\hskip 14.22636ptz\rightarrow\infty$.

  * 6.

As $z\rightarrow\varkappa_{l}=e^{\frac{i\pi(l-1)}{3}},l=1,...,6$, the limit of
$M^{R}(z)$ have the pole singularities

$\displaystyle M^{R}(z)=\frac{1}{z\mp
1}\left(\begin{array}[]{ccc}\alpha^{R}_{\pm}&\alpha^{R}_{\pm}&\beta^{R}_{\pm}\\\
-\alpha^{R}_{\pm}&-\alpha^{R}_{\pm}&-\beta^{R}_{\pm}\\\
0&0&0\end{array}\right)+\mathcal{O}(1),\ z\to\pm 1,$ $\displaystyle
M^{R}(z)=\frac{1}{z\mp\omega^{2}}\left(\begin{array}[]{ccc}0&0&0\\\
\beta^{R}_{\pm}&\alpha^{R}_{\pm}&\alpha^{R}_{\pm}\\\
-\beta^{R}_{\pm}&-\alpha^{R}_{\pm}&-\alpha^{R}_{\pm}\end{array}\right)+\mathcal{O}(1),\
z\to\pm\omega^{2},$ $\displaystyle
M^{R}(z)=\frac{1}{z\mp\omega}\left(\begin{array}[]{ccc}-\alpha^{R}_{\pm}&-\beta^{R}_{\pm}&-\alpha^{R}_{\pm}\\\
0&0&0\\\
\alpha^{R}_{\pm}&\beta^{R}_{\pm}&\alpha^{R}_{\pm}\end{array}\right)+\mathcal{O}(1),\
z\to\pm\omega,$

with $\alpha^{R}_{\pm}=\alpha^{R}_{\pm}(y,t)=-\bar{\alpha}^{R}_{\pm}$,
$\beta^{R}_{\pm}=\beta^{R}_{\pm}(y,t)=-\bar{\beta}^{R}_{\pm}$ and
$M^{R}(z)^{-1}$ has same specific matrix structure with $\alpha^{R}_{\pm}$,
$\beta^{R}_{\pm}$ replaced by $\tilde{\alpha}^{R}_{\pm}$,
$\tilde{\beta}^{R}_{\pm}$.

  * 7.

$M^{R}(z)$ has the simple poles at each point $\zeta_{n}$ and
$\bar{\zeta}_{n}$ for $n\in\lozenge$ with

$\displaystyle\operatorname*{Res}_{z=\zeta_{n}}M^{R}(z)=\lim_{z\to\zeta_{n}}M^{R}(z)\left[T^{-1}(z)B_{n}T(z)\right].$

Observing the above problem, we find that both the discrete spectrum and the
jump discontinuities make distinct contributions to this RH problem. Moreover,
by (5.21), $V^{(3)}(z)\to I$ as $t\to\infty$ uniformly, expect for the
neighborhoods of the saddle points $\pm p_{0},\pm 1/p_{0}$. For $l=0,1,2$, let

$\displaystyle U_{a}^{l}=\\{z\in\mathbb{C}:|z-\omega^{l}|\leq c_{0}\\},\quad
U_{b}^{l}=\\{z\in\mathbb{C}:|z+\omega^{l}|\leq c_{0}\\},$

be two small disks around $z=1$ and $z=-1$, respectively, where

$\displaystyle c_{0}=\min\\{1/2,2(p_{0}-1)t^{\delta_{2}}\\},$

which is a constant depending on $t$ with $\delta_{2}\in(1/9,1/6)$. For $t$