# Long-time asymptotic analysis for defocusing Ablowitz-Ladik system with
initial value in lower regularity

Meisen Chen Corresponding author<EMAIL_ADDRESS>Institute for Advanced
Study, Shenzhen University, Shenzhen, 518060, China College of Physics and
Optoelectronic Engineering, Shenzhen University, Shenzhen, 518060, China
Engui Fan  School of Mathematical Sciences, Fudan University, Shanghai,
200433, China  Jingsong He  Institute for Advanced Study, Shenzhen University,
Shenzhen, 518060, China

###### Abstract

Recently, we have given the $l^{2}$ bijectivity for defocusing Ablowitz-Ladik
systems in the discrete Sobolev space $l^{2,1}$ by inverse spectral method.
Based on these results, the goal of this article is to investigate the long-
time asymptotic property for the initial-valued problem of the defocusing
Ablowitz-Ladik system with initial potential in lower regularity. The main
idea is to perform proper deformations and analysis to the corespondent
Riemann-Hilbert problem with the unit circle as the jump contour $\Sigma$. As
a result, we show that when $|\frac{n}{2t}|\leq 1<1$, the solution admits
Zakharov-Manakov type formula, and when $|\frac{n}{2t}|\geq 1>1$, the solution
decays fast to zero.

Key Words: defocusing Ablowitz-Ladik system, inverse spectral method, long-
time asymptotic property, Riemann-Hilbert problem
2010 Mathematics Subject Classification Numbers: 37K15, 35Q15, 35Q55

###### Contents

  1. 1 Introduction
    1. 1.1 Main results
  2. 2 Riemann-Hilbert problem and the solution for defoxusing Ablowitz-Ladik systems
  3. 3 Long-time asymptotic analysis on $-1<-V_{0}\leq\xi\leq V_{0}<1$
    1. 3.1 Deformation on the jump contour
    2. 3.2 Split the circle
    3. 3.3 Deformation of the $\bar{\partial}$-RH problem
    4. 3.4 RH problems at stationary phase points
    5. 3.5 Scaling and rotation
    6. 3.6 Analysis on $\bar{\partial}$-problem
    7. 3.7 the long-time asymptotics
    8. 3.8 The invertibility of operators
  4. 4 Long-time asymptotic analysis on $\xi\geq V_{0}>1$
  5. 5 Long-time asymptotic analysis on $\xi\leq-V_{0}<-1$

##  1 Introduction

It’s well known that inverse scattering transform is an effective method when
solving the integrable system. In 1967, the inverse scattering transform is
firstly introduced when solving the KdV equation by Gardne, Greene, Kruskal
and Miura [1]. This method is also applied to ZS-AKNS systems [2]. More
literatures for solving the continuous and discrete integrable systems refers
to [3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19], Except for
solving initial-valued problem, this method also have a great number of
important results in mathematics and physics. Particularly, Deift and Zhou
apply this method to the nonlinear Schrödinger equations [20, 21] and modified
KdV equations [22] to obtain solutions, and further develop a nonlinear
steepest descent method to study the long-time asymptotic analysis with
potentials in Schwartz space. This method also has been applied to numerous
integrable systems for the long-time asymptotic analysis [23, 24, 25, 26, 27,
28, 29, 30, 31, 32, 33]. Recent years, people become interested in extending
the long-time asymptotic analysis for the integrable system with lower
regularity. In particular, for nonlinear Schrödinger equation as one of the
most important integrable systems, based on the $L^{2}$-Sobolev bijectivity
for the inverse scattering transform [34], with a dbar steepest descent
method, people investigate the long-time asymptotic property when the initial
potential belongs to a weighted Sobolev space:

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

In this paper, we focus on defocusing Ablowitz-Ladik systems

$\displaystyle
i\partial_{t}q_{n}(t)=q_{n+1}(t)-2q_{n}(t)+q_{n-1}(t)-|q_{n}(t)|^{2}(q_{n+1}(t)+q_{n-1}(t)),$
(1)

that is integrable systems introduced by Ablowitz and Ladik [35, 36] in
1975-1976, where $n\in\mathbb{Z}$ is the discrete spatial variable and
$t\in\mathbb{R}^{+}$ is the continuous variable. It’s shown as the spatial
integrable discretization of the defocusing nonlinear Schrödinger equation:

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

and there are many important researches for it in aspects of mathematics and
physics. In 2005, Nenciu [37] constructs the Lax pair for defocusing Ablowitz-
Ladik systems by the connection between defocusing Ablowitz-Ladik systems and
the orthogonal polynomials on the unit circle. In 2006, Chow et al [38] show
the analytic doubly periodic waves pattern for Ablowitz-Ladik systems. Using
methods of algebraic geometry, Miller et al [39] obtain the finite genus
solutions for Ablowtiz-Ladik systems. Recently, Yamane [40] studies the long-
time asymptotic behavior by Deift-Zhou method with initial data bounded by
$\sup_{n\in\mathbb{Z}}|q_{n}|<1$ and in the discrete Schwartz space:

$\displaystyle\\{q_{n}\\}_{n=-\infty}^{\infty}\in\left\\{\\{a_{n}\\}_{n=-\infty}^{\infty}:\sum_{n=-\infty}^{\infty}(1+n^{2})^{k}a_{n}^{2}<\infty,\
k\in\mathbb{N}^{+}\right\\}.$

Here, we present the long-time asymptotic analysis for initial-valued problem
of (1) with initial potential in lower regularity, and it satisfies

$\displaystyle\\{q_{n}\\}_{-\infty}^{\infty}\in
l^{2,1}=\left\\{\\{a_{n}\\}_{-\infty}^{\infty}:\sum_{n=-\infty}^{\infty}(1+n^{2})a_{n}^{2}<\infty\right\\},\quad\sup_{n\in\mathbb{Z}}|q_{n}|<1,$
(2)

where $l^{2,1}$ is a discrete weighted Sobolev space.

For the initial potential satisfies (2), recently, we prove in [41] that the
direct scattering mapping maps these potentials to reflection coefficients
$r(\lambda)$ which belongs to a Sobolev space $H^{1}_{\theta}(\Sigma)$ and is
bounded by $\parallel{r}\parallel_{L^{\infty}(\Sigma)}<1$, where $\Sigma$ is
the jump contour shown in Figure 1 and $\theta\in[0,2\pi]$ is the parameter of
it. Reversely, by the inverse scattering mapping, if reflection coefficients
belong to $H^{1}_{\theta}(\Sigma)$ and is bounded by
$\parallel{r}\parallel_{L^{\infty}(\Sigma)}<1$, potentials also belongs to the
discrete weighted Sobolev space $l^{2,1}$. In fact, by the argument in [41],
these two mappings is of Lipschitz continuity. Moreover, for Ablowitz-Ladik
systems, if we denote $r(\lambda,t)$ the reflection coefficient for
$q_{n}(t)$, the time flow

$\displaystyle r(\lambda)\equiv r(\lambda,0)\mapsto
r(\lambda,t)=r(\lambda)e^{2i(\cos\theta-1)t},\quad\lambda=e^{i\theta}\in\Sigma,$

persists the reflection coefficient in $H^{1}_{\theta}(\Sigma)$ and bounded by
$\parallel r(\cdot,t)\parallel_{L^{\infty}(\Sigma)}<1$. In (2), the weighted
Sobolev space $l^{2,1}$ is the minimal condition for the solution solved by
the inverse scattering transform.

Figure 1: $\Sigma$: jump contour for RH problem 2.1

$D_{+}/D_{-}$: the region outside/inside of $\Sigma$.

Since $r\in H^{1}_{\theta}(\Sigma)$, we obtain Riemann-Hilbert (RH) problem
2.1 and it couldn’t applied proper rational approximation to $r(\lambda)$;
therefore, it’s hard to extend the jump contour and make the RH transform like
that in [40]. For $r\in H^{1}_{\theta}(\Sigma)$, by Sobolev embedding theorem,
it’s also $\frac{1}{2}$-Hölder continuous on $\Sigma$; as a result, we deform
RH problem 2.1 properly into a $\bar{\partial}$-RH problem; moreover, solution
for this $\bar{\partial}$-RH problem can be factorized into a product of
solutions for an RH problem and a $\bar{\partial}$ problem; and, we further
analyze this two problems separately and finally obtain the long-time
asymptotic property. This idea is generalized from the dbar steepest descent
method that people applied to orthogonal polynomials and nonlinear Schrödinger
equations [42, 43]. This method make it eligible to analyze the long-time
asymptotic property under the condition $r\in H^{1}_{\theta}(\Sigma)$. In our
method, the jump contour is a unit circle and there are two separate first-
order stationary phase points, which make the deformation of RH problem more
tricky.

In this paper, for Hilbert spaces $P_{1}$ and $P_{2}$, we denote
$\mathcal{B}(P_{1},P_{2})$ as the normed linear space consisting of all linear
bounded operators from $P_{1}$ to $P_{2}$, and simply write
$\mathcal{B}(P_{1},P_{1})$ as $\mathcal{B}(P_{1})$.

The article is organized as follows. In Section 2, according to Lax pair, we
give the direct scattering including Jost solutions, modified Jost solution
and the reflection coefficient, and then perform inverse scattering transform
to obtain the reconstructed formula by constructing the correspondent RH
problem. In Section 3, we analyze the long-time asymptotic behavior on the
region $-1<-V_{0}\leq\xi\overset{\mathrm{def}}{=}\frac{n}{2t}\leq V_{9}<1$. In
Section 4 and 5, we analyze the long-time aaymptotics for $|\xi|\geq V_{0}>1$.

###  1.1 Main results

In this paper, we study the long-time asymptotic analysis on three regions as
shown in Figure 2.

Figure 2: Regions on the half plane divided by rays $\frac{n}{2t}=\pm 1$.

When $-1<-V_{0}\leq\xi\leq V_{0}<1$, we see that first-order stationary phase
points $S_{j}=i\xi+\sqrt{1-\xi^{2}}$ are on the jump contour $\Sigma$. Since
these stationary phase points appear on $\Sigma$, the reflection coefficient
at $S_{j}$ has impact to the solution; so, we come to investigate the local RH
problem at $S_{j}$ on Section 3.4, and make scaling transform for them to
obtain the model RH problem that is related to parabolic cylinder functions,
which refers to [44]. By the model RH problem, we obtain the oscillatory
leading term that is $\mathcal{O}(t^{-\frac{1}{2}})$ as shown in (3), and we
obtain that as $t\to+\infty$, the solution in Zakharov-Manakov type formulas
(3). It’s notable that under the condition
$\\{q_{n}\\}_{n=-\infty}^{\infty}\in L^{2,1}$, the decaying rate is
$\mathcal{O}(t^{-1})$.

When $|\xi|\geq V_{0}>1$, stationary phase points are pure imaginary number
and off the jump contour $\Sigma$; in this case, the leading term in (3)
vanishes, and $q_{n}(t)$ decays fast when $t\to+\infty$, and we call these
regions as fast decaying region.

######  Theorem 1.1.

For the initial-valued problem (1)-(2), the solution $q_{n}(t)$ admits the
following long-time asymptotic formula in the sectors obtained by dividing the
half plane by rays $\frac{n}{2t}=\pm 1$, which is shown in Figure 2.

  1. a)

The Zakharov-Manakov region $-1<-V_{0}\leq\xi\leq V_{0}<1$ with some positive
constant $V_{0}$: as $t\to+\infty$, the solution admits Zakharov-Manakov type
formulas:

$\displaystyle
q_{n}(t)=\frac{i}{2}t^{-\frac{1}{2}}(1-\xi^{2})^{-\frac{1}{4}}\delta^{-1}(0)\sum_{j=1}^{2}\delta_{j0}^{2}[M^{L,j}_{1}]_{1,2}+\mathcal{O}(t^{-\frac{3}{4}}),$
(3)
$\displaystyle\quad[M_{1}^{L,j}]_{1,2}=-i\frac{(2\pi)^{\frac{1}{2}}e^{\frac{\pi
i}{4}}e^{\frac{-\pi\nu_{j}}{2}}}{r(S_{j})\Gamma(-i\nu_{j})},\quad j=1,2,$

where $r(S_{j})$ is the value of reflection coefficients at stationary phase
points $S_{j}$, $j=1,2$, $\int^{S_{1}}_{S_{2}}$ denotes the integral on
$\Sigma$ from $S_{2}$ to $S_{1}$, $\Gamma(\lambda)$ is the Euler’s gamma
function,

$\displaystyle
S_{j}=-i\xi+(-1)^{j}\sqrt{1-\xi^{2}},\quad\beta_{j}=\frac{it^{-\frac{1}{2}}}{2}(1-\xi^{2})^{-\frac{1}{4}}S_{j},\quad\nu_{j}=-\frac{1}{2\pi}\ln(1-|r(S_{j})|^{2})$
$\displaystyle\delta^{-1}(0)=e^{\frac{1}{2\pi
i}\int_{S_{2}}^{S_{1}}s^{-1}\ln(1-|r(s)|^{2})\mathrm{d}s},\quad\phi(S_{j})=2((-1)^{j}\sqrt{1-\xi^{2}}-\xi\arg
S_{j}-1),$ $\displaystyle\alpha_{j}(S_{j})=\frac{1}{2\pi
i}\int_{S_{2}}^{S_{1}}\frac{\ln(1-|r(s)|^{2})-\ln(1-|r(S_{j})|^{2})}{s-S_{j}}\mathrm{d}s,$
$\displaystyle\delta_{j0}=e^{\alpha_{j}(S_{j})-\frac{it}{2}\phi(S_{j})}\left(\frac{(-1)^{j-1}\beta_{j}}{S_{1}-S_{2}}\right)^{(-1)^{j-1}i\nu_{j}}.$

  2. b)

The fast decaying region $|\xi|\geq V_{0}>1$: as $t\to+\infty$, the solution
decays to zero

$\displaystyle q_{n}(t)\sim\mathcal{O}(t^{-1}).$

##  2 Riemann-Hilbert problem and the solution for defoxusing Ablowitz-Ladik
systems

In this section, we investigate Jost solutions: $X^{\pm}$ for the Lax pair
corespondent to the initial-valued problem, make a transform for Jost
solutions to obtain a modified Jost solution: $X^{\pm}\rightsquigarrow
Y^{\pm}$, and then construct the RH problem. The Lax pair for (1) is:

$\displaystyle X(z,n+1,t)=\left[\begin{matrix}z&q_{n}(t)\\\
\overline{q_{n}(t)}&z^{-1}\end{matrix}\right]X(z,n,t),$ (4a)
$\displaystyle\partial_{t}X(z,n,t)=i\left[\begin{matrix}-\frac{1}{2}(z-z^{-1})^{2}+q_{n}(t)\overline{q_{n-1}(t)}&q_{n-1}(t)z^{-1}-q_{n}(t)z\\\
\overline{q_{n}(t)}z^{-1}-\overline{q_{n-1}(t)}z&\frac{1}{2}(z-z^{-1})^{2}-q_{n-1}(t)\overline{q_{n}(t)}\end{matrix}\right]$
$\displaystyle\quad\times X(z,n,t).$ (4b)

where $z$ is the spectral parameter and the Lax pair admits the Jost solution,

$\displaystyle X^{\pm}\equiv X^{\pm}(z,n,t)\sim
z^{n\sigma_{3}}e^{-\frac{it}{2}(z-z^{-1})^{2}\sigma_{3}},\quad n\to\pm\infty.$
(5)

Seeing from the spatial problem (4a), we derive that

$\displaystyle\det X^{-}=\prod_{k=-\infty}^{n-1}(1-|q_{k}(t)|^{2}),\quad\det
X^{+}=\prod_{k=n}^{\infty}(1-|q_{k}(t)|^{2})^{-1}.$ (6)

Introducing

$\displaystyle c_{n}(t)=\prod_{k=n}^{\infty}(1-|q_{k}(t)|^{2}),\quad
n\in\mathbb{Z}\cup\\{-\infty\\},$

because of assumption (2), from the result of [41], we learn that
$c_{-\infty}=\lim_{n\to-\infty}c_{n}(t)$ is nonzero and independent on $t$;
therefore, for any fixed $t\leq 0$,

$\displaystyle 1-|q_{n}(t)|^{2}\neq 0,\quad n\in\mathbb{Z},$

which combined with (4a) deduces that $X^{\pm}$ are invertible and fundamental
solutions for the Lax pair; moreover, by the uniqueness of solution, there is
a unique $2\times 2$ matrix-valued function depending only on $z$ such that

$\displaystyle X^{-}(z,n,t)=X^{+}(z,n,t)S(z),\quad
S(z)=\left[\begin{matrix}a(z)&\breve{b}(z)\\\
b(z)&\breve{a}(z)\end{matrix}\right].$ (7)

$S(z)$ is also known as the scattering matrix. Since
$\left[\begin{matrix}0&q_{n}(t)\\\ \overline{q_{n}(t)}&0\end{matrix}\right]$
is Hermitian, we learn from the spatial problem (4a) that $X^{\pm}$ admit the
following symmetric property

$\displaystyle
X^{\pm}(z,n,t)=\sigma_{1}\overline{X^{\pm}(\bar{z}^{-1},n,t)}\sigma_{1},\quad\sigma_{1}=\left[\begin{matrix}0&1\\\
1&0\end{matrix}\right],$

which combined with (7) deduces symmetry properties of $S(z)$:

$\displaystyle\breve{a}(z)=\overline{a(\bar{z}^{-1})},\quad\breve{b}(z)=\overline{b(\bar{z}^{-1})},\quad
z\in\mathbb{C}.$ (8)

Making the transformation

$\displaystyle Y^{\pm}(z,n,t)=\left[\begin{matrix}1&0\\\
0&z\end{matrix}\right]X^{\pm}(z,n,t)z^{-n\sigma_{3}}e^{\frac{it}{2}(z-z^{-1})^{2}\sigma_{3}}\left[\begin{matrix}1&0\\\
0&z^{-1}\end{matrix}\right],$ (9)

from property (5), we learn that $Y^{\pm}(z,n,t)$ satisfy the normalization
property

$\displaystyle Y^{\pm}(z,n,t)\sim I,\quad n\to\pm\infty.$ (10)

Making the spectral parameter transform: $\lambda=z^{2}$, seeing (2) and (8),
we deduce that $Y^{\pm}(z,n,t)=Y^{\pm}(-z,n,t)$; thus, we simply write
$Y^{\pm}(\lambda,n,t)=Y^{\pm}(z,n,t)$. Seeing from [41], we learn that
$Y^{+}_{1}$ and $Y^{-}_{2}$ are both holomorphic on $D^{-}=\\{|\lambda|<1\\}$
and continuously extended to $D^{-}\cup\Sigma$, $Y^{-}_{1}$ and $Y^{+}_{2}$
are both holomorphic on $D^{+}=\\{|\lambda|>1\\}$ and continously extended to
$D^{+}\cup\Sigma$. By (7) and (9), we derive that

$\displaystyle a(z)=c_{n}(t)\det[Y^{-}_{1},Y^{+}_{2}](\lambda,n,t),\quad
zb(z)=c_{n}(t)e^{-it\phi(\lambda,n,t)}\det[Y^{+}_{1},Y^{-}_{1}](\lambda,n,t),$
(11)

where

$\displaystyle\phi(\lambda,n,t)=\lambda+\lambda^{-1}+2i\xi\log\lambda-2;$

since then, we write $a(\lambda=z^{2})$ instead of $a(z)$. Because
$[Y^{+}_{1},Y^{-}_{2}]$ is holomorphic on $D^{-}$ and continuously extended to
$D^{-}\cup\Sigma$, by (11), $a(\lambda)$ is holomorphic on $D^{-}$ and
continuously extended to $D^{+}\cup\Sigma$, too; moreover, because of the
continuity of $[Y^{+}_{1},Y^{-}_{1}]$ on $\Sigma$, $zb(z)$ is also continuous
on $\Sigma$; therefore, we define the reflection coefficients on the circle
$\lambda=e^{i\theta}\in\Sigma$:

$\displaystyle r(\lambda)=\frac{zb(z)}{a(z)},$

which is well-defined and belongs to $H^{1}_{\theta}(\Sigma)$ according to
[41]. Sometimes, we denote $r(\theta)=r(e^{i\theta})$ without confusion of
notation on the jump contour $\Sigma$. Since $r\in H^{1}_{\theta}(\Sigma)$, by
Sobolev embedding theory, we learn that $r(\theta)$ is $\frac{1}{2}$-Hölder
continuous and boundend on $\Sigma$. By (6), (7) and (8), we see that

$\displaystyle
1-|r(\theta)|^{2}=\frac{c_{-\infty}}{|a(e^{i\theta})|}>0,\quad\theta\in[0,2\pi].$
(12)

Seeing from [41] and setting $Y^{\pm}=[Y^{\pm}_{1},Y^{\pm}_{2}]$, we learn
that as $\lambda\to\infty$ and $\lambda\to 0$, $Y^{\pm}(z,n,t)$ admit the
following asymptotic property:

$\displaystyle[Y^{+}_{1},Y^{-}_{2}](\lambda,n,t)\sim\left[\begin{matrix}c_{n}^{-1}(t)+O(\lambda)&q_{n-1}(t)+O(\lambda)\\\
-c_{n}^{-1}(t)\overline{q_{n}(t)}\lambda+O(\lambda^{2})&1+O(\lambda)\end{matrix}\right],\quad\lambda\to
0,$ (13a)
$\displaystyle[Y^{-}_{1},Y^{+}_{2}](\lambda,n,t)\sim\left[\begin{matrix}1+O(\lambda^{-1})&-c_{n}^{-1}(t)q_{n}(t)\lambda^{-1}+O(\lambda^{-2})\\\
\overline{q_{n-1}(t)}+O(\lambda^{-1})&c_{n}^{-1}(t)+O(\lambda^{-1})\end{matrix}\right],\quad\lambda\to\infty.$
(13b)

From (11) and (13b), we learn that

$\displaystyle\lim_{\lambda\to\infty}a(\lambda)=1.$ (14)

We come to the construction of corespondent RH problems. Set a $2\times 2$
matrix-valued function

$\displaystyle M\equiv M(\lambda,n,t)=\begin{cases}\left[\begin{matrix}1&0\\\
-\overline{q_{n-1}(t)}c_{n}(t)&c_{n}(t)\end{matrix}\right]\left[Y_{1}^{+},\frac{Y_{2}^{-}}{\breve{a}}\right](\lambda,n,t)&|\lambda|<1,\\\
\left[\begin{matrix}1&0\\\
-\overline{q_{n-1}(t)}c_{n}(t)&c_{n}(t)\end{matrix}\right]\left[\frac{Y_{1}^{-}}{a},Y_{2}^{+}\right](\lambda,n,t)&|\lambda|>1;\end{cases}$
(15)

then, we clain that $M$ admits the following RH problem. The first item comes
from that of $Y^{\pm}(\lambda,n,t)$ and $a(\lambda)$. The second item
naturally follows after the definition of $M(\lambda,n,t)$. (9) and (15)
deduce the third item. Here, the orientation of jump contour is clockwise, and
as general notation, we call the left/right side of the jump contour as $+/-$
side. As follows, we obtain the reconstructed formula from (8), (13a), (14)
and (15) that the reconstructed formula is

$\displaystyle q_{n}(t)=M_{1,2}(0,n+1,t).$ (16)

######  RH problem 2.1.

Find a $2\times 2$ matrix-valued function $M$ such that

  * •

$M$ is analytic on $\Sigma$.

  * •

As $\lambda\to\infty$, $M\sim I+O(\lambda^{-1})$.

  * •

On $\lambda\in\Sigma$,

$\displaystyle M_{+}=M_{-}V,\quad V\equiv
V(\lambda,n,t)=\left[\begin{matrix}1-|r(\lambda)|^{2}&-\overline{r(\lambda)}e^{-it\phi(\lambda,n,t)}\\\
r(\lambda)e^{it\phi(\lambda,n,t)}&1\end{matrix}\right].$

##  3 Long-time asymptotic analysis on $-1<-V_{0}\leq\xi\leq V_{0}<1$

In this section, we study the case for the region $-1<-V_{0}\leq\xi\leq
V_{0}<1$. In this case, we derive that there are two first-order stationary
phase points $S_{1}$ and $S_{2}$ on the unit circle $\Sigma$: therefore, we
obtain that

$\displaystyle\phi(\lambda,n,t)-\phi(S_{j},n,t)\sim\frac{\phi^{\prime\prime}(S_{j},n,t)}{2}(\lambda-
S_{j})^{2}+\mathcal{O}(\lambda-S_{j})^{3},$ (17)
$\displaystyle\quad\phi^{\prime\prime}(S_{j},n,t)=(-1)^{j}S_{j}^{-2}\sqrt{1-\xi^{2}}.$

Figure 3: The sign of $\mathrm{Im}\phi$ in region I; $S_{1}$, $S_{2}$ the
stationary phase point

###  3.1 Deformation on the jump contour

In this part, we study an RH problem transform: $M\rightsquigarrow
M^{(1)}=M\delta^{-\sigma_{3}}$; and the new jump matrix admits a proper
factorization, seeing in RH problem 3.3.

Introducing the scalar function

$\displaystyle\delta\equiv\delta(\lambda)=e^{\frac{1}{2\pi
i}\int_{S_{2}}^{S_{1}}\frac{\ln(1-|r(s)|^{2})}{s-\lambda}\mathrm{d}s},$ (18)

where the integral is along the lower-half arc on $\Sigma$:
$\wideparen{S_{2}S_{1}}$ and this arc’s orientation is from $S_{2}$ to $S_{1}$
along $\Sigma$, we derive that $\delta$ admits properties shown in Proposition
3.1

######  Proposition 3.1.

Since $r\in H^{1}_{\theta}(\Sigma)$, the scalar function $\delta$ satisfies:

  1. (a)

$\delta$ is analytic on $\mathbb{C}\setminus\wideparen{S_{2}S_{1}}$.

  2. (b)

As $\lambda\to\infty$, $\delta(\lambda)\sim 1+O(\lambda^{-1})$.

  3. (c)

On the arc $\lambda\in\wideparen{S_{2}S_{1}}$.
$\delta_{+}(\lambda)=\delta_{-}(\lambda)(1-|r(\lambda|^{2}))$.

  4. (d)

On $\lambda\in\mathbb{C}\setminus\wideparen{S_{2}S_{1}}$, $\delta$ admits the
symmetry:

$\displaystyle\delta(\lambda)=\overline{\delta(0)/\delta(\bar{\lambda}^{-1})}.$

  5. (e)

On the neighborhood of $S_{1}$ and $S_{2}$, $\delta$ admits following
asymptotic properties:

$\displaystyle\delta(\lambda)\sim\left(\frac{\lambda-S_{1}}{\lambda-
S_{2}}\right)^{i\nu_{j}}e^{\alpha_{j}(S_{j})}+O(|\lambda-
S_{j}|^{\frac{1}{2}}),\quad\lambda\to S_{j},$
$\displaystyle\quad\alpha_{j}(\lambda)=\frac{1}{2\pi
i}\int_{S_{2}}^{S_{1}}\frac{\ln(1-|r(s)|^{2})-\ln(1-|r(S_{j})|^{2})}{s-\lambda}\mathrm{d}s,$
$\displaystyle\quad\nu_{j}=-\frac{\ln(1-|r(S_{j})|^{2})}{2\pi},\quad\quad
j=1,2.$

###### Proof.

(a) comes from the analyticity of $Y^{\pm}(\lambda,n,t)$, $a(\lambda)$ and
(15). (b) comes from (13b) and (15). (c) comes from (7) and (9). Since
$\delta(\lambda)$ satisfies (a-c) which determine an RH problem with the jump
contour $\wideparen{S_{2}S_{1}}$, we also verify by computation that
$\overline{\delta(0)/\delta(\bar{\lambda}^{-1})}$ is also the solution of this
RH problem; then, by uniqueness of solution, we derive (d). Changing the
variable on $\in\Sigma$: $\lambda\to\lambda^{\prime}$, $s\to s^{\prime}$,

$\displaystyle\lambda=\sqrt{S_{1}S_{2}}\frac{\sqrt{S_{1}}+\sqrt{S_{2}}\lambda^{\prime}}{\sqrt{S_{2}}+\sqrt{S_{1}}\lambda^{\prime}},\quad
s=\sqrt{S_{1}S_{2}}\frac{\sqrt{S_{1}}+\sqrt{S_{2}}s^{\prime}}{\sqrt{S_{2}}+\sqrt{S_{1}}s^{\prime}},$

we obtain that

$\displaystyle\int_{S_{2}}^{S_{1}}\frac{f(s)-f(S_{1})}{(2\pi
i)(s-\lambda)}\mathrm{d}s=\frac{\sqrt{S_{2}}+\sqrt{S_{1}}\lambda^{\prime}}{2\pi
i}\int_{-\infty}^{0}\frac{f(s)-f(S_{1})\mathrm{d}s^{\prime}}{(\sqrt{S_{2}}+\sqrt{S_{1}}s^{\prime})(s^{\prime}-\lambda^{\prime})},$
(19)

where $f(s)=\ln(1-|r(s)|^{2})$. Since $r(s)\in H^{1}_{\theta}(\Sigma)\subset
L^{\infty}(\Sigma)$, it naturally follows that

$\displaystyle
F(s^{\prime})=\begin{cases}\frac{f(s)-f(S_{1})}{\sqrt{S_{2}}+\sqrt{S_{1}}s^{\prime}},&s^{\prime}\leq
0,\\\ 0,&s^{\prime}>0,\end{cases}$ (20)

belongs to $H^{1}(\mathbb{R})$ and

$\displaystyle F(s=0)=0;$

therefore, we apply Lemma 23.3 in [45] and get that for any
$\lambda\in\mathbb{C}\setminus\mathbb{R}$

$\displaystyle\Big{|}\int_{-\infty}^{+\infty}\frac{F(s^{\prime})}{(2\pi
i)(s^{\prime}-\lambda^{\prime})}\mathrm{d}s^{\prime}-\int_{-\infty}^{+\infty}\frac{F(s^{\prime})}{2\pi
is^{\prime}}\mathrm{d}s^{\prime}\Big{|}\lesssim\parallel
F\parallel_{H^{1}}|\lambda^{\prime}|^{\frac{1}{2}}$ (21)

Considering (19), (20) and (21), we obtain (e) for $j=1$, and the proof for
$j=2$ is parallel. We confirm the result. ∎

######  Remark 3.2.

Seeing property (c) and (d) in Proposition 3.1, we learn that for
$\lambda\in\wideparen{S_{2}S_{1}}$,

$\displaystyle|\delta_{+}(\lambda)|^{2}=(1-|r(s)|^{2})\delta(0),\quad|\delta_{-}(\lambda)|^{2}=\delta(0)(1-|r(\lambda)|^{2})^{-1}.$

which are nonzero and finite by (12) and (19); moreover, $\delta(\lambda)\to
1$ as $\lambda\to\infty$; so, $\delta$ and $\delta^{-1}$ are both bounded
function by (a) in Proposition 3.1 and the maximal modular theorem.

Here, we confirm that a new $2\times 2$ matrix-valued function

$\displaystyle M^{(1)}=M\delta^{-\sigma_{3}},$ (22)

satisfies RH problem 3.3, which is the natural consequence of RH problem 2.1,
(a-c) in Proposition 3.1 and (22).

######  RH problem 3.3.

Find a $2\times 2$ matrix-valued function such that

  * •

$M^{(1)}$ is analytic on $\mathbb{C}\setminus\Sigma$.

  * •

As $\lambda\to\infty$, $M^{(1)}(\lambda,n,t)\sim I+O(\lambda^{-1})$.

  * •

On $\lambda\in\Sigma$,

$\displaystyle M^{(1)}_{+}=M^{(1)}_{-}V^{(1)},\quad$

where $V^{(1)}\equiv V^{(1)}(\lambda,n,t)$ is the jump matrix

$\displaystyle
V^{(1)}=\begin{cases}\left[\begin{matrix}1&-\delta_{+}^{2}\frac{\overline{r(\lambda)}}{1-|r(\lambda)|^{2}}e^{-it\phi(\lambda,n,t)}\\\
\delta_{-}^{-2}\frac{r(\lambda)}{1-|r(\lambda)|^{2}}e^{it\phi(\lambda,n,t)}&1-|r(\lambda)|^{2}\end{matrix}\right]\\\
\left[\begin{matrix}1-|r(\lambda)|^{2}&-\overline{r(\lambda)}\delta^{2}(\lambda)e^{-it\phi(\lambda,n,t)}\\\
r(\lambda)\delta^{-2}(\lambda)e^{it\phi(\lambda,n,t)}&1\end{matrix}\right]\end{cases}$
$\displaystyle\quad=\begin{cases}\left[\begin{matrix}1&0\\\
\frac{\delta_{-}^{-2}r(\lambda)}{1-|r(\lambda)|^{2}}e^{it\phi(\lambda,n,t)}&1\end{matrix}\right]\left[\begin{matrix}1&-\frac{\delta_{+}^{2}\overline{r(\lambda)}}{1-|r(\lambda)|^{2}}e^{-it\phi(\lambda,n,t)}\\\
0&1\end{matrix}\right]&\lambda\in\wideparen{S_{2}S_{1}},\\\
\left[\begin{matrix}1&-\overline{r(\lambda)}\delta^{2}(\lambda)e^{-it\phi(\lambda,n,t)}\\\
0&1\end{matrix}\right]\left[\begin{matrix}1&0\\\
r(\lambda)\delta^{-2}(\lambda)e^{it\phi(\lambda,n,t)}&1\end{matrix}\right]&\lambda\in\Sigma\setminus\wideparen{S_{2}S_{1}}.\end{cases}$

###  3.2 Split the circle

In this part, we introduce a new RH problem transform:
$M^{(1)}\rightsquigarrow M^{(2)}$. As a result, the jump contour $\Sigma$ is
deformed into $\Sigma^{(2)}$, where the jump matrix is decaying.

Introducing the jump contour $\Sigma^{(2)}=L\cup\tilde{L}$ as shown in Figure
4, we see that $L$ and $\tilde{L}$ are both closed curves consisting of
straight line segments and arcs centering at the origin. As for orientation of
the jump contour, we denote it in Figure 4. Again seeing Figure 4, the complex
plane is divided into six parts by $\Sigma^{(2)}$ and $\Sigma$: $\Omega_{j}$,
$j=1,\dots,6$.

Figure 4: Jump contour $\Sigma^{(2)}=L\cup\tilde{L}$.

Since we have constructed the jump contour, introduce a $2\times 2$ matrix-
valued function

$\displaystyle M^{(2)}=M^{(1)}\mathcal{P},$ (23)

where the $2\times 2$ matrix-valued function $\mathcal{P}$ is invertible and
written as

$\displaystyle\mathcal{P}\equiv\mathcal{P}(\lambda,n,t)=\begin{cases}\left[\begin{matrix}1&0\\\
-R_{1}(\lambda)\delta^{-2}(\lambda)e^{it\phi(\lambda,n,t)}&1\end{matrix}\right]&\lambda\in\Omega_{1},\\\
\left[\begin{matrix}1&R_{3}(\lambda)\delta^{2}(\lambda)e^{-it\phi(\lambda,n,t)}\\\
0&1\end{matrix}\right]&\lambda\in\Omega_{3},\\\ \left[\begin{matrix}1&0\\\
R_{4}(\lambda)\delta^{-2}(\lambda)e^{it\phi(\lambda,n,t)}&1\end{matrix}\right]&\lambda\in\Omega_{4},\\\
\left[\begin{matrix}1&-R_{6}(\lambda)\delta^{2}(\lambda)e^{-it\phi(\lambda,n,t)}\\\
0&1\end{matrix}\right]&\lambda\in\Omega_{6},\\\
I&\lambda\in\Omega_{2}\cup\Omega_{5}\end{cases}$ (24)

and we denote $\lambda=\rho e^{i\theta}$, $\rho\geq 0$, $\theta\in[0,2\pi]$,
and

$\displaystyle R_{1}(\lambda)=r(\theta),\quad
R_{3}(\lambda)=\frac{\overline{r(\theta)}}{1-|r(\theta)|^{2}},\quad
R_{4}(\lambda)=\frac{r(\theta)}{1-|r(\theta)|^{2}},\quad
R_{6}(\lambda)=\overline{r(\theta)}.$ (25)

Recalling (12) and that the reflection coefficients $r$ is bounded on
$\Sigma$, we see by (25) that $R_{1}(\lambda)$, $R_{3}(\lambda)$,
$R_{4}(\lambda)$ and $R_{6}(\lambda)$ are bounded on $\mathbb{C}$. Then, as a
consequence of RH problem 3.3 and (23), we deduce that $M^{(2)}$ satisfies
$\bar{\partial}$-RH problem 3.4, and the jump matrix admits the lower/upper
triangular factorization:

$\displaystyle V^{(2)}=(1-w_{-})^{-1}(1+w_{+}),$

where we denote

$\displaystyle w=w_{+}+w_{-},$ (26) $\displaystyle
w_{+}=\begin{cases}\textbf{0}&L,\\\
\left[\begin{matrix}0&-\delta^{2}(\lambda)R_{3}(\lambda)e^{-it\phi(\lambda,n,t)}\\\
0&0\end{matrix}\right]&\tilde{L}\cap D_{+},\\\
\left[\begin{matrix}0&-\delta^{2}(\lambda)R_{6}(\lambda)e^{-it\phi(\lambda,n,t)}\\\
0&0\end{matrix}\right]&\tilde{L}\cap D_{-},\end{cases}$ (27) $\displaystyle
w_{-}=\begin{cases}\left[\begin{matrix}0&0\\\
\delta^{-2}(\lambda)R_{1}(\lambda)e^{it\phi(\lambda,n,t)}&0\end{matrix}\right]&L\cap
D_{+},\\\ \left[\begin{matrix}0&0\\\
\delta^{-2}(\lambda)R_{4}(\lambda)e^{it\phi(\lambda,n,t)}&0\end{matrix}\right]&L\cap
D_{-},\\\ \textbf{0}&\tilde{L}.\end{cases}$ (28)

and it is easy to check that $w_{\pm}$ are $2\times 2$ nilpotent matrices.

######  $\bar{\partial}$-RH problem 3.4.

Find a $2\times 2$ matrix-valued function $M^{(2)}$ such that

  * •

$M^{(2)}$ belongs to $C^{0}(\mathbb{C}\setminus(L\cup\tilde{L}))$ and its
first-order partial derivatives are continuous on
$\mathbb{C}\setminus(\Sigma\cup L\cup\tilde{L})$.

  * •

As $\lambda\to\infty$, $M^{(2)}\sim I+\mathcal{O}(\lambda^{-1})$.

  * •

On $\lambda\in\Sigma$,

$\displaystyle M^{(2)}_{+}=M^{(2)}_{-}V^{(2)},$

where

$\displaystyle V^{(2)}\equiv
V^{(2)}(\lambda,n,t)=\begin{cases}\left[\begin{matrix}1&0\\\
R_{1}(\lambda)\delta^{-2}(\lambda)e^{it\phi(\lambda,n,t)}&1\end{matrix}\right]&\lambda\in
L\cap D_{+},\\\
\left[\begin{matrix}1&-R_{3}(\lambda)\delta^{2}(\lambda)e^{-it\phi(\lambda,n,t)}\\\
0&1\end{matrix}\right]&\lambda\in\tilde{L}\cap D_{+},\\\
\left[\begin{matrix}1&0\\\
R_{4}(\lambda)\delta^{-2}(\lambda)e^{it\phi(\lambda,n,t)}&1\end{matrix}\right]&\lambda\in
L\cap D_{-},\\\
\left[\begin{matrix}1&-R_{6}(\lambda)\delta^{2}(\lambda)e^{-it\phi(\lambda,n,t)}\\\
0&1\end{matrix}\right]&\lambda\in\tilde{L}\cap D_{-}.\end{cases}$

  * •

On $\mathbb{C}\setminus(\Sigma\cup L\cup\tilde{L})$,

$\displaystyle\bar{\partial}M^{(2)}=M^{(2)}\bar{\partial}\mathcal{P},$

where $\bar{\partial}\mathcal{P}$ is a nilpotent matrix and

$\displaystyle\bar{\partial}\mathcal{P}(\lambda,n,t)=\begin{cases}\left[\begin{matrix}0&0\\\
-\bar{\partial}R_{1}(\lambda)\delta^{-2}(\lambda)e^{it\phi(\lambda,n,t)}&0\end{matrix}\right]&\lambda\in\Omega_{1},\\\
\left[\begin{matrix}0&\bar{\partial}R_{3}(\lambda)\delta^{2}(\lambda)e^{-it\phi(\lambda,n,t)}\\\
0&0\end{matrix}\right]&\lambda\in\Omega_{3},\\\ \left[\begin{matrix}0&0\\\
\bar{\partial}R_{4}(\lambda)\delta^{-2}(\lambda)e^{it\phi(\lambda,n,t)}&0\end{matrix}\right]&\lambda\in\Omega_{4},\\\
\left[\begin{matrix}0&-\bar{\partial}R_{6}(\lambda)\delta^{2}(\lambda)e^{-it\phi(\lambda,n,t)}\\\
0&0\end{matrix}\right]&\lambda\in\Omega_{6},\\\
\textbf{0}&\lambda\in\Omega_{2}\cup\Omega_{5}.\end{cases}$

###  3.3 Deformation of the $\bar{\partial}$-RH problem

In this part, we deform the solution for $\bar{\partial}$-RH problem 3.4 into
the product of solutions for RH problem 3.5 and $\bar{\partial}$-problem 3.6:

$\displaystyle M^{(2)}=M^{(2,D)}M^{(2,R)},$ (29)

where $M^{(2,R)}$ admits the same jump condition as $M^{(2)}$’s, and
$M^{(2,D)}$ is continuous over complex plane $\mathbb{C}$. Seeing from RH
problem 3.5, $M^{(2,R)}$ has no pole and on $\lambda\in L\cup\tilde{L}$,

$\displaystyle\det M^{(2,R)}_{+}(\lambda,n,t)=\det
M^{(2,R)}_{-}(\lambda,n,t);$

so, it is analytic in the whole complex plane; moreover, we see that as
$\lambda\to\infty$,

$\displaystyle\det M^{(2,R)}(\lambda,n,t)\sim 1+\mathcal{O}(\lambda^{-1}),$

which deduce by Liouville’s Theorem that

$\displaystyle\det M^{(2,R)}\equiv 1,$

and $M^{(2,R)}$ is invertible. Since $M^{(2,R)}$ is invertible, we obtain that
(29) is well-defined.

######  RH problem 3.5.

Find a $2\times 2$ matrix-valued function $M^{(2,R)}$ such that:

  * •

$M^{(2,R)}$ is analytic on $\mathbb{C}\setminus L\cap\tilde{L}$.

  * •

As $\lambda\to\infty$, $M^{(2,R)}\sim I+\mathcal{O}(\lambda^{-1})$.

  * •

On $\lambda\in L\cap\tilde{L}$, $M^{(2,R)}_{+}=M^{(2,R)}_{-}V^{(2)}$.

######  $\bar{\partial}$-problem 3.6.

Find a $2\times 2$ matrix-valued function $M^{(2,D)}$ such that

  * •

$M^{(2,D)}$ belongs to $C^{0}(\mathbb{C})$ and its first-order partial
derivatives are continuous on $\mathbb{C}\setminus(\Sigma\cap
L\cap\tilde{L})$.

  * •

As $\lambda\to\infty$, $M^{(2,D)}\sim I+\mathcal{O}(\lambda^{-1})$.

  * •

On $\lambda\in L\cap\tilde{L}$,
$\bar{\partial}M^{(2,D)}=M^{(2,D)}\tilde{\mathcal{P}}$, where
$\tilde{\mathcal{P}}=M^{(2,R)}\bar{\partial}\mathcal{P}(M^{(2,R)})^{-1}$.

Introducing a Cauchy-like operator $C_{w}$, for a $2\times 2$ matrix-valued
function $f$:

$\displaystyle C_{w}f=C_{+}(fw_{-})+C_{-}(fw_{+}),\quad w=w_{+}+w_{-},$
$\displaystyle
C_{\pm}f(\lambda)=\lim_{\lambda^{\prime}\to\lambda,\lambda^{\prime}\text{ on
$\pm$ side of
$\Sigma^{(2)}$}}\int_{\Sigma^{(2)}}\frac{f(s)\mathrm{d}s}{s-\lambda^{\prime}},$

for RH problem 3.3, since $(1-C_{w})^{-1}\in\mathcal{B}(L^{2}(\Sigma^{(2)}))$,
which is shown in Section 3.8, we obtain Beals-Coifman solutions:

$\displaystyle M^{(2,R)}(\lambda,n,t)=I+\frac{1}{2\pi
i}\int_{\Sigma}\frac{((1-C_{w})^{-1}Iw)(s,n,t)\mathrm{d}s}{s-\lambda}.$ (30)

###  3.4 RH problems at stationary phase points

In this part, we introduce two crosses contained in $\Sigma^{(2)}$ associated
to the two stationary phase points. And then, by these crosses, we analyze the
solution of RH problem 3.5 at $\lambda=0$. We find that as $t\to+\infty$, the
leading terms is related to RH problems on each of the crosses, and the
remaining part decays not faster than $\mathcal{O}(t^{-1})$, i.e. the result
in (61).

Figure 5: The jump contour $\Sigma^{\prime}=\Sigma_{1}\cup\Sigma_{2}$.

Set $\epsilon_{0}>0$ fixed and introduce contours

$\displaystyle\Sigma_{1}=\cup_{l=1}^{4}\Sigma_{1l},\quad\Sigma_{1l}=\left\\{S_{1}(1+xe^{\frac{2l-3}{4}\pi
i}):x\in[0,\epsilon_{0})\right\\},$
$\displaystyle\Sigma_{2}=\cup_{l=1}^{4}\Sigma_{2l},\quad\Sigma_{2l}=\left\\{S_{2}(1+xe^{\frac{2l-3}{4}\pi
i}):x\in[0,\epsilon_{0})\right\\}.$

We choose $\epsilon_{0}$ sufficiently small such that
$\Sigma^{\prime}\overset{\mathrm{def}}{=}\Sigma_{1}\cup\Sigma_{2}\subset\Sigma^{(2)}$,
and on $\lambda\in\Sigma^{\prime}$, (17) guarantees that

$\displaystyle\mathrm{Im}\phi(\lambda,n,t)\geq
x^{2}\sqrt{1-\xi^{2}},\quad\lambda=S_{j}(1+xe^{(-1)^{j}\frac{\pi i}{4}}),\quad
x\in(-\epsilon_{0},\epsilon_{0}),$ (31a)
$\displaystyle\mathrm{Im}\phi(\lambda,n,t)\leq-x^{2}\sqrt{1-\xi^{2}},\quad\lambda=S_{j}(1+xe^{(-1)^{j-1}\frac{\pi
i}{4}}),\quad x\in(-\epsilon_{0},\epsilon_{0}).$ (31b)

We also reasonably restrict $\epsilon_{0}$ such that for
$\lambda\in\Sigma^{\prime}$,

$\displaystyle|\mathrm{Re}\lambda|\geq\frac{1}{2}|\mathrm{Re}S_{1}|=\frac{\sqrt{1-\xi^{2}}}{2}\geq\frac{\sqrt{1-V_{0}^{2}}}{2}.$
(32)

Set $2\times 2$ matrix-valued functions supported on $\Sigma^{\prime}$:

$\displaystyle w^{\prime}=w_{-}^{\prime}+w_{+}^{\prime},\quad
w_{\pm}^{\prime}=\begin{cases}w_{\pm}&\lambda\in\Sigma^{\prime},\\\
0&\lambda\in\Sigma^{(2)}\setminus\Sigma^{\prime}.\end{cases}$ (33)

Similar to $C_{w}$, we also define a Cauchy-like operator $C_{w^{\prime}}$,
and in Section 3.8, we show that
$(1-C_{w^{\prime}})^{-1}\in\mathcal{B}(L^{2}(\Sigma^{(2)}))$ for sufficiently
large $t$. Taking $\lambda=0$ in (30), since $1-C_{w}$ and $1-C_{w^{\prime}}$
is invertible on $L^{2}(\Sigma^{(2)})$, we have by second resolvent identity
that

$\displaystyle M^{(2,R)}(0,n,t)=I+\frac{1}{2\pi
i}\int_{\Sigma^{(2)}}\frac{((1-C_{w})^{-1}Iw)(s,n,t)\mathrm{d}s}{s}$
$\displaystyle\quad=I+\frac{1}{2\pi
i}\int_{\Sigma^{(2)}}\frac{((1-C_{w^{\prime}})^{-1}Iw^{\prime})(s,n,t)\mathrm{d}s}{s}+I_{1}+I_{2},$
(34)

where

$\displaystyle I_{1}=\frac{1}{2\pi
i}\int_{\Sigma^{(2)}}s^{-1}[(1-C_{w})^{-1}I(w-w^{\prime})](s,n,t)\mathrm{d}s,$
$\displaystyle I_{2}=\frac{1}{2\pi
i}\int_{\Sigma^{(2)}}s^{-1}[(1-C_{w})^{-1}C_{w-w^{\prime}}(1-C_{w^{\prime}})^{-1}Iw^{\prime}](s,n,t)\mathrm{d}s.$

######  Lemma 3.7.

Since $r\in H^{1}_{\theta}(\Sigma)$, $w_{\pm}$ and $w$ are bounded functions
on $\Sigma^{(2)}$. Moreover, setting $w^{r}=w-w^{\prime}$ and
$w_{\pm}^{r}=w_{\pm}-w_{\pm}^{\prime}$ on $\Sigma^{(2)}$, we obtain that there
is a positive constant $C>0$ such that for $\lambda\in\Sigma^{(2)}$:

$\displaystyle|w_{\pm}^{r}(\lambda)|\lesssim
e^{-Ct},\quad|w^{r}(\lambda)|\lesssim e^{-Ct}.$

###### Proof.

Recalling that $\delta$, $\delta^{-1}$, $R_{j}$ are bounded functions on
$\Sigma^{(2)}$, by (27) and (28), we only have to check the boundedness of
$\mathrm{Im}\phi$; moreover, seeing Figure 3, since $\mathrm{Im}\phi>0$ on $L$
and $\mathrm{Im}\phi<0$ on $\tilde{L}$, we obtain that $w_{\pm}$ are bounded
on $\Sigma^{(2)}$. As for the estimates of $w^{r}_{\pm}$, seeing Figure 3,
$L\setminus\Sigma^{\prime}$ is a compact on $\\{\mathrm{Im}\phi>0\\}$, then
there is a constant $C>0$ such that

$\displaystyle\mathrm{Im}\phi\big{|}_{L\setminus\Sigma^{\prime}}\geq C;$ (35)

similarly, choosing proper $C$, we also have

$\displaystyle\mathrm{Im}\phi\big{|}_{\tilde{L}\setminus\Sigma^{\prime}}\leq-C;$
(36)

therefore, we confirm results for $w^{r}_{\pm}$ and $w^{r}$. ∎

The next task is to prove that as $t\to+\infty$

$\displaystyle I_{1},I_{2}\sim\mathcal{O}(t^{-1}),$ (37)

By Schwartz’s inequality, the boundedness of $(1-C_{w})^{-1}$ and
$(1-C_{w^{\prime}})^{-1}$ shown in Section 3.8, the fact that $\Sigma^{(2)}$
is a compact set contained in $\mathbb{C}\setminus\\{0\\}$ and Lemma 3.7, we
obtain that as $t\to+\infty$

$\displaystyle|I_{1}|\lesssim\parallel
w^{r}\parallel_{L^{2}(\Sigma^{(2)})}\lesssim\parallel
w^{r}\parallel_{L^{\infty}(\Sigma^{(2)})}\sim\mathcal{O}(t^{-1}).$ (38)

Similarly for $|I_{2}|$, also by Schwartz’s inequality, the result in Section
3.8 and Lemma 3.7, we have that as $t\to+\infty$,

$\displaystyle|I_{2}|\lesssim\parallel C_{w^{r}}\parallel_{L^{2}\to
L^{2}}\parallel I\parallel_{L^{2}(\Sigma^{(2)})}\parallel
w^{\prime}\parallel_{L^{2}(\Sigma^{(2)})}$
$\displaystyle\quad\lesssim\parallel C_{w^{r}}\parallel_{L^{2}\to
L^{2}}\lesssim\parallel
w^{r}\parallel_{L^{\infty}(\Sigma^{(2)})}\sim\mathcal{O}(t^{-1}).$ (39)

In (3.4), the second inequality is confirmed by the fact that $I$ and
$w^{\prime}$ are bounded on $\Sigma^{(2)}$, where the boundedness of
$w^{\prime}$ is supported by Lemma 3.7 and (33); the third inequality is also
correct because for a $2\times 2$ matrix function $f$ on $\Sigma^{(2)}$,

$\displaystyle\parallel C_{w^{r}}f\parallel_{L^{2}(\Sigma^{(2)})}=\parallel
C_{+}(fw_{-}^{r})+C_{-}(fw_{+}^{r})\parallel_{L^{2}(\Sigma^{(2)})}$
$\displaystyle\quad\lesssim\parallel
fw_{+}\parallel_{L^{2}(\Sigma^{(2)})}+\parallel
fw_{-}\parallel_{L^{2}(\Sigma^{(2)})}\lesssim\parallel
f\parallel_{L^{2}(\Sigma^{(2)})}\parallel
w^{r}\parallel_{L^{\infty}(\Sigma^{(2)})},$ (40)

by the fact that Cauchy integral operators $C_{\pm}$ are bounded on the
$L^{2}$ space. Seeing (3.4) and (37), we derive that as $t\to+\infty$,

$\displaystyle M^{(2,R)}(0,n,t)$ $\displaystyle=I+\frac{1}{2\pi
i}\int_{\Sigma\textquoteright}\frac{((1-C_{w^{\prime}})^{-1}Iw^{\prime})(s,n,t)\mathrm{d}s}{s}+\mathcal{O}(t^{-1}).$
(41)

The remaining work is to separate the contribution of each cross in
$\Sigma^{\prime}$. Introducing

$\displaystyle w^{j}=w^{j}_{-}+w^{j}_{+}\quad j=1,2,$ $\displaystyle
w^{j}_{\pm}=(\lambda,n,t)\begin{cases}w_{\pm}(\lambda,n,t)&\lambda\in\Sigma_{j},\\\
0&\lambda\in\Sigma^{(2)}\setminus\Sigma_{j},\end{cases}$

we deduce that

$\displaystyle w_{\pm}^{\prime}=w^{1}_{\pm}+w^{2}_{\pm}.$

and also define correspondent Cauchy-like operators $C_{w^{j}}$, $j=1,2$. By
(3.17) in [22], we obtain

$\displaystyle(1-C_{w^{\prime}})(1+C_{w^{1}}(1-C_{w^{1}})^{-1}+C_{w^{2}}(1-C_{w^{2}})^{-1})$
$\displaystyle\quad=(1-C_{w^{2}}C_{w^{1}}(1-C_{w^{1}})^{-1}-C_{w^{1}}C_{w^{2}}(1-C_{w^{2}})^{-1}).$
(42)

Considering (41) and (3.4), we obtain that

$\displaystyle M^{(2,R)}(0,n,t)=I+\sum_{j=1,2}\frac{1}{2\pi
i}\int_{\Sigma_{j}}s^{-1}[(1-C_{w^{j}})^{-1}Iw^{j}](s,n,t)\mathrm{d}s$
$\displaystyle\quad+I_{3}+I_{4}+I_{5}+\mathcal{O}(t^{-1}).$ (43)

where

$\displaystyle I_{3}=\frac{1}{2\pi i}\int_{\Sigma^{\prime}}$
$\displaystyle[s^{-1}(1+C_{w^{1}}(1-C_{w^{1}})^{-1}+C_{w^{2}}(1-C_{w^{2}})^{-1})$
$\displaystyle(1-C_{w^{2}}C_{w^{1}}(1-C_{w^{1}})^{-1}-C_{w^{1}}C_{w^{2}}(1-C_{w^{2}})^{-1})^{-1}$
$\displaystyle(C_{w^{2}}C_{w^{1}}(1-C_{w^{1}})^{-1}+C_{w^{1}}C_{w^{2}}(1-C_{w^{2}})^{-1})Iw^{\prime}](s,n,t)\mathrm{d}s,$
(44a) $\displaystyle I_{4}=\frac{1}{2\pi i}\int_{\Sigma^{\prime}}$
$\displaystyle s^{-1}[(1-C_{w^{2}})^{-1}C_{w^{2}}Iw^{1}](s,n,t)\mathrm{d}s,$
(44b) $\displaystyle I_{5}=\frac{1}{2\pi i}\int_{\Sigma^{\prime}}$
$\displaystyle s^{-1}[(1-C_{w^{1}})^{-1}C_{w^{1}}Iw^{2}](s,n,t)\mathrm{d}s.$
(44c)

For $I_{3}$, considering
$(1-C_{w^{j}})^{-1}\in\mathcal{B}(L^{2}(\Sigma^{(2)}))$ for $j=1,2$ in Section
3.8, by Lamma 3.9,

$\displaystyle(1-C_{w^{2}}C_{w^{1}}(1-C_{w^{1}})^{-1}-C_{w^{1}}C_{w^{2}}(1-C_{w^{2}})^{-1})^{-1}\in\mathcal{B}(L^{2}(\Sigma^{(2)}));$

also, applying the technique used in (3.4) on $C_{w^{j}}$, we claim that
$C_{w^{j}}$ belongs to $\mathcal{B}(L^{2}(\Sigma^{(2)}))$, $j=1,2$; thus, we
obtain by (44a) and Schwartz inequality that

$\displaystyle|I_{3}|$
$\displaystyle\lesssim\parallel(C_{w^{2}}C_{w^{1}}(1-C_{w^{1}})^{-1}+C_{w^{1}}C_{w^{2}}(1-C_{w^{2}})^{-1})I\parallel_{L^{2}(\Sigma^{(2)})}\parallel
w^{\prime}\parallel_{L^{2}(\Sigma^{(2)})},$ $\displaystyle\leq(\parallel
C_{w^{2}}C_{w^{1}}(1-C_{w^{1}})^{-1}I\parallel_{L^{2}(\Sigma^{(2)})}+\parallel
C_{w^{1}}C_{w^{2}}(1-C_{w^{2}})^{-1}I\parallel_{L^{2}(\Sigma^{(2)})})$
$\displaystyle\quad\times\parallel w^{\prime}\parallel_{L^{2}(\Sigma^{(2)})}$
(45)

By computation, applying (56a) in Lemma 3.9, Lemma 3.8,
$(1-C_{w^{1}})^{-1}\in\mathcal{B}(L^{2}(\Sigma^{(2)}))$ shown in Section 3.8
and the fact that

$\displaystyle\parallel C_{w^{1}}I\parallel_{L^{2}(\Sigma^{(2)})}\leq\parallel
C_{+}(w^{1}_{-})\parallel_{L^{2}(\Sigma^{(2)})}+\parallel
C_{-}(w^{1}_{+})\parallel_{L^{2}(\Sigma^{(2)})}$
$\displaystyle\quad\leq\parallel
w^{1}_{-}\parallel_{L^{2}(\Sigma^{(2)})}+\parallel
w^{1}_{+}\parallel_{L^{2}(\Sigma^{(2)})}\lesssim t^{-\frac{1}{4}},$ (46)

we obtain estimates:

$\displaystyle\parallel
C_{w^{2}}C_{w^{1}}(1-C_{w^{1}})^{-1}I\parallel_{L^{2}(\Sigma^{(2)})}\leq\parallel
C_{w^{2}}C_{w^{1}}(1-C_{w^{1}})^{-1}C_{w^{1}}I\parallel_{L^{2}(\Sigma^{(2)})}$
$\displaystyle\quad+\parallel
C_{w^{2}}C_{w^{1}}I\parallel_{L^{2}(\Sigma^{(2)})}\leq\parallel
C_{w^{2}}C_{w^{1}}\parallel_{L^{2}\to
L^{2}}\parallel(1-C_{w^{1}})^{-1}\parallel_{L^{2}\to L^{2}}\parallel
C_{w^{1}}I\parallel_{L^{2}(\Sigma^{(2)})}$ $\displaystyle\quad+\parallel
C_{w^{2}}C_{w^{1}}\parallel_{L^{\infty}\to L^{2}}\parallel
I\parallel_{L^{\infty}(\Sigma^{(2)})}\sim\mathcal{O}(t^{-\frac{3}{4}}).$ (47)

Similarly, we obtain that

$\displaystyle\parallel
C_{w^{2}}C_{w^{1}}(1-C_{w^{1}})^{-1}I\parallel_{L^{2}(\Sigma^{(2)})}\sim\mathcal{O}(t^{-\frac{3}{4}}).$
(48)

By (3.4), (3.4), (48) and Lemma 3.8, we confirm that

$\displaystyle I_{3}\sim\mathcal{O}(t^{-1}).$ (49)

For $I_{4}$, we split the integral into two parts

$\displaystyle
I_{4}=\int_{\Sigma^{\prime}}\frac{[C_{w^{2}}Iw^{1}](s,n,t)}{2\pi
is}\mathrm{d}s+\int_{\Sigma^{\prime}}\frac{[C_{w^{2}}(1-C_{w^{2}})^{-1}C_{w^{2}}Iw^{1}](s,n,t)}{2\pi
is}\mathrm{d}s;$ (50)

estimate the first part by Lemma 3.8,

$\displaystyle\Big{|}\int_{\Sigma^{\prime}}\frac{[C_{w^{2}}Iw^{1}](s,n,t)}{2\pi
is}\mathrm{d}s\Big{|}\leq\frac{1}{2\pi}\int_{\Sigma_{1}}\int_{\Sigma_{2}}\Big{|}\frac{w^{2}(s^{\prime},n,t)w^{1}(s,n,t)}{s(s^{\prime}-s)}\Big{|}\mathrm{d}s^{\prime}\mathrm{d}s$
$\displaystyle\quad\leq\frac{1}{2\pi d\inf_{s\in\Sigma_{1}}|s|}\parallel
w_{1}\parallel_{L^{1}(\Sigma_{1})}\parallel
w_{2}\parallel_{L^{1}(\Sigma_{2})}\lesssim t^{-1};$ (51)

estimate the second part by Schwartz inequality, the boundedness of
$(1-C_{w^{2}})^{-1}$, Lemma 3.8 and (3.4)

$\displaystyle\Big{|}\int_{\Sigma^{\prime}}\frac{[C_{w^{2}}(1-C_{w^{2}})^{-1}C_{w^{2}}Iw^{1}](s,n,t)}{2\pi
is}\mathrm{d}s\Big{|}$
$\displaystyle\quad\leq\frac{1}{2\pi}\int_{\Sigma_{1}}\int_{\Sigma_{2}}\Big{|}\frac{[(1-C_{w^{2}})^{-1}C_{w^{2}}Iw^{2}](s^{\prime},n,t)w^{1}(s,n,t)}{s(s^{\prime}-s)}\Big{|}\mathrm{d}s^{\prime}\mathrm{d}s$
$\displaystyle\quad\leq\frac{1}{2\pi
d\inf_{s\in\Sigma_{1}}|s|}\parallel(1-C_{w^{2}})^{-1}C_{w^{2}}Iw^{2}\parallel_{L^{1}(\Sigma_{2})}\parallel
w^{1}\parallel_{L^{1}(\Sigma_{1})}$ $\displaystyle\quad\lesssim\parallel
C_{w^{2}}I\parallel_{2}\parallel w^{2}\parallel_{L^{2}(\Sigma_{2})}\parallel
w^{1}\parallel_{L^{1}(\Sigma_{1})}\lesssim t^{-1};$ (52)

thus, by (50), (3.4) and (3.4), we have

$\displaystyle I_{4}\sim\mathcal{O}(t^{-1}).$ (53)

For $I_{5}$, we claim that

$\displaystyle I_{5}\sim\mathcal{O}(t^{-1}),$ (54)

and the technique is parallel to that for $I_{4}$. By (3.4), (49), (53) and
(54), we get

$\displaystyle M^{(2,R)}(0,n,t)=$ $\displaystyle I+\sum_{j=1}^{2}\frac{1}{2\pi
i}\int_{\Sigma_{j}}s^{-1}[(1-C_{w^{j}})^{-1}Iw^{j}](s,n,t)\mathrm{d}s+\mathcal{O}(t^{-1}).$
(55)

######  Lemma 3.8.

For $r\in H^{1}_{\theta}(\Sigma)$, we obtain these estimates for $w^{j}$:

$\displaystyle\parallel w^{j}\parallel_{L^{1}(\Sigma^{(2)})}\lesssim
t^{-\frac{1}{2}},\quad\parallel w^{j}\parallel_{L^{2}(\Sigma^{(2)})}\lesssim
t^{-\frac{1}{4}}.$

###### Proof.

Recalling that $\delta$, $\delta^{-1}$, $R_{j}$ are bounded functions, by
(27), (28) and (31), we obtain that on $\lambda\in\Sigma_{j}$,

$\displaystyle|w^{j}(\lambda,n,t)|\lesssim e^{-t\sqrt{1-\xi^{2}}x^{2}},\quad
x=|\lambda-S_{j}|,$

by which we confirm the result since $w^{j}$ is supported only on
$\Sigma_{j}$. ∎

######  Lemma 3.9.

For $r\in H^{1}_{\theta}(\theta)$, we claim that $C_{w^{1}}C_{w^{2}}$,
$C_{w^{1}}C_{w^{2}}$ belong to $\mathcal{B}(L^{2}(\Sigma^{(2)}))$ and
$\mathcal{B}(L^{\infty}(\Sigma^{(2)}),L^{2}(\Sigma^{(2)}))$, and we have the
following estimate

$\displaystyle\parallel C_{w^{1}}C_{w^{2}}\parallel_{L^{2}\to L^{2}}\lesssim
t^{-\frac{1}{2}},\quad\parallel C_{w^{1}}C_{w^{2}}\parallel_{L^{\infty}\to
L^{2}}\lesssim t^{-\frac{3}{4}},$ (56a) $\displaystyle\parallel
C_{w^{2}}C_{w^{1}}\parallel_{L^{2}\to L^{2}}\lesssim
t^{-\frac{1}{2}},\quad\parallel C_{w^{2}}C_{w^{1}}\parallel_{L^{\infty}\to
L^{2}}\lesssim t^{-\frac{3}{4}}.$ (56b)

###### Proof.

For any $2\times 2$ matrix-valued function $f\in L^{2}(\Sigma^{(2)})$, we
obtain that

$\displaystyle
C_{w^{1}}C_{w^{2}}f=C_{+}(C_{-}(fw^{2}_{+})w^{1}_{-})+C_{-}(C_{+}(fw^{2}_{-})w^{1}_{+}).$
(57)

By the formula

$\displaystyle(C_{-}(fw^{2}_{+})w^{1}_{-})(s_{1})=\frac{1}{2\pi
i}\int_{\Sigma_{2}}\frac{f(s_{2})w^{2}_{+}(s_{2})w^{1}_{-}(s_{1})}{s_{2}-s_{1}}\mathrm{d}s_{2}$

ope and Lemma 3.8, the $L^{2}$-norm of the first term in (57) satisfies that

$\displaystyle\parallel
C_{+}(C_{-}(fw^{2}_{+})w^{1}_{-})\parallel_{L^{2}(\Sigma^{(2)})}\leq\frac{1}{2\pi}\left(\int_{\Sigma_{1}}\Big{|}\int_{\Sigma_{2}}\frac{f(s_{2})w^{2}_{+}(s_{2})w^{1}_{-}(s_{1})}{s_{2}-s_{1}}\mathrm{d}s_{2}\Big{|}^{2}\mathrm{d}s_{1}\right)^{\frac{1}{2}}$
$\displaystyle\quad\leq\frac{1}{2\pi d}\parallel
fw^{2}_{+}\parallel_{L^{1}(\Sigma^{(2)})}\parallel
w^{1}_{-}\parallel_{L^{2}(\Sigma^{(2)})}$
$\displaystyle\quad\lesssim\begin{cases}\parallel
f\parallel_{L^{\infty}(\Sigma^{(2)})}\parallel
w^{2}_{+}\parallel_{L^{1}(\Sigma^{(2)})}\parallel
w^{1}_{-}\parallel_{L^{2}(\Sigma^{(2)})}\lesssim t^{-\frac{3}{4}}\parallel
f\parallel_{L^{\infty}(\Sigma^{(2)})},\\\ \parallel
f\parallel_{L^{2}(\Sigma^{(2)})}\parallel
w^{2}_{+}\parallel_{L^{2}(\Sigma^{(2)})}\parallel
w^{1}_{-}\parallel_{L^{2}(\Sigma^{(2)})}\lesssim t^{-\frac{1}{2}}\parallel
f\parallel_{L^{2}(\Sigma^{(2)})},\end{cases}$ (58)

where $d$ denotes the distance of $\Sigma_{1}$ and $\Sigma_{2}$. For the
second term on the right of (57), we obtain the similar estimate,

$\displaystyle\parallel
C_{+}(C_{-}(fw^{2}_{+})w^{1}_{-})\parallel_{L^{2}(\Sigma^{(2)})}\lesssim\begin{cases}t^{-\frac{3}{4}}\parallel
f\parallel_{L^{\infty}(\Sigma^{(2)})},\\\ t^{-\frac{1}{2}}\parallel
f\parallel_{L^{2}(\Sigma^{(2)})}.\end{cases}$ (59)

By (57), (3.4) and (59), we confirm the result of $C_{w^{1}}C_{w^{2}}$, and
that of $C_{w^{2}}C_{w^{1}}$ is similarly checked. ∎

Figure 6: The infinite cross $\Sigma_{1}^{\prime}$.

Define infinite crosses

$\displaystyle\Sigma_{j}^{\prime}=S_{j}((1+e^{\frac{\pi
i}{4}}\mathbb{R})\cup(1+e^{-\frac{\pi i}{4}}\mathbb{R}))\quad j=1,2.$

Here, we give the infinite crosses the orientation consisting with that of
$\Sigma_{j}$, example as shown in Figure 6. Set $\check{w}^{j}_{\pm}$ the zero
extension of $w^{j}_{\pm}\big{|}_{\Sigma_{j}}$ on $\Sigma_{j}^{\prime}$ and
$\check{w}^{j}=\check{w}^{j}_{+}+\check{w}^{j}_{-}$; then, define Cauchy-like
integral operators $A_{j}$: $L^{2}(\Sigma_{j}^{\prime})\to
L^{2}(\Sigma_{j}^{\prime})$, $f\mapsto A_{j}f$,

$\displaystyle
A_{j}f=C_{+}^{\Sigma_{j}^{\prime}}(f\check{w}_{-})+C_{-}^{\Sigma_{j}^{\prime}}(f\check{w}_{+}),$
(60) $\displaystyle\quad
C_{\pm}^{\Sigma_{j}^{\prime}}f(\lambda)=\lim_{\lambda^{\prime}\to\lambda,\lambda^{\prime}\text{
on $\pm$ sides of
}\Sigma_{j}^{\prime}}\int_{\Sigma_{j}^{\prime}}\frac{f(s)\mathrm{d}s}{s-\lambda^{\prime}};$

combining (55) and (60), we claim that

$\displaystyle M^{(2,R)}(0,n,t)=$ $\displaystyle I+\sum_{j=1}^{2}\frac{1}{2\pi
i}\int_{\Sigma_{j}^{\prime}}s^{-1}[(1-A_{j})^{-1}I\check{w}^{j}](s,n,t)\mathrm{d}s+\mathcal{O}(t^{-1}).$
(61)

###  3.5 Scaling and rotation

Here, we firstly introduce the scaling operator $N_{j}$ according to each
cross, $j=1,2$; secondly, we apply these scaling operators to some functions
and obtain the oscillatory part $\delta_{j0}$; thirdly, by these scaling
operators, we obtain the Cauchy-like integral operators $\tilde{A}_{j}$ and
their limits as $t\to+\infty$: $\tilde{A}_{j}^{\infty}$; finally, by proper
evaluations and estimates, we rewrite formula (61) by
$\tilde{A}_{j}^{\infty}$, which results in a long-time asymptotic formula (81)
related to $M^{(2,R)}$. From (81), we see that the leading term in the long-
time asymptotic formula is determined by the model RH problem at the
stationary phase point. The model RH problem is shown in the later part.

Define infinite crosses

$\displaystyle\tilde{\Sigma}_{j}=e^{\frac{\pi i}{4}}\mathbb{R}\cup
e^{-\frac{\pi i}{4}}\mathbb{R},\quad j=1,2,$

and scaling mapping

$\displaystyle N_{j}:f(\lambda)\mapsto N_{j}f(\zeta)=f(\beta_{j}\zeta+S_{j}).$
(62)

It follows from direct computation that $N_{j}$ belongs to
$\mathcal{B}(L^{2}(\Sigma_{j}^{\prime}),L^{2}(\tilde{\Sigma}_{j}))$ and
invertible; moreover, its norm satisfies that

$\displaystyle\parallel N_{j}\parallel_{L^{2}\to
L^{2}}=|\beta_{j}|^{-\frac{1}{2}}=t^{\frac{1}{4}}(1-\xi^{2})^{\frac{1}{8}}.$

Seeing the definition of $\Sigma_{j}^{\prime}$ and $\tilde{\Sigma}_{j}$, we
have the relationship

$\displaystyle\Sigma_{j}^{\prime}=\beta_{j}\tilde{\Sigma}_{j}+S_{j},$

and also set the orientation of $\tilde{\Sigma}_{j}$ consisting with that of
$\Sigma_{j}^{\prime}$ by mapping: $\tilde{\Sigma}_{j}\to\Sigma_{j}^{\prime}$,
$\zeta\mapsto\beta_{j}\zeta+S_{j}$.

By direct computation, we obtain that

$\displaystyle
N_{j}(\delta^{2}e^{-it\phi})(\zeta)=\zeta^{(-1)^{j-1}2i\nu_{j}}e^{(-1)^{j}\frac{i}{2}\zeta^{2}}\delta^{2}_{j0}\delta_{j1}(\zeta),$
(63a)
$\displaystyle\quad\delta_{j0}=e^{\alpha_{j}(S_{j})-\frac{it}{2}\phi(S_{j})}\left(\frac{(-1)^{j-1}\beta_{j}}{S_{1}-S_{2}}\right)^{(-1)^{j-1}i\nu_{j}},$
(63b)
$\displaystyle\quad\delta_{j1}(\zeta)=\left(\frac{S_{1}-S_{2}}{(-1)^{j-1}\beta_{1}\zeta+S_{1}-S_{2}}\right)^{2(-1)^{j-1}i\nu_{1}}$
$\displaystyle\quad\quad\times
e^{2(\alpha_{j}(\beta_{j}\zeta+S_{j})-\alpha_{j}(S_{j}))-it(\phi(\beta_{j}\zeta+S_{j},n,t)-\phi(S_{j},n,t))+(-1)^{j-1}\frac{i}{2}\zeta^{2}},$
(63c)

where $\delta_{j0}$ is the oscillatory part and $\delta_{j1}(\zeta)$ admits
limits at $\zeta\to 0$ as shown in Lemma 3.10.

######  Lemma 3.10.

For $r\in H^{1}_{\theta}(\Sigma)$, we have $\delta_{11}(\zeta)\to 1$ as
$\zeta\to 0$, and on $\zeta\in\beta_{j}^{-1}(\Sigma_{j}-S_{j})$

$\displaystyle|\delta_{j1}(\zeta)-1|\lesssim
t^{-\frac{1}{4}}|\zeta|^{\frac{1}{2}},\quad|\delta_{j1}^{-1}(\zeta)-1|\lesssim
t^{-\frac{1}{4}}|\zeta|^{\frac{1}{2}}.$ (64)

###### Proof.

We first detail the proof for the first formula in (64) with $j=1$ and
$\zeta\in\beta_{1}^{-1}(\Sigma_{11}-S_{1})$. Since
$\left(\frac{S_{1}-S_{2}}{\lambda-S_{2}}\right)^{2i\nu_{1}}$ and
$it(\phi(\lambda,n,t)-\phi(S_{1},n,t))-\frac{\phi^{\prime\prime}(S_{j},n,t)}{2}(\lambda-
S_{j})^{2}$ are both holomorphic on $\lambda\in\Sigma_{11}$, we deduce that

$\displaystyle\Big{|}\left(\frac{S_{1}-S_{2}}{\lambda-
S_{2}}\right)^{2i\nu_{1}}-1\Big{|}\lesssim|\lambda-S_{1}|,$ (65a)
$\displaystyle|it(\phi(\lambda,n,t)-\phi(S_{1},n,t))-2\phi^{\prime\prime}(S_{1},n,t)(\lambda-
S_{1})^{2}|\lesssim t|\lambda-S_{1}|^{3};$ (65b) moreover, seeing (19) and
(21), we derive that on $\lambda\in\Sigma_{11}$,
$\displaystyle|\alpha_{1}(\lambda)-\alpha_{1}(S_{1})|\lesssim|\lambda-
S_{1}|^{\frac{1}{2}};$ (65c)

thus, by (62), (63c) and (65), we obtain that

$\displaystyle|\delta_{11}(\zeta)-1|\lesssim
t^{-\frac{1}{4}}|\zeta|^{\frac{1}{2}},$

Since we have proven the first formula in (64) with $j=1$ and
$\zeta\in\beta_{1}^{-1}(\Sigma_{11}-S_{1})$, we generalize similarly this
proof to the case for $\delta_{j1}(\zeta)$ on
$\zeta\in\beta_{j}^{-1}(\Sigma_{j}-S_{j})$, $j=1,2$. The proof for the second
formula is similarly obtained. Thus, we confirm the result in this lemma. ∎

Define operator $\Delta_{j0}:f\mapsto f\delta_{j0}^{\sigma_{3}}$, where $f$ is
a $2\times 2$ matrix-valued function; by the definition of $\delta_{j0}$, it
and its inverse operator $\Delta_{j0}^{-1}$ both belong to
$\mathcal{B}(L^{2}(\tilde{\Sigma}_{j}))$. Define a Cauchy-like integral
operator $\tilde{A}_{j}$: $L^{2}(\tilde{\Sigma}_{j})\to
L^{2}(\tilde{\Sigma}_{j})$, $f\mapsto\tilde{A}_{j}f$,

$\displaystyle\tilde{A}_{j}f=C_{+}^{\tilde{\Sigma}_{j}}(f\delta_{j0}^{-\sigma_{3}}N_{j}\check{w}^{j}_{-}\delta_{j0}^{\sigma_{3}})+C_{-}^{\tilde{\Sigma}_{j}}(f\delta_{j0}^{-\sigma_{3}}N_{j}\check{w}^{j}_{+}\delta_{j0}^{\sigma_{3}}),$
(66) $\displaystyle\quad
C_{\pm}^{\tilde{\Sigma}_{j}}f(\lambda)=\lim_{\lambda^{\prime}\to\lambda,\lambda^{\prime}\text{
on $\pm$ sides of
}\tilde{\Sigma}_{j}}\int_{\tilde{\Sigma}_{j}}\frac{f(s)\mathrm{d}s}{s-\lambda^{\prime}}.$

By the relations (60) and (66), we have these formulas

$\displaystyle\tilde{A}_{j}=\Delta_{j0}N_{j}A_{j}N_{j}^{-1}\Delta_{j0}^{-1},\quad
j=1,2.$ (67)

Write $\tilde{w}^{j}=\tilde{w}_{+}^{j}+\tilde{w}_{-}^{j}$ on
$\tilde{\Sigma}_{j}$,

$\displaystyle\tilde{w}_{\pm}^{j}\equiv\tilde{w}_{\pm}^{j}(\zeta,n,t)=\delta_{j0}^{-\sigma_{3}}N_{j}\check{w}^{j}_{\pm}(\zeta,n,t)\delta_{j0}^{\sigma_{3}}.$
(68)

By the definition of $w_{\pm}^{j}$ and (63), we deduce that

$\displaystyle\tilde{w}_{+}^{1}(\zeta,n,t)=\begin{cases}\left[\begin{matrix}0&-N_{1}R_{3}(\zeta)\zeta^{2i\nu_{j}}e^{-\frac{i}{2}\zeta^{2}}\delta_{11}(\zeta,n,t)\\\
0&0\end{matrix}\right]&\zeta\in e^{\frac{3\pi
i}{4}}(0,|\beta_{j}^{-1}|\epsilon_{0}),\\\
\left[\begin{matrix}0&-N_{1}R_{6}(\zeta)\zeta^{2i\nu_{j}}e^{-\frac{i}{2}\zeta^{2}}\delta_{11}(\zeta,n,t)\\\
0&0\end{matrix}\right]&\zeta\in e^{\frac{3\pi
i}{4}}(-|\beta_{j}^{-1}|\epsilon_{0},0),\\\
\textbf{0}&\text{otherwise},\end{cases}$
$\displaystyle\tilde{w}_{-}^{1}(\zeta,n,t)=\begin{cases}\left[\begin{matrix}0&0\\\
N_{1}R_{1}(\zeta)\zeta^{-2i\nu_{j}}e^{\frac{i}{2}\zeta^{2}}\delta_{11}^{-1}(\zeta,n,t)&0\end{matrix}\right]&\zeta\in
e^{\frac{\pi i}{4}}(0,|\beta_{j}^{-1}|\epsilon_{0}),\\\
\left[\begin{matrix}0&0\\\
N_{1}R_{4}(\zeta)\zeta^{-2i\nu_{j}}e^{\frac{i}{2}\zeta^{2}}\delta_{11}^{-1}(\zeta,n,t)&0\end{matrix}\right]&\zeta\in
e^{\frac{\pi i}{4}}(-|\beta_{j}^{-1}|\epsilon_{0},0),\\\
\textbf{0}&\text{otherwise},\end{cases}$
$\displaystyle\tilde{w}_{+}^{2}(\zeta,n,t)=\begin{cases}\left[\begin{matrix}0&-N_{2}R_{3}(\zeta)\zeta^{-2i\nu_{j}}e^{\frac{i}{2}\zeta^{2}}\delta_{21}(\zeta,n,t)\\\
0&0\end{matrix}\right]&\zeta\in e^{\frac{\pi
i}{4}}(0,|\beta_{j}^{-1}|\epsilon_{0}),\\\
\left[\begin{matrix}0&-N_{2}R_{6}(\zeta)\zeta^{-2i\nu_{j}}e^{\frac{i}{2}\zeta^{2}}\delta_{21}(\zeta,n,t)\\\
0&0\end{matrix}\right]&\zeta\in e^{\frac{\pi
i}{4}}(-|\beta_{j}^{-1}|\epsilon_{0},0),\\\
\textbf{0}&\text{otherwise},\end{cases}$
$\displaystyle\tilde{w}_{-}^{2}(\zeta,n,t)=\begin{cases}\left[\begin{matrix}0&0\\\
N_{2}R_{1}(\zeta)\zeta^{2i\nu_{j}}e^{-\frac{i}{2}\zeta^{2}}\delta_{21}^{-1}(\zeta,n,t)&0\end{matrix}\right]&\zeta\in
e^{\frac{3\pi i}{4}}(0,|\beta_{j}^{-1}|\epsilon_{0}),\\\
\left[\begin{matrix}0&0\\\
N_{2}R_{4}(\zeta)\zeta^{2i\nu_{j}}e^{-\frac{i}{2}\zeta^{2}}\delta_{21}^{-1}(\zeta,n,t)&0\end{matrix}\right]&\zeta\in
e^{\frac{3\pi i}{4}}(-|\beta_{j}^{-1}|\epsilon_{0},0),\\\
\textbf{0}&\text{otherwise}.\end{cases}$

Taking limits of $\tilde{w}_{\pm}^{j}$ and denoting them as
$\tilde{w}_{\pm}^{j,\infty}\equiv\tilde{w}_{\pm}^{j,\infty}(\zeta)$ when
$t\to+\infty$, by direct computation, we obtain that

$\displaystyle\tilde{w}_{+}^{j,\infty}(\zeta)=\begin{cases}\textbf{0}&\zeta\in
e^{\frac{(2+(-1)^{j})\pi i}{4}}\mathbb{R},\\\
\left[\begin{matrix}0&-R_{3}(S_{j})\zeta^{2(-1)^{j-1}i\nu_{j}}e^{(-1)^{j}\frac{i}{2}\zeta^{2}}\\\
0&0\end{matrix}\right]&\zeta\in e^{\frac{(2-(-1)^{j})\pi
i}{4}}\mathbb{R}^{+},\\\
\left[\begin{matrix}0&-R_{6}(S_{j})\zeta^{2(-1)^{j-1}i\nu_{j}}e^{(-1)^{j}\frac{i}{2}\zeta^{2}}\\\
0&0\end{matrix}\right]&\zeta\in e^{\frac{(2-(-1)^{j})\pi
i}{4}}\mathbb{R}^{-},\end{cases}$ (69a)
$\displaystyle\tilde{w}_{-}^{j,\infty}(\zeta)=\begin{cases}\textbf{0}&\zeta\in
e^{\frac{(2-(-1)^{j})\pi i}{4}}\mathbb{R},\\\ \left[\begin{matrix}0&0\\\
R_{1}(S_{j})\zeta^{2(-1)^{j}i\nu_{j}}e^{(-1)^{j-1}\frac{i}{2}\zeta^{2}}&0\end{matrix}\right]&\zeta\in
e^{\frac{(2+(-1)^{j})\pi i}{4}}\mathbb{R^{+}},\\\ \left[\begin{matrix}0&0\\\
R_{4}(S_{j})\zeta^{2(-1)^{j}i\nu_{j}}e^{(-1)^{j-1}\frac{i}{2}\zeta^{2}}&0\end{matrix}\right]&\zeta\in
e^{\frac{(2+(-1)^{j})\pi i}{4}}\mathbb{R^{-}}.\end{cases}$ (69b)

Figure 7: The jump contour $\tilde{\Sigma}_{1}$ and regions:
$\tilde{\Omega}_{1},\dots,\tilde{\Omega}_{6}$.

######  Proposition 3.11.

If $r\in H^{1}_{\theta}(\Sigma)$, we obtain that as $t\to+\infty$, for
$j=1,2$,

$\displaystyle\parallel\tilde{w}^{j}-\tilde{w}^{j,\infty}\parallel_{L^{\infty}(\tilde{\Sigma}_{j})\cap
L^{2}(\tilde{\Sigma}_{j})\cap L^{1}(\tilde{\Sigma}_{j})}\lesssim
t^{-\frac{1}{4}}.$

###### Proof.

In this proof, we claim that for $\zeta\in\tilde{\Sigma}_{j}$, $\tilde{w}^{j}$
and $\tilde{w}^{j,\infty}$ satisfy

$\displaystyle|\tilde{w}^{j}-\tilde{w}^{j,\infty}|(\zeta,n,t)\lesssim
t^{-\frac{1}{4}}|\zeta|^{\frac{1}{4}}e^{-\frac{|\zeta|^{2}}{2}},$ (70)

and then, this proposition’s results naturally follow. In the following, we
will look into the case of $\zeta\in e^{\frac{\pi i}{4}}\mathbb{R}^{+}$ and
$j=1$; then, proofs for other cases are parallel to omit. For
$\zeta=xe^{\frac{\pi i}{4}}\in\beta_{j}^{-1}(\Sigma_{11}-S_{1})\subset
e^{\frac{\pi i}{4}}\mathbb{R}$, we obtain the estimate

$\displaystyle|N_{1}R_{1}(\zeta)-R_{1}(S_{1})|=|r(\theta)-r(\theta_{1})|=\Big{|}\int_{\theta_{1}}^{\theta}r^{\prime}(\check{\theta})\mathrm{d}\check{\theta}\Big{|}\leq|\theta-\theta_{1}|^{\frac{1}{2}}\parallel
r\parallel_{H^{1}},$ (71)
$\displaystyle\quad\quad\theta=\arg(\beta_{1}\zeta+S_{1}),\quad\theta_{1}=\arg
S_{1}\in[0,2\pi];$

seeing the relationship between $\zeta$ and $\theta$ in Figure 8, since
$\zeta\in\beta_{j}^{-1}(\Sigma_{11}-S_{1})$, we get the following estimate by
triangular knowledge

$\displaystyle|\theta_{1}-\theta|\lesssim|\beta_{1}\zeta|=t^{-\frac{1}{2}}(1-\xi^{2})^{-\frac{1}{4}}x;$
(72)

therefore, by (71) and (72), recalling $|\xi|\leq V_{0}$, we obtain that

$\displaystyle|N_{1}R_{1}(\zeta)-R_{1}(S_{1})|\lesssim
t^{-\frac{1}{4}}x^{\frac{1}{2}}.$ (73)

Figure 8: the relation between parameters $\theta$ and $\zeta$ for
$\beta_{1}\zeta+S_{1}\in\Sigma_{11}$.

By definition of $\tilde{w}^{j}$, (69), Lemma 3.10 and (73), we estimate that

$\displaystyle|\tilde{w}^{j}-\tilde{w}^{j,\infty}|(\zeta,n,t)\leq\Big{|}\zeta^{-2i\nu_{1}}e^{\frac{i\zeta^{2}}{2}}\Big{|}\big{|}N_{1}R_{1}(\zeta)\delta_{11}^{-1}(\zeta)-R_{1}(S_{1})\big{|}$
$\displaystyle\quad\lesssim
e^{-\frac{x^{2}}{2}}(|R_{1}(\beta_{1}\zeta+S_{1})||\delta_{11}^{-1}(\zeta)-1|+|R_{1}(\beta_{1}\zeta+S_{1})-R_{1}(S_{1})|)$
$\displaystyle\quad\lesssim e^{-\frac{x^{2}}{2}}(\parallel
r\parallel_{\infty}t^{-\frac{1}{4}}x^{\frac{1}{2}}+t^{-\frac{1}{4}}x^{\frac{1}{2}})\lesssim
t^{-\frac{1}{4}}x^{\frac{1}{2}}e^{-\frac{x^{2}}{2}}.$

For $\zeta=xe^{\frac{i\pi}{4}}\in e^{\frac{\pi
i}{4}}\mathbb{R}\setminus\beta_{j}^{-1}(\Sigma_{11}-S_{1})$, we obtain by (69)
that

$\displaystyle|\tilde{w}^{j}-\tilde{w}^{j,\infty}|(\zeta,n,t)=|\tilde{w}^{j,\infty}|(\zeta,n,t)=|R_{1}(S_{j})||(xe^{\frac{\pi
i}{4}})^{i\nu_{1}}|e^{-\frac{x^{2}}{2}}$ $\displaystyle\quad\lesssim
e^{-\frac{x^{2}}{2}}\leq\Big{|}\frac{\beta_{1}\zeta}{\epsilon_{0}}\Big{|}^{\frac{1}{2}}e^{-\frac{x^{2}}{2}}\lesssim
t^{-\frac{1}{4}}x^{\frac{1}{2}}e^{-\frac{x^{2}}{2}}.$

We complete the proof. ∎

Here, we come to estimates of integral parts in (61). By direct computation,
(67), (68) and (69), we obtain that

$\displaystyle\frac{1}{2\pi
i}\int_{\Sigma_{j}^{\prime}}s^{-1}[(1-A_{j})^{-1}I\check{w}^{j}](s,n,t)\mathrm{d}s$
$\displaystyle\quad=\frac{1}{2\pi
i}\int_{\Sigma_{j}^{\prime}}s^{-1}[N_{j}^{-1}\Delta_{j0}^{-1}(1-\tilde{A})^{-1}\Delta_{j0}N_{j}I\delta_{j0}^{\sigma_{3}}N_{j}^{-1}\tilde{w}^{j}\delta_{j0}^{-\sigma_{3}}](s,n,t)\mathrm{d}s$
$\displaystyle\quad=\frac{\beta_{j}}{2\pi
i}\int_{\tilde{\Sigma}_{j}}(\beta_{j}s+S_{j})^{-1}[(1-\tilde{A}_{j})^{-1}\delta_{j0}^{\sigma_{3}}\tilde{w}^{j}](s,n,t)\delta_{j0}^{-\sigma_{3}}\mathrm{d}s$
$\displaystyle\quad=\frac{\beta_{j}}{2\pi
i}\delta_{j0}^{\sigma_{3}}\int_{\tilde{\Sigma}_{j}}(\beta_{j}s+S_{j})^{-1}[(1-\tilde{A}_{j})^{-1}I\tilde{w}^{j}](s,n,t)\mathrm{d}s\delta_{j0}^{-\sigma_{3}}$
$\displaystyle\quad=\frac{\beta_{j}}{2\pi
i}S_{j}^{-1}\delta_{j0}^{\sigma_{3}}\int_{\tilde{\Sigma}_{j}}[(1-\tilde{A}_{j}^{\infty})^{-1}I\tilde{w}^{j,\infty}](s)\mathrm{d}s\delta_{j0}^{-\sigma_{3}}+\frac{\beta_{j}}{2\pi
i}\delta_{j0}^{\sigma_{3}}(I_{6}+I_{7})\delta_{j0}^{-\sigma_{3}},$ (74)

where

$\displaystyle
I_{6}=\int_{\tilde{\Sigma}_{j}}(\beta_{j}s+S_{j})^{-1}[((1-\tilde{A}_{j})^{-1}-(1-\tilde{A}_{j}^{\infty})^{-1})I\tilde{w}^{j}](s,n,t)\mathrm{d}s,$
(75a) $\displaystyle
I_{7}=\int_{\tilde{\Sigma}_{j}}[(1-\tilde{A}_{j}^{\infty})^{-1}I((\beta_{j}s+S_{j})^{-1}\tilde{w}^{j}-S_{j}^{-1}\tilde{w}^{j,\infty})](s,n,t)\mathrm{d}s.$
(75b)

By Schwartz inequality,

$\displaystyle|I_{6}|=\Big{|}\int_{\tilde{\Sigma}_{j}}[(\beta_{j}s+S_{j})^{-1}((1-\tilde{A}_{j})^{-1}(\tilde{A}_{j}-\tilde{A}_{j}^{\infty})(1-\tilde{A}_{j}^{\infty})^{-1})I\tilde{w}^{j}](s,n,t)\mathrm{d}s\Big{|}$
$\displaystyle\quad\leq\parallel(1-\tilde{A}_{j})^{-1}(\tilde{A}_{j}-\tilde{A}_{j}^{\infty})(1-\tilde{A}_{j}^{\infty})^{-1}I\parallel_{L^{2}(\tilde{\Sigma}_{j})}\parallel(\beta_{j}\cdot+S_{j})^{-1}\tilde{w}^{j}\parallel_{L^{2}(\tilde{\Sigma}_{j})},$
(76a)
$\displaystyle|I_{7}|=\int_{\tilde{\Sigma}_{j}}[(1-\tilde{A}_{j}^{\infty})^{-1}I((\beta_{j}s+S_{j})^{-1}\tilde{w}^{j}-S_{j}^{-1}\tilde{w}^{j,\infty})](s,n,t)\mathrm{d}s$
$\displaystyle\quad=\int_{\tilde{\Sigma}_{j}}[(1-\tilde{A}_{j}^{\infty})^{-1}\tilde{A}_{j}I((\beta_{j}s+S_{j})^{-1}\tilde{w}^{j}-S_{j}^{-1}\tilde{w}^{j,\infty})](s,n,t)\mathrm{d}s$
$\displaystyle\quad\quad+\int_{\tilde{\Sigma}_{j}}((\beta_{j}\cdot+S_{j})^{-1}\tilde{w}^{j}-S_{j}^{-1}\tilde{w}^{j,\infty})(s,n,t)\mathrm{d}s$
$\displaystyle\quad\leq\parallel(1-\tilde{A}_{j}^{\infty})^{-1}\parallel_{L^{2}\to
L^{2}}\parallel\tilde{A}_{j}I\parallel_{L^{2}(\tilde{\Sigma}_{j})}\parallel(\beta_{j}\cdot+S_{j})^{-1}\tilde{w}^{j}-S_{j}^{-1}\tilde{w}^{j,\infty}\parallel_{L^{2}(\tilde{\Sigma}_{j})}$
$\displaystyle\quad\quad+\parallel(\beta_{j}\cdot+S_{j})^{-1}\tilde{w}^{j}-S_{j}^{-1}\tilde{w}^{j,\infty}\parallel_{L^{1}(\tilde{\Sigma}_{j})}.$
(76b)

For $I_{6}$, estimating by Lemma 3.10 and $L^{2}$ boundedness of
$\tilde{w}^{j,\infty}$ on $L^{2}(\tilde{\Sigma}_{j})$, we get that

$\displaystyle\parallel(1-\tilde{A}_{j})^{-1}(\tilde{A}_{j}-\tilde{A}_{j}^{\infty})(1-\tilde{A}_{j}^{\infty})^{-1}I\parallel_{L^{2}(\tilde{\Sigma}_{j})}$
$\displaystyle\quad=\parallel(1-\tilde{A}_{j})^{-1}(\tilde{A}_{j}-\tilde{A}_{j}^{\infty})I+(1-\tilde{A}_{j})^{-1}(\tilde{A}_{j}-\tilde{A}_{j}^{\infty})(1-\tilde{A}_{j}^{\infty})^{-1}\tilde{A}_{j}^{\infty}I\parallel_{L^{2}(\tilde{\Sigma}_{j})}$
$\displaystyle\quad\leq\parallel(\tilde{A}_{j}-\tilde{A}_{j}^{\infty})I\parallel_{L^{2}(\tilde{\Sigma}_{j})}+\parallel\tilde{A}_{j}-\tilde{A}_{j}^{\infty}\parallel_{L^{2}\to
L^{2}}\parallel\tilde{A}_{j}^{\infty}I\parallel_{L^{2}(\tilde{\Sigma}_{j})}$
$\displaystyle\quad\lesssim\parallel\tilde{w}^{j}-\tilde{w}^{j,\infty}\parallel_{L^{2}(\tilde{\Sigma}_{j})}+\parallel\tilde{w}^{j}-\tilde{w}^{j,\infty}\parallel_{L^{\infty}(\tilde{\Sigma}_{j})}\parallel\tilde{w}^{j,\infty}\parallel_{L^{2}(\tilde{\Sigma}_{j})}$
$\displaystyle\quad\lesssim
t^{-\frac{1}{4}}(1+\parallel\tilde{w}^{j,\infty}\parallel_{L^{2}(\tilde{\Sigma}_{j})})\lesssim
t^{-\frac{1}{4}};$ (77)

estimate the $L^{2}$-norm by direct computation, the boundedness of
$(\beta_{j}\zeta+S_{j})^{-1}$ on $\zeta\in\tilde{\Sigma}_{j}$, (62), (68),
Lemma 3.8 and definition of $\beta_{j}$

$\displaystyle\parallel(\beta_{j}\cdot+S_{j})^{-1}\tilde{w}^{j}\parallel_{L^{2}(\tilde{\Sigma}_{j})}\lesssim\parallel\tilde{w}^{j}\parallel_{L^{2}(\tilde{\Sigma}_{j})}=\parallel\delta_{j0}^{-\sigma_{3}}N_{j}\check{w}^{j}_{-}\delta_{j0}^{\sigma_{3}}\parallel_{L^{2}(\tilde{\Sigma}_{j})}$
$\displaystyle\quad\lesssim\parallel
N_{j}\check{w}^{j}\parallel_{L^{2}(\tilde{\Sigma}_{j})}=\parallel
w^{j}\parallel_{L^{2}(\Sigma_{j})}|\beta_{j}|^{-\frac{1}{2}}\lesssim 1;$ (78)

therefore, we obtain by (76a), (3.5) and (3.5) that

$\displaystyle|I_{6}|\lesssim t^{-\frac{1}{4}}.$ (79)

For $I_{7}$, similar to Lemma 3.10, we obtain that for $j=1,2$,

$\displaystyle\parallel(\beta_{j}\cdot+S_{j})^{-1}\tilde{w}^{j}-S_{j}^{-1}\tilde{w}^{j,\infty}\parallel_{L^{1}(\tilde{\Sigma}_{j})\cap
L^{2}(\tilde{\Sigma}_{j})}\lesssim t^{-\frac{1}{4}};$

which deduces that

$\displaystyle|I_{7}|\lesssim(\parallel(1-\tilde{A}_{j}^{\infty})^{-1}\parallel_{L^{2}\to
L^{2}}\parallel\tilde{A}_{j}I\parallel_{L^{2}(\tilde{\Sigma}_{j})}+1)t^{-\frac{1}{4}}\lesssim
t^{-\frac{1}{4}},$ (80)

by (76b), the boundedness of $(1-\tilde{A}_{j}^{\infty})^{-1}$ shown in
Section 3.8 and

$\displaystyle\parallel\tilde{A}_{j}I\parallel_{L^{2}(\tilde{\Sigma}_{j})}\lesssim\parallel\tilde{w}^{j}\parallel_{2}\lesssim|\beta_{j}|^{-\frac{1}{2}}\parallel
w^{j}\parallel_{L^{2}(\Sigma_{j})}\lesssim 1.$

Considering (61), (3.5), (79) and (80), we obtain the asymptotic formula

$\displaystyle M^{(2,R)}(0,n,t)=$ $\displaystyle
I+\sum_{j=1}^{2}\frac{\beta_{j}}{2\pi
i}\delta_{j0}^{\sigma_{3}}\int_{\tilde{\Sigma}_{j}}S_{j}^{-1}[(1-\tilde{A}_{j}^{\infty})^{-1}I\tilde{w}^{j,\infty}](s)\mathrm{d}s\delta_{j0}^{-\sigma_{3}}+\mathcal{O}(t^{-\frac{3}{4}}).$
(81)

######  RH problem 3.12.

Find a $2\times 2$ matrix-valued function $M^{L,j}\equiv M^{L,j}(\zeta,n,t)$
such that

  * •

$M^{L,j}(\zeta,n,t)$ is analytic on $\mathbb{C}\setminus\tilde{\Sigma}_{j}$.

  * •

As $\zeta\to\infty$, $M^{L,j}(\zeta,n,t)\sim I$.

  * •

On $\zeta\in\tilde{\Sigma}_{j}$,
$M^{L,j}_{+}(\zeta)=M^{L,j}_{-}(\zeta)V^{L,j}(\zeta)$, where
$V^{L,j}=(I-\tilde{w}^{j,\infty}_{-})^{-1}(I+\tilde{w}^{j,\infty}_{+})$.

We see that RH problem 3.12 is a model RH problem that related to parabolic
cylinder functions, seeing [44]. And we see that the solution of RH problem
3.12 is written in the Beals-Coifman solution:

$\displaystyle M^{L,j}(\zeta)=I+\frac{1}{2\pi
i}\int_{\tilde{\Sigma}_{j}}\frac{[(1-\tilde{A}_{j}^{\infty})^{-1}I\tilde{w}^{j,\infty}](s)}{s-\zeta}\mathrm{d}s$
(82)

If write $M^{L,j}(\zeta,n,t)$ by its asymptotic expansion at $\zeta\to\infty$,

$\displaystyle M^{L,j}(\zeta)=I+\zeta^{-1}M^{L,j}_{1}+\dots,$ (83)

then, we derive that

$\displaystyle M^{L,j}_{1}=\frac{1}{2\pi
i}\int_{\tilde{\Sigma}_{j}}[(1-\tilde{A}_{j}^{\infty})^{-1}I\tilde{w}^{j,\infty}](s)\mathrm{d}s;$
(84)

recalling (81), we obtain that

$\displaystyle M^{(2,R)}(0,n,t)=$ $\displaystyle
I+\sum_{j=1}^{2}\beta_{j}S_{j}^{-1}\delta_{j0}^{\sigma_{3}}M^{L,j}_{1}\delta_{j0}^{-\sigma_{3}}+\mathcal{O}(t^{-\frac{3}{4}}),$
(85)

which means that the solution of RH problem 3.12 describes the leading terms
of solutions of $M^{(2,R)}(0,n,t)$. Seeing [44, 20] and solving RH problem
3.12, we learn that $M^{L,j}_{1}$ is explicitly obtain that

$\displaystyle[M_{1}^{L,j}]_{1,2}=-i\frac{(2\pi)^{\frac{1}{2}}e^{\frac{\pi
i}{4}}e^{\frac{-\pi\nu_{j}}{2}}}{r(S_{j})\Gamma(-i\nu_{j})}.$ (86)

where $\Gamma(\lambda)$ is Euler’s Gamma functions.

###  3.6 Analysis on $\bar{\partial}$-problem

Since we have obtain the long-time asymptotic formula (85), in this part, we
study the solution of $\bar{\partial}$-problem 3.6 as $t\to+\infty$.

The solution of $\bar{\partial}$-problem 3.6 can be written as

$\displaystyle
M^{(2,D)}(\lambda,n,t)=I-\frac{1}{\pi}\iint_{\mathbb{C}}\frac{[M^{(2,D)}\tilde{\mathcal{P}}](s,n,t)}{s-\lambda}\mathrm{d}s.$
(87)

(87) can also be written by operator $S$: $f\mapsto Sf$

$\displaystyle(1-S)M^{(2,D)}=I,\quad
Sf(\lambda)=-\frac{1}{\pi}\iint_{\mathbb{C}}\frac{f(s)\tilde{\mathcal{P}}(s)}{s-\lambda}\mathrm{d}L(s),$
(88)

where $L(s)$ is the Lebesgue distribution on complex plane $\mathbb{C}$. We
prove in the following that $S\in\mathcal{B}(L^{\infty}(\mathbb{C}))$ and as
$t$ is sufficiently large, $\parallel S\parallel_{L^{\infty}\to L^{\infty}}$
decays.

######  Proposition 3.13.

If $r\in H^{1}_{\theta}(\Sigma)$, the operator
$S\in\mathcal{B}(L^{\infty}(\mathbb{C}))$, and as $t\to+\infty$, the norm of
$S$ satisfies

$\displaystyle\parallel S\parallel_{L^{\infty}\to L^{\infty}}\lesssim
t^{-\frac{1}{4}}.$

###### Proof.

As the definition of $\tilde{\mathcal{P}}$, it is supported on
$\cup_{k=1,3,4,6}\Omega_{k}$; therefore, the integral in (88) is only
supported on $\cup_{k=1,3,4,6}\Omega_{6}$. To make the proof concise, we
detail the case for functions supported on
$\Omega_{1}^{-}\overset{\mathrm{def}}{=}\Omega_{1}\cap\\{\mathrm{Re}\lambda<0\\}$;
and the other cases’ proofs are parallel. For $2\times 2$ matrix-valued
functions $f\equiv f(\lambda)$ supported on $\Omega_{1}^{-}$, by (24), (25)
and classical computation, we obtain a series of estimates:

$\displaystyle|Sf(\lambda,n,t)|=\Big{|}\frac{1}{\pi}\iint_{\Omega_{1}^{-}}\frac{f(s)[M^{(2,R)}\bar{\partial}\mathcal{P}(M^{(2,R)})^{-1}](s,n,t)}{s-\lambda}\mathrm{d}L(s)\Big{|}$
$\displaystyle\quad\leq\parallel
f\parallel_{L^{\infty}(\Omega_{1}^{-})}\parallel
M^{(2,R)}\parallel_{L^{\infty}(\Omega_{1}^{-})}\parallel(M^{(2,R)})^{-1}\parallel_{L^{\infty}(\Omega_{1}^{-})}\parallel\delta^{-2}\parallel_{L^{\infty}(\Omega_{1}^{-})}$
$\displaystyle\quad\quad\iint_{\Omega_{1}\cap\\{\mathrm{Re}\lambda>0\\}}\frac{|\bar{\partial}R_{1}(s)|e^{-t\mathrm{Im}\phi(s,n,t)}\mathrm{d}L(s)}{|s-\lambda|}$
$\displaystyle\quad\overset{\leavevmode\hbox to8.59pt{\vbox
to8.59pt{\pgfpicture\makeatletter\hbox{\hskip 4.29594pt\lower-4.29594pt\hbox
to0.0pt{\pgfsys@beginscope\pgfsys@invoke{
}\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{
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}\pgfsys@setlinewidth{0.4pt}\pgfsys@invoke{ }\nullfont\hbox
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} }{{{{}}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@transformcm{1.0}{0.0}{0.0}{1.0}{-2.25pt}{-2.9pt}\pgfsys@invoke{
}\hbox{{\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{
}\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{ }\hbox{{1}}
}}\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}}
\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}} }
\pgfsys@invoke{\lxSVG@closescope
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}\pgfsys@endscope\hss}}\lxSVG@closescope\endpgfpicture}}}{\lesssim}\parallel
f\parallel_{L^{\infty}(\Omega_{1}^{-})}\iint_{\Omega_{1}\cap\\{\mathrm{Re}\lambda>0\\}}\frac{|\bar{\partial}R_{1}(s)|e^{-t\mathrm{Im}\phi(s,n,t)}\mathrm{d}L(s)}{|s-\lambda|}$
$\displaystyle\quad\overset{\leavevmode\hbox to8.59pt{\vbox
to8.59pt{\pgfpicture\makeatletter\hbox{\hskip 4.29594pt\lower-4.29594pt\hbox
to0.0pt{\pgfsys@beginscope\pgfsys@invoke{
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}\pgfsys@setlinewidth{0.4pt}\pgfsys@invoke{ }\nullfont\hbox
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{{}}\hbox{\hbox{{\pgfsys@beginscope\pgfsys@invoke{
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} }{{{{}}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@transformcm{1.0}{0.0}{0.0}{1.0}{-2.25pt}{-2.9pt}\pgfsys@invoke{
}\hbox{{\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{
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\pgfsys@invoke{\lxSVG@closescope
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}\pgfsys@endscope\hss}}\lxSVG@closescope\endpgfpicture}}}{=}\parallel
f\parallel_{L^{\infty}(\Omega_{1}^{-})}\int_{1}^{\rho_{0}}\int_{\frac{\pi}{2}}^{\theta_{\rho}}\frac{|r^{\prime}(\theta)|e^{-t((\rho-\rho^{-1})\sin\theta+2\xi\log\rho)}\rho\mathrm{d}\theta\mathrm{d}\rho}{|\rho
e^{i\theta}-\lambda|}$ $\displaystyle\quad\overset{\leavevmode\hbox
to8.59pt{\vbox to8.59pt{\pgfpicture\makeatletter\hbox{\hskip
4.29594pt\lower-4.29594pt\hbox to0.0pt{\pgfsys@beginscope\pgfsys@invoke{
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} }{{{{}}\pgfsys@beginscope\pgfsys@invoke{
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}\hbox{{\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{
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}\pgfsys@endscope\hss}}\lxSVG@closescope\endpgfpicture}}}{\leq}\parallel
f\parallel_{L^{\infty}(\Omega_{1}^{-})}\int_{1}^{\rho_{0}}\int_{\frac{\pi}{2}}^{\theta_{\rho}}\frac{|r^{\prime}(\theta)|e^{-t((\rho-\rho^{-1})\sin\theta_{\rho}+2\xi\log\rho)}\rho_{0}\mathrm{d}\theta\mathrm{d}\rho}{|\rho
e^{i\theta}-\lambda|}$ $\displaystyle\quad=\parallel
f\parallel_{L^{\infty}(\Omega_{1}^{-})}\int_{1}^{\rho_{0}}e^{-t\mathrm{Im}\phi(\rho
e^{i\theta_{\rho}},n,t)}\rho_{0}\mathrm{d}\rho\int_{\frac{\pi}{2}}^{\theta_{\rho}}\frac{|r^{\prime}(\theta)|}{|\rho
e^{i\theta}-\lambda|}\mathrm{d}\theta$
$\displaystyle\quad\overset{\leavevmode\hbox to8.59pt{\vbox
to8.59pt{\pgfpicture\makeatletter\hbox{\hskip 4.29594pt\lower-4.29594pt\hbox
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} }{{{{}}\pgfsys@beginscope\pgfsys@invoke{
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}\pgfsys@endscope\hss}}\lxSVG@closescope\endpgfpicture}}}{\lesssim}\parallel
f\parallel_{L^{\infty}(\Omega_{1}^{-})}\int_{1}^{\rho_{0}}\frac{e^{-t\mathrm{Im}\phi(\rho
e^{i\theta_{\rho}},n,t)}}{|\rho-|\lambda||^{\frac{1}{2}}}\mathrm{d}\rho\overset{\leavevmode\hbox
to8.59pt{\vbox to8.59pt{\pgfpicture\makeatletter\hbox{\hskip
4.29594pt\lower-4.29594pt\hbox to0.0pt{\pgfsys@beginscope\pgfsys@invoke{
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{{}}\hbox{\hbox{{\pgfsys@beginscope\pgfsys@invoke{
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} }{{{{}}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@transformcm{1.0}{0.0}{0.0}{1.0}{-2.25pt}{-2.9pt}\pgfsys@invoke{
}\hbox{{\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{
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}\pgfsys@endscope\hss}}\lxSVG@closescope\endpgfpicture}}}{\leq}\parallel
f\parallel_{L^{\infty}(\Omega_{1}^{-})}\int_{1}^{\rho_{0}}\frac{e^{-\frac{t\sqrt{1-\xi^{2}}}{2}(\rho-1)^{2}}}{|\rho-|\lambda||^{\frac{1}{2}}}\mathrm{d}\rho$
$\displaystyle\quad\overset{\leavevmode\hbox to8.59pt{\vbox
to8.59pt{\pgfpicture\makeatletter\hbox{\hskip 4.29594pt\lower-4.29594pt\hbox
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} }{{{{}}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@transformcm{1.0}{0.0}{0.0}{1.0}{-2.25pt}{-2.9pt}\pgfsys@invoke{
}\hbox{{\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{
}\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{ }\hbox{{6}}
}}\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}}
\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}} }
\pgfsys@invoke{\lxSVG@closescope
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}\pgfsys@endscope\hss}}\lxSVG@closescope\endpgfpicture}}}{\lesssim}t^{-\frac{1}{4}}\parallel
f\parallel_{L^{\infty}(\Omega_{1}^{-})}.$ (89)

In (89), recalling the boundedness of $\parallel
M^{(2,R)}\parallel_{L^{\infty}(\Omega_{1}^{-})}$,
$\parallel(M^{(2,R)})^{-1}\parallel_{L^{\infty}(\Omega_{1}^{-})}$,
$\parallel\delta^{-1}\parallel_{L^{\infty}(\Omega_{1}^{-})}$, we obtain
estimate 1. For equality 2, $\theta_{\rho}$ is the unique argument such that
$\rho e^{i\theta_{\rho}}\in\Sigma_{11}$; $\rho_{0}$ denotes the modular of $s$
belonging to the arc contained in $L\cap D_{+}$; moreover, for $s=\rho
e^{i\theta}$, we obtain that

$\displaystyle\mathrm{d}L(s)=\rho\mathrm{d}\theta\mathrm{d}\rho,\quad\bar{\partial}=e^{i\theta}(\partial_{\rho}+i\rho^{-1}\partial_{\theta})/2.$

For the triangular function, it follows that

$\displaystyle\sin\theta\geq\sin\theta_{\rho},\quad\theta\in[\frac{\pi}{2},\theta_{\rho}],$

and then, as a consequence, we obtain 3. 4 is the consequence of Schwartz
inequalities, the boundedness of $\parallel r\parallel_{H^{1}}$ and the
elementary fact that

$\displaystyle\int_{\frac{\pi}{2}}^{\theta_{\rho}}\frac{1}{|\rho
e^{i\theta}-\lambda|^{2}}\mathrm{d}\theta\leq\int_{-\pi}^{\pi}\frac{1}{|\rho
e^{i\theta}-\lambda|^{2}}\mathrm{d}\theta=\int_{-\pi}^{\pi}\frac{\mathrm{d}\theta}{(\rho-|\lambda|)^{2}+4\rho|\lambda|\sin^{2}\frac{\theta}{2}}$
$\displaystyle\quad\leq\int_{-\pi}^{\pi}\frac{\mathrm{d}\theta}{(\rho-|\lambda|)^{2}+\frac{4\rho|\lambda|}{\pi^{2}}\theta^{2}}\lesssim|\rho-|\lambda||^{-1}.$

5 is obtained by (31) and the fact that $\rho
e^{i\theta_{\rho}}\in\Sigma_{11}$. We derive 6 by the proof found in [42]. We
complete the proof. ∎

Since $\parallel S\parallel_{L^{\infty}\to L^{\infty}}$ decays as
$t\to+\infty$, for sufficiently large $t$, $(1-S)^{-1}$ belongs to
$\mathcal{B}(L^{\infty}(\mathbb{C}))$: therefore, by (88), we learn that the
solution $M^{(2,D)}$ exists and (87) is written as

$\displaystyle
M^{(2,D)}(\lambda,n,t)=I-\frac{1}{\pi}\iint_{\mathbb{C}}\frac{[(1-S)^{-1}I\mathcal{P}](s,n,t)}{s-\lambda}\mathrm{d}s.$

Estimate $M^{(2,D)}(\lambda,n,t)$ at $\lambda=0$,

$\displaystyle
M^{(2,D)}(0,n,t)=I-\frac{1}{\pi}\iint_{\mathbb{C}}\frac{[(1-S)^{-1}I\mathcal{P}](s,n,t)}{s}\mathrm{d}s.$
(90)

For sufficiently large $t$, using the technique applied on (89), we derive a
series of estimates

$\displaystyle\Big{|}\frac{1}{\pi}\iint_{\Omega_{1}^{-}}\frac{[(1-S)^{-1}I\tilde{\mathcal{P}}](s,n,t)}{s}\mathrm{d}L(s)\Big{|}$
$\displaystyle\quad\leq\frac{\parallel(1-S)^{-1}I\parallel_{L^{\infty}(\Omega_{1}^{-})}}{\pi}\iint_{\Omega_{1}^{-}}\frac{|\tilde{\mathcal{P}}|(s,n,t)\mathrm{d}L(s)}{|s|}\overset{\leavevmode\hbox
to8.59pt{\vbox to8.59pt{\pgfpicture\makeatletter\hbox{\hskip
4.29594pt\lower-4.29594pt\hbox to0.0pt{\pgfsys@beginscope\pgfsys@invoke{
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}\pgfsys@setlinewidth{0.4pt}\pgfsys@invoke{ }\nullfont\hbox
to0.0pt{\pgfsys@beginscope\pgfsys@invoke{ }{
{{}}\hbox{\hbox{{\pgfsys@beginscope\pgfsys@invoke{
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} }{{{{}}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@transformcm{1.0}{0.0}{0.0}{1.0}{-2.25pt}{-2.9pt}\pgfsys@invoke{
}\hbox{{\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{
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}}\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}}
\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}} }
\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope{{{}}}{}{}\hss}\pgfsys@discardpath\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope\hss}}\lxSVG@closescope\endpgfpicture}}}{\lesssim}\iint_{\Omega_{1}^{-}}\frac{|\bar{\partial}\mathcal{P}|(s,n,t)}{|s|}\mathrm{d}L(s)$
$\displaystyle\quad=\int_{1}^{\rho_{0}}\int_{\frac{\pi}{2}}^{\theta_{\rho}}\frac{|r^{\prime}(\theta)|e^{-t\mathrm{Im}\phi(\rho
e^{i\theta},n,t)}}{|\delta^{2}(\rho
e^{i\theta})|}\mathrm{d}\theta\mathrm{d}\rho\lesssim\int_{1}^{\rho_{0}}\int_{\frac{\pi}{2}}^{\theta_{\rho}}|r^{\prime}(\theta)|e^{-t\mathrm{Im}\phi(\rho
e^{i\theta},n,t)}\mathrm{d}\theta\mathrm{d}\rho$
$\displaystyle\quad\overset{\leavevmode\hbox to8.59pt{\vbox
to8.59pt{\pgfpicture\makeatletter\hbox{\hskip 4.29594pt\lower-4.29594pt\hbox
to0.0pt{\pgfsys@beginscope\pgfsys@invoke{
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{{}}\hbox{\hbox{{\pgfsys@beginscope\pgfsys@invoke{
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} }{{{{}}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@transformcm{1.0}{0.0}{0.0}{1.0}{-2.25pt}{-2.9pt}\pgfsys@invoke{
}\hbox{{\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{
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}\pgfsys@endscope\hss}}\lxSVG@closescope\endpgfpicture}}}{\leq}\int_{1}^{\rho_{0}}e^{-t\mathrm{Im}\phi(\rho
e^{i\theta_{\rho}},n,t)}\int_{\frac{\pi}{2}}^{\theta_{\rho}}|r^{\prime}(\theta)|e^{-t(\rho-\rho^{-1})(\theta-\theta_{\rho})\frac{\cos\theta_{\rho}}{2}}\mathrm{d}\theta\mathrm{d}\rho$
$\displaystyle\quad\lesssim
t^{-\frac{1}{2}}\int_{1}^{\rho_{0}}\frac{e^{-\frac{t\sqrt{1-\xi^{2}}}{2}(\rho-1)^{2}}\mathrm{d}\rho}{((1-\rho^{-2})\rho\cos\theta_{\rho})^{\frac{1}{2}}}\overset{\leavevmode\hbox
to8.59pt{\vbox to8.59pt{\pgfpicture\makeatletter\hbox{\hskip
4.29594pt\lower-4.29594pt\hbox to0.0pt{\pgfsys@beginscope\pgfsys@invoke{
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{{}}\hbox{\hbox{{\pgfsys@beginscope\pgfsys@invoke{
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} }{{{{}}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@transformcm{1.0}{0.0}{0.0}{1.0}{-2.25pt}{-2.9pt}\pgfsys@invoke{
}\hbox{{\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{
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}\pgfsys@endscope\hss}}\lxSVG@closescope\endpgfpicture}}}{\lesssim}t^{-\frac{1}{2}}\int_{1}^{\rho_{0}}\frac{e^{-\frac{t\sqrt{1-\xi^{2}}}{2}(\rho-1)^{2}}\mathrm{d}\rho}{((1-\rho^{-2})\frac{\sqrt{1-V_{0}^{2}}}{2})^{\frac{1}{2}}}\lesssim
t^{-\frac{3}{4}}.$ (91)

In addition to proof for (3.6), the sufficiently large $t$ and Proposition
3.13 guarantee the boundedness of
$\parallel(1-S)^{-1}I\parallel_{L^{\infty}(\mathbb{C})}$, which combined with
the boundedness of $\parallel
M^{(2,R)}\parallel_{L^{\infty}(\Omega_{1}^{-})}$,
$\parallel(M^{(2,R)})^{-1}\parallel_{L^{\infty}(\Omega_{1}^{-})}$,
$\parallel\delta^{-1}\parallel_{L^{\infty}(\Omega_{1}^{-})}$ deduces 7; by the
elementary fact that for $\theta\in[\frac{\pi}{2},\theta_{\rho}]$

$\displaystyle\sin\theta\geq\sin\theta_{\rho}+\frac{\cos\theta_{\rho}}{2}(\theta-\theta_{\rho}),$

we obtain 8; recalling $\rho e^{i\theta_{\rho}}\in\Sigma_{11}$ and (32), we
obtain that

$\displaystyle\rho\cos\theta_{\rho}=\mathrm{Re}(\rho
e^{i\theta_{\rho}})>\frac{\sqrt{1-V_{0}^{2}}}{2}>0,$ (92)

which deduces 9. We extend the result in (3.6) to that for the case when
integral regions is $\mathbb{C}$

$\displaystyle\Big{|}\frac{1}{\pi}\iint_{\mathbb{C}}\frac{[(1-S)^{-1}I\tilde{\mathcal{P}}](s,n,t)}{s}\mathrm{d}L(s)\Big{|}\lesssim
t^{-\frac{3}{4}};$ (93)

therefore, combining (90) and (93), we obtain that as $t\to+\infty$,

$\displaystyle M^{(2,D)}\sim I+\mathcal{O}(t^{-\frac{3}{4}}).$ (94)

###  3.7 the long-time asymptotics

Considering (16), (22), (23), (29), (85), (86) and (94), we obtain that

$\displaystyle
q_{n}(t)=\delta^{-1}(0)\sum_{j=1}^{2}\beta_{j}S_{j}^{-1}\delta_{j0}^{2}[M^{L,j}_{1}]_{1,2}+\mathcal{O}(t^{-\frac{3}{4}}),$
$\displaystyle\quad[M_{1}^{L,j}]_{1,2}=-i\frac{(2\pi)^{\frac{1}{2}}e^{\frac{\pi
i}{4}}e^{\frac{-\pi\nu_{1}}{2}}}{r(S_{j})\Gamma(-a)},\quad j=1,2.$

###  3.8 The invertibility of operators

In this part, we come to accomplish this section by proving the boundedness of
some operators. We check the boundedness of these operators one-by-one in this
order:

$\displaystyle(1-\tilde{A}_{j}^{\infty})^{-1}\rightharpoonup(1-\tilde{A}_{j})^{-1}\rightharpoonup(1-A_{j})^{-1}$
$\displaystyle\quad\rightharpoonup(1-C_{w^{j}})^{-1}\rightharpoonup(1-C_{w^{\prime}})^{-1}\rightharpoonup(1-C_{w})^{-1},\quad
j=1,2.$

By taking $\tilde{A}_{j}^{\infty}$ in place of $C_{w^{\infty}}$ in Step 7 of
[20], we obtain that
$(1-\tilde{A}_{j}^{\infty})^{-1}\in\mathcal{B}(L^{2}(\tilde{\Sigma}_{j}))$. By
Proposition 3.11, we learn that

$\displaystyle\parallel\tilde{A}_{j}-\tilde{A}_{j}^{\infty}\parallel_{L^{2}\to
L^{2}}\lesssim\parallel\tilde{w}^{j}-\tilde{w}^{j,\infty}\parallel_{L^{\infty}(\tilde{\Sigma}_{j})}\lesssim
t^{-\frac{1}{4}},$

which combined with the boundedness of $(1-\tilde{A}_{j}^{\infty})^{-1}$
deduces that for sufficiently large $t$,

$\displaystyle(1-\tilde{A}_{j})^{-1}=(1-\tilde{A}_{j}^{\infty})^{-1}(1-(\tilde{A}_{j}-\tilde{A}_{j}^{\infty})(1-\tilde{A}_{j}^{\infty})^{-1})^{-1}\in\mathcal{B}(L^{2}(\tilde{\Sigma}_{j})).$

For $(1-A_{j})^{-1}$, by (67), we obtain htat

$\displaystyle(1-A_{j})^{-1}=N_{j}^{-1}\Delta_{j0}^{-1}(1-\tilde{A}_{j})^{-1}\Delta_{j0}N_{j}\in\mathcal{B}(L^{2}(\Sigma_{j}^{\prime})),\quad
j=1,2,$

because $(1-\tilde{A}_{j})^{-1}\in\mathcal{B}(L^{2}(\tilde{\Sigma}_{j}))$. By
Lemma 2.56 in [22], we learn that since $w^{j}$ is supported on
$\Sigma_{j}\subset\Sigma^{(2)}\cap\Sigma_{j}^{\prime}$, that
$(1-C_{w^{j}})^{-1}\in\mathcal{B}(L^{2}(\Sigma^{(2)}))$ are equivalent to that
$(1-A_{j})^{-1}\in\mathcal{B}(L^{2}(\Sigma_{j}^{\prime}))$. By (3.4),
boundedness of $(1-C_{w^{j}})^{-1}$ and Lemma 3.9, we derive that for
sufficiently large $t$,

$\displaystyle(1-C_{w^{\prime}})^{-1}=(1+C_{w^{1}}(1-C_{w^{1}})^{-1}+C_{w^{2}}(1-C_{w^{2}})^{-1})$
$\displaystyle\quad\times(1-C_{w^{2}}C_{w^{1}}(1-C_{w^{1}})^{-1}-C_{w^{1}}C_{w^{2}}(1-C_{w^{2}})^{-1})^{-1}\in\mathcal{B}(L^{2}(\Sigma^{(2)})).$

By direct computation and Lemma 3.7, we derive that

$\displaystyle\parallel C_{w}-C_{w^{\prime}}\parallel_{L^{2}\to
L^{2}}=\parallel C_{w^{r}}\parallel_{L^{2}\to L^{2}}\lesssim\parallel
w^{r}\parallel_{L^{\infty}(\Sigma^{(2)})}\lesssim t^{-1}.$ (95)

For sufficiently large $t$, by boundedness of $(1-C_{w^{\prime}})^{-1}$ and
(95), it follows that

$\displaystyle(1-C_{w})^{-1}=(1-C_{w^{\prime}})^{-1}(1-(C_{w}-C_{w^{\prime}})(1-C_{w^{\prime}})^{-1})^{-1}\in\mathcal{B}(L^{2}(\Sigma^{(2)})).$

We complete the proof of this part.

##  4 Long-time asymptotic analysis on $\xi\geq V_{0}>1$

In this section, we analyzing asymptotic properties for the case when
$\frac{n}{2t}=\xi\geq V_{0}>1$. In this case, we set
$\epsilon_{0}=\frac{\sqrt{V_{0}^{2}-1}+V_{0}-1}{2}$. We see that the jump
matrix $V$ admits a upper/lower triangular factorization shown in Jump
condition of RH problem. The exponential part in the lower/upper triangular
matrix decays fast as $t\to+\infty$ on the $-/+$ side of the jump contour,
respectively, seeing (a) in Figure 9. According to this factorization,
firstly, we deform the RH problem: $M\rightsquigarrow M^{(3)}$, such that it
admits $\bar{\partial}$-RH problem 4.1 and the new jump matrix decay to $I$ as
$t\to+\infty$; secondly, we deform the solution for $\bar{\partial}$-RH
problem into the product of solutions for RH problem 4.2 and
$\bar{\partial}$-problem 4.3, and analyze long-time asymptotic properties of
these solutions separately. Finally, we obtain the asymptotic property (110)
by the reconstructed formula.

(a)

(b)

Figure 9: The sign of $\mathrm{Im}\phi$ in region II and III.

Introducing a $2\times 2$ matrix-valued function $M^{(3)}\equiv
M^{(3)}(\lambda,n,t)$

$\displaystyle
M^{(3)}=M\mathcal{P}^{1},\quad\mathcal{P}^{1}(\lambda,n,t)=\begin{cases}I&\lambda\in\Omega_{7}\cup\Omega_{10},\\\
\left[\begin{matrix}1&0\\\
R_{8}(\lambda)e^{it\phi(\lambda,n,t)}&1\end{matrix}\right]&\lambda\in\Omega_{8},\\\
\left[\begin{matrix}1&R_{9}(\lambda)e^{-it\phi(\lambda,n,t)}\\\
0&1\end{matrix}\right]&\lambda\in\Omega_{9},\end{cases}$ (96a)
$\displaystyle\quad R_{8}(\lambda)=-r(\theta),\quad
R_{9}(\lambda)=-\overline{r(\theta)},\quad\lambda=|\lambda|e^{i\theta},$ (96b)

and

$\displaystyle w^{(3)}_{+}\equiv
w^{(3)}_{+}(\lambda,n,t)=\begin{cases}\textbf{0},&\lambda\in\Sigma_{-\epsilon_{0}}\\\
\left[\begin{matrix}0&0\\\
-R_{8}(\lambda)e^{it\phi(\lambda,n,t)}&0\end{matrix}\right]&\lambda\in\Sigma_{\epsilon_{0}},\end{cases}$
(97a) $\displaystyle w^{(3)}_{-}\equiv
w^{(3)}_{-}(\lambda,n,t)=\begin{cases}\left[\begin{matrix}0&R_{9}(\lambda)e^{-it\phi(\lambda,n,t)}\\\
0&0\end{matrix}\right]&\lambda\in\Sigma_{-\epsilon_{0}},\\\
\textbf{0}&\lambda\in\Sigma_{\epsilon_{0}},\end{cases}$ (97b)

we obtain a $\bar{\partial}$-RH problem:

######  $\bar{\partial}$-RH problem 4.1.

Find a $2\times 2$ matrix-valued function such that

  * •

$M^{(3)}$ belongs to
$C^{0}(\mathbb{C}\setminus(\Sigma_{-\epsilon_{0}}\cup\Sigma_{\epsilon_{0}}))$
and its first-order partial derivatives are continuous on
$\mathbb{C}\setminus(\Sigma\cup\Sigma_{-\epsilon_{0}}\cup\Sigma_{\epsilon_{0}})$.

  * •

As $\lambda\to\infty$, $M^{(3)}(\lambda,n,t)\sim I+\mathcal{O}(\lambda^{-1})$.

  * •

On $\lambda\in(\Sigma_{\epsilon_{0}}\cup\Sigma_{-\epsilon_{0}})$,
$M^{(3)}_{+}=M^{(3)}_{-}V^{(3)}$,
$V^{(3)}=(1-w^{(3)}_{-})^{-1}(1+w^{(3)}_{+})$.

  * •

On
$\lambda\in\mathbb{C}\setminus(\Sigma\cup\Sigma_{-\epsilon_{0}}\cup\Sigma_{\epsilon_{0}})$,

$\displaystyle\bar{\partial}M^{(3)}=M^{(3)}\bar{\partial}\mathcal{P}^{1},$
$\displaystyle\bar{\partial}\mathcal{P}^{1}(\lambda,n,t)=\begin{cases}\textbf{0}&\lambda\in\Omega_{7}\cup\Omega_{10},\\\
\left[\begin{matrix}0&0\\\
\bar{\partial}R_{8}(\lambda)e^{it\phi(\lambda,n,t)}&0\end{matrix}\right]&\lambda\in\Omega_{8},\\\
\left[\begin{matrix}0&\bar{\partial}R_{9}(\lambda)e^{-it\phi(\lambda,n,t)}\\\
0&0\end{matrix}\right]&\lambda\in\Omega_{9}.\end{cases}$

Figure 10: The jump contour:
$\Sigma=\Sigma_{-\epsilon_{0}}\cup\Sigma_{\sigma_{0}}$.

For sufficiently large $t>0$, we deform the solution for $\bar{\partial}$-RH
problem 4.1 into the product of solutions for RH problem 4.2 and
$\bar{\partial}$ problem 4.3:

$\displaystyle M^{(3)}=M^{(3,D)}M^{(3,R)},$ (98)

where $M^{(3,R)}\equiv M^{(3,R)}(\lambda,n,t)$ admits the same jump condition
as $M^{(3)}$’s and $M^{(3,D)}\equiv M^{(3,D)}(\lambda,n,t)$ is continuous over
complex plane $\mathbb{C}$; then, in the remaining part of this section, by
analyzing long-time asymptotic properties of these solutions, we claim that

$\displaystyle M^{(3)}(0,n,t)\sim I+\mathcal{O}(t^{-1}).$ (99)

Below, we study the following problems.

######  RH problem 4.2.

Find a $2\times 2$ matrix-valued function $M^{(3,R)}$ such that

  * •

$M^{(3,R)}$ is holomorphic on
$\mathbb{C}\setminus(\Sigma_{-\epsilon_{0}}\cup\Sigma_{\epsilon_{0}})$.

  * •

As $\lambda\to\infty$, $M^{(3,R)}(\lambda,n,t)\sim
I+\mathcal{O}(\lambda^{-1})$.

  * •

On $\lambda\in\Sigma_{-\epsilon_{0}}\cup\Sigma_{\epsilon_{0}}$,
$M^{(3,R)}_{+}=M^{(3,R)}_{-}V^{(3)}$.

######  $\bar{\partial}$-problem 4.3.

Find a $2\times 2$ matrix-valued function $M^{(3,D)}$ such that

  * •

$M^{(3,D)}$ is continuous on $\mathbb{C}$, and its first-order partial
derivatives are continuous on
$\mathbb{C}\setminus(\Sigma\cup\Sigma_{-\epsilon_{0}}\cup\Sigma_{\epsilon_{0}})$.

  * •

As $\lambda\to\infty$, $M^{(3,D)}\sim I+\mathcal{O}(\lambda^{-1})$.

  * •

On
$\lambda\in\mathbb{C}\setminus(\Sigma\cup\Sigma_{-\epsilon_{0}}\cup\Sigma_{\epsilon_{0}})$,

$\displaystyle\bar{\partial}M^{(3,D)}=M^{(3,D)}\tilde{\mathcal{P}}^{1},\quad\tilde{\mathcal{P}}^{1}=M^{(3,R)}\bar{\partial}\mathcal{P}^{1}(M^{(3,R)})^{-1}.$

For RH problem 4.2, like what we have done to RH problem 3.5, we obtain the
Beals-Coifman solution for RH problem 4.2,

$\displaystyle M^{(3,R)}(\lambda,n,t)=I+\frac{1}{2\pi
i}\int_{\Sigma^{(3)}}\frac{[(1-C_{w^{(3)}}^{\Sigma^{(3)}})^{-1}Iw^{(3)}](s,n,t)}{s-\lambda}\mathrm{d}s,$
(100) $\displaystyle\quad
C_{w^{(3)}}^{\Sigma^{(3)}}=C_{+}^{\Sigma^{(3)}}(\cdot
w^{(3)}_{-})+C_{-}^{\Sigma^{(3)}}(\cdot w^{(3)}_{+}),\quad
w^{(3)}=w_{+}^{(3)}+w^{(3)}_{-},$ $\displaystyle\quad
C_{\pm}^{\Sigma^{(3)}}f(\lambda)=\lim_{\lambda^{\prime}\to\lambda,\lambda\text{
on the $\pm$ side of $\Sigma^{(3)}$}}\int_{\Sigma^{(3)}}\frac{f(s)}{2\pi
i(s-\lambda^{\prime})}\mathrm{d}s.$

Since $\Sigma_{\epsilon_{0}}$, $\Sigma_{-\epsilon_{0}}$ are compact curves and
contained in the region $\\{\mathrm{Im}\phi>0\\}$, $\\{\mathrm{Im}\phi<0\\}$,
respectively, there is a positive constant $C$ such that

$\displaystyle\mathrm{Im}\phi\big{|}_{\Sigma_{\epsilon_{0}}}\geq
C,\quad\mathrm{Im}\phi\big{|}_{\Sigma_{-\epsilon_{0}}}\leq-C;$ (101)

therefore, by the fact that $r\in H^{1}_{\theta}(\Sigma)\subset
L^{\infty}(\Sigma)$, (96b) and (97), we obtain that as $t\to+\infty$,

$\displaystyle\parallel
w^{(3)}_{\pm}\parallel_{L^{\infty}(\Sigma^{(3)})}\lesssim e^{-Ct};$ (102)

as a result, using the technique applied in (3.4), we deduce that

$\displaystyle\parallel C^{(\Sigma^{(3)})}_{w^{(3)}}\parallel_{L^{2}\to
L^{2}}\lesssim\parallel w^{(3)}\parallel_{L^{\infty}(\Sigma^{(3)})}\lesssim
e^{-Ct}.$ (103)

Considering (100) and (103), we obtain that as $t\to+\infty$,

$\displaystyle M^{(3,R)}(0,n,t)\sim I+\mathcal{O}(e^{-Ct}).$ (104)

For the boundedness of $M^{(3,R)}$, by RH problem 4.2, we find that $\det
M^{(3,R)}$ admits no jump on $\mathbb{C}$, and as $\lambda\to\infty$,

$\displaystyle\lim_{\lambda\to\infty}\det M^{(3,R)}\sim
1+\mathcal{O}(\lambda^{-1});$

therefore, by Liouville’s Theorem, we derive that

$\displaystyle\det M^{(3,R)}\equiv 1;$

moreover, since $M^{(3,R)}$ admits no pole, both $M^{(3,R)}$ and
$(M^{(3,R)})^{-1}$ are bounded on $\mathbb{C}$.

For $\bar{\partial}$-problem 4.3, Like that in Section 3.6, we also have the
solution for $\bar{\partial}$-problem 4.3:

$\displaystyle M^{(3,D)}(\lambda,n,t)=I+[S^{1}M^{(3,D)}](\lambda,n,t),$ (105)

where $S^{1}$: $f\mapsto S^{1}f$ is a linear operator and belongs to
$\mathcal{B}(L^{\infty}(\mathbb{C}))$,

$\displaystyle
S^{1}f(\lambda,n,t)=-\frac{1}{\pi}\iint_{\mathbb{C}}\frac{f(s)\tilde{\mathcal{P}}^{1}(s,n,t)}{s-\lambda}\mathrm{d}L(s).$
(106)

For this operator $S^{1}$, we find it decays as $t\to+\infty$ on
$L^{\infty}(\mathbb{C})$ as shown in Proposition 4.4.

######  Proposition 4.4.

For $r\in H^{1}_{\theta}(\Sigma)$, the integral operator $S^{1}$ belongs to
$\mathcal{B}(L^{\infty}(\mathbb{C}))$, and it satisfies that

$\displaystyle\parallel S^{1}\parallel_{L^{\infty}\to L^{\infty}}\lesssim
t^{-\frac{1}{4}}.$

###### Proof.

We detail the case when functions $f(s)$ are only supported on $\Omega_{8}$.
Like the proof in Proposition 3.13, we obtain a series of estimates:

$\displaystyle|S^{1}f(\lambda)|\leq\frac{\parallel
f\parallel_{L^{\infty}(\Omega_{8})}}{\pi}\parallel
M^{(3,R)}\parallel_{L^{\infty}(\Omega_{8})}\parallel(M^{(3,R)})^{-1}\parallel_{L^{\infty}(\Omega_{8})}$
$\displaystyle\quad\quad\iint_{\Omega_{8}}\frac{|\bar{\partial}R_{8}(s)|e^{-t\mathrm{Im}\phi(s,n,t)}}{|s-\lambda|}\rho\mathrm{d}L(s)$
$\displaystyle\quad\overset{\leavevmode\hbox to11.94pt{\vbox
to11.94pt{\pgfpicture\makeatletter\hbox{\hskip 5.9688pt\lower-5.9688pt\hbox
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{{}}\hbox{\hbox{{\pgfsys@beginscope\pgfsys@invoke{
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}\pgfsys@endscope\hss}}\lxSVG@closescope\endpgfpicture}}}{\lesssim}\parallel
f\parallel_{L^{\infty}(\Omega_{8})}\int_{1}^{1+\epsilon_{0}}\int_{0}^{2\pi}\frac{|\bar{\partial}R_{8}(s)|e^{-t(2\xi\ln\rho-(\rho-\rho^{-1}))}}{|\rho
e^{i\theta}-\lambda|}\rho\mathrm{d}\theta\mathrm{d}\rho$
$\displaystyle\quad=\parallel
f\parallel_{L^{\infty}(\Omega_{8})}\int_{1}^{1+\epsilon_{0}}\Big{|}\frac{r^{\prime}(\theta)}{\rho
e^{i\theta}-\lambda}\Big{|}\mathrm{d}\theta\int_{0}^{2\pi}e^{-t(2\xi\ln\rho-(\rho-\rho^{-1}))}\rho\mathrm{d}\rho$
$\displaystyle\quad\overset{\leavevmode\hbox to11.94pt{\vbox
to11.94pt{\pgfpicture\makeatletter\hbox{\hskip 5.9688pt\lower-5.9688pt\hbox
to0.0pt{\pgfsys@beginscope\pgfsys@invoke{
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to0.0pt{\pgfsys@beginscope\pgfsys@invoke{ }{
{{}}\hbox{\hbox{{\pgfsys@beginscope\pgfsys@invoke{
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} }{{{{}}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@transformcm{1.0}{0.0}{0.0}{1.0}{-4.50002pt}{-2.9pt}\pgfsys@invoke{
}\hbox{{\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{
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}}\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}}
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\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope{{{}}}{}{}\hss}\pgfsys@discardpath\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope\hss}}\lxSVG@closescope\endpgfpicture}}}{\leq}\parallel
f\parallel_{L^{\infty}(\Omega_{8})}\parallel
r^{\prime}\parallel_{L^{2}_{\theta}(\sigma)}\int_{0}^{2\pi}\parallel\frac{1}{\rho
e^{i\theta}-\lambda}\parallel_{L^{2}(\\{|s|=\rho\\})}e^{-2t(V_{0}\frac{\ln(1+\epsilon_{0})}{\epsilon_{0}}-1)(\rho-1)}\rho\mathrm{d}\rho$
$\displaystyle\quad\lesssim\parallel
f\parallel_{L^{\infty}(\Omega_{8})}\int_{0}^{2\pi}\frac{e^{-2t(V_{0}\frac{\ln(1+\epsilon_{0})}{\epsilon_{0}}-1)(\rho-1)}}{|\rho-|\lambda||^{\frac{1}{2}}}\rho\mathrm{d}\rho\lesssim
t^{-\frac{1}{4}}\parallel f\parallel_{L^{\infty}(\Omega_{8})}.$ (107)

In (4), by the fact that for $s=\rho e^{i\theta}$,

$\displaystyle\mathrm{Im}\phi(s,n,t)=(\rho-\rho^{-1})\sin\theta+2\xi\ln\rho\geq
2\xi\ln\rho-(\rho-\rho^{-1}),$

which with the boundedness of $\parallel
M^{(3,R)}\parallel_{L^{\infty}(\Omega_{8})}$ and
$\parallel(M^{(3,R)})^{-1}\parallel_{L^{\infty}(\Omega_{8})}$ deduces 10; 11
is by Schwartz Inequalities and the fact that on the region $\Omega_{8}$,
$\rho>1$ such that

$\displaystyle
2\xi\ln\rho-\rho+\rho^{-1}=\left(\frac{2\xi\ln\rho}{\rho-1}-\left(1+\frac{1}{\rho}\right)\right)(\rho-1)\geq
2(\rho-1)\left(\frac{V_{0}\ln(1+\epsilon_{0})}{\epsilon_{0}}-1\right)$

where $\frac{V_{0}\ln(1+\epsilon_{0})}{\epsilon_{0}}-1$ is a positive number
because of the choosing of $\epsilon_{0}$ and $V_{0}>1$. We complete the proof
for $f(s)$ only supported on $\Omega_{8}$, and the proof for general $f(s)\in
L^{\infty}(\mathbb{C})$ is similarly obtained. ∎

From Proposition 4.4, we see that for sufficiently large $t$,
$(1-S^{1})^{-1}I$ exists and is bounded on $L^{\infty}(\mathbb{C})$;
therefore, we obtain that

$\displaystyle\Big{|}\frac{1}{\pi}\iint_{\Omega_{8}}\frac{[(1-S)^{-1}I\tilde{\mathcal{P}}^{1}](s,n,t)}{s}\mathrm{d}L(s)\Big{|}$
$\displaystyle\quad\leq\frac{1}{\pi}\parallel(1-S)^{-1}I\parallel_{L^{\infty}(\mathbb{C})}\parallel
M^{(3,R)}\parallel_{L^{\infty}(\Omega_{8})}\parallel(M^{(3,R)})^{-1}\parallel_{L^{\infty}(\Omega_{8})}$
$\displaystyle\quad\quad\iint_{\Omega_{8}}\frac{|\bar{\partial}R_{8}(s)|e^{-t\mathrm{Im}\phi(s,n,t)}}{|s|}\rho\mathrm{d}L(s)$
$\displaystyle\quad\lesssim\iint_{\Omega_{8}}\frac{|\bar{\partial}R_{8}(s)|e^{-t\mathrm{Im}\phi(s,n,t)}}{|s|}\rho\mathrm{d}L(s)=\int_{1}^{1+\epsilon_{0}}\int_{0}^{2\pi}\frac{|r^{\prime}(\theta)|\mathrm{d}\theta\mathrm{d}\rho}{e^{t((\rho-\frac{1}{\rho})\sin\theta+2\xi\ln\rho)}}$
$\displaystyle\quad\leq\parallel
r^{\prime}\parallel_{L^{1}(\Sigma)}\int_{1}^{1+\epsilon_{0}}e^{-t(2\xi\ln\rho-\rho+\rho^{-1})}\mathrm{d}\rho$
$\displaystyle\quad\leq\parallel
r\parallel_{L^{1}(\Sigma)}\int_{1}^{1+\epsilon_{0}}e^{-2t(V_{0}\frac{\ln(1+\epsilon_{0})}{\epsilon_{0}}-1)(\rho-1)}\mathrm{d}\rho\lesssim
t^{-1};$ (108)

similar to (4), we obtain the result for integral region $\mathbb{C}$ that

$\displaystyle\Big{|}\frac{1}{\pi}\iint_{\mathbb{C}}\frac{[(1-S)^{-1}I\tilde{\mathcal{P}}^{1}](s,n,t)}{s}\mathrm{d}L(s)\Big{|}\lesssim
t^{-1}.$ (109)

Considering (98), (104), (105), (106) and (109), we prove (99); therefore, by
(16), (96a) and (99), we obtain that

$\displaystyle q_{n}(t)\sim\mathcal{O}(t^{-1}).$

##  5 Long-time asymptotic analysis on $\xi\leq-V_{0}<-1$

In the last section, we analyze the case of $\xi\leq-V_{0}<-1$. Introducing a
$2\times 2$ matrix-valued function

$\displaystyle
M^{(4)}=M\tilde{\delta}^{-\sigma_{3}},\quad\tilde{\delta}\equiv\tilde{\delta}(\lambda)=e^{\frac{1}{2\pi
i}\int_{\Sigma}\frac{\ln(1-|r(s)|^{2})\mathrm{d}s}{s-\lambda}}$ (110)

we claim that $\tilde{\delta}$ satisfies Proposition 5.1; moreover, by
Proposition 5.1 and RH problem 2.1, we obtain that $M^{(4)}$ admits RH problem
5.2. We see in RH problem 5.2 that on $\lambda\in\Sigma$, the jump matrix
admits a lower/upper triangular factorization

$\displaystyle V^{(4)}=(I-w^{(4)}_{-})^{-1}(I-w^{(4)}_{+}),$ (111)
$\displaystyle\quad w^{(4)}_{-}\equiv
w^{(4)}_{-}(\lambda,n,t)=\left[\begin{matrix}0&0\\\
\frac{\delta_{-}^{-2}(\lambda)r(\lambda)}{1-|r(\lambda)|^{2}}e^{it\phi(\lambda,n,t)}&0\end{matrix}\right],$
$\displaystyle\quad w^{(4)}_{+}\equiv
w^{(4)}_{+}(\lambda,n,t)=\left[\begin{matrix}0&-\frac{\delta_{+}^{2}(\lambda)\overline{r(\lambda)}}{1-|r(\lambda)|^{2}}e^{-it\phi(\lambda,n,t)}\\\
0&0\end{matrix}\right],$

and the exponential part of $w^{(4)}_{+}$, $w^{(4)}_{-}$ decay exponentially
on regions to the $+$, $-$ side of $\Sigma$ as $t\to+\infty$, respectively,
seeing (b) in Figure 9; therefore, similar to the case of $\xi\geq V_{0}>1$ we
obtain the long-time asymptotics when $\xi\leq-V_{0}<-1$, and that

$\displaystyle q_{n}(t)\sim\mathcal{O}(t^{-1}).$

######  Proposition 5.1.

Since $r\in H^{1}_{\theta}(\Sigma),$the scalar function $\tilde{\delta}$
satisfies:

  1. (a)

$\tilde{\delta}$ is analytic on $\mathbb{C}\setminus\Sigma$.

  2. (b)

As $\lambda\to\infty$,

$\displaystyle\tilde{\delta}(\lambda)\sim 1+O(\lambda^{-1}).$

  3. (c)

On the unit circle $\lambda\in\Sigma$,

$\displaystyle\tilde{\delta}_{+}(\lambda)=\tilde{\delta}_{-}(\lambda)(1-|r(\lambda|^{2})).$

  4. (d)

$\tilde{\delta}$ admits the symmetry:

$\displaystyle\tilde{\delta}(\lambda)=\overline{\tilde{\delta}(0)/\tilde{\delta}(\bar{\lambda}^{-1})}.$

###### Proof.

This proof is parallel to that for Proposition 3.1. ∎

######  RH problem 5.2.

Find a $2\times 2$ matrix-valued function $M^{(4)}$ such that

  * •

Analyticity: $M^{(1)}$ is analytic on $\mathbb{C}\setminus\Sigma$.

  * •

Normalization: As $\lambda\to\infty$,

$\displaystyle M^{(4)}(\lambda,n,t)\sim I+O(\lambda^{-1}).$

  * •

Jump condition: On $\lambda\in\Sigma$,

$\displaystyle M^{(4)}_{+}=M^{(4)}_{-}V^{(4)},$ $\displaystyle\quad
V^{(4)}\equiv
V^{(3)}(\lambda,n,t)=\left[\begin{matrix}1&-\frac{\tilde{\delta}_{+}^{2}(\lambda)\overline{r(\lambda)}}{1-|r(\lambda)|^{2}}e^{-it\phi(\lambda,n,t)}\\\
\frac{\tilde{\delta}_{-}^{-2}(\lambda)r(\lambda)}{1-|r(\lambda)|^{2}}e^{it\phi(\lambda,n,t)}&1-|r(\lambda)|^{2}\end{matrix}\right].$

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