# Higher order Airy and Painlevé asymptotics
for the mKdV Hierarchy

Lin Huang† and Lun Zhang‡ †School of Science, Hangzhou Dianzi University,
Zhejiang 310018, China.
‡School of Mathematical Sciences and Shanghai Key Laboratory for Contemporary
Applied Mathematics, Fudan University,
Shanghai 200433, China<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

In this paper, we consider Cauchy problem for the modified Korteweg-de Vries
hierarchy on the real line with decaying initial data. Using the
Riemann–Hilbert formulation and nonlinear steepest descent method, we derive a
uniform asymptotic expansion to all orders in powers of $t^{-1/(2n+1)}$ with
smooth coefficients of the variable $(-1)^{n+1}x((2n+1)t)^{-1/(2n+1)}$ in the
self-similarity region for the solution of $n$-th member of the hierarchy. It
turns out that the leading asymptotics is described by a family of special
solutions of the Painlevé II hierarchy, which generalize the classical
Ablowitz-Segur solution for the Painelvé II equation and appear in a variety
of random matrix and statistical physics models. We establish the connection
formulas for this family of solutions. In the special case that the reflection
coefficient vanishes at the origin, the solutions of Painlevé II hierarchy in
the leading coefficient vanishes as well, the leading and subleading terms in
the asymptotic expansion are instead given explicitly in terms of derivatives
of the generalized Airy function.

AMS Subject Classification (2010): 37K15, 41A60, 35Q15, 35Q53.

Keywords: Long-time asymptotics, modified Korteweg-de Vries hierarchy,
Painlevé II transcendents, Riemann–Hilbert problems, nonlinear steepest
descent method.

###### Contents

  1. 1 Introduction and main results
  2. 2 Preliminaries
  3. 3 Asymptotic analysis of the RH problem for $\Psi$
  4. 4 Asymptotic analysis of the RH problem for $m$: Case I
  5. 5 Asymptotic analysis of the RH problem for $m$: Case II
  6. 6 Proofs of main results
  7. A Model RH problems

##  1\. Introduction and main results

In a seminal work [18], Deift and Zhou introduced the celebrated nonlinear
steepest descent method to analyze asymptotics of oscillatory Riemann–Hilbert
(RH) problems. This approach deals with the RH problem directly and consists
of various contour deformations, following the spirit of classical steepest
descent method. Compared with the inverse scattering method [38], it does not
require any priori ansatz for the asymptotic form of the solution. Since the
solutions for a variety of integrable nonlinear differential equations are
closely related to RH problems, the nonlinear steepest descent method and its
variants has been applied successfully to resolve many asymptotic problems
arising from integrable systems.

The illustrative example used in [18] is the modified Korteweg-de Vries (mKdV)
equation

$u_{t}-6u^{2}u_{x}+u_{xxx}=0,\quad x\in\mathbb{R},\quad t>0,$ (1.1)

with Schwartz space initial data. Long-time asymptotics of the solution, which
depends explicitly on the reflection coefficient associated with the initial
data, is presented in six regions of the $(x,t)$-plane. In the similarity
region $-M_{1}\leq x/t\leq-M_{2}$, $M_{1,2}>0$, the mKdV equation can be
solved to any fixed order $\mathcal{O}(1/t^{n})$,
$n\in\mathbb{N}=\\{1,2,\ldots\\}$, and the leading asymptotics is described by
a slowly decaying modulated sine wave. The full asymptotic expansion in this
region is later derived in [17], and each higher order coefficient therein can
be calculated recursively. In the self-similarity region $|x|\leq Mt^{1/3}$,
$M>0$, leading asymptoics of the solution, however, is given in terms of the
Ablowitz-Segur solution [2, 36] for the homogeneous Painelvé II equation

$q^{\prime\prime}(x)-2q(x)^{3}=xq(x),$ (1.2)

which is determined by the reflection coefficient. This result has recently
been improved in a paper of Charlier and Lenells [9] by showing that the
solution actually admits a uniform expansion to all orders in powers of
$t^{-1/3}$ with smooth coefficients. Moreover, if the reflection coefficient
vanishes at the origin, they derived the leading and subleading terms in the
expansion explicitly with the aid of the classical Airy function. Besides
these studies of mKdV equation on the real line, we also refer to [6, 29] for
the asymptotic results on the half-line.

In this paper, we are concerned with the mKdV hierarchy [12] which is defined
by

$\displaystyle u_{t}+\frac{\partial}{\partial
x}\bigg{(}\frac{\partial}{\partial
x}+2u\bigg{)}\mathcal{L}_{n}[u_{x}-u^{2}]=0,\quad n\in\mathbb{N},$ (1.3)

where the operator $\mathcal{L}_{n}$ satisfies the Lenard recursion relation
[28]:

$\displaystyle\begin{cases}\frac{d}{dx}\mathcal{L}_{j+1}f=\bigg{(}\frac{d^{3}}{dx^{3}}+4f\frac{d}{dx}+2f_{x}\bigg{)}\mathcal{L}_{j}f,\\\
\mathcal{L}_{0}f=\frac{1}{2},\quad\mathcal{L}_{j}0=0,\quad
j=1,2,\cdots.\end{cases}$ (1.4)

If $n=1$ in (1.3), one recovers (1.1), and the equation for $n=2$ reads

$\displaystyle
u_{t}-10u^{2}u_{xxx}-40uu_{x}u_{xx}-10u_{x}^{3}+30u^{4}u_{x}+u_{xxxxx}=0.$

Following the spirit in [9], emphasis will be put on the higher order
Painlevé-type asymptotics for the mKdV hierarchy in the self-similarity region
with initial data $u(x,0)=u_{0}(x)\in\mathcal{S}({\mathbb{R}})$, where
$\mathcal{S}({\mathbb{R}})$ is the Schwartz class of smooth rapidly decaying
functions. As we will show later, the role played by the Ablowitz-Segur
solution and Airy function in the mKdV equation will be replaced by their
higher-order generalizations. In the literatures, we note that the Painlevé
transcendents and their higher-order analogues are crucial in asymptotic
analysis of many integrable nonlinear differential equations, as can be seen
from their appearances in the focusing nonlinear Schrödinger equation [4, 5],
in critical asymptotics for Hamiltonian perturbations of hyperbolic and
elliptic systems [10], in the Camassa-Holm equation [7], in the Sasa-Satsuma
equation [24], in an extended mKdV equation [30, 31] and in the sine-Gordon
equation [32]. The higher order asymptotics in similarity region for other
integrable equations can be found in [26, 37].

Main results of this paper are stated in what follows.

###  1.1. Main Results

To state our results, we start with the Painlevé II hierarchy, which is a
sequence of ordinary differential equations obtained from equations of the
mKdV hierarchy via self-similar reduction [20]; see also [13, 27, 33]. The
$n$-th member of the Painlevé II hierarchy is a non-linear ordinary
differential equation of order $2n$ for $q=q(x)$, and reads as

$\left(\frac{d}{dx}+2q\right)\mathcal{L}_{n}[q_{x}-q^{2}]=xq,\qquad
n\in\mathbb{N},$ (1.5)

where the Lenard operators are defined in (1.4). The Painlevé II equation
(1.2) corresponds to $n=1$ in (1.5), while for $n=2$, we have

$q^{\prime\prime\prime\prime}(x)-10q(x)(q^{\prime}(x))^{2}-10q(x)^{2}q^{\prime\prime}(x)+6q(x)^{5}=xq(x).$

In the most general case, the Painlevé II hierarchy depends on several
parameters, and the one in (1.5) corresponds to the case that all the
parameters are taken to be zero.

As is well-known, the solutions to Painlevé equations and hierarchy are
transcendental in general, hence, we cannot expect simple closed forms for the
solutions. By assuming that $q((-1)^{n+1}x)$ tends to zero exponentially fast
as $x\to+\infty$, it is readily seen that the $n$-th member of the Painlevé II
hierarchy (1.5) is approximated by the generalized Airy equation

$\frac{d^{2n}}{dx^{2n}}y(x)=xy(x).$ (1.6)

If $n=1$, the above equation is nothing but the classical Airy equation. It is
straightforward to check that the function $\text{Ai}_{2n+1}((-1)^{n+1}x)$
with

$\text{Ai}_{2n+1}(x):=\frac{(-1)^{n+1}}{2\pi
i}\int_{\gamma}e^{(-1)^{n}\frac{s^{2n+1}}{2n+1}+xs}ds$ (1.7)

solves (1.6), where $\gamma$ is a curve in the left half of the complex plane
that is asymptotic to straight lines with arguments $\pm\frac{n+1}{2n+1}\pi$
at infinity with the orientation from the bottom to the top. Note that
$\text{Ai}_{3}(x)=\frac{1}{2\pi i}\int_{\gamma}e^{-\frac{s^{3}}{3}+xs}ds$ is
the standard Airy function $\text{Ai}(x)$ [35].

Let $-1<\rho<1$ be a real number, it has recently been shown in [8] that each
of the $n$-th member of the Painlevé II hierarchy (1.5) admits a one-parameter
family of real solutions111In [8], the authors only considered the case that
$0<\rho<1$, but it is clear that the arguments therein can be extended to
$-1<\rho<1$. denoted by $q_{\textrm{AS},n}(x;\rho)$ that are pole-free on the
real line with the asymptotics

$q_{\textrm{AS},n}((-1)^{n+1}x;\rho)\sim\rho\text{Ai}_{2n+1}(x),\qquad
x\to+\infty,$

where $\text{Ai}_{2n+1}(x)$ is defined in (1.7). This family of solutions
plays an important role in multicritical edge statistics for the momenta of
fermions in nonharmonic traps [19] and other statistical physics model [3].
They are natural generalizations of the classical real Ablowitz-Segur
solutions for the Painelvé II equation, and are determined by the Stokes
multipliers

$s_{1}=-s_{2n+1}=\rho,\qquad s_{2}=\cdots=s_{2n}=0.$ (1.8)

Since $q_{\textrm{AS},n}$ will also appear in long-time asymptotics of the
mKdV hierarchy, our first result is the following asymptotics of
$q_{\textrm{AS},n}((-1)^{n+1}x;\rho)$ as $x\to-\infty$. This particularly
establishes the so-called connection formulas for this special family of
solutions.

######  Theorem 1.1.

For each of the $n$-th member of the Painlevé II hierarchy (1.5), there exists
a one-parameter family of real solutions $q_{\textrm{AS},n}(x;\rho)$ such that
as $x\to+\infty$,

$q_{\textrm{AS},n}((-1)^{n+1}x;\rho)=\rho\text{Ai}_{2n+1}(x)(1+o(1)),\qquad-1<\rho<1,$
(1.9)

where $\text{Ai}_{2n+1}(x)$ is defined in (1.7), and as $x\to-\infty$,

$q_{\textrm{AS},n}((-1)^{n+1}x;\rho)=\frac{\,\mathrm{d}}{\sqrt{n}(-x)^{\frac{2n-1}{4n}}}\cos\left(\frac{2n}{2n+1}(-x)^{\frac{2n+1}{2n}}-\frac{2n+1}{4n}\,\mathrm{d}^{2}\ln(-x)+\varphi\right)\\\
+\mathcal{O}((-x)^{-1}),$ (1.10)

where the constants $\,\mathrm{d}$ and $\varphi$ are related to the parameter
$\rho$ through the connection formulas

$\displaystyle\,\mathrm{d}$
$\displaystyle=\sqrt{-\frac{\ln(1-\rho^{2})}{\pi}},$ (1.11)
$\displaystyle\varphi$
$\displaystyle=-\frac{\,\mathrm{d}^{2}}{2}\ln(8n)+\arg\Gamma\left(\frac{\,\mathrm{d}^{2}}{2}i\right)+\frac{\pi}{2}\mathrm{sgn}(\rho)-\frac{\pi}{4},$
(1.12)

with $\Gamma(z)$ being the Gamma function.

If $n=1$, Theorem 1.1 is due to Ablowitz and Segur [2, 36]; see also [14, 16,
23] for rigorous derivations using different methods.

We are now ready to state long-time asymptotics of the mKdV hierarchy. It
comes out that, in the self-similarity region $|x|<Ct^{1/(2n+1)}$, where
$C=C(n)$ is a positive constant for fixed $n$, the solution of the mKdV
hierarchy (1.3) with Schwartz space initial data admits a uniform expansion to
all orders in powers of $t^{-1/(2n+1)}$ with smooth coefficients. Furthermore,
the leading coefficient is described by the generalized Ablowitz-Segur
solution $q_{\textrm{AS},n}$ with the parameter explicitly determined by the
reflection coefficient according to the inverse scattering transform on the
real line.

######  Theorem 1.2.

Let $u(x,t)$ be the solution for each of the $n$-th member of the mKdV
hierarchy (1.3) with initial condition
$u_{0}(x)=u(x,0)\in\mathcal{S}({\mathbb{R}})$. As $t\to\infty$, we have

$\displaystyle
u(x,t)=\sum_{j=1}^{N}\frac{u_{j}(y)}{t^{\frac{j}{2n+1}}}+\mathcal{O}\big{(}t^{-\frac{N+1}{2n+1}}\big{)},\qquad
y\doteq(-1)^{n+1}\frac{x}{((2n+1)t)^{\frac{1}{2n+1}}},$ (1.13)

uniformly for $|x|\leq Ct^{1/(2n+1)}$ with fixed $C>0$ and $N\geq 1$, where
$\\{u_{j}(y)\\}_{j\in{\mathbb{N}}}$ are smooth functions of $y\in{\mathbb{R}}$
and

$\displaystyle
u_{1}(y)=(2n+1)^{-\frac{1}{2n+1}}q_{\textrm{AS},n}((-1)^{n+1}y,ir(0)).$ (1.14)

In (1.14), $q_{\textrm{AS},n}$ is the generalized Ablowitz-Segur solution of
the Painlevé II hierarchy (1.5) as stated in Theorem 1.1, and $r(\lambda)$ is
the reflection coefficient associated with the initial date $u_{0}$.

We shall see existence of the solution for each $n$-th member of the mKdV
hierarchy with Schwartz class initial data from Lemma 2.3 below.

By substituting (1.13) into (1.3) and comparing the coefficients of powers of
$t^{-1/(2n+1)}$, it follows that the coefficients $u_{j}(y)$, $j=2,3,\ldots$,
in (1.13) satisfy coupled differential equations, which in general cannot be
solved explicitly. If the reflection coefficient vanishes at the origin,
however, one can calculate the first few terms as shown in the following
theorem.

######  Theorem 1.3.

If $r(0)=0$, then the asymptotic formula (1.13) still holds with

$\displaystyle u_{1}(y)\equiv 0,\quad
u_{2}(y)=\frac{r^{\prime}(0)}{2\times(2n+1)^{\frac{2}{2n+1}}}\text{Ai}^{\prime}_{2n+1}(y),$
(1.15)

and

$\displaystyle
u_{3}(y)=-\frac{ir^{\prime\prime}(0)}{8\times(2n+1)^{\frac{3}{2n+1}}}\text{Ai}^{{}^{\prime\prime}}_{2n+1}(y),$
(1.16)

where the $\text{Ai}_{2n+1}$ is the generalized Airy function defined in
(1.7).

Theorems 1.2 and 1.3 extend the results for the mKdV equation (1.1) in [9] to
the mKdV hierarchy (1.3). For $n=2$, the leading asymptotics in (1.13) is also
known in [31].

###  1.2. Organization of the paper and notation

The rest of this paper is devoted to the proofs of our main results, which
rely on the Deift/Zhou steepest descent analysis of the associated RH problems
and a technique introduced by Charlier and Lenells in [9] to derive the higher
order asymptotic expansion. In Section 2, we present RH representations for
the mKdV hierarchy and the Painlevé II hierarchy, respectively. Asymptotic
analysis of these RH problems are scattered in Sections 3–5. To analyse the RH
problem for the mKdV hierarchy, it is necessary to divide the discussion into
several different cases. The asymptotic outcomes will finally lead to the
proofs of our main results, i.e., Theorems 1.1–1.3, as shown in Section 6.

We conclude this section with some notation used throughout this paper.

  * •

If $A$ is an $n\times m$ matrix, the $A_{ij}$ stands for the $(i,j)$-th entry
of $A$. We define $|A|\geq 0$ by $|A|^{2}=\sum_{i,j}|A_{ij}|^{2}$. It is then
easily seen that $|A+B|\leq|A|+|B|$ and $|AB|\leq|A||B|$.

  * •

For a (piecewise smooth) contour $\gamma\subset{\mathbb{C}}$ and $1\leq
p\leq\infty$, we write $A\in L^{p}(\gamma)$ if $|A|$ belongs to
$L^{p}(\gamma)$. $A\in L^{p}(\gamma)$ if and only if each entry $A_{ij}$
belongs to $L^{p}(\gamma)$. We also define
$\|A\|_{L^{p}(\gamma)}:=\||A|\|_{L^{p}(\gamma)}$.

  * •

For a complex-valued function $f(k)$ of $k\in{\mathbb{C}}$, we use

$f^{*}(k):=\overline{f(\bar{k})}$ (1.17)

to denote its Schwartz conjugate.

  * •

As usual, the three Pauli matrices $\\{\sigma_{j}\\}_{j=1}^{3}$ are defined by

$\sigma_{1}=\begin{pmatrix}0&1\\\
1&0\end{pmatrix},\quad\sigma_{2}=\begin{pmatrix}0&-i\\\
i&0\end{pmatrix},\quad\sigma_{3}=\begin{pmatrix}1&0\\\ 0&-1\end{pmatrix}.$
(1.18)

  * •

Let $D$ be an open connected subset of ${\mathbb{C}}$ bounded by a piecewise
smooth curve $\gamma\subset\hat{{\mathbb{C}}}:={\mathbb{C}}\cup\\{\infty\\}$
and $z_{0}\in{\mathbb{C}}\setminus\bar{D}$. We use $\dot{E}^{p}(D)$, $1\leq
p<\infty$, to denote the space of all analytic functions $f:D\to{\mathbb{C}}$
with the property that there exist piecewise smooth curves
$\\{C_{n}\\}_{n=1}^{\infty}$ in $D$ tending to $\gamma$ in the sense that
$C_{n}$ eventually surrounds each compact subset of $D$ and such that

$\sup_{n\geq 1}\int_{C_{n}}|z-z_{0}|^{p-2}|f(z)|^{p}|dz|<\infty.$

If $D=D_{1}\cup\cdots\cup D_{n}$ is a finite union of such open subsets, then
$\dot{E}^{p}(D)$ denotes the space of analytic functions $f:D\to{\mathbb{C}}$
such that $f|_{D_{j}}\in\dot{E}^{p}(D_{j})$ for each $j=1,\ldots,n$.

  * •

For a (piecewise smooth) oriented contour $\gamma\subset\hat{{\mathbb{C}}}$
and a function $h$ defined on $\gamma$, the Cauchy transform of $h$ is defined
by

$\displaystyle(\mathcal{C}h)(z)=\frac{1}{2\pi
i}\int_{\gamma}\frac{h(z^{\prime})dz^{\prime}}{z^{\prime}-z},\qquad
z\in{\mathbb{C}}\setminus\gamma,$

whenever the integral converges. If $h\in L^{2}(\gamma)$, the left and right
non-tangential boundary values of $\mathcal{C}h$ exist a.e. on $\gamma$ and
belong to $L^{2}(\gamma)$, which we denote by $\mathcal{C}_{+}h$ and
$\mathcal{C}_{-}h$, respectively. Moreover,
$\mathcal{C}_{\pm}\in\mathcal{B}(L^{2}(\gamma))$, where
$\mathcal{B}(L^{2}(\gamma))$ is the space of bounded linear operators on
$L^{2}(\gamma)$, and by the Sokhotski-Plemelj relation, it follows that
$\mathcal{C}_{+}-\mathcal{C}_{-}=I$,

  * •

In what follows, it is understood that most of the quantities encountered
should depend on a parameter $n$, which corresponds to the $n$-th member of
the mKdV hierarchy or the Painlevé II hierarchy. For simplicity, we omit this
dependence unless otherwise specified.

##  2\. Preliminaries

###  2.1. Lax Pair for the mKdV hierarchy

According to [1] (see also [12, Proposition 1]), the Lax pair for $n$-th
equation of the mKdV hierarchy (1.3) is given by

$\displaystyle\begin{cases}\phi_{x}=\begin{pmatrix}-i\lambda&u\\\
u&i\lambda\end{pmatrix}\phi,\\\ \phi_{t}=\begin{pmatrix}A&B\\\
D&-A\end{pmatrix}\phi,\end{cases}$ (2.1)

where

$A=\sum_{j=0}^{2n+1}A_{j}(i\lambda)^{j},\quad
B=\sum_{j=0}^{2n}B_{j}(i\lambda)^{j},\quad
D=\sum_{j=0}^{2n}D_{j}(i\lambda)^{j},$

with

$\displaystyle A_{2n+1}$ $\displaystyle=4^{n},\quad
A_{2k}=0,~{}~{}k=0,1,\ldots,n,$ $\displaystyle A_{2k+1}$
$\displaystyle=\frac{4^{k+1}}{2}\left\\{\mathcal{L}_{n-k}[u_{x}-u^{2}]-\frac{\partial}{\partial
x}\left(\frac{\partial}{\partial
x}+2u\right)\mathcal{L}_{n-k-1}[u_{x}-u^{2}]\right\\},~{}~{}k=0,1,\ldots,n-1,$
$\displaystyle B_{2k+1}$
$\displaystyle=\frac{4^{k+1}}{2}\left\\{\frac{\partial}{\partial
x}\left(\frac{\partial}{\partial
x}+2u\right)\mathcal{L}_{n-k-1}[u_{x}-u^{2}]\right\\},~{}~{}k=0,1,\ldots,n-1,$
$\displaystyle B_{2k}$ $\displaystyle=-4^{k}\bigg{(}\frac{\partial}{\partial
x}+2u\bigg{)}\mathcal{L}_{n-k}[u_{x}-u^{2}],~{}~{}k=0,1,\ldots,n,$
$\displaystyle D_{k}$ $\displaystyle=(-1)^{k}B_{k},~{}~{}k=0,1,\ldots,2n,$
(2.2)

and $\mathcal{L}_{k}$, $k=0,1,\ldots,n$, being the Lenard operators defined in
(1.4).

######  Remark 2.1.

If the initial data $u_{0}(x)\in\mathcal{S}({\mathbb{R}})$, then
$\begin{pmatrix}B(x,0)\\\ D(x,0)\end{pmatrix}\to\begin{pmatrix}0\\\
0\end{pmatrix}$ as $x\to\infty$.

By introducing the matrix-valued functions

$\displaystyle Q(x,t)=$ $\displaystyle\begin{pmatrix}0&u\\\ u&0\end{pmatrix},$
$\displaystyle\tilde{Q}(x,t,\lambda)=$
$\displaystyle\begin{pmatrix}A(x,t,\lambda)-2^{2n}(i\lambda)^{2n+1}&B(x,t,\lambda)\\\
D(x,t,\lambda)&-A(x,t,\lambda)+2^{2n}(i\lambda)^{2n+1}\end{pmatrix},$

we could rewrite the Lax pair (2.1) as

$\displaystyle\begin{cases}\psi_{x}=(i(-1)^{n}\lambda\sigma_{3}+Q(x,t))\psi\doteq{\bf
U}(x,t,\lambda)\psi,\\\
\psi_{t}=(-i2^{2n}\lambda^{2n+1}\sigma_{3}+\tilde{Q}(x,t,(-1)^{n+1}\lambda))\psi\doteq{\bf
V}(x,t,\lambda)\psi,\end{cases}$ (2.3)

where $\sigma_{3}$ is the Pauli matrix defined in (1.18). Equivalently, let

$\displaystyle\psi=\Phi e^{-i((-1)^{n+1}\lambda
x+2^{2n}\lambda^{2n+1}t)\sigma_{3}},$

we have

$\displaystyle\begin{cases}\Phi_{x}+i(-1)^{n+1}\lambda[\sigma_{3},\Phi]=Q(x,t)\Phi,\\\
\Phi_{t}+i2^{2n}\lambda^{2n+1}[\sigma_{3},\Phi]=\tilde{Q}(x,t,(-1)^{n+1}\lambda)\Phi.\end{cases}$

From (2.2), it follows that ${\bf U}(x,t,\lambda)$ and ${\bf V}(x,t,\lambda)$
in (2.3) satisfy the following symmetry relations:

$\displaystyle{\bf U}(x,t,-\lambda)$ $\displaystyle=\overline{{\bf
U}(x,t,\bar{\lambda})},\qquad\,{\bf V}(x,t,-\lambda)=\overline{{\bf
V}(x,t,\bar{\lambda})},$ $\displaystyle\sigma_{1}\overline{{\bf
U}(x,t,\bar{\lambda})}\sigma_{1}$ $\displaystyle={\bf
U}(x,t,\lambda),\quad\sigma_{1}\overline{{\bf
V}(x,t,\bar{\lambda})}\sigma_{1}={\bf V}(x,t,\lambda).$

where $\sigma_{1}$ is given in (1.18). This, in turn, implies that

$\displaystyle\Phi(x,t,-\lambda)=\overline{\Phi(x,t,\bar{\lambda})},\qquad\sigma_{1}\overline{\Phi(x,t,\bar{\lambda})}\sigma_{1}=\Phi(x,t,\lambda).$

###  2.2. RH problem for the mKdV hierarchy

For the Cauchy problem of the mKdV hierarchy with initial data $u_{0}(x)$, we
see from the nonlinear Fourier transform formalism [34] and the standard
unified method introduced by Fokas [21] that the associated reflection
coefficient $r$ is defined by

$\displaystyle r(\lambda)=\frac{b^{\ast}(\lambda)}{a(\lambda)},$ (2.4)

where $f^{\ast}$ stands for its Schwartz conjugate (1.17), and the spectral
functions $a(\lambda)$ and $b(\lambda)$ constitute the scattering matrix

$\displaystyle\begin{pmatrix}a^{\ast}(\lambda)&b(\lambda)\\\
b^{\ast}(\lambda)&a(\lambda)\end{pmatrix}.$

We note that $r$ satisfies the symmetry relation

$\displaystyle
r(\lambda)=-\overline{r(-\lambda)},\quad\lambda\in{\mathbb{R}},$ (2.5)

and if $u_{0}(x)\in\mathcal{S}({\mathbb{R}})$, then

$\displaystyle\sup_{\lambda\in{\mathbb{R}}}|r(\lambda)|<1.$

With the reflection coefficient $r$ in (2.4), we define a $2\times 2$ matrix-
valued function $v(x,t,\lambda)$ by

$v(x,t,\lambda)=\begin{pmatrix}1-|r(\lambda)|^{2}&-\overline{r(\lambda)}e^{-t\Theta(\xi,\lambda)}\\\
r(\lambda)e^{t\Theta(\xi,\lambda)}&1\end{pmatrix},$ (2.6)

where

$\Theta(\xi,\lambda):=2i((-1)^{n+1}\xi\lambda+2^{2n}\lambda^{2n+1}),\qquad\xi:=\frac{x}{t}.$
(2.7)

We then formulate the following RH problem:

######  RH problem 2.2.

  1. _(a)_

$m(x,t,\lambda)$ is analytic for
$\lambda\in{\mathbb{C}}\setminus{\mathbb{R}}$.

  2. _(b)_

For a.e. $\lambda\in{\mathbb{R}}$, the limiting values

$m_{+/-}(x,t,\lambda):=\lim_{\begin{subarray}{c}\lambda^{\prime}\to\lambda\\\
\lambda^{\prime}\textrm{ on the upper/lower half-
plance}\end{subarray}}m(x,t,\lambda^{\prime})$

exist, and satisfy the jump condition

$\displaystyle m_{+}(x,t,\lambda)=m_{-}(x,t,\lambda)v(x,t,\lambda),$ (2.8)

where the jump matrix $v$ is given in (2.6).

  3. _(c)_

As $\lambda\to\infty$, we have $m(x,t,\lambda)=I+\mathcal{O}(\lambda^{-1})$.

From the standard argument (cf. [24, Theorem 1]), the relation between the
above RH problem and the solution of the mKdV hierarchy is given in the
following lemma.

######  Lemma 2.3.

The RH problem 2.2 for $m$ has a unique solution for the each
$(x,t)\in{\mathbb{R}}^{2}$ and the limit $\lim_{\lambda\to\infty}(\lambda
m(x,t,\lambda))_{21}$ exists. Moreover, the function $u(x,t)$ defined by

$\displaystyle u(x,t)=2\lim_{\lambda\to\infty}(\lambda m(x,t,\lambda))_{21},$
(2.9)

is a smooth function with rapid decay as $|x|\to\infty$, which satisfies the
$n$-th member of the mKdV hierarchy (1.3) with Schwartz class initial data.

######  Remark 2.4.

By (2.5), it is clear that $r(0)$ is purely imaginary and the jump matrix $v$
defined in (2.6) satisfies

$v(x,t,\lambda)=\sigma_{1}\overline{v(x,t,\bar{\lambda})}^{-1}\sigma_{1}=\sigma_{1}\sigma_{3}v(x,t,-\lambda)^{-1}\sigma_{3}\sigma_{1},\qquad\lambda\in{\mathbb{R}}.$

Thus, by uniqueness of the solution of the RH problem 2.2, it follows that

$\displaystyle
m(x,t,\lambda)=\sigma_{1}\overline{m(x,t,\bar{\lambda})}\sigma_{1}=\sigma_{1}\sigma_{3}m(x,t,-\lambda)\sigma_{3}\sigma_{1},\qquad\lambda\in{\mathbb{C}}\setminus{\mathbb{R}}.$
(2.10)

We will perform asymptotic analysis of the RH problem 2.2 for $m$ as
$t\to\infty$. Based on the parity of $n$ and the range of $x$, we split the
analysis into four cases, namely,

Case I $\displaystyle\doteq\\{(x,t)\mid 0\leq{x}\leq Mt^{\frac{1}{2n+1}}\\},$
$\displaystyle\quad n\text{ is odd},$ Case II $\displaystyle\doteq\\{(x,t)\mid
0\leq{x}\leq Mt^{\frac{1}{2n+1}}\\},$ $\displaystyle\quad n\text{ is even},$
Case III $\displaystyle\doteq\\{(x,t)\mid-Mt^{\frac{1}{2n+1}}\leq{x}\leq
0\\},$ $\displaystyle\quad n\text{ is even},$ Case IV
$\displaystyle\doteq\\{(x,t)\mid-Mt^{\frac{1}{2n+1}}\leq x\leq 0\\},$
$\displaystyle\quad n\text{ is odd},$

where $M$ is a positive constant. Since the analysis for Case III and Case IV
is similar to that for Case I and Case II (see Remark 5.7 below for a brief
comment), it suffices to focus on the first two cases, which is presented in
Sections 4 and 5 below, respectively.

###  2.3. RH problem for the Painlevé II hierarchy

We finally give an RH characterization of the generalized Ablowitz-Segur
solution $q_{\textrm{AS},n}$ of the Painlevé II hierarchy (1.5). As
aforementioned, the RH problem below is obtained from the general one (cf.
[11]) by choosing the specified Stokes multipliers (1.8).

######  RH problem 2.5.

  1. _(a)_

$\Psi(\zeta)=\Psi(x,\rho,\zeta)$ is defined and analytic in
$\mathbb{C}\setminus\Upsilon$, where

$\Upsilon:=\Upsilon_{1}\cup\Upsilon_{2n+1}\cup\Upsilon_{2n+2}\cup\Upsilon_{4n+2}$
(2.11)

with

$\Upsilon_{j}:=\left\\{\zeta\in\mathbb{C}~{}\bigg{|}~{}\arg\zeta=\frac{2j-1}{4n+2}\pi\right\\},\qquad
j=1,\ldots,4n+2.$ (2.12)

  2. _(b)_

For $\zeta\in\Upsilon$, we have

$\Psi_{+}(\zeta)=\Psi_{-}(\zeta)J_{\Psi}(\zeta),$

where

$J_{\Psi}(\zeta):=\left\\{\begin{array}[]{ll}\begin{pmatrix}1&0\\\
\rho&1\end{pmatrix},&\qquad\hbox{$\zeta\in\Upsilon_{1}$,}\\\
\begin{pmatrix}1&0\\\
-\rho&1\end{pmatrix},&\qquad\hbox{$\zeta\in\Upsilon_{2n+1}$,}\\\
\begin{pmatrix}1&\rho\\\
0&1\end{pmatrix},&\qquad\hbox{$\zeta\in\Upsilon_{2n+2}$,}\\\
\begin{pmatrix}1&-\rho\\\
0&1\end{pmatrix},&\qquad\hbox{$\zeta\in\Upsilon_{4n+2}$,}\end{array}\right.$

and the orientation of $\Upsilon$ is shown in Figure 1.

  3. _(c)_

As $\zeta\to\infty$, we have

$\Psi(\zeta)=\left(I+\frac{\Psi_{1}(x)}{\zeta}+\mathcal{O}(\zeta^{-2})\right)e^{-i\Xi(\zeta)\sigma_{3}}$
(2.13)

for some function $\Psi_{1}$, where

$\Xi(\zeta)=\Xi(x,\zeta):=\frac{(2\zeta)^{2n+1}}{4n+2}+x\zeta.$ (2.14)

  4. _(d)_

$\Psi(\zeta)$ is bounded near the origin.

$0$$\Upsilon_{1}$$\Upsilon_{4n+2}$$\Upsilon_{2n+1}$$\Upsilon_{2n+2}$ Figure 1.
The jump contour $\Upsilon$ for the RH problem 2.5 for $\Psi$.

Let $\Psi$ be a solution of the above RH problem, by [8, Proposition 2.3], it
follows that the function

$q_{\textrm{AS},n}(x;\rho):=\left\\{\begin{array}[]{ll}2i(\Psi_{1})_{12}(x)=-2i(\Psi_{1})_{21}(x),&\hbox{$n$
odd,}\\\ 2i(\Psi_{1})_{12}(-x)=-2i(\Psi_{1})_{21}(-x),&\hbox{$n$
even,}\end{array}\right.$ (2.15)

is real for $x\in\mathbb{R}$ and $-1<\rho<1$, where $\Psi_{1}(x)$ is given in
(2.13). Moreover, it satisfies the Painlevé II hierarchy (1.5) and the
boundary condition (1.9). We will prove Theorem 1.1 by analysing RH problem
2.5 for large negative $x$ in Section 3 below.

######  Remark 2.6.

One can strengthen the asymptotic behavior (2.13) to be

$\displaystyle\Psi(\zeta)=\left(I+\sum_{j=1}^{N}\frac{\Psi_{j}(x)}{\zeta^{j}}+\mathcal{O}(\zeta^{-N-1})\right)e^{-i\Xi(\zeta)\sigma_{3}},\qquad\zeta\to\infty,$
(2.16)

uniformly for $x$ in compact subsets of ${\mathbb{C}}\setminus\Upsilon$, where
$\Psi_{j}(x)$, $j=1,\ldots,N$, are smooth functions. If $\rho=0$, we have
$\Psi(\zeta)\equiv e^{-i\Xi(\zeta)\sigma_{3}}$, which implies that
$q_{\textrm{AS},n}\equiv 0$. Moreover, since it is readily seen that both
$\sigma_{3}\Psi(x,\rho,\zeta)\sigma_{3}$ and $\Psi(x,-\rho,\zeta)$ satisfy the
same RH problem, by (2.15), it follows that

$q_{\textrm{AS},n}(x;-\rho)=-q_{\textrm{AS},n}(x;\rho).$ (2.17)

##  3\. Asymptotic analysis of the RH problem for $\Psi$

In this section, we perform a Deift-Zhou steepest descent analysis [18] to the
RH problem 2.5 for $\Psi$ as $x\to-\infty$. It consists of a series of
explicit and invertible transformations which leads to an RH problem tending
to the identity matrix for large negative $x$.

###  3.1. First transformation: $\Psi\to X$

The first transformation is a rescaling and normalization of the RH problem
for $\Psi$, which is defined by

$X(\zeta)=\Psi(|x|^{\frac{1}{2n}}\zeta)e^{i|x|^{\frac{2n+1}{2n}}\widetilde{\Xi}(\zeta)\sigma_{3}},$
(3.1)

where

$\widetilde{\Xi}(\zeta):=\frac{2^{2n}}{2n+1}\zeta^{2n+1}-\zeta.$ (3.2)

It is then straightforward to check that $X$ satisfies the following RH
problem.

######  RH problem 3.1.

  1. _(a)_

$X(\zeta)$ is defined and analytic in $\mathbb{C}\setminus\Upsilon$, where
$\Upsilon$ is defined in (2.11).

  2. _(b)_

For $\zeta\in\Upsilon$, we have

$X_{+}(\zeta)=X_{-}(\zeta)J_{X}(\zeta),$

where

$\displaystyle J_{X}(\zeta)$
$\displaystyle=e^{-i|x|^{\frac{2n+1}{2n}}\widetilde{\Xi}(\zeta)\sigma_{3}}J_{\Psi}(\zeta)e^{i|x|^{\frac{2n+1}{2n}}\widetilde{\Xi}(\zeta)\sigma_{3}}$
$\displaystyle=\left\\{\begin{array}[]{ll}\begin{pmatrix}1&0\\\ \rho
e^{2i|x|^{\frac{2n+1}{2n}}\widetilde{\Xi}(\zeta)}&1\end{pmatrix},&\qquad\hbox{$\zeta\in\Upsilon_{1}$,}\\\
\begin{pmatrix}1&0\\\ -\rho
e^{2i|x|^{\frac{2n+1}{2n}}\widetilde{\Xi}(\zeta)}&1\end{pmatrix},&\qquad\hbox{$\zeta\in\Upsilon_{2n+1}$,}\\\
\begin{pmatrix}1&\rho e^{-2i|x|^{\frac{2n+1}{2n}}\widetilde{\Xi}(\zeta)}\\\
0&1\end{pmatrix},&\qquad\hbox{$\zeta\in\Upsilon_{2n+2}$,}\\\
\begin{pmatrix}1&-\rho e^{-2i|x|^{\frac{2n+1}{2n}}\widetilde{\Xi}(\zeta)}\\\
0&1\end{pmatrix},&\qquad\hbox{$\zeta\in\Upsilon_{4n+2}$.}\end{array}\right.$

  3. _(c)_

As $\zeta\to\infty$, we have

$X(\zeta)=I+\frac{\Psi_{1}(x)}{|x|^{\frac{1}{2n}}\zeta}+\mathcal{O}(\zeta^{-2}),$

where $\Psi_{1}(x)$ is given in (2.13).

  4. _(d)_

$X(\zeta)$ is bounded near the origin.

###  3.2. Second transformation: $X\to Y$

In the second transformation we apply contour deformations. The four rays
$\Upsilon_{j}$, $j=1,2n+1,2n+2,4n+2$, are replaced by their parallel lines
emanating from some special points on the real line. More precisely, we
replace $\Upsilon_{1}$ and $\Upsilon_{4n+2}$ by their parallel rays
$\widetilde{\Upsilon}_{1}$ and $\widetilde{\Upsilon}_{4n+2}$ emanating from
the point 1/2, and replace $\Upsilon_{2n+1}$ and $\Upsilon_{2n+2}$ by their
parallel rays $\widetilde{\Upsilon}_{2n+1}$ and $\widetilde{\Upsilon}_{2n+2}$
emanating from the point $-1/2$; see Figure 2 for an illustration.

0.8(40,40)(10,30) 0.8(40,40)(10,50) 0.8(40,40)(70,30) 0.8(40,40)(70,50)
$0$$-1/2$$1/2$$\Upsilon_{1}$$\Upsilon_{4n+2}$$\Upsilon_{2n+1}$$\Upsilon_{2n+2}$$\widetilde{\Upsilon}_{1}$$\widetilde{\Upsilon}_{2n+1}$$\widetilde{\Upsilon}_{2n+2}$$\widetilde{\Upsilon}_{4n+2}$
Figure 2. The jump contour $\Sigma_{Y}$ for the RH problem 3.2 for $Y$.

The second transformation is defined as follows.

$Y(\zeta)=\left\\{\begin{array}[]{ll}X(\zeta)\begin{pmatrix}1&0\\\ \rho
e^{2i|x|^{\frac{2n+1}{2n}}\widetilde{\Xi}(\zeta)}&1\end{pmatrix},&\text{$\zeta$
between $\Upsilon_{1}$ and $\widetilde{\Upsilon}_{1}$}\\\ &\text{and $\zeta$
between $\Upsilon_{2n+1}$ and $\widetilde{\Upsilon}_{2n+1}$, }\\\
X(\zeta)\begin{pmatrix}1&\rho
e^{-2i|x|^{\frac{2n+1}{2n}}\widetilde{\Xi}(\zeta)}\\\
0&1\end{pmatrix},&\text{$\zeta$ between $\Upsilon_{2n+2}$ and
$\widetilde{\Upsilon}_{2n+2}$}\\\ &\text{and $\zeta$ between $\Upsilon_{4n+2}$
and $\widetilde{\Upsilon}_{4n+2}$,}\\\
X(\zeta),&\hbox{elsewhere.}\end{array}\right.$ (3.3)

In view of the RH problem 3.1 for $X$ and (3.3), it is readily seen that $Y$
satisfies the following RH problem.

######  RH problem 3.2.

  1. _(a)_

$Y(\zeta)$ is defined and analytic in $\mathbb{C}\setminus\Sigma_{Y}$, where

$\Sigma_{Y}:=\widetilde{\Upsilon}_{1}\cup\widetilde{\Upsilon}_{2n+1}\cup\widetilde{\Upsilon}_{2n+2}\cup\widetilde{\Upsilon}_{4n+2}\cup[-1/2,1/2];$

see the solid lines in Figure 2.

  2. _(b)_

For $\zeta\in\Sigma_{Y}$, we have

$Y_{+}(\zeta)=Y_{-}(\zeta)J_{Y}(\zeta),$

where

$J_{Y}(\zeta)=\left\\{\begin{array}[]{ll}\begin{pmatrix}1&0\\\ \rho
e^{2i|x|^{\frac{2n+1}{2n}}\widetilde{\Xi}(\zeta)}&1\end{pmatrix},&\qquad\hbox{$\zeta\in\widetilde{\Upsilon}_{1}$,}\\\
\begin{pmatrix}1&0\\\ -\rho
e^{2i|x|^{\frac{2n+1}{2n}}\widetilde{\Xi}(\zeta)}&1\end{pmatrix},&\qquad\hbox{$\zeta\in\widetilde{\Upsilon}_{2n+1}$,}\\\
\begin{pmatrix}1&\rho e^{-2i|x|^{\frac{2n+1}{2n}}\widetilde{\Xi}(\zeta)}\\\
0&1\end{pmatrix},&\qquad\hbox{$\zeta\in\widetilde{\Upsilon}_{2n+2}$,}\\\
\begin{pmatrix}1&-\rho e^{-2i|x|^{\frac{2n+1}{2n}}\widetilde{\Xi}(\zeta)}\\\
0&1\end{pmatrix},&\qquad\hbox{$\zeta\in\widetilde{\Upsilon}_{4n+2}$,}\\\
\begin{pmatrix}1-\rho^{2}&-\rho
e^{-2i|x|^{\frac{2n+1}{2n}}\widetilde{\Xi}(\zeta)}\\\ \rho
e^{2i|x|^{\frac{2n+1}{2n}}\widetilde{\Xi}(\zeta)}&1\end{pmatrix},&\qquad\hbox{$\zeta\in(-\frac{1}{2},\frac{1}{2})$.}\end{array}\right.$
(3.4)

  3. _(c)_

As $\zeta\to\infty$, we have

$Y(\zeta)=I+\frac{\Psi_{1}(x)}{|x|^{\frac{1}{2n}}\zeta}+\mathcal{O}(\zeta^{-2}),$

where $\Psi_{1}(x)$ is given in (2.13).

  4. _(d)_

$Y(\zeta)$ is bounded near $\zeta=\pm 1/2$.

###  3.3. Third transformation: $Y\to T$

As $x\to-\infty$, it comes out that $J_{Y}(\zeta)$ tends to the identity
matrix exponentially fast except for $\zeta\in(-1/2,1/2)$, as evidenced in
Figure 3. Since $\text{Im\,}\widetilde{\Xi}(\zeta)=0$ for
$\zeta\in(-1/2,1/2)$, we have that $J_{Y}(\zeta)$ is highly oscillatory for
large negative $x$. The third transformation then involves the so-called lens
opening, which is based on the following factorization:

$J_{Y}(\zeta)=\begin{pmatrix}1&0\\\
\frac{\rho}{1-\rho^{2}}e^{2i|x|^{\frac{2n+1}{2n}}\widetilde{\Xi}(\zeta)}&1\end{pmatrix}\begin{pmatrix}1-\rho^{2}&0\\\
0&\frac{1}{1-\rho^{2}}\end{pmatrix}\begin{pmatrix}1&-\frac{\rho}{1-\rho^{2}}e^{-2i|x|^{\frac{2n+1}{2n}}\widetilde{\Xi}(\zeta)}\\\
0&1\end{pmatrix}$

for $\zeta\in(-1/2,1/2)$.

\begin{overpic}[width=142.26378pt]{tildeXi.pdf}\vspace{1pt}
\put(50.0,38.0){\tiny$\Sigma_{2}$} \put(50.0,59.0){\tiny$\Sigma_{1}$}
\put(24.0,42.5){\tiny$-1/2$} \put(59.0,42.5){\tiny$1/2$}
\put(-3.0,40.5){\tiny$\widetilde{\Upsilon}_{2n+2}$}
\put(-3.0,58.5){\tiny$\widetilde{\Upsilon}_{2n+1}$}
\put(95.0,40.5){\tiny$\widetilde{\Upsilon}_{4n+2}$}
\put(95.0,58.5){\tiny$\widetilde{\Upsilon}_{1}$}
\put(34.0,50.0){\circle*{2.0}} \put(65.8,50.0){\circle*{2.0}} \end{overpic}

(a) $n=2$

(b) $n=3$

Figure 3.  The jump contour $\Sigma_{T}$ for the RH problem 3.3 for $T$. The
shaded areas indicate the regions where $\text{Re\,}i\widetilde{\Xi}>0$ for
$n=2$ (left) and $n=3$ (right).

By opening lens around $(-1/2,1/2)$ as shown in Figure 3 so that
$\text{Re\,}i\widetilde{\Xi}(\zeta)>0$ for $\zeta\in\Sigma_{1}$ and
$\text{Re\,}i\widetilde{\Xi}(\zeta)<0$ for $\zeta\in\Sigma_{2}$, the third
transformation is defined by

$T(\zeta)=\begin{cases}Y(\zeta),&\text{$\zeta$ outside the lens},\\\
Y(\zeta)\begin{pmatrix}1&\frac{\rho}{1-\rho^{2}}e^{-2i|x|^{\frac{2n+1}{2n}}\widetilde{\Xi}(\zeta)}\\\
0&1\end{pmatrix},&\text{$\zeta$ in the upper part of the lens},\\\
Y(\zeta)\begin{pmatrix}1&0\\\
\frac{\rho}{1-\rho^{2}}e^{2i|x|^{\frac{2n+1}{2n}}\widetilde{\Xi}(\zeta)}&1\end{pmatrix},&\text{$\zeta$
in the lower part of the lens.}\end{cases}$ (3.5)

It is then straightforward to check that $T$ satisfies the following RH
problem

######  RH problem 3.3.

  1. _(a)_

$T(\zeta)$ is defined and analytic in $\mathbb{C}\setminus\Sigma_{T}$, where

$\Sigma_{T}:=\widetilde{\Upsilon}_{1}\cup\widetilde{\Upsilon}_{2n+1}\cup\widetilde{\Upsilon}_{2n+2}\cup\widetilde{\Upsilon}_{4n+2}\cup[-1/2,1/2]\cup\Sigma_{1}\cup\Sigma_{2};$
(3.6)

see Figure 3 for an illustration.

  2. _(b)_

For $\zeta\in\Sigma_{T}$, we have

$T_{+}(\zeta)=T_{-}(\zeta)J_{T}(\zeta),$

where

$J_{T}(\zeta)=\left\\{\begin{array}[]{ll}\begin{pmatrix}1&-\frac{\rho}{1-\rho^{2}}e^{-2i|x|^{\frac{2n+1}{2n}}\widetilde{\Xi}(\zeta)}\\\
0&1\end{pmatrix},&\qquad\hbox{$\zeta\in\Sigma_{1}$,}\\\ \begin{pmatrix}1&0\\\
\frac{\rho}{1-\rho^{2}}e^{2i|x|^{\frac{2n+1}{2n}}\widetilde{\Xi}(\zeta)}&1\end{pmatrix},&\qquad\hbox{$\zeta\in\Sigma_{2}$,}\\\
\begin{pmatrix}1-\rho^{2}&0\\\
0&\frac{1}{1-\rho^{2}}\end{pmatrix},&\qquad\hbox{$\zeta\in(-\frac{1}{2},\frac{1}{2})$,}\\\
J_{Y}(\zeta),&\qquad\hbox{elsewhere,}\\\ \end{array}\right.$ (3.7)

and where $J_{Y}(\zeta)$ is given in (3.4).

  3. _(c)_

As $\zeta\to\infty$, we have

$T(\zeta)=I+\frac{\Psi_{1}(x)}{|x|^{\frac{1}{2n}}\zeta}+\mathcal{O}(\zeta^{-2}),$

where $\Psi_{1}(x)$ is given in (2.13).

  4. _(d)_

$T(\zeta)$ is bounded near $\zeta=\pm 1/2$.

###  3.4. Global parametrix

It is now easily seen that the jump matrix $J_{T}$ tends to the identity
matrix except for $\zeta\in(-1/2,1/2)$. We are then led to consider the
following RH problem for the global parametrix $P^{(\infty)}$.

######  RH problem 3.4.

  1. _(a)_

$P^{(\infty)}(\zeta)$ is defined and analytic in
$\mathbb{C}\setminus[-1/2,1/2]$.

  2. _(b)_

For $\zeta\in(-1/2,1/2)$, we have

$P^{(\infty)}_{+}(\zeta)=P^{(\infty)}_{-}(\zeta)\begin{pmatrix}1-\rho^{2}&0\\\
0&\frac{1}{1-\rho^{2}}\end{pmatrix}.$

  3. _(c)_

As $\zeta\to\infty$, we have

$P^{(\infty)}(\zeta)=I+\mathcal{O}(\zeta^{-1}).$

One can check that the solution to the above RH problem is explicitly given by

$P^{(\infty)}(\zeta)=\left(\frac{\zeta+1/2}{\zeta-1/2}\right)^{\nu\sigma_{3}},\qquad\nu=-\frac{1}{2\pi
i}\ln(1-\rho^{2}),$ (3.8)

where the branch cut of $\left(\frac{\zeta+1/2}{\zeta-1/2}\right)^{\nu}$ is
taken along the interval $[-1/2,1/2]$ such that
$\left(\frac{\zeta+1/2}{\zeta-1/2}\right)^{\nu}\to 1$ as $\zeta\to\infty$.

###  3.5. Local parametrices near $\zeta=\pm 1/2$

Since the convergence of $J_{T}$ to the identity matrix is not uniform near
$\zeta=\pm 1/2$, we have to build local parametrices around these two points.
Denote by $D(z_{0},\delta)$ a fixed open disc centered at $z_{0}$ with radius
$\delta>0$, the local parametrix near $\zeta=1/2$ reads as follows

######  RH problem 3.5.

  1. _(a)_

$P^{(\frac{1}{2})}(\zeta)$ is defined and analytic in
$D(1/2,\delta)\setminus\Sigma_{T}$, where $\Sigma_{T}$ is defined (3.6).

  2. _(b)_

For $\zeta\in D(1/2,\delta)\cap\Sigma_{T}$, we have

$P^{(\frac{1}{2})}_{+}(\zeta)=P^{(\frac{1}{2})}_{-}(\zeta)J_{T}(\zeta),$

where $J_{T}(\zeta)$ is defined in (3.7).

  3. _(c)_

As $x\to-\infty$, $P^{(\frac{1}{2})}(\zeta)$ matches $P^{(\infty)}(\zeta)$ on
the boundary $\partial D(1/2,\delta)$ of $D(1/2,\delta)$, i.e.,

$P^{(\frac{1}{2})}(\zeta)=(I+\mathcal{O}(|x|^{-\frac{2n+1}{4n}}))P^{(\infty)}(\zeta),$
(3.9)

where $P^{(\infty)}(\zeta)$ is given in (3.8).

We can construct $P^{(\frac{1}{2})}(\zeta)$ explicitly by using the parabolic
cylinder parametrix $\Psi^{(\text{PC})}$ introduced in Appendix A.1, following
the strategy in [22, Section 9.4]. To proceed, we define

$\eta(\zeta):=2(-i\widetilde{\Xi}(\zeta)+i\widetilde{\Xi}(1/2))^{\frac{1}{2}},$
(3.10)

where $\widetilde{\Xi}(\zeta)$ is defined in (3.2) and the branch cut of
$(\cdot)^{1/2}$ is chosen such that $\arg(\zeta-1/2)\in(-\pi,\pi)$. It is
readily seen that

$\eta(\zeta)\sim e^{\frac{3\pi
i}{4}}2\sqrt{2n}(\zeta-\frac{1}{2}),\qquad\zeta\to\frac{1}{2}.$ (3.11)

Let $\Psi^{(\text{PC})}(\zeta;\nu)$ be the parabolic cylinder parametrix with
$\nu$ given in (3.8) (see Appendix A.1 below), we set, for $\zeta\in
D(\frac{1}{2},\delta)\setminus\Sigma_{T}$,

$P^{(\frac{1}{2})}(\zeta)=E(\zeta)\Psi^{(\text{PC})}(|x|^{\frac{2n+1}{4n}}\eta(\zeta);\nu)e^{i|x|^{\frac{2n+1}{2n}}\widetilde{\Xi}(\zeta)\sigma_{3}}\left(i\frac{h_{1}}{\rho}\right)^{\sigma_{3}/2}\begin{pmatrix}1&0\\\
0&-i\end{pmatrix}$ (3.12)

where $\eta(\zeta)$ is defined in (3.10),
$h_{1}=\frac{\sqrt{2\pi}}{\Gamma(-\nu)}e^{i\pi\nu}$, and

$E(\zeta):=\begin{pmatrix}1&0\\\
0&i\end{pmatrix}(\beta(\zeta))^{\sigma_{3}}\left(i\frac{2h_{1}}{\rho}\right)^{-\sigma_{3}/2}e^{in|x|^{\frac{2n+1}{2n}}\sigma_{3}/(2n+1)}\begin{pmatrix}|x|^{\frac{2n+1}{4n}}\eta(\zeta)&1\\\
1&0\end{pmatrix}$

and where

$\beta(\zeta):=\left(|x|^{\frac{2n+1}{4n}}\eta(\zeta)\frac{\zeta+1/2}{\zeta-1/2}\right)^{\nu}.$
(3.13)

By (3.11), it follows that $\beta(\zeta)$ is analytic near $\zeta=1/2$ with

$\beta(1/2)=(2\sqrt{2n}|x|^{\frac{2n+1}{4n}})^{\nu}e^{3\nu\pi i/4},$

which also implies that $E(\zeta)$ is an analytic prefactor in
$D(\frac{1}{2},\delta)$. From the RH problem A.1 for $\Psi^{(\text{PC})}$, it
is straightforward to show (cf. [22] for the case $k=1$) that
$P^{(\frac{1}{2})}(\zeta)$ defined in (3.12) indeed solves the RH problem 3.5.
Moreover, the matching condition (3.9) now reads

$P^{(\frac{1}{2})}(\zeta)=\Bigg{(}I+\begin{pmatrix}0&-\frac{\rho\nu}{h_{1}}e^{2n|x|^{\frac{2n+1}{2n}}\frac{i}{(2n+1)}}\frac{\beta(\zeta)^{2}}{\eta(\zeta)}\\\
-\frac{h_{1}}{\rho}e^{-2n|x|^{\frac{2n+1}{2n}}\frac{i}{(2n+1)}}\frac{1}{\beta(\zeta)^{2}\eta(\zeta)}&0\end{pmatrix}|x|^{-\frac{2n+1}{4n}}\\\
+\mathcal{O}(|x|^{-\frac{2n+1}{2n}})\Bigg{)}P^{(\infty)}(\zeta),$ (3.14)

as $x\to-\infty$, uniformly for $\zeta\in\partial D(1/2,\delta)$.

Similarly, near $\zeta=-1/2$, we intend to find a function
$P^{(-\frac{1}{2})}$ satisfying the following RH problem.

######  RH problem 3.6.

  1. _(a)_

$P^{(-\frac{1}{2})}(\zeta)$ is defined and analytic in
$D(-1/2,\delta)\setminus\Sigma_{T}$, where $\Sigma_{T}$ is defined (3.6).

  2. _(b)_

For $\zeta\in D(-1/2,\delta)\cap\Sigma_{T}$, we have

$P^{(-\frac{1}{2})}_{+}(\zeta)=P^{(-\frac{1}{2})}_{-}(\zeta)J_{T}(\zeta),$

where $J_{T}(\zeta)$ is defined in (3.7).

  3. _(c)_

As $x\to-\infty$, we have, for $\zeta\in\partial D(-1/2,\delta)$

$P^{(-\frac{1}{2})}(\zeta)=(I+\mathcal{O}(|x|^{-\frac{2n+1}{4n}}))P^{(\infty)}(\zeta),$
(3.15)

where $P^{(\infty)}(\zeta)$ is given in (3.8).

From the symmetry of $J_{T}$, one can check directly that

$P^{(-\frac{1}{2})}(\zeta)=\sigma_{1}P^{(\frac{1}{2})}(-\zeta)\sigma_{1}$
(3.16)

with $\sigma_{1}$ and $P^{(\frac{1}{2})}(\zeta)$ given (1.18) and (3.12)
solves the above RH problem.

###  3.6. Final transformation

The final transformation is defined by

$R(\zeta)=\left\\{\begin{array}[]{ll}T(\zeta)P^{(\infty)}(\zeta)^{-1},&\hbox{$\zeta\in\mathbb{C}\setminus\\{D(\frac{1}{2},\delta)\cup
D(-\frac{1}{2},\delta)\cup\Sigma_{T}\\}$,}\\\
T(\zeta)P^{(\frac{1}{2})}(\zeta)^{-1},&\hbox{$\zeta\in
D(\frac{1}{2},\delta)$,}\\\
T(\zeta)P^{(-\frac{1}{2})}(\zeta)^{-1},&\hbox{$\zeta\in
D(-\frac{1}{2},\delta)$.}\end{array}\right.$ (3.17)

It is then easily seen that $R$ satisfies the following RH problem.

######  RH problem 3.7.

  1. _(a)_

$R(\zeta)$ is defined and analytic in $\mathbb{C}\setminus\Sigma_{R}$, where

$\Sigma_{R}:=\Sigma_{T}\cup\partial D(\frac{1}{2},\delta)\cup\partial
D(-\frac{1}{2},\delta)\setminus\\{(-\frac{1}{2},\frac{1}{2})\cup
D(\frac{1}{2},\delta)\cup D(-\frac{1}{2},\delta)\\}.$

  2. _(b)_

For $\zeta\in\Sigma_{R}$, we have

$R_{+}(\zeta)=R_{-}(\zeta)J_{R}(\zeta),$ (3.18)

where

$J_{R}(\zeta)=\left\\{\begin{array}[]{ll}P^{(\frac{1}{2})}(\zeta)P^{(\infty)}(\zeta)^{-1},&\hbox{$\zeta\in\partial
D(\frac{1}{2},\delta)$,}\\\
P^{(-\frac{1}{2})}(\zeta)P^{(\infty)}(\zeta)^{-1},&\hbox{$\zeta\in\partial
D(-\frac{1}{2},\delta)$,}\\\
P^{(\infty)}(\zeta)J_{T}(\zeta)P^{(\infty)}(\zeta)^{-1},&\hbox{$\zeta\in\Sigma_{R}\setminus\left\\{\partial
D(\frac{1}{2},\delta)\cup\partial
D(-\frac{1}{2},\delta)\right\\}$,}\end{array}\right.$ (3.19)

and where $J_{T}(\zeta)$ is defined in (3.7) and the orientation of $\partial
D(\pm 1/2,\delta)$ is taken in a clock-wise manner.

  3. _(c)_

As $\zeta\to\infty$, we have,

$R(\zeta)=I+\mathcal{O}(\zeta^{-1}).$

From (3.7), (3.9) and (3.15), it is readily seen from (3.18) that, as
$x\to-\infty$,

$J_{R}(\zeta)=\left\\{\begin{array}[]{ll}I+\mathcal{O}(|x|^{-\frac{2n+1}{4n}})),&\hbox{$\zeta\in\partial
D(\frac{1}{2},\delta)\cup\partial D(-\frac{1}{2},\delta)$,}\\\
I+\mathcal{O}(e^{-c|x|^{-\frac{2n+1}{2n}})}),&\hbox{$\zeta\in\Sigma_{R}\setminus\\{\partial
D(\frac{1}{2},\delta)\cup\partial
D(-\frac{1}{2},\delta)\\}$,}\end{array}\right.$ (3.20)

for some $c>0$. An appeal to the standard small norm arguments (cf. [18, 15])
then shows that

$R(\zeta)=I+\mathcal{O}(|x|^{-\frac{2n+1}{4n}}),\qquad x\to-\infty,$ (3.21)

uniformly for $\zeta\in\mathbb{C}\setminus\Sigma_{R}$.

For later use, it is worth noting that $R$ is also characterized by the
following integral equation

$R(\zeta)=I+\frac{1}{2\pi
i}\int_{\Sigma_{R}}\frac{R_{-}(s)(J_{R}(s)-I)}{s-\zeta}ds.$ (3.22)

##  4\. Asymptotic analysis of the RH problem for $m$: Case I

In this section, we analyse the RH problem 2.2 for $m$ as $t\to\infty$ in Case
I, that is,

$\displaystyle\\{(x,t)\mid 0\leq x\leq Mt^{\frac{1}{2n+1}}\\},\qquad n\text{
is odd},$

where $M>0$. In this case, the critical points of phase function $\Theta$ in
(2.7) are

$\lambda_{0}^{(n,j)}=\sqrt[2n]{\frac{\xi}{(2n+1)2^{2n}}}e^{i\frac{(2j-1)\pi}{2n}},\qquad
j=1,2,\ldots,2n,$

which are not on the real line and approach $0$ at least as fast as
$t^{-1/(2n+1)}$ as $t\to\infty$. Thus, it follows that
$|\lambda_{0}^{(n,j)}|\leq Ct^{-1/(2n+1)}$ for some $C>0$. For $n=3$ and
$\xi=1$, an illustration of the six critical points and signature of
$\text{Re\,}\Theta$ are shown in Figure 4.

\begin{overpic}[width=241.81003pt]{casesI-phasefunction.pdf}
\put(92.0,45.0){\small${\mathbb{R}}$} \put(50.0,30.0){\circle*{2.0}}
\put(50.0,61.0){\circle*{2.0}} \put(39.0,54.0){\circle*{2.0}}
\put(39.0,37.0){\circle*{2.0}} \put(61.0,37.0){\circle*{2.0}}
\put(61.0,54.0){\circle*{2.0}} \par\put(42.0,51.0){\tiny$\text{Re\,}\Theta<0$}
\put(42.0,40.0){\tiny$\text{Re\,}\Theta>0$}
\put(72.0,67.0){\tiny$\text{Re\,}\Theta>0$}
\put(72.0,21.0){\tiny$\text{Re\,}\Theta<0$}
\put(44.0,80.0){\tiny$\text{Re\,}\Theta>0$}
\put(43.0,8.0){\tiny$\text{Re\,}\Theta<0$}
\put(16.0,67.0){\tiny$\text{Re\,}\Theta>0$}
\put(16.0,21.0){\tiny$\text{Re\,}\Theta<0$} \end{overpic} Figure 4.  The
critical points $\lambda^{(3,j)}_{0}$, $j=1,\ldots,6$, and signature of
$\text{Re\,}\Theta$ in the complex $\lambda$-plane for $\xi=1$ in Case I.

###  4.1. First transformation: $m\to m^{(1)}$

The first transformation involves deformation of the jump matrix along the
real axis and we need to decompose the reflection coefficient $r$ into an
analytic part $r_{a}$ and a small remainder $r_{r}$. To proceed, recall the
rays $\Upsilon_{j}$, $j=1,2n+1,2n+2,4n+2$, defined in (2.12) with the
orientations shown in Figure 1. By reversing the orientations of
$\Upsilon_{j}$, $j=2n+1,2n+2$, we obtain two new rays denoted by
$\Upsilon_{j}^{*}$. Clearly, the three curves
$\Upsilon_{1}\cup\Upsilon_{2n+1}^{*}$, $\mathbb{R}$ and
$\Upsilon_{2n+2}^{*}\cup\Upsilon_{4n+1}$ formulate the boundaries of two open
subsets $V$ and $V^{*}$, as shown in Figure 5. The decomposition of $r$ is
then given in the following lemma. Since the proof is similar to that of [9,
Lemma 2.1] or [25, Lemma 7.1], we omit the details here.

\begin{overpic}[width=241.81003pt]{casesI-phasefunction-CasesII.pdf}
\put(50.0,30.0){\circle*{2.0}} \put(50.0,61.0){\circle*{2.0}}
\put(52.5,45.5){\circle*{3.0}} \put(39.0,54.0){\circle*{2.0}}
\put(39.0,37.0){\circle*{2.0}} \put(61.0,37.0){\circle*{2.0}}
\put(61.0,54.0){\circle*{2.0}} \par\put(42.0,51.0){\tiny$\text{Re\,}\Theta<0$}
\put(42.0,38.0){\tiny$\text{Re\,}\Theta>0$}
\put(72.0,67.0){\tiny$\text{Re\,}\Theta>0$}
\put(72.0,21.0){\tiny$\text{Re\,}\Theta<0$}
\put(44.0,80.0){\tiny$\text{Re\,}\Theta>0$}
\put(43.0,8.0){\tiny$\text{Re\,}\Theta<0$}
\put(16.0,67.0){\tiny$\text{Re\,}\Theta>0$}
\put(16.0,21.0){\tiny$\text{Re\,}\Theta<0$}
\put(5.0,57.0){\tiny$\Upsilon_{2n+1}^{*}$}
\put(5.0,30.0){\tiny$\Upsilon_{2n+2}^{*}$}
\put(90.0,58.0){\tiny$\Upsilon_{1}$} \put(90.0,31.0){\tiny$\Upsilon_{4n+2}$}
\put(20.0,50.0){\tiny$V$} \put(80.0,50.0){\tiny$V$}
\put(20.0,38.0){\tiny$V^{*}$} \put(80.0,38.0){\tiny$V^{*}$} \end{overpic}
Figure 5.  The open subsets $V$, $V^{*}$ in Case I and the contour
$\Gamma^{(1)}$ for the RH problem 4.2 for $m^{(1)}$.

######  Lemma 4.1 (Analytic approximation for $\xi\geq 0$).

Let $r$ be the reflection coefficient defined in (2.4), we have

$r(\lambda)=r_{a}(x,t,\lambda)+r_{r}(x,t,\lambda),\qquad t\geq
C(n),\quad\lambda\in{\mathbb{R}},$ (4.1)

where $C(n)$ is a positive constant depending on $n$, and for a fixed
$N\in{\mathbb{N}}$ the functions $r_{a}$ and $r_{r}$ have the following
properties:

  1. _(a)_

For each $t\geq C(n)$, $r_{a}(t,\lambda)$ is defined and continuous for
$\lambda\in\bar{V}$ and analytic for $\lambda\in V$, where $\bar{V}$ denotes
the closure of $V$.

  2. _(b)_

There exists a constant $C>0$ such that

$\displaystyle|r_{a}(x,t,\lambda)|\leq\frac{C}{1+|\lambda|}e^{\frac{t}{4}|\text{Re\,}\Theta(\xi,\lambda)|},\qquad\lambda\in\bar{V},$

and

$\displaystyle\bigg{|}r_{a}(x,t,\lambda)-\sum_{j=0}^{N}\frac{r^{(j)}(0)}{j!}\lambda^{j}\bigg{|}\leq
C|\lambda|^{N+1}e^{\frac{t}{4}|\text{Re\,}\Theta(\xi,\lambda)|},\qquad\lambda\in\bar{V},$
(4.2)

for each $\xi\geq 0$ and $t\geq C(n)$.

  3. _(c)_

As $t\to\infty$, the $L^{1}$ and $L^{\infty}$ norms of $r_{r}(x,t,\cdot)$ on
${\mathbb{R}}$ are $\mathcal{O}(t^{-N})$.

  4. _(d)_

The functions $r_{a}$ and $r_{r}$ satisfy the following symmetries:

$\displaystyle\begin{cases}r_{a}(x,t,\lambda)=-r_{a}^{*}(x,t,-\lambda),&\lambda\in\bar{V},\\\
r_{r}(x,t,\lambda)=-r_{r}^{*}(x,t,-\lambda),&\lambda\in{\mathbb{R}}.\end{cases}$

With the aid of decomposition (4.1), we define a sectionally analytic matrix-
valued function $m^{(1)}$ by

$\displaystyle m^{(1)}(x,t,\lambda)=m(x,t,\lambda)G(x,t,\lambda),$ (4.3)

where

$\displaystyle G(x,t,\lambda):=\begin{cases}\begin{pmatrix}1&0\\\
-r_{a}e^{t\Theta}&1\end{pmatrix},&\lambda\in V,\\\
\begin{pmatrix}1&-r_{a}^{*}e^{-t\Theta}\\\ 0&1\end{pmatrix},&\lambda\in
V^{*},\\\ I,&\text{elsewhere}.\end{cases}$ (4.4)

By Lemma 4.1 and the notation $E^{p}(D)$ introduced at the end of Section 1.2,
it is readily seen that

$G(x,t,\cdot)\in I+(\dot{E}^{2}\cap E^{\infty})(V\cup V^{*}),$

and $m$ satisfies the RH problem 2.2 if and only if $m^{(1)}$ defined in (4.3)
solves the following RH problem.

######  RH problem 4.2.

  1. _(a)_

$m^{(1)}(x,t,\cdot)\in I+\dot{E}^{2}({\mathbb{C}}\setminus\Gamma^{(1)})$,
where

$\Gamma^{(1)}:={\mathbb{R}}\cup\Upsilon_{1}\cup\Upsilon_{2n+1}^{*}\cup\Upsilon_{2n+2}^{*}\cup\Upsilon_{4n+2};$
(4.5)

see Figure 5 for an illustration and the orientation.

  2. _(b)_

For a.e. $\lambda\in\Gamma^{(1)}$, we have

$m^{(1)}_{+}(x,t,\lambda)=m^{(1)}_{-}(x,t,\lambda)v^{(1)}(x,t,\lambda),$

where

$\displaystyle v^{(1)}(x,t,\lambda)=\begin{cases}\begin{pmatrix}1&0\\\
r_{a}e^{t\Theta}&1\end{pmatrix},&\lambda\in\Upsilon_{1}\cup\Upsilon_{2n+1}^{*},\\\
\begin{pmatrix}1&-r_{a}^{*}e^{-t\Theta}\\\
0&1\end{pmatrix},&\lambda\in\Upsilon_{2n+2}^{*}\cup\Upsilon_{4n+2},\\\
\begin{pmatrix}1-|r_{r}|^{2}&-r_{r}^{*}e^{-t\Theta}\\\
r_{r}e^{t\Theta}&1\end{pmatrix},&\lambda\in{\mathbb{R}}.\end{cases}$ (4.6)

###  4.2. Local parametrix near the origin

As $t\to\infty$, it is easily seen from the signature of $\text{Re\,}\Theta$
(c.f. Figure 4) and Item (c) of Lemma 4.1 that $v^{(1)}$ in (4.6) tends to the
identity matrix. The convergence, however, is not uniform for
$\lambda\in\Gamma^{(1)}\setminus{\mathbb{R}}$ and close to the origin, which
means we have to construct a local parametrix in a small neighborhood of the
origin $D(0,\epsilon)$ with $\epsilon>0$ being small and fixed.

To formulate the local parametrix, we make a local change of variable for
$\lambda$ near origin and introduce two new variables $y$ and $z$ by

$\displaystyle y:=((2n+1)t)^{-\frac{1}{2n+1}}x,\qquad
z:=((2n+1)t)^{{\frac{1}{2n+1}}}\lambda,$ (4.7)

so that

$\displaystyle
t\Theta(\xi,\lambda)=2i\left(yz+\frac{(2z)^{2n+1}}{4n+2}\right).$

Since $0\leq{x}\leq Mt^{1/(2n+1)}$, we have that $y$ is bounded and the map
$\lambda\mapsto z$ maps $D(0,\epsilon)$ onto the open disk
$D(0,((2n+1)t)^{{\frac{1}{2n+1}}}\epsilon)$ in the complex $z$-plane. The idea
now is, in view of Item (b) of Lemma 4.1, to replace the analytic part $r_{a}$
of $r$ in (4.6) by its $N$-th order polynomial approximation

$\displaystyle
p_{N}(t,z):=\sum_{j=0}^{N}\frac{r^{(j)}(0)}{j!}\lambda^{j}=\sum_{j=0}^{N}\frac{r^{(j)}(0)}{j!(2n+1)^{\frac{j}{2n+1}}}\frac{z^{j}}{t^{\frac{j}{2n+1}}}.$
(4.8)

and approximate $v^{(1)}$ by

$\displaystyle
v_{0}(x,t,\lambda):=\left\\{\begin{array}[]{ll}\begin{pmatrix}1&0\\\
p_{N}(t,z)e^{t\Theta}&1\end{pmatrix},&\hbox{$\lambda\in
D(0,\epsilon)\cap\\{\Upsilon_{1}\cup\Upsilon_{2n+1}^{*}\\},$}\\\
\begin{pmatrix}1&-p_{N}^{*}(t,z)e^{-t\Theta}\\\
0&1\end{pmatrix},&\hbox{$\lambda\in
D(0,\epsilon)\cap\\{\Upsilon_{2n+2}^{*}\cup\Upsilon_{4n+2}\\}$.}\end{array}\right.$
(4.11)

Indeed, for $T>0$ large enough, we define

${\bf Case~{}~{}I^{T}}\doteq{\bf Case~{}~{}I}\cap\\{t\geq T\\},$

and the following estimate holds.

######  Lemma 4.3.

For $(x,t)\in{\bf Case~{}~{}I^{T}}$, we have

$\displaystyle\|v^{(1)}-v_{0}\|_{L^{p}(D(0,\epsilon)\cap\Gamma^{(1)}\setminus{\mathbb{R}})}\leq
Ct^{-\frac{N+1}{2n+1}},$ (4.12)

for each $1\leq p\leq\infty$ and some $C>0$, where $v^{(1)}$ and $v_{0}$ are
defined in (4.6) and (4.11), respectively.

###### Proof.

It is readily seen from (4.6) and (4.11) that

$\displaystyle v^{(1)}-v_{0}=\begin{cases}\begin{pmatrix}0&0\\\
(r_{a}(x,t,\lambda)-p_{N}(t,z))e^{t\Theta}&0\end{pmatrix},&\lambda\in
D(0,\epsilon)\cap\\{\Upsilon_{1}\cup\Upsilon_{2n+1}^{*}\\},\\\
\begin{pmatrix}0&-(r_{a}^{*}(x,t,\lambda)-p_{N}^{*}(t,z))e^{-t\Theta}\\\
0&0\end{pmatrix},&\lambda\in
D(0,\epsilon)\cap\\{\Upsilon_{2n+2}^{*}\cup\Upsilon_{4n+2}\\}.\end{cases}$
(4.13)

We will only prove (4.12) for $\lambda\in D(0,\epsilon)\cap\Upsilon_{1}$,
since the proofs for the other contours follow from similar arguments. Let
$\lambda\in D(0,\epsilon)\cap\Upsilon_{1}$ and $(x,t)\in{\bf
Case~{}~{}I^{T}}$, for $\xi\geq 0$, we have from (2.7) that

$\text{Re\,}\Theta(\xi,\lambda)=\text{Re\,}\Theta(\xi,|\lambda|e^{\frac{\pi
i}{4n+2}})\\\
=-2^{2n+1}|\lambda|^{2n+1}-2\xi|\lambda|\sin\left(\frac{\pi}{4n+2}\right)\leq-2^{2n+1}|\lambda|^{2n+1}.$
(4.14)

In particular,

$\displaystyle e^{-\frac{3t}{4}|\text{Re\,}\Theta|}\leq
Ce^{-\frac{3}{4}t2^{2n+1}|\lambda|^{2n+1}}\leq
Ce^{-{2^{2n-1}}|z|^{2n+1}},\qquad\lambda\in D(0,\epsilon)\cap\Upsilon_{1},$

for some $C>0$. Thus, by (4.13), (4.2) and (4.8), it follows that

$\displaystyle|v^{(1)}-v_{0}|$ $\displaystyle\leq
C|r_{a}(t,\lambda)-p_{N}(t,z)|e^{t\text{Re\,}\Theta}\leq
C|zt^{-\frac{1}{2n+1}}|^{N+1}e^{-\frac{3t}{4}|\text{Re\,}\Theta|}$
$\displaystyle\leq
C|zt^{-\frac{1}{2n+1}}|^{N+1}e^{-{2^{2n-1}}|z|^{2n+1}},\qquad\lambda\in
D(0,\epsilon)\cap\Upsilon_{1}.$

Consequently, writing $\varrho=|z|$,

$\displaystyle\|v^{(1)}-v_{0}\|_{L^{\infty}(D(0,\epsilon)\cap\Upsilon_{1})}\leq
C\sup_{0\leq\varrho<\infty}(\varrho
t^{-\frac{1}{2n+1}})^{N+1}e^{-{2^{2n-1}}\rho^{2n+1}}\leq
Ct^{-\frac{N+1}{2n+1}}$

and

$\displaystyle\|v^{(1)}-v_{0}\|_{L^{1}(D(0,\epsilon)\cap\Upsilon_{1})}\leq
C\int_{0}^{\infty}(\varrho
t^{-\frac{1}{2n+1}})^{N+1}e^{-{2^{2n-1}}\varrho^{2n+1}}\frac{d\varrho}{t^{\frac{1}{2n+1}}}\leq
Ct^{-\frac{N+2}{2n+1}},$

which gives us (4.12). ∎

In the virtue of Lemma 4.3, we are then led to consider the following local
parametrix.

######  RH problem 4.4.

  1. _(a)_

$m_{0}(x,t,\lambda)$ is analytic for
$D(0,\epsilon)\cap\Gamma^{(1)}\setminus{\mathbb{R}}$.

  2. _(b)_

For a.e. $\lambda\in D(0,\epsilon)\cap\Gamma^{(1)}\setminus{\mathbb{R}}$, we
have

$m_{0,+}(x,t,\lambda)=m_{0,-}(x,t,\lambda)v_{0}(x,t,\lambda),$

where $v_{0}(x,t,\lambda)$ is defined in (4.11).

  3. _(c)_

For $\lambda\in\partial D(0,\epsilon)$, we have $m_{0}(x,t,\lambda)\to I$ as
$t\to\infty$.

We can solve the above RH problem by using the solution of model RH problem
introduced in Section A.2. Let $m^{\Upsilon^{*}}(y,t,z)$ be the solution of RH
problem A.2 with the polynomial (A.3) given by (4.8), i.e., the parameters $s$
and $p_{j}$, $j=1,\ldots,N$, therein are chosen to be

$\displaystyle s=r(0)\in i{\mathbb{R}},\qquad
p_{j}=\frac{r^{(j)}(0)}{j!(2n+1)^{\frac{j}{2n+1}}}.$ (4.15)

Due to the symmetry of reflection coefficient in (2.5), the polynomial
$p_{N}(t,z)$ satisfies the symmetry relation

$p_{N}(t,z)=-\overline{p_{N}(t,-\bar{z})},$

which implies that $p_{1}\in{\mathbb{R}}$ and $p_{2}\in i{\mathbb{R}}$. We
then define

$\displaystyle m_{0}(x,t,\lambda)=m^{\Upsilon^{*}}(y,t,z),\qquad\lambda\in
D(0,\epsilon),$ (4.16)

where $y$ and $z$ are defined in (4.7). By Lemma A.3, it follows that we can
choose $T\geq C(n)$ such that $m_{0}$ in (4.16) is well-defined whenever
$(x,t)\in{\bf Case~{}~{}I^{T}}$ and indeed solves the RH problem 4.4. More
properties of $m_{0}$ are collected in the following lemma for later use.

######  Lemma 4.5.

For each $(x,t)\in{\bf Case~{}~{}I^{T}}$, we have

$\displaystyle|m_{0}(x,t,\lambda)|\leq C,\qquad\lambda\in
D(0,\epsilon)\cap\Gamma^{(1)}\setminus{\mathbb{R}},$ (4.17)

for some $C>0$. Moreover,

$\displaystyle
m_{0}(x,t,\lambda)^{-1}=I+\sum_{j=1}^{N}\sum_{l=0}^{N}\frac{\Phi^{(0)}_{jl}(y)}{\lambda^{j}t^{\frac{j+l}{2n+1}}}+\mathcal{O}(t^{-\frac{N+1}{2n+1}})$
(4.18)

uniformly for $(x,t)\in{\bf Case~{}~{}I^{T}}$ and $\lambda\in\partial
D(0,\epsilon)$, where the coefficients $\Phi^{(0)}_{jl}(y)$ are smooth
functions of $y\in[0,\infty)$. In particular,

$||m_{0}(x,t,\cdot)^{-1}-I||_{L^{\infty}(\partial
D(0,\epsilon))}=\mathcal{O}(t^{-\frac{1}{2n+1}}),$ (4.19)

and

$\displaystyle\frac{1}{2\pi i}\int_{\partial
D(0,\epsilon)}(m_{0}(x,t,\lambda)^{-1}-I)d\lambda=\sum_{l=1}^{N}\frac{g_{l}(y)}{t^{\frac{l}{2n+1}}}+\mathcal{O}(t^{-\frac{N+1}{2n+1}}),$
(4.20)

uniformly for $(x,t)\in{\bf Case~{}~{}I^{T}}$, where
$g_{l+1}(y)=\Phi^{(0)}_{1l}(y)$ for each $l\geq 0$ and

$\displaystyle(g_{1})_{12}(y)=(g_{1})_{21}(y)=-\frac{(2n+1)^{-\frac{1}{2n+1}}}{2}q_{AS,n}(y;ir(0))$
(4.21)

with $q_{\textrm{AS},n}$ being the generalized Ablowitz-Segur solution of the
Painlevé II hierarchy (1.5). If $r(0)=0$, we have

$\displaystyle g_{1}(y)$ $\displaystyle=0,$ (4.22) $\displaystyle g_{2}(y)$
$\displaystyle=-\frac{r^{\prime}(0)}{4\times(2n+1)^{\frac{2}{2n+1}}}\text{Ai}^{\prime}_{2n+1}(y)\sigma_{1},$
(4.23) $\displaystyle g_{3}(y)$
$\displaystyle=\frac{ir^{\prime}(0)^{2}}{8\times(2n+1)^{\frac{3}{2n+1}}}\left(\int_{y}^{\infty}(\text{Ai}^{\prime}_{2n+1}(y^{\prime}))^{2}dy^{\prime}\right)\sigma_{3}+\frac{ir^{\prime\prime}(0)}{16\times(2n+1)^{\frac{3}{2n+1}}}\text{Ai}^{{}^{\prime\prime}}_{2n+1}(y)\sigma_{1}.$
(4.24)

###### Proof.

The proof essentially follows from Lemma A.3. By (A.5), we obtain the bound
(4.17). To show (4.18), we note that if $|\lambda|=\epsilon$, the variable
$z=((2n+1)t)^{{\frac{1}{2n+1}}}\lambda$ satisfies
$|z|=((2n+1)t)^{{\frac{1}{2n+1}}}\epsilon$ and $z\to\infty$ if $t\to\infty$.
Thus equation (A.4) yields

$\displaystyle
m_{0}(x,t,\lambda)=I+\sum_{j=1}^{N}\sum_{l=0}^{N}\frac{\Phi_{jl}^{\Upsilon^{*}}(y)}{(((2n+1)t)^{{\frac{1}{2n+1}}}\lambda)^{j}t^{\frac{l}{2n+1}}}+\mathcal{O}\big{(}t^{-\frac{N+1}{2n+1}}\big{)},$

uniformly for $(x,t)\in{\bf Case~{}~{}I^{T}}$ and $\lambda\in\partial
D(0,\epsilon)$. It is then straightforward to check that the expansion (4.18)
exists where the coefficients $\Phi^{(0)}_{jl}(y)$ can be expressed in terms
of $\Phi_{jl}^{\Upsilon^{*}}(y)$. In particular, we have
$\Phi^{(0)}_{10}(y)=-(2n+1)^{-\frac{1}{2n+1}}\Phi_{10}^{\Upsilon^{*}}(y)$. By
(4.18), the estimate (4.19) holds and we obtain from the Cauchy’s formula that

$\displaystyle\frac{1}{2\pi i}\int_{\partial
D(0,\epsilon)}(m_{0}^{-1}-I)d\lambda=\sum_{l=0}^{N}\frac{\Phi^{(0)}_{1l}(y)}{t^{\frac{l+1}{2n+1}}}+\mathcal{O}(t^{-\frac{N+2}{2n+1}}),$

which is (4.20).

Finally, note that

$g_{1}(y)=\Phi_{10}^{(0)}(y)=-(2n+1)^{-\frac{1}{2n+1}}\Phi_{10}^{\Upsilon^{*}}(y),$

we then obtain from (A.8) and (2.17) that

$\displaystyle(g_{1})_{12}(y)=(g_{1})_{21}(y)=\frac{(2n+1)^{-\frac{1}{2n+1}}}{2}q_{AS,n}(y;-ir(0))=-\frac{(2n+1)^{-\frac{1}{2n+1}}}{2}q_{AS,n}(y;ir(0)),$

which is (4.21). If $r(0)=0$, a combination of the fact that

$g_{l+1}(y)=\Phi^{(0)}_{1l}(y)=-(2n+1)^{-\frac{1}{2n+1}}\Phi_{1l}^{\Upsilon^{*}}(y),\quad
l=0,1,2,$

(4.15) and (A.9)–(A.11) gives us (4.22)–(4.24). ∎

###  4.3. Final transformation

The final transformation is defined by

$\displaystyle\hat{m}(x,t,\lambda)=\begin{cases}m^{(1)}(x,t,\lambda)m_{0}(x,t,\lambda)^{-1},&\lambda\in
D(0,\epsilon),\\\ m^{(1)}(x,t,\lambda),&\lambda\in{\mathbb{C}}\setminus
D(0,\epsilon),\end{cases}$ (4.25)

and it is readily seen that $\hat{m}$ satisfies the following RH problem.

######  RH problem 4.6.

  1. _(a)_

$\hat{m}(x,t,\cdot)\in I+\dot{E}^{2}({\mathbb{C}}\setminus\hat{\Gamma})$,
where $\hat{\Gamma}:=\Gamma^{(1)}\cup\partial D(0,\epsilon)$ with
$\Gamma^{(1)}$ defined in (4.5); see Figure 6 for an illustration.

  2. _(b)_

For a.e. $\lambda\in\hat{\Gamma}$, we have

$\hat{m}_{+}(x,t,\lambda)=\hat{m}_{-}(x,t,\lambda)\hat{v}(x,t,\lambda),$
(4.26)

where

$\displaystyle\hat{v}=\begin{cases}m_{0,-}v^{(1)}m_{0,+}^{-1},&\lambda\in\hat{\Gamma}\cap
D(0,\epsilon),\\\ m_{0}^{-1},&\lambda\in\partial D(0,\epsilon),\\\
v^{(1)},&\lambda\in\hat{\Gamma}\setminus\overline{D(0,\epsilon)},\end{cases}$

and where $v^{(1)}$ is defined in (4.6).

\begin{overpic}[width=241.81003pt]{casesI-phasefunction-CasesII-2.pdf}
\put(99.0,48.0){\small$\hat{\Gamma}$} \put(83.0,46.5){\small$\epsilon$}
\put(15.5,46.5){\small$-\epsilon$} \end{overpic} Figure 6.  The contour
$\hat{\Gamma}$ for the RH problem 4.6 for $\hat{m}$.

The jump matrix $\hat{v}$ in (4.26) tends to the identity matrix as
$t\to\infty$. More precisely, let

$\hat{w}:=\hat{v}-I,$ (4.27)

we have the following estimates.

######  Lemma 4.7.

For each $1\leq p\leq\infty$, the following estimates hold uniformly for
$(x,t)\in{\bf Case~{}~{}I^{T}}$:

$\displaystyle\|\hat{w}\|_{L^{p}(\partial D(0,\epsilon))}\leq
Ct^{-\frac{1}{2n+1}},$ (4.28a)
$\displaystyle\|\hat{w}\|_{L^{p}(D(0,\epsilon)\cap\Gamma^{(1)}\setminus{\mathbb{R}})}\leq
Ct^{-\frac{N+1}{2n+1}},$ (4.28b)
$\displaystyle\|\hat{w}\|_{L^{p}({\mathbb{R}})}\leq Ct^{-N},$ (4.28c)
$\displaystyle\|\hat{w}\|_{L^{p}(\Gamma^{(1)}\setminus\\{{\mathbb{R}}\cup\overline{D(0,\epsilon)}\\})}\leq
Ce^{-ct},$ (4.28d) for some positive constants $c$ and $C$.

###### Proof.

The estimate (4.28a) follows from (4.19). For $\lambda\in
D(0,\epsilon)\cap\Gamma^{(1)}\setminus{\mathbb{R}}$, we have

$\hat{w}=m_{0,-}(v^{(1)}-v_{0})m_{0,+}^{-1}.$

This, together with (4.12) and (4.17), gives us (4.28b). According to Item (c)
of Lemma 4.1 and the boundedness (4.17) of $m_{0}$, one has (4.28c). Finally,
(4.28d) follows because $e^{-t|\text{Re\,}\Theta|}\leq Ce^{-ct}$ uniformly on
$\Gamma^{(1)}\setminus\\{{\mathbb{R}}\cup\overline{D(0,\epsilon)}\\}$. ∎

##  5\. Asymptotic analysis of the RH problem for $m$: Case II

In this section, we analyse the RH problem 2.2 for $m$ as $t\to\infty$ in Case
II, that is,

$\displaystyle\\{(x,t)\mid 0\leq{x}\leq Mt^{\frac{1}{2n+1}}\\},$
$\displaystyle\qquad n\text{ is even},$

where $M>0$. In this case, the critical points of phase function $\Theta$ in
(2.7) are

$\lambda^{(n,j)}_{0}=\sqrt[2n]{\frac{\xi}{2^{2n}(2n+1)}}e^{j\frac{\pi
i}{n}},\qquad j=1,2,\cdots,2n,$

which approach $0$ at least as fast as $t^{-1/(2n+1)}$ as $t\to\infty$. Thus,
it follows that $|\lambda^{(n,j)}_{0}|\leq Ct^{-1/2n+1}$ for some $C>0$. The
difference between the present case and Case I lies in the fact that there are
two critical points lying on the real line, which we denote by

$\pm\lambda_{0}:=\pm\sqrt[2n]{\frac{\xi}{2^{2n}(2n+1)}}.$ (5.1)

For $n=2$ and $\xi=1$, an illustration of the four critical points and
signature of $\text{Re\,}\Theta$ are shown in Figure 7.

During our analysis performed in what follows, the same notations
$m^{(1)},V,\ldots$, will be used to emphasize the analogies with the previous
section, and we believe this will not cause any confusion.

\begin{overpic}[width=241.81003pt]{phasefunction-CasesII-even.pdf}
\put(95.0,75.0){\small$\Gamma^{(1)}$}
\put(84.0,70.0){\small$\Upsilon_{1,\lambda_{0}}$}
\put(1.0,70.0){\small$\Upsilon_{2n+1,\lambda_{0}}^{*}$}
\put(84.0,40.0){\small$\Upsilon_{4n+2,\lambda_{0}}$}
\put(1.0,40.0){\small$\Upsilon_{2n+2,\lambda_{0}}^{*}$}
\put(10.0,58.0){\small$V$} \put(10.0,48.0){\small$V^{*}$}
\put(85.0,58.0){\small$V$} \put(85.0,48.0){\small$V^{*}$}
\put(74.0,88.0){\tiny$\text{Re\,}\Theta>0$} \put(49.0,41.3){\circle*{2.0}}
\put(49.0,69.3){\circle*{2.0}} \put(34.8,55.3){\circle*{2.0}}
\put(31.0,50.3){\small$-\lambda_{0}$} \put(63.0,55.3){\circle*{2.0}}
\put(60.0,50.3){\small$\lambda_{0}$} \end{overpic} Figure 7.  The contour
$\Gamma^{(1)}$ for the RH problem 5.2 for $m^{(1)}$ and the sets $V$ and
$V^{*}$ in ${\bf Case~{}~{}II}$. The region where $\text{Re\,}\Theta>0$ is
shaded and the dark points stand for the critical points
$\lambda_{0}^{(2,j)}$, $j=1,\ldots,4$, for $\xi=1$.

###  5.1. First transformation: $m\to m^{(1)}$

As in Section 4.1, we need to decompose the reflection coefficient $r$ into an
analytic part and a small remainder to define the first transformation. In
this case, we define

$\Upsilon_{i,\lambda_{0}}:=\lambda_{0}+\Upsilon_{i},\quad
i=1,4n+2,\qquad\Upsilon_{i,\lambda_{0}}^{*}:=-\lambda_{0}+\Upsilon_{i}^{*},\quad
i=2n+1,2n+2.$

These four rays, together with
$(-\infty,-\lambda_{0}]\cup[\lambda_{0},\infty)$, formulate the boundaries of
open subsets $V\equiv V(\xi)$ and $V^{*}\equiv V^{*}(\xi)$, as illustrated in
Figure 7. Similar to [9, Lemma 5.7], [24, Lemma 5.2] and [25, Lemma 7.1], the
decomposition of $r$ in Case II reads as follows and we again omit the proof.

######  Lemma 5.1 (Analytic approximation for $0\leq\xi\leq A$).

Let $r$ be the reflection coefficient defined in (2.4), we have

$r(\lambda)=r_{a}(x,t,\lambda)+r_{r}(x,t,\lambda),\qquad\lambda\in(-\infty,-\lambda_{0})\cup(\lambda_{0},\infty),$
(5.2)

and for a fixed $N\in{\mathbb{N}}$ the functions $r_{a}$ and $r_{r}$ have the
following properties:

  1. _(a)_

For each $\xi\in[0,A]$ and $t\geq C(n)$ with positive $A$ and $C(n)$,
$r_{a}(x,t,\lambda)$ is defined and continuous for $\lambda\in\bar{V}$ and
analytic for $\lambda\in V$.

  2. _(b)_

There exists a constant $C>0$ such that

$\displaystyle|r_{a}(x,t,\lambda)|\leq\frac{C}{1+|\lambda|}e^{\frac{t}{4}|\text{Re\,}\Theta(\xi,\lambda)|},\qquad\lambda\in\bar{V},$

and

$\bigg{|}r_{a}(x,t,\lambda)-\sum_{j=0}^{N}\frac{r^{(j)}(\lambda_{0})}{j!}(\lambda-\lambda_{0})^{j}\bigg{|}\leq
C|\lambda-\lambda_{0}|^{N+1}e^{\frac{t}{4}|\text{Re\,}\Theta(\xi,\lambda)|},\qquad\lambda\in\bar{V},$
(5.3)

uniformly for $\xi\in[0,A]$ and $t\geq C(n)$.

  3. _(c)_

The $L^{1}$ and $L^{\infty}$ norms of $r_{r}(x,t,\cdot)$ on
$(-\infty,-\lambda_{0})\cup(\lambda_{0},\infty)$ are $\mathcal{O}(t^{-N})$ as
$t\to\infty$ uniformly with respect to $\xi\in[0,A]$.

  4. _(d)_

The functions $r_{a}$ and $r_{r}$ satisfy the following symmetries:

$\displaystyle\begin{cases}r_{a}(x,t,\lambda)=-r_{a}^{*}(x,t,-\lambda),&\lambda\in\bar{V},\\\
r_{r}(x,t,\lambda)=-r_{r}^{*}(x,t,-\lambda),&\lambda\in(-\infty,-\lambda_{0})\cup(\lambda_{0},\infty).\end{cases}$

We now define $m^{(1)}$ by (4.3) with $G(x,t,\lambda)$ given by (4.4) and
emphasize that the function $r_{a}$ therein appears in the decomposition
(5.2). By Lemma 5.1, one has

$G(x,t,\cdot)\in I+(\dot{E}^{2}\cap E^{\infty})(V\cup V^{*}),$

and $m$ satisfies the RH problem 2.2 if and only if $m^{(1)}$ solves the
following RH problem.

######  RH problem 5.2.

  1. _(a)_

$m^{(1)}(x,t,\cdot)\in I+\dot{E}^{2}({\mathbb{C}}\setminus\Gamma^{(1)})$,
where

$\Gamma^{(1)}:={\mathbb{R}}\cup\Upsilon_{1,\lambda_{0}}\cup\Upsilon_{2n+1,\lambda_{0}}^{*}\cup\Upsilon_{2n+2,\lambda_{0}}^{*}\cup\Upsilon_{4n+2,\lambda_{0}};$
(5.4)

see Figure 7 for an illustration and the orientation.

  2. _(b)_

For a.e. $\lambda\in\Gamma^{(1)}$, we have

$m^{(1)}_{+}(x,t,\lambda)=m^{(1)}_{-}(x,t,\lambda)v^{(1)}(x,t,\lambda),$

where

$\displaystyle v^{(1)}(x,t,\lambda)=\begin{cases}\begin{pmatrix}1&0\\\
r_{a}e^{t\Theta}&1\end{pmatrix},&\lambda\in\Upsilon_{1,\lambda_{0}}\cup\Upsilon_{2n+1,\lambda_{0}}^{*},\\\
\begin{pmatrix}1&-r_{a}^{*}e^{-t\Theta}\\\
0&1\end{pmatrix},&\lambda\in\Upsilon_{2n+2,\lambda_{0}}^{*}\cup\Upsilon_{4n+2,\lambda_{0}},\\\
\begin{pmatrix}1-|r_{r}|^{2}&-r_{r}^{*}e^{-t\Theta}\\\
r_{r}e^{t\Theta}&1\end{pmatrix},&\lambda\in(-\infty,-\lambda_{0})\cup(\lambda_{0},\infty),\\\
\begin{pmatrix}1-|r|^{2}&-r^{*}e^{-t\Theta}\\\
re^{t\Theta}&1\end{pmatrix},&\lambda\in(-\lambda_{0},\lambda_{0}).\end{cases}$
(5.5)

###  5.2. Local parametrix near the origin

As $t\to\infty$, it is easily seen from the signature of $\text{Re\,}\Theta$
(c.f. Figure 7) and Item (c) of Lemma 5.1 that $v^{(1)}$ in (5.5) tends to the
identity matrix. The convergence is not uniform for
$\lambda\in\Gamma^{(1)}\setminus(-\infty,-\lambda_{0}]\cup[\lambda_{0},\infty)$
and close to $\pm\lambda_{0}$. Since $\lambda_{0}=\mathcal{O}(t^{-1/(2n+1)})$,
we need to build a local parametrix in a small neighborhood of the origin
$D(0,\epsilon)$.

To formulate the local parametrix, as in Case I, we introduce the new
variables $y$ and $z$ by

$\displaystyle y:=-((2n+1)t)^{-\frac{1}{2n+1}}x,\qquad
z:=((2n+1)t)^{{\frac{1}{2n+1}}}\lambda,$ (5.6)

so that

$\displaystyle
t\Theta(\xi,\lambda)=2i\left(yz+\frac{(2z)^{2n+1}}{4n+2}\right).$

Since $0\leq{x}\leq Mt^{1/(2n+1)}$, it is easily seen that $-C\leq y\leq 0$
for some $C>0$. Moreover, the critical point $\lambda_{0}$ is mapped to the
point

$z_{0}:=((2n+1)t)^{\frac{1}{2n+1}}\lambda_{0}=\sqrt[2n]{|y|}/2$ (5.7)

on the $z$-plane. Following the strategy used in Section 4.2, we approximate
$r_{a}$ and $r$ by the $N$-th order polynomial $p_{N}$ given in (4.8) for
large $t$, which also leads to the approximation of $v^{(1)}$ in (5.5) by

$\displaystyle
v_{0}(x,t,\lambda):=\left\\{\begin{array}[]{ll}\begin{pmatrix}1&0\\\
p_{N}(t,z)e^{t\Theta}&1\end{pmatrix},&\hbox{$\lambda\in
D(0,\epsilon)\cap\\{\Upsilon_{1,\lambda_{0}}\cup\Upsilon_{2n+1,\lambda_{0}}^{*}\\},$}\\\
\begin{pmatrix}1&-p_{N}^{*}(t,z)e^{-t\Theta}\\\
0&1\end{pmatrix},&\hbox{$\lambda\in
D(0,\epsilon)\cap\\{\Upsilon_{2n+2,\lambda_{0}}^{*}\cup\Upsilon_{4n+2,\lambda_{0}}\\}$,}\\\
\begin{pmatrix}1-|p_{N}(t,z)|^{2}&-p_{N}^{*}(t,z)e^{-t\Theta}\\\
p_{N}(t,z)e^{t\Theta}&1\end{pmatrix},&\lambda\in(-\lambda_{0},\lambda_{0}).\end{array}\right.$
(5.11)

For $T>0$ large enough, by setting

${\bf Case~{}~{}II^{T}}\doteq{\bf Case~{}~{}II}\cap\\{t\geq T\\},$

the following estimate holds.

######  Lemma 5.3.

For $(x,t)\in{\bf Case~{}~{}II^{T}}$, we have

$\displaystyle\|v^{(1)}-v_{0}\|_{L^{p}(D(0,\epsilon)\cap\Gamma^{(1)}\setminus\\{(\infty,-\lambda_{0}]\cup[\lambda_{0},\infty)\\})}\leq
Ct^{-\frac{N+1}{2n+1}},$ (5.12)

for each $1\leq p\leq\infty$ and some $C>0$, where $v^{(1)}$ and $v_{0}$ are
defined in (5.5) and (5.11), respectively.

###### Proof.

It is readily seen from (5.5) and (5.11) that

$\displaystyle v^{(1)}-v_{0}$
$\displaystyle=\begin{cases}\begin{pmatrix}0&0\\\
(r_{a}(x,t,\lambda)-p_{N}(t,z))e^{t\Theta}&0\end{pmatrix},&\lambda\in
D(0,\epsilon)\cap\\{\Upsilon_{1,\lambda_{0}}\cup\Upsilon_{2n+1,\lambda_{0}}^{*}\\},\\\
\begin{pmatrix}0&-(r_{a}^{*}(x,t,\lambda)-p_{N}^{*}(t,z))e^{-t\Theta}\\\
0&0\end{pmatrix},&\lambda\in
D(0,\epsilon)\cap\\{\Upsilon_{2n+2,\lambda_{0}}^{*}\cup\Upsilon_{4n+2,\lambda_{0}}\\},\\\
\begin{pmatrix}-|r|^{2}+|p_{N}|^{2}&-(r^{*}(\lambda)-p_{N}^{*}(t,z))e^{-t\Theta}\\\
(r(\lambda)-p_{N}(t,z))e^{t\Theta}&0\end{pmatrix},&\lambda\in(-\lambda_{0},\lambda_{0}).\end{cases}$
(5.13)

We will prove (5.12) for $\lambda\in(-\lambda_{0},\lambda_{0})$ and
$\lambda\in D(0,\epsilon)\cap\Upsilon_{1,\lambda_{0}}$, since the proofs for
the other contours follow from similar arguments.

By expanding $r^{(j)}(\lambda_{0})$, $j=1,\ldots,N$, around the origin, it is
easily seen that

$\displaystyle\sum_{j=0}^{N}\frac{r^{(j)}(\lambda_{0})}{j!}(\lambda-\lambda_{0})^{j}$
$\displaystyle=\sum_{j=0}^{N}\sum_{l=0}^{N}\frac{r^{(j+l)}(0)}{j!l!}\lambda_{0}^{l}(\lambda-\lambda_{0})^{j}+\mathcal{O}(\lambda_{0}^{N+1})$
$\displaystyle=\sum_{l=0}^{N}\sum_{\kappa=l}^{N+l}\frac{r^{(\kappa)}(0)}{(\kappa-l)!l!}\lambda_{0}^{l}(\lambda-\lambda_{0})^{\kappa-l}+\mathcal{O}(\lambda_{0}^{N+1})$
$\displaystyle=\sum_{\kappa=0}^{N}\frac{r^{(\kappa)}(0)}{\kappa!}\lambda^{\kappa}+\mathcal{O}(|\lambda|^{N+1})=P_{N}(t,z)+\mathcal{O}(|\lambda|^{N+1}),$
(5.14)

uniformly for $0\leq\lambda_{0}\leq C$ and $\lambda\in
D(0,\epsilon)\cap\Gamma^{(1)}\setminus\\{(\infty,-\lambda_{0}]\cup[\lambda_{0},\infty)\\})$.
If $\lambda\in(-\lambda_{0},\lambda_{0})$, we have $\text{Re\,}\Theta=0$. A
combination of (5.13), (5.14) and Lemma 5.1 shows that

$\displaystyle|v^{(1)}-v_{0}|\leq C|r(\lambda)-p_{N}(t,z)|\leq
C|\lambda|^{N+1}\leq Ct^{-\frac{N+1}{2n+1}},$

which gives us (5.12) for $\lambda\in(-\lambda_{0},\lambda_{0})$. If
$\lambda=\lambda_{0}+\varrho e^{\frac{\pi i}{4n+2}}\in
D(0,\epsilon)\cap\Upsilon_{1,\lambda_{0}}$, we see from the fact
$\xi=(2n+1)(2\lambda_{0})^{2n}>0$ and (2.7) that

$\displaystyle\text{Re\,}\Theta(\xi,\lambda)$
$\displaystyle=\text{Re\,}\Theta(\xi,\lambda_{0}+\varrho e^{\frac{\pi
i}{4n+2}})$ $\displaystyle=\text{Re\,}2i(-\xi(\lambda_{0}+\varrho e^{\frac{\pi
i}{4n+2}})+2^{2n}(\lambda_{0}+\varrho e^{\frac{\pi i}{4n+2}})^{2n+1})$
$\displaystyle=-2^{2n+1}\bigg{(}\begin{pmatrix}2n+1\\\
2\end{pmatrix}\lambda_{0}^{2n-1}\varrho^{2}\sin\frac{n}{2n+1}\pi+\cdots+\varrho^{2n+1}\bigg{)}$
$\displaystyle\leq-2^{2n+1}\varrho^{2n+1}=-2^{2n+1}|\lambda-\lambda_{0}|^{2n+1},$

where $\begin{pmatrix}n\\\ k\end{pmatrix}$ is the binomial number. Thus,

$\displaystyle e^{-\frac{3t}{4}|\text{Re\,}\Theta|}\leq
e^{-\frac{3}{4}t2^{2n+1}\varrho^{2n+1}}\leq$ $\displaystyle
Ce^{-\frac{3}{4}t|\lambda|^{2n+1}}\leq Ce^{-\frac{3}{4(2n+1)}|z|^{2n+1}}.$
(5.15)

It then follows from (5.3), (5.13), (5.14) and (5.15) that

$\displaystyle|v^{(1)}-v_{0}|$ $\displaystyle\leq
C|r_{a}(x,t,\lambda)-p_{N}(t,z)|e^{t\text{Re\,}\Theta}$ $\displaystyle\leq
C\bigg{|}r_{a}(x,t,\lambda)-\sum_{j=0}^{N}\frac{r^{(j)}(\lambda_{0})}{j!}(\lambda-\lambda_{0})^{j}\bigg{|}e^{t\text{Re\,}\Theta}$
$\displaystyle\quad+C\bigg{|}\sum_{j=0}^{N}\frac{r^{(j)}(\lambda_{0})}{j!}(\lambda-\lambda_{0})^{j}-\sum_{j=0}^{N}\frac{r^{(j)}(0)}{j!}\lambda^{j}\bigg{|}e^{t\text{Re\,}\Theta}$
$\displaystyle\leq
C|\lambda-\lambda_{0}|^{N+1}e^{-\frac{3}{4}t|\text{Re\,}\Theta|}+C|\lambda|^{N+1}e^{-t|\text{Re\,}\Theta|}$
$\displaystyle\leq
C\big{|}zt^{-\frac{1}{2n+1}}\big{|}^{N+1}e^{-\frac{3}{4(2n+1)|z|^{2n+1}}}.$

As in the proof of Lemma 4.5, this implies

$\|v^{(1)}-v_{0}\|_{(L^{1}\cap
L^{\infty})(D(0,\epsilon)\cap\Upsilon_{1,\lambda_{0}})}\leq
Ct^{-\frac{N+1}{2n+1}},$

as required. ∎

In the virtue of Lemma 5.3, we are then led to consider the following local
parametrix.

######  RH problem 5.4.

  1. _(a)_

$m_{0}(x,t,\lambda)$ is analytic for
$D(0,\epsilon)\cap\Gamma^{(1)}\setminus\\{(-\infty,-\lambda_{0}]\cup[\lambda_{0},\infty\\}$,
where $\Gamma^{(1)}$ is defined in (5.4).

  2. _(b)_

For a.e. $\lambda\in
D(0,\epsilon)\cap\Gamma^{(1)}\setminus\\{(-\infty,-\lambda_{0}]\cup[\lambda_{0},\infty\\}$,
we have

$m_{0,+}(x,t,\lambda)=m_{0,-}(x,t,\lambda)v_{0}(x,t,\lambda),$ (5.16)

where $v_{0}(x,t,\lambda)$ is defined in (5.11).

  3. _(c)_

For $\lambda\in\partial D(0,\epsilon)$, we have $m_{0}(x,t,\lambda)\to I$ as
$t\to\infty$.

We can solve the above RH problem with the aid of the solution of model RH
problem introduced in Section A.3. Let $m^{Z}(y,t,z_{0},z)$ be solution of RH
problem A.6 with the polynomial (A.3) in the jump condition given by (4.8), we
define

$\displaystyle m_{0}(x,t,\lambda):=m^{Z}(y,t,z_{0},z),\qquad\lambda\in
D(0,\epsilon),$ (5.17)

where $y$, $z$ and $z_{0}$ are defined in (5.6) and (5.7), respectively. If
$(x,t)\in{\bf Case~{}~{}II^{T}}$, then $(y,t,z_{0})\in\mathcal{P}_{T}$, where
$\mathcal{P}_{T}$ defined in (A.39). By Lemma A.7, we can choose suitable $T$
such that $m_{0}$ in (5.17) is well-defined and check that it indeed solves
the RH problem 5.4. The function $m_{0}$ in this case has similar properties
as that in ${\bf Case~{}~{}I}$, which are collected in the following lemma and
follow directly from Lemma A.7.

######  Lemma 5.5.

For each $(x,t)\in{\bf Case~{}~{}II^{T}}$, the function $m_{0}(x,t,\lambda)$
defined in (5.17) is uniformly bounded, i.e.,

$\displaystyle|m_{0}(x,t,\lambda)|\leq C,\qquad\lambda\in
D(0,\epsilon)\cap\Gamma^{(1)}\setminus\\{(-\infty,-\lambda_{0}]\cup[\lambda_{0},\infty\\}.$
(5.18)

Moreover,

$\displaystyle
m_{0}(x,t,\lambda)^{-1}=I+\sum_{j=1}^{N}\sum_{l=0}^{N}\frac{\Phi^{(0)}_{jl}(y)}{\lambda^{j}t^{\frac{j+l}{2n+1}}}+O(t^{-\frac{N+1}{2n+1}})$
(5.19)

uniformly for $(x,t)\in{\bf Case~{}~{}II^{T}}$ and $\lambda\in\partial
D(0,\epsilon)$. The function $m_{0}(x,t,\lambda)$ satisfies (4.19) and (4.20)
uniformly for $(x,t)\in{\bf Case~{}~{}II^{T}}$, where the coefficients
$\\{g_{j}(y)\\}_{1}^{N}$ are smooth functions of $y\in{\mathbb{R}}$ with

$\displaystyle(g_{1})_{12}(y)=(g_{1})_{21}(y)=-\frac{(2n+1)^{-\frac{1}{2n+1}}}{2}q_{AS,n}(-y;ir(0)).$
(5.20)

If $r(0)=0$, we still have (4.22)–(4.24) for the first few terms in (5.19).

###  5.3. Final transformation

As in ${\bf Case~{}~{}I}$, we define $\hat{m}$ by (4.25) in the final
transformation, but it is understood that the functions $m^{(1)}$ and $m_{0}$
therein solve RH problems 5.2 and 5.4, respectively. Thus, $\hat{m}$ satisfies
RH problem 4.6 with the contour $\hat{\Gamma}:=\Gamma^{(1)}\cup\partial
D(0,\epsilon)$ illustrated in Figure 8.

\begin{overpic}[width=241.81003pt]{phasefunction-CasesII-even-epsilon.pdf}
\put(92.0,60.0){\small$\hat{\Gamma}$} \put(90.0,50.0){\small$\epsilon$}
\put(3.0,50.0){\small$-\epsilon$} \end{overpic} Figure 8.  The contour
$\hat{\Gamma}$ in ${\bf Case~{}~{}II}$.

The estimates of $\hat{w}$ in (4.27) for Case II are given in the following
lemma, which can be proved in a manner similar to that of Lemma 4.7.

######  Lemma 5.6.

For each $1\leq p\leq\infty$, the following estimates hold uniformly for
$(x,t)\in{\bf Case~{}~{}II^{T}}$:

$\displaystyle\|\hat{w}\|_{L^{p}(\partial D(0,\epsilon))}\leq
Ct^{-\frac{1}{2n+1}},$ (5.21a)
$\displaystyle\|\hat{w}\|_{L^{p}(D(0,\epsilon)\cap\Gamma^{(1)}\setminus\\{(-\infty,-\lambda_{0}]\cup[\lambda_{0},\infty\\})}\leq
Ct^{-\frac{N+1}{2n+1}},$ (5.21b)
$\displaystyle\|\hat{w}\|_{L^{p}({\mathbb{R}}\setminus[-\lambda_{0},\lambda_{0}])}\leq
Ct^{-N},$ (5.21c)
$\displaystyle\|\hat{w}\|_{L^{p}(\Gamma^{(1)}\setminus\\{{\mathbb{R}}\cup\overline{D(0,\epsilon)}\\})}\leq
Ce^{-ct},$ (5.21d) for some positive constants $c$ and $C$.

######  Remark 5.7.

We have completed asymptotic analysis of the RH problem 2.2 for $m$ in Case I
and Case II. For Case III, critical points of the phase function $\Theta$ in
(2.7) are not on the real line; see the left picture in Figure 9. Therefore,
the asymptotic analysis is analogous to that in Case I and local analysis near
the origin is related to the model RH problem A.2. For Case IV, the phase
function $\Theta$ has two critical points lying on the real line; see the
right picture in Figure 9, which implies that the analysis can be carried out
in a way similar to that in Case II and the model RH problem A.6 will play a
crucial role in the local analysis.

\begin{overpic}[width=142.26378pt]{caseIII.pdf}\vspace{1pt}
\put(39.0,65.0){\circle*{2.0}} \put(39.0,35.0){\circle*{2.0}}
\put(61.0,35.0){\circle*{2.0}} \put(61.0,65.0){\circle*{2.0}}
\put(41.0,88.0){\tiny$\text{Re\,}\Theta<0$}
\put(41.0,8.0){\tiny$\text{Re\,}\Theta>0$} \end{overpic}

(a) $n=2$

\begin{overpic}[width=142.26378pt]{caseIV.pdf}\vspace{1pt}
\put(44.0,31.0){\circle*{2.0}} \put(44.0,66.0){\circle*{2.0}}
\put(33.0,47.0){\circle*{2.0}} \put(68.0,47.0){\circle*{2.0}}
\put(61.0,31.0){\circle*{2.0}} \put(61.0,66.0){\circle*{2.0}}
\par\put(42.0,55.0){\tiny$\text{Re\,}\Theta>0$}
\put(42.0,38.0){\tiny$\text{Re\,}\Theta<0$} \end{overpic}

(b) $n=3$

Figure 9.  The critical points and signature of $\text{Re\,}\Theta$ in the
complex $\lambda$-plane with $\xi=-1$ for Case III and Case IV. The shaded
areas indicate the regions where $\text{Re\,}\Theta>0$ for $n=2$ (left) and
$n=3$ (right).

##  6\. Proofs of main results

###  6.1. Proof of Theorem 1.1

The asymptotics (1.9) is given in [8] by analysing the RH problem 2.5 for
$\Psi$ as $x\to\infty$ and we only need to show (1.10). By tracing back the
transformations $\Psi\to X\to Y\to T\to R$ given in (3.1), (3.3), (3.5) and
(3.17), it is readily seen from (2.15) and (3.22) that

$\displaystyle q_{\textrm{AS},n}((-1)^{n+1}x;\rho)=2i(\Psi_{1})_{12}(x)$
$\displaystyle=2|x|^{\frac{1}{2n}}i\lim_{\zeta\to\infty}\zeta
R_{12}(\zeta)=-\frac{|x|^{\frac{1}{2n}}}{\pi}\int_{\Sigma_{R}}\bigg{(}R_{-}(s)(J_{R}(s)-I)\bigg{)}_{12}ds.$

This, together with (3.20) and (3.21), implies that, as $x\to-\infty$,

$q_{\textrm{AS},n}((-1)^{n+1}x;\rho)\\\
=-\frac{|x|^{\frac{1}{2n}}}{\pi}\left(\int_{\partial D(1/2,\delta)\cup\partial
D(-1/2,\delta)}(J_{R})_{12}(s)ds+\mathcal{O}(|x|^{-\frac{2n+1}{2n}})\right).$
(6.1)

In view of (3.19), we obtain from (3.14) and (3.16) that, for large negative
$x$,

$(J_{R})_{12}(\zeta)=\left\\{\begin{array}[]{ll}-\frac{\rho\nu}{h_{1}}e^{2n|x|^{\frac{2n+1}{2n}}i/(2n+1)}\frac{\beta(\zeta)^{2}}{\eta(\zeta)}|x|^{-\frac{2n+1}{4n}}+\mathcal{O}(|x|^{-\frac{2n+1}{2n}}),&\hbox{$\zeta\in\partial
D(1/2,\delta)$,}\\\
-\frac{h_{1}}{\rho}e^{-2n|x|^{\frac{2n+1}{2n}}i/(2n+1)}\frac{1}{\beta(-\zeta)^{2}\eta(-\zeta)}|x|^{-\frac{2n+1}{4n}}+\mathcal{O}(|x|^{-\frac{2n+1}{2n}}),&\hbox{$\zeta\in\partial
D(-1/2,\delta)$.}\end{array}\right.$

Inserting the above estimates into (6.1), it is readily seen that

$q_{\textrm{AS},n}((-1)^{n+1}x;\rho)=\frac{1}{\pi|x|^{\frac{2n-1}{4n}}}\Bigg{(}\frac{\rho\nu}{h_{1}}e^{2|x|^{\frac{2n+1}{2n}}i/3}\int_{\partial
D(1/2,\delta)}\frac{\beta(s)^{2}}{\eta(s)}ds\\\
+\frac{h_{1}}{\rho}e^{-2|x|^{\frac{2n+1}{2n}}i/3}\int_{\partial
D(-1/2,\delta)}\frac{1}{\beta(-s)^{2}\eta(-s)}ds\Bigg{)}+\mathcal{O}(|x|^{-1}).$

By (3.11) and (3.13), we have

$\displaystyle\int_{\partial D(1/2,\delta)}\frac{\beta(s)^{2}}{\eta(s)}ds$
$\displaystyle=\frac{\pi}{\sqrt{2n}}e^{\frac{3}{2}\nu\pi i+\frac{3}{4}\pi
i}\left(8n|x|^{\frac{2n+1}{2n}}\right)^{\nu},$ $\displaystyle\int_{\partial
D(-1/2,\delta)}\frac{1}{\beta(-s)^{2}\eta(-s)}ds$
$\displaystyle=-\frac{\pi}{\sqrt{2n}}e^{-\frac{3}{2}\nu\pi i+\frac{3}{4}\pi
i}\left(8n|x|^{\frac{2n+1}{2n}}\right)^{-\nu}.$

Thus,

$\displaystyle q_{\textrm{AS},n}((-1)^{n+1}x;\rho)$
$\displaystyle=\frac{e^{\frac{3}{4}\pi
i}}{\sqrt{2n}|x|^{\frac{2n-1}{4n}}}\Bigg{(}\frac{\rho\nu}{h_{1}}e^{\frac{2n|x|^{(2n+1)/(2n)}}{2n+1}i+\frac{3}{2}\nu\pi
i}\left(8n|x|^{\frac{2n+1}{2n}}\right)^{\nu}$
$\displaystyle\quad-\frac{h_{1}}{\rho}e^{-\frac{2n|x|^{(2n+1)/(2n)}}{2n+1}i-\frac{3}{2}\nu\pi
i}\left(8n|x|^{\frac{2n+1}{2n}}\right)^{-\nu}\Bigg{)}+\mathcal{O}(|x|^{-1})$
$\displaystyle=\sqrt{\frac{\pi}{n}}\frac{1}{\rho|x|^{\frac{2n-1}{4n}}}\Bigg{(}\frac{1}{\Gamma(\nu)}e^{\frac{2n|x|^{(2n+1)/(2n)}}{2n+1}i-\frac{\nu}{2}\pi
i+\nu\ln\left(8n|x|^{\frac{2n+1}{2n}}\right)+\frac{1}{4}\pi i}$
$\displaystyle\quad+\frac{1}{\Gamma(-\nu)}e^{-\frac{2n|x|^{(2n+1)/(2n)}}{2n+1}i-\frac{\nu}{2}\pi
i-\nu\ln\left(8n|x|^{\frac{2n+1}{2n}}\right)-\frac{1}{4}\pi
i}\Bigg{)}+\mathcal{O}(|x|^{-1})$
$\displaystyle=\frac{2}{\sqrt{n}|x|^{\frac{2n-1}{4n}}}\text{Re\,}\left[\frac{\sqrt{\pi}}{\rho\Gamma(\nu)}e^{-\frac{\nu}{2}\pi
i+\frac{2n|x|^{(2n+1)/(2n)}}{2n+1}i+\nu\ln\left(8n|x|^{\frac{2n+1}{2n}}\right)+\frac{1}{4}\pi
i}\right]+\mathcal{O}(|x|^{-1}),$ (6.2)

where we have made use of the facts that
$h_{1}=\frac{\sqrt{2\pi}}{\Gamma(-\nu)}e^{i\pi\nu}$,
$\nu=-\frac{\ln(1-\rho^{2})}{2\pi i}$ and

$\Gamma(\nu)\Gamma(-\nu)=-\frac{\pi}{\nu\sin(\nu\pi)}=\frac{2\pi}{\rho^{2}\nu
i}e^{-\nu\pi i}$ (6.3)

in the second equality, and since $\nu$ is purely imaginary, the third
equality follows immediately. To this end, we note that
$|\Gamma(\nu)|^{2}=\Gamma(\nu)\Gamma(-\nu)$, it is then easily seen from (6.3)
that

$\Gamma(\nu)^{-1}=\frac{|\rho|}{\sqrt{2\pi}}\sqrt{-\ln(1-\rho^{2})}e^{\frac{\nu}{2}\pi
i+i\arg\Gamma(-\nu)}.$

Substituting this formula into (6.1), we then obtain the asymptotics
(1.10)–(1.12) by straightforward calculations, which completes the proof of
Theorem 1.1. ∎

###  6.2. Proofs of Theorems 1.2 and 1.3

We will prove Theorems 1.2 and 1.3 for Case I and Case II in a unified way.
Thus, it is always assumed that $(x,t)\in{\bf Case~{}~{}I^{T}}\cup{\bf
Case~{}~{}II^{T}}$ for some $T$ and the same notations might have different
definitions for different cases.

To proceed, we observe from Lemmas 4.7 and 5.6 that

$\displaystyle\|\hat{w}\|_{(L^{1}\cap L^{\infty})(\hat{\Gamma})}\leq
Ct^{-\frac{1}{2n+1}}.$ (6.4)

In particular, $\|\hat{w}\|_{L^{\infty}(\hat{\Gamma})}\to 0$ uniformly as
$t\to\infty$, which implies that there exist a suitable constant $T$ such that

$\|\hat{\mathcal{C}}_{\hat{w}}\|_{\mathcal{B}(L^{2}(\hat{\Gamma}))}\leq
C\|\hat{w}\|_{L^{\infty}(\hat{\Gamma})}\leq 1/2$ (6.5)

for all $(x,t)$, where

$\hat{\mathcal{C}}_{\hat{w}}f:=\hat{\mathcal{C}}_{-}(f\hat{w})$

with $\hat{\mathcal{C}}$ being the Cauchy transform associated with the
contour $\hat{\Gamma}$; see Section 1.2. By (6.5), the operator
$I-\hat{\mathcal{C}}_{\hat{w}}\in\mathcal{B}(L^{2}(\hat{\Gamma}))$ is
invertible and the RH problem 4.6 has a unique solution given by

$\displaystyle\hat{m}(x,t,\lambda)=I+\hat{\mathcal{C}}(\hat{\mu}\hat{w})=I+\frac{1}{2\pi
i}\int_{\hat{\Gamma}}(\hat{\mu}\hat{w})(x,t,s)\frac{ds}{s-\lambda},$ (6.6)

where $\hat{\mu}(x,t,\lambda)\in I+L^{2}(\hat{\Gamma})$ is defined by
$\hat{\mu}=I+(I-\hat{\mathcal{C}}_{\hat{w}})^{-1}\hat{\mathcal{C}}_{\hat{w}}I$.
Also, it is readily seen from (6.4) that

$\displaystyle\|\hat{\mu}(x,t,\cdot)-I\|_{L^{2}(\hat{\Gamma})}\leq\frac{C\|\hat{w}\|_{L^{2}(\hat{\Gamma})}}{1-\|\hat{\mathcal{C}}_{\hat{w}}\|_{\mathcal{B}(L^{2}(\hat{\Gamma}))}}\leq
Ct^{-\frac{1}{2n+1}}.$

For $\lambda\in\hat{\Gamma}$, it is easily seen from (4.18), (5.19), Lemmas
4.7 and 5.6 that as $t\to\infty$,

$\hat{w}(x,t,\lambda)=\frac{\hat{w}_{1}(y,\lambda)}{t^{\frac{1}{2n+1}}}+\frac{\hat{w}_{2}(y,\lambda)}{t^{\frac{2}{2n+1}}}+\cdots+\frac{\hat{w}_{N}(y,\lambda)}{t^{\frac{N}{2n+1}}}+\frac{\hat{w}_{err}(x,t,\lambda)}{t^{\frac{N+1}{2n+1}}},$
(6.7)

where the coefficients $\\{\hat{w}_{j}\\}_{j=1}^{N}$ are nonzero only for
$\lambda\in\partial D(0,\epsilon)$, and for $1\leq p\leq\infty$,

$\displaystyle\begin{cases}\|\hat{w}_{j}(y,\cdot)\|_{L^{p}(\hat{\Gamma})}\leq
C,\qquad j=1,\dots,N,\\\
\|\hat{w}_{err}(x,t,\cdot)\|_{L^{p}(\hat{\Gamma})}\leq C.\end{cases}$

As a consequence, we also have

$\hat{\mathcal{C}}_{\hat{w}}=\frac{\hat{\mathcal{C}}_{\hat{w}_{1}}}{t^{\frac{1}{2n+1}}}+\frac{\hat{\mathcal{C}}_{\hat{w}_{2}}}{t^{\frac{2}{2n+1}}}+\cdots+\frac{\hat{\mathcal{C}}_{\hat{w}_{N}}}{t^{\frac{N}{2n+1}}}+\frac{\hat{\mathcal{C}}_{\hat{w}_{err}}}{t^{\frac{N+1}{2n+1}}}.$

Since
$\hat{\mu}=\sum_{j=0}^{N}\hat{\mathcal{C}}_{\hat{w}}^{j}I+(I-\hat{\mathcal{C}}_{\hat{w}})^{-1}\hat{\mathcal{C}}_{\hat{w}}^{N+1}I$,
it is readily seen from the above expansion that

$\hat{\mu}(x,t,\lambda)=I+\frac{\hat{\mu}_{1}(y,\lambda)}{t^{\frac{1}{2n+1}}}+\cdots+\frac{\hat{\mu}_{N}(y,\lambda)}{t^{\frac{N}{2n+1}}}+\frac{\hat{\mu}_{err}(x,t,\lambda)}{t^{\frac{N+1}{2n+1}}},$
(6.8)

where the coefficients $\\{\hat{\mu}_{j}(y,\lambda)\\}_{j=1}^{N}$ are smooth
functions of $y\in[0,\infty)$ and

$\displaystyle\begin{cases}\|\hat{\mu}_{j}(y,\cdot)\|_{L^{2}(\hat{\Gamma})}\leq
C,\qquad j=1,\dots,N,\\\
\|\hat{\mu}_{err}(x,t,\cdot)\|_{L^{2}(\hat{\Gamma})}\leq C.\end{cases}$

By inverting the transformations (4.3) and (4.25), we obtain from (6.6), (6.7)
and (6.8) that

$\displaystyle\lim_{\lambda\to\infty}\lambda(m(x,t,\lambda)-I)=\lim_{\lambda\to\infty}\lambda(\hat{m}(x,t,\lambda)-I)$
$\displaystyle=-\frac{1}{2\pi
i}\int_{\hat{\Gamma}}\hat{\mu}(x,t,\lambda)\hat{w}(x,t,\lambda)d\lambda=-\sum_{j=1}^{N}\frac{h_{j}(y)}{t^{\frac{j}{2n+1}}}+\mathcal{O}\big{(}t^{-\frac{N+1}{2n+1}}\big{)},\quad
t\to\infty,$

uniformly for $(x,t)$, where $\\{h_{j}(y)\\}_{j=1}^{N}$ are smooth functions
with $h_{1}(y)=g_{1}(y)$. In view of (2.9), this leads to the asymptotic
formula

$\displaystyle
u(x,t)=2\lim_{\lambda\to\infty}\lambda(m(x,t,\lambda))_{21}=\sum_{j=1}^{N}\frac{u_{j}(y)}{t^{\frac{j}{2n+1}}}+\mathcal{O}\big{(}t^{-\frac{N+1}{2n+1}}\big{)},$

where

$\displaystyle u_{j}(y)=-2(h_{j})_{21}(y),\qquad j=1,\dots,N,$ (6.9)

are smooth functions of $y\in[0,\infty)$. Thus, we have proved (1.13).

Since $h_{1}(y)=g_{1}(y)$, we obtain (1.14) from (4.21) and (5.20). If it is
further assumed that $r(0)=0$, then $\hat{w}_{1}=\hat{\mu}_{1}=0$, and so
$h_{j}(y)=g_{j}(y)$ for $j=1,2,3$. This, together with (6.9) and
(4.22)–(4.24), implies (1.15) and (1.16). We thus complete the proof of
Theorems 1.2 and 1.3 for ${\bf Case~{}~{}I}$ and ${\bf Case~{}~{}II}$. ∎

##  Appendix A Model RH problems

###  A.1. The parabolic cylinder parametrix

The parabolic cylinder parametrix
$\Psi^{(\text{PC})}(\zeta)=\Psi^{(\text{PC})}(\zeta;\nu)$ with $\nu$ being a
real or complex parameter is a solution of the following RH problem; see [22,
Chapter 9].

######  RH problem A.1.

  1. _(a)_

$\Psi^{(\text{PC})}(\zeta)$ is analytic in
$\mathbb{C}\setminus\\{\cup^{4}_{j=0}\widehat{\Sigma}_{j}\cup\\{0\\}\\}$,
where the contours $\widehat{\Sigma}_{j}$, $j=0,\ldots,4,$ are indicated in
Figure 10.

$0$$\widehat{\Sigma}_{0}$$\widehat{\Sigma}_{1}$$\widehat{\Sigma}_{2}$$\widehat{\Sigma}_{3}$$\widehat{\Sigma}_{4}$
Figure 10. The jump contours for the RH problem A.1 for $\Psi^{(\text{PC})}$.

  2. _(b)_

$\Psi^{(\text{PC})}$ satisfies the following jump condition:

$\Psi^{(\text{PC})}_{+}(\zeta)=\Psi^{(\text{PC})}_{-}(\zeta)\left\\{\begin{array}[]{ll}H_{j},&\hbox{$\zeta\in\widehat{\Sigma}_{j}$,
$j=0,\ldots,3$,}\\\ e^{2\nu\pi
i\sigma_{3}},&\hbox{$\zeta\in\widehat{\Sigma}_{4}$,}\end{array}\right.$ (A.1)

where

$H_{0}=\begin{pmatrix}1&0\\\ h_{0}&1\end{pmatrix},\quad
H_{1}=\begin{pmatrix}1&h_{1}\\\ 0&1\end{pmatrix},\quad
H_{i+2}(z)=e^{i\pi(\nu+\frac{1}{2})\sigma_{3}}H_{i}e^{-i\pi(\nu+\frac{1}{2})\sigma_{3}},~{}~{}i=0,1,$

with

$h_{0}=-i\frac{\sqrt{2\pi}}{\Gamma(\nu+1)},\qquad
h_{1}=\frac{\sqrt{2\pi}}{\Gamma(-\nu)}e^{i\pi\nu}.$

  3. _(c)_

As $\zeta\to\infty$, we have

$\Psi^{(\text{PC})}(\zeta)=\frac{\zeta^{-\sigma_{3}/2}}{\sqrt{2}}\left(\begin{pmatrix}1&1\\\
1&-1\end{pmatrix}+\frac{1}{2\zeta^{2}}\begin{pmatrix}(\nu+1)(\nu+2)&-\nu(\nu-1)\\\
(\nu+1)(\nu-2)&\nu(\nu+3)\end{pmatrix}+\mathcal{O}(\zeta^{-4})\right)\\\
\times e^{(\frac{\zeta^{2}}{4}-(\nu+\frac{1}{2})\ln\zeta)\sigma_{3}}.$

From [22, Section 9.4], it follows that the above RH problem can be solved
explicitly in the following way. For $z$ belonging to the region bounded by
the rays $\widehat{\Sigma}_{4}$ and $\widehat{\Sigma}_{0}$,

$\Psi^{(\text{PC})}(\zeta)=2^{-\frac{\sigma_{3}}{2}}\begin{pmatrix}D_{-\nu-1}(i\zeta)&D_{\nu}(\zeta)\\\
\frac{d}{d\zeta}D_{-\nu-1}(i\zeta)&\frac{d}{d\zeta}D_{\nu}(\zeta)\end{pmatrix}\begin{pmatrix}e^{\frac{(\nu+1)\pi}{2}i}&0\\\
0&1\end{pmatrix},$

where $D_{\nu}$ stands for the parabolic cylinder function (cf. [35, Chapter
12]). The explicit formula of $\Psi^{(\text{PC})}(\zeta)$ in the other sectors
is then determined by using the jump condition (A.1).

###  A.2. The Model RH problem for Case I and Case III

The long-time asymptotics of $m$ in Case I and Case III is related to the
solution $m^{\Upsilon^{*}}$ of the following model RH problem parameterized by
$(y,t)$.

######  RH problem A.2.

  1. _(a)_

$m^{\Upsilon^{*}}(y,t,\cdot)\in
I+\dot{E}^{2}({\mathbb{C}}\setminus\Upsilon^{*})$, where

$\Upsilon^{*}:=\Upsilon_{1}\cup\Upsilon_{2n+1}^{*}\cup\Upsilon_{2n+2}^{*}\cup\Upsilon_{4n+2},$

and where the contours $\Upsilon_{1},\Upsilon_{2n+1}^{*},\Upsilon_{2n+2}^{*}$
and $\Upsilon_{4n+2}$ are illustrated in Figure 4.

  2. _(b)_

For a.e. $\lambda\in\Upsilon^{*}$, we have

$m_{+}^{\Upsilon^{*}}(y,t,z)=m_{-}^{\Upsilon^{*}}(y,t,z)v^{\Upsilon^{*}}(y,t,z),$

where

$v^{\Upsilon^{*}}(y,t,z)=\begin{cases}\begin{pmatrix}1&0\\\
p_{N}(t,z)e^{2i\left(yz+\frac{(2z)^{2n+1}}{4n+2}\right)}&1\end{pmatrix},&z\in\Upsilon_{1}\cup\Upsilon_{2n+1}^{*},\\\
\begin{pmatrix}1&-p_{N}^{*}(t,z)e^{-2i\left(yz+\frac{(2z)^{2n+1}}{4n+2}\right)}\\\
0&1\end{pmatrix},&z\in\Upsilon_{2n+2}^{*}\cup\Upsilon_{4n+2},\end{cases}$
(A.2)

and where

$\displaystyle
p_{N}(t,z)=s+\sum_{j=1}^{N}\frac{p_{j}z^{j}}{t^{\frac{j}{2n+1}}},\qquad
n\in{\mathbb{N}},$ (A.3)

is a polynomial of degree $N$ in $zt^{-1/(2n+1)}$ with coefficients $s=i\rho$,
$-1<\rho<1$, and $\\{p_{j}\\}_{1}^{N}\subset{\mathbb{C}}$.

Note that if $p_{j}$, $j=1,\ldots,N$, in (A.3) all vanishes, the above RH
problem reduces to the RH problem 2.5 for the Painlevé II hierarchy, up to a
conjugation. For general $p_{j}$, we have the following lemma about the
properties of $m^{\Upsilon^{*}}$.

######  Lemma A.3.

There exists a suitable $T$ such that the RH problem A.2 has a unique solution
$m^{\Upsilon^{*}}(y,t,z)$ whenever $y\geq 0$ and $t\geq T$. Moreover, the
following properties of $m^{\Upsilon^{*}}$ hold.

  1. _(a)_

For each integer $N\geq 1$, there are smooth functions
$\\{\Phi_{jl}^{\Upsilon^{*}}(y)\\}$ of $y\in[0,\infty)$ such that, as
$z\to\infty$,

$m^{\Upsilon^{*}}(y,t,z)=I+\sum_{j=1}^{N}\sum_{l=0}^{N}\frac{\Phi_{jl}^{\Upsilon^{*}}(y)}{z^{j}t^{\frac{l}{2n+1}}}+\mathcal{O}\left(\frac{t^{-\frac{N+1}{2n+1}}}{|z|}+\frac{1}{|z|^{N+1}}\right),$
(A.4)

uniformly with respect to $\arg z\in[0,2\pi]$, $y\geq 0$, and $t\geq T$.

  2. _(b)_

$m^{\Upsilon^{*}}$ is uniformly bounded, i.e.,

$\displaystyle\sup_{y\geq 0}\sup_{t\geq
T}\sup_{z\in{\mathbb{C}}\setminus\Upsilon^{*}}|m^{\Upsilon^{*}}(y,t,z)|<\infty,$
(A.5)

and satisfies the symmetry relation

$\displaystyle
m^{\Upsilon^{*}}(y,t,z)=\sigma_{1}\overline{m^{\Upsilon^{*}}(y,t,\bar{z})}\sigma_{1}.$
(A.6)

If $p_{N}(t,z)=-\overline{p_{N}(t,-\bar{z})}$, we also have

$\displaystyle
m^{\Upsilon^{*}}(y,t,z)=\sigma_{1}\sigma_{3}m^{\Upsilon^{*}}(y,t,-z)\sigma_{3}\sigma_{1}.$
(A.7)

  3. _(c)_

For the leading coefficient in (A.4), we have

$\displaystyle(\Phi_{10}^{\Upsilon^{*}})_{12}(y)=(\Phi_{10}^{\Upsilon^{*}})_{21}(y)=-\frac{1}{2}q_{AS,n}((-1)^{n+1}y;\rho),$
(A.8)

where $q_{\textrm{AS},n}$ is the generalized Ablowitz-Segur solution of the
Painlevé II hierarchy (1.5) as stated in Theorem 1.1. Furthermore, if $s=0$,
$p_{1}\in{\mathbb{R}}$, and $p_{2}\in i{\mathbb{R}}$, the first few terms in
(A.4) are given by

$\displaystyle\Phi_{10}^{\Upsilon^{*}}(y)=0,$ (A.9)
$\displaystyle\Phi_{11}^{\Upsilon^{*}}(y)=\frac{p_{1}}{4}\text{Ai}_{2n+1}^{\prime}(y)\sigma_{1},$
(A.10)
$\displaystyle\Phi_{12}^{\Upsilon^{*}}(y)=\frac{p_{1}^{2}}{8i}\bigg{(}\int_{y}^{\infty}(\text{Ai}_{2n+1}^{\prime}(y^{\prime}))^{2}dy^{\prime}\bigg{)}\sigma_{3}+\frac{p_{2}}{8i}\text{Ai}_{2n+1}^{\prime\prime}(y)\sigma_{1},$
(A.11)
$\displaystyle\Phi_{21}^{\Upsilon^{*}}(y)=-\frac{p_{1}}{8i}\text{Ai}_{2n+1}^{\prime\prime}(y)\sigma_{3}\sigma_{1}.$
(A.12)

###### Proof.

We follow the strategy in [9]. Define

$\displaystyle m^{P}(z):=\begin{pmatrix}1&0\\\
0&i\end{pmatrix}\Psi(y,\rho,z)e^{i\Xi(y,z)\sigma_{3}}\begin{pmatrix}1&0\\\
0&i\end{pmatrix}^{-1},$ (A.13)

where $\Psi$ solves the RH problem 2.5 and $\Xi$ is given in (2.14). It is
then readily seen that $m^{P}$ satisfies the following RH problem.

######  RH problem A.4.

  1. _(a)_

$m^{P}(z)=m^{P}(y,\rho,z)$ is defined and analytic in
$\mathbb{C}\setminus\Upsilon^{*}$.

  2. _(b)_

For $z\in\Upsilon^{*}$, we have

$m^{P}_{+}(z)=m^{P}_{-}(z)v^{P},$ (A.14)

where

$v^{P}(z):=\left\\{\begin{array}[]{ll}\begin{pmatrix}1&0\\\ i\rho
e^{2i\left(yz+\frac{(2z)^{2n+1}}{4n+2}\right)}&1\end{pmatrix},&\qquad\hbox{$z\in\Upsilon_{1}\cup\Upsilon^{*}_{2n+1}$,}\\\
\begin{pmatrix}1&i\rho e^{-2i\left(yz+\frac{(2z)^{2n+1}}{4n+2}\right)}\\\
0&1\end{pmatrix},&\qquad\hbox{$z\in\Upsilon^{*}_{2n+2}\cup\Upsilon_{4n+2}$.}\end{array}\right.$
(A.15)

  3. _(c)_

As $z\to\infty$, we have

$m^{P}(z)=I+\sum_{j=1}^{N}\frac{m^{P}_{j}(y)}{z^{j}}+\mathcal{O}(z^{-N-1}),$
(A.16)

where $m^{P}_{j}=\begin{pmatrix}1&0\\\
0&i\end{pmatrix}\Psi_{j}\begin{pmatrix}1&0\\\ 0&i\end{pmatrix}^{-1}$ with
$\Psi_{j}$ given in (2.16) for $j=1,\ldots,N$.

  4. _(d)_

$m^{P}(z)$ is bounded near the origin.

By (A.13) and (2.15), we have

$q_{\textrm{AS},n}(y;\rho):=\left\\{\begin{array}[]{ll}-2(m^{P}_{1})_{12}(y)=-2(m^{P}_{1})_{21}(y),&\hbox{$n$
odd,}\\\ -2(m^{P}_{1})_{12}(-y)=-2(m^{P}_{1})_{21}(-y),&\hbox{$n$
even,}\end{array}\right.$ (A.17)

and

$\displaystyle\sup_{y\geq-
C_{1}}\sup_{z\in{\mathbb{C}}\setminus\Upsilon^{*}}|m^{P}(y,z)|<\infty.$ (A.18)

for each $C_{1}>0$.

It is straightforward to check that the matrix-valued function
$m^{\Upsilon^{*}}$ satisfies RH problem A.2 if and only if

$\displaystyle\hat{m}^{\Upsilon^{*}}:=m^{\Upsilon^{*}}(m^{{P}})^{-1}$ (A.19)

satisfies the following RH problem.

######  RH problem A.5.

  1. _(a)_

$\hat{m}^{\Upsilon^{*}}(y,t,\cdot)\in
I+\dot{E}^{2}({\mathbb{C}}\setminus\Upsilon^{*})$.

  2. _(b)_

For a.e. $\lambda\in\Upsilon^{*}$, we have

$m_{+}^{\Upsilon^{*}}(y,t,z)=m_{-}^{\Upsilon^{*}}(y,t,z)\hat{v}^{\Upsilon^{*}}(y,t,z),$

where

$\hat{v}^{\Upsilon^{*}}=m_{-}^{P}v^{\Upsilon^{*}}(m_{+}^{P})^{-1},$

and where $v^{\Upsilon^{*}}$ is given in (A.2).

By setting

$\hat{w}^{\Upsilon^{*}}:=\hat{v}^{\Upsilon^{*}}-I=m_{-}^{P}(v^{\Upsilon^{*}}-v^{P})(m_{+}^{P})^{-1},$

we have

$\displaystyle\hat{w}^{\Upsilon^{*}}(y,t,z)=\frac{\hat{w}_{1}^{\Upsilon^{*}}(y,z)}{t^{\frac{1}{2n+1}}}+\cdots+\frac{\hat{w}_{{N}}^{\Upsilon^{*}}(y,z)}{t^{\frac{N}{2n+1}}},$
(A.20)

where, for $j=1,\ldots,N$,

$\displaystyle\hat{w}_{j}^{\Upsilon^{*}}(y,z)=\begin{cases}m_{-}^{P}\begin{pmatrix}0&0\\\
p_{j}z^{j}e^{2i\left(yz+\frac{(2z)^{2n+1}}{4n+2}\right)}&0\end{pmatrix}(m_{+}^{P})^{-1},&z\in\Upsilon_{1}\cup\Upsilon^{*}_{2n+1},\\\
m_{-}^{P}\begin{pmatrix}0&-{p}^{*}_{j}z^{j}e^{-2i\left(yz+\frac{(2z)^{2n+1}}{4n+2}\right)}\\\
0&0\end{pmatrix}(m_{+}^{P})^{-1},&z\in\Upsilon^{*}_{2n+2}\cup\Upsilon_{4n+2}.\end{cases}$
(A.21)

Since $y\geq 0$, it follows that

$\displaystyle\begin{cases}\left|e^{2i\left(yz+\frac{(2z)^{2n+1}}{4n+2}\right)}\right|\leq
e^{-\frac{2^{2n+1}}{2n+1}|z|^{2n+1}},\quad&z\in\Upsilon_{1}\cup\Upsilon^{*}_{2n+1},\\\
\left|e^{-2i\left(yz+\frac{(2z)^{2n+1}}{4n+2}\right)}\right|\leq
e^{-\frac{2^{2n+1}}{2n+1}|z|^{2n+1}},\quad&z\in\Upsilon^{*}_{2n+2}\cup\Upsilon_{4n+2}.\end{cases}$

This, together with (A.21), implies that for any integer $m\geq 0$,

$\displaystyle|z^{m}\hat{w}_{j}^{\Upsilon^{*}}(y,z)|\leq$
$\displaystyle\;C|z|^{m+j}e^{-\frac{2^{2n+1}}{2n+1}|z|^{2n+1}}\leq
C(n)e^{-c(n)|z|^{2n+1}},\quad z\in\Upsilon^{*},$

and hence, for any $1\leq p\leq\infty$,

$\displaystyle\begin{cases}\|z^{m}\hat{w}_{j}^{\Upsilon^{*}}(y,z)\|_{L^{p}(\Upsilon^{*})}\leq
C,\\\ \|z^{m}\hat{w}^{\Upsilon^{*}}(y,t,z)\|_{L^{p}(\Upsilon^{*})}\leq
Ct^{-\frac{1}{2n+1}},\quad&t\geq 1.\end{cases}$ (A.22)

We then particularly have

$\|\mathcal{C}_{\hat{w}^{\Upsilon^{*}}(y,t,\cdot)}^{\Upsilon^{*}}\|_{\mathcal{B}(L^{2}(\Upsilon^{*})}\leq
C\|\hat{w}^{\Upsilon^{*}}\|_{L^{\infty}(\Upsilon^{*})},\quad t\geq 1.$

Therefore, there exists a suitable $T\geq 1$ such that the RH problem A.5 has
a unique solution which is given by

$\displaystyle\hat{m}^{\Upsilon^{*}}(y,t,z)$
$\displaystyle=I+\mathcal{C}^{\Upsilon^{*}}(\hat{\mu}^{\Upsilon^{*}}\hat{w}^{\Upsilon^{*}})$
$\displaystyle=I+\frac{1}{2\pi
i}\int_{\Upsilon^{*}}(\hat{\mu}^{\Upsilon^{*}}\hat{w}^{\Upsilon^{*}})(y,t,s)\frac{ds}{s-z},\quad
t\geq T,$ (A.23)

where $\hat{\mu}^{\Upsilon^{*}}(y,t,\cdot)\in I+L^{2}(\Upsilon^{*})$ is
defined by

$\displaystyle\hat{\mu}^{\Upsilon^{*}}=I+\mathcal{C}^{\Upsilon^{*}}_{\hat{w}^{\Upsilon^{*}}}\hat{\mu}^{\Upsilon^{*}}=I+(I-\mathcal{C}^{\Upsilon^{*}}_{\hat{w}^{\Upsilon^{*}}})^{-1}\mathcal{C}^{\Upsilon^{*}}_{\hat{w}^{\Upsilon^{*}}}I.$
(A.24)

Note that

$\mathcal{C}^{\Upsilon^{*}}_{\hat{w}^{\Upsilon^{*}}}=\frac{\mathcal{C}^{\Upsilon^{*}}_{\hat{w}_{1}^{\Upsilon^{*}}}}{t^{\frac{1}{2n+1}}}+\frac{\mathcal{C}^{\Upsilon^{*}}_{\hat{w}_{2}^{\Upsilon^{*}}}}{t^{\frac{2}{2n+1}}}+\cdots+\frac{\mathcal{C}^{\Upsilon^{*}}_{\hat{w}_{N}^{\Upsilon^{*}}}}{t^{\frac{N}{2n+1}}},\qquad
N\in{\mathbb{N}},$

it follows from (A.24) that

$\displaystyle\hat{\mu}^{\Upsilon^{*}}(y,t,z)$
$\displaystyle=\sum_{r=0}^{N}(\mathcal{C}^{\Upsilon^{*}}_{\hat{w}^{\Upsilon^{*}}})^{r}I+(I-\mathcal{C}^{\Upsilon^{*}}_{\hat{w}^{\Upsilon^{*}}})^{-1}(\mathcal{C}^{\Upsilon^{*}}_{\hat{w}^{\Upsilon^{*}}})^{N+1}I$
$\displaystyle=I+\sum_{j=1}^{N}\frac{\hat{\mu}_{j}^{\Upsilon^{*}}(y,z)}{t^{\frac{j}{2n+1}}}+\frac{\hat{\mu}_{err}^{\Upsilon^{*}}(y,t,z)}{t^{\frac{N+1}{2n+1}}},$
(A.25)

where $\hat{\mu}_{j}^{\Upsilon^{*}}$ is a sum of terms of the form
$\prod_{i=1}^{r}\mathcal{C}^{\Upsilon^{*}}_{\hat{w}_{j_{i}}^{\Upsilon^{*}}}I$
with $\Sigma_{i=1}^{r}j_{i}=j$, and the $\hat{\mu}_{err}^{\Upsilon^{*}}$
involves terms of the same form (but with $j_{1}+\cdots j_{r}\geq N+1$). By
(A.22), it follows that

$\displaystyle\begin{cases}\|\hat{\mu}_{j}^{\Upsilon^{*}}(y,\cdot)\|_{L^{2}(\Upsilon^{*})}\leq
C,&j=1,\dots,N,\\\
\|\hat{\mu}_{err}^{\Upsilon^{*}}(y,t,\cdot)\|_{L^{2}(\Upsilon^{*})}\leq
C,\quad&t\geq T,\end{cases}$ (A.26)

for some $C>0$.

We now plug (A.20) and (A.25) back into (A.23), it is readily seen from (A.22)
and (A.26) that, as $z\to\infty$,

$\displaystyle\hat{m}^{\Upsilon^{*}}(y,t,z)=$
$\displaystyle\;I-\sum_{j=1}^{N}\frac{1}{2\pi
iz^{j}}\bigg{\\{}\sum_{l=1}^{N}t^{-\frac{l}{2n+1}}\int_{\Upsilon^{*}}s^{j-1}\bigg{(}\hat{w}_{l}^{\Upsilon^{*}}+\sum_{i=1}^{l-1}\hat{\mu}_{l-i}^{\Upsilon^{*}}\hat{w}_{i}^{\Upsilon^{*}}\bigg{)}ds+\mathcal{O}\big{(}t^{-\frac{N+1}{2n+1}}\big{)}\bigg{\\}}$
$\displaystyle+\mathcal{O}\big{(}|z|^{-N-1}t^{-\frac{1}{2n+1}}\big{)},$ (A.27)

uniformly for $y\geq 0$ and $t\geq T$. Denote by

$\hat{\Phi}_{jl}^{\Upsilon^{*}}(y)=-\frac{1}{2\pi
i}\int_{\Upsilon^{*}}s^{j-1}\bigg{(}\hat{w}_{l}^{\Upsilon^{*}}+\sum_{i=1}^{l-1}\hat{\mu}_{l-i}^{\Upsilon^{*}}\hat{w}_{i}^{\Upsilon^{*}}\bigg{)}(y,s)ds,\qquad
1\leq j,l\leq N.$ (A.28)

we can rewrite (A.27) as

$\displaystyle\hat{m}^{\Upsilon^{*}}(y,t,z)=I+\sum_{j=1}^{N}\sum_{l=1}^{N}\frac{\hat{\Phi}_{jl}^{\Upsilon^{*}}(y)}{z^{j}t^{\frac{l}{2n+1}}}+\mathcal{O}\biggl{(}\frac{t^{-\frac{N+1}{2n+1}}}{|z|}+\frac{t^{-\frac{1}{2n+1}}}{|z|^{N+1}}\biggr{)},$
(A.29)

uniformly with respect to $\arg z\in[0,2\pi]$, $y\geq 0$ and $t\geq T$.
Moreover, from (A.28), (A.21) and (A.25), we obtain the smoothness of
$\hat{\Phi}_{jl}^{\Upsilon^{*}}(y)$. A combination of (A.16), (A.19) and
(A.29) gives us (A.4) with

$\Phi_{j0}^{\Upsilon^{*}}(y)=m_{j}^{P}(z),\qquad j=1,\ldots,N.$ (A.30)

The bound (A.5) follows from (A.18), (A.29) and the fact that the contour can
be deformed. The symmetries (A.6) and (A.7) follow from the analogous
symmetries for the jump $v^{\Upsilon^{*}}$.

We finally prove Item (c) of the lemma. The relation (A.8) follows directly
from (A.17) and (A.30). If $s=0$, $p_{1}\in{\mathbb{R}}$, and $p_{2}\in
i{\mathbb{R}}$, it is easily seen from (A.13) and Remark 2.6 that $m^{P}\equiv
I$, and by (A.21), we have

$\displaystyle\hat{w}_{1}^{\Upsilon^{*}}$
$\displaystyle=\begin{pmatrix}0&-p_{1}ze^{-2i\left(yz+\frac{(2z)^{2n+1}}{4n+2}\right)}1_{\Upsilon_{2n+2}^{*}\cup\Upsilon_{4n+2}}(z)\\\
p_{1}ze^{2i\left(yz+\frac{(2z)^{2n+1}}{4n+2}\right)}1_{\Upsilon_{1}\cup\Upsilon_{2n+1}^{*}}(z)&0\end{pmatrix},$
(A.31) $\displaystyle\hat{w}_{2}^{\Upsilon^{*}}$
$\displaystyle=\begin{pmatrix}0&p_{2}z^{2}e^{-2i\left(yz+\frac{(2z)^{2n+1}}{4n+2}\right)}1_{\Upsilon_{2n+2}^{*}\cup\Upsilon_{4n+2}}(z)\\\
p_{2}z^{2}e^{2i\left(yz+\frac{(2z)^{2n+1}}{4n+2}\right)}1_{\Upsilon_{1}\cup\Upsilon^{*}_{2n+1}}(z)&0\end{pmatrix},$
(A.32)

where $1_{A}(z)$ denotes the characteristic function of the set
$A\subset{\mathbb{C}}$. In view of the function $\text{Ai}_{2n+1}(y)$ defined
in (1.7), a change of variable shows that

$\displaystyle\int_{\Upsilon_{1}\cup\Upsilon_{2n+1}^{*}}e^{2i\left(yz+\frac{(2z)^{2n+1}}{4n+2}\right)}dz=\int_{\Upsilon_{2n+2}^{*}\cup\Upsilon_{4n+2}}e^{-2i\left(yz+\frac{(2z)^{2n+1}}{4n+2}\right)}dz=\pi\text{Ai}_{2n+1}(y).$

By differentiating the above formula $j$ times with respect to $y$, it follows
that

$\int_{\Upsilon_{1}\cup\Upsilon_{2n+1}^{*}}z^{j}e^{2i\left(yz+\frac{(2z)^{2n+1}}{4n+2}\right)}dz=(-1)^{j}\int_{\Upsilon_{2n+2}^{*}\cup\Upsilon_{4n+2}}z^{j}e^{-2i\left(yz+\frac{(2z)^{2n+1}}{4n+2}\right)}dz\\\
=\frac{\pi\text{Ai}_{2n+1}^{(j)}(y)}{(2i)^{j}},$ (A.33)

for each integer $j\geq 0$. A combination of (A.28) and (A.30)–(A.33) then
implies that

$\displaystyle\Phi_{10}^{\Upsilon^{*}}(y)$ $\displaystyle=0,$
$\displaystyle\Phi_{11}^{\Upsilon^{*}}(y)$
$\displaystyle=\hat{\Phi}_{11}^{\Upsilon^{*}}(y)=-\frac{1}{2\pi
i}\int_{\Upsilon^{*}}\hat{w}_{1}^{\Upsilon^{*}}dz$
$\displaystyle=-\frac{p_{1}}{2\pi
i}\begin{pmatrix}0&-\int_{\Upsilon_{2n+2}^{*}\cup\Upsilon_{4n+2}}ze^{-2i\left(yz+\frac{(2z)^{2n+1}}{4n+2}\right)}dz\\\
\int_{\Upsilon_{1}\cup\Upsilon_{2n+1}^{*}}ze^{2i\left(yz+\frac{(2z)^{2n+1}}{4n+2}\right)}dz&0\end{pmatrix}$
$\displaystyle=-\frac{p_{1}}{2\pi
i}\frac{\pi\text{Ai}^{\prime}_{2n+1}(y)}{2i}\begin{pmatrix}0&1\\\
1&0\end{pmatrix},$ $\displaystyle\Phi_{21}^{\Upsilon^{*}}(y)$
$\displaystyle=\hat{\Phi}_{21}^{\Upsilon^{*}}(y)=-\frac{1}{2\pi
i}\int_{\Upsilon^{*}}z\hat{w}_{1}^{\Upsilon^{*}}dz$
$\displaystyle=-\frac{p_{1}}{2\pi
i}\begin{pmatrix}0&-\int_{\Upsilon_{2n+2}^{*}\cup\Upsilon_{4n+2}}z^{2}e^{-2i\left(yz+\frac{(2z)^{2n+1}}{4n+2}\right)}dz\\\
\int_{\Upsilon_{1}\cup\Upsilon_{2n+1}^{*}}z^{2}e^{2i\left(yz+\frac{(2z)^{2n+1}}{4n+2}\right)}dz&0\end{pmatrix}$
$\displaystyle=\frac{p_{1}}{2\pi
i}\frac{\pi\text{Ai}_{2n+1}^{\prime\prime}(y)}{4}\begin{pmatrix}0&-1\\\
1&0\end{pmatrix},$

which are (A.9), (A.10) and (A.12). It remains to show (A.11). By (A.28) and
(A.25), we have

$\Phi_{12}^{\Upsilon^{*}}(y)=-\frac{1}{2\pi
i}\int_{\Upsilon^{*}}(\hat{w}_{2}^{\Upsilon^{*}}(y,z)+\hat{\mu}_{1}^{\Upsilon^{*}}(y,z)\hat{w}_{1}^{\Upsilon^{*}}(y,z))dz,$
(A.34)

and

$\displaystyle\hat{\mu}_{1}^{\Upsilon^{*}}(y,z)$
$\displaystyle=\mathcal{C}^{\Upsilon^{*}}_{\hat{w}_{1}^{\Upsilon^{*}}}I=\mathcal{C}^{\Upsilon^{*}}_{-}(\hat{w}_{1}^{\Upsilon^{*}})=\frac{1}{2\pi
i}\int_{\Upsilon^{*}}\frac{\hat{w}_{1}^{\Upsilon^{*}}(y,s)ds}{s-z_{-}}$
$\displaystyle=\frac{p_{1}}{2\pi
i}\begin{pmatrix}0&-\int_{\Upsilon_{2n+2}^{*}\cup\Upsilon_{4n+2}}\frac{se^{-2i\left(yz+\frac{(2z)^{2n+1}}{4n+2}\right)}}{s-z_{-}}ds\\\
\int_{\Upsilon_{1}\cup\Upsilon_{2n+1}^{*}}\frac{se^{2i\left(yz+\frac{(2z)^{2n+1}}{4n+2}\right)}}{s-z_{-}}ds&0\end{pmatrix}.$

We now define

$\displaystyle
F(y)=\int_{\Upsilon^{*}}\hat{\mu}_{1}^{\Upsilon^{*}}\hat{w}_{1}^{\Upsilon^{*}}dz=-\frac{p_{1}^{2}}{2\pi
i}\begin{pmatrix}F_{1}(y)&0\\\ 0&F_{2}(y)\end{pmatrix},$

where

$\displaystyle F_{1}(y)$
$\displaystyle=\int_{\Upsilon_{1}\cup\Upsilon_{2n+1}^{*}}\bigg{(}\int_{\Upsilon_{2n+2}^{*}\cup\Upsilon_{4n+2}}\frac{sze^{-2i\left(ys+\frac{(2s)^{2n+1}}{4n+2}\right)}e^{2i\left(yz+\frac{(2z)^{2n+1}}{4n+2}\right)}}{s-z}ds\bigg{)}dz,$
$\displaystyle F_{2}(y)$
$\displaystyle=\int_{\Upsilon_{2n+2}^{*}\cup\Upsilon_{4n+2}}\bigg{(}\int_{\Upsilon_{1}\cup\Upsilon_{2n+1}^{*}}\frac{sze^{2i\left(ys+\frac{(2s)^{2n+1}}{4n+2}\right)}e^{-2i\left(yz+\frac{(2z)^{2n+1}}{4n+2}\right)}}{s-z}ds\bigg{)}dz.$

Since $F_{2}(y)=-F_{1}(y)$ (by Fubini’s theorem), it follows that

$\displaystyle F(y)=-\frac{p_{1}^{2}}{2\pi i}F_{1}(y)\sigma_{3}.$

Differentiating both sides of the above two equalities with respect to $y$, we
observe from (A.33) that

$\displaystyle F^{\prime}(y)$
$\displaystyle=\frac{p_{1}^{2}}{\pi}\bigg{(}\int_{\Upsilon_{1}\cup\Upsilon_{2n+1}^{*}}ze^{2i\left(yz+\frac{(2z)^{2n+1}}{4n+2}\right)}dz\bigg{)}\bigg{(}\int_{\Upsilon_{2n+2}^{*}\cup\Upsilon_{4n+2}}se^{-2i\left(ys+\frac{(2s)^{2n+1}}{4n+2}\right)}ds\bigg{)}\sigma_{3}$
$\displaystyle=\frac{p_{1}^{2}}{\pi}\bigg{(}\frac{\pi\text{Ai}_{2n+1}^{\prime}(y)}{2i}\bigg{)}\bigg{(}-\frac{\pi\text{Ai}_{2n+1}^{\prime}(y)}{2i}\bigg{)}\sigma_{3}=\frac{\pi
p_{1}^{2}}{4}(\text{Ai}_{2n+1}^{\prime}(y))^{2}\sigma_{3}.$

Note that $F(y)\to 0$ as $y\to\infty$, we have

$\int_{\Upsilon^{*}}\hat{\mu}_{1}^{\Upsilon^{*}}(y,z)\hat{w}_{1}^{\Upsilon^{*}}(y,z)dz=F(y)=\int_{+\infty}^{y}F^{\prime}(y^{\prime})dy^{\prime}=\frac{\pi
p_{1}^{2}}{4}\left(\int_{+\infty}^{y}(\text{Ai}_{2n+1}^{\prime}(y^{\prime}))^{2}dy^{\prime}\right)\sigma_{3}.$

Combining the above formula and (A.32)–(A.34), we arrive at

$\displaystyle\Phi_{12}^{\Upsilon^{*}}(y)$ $\displaystyle=-\frac{p_{2}}{2\pi
i}\begin{pmatrix}0&\int_{\Upsilon_{2n+2}^{*}\cup\Upsilon_{4n+2}}z^{2}e^{-2i\left(yz+\frac{(2z)^{2n+1}}{4n+2}\right)}dz\\\
\int_{\Upsilon_{1}\cup\Upsilon_{2n+1}^{*}}z^{2}e^{2i\left(yz+\frac{(2z)^{2n+1}}{4n+2}\right)}dz&0\end{pmatrix}$
$\displaystyle\quad-\frac{1}{2\pi i}\frac{\pi
p_{1}^{2}}{4}\left(\int_{+\infty}^{y}(\text{Ai}_{2n+1}^{\prime}(y^{\prime}))^{2}dy^{\prime}\right)\sigma_{3}$
$\displaystyle=\frac{p_{2}}{2\pi
i}\frac{\pi}{4}\text{Ai}_{2n+1}^{\prime\prime}(y)\sigma_{1}+\frac{p_{1}^{2}}{8i}\left(\int_{y}^{\infty}(\text{Ai}_{2n+1}^{\prime}(y^{\prime}))^{2}dy^{\prime}\right)\sigma_{3},$

which is (A.11).

This completes the proof of Lemma A.3. ∎

###  A.3. The Model RH problem for Case II and Case IV

For each $z_{0}\geq 0$, we define

$Z_{1}=z_{0}+\Upsilon_{1},\quad Z_{2}=-z_{0}+\Upsilon_{2n+1}^{*},\quad
Z_{3}=-z_{0}+\Upsilon_{2n+2}^{*},\quad Z_{4}=z_{0}+\Upsilon_{4n+2},$ (A.35)

where the contours $\Upsilon_{1},\Upsilon_{2n+1}^{*},\Upsilon_{2n+2}$ and
$\Upsilon_{4n+2}$ are shown in Figure 4. The long-time asymptotics of $m$ in
Case II and Case IV is related to the solution $m^{Z}$ of the following model
RH problem parameterized by $y\leq 0$, $t\geq 0$, and $z_{0}\geq 0$.

######  RH problem A.6.

  1. _(a)_

$m^{Z}(y,t,z_{0},\cdot)\in I+\dot{E}^{2}({\mathbb{C}}\setminus Z)$, where

$Z:=\cup_{j=1}^{4}Z_{j}\cup(-z_{0},z_{0});$ (A.36)

see Figure 11 for an illustration and the orientation.

  2. _(b)_

For a.e. $\lambda\in Z$, we have

$m_{+}^{Z}(y,t,z_{0},z)=m_{-}^{Z}(y,t,z_{0},z)v^{Z}(y,t,z_{0},z),$ (A.37)

where

$\displaystyle v^{Z}(y,t,z_{0},z)$
$\displaystyle=\begin{cases}\begin{pmatrix}1&0\\\
p_{N}(t,z)e^{2i\left(yz+\frac{(2z)^{2n+1}}{4n+2}\right)}&1\end{pmatrix},&z\in
Z_{1}\cup Z_{2},\\\
\begin{pmatrix}1&-p_{N}^{*}(t,z)e^{-2i\left(yz+\frac{(2z)^{2n+1}}{4n+2}\right)}\\\
0&1\end{pmatrix},&z\in Z_{3}\cup Z_{4},\\\
\begin{pmatrix}1-|p_{N}(t,z)|^{2}&-p_{N}^{*}(t,z)e^{-2i\left(yz+\frac{(2z)^{2n+1}}{4n+2}\right)}\\\
p_{N}(t,z)e^{2i\left(yz+\frac{(2z)^{2n+1}}{4n+2}\right)}&1\end{pmatrix},&z\in(-z_{0},z_{0}),\end{cases}$
(A.38)

with $p_{N}(t,z)$ given in (A.3).

\begin{overpic}[width=265.99251pt]{Z.pdf} \put(95.0,36.0){\small$Z_{1}$}
\put(2.0,36.0){\small$Z_{2}$} \put(2.5,6.0){\small$Z_{3}$}
\put(95.0,6.0){\small$Z_{4}$} \put(62.0,36.0){\small$\Upsilon_{1}$}
\put(34.0,36.0){\small$\Upsilon_{2n+1}^{*}$}
\put(62.0,6.0){\small$\Upsilon_{4n+2}$}
\put(34.0,6.0){\small$\Upsilon_{2n+2}^{*}$} \put(62.0,18.0){\small$z_{0}$}
\put(34.0,18.0){\small$-z_{0}$} \end{overpic} Figure 11.  The contour $Z$ for
the RH problem A.6 for $m^{Z}$.

Define the parameter subset $\mathcal{P}_{T}$ of ${\mathbb{R}}^{3}$ by

$\displaystyle\mathcal{P}_{T}=\\{(y,t,z_{0})\in{\mathbb{R}}^{3}\,|\,-C_{1}\leq
y\leq 0,\,t\geq T,\,\sqrt[2n]{|y|}/2\leq z_{0}\leq C_{2}\\},$ (A.39)

where $C_{1},C_{2}>0$ are constants, we have the following lemma about the
properties of $m^{Z}$, which serves as the counterpart of Lemma A.3 in Case I
and Case III.

######  Lemma A.7.

There exists a suitable $T\geq 1$ such that the RH problem (A.6) has a unique
solution $m^{Z}(y,t,z_{0},z)$ whenever $(y,t,z_{0})\in\mathcal{P}_{T}$.
Moreover, the following properties of $m^{Z}$ hold.

  1. _(a)_

For each integer $N\geq 1$,

$\displaystyle
m^{Z}(y,t,z_{0},z)=I+\sum_{j=1}^{N}\sum_{l=0}^{N}\frac{\Phi_{jl}^{\Upsilon}(y)}{z^{j}t^{\frac{l}{2n+1}}}+\mathcal{O}\left(\frac{t^{-\frac{N+1}{2n+1}}}{|z|}+\frac{1}{|z|^{N+1}}\right),$

uniformly with respect to $\arg z\in[0,2\pi]$ and
$(y,t,z_{0})\in\mathcal{P}_{T}$ as $z\to\infty$, where
$\\{\Phi_{jl}^{\Upsilon}(y)\\}$ are smooth functions of $y\in{\mathbb{R}}$
which coincide with the functions in (A.4) for $y\geq 0$ and satisfy the
properties indicated in Item (c) of Lemma A.3.

  2. _(b)_

$m^{Z}$ is uniformly bounded, i.e.,

$\displaystyle\sup_{(y,t,z_{0})\in\mathcal{P}_{T}}\sup_{z\in{\mathbb{C}}\setminus
Z}|m^{Z}(y,t,z_{0},z)|<\infty,$

and satisfies the symmetry relation

$\displaystyle
m^{Z}(y,t,z_{0},z)=\sigma_{1}\overline{m^{Z}(y,t,z_{0},\bar{z})}\sigma_{1}.$
If $p_{N}(t,z)=-\overline{p_{N}(t,-\bar{z})}$, it also follows that
$\displaystyle
m^{Z}(y,t,z_{0},z)=\sigma_{1}\sigma_{3}m^{Z}(y,t,z_{0},-z)\sigma_{3}\sigma_{1}.$

###### Proof.

Proceeding as in Section 3.2, we introduce a matrix-valued function
$m^{P1}(y,z)$ by

$m^{P1}(y,z)=\left\\{\begin{array}[]{ll}m^{P}(z)\begin{pmatrix}1&0\\\ i\rho
e^{2i\left(yz+\frac{(2z)^{2n+1}}{4n+2}\right)}&1\end{pmatrix},&\text{$z$
between $Z_{1}$ and $\Upsilon_{1}$}\\\ &\text{and $z$ between
$\Upsilon_{2n+1}^{*}$ and $\widetilde{Z}_{2}$, }\\\
m^{P}(z)\begin{pmatrix}1&-i\rho
e^{-2i\left(yz+\frac{(2z)^{2n+1}}{4n+2}\right)}\\\ 0&1\end{pmatrix},&\text{$z$
between $Z_{3}$ and $\Upsilon_{2n+2}^{*}$}\\\ &\text{and $z$ between $Z_{4}$
and $\Upsilon_{4n+2}$,}\\\ m^{P}(z),&\hbox{elsewhere,}\end{array}\right.$

where $m^{P}$ is defined in (A.13). In view of the RH problem A.4 for $m^{P}$,
it can be easily checked that $m^{P1}$ satisfies the following RH problem.

######  RH problem A.8.

  1. _(a)_

$m^{P1}(y,t,z_{0},\cdot)\in I+\dot{E}^{2}({\mathbb{C}}\setminus Z)$, where the
contour $Z$ is defined in (A.36).

  2. _(b)_

For a.e. $\lambda\in Z$, we have

$m_{+}^{P1}(y,t,z_{0},z)=m_{-}^{P1}(y,t,z_{0},z)v^{P1}(y,t,z_{0},z),$

where

$\displaystyle v^{P1}(y,t,z_{0},z)$
$\displaystyle=\begin{cases}\begin{pmatrix}1&0\\\ i\rho
e^{2i\left(yz+\frac{(2z)^{2n+1}}{4n+2}\right)}&1\end{pmatrix},&z\in Z_{1}\cup
Z_{2},\\\ \begin{pmatrix}1&i\rho
e^{-2i\left(yz+\frac{(2z)^{2n+1}}{4n+2}\right)}\\\ 0&1\end{pmatrix},&z\in
Z_{3}\cup Z_{4},\\\ \begin{pmatrix}1-|\rho|^{2}&i\rho
e^{-2i\left(yz+\frac{(2z)^{2n+1}}{4n+2}\right)}\\\ i\rho
e^{2i\left(yz+\frac{(2z)^{2n+1}}{4n+2}\right)}&1\end{pmatrix},&z\in(-z_{0},z_{0}).\end{cases}$

By defining

$\displaystyle\hat{m}^{Z}:=m^{Z}(m^{P1})^{-1},$

one can show that, analogous to the arguments used in the proof of Lemma A.3,
that $\hat{m}^{Z}$ admits an asymptotic expansion like (A.29), which finally
leads to the statements of Lemma A.7. We omit the details here. ∎

Acknowledgements Lin Huang was partially supported by National Natural Science
Foundation of China under grant number 11901141. Lun Zhang was partially
supported by National Natural Science Foundation of China under grant number
11822104, by The Program for Professor of Special Appointment (Eastern
Scholar) at Shanghai Institutions of Higher Learning, and by Grant EZH1411513
from Fudan University.

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