# Asymptotics of the determinant of the modified Bessel functions and the
second Painlevé equation

Yu Chen  Department of Mathematics, Sun Yat-sen University, Guangzhou, 510275,
China  Shuai-Xia Xu  Institut Franco-Chinois de l’Energie Nucléaire, Sun Yat-
sen University, Guangzhou, 510275, China  Yu-Qiu Zhao  Department of
Mathematics, Sun Yat-sen University, Guangzhou, 510275, China

###### Abstract

In the paper, we consider the extended Gross-Witten-Wadia unitary matrix model
by introducing a logarithmic term in the potential. The partition function of
the model can be expressed equivalently in terms of the Toeplitz determinant
with the $(i,j)$-entry being the modified Bessel functions of order $i-j-\nu$,
$\nu\in\mathbb{C}$. When the degree $n$ is finite, we show that the Toeplitz
determinant is described by the isomonodromy $\tau$-function of the Painlevé
III equation. As a double scaling limit, we establish an asymptotic
approximation of the logarithmic derivative of the Toeplitz determinant,
expressed in terms of the Hastings-McLeod solution of the inhomogeneous
Painlevé II equation with parameter $\nu+\frac{1}{2}$. The asymptotics of the
leading coefficient and recurrence coefficient of the associated orthogonal
polynomials are also derived. We obtain the results by applying the Deift-Zhou
nonlinear steepest descent method to the Riemann-Hilbert problem for
orthogonal polynomials on the Hankel loop. The main concern here is the
construction of a local parametrix at the critical point $z=-1$, where the
$\psi$-function of the Jimbo-Miwa Lax pair for the inhomogeneous Painlevé II
equation is involved.

2020 mathematics subject classification: 33E17; 34M55; 41A60
Keywords and phrases: Random Matrices; Toeplitz determinants; Painlevé
equations; Asymptotics

###### Contents

  1. 1 Introduction
    1. 1.1 Statement of results
  2. 2 Riemann-Hilbert problem for the orthogonal polynomials
    1. 2.1 Riemann-Hilbert problem for orthogonal polynomials and differential identity
    2. 2.2 Proof of Theorem 1.1
  3. 3 Asymptotics of the Riemann-Hilbert problem for orthogonal polynomials
    1. 3.1 Normalization: $\widehat{Y}\to T$
    2. 3.2 Deformation: $T\to S$
    3. 3.3 Global parametrix: $N$
    4. 3.4 Local parametrix and the model Riemann-Hilbert problem for $\Psi$
    5. 3.5 Final transformation $S\to R$ and error estimate
  4. 4 Asymptotics for $\Psi$ as $x\rightarrow+\infty$
    1. 4.1 Global parametrix on $[-i,i]$
      1. 4.1.1 Local parametrices near $\pm i$
    2. 4.2 Final transformation
    3. 4.3 Asymptotics for $H$ and $u$ as $x\rightarrow+\infty$
  5. 5 Asymptotics for $\Psi$ as $x\rightarrow-\infty$
    1. 5.1 Asymptotics for $\Psi$ as $x\rightarrow-\infty$ with $\nu\in\mathbb{C}\setminus\\{0\\}$
      1. 5.1.1 Global parametrix
      2. 5.1.2 Local parametrices near $z_{-}=-\frac{\sqrt{2}}{2}$
      3. 5.1.3 Final transformation
    2. 5.2 Asymptotics of $u(x)$ and $H(x)$ as $x\to-\infty$ for $\nu\in\mathbb{C}\setminus\\{0\\}$
    3. 5.3 Asymptotics for $\Psi$ as $x\rightarrow-\infty$ for $\nu=0$
  6. 6 Proof of Theorem 1.5
  7. A Local parametrix models
    1. A.1 Airy parametrix
    2. A.2 Parabolic cylinder parametrix

##  1 Introduction

Consider the following partition function of the matrix model

$Z_{n,\nu}(t)=\frac{1}{n!}\left(\prod_{k=1}^{n}\int_{\Gamma}\frac{ds_{k}}{2\pi
is_{k}}\right)\Delta(\mathbf{s})\Delta(\mathbf{s}^{-1})e^{\frac{t}{2}\sum\limits_{k=1}^{n}(s_{k}+s_{k}^{-1})+\nu\log
s_{k}},$ (1.1)

where $\nu\in\mathbb{C}$, $t>0$, $\Delta(\mathbf{s})=\prod_{i<j}(s_{i}-s_{j})$
is the Vandermonde determinant and the branch of $\log s_{k}$ is chosen such
that $\mathop{\rm arg}\nolimits s_{k}\in(-\pi,\pi)$. Here, the integral path
$\Gamma$ denotes the Hankel loop as shown in Figure 1, which starts at
$-\infty$, encircles the origin once in the positive direction and returns to
$-\infty$.

Figure 1: The Hankel loop $\Gamma$

The matrix model has been obtained recently in the studies of irregular
conformal blocks [27, 28, 29]. The matrix model (1.1) has also been considered
in the studies of random matrix average over unitary group with Haar measure
earlier in [23, 24]. When $\nu$ is an integer, the contour is reduced to the
positively oriented unit circle around the origin, and the model (1.1) becomes
the unitary matrix model extending the classical Gross-Witten-Wadia unitary
matrix model [26, 35] by introducing the extra logarithmic term in the
potential [24].

The partition function (1.1) can be expressed equivalently in terms of the
Toeplitz determinant and in terms of certain orthogonal polynomials. To this
aim, we introduce the systems of monic orthogonal polynomials
$\\{\pi_{n}\\}_{n\in\mathbb{N}}$ and $\\{\tilde{\pi}_{n}\\}_{n\in\mathbb{N}}$
defined by the orthogonality on the Hankel loop depicted in Figure 1:

$\int_{\Gamma}\pi_{n}(s)\tilde{\pi}_{m}(s^{-1})w(s)\frac{ds}{2\pi
is}=h_{n}\delta_{n,m},$ (1.2)

where $h_{n}$ is a constant, and

$w(z)=e^{\frac{t}{2}(z+\frac{1}{z})+\nu\log z},\quad\nu\in\mathbb{C},$ (1.3)

with the branch of $\log z$ chosen such that $\mathop{\rm arg}\nolimits
z\in(-\pi,\pi)$. For integer $\nu$, the orthogonality (1.2) is equivalent to a
non-Hermitian orthogonality on the unit circle oriented counterclockwise,
which was studied in [24]. We denote

$m_{k}=m_{k}(t)=\int_{\Gamma}s^{k}w(s)\frac{ds}{2\pi is},\quad
k\in\mathbb{Z}.$

The Toeplitz determinant associated with the weight function (1.3) can be
defined by

$D_{n,\nu}(t)=\det\left(m_{j-i}\right)_{i,j=0}^{n-1}.$ (1.4)

Then, the partition function (1.1) can be expressed in terms of the Toeplitz
determinant

$Z_{n,\nu}(t)=D_{n,\nu}(t).$ (1.5)

From the integral representation of the $I$ Bessel function [1, 34], we have

$m_{k}(t)=I_{-k-\nu}(t),$ (1.6)

and

$Z_{n,\nu}(t)=D_{n,\nu}(t)=\det\left(I_{i-j-\nu}(t)\right)_{i,j=0}^{n-1}.$
(1.7)

Here $I_{-\nu}(t)$ denotes the $I$ Bessel function of order
$-\nu\in\mathbb{C}$:

$I_{-\nu}(t)=\int_{\Gamma}w(s)\frac{ds}{2\pi is},$

where $w(z)$ is defined in (1.3) and $\Gamma$ is the Hankel loop depicted in
Figure 1; see [1, (9.6.20)].

Suppose the determinant $D_{n,\nu}(t)$ ($D_{n,\nu}$, for short) does not
vanish, then the orthogonal polynomials can be constructed explicitly as
follows:

$\pi_{n}(z)=\frac{1}{D_{n,\nu}}\left|\begin{array}[]{cccc}m_{0}&m_{1}&\cdots&m_{n}\\\
m_{-1}&m_{0}&\cdots&m_{-1+n}\\\ \vdots&\vdots&\vdots&\vdots\\\
m_{-n+1}&m_{-n+2}&\cdots&m_{1}\\\ 1&z&\cdots&z^{n}\end{array}\right|,$ (1.8)

and

$\tilde{\pi}_{n}(z)=\frac{1}{D_{n,\nu}}\left|\begin{array}[]{cccc}m_{0}&m_{-1}&\cdots&m_{-n}\\\
m_{1}&m_{0}&\cdots&m_{1-n}\\\ \vdots&\vdots&\vdots&\vdots\\\
m_{n-1}&m_{n-2}&\cdots&m_{-1}\\\ 1&z&\cdots&z^{n}\end{array}\right|.$ (1.9)

The equations (1.2) and (1.8) then imply the relation

$h_{n}=\frac{D_{n+1,\nu}}{D_{n,\nu}}.$ (1.10)

The monic orthogonal polynomials $\pi_{n}(z)$ and $\widetilde{\pi}_{n}(z)$
satisfy the recurrence relations as follows:

$\pi_{n+1}(z)=z\pi_{n}(z)+\pi_{n+1}(0)\,\widetilde{\pi}^{*}_{n}(z),\quad\widetilde{\pi}^{*}_{n+1}(z)=\widetilde{\pi}^{*}_{n}(z)+\widetilde{\pi}_{n+1}(0)z\pi_{n}(z)$
(1.11)

where $\widetilde{\pi}^{*}_{n}(z)=z^{n}\widetilde{\pi}_{n}(z^{-1})$ is the
reversed polynomial associated with $\widetilde{\pi}_{n}(z)$.

The Toeplitz determinant (1.7) has important applications in random matrix
theory. According to Gessel’s formula [25], the Toeplitz determinant (1.7)
with the parameter $\nu=0$ can be used to represent the distribution of the
length of the longest increasing subsequence of random permutations of the
numbers $1,2,\cdots,n$. In the seminal work [2], Baik, Deift, and Johansson
show that as $n\to\infty$ the scaled distribution converges to the Tracy-Widom
distribution

$F(s)=\exp\left(-\int_{s}^{+\infty}(x-s)u_{HM}(x)^{2}dx\right),$ (1.12)

where $u_{HM}(x)$ is the Hastings-McLeod solution of the homogeneous Painlevé
II equation

$u_{HM}^{\prime\prime}=2u_{HM}^{3}+xu_{HM},\qquad u_{HM}(x)\sim{\rm
Ai}(x),\quad x\to\infty.$ (1.13)

They obtain the above celebrated result by exploring the asymptotics of the
Toeplitz determinant, which is shown to be described by the Hastings-McLeod
solution of the homogeneous Painlevé II equation in a certain double scaling
limit by using the Riemann-Hilbert method.

For general parameter $\nu$, the Toeplitz determinant with entries given in
terms of the $I$ Bessel functions (1.7) has been considered in the studies of
random matrix average over unitary groups [21, 22, 23, 24]. The Toeplitz
determinant is identified therein as the $\tau$-function of the Painlevé III
equation by using Okamoto’s development of the theory of Painlevé equations
[33]. Recently, the determinant with the entries expressed in terms of the $K$
Bessel functions, instead of $I$ Bessel functions, appears in the studies of
linear statistic of the inverse eigenvalues of Laguerre unitary ensemble and
is related to the Painlevé III equation both for finite dimension $n$ and in
the large-$n$ limit [11]; see also [6, 36, 37].

Recently, the matrix model (1.1) is investigated in [27, 28, 29], motivated by
the applications in the studies of irregular conformal blocks. For general
integer parameter $\nu$, they first show that for fixed $n$ the partition
function (1.1) is identified with the $\tau$-function of the Painlevé III
equation. Then, by conducting appropriate double scaling analysis of the
associated discrete Painlevé system, they formally derive the inhomogeneous
Painlevé II equation for a certain scaling function as $n$ tends to infinity.
It is worth mentioning that the solution of the inhomogeneous Painlevé II
equation has also arisen in the studies of multi-critical Hermitian random
matrix ensembles when the spectral singularity collides with the interior
singularity where the global eigenvalue density function vanishes
quadratically; see [10, 9].

In recent years, there has been great interest in studying the asymptotic
behavior of Toeplitz determinants. The asymptotics of Toeplitz determinants
associated with general weight functions on the unit circle perturbed by
several Fisher-Hartwig singularities have been derived in [14, 13]. When the
location of the Fisher-Hartwig singularities varies with the size of the
determinants, the transition asymptotics of the Toeplitz determinants are
established in several different situations, such as the emergence of a
Fisher-Hartwig singularity [7], the merging of two Fisher-Hartwig
singularities [8], the emergence of a gap on the unit circle [5] and the
closing of a gap to which a Fisher-Hartwig singularity belongs [38].

In the present paper, we are concerned with the partition function (1.1) or
equivalently the Toeplitz determinant (1.7) with entries given in terms of $I$
Bessel functions and general parameter $\nu\in\mathbb{C}$ therein. When the
degree $n$ is finite, we show that the Toeplitz determinant is described by
the isomonodromy $\tau$-function of the Painlevé III equation. In the double
scaling limit as the degree $n$ tends to infinity and $\frac{t}{n}\rightarrow
1$, we establish an asymptotic approximation of the Toeplitz determinant,
expressed in terms of the Hasting-McLeod solution of the inhomogeneous
Painlevé II equation. The asymptotics of the leading coefficient and
recurrence coefficient of the associated orthogonal polynomials are also
derived. We obtain the results by applying the Deift-Zhou nonlinear steepest
descent method developed in [12, 15, 16, 17] to the Riemann-Hilbert problem
for orthogonal polynomials defined by the orthogonality on the Hankel loop
(1.2). The main focus is the construction of a local parametrix at the
critical point $z=-1$, where the $\psi$-function of the Jimbo-Miwa Lax pair
for the inhomogeneous Painlevé II equation is involved.

###  1.1 Statement of results

Our first result shows that, for fixed degree $n$, the ratio of the contiguous
recurrence coefficients satisfies the Painlevé III equation, and the Toeplitz
determinant equals to the isomonodromy $\tau$-function of the Painlevé III
equation up to a constant factor.

######  Theorem 1.1.

Let $\nu\in\mathbb{C}$ and $\pi_{n}$ be the $n$-th monic orthogonal
polynomials defined by (1.2), and

$a_{n}\left(\frac{t}{2}\right)=-\frac{\pi_{n+1}(0)}{\pi_{n}(0)}.$ (1.14)

Then $a_{n}(t)$ satisfies the Painlevé III equation

$a^{\prime\prime}_{n}=\frac{(a^{\prime}_{n})^{2}}{a_{n}}-\frac{a^{\prime}_{n}}{t}+\frac{4}{t}\left((\nu-n)a_{n}^{2}+n+\nu+1\right)+4a_{n}^{3}-\frac{4}{a_{n}}.$
(1.15)

Moreover, we have

$D_{n,\nu}(t)=c~{}e^{\frac{1}{8}t^{2}}\tau\left(\frac{t}{2}\right),$ (1.16)

where $c$ is a constant and $\tau(t)$ is the Jimbo-Miwa-Ueno isomonodromy
$\tau$-function of the Painlevé III equation (1.15).

######  Remark 1.2.

The identification of the Toeplitz determinant with the $\tau$-function of the
Painlevé III equation (1.16) is derived earlier in [23, Proposition 2] and
[24, Proposition 3.2] by using Okamoto’s development of the theory of Painlevé
equations. According to Okamoto’s theory of Painlevé III equation, there
exists a sequence of $\tau$-functions expressed in terms of the Toeplitz
determinant of a certain linear combination of the modified Bessel functions;
see [11, 23, 24] and the references therein. Setting
$\sigma(t)=t\frac{d}{dt}\log\tau(t)$, then $\sigma(t)$ satisfies the
$\sigma$-form of Painlevé III equation [30, (C.29)]:

$\left(t\sigma^{\prime\prime}-\sigma^{\prime}\right)^{2}=4(2\sigma-t\sigma^{\prime})((\sigma^{\prime})^{2}-4t^{2})+2(\theta_{0}^{2}+\theta_{\infty}^{2})((\sigma^{\prime})^{2}+4t^{2})-16\theta_{0}\theta_{\infty}t\sigma^{\prime},$
(1.17)

with $\theta_{0}=\nu-n$ and $\theta_{\infty}=-(n+\nu)$. Therefore, the
logarithmic derivative of $D_{n,\nu}(t)$ given in (1.7) provides a classical
solution of the differential equation (1.17).

To state our result on the asymptotics of the Toeplitz determinant
$D_{n,\nu}(t)$ and the recurrence coefficient of the orthogonal polynomials,
we need a family of solutions of the general Painlevé II equation

$u^{\prime\prime}=2u^{3}+xu+\left(\nu+\frac{1}{2}\right),\qquad\nu\in\mathbb{C},$
(1.18)

and the Hamiltonian associated with the solutions

$H(x;\nu)=\frac{1}{2}z(z+x)+zu^{2}-\nu u,$ (1.19)

with

$z=u^{\prime}-u^{2}-\frac{x}{2}.$ (1.20)

The asymptotic behaviors of this family of solutions and the Hamiltonians are
stated in the following theorem.

######  Theorem 1.3.

There exist one parameter family of solutions to the general Painlevé II
equation (1.18) with the parameter $\nu\in\mathbb{C}$ such that

$u(x;\nu)=-\frac{\nu+\frac{1}{2}}{x}+O\left(x^{-\frac{5}{2}}\right),\quad
x\to+\infty$ (1.21)

and

$u(x;\nu)=\left\\{\begin{aligned}
&\sqrt{-\frac{x}{2}}+\frac{\nu+\frac{1}{2}}{2x}+O\left((-x)^{-\frac{5}{2}}\right),\quad&\nu\notin\mathbb{N},\quad&x\rightarrow-\infty,\\\
&-\sqrt{-\frac{x}{2}}+\frac{\nu+\frac{1}{2}}{2x}+O\left((-x)^{-\frac{5}{2}}\right),\quad&\nu\in\mathbb{N},\quad&x\rightarrow-\infty.\\\
\end{aligned}\right.$ (1.22)

Moreover, the Hamiltonians associated with these solutions satisfy the
asymptotic behaviors

$H(x;\nu)=-\frac{1}{8}x^{2}+\frac{\nu^{2}-\frac{1}{4}}{2x}+O\left(x^{-3}\right),\quad
x\to+\infty,$ (1.23)

and

$H(x;\nu)=\left\\{\begin{aligned}
&-\nu\sqrt{-\frac{x}{2}}+O\left((-x)^{-1}\right),\quad&\nu\notin\mathbb{N},\quad&x\rightarrow-\infty,\\\
&\nu\sqrt{-\frac{x}{2}}+O\left((-x)^{-1}\right),\quad&\nu\in\mathbb{N},\quad&x\rightarrow-\infty.\\\
\end{aligned}\right.$ (1.24)

######  Remark 1.4.

The solution $u(x;\nu)$ determined by the asymptotic behaviors (1.21) and
(1.22) is known as the Hastings-McLeod solution of the general Painlevé II
equation (1.18); see [19, 18, Chapter 11]. The Hastings-McLeod solution is
meromorphic in the complex plane. For $\nu<0$, it is shown in [10] that the
Hastings-McLeod solution $u(x;\nu)$ is pole free on the real axis. By the
uniqueness of solution, it is seen from (1.21) and (1.22) that for
$\nu\in\mathbb{N}$

$u(x;\nu)=-u(x;-\nu-1).$ (1.25)

Therefore, the solution $u(x;\nu)$, $\nu\in\mathbb{N}$ is also pole free on
the real axis.

Now, we consider the double scaling limit when $n\rightarrow\infty$ and
$\tau=\frac{t}{n}\rightarrow 1$ such that $\tau-1=O\left(n^{-2/3}\right)$. We
show that the asymptotics of the logarithmic derivative of the Toeplitz
determinant $D_{n,\nu}(t)$ generated by (1.3), and equivalently the Toeplitz
determinant (1.7) with entries given in terms of $I$ Bessel functions, can be
expressed in terms of the Hamiltonian of the Painlevé II equation (1.18). The
asymptotics of the recurrence coefficient in (1.11) and the leading
coefficient of the orthonormal polynomials with respect to the weight function
(1.3) are also derived.

######  Theorem 1.5.

Let $n\rightarrow\infty$ and $\tau=\frac{t}{n}\rightarrow 1$, such that
$\tau-1=O(n^{-2/3})$. We have the asymptotic approximation of the logarithmic
derivative of the Toeplitz determinant associated with (1.3)

$\frac{d}{dt}\log
D_{n,\nu}(t)=\frac{t}{2}+{2^{\frac{2}{3}}}{n^{-\frac{1}{3}}}H\left(2^{\frac{2}{3}}n^{\frac{2}{3}}(\tau-1);\nu\right)+O\left(n^{-\frac{2}{3}}\right).$
(1.26)

Moreover, we have the asymptotics of the leading coefficient
$\gamma_{n}=h_{n}^{-1/2}$ of the $n$-th orthonormal polynomials associated
with (1.3)

$\gamma_{n}=1+{2^{-\frac{1}{3}}}{n^{-\frac{1}{3}}}H\left(2^{\frac{2}{3}}n^{\frac{2}{3}}(\tau-1);\nu\right)+O\left(n^{-\frac{2}{3}}\right),$
(1.27)

and the logarithmic derivative of the recurrence coefficient in (1.11)

$\frac{d}{dt}\log\pi_{n}(0)=-{2^{\frac{2}{3}}}{n^{-\frac{1}{3}}}u\left(2^{\frac{2}{3}}n^{\frac{2}{3}}(\tau-1);\nu\right)+O\left(n^{-\frac{2}{3}}\right).$
(1.28)

Here, $u(s;\nu)$ and $H(s;\nu)$ are respectively the solution of the Painlevé
II equation and the associated Hamiltonian with the asymptotic behaviors given
in Theorem 1.3. For $\nu<0$ or $\nu\in\mathbb{N}$, $u(s;\nu)$ and $H(s;\nu)$
are pole free on the real axis and the error terms are uniform for
$n^{{2}/{3}}(\tau-1)$ in any compact subsets of the real axis. Otherwise, the
error terms are uniform for $n^{3/2}(\tau-1)$ bounded and bounded away from
the poles of $u(s;\nu)$ on the real axis.

######  Remark 1.6.

From (1.26), we see that there is a sign difference in the asymptotics of the
Hamiltonian (1.24). It is consistent with the fact that $H(s;\nu)=H(s;-\nu)$
for $\nu\in\mathbb{N}$ as shown later in (5.38), and the symmetry relation

$D_{n,\nu}(t)=D_{n,-\nu}(t)$ (1.29)

for $\nu\in\mathbb{N}$. The sign difference in (1.22) and (1.24) may indicate
that the asymptotic behaviors of $u(s;\nu)$ and $H(s;\nu)$ are sensitive to
the parameter $\nu$ near $\mathbb{N}$. Similar phenomenon has also been
observed in the study of the asymptotics of the increasing tritronquée
solutions of the second Painlevé equation (1.18) where the asymptotic
behaviors change dramatically for the parameter $\nu$ near $\mathbb{Z}$; see
[32, Theorem 1.2] and the discussion after Theorem 1.4 in [32]. It would be
interesting to consider the transition asymptotics of $u(s;\nu)$ and
$H(s;\nu)$ as $x\to-\infty$ and $\nu\to k$ for given $k\in\mathbb{N}$ in an
interrelated manner.

The rest of the paper is arranged as follows. In Section 2, we formulate a
Riemann-Hilbert (RH, for short) problem $Y$ for the orthogonal polynomials
with respect to the weight function (1.3), and derive the differential
identity for the Toeplitz determinant. We then prove Theorem 1.1 at the end of
this Section by relating the RH problem to that of the Painlevé III equation.
In Section 3, we perform the Deift-Zhou nonlinear steepest descent analysis of
the RH problem for $Y$. The main concern is the construction of a local
parametrix at the critical point $z=-1$, where the $\psi$-function of the
Jimbo-Miwa Lax pair for the Painlevé II equation is involved. The construction
is different from [9, 10] where the Flaschka-Newell Lax pair for the Painlevé
II equation was used. In Sections 4 and 5, we derive the asymptotics of the
one parameter family of solutions of the Painlevé II equation and the
associated Hamiltonians as stated in Theorem 1.3 by carrying out a nonlinear
steepest descent analysis of the RH problem for the Jimbo-Miwa Lax pair of the
Painlevé II equation. Then, the proof of Theorem 1.5 is given in Section 6.
For the convenience of the reader, we collect the Airy and parabolic cylinder
parametrix models in the Appendix.

##  2 Riemann-Hilbert problem for the orthogonal polynomials

###  2.1 Riemann-Hilbert problem for orthogonal polynomials and differential
identity

######  RH problem 2.1.

We look for a $2\times 2$ matrix-valued function $Y(z;n)$ ($Y(z)$, for short)
satisfying the properties.

  * (1)

$Y(z)$ is analytic in $\mathbb{C}\setminus\Gamma$; where the contour $\Gamma$
is shown in Figure 1.

  * (2)

$Y(z)$ satisfies the jump condition

$Y_{+}(z)=Y_{-}(z)\begin{pmatrix}1&\frac{w(z)}{z^{n}}\\\
0&1\end{pmatrix},\qquad z\in\Gamma,$ (2.1)

where the weight function is defined in (1.3).

  * (3)

As $z\to\infty$, we have

$Y(z)=\left(I+\frac{Y_{-1}}{z}+O\left(\frac{1}{z^{2}}\right)\right)\begin{pmatrix}z^{n}&0\\\
0&z^{-n}\end{pmatrix}.$ (2.2)

According to [20], if there exists a solution to the RH problem for $Y$, the
solution is unique and given by

$Y(z)=Y(z;n)=\begin{pmatrix}\pi_{n}(z)&\frac{1}{2\pi
i}\int_{\Gamma}\frac{\pi_{n}(x)w(x)dx}{x^{n}(x-z)}\\\\[8.5359pt]
-h_{n-1}^{-1}\tilde{\pi}^{*}_{n-1}(z)&-\frac{h_{n-1}^{-1}}{2\pi
i}\int_{\Gamma}\frac{\tilde{\pi}^{*}_{n-1}(x)w(x)dx}{x^{n}(x-z)}\end{pmatrix},$
(2.3)

where $\widetilde{\pi}^{*}_{n-1}(z)=z^{n-1}\widetilde{\pi}_{n-1}(z^{-1})$ and
$\pi_{n}$, $\tilde{\pi}_{n-1}$ and $h_{n-1}$ are defined by (1.2).

######  Remark 2.2.

In the asymptotic analysis of the RH problem for $Y$ given in the next
section, we will transform the RH problem for $Y$ to a small-norm RH problem
for $R$ by a series of invertible transformations. The RH problem for $R$ is
solvable for sufficiently large $n$ and $\tau=\frac{t}{n}\to 1$ such that
$\xi=n^{2/3}(\tau-1)=O(1)$ and $x=\xi$ is not a pole of the Hasting-McLeod
solution $u(x;\nu)$ of Painlevé II equation (1.18). As shown in Remark 1.4,
the Hastings-McLeod solution $u(x;\nu)$ is pole free on the real line for
$\nu<0$ or $\nu\in\mathbb{N}$, and may have poles on the real line for other
parameter $\nu\in\mathbb{C}$. Therefore, the RH problem for $Y$ is solvable
under the same conditions as that for $R$ by tracing back the series of
invertible transformations.

For later use, we derive a differential identity for the logarithmic
derivative of the Toeplitz determinant associated with (1.3).

######  Proposition 1.

We have the following differential identity

$\frac{d}{dt}\log
D_{n,\nu}=-\frac{1}{2}\left((Y_{-1})_{11}+\frac{Y_{21}^{\prime}(0;n+1)}{Y_{21}(0;n+1)}\right),$
(2.4) $h_{n}=Y_{12}(0;n),$ (2.5)

where $Y_{-1}$ is given in (2.2) and $Y_{21}^{\prime}$ denotes the derivative
of $Y_{21}$ with respect to $z$.

###### Proof.

The relation (2.5) follows directly from (2.3). To prove (2.4), we first
derive from (1.10) that

$\frac{d}{dt}\log D_{n,\nu}=\sum_{k=0}^{n-1}\frac{1}{h_{k}}\frac{d}{dt}h_{k}.$
(2.6)

Next, we calculate the derivative of $h_{k}$ by using the integral
representation (1.2)

$\displaystyle\frac{d}{dt}h_{k}=$
$\displaystyle\frac{d}{dt}\int_{\Gamma}\pi_{k}(s)\tilde{\pi}_{k}(s^{-1})w(s)\frac{ds}{2\pi
is}$ $\displaystyle=$
$\displaystyle\int_{\Gamma}\frac{d}{dt}(\pi_{k}(s))\tilde{\pi}_{k}(s^{-1})w(s)\frac{ds}{2\pi
is}+\int_{\Gamma}\pi_{k}(s)\frac{d}{dt}(\tilde{\pi}_{k}(s^{-1}))w(s)\frac{ds}{2\pi
is}$ (2.7)
$\displaystyle+\int_{\Gamma}\pi_{k}(s)\tilde{\pi}_{k}(s^{-1})\frac{d}{dt}w(s)\frac{ds}{2\pi
is}$ $\displaystyle=$
$\displaystyle\frac{1}{2}\int_{\Gamma}\pi_{k}(s)s\tilde{\pi}_{k}(s^{-1})w(s)\frac{ds}{2\pi
is}+\frac{1}{2}\int_{\Gamma}\pi_{k}(s)\tilde{\pi}_{k}(s^{-1})s^{-1}w(s)\frac{ds}{2\pi
is}.$ (2.8)

Denote

$\pi_{n}(z)=z^{n}+a_{n,n-1}z^{n-1}+\cdots,\quad\tilde{\pi}_{n}(z)=z^{n}+\tilde{a}_{n,n-1}z^{n-1}+\cdots.$
(2.9)

We have

$z\pi_{n}(z)=\pi_{n+1}(z)+(a_{n,n-1}-a_{n+1,n})\pi_{n}(z)+\cdots,$ (2.10)
$z\tilde{\pi}_{n}(z)=\tilde{\pi}_{n+1}(z)+(\tilde{a}_{n,n-1}-\tilde{a}_{n+1,n})\tilde{\pi}_{n}(z)+\cdots.$
(2.11)

Substituting (2.10) and (2.11) into (2.8) yields

$\int_{\Gamma}\pi_{k}(s)s\tilde{\pi}_{k}(s^{-1})w(s)\frac{ds}{2\pi
is}=(a_{k,k-1}-a_{k+1,k})h_{k},$ (2.12)

and

$\int_{\Gamma}\pi_{k}(s)\tilde{\pi}_{k}(s^{-1})s^{-1}w(s)\frac{ds}{2\pi
is}=(\tilde{a}_{k,k-1}-\tilde{a}_{k+1,k})h_{k}.$ (2.13)

Thus, we have

$\frac{d}{dt}\log D_{n,\nu}=-\frac{1}{2}(a_{n,n-1}+\tilde{a}_{n,n-1}).$ (2.14)

Comparing (2.3) and (2.9), we have

$a_{n,n-1}=(Y_{-1})_{11},\quad\tilde{a}_{n,n-1}=(Y)_{21}^{\prime}(0)/(Y)_{21}(0),$
(2.15)

where $Y_{21}^{\prime}$ denotes the derivative of $Y_{21}$ with respect to
$z$. Therefore, we obtain (2.4) by substituting (2.15) in (2.14). This
completes the proof of the proposition.

∎

Next, we relate the RH problem for $Y$ to the known one in the literature for
Painlevé III equation; see [19, Chapter 5.3]. For this purpose, we introduce
the new independent variables

$x=\frac{t}{2}\quad\text{and}\quad\lambda=-iz,$

and define

$\Phi(\lambda,x)=e^{-\frac{(n+\nu)\pi i}{4}{\sigma_{3}}}Y(e^{\frac{1}{2}\pi
i}\lambda)e^{\frac{ix}{2}(\lambda-\frac{1}{\lambda})\sigma_{3}}(e^{\frac{1}{2}\pi
i}\lambda)^{-(\frac{n-\nu}{2})\sigma_{3}},$ (2.16)

where the branch of the function $\lambda^{-(\frac{n-\nu}{2})}$ is chosen such
that $\mathop{\rm arg}\nolimits\lambda\in(-\frac{3}{2}\pi,\frac{1}{2}\pi)$.
Then, $\Phi(\lambda)=\Phi(\lambda,x)$ solves the following RH problem.

######  RH problem 2.3.

(1) $\Phi(\lambda)$ is analytic for
$\lambda\in\mathbb{C}\setminus\\{-i\Gamma\cup(0,i\infty)\\}$, where $\Gamma$
is shown in Figure 1.

(2) $\Phi(\lambda)$ satisfies the following jump condition

$\Phi_{+}(\lambda)=\Phi_{-}(\lambda)\begin{pmatrix}1&1\\\
0&1\end{pmatrix},\quad\lambda\in-i\Gamma~{}~{}\mbox{and}~{}~{}\Phi_{+}(\lambda)=\Phi_{-}(\lambda)e^{-\pi
i(n+\nu)\sigma_{3}},\quad\lambda\in(0,i\infty).$ (2.17)

(3) As $\lambda\to\infty$, we have

$\Phi(\lambda)=\left(I+\sum_{k=1}^{\infty}\frac{\Phi_{-k}}{\lambda^{k}}\right)e^{\frac{ix\lambda}{2}\sigma_{3}}\lambda^{\frac{n+\nu}{2}\sigma_{3}},$
(2.18)

where the branch is chosen such that $\mathop{\rm
arg}\nolimits\lambda\in(-\frac{3}{2}\pi,\frac{1}{2}\pi)$ and the coefficient

$\Phi_{-1}(x)=-ie^{-\frac{(n+\nu)\pi
i}{4}{\sigma_{3}}}\begin{pmatrix}a_{n,n-1}+\frac{x}{2}&\pi_{n+1}(0)h_{n}\\\
-\widetilde{\pi}_{n-1}(0)h^{-1}_{n-1}&-a_{n,n-1}-\frac{x}{2}\end{pmatrix}e^{\frac{(n+\nu)\pi
i}{4}{\sigma_{3}}}.$ (2.19)

(4) As $\lambda\to 0$, we have

$\Phi(\lambda)=\Phi(0)\left(I+\sum_{k=1}^{\infty}{\Phi_{k}}{\lambda^{k}}\right)e^{-\frac{ix}{2\lambda}\sigma_{3}}\lambda^{\frac{\nu-n}{2}\sigma_{3}},$
(2.20)

where the branch is chosen such that $\mathop{\rm
arg}\nolimits\lambda\in(-\frac{3}{2}\pi,\frac{1}{2}\pi)$ and the coefficient

$\Phi(0)=e^{-\frac{(n+\nu)\pi i}{4}{\sigma_{3}}}\begin{pmatrix}1&p_{n}\\\
-q_{n}&1-p_{n}q_{n}\end{pmatrix}\pi_{n}(0)^{\sigma_{3}}e^{-\frac{(n-\nu)\pi
i}{4}{\sigma_{3}}},$

with

$p_{n}=\pi_{n}(0)h_{n},\quad q_{n}=\frac{1}{h_{n-1}\pi_{n}(0)}.$

It is seen that $\Phi$ satisfies the same RH problem for Painlevé III equation
as given in [19, Chapter 5.3]. From the RH problem 2.3, we have the following
Lax pair

$\Phi_{\lambda}=A(\lambda,x)\Phi\quad\text{and}\quad\Phi_{x}=B(\lambda,x)\Phi,$
(2.21)

where

$A(\lambda,x)=\frac{ix}{2}\sigma_{3}+\frac{A_{-1}}{\lambda}+\frac{A_{-2}}{\lambda^{2}},\quad
B(\lambda,x)=\frac{i}{2}\lambda\sigma_{3}+B_{0}+\frac{B_{-1}}{\lambda}.$
(2.22)

The coefficients in the above formula are given by

$A_{-1}=\begin{pmatrix}\frac{n+\nu}{2}&-xe^{-\frac{(n+\nu)\pi
i}{2}}\pi_{n+1}(0)h_{n}\\\ -xe^{\frac{(n+\nu)\pi
i}{2}}\widetilde{\pi}_{n-1}(0)h^{-1}_{n-1}&-\frac{(n+\nu)}{2}\end{pmatrix},$
(2.23)

$A_{-2}=\frac{ix}{2}\begin{pmatrix}1-2p_{n}q_{n}&-2e^{-\frac{(n+\nu)\pi
i}{2}}p_{n}\\\ 2e^{\frac{(n+\nu)\pi
i}{2}}q_{n}(p_{n}q_{n}-1)&2p_{n}q_{n}-1\end{pmatrix},$ (2.24)

and

$B_{0}=\frac{1}{x}\left(A_{-1}-\frac{n+\nu}{2}\sigma_{3}\right),\quad
B_{-1}=-\frac{1}{x}A_{-2}.$ (2.25)

To complete this section, we shall identify the Toeplitz determinant with the
$\tau$-function of the Painlevé III equation, as stated in Theorem 1.1.

###  2.2 Proof of Theorem 1.1

According to (2.23), (2.24) and [19, (5.3.4), (5.3.7)], we find that

$a_{n}(x)=-i(A_{-1})_{12}/(A_{-2})_{12}=-\frac{\pi_{n+1}(0)}{\pi_{n}(0)}$
(2.26)

with $t=2x$, solves the Painlevé III equation (1.15).

Next, we consider the Toeplitz determinant $D_{n,\nu}$. To this end, we derive
from (1.11), (2.3) and (2.16) the coefficient $\Phi_{1}$ in the expansion
(2.20)

$\Phi_{1}=i\left(\widetilde{a}_{n,n-1}+\frac{x}{2}\right)\sigma_{3}+\begin{pmatrix}0&*\\\
*&0\end{pmatrix}.$ (2.27)

According to the general theory of Jimbo-Miwa-Ueno[31, (1.11)], the
isomonodromy $\tau$-function for the Lax pair in (2.21)-(2.25) is defined by

$\displaystyle d\log\tau(x)$ $\displaystyle=$ $\displaystyle-\mathop{\rm
Res}\nolimits\limits_{\lambda=0}\;{\rm{tr}}\left(\Phi^{-1}_{0}(\lambda)\frac{\partial\,\Phi_{0}(\lambda)}{\partial\lambda}d\,T_{0}(\lambda)\right)$
(2.28) $\displaystyle-\mathop{\rm
Res}\nolimits\limits_{\lambda=\infty}\;{\rm{tr}}\left(\Phi^{-1}_{\infty}(\lambda)\frac{\partial\,\Phi_{\infty}(\lambda)}{\partial\lambda}d\,T_{\infty}(\lambda)\right),$

where

$d\,T_{0}(\lambda)=-\frac{i}{2\lambda}\sigma_{3}dx,\quad
d\,T_{\infty}(\lambda)=\frac{i\lambda}{2}\sigma_{3}dx.$

Substituting (2.19) and (2.27) into (2.28), we arrive at the equation

$\displaystyle\frac{d}{dx}\log\tau(x)$ $\displaystyle=$
$\displaystyle\frac{i}{2}\;{\rm{tr}}(\Phi_{1}\sigma_{3}-\Phi_{-1}\sigma_{3})$
(2.29) $\displaystyle=$
$\displaystyle\frac{i}{2}(2i\widetilde{a}_{n,n-1}+2ia_{n,n-1}+2ix)$
$\displaystyle=$ $\displaystyle-\widetilde{a}_{n,n-1}-a_{n,n-1}-x.$

From (2.14) and (2.29), we obtain (1.16). This completes the proof of Theorem
1.1.

##  3 Asymptotics of the Riemann-Hilbert problem for orthogonal polynomials

In this section, we perform the Deift-Zhou nonlinear steepest descent analysis
of the RH problem for $Y$. The analysis includes a series of invertible
transformations $Y\rightarrow\widehat{Y}\rightarrow T\rightarrow S\rightarrow
R$ such that the jump matrices in the final RH problem for $R$ are uniformly
close to the identity matrix; [12, 15, 16, 17].

We first modify the contour in RH problem for $Y$ and define

$\widehat{Y}(z)=\left\\{\begin{array}[]{ll}Y(z),&\quad
z\in\Omega_{-}\cup\Omega_{0}\\\ Y(z)\begin{pmatrix}1&-\frac{w(z)}{z^{n}}\\\
0&1\end{pmatrix},&\quad z\in\Omega_{+}\backslash\Omega_{0},\end{array}\right.$
(3.1)

where the regions and contours are shown in Figure 2. Then we obtain the
following RH problem for $\widehat{Y}$.

Figure 2: Contours for the transformation $Y\to\widehat{Y}$

######  RH problem 3.1.

$\widehat{Y}(z)$ is analytic in $\mathbb{C}\setminus\Sigma_{\widehat{Y}}$,
where the contour $\Sigma_{\widehat{Y}}=(-1,-\infty)\cup C$ are depicted in
Figure 3, of which $C$ is the unit circle centered at the origin.

$\widehat{Y}(z)$ satisfies
$\widehat{Y}_{+}(z)=\widehat{Y}_{-}(z)J_{\widehat{Y}}$, where

$J_{\widehat{Y}}(z)=\left\\{\begin{aligned}
&\begin{pmatrix}1&\frac{w(z)}{z^{n}}\\\ 0&1\end{pmatrix},\quad&z&\in C,\\\
&\begin{pmatrix}1&\frac{w_{+}(z)-w_{-}(z)}{z^{n}}\\\
0&1\end{pmatrix},\quad&z&\in(-1,-\infty),\end{aligned}\right.$ (3.2)

with $w_{+}(z)-w_{-}(z)=|z|^{\nu}e^{\frac{t}{2}(z+\frac{1}{z})}(e^{\pi
i\nu}-e^{-\pi i\nu})$.

As $z\to\infty$, we have

$\widehat{Y}(z)=\left(I+\frac{\widehat{Y}_{-1}}{z}+O\left(\frac{1}{z^{2}}\right)\right)z^{n\sigma_{3}}.$
(3.3)

Figure 3: The contours $\Sigma_{\widehat{Y}}$ of RH problem for $\widehat{Y}$

###  3.1 Normalization: $\widehat{Y}\to T$

To normalize the large-$z$ behavior of $\widehat{Y}(z)$, we introduce the
transformation

$T(z)=\left\\{\begin{array}[]{ll}\widehat{Y}(z)e^{\frac{t}{2z}\sigma_{3}}z^{-n\sigma_{3}},&\hbox{$|z|>1$,}\\\
\widehat{Y}(z)e^{\frac{t}{2}z\sigma_{3}},&\hbox{$|z|<1$.}\end{array}\right.$
(3.4)

######  RH problem 3.2.

The function $T(z)$ defined in (3.4) satisfies the following RH problem.

  * (1)

$T(z)$ is analytic in $\mathbb{C}\setminus\Sigma_{\widehat{Y}}$.

  * (2)

$T(z)$ satisfies the jump condition

$T_{+}(z)=T_{-}(z)\left\\{\begin{aligned}
&\begin{pmatrix}e^{n\phi(z)}&z^{\nu}\\\
0&e^{-n\phi(z)}\end{pmatrix},\quad&z&\in C,\\\
&\begin{pmatrix}1&|z|^{\nu}e^{n\phi(z)}(e^{\pi i\nu}-e^{-\pi i\nu})\\\
0&1\end{pmatrix},\quad&z&\in(-1,-\infty),\end{aligned}\right.$ (3.5)

where

$\phi(z)=\frac{t}{2n}(z-z^{-1})+\log z.$ (3.6)

  * (3)

As $z\to\infty$, we have

$T(z)=I+O\left(\frac{1}{z}\right).$ (3.7)

###  3.2 Deformation: $T\to S$

It is seen from (3.5) that the diagonal entries of the jump matrix for $T$ are
highly oscillating for $n$ large. To transform the oscillating entries to
exponential decay ones on certain contours, we introduce the second
transformation $T\to S$. The transformation is based on the following
factorization of jump matrix

$\begin{pmatrix}e^{n\phi(z)}&z^{\nu}\\\
0&e^{-n\phi(z)}\end{pmatrix}=\begin{pmatrix}1&0\\\
z^{-\nu}e^{-n\phi(z)}&1\end{pmatrix}\begin{pmatrix}0&z^{\nu}\\\
-z^{-\nu}&0\end{pmatrix}\begin{pmatrix}1&0\\\
z^{-\nu}e^{n\phi(z)}&1\end{pmatrix}.$ (3.8)

We introduce the transformation

$S(z)=\left\\{\begin{array}[]{lll}T(z)\begin{pmatrix}1&0\\\
z^{-\nu}e^{-n\phi(z)}&1\end{pmatrix},&\hbox{for
$z\in\Omega_{E}$,}\\\\[14.22636pt] T(z)\begin{pmatrix}1&0\\\
-z^{-\nu}e^{n\phi(z)}&1\end{pmatrix},&\hbox{for $z\in\Omega_{I}$,}\\\
T(z),&\hbox{otherwise.}\\\ \end{array}\right.$ (3.9)

Here $\Omega_{E}$ and $\Omega_{I}$ denote some lens-shaped regions outside and
inside the unit circle, which are shown in Figure 4.

######  RH problem 3.3.

The function $S(z)$ defined in (3.9) satisfies the following RH problem.

  * (1)

$S(z)$ is analytic in $\mathbb{C}\setminus\Sigma$, where
$\Sigma=\Sigma_{E}\cup C\cup\Sigma_{I}\cup(-1,-\infty)$, with $\Sigma_{E}$ and
$\Sigma_{I}$ the boundary of the lens shaped regions $\Omega_{E}$ and
$\Omega_{I}$ as indicated in Figure 4.

  * (2)

$S(z)$ satisfies the jump condition

$S_{+}(z)=S_{-}(z)\left\\{\begin{array}[]{lll}\begin{pmatrix}1&0\\\
z^{-\nu}e^{-n\phi(z)}&1\end{pmatrix},&\hbox{for
$z\in\Sigma_{E}$,}\\\\[14.22636pt] \begin{pmatrix}0&z^{\nu}\\\
-z^{-\nu}&0\end{pmatrix},&\hbox{for $z\in C$,}\\\\[14.22636pt]
\begin{pmatrix}1&0\\\ z^{-\nu}e^{n\phi(z)}&1\end{pmatrix},&\hbox{for
$z\in\Sigma_{I}$,}\\\\[14.22636pt]
\begin{pmatrix}1&|z|^{\nu}e^{n\phi(z)}(e^{\pi i\nu}-e^{-\pi i\nu})\\\
0&1\end{pmatrix},&\hbox{for $z\in(-1,-\infty)$},\\\ \end{array}\right.$ (3.10)

with $\phi(z)$ defined in (3.6).

  * (3)

As $z\to\infty$, we have

$S(z)=I+O\left(\frac{1}{z}\right).$ (3.11)

Figure 4: Contours and regions for the RH problem for $S(z)$

For $t=n\tau$, $\tau\to 1$, it follows from (3.6) that the $\phi$-function
possesses the following properties

$\left\\{\begin{array}[]{lll}\mathop{\rm Re}\nolimits\phi(z)=0,&z\in C,&\\\
\mathop{\rm Re}\nolimits\phi(z)>0,&z\in\Sigma_{E},&|z+1|>\delta,\\\
\mathop{\rm
Re}\nolimits\phi(z)<0,&z\in\Sigma_{I},&|z+1|>\delta;\end{array}\right.$ (3.12)

see [2].

###  3.3 Global parametrix: $N$

It is readily seen from (3.12) that on $\Sigma_{I}\cup\Sigma_{E}$ and bounded
away from $z=-1$, the jump matrices for $S$ tend to the identity matrix
exponentially fast as $n\to\infty$. Then, we arrive at the following
approximate RH problem for $n$ large.

######  RH problem 3.4.

We look for a $2\times 2$ matrix-valued function $N(z)$ satisfying the
following properties.

  * (1)

$N(z)$ is analytic in $\mathbb{C}\setminus C$, where $C$ is the unit circle
oriented counterclockwise.

  * (2)

$N(z)$ satisfies the jump condition

$N_{+}(z)=N_{-}(z)\begin{pmatrix}0&z^{\nu}\\\ -z^{-\nu}&0\end{pmatrix},\qquad
z\in C.$ (3.13)

  * (3)

As $z\to\infty$, we have

$N(z)=I+O\left(\frac{1}{z}\right).$ (3.14)

The solution to the RH problem for $N(z)$ can be constructed by elementary
functions as follows:

$N(z)=\left\\{\begin{array}[]{ll}\left(\frac{z+1}{z}\right)^{\nu\sigma_{3}},&\hbox{$|z|>1$,}\\\
(z+1)^{\nu\sigma_{3}}\begin{pmatrix}0&1\\\
-1&0\end{pmatrix},&\hbox{$|z|<1$.}\end{array}\right.$ (3.15)

Here, the branches for the power functions are chosen such that $\mathop{\rm
arg}\nolimits(z)\in(-\pi,\pi)$ and $\mathop{\rm
arg}\nolimits(z+1)\in(-\pi,\pi)$.

###  3.4 Local parametrix and the model Riemann-Hilbert problem for $\Psi$

In this subsection, we intend to construct a local parametrix in a
neighborhood of $z=-1$.

######  RH problem 3.5.

We look for a $2\times 2$ matrix-valued function $P(z)$ satisfying the
following properties.

  * (1)

$P(z)$ is analytic in $U(-1,\delta)\setminus\Sigma$, where $U(-1,\delta)$ is a
neighborhood of $z=-1$.

  * (2)

$P(z)$ satisfies the same jump condition as $S(z)$ on
$U(-1,\delta)\cap\Sigma$.

  * (3)

On the boundary of $U(-1,\delta)$, $P(z)$ satisfies the matching condition

$P(z)N^{-1}=n^{-\frac{\nu}{3}\sigma_{3}}\left(I+O(1/n^{1/3})\right)n^{\frac{\nu}{3}\sigma_{3}}.$
(3.16)

Let

$\tilde{P}(z)=\left\\{\begin{array}[]{lll}P(z),&z\in U(-1,\delta),&|z|>1,\\\
P(z)\begin{pmatrix}0&-z^{\nu}\\\ z^{-\nu}&0\end{pmatrix},&z\in
U(-1,\delta),&|z|<1.\end{array}\right.$ (3.17)

It is seen from (3.10) that $\tilde{P}(z)$ satisfies the jump relations

$\tilde{P}_{+}(z)=\tilde{P}_{-}(z)\left\\{\begin{array}[]{lll}\begin{pmatrix}1&0\\\
z^{-\nu}e^{-n\phi(z)}&1\end{pmatrix},&\hbox{for $z\in
U(-1,\delta)\cap\Sigma_{E}$,}\\\\[14.22636pt]
\begin{pmatrix}1&-z^{\nu}e^{n\phi(z)}\\\ 0&1\end{pmatrix},&\hbox{for $z\in
U(-1,\delta)\cap\Sigma_{I}$,}\\\\[14.22636pt]
\begin{pmatrix}1&|z|^{\nu}e^{n\phi(z)}(e^{\pi i\nu}-e^{-\pi i\nu})\\\
0&1\end{pmatrix},&\hbox{for $z\in
U(-1,\delta)\cap(-1,-\infty)$},\\\\[14.22636pt] \begin{pmatrix}e^{-2\pi
i\nu}&0\\\ 0&e^{2\pi i\nu}\end{pmatrix},&\hbox{for $z\in
U(-1,\delta)\cap(-1,0)$}.\\\ \end{array}\right.$ (3.18)

We seek a solution to the RH problem for $P$ of the following form for
$|z+1|<\delta$

$P(z)=E(z)\Psi(n^{1/3}f(z),n^{2/3}s(t,z))\left\\{\begin{array}[]{ll}z^{-\frac{\nu}{2}\sigma_{3}}e^{-\frac{1}{2}\pi
i\nu\sigma_{3}}e^{-\frac{n}{2}\phi(z)\sigma_{3}}M(z),&\mathop{\rm
arg}\nolimits z\in(0,\pi),\\\\[5.69046pt]
z^{-\frac{\nu}{2}\sigma_{3}}e^{\frac{1}{2}\pi
i\nu\sigma_{3}}e^{-\frac{n}{2}(\phi(z)+2\pi i)\sigma_{3}}M(z),&\mathop{\rm
arg}\nolimits z\in(-\pi,0),\end{array}\right.$ (3.19)

where

$M(z)=\left\\{\begin{array}[]{lll}I,&z\in U(-1,\delta),&|z|>1,\\\
\begin{pmatrix}0&z^{\nu}\\\ -z^{-\nu}&0\end{pmatrix},&z\in
U(-1,\delta),&|z|<1,\end{array}\right.$ (3.20)

and the branch cut for $z^{\nu/2}$ is chosen such that $\mathop{\rm
arg}\nolimits z\in(-\pi,\pi)$. The function $f(z)$ is defined as

$f(z)=\left(\frac{3}{2}\phi_{0}(z)\right)^{1/3}\sim 2^{-2/3}(z+1),\quad
z\to-1,$ (3.21)

and serves as a conformal mapping at $z=-1$, where

$\phi_{0}(z)=\frac{1}{2}(z-z^{-1})+\log(z)\mp\pi
i~{}~{}~{}~{}\mbox{for}~{}~{}\pm\mathop{\rm arg}\nolimits z\in(-\pi,\pi).$

######  RH problem 3.6.

The function $\Psi(z)=\Psi(z,x)$ satisfies the following model RH problem.

  * (1)

$\Psi(z)$ is analytic in $\mathbb{C}\setminus\cup_{k=1}^{5}~{}\Gamma_{k}$,
where

$\Gamma_{1}=e^{\frac{\pi}{3}i}\mathbb{R}^{+},~{}\Gamma_{2}=e^{\frac{2\pi}{3}i}\mathbb{R}^{+},~{}\Gamma_{3}=e^{\pi
i}\mathbb{R}^{+},~{}\Gamma_{4}=e^{-\frac{2\pi}{3}i}\mathbb{R}^{+},~{}\Gamma_{5}=e^{-\frac{\pi}{3}i}\mathbb{R}^{+},$

as depicted in Figure 5.

  * (2)

For $z\in\cup_{k=1}^{5}~{}\Gamma_{k}$, $\Psi(z)$ satisfies the jump condition

$\Psi_{+}(z)=\Psi_{-}(z)J_{\Psi}(z),$ (3.22)

where

$J_{\Psi}(z)=\left\\{\begin{array}[]{lll}\begin{pmatrix}1&e^{-\pi i\nu}\\\
0&1\end{pmatrix},&z\in\Gamma_{1},\\\\[14.22636pt] \begin{pmatrix}1&0\\\
-e^{\pi i\nu}&1\end{pmatrix},&z\in\Gamma_{2},\\\\[14.22636pt]
\begin{pmatrix}e^{-2\pi i\nu}&e^{\pi i\nu}-e^{-\pi i\nu}\\\ 0&e^{2\pi
i\nu}\end{pmatrix},&z\in\Gamma_{3},\\\\[14.22636pt] \begin{pmatrix}1&0\\\
e^{-\pi i\nu}&1\end{pmatrix},&z\in\Gamma_{4},\\\\[14.22636pt]
\begin{pmatrix}1&-e^{\pi i\nu}\\\
0&1\end{pmatrix},&z\in\Gamma_{5}.\\\\[14.22636pt] \end{array}\right.$ (3.23)

  * (3)

As $z\to\infty$, we have

$\Psi(z)=\left(I+\frac{\Psi_{1}(x;\nu)}{z}+O\left(\frac{1}{z^{2}}\right)\right)\exp\left(\left(\frac{1}{3}z^{3}+\frac{x}{2}z\right)\sigma_{3}\right)z^{\nu\sigma_{3}},$
(3.24)

with $\mathop{\rm arg}\nolimits z\in(-\pi,\pi)$.

######  Remark 3.7.

The solution $\Psi(z)$ of the model RH problem can be constructed by using the
$\psi$-functions of the Jimbo-Miwa-Garnier Lax pair for the general Painlevé
II equation (1.18); see [19, Proposition 5.3] and [30]. It is worth mentioning
that the $\psi$-functions are entire functions in the variable $z$.
Conversely, the solution $u(x;\nu)$ of (1.18) and the associated Hamiltonian
$H(x;\nu)$ can be expressed in terms of the solution of the model RH problem

$H(x;\nu)=-(\Psi_{1})_{11}(x;\nu)=(\Psi_{1})_{22}(x;\nu),\quad
u(x;\nu)=-\frac{d}{dx}\log(\Psi_{1})_{12}(x;\nu)$ (3.25)

with $\Psi_{1}(x;\nu)$ given in (3.24). Moreover, the model RH problem
$\Psi(z,x)$ is solvable if and only if $x$ is not a pole of $u(x;\nu)$.

Figure 5: Contours and regions for the model RH problem

To fulfill the matching condition (3.16), we define

$s(t,z)=\frac{\phi(z)-\phi_{0}(z)\mp\pi
i}{f(z)}=\frac{\tau-1}{2}\frac{z-z^{-1}}{f(z)}=2^{2/3}(\tau-1)\left(1+\sum_{k=1}^{\infty}c_{k}(z+1)^{k}\right),$
(3.26)

for $\pm\mathop{\rm arg}\nolimits z\in(0,\pi)$ with $\tau=\frac{t}{n}$. Now we
choose

$E(z)=n^{-\frac{\nu}{3}\sigma_{3}}\left(\frac{z+1}{f(z)}\right)^{\nu\sigma_{3}}e^{\frac{n\pi
i}{2}\sigma_{3}}z^{-\frac{\nu}{2}\sigma_{3}}e^{\pm\frac{1}{2}\pi
i\nu\sigma_{3}},\quad\pm\mathop{\rm arg}\nolimits z\in(0,\pi).$ (3.27)

######  Proposition 2.

Let $s(t,z)$ be given by (3.26) and $n^{2/3}s(t,z)$ be bounded away from the
poles of $u(x;\nu)$, the matching condition (3.16) is fulfilled for general
parameter $\nu\in\mathbb{C}$.

###### Proof.

According to Remark 3.7, the solution
$\Psi\left(n^{1/3}f(z),n^{2/3}s(t,z)\right)$ exists for $n^{2/3}s(t,z)$
bounded away from the poles of $u(x;\nu)$. It follows from (3.26) that

$\frac{1}{3}f(z)^{3}+\frac{1}{2}f(z)s(t,z)=\frac{1}{2}(\phi(z)\mp\pi
i),~{}\pm\mathop{\rm arg}\nolimits z\in(0,\pi).$ (3.28)

This, together with (3.24) and (3.27), implies that

$P(z)N^{-1}=n^{-\frac{\nu}{3}\sigma_{3}}\left(I+\frac{G(z)\Psi_{1}\left(n^{2/3}s(t,z)\right)G(z)^{-1}}{n^{1/3}f(z)}+O\left(1/n^{2/3}\right)\right)n^{\frac{\nu}{3}\sigma_{3}},$
(3.29)

where

$G(z)=z^{-\frac{\nu}{2}\sigma_{3}}\left(\frac{z+1}{f(z)}\right)^{\nu\sigma_{3}}e^{\frac{n\pi
i}{2}\sigma_{3}}e^{\pm\frac{1}{2}\pi i\nu\sigma_{3}}$ (3.30)

for $\pm\mathop{\rm arg}\nolimits z\in(0,\pi)$. Therefore, we have the
matching condition (3.16). ∎

###  3.5 Final transformation $S\to R$ and error estimate

We define

$R(z)=\left\\{\begin{array}[]{ll}n^{\frac{\nu}{3}\sigma_{3}}S(z)N(z)^{-1}n^{-\frac{\nu}{3}\sigma_{3}},&\hbox{$|z+1|>\delta$,}\\\
n^{\frac{\nu}{3}\sigma_{3}}S(z)P(z)^{-1}n^{-\frac{\nu}{3}\sigma_{3}},&\hbox{$|z+1|<\delta$.}\end{array}\right.$
(3.31)

Using (3.16) and (3.31), we see that $R(z)$ satisfies the following RH
problem.

######  RH problem 3.8.

The function $R(z)$ defined by (3.31) satisfies the following properties.

  * (1)

$R(z)$ is analytic in $\mathbb{C}\setminus\Sigma_{R}$, where the remaining
contour $\Sigma_{R}$ is illustrated in Figure 6.

  * (2)

$R(z)$ satisfies the same jump condition

$R_{+}(z)=R_{-}(z)J_{R}(z),$ (3.32)

with

$J_{R}(z)=I+O(1/n^{1/3}),\quad|z+1|=\delta,$ (3.33)

and

$J_{R}(z)=I+O(e^{-cn}),$ (3.34)

for z on the other jump contours.

  * (3)

Near infinity, we have

$R(z)=I+O(1/z).$ (3.35)

Figure 6:  Contours for the RH problem for $R(z)$

######  Proposition 3.

For $|\tau-1|=O(n^{-2/3})$, we have the estimate

$R(z)=I+\frac{R_{1}}{n^{1/3}}+O\left(\frac{1}{n^{2/3}}\right),$ (3.36)

where the error term is uniform for $z$ in the complex plane. Here

$R_{1}(z)=\frac{2^{\frac{2}{3}}}{1+z}e^{\frac{n\pi
i}{2}\sigma_{3}}2^{\frac{2}{3}\nu\sigma_{3}}\Psi_{1}(2^{\frac{2}{3}}n^{2/3}(\tau-1))2^{-\frac{2}{3}\nu\sigma_{3}}e^{-\frac{n\pi
i}{2}\sigma_{3}}$ (3.37)

for $|1+z|>\delta$.

###### Proof.

It is seen from (3.29) and (3.31) that

$J_{R}(z)=I+\frac{J_{1}(z)}{n^{1/3}}+O\left(\frac{1}{n^{2/3}}\right),$ (3.38)

for $|1+z|=\delta$. Here

$J_{1}(z)=G(z)\frac{\Psi_{1}(n^{2/3}s(t,z))}{f(z)}G(z)^{-1},$ (3.39)

with $G(z)$ given in (3.30). Accordingly we have the expansion (3.36).
Substituting (3.36) and (3.38) into (3.32), we obtain

$R_{1+}(z)=R_{1-}(z)+J_{1}(z).$ (3.40)

This, together with the behavior $R_{1}(z)=O(1/z)$, implies

$R_{1}(z)=\frac{1}{2\pi i}\oint_{|x+1|=\delta}\frac{J_{1}(x)}{x-z}dx,$ (3.41)

where the integral contour is oriented clockwise. Therefore, we have (3.37) by
(3.29), (3.39) and (3.41). This completes the proof of the proposition. ∎

##  4 Asymptotics for $\Psi$ as $x\rightarrow+\infty$

We begin with the following re-scaling of variable

$\Phi(z)=x^{-\frac{\nu}{2}\sigma_{3}}\Psi(x^{\frac{1}{2}}z;x)\left\\{\begin{array}[]{ll}I,&\mathop{\rm
arg}\nolimits z\in(-\frac{1}{2}\pi,\pi),\\\ e^{2\pi
i\nu\sigma_{3}},&\mathop{\rm arg}\nolimits
z\in(-\pi,-\frac{1}{2}\pi).\end{array}\right.$ (4.1)

It is seen from the RH problem 3.6 that $\Phi(z)$ satisfies the following RH
problem.

######  RH problem 4.1.

The function $\Phi(z)$ defined by (4.1) satisfies the following properties.

  * (1)

$\Phi(z)$ is analytic in $\mathbb{C}\setminus\cup_{k=1}^{6}\pi_{k}$, where the
contours $\cup_{k=1}^{6}\pi_{k}$ are shown in Figure 7.

  * (2)

$\Phi(z)$ satisfies $\Phi_{+}(z)=\Phi_{-}(z)J_{\Phi}(z)$, for
$z\in\cup_{k=1}^{6}\pi_{k}$, where

$J_{\Phi}(z)=\left\\{\begin{array}[]{lll}\begin{pmatrix}1&e^{-\pi i\nu}\\\
0&1\end{pmatrix},&z\in\pi_{1},\\\\[14.22636pt] \begin{pmatrix}1&0\\\ -e^{\pi
i\nu}&1\end{pmatrix},&z\in\pi_{2},\\\\[14.22636pt] \begin{pmatrix}1&e^{-\pi
i\nu}-e^{-3\pi i\nu}\\\ 0&1\end{pmatrix},&z\in\pi_{3},\\\\[14.22636pt]
\begin{pmatrix}1&0\\\ e^{3\pi
i\nu}&1\end{pmatrix},&z\in\pi_{4},\\\\[14.22636pt] \begin{pmatrix}1&-e^{\pi
i\nu}\\\ 0&1\end{pmatrix},&z\in\pi_{5},\\\\[14.22636pt]
\begin{pmatrix}e^{-2\pi i\nu}&0\\\ 0&e^{2\pi
i\nu}\end{pmatrix},&z\in\pi_{6}.\\\\[14.22636pt] \end{array}\right.$ (4.2)

  * (3)

As $z\to\infty$, we have

$\Phi(z)=\left(I+\frac{\Phi_{1}}{z}+O\left(\frac{1}{z^{2}}\right)\right)e^{t\theta(z)\sigma_{3}}z^{\nu\sigma_{3}},$
(4.3)

where $t=x^{\frac{3}{2}},\theta(z)=\frac{1}{3}z^{3}+\frac{1}{2}z$ and
$\mathop{\rm arg}\nolimits z\in(-\frac{1}{2}\pi,\frac{3}{2}\pi)$.

Figure 7: Contours and regions of the RH problem for $\Phi$

To normalize the asymptotic behavior of $\Phi(z)$ at infinity, we introduce
the $g$-function

$g(z)=\frac{1}{3}(z^{2}+1)^{\frac{3}{2}},$ (4.4)

where the branches are taken such that $\mathop{\rm arg}\nolimits(z\pm
i)\in(-\frac{\pi}{2},\frac{3\pi}{2})$. Therefore, we have

$g_{+}(z)+g_{-}(z)=0,\quad z\in(-i,i).$

A straightforward computation gives

$g(z)=\frac{1}{3}z^{3}+\frac{1}{2}z+\frac{1}{8}z^{-1}+O(z^{-3}),\quad
z\rightarrow\infty.$ (4.5)

Figure 8: The jump curves $\Sigma_{{U}}$ of the RH problem for ${U}(z)$

The second transformation is now defined as

$U(z)=2^{\nu\sigma_{3}}\Phi(z)e^{-tg(z)\sigma_{3}}\varphi(z)^{-\nu\sigma_{3}},$
(4.6)

where $\varphi(z)=z+\sqrt{z^{2}+1}$ is a conformal mapping from
$\mathbb{C}\setminus[-i,i]$ onto $|\varphi(z)|>1$, and the branches are taken
such that $\mathop{\rm arg}\nolimits(z\pm
i)\in(-\frac{\pi}{2},\frac{3\pi}{2})$ and $\mathop{\rm
arg}\nolimits\varphi(z)\in(-\frac{\pi}{2},\frac{3\pi}{2})$. This implies that

$\left\\{\begin{aligned}
&\left(\varphi(z)^{\nu}\right)_{+}\left(\varphi(z)^{\nu}\right)_{-}=e^{\pi
i\nu},\quad z\in(-i,i),\\\
&\left(\varphi(z)^{\nu}\right)_{+}=\left(\varphi(z)^{\nu}\right)_{-}e^{2\pi
i\nu},\quad z\in(-i\infty,-i).\end{aligned}\right.$ (4.7)

######  RH problem 4.2.

$U(z)$ satisfies the following RH problem.

  * (1)

$U(z)$ is analytic in $\mathbb{C}\setminus\Sigma_{U}$, where $\Sigma_{{U}}$ is
shown in Figure 8, and where we have deformed $\pi_{k},k=1,2,4,5$ into the
steepest descent curves $\widetilde{\pi}_{k},k=1,2,4,5$ of the $g$-function
$g(z)$ .

  * (2)

${U}(z)$ satisfies ${U}_{+}(z)={U}_{-}(z)J_{{U}}(z)$, where

$J_{{U}}(z)=\left\\{\begin{aligned}
&\begin{pmatrix}1&\varphi(z)^{2\nu}e^{2tg(z)-\pi i\nu}\\\
0&1\end{pmatrix},\quad&z&\in\widetilde{\pi}_{1},\\\ &\begin{pmatrix}1&0\\\
-\varphi(z)^{-2\nu}e^{-2tg(z)+\pi
i\nu}&1\end{pmatrix},\quad&z&\in\widetilde{\pi}_{2},\\\
&\begin{pmatrix}1&(e^{-\pi i\nu}-e^{-3\pi i\nu})\varphi(z)^{2\nu}e^{2tg(z)}\\\
0&1\end{pmatrix},\quad&z&\in\pi_{3}.\\\ &\begin{pmatrix}1&0\\\
\varphi(z)^{-2\nu}e^{-2tg(z)+3\pi
i\nu}&1\end{pmatrix},\quad&z&\in\widetilde{\pi}_{4},\\\
&\begin{pmatrix}1&-\varphi(z)^{2\nu}e^{2tg(z)+\pi i\nu}\\\
0&1\end{pmatrix},\quad&z&\in\widetilde{\pi}_{5},\\\ &\begin{pmatrix}0&1\\\
-1&e^{-\pi
i\nu}\varphi(z)^{2\nu}_{+}e^{2tg_{+}(z)}\end{pmatrix},\quad&z&\in\pi_{+},\\\
&\begin{pmatrix}e^{-3\pi i\nu}\varphi(z)^{2\nu}_{+}e^{2tg_{+}(z)}&-1\\\
1&0\end{pmatrix},\quad&z&\in\pi_{-}.\end{aligned}\right.$ (4.8)

  * (3)

As $z\to\infty$, we have

$U(z)=I+O\left(\frac{1}{z}\right).$ (4.9)

According to the signature of $\mathop{\rm Re}\nolimits g(z)$, it is readily
seen that the jump matrices on $\widetilde{\pi}_{k},k=1,2,4,5$ and $\pi_{3}$
tend to the identity matrix exponentially fast as $t\rightarrow+\infty$. The
next task is to construct a global parametrix with constant jump on the
segment $[-i,i]$ and two local parametrices near the critical points $z=\pm
i$.

###  4.1 Global parametrix on $[-i,i]$

######  RH problem 4.3.

We look for a $2\times 2$ matrix-valued function $\widetilde{P}^{(\infty)}(z)$
satisfying the following properties.

  * (1)

$\widetilde{P}^{(\infty)}(z)$ is analytic in $\mathbb{C}\setminus[-i,i]$.

  * (2)

$\widetilde{P}^{(\infty)}(z)$ satisfies the jump condition

$\widetilde{P}^{(\infty)}_{+}(z)=\widetilde{P}^{(\infty)}_{-}(z)\begin{pmatrix}0&1\\\
-1&0\end{pmatrix},\qquad z\in[-i,i].$ (4.10)

  * (3)

As $z\to\infty$, we have

$\widetilde{P}^{(\infty)}(z)=I+O\left(\frac{1}{z}\right).$ (4.11)

A solution of the above RH problem is given by

$\widetilde{P}^{(\infty)}(z)=\begin{pmatrix}1&1\\\
i&-i\end{pmatrix}w(z)^{\sigma_{3}}\begin{pmatrix}1&1\\\
i&-i\end{pmatrix}^{-1},\quad w(z)=\left(\frac{z-i}{z+i}\right)^{\frac{1}{4}},$
(4.12)

where the branches are taken such that $\mathop{\rm arg}\nolimits(z\pm
i)\in(-\frac{\pi}{2},\frac{3\pi}{2})$.

####  4.1.1 Local parametrices near $\pm i$

In this subsection, we seek two parametrices $P^{(\pm i)}(z)$ satisfying the
same jumps on the jump contours as $U(z)$ respectively in the neighborhoods
$U(\pm i,\delta)$ of the saddle points $\pm i$, and matching with
$\widetilde{P}^{(\infty)}(z)$ on the boundaries $\partial U(\pm i,\delta)$.

######  RH problem 4.4.

We look for a $2\times 2$ matrix-valued function $P^{(i)}(z)$ satisfying the
following properties.

  * (1)

$P^{(i)}(z)$ is analytic in $U(i,\delta)\setminus\Sigma_{U}$.

  * (2)

$P^{(i)}(z)$ satisfies the same jump condition as $U(z)$ on
$U(i,\delta)\cap\Sigma_{U}$.

  * (3)

On $\partial U(i,\delta)$, we have

$P^{(i)}(z)=\left(I+O(t^{-1})\right)\widetilde{P}^{(\infty)}(z),\quad\mathrm{as}\quad
t\to\infty.$ (4.13)

To construct a solution to the above RH problem, we first define a conformal
mapping

$\lambda(z)=\left(\frac{3}{2}g(z)\right)^{\frac{2}{3}}=e^{\frac{\pi}{2}i}2^{\frac{1}{3}}(z-i)(1+o(1)),\quad
z\rightarrow i.$ (4.14)

Then, the solution to the above RH problem can be built out of the Airy
function as follows:

$P^{(i)}(z)=E^{(i)}(z)\Phi^{\mathrm{(Ai)}}\left(t^{\frac{2}{3}}\lambda(z)\right)e^{\frac{1}{2}\pi
i\nu\sigma_{3}}\sigma_{3}\varphi(z)^{-\nu\sigma_{3}}e^{-tg(z)\sigma_{3}},$
(4.15)

where $\Phi^{\mathrm{(Ai)}}$ denotes the Airy parametrix given in Appendix A.1
and $E^{(i)}(z)$ is given by

$E^{(i)}(z)=\widetilde{P}^{(\infty)}(z)\varphi(z)^{\nu\sigma_{3}}\sigma_{3}e^{-\frac{1}{2}\pi
i\nu\sigma_{3}}\begin{pmatrix}-i&i\\\
1&1\end{pmatrix}\left(t^{\frac{2}{3}}\lambda(z)\right)^{\frac{\sigma_{3}}{4}}.$
(4.16)

Here, the branch of the function $\lambda(z)^{\frac{1}{4}}$ is chosen such
that $\mathop{\rm arg}\nolimits\lambda(z)\in(0,2\pi)$. Particularly, we have
on the segment $[-i,i]$

$\left(\lambda(z)^{\frac{\sigma_{3}}{4}}\right)_{+}=\left(\lambda(z)^{\frac{\sigma_{3}}{4}}\right)_{-}e^{\frac{\pi
i}{2}\sigma_{3}}.$ (4.17)

From (4.10) and (4.17), it is straightforward to check that $E^{(i)}(z)$ is
analytic in the neighborhood $U(i,\delta)$. Furthermore, using (4.12), (4.15)
and (A.2), we arrive at the following matching condition

$\displaystyle P^{(i)}(z)\widetilde{P}^{(\infty)}(z)^{-1}$ (4.18)
$\displaystyle=W^{(i)}(z)\left(I+\frac{1}{48\lambda(z)^{\frac{3}{2}}t}\begin{pmatrix}-1&6i\\\
6i&1\end{pmatrix}+O(t^{-2})\right)W^{(i)}(z)^{-1},$
$\displaystyle=I+Q^{(i)}(z)t^{-1}+O\left(t^{-2}\right),$

where

$W^{(i)}(z)=\widetilde{P}^{(\infty)}(z)\varphi(z)^{\nu\sigma_{3}}\sigma_{3}e^{-\frac{1}{2}\pi
i\nu\sigma_{3}},$ (4.19)

and

$Q^{(i)}(z)=\frac{1}{96\lambda(z)^{\frac{3}{2}}}\begin{pmatrix}c_{1}(z)&*\\\
*&*\end{pmatrix},$ (4.20)

with

$c_{1}(z)=\left(-1-3\varphi(z)^{-2\nu}e^{\pi i\nu}-3\varphi(z)^{2\nu}e^{-\pi
i\nu}\right)w(z)^{2}+\left(-1+3\varphi(z)^{-2\nu}e^{\pi
i\nu}+3\varphi(z)^{2\nu}e^{-\pi i\nu}\right)w(z)^{-2}.\\\ $

######  RH problem 4.5.

We look for a $2\times 2$ matrix-valued function $P^{(-i)}(z)$ satisfying the
following properties.

  * (1)

$P^{(-i)}(z)$ is analytic in $U(-i,\delta)\setminus\Sigma_{U}$.

  * (2)

$P^{(-i)}(z)$ satisfies the same jump condition as $U(z)$ on
$U(-i,\delta)\cap\Sigma_{U}$.

  * (3)

On $\partial U(-i,\delta)$, we have

$P^{(-i)}(z)=\left(I+O(t^{-1})\right)\widetilde{P}^{(\infty)}(z),\quad\mathrm{as}\quad
t\to\infty.$ (4.21)

Similarly, the solution to the above RH problem can be built out of the Airy
function as follows:

$P^{(-i)}(z)=\left\\{\begin{aligned}
&E^{(-i)}(z)\Phi^{\mathrm{(Ai)}}\left(t^{\frac{2}{3}}\lambda(-z)\right)\sigma_{2}e^{\frac{3}{2}\pi
i\nu\sigma_{3}}\varphi(z)^{-\nu\sigma_{3}}e^{-tg(z)\sigma_{3}},&\mathop{\rm
Re}\nolimits z<0,\\\
&E^{(-i)}(z)\Phi^{\mathrm{(Ai)}}\left(t^{\frac{2}{3}}\lambda(-z)\right)\sigma_{2}e^{-\frac{1}{2}\pi
i\nu\sigma_{3}}\varphi(z)^{-\nu\sigma_{3}}e^{-tg(z)\sigma_{3}},&\mathop{\rm
Re}\nolimits z>0,\\\ \end{aligned}\right.$ (4.22)

where $\Phi^{\mathrm{(Ai)}}$ is the Airy parametrix given in Appendix A.1,
$\lambda(z)$ is defined in (4.17) and $E^{(-i)}(z)$ is given by

$E^{(-i)}(z)=\left\\{\begin{aligned}
&\widetilde{P}^{(\infty)}(z)\varphi(z)^{\nu\sigma_{3}}e^{-\frac{3}{2}\pi
i\nu\sigma_{3}}\sigma_{2}\begin{pmatrix}-i&i\\\
1&1\end{pmatrix}\left(t^{\frac{2}{3}}\lambda(-z)\right)^{\frac{\sigma_{3}}{4}},&\mathop{\rm
Re}\nolimits z<0,\\\
&\widetilde{P}^{(\infty)}(z)\varphi(z)^{\nu\sigma_{3}}e^{\frac{1}{2}\pi
i\nu\sigma_{3}}\sigma_{2}\begin{pmatrix}-i&i\\\
1&1\end{pmatrix}\left(t^{\frac{2}{3}}\lambda(-z)\right)^{\frac{\sigma_{3}}{4}},&\mathop{\rm
Re}\nolimits z>0.\\\ \end{aligned}\right.$ (4.23)

It is readily seen that $E^{(-i)}(z)$ is analytic in $U(-i,\delta)$ and the
function $P^{(-i)}(z)$ in (4.22) satisfies the matching condition (4.21). It
follows from (4.12), (4.22) and (A.2) that

$\displaystyle P^{(-i)}(z)\widetilde{P}^{(\infty)}(z)^{-1}$ (4.24)
$\displaystyle=W^{(-i)}(z)\left(I+\frac{1}{48\lambda(-z)^{\frac{3}{2}}t}\begin{pmatrix}-1&6i\\\
6i&1\end{pmatrix}+O\left(t^{-2}\right)\right)W^{(-i)}(z)^{-1},$
$\displaystyle=I+Q^{(-i)}(z)t^{-1}+O\left(t^{-2}\right),$

where

$W^{(-i)}(z)=\left\\{\begin{aligned}
&\widetilde{P}^{(\infty)}(z)\varphi(z)^{\nu\sigma_{3}}e^{-\frac{3}{2}\pi
i\nu\sigma_{3}}\sigma_{2},\quad\mathop{\rm Re}\nolimits z<0,\\\
&\widetilde{P}^{(\infty)}(z)\varphi(z)^{\nu\sigma_{3}}e^{\frac{1}{2}\pi
i\nu\sigma_{3}}\sigma_{2},\quad\mathop{\rm Re}\nolimits z>0,\\\
\end{aligned}\right.$ (4.25)

and

$Q^{(-i)}(z)=\frac{1}{96\lambda(-z)^{\frac{3}{2}}}\begin{pmatrix}c_{2}(z)&*\\\
*&*\end{pmatrix},$ (4.26)

with

$\displaystyle c_{2}(z)=(1-3\varphi(z)^{-2\nu}e^{-\pi
i\nu}-3\varphi(z)^{2\nu}e^{\pi i\nu})w(z)^{2}+(1+3\varphi(z)^{-2\nu}e^{-\pi
i\nu}+3\varphi(z)^{2\nu}e^{\pi i\nu})w(z)^{-2}.$

###  4.2 Final transformation

The final transformation is defined as

$\widetilde{R}(z)=\left\\{\begin{aligned} &U(z)P^{(\pm i)}(z)^{-1},\quad&z&\in
U(\pm i,\delta)\setminus\Sigma_{U},\\\
&U(z)\widetilde{P}^{(\infty)}(z)^{-1},&\mathrm{e}&\mathrm{lsewhere}.\\\
\end{aligned}\right.$ (4.27)

Figure 9: The jump contours $\Sigma_{\widetilde{R}}$ of the RH problem for
$\widetilde{R}(z)$

######  RH problem 4.6.

$\widetilde{R}(z)$ satisfies the following properties.

(1) $\widetilde{R}(z)$ is analytic for
$z\in\mathbb{C}\setminus\Sigma_{\widetilde{R}}$, where the contour
$\Sigma_{\widetilde{R}}$ is illustrated in Figure 9.

(2) For $z\in\Sigma_{\widetilde{R}}$, we have
$\widetilde{R}_{+}(z)=\widetilde{R}_{-}(z)J_{\widetilde{R}}(z)$, where

$J_{\widetilde{R}}(z)=\left\\{\begin{aligned} &P^{(\pm
i)}(z)\widetilde{P}^{(\infty)}(z)^{-1},\quad&z&\in\partial U(\pm i,\delta),\\\
&\widetilde{P}^{(\infty)}(z)J_{U}(z)\widetilde{P}^{(\infty)}(z)^{-1},\quad&\mathrm{e}&\mathrm{lsewhere}.\end{aligned}\right.$
(4.28)

(3) As $z\rightarrow\infty$, we have

$\widetilde{R}(z)=I+\frac{\widetilde{R}_{1}}{z}+O\left(\frac{1}{z^{2}}\right).$
(4.29)

Based on the matching conditions (4.13) and (4.21), we have the following
estimates as $t\rightarrow\infty$

$J_{\widetilde{R}}(z)=\left\\{\begin{aligned}
&I+O\left(t^{-1}\right),&z&\in\partial U(\pm i,\delta),\\\
&I+O\left(e^{-c_{1}t}\right),&\mathrm{e}&\mathrm{lsewhere},\end{aligned}\right.$
(4.30)

where $c_{1}$ is a positive constant. Consequently, we have the asymptotic
approximation

$\widetilde{R}(z)=I+O\left(t^{-1}\right),\quad\mathrm{as}\quad
t\rightarrow\infty,$ (4.31)

where the error term is uniform for
$z\in\mathbb{C}\setminus\Sigma_{\widetilde{R}}$.

###  4.3 Asymptotics for $H$ and $u$ as $x\rightarrow+\infty$

According to our previous RH analysis $\widetilde{\Psi}\mapsto\Phi\mapsto
U\mapsto\widetilde{R}$, for large $z$, we have

$2^{\nu\sigma_{3}}\Phi(z)e^{-tg(z)\sigma_{3}}\varphi(z)^{-\nu\sigma_{3}}=\widetilde{R}(z)\widetilde{P}^{(\infty)}(z).$
(4.32)

It follows from the large $z$ asymptotics (4.3) and (4.5) that

$2^{\nu\sigma_{3}}\Phi(z)e^{-tg(z)\sigma_{3}}\varphi(z)^{-\nu\sigma_{3}}=I+\frac{2^{\nu\sigma_{3}}\Phi_{1}2^{-\nu\sigma_{3}}-\frac{1}{8}t\sigma_{3}}{z}+O\left(\frac{1}{z^{2}}\right).$
(4.33)

The asymptotic approximation of $\widetilde{P}^{(\infty)}(z)$ is obtained by
expanding (4.12),

$\widetilde{P}^{(\infty)}(z)=I+\frac{\widetilde{P}^{(\infty)}_{1}}{z}+O\left(\frac{1}{z^{2}}\right),\quad\mathrm{as}~{}z\rightarrow\infty,$
(4.34)

where

$\widetilde{P}^{(\infty)}_{1}=\begin{pmatrix}0&-\frac{1}{2}\\\
\frac{1}{2}&0\end{pmatrix}.$ (4.35)

From (4.33) and the asymptotic behaviors (4.29) and (4.34), we have

$\Phi_{1}=2^{-\nu\sigma_{3}}\left(\widetilde{P}^{(\infty)}_{1}+\widetilde{R}_{1}\right)2^{\nu\sigma_{3}}+\frac{1}{8}t\sigma_{3},$
(4.36)

which implies

$(\Phi_{1})_{12}=4^{-\nu}\left((\widetilde{P}^{(\infty)}_{1})_{12}+(\widetilde{R}_{1})_{12}\right),$
(4.37)

and

$(\Phi_{1})_{11}=(\widetilde{P}^{(\infty)}_{1})_{11}+(\widetilde{R}_{1})_{11}+\frac{1}{8}t.$
(4.38)

In virtue of (4.18), (4.24) and (4.28), we have as $t\rightarrow\infty$

$\widetilde{R}(z)=I+\frac{\widetilde{R}^{(1)}(z)}{t}+O\left(\frac{1}{t^{2}}\right),\quad
z\in\partial U(\pm i,\delta),$ (4.39)

where the coefficient $\widetilde{R}^{(1)}(z)=O(z^{-1})$ as
$z\rightarrow\infty$, and $\widetilde{R}^{(1)}(z)$ satisfies the jump relation

$\widetilde{R}^{(1)}_{+}(z)-\widetilde{R}^{(1)}_{-}(z)=Q^{(\pm i)}(z),\quad
z\in\partial U(\pm i,\delta).$ (4.40)

The solution to the above RH problem is explicitly given by

$\widetilde{R}^{(1)}(z)=\left\\{\begin{aligned}
&\frac{C_{+}}{z-i}+\frac{C_{-}}{z+i}+\frac{D_{+}}{(z-i)^{2}}+\frac{D_{-}}{(z+i)^{2}},&z\notin\overline{U(i,\delta)}\cup\overline{U(-i,\delta)},\\\
&\frac{C_{+}}{z-i}+\frac{C_{-}}{z+i}+\frac{D_{+}}{(z-i)^{2}}+\frac{D_{-}}{(z+i)^{2}}-Q^{(\pm
i)}(z),&z\in U(i,\delta)\cup U(-i,\delta),\end{aligned}\right.$ (4.41)

where $C_{\pm}=\mathop{\rm Res}\nolimits(Q^{(\pm i)}(z),z=\pm i)$. Expanding
this as $z\rightarrow\infty$, we obtain

$\widetilde{R}_{1}=\frac{\widetilde{R}^{(1)}_{1}}{t}+O\left(\frac{1}{t^{2}}\right),\quad
t\rightarrow\infty,$ (4.42)

where $\widetilde{R}^{(1)}_{1}=C_{+}+C_{-}$, after a direct computation by
expanding $\varphi(z)$, $w(z)$ and $\lambda(\pm z)^{\frac{3}{2}}$ as
$z\rightarrow\pm i$ , it follows that

$(\widetilde{R}^{(1)}_{1})_{11}=(C_{+}+C_{-})_{11}=-\frac{1}{2}\nu^{2}+\frac{1}{8}.$
(4.43)

Substituting (4.35) and (4.42) into (4.37) and (4.38), and using the formula
(3.25) and (4.1), we obtain (1.21) and (1.23).

##  5 Asymptotics for $\Psi$ as $x\rightarrow-\infty$

###  5.1 Asymptotics for $\Psi$ as $x\rightarrow-\infty$ with
$\nu\in\mathbb{C}\setminus\\{0\\}$

First, we introduce the following re-scaling of variable

$\widehat{\Phi}(z)=(-x)^{-\frac{\nu}{2}\sigma_{3}}\Psi\left((-x)^{\frac{1}{2}}z;x\right).$
(5.1)

######  RH problem 5.1.

$\widehat{\Phi}(z)$ solves the following RH problem.

  * (1)

$\widehat{\Phi}(z)$ is analytic in
$\mathbb{C}\setminus\cup_{k=1}^{5}~{}\Gamma_{k}$; see Figure 5.

  * (2)

$\widehat{\Phi}(z)$ satisfies
$\widehat{\Phi}_{+}(z)=\widehat{\Phi}_{-}(z)J_{\Psi}$,
$z\in\cup_{k=1}^{5}~{}\Gamma_{k}$ with $J_{\Psi}$ given in(3.23).

  * (3)

As $z\to\infty$, we have

$\widehat{\Phi}(z)=\left(I+\frac{\widehat{\Phi}_{1}}{z}+O\left(\frac{1}{z^{2}}\right)\right)e^{t\widehat{\theta}(z)\sigma_{3}}z^{\nu\sigma_{3}},$
(5.2)

where $t=|x|^{\frac{3}{2}}$,
$\widehat{\theta}(z)=\frac{1}{3}z^{3}-\frac{1}{2}z$, and $\mathop{\rm
arg}\nolimits z\in(-\pi,\pi)$.

We introduce the $g$-function

$\widehat{g}(z)=\widehat{\theta}(z)-\widehat{\theta}\left(-{\sqrt{2}}/{2}\right).$
(5.3)

It is direct to see that $\widehat{g}(z)$ has two saddle points at
$z_{\pm}=\pm\frac{\sqrt{2}}{2}$. The second transformation is defined by

$\widehat{U}(z)=e^{-t\widehat{\theta}\left(-\frac{\sqrt{2}}{2}\right)\sigma_{3}}\widehat{\Phi}(z)e^{-t\widehat{g}(z)\sigma_{3}}.$
(5.4)

We also deform the jump contours $\Gamma_{k}$ into the steepest descent curves
$\widehat{\Gamma}_{k}$, $k=1,2,4,5$ of the function $\widehat{g}(z)$. Then, we
have

$\mathop{\rm Re}\nolimits\widehat{g}(z)<0,\quad
z\in\widehat{\Gamma}_{1}\cup\widehat{\Gamma}_{3}\cup\widehat{\Gamma}_{5}\cup\widehat{\Gamma}_{6},~{}~{}\mbox{and}~{}~{}\mathop{\rm
Re}\nolimits\widehat{g}(z)>0,\quad
z\in\widehat{\Gamma}_{2}\cup\widehat{\Gamma}_{4},$ (5.5)

where the curves $\widehat{\Gamma}_{k}$, $k=1,2,3,4,5,6$ are illustrated in
Figure 10.

######  RH problem 5.2.

$\widehat{U}(z)$ satisfies the following RH problem.

  * (1)

$\widehat{U}(z)$ is analytic in $\mathbb{C}\setminus\Sigma_{\widehat{U}}$,
where $\Sigma_{\widehat{U}}$ is indicated in Figure 10.

  * (2)

$\widehat{U}(z)$ satisfies
$\widehat{U}_{+}(z)=\widehat{U}_{-}(z)J_{\widehat{U}}(z)$, where

$J_{\widehat{U}}(z)=\left\\{\begin{aligned}
&\begin{pmatrix}1&e^{2t\widehat{g}(z)-\pi i\nu}\\\
0&1\end{pmatrix},\quad&z&\in\widehat{\Gamma}_{1},\\\ &\begin{pmatrix}1&0\\\
-e^{-2t\widehat{g}(z)+\pi
i\nu}&1\end{pmatrix},\quad&z&\in\widehat{\Gamma}_{2},\\\
&\begin{pmatrix}e^{-2\pi i\nu}&(e^{\pi i\nu}-e^{-\pi
i\nu})e^{2t\widehat{g}(z)}\\\ 0&e^{2\pi
i\nu}\end{pmatrix},\quad&z&\in\widehat{\Gamma}_{3},\\\ &\begin{pmatrix}1&0\\\
e^{-2t\widehat{g}(z)-\pi
i\nu}&1\end{pmatrix},\quad&z&\in\widehat{\Gamma}_{4},\\\
&\begin{pmatrix}1&-e^{2t\widehat{g}(z)+\pi i\nu}\\\
0&1\end{pmatrix},\quad&z&\in\widehat{\Gamma}_{5},\\\
&\begin{pmatrix}1&(e^{-\pi i\nu}-e^{\pi i\nu})e^{2t\widehat{g}(z)}\\\
0&1\end{pmatrix},\quad&z&\in\widehat{\Gamma}_{6}.\end{aligned}\right.$ (5.6)

  * (3)

As $z\to\infty$, we have

$\widehat{U}(z)=\left(I+\frac{\widehat{U}_{1}}{z}+O\left(\frac{1}{z^{2}}\right)\right)z^{\nu\sigma_{3}},$
(5.7)

where $\mathop{\rm arg}\nolimits z\in(-\pi,\pi)$.

Figure 10:  The jump contours $\Sigma_{\widehat{U}}$ of RH problem for
$\widehat{U}(z)$

It is seen from (5.5) that the jump matrices on
$\widehat{\Gamma}_{k},k=1,2,4,5,6$ tend to the identity matrix exponentially
fast as $t\rightarrow+\infty$. While, the jump on $\widehat{\Gamma}_{3}$ tends
to $e^{-2\pi i\nu\sigma_{3}}$ exponentially fast as $t\rightarrow+\infty$. In
the following two sections, we construct a global parametrix with the
remaining constant jump $e^{-2\pi i\nu\sigma_{3}}$ on $\widehat{\Gamma}_{3}$
and a local parametrix near the saddle points $z=-\frac{\sqrt{2}}{2}$.

####  5.1.1 Global parametrix

######  RH problem 5.3.

We look for a $2\times 2$ matrix-valued function $\widehat{P}^{(\infty)}(z)$
satisfying the following properties.

  * (1)

$\widehat{P}^{(\infty)}(z)$ is analytic in
$\mathbb{C}\setminus(-\infty,z_{-}]$.

  * (2)

$\widehat{P}^{(\infty)}(z)$ satisfies the jump condition

$\widehat{P}^{(\infty)}_{+}(z)=\widehat{P}^{(\infty)}_{-}(z)J_{\infty}(z)$
(5.8)

where

$J_{\infty}(z)=\begin{aligned} &\begin{pmatrix}e^{2\pi i\nu}&0\\\ 0&e^{-2\pi
i\nu}\end{pmatrix},\qquad z\in(-\infty,z_{-}]\\\ \end{aligned}.$

  * (3)

As $z\to\infty$, we have

$\widehat{P}^{(\infty)}(z)=\left(I+O\left(\frac{1}{z}\right)\right)z^{\nu\sigma_{3}},$
(5.9)

where $\mathop{\rm arg}\nolimits z\in(-\pi,\pi)$.

The solution of the above RH problem is explicitly given by

$\widehat{P}^{(\infty)}(z)=\left(z+\frac{\sqrt{2}}{2}\right)^{\nu\sigma_{3}},$
(5.10)

where $\mathop{\rm
arg}\nolimits\left(z+\frac{\sqrt{2}}{2}\,\right)\in(-\pi,\pi)$.

####  5.1.2 Local parametrices near $z_{-}=-\frac{\sqrt{2}}{2}$

In this subsection, we seek a function $P^{(z_{-})}(z)$ satisfying the same
jumps as $\widehat{U}(z)$ on the jump contours $\Sigma_{\widehat{U}}$ in the
neighborhood $U(z_{-},\delta)$ of the saddle points
$z_{-}=-\frac{\sqrt{2}}{2}$ and matching with $\widehat{P}^{(\infty)}(z)$ on
the boundary $\partial U(z_{-},\delta)$.

######  RH problem 5.4.

We look for a $2\times 2$ matrix-valued function $P^{(z_{-})}(z)$ satisfying
the following properties.

  * (1)

$P^{(z_{-})}(z)$ is analytic in
$U(z_{-},\delta)\setminus\Sigma_{\widehat{U}}$.

  * (2)

$P^{(z_{-})}(z)$ satisfies the same jump condition as $\widehat{U}(z)$ on
$U(z_{-},\delta)\cap\Sigma_{\widehat{U}}$.

  * (3)

On $\partial U(z_{-},\delta)$, we have

$P^{(z_{-})}(z)=t^{-\frac{\nu}{2}\sigma_{3}}\left(I+O\left(t^{-\frac{1}{2}}\right)\right)t^{\frac{\nu}{2}\sigma_{3}}\widehat{P}^{(\infty)}(z),\quad\mathrm{as}\quad
t\to\infty.$ (5.11)

In the case $\nu\in\mathbb{C}\setminus\mathbb{N}$, we define the following
conformal mapping

$\eta(z)=2\sqrt{-\widehat{g}(z)}=2^{\frac{3}{4}}\left(z+\frac{\sqrt{2}}{2}\right)\left(1+o(1)\right),\quad
z\rightarrow-\frac{\sqrt{2}}{2}.$ (5.12)

Let $\Phi^{(PC)}$ be the parabolic cylinder parametrix given in Appendix A.2.
Then the parametrix near $z=z_{-}$ can be constructed as follows:

$P^{(z_{-})}(z)=E^{(z_{-})}(z)\Phi^{(PC)}\left(t^{\frac{1}{2}}\eta(z)\right)\left(-e^{-\pi
i\nu}h_{1}\right)^{\frac{\sigma_{3}}{2}}D(z)\sigma_{1}e^{-t\widehat{g}(z)\sigma_{3}},$
(5.13)

where $h_{1}=\frac{\sqrt{2\pi}}{\Gamma(-\nu)}e^{i\pi\nu}$ for
$\nu\in\mathbb{C}\setminus\mathbb{N}$, as defined in (A.3), and
$E^{(z_{-})}(z)$ is defined as

$E^{(z_{-})}(z)=\widehat{P}^{(\infty)}(z)\sigma_{1}\eta(z)^{\nu\sigma_{3}}t^{\frac{1}{2}\nu\sigma_{3}}\left(-e^{-\pi
i\nu}h_{1}\right)^{-\frac{\sigma_{3}}{2}}2^{-\frac{\sigma_{3}}{2}}\begin{pmatrix}t^{\frac{1}{2}}\eta(z)&1\\\
1&0\end{pmatrix},$ (5.14)

with

$D(z)=\left\\{\begin{aligned} &I,\quad&\mathop{\rm arg}\nolimits
z&\in(-\frac{\pi}{4},\pi),\\\ &e^{2\pi i\nu\sigma_{3}},\quad&\mathop{\rm
arg}\nolimits z&\in(\pi,\frac{7\pi}{4}).\end{aligned}\right.$ (5.15)

Here, the branch of the function $\eta(z)^{\nu}$ is chosen such that
$\mathop{\rm arg}\nolimits\eta(z)\in(-\pi,\pi)$. It is easily seen that
$E^{(z_{-})}(z)$ is analytic in the neighborhood $U(z_{-},\delta)$.
Furthermore, a combination of (5.10), (5.13), (5.14) and (A.4) gives

$\displaystyle P^{(z_{-})}(z)\widehat{P}^{(\infty)}(z)^{-1}$ (5.16)
$\displaystyle=W^{(z_{-})}(z)\begin{pmatrix}1+\frac{\nu(\nu+1)}{2t\eta(z)^{2}}+O(t^{-2})&\frac{\nu}{t^{\frac{1}{2}}\eta(z)}+O(t^{-\frac{3}{2}})\\\
\frac{1}{t^{\frac{1}{2}}\eta(z)}+O(t^{-\frac{3}{2}})&1-\frac{\nu(\nu-1)}{2t\eta(z)^{2}}+O(t^{-2})\end{pmatrix}W^{(z_{-})}(z)^{-1},$
$\displaystyle=I+Q^{(z_{-})}(z)t^{-\frac{1}{2}}+O(t^{-1}),$

where

$W^{(z_{-})}(z)=\widehat{P}^{(\infty)}(z)\sigma_{1}\eta(z)^{\nu\sigma_{3}}t^{\frac{1}{2}\nu\sigma_{3}}\left(-e^{-\pi
i\nu}h_{1}\right)^{-\frac{\sigma_{3}}{2}},$ (5.17)

and $Q^{(z_{-})}(z)$ is given by

$Q^{(z_{-})}(z)=t^{-\frac{1}{2}\nu\sigma_{3}}\begin{pmatrix}0&-\frac{\left(z+\frac{\sqrt{2}}{2}\right)^{2\nu}h_{1}e^{-\pi
i\nu}}{\eta(z)^{2\nu+1}}\\\ -\frac{\nu\eta(z)^{2\nu-1}e^{\pi
i\nu}}{h_{1}\left(z+\frac{\sqrt{2}}{2}\right)^{2\nu}}&0\end{pmatrix}t^{\frac{1}{2}\nu\sigma_{3}}.$
(5.18)

Figure 11: The jump contours $\Sigma_{\widehat{R}}$ of the RH problem for
$\widehat{R}(z)$

####  5.1.3 Final transformation

The final transformation is defined as

$\widehat{R}(z)=\left\\{\begin{aligned}
&t^{\frac{1}{2}\nu\sigma_{3}}\widehat{U}(z)P^{(z_{-})}(z)^{-1}t^{-\frac{1}{2}\nu\sigma_{3}},\quad&z&\in
U(z_{-},\delta)\setminus\Sigma_{\widehat{U}},\\\
&t^{\frac{1}{2}\nu\sigma_{3}}\widehat{U}(z)\widehat{P}^{(\infty)}(z)^{-1}t^{-\frac{1}{2}\nu\sigma_{3}},&\mathrm{e}&\mathrm{lsewhere}.\\\
\end{aligned}\right.$ (5.19)

Then $\widehat{R}(z)$ satisfies the following RH problem.

######  RH problem 5.5.

(1) $\widehat{R}(z)$ is analytic for
$z\in\mathbb{C}\setminus\Sigma_{\widehat{R}}$, where the contour
$\Sigma_{\widehat{R}}$ is illustrated in Figure 11.

(2) On the contour $\Sigma_{\widehat{R}}$, we have
$\widehat{R}_{+}(z)=\widehat{R}_{-}(z)J_{\widehat{R}}(z)$, where

$J_{\widehat{R}}(z)=\left\\{\begin{aligned}
&t^{\frac{1}{2}\nu\sigma_{3}}P^{(z_{-})}(z)\widehat{P}^{(\infty)}(z)^{-1}t^{-\frac{1}{2}\nu\sigma_{3}},\quad&z&\in\partial
U(z_{-},\delta),\\\
&t^{\frac{1}{2}\nu\sigma_{3}}\widehat{P}^{(\infty)}(z)J_{\widehat{U}}(z)\widehat{P}^{(\infty)}(z)^{-1}t^{-\frac{1}{2}\nu\sigma_{3}},\quad&\mathrm{e}&\mathrm{lsewhere}.\\\
\end{aligned}\right.$ (5.20)

(3) As $z\rightarrow\infty$, we have

$\widehat{R}(z)=I+\frac{\widehat{R}_{1}}{z}+O\left(\frac{1}{z^{2}}\right).$
(5.21)

By virtue of the matching condition (5.11), we have the following estimates

$J_{\widehat{R}}(z)=\left\\{\begin{aligned}
&I+O(t^{-\frac{1}{2}}),&z&\in\partial U(z_{-},\delta),\\\
&I+O(e^{-c_{2}t}),\quad&\mathrm{e}&\mathrm{lsewhere},\end{aligned}\right.$
(5.22)

where $c_{2}$ is a positive constant. Consequently, we have

$\widehat{R}(z)=I+O(t^{-\frac{1}{2}}),\quad\text{as}\quad t\rightarrow\infty,$
(5.23)

uniformly for $z\in\mathbb{C}\setminus\Sigma_{\widehat{R}}$. This completes
the RH analysis for the case $\nu\in\mathbb{C}\setminus\mathbb{N}$.

For $\nu\in\mathbb{N}\setminus\\{0\\}$, the asymptotics of $\Psi(z,x;\nu)$ can
be obtained from that of $\Psi(z,x;-\nu)$ by using the symmetry relation

$\Psi(z,x;\nu)=(-1)^{\nu}\sigma_{1}\Psi(-z,x;-\nu)\sigma_{1},\quad\nu\in\mathbb{N}.$
(5.24)

###  5.2 Asymptotics of $u(x)$ and $H(x)$ as $x\to-\infty$ for
$\nu\in\mathbb{C}\setminus\\{0\\}$

First, we consider the case $\nu\in\mathbb{C}\setminus\mathbb{N}$. According
to our previous RH analysis
$\Psi\mapsto\widehat{\Phi}\mapsto\widehat{U}\mapsto\widehat{R}$, we have that,
for large $z$,

$e^{-t\widehat{\theta}(-\frac{\sqrt{2}}{2})\sigma_{3}}\widehat{\Phi}(z)e^{-t\widehat{g}(z)\sigma_{3}}=t^{-\frac{1}{2}\nu\sigma_{3}}\widehat{R}(z)t^{\frac{1}{2}\nu\sigma_{3}}\widehat{P}^{(\infty)}(z).$
(5.25)

The asymptotic approximation of $\widehat{P}^{(\infty)}(z)$ can be obtained by
expanding the explicit representation (5.10).

$\widehat{P}^{(\infty)}(z)=\left(I+\frac{\widehat{P}^{(\infty)}_{1}}{z}+O\left(\frac{1}{z^{2}}\right)\right)z^{\nu\sigma_{3}},\quad\mathrm{as}\quad
z\rightarrow\infty,$ (5.26)

where

$\widehat{P}^{(\infty)}_{1}=\begin{pmatrix}\frac{\sqrt{2}}{2}\nu&0\\\
0&-\frac{\sqrt{2}}{2}\nu\end{pmatrix}.$ (5.27)

From (5.25) and the asymptotic behaviors (5.2), (5.21), (5.26) and (5.27), we
have

$e^{-\frac{\sqrt{2}}{3}t}(\widehat{\Phi}_{1})_{12}=t^{-\nu}(\widehat{R}_{1})_{12},\quad
e^{\frac{\sqrt{2}}{3}t}(\widehat{\Phi}_{1})_{21}=t^{\nu}(\widehat{R}_{1})_{21},$
(5.28)

and

$(\widehat{\Phi}_{1})_{11}=(\widehat{P}^{(\infty)}_{1})_{11}+(\widehat{R}_{1})_{11}.$
(5.29)

In virtue of (5.16), (5.20) and (5.21), we easily get

$\widehat{R}(z)=I+\frac{\widehat{R}^{(1)}(z)}{t^{{1}/{2}}}+O\left(\frac{1}{t}\right),\quad\text{for}\quad
z\in\partial U(z_{-},\delta),$ (5.30)

where $\widehat{R}^{(1)}(z)=O(z^{-1})$ as $z\rightarrow\infty$ and
$\widehat{R}^{(1)}(z)$ satisfies the jump relation

$\widehat{R}^{(1)}_{+}(z)-\widehat{R}^{(1)}_{-}(z)=Q^{(z_{-})}(z),\quad
z\in\partial U(z_{-},\delta).$ (5.31)

The solution to the above RH problem can be constructed explicitly as follows:

$\widehat{R}^{(1)}(z)=\left\\{\begin{aligned}
&\frac{L_{-}}{z+\frac{\sqrt{2}}{2}},&z&\notin\overline{U(z_{-},\delta)},\\\
&\frac{L_{-}}{z+\frac{\sqrt{2}}{2}}-Q^{(z_{-})}(z),&z&\in
U(z_{-},\delta),\end{aligned}\right.$ (5.32)

where $L_{-}=\mathop{\rm
Res}\nolimits\left(Q^{(z_{-})}(z),z=-\frac{\sqrt{2}}{2}\right)$. Expanding
this as $z\rightarrow\infty$, we obtain the asymptotics for $\widehat{R}_{1}$
in the expansion (5.21)

$\widehat{R}_{1}=\frac{\widehat{R}^{(1)}_{1}}{t^{{1}/{2}}}+O\left(\frac{1}{t}\right),\quad\mathrm{as}\quad
t\rightarrow\infty,$ (5.33)

where

$\widehat{R}^{(1)}_{1}=L_{-}=\begin{pmatrix}0&-2^{-\frac{3}{2}\nu-\frac{3}{4}}h_{1}e^{-\pi
i\nu}\\\ -\nu 2^{\frac{3}{2}\nu-\frac{3}{4}}e^{\pi
i\nu}h^{-1}_{1}&0\end{pmatrix}.$ (5.34)

Substituting (5.33) and (5.27) into (5.28) and (5.29), and using (3.25) and
(5.1), we obtain for $\nu\in\mathbb{C}\setminus\mathbb{N}$

$u(x;\nu)=\sqrt{-\frac{x}{2}}+\frac{\nu+\frac{1}{2}}{2x}+O\left((-x)^{-\frac{5}{2}}\right),\quad
x\rightarrow-\infty,$ (5.35)

and

$H(x;\nu)=-\nu\sqrt{-\frac{x}{2}}+O\left((-x)^{-1}\right),\quad
x\rightarrow-\infty.$ (5.36)

Next, we consider the case $\nu\in\mathbb{N}\setminus\\{0\\}$. From (3.25) and
(5.24), we have

$u(x;\nu)=-\frac{d}{dx}\log(\Psi_{1})_{21}(x;-\nu),$ (5.37)

and

$H(x;\nu)=H(x;-\nu)$ (5.38)

for $\nu\in\mathbb{N}\setminus\\{0\\}$. Therefore, we obtain from (5.28),
(5.36), (5.37) and (5.38) that

$u(x;\nu)=-\sqrt{-\frac{x}{2}}+\frac{\nu+\frac{1}{2}}{2x}+O\left((-x)^{-\frac{5}{2}}\right),\quad
x\rightarrow-\infty,$ (5.39)

and

$H(x;\nu)=\nu\sqrt{-\frac{x}{2}}+O\left((-x)^{-1}\right),\quad
x\rightarrow-\infty,$ (5.40)

for $\nu\in\mathbb{N}\setminus\\{0\\}$.

###  5.3 Asymptotics for $\Psi$ as $x\rightarrow-\infty$ for $\nu=0$

In the section, we consider the remaining case $\nu=0$. Still, we carry out a
nonlinear steepest descent analysis of the Riemann-Hilbert problem for $\Psi$.
The first transformation is the same as (5.1). While the second transformation
is now defined by

$\widetilde{U}(z)=\widehat{\Phi}(z)e^{-t\widehat{\theta}(z)\sigma_{3}},\quad\mbox{with}\quad\widehat{\theta}(z)=\frac{1}{3}z^{3}-\frac{1}{2}z.$
(5.41)

We also deform the original contour $\Gamma_{k},~{}k=1,2,4,5$ into the anti-
Stokes curves of $\widehat{\theta}(z)$ as shown in Figure 12. Then
$\widetilde{U}(z)$ solves the following RH problem.

Figure 12: The jump curves $\Sigma_{\widetilde{U}}$ of the RH problem for
$\widetilde{U}(z)$

######  RH problem 5.6.

(1) $\widetilde{U}(z)$ is analytic in
$\mathbb{C}\setminus\Sigma_{\widetilde{U}}$.

(2) For $z\in\Sigma_{\widetilde{U}}$, we have
$\widetilde{U}_{+}(z)=\widetilde{U}_{-}(z)J_{\widetilde{U}}(z)$, where

$J_{\widetilde{U}}(z)=\left\\{\begin{aligned}
&\begin{pmatrix}1&e^{2t\widehat{\theta}(z)}\\\
0&1\end{pmatrix},\quad&z&\in\Gamma_{+},\\\ &\begin{pmatrix}1&0\\\
e^{-2t\widehat{\theta}(z)}&1\end{pmatrix},\quad&z&\in\Gamma_{-}.\end{aligned}\right.$
(5.42)

(3) As $z\to\infty$, we have

$\widetilde{U}(z)=\left(I+\frac{\widetilde{U}_{1}}{z}+O\left(\frac{1}{z^{2}}\right)\right).$
(5.43)

From (5.42), we have

$J_{\widetilde{U}}(z)-I=O\left(e^{-2t\widehat{\theta}(z)}\right),$ (5.44)

where $z\in\Sigma_{\widetilde{U}}$. Therefore, we have the estimate

$||J_{\widetilde{U}}(z)-I||_{L^{2}\cap
L^{\infty}(\Gamma_{+}\cup\Gamma_{-1})}=O\left(e^{-\frac{\sqrt{2}}{3}t}\right).$
(5.45)

This ensures the solvability of the RH problem for $\widetilde{U}(z)$, and the
solution can be given by the integral formula

$\widetilde{U}(z)=I+\frac{1}{2\pi
i}\int_{\Gamma_{+}\cup\Gamma_{-}}\frac{\widetilde{U}_{-}(s)(J_{\widetilde{U}}(s)-I)}{s-z}ds.$
(5.46)

This expression implies the large-$z$ expansion

$\widetilde{U}(z)=I+\frac{\widetilde{U}_{1}(s)}{z}+O\left(\frac{1}{z^{2}}\right),$
(5.47)

where

$\widetilde{U}_{1}(t)=-\frac{1}{2\pi
i}\int_{\Gamma_{+}\cup\Gamma_{-}}\widetilde{U}_{-}(s)\left(J_{\widetilde{U}}(s)-I\right)ds.$
(5.48)

To continue, we compute by using (5.42) and (5.45)

$\left(\widetilde{U}_{1}\right)_{12}=-\frac{1}{2\pi
i}\int_{\Gamma_{+}\cup\Gamma_{-}}\left(J_{\widetilde{U}}(s)-I\right)_{12}ds+O\left(e^{-\frac{2\sqrt{2}}{3}t}\right)=c\,e^{-\frac{\sqrt{2}}{3}t}t^{-\frac{1}{2}}\left(1+O\left(t^{-1}\right)\right),$
(5.49)

for some nonzero constant $c$. We also have from (5.42)

$\displaystyle\left(\widetilde{U}_{1}\right)_{11}$
$\displaystyle=-\frac{1}{2\pi
i}\int_{\Gamma_{+}\cup\Gamma_{-}}\left((\widetilde{U}_{-}(s)-I)(J_{\widetilde{U}}(s)-I)\right)_{11}ds$
(5.50) $\displaystyle=-\frac{1}{2\pi
i}\int_{\Gamma_{-}}(\widetilde{U}_{-}(s)-I)_{12}e^{-t\widehat{\theta}(s)}ds$
(5.51) $\displaystyle=O\left(e^{-\frac{2\sqrt{2}}{3}t}\right).$ (5.52)

Recalling the relation $\Psi_{1}=(-x)^{\frac{1}{2}}\widetilde{U}_{1}$, we
obtain from (3.25) and (5.49) that

$u(x;0)=-\sqrt{-\frac{x}{2}}+\frac{1}{4x}+O\left((-x)^{-\frac{5}{2}}\right),\quad
x\rightarrow-\infty.$ (5.53)

From (3.25) and (5.50), we have

$H(x;0)=O\left((-x)^{1/2}e^{-\frac{2\sqrt{2}}{3}(-x)^{3/2}}\right),\quad
x\rightarrow-\infty.$ (5.54)

Now combining (5.35)-(5.40), (5.53) and (5.54), we further have (1.22) and
(1.24).

##  6 Proof of Theorem 1.5

Tracing back the series of invertible transformations $Y\to\widehat{Y}\to T\to
S\to R$, we obtain

$Y(z)=n^{-\frac{\nu}{3}\sigma_{3}}R(z)n^{\frac{\nu}{3}\sigma_{3}}\left(\frac{z+1}{z}\right)^{\nu\sigma_{3}}e^{-\frac{t}{2z}\sigma_{3}}z^{n\sigma_{3}}\quad\mbox{for}\quad|z|>1+\delta,$
(6.1)

and

$Y(z)=n^{-\frac{\nu}{3}\sigma_{3}}R(z)n^{\frac{\nu}{3}\sigma_{3}}(z+1)^{\nu\sigma_{3}}i\sigma_{2}e^{-\frac{t}{2}z\sigma_{3}}\quad\mbox{for}\quad|z|<\delta,$
(6.2)

where

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

is one of the Pauli matrices.

Thus, we get from (3.37) and (6.1) that

$(Y_{-1})_{11}=-\frac{t}{2}+\nu-\frac{1}{n^{1/3}}2^{\frac{2}{3}}H\left(2^{\frac{2}{3}}n^{2/3}(\tau-1);\nu\right)+O\left(\frac{1}{n^{2/3}}\right),$
(6.3)

where the Hamiltonian $H(s;\nu)=-(\Psi_{1})_{11}(s)$; cf. (3.25). Similarly,
using (3.37) and (6.2), we obtain

$\displaystyle\frac{d}{dz}\log(Y)_{21}(z)|_{z=0}$
$\displaystyle=-\frac{t}{2}-\nu+\frac{d}{dz}\log(R)_{22}(z)|_{z=0}$
$\displaystyle=-\frac{t}{2}-\nu-2^{\frac{2}{3}}n^{-\frac{1}{3}}H\left(2^{\frac{2}{3}}n^{\frac{2}{3}}(\tau-1);\nu\right)+O\left(n^{-\frac{2}{3}}\right).$
(6.4)

From the differential identity (2.4), (6.3) and (6.4), we have (1.26). In
virtue of (2.5), (3.37) and (6.2), we have

$h_{n}=Y_{12}(0)=1-2^{\frac{2}{3}}n^{-\frac{1}{3}}H\left(2^{\frac{2}{3}}n^{\frac{2}{3}}(\tau-1);\nu\right)+O\left(n^{-\frac{2}{3}}\right),$
(6.5)
$\pi_{n}(0)=Y_{11}(0)=-2^{\frac{2}{3}+\frac{4}{3}\nu}n^{-\frac{1}{3}-\frac{2}{3}\nu}e^{n\pi
i}\left((\Psi_{1})_{12}\left(2^{\frac{2}{3}}n^{\frac{2}{3}}(\tau-1);\nu\right)+O\left(n^{-\frac{1}{3}}\right)\right).$
(6.6)

Recalling (3.25), we obtain (1.27) and (1.28). This completes the proof of
Theorems 1.5.

## Acknowledgements

The authors are grateful to the reviewers for their constructive comments and
suggestions. The work of Shuai-Xia Xu was supported in part by the National
Natural Science Foundation of China under grant numbers 11971492 and 12371257,
and by Guangdong Basic and Applied Basic Research Foundation (Grant No.
2022B1515020063). Yu-Qiu Zhao was supported in part by the National Natural
Science Foundation of China under grant numbers 11971489 and 12371077.

##  Appendix A Local parametrix models

###  A.1 Airy parametrix

Define

$\Phi^{(\mathrm{Ai})}(z)=M\left\\{\begin{aligned}
&\begin{pmatrix}\mathrm{Ai}(e^{-\frac{2}{3}\pi i}z)&\mathrm{Ai}(z)\\\
e^{-\frac{2}{3}\pi i}\mathrm{Ai}^{\prime}(e^{-\frac{2}{3}\pi
i}z)&\mathrm{Ai}^{\prime}(z)\end{pmatrix}e^{i\frac{\pi}{6}\sigma_{3}},&z&\in\mathrm{I},\\\
&\begin{pmatrix}\mathrm{Ai}(e^{-\frac{2}{3}\pi i}z)&\mathrm{Ai}(z)\\\
e^{-\frac{2}{3}\pi i}\mathrm{Ai}^{\prime}(e^{-\frac{2}{3}\pi
i}z)&\mathrm{Ai}^{\prime}(z)\end{pmatrix}e^{i\frac{\pi}{6}\sigma_{3}}\begin{pmatrix}1&-1\\\
0&1\end{pmatrix},&z&\in\mathrm{II},\\\
&\begin{pmatrix}\mathrm{Ai}(e^{-\frac{2}{3}\pi i}z)&\mathrm{Ai}(z)\\\
e^{-\frac{2}{3}\pi i}\mathrm{Ai}^{\prime}(e^{-\frac{2}{3}\pi
i}z)&\mathrm{Ai}^{\prime}(z)\end{pmatrix}e^{i\frac{\pi}{6}\sigma_{3}}\begin{pmatrix}0&-1\\\
1&1\end{pmatrix},&z&\in\mathrm{III},\end{aligned}\right.$ (A.1)

where $\mathrm{Ai}(z)$ is the Airy function,

$M=\sqrt{2\pi}\begin{pmatrix}e^{\frac{1}{6}\pi i}&0\\\ 0&-e^{\frac{1}{6}\pi
i}\end{pmatrix},$

and the regions I-III are shown in Figure 13. Then, $\Phi^{(\mathrm{Ai})}(z)$
satisfies the following RH problem [4, Section 3.7], which is related to the
standard one in [12, Chapter 7] and [16] after some minor modifications.

Figure 13: The jump contours and regions for $\Phi^{(\mathrm{Ai})}$

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

$\Phi^{\mathrm{(Ai)}}(z)$ is analytic for
$s\in\mathbb{C}\setminus\cup^{3}_{k=1}\gamma_{k}$, see Figure 13.

$\Phi^{\mathrm{(Ai)}}(z)$ satisfies the jump relations
$\Phi^{\mathrm{(Ai)}}_{+}(z)=\Phi^{\mathrm{(Ai)}}_{-}(z)J^{(\mathrm{Ai})}_{k}(z)$,
$z\in\gamma_{k}$, $k=1,2,3$, where

$J^{(\mathrm{Ai})}_{1}(z)=\begin{pmatrix}1&1\\\ -1&0\end{pmatrix},\
J^{(\mathrm{Ai})}_{2}(z)=\begin{pmatrix}1&1\\\ 0&1\end{pmatrix},\
J^{(\mathrm{Ai})}_{3}(z)=\begin{pmatrix}1&0\\\ -1&1\end{pmatrix}.$

As $z\to\infty$, we have

$\Phi^{(\mathrm{Ai})}(z)=z^{-\frac{\sigma_{3}}{4}}\frac{1}{\sqrt{2}}\begin{pmatrix}i&1\\\
-i&1\end{pmatrix}\left(I+\frac{1}{48z^{\frac{3}{2}}}\begin{pmatrix}-1&6i\\\
6i&1\end{pmatrix}+O\left(z^{-3}\right)\right)e^{\frac{2}{3}z^{\frac{3}{2}}\sigma_{3}},$
(A.2)

where $\mathop{\rm arg}\nolimits z\in(0,2\pi)$.

###  A.2 Parabolic cylinder parametrix

Define

$Z_{0}(z)=2^{-\frac{\sigma_{3}}{2}}\begin{pmatrix}D_{-\nu-1}(iz)&D_{\nu}(z)\\\
D_{-\nu-1}^{\prime}(iz)&D_{\nu}^{\prime}(z)\end{pmatrix}\begin{pmatrix}e^{i\frac{\pi}{2}(\nu+1)}&0\\\
0&1\end{pmatrix}$

and

$Z_{n+1}(z)=Z_{n}(z)H_{n},\quad n=0,1,2,3,$

where $D_{-\nu-1}$, $D_{\nu}$ are the standard parabolic cylinder functions
(cf. [34, Chapter 12]),

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

with

$h_{0}=-i\frac{\sqrt{2\pi}}{\Gamma(\nu+1)},\quad
h_{1}=\frac{\sqrt{2\pi}}{\Gamma(-\nu)}e^{i\pi\nu},\quad 1+h_{0}h_{1}=e^{2\pi
i\nu}.$ (A.3)

Figure 14: The jump contours and jump matrices for $\Phi^{(\mathrm{PC})}$

Let

$\Phi^{(\mathrm{PC})}(z)=\left\\{\begin{aligned} Z_{0}(z),\quad&\mathop{\rm
arg}\nolimits z\in\left(-\frac{\pi}{4},0\right),\\\ Z_{1}(z),\quad&\mathop{\rm
arg}\nolimits z\in\left(0,\frac{\pi}{2}\right),\\\ Z_{2}(z),\quad&\mathop{\rm
arg}\nolimits z\in\left(\frac{\pi}{2},\pi\right),\\\
Z_{3}(z),\quad&\mathop{\rm arg}\nolimits
z\in\left(\pi,\frac{3\pi}{2}\right),\\\ Z_{4}(z),\quad&\mathop{\rm
arg}\nolimits
z\in\left(\frac{3\pi}{2},\frac{7\pi}{4}\right),\end{aligned}\right.$

then $\Phi^{(\mathrm{PC})}(z)$ solves the following RH problem (cf. [3, 19]).

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

$\Phi^{\mathrm{(PC)}}(z)$ is analytic for all
$z\in\mathbb{C}\setminus\bigcup^{5}_{k=1}\Sigma_{k}$, where
$\Sigma_{k}=\\{z\in\mathbb{C}:\mathop{\rm arg}\nolimits z=\frac{k\pi}{2}\\}$,
$k=1,2,3,4$ and $\Sigma_{5}=\\{z\in\mathbb{C}:\mathop{\rm arg}\nolimits
z=-\frac{\pi}{4}\\}$; see Figure 14.

$\Phi^{\mathrm{(PC)}}(z)$ satisfies the jump conditions as shown in Figure 14.

As $z\to\infty$, we have

$\displaystyle\Phi^{\mathrm{(PC)}}(z)=\begin{pmatrix}0&1\\\
1&-z\end{pmatrix}2^{\frac{\sigma_{3}}{2}}\begin{pmatrix}1+O\left(\frac{1}{z^{2}}\right)&\frac{\nu}{z}+O\left(\frac{1}{z^{3}}\right)\\\
\frac{1}{z}+O\left(\frac{1}{z^{3}}\right)&1+O\left(\frac{1}{z^{2}}\right)\end{pmatrix}e^{\left(\frac{z^{2}}{4}-\nu\log
z\right)\sigma_{3}}.$ (A.4)

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